Machine Design

STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS

1

Chapter 1 1-1 Calculate the mass m of a body that weighs 600 lb at the surface of the earth. SOLUTION

slug63.182.32

600g

Wm ....................................................Ans.

1-2 Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg. SOLUTION

kN62.6N1062.681.9675 3mgW ..................................Ans.

1-3 If a man weighs 180 lb at sea level, determine the weight W of the man (a) At the top of Mt. McKinley (20,320 ft above sea level). (b) At the top of Mt. Everest (29,028 ft above sea level). SOLUTION

2e

e

rmGmW

Therefore: mGmrWrW ehh22

00

where ft10090.2 70r

(a) ft10092032.2100320.210090.2 7470 hrrh

lb7.17910092032.210090.2180

27

27

2

200

hh r

rWW ......................................Ans.

(b) ft100929028.2109028.210090.2 7470 hrrh

lb5.179100929028.2

10090.218027

27

2

200

hh r

rWW ....................................Ans.

1-4 Calculate the weight W of a navigation satellite at a distance of 20,200 km above the earth’s surface if the satellite weighs 9750 N at the earth’s surface.

SOLUTION

2e

e

rmGmW

Therefore: mGmrWrW ehh22

00

where 60 6.370 10 mr

(a) 0 6370 20,200 26,570 kmhr r h

22

0 022

9750 6370560 N

26,570h

h

W rWr

...........................................Ans.


2

1-5 Compute the gravitational force acting between two spheres that are touching each other if each sphere weighs 1125 lb and has a diameter of 20 in.

SOLUTION

2

2

2221

s

s

ss

ss

dGm

rrmGm

rmGmF

28

62

11253.439 10 32.2 15.11 10 lb20

12

..................................Ans.

1-6 Two spherical bodies have masses of 60 kg and 80 kg, respectively. Determine the gravitational force of attraction between the spheres if the distance from center to center is 600 mm.

SOLUTION

11

61 222

6.673 10 60 800.890 10 N

0.600Gm mF

r........................Ans.

1-7 Determine the weight W of a satellite when it is in orbit 8500 mi above the surface of the earth if the satellite weighs 7600 lb at the earth’s surface.

SOLUTION

221

rmGmW

Therefore: behh mGmrWrW 2200

where ft10090.2 70r

ft10578.6)5280(850010090.2 77hrr oh

272

22 7

7600 2.090 10768 lb

6.578 10o o

hh

W rWr

.....................................Ans.

1-8 Determine the weight W of a satellite when it is in orbit 20.2(106) m above the surface of the earth if the satellite weighs 8450 N at the earth’s surface.

SOLUTION

221

rmGmW

Therefore: behh mGmrWrW 2200

where 60 6.370 10 mr

6 6 66.370 10 20.2 10 26.570 10 mh or r h

262

22 6

8450 6.370 10486 N

26.570 10o o

hh

W rWr

.....................................Ans.


3

1-9 If a woman weighs 135 lb when standing on the surface of the earth, how much would she weigh when standing on the surface of the moon?

SOLUTION

221

rmGmW

Therefore on the surface of the earth where 234.095 10 slugsem and 3960 mir

23

2

4.095 10135

3960 5280

G m

7 21.4413 10 lb ft /slugGm

Then on the surface of the moon where 215.037 10 slugsmm and 1080 mir

7 21

2

1.4413 10 5.037 1022.33 lb

1080 5280W ...................................Ans.

1-10 Determine the weight W of a body that has a mass of 1000 kg (a.) At the surface of the earth. (b.) At the top of Mt. McKinley (6193 m above sea level). (c.) In a satellite at an altitude of 250 km. SOLUTION

221

rmGmW

(a) 11 24

23

6.673 10 5.976 10 10009830 N

6370 10W .............................Ans.

(b) 11 24

23

6.673 10 5.976 10 10009810 N

6370 10 6193W .............................Ans.

(c) 11 24

23 3

6.673 10 5.976 10 10009100 N

6370 10 250 10W .............................Ans.

1-11 If a man weighs 210 lb at sea level, determine the weight W of the man (a.) At the top of Mt. Everest (29,028 ft above sea level). (b.) In a satellite at an altitude of 200 mi. SOLUTION

221

rmGmW

Therefore 22103960 5280

eGm m

16 29.181 10 lb fteGm m


4

(a) 16

29.181 10 209.4 lb

3960 5280 29,028W .......................................Ans.

(b) 16

29.181 10 190.3 lb

3960 200 5280W ........................................Ans.

1-12 A space traveler weighs 800 N on earth. A planet having a mass of 5(1025) kg and a diameter of 30(106) m orbits a distant star. Determine the weight W of the traveler on the surface of this planet.

SOLUTION

221

rmGmW

Therefore on the surface of the earth where 245.976 10 kgem and 6370 kmr

24

23

5.976 10800

6370 10

G m

9 25.432 10 N m /kgGm

Then on the surface of the planet where 255 10 kgm and 615 10 mr

9 25

26

5.432 10 5 101207 N

15 10W .........................................Ans.

1-13 The planet Jupiter has a mass of 1.302(1026) slug and a visible diameter (top of the cloud layer) of 88,700 mi. Determine the gravitational acceleration g

(a.) At a point 100,000 miles above the top of the clouds. (b.) At the top of the cloud layers. SOLUTION

221

rmGmW

(a) 8 26

2

3.439 10 1.302 10

44,350 100,000 5280

mW mg

27.71 ft/sg .................................................................Ans.

(b) 8 26

2

3.439 10 1.302 10

44,350 5280

mW mg

281.7 ft/sg ................................................................Ans.

1-14 The planet Saturn has a mass of 5.67(1026) kg and a visible diameter (top of the cloud layer) of 120,000 km. The weight W of a planetary probe on earth is 4.50 kN. Determine

(a.) The weight of the probe when it is 600,000 km above the top of the clouds. (b.) The weight of the probe as it begins its penetration of the cloud layers. SOLUTION

221

rmGmW


5

Therefore on the surface of the earth where 245.976 10 kgem and 6370 kmr

24

23

5.976 104500

6370 10

G m

8 23.055 10 N m /kgGm

(a) 8 26

23

3.055 10 5.67 1039.8 N

60,000 600,000 10W .....................................Ans.

(b) 8 26

23

3.055 10 5.67 104810 N

60,000 10W ......................................Ans.

1-15 The first U.S. satellite, Explorer 1, had a mass of approximately 1 slug. Determine the force exerted on the satellite by the earth at the low and high points of its orbit, which were 175 mi and 2200 mi, respectively, above the surface of the earth.

SOLUTION

1 22

Gm mFr

8 23

2

3.439 10 4.095 10 129.5 lb

3960 175 5280F ...................................Ans.

8 23

2

3.439 10 4.095 10 113.31 lb

3960 2200 5280F ..................................Ans.

1-16 A neutron star has a mass of 2(1030) kg and a diameter of 10 km. Determine the gravitational force of attraction on a 10-kg space probe

(a.) When it is 1000 km from the center of the star. (b.) At the instant of impact with the surface of the star. SOLUTION

1 22

Gm mFr

11 30

923

6.673 10 2 10 101.335 10 N

1000 10F ...............................Ans.

11 30

1323

6.673 10 2 10 105.34 10 N

5 10F ...............................Ans.

1-17 Determine the weight W, in U.S. customary units, of a 75-kg steel bar under standard conditions (sea level at a latitude of 45 degrees).

SOLUTION

75 9.81 735.75 N 0.2248 lb/N 165.4 lbW mg ......................Ans.


6

1-18 Determine the mass m, in SI units, for a 500-lb steel beam under standard conditions (sea level at a latitude of 45 degrees).

SOLUTION W mg

500 15.528 slug 14.59 kg/slug 227 kg32.2

Wmg

........................Ans.

1-19 An automobile has a 440 cubic inch engine displacement. Determine the engine displacement in liters. SOLUTION

33 3 3 3 1 3 3440 in. 16.39 10 mm /in. 10 cm/mm 10 L/cmV

7.21 LV ...................................................................Ans.

1-20 How many barrels of oil are contained in 100 kL of oil? One barrel (petroleum) equals 42.0 gal. SOLUTION

3100 10 L 0.2642 gal/L 1 barrel 42 gal 629 barrelV ..................Ans.

1-21* Express the density, in SI units, of a specimen of material that has a specific weight of 0.025 lb/in.3 SOLUTION g

3 3

3 3 3

0.025 slug 14.59 kg in. 1000 mm32.2 in. slug 16.39 10 mm mg

3691 kg/m ...............................................................Ans.

1-22 The viscosity of crude oil under conditions of standard temperature and pressure is 7.13(10-3) N-s/m2. Determine the viscosity of crude oil in U.S. Customary units.

SOLUTION

2

3 32 2 2

N s 0.2248 lb 0.0929 m lb s7.13 10 0.1489 10 m N ft ft

...........Ans.

1-23 One acre equals 43,560 ft2. One gallon equals 231 in.3 Determine the number of liters of water required to cover 2000 acres to a depth of 1 foot.

SOLUTION

32

3

43,560 ft 12 in gal 3.785 L2000 acre ftacre ft 231 in. gal

V

92.47 10 LV ..............................................................Ans.

1-24 The stress in a steel bar is 150 MPa. Express the stress in appropriate U.S. Customary units (ksi) by using the values listed in Table 1-6 for length and force as defined values.

SOLUTION

2

6 2 2 2 ftstress 150 10 N/m 0.2248 lb/N 0.0929 m /ft12 in.

3 2stress 21.75 10 lb/in. 21.75 ksi .........................................Ans.


7

1-25 By definition, 1 hp = 33,000 ft-lb/min and 1 W = 1 N-m/s. Verify the conversion factors listed in Table 1-6 for converting power from U.S. Customary units to SI units by using the values listed for length and force as defined values.

SOLUTION

ft lb 4.448 N 0.3048 m min N m1 hp 33,000 745.7 min lb ft 60 s s

.............Ans.

1-26 The specific heat of air under standard atmospheric pressure, in SI units, is 1003 N-m/kg- K. Determine the specific heat of air under standard atmospheric pressure in U.S. customary units (ft-lb/slug- R).

SOLUTION

0.2248 lb 3.281 ft 14.59 kg 5 KN m1003 kg K N m slug 9 R

lb ft6000 slug R ..........................................................Ans.

1-27 Newton’s law of gravitation can be expressed in equation form as

221

rmmGF

If F is a force, m1 and m2 are masses, and r is a distance, determine the dimensions of G. SOLUTION

222 3

21 2

ML T LFr LGm m M M MT

..........................................Ans.

1-28 The elongation of a bar of uniform cross section subjected to an axial force is given by the equation EAPL . What are the dimensions of E if and L are lengths, P is a force, and A is an area?

SOLUTION

2

2 22

ML T LPL M FEA LT LL L

......................................Ans.

1-29 The period of oscillation of a simple pendulum is given by the equation gLkT , where T is in seconds, L is in feet, g is the acceleration due to gravity, and k is a constant. What are the dimensions of k for dimensional homogeneity?

SOLUTION

1 22

1 (dimensionless)L Tk T g L TL

...............................Ans.

1-30 An important parameter in fluid flow problems involving thin films is the Weber number (We) which can be expressed in equation form as

Lv2

We

where is the density of the fluid, v is a velocity, L is a length, and is the surface tension of the fluid. If the Weber number is dimensionless, what are the dimensions of the surface tension ?

SOLUTION

232

21e

M L L T Lv L M FW T L

................................Ans.


8

1-31 In the dimensionally homogeneous equation

I

McAP

is a stress, A is an area, M is a moment of a force, and c is a length. Determine the dimensions of P and I. SOLUTION

2 2

2 2

ML T LM PLT IL

Therefore

22 2

M MLP L FLT T

...............................................Ans.

2 2

42

ML T LI L

M LT......................................................Ans.

1-32 In the dimensionally homogeneous equation 2 21 1

2 2Pd mv I

d is a length, m is a mass, v is a linear velocity, and is an angular velocity. Determine the dimensions of P and I.

SOLUTION

2

2

1LP L M IT T

Therefore

2 2

2

ML T MLP FL T

................................................Ans.

2

2 22

MLI T MLT

......................................................Ans.


bI

VQJ

Tr

is a stress, T is a torque (moment of a force), V is a force, r and b are lengths and I is a second moment of an area. Determine the dimensions of J and Q.

SOLUTION

2 2 2

2 4

ML T L ML T QMLT J L L

Therefore

2 2

42

ML T LJ L

M LT.....................................................Ans.


9

2 5

32

M LT LQ L

ML T.....................................................Ans.


J

TrAP

is a stress, A is an area, T is a torque (moment of a force), and r is a length. Determine the dimensions of P and J.

SOLUTION

2 2

2 2

ML T LM PLT JL

Therefore

22 2

M MLP L FLT T

...............................................Ans.

2 2

42

ML T LJ L

M LT.....................................................Ans.

1-35 The equation atAex bt sin is dimensionally homogeneous. If A is a length and t is time, determine the dimensions of x, a, b, and .

SOLUTION

sinT bx L e a T

Therefore

1 1x L L ...........................................................Ans.

b T ......................................................................Ans.

1a T ....................................................................Ans.

1 (dimensionless) ........................................................Ans.

1-36 In the dimensionally homogeneous equation xbabxaxxw 223 , if x is a length, what are the dimensions of a, b, and w?

SOLUTION

If x L , then each term has the dimension 3L . Therefore

3w L .....................................................................Ans.

3

2

La L

L...............................................................Ans.

3

2L

b LL

...............................................................Ans.

Using the last term as a check,


10

2 22

3L La b L

x L

1-37 Determine the dimensions of a, b, c, and y in the dimensionally homogeneous equation

cbtaAey bt 21cos

in which A is a length and t is time. SOLUTION

2cos 1b Ty L e a b T c

Therefore

1 1y L L ..........................................................Ans.

1b T 1b T ....................................................................Ans.

1 (dimensionless)a ........................................................Ans.

1 (dimensionless)c ........................................................Ans.

1-38 Determine the dimensions of c, , k and P in the differential equation

tPkxdtdxc

dtxdm cos2

2

in which m is a mass, x is a length, and t is time. SOLUTION

2cos

M L Lc k L P T

TT

Therefore

2ML T Mc

L T T.........................................................Ans.

2

2

ML T M FkL T L

..................................................Ans.

2

MLP FT

.............................................................Ans.

1 T ....................................................................Ans.

1-39 Round off the following numbers to two significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.015362 (b) 55.33682 (c) 63,746.27 SOLUTION

(a) 0.015 0.015362 100 2.36 %

0.015362.............................................Ans.

(b) 55 55.33682 100 0.609 %

55.33682...............................................Ans.


11

(c) 64,000 63,746.27 100 0.398 %

63,746.27.........................................Ans.

1-40 Round off the following numbers to two significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.837482 (b) 374.9371 (c) 937,284.9 SOLUTION

(a) 0.84 0.837482 100 0.301%

0.837482.............................................Ans.

(b) 370 374.9371 100 1.317 %

374.9371.............................................Ans.

(c) 940,000 937,284.9 100 0.290%

937,284.9.........................................Ans.

1-41 Round off the following numbers to three significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.034739 (b) 26.39473 (c) 55,129.92 SOLUTION

(a) 0.0347 0.034739 100 0.1123 %

0.034739.........................................Ans.

(b) 26.4 26.39473 100 0.01997 %

26.39473..........................................Ans.

(c) 55,100 55,129.92 100 0.0543 %

55,129.92.........................................Ans.

1-42 Round off the following numbers to three significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.472916 (b) 826.4836 (c) 339,872.8 SOLUTION

(a) 0.473 0.472916 100 0.01776 %

0.472916.........................................Ans.

(b) 826 826.4836 100 0.0585 %

826.4836............................................Ans.

(c) 340,000 339,872.8 100 0.0374 %

339,872.8.......................................Ans.

1-43 Round off the following numbers to four significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.056623 (b) 74.82917 (c) 27,382.84 SOLUTION

(a) 30.05662 0.056623 100 5.30 10 %0.056623

.....................................Ans.

(b) 374.83 74.82917 100 1.109 10 %74.82917

......................................Ans.


12

(c) 327,380 27,382.84 100 10.37 10 %27,382.84

....................................Ans.

1-44 Round off the following numbers to four significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference.

(a) 0.664473 (b) 349.3378 (c) 274,918.2 SOLUTION

(a) 30.6645 0.664473 100 4.06 10 %0.664473

......................................Ans.

(b) 3349.3 349.3378 100 10.82 10 %349.3378

......................................Ans.

(c) 3274,900 274,918.2 100 6.62 10 %274,918.2

....................................Ans.

1-45 The weight of the first Russian satellite, Sputnik I, was 184 lb on the surface of the earth. Determine the force exerted on the satellite by the earth at the low and high points of its orbit which were 149 mi and 597 mi, respectively, above the surface of the earth.

SOLUTION

1 22

Gm mFr

Therefore 21843960 5280

eGm m

16 28.044 10 lb fteGm m

16

28.044 10 170.9 lb

3960 149 5280F .........................................Ans.

16

28.044 10 138.9 lb

3960 597 5280F .........................................Ans.

1-46 The planet Jupiter has a mass of 1.90(1027) kg and a radius of 7.14(107) m. Determine the force of attraction between the earth and Jupiter when the minimum distance between the two planets is 6(1011) m.

SOLUTION

1 22

Gm mFr

11 24 27

18211

6.673 10 5.976 10 1.90 102.10 10 N

6 10F ....................Ans.

1-47 Convert 640 acres (1 square mile) to hectares if 1 acre equals 4840 yd2 and 1 hectare equals 104 m2. SOLUTION

22 2

2 4

4840 yd 3 ft 0.0929 m hect640 acres 259 hectareacre yd ft 10 m

............Ans.


13

1-48 Determine the dimension of c in the dimensionally homogeneous equation

mctec

mgv 1

in which v is a velocity, m is a mass, t is time, and g is the gravitational acceleration. SOLUTION

2

1 c T MM L TL eT c

Therefore

2ML T Mc

L T T.........................................................Ans.

As a check, 1 (dimensionless)M T Tct

m M

1-49 Develop an expression for the change in gravitational acceleration g between the surface of the earth and a height h when h << Re.

SOLUTION

2eGm mW mg

r

Therefore

2e e emg Gm m r

2

e emg Gm m r h

2 2

2 2 22 2

22 2

1 1

2

2

e eee

ee e e

e e

ee

e e

g g g Gmrr h

Gm r r r h hr h r

Gm r h hr h r

4 3

22e ee

ee

Gm Gm hg r hrr

..........................................................Ans.


14

Chapter 2 2-1 Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the

following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-1.

SOLUTION From the law of cosines

2 2 290 120 2 90 120 cos90R

150.0 lbR From the law of sines

sin sin 9090 R

1 90sin 90sin 36.87150.0

150 lbR 36.87 .......................................................................................................Ans.

2-2 Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-2.


2 2 2250 200 2 250 200 cos50R

408.386 NR From the law of sines

sin sin130200 R

1 200sin130sin 22.03408.386

408 NR 36.03 .......................................................................................................Ans.



2 2 2600 800 2 600 800 cos75R

866.910 lbR From the law of sines

sin sin 75800 R

1 800sin 75sin 63.05866.910

867 lbR 86.95 .......................................................................................................Ans.


15



2 2 210 25 2 10 25 cos120R

31.225 kNR From the law of sines

sin sin12025 R

1 25sin120sin 43.9031.225

31.2 kNR 19.90 ....................................................................................................Ans.

2-5 The support ring of the traffic light shown in Fig. P2-5 is acted on by three forces – the weight of the traffic light (220 lb), a force in cable A (280 lb), and a force in cable B (FB). If the resultant of the three forces is zero, determine the magnitude and direction of FB.


2 2 212 220 280 2 220 280 cos70R

12 290.969 lbR

From the law of sines

12

12

sin sin 70220 R

112

220sin 70sin 45.28290.969

12 291 lbR 25.28

But 12 BR F 0

Therefore 291 lbBF R 25.28 ..........................................................................................Ans.

2-6 Determine the resultant of the three forces shown in Fig. P2-6. Locate the resultant with respect to the x-axis shown.

SOLUTION

2 212 3 4 5 kNR

112

3tan 36.8704

Then, using the law of cosines

2 2 25 8 2 5 8 cos113.130R

10.974 kNR From the law of sines

sin sin113.130

5 R


16

1 5sin113.130sin 24.7710.974

10.97 kNR 5.23 ....................................................................................................Ans.

2-7 A stalled automobile is being pulled by the two forces shown in Fig. P2-7. If the resultant pull is to be 120 lb in the x-direction, determine the magnitude and direction of the force P.


2 2 280 120 2 80 120 cos 20P

52.516 lbP From the law of sines

sin sin 20120 P

1 120sin 20sin 128.6052.516

52.5 lbP 31.40 ......................................................................................................Ans.

2-8 The eye bolt shown in Fig. P2-8 is subjected to a 2700 N force and an unknown force P. If the resultant pull is 2000 N in the x-direction, determine the magnitude and direction of the force P.


2 2 22000 2700 2 2000 2700 cos 60P

2426.932 kNP From the law of sines

sin sin 602000 P

1 2000sin 60sin 45.542426.932

2430 kNP 74.46 ..................................................................................................Ans.

2-9 Two forces act on the bracket shown in Fig. P2-9. Determine the angle that will make the vertical component of the resultant of these two forces zero.


2 2 2145 175 2 175 cos50R R

2 224.976 9600 0R R

167.747 lbR or 57.229 lbR From the law of sines

sin sin 50175 145

1 175sin 50sin145

67.60 or 112.40 .............................................................................................Ans.


17

167.7 lb R when 67.60

57.2 lb R when 112.40

2-10 A 270-N force and a 400-N force act at point B of the truss shown in Fig. P2-10. A third force F is to be applied at point B so that the resultant of the three forces is zero. Determine the magnitude of F and its orientation with respect to the 400-N force.

SOLUTION Using the law of cosines

2 2 212 270 400 2 270 400 cos50R

12 306.689 NR


12

12

sin sin 50270 R

112

270sin 50sin 42.41306.689

12 307 NR 42.41

But 12 3R R R 0

Therefore 3 12 307 NR R 42.41 ........................................................................................Ans.

2-11 Three forces are applied to a bracket mounted on a post as shown in Fig. P2-11. Determine (a) The magnitude and direction (angle x) of the resultant R of the three forces. (b) The magnitudes of two other forces Fx and Fy that would have the same resultant. SOLUTION (a) For the system of forces,

1 2 3 12 3R F F F R F

From the parallelogram of forces constructed using F1 and F2

2 2

12 120 140 2 120 140 cos105

206.631 lb

R

112

140sin105sin 40.878206.631

12 206.631 lbR 4.122

For the second parallelogram of forces

2 2100 206.631 2 100 206.631 cos85.878

222.993 lb

R

1 206.632sin85.878sin 67.55222.993

223 lbR 22.45 ........................................................................................................Ans. (b) From the definition of sine and cosine

223.993 cos 22.45 207.0 lb xR ......................................................................Ans.

223.993 sin 22.45 85.5 lb yR ...........................................................................Ans.


18

2-12 Four forces act on a small airplane in flight, as shown in Fig. P2-12; its weight (25 kN), the thrust provided by the engine (10 kN), the lift provided by the wings (24 kN), and the drag resulting from its motion through the air (3 kN). Determine the resultant of the four forces and its line of action with respect to the axis of the plane.

SOLUTION For the system of forces,

1 2 3 4

12 3 4

123 4

R F F F FR F FR F


2 2

12 3 24 2 3 24 cos90

24.1868 kN

R

11

3sin 90sin 7.12524.1868

12 24.1868 kNR 73.875

From the parallelogram of forces constructed using R12 and F3

2 2

123 12 1210 2 10 cos82.875

25.000 kN

R R R

1 122

sin 82.875sin 73.74025.000

R

123 25.000 kNR 83.740

Finally, from the parallelogram of forces constructed using R123 and F4

2 2

123 12325 2 25 cos 6.260

2.730 kN

R R R

1 1233

sin 6.260sin 86.9022.730

R

2.730 kNR 3.088 ..................................................................................................Ans.

13.088 below the -axisx .........................................................................................Ans.

2-13 Determine the resultant force R of the three forces applied to the gusset plate shown in Fig. P2-13. SOLUTION For the system of forces,

1 2 3 12 3R F F F R F


2 2

12 11,000 4500 2 11,000 4500 cos135

14,534.565 lb

R

112

4500sin135sin 12.64614,534.565

12 14,534.565 lbR 77.354


19


2 2

12 129000 2 9000 cos137.354

22,015.69 lb

R R R

1 9000sin137.354sin 16.07822,015.69

22,020 lbR 86.57 ............................................................................Ans.

2-14 Determine the magnitude and direction of the resultant of the three forces shown in Fig. P2-14. SOLUTION For the system of forces,

1 2 3 12 3R F F F R F


2 2

12 5 7.5 2 5 7.5 cos130

11.3780 kN

R

112

7.5sin130sin 30.32811.3780

12 11.3780 kNR 20.328


2 2

12 1210 2 10 cos65.328

11.5961 kN

R R R

1 10sin 65.328sin 51.59511.5961

11.60 kNR 71.92 ..................................................................................................Ans.

2-15 Two forces A and B are applied to an eyebolt as shown in Fig. P2-15. If the magnitudes of the two forces are A = 50 lb and B = 100 lb, calculate and plot the magnitude of the resultant R as a function of the angle A (0

A 180 ). Also calculate and plot the angle R that the resultant makes with the force B as a function of the angle A. When is the resultant a maximum? When is the resultant a minimum? When is the angle R a maximum? Repeat for A = 100 lb and B = 50 lb.


2 2 2 cos AR A B AB


sin 180sin AB

A R

1 sinsin AB

AR

For A = 50 lb and B = 100 lb


20

2 250 100 2 50 100 cos

100 1.25 cos lb

A

A

R

1

1

50sinsin100 1.25 cos

sinsin2 1.25 cos

AB

A

A

A

For A = 100 lb and B = 50 lb

2 2100 50 2 100 50 cos

100 1.25 cos lb

A

A

R

1

1

100sinsin100 1.25 cos

sinsin1.25 cos

AB

A

A

A

2-16 Two forces A and B are applied to a bracket as shown in Fig. P2-16. If the magnitude of the force B is 325 N, calculate and plot the magnitude of the resultant R as a function of the magnitude of the force A (0 A 900 N).


2 2 2325 2 325 cos 75R A A

2105,625 168.2324 NR A A


21

2-17 Determine the magnitudes of the u- and v-components of the 1000-lb force shown in Fig. P2-17. SOLUTION From the law of sines

1000

sin 35 sin100 sin 45u vF F

582 lbuF ........................................................................Ans.

718 lbvF .........................................................................Ans.

2-18 Determine the components of the 3000-N force in the directions of members AB and BC of the truss shown in Fig. P2-18 when = 45 .

SOLUTION From the law of sines

3000

sin15 sin 90 sin 75BCAB FF

776 NABF ......................................................................Ans.

2900 NBCF ....................................................................Ans.

2-19 Two cables are used to support a stop light as shown in Fig. P2-19. The resultant R of the cable forces Fu and Fv has a magnitude of 300 lb and its line of action is vertical. Determine the magnitudes of the forces Fu and Fv.


300

sin 36.87 sin 98.13 sin 45u vF F

181.8 lbuF .....................................................................Ans.

214.3 lbvF .....................................................................Ans.

2-20 Two ropes are used to tow a boat upstream as shown in Fig. P2-20. The resultant R of the rope forces Fu and Fv has a magnitude of 1500 N and its line of action is directed along the axis of the boat. Determine the magnitudes of the forces Fu and Fv.


1500

sin 40 sin110 sin 30u vF F

1026 NuF .......................................................................Ans.

798 NvF .........................................................................Ans.

2-21 Determine the u- and v-components of the 5200-lb force acting on the bracket shown in Fig. P2-21. SOLUTION From the law of sines

5200

sin 22.38 sin135 sin 22.62u vF F

2800 lbuF ...............................................................Ans.

2830 lbvF ...............................................................Ans.


22

2-22 A 5000-N force acts in the vertical direction on the block shown in Fig. P2-22. Determine the components of the force perpendicular to and parallel to the line AB.


5000

sin 60 sin 90 sin 30pa FF

4330 NaF .................................................................................... Ans.

2500 NpF .................................................................................... Ans.

2-23 Two forces Fu and Fv are applied to a bracket as shown in Fig. P2-23. If the resultant R of the two forces has a magnitude of 725 lb and a direction as shown on the figure, determine the magnitudes of the forces Fu and Fv.


725

sin 47.73 sin 49.40 sin82.88u vF F

707 lbuF .................................................................Ans.

948 lbvF ..................................................................Ans.

2-24 A 20-kN force acts on the member shown in Fig. P2-24. Determine the components of the force in the horizontal and vertical directions (FAC and FBC).


20

sin 53.13 sin 90 sin 36.87AC BCF F

16.00 NACF ................................................................................. Ans.

12.00 NBCF ................................................................................. Ans.

2-25 A 500-lb force acts along the line AB shown in Fig. P2-25. Determine the magnitude of the components in the direction of lines AC and AD.


500

sin 40.60 sin108.44 sin 30.96ACAD FF

271 lbACF ........................................................Ans.

343 lbADF ........................................................Ans.


23

2-26 Two forces Fu and Fv are applied to a bracket as shown in Fig. P2-26. If the resultant R of the two forces has a magnitude of 375 N and a direction as shown on the figure, determine the magnitudes of the forces Fu and Fv.


375

sin 70.35 sin 67.38 sin 42.27u vF F

383 NuF ...........................................................Ans.

273 NvF ...........................................................Ans.

2-27 Three forces are applied to a bracket as shown in Fig. P2-27. The magnitude of the resultant R of the three forces is 50 kip. If the force F1 has a magnitude of 30 kip, determine the magnitudes of the forces F2 and F3.


1 2 3 1 23R F F F F R

From the parallelogram of forces constructed using F1 and R23

2 2

23 30 50 2 30 50 cos75

51.2205 kip

R

12

50sin 75sin 70.54651.2205

23 51.2205 kipR 27.546

For the second parallelogram of forces 12 27.546 39.546

38 27.546 10.454

180 130

3 251.2205sin sin sinF F

2 42.6 kipF ....................................................................................................................... Ans.

3 12.13 kipF ..................................................................................................................... Ans.

2-28 A homogeneous cylinder is subjected to the three forces shown in Fig. P2-28. If the resultant of the three forces is zero, determine the magnitude of N1 and the angle of the surface to which N1 is normal.

SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines and the law of cosines gives

2 2

1 2220 1940 2 2220 1940 cos15

609.837 N

N

1 1940sin15sin 55.42609.837

1 610 NN ........................................................................Ans.

55.42 ..........................................................................Ans.


24

2-29 Four forces act on the machine component shown in Fig. P2-29. If the resultant of F1, F2, and F3 is horizontal to the left and has a magnitude of 400 lb, determine the magnitudes of the forces F2 and F3.


1 2 3 1 23 400 lb F F F F R

From the force triangle constructed using F1 and R23

2 223 400 800 894.427 lbR

12

400tan 26.565800

23 894.427 lbR 63.435


32 894.427sin 33.435 sin 75 sin 71.565

FF

2 510 lbF .......................................................................................................................... Ans.

3 878 lbF ........................................................................................................................... Ans.

2-30 A 140-N light is supported by two cables as shown in Fig. P2-30. If the resultant of the three forces is zero, determine the force T1 and the angle .

SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines and the law of cosines gives

2 2

1 140 155 2 140 155 cos50

125.411 N

T

1

sin 40 sin 50140 T

1 140sin 5040 sin 58.777125.411

1 125.4 NT ........................................................................................................................ Ans.

18.77 ............................................................................................................................ Ans.

2-31 Two forces A and B are applied to an eye bolt using ropes as shown in Fig. P2-31. The resultant R of the two forces has a magnitude R = 4000 lb and makes an angle of 30 with the force A as shown. If both of the forces pull on the eye bolt as shown (ropes cannot push on the eye bolt), what is the range of angles ( min B

max) for which this problem has a solution? Calculate and plot the required magnitudes A and B as functions of the angle angles B ( min B max). Why is the magnitude of B a minimum when B = 90 ?


4000

sin sin sin 30A B

180 B

180 30 30B


25

4000sin 30

lbsin 180

B

B

A

4000sin 30 lb

sin 180 B

B

30B or else 0A

180B or else 0B

The magnitude of B is a minimum when B is vertical because that results in the shortest distance from the end of the 4000 lb force to the horizontal.

2-32 Three forces A, B, and C are applied to an eye bolt using ropes as shown in Fig. P2-32. Force A has a magnitude A = 50 N, and the resultant of the three forces is zero. If all of the forces pull on the eye bolt as shown (ropes cannot push on the eye bolt), what is the range of angles ( min C max) for which this problem has a solution? Calculate and plot the required magnitudes B and C as functions of the angle C ( min C max). Why is the magnitude of C a minimum when C = 90 ?

SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines gives

50

sin sin sin 40B C

180 C

180 40 40C

50sin 40

kNsin 180

C

C

B

50sin 40 kN

sin 180 C

C

40C or else 0B

180C or else B C

The magnitude of C is a minimum when C is vertical because that results in the shortest distance from the end of the 50 kN force to the horizontal.

2-33 Determine the x- and y-components of the 1000-lb force shown in Fig. P2-33. SOLUTION

1000cos30 866 lbxF ...............................................................................................Ans.

1000sin 30 500 lbyF ...............................................................................................Ans.

2-34 Determine the x- and y-components of the 800-N force shown in Fig. P2-34. SOLUTION

800sin 25 338 NxF .................................................................................................Ans.

800cos 25 725 NyF .................................................................................................Ans.


26

2-35 Determine the x- and y-components of each force shown in Fig. P2-35. SOLUTION For the 600 lb force

600sin 60 520 lbxF .................................................................................................Ans.

600cos60 300 lbyF .................................................................................................Ans.

For the 800 lb force

800sin 45 566 lbxF ............................................................................................Ans.

800cos 45 566 lbyF .................................................................................................Ans.

2-36 Determine the x- and y-components of each force shown in Fig. P2-36. SOLUTION

1 3 5 950 570 NxF ................................................................................................Ans.

1 4 5 950 760 NyF ...............................................................................................Ans.

2 1 2 800 566 NxF ......................................................................................Ans.

2 1 2 800 566 NyF ............................................................................................Ans.

2-37 Two forces are applied to a post as shown in Fig. P2-37. Determine (a) The x- and y-components of each force. (b) The x’- and y’-components of each force. SOLUTION

(a) 1 500cos70 171.0 lbxF ............................................................................................Ans.

1 500sin 70 470 lbyF ................................................................................................Ans.

2 750cos30 650 lbxF ...............................................................................................Ans.

2 750sin 30 375 lbyF ..........................................................................................Ans.

(b) 1 500sin 25 211.3 lbxF .......................................................................................Ans.

1 500cos 25 453 lbyF ...............................................................................................Ans.

2 750cos15 724 lbxF ...............................................................................................Ans.

2 750sin15 194.1 lbyF .............................................................................................Ans.

2-38 For the 900-N force shown in Fig. P2-38 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. SOLUTION

(a) 900cos35 737.237 NxyF

cos60 369 Nx xyF F ..................................................................................................Ans.

sin 60 638 Ny xyF F ...................................................................................................Ans.

900sin 35 516 NzF ..................................................................................................Ans.

(b) 369 638 516 NF i j k ..........................................................................................Ans.


27

2-39 As an automobile rounds a curve, a force F of magnitude 600 lb is exerted on one of the tires, as shown in Fig. P2-39. Determine the x, y, and z scalar components of the force. The xy plane is parallel to the roadway.

SOLUTION

2 2 2

2 2 12600 97.3 97.3 584 lb2 2 12

F i j kF e i j k

97.3 lbxF ....................................................................................................................... Ans.

97.3 lbyF ......................................................................................................................... Ans.

584 lbzF .......................................................................................................................... Ans.

2-40 A 50-kN force is applied to an eye bolt as shown in Fig. P2-40. (a) Determine the direction angles x, y, and z. (b) Determine the x, y, and z scalar components of the force. (c) Express the force in Cartesian vector form. SOLUTION

(c) 2 2 2

3 2 250 36.38 24.25 24.25 kN3 2 2

F i j kF e i j k

36.4 24.3 24.3 kNF i j k ..................................................................................Ans.

(b) 36.4 kNxF ..................................................................................................................... Ans.

24.3 kNyF ..................................................................................................................... Ans.

24.3 kNzF ....................................................................................................................... Ans.

(a) 1 36.38cos 136.6950x ............................................................................................Ans.

1 24.25cos 119.0150y ............................................................................................Ans.

1 24.25cos 60.9950z ................................................................................................Ans.

2-41 Two forces are applied at a point on a body as shown in Fig. P2-41. Determine (a) The x- and y- components of each force. (b) The x’- and y’- components of each force. SOLUTION

(a) 1 800sin 40 514 lbxF ...........................................................................................Ans.

1 800cos 40 613 lbyF ...............................................................................................Ans.

2 1000cos 20 940 lbxF .............................................................................................Ans.

2 1000sin 20 342 lbyF .............................................................................................Ans.

(b) 1 800sin 70 752 lbxF ..........................................................................................Ans.

1 800cos70 274 lbyF ...............................................................................................Ans.

2 1000cos50 643 lbxF .............................................................................................Ans.

2 1000sin 50 766 lbyF .............................................................................................Ans.


28

2-42 Two forces are applied to an eye bolt as shown in Fig. P2-42. (a) Determine the x, y, and z scalar components of the 30-kN force F1. (b) Express the 30-kN force F1 in Cartesian vector form. (c) Determine the magnitude of the rectangular component of the 30-kN force F1 along the line of action of

the 50-kN force F2. (d) Determine the angle between the two forces. SOLUTION

(a) 1 1 1 2 2 2

5 330 25.355 15.213 5.071 kN5 3 1

F i j kF e i j k

1 25.4 kNxF ...................................................................................................................... Ans.

1 15.21 kNyF .................................................................................................................Ans.

1 5.07 kNzF ...................................................................................................................... Ans.

(b) 1 25.4 15.21 5.07 kNF i j k ..................................................................................Ans.

(c) 2 2 2 2

2 3 2 0.48507 0.72761 0.485072 3 2i j ke i j k

12 1 2 25.355 0.48507 15.213 0.72761 5.071 0.4850725.83 kN

F F e

12 25.8 kNF ...................................................................................................................... Ans.

(d) 1 2 125.83 kN cosFF e

1 25.83cos 30.5730

.................................................................................................Ans.

2-43 A wire is stretched between two pylons, one of which is shown in Fig. P2-43. The 250-lb force in the wire is parallel to the xy-plane and makes an angle of 30 with the y-axis. Point A lies in the xz-plane and point C lies in the yz-plane. Determine

(a) The magnitude of the rectangular component of the 250-lb force in the direction of member AD. (b) The magnitude of the rectangular component of the 250-lb force in the direction of member CD. (c) The angle between members AD and CD. SOLUTION

(a) 250sin 30 250cos30 125.00 216.51 lbFF e i j i j

2 2

12 18 0.55470 0.8320512 18

ADi ke i k

2 2

9 18 0.44721 0.894439 18

CDj ke j k

125 0.55470 216.51 0 0 0.83205AD ADF F e

69.3 lbADF ....................................................................................................................... Ans.

(b) 125 0 216.51 0.44721 0 0.89443CD CDF F e

96.8 lbCDF ....................................................................................................................... Ans.

(c) 1cos 41.91AD CDe e ..........................................................................................Ans.


29

2-44 The hot-air balloon shown in Fig. P2-44 is tethered with three mooring cables. The force TA in cable AD has a magnitude of 1860 N.

(a) Express TA in Cartesian vector form. (b) Determine the magnitude of the rectangular component of the force TA along BD. (c) Determine the angle between cables AD and BD. SOLUTION

(a) 2 2 2

20 30 501860 603 905 1509 N20 30 50

A A AT i j kT e i j k ..........................Ans.

(b) 2 2 2

16 25 50 0.27517 0.42995 0.8599016 25 50

BDi j ke i j k

/ 603 0.27517 905 0.42995 1509 0.85990A BD A BDT T e

/ 1074 NA BDT .................................................................................................................... Ans.

(c) 1 1 1 1074cos cos cos 54.721860

A BDAD BD

ATT ee e ..........................Ans.

2-45 A system of three cables supports the cylinder shown in Fig. P2-45. The magnitude of the force T1 in cable AO is 2000 lb. Determine

(a) The magnitude of the rectangular component of the force T1 along the line OB. (b) The angle between the force T1 and the line OB. SOLUTION

(a) 1 1 1 2 2 2

4 3 62000 1024.30 768.22 1536.44 lb4 3 6

T i j kT e i j k

2 2 2

4 6 4 0.48507 0.72761 0.485074 6 4

OBi j ke i j k

1 807.39 lb 807 lbOB OBT T e .................................................................................Ans.

(b) 1 807.39 lb 2000 1 cosOBT e

1 807.39cos 66.192000

..........................................................................................Ans.

2-46 A 2000-N force F acts on a machine component as shown in Fig. P2-46. (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. (c) Determine the angle between the force and the line AB. SOLUTION

(a) 2 2 2

8 14 102000 843.27 1475.73 1054.09 N8 14 10

F i j kF e i j k

843 NxF ........................................................................................................................... Ans.

1476 NyF ........................................................................................................................ Ans.

1054 NzF ...................................................................................................................... Ans.

(b) 843 1476 1054 NF i j k ......................................................................................Ans.


30

(c) 2 2

150 130 0.75569 0.65493150 130

ABi je i j

1603.75 N 2000 1 cosAB ABF F e

1 1603.75cos 36.692000

.............................................................................................Ans.

2-47 Two flower pots are supported with wires as shown in Fig. P2-47. If = 6 , the tension force T1 has a magnitude of 13 lb, and the tension force T2 has a magnitude of 9 lb, determine the magnitude and orientation of the resultant of the forces T1 and T2.

SOLUTION

1 13cos 45 13sin 45 9.1924 9.1924 lbT i j i j

2 9cos6 9sin 6 8.9507 0.94076 lbT i j i j

1 2 9.1924 8.9507 9.1924 0.94076

0.2417 10.1332 lb

R T T i j

i j

10.14 lbR 88.63 .....................................................................................................Ans.

2-48 Two forces are applied to an eye bolt as shown in Fig. P2-48. Determine the magnitude of the resultant R of the two forces and the angle x between the line of action of the resultant and the x-axis.

SOLUTION

1 500cos60 500sin 60 250.00 433.01 NF i j i j

2 375cos30 375sin 30 324.76 187.50 NF i j i j

1 2 250 324.76 433.01 187.50

574.76 620.51 N

R F F i j

i j

846 NR 47.19 ........................................................................................................Ans.

2-49 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P2-49. The forces T1 and T2 in the two segments of the cable each have a magnitude of 650 lb. If the resultant of the three forces T1, T2, and P is to be zero, determine the magnitude of P.

SOLUTION

1 650cos5 650sin 5 647.53 56.65 lbT i j i j

2 650cos5 650sin 5 647.53 56.65 lbT i j i j

1 2 113.30R T T P j P 0

113.3 lbP j

113.3 lbP ......................................................................................................................... Ans.

2-50 Three forces act on the structural member shown in Fig. P2-50. Determine the resultant of the forces and express the result in Cartesian vector form.

SOLUTION

2500cos30 900 1265 NxR

2000 2500sin 30 3250 NyR

1265 3250 NR i j ....................................................................................................Ans.


31

2-51 Express the resultant of the two forces shown in Fig. P2-51 in Cartesian vector form, and determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes.

SOLUTION

1 100cos 44 71.934 lbxF

1 0 lbyF

1 100sin 44 69.466 lbzF

2 200cos53 120.363 lbxyF

2 2 cos70 41.167 lbx xyF F

2 2 sin 70 113.104 lby xyF F

2 200sin 53 159.727 lbzF

1 2 113.101 113.104 229.193 lbR F F i j k

113.1 113.1 229 lbR i j k ....................................................................................Ans.

279.49 lbR

1 113.101cos 66.13279.49x ............................................................................................Ans.

1 113.104cos 66.13279.49y ............................................................................................Ans.

1 229.193cos 34.91279.49z ............................................................................................Ans.

2-52 Determine the magnitude R of the resultant of the two forces shown in Fig. P2-52. Also determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes.

SOLUTION

1 10cos67 3.9073 kNxF

1 0 kNyF

1 10sin 67 9.2051 kNzF

2 0 kNxF

2 20cos60 10.0000 kNyF

2 20sin 60 17.3205 kNzF

1 2 3.907 10.000 26.526 kNR F F i j k

28.616 28.6 kNR .......................................................................................................Ans.

1 3.907cos 82.1528.616x ..............................................................................................Ans.

1 10cos 69.5528.616y .............................................................................................Ans.

1 26.526cos 22.0328.616z .............................................................................................Ans.


32

2-53 Four forces are applied to the block shown in Fig. P2-53. Determine the magnitude of the resultant and the angle between the resultant and the x-axis.

SOLUTION

15 700 649.93 lb29xF 1

2 700 259.97 lb29yF

23 300 154.35 lb34xF 2

5 300 257.25 lb34yF

32 600 222.83 lb29xF 3

5 600 557.09 lb29yF

42 900 804.98 lb5xF 4

1 900 402.49 lb5yF

223.54 lbxR 1476.80 lbyR

1494 lbR 81.39 ......................................................................................................Ans.

2-54 Determine the magnitude and orientation of the resultant of the two forces applied to the eye bolt shown in Fig. P2-54.

SOLUTION

1 2 2

120 100 44.7214 N120 240

xF 1 2 2

240 100 89.4427 N120 240

yF

1 0 NzF 2 0 NxF

2 2 2

90 75 33.5410 N90 180

yF 2 2 2

180 75 67.0820 N90 180

zF

1 2 44.721 122.984 67.082 NR F F i j k

147.054 147.1 NR .....................................................................................................Ans.

1 44.721cos 72.30147.054x ............................................................................................Ans.

1 122.984cos 33.25147.054y ............................................................................................Ans.

1 67.082cos 62.86147.054z ............................................................................................Ans.

2-55 The three forces acting on the machine component shown in Fig. P2-55 have magnitudes F1 = 500 lb, F2 = 300 lb, and F3 = 200 lb. Determine the resultant R of the three forces and express the resultant in Cartesian vector form. Also, determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes.

SOLUTION

1 1 1 2 2 2

2 35 40500 18.80 329.02 376.02 lb2 35 40

F i j kF e i j k

2 2 2 2 2 2

9 22 20300 86.92 212.46 193.15 lb9 22 20

F i j kF e i j k

3 200 lbF j


33

1 2 3 105.72 83.44 569.17 lbR F F F i j k

105.7 83.4 569 lbR i j k ..................................................................................Ans.

584.89 lbR

1 105.72cos 100.41584.89x .........................................................................................Ans.

1 83.44cos 98.20584.89y .............................................................................................Ans.

1 569.17cos 166.69584.89z .........................................................................................Ans.

2-56 Determine the magnitude R of the resultant of the three forces shown in Fig. P 2-56. Also determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes.

SOLUTION

1 10cos 26 cos 42 6.679 kNxF 1 10sin 26 4.384 kNzF

1 10cos 26 sin 42 6.014 kNyF

2 16cos 40 sin 35 7.030 kNxF 2 16sin 40 10.285 kNzF

2 16cos 40 cos35 10.040 kNyF

3 24cos50 cos60 7.713 kNxF 3 24sin 50 18.385 kNzF

3 24cos50 sin 60 13.360 kNyF

1 2 3 7.362 2.694 33.054 kNR F F F i j k

33.971 34.0 kNR .......................................................................................................Ans.

1 7.362cos 77.4833.971x ..............................................................................................Ans.

1 2.694cos 94.5533.971y .............................................................................................Ans.

1 33.054cos 13.3433.971z ..............................................................................................Ans.

2-57 Three forces are applied at the corner of the box shown in Fig. P2-57. Determine the magnitude R of the resultant and the angles x, y, and z between the line of action of of the resultant and the positive coordinate axes.

SOLUTION

1 1 1 2 2 2

8 15 2010 3.0478 5.7145 7.6194 kip8 15 20

F i j kF e i j k

2 2 2 2 2 2

16 15 2020 10.7811 10.1073 13.4763 kip16 15 20

F i j kF e i j k

3 3 3 2 2 2

16 15 525 17.7822 16.6708 5.5569 kip16 15 5

F i j kF e i j k


34

1 2 3 31.611 32.493 26.653 kipR F F F i j k

52.587 52.6 kipR .......................................................................................................Ans.

1 31.611cos 53.0552.587x ..............................................................................................Ans.

1 32.493cos 128.1652.587y .........................................................................................Ans.

1 26.653cos 59.5552.587z ..............................................................................................Ans.

2-58 The pin A shown in Fig. P2-58 supports a load F of magnitude 1250 N, and is held in place by a wire AD and compression members AB and AC. If the magnitudes of the forces CB and CC in the compression members are 995 N and 700 N, respectively, and the resultant of the four forces is zero, determine the magnitude of the force TD.

SOLUTION

2 2 2

2 6 2995 300.00 900.01 300.00 N2 6 2

Bi j kC i j k

2 2 2

3 6 2700 300.00 600.00 200.00 N3 6 2

Ci j kC i j k

2 2

6 3 0.89443 0.44721 N6 3

D D D DT T Tj kT j k

1250 NF k

0xR

1500.01 0.89443 0y DR T

750.00 0.44721 0z DR T

1677 NDT ........................................................................................................................ Ans.

2-59 Three forces are applied to an eye bolt as shown in Fig. P2-59. Determine the magnitude R of the resultant of the forces and the angle x between the line of action of the resultant and the x-axis.

SOLUTION

12 5000 4472.136 lb5xF 1

1 5000 2236.068 lb5yF

21 2000 894.427 lb5xF 2

2 2000 1788.854 lb5yF

32 1000 894.427 lb5xF 3

1 1000 447.214 lb5yF

2683.28 lbxR 4472.14 lbyR

5220 lbR 59.04 ......................................................................................................Ans.


35

2-60 Two forces, F1 and F2, are applied to an eye bolt as shown in Fig. P2-60. Determine (a) The magnitude and direction (angle x) of the resultant R of the two forces. (b) The magnitudes of two other forces Fu and Fv (along the axes u and v) that would have the same

resultant. SOLUTION

(a) 1 80cos15 77.2741 NxF 1 80sin15 20.7055 NyF

2 60cos 26 53.9276 NxF 2 60sin 26 26.3023 NyF

23.3465 NxR 47.0078 NyR

52.5 NR 63.59 .......................................................................................................Ans.

(b) 63.5886 45 18.5886

70 45 115

180 46.4114

52.4861

sin sin sinu vF F

41.9 NuF ......................................................................................................................... Ans.

18.46 NvF ........................................................................................................................ Ans.

2-61 Three forces are applied at a point on a body as shown in Fig. P2-61. Determine the resultant R of the three forces and the angles x, y, and z between the line of action of the resultant and the positive x-, y-, and z-coordinate axes.

SOLUTION

1 2 2

2 2500 353.553 353.553 lb2 2j kF j k

2 2 2

4 4800 565.685 565.685 lb4 4i jF i j

3 2 2

2 2700 494.975 494.975 lb2 2i kF i k

1 2 3 1060.660 919.238 848.528 lbR F F F i j k

1061 919 849 lbR i j k .........................................................................................Ans.

1640.121 lbR

1 1060.660cos 49.711640.121x ..........................................................................................Ans.

1 919.238cos 55.911640.121y ..........................................................................................Ans.

1 848.528cos 58.841640.121z ..........................................................................................Ans.


36

2-62 Three forces are applied at a point on a body as shown in Fig. P2-62. Determine the resultant R of the three forces and the angles x, y, and z between the line of action of the resultant and the positive x-, y-, and z-coordinate axes.

SOLUTION

1 2 2 2

3 2 330 19.1881 12.7920 19.1881 kN3 2 3i j kF i j k

2 2 2 2

3 4 1.520 11.4939 15.3252 5.7470 kN3 4 1.5i j kF i j k

3 2 2

1.5 425 8.7781 23.4082 kN1.5 4

i jF i j

1 2 3 39.4601 51.5254 24.9351 kNR F F F i j k

39.5 51.5 24.9 kNR i j k .................................................................................Ans.

69.5250 kNR

1 39.4601cos 124.5869.5250x .......................................................................................Ans.

1 51.5254cos 42.1769.5250y ...........................................................................................Ans.

1 24.9351cos 68.9869.5250z ............................................................................................Ans.

2-63 Three forces are applied to a stalled automobile as shown in Fig. P2-63. Determine the magnitude of the force F3 and the magnitude of the resultant R if the line of action of the resultant is along the x-axis.

SOLUTION

1 50cos 48 33.4565 lbxF 1 50sin 48 37.1572 lbyF

2 30cos 23 27.6152 lbxF 2 30sin 23 11.7219 lbyF

3 3 3cos37 0.79864 lbxF F F 3 3 3sin 37 0.60182 lbyF F F

333.4565 27.6152 0.79864 lbxR F R

337.1572 11.7219 0.60182 0 lbyR F

3 81.2 lbF .......................................................................................................................... Ans.

125.9 lbR ......................................................................................................................... Ans. 2-64 Three cables are used to drag a heavy crate along a horizontal surface as shown in Fig. P2-64. The resultant

R of the forces has a magnitude of 2800 N and its line of action is directed along the x-axis. Determine the magnitudes of the forces F1 and F3.

SOLUTION

1 1 1cos30 0.86603 NxF F F 1 1 1sin 30 0.50000 NyF F F

2 1600cos10 1575.6924 NxF 2 1600sin10 277.8371 NyF

3 3 3cos 45 0.70711 NxF F F 3 3 3sin 45 0.70711 NyF F F

1 30.86603 1575.6924 0.70711 2800 NxR F F


37

1 30.50000 277.8371 0.70711 0 NyR F F

1 1100 NF ......................................................................................................................... Ans.

3 385 NF ........................................................................................................................... Ans.

2-65 Two forces are applied at a point in a body as shown in Fig. P2-65. Determine (a) The magnitude and direction (angles x, y, and z) of the resultant R of the two forces. (b) The magnitude of the rectangular component of the force F1 along the line of action of the force F2. (c) The angle between forces F1 and F2. SOLUTION

(a) 1 1 1 2 2 2

1.5 6 4.5150 29.4174 117.6697 88.2523 lb1.5 6 4.5

F i j kF e i j k

21.5120cos60 36.0000 lb2.5xF

22120cos60 48.0000 lb

2.5yF

2 120sin 60 103.9231 lbzF

65.4174 69.6697 192.1754 lbR i j k

214.6269 lb 215 lbR ................................................................................................Ans.

1 65.4174cos 72.25214.6269x .........................................................................................Ans.

1 69.6697cos 71.06214.6269y .........................................................................................Ans.

1 192.1754cos 26.44214.6269z .........................................................................................Ans.

(b) 1 21/ 2

2

4582.333 38.2 lb120

FF

F F................................................................................Ans.

(c) 1 11 2

2

4582.333cos cos 75.25150 120F

F F.............................................................Ans.

2-66 Three forces are applied with cables to the anchor block shown in Fig. P2-66. Determine (a) The magnitude and direction (angles x, y, and z) of the resultant R of the three forces. (b) The magnitude of the rectangular component of the force F1 along the line of action of the force F2. (c) The angle between forces F1 and F3. SOLUTION

(a) 1 2 2 2

2.4 2.7 3.6136 64.000 72.000 96.000 N2.4 2.7 3.6

i j kF i j k

2 2 2 2

0.6 1.8 2.7250 45.455 136.364 204.545 N0.6 1.8 2.7

i j kF i j k

3 2 2 2

3.6 1.2 0.9325 300.000 100.000 75.000 N3.6 1.2 0.9

i j kF i j k

409.455 164.364 375.545 NR i j k


38

579.399 N 579 NR ...................................................................................................Ans.

1 409.455cos 45.03579.399x ...........................................................................................Ans.

1 164.364cos 106.48579.399y .......................................................................................Ans.

1 375.545cos 49.60579.399z ............................................................................................Ans.

(b) 1 21/ 2

2

12,727.232 50.9 N250

FF

F F............................................................................Ans.

(c) 1 11 3

1 3

19, 200cos cos 64.25136 325FF

F F............................................................Ans.


40

3-4 A 750-kg body is supported by the flexible cable system shown in Fig. P3-4. Determine the tensions in cables AC, BC, and CD.

SOLUTION From a free-body diagram for the body

0 :yF 750 9.81 0CDT

7357.50 7360 NCDT .......................... Ans.

From a free-body diagram of the ring, the equations of equilibrium

0 :xF cos60 cos30 0BC ACT T

0 :yF sin 60 sin 30 7357.50 0BC ACT T

are solved to get

1.73205BC ACT T

7360 NACT ..................................................................................................................................Ans.

12,740 NBCT ..............................................................................................................................Ans.

3-5 An 800-lb homogeneous cylinder is supported by two rollers as shown in Fig. P3-5. Determine the forces exerted by the rollers on the cylinder. All surfaces are smooth (frictionless).

SOLUTION The equations of equilibrium

0 :xF sin 30 sin 30 0A BN N

0 :yF cos30 cos30 800 0A BN N

are solved to get

A BN N

462 lbAN ..............................................................................Ans.

462 lbBN ..............................................................................Ans.

3-6 A worker is using a hoist and cable to lift a 175-kg engine from a car as shown in Fig. P3-6. Determine the forces in the three cables attached to the ring.

SOLUTION From a free-body diagram of the ring, the equations of equilibrium

0 :xF 2 1cos10 sin10 0T T

0 :yF 1 2cos10 sin10 175 9.81 0T T

are solved to get

1 25.67128T T

1 1799 NT ...................................................................................... Ans.

2 317 NT ....................................................................................... Ans.

3-7 The lightweight collar A shown in Fig. P3-7 is free to slide on the smooth rod BC. Determine the forces exerted on the collar by the cable and by the rod when the 900 lb downward force F is applied to the collar.

SOLUTION From a free-body diagram of the collar, the equations of equilibrium


41

0 :xF cos 20 sin 30 0T N

0 :yF sin 20 cos30 900 0T N

are solved to get 1.87939N T

700 lbT ............................................................................................... Ans.

1316 lbN ............................................................................................ Ans.

3-8 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P3-8. When a 500-N sideways force P is applied to the cable, the cable is pulled 5 to the side as shown. For this position, determine the x- and y-components of the cable force being applied to the automobile.


0 :xF 2 1cos5 cos5 0T T

0 :yF 1 2500 sin 5 sin 5 0T T

are solved to get

1 2 2868.43 NT T

The x- and y-components of the cable force being applied to the automobile are then

1 1 cos5 2860 NxT T ..................................................Ans.

1 1 sin 5 250 NyT T .....................................................Ans.

3-9 Two flower pots are supported with cables as shown in Fig. P3-9. If pot A weighs 10 lb and pot B weighs 8 lb, determine the tension in each of the cables and the slope of cable BC.

SOLUTION From a free-body diagram of the upper ring

0 :xF cos 45 cos 0CD BCT T (a)

0 :yF sin 45 sin 8 0CD BCT T (b)

From a free-body diagram for the lower ring

0 :xF cos cos 45 0BC ABT T (c)

0 :yF sin 45 sin 10 0AB BCT T (d)

Adding equations (a) and (c) gives

cos 45 cos 45 0CD ABT T

CD ABT T

Then adding Eqs. (b) and (d) gives

sin 45 sin 45 18 lbCD ABT T

12.7279 12.73 lbCD ABT T ............... Ans.

Now Eqs. (b) and (a) can be written

sin 1.0000BCT (e)

cos 9.0000BCT (f)

Dividing Eq. (e) by Eq. (f) gives


42

tan 1 9

6.340 .................................................... 9.06 lbBCT ........................................................Ans.

3-10 Three smooth homogeneous cylinders A, B, and C are stacked in a box as shown in Fig. P3-10. Each cylinder has a diameter of 250 mm and a mass of 245 kg. Determine

(a) The force exerted by cylinder B on cylinder A. (b) The forces exerted on cylinder B by the vertical and horizontal surfaces at D and E. SOLUTION

245 9.81 2403.45 NW

(a) From a free-body diagram of cylinder A, the equations of equilibrium

0 :xF cos 40 cos 40 0AB ACN N

0 :yF sin 40 sin 40 2403.45 0AB ACN N

are solved to get

1869.55 1870 NAB ACN N ............. Ans.

(b) From a free-body diagram of cylinder B, the equations of equilibrium

0 :xF cos 40 0D ABN N

0 :yF sin 40 2403.45 0E ABN N

are solved to get

1432 NDN .....................................................................Ans.

3610 NEN .....................................................................Ans.

3-11 Three smooth homogeneous cylinders A, B, and C are stacked in a V-shaped trough as shown in Fig. P3-11. Each cylinder weighs 100 lb and has a diameter of 5 in. Determine the minimum angle for equilibrium.

SOLUTION From a free-body diagram of cylinder A, the equations of equilibrium

0 :xF sin 30 sin 30 0AB ACN N

0 :yF cos30 cos30 100 0AB ACN N

are solved to get

57.735 lbAB ACN N

From a free-body diagram of cylinder B, the equations of equilibrium are

0 :xF sin 57.735sin 30 0B BCF N

0 :yF cos 100 57.735cos30 0BF

But the contact force between cylinders B and C cannot be negative. Therefore, the minimum angle corresponds to 0BCN and

sin 28.868 lbBF

cos 150.000 lbBF

sin 28.868tan 0.19245cos 150.000

B

B

FF

10.89 ........................................................................................................................................Ans.


43

3-12 A 250-kg body is supported by the flexible cable system shown in Fig. P3-12. Determine the tensions in cables A, B, C, and D.

SOLUTION From a free-body diagram of the lower ring, the equations of equilibrium

0 :xF cos60 0C DT T

0 :yF sin 60 250 9.81 0CT

are solved to get

2831.90 2830 NCT ..................................................Ans.

1415.95 1416 NDT ...................................................Ans.

From a free-body diagram of the upper ring, the equations of equilibrium

0 :xF cos 40 cos30 2831.90cos 60 0A BT T

0 :yF sin 40 sin 30 2831.90sin 60 0A BT T

are solved to get

1507 NAT .......................................................................Ans.

2970 NBT .......................................................................Ans.

3-13 A 500-lb lawn roller is to be pulled over a curb as shown in Fig. P3-13. Determine the minimum pulling force that must be applied by the man to just start the 3-ft diameter roller over the curb. Also determine the angle which gives the minimum pulling force. Assume that the pulling force is along the handle, which makes an angle of with the horizontal.

SOLUTION From a free-body diagram of the roller, the equations of equilibrium are

0 :xF cos cos 0CT N (a)

0 :yF sin sin 500 0F CN N T (b)

where

1sin 12 18 41.810

The roller just begins to roll over the curb when 0FN . Therefore, Eqs. (a) and (b) become

sin 500 0.66667 CT N (c)

cos 0.74536 CT N (d)

Dividing Eq. (c) by Eq. (d) gives

500 0.66667tan

0.74536C

C

NN

.......................................(e)

Solving Eqs. (d) and (e) by trial and error for the angle that makes the tension a minimum NC T 40 386.968 376.519 45 354.101 373.257 48.19 333.332 372.679 50 321.553 372.865 55 288.825 375.327

min 373 lb @ 48.19T ........................................................................................................... Ans.


44

3-14 In order to hold a 130-kg crate in a stationary position, a worker exerts a force P at an angle on a rope as shown in Fig. P3-14. Determine the force exerted by the worker when = 20 .


0 :xF sin 8 cos 20 0ABT P

0 :yF cos8 sin 20 130 9.81 0ABT P

are solved to get

6.75197ABT P

201.0 NP ..............................................................................Ans.

1357 NABT

3-15 A farmer is extracting a post from the ground using the structure shown in Fig. P3-15. What force must the farmer apply to the cable system if the force required to remove the post is 2000 lb?

SOLUTION From a free-body diagram of the right ring, the equations of equilibrium

0 :xF sin15 0CE BCT T

0 :yF cos15 2000 0CET

are solved to get

535.898 lbBCT

2070.552 lbCET

From a free-body diagram of the left ring, the equations of equilibrium

0 :xF 535.898 cos15 0ABT

0 :yF sin15 0ABT P

are solved to get

554.803 lbABT 143.6 lbP ....................................................................................................................................Ans.

3-16 A continuous cable is used to support two blocks as shown in Fig. P3-16. Block A is supported by a small wheel that is free to roll on the cable. Determine the displacement y of block A for equilibrium if the masses of blocks A and B are 22 kg and 34 kg, respectively.

SOLUTION From a free-body diagram of the hanging weight, the equations of equilibrium

0 :yF 34 9.81 0T

are solved to get

333.54 NT Since the pulleys are free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 22 kg weight, the equations of equilibrium

0 :xF 2 1cos cos 0T T


46

Adding Eqs. (a) and (c) together gives

sin 60 sin 30 0 NB AN N (e)

while adding Eqs. (b) and (d) together gives

cos60 cos30 1000 NB AN N (f)

Solving Eqs. (e) and (f) gives

866.03 N 866 NAN ........................... Ans.

500.00 N 500 NBN ........................... Ans.

Then Eqs. (b) and (a) can be written

sin 50 NT (g)

cos 433.013 NT (h) Dividing Eq. (g) by Eq. (h) gives

50tan

433.013

6.587 .......................................................................................................................................Ans.

436 NT .......................................................................................................................................Ans.

3-19 Concrete is to be moved from a mixer to the second floor of a building under construction using a container as shown in Fig. P3-19. The container and its contents weigh 3000 lb, and it is supported by three cables equally spaced around the top of the 4-ft diameter container. Determine the force in each cable.

SOLUTION The coordinates of points A, B, C, and D are:

A: 2cos30 1.73205 ftx

2sin 30 1.0000 fty

B: 0 ftx

2.0000 fty

C: 2cos30 1.73205 ftx

2sin 30 1.0000 fty

D: 0 ftx y

4.5sinz h where

1cos 2 4.5 63.612

Therefore

4.0311 fth

2 2 2

1.73205 4.0311 0.38490 0.22222 0.895811.73205 1 4.0311

A A A A AT T T Ti j kT i j k

0.38490 0.22222 0.89581C C C CT T TT i j k

2 2

2 4.0311 0.44445 0.895812 4.0311

B B B BT T Tj kT j k

Then, the equations of equilibrium

0 :xF 0.38490 0.38490 0A CT T


47

0 :yF 0.22222 0.44445 0.22222 0A B CT T T

0 :zF 0.89581 0.89581 0.89581 3000 0A B CT T T

are solved simultaneously to get

1116 lbA B CT T T .................................................................................................................Ans.

3-20 A mass m is to be supported by two cables (A and B) as shown Fig. P3-20. If the maximum force that the cables can withstand is 15 kN, determine the maximum mass m that can be supported.

SOLUTION From a free-body diagram of the ring, the equations of equilibrium are

0 :xF cos 25 cos 40 0B AT T

0 :yF sin 40 sin 25 0A BT T mg

If max 15 kNA AT T , then

12.6785 kNBT

which is okay since 12.6785 kN 15 kNBT . Then

sin 40 sin 25 15 kNA Bmg T T

1529 kgm ....................................................................................................................................Ans.

3-21 The hot-air balloon shown in Fig. P3-21 is tethered with three mooring cables. If the net lift of the balloon is 900 lb, determine the force exerted on the balloon by each of the three cables.

SOLUTION

2 2 2

20 30 5020 30 50

0.32444 0.48666 0.81111

A A

A A A

T

T T T

i j kT

i j k

2 2 2

16 25 5016 25 50

0.27517 0.42995 0.85990

B B

B B B

T

T T T

i j kT

i j k

2 2 2

25 15 5025 15 50

0.43193 0.25916 0.86387

C C

C C C

T

T T T

i j kT

i j k


0 :xF 0.32444 0.27517 0.43193 0A B CT T T

0 :yF 0.48666 0.42995 0.25916 0A B CT T T

0 :zF 0.81111 0.85990 0.86387 900 0A B CT T T


418 lbAT ...........................................................................................................................Ans.

205 lbBT ...........................................................................................................................Ans.

445 lbCT ...........................................................................................................................Ans.


48

3-22 A 100-kg traffic light is supported by a system of cables as shown in Fig. P3-22. Determine the tensions in each of the three cables.

SOLUTION

2 2 2

4 8 5 0.39036 0.78072 0.487954 8 5


2 2 2

6 8 5 0.53666 0.71554 0.447216 8 5

B B B B BT T T Ti j kT i j k

2 2

8 5 0.84800 0.530008 5

C C C CT T Tj kT j k


0 :xF 0.39036 0.53666 0A BT T

0 :yF 0.78072 0.71554 0.84800 0A B CT T T

0 :zF 0.48795 0.44721 0.53000 100 9.81 0A B CT T T


603 NAT ...........................................................................................................................Ans.

439 NBT ...........................................................................................................................Ans.

925 NCT ...........................................................................................................................Ans.

3-23 A 250-lb force is applied to the joint at the top of the structure shown in Fig. P3-23. The joint is held in position by the slender members AD, BD, and CD, which can only exert forces that act along the members. If the 250-lb force is in the xy-plane, determine the forces in members AD, BD, and CD.

SOLUTION

250sin 30 250cos30 125 216.506 lbF i j i j

2 2

12 18 0.55470 0.8320512 18

AD AD AD ADT T Ti kT i k

BD BDTT k

2 2

9 18 0.44721 0.894439 18

CD CD CD CDT T Tj kT j k


0 :xF 0.55470 125 0ADT

0 :yF 0.44721 216.506 0CDT

0 :zF 0.83205 0.89443 0AD BD CDT T T


225 lbADT .......................................................................................................................Ans.

621 lbBDT .........................................................................................................................Ans.

484 lbCDT .......................................................................................................................Ans.


49

3-24 Three cables are used to support a 250-kg homogeneous plate as shown in Fig. P3-24. Determine the force in each of the three cables.

SOLUTION

2 2 2

0.5 0.6 1.5 0.29566 0.35479 0.886970.5 0.6 1.5


2 2 2

0.5 0.6 1.5 0.29566 0.35479 0.886970.5 0.6 1.5


2 2

0.5 1.5 0.31623 0.948680.5 1.5

C C C CT T Ti kT i k


0 :xF 0.29566 0.29566 0.31623 0A B CT T T

0 :yF 0.35479 0.35479 0A BT T

0 :zF 0.88697 0.88697 0.94868 250 9.81 0A B CT T T


691 NAT ............................................................................................................................Ans.

691 NBT ............................................................................................................................Ans.

1293 NCT .........................................................................................................................Ans.

3-25 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-25. Determine the magnitude and the coordinate direction angles of the unknown force F4.

SOLUTION

1 4 5 80 3 5 80 64 48 lbF i j i j

2 150 lbF k

3 200cos30 200sin 30 173.205 100.000 lbF j k j k


0 :xF 464 0xF

0 :yF 448 173.205 0yF

0 :zF 4150 100 0zF


4 64.000 lbxF 4 125.205 lbyF 4 50.000 lbzF

4 149.239 lb 149.2 lbF ...............................................................................................Ans.

1 64cos 115.39149.239x ..........................................................................................Ans.

1 125.205cos 147.03149.239y .......................................................................................Ans.

1 50cos 70.43149.239z ............................................................................................Ans.


50

3-26 A pair of steel pipes is stacked in a box as shown in Fig. P3-26. The masses and diameters of the smooth pipes are mA = 5 kg, mB = 20 kg, dA = 100 mm, and dB = 200 mm. Plot the two forces exerted on pipe A (by pipe B and by the side wall) as a function of the distance b between the walls of the box (200 mm b 300 mm). Determine the range of b for which

(a) The force at the side wall is less than WA, the weight of pipe A. (b) Neither of the two forces exceed 2WA. (c) Neither of the two forces exceed 4WA. SOLUTION From a free-body diagram of pipe A, the equations of equilibrium are

0 :xF sin 0A ABN F

0 :yF cos 5 9.81 0ABF

where

1 150sin150b

Therefore

49.05 NcosABF ............................................ Ans.

sin NA ABN F ........................................ Ans.

(a) 50 NAN for 260 mmb ..................... Ans.

(b) 100 NABF for 280 mmb .................. Ans.

(c) 200 NABF for 295 mmb .................. Ans.

3-27 A 75-lb stoplight is suspended between two poles as shown in Fig. P3-27. Neglect the weight of the flexible cables and plot the tension in both cables as a function of the sag distance d (0 d 8 ft). Determine the minimum sag d for which both tensions are less than

(a) 100 lb (b) 250 lb (c) 500 lb SOLUTION From a free-body diagram of the ring, the equations of equilibrium are

0 :xF cos cos 0B B A AT T

0 :yF sin sin 75 0A A B BT T

where

1tan20Ad

1tan10Bd

Therefore

cos

cosA A

BB

TT

75cos 75cos lb

sin cos sin cos sinB B

AA B B A A B

T ........................................................Ans.


51

75cos lb

sinA

BA B

T ................................................................................................................Ans.

(a) 100 lbBT for 5.6 ftd ......................... Ans.

(b) 250 lbBT for 2.0 ftd ......................... Ans.

(c) 500 lbBT for 1.0 ftd ......................... Ans.

3-28 A 50-kg load is suspended from a pulley as shown in Fig. P3-28. The tension in the flexible cable does not change as it passes around the small frictionless pulleys, and the weight of the cable may be neglected. Plot the force P required for equilibrium as a function of the sag distance d (0 d 1 m). Determine the minimum sag d for which P is less than

(a) Twice the weight of the load. (b) Four times the weight of the load. (c) Eight times the weight of the load. SOLUTION From a free-body diagram of pulley supporting the load, the equations of equilibrium

0 :xF cos cos 0C AT T

0 :yF sin sin 50 9.81 0A CT T

give

A C

49.05 N

2sin A

P T .................................... Ans.

Since the two angles are equal, the pulley is midway between the supports, and

1tan1.5Ad

(a) 1000 NP for 380 mmd ................... Ans.

(b) 2000 NP for 185 mmd .................... Ans.

(c) 4000 NP for 92 mmd ...................... Ans.


52

3-29 A worker positions a 250-lb crate by pulling on the rope BD as shown in Fig. P3-29. The 3-ft long rope BD is horizontal ( = 0) when the 5-ft long rope AB is vertical ( = 0).

(a) What is the maximum distance bmax that the crate can be pulled to the side using this arrangement? (b) Calculate and plot the forces in ropes AB and BD as a function of the distance b for 0 b bmax. (c) How could the worker pull the crate to the side more than the bmax calculated in part a? SOLUTION From a free-body diagram of the ring, the equations of equilibrium are

0 :xF sin cos 0AB BDT T

0 :yF cos sin 250 0AB BDT T

where

1sin5b

1 5 1 cossin

3

Therefore

cos

sinBD

ABTT

250sin lb

cos cos sin sinBDT ............Ans.

250cos lb

cos cos sin sinABT ............Ans.

(a) maxb occurs when BDT goes negative (after it goes to infinity);

max 3.90 ftb ..................................... Ans.

(c) To pull further to the side, the worker needs a longer rope to pull on or he needs to attach his rope lower - closer to the crate.................... Ans.

3-30 Two small wheels are connected by a light-weight rigid rod as shown in Fig. P3-30. Plot the angle (between the rod and the horizontal) as a function of the weight W1 (0.25W2 W1 10W2). Determine the weight W1 for which

(a) = 50 (b) = 10 (c) = 25 SOLUTION From a free-body diagram of body 1, the equations of equilibrium are

0 :xF 1cos sin 30 0T N (a)

0 :yF 1 1cos30 sin 0N T W (b)

From a free-body diagram of body 2, the equations of equilibrium are

0 :xF 2 sin 60 cos 0N T (c)

0 :yF 2 cos60 sin 5 9.81 0N T (d)

Adding Eqs. (a) and (c) together gives

2 1sin 60 sin 30 0N N (e)

while adding Eqs. (b) and (d) together gives


53

2 1 1cos60 cos30 49.05N N W (f)

Solving Eqs. (e) and (f) gives

1 21.73205N N

12

49.052

WN

Then Eqs. (b) and (a) can be written

2sin cos 60 49.05T N (g)

2cos sin 60T N (h)

Dividing Eq. (g) by Eq. (h) gives

2

2

cos60 49.05tansin 60

NN

(a) 50 for 1 14.98 NW ....................... Ans.

(b) 10 for 1 233 NW .............................. Ans.

(c) 25 for 1 971 NW .............................. Ans.

3-31 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P3-31. If the side force P has a magnitude of 150 lb,

(a) Calculate and plot FC, the force applied to the car as a function of the angle (0 45 ). (b) What is the maximum angle for which this method is effective (that is, for which P FC)? SOLUTION From a free-body diagram of the knot, the equations of equilibrium are

0 :xF 1 1 2 2cos cos 0T T

0 :yF 1 1 2 2sin sin 0P T T

(a) If 1 2 , then 1 2 CT T F and

150 lb

2sin 2sinCPF .......................... Ans.

(b) CP F when

30 .......................................................... Ans.


54

3-32 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-32. Determine the magnitude and the direction angle of the unknown force F4.

SOLUTION

1 500cos63 226.995 NxF 1 500sin 63 445.503 NyF

2 750cos30 649.519 NxF 2 750sin 30 375.000 NyF

3 1000cos60 500.000 NxF 3 1000sin 60 866.025 NyF


0 :xF 4226.995 649.519 500 0xF

0 :yF 4445.503 375 866.025 0yF

give

4 1376.514 NxF 4 45.522 NyF

4 1377 NF 1.89 ........................................................................................................Ans.

3-33 Two 10-in.-diameter pipes and a 6-in.-diameter pipe are supported in a pipe rack as shown in Fig. P3-33. The 10-in.-diameter pipes each weigh 300 lb and the 6-in.-diameter pipe weighs 175 lb. Determine the forces exerted on the pipes by the supports at the contact surfaces A, B, and C. Assume all surfaces to be smooth.

SOLUTION

1 6cos 41.4108AB BC

From a free-body diagram of cylinder B, the equations of equilibrium

0 :xF cos 0B BCN F

0 :yF sin 300 0BCF

are solved to get

453.557 lbBCF

340 lbBN ................................................. Ans.

From a free-body diagram of cylinder C, the equations of equilibrium

0 :xF cos cos 0BC AB CF F N

0 :yF sin sin 175 0AB BCF F

are solved to get

718.132 lbABF

879 lbCN ................................................. Ans.

Finally, from a free-body diagram of cylinder A, the equations of equilibrium

0 :xF cos 0A ABN F

0 :yF 300 sin 0D ABN F

are solved to get

539 lbAN ................................................. Ans.


55

3-34 The 250-kg block A of Fig. P3-34 is supported by a small wheel that is free to roll on the continuous cable between supports B and C. If the length of the cable is 42 m, determine the distance x and the tension T in the cable when the system is in equilibrium.

SOLUTION Since the pulley is free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 250 kg weight, the equations of equilibrium


0 :yF 1 2sin sin 250 9.81 0T T

are solved to get

1 2

1226.25 N

sinT (a)

From the geometry of the cable

1 2 42 m (b)

1 2cos cos 40 m (c)

1 2sin sin 6 m (d)

Therefore, from Eqs. (b) and (c) 42cos 40

17.753

4020 NT .................................................. Ans. Finally, Eq. (d) can be written

1 2 19.6779 m (e)

and adding Eqs. (b) and (e) gives

1 30.839 m

1 cos 29.4 mx ................................. Ans.

3-35 A joint in a bridge truss is subjected to the forces shown in Fig. P3-35. Determine the forces C and T required for equilibrium.


0 :xF 9 cos 45 9cos 60 12 0T

0 :yF sin 45 9sin 60 0T C

are solved to give

10.61 kipT .......................................................................................................................Ans.

15.29 kipC .......................................................................................................................Ans.


56

3-36 Four forces act on a small airplane in flight, as shown in Fig. P3-36; the weight W (25 kN), the thrust provided by the engine FT (10 kN), the lift provided by the wings FL, and the drag resulting from motion through the air FD. If the airplane is flying with a constant velocity in the x’ direction, determine the magnitudes of the lift and drag forces.


0 :xF cos10 sin10 10cos10 0D LF F

0 :yF cos10 10sin10 sin10 25 0L DF F

are solved to give

24.6 kNLF .......................................................................................................................Ans.

5.66 kNDF .......................................................................................................................Ans.

3-37 A 500-lb block is supported by three cables as shown in Fig. P3-37. Determine the tensions in the cables AB, AC, and AD.

SOLUTION

2 2

6 12 0.44721 0.894436 12

AB AB AB ABT T Tj kT j k

2 2 2

4 3 12 0.30769 0.23077 0.923084 3 12

AC AC AC AC ACT T T Ti j kT i j k

2 2 2

4 8 12 0.26726 0.53452 0.801784 8 12

AD AD AD AD ADT T T Ti j kT i j k


0 :xF 0.30769 0.26726 0AC ADT T

0 :yF 0.44721 0.23077 0.53452 0AB AC ADT T T

0 :zF 0.89443 0.92308 0.80178 500 0AB AC ADT T T


267 lbABT .........................................................................................................................Ans.

141.3 lbACT ......................................................................................................................Ans.

162.7 lbADT ......................................................................................................................Ans.

3-38 A 1250-N force F is supported by a cable AD and by struts AB and AC, as shown in Fig. P3-38. If the struts can transmit only axial tensile or compressive forces, determine the forces in the struts and the tension in the cable.

SOLUTION

2 2

6 3 0.89443 0.447216 3

AD AD AD ADT T Tj kT j k

2 2 2

2 6 2 0.30151 0.90453 0.301512 6 2

AB AB AB AB ABF F F Fi j kF i j k

2 2 2

3 6 2 0.42857 0.85714 0.285713 6 2

AC AC AC AC ACF F F Fi j kF i j k



57

0 :xF 0.30151 0.42857 0AB ACF F

0 :yF 0.89443 0.90453 0.85714 0AD AB ACT F F

0 :zF 0.44721 0.30151 0.28571 1250 0AD AB ACT F F


1677 NADT .......................................................................................................................Ans.

995 NABF .........................................................................................................................Ans.

700 NACF .........................................................................................................................Ans.

3-39 A 75-lb force F is supported by a tripod as shown in Fig. P3-39. If the legs can transmit only axial tensile or compressive forces, determine the forces in the legs AB, AC, and AD.

SOLUTION

2 2 2

25 20 54 0.39823 0.31859 0.8601825 20 54

AB AB AB AB ABF F F Fi j kF i j k

2 2 2

25 40 54 0.34867 0.55787 0.7531325 40 54

AC AC AC AC ACF F F Fi j kF i j k

2 2 2

25 6 54 0.41800 0.10032 0.9028925 6 54

AD AD AD AD ADF F F Fi j kF i j k


0 :xF 0.39823 0.34867 0.41800 0AB AC ADF F F

0 :yF 0.31859 0.55787 0.10032 0AB AC ADF F F

0 :zF 0.86018 0.75313 0.90289 75 0AB AC ADF F F


33.4 lbABF ........................................................................................................................Ans.

11.62 lbACF .....................................................................................................................Ans.

41.5 lbADF .......................................................................................................................Ans.

3-40 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-40. Determine the magnitude of the unknown forces F1, F2, and F3.

SOLUTION

1 1 1 1 12 2 2

3 2.5 3.5 0.57208 0.47673 0.667423 2.5 3.5

F F F Fi j kF i j k

2 2 2 2 22 2 2

3 4.5 3.5 0.46569 0.69854 0.543313 4.5 3.5

F F F Fi j kF i j k

3 3 3 3 32 2 2

3 5 4 0.42426 0.70711 0.565693 5 4

F F F Fi j kF i j k


0 :xF 1 2 30.57208 0.46569 0.42426 0F F F

0 :yF 1 2 30.47673 0.69854 0.70711 0F F F


58

0 :zF 1 2 30.66742 0.54331 0.56569 50 0F F F


1 2.34 kNF ........................................................................................................................Ans.

2 43.1 kNF ........................................................................................................................Ans.

3 44.2 kNF ........................................................................................................................Ans.


59

Chapter 4 4-1 An aluminum tube with an outside diameter of 1.000 in. will be used to support a 10-kip load. If the axial

stress in the member must be limited to 30 ksi T or C, determine the wall thickness required for the tube. SOLUTION

10,000 30,000 psiP

A A

2 2 210,0001 in.4 30,000iA d

0.75867 in. 1 2id t

0.1207 in.t .......................................................................................................................... Ans.

4-2 Three steel bars with 25 15-mm cross sections are welded to a gusset plate as shown in Fig. P4-2. Determine the normal stresses in the bars when the forces shown are being applied to the plate.

SOLUTION

6 240,000 106.7 10 N/m 106.7 MPa0.025 0.015A

PA

........................................Ans.

6 250,000 133.3 10 N/m 133.3 MPa0.025 0.015B

PA

.........................................Ans.

6 220,000 53.3 10 N/m 53.3 MPa0.025 0.015C

PA

.............................................Ans.

4-3 Two ¼-in. diameter steel cables A and B are used to support a 220-lb traffic light as shown in Fig. P4-3. Determine the normal stress in each of the cables.


0 :xF cos 25 cos 20 0B AT T

0 :yF sin 20 sin 25 220 0A BT T

are solved to get

1.03684B AT T

281.977 lbAT 292.365 lbBT

2281.977 5740 psi0.25 4

AA

A

TA

......................................................................................Ans.

2292.365 5960 psi0.25 4

BB

B

TA

......................................................................................Ans.

4-4 A system of steel pipes is loaded and supported as shown in Fig. P4-4. If the normal stress in each pipe must not exceed 150 MPa, determine the cross-sectional areas required for each of the sections.

SOLUTION From free-body diagrams of the pipes, the equations of equilibrium give

0 :yF 650 0AP 650 kNAP

0 :yF 650 850 0BP 1500 kNBP


60

0 :yF 650 850 1500 0CP 3000 kNCP

3

6 2 26

650 10 4333 10 m 4330 mm150 10

AA

A

PA .......................................................Ans.

3

6 2 26

1500 10 10,000 10 m 10,000 mm150 10

BB

B

PA ..............................................Ans.

3

6 2 26

3000 10 20,000 10 m 20,000 mm150 10

CC

C

PA ............................................Ans.

4-5 Two 1-in. diameter steel bars are welded to a gusset plate as shown in Fig. P4-5. Determine the normal stresses in the bars when forces F1 = 550 lb and F2 = 750 lb are applied to the plate.

SOLUTION

12

550 700 psi1 4A

A

FA

...............................................................................................Ans.

22

750 955 psi1 4B

B

FA

...............................................................................................Ans.

4-6 Two strips of a plastic material are bonded together as shown in Fig. P4-6. The average shearing stress in the glue must be limited to 950 kPa. What length L of splice plate is needed if the axial load carried by the joint is 50 kN?

SOLUTION From a free-body diagram of the middle plate, the equations of equilibrium

0 :xF 32 50 10 0V

gives

3

3

25 10 N

950 10 0.3 2

V A

L

0.1754 m 175.4 mmL .....................................................................................................Ans.

4-7 A coupling is used to connect a 2-in. diameter plastic rod to a 1.5-in. diameter rod, as shown in Fig. P4-7. If the average shearing stress in the adhesive must be limited to 500 psi, determine the minimum lengths L1 and L2 required for the joint if the applied axial load P is 8000 lb.

SOLUTION From a free-body diagram of the rod, the equation of equilibrium

0 :xF 8000 0V

gives

1

2

8000 lb500 2

500 1.5

V AL

L

1 2.55 in.L ................................................................Ans.

2 3.40 in.L ................................................................Ans.


61

4-8 Three plates are joined with a 12-mm diameter pin as shown in Fig. P4-8. Determine the maximum load P that can be transmitted by the joint if

(a) The maximum normal stress on a cross section at the pin must be limited to 350 MPa. (b) The maximum bearing stress between a plate and the pin must be limited to 650 MPa. (c) The maximum shearing stress on a cross section of the pin must be limited to 240 MPa. (d) The punching shear resistance of the material in the top and bottom plates is 300 MPa. SOLUTION (a) For the outside plates

2 6 230 12 10 180 mm 180 10 mnA

6 6 32 2 350 10 180 10 126 10 NnP A

For the middle plate

2 6 225 12 25 325 mm 325 10 mnA

6 6 3350 10 325 10 113.7 10 NnP A

Therefore

max 113.7 kNP ......................................................................................................................... Ans.

(b) For the bearing stress on the hole in the middle plate

2 6 212 25 300 mm 300 10 mbA dt

6 6 3650 10 300 10 195 10 NbP A

For the outer plates

2 6 212 10 120 mm 120 10 mbA dt

6 6 32 2 650 10 120 10 156 10 NbP A

Therefore

max 156 kNP ............................................................................................................................. Ans.

(c) For the shearing stress on the cross section of the pin (which is in double shear)

22

2 6 2

4 12 4

113.10 mm 113.10 10 msA d

6 62 2 240 10 113.10 10sP A

354.3 10 N 54.3 kN .....................................................................................................Ans. (d) For the punching shear stress in the outside plates

2 6 2

2 2 18 10

360 mm 360 10 msA Lt

6 62 2 300 10 360 10sP A

3216 10 N 216 kN ..........................................................................................................Ans.


62

4-9 A 100-ton hydraulic punch press is used to punch holes in a 0.50-in.-thick steel plate, as illustrated schematically in Fig. P4-9. If the average punching shear resistance of the steel plate is 40 ksi, determine the maximum diameter hole that can be punched.

SOLUTION

340 10 0.5

100 2000 lb

sP A dt

d

3.18 in.d ............................................................................................. Ans.

4-10 A body with a mass of 250 kg is supported by five 15-mm diameter cables, as shown in Fig. P4-10. Determine the normal stress in each of the cables.

SOLUTION The tension in cable E is equal to the weight of the hanging body. The normal stress in cable E is then

6 22

250 9.8113.88 10 N/m 13.88 MPa

0.015 4E

ETA

...........................................Ans.

From a free-body diagram of the lower ring, the equations of equilibrium

0 :xF cos60 0C DT T

0 :yF sin 60 250 9.81 0CT

give

22831.90 N 0.015 4C CT

6 216.03 10 N/m 16.03 MPaC ....................................................................................Ans.

21415.95 N 0.015 4D DT

6 28.01 10 N/m 8.01 MPaD ........................................................................................Ans.

Then, from a free-body diagram of the upper ring, the equations of equilibrium

0 :xF cos30 cos 40 cos 60 0B A CT T T

0 :yF sin 30 sin 40 sin 60 0B A CT T T

give

21506.83 N 0.015 4A AT

6 28.53 10 N/m 8.53 MPaA ........................................................................................Ans.

22967.85 N 0.015 4B BT

6 216.79 10 N/m 16.79 MPaB ...................................................................................Ans.


63

4-11 The joint shown in Fig. P4-11 is used in a steel tension member which has a 2 1-in. rectangular cross section. If the allowable normal, bearing, and punching-shearing stresses in the joint are 13.5 ksi, 18.0 ksi, and 6.50 ksi, respectively, determine the maximum load P that can be carried by the joint.

SOLUTION For normal stress at the narrowest section

20.5 1 0.5 in.nA

313.5 10 0.5 6750 lbnP A

For bearing stress

22 3 8 1 0.75 in.bA

318.0 10 0.75 13,500 lbbP A

For punching shearing stress

22 3 8 1 0.75 in.sA

36.5 10 0.75 4880 lbsP A

Therefore

max 4880 lbP ............................................................................................................................ Ans.

4-12 A vertical shaft is supported by a thrust collar and bearing plate, as shown in Fig. P4-12. Determine the maximum axial load that can be applied to the shaft if the average punching shear stress in the collar and the average bearing stress between the collar and the plate are limited to 75 and 100 MPa, respectively.

SOLUTION For bearing stress between the collar and the plate

2 2 3 20.15 0.10 4 9.8175 10 mbA

6 3 3100 10 9.8175 10 982 10 NbP A

For punching shearing stress in the collar

3 20.10 0.025 7.854 10 msA dL

6 3 375 10 7.854 10 589 10 NsP A

Therefore

max 589 kNP ............................................................................................................................ Ans.

4-13 A device for determining the shearing strength of wood is shown in Fig. P4-13. The dimensions of the wood specimen are 6-in. wide by 8-in. high by 2-in. thick. If a load of 16,800 lb is required to fail the specimen, determine the shearing strength of the wood.

SOLUTION

16,800 1050 psi

8 2s

VA

...................................................................................................Ans.


64

4-14 The 75-kg traffic light shown in Fig. P4-14 is supported by three cables of equal diameter. Determine the minimum diameter required if the normal stress in any cable must not exceed 160 MPa.

SOLUTION

2 2 2

4 8 5 0.39036 0.78072 0.487954 8 5


2 2 2

6 8 5 0.53666 0.71554 0.447216 8 5


2 2

8 5 0.84800 0.530008 5

C C C CT T Tj kT j k

The equations of equilibrium

0 :xF 0.39036 0.53666 0A BT T

0 :yF 0.78072 0.71554 0.84800 0A B CT T T

0 :zF 0.48795 0.44721 0.53000 75 9.81 0A B CT T T

give

1.37478A BT T 2.10950C BT T

6 2452.353 N 160 10 4A AT d 0.00190 mAd

6 2329.037 N 160 10 4B BT d 0.00162 mBd

6 2694.103 N 160 10 4C CT d 0.00235 mCd

Therefore

min 2.35 mmd .......................................................................................................................... Ans.

4-15 A 1000-lb load is securely fastened to a hoisting rope as shown in Fig. P4-15. The force in the weightless flexible cable does not change as it passes around the small frictionless pulley at support C. The sag distance d is 4 ft. Cables AB and BC have the same diameter. The normal stresses in these cables must not exceed 24 ksi, and the shearing stress in pin A (double shear) must not exceed 12 ksi. Determine

(a) The minimum diameter of cables AB and BC. (b) The minimum diameter of the pin at A. SOLUTION

11

4sin 23.57810

110cos 9.16515 fta

30 20.83485 ftb a

12

4tan 10.868b

Then, from a free-body diagram of the ring, the equations of equilibrium

0 :xF 2 1cos cos 0AP T

0 :yF 2 1sin sin 1000 0AP T

give


65

1.07152AT P

1620.35 lbP 1736.24 lbAT

(a) For the normal stress in the cables

3 21620.35 lb 24 10 4BCP d

0.293 in.BCd ........................................................................................................................... Ans.

3 21736.24 lb 24 10 4A ABT d

0.304 in.ABd ........................................................................................................................... Ans.

(b) For the pin at A, which is in double shear,

3 21736.24 lb 12 10 2 4A pT d

0.304 in.pd ............................................................................................................................. Ans.

4-16 A flat steel bar 100 mm wide by 25 mm thick has axial loads applied with 40 mm-diameter pins in double shear at points A, B, C, and D as shown in Fig. P4-16. Determine

(a) The axial stress in the bar on a cross section at pin B. (b) The average bearing stress on the bar at pin B. (c) The shearing stress on the pin at A. SOLUTION (a) For axial stress in the bar

3

6 2350 10 233 10 N/m 233 MPa0.06 0.025n

n

PA

....................................................Ans.

(b) For bearing stress at the pin

3

6 2250 10 250 10 N/m 250 MPa0.04 0.025b

b

PA

...................................................Ans.

(c) For shearing stress on the pin (which is in double shear)

3

6 22

350 10 139.3 10 N/m 139.3 MPa2 2 0.04 4s

VA

.......................................Ans.

4-17 Two flower pots are supported with steel wires of equal diameter. Pot A weighs 10 lb and pot B weighs 8 lb. Determine the minimum required diameter of the wires if the normal stress in the wires must not exceed 18 ksi.

SOLUTION From a free-body diagram of the upper ring

0 :xF cos 45 cos 0CD BCT T (a)

0 :yF sin 45 sin 8 0CD BCT T (b)

From a free-body diagram for the lower ring

0 :xF cos cos 45 0BC ABT T (c)

0 :yF sin 45 sin 10 0AB BCT T (d)

Adding equations (a) and (c) gives

cos 45 cos 45 0CD ABT T


66

CD ABT T

Then adding Eqs. (b) and (d) gives

sin 45 sin 45 18 lbCD ABT T

12.7279 lbCD ABT T

Now Eqs. (b) and (a) can be written

sin 1.0000BCT (e)

cos 9.0000BCT (f)

Dividing Eq. (e) by Eq. (f) gives

tan 1 9 6.340

9.0554 lbBCT

3max 2

12.7279 18 10 psi4

PA d

0.0300 in.d ............................................................................................................................ Ans.

4-18 A steel pipe will be used to support a 40-kN load. If the wall thickness of the pipe is 10 percent of the pipe’s outside diameter do, calculate and plot the normal stress in the pipe as a function of the diameter do, (20 mm

do 75 mm). If the axial stress in the pipe must be limited to 250 MPa, what is the smallest size standard steel pipe (see Appendix A) that could be used.

SOLUTION

2 0.1 0.8i o o od d d d

3

222 2

40 10 141, 471.06 N/m4 oo i

PA dd d

For standard steel pipe,

3

6 240 10 250 10 N/mPA A

6 2 2160 10 m 160 mmA

13 mmd .................................................... Ans.

4-19 An aluminum tube with an outside diameter do will be used to support a 10-kip load. If the axial stress in the tube must be limited to 30 ksi T or C, calculate and plot the required wall thickness t as a function of do, (0.75 in. do 4 in.). What diameter would be required for a solid aluminum shaft?

SOLUTION

2 2

10,000 30,000 psi4o i

PA d d

2 0.42441 2i o od d d t

2 0.424412

o od dt

For a solid aluminum shaft


67

2

10,000 30,000 psi4

PA d

0.65 in.d .................................................................................................................................Ans.

4-20 The steel pipe column shown in Fig. P4-20a has an outside diameter of 150 mm and a wall thickness of 15 mm. The load imposed on the column by the timber beam is 150 kN. If the bearing stress between the circular steel bearing plate and the timber beam is not to exceed 3.25 MPa, determine the minimum diameter bearing plate that must be used between the column and the beam. Assume that the bearing stress is uniformly distributed over the surface of the plate.

If the bearing plate is not rigid, the stress between the bearing plate and the timber beam will not be uniform. If the stress varies as shown in Fig. P4-20b (a uniform value of max above the column and decreasing linearly to max/5 at the outside edge rp of the bearing plate), calculate and plot max versus the radius rp of the bearing plate (75 mm rp 500 mm). Now what minimum diameter bearing plate must be used if the bearing stress must not exceed 3.25 MPa? What is the percent decrease in max for a 400 mm-diameter bearing plate compared with a 150 mm-diameter bearing plate? For a 600 mm-diameter bearing plate compared with a 150 mm-diameter bearing plate?

SOLUTION For uniform stress over a rigid bearing plate

3

6 22

150 10 3.25 10 N/m4

PA d

0.242 m 242 mmd ...........................................................................................................Ans. For the non-rigid bearing plate

1 2max

c r c

where the constants c1 and c2 are chosen such that

1 21 0.075c c

1 20.2 pc r c

Therefore,

10.8

0.075 p

cr

2

0.8 0.0751

0.075 p

cr

Integrating the stress to get the axial force

F dA

gives

0.0753max max 1 20 0.075

3 22 3 2

max 1 2

150 10 2 2

0.075 0.075 0.07522 3 3 2 2

pr

p p

r dr c r c r dr

r rc c


68

3

2max 3 22 3 2

1 2

150 10 N/m0.075 0.075 0.0752

2 3 3 2 2p pr r

c c

When 0.075 mpr 6 2max 8.49 10 N/m 8.49 MPa ......................................Ans.

When 0.2 mpr max 1.976 MPa (a 76.7% decrease) ................................Ans.

When 0.3 mpr max 0.965 MPa (an 88.6% decrease) ..............................Ans.

4-21 The tie rod shown in Fig. P4-21a has a diameter of 1.50 in and is used to resist the lateral pressure against the walls of a grain bin. The force imposed on the wall by the rod is 18,000 lb. If the bearing stress between the washer and the wall must not exceed 400 psi, determine the minimum diameter washer that must be used between the head of the bolt and the grain bin wall. Assume that the bearing stress is uniformly distributed over the surface of the washer.

If the washer is not rigid, the stress between the washer and the wall will not be uniform. If the stress varies as shown in Fig. P4-21b (a uniform value of max under the 2.4-in.-diameter restraining nut and decreasing as 1/r to the outside edge rw of the washer), calculate and plot max versus the radius rw of the washer (1 in. rw 8 in.). Now what minimum diameter washer must be used if the bearing stress must not exceed 400 psi? What is the percent decrease in max for an 8-in.-diameter washer compared with a 4-in.-diameter washer? For a 12-in.-diameter washer compared with an 8-in.-diameter washer?

SOLUTION For uniform stress over a rigid washer

3

2 2

18 10 400 psi1.5 4

PA d

7.72 in.d .................................................. Ans. For the non-rigid washer

max1.2r

Integrating the stress to get the axial force

F dA

gives

1.23 maxmax0.75 1.2

2 2max

max

1.218 10 2 2

1.2 0.75 2.4 1.2

0.87750 2.4 1.2

wr

w

w

r dr r drrr

r

3

max18 10 psi

0.87750 2.4 1.2wr

When max 400 psi 2 13.61 in.wd r ......................................................................Ans.

When 2 2.4 in.wd r max 6529 psi

When 2 8 in.wd r max 754 psi an 88.5% decrease.................................. Ans.

When 2 12 in.wd r max 462 psi a 38.7% decrease.................................... Ans.


69

4-22 The steel bar shown in Fig. P4-22 will be used to carry an axial tensile load of 400 kN. If the thickness of the bar is 45 mm, determine the normal and shearing stresses on plane AB.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium

0 :nF 400cos37 0N

0 :tF 400sin 37 0V

give

319.45 kNN 240.73 kNV The areas are related by

0.075 0.045 cos37nA A 3 24.226 10 mnA

Therefore

3

6 23

319.45 10 75.6 10 N/m 75.6 MPa4.226 10n

NA

...................................................Ans.

3

6 23

240.73 10 57.0 10 N/m 57.0 MPa4.226 10n

VA

....................................................Ans.

4-23 An axial load P is applied to a timber block with a 4 4-in. square cross section, as shown in Fig. P4-23. Determine the normal and shear stresses on the planes of the grain if P = 5000 lb.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium

0 :nF 5000sin14 0N

0 :tF 5000cos14 0V

give

1209.61 lbN

4851.48 lbV The areas are related by

4 4 sin14nA A

266.137 in.nA

Therefore

1209.61 18.29 psi66.137n

NA

...............................................................................................Ans.

4851.48 73.4 psi66.137n

VA

.................................................................................................Ans.

4-24 A structural steel bar with a 20 25-mm rectangular cross section is subjected to an axial tensile load of 55 kN. Determine the maximum normal and shear stresses in the bar.

SOLUTION

3

6 2max

55 10 110.0 10 N/m 110.0 MPa0.02 0.025

PA

...........................................Ans.

max 55.0 MPa2 2PA

....................................................................................................Ans.


70

4-25 A steel rod of circular cross section will be used to carry an axial tensile load of 50 kip. The maximum stresses in the rod must be limited to 25 ksi in tension and 15 ksi in shear. Determine the minimum diameter d for the rod.

SOLUTION

3

3max 2

50 10 25 10 psi4

PA d

1.596 in.d

3

3max 2

50 10 15 10 psi2 2 4PA d

1.457 in.d

min 1.596 in.d ..........................................................................................................................Ans.

4-26 Determine the maximum axial load P that can be applied to the wood compression block shown in Fig. P4-26 if specifications require that the shear stress parallel to the grain not exceed 5.25 MPa, the compressive stress perpendicular to the grain not exceed 13.60 MPa, and the maximum shear stress in the block not exceed 8.75 MPa.

SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 60 0P N

0 :tF cos 60 0P V

The areas are related by

0.2 0.12 sin 60nA A 20.027713 mnA

Therefore

613.60 10 0.027713

sin 60 sin 60 sin 60nANP 3435 10 NP

65.25 10 0.027713

cos60 cos 60 cos 60nAVP 3291 10 NP

max 291 kNP .............................................................................................................................Ans.

4-27 A steel bar with a 4 1-in. rectangular cross section is being used to transmit an axial tensile load, as shown in Fig. P4-27. Normal and shear stresses on plane AB of the bar are 12 ksi tension and 9 ksi shear. Determine the angle and the applied load P.


0 :nF cos 0N P

0 :tF sin 0V P


4 1 cosnA A 24 cos in.nA

Therefore

3cos 12 10 4 cosnN P A

3sin 9 10 4 cosnV P A

Therefore


71

sin 9tancos 12PP

36.870 .................................................................................................................................Ans.

375 10 lb 75 kipP ...........................................................................................................Ans.

4-28 A steel bar with a butt-welded joint, as shown in Fig. P4-28 will be used to carry an axial tensile load of 400 kN. If the normal and shear stresses on the plane of the weld must be limited to 70 MPa and 45 MPa, respectively, determine the minimum thickness t required for the bar.


0 :nF 400sin 57 0N

0 :tF 400cos57 0V


0.1 sin 57nA t A 20.11924 mnA t

Therefore

3

6 2400 10 sin 57

70 10 N/m0.11924n

NA t

0.0402 mt

3

6 2400 10 cos57

45 10 N/m0.11924n

VA t

0.0406 mt

min 40.6 mmt ...........................................................................................................................Ans.

4-29 The shearing stress on plane AB of the 4 8-in. rectangular block shown in Fig. P4-29 is 2 ksi when the axial load P is applied. If the angle is 35 , determine

(a) The load P. (b) The normal stress on plane AB. (c) The maximum normal and shearing stresses in the block. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 35 0P N

0 :tF cos35 0P V


4 8 sin 35nA A

255.790 in.nA

(a) Solving the second equation for P gives

32 10 55.790

136, 215 lb 136.2 kipcos35 cos35 cos35

nAVP ........................Ans.

(b) Solving the first equation for N gives

sin 35 78,129.6 lb nN P A

1400 psi ...............................................................................................................................Ans.

(c) Then the maximum normal and shearing stresses are given by


72

max136, 215 4260 psi

4 8PA

............................................................................................Ans.

maxmax 2130 psi

2 2PA

.................................................................................................Ans.

4-30 A wood tension member with a 50 100-mm rectangular cross section will be fabricated with an inclined glued joint (45 90 ) at its midsection, as shown in Fig. P4-30. If the allowable stresses for the glue are 5 MPa in tension and 3 MPa in shear, determine

(a) The optimum angle for the joint. (b) The maximum safe load P for the member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 0N P

0 :tF cos 0V P


0.05 0.1 sinnA A

20.005 sin mnA

Therefore

6sin 5 10 0.005 sinnN P A

6cos 3 10 0.005 sinnV P A

and dividing the first equation by the second gives

sin 5tan 1.66667cos 3PP

59.04 ...................................................................................................................................Ans.

3max 34.0 10 N 34.0 kNP ...............................................................................................Ans.

4-31 The two parts of the eyebar shown in Fig. P4-31 are connected with two ½-in. diameter bolts (one on each side). Specifications for the bolts require that the axial tensile stress not exceed 12.0 ksi and that the shearing stress not exceed 8.0 ksi. Determine the maximum load P that can be applied to the eyebar without exceeding either specification.


0 :nF 2 cos30 0N P

0 :tF 2 sin 30 0V P

The cross sectional areas of the bolts are

2 20.5 4 0.19635 in.bA

The first equation gives

32 12 10 0.1963522 5441 lb

cos30 cos30 cos30bANP

while the second equation gives


73

32 8 10 0.1963522 6283 lb

sin 30 sin 30 sin 30bAVP

Therefore

max 5441 lb 5440 lbP .........................................................................................................Ans.

4-32 The bar shown in Fig. P4-32 has a 200 100-mm rectangular cross section. Determine (a) The normal and shearing stresses on plane a-a. (b) The maximum normal and shearing stresses in the bar. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF 500cos30 0N

0 :tF 500sin 30 0V


0.2 0.1 cos30nA A

3 223.094 10 mnA

Therefore

3500 10 cos30 433.01 kN nN A

3500 10 sin 30 250.00 kN nV A

6 218.75 10 N/m 18.75 MPa ......................................................................................Ans.

6 210.83 10 N/m 10.83 MPa .......................................................................................Ans.

3

6 2max

500 10 25.0 10 N/m 25.0 MPa0.2 0.1

PA

.....................................................Ans.

maxmax 12.50 MPa

2 2PA

.............................................................................................Ans.

4-33 A steel eyebar with a 4 1-in. rectangular cross section has been designed to transmit an axial tensile load. The length of the eyebar must be increased by welding a new center section in the bar (45 90 ), as shown in Fig. P4-33. The stresses in the weld material must be limited to 12 ksi in tension and 9 ksi in shear. Determine

(a) The optimum angle for the joint. (b) The maximum safe load P for the redesigned member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are

0 :nF sin 0N P

0 :tF cos 0V P


4 1 sinnA A 24 sin in.nA

Therefore

3sin 12 10 4 sinnN P A

3cos 9 10 4 sinnV P A


74

and dividing the first equation by the second gives

sin 12tan 1.33333cos 9PP

53.13 .................................................Ans.

max 75,000 lbP ........................................................................................................................Ans.

4-34 A steel eyebar with a 100 25-mm rectangular cross section has been designed to transmit an axial tensile load P. The length of the eyebar must be increased by welding a new center section in the bar, as shown in Fig. P4-33. If P = 250 kN, calculate and plot the normal stress n and the shear stress n in the weld material for weld angles (30 90 ). If the stresses in the weld material must be limited to 80 MPa in tension and 60 MPa in shear, what ranges of would be acceptable for the joint? Repeat for P = 305 kN and for P = 350 kN. Are weld angles < 30 reasonable? Why or why not?


0 :nF sin 0N P

0 :tF cos 0V P


0.1 0.025 sinnA A

20.0025 sin mnA

Therefore

2

2sin sin N/m0.0025 sin 0.0025n

N P PA

............................................................................Ans.

2cos sin cos N/m0.0025 sin 0.0025n

V P PA

.....................................................................Ans.

When 250 kNP 0 64 ..................................................................................Ans.

When 305 kNP 0 40 or 50 54 ..................................................Ans.

When 350 kNP 0 30 ..................................................................................Ans.

Angles less than about 30 are not very practical since the weld gets long quickly as the angle decreases.................................................................................................................................Ans.

4-35 Specifications for the rectangular (3 3 21-in.) block shown in Fig. P4-35 require that the normal and shearing stresses on plane A-A not exceed 800 psi and 500 psi, respectively. If the plane A-A makes an angle

= 37 with the horizontal, calculate and plot the ratios / max and / max as a function of the load P (0 P 13 kip). What is the maximum load Pmax that can be applied to the block? Which condition controls what the maximum load can be? Repeat for = 25 . For what angle will the normal stress and the shear stress both reach their limiting values at the same time?


0 :nF cos 0N P


75

0 :tF sin 0V P


3 3 cosnA A

29 cos in.nA

Therefore

2

max max

cos 9 psi800

nN A P.................................................. Ans.

max max

sin cos 9 psi500

nV A P......... Ans.

When 37 , shear stress controls and

max 9360 lbP ............................................. Ans.

When 25 , normal stress controls and

max 8770 lbP ............................................. Ans.

The normal stress and the shear stress will reach their maximum values at the same time if

max

max

500tan 0.62500800

nopt

n

AVN A

32.01opt ................................................................................................................................Ans.

4-36 Compression tests of concrete indicate that concrete fails when the axial compressive strain is 1200 m/m. Determine the maximum change in length that a 200-mm diameter 400-mm long concrete test specimen can tolerate before failure occurs.

SOLUTION

61200 10 400 0.480 mmL ..........................................................................Ans.

4-37 The 0.5 2 4-in. rubber mounts shown in Fig. P4-37 are used to isolate the vibrational motion of a machine from its supports. Determine the average shearing strain in the rubber mounts if the rigid frame displaces 0.01 in. vertically relative to the support.

SOLUTION

0.01 0.02 in./in. 0.02 rad0.5L

.................................................................................Ans.

4-38 A structural steel bar was loaded in tension to fracture. A 200 mm-length of the bar was marked off in 25-mm lengths before loading. After the rod broke, the 25-mm segments were found to have lengthened to 30.0, 30.5, 31.5, 34.0, 44.5, 32.0, 31.0, and 30.0 mm, consecutively. Determine

(a) The average strain over the 200-mm length. (b) The maximum average strain over any 50-mm length. SOLUTION

263.5 200 0.317 mm/mm

200f i

i

L LL L

.............................................................Ans.

max78.5 50 0.570 mm/mm

50........................................................................................Ans.


76

4-39 Mutually perpendicular axes in an unstressed member were found to be oriented at 89.92 when the member was stressed. Determine the shearing strain associated with these axes in the stressed member.

SOLUTION

390 89.92 1.396 10 rad2 180

..........................................................Ans.

4-40 A thin triangular plate is uniformly deformed as shown in Fig. P4-40. Determine the shearing strain at P associated with the two edges (PQ and PR) that were orthogonal in the undeformed plate.

SOLUTION

500 mmP QL

500cos 45 353.5534 mmMQ MPL L

10 363.5534 mmMP MPL L

1tan 44.20107MQ

MP

LL

2

2 2

90 88.40213180

327.9 10 rad ..................................................................................................................... Ans.

4-41 The sanding-drum mandrel shown in Fig. P4-41 is made for use with a hand drill. The mandrel is made from a rubber-like material which expands when the nut is tightened to secure the sanding drum placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine

(a) The average normal strain along a diameter of the mandrel. (b) The circumferential strain at the outside surface of the mandrel. SOLUTION

(a) 32.15 2.00 75.0 10 in./in.2.00D ...................................................................................Ans.

(b) 32.15 2.0075.0 10 in./in.

2.00C .....................................................................Ans.

4-42 A thin rectangular plate is uniformly deformed as shown by PRSQ in Fig. P4-42. Determine the shearing stain xy at P.

SOLUTION

11

0.380tan 0.04354500

12

0.200tan 0.04584250

1 220.08938

31.560 10 rad ................................................................................................................... Ans.


77

4-43 A steel sleeve is connected to a steel shaft with a flexible rubber insert, as shown in Fig. P4-43. The insert has an inside diameter of 3.3 in. and an outside diameter of 4.3 in. When the unit is subjected to a torque T, the shaft rotates 1.5 with respect to the sleeve. Assume that radial lines in the unstressed state remain straight as the rubber deforms. Determine the shearing strain r in the rubber insert

(a) At the inside surface. (b) At the outside surface. SOLUTION

3.3 1.5 0.0431969 in.2 180

b r

(b) 11 tan 4.93774

0.5b

386.2 10 rad outside ..................................................Ans.

(a) 2 1 1.5 6.43774

3112.4 10 rad inside .................................................Ans.

4-44 A steel rod is subjected to a nonuniform heating that produces an extensional (axial) strain that is proportional to the square of the distance from the unheated end ( = kx2). If the strain is 1250 m/m at the midpoint of a 3.00-m rod, determine

(a) The change in length of the rod. (b) The average axial strain over the length L of the rod. (c) The maximum axial strain in the rod. SOLUTION

2kx 261250 10 1.5k 6555.556 10k

2d kxdx

(a)

333 2 3

00

9 5.00 10 m 5.00 mm3kxd dx k x dx k ...................Ans.

(b) 30.005 1.667 10 m/m 1667 m/m3avg .................................................................Ans.

(c) 2 3

max 3 3 9 5.00 10 m/m 5000 m/mk k ................................................Ans.

4-45 The axial strain in a suspended bar of material of varying cross section due to its own weight, as shown in Fig. P4-45, is given by the expression y/3E, where is the specific weight of the material, y is the distance from the free (bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E,

(a) The change in length of the bar due to its own weight. (b) The average axial strain over the length L of the bar. (c) The maximum axial strain in the bar. SOLUTION

3y dE dy

(a) 2 2

003 3 2 6

LL y Ld y dy

E E E..........................................................................Ans.


78

(b) in./in.6avgL

L E................................................................................................................Ans.

(c) max in./in.3LLE

...............................................................................................................Ans.

4-46 A steel cable is used to tether an observation balloon. The force exerted on the cable by the balloon is sufficient to produce a uniform strain of 500 m/m in the cable. In addition, at each point in the cable, the weight of the cable reduces the axial strain by an amount that is proportional to the length of the cable between the balloon and the point. When the balloon is directly overhead at an elevation of 300 m, the axial strain at the midlength of the cable is 350 m/m. Determine

(a) The total elongation of the cable. (b) The maximum height that the balloon could achieve SOLUTION

6500 10 cy 6 6350 10 500 10 150c 61 10c

6 6500 10 1 10d ydy

2

6 6 2 6

00

10 500 10 500 500 0.5 102

LL yd y dy y L L

(a) If 300 mL

0.1050 m 105.0 mm .....................................................................................................Ans.

(b) If 500 my , then 0 . If 1000 mL , then 0 . Neither is possible. Therefore

max 500 my .............................................................................................................................. Ans.

4-47 A steel cable is used to support an elevator cage at the bottom of a 2000 ft-deep mine shaft. A uniform axial strain of 250 in./in. is produced in the cable by the weight of the cage. At each point the weight of the cable produces an additional axial strain that is proportional to the length of the cable below the point. If the total axial strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine

(a) The strain in the cable at a depth of 500 ft. (b) The total elongation of the cable. SOLUTION

6250 10 cy 6 6700 10 250 10 2000c 60.22500 10c

(a) 6 6500 250 10 0.22500 10 1500

6587 10 in./in. 587 in./in. ..................................................................................Ans.

(b) 6 6250 10 0.22500 10d ydy

2000220006 6

00

2 6

0.22510 250 0.225 10 2502

250 2000 0.1125 2000 10

yd y dy y

0.9500 ft 11.40 in. .........................................................................................................Ans.


79

4-48 At the proportional limit, a 200 mm-gage length of a 15 mm-diameter alloy bar has elongated 0.90 mm and the diameter has been reduced 0.022 mm. The total axial load carried was 62.6 kN. Determine the modulus of elasticity, Poisson’s ratio, and the proportional limit for the material.

SOLUTION E

3

262.6 10 0.90

2000.015 4E

9 278.7 10 N/m 78.7 GPaE ......................................................................................Ans.

0.022 15 0.3260.9 200

t

a

.........................................................................................Ans.

3

6 22

62.6 10 354 10 N/m 354 MPa0.015 4p

PA

...............................................Ans.

4-49 A 1.50 in.-diameter rod 20 ft long elongates 0.48 in. under a load of 53 kip. The diameter of the rod decreases 0.001 in. during the loading. Determine the modulus of elasticity, Poisson’s ratio, and the modulus of rigidity for the material.

SOLUTION E

3

253 10 0.48

20 121.5 4E

615.00 10 psi 15,000 ksiE ........................................................................................Ans.

0.001 1.5 0.333

0.48 20 12t

a

..................................................................................Ans.

15,000 5625 ksi

2 1 2 1 0.333EG ..........................................................................Ans.

4-50 A tensile test specimen having a diameter of 5.64 mm and a gage length of 50 mm was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-50. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if Poisson’s ratio = 0.30 remains constant. (h) The tangent modulus at a stress level of 400 MPa. (i) The secant modulus at a stress level of 400 MPa. SOLUTION From the stress-strain diagram:

(a) 3140 70.0 10 MPa 70.0 GPa0.002

E ...........................................................................Ans.

(b) 150 MPapl ........................................................................................................................... Ans.

(c) 450 MPault .......................................................................................................................... Ans.


80

(d) .05 220 MPa ............................. Ans.

(e) .2 275 MPa .............................. Ans.

(f) 450 MPaf .............................. Ans.

(g) 0.3 0.115t add

0.19458 mmd

5.64 0.19458 5.4454 mmfd

2

2

450 5.64 4483 MPa

5.4454 4f f i

tff f

P AA A

......................................................Ans.

(h) 430 350 1000 MPa 1.000 GPa0.08 0tE ..........................................................................Ans.

(i) 400 8890 MPa 8.89 GPa

0.045sE ...................................................................................Ans.

4-51 A tensile test specimen having a diameter of 0.250 in. and a gage length of 2.000 in. was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-51. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.212 in. (h) The tangent modulus at a stress level of 56 ksi. (i) The secant modulus at a stress level of 56 ksi. SOLUTION From the stress-strain diagram:

(a) 3

628 10 28.0 10 psi 28,000 ksi0.001

E ........................................................................Ans.

(b) 34 ksipl ................................... Ans.

(c) 74 ksiult ................................... Ans.

(d) .05 43 ksi ................................... Ans.

(e) .2 43 ksi .................................... Ans.

(f) 65 ksif .................................... Ans.

(g)

2

2

65 0.25 4

0.212 4f f i

tff f

P AA A

90.4 ksi ............................................................................................................................. Ans.

(h) 78 35 537 ksi0.08 0tE ............................................................................................................Ans.


81

(i) 56 1400 ksi

0.04sE ................................................................................................................Ans.

4-52 A tensile test specimen having a diameter of 11.28 mm and a gage length of 50 mm was tested to fracture. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 9.50 mm. (h) The tangent modulus at a stress level of 315 MPa. (i) The secant modulus at a stress level of 315 MPa. SOLUTION From the stress-strain diagram:

(a) 3222.1 185 10 MPa0.0012

E

185 GPaE ...................Ans.

(b) 280 MPapl ...............Ans.

(c) 507 MPault ...............Ans.

(d) .05 300 MPa ...............Ans.

(e) .2 330 MPa ................Ans.

(f) 450 MPaf ................Ans.

(g) 3

6 22

45.1 10 636 10 N/m 636 MPa0.0095 4

ftf

f

PA

..............................................Ans.

(h) 3320 306 17.5 10 MPa 17.5 GPa0.0032 0.0024tE ........................................................Ans.

(i) 3315 105 10 MPa 105 GPa0.003sE .............................................................................Ans.

4-53 A tensile test specimen having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Determine

(a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.425 in. (h) The tangent modulus at a stress level of 46,000 psi. (i) The secant modulus at a stress level of 46,000 psi. SOLUTION From the stress-strain diagram:

(a) 3

632 10 26.7 10 psi 26,700 ksi0.0012

E ........................................................................Ans.


82

(b) 40 ksipl .....................Ans.

(c) 73 ksiult .....................Ans.

(d) .05 44 ksi .....................Ans.

(e) .2 47 ksi ......................Ans.

(f) 65 ksif ......................Ans.

(g) 3

213 100.425 4

ftf

f

PA

391.6 10 psi 91.6 ksi ..........................................................................Ans.

(h) 47.43 45.93 1875 ksi

0.004 0.0032tE ........................................................................Ans.

(i) 46 14,370 ksi

0.0032sE ..................................................................................Ans.

4-54 A cast iron pipe has an inside diameter of 70 mm and an outside diameter of 105 mm. The length of the pipe is 2.5 m. The coefficient of thermal expansion for cast iron is = 12.1(10 6)/ C. Determine the dimension changes caused by

(a) An increase in temperature of 70 C. (b) A decrease in temperature of 85 C. SOLUTION

(a) 612.1 10 70 2500 2.12 mmL ..........................................................................Ans.

612.1 10 70 105 0.0889 mmo .........................................................................Ans.

612.1 10 70 70 0.0593 mmi ............................................................................Ans.

(b) 612.1 10 85 2500 2.57 mmL .....................................................................Ans.

612.1 10 85 105 0.108 mmo ......................................................................Ans.

612.1 10 85 70 0.072 mmi .........................................................................Ans.

4-55 A large cement kiln has a length of 225 ft and a diameter of 12 ft. Determine the change in length and diameter of the structural steel shell [ = 6.5(10 6)/ F] caused by an increase in temperature of 250 F.

SOLUTION

66.5 10 250 225 12 4.39 in.L .......................................................................Ans.

66.5 10 250 12 12 0.234 in.d ........................................................................Ans.

4-56 An airplane has a wing span of 40 m. Determine the change in length of the aluminum alloy [ = 22.5(10 6)/ C] wing spar if the plane leaves the ground at a temperature of 40 C and climbs to an altitude where the temperature is 40 C.

SOLUTION

6 322.5 10 80 40 72.0 10 m 72.0 mm ...........................................Ans.


83

4-57 Determine the movement of the pointer of Fig. P4-57 with respect to the scale zero when the temperature increases 80 F. The coefficients of thermal expansion are 6.6(10 6)/ F for the steel and 12.5(10 6)/ F for the aluminum.

SOLUTION The steel post and the steel scale each stretch the same amount

66.6 10 80 20 0.01056 in.s o

The aluminum post stretches

612.5 10 80 20 0.02000 in.a

Therefore, the motion of the pointer relative to the zero on the steel scale is

5 1p a s

0.0472 in. p ..................................................................................................................... Ans.

4-58 A bronze [ B = 16.9(10 6)/ C] sleeve with an inside diameter of 99.8 mm is to be placed over a solid steel [ S = 11.9(10 6)/ C] cylinder, which has an outside diameter of 100 mm. If the temperatures of the cylinder and sleeve remain equal, how much must the temperature be increased in order for the bronze sleeve to slip over the steel cylinder?

SOLUTION

6

6

100 11.9 10 100

99.8 16.9 10 99.8

d T

T

403 CT ................................................................................................................................Ans.

4-59 A steel [E = 30,000 ksi and = 6.5(10 6)/ F] surveyor’s tape ½-in. wide 1/32-in. thick is exactly 100 ft long at 72 F and under a pull of 10 lb. What correction should be introduced if the tape is used to make a 100-ft measurement at a temperature of 100 F and under a pull of 25 lb.

SOLUTION

1L L LTE L L L

where L is the length at 72 F and zero force. When 10 lbP , 10 640 psi

1 2 1 32, 100 ftL , and

6

100 6401 030 10L

99.99787 ftL

Then, when 25 lbP , 25 1600 psi

1 2 1 32, and 28 FT

66

16001 6.5 10 2830 10

LL

100.02140 ftL

0.02140 ft 0.257 in.L

correction 0.257 in. ............................................................................................................Ans.


84

4-60 A 25-mm diameter aluminum [ = 22.5(10 6)/ C, E = 73 GPa, and = 0.33] rod hangs vertically while suspended from one end. A 2500-kg mass is attached at the other end. After the load is applied, the temperature decreases 50 C. Determine

(a) The axial stress in the rod. (b) The axial strain in the rod. (c) The change in diameter of the rod. SOLUTION

(a) 6 22

2500 9.8149.962 10 N/m 50.0 MPa

0.025 4......................................................Ans.

(b) 6

6 69

49.962 10 22.5 10 50 441 10 m/m73 10a T

E......................Ans.

(c) 6

6 69

49.962 100.33 22.5 10 50 1350.86 10 m/m73 10t T

E

61350.86 10 25 0.0338 mmd td ...........................................................Ans.

4-61 The rigid yokes B and C of Fig. P4-61 are securely fastened to the 2-in. square steel (E 30,000 ksi) bar AD. Determine

(a) The maximum normal stress in the bar. (b) The change in length of segment AB. (c) The change in length of segment BC. (d) The change in length of the complete bar. SOLUTION

(a) 3

maxmax

82 10 20,500 psi 20.5 ksi2 2

PA

.................................................................Ans.

(b) 3

6

82 10 8 120.0656 in.

30 10 2 2ABPLEA

........... Ans.

(c) 3

6

12 10 5 120.00600 in.

30 10 2 2BC .............. Ans.

3

6

45 10 4 120.01800 in.

30 10 2 2CD

(d) 0.0776 in.total AB BC CD ......................................................................................Ans.

4-62 A structural tension member of aluminum alloy (E = 70 GPa) has a rectangular cross section 25 75 mm and is 2 m long. Determine the maximum axial load that may be applied if the axial stress is not to exceed 100 MPa and the total elongation is not to exceed 4 mm.

SOLUTION

9

20004 mm

70 10 0.025 0.075PPL

EA 262,500 NP

6100 10 0.025 0.075 187,500 NP A

max 187.5 kNP ......................................................................................................................... Ans.


85

4-63 The tension member of Fig. P4-63 consists of a steel (E = 30,000 ksi) pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid aluminum alloy (E = 10,600 ksi) bar B, which has a diameter of 4 in. Determine the overall elongation of the member.

SOLUTION

120 kipA BP P

3 3

6 2 2 6 2

120 10 3 12 120 10 4 12

30 10 6 4.5 4 10.6 10 4 4A B

0.01164 0.04324 0.0549 in. .....................................................................................Ans.

4-64 A steel (E = 200 GPa) rod, which has a diameter of 30 mm and a length of 1.0 m, is attached to the end of a Monel (E = 180 GPa) tube, which has an internal diameter of 40 mm, a wall thickness of 10 mm, and a length of 2.0 m, as shown in Fig. P4-64. Determine the load required to stretch the assembly 3.00 mm.

SOLUTION

S MP P P

9 2 9 2 2

1000 20003.00 mm

200 10 0.030 4 180 10 0.060 0.040 4P P

3212 10 N 212 kNP ......................................................................................................Ans.

4-65 The floor of a warehouse is supported by an air-dried, red-oak, timber column (see Appendix A for properties), as shown in Fig. P4-65. Contents of the warehouse subject the 12 12-in. column to an axial load of 200 kip. If the column is 30-in. long, determine

(a) The deformation of the column. (b) The normal stress in the column. (c) The bearing stress between the column and the lower bearing plate. (d) The maximum shearing stress in the column. SOLUTION 1800 ksiE

(a) 3

3

200 10 300.0231 in.

1800 10 12 12PLEA

.....................................................................Ans.

(b) 3200 10 1389 psi

12 12PA

................................................................................................Ans.

(c) 3200 10 1389 psi

12 12b .......................................................................................................Ans.

(d) 3

max200 10 694 psi

2 2 12 12PA

..........................................................................................Ans.

4-66 The roof and second floor of a building are supported by the column shown in Fig. P4-66. The column is a structural steel (see Appendix A for properties) section having a cross sectional area of 5700 mm2. The roof and floor subject the column to the axial forces shown. Determine

(a) The amount that the first floor will settle. (b) The amount that the roof will settle. SOLUTION

200 GPaE 1 1030 kN (C)P 2 380 kN (C)P


86

(a) 3

1 9 6

1030 10 35003.1623 3.16 mm

200 10 5700 10PLEA

.............................................Ans.

(b) 3

1 9 6

380 10 35003.1623 1.1667 4.33 mm

200 10 5700 10r ...............................Ans.

4-67 The tension member of Fig. P4-67 consists of a structural steel pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid 2014-T4 aluminum alloy bar B, which has a diameter of 4 in. (see Appendix A for properties). Determine

(a) The change in length of the steel pipe. (b) The overall deflection of the member. (c) The maximum normal and shearing stresses in the aluminum bar. SOLUTION

29,000 ksistE 10,600 ksialE

205 kip (T)AP 120 kip (T)BP

(a) 3

6 2 2

205 10 500.0286 in.

29 10 6 4.5 4A

PLEA

......................................................Ans.

(b) 3

6 2

120 10 400.0360 in.

10.6 10 4 4B

0.0646 in.total A B ....................................................................................................Ans.

(c) 3

max 2120 10 9550 psi

4 4PA

...........................................................................................Ans.

maxmax 4770 psi

2 2PA

.................................................................................................Ans.

4-68 An aluminum alloy (E = 73 GPa) tube A with an outside diameter of 75 mm is used to support a 25-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-68. Determine the minimum thickness t required for the tube if the maximum deflection of the loaded end of the rod must be limited to 0.40 mm.

SOLUTION

total A B

3 3

29 9

35 10 300 35 10 9000.40 mm

73 10 200 10 0.025 4A

6 2 2

2 2

1817.40 10 m 1817.40 mm

75 4i

A

d

57.54 mm 75 2id t

8.73 mmt ................................................................................................................................Ans.


87

4-69 A structural steel (see Appendix A for properties) bar of rectangular cross section consists of uniform and tapered sections as shown in Fig. P4-69. The width of the tapered section varies linearly from 2 in. at the bottom to 5 in. at the top. The bar has a constant thickness of ½ in. Determine the elongation of the bar resulting from application of the 30-kip load P. Neglect the weight of the bar.

SOLUTION 29,000 ksiE

32 in.60yb 0 60 in.y

3 3 60

66 0

6030

3

30 10 25 30 10329 1029 10 2 0.5 2 0.560

0.02586 1.03448 10 40ln 40

0.02586 41.37931 10 4.60517 3.68888

dyy

y

0.0638 in. ............................................................................................................................ Ans.

4-70 Determine the change in length of the homogeneous conical bar of Fig. P4-70 due to its own weight. Express the results in terms of L, E, and the specific weight of the material. The taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

SOLUTION

0 :F 0A W

2

2

3R yR W Vol

3y

ryRL

2 2

003 3 2 6

LLP dy y dy y Ldy

EA E E E E........................................................Ans.

4-71 The bar shown in Fig. P4-71 is made of annealed bronze (see Appendix A for properties). In addition to its own weight, the bar is subjected to an axial tensile load P of 5000 lb at its lower end. Determine the elongation of the bar due to the combined effects of its weight and the load P. Let r = 4 in. and L = 60 in.

SOLUTION

15,000 ksiE 30.320 lb in

60 604 in.

60 15y yR

22 4 6060

3 3R y

W Vol

0 :F 5000 0A W

2 25000 60 9603

R R y


88

2

60

20

5000 60 9603

R yP dy dy dyEA E E R

60 60

26 60 0

60

26 0

5000 22560

15 10 3 15 1060

960 225

3 15 10 60

dy y dyy

dyy

6060 26

0 060

6

0

6010.02387 0.00711 1060 2

1 1536.00 1060

yy

y

3 3 30.19892 10 0.03839 10 0.01280 10

30.225 10 in. ................................................................................................................... Ans.

4-72 A hollow brass (E = 100 GPa) tube A with an outside diameter of 100 mm and an inside diameter of 50 mm is fastened to a 50-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-72. The supports at the top and bottom of the assembly and the collar C used to apply the 500-kN load P are rigid. Determine

(a) The normal stresses in each of the members. (b) The deflection of the collar C. SOLUTION (a) From equilibrium

0 :F 500 0B AT P

2 2 2 30.05 0.1 0.05 500 10 N

4 4B A

where B is a tension stress and A is a compressive stress. Then, expressing that the stretch of the rod is equal to the shrink of the tube

B A in terms of the stresses gives

9 9

2000 1500200 10 100 10B A

1.5B A

Substituting back into the equilibrium equation gives

256.588 N/m 56.6 MPaA ..............................................................................Ans.

1.5 84.9 MPaB A ............................................................................................Ans.

(b) 6

9

56.588 10 15000.849 mm

100 10C A ................................................Ans.


89

4-73 The 7.5 7.5 20-in. oak (E = 1800 ksi) block shown in Fig. P4-73 was reinforced by bolting two 2 7.5 20-in. steel (E = 29,000 ksi) plates to opposite sides of the block. If the stresses in the wood and the steel are to be limited to 4.6 ksi and 22 ksi, respectively, determine

(a) The maximum axial compressive load P that can be applied to the reinforced block. (b) The shortening of the block when the load of part (a) is applied. SOLUTION (a) From equilibrium

0 :F 2 0w sP P P

7.5 7.5 2 2 7.5w s P

where w and s are both compressive stresses. Expressing that

the shrink of the steel is equal to the shrink of the wood s w in terms of the stresses gives

3 329,000 10 1800 10s wL L

16.111s w

If 22 ksis , then 1.36552 ksi 4.6 ksiw , and

1.36552 7.5 7.5 2 22 2 7.5 737 kipP .......................................Ans.

(b) 3

3

22 10 200.01517 in.

29,000 10s .............................................................Ans.

4-74 Five 25-mm diameter steel (E = 200 GPa) reinforcing bars will be used in a 1-m long concrete (E = 31 GPa) pier with a square cross section, as shown in Fig. P4-74. The allowable strengths in compression for steel and concrete are 130 MPa and 9.5 MPa, respectively. Determine the minimum size of pier required to support a 900-kN axial load.

SOLUTION From equilibrium

0 :F 5 900 0s cP P

2 35 0.025 4 900 10 Ns c cA

where s and c are both compressive stresses. Expressing that the

shrink of the steel rods is equal to the shrink of the concrete s c in terms of the stresses gives

9 9200 10 31 10s cL L

6.4516s c

If 9.5 MPac , then 61.29 MPa 130 MPas , and

6 2 2

22

78,902 10 m 78,902 mm

5 25 4cA

b

285 mmb ................................................................................................................................Ans.


90

4-75 A hollow steel (E = 30,000 ksi) tube A with an outside diameter of 2.5 in. and an inside diameter of 2 in. is fastened to an aluminum (E = 10,000 ksi) bar B that has a 2-in. diameter over one-half of its length and a 1-in. diameter over the other half. The assembly is attached to unyielding supports at the left and right ends and is loaded as shown in Fig. P4-75. Determine

(a) The normal stresses in all parts of the bar. (b) The deflection of cross-section a-a. SOLUTION (a) From equilibrium

40 kipB AP P 10 kipC AP P

where AP , BP , and CP are all tension forces. Since the total length of the assembly cannot change, the total stretch must be zero 0A B C . Therefore

3 2 2 3 2 3 2

20 40 24 10 240

30 10 2.5 2 4 10 10 2 4 10 10 1 4A A AP P P

0 kipAP 0 ksiA .......................................................................Ans.

2

0 4012.73 ksi (T)

2 4B ....................................................................................Ans.

2

0 1012.73 ksi 12.73 ksi (C)

1 4C ............................................................Ans.

(b) 3

26

10 10 240.03056 in.

10 10 1 4C

Therefore section a-a moves 0.03056 in. ............................................................................Ans.

4-76 A 150-mm diameter 200-mm long polymer (E = 2.10 GPa) cylinder will be attached to a 45-mm diameter 400-mm long brass (E = 100 GPa) rod by using the flange type of connection shown in Fig. P4-76. A 0.15-mm clearance exists between the parts as a result of a machining error. If the bolts are inserted and tightened, determine

(a) The normal stresses produced in each of the members. (b) The final position of the flange-polymer interface after assembly, with respect to the left support. SOLUTION (a) From equilibrium

p bT T

where pT and bT are both tension forces. When the bolts are tightened, the stretch of the two pieces closes the gap,

0.15 mmp b . Therefore

2 29 9

200 4000.15 mm

2.1 10 0.15 4 100 10 0.045 4p bT T

18,976.74 Np bT T


91

6 22

18,976.741.074 10 N/m 1.074 MPa (T)

0.15 4p ......................................Ans.

6 22

18,976.7411.93 10 N/m 11.93 MPa (T)

0.045 4b ....................................Ans.

(b) 29

18,976.74 2000.10227 mm

2.1 10 0.15 4p

Therefore the interface will be located at 200.1023 mm ...........................................................Ans.

4-77 The assembly shown in Fig. P4-77 consists of a steel bar A (Es = 30,000 ksi and As = 1.25 in.2), a rigid bearing plate C that is securely fastened to bar A, and a bronze bar B (EB = 15,000 ksi and AB = 3.75 in.2). A clearance of 0.015 in. exists between the bearing plate C and bar B before the assembly is loaded. After a load P of 95 kip is applied to the bearing plate, determine

(a) The normal stresses in bars A and B. (b) The vertical displacement of the bearing plate C. SOLUTION (a) From equilibrium (assuming that the gap is closed and the bearing plate

pushes on the brass bar,

0 :F 95 0A BT P

or in terms of stresses

31.25 3.75 95 10 lbA B

in which A is a tension stress and B is a compression stress. If the bearing plate presses against the brass bar, the stretch of the steel rod will exceed the shrink of the brass bar by the initial gap, 0.015 in.A B Therefore

6 6

6 12 2 120.015 in.

30 10 15 10A B

2 3 6250 psiA B

Combining the equilibrium equation and the deformation equation gives

18,932 psi 18.93 ksi (T)A .................................................................................Ans.

19,023 psi 19.02 ksi (C)B .................................................................................Ans.

(b) 6

18,932 6 120.0454 in.

30 10C A .......................................................................Ans.

4-78 A column similar to Fig. P4-74 is being designed to carry a load of 4450 kN. The column, which will have a 500 500 mm square cross section, will be made of concrete (E = 20 GPa) and will be reinforced with 50-mm diameter steel (E = 200 GPa) bars. If the allowable stresses are 120 MPa in the steel and 8 MPa in the concrete, determine

(a) The number of steel bars required. (b) The stresses in the steel and concrete when the bars of part (a) are used. (c) The change in length of a 3-m long column when the bars of part (a) are used. SOLUTION (a) From equilibrium


92

0 :F 4450 0c sP nP

4450 kNc sP nP

Expressing that the steel rods and the concrete must shrink the same amount, c s , in terms of stresses gives

9 920 10 200 10c sL L

10s c

If max 8 MPac , then 80 MPa 120 MPas , and

2 26 2 6 38 10 0.5 0.05 4 80 10 0.05 4 4450 10n n

17.33n

Therefore the required number of rods is 18n ..........................................................................Ans.

(b) If 18n , then

2 22 30.5 18 0.05 4 18 10 0.05 4 4450 10c c

6 27.83 10 N/m 7.83 MPac .............................................................................Ans.

10 78.3 MPas c ..................................................................................................Ans.

(c) 6

9 9

78.3 10 300030001.175 mm

200 10 200 10s

s ........................................Ans.

4-79 A ½-in. diameter alloy-steel bolt (E = 30,000 ksi) passes through a cold-rolled brass sleeve (E = 15,000 ksi) as shown in Fig. P4-79. The cross-sectional area of the sleeve is 0.375 in.2 Determine the normal stresses produced in the bolt and sleeves by tightening the nut ¼ turn (0.020 in.).


st brT F


20.5 4 0.375st br

1.90986st br

in which st is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of the

brass sleeve, and the movement of the nut are related by 0.02st br . Therefore

6 6

6 60.02

30 10 15 10st br 100,000 2st br


25,576 psi 25.6 ksi (C)br ..................................................................................Ans.

48,847 psi 48.8 ksi (T)st ..................................................................................Ans.


93

4-80 The two faces of the clamp shown in Fig. P4-80 are 250 mm apart when the two stainless-steel (Es = 190 GPa) bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy (Ea = 73 GPa) bar with a length of 251 mm can be inserted as shown. Each of the bolts has a cross-sectional area of 120 mm2 and the bar has a cross-sectional area of 625 mm2. After the load P is removed, determine

(a) The axial stresses in the bolts and in the bar. (b) The change in length of the aluminum alloy bar. SOLUTION (a) Expressing the equilibrium equation

2a sF T

in terms of stresses gives

6 6625 10 2 120 10a s

2.60417s a

in which s is a tension stress and a is a compression stress. The stretch of the steel bolts and the shrink

of the aluminum bar are related by 1s a Therefore

9 9

330 2511

190 10 73 10s a 6575.76 10 1.97966s a


6 2327 10 N/m 327 MPas ................................................................................Ans.

6 2125.6 10 N/m 125.6 MPaa .........................................................................Ans.

(b) 6

9

125.6 10 2510.432 mm

73 10a .......................................................................Ans.

4-81 A high-strength steel bolt (Es = 30,000 ksi and As = 0.785 in.2) passes through a brass sleeve (Eb = 15,000 ksi and Ab = 1.767 in.2), as shown in Fig. P4-81. As the nut is tightened, it advances a distance of 0.125 in. along the bolt for each complete turn of the nut. Compute and plot

(a) The axial stresses s (in the steel bolt) and b (in the brass sleeve) as functions of the angle of twist of the nut (0 180 ).

(b) The elongations s (of the steel bolt) and b (of the brass sleeve) as function of (0 180 ). (c) The distance L between the two washers as a function of (0 180 ). SOLUTION (a) From equilibrium

st brT F


0.785 1.767st br

2.25096st br

in which st is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of

the brass sleeve, and the movement of the nut are related by 0.125 360st br in which is in degrees. Therefore

6 6

14 120.12530 10 360 15 10st br 744.0476 1.71429st br



94

422.374 psi (T)st ................... Ans.

187.642 psi (C)br ................... Ans.

(b) 36

140.19711 10 in.

30 10st

st .......................................................................Ans.

36

120.15011 10 in.

15 10br

br ......................................................................Ans.

(c) 312 12 0.15011 10 in.brL ..................................................................Ans.

4-82 The short pier shown in Fig. P4-82 is reinforced with nine steel (E = 210 GPa) reinforcing bars. An axial compressive load P is applied to the pier through the rigid capping plate. The axial load carried by the matrix material is a function of R, the percentage of the cross section taken up by the steel reinforcement bars. The load is also a function of the modulus ratio ER/EM where ER and EM are the modulus of elasticity for the reinforcement material and the matrix material, respectively. For the three matrix-reinforcement combinations listed, compute and plot the percentage of the load carried by the matrix as a function of R (0 R 100 %).


0 :F 9 0M sP P P

9M sP P P

Since the steel rods and the pier must shrink the same amount, M s ,

sM

M M s s

P LP LE A E A

s ss M

M M

E AP PE A


95


1 9 s sM

M M

E AP PE A

Then, since

9 9 100100 100

9 9 1s s

M s M s

A ARA A A A A

100 1

9M

s

AA R

9 1

100 1 100s

M

A RA R R

and

1

1100

M

s

M

PE RPE R

..............Ans.

4-83 A 3-in. diameter 80-in. long aluminum alloy bar is stress free after being attached to rigid supports, as shown in Fig P4-83. Determine the normal stress in the bar after the temperature drops 100 F. Use E = 10,600 ksi and = 12.5(10 6)/ F.

SOLUTION

66 12.5 10 100 0

10.6 10T

E

313.25 10 psi 13.25 ksi (T) ............................................................................Ans.

4-84 A 6-m long 50-mm diameter rod of aluminum alloy [E = 70 GPa, = 0.346, and = 22.5(10 6)/ C] is attached at the ends to supports that yield to permit a change in length of 1.00 mm in the rod when stressed. When the temperature is 35 C, there is no stress in the rod. After the temperature of the rod drops to 20 C, determine

(a) The normal stress in the rod. (b) The change in diameter of the rod. SOLUTION

(a) 69

600022.5 10 55 6000 1 mm

70 10L T LE

6 274.958 10 N/m 75.0 MPa (T) ...................................................................Ans.

(b) dL T LE

6

69

74.958 10 500.346 22.5 10 55 50

70 10

0.0804 mm 0.0804 mm (shrink) ...............................................................Ans.


96

4-85 A bar consists of 3-in. diameter aluminum alloy [E = 10,600 ksi, = 0.33, and = 12.5(10 6)/ F] and 4-in. diameter steel [E = 30,000 ksi, = 0.30, and = 6.6(10 6)/ F] parts, as shown in Fig. P4-85. If end supports are rigid and the bar is stress free at 0 F, determine

(a) The normal stress in both parts of the bar at 80 F. (b) The change in diameter of the steel part of the bar. SOLUTION (a) From equilibrium

s aP P

in which both sP and aP are compressive forces. Since the end supports are rigid, the total stretch of the rod must be zero 0s a

626

626

206.6 10 80 20

30 10 4 4

30 12.5 10 80 30 0

10.6 10 3 4

s

a

P

P

89, 449 lb (C)s aP P

289, 449 7118 psi 7.12 ksi

4 4s ........................................................................Ans.

289, 449 12,654 psi 12.65 ksi

3 4a ...................................................................Ans.

(b) dsL T LE

66

7118 40.3 6.6 10 80 4 0.00240 in.

30 10..........................Ans.

4-86 A steel tie rod containing a rigid turnbuckle (see Fig. P4-86) has its ends attached to rigid walls. During the summer when the temperature is 30 C, the turnbuckle is tightened to produce a stress in the rod of 15 MPa. Determine the normal stress in the rod in the winter when the temperature is 10 C. Use E = 200 GPa and = 11.9(10 6)/ C.

SOLUTION

30 10 L T LE

6

69 9

15 100 11.9 10 40

200 10 200 10L L L

6 2110.2 10 N/m 110.2 MPa ...........................................................................Ans.

4-87 Nine ¾-in. diameter steel (E = 30,000 ksi) reinforcing bars were used when the short concrete (E = 4500 ksi) pier shown in Fig. P4-87 was constructed. After a load P of 150 kip was applied to the pier, the temperature increased 100 F. The coefficients of thermal expansion for steel and concrete are 6.6(10 6)/ F and 6.0(10 6)/ F, respectively. Determine

(a) The normal stresses in the concrete and in the steel bars after the temperature increases.


97

(b) The change in length of the pier resulting from the combined effects of the temperature change and the load.

SOLUTION (a) The equilibrium equation

0 :F 9 150 0c sP P

is written in terms of stresses

2 2 3100 9 0.75 4 9 0.75 4 150 10 lbc s

24.1504 37,725.62 psis c

in which c and s are both compressive stresses. Expressing that

the steel rods and the concrete must stretch the same amount, c s , in terms of stresses gives

6 66 66.0 10 100 6.6 10 100

4.5 10 30 10c sL LL L

6.6667 1800.00 psis c


1165.770 psi 1.166 ksi (C)c .............................................................................Ans.

9571.805 psi 9.57 ksi (C)s ...............................................................................Ans.

(b) 66

9571.805 246.6 10 100 24 0.00818 in.

30 10............................Ans.

4-88 The assembly shown in Fig. P4-88 consists of a steel (E = 210 GPa) cylinder A, a rigid bearing plate C, and an aluminum alloy (E = 71 GPa) bar B. Cylinder A has a cross-sectional area of 1850 mm2, and bar B has a cross-sectional area of 2500 mm2. After an axial load of 600 kN is applied, the temperature of cylinder A decreases 50 C and the temperature of bar B increases 25 C. The coefficients of thermal expansion are 11.9(10 6)/ C for the steel and 22.5(10 6)/ C for the aluminum. Determine

(a) The normal stresses in the cylinder and in the bar after the load is applied and the temperatures change. (b) The displacement of plate C after the load is applied and the temperatures change. SOLUTION (a) The equilibrium equation

0 :F 600 0B AT T

can be written in terms of stresses

6 6 31850 10 2500 10 600 10 NA B

6 21.35135 324.324 10 N/mA B

in which A and B are both tension stresses. The total stretch of the assembly must be zero,

0A B , therefore

6 69 9

750 60011.9 10 50 750 22.5 10 25 600 0

210 10 71 10A B

6 22.36620 30.450 10 N/mA B


6 2217.5 10 N/m 217 MPa (T)A ....................................................................Ans.


98

6 279.051 10 N/m 79.1 MPa (C)B ...............................................................Ans.

(b) 6

69

217.5 10 75011.9 10 50 750

210 10C A

0.331 mm C .........................................................................................................Ans.

4-89 A high-strength steel [Es = 30,000 ksi, As = 0.785 in.2, and s = 6.6(10 6)/ F] bolt passes through a brass [Eb = 15,000 ksi, Ab = 1.767 in.2, and b = 9.8(10 6)/ F] sleeve, as shown in Fig. P4-89. After the unit is assembled at 40 F, the temperature is increased to 100 F. If the unit is free of stress at 40 F, determine the normal stresses in the bolt and in the sleeve at 100 F.

SOLUTION Writing the equilibrium equation

st brT F


0.785 1.767st br

in which st is a tension stress and br is a compressive stress. Expressing that the steel bolt and the brass sleeve must stretch the same amount, st br , in terms of stresses gives

6 66 6

14 126.6 10 60 14 9.8 10 60 12

30 10 15 10st br

14 24 45,360 psist br


1839.29 psi 1839 psi (T)st ................................................................................Ans.

817.10 psi 817 psi (C)br ...................................................................................Ans.

4-90 The two faces of the clamp of Fig. P4-90 are 250 mm apart when the two stainless-steel [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy [Ea = 73 GPa, Aa = 625 mm2, and a = 22.5(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature is raised 100 C. Determine the normal stresses in the bolts and in the bar, and the distance between the faces of the clamps.


2a sF T


625 2 115a s

in which s is a tension stress and a is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the aluminum bar must stretch, 0.5s a , in terms of stresses gives


99

69

69

33017.3 10 100 330

190 10

250.5 0.5 22.5 10 100 250.5

73 10

s

a

6 21.73684 3.43151 492.725 10 N/ms a


6 260.448 10 N/m 60.4 MPa (C)a .................................................................Ans.

6 2164.262 10 N/m 164.3 MPa (T)s .............................................................Ans.

Then, the stretch of the aluminum bar is

6

69

60.448 10 250.522.5 10 100 250.5 0.35620 mm

73 10a

and the distance between the faces of the clamp is

250.5 250.856 mmaL ....................................................................................Ans.

4-91 A prismatic bar [E = 10,000 ksi and = 12.5(10 6)/ F], free of stress at room temperature, is fastened to rigid walls at its ends. One end of the bar is heated 200 F above room temperature while the other end is maintained at room temperature. The change in temperature T along the bar is proportional to the square of the distance from the unheated end. Determine the normal stress in the bar after the change in temperature.

SOLUTION

dxd T dxE

2

2

200xTL

2

20 0

200 0L Ldx xd dxE L

3

2

200 03

L LE L

6 6200 12.5 10 10 10200

3 3E

8333 psi 8.33 ksi (C) ......................................................................................Ans.

4-92 The two faces of the clamp shown in Fig. P4-90 are 250 mm apart when the two steel bolts [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an brass bar [Eb = 100 GPa, Ab = 625 mm2, and b = 17.6(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature of the system is slowly raised. Compute and plot

(a) The axial stress s in the steel bolts and the axial stress b in the brass bar as functions of the temperature rise T (0 T 100 C).

(b) The elongation s of the steel bolts and the elongation b of the brass bar as functions of the temperature rise T (0 T 100 C).

(c) The distance L between the faces of the clamp as a function of the temperature rise T (0 T 100 C).

SOLUTION (a) Writing the equilibrium equation


100

2b sP T


625 2 115b s

in which s is a tension stress and b is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the brass bar must stretch, 0.5s b , in terms of stresses gives

69

69

33017.3 10 330

190 10

250.5 0.5 17.6 10 250.5

100 10

s

b

T

T

6 6 21.73684 2.5050 500 10 1.3002 10 T N/ms b


6 269.207 0.17997 10 N/m (C)b T .........................................................Ans.

6 2188.063 0.48904 10 N/m (T)s T ........................................................Ans.

(b) Then, the elongation of the steel bolts and the brass bar are

6

69

188.063 0.48904 10 33017.3 10 330

190 10sT

T

30.32664 4.860 10 mmT .............................................................................Ans.

6

69

69.207 0.17997 10 250.517.6 10 250.5

100 10bT

T

30.17336 4.860 10 mmT Ans.

(both stretches). (c) Finally, the distance between the faces of

the clamp is

250.5 bL

6250.32664 4.860 10 mmT ..........................................................................Ans.


101

4-93 An aluminum (Eal = 10,000 ksi, al = 12.5 10-6/ F, Aal = 1.4 in.2) bolt passes through a steel (Est = 30,000 ksi, st = 6.6 10-6/ F, Ast = 0.4 in.2) sleeve as shown in Fig. P4-89. Initially, the nut is tightened against the

washer at room temperature until the bolt has a tensile force of 3500 lb. Then the temperature of the assembly is slowly raised. Calculate and plot:

(a) The stress al in the aluminum bolt and the stress st in the steel sleeve as a function of the temperature increase T (0 F < T < 100 F).

(b) The change in length of the aluminum bolt al and the change in length of the steel sleeve st as a function of the temperature increase T (0 F < T < 100 F).

SOLUTION (a) Writing the equilibrium equation

al stT P


1.4 0.4al st

in which al is a tension stress and st is a compressive stress. The aluminum bolt must stretch more than

the steel sleeve stretches by the amount that the nut moves, al st nut . In terms of stresses this gives

6 66 6

14 1212.5 10 14 6.6 10 12

10 10 30 10al st

nutT T

When 0T the tension in the bolt and the compression in the sleeve are each equal to 3500 lb and

3500 2500 psi1.4al

3500 8750 psi0.4st

and the initial movement of the nut is

0.00700 in.nut

Therefore, the deformation equation can be written

14 4 70,000 958.00 psial st T


2500 34.214 psi (T)al T ..............................................................................Ans.

8750 119.750 psi (C)st T ............................................................................Ans.

If the assembly is not welded together, neither stress can be negative. Therefore, the above equations are valid only for T less than about 73 . For T greater than about 73 , the bolt and the sleeve will separate, and both stresses will be zero

0al st for 73T ..................................................................................Ans.


102

(b) Then, the change in length of the aluminum bolt and the steel sleeve are

66

1412.5 10 14 in. (stretch)

10 10al

al T .............................................Ans.

66

126.6 10 12 in. (stretch)

30 10st

st T .............................................Ans.

4-94 A short standard-weight steel pipe (see Appendix A) is used to support an axial compressive load of 100 kN. If yielding ( y = 250 MPa) should not occur, and the factor of safety is to be 1.6, determine the smallest nominal diameter pipe that may be used to support the load.

SOLUTION

3 6100 10 250 10

1.6A 6 2640 10 mA

Use 51 mmd (for which 2693.5 mmA ) ................................................................Ans.

4-95 A short column made of structural steel is used to support the floor beams of a building, as shown in Fig. P4-95. Each floor beam (A and B) transmits a force of 40 kip to the column. The column has the shape of a wide-flange (W) section (see Appendix A). The factor of safety based on failure by yielding is 3.0. Select the lightest wide-flange section that will support the given loads.

SOLUTION

36 ksiy 3.0FS

3 380 10 36 10

3.0A 26.667 in.A

W6 25 , W8 24 , W10 30 , W12 30 , etc. all have sufficiently large area. The lightest section is

W8 24 ....................................................................................................................................Ans.

4-96 The two structural steel (see Appendix A) rods A and B shown in Fig. P4-96 are used to support a mass m = 2000 kg. If failure is by yielding and a factor of safety of 1.75 is specified, determine the diameters of the rods (to the nearest 1 mm) that must be used to support the mass. Both rods are to have the same diameter.

SOLUTION

250 MPay

1.75FS From a free body diagram of the connection, the equilibrium equations

0 :xF cos30 cos50 0B AT T

0 :yF sin 50 sin 30 2000 9.81 0A BT T

give

6250 1017,254 N

1.75AT A 6 2120.8 10 mA

6250 1012,806 N

1.75BT A 6 289.6 10 mA

If the rods are made from the same material and diameter, then

2

2120.8 mm4d A 12.4 mmd

Use 13 mmd ........................................................................................................................ Ans.


103

4-97 The machine component shown in Fig. P4-97 is made of hot-rolled Monel. The forces at B are applied to the component with a rigid collar that is firmly attached to the component. If the mode of failure is yielding and the factor of safety is 1.5, determine the minimum permissible diameter of each segment of the machine component.

SOLUTION

50 ksiy 1.5FS

20 kip (T)ABF 3 3

2

20 10 50 104 1.5d

0.874 in.ABd ................................................................................................................Ans.

30 kip (C)BCF 3 3

2

30 10 50 104 1.5d

1.070 in.BCd .................................................................................................................Ans.

4-98 An axial load P = 1000 kN is applied to the rigid steel bearing plate on the top of the short column shown in Fig. P4-98. The outside segment of the column is made of structural steel. The inside core is made of fairly high strength concrete. Both segments are square. The failure modes are yielding for the steel and fracture for the concrete. The factor of safety is to be 1.4. If the area of the concrete is to be 10 times the area of the steel, determine the required dimensions.

SOLUTION

34 MPafc 250 MPays

1.4FS 10c sA A

Writing the equilibrium equation

1000 kNs cF F


31000 10 Ns s c cA A

in which c and s are both compressive stresses. Expressing that the steel and the concrete must stretch the

same amount, c s , in terms of stresses gives

9 9200 10 31 10s cL L

6.4516s c

If 34 MPac fc , then

6.4516 219.355 MPa 250 MPas c

Therefore

6 6 3219.355 10 34 10 10 1000 10 Ns sA A

6 21787.77 10 msA 6 210 17,877.7 10 mc sA A

and the required sizes are

concrete 133.7 mm 133.7 mm .............................................................................Ans.

2 17,877.7 1787.77b 140.2 mmb

steel 140.2 mm 140.2 mm .............................................................................Ans.


104

4-99 Four axial forces are applied to the 1-in. thick, 0.4% C hot-rolled steel bar, as shown in Fig. P4-99. The factor of safety for failure by yielding is 1.75. Determine the minimum width w of the constant cross-sectional area bar.

SOLUTION

53 ksiy 1.75FS

20 kip (T)ABF 3 320 10 53 10

1 1.75w 0.660 in.w

10 kip (C)BCF 3 310 10 53 10

1 1.75w 0.330 in.w

50 kip (T)BCF 3 350 10 53 10

1 1.75w 1.651 in.w

1.651 in.w .....................................................................................................................Ans.

4-100 The two parts of the eyebar shown in Fig. P4-100 are connected by two bolts (one on each side of the eyebar). The bolts are made of a grade of steel with a tensile yield strength of 1035 MPa and a shear yield strength of 620 MPa. The eyebar is subjected to the forces P = 85 kN. Determine the minimum bolt diameter required to safely support the forces if the mode of failure is yielding and the factor of safety is 1.5.

SOLUTION Since there are two bolts, there are two normal forces and two shear forces shown on the free-body diagram. The equilibrium equations gives

0 :nF 2 85cos 30 0N

0 :tF 2 85sin 30 0V

give

3 6 2

385 10 1035 10cos30 36.806 102 1.5 4

dN 0.00824 md

3 6 2

385 10 620 10sin 30 21.25 102 1.5 4

dV 0.00809 md

min 8.24 mmd ...............................................................................................................Ans.

4-101 The two solid rods shown in Fig. P4-101 are pin-connected at the ends and support a weight of 10 kip. The rods are made of SAE 4340 heat-treated steel. The factor of safety for failure by yielding is to be 1.5. For a minimum weight of rod design, determine

(a) The optimum angle . (b) The required diameter for the rods. (c) The weight of each rod. Is it reasonable to neglect the weight of the rods in the design? SOLUTION

132 ksiy 1.5FS

30.283 lb/in.

(a) From a free body diagram of the pin connection, the equilibrium equations


0 :yF 1 2sin sin 10,000 0T T


105

give

3

1 2

132 1010,0002sin 1.5

AT T

3256.818 10 in.

sinA

The optimum angle is the angle that makes the volume (and therefore the weight and cost) of the rods a minimum. The volume of each rod is

356.818 10 25 12 17.04545 34.0909

sin cos sin cos sin 2V AL

The minimum volume occurs when sin 2 1

45 .............................................................................................................................. Ans.

(b) When 45

2 3

256.818 10 in.4 sin 45dA

0.31986 in. 0.320 in.d ..........................................................................................Ans.

(c) When 45

34.09090.283 9.65 lbsin 90

W V ....................................................................Ans.

Yes, it is safe to neglect the weight of the rods.....................................................................Ans.

4-102 A tension member consists of a 50-mm diameter brass (E = 100 GPa) bar connected to a 32-mm diameter stainless steel (E = 190 GPa) bar, as shown in Fig. P4-102. For an applied load P = 50 kN, determine

(a) The normal stresses in each segment of the member. (b) The elongation of the member. SOLUTION

50 kNAB BCF F

(a) 3

6 22

50 10 25.5 10 N/m 25.5 MPa0.05 4AB ...............................................Ans.

3

6 22

50 10 62.2 10 N/m 62.2 MPa0.032 4BC .............................................Ans.

(b) 3 3

2 29 9

50 10 1500 50 10 10000.709 mm

100 10 0.05 4 190 10 0.032 4......Ans.

4-103 An alloy steel (E = 30,000 ksi) bar is loaded and supported as shown in Fig. P4-103. The loading collar at B is free to slide on section BC. The diameters of sections AB, BC, and CD are 2.50 in., 1.50 in., and 1.00 in., respectively. The lengths of all three segments are 15 in. Determine the normal stresses in each section and the overall change in length of the bar.

SOLUTION

20 kipABT

60 kipBCT

10 kipCDT


106

3

220 10 4074 psi 4.07 ksi (C)2.5 4AB ........................................................Ans.

3

260 10 33,950 psi 33.9 ksi (T)1.5 4BC .....................................................Ans.

3

210 10 12,730 psi 12.73 ksi (T)

1 4CD .....................................................Ans.

3 3 3

2 2 26 6 6

20 10 15 60 10 15 10 10 15

30 10 2.5 4 30 10 1.5 4 30 10 1 4

0.0213 in. .................................................................................................................. Ans.

4-104 The floor beams of a storage shed are supported as shown in Fig. P4-104. Each of the floor beams B and C transmits a 50 kN load to post A. Post A, the baseplate, and the footing have cross-sectional areas of 15,000 mm2, 30,000 mm2, and 260,000 mm2, respectively. Determine

(a) The normal stress in post A. (b) The bearing stress between the post and the baseplate. (c) The bearing stress between the baseplate and the footing. (d) The bearing stress between the footing and the ground. SOLUTION

(a) 3

6 26

100 10 6.67 10 N/m 6.67 MPa15,000 10n ...............................................Ans.

(b) 3

6 26

100 10 6.67 10 N/m 6.67 MPa15,000 10b ...............................................Ans.

(c) 3

6 26

100 10 3.33 10 N/m 3.33 MPa30,000 10b ...............................................Ans.

(d) 3

3 26

100 10 385 10 N/m 385 kPa260,000 10b .................................................Ans.

4-105 A 1-in.-diameter steel [ = 6.5(10 6)/ F, E = 30,000 ksi, and = 0.30] bar is subjected to a temperature decrease of 150 F. The ends of the bar are supported by two walls that displace a small amount during the temperature change. If the measured strain in the bar is 600 in./in. after the temperature change, determine the load being transmitted to the walls.

SOLUTION

6600 10TE

6 66 6.5 10 150 600 10

30 10

11,250 psi 11.25 ksi (T)

211.25 1 4 8.84 kipP A ...................................................................Ans.


107

4-106 A 90-mm diameter brass (E = 100 GPa) bar is securely fastened to a 50-mm diameter steel (E = 200 GPa) bar. The ends of the composite bar are then attached to rigid supports, as shown in Fig. P4-106. Determine the stresses in the brass and the steel after a temperature drop of 70 C occurs. The thermal coefficients of expansion for the brass and the steel are 17.6(10 6)/ C and 11.9(10 6)/ C, respectively.


b sF F


2 290 4 50 4b s 3.24s b

in which s and b are both tensile stresses. Since the supports are rigid, the total stretch of the bar must be zero,

0b s . In terms of stresses

6 69 9

800 48017.6 10 70 800 11.9 10 70 480 0

100 10 200 10b s

9 28 2.4 1.38544 10 N/mb s


6 287.82 10 N/m 87.8 MPa (T)b ....................................................................Ans.

6 2284.5 10 N/m 285 MPa (T)s .....................................................................Ans.

4-107 A steel (E = 30,000 ksi) pipe column with an outside diameter of 3 in. and an inside diameter of 2.5 in. is attached to unyielding supports at the top and bottom as shown in Fig. P4-107. A rigid collar C is used to apply a 50-kip load P. Determine

(a) The normal stresses in the top and bottom portions of the pipe. (b) The deflection of the collar C. SOLUTION

2 2 23 2.5 4 2.15984 in.A

(a) Writing the equilibrium equation

50 0B AP T


32.15984 2.15984 50 10 lbB A

in which A is a tension stress and B is a compressive stress. Since the supports are rigid, the stretch of the top section must be the same as the shrink of the bottom section, A B . In terms of stresses

6 6

7 12 4 1230 10 30 10A B 1.75B A


8418 psi 8.42 ksi (T)A .......................................................................................Ans.

14,732 psi 14.73 ksi (C)B .................................................................................Ans.

(b) 6

8418 7 120.0236 in.

30 10C A ................................................................Ans.


108

4-108 A 3-mm diameter cord (E = 7 GPa) that is covered with a 0.5-mm thick plastic sheath (E = 14 GPa) is subjected to an axial tensile load P, as shown in Fig. P4-108. The load is transferred to the cord and sheath by rigid blocks attached to the ends of the assembly. The yield strengths for the cord and sheath are 15 MPa and 56 MPa, respectively. Determine the maximum allowable load if a factor of safety of 3 with respect to failure by yielding is specified.

SOLUTION Equilibrium gives that the total force is the sum of the force carried by the cord and the force carried by the sheath

C SP P P

The cord and the sheath must both stretch the same amount, C S . In terms of stresses

9 97 10 14 10C SL L

2S C

If max 15 3 5 MPaC C , then

562 10 MPa 18.667 MPa3S C

Therefore

26 6 2 2

max 5 10 0.003 4 10 10 0.004 0.003 4P

max 90.3 NP ................................................................................................................... Ans.

4-109 The assembly shown in Fig. P4-109 consists of a steel (Es = 30,000 ksi, As = 1.25 in.2) bar A, a rigid bearing plate C that is securely fastened to bar A, and a bronze (Eb = 15,000 ksi, Ab = 3.75 in.2) bar B. A clearance of 0.025 in. exists before the assembly is loaded by a force P = 15 kip. Determine, for each segment of the assembly

(a) The normal stress. (b) The change in length. SOLUTION (a) Writing the equilibrium equation

15 0A BT P


31.25 3.75 15 10 lbA B

in which A is a tension stress and B is a compressive stress. Assuming that the force is sufficient to close the gap, the stretch of bar A must be greater than the shrink of bar B by the amount of the clearance,

0.025 in.A B In terms of stresses

6 6

6 12 2 120.025 in.

30 10 15 10A B

372 48 750 10 psiA B


310.7045 10 psi 10.70 ksi (T)A .....................................................................Ans.

431.8 psi 0.432 ksi (C)B ...................................................................................Ans.

(b) 3

6

10.7045 10 6 120.0257 in. (stretch)

30 10A ..............................................Ans.


109

6

431.8 2 120.000691 in. (shrink)

15 10B .........................................................Ans.

4-110 An 80-kN force P is applied to the 150 180-mm wood block shown in Fig. P4-110. Determine the normal stress perpendicular to the grain of the wood and the shearing stress parallel to the grain of the wood.

SOLUTION From a free-body diagram, the equilibrium equations

0 :nF 80sin 25 0N

0 :tF 80cos 25 0V

give

80sin 25 n nN A

80cos 25 n nV A

where the areas are related by

20.150 0.180 mm sin 25nA A

Therefore

3 2

3 280 10 sin 25

529 10 N/m 0.529 MPa0.150 0.180n .....................................Ans.

3

6 280 10 sin 25 cos 25

1.135 10 N/m 1.135 MPa0.150 0.180n .......................Ans.

4-111 A 4000-lb force P is applied to the square structural steel block shown in Fig. P4-111. Determine (a) The change in length of the block. (b) The maximum normal stress in the block. (c) The maximum shearing stress in the block. (d) The normal stress perpendicular to the plane A-A. (e) The shearing stress parallel to the plane A-A. SOLUTION

sin 3 5 cos 4 5

3 3 cosnA A 211.25 in.nA

(a) 36

4000 210.322 10 in.

29 10 3 3PLEA

.........................................................Ans.

(b) max4000 444 psi3 3

PA

..........................................................................................Ans.

(c) maxmax 222 psi

2 2PA

.......................................Ans.

From a free-body diagram, the equilibrium equations

0 :nF 4000cos 0N

0 :tF 4000sin 0V

give

4000 4 5 11.25nN


110

(d) 284 psin ..................................................................................................................... Ans.

4000 3 5 11.25nV

(e) 213 psin ...................................................................................................................... Ans.


111

.Chapter 5 5-1 Two forces are applied to a bracket as shown in Fig. P5-1. Determine (a) The moment of force F1 about point O. (b) The moment of force F2 about point O. SOLUTION

(a) ( )1 10 10 100 lb in. = = ⋅M .........................................................................................Ans.

(b) ( )2 25 21 525 lb in. = = ⋅M ........................................................................................Ans.

5-2 Determine the moments of the 225-N force shown in Fig. P5-2 about points A, B, and C. SOLUTION

( )225 0.6 135 N m A = = ⋅M .....................................................................................Ans.

( )225 0.4 90 N m B = = ⋅M .......................................................................................Ans.

( )225 0.8 180 N m C = = ⋅M .....................................................................................Ans.

5-3 Two forces are applied at a point in the plane of a rigid steel plate as shown in Fig. P5-3. Determine the moments of

(a) The 500-lb force about points A and B. (b) The 300-lb force about points B and C. SOLUTION

(a) ( )500 30 15,000 lb in. A = = ⋅M ...............................................................................Ans.

(b) ( )500 20 10,000 lb in. B = = ⋅M ...............................................................................Ans.

(c) ( )300 30 9000 lb in. B = = ⋅M ..................................................................................Ans.

(d) ( )300 25 7500 lb in. C = = ⋅M ..................................................................................Ans.

5-4 Two forces are applied to the bridge truss shown in Fig. P5-4. Determine the moments of (a) The 3.6-kN force about points A and D. (b) The 2.7-kN force about point A. SOLUTION

(a) ( )3.6 3 10.80 kN m A = = ⋅M ....................................................................................Ans.

( )3.6 3 10.80 kN m D = = ⋅M ....................................................................................Ans.

(b) ( )2.7 3 8.1 kN m A = = ⋅M .........................................................................................Ans.

5-5 A 50-lb force is applied to the handle of a lug wrench, which is being used to tighten the nuts on the rim of an automobile tire as shown in Fig. P5-5. The diameter of the bolt circle is 5.5 in. Determine the moments of the force about the axle of the wheel (point O) and about the point of contact of the wheel with the pavement (point A).

SOLUTION

( )( )( )( )50cos 20 20cos 20

50sin 20 2.75 20sin 20O = ° °

+ ° + °

M

1047 lb in. = ⋅ ...............................................................Ans.


112

( )( )( )( )50cos 20 20cos 20

50sin 20 16.75 20sin 20A = ° °

+ ° + °

M

1286 lb in. = ⋅ ....................................................................................................Ans.

5-6 A 160-N force is applied to the handle of a door as shown in Fig. P5-6. Determine the moments of the force about the hinges A and B.

SOLUTION

( )( ) ( )( )160cos 45 0.6 160sin 45 0.8 158.4 N m A = ° + ° = ⋅M ......................Ans.

( )( ) ( )( )160sin 45 0.8 160cos 45 0.5 33.9 N m B = ° − ° = ⋅M ........................Ans.

5-7 Determine the moment of the 425-lb force shown in Fig. P5-7 about point B. SOLUTION

( )( ) ( )( )425cos35 16 425sin 35 16 9470 lb in. B = ° + ° = ⋅M ........................Ans.

5-8 A pry bar is used to extract a nail from a board as shown in Fig. P5-8. Determine the moment of the 120-N force

(a) About point A. (b) About point B. SOLUTION

(a) ( )( )( )( )120cos 20 0.75

120sin 20 0.46A = °

+ °

M

103.5 N m = ⋅ ..........................................Ans.

(b) ( )( )( )( )120cos 20 0.65

120sin 20 0.460B = °

+ °

M

92.2 N m= ⋅ .............................................Ans.

5-9 A man exerts a force P to hold a 60-ft pole in the position shown in Fig. P5-9. If the moment of the force P about point A is 4000 lb ft⋅ , determine the magnitude of the force P.

SOLUTION 60cos50 38.567 ftb = ° =

60sin 50 45.963 fth = ° =

1tan 23.72866hb

φ −= = °+

( )( ) ( )( )cos 45.963 sin 38.567 4000 lb ftA P Pφ φ= − = ⋅M

150.6 lbP = .......................................................................................................................Ans.

5-10 Two forces act on a truss as shown in Fig. P5-10. Member AC of the truss is perpendicular to members BC and CD. Determine the moment of

(a) The 4-kN force about point A. (b) The 3-kN force about point D. SOLUTION

(a) ( )4 2sin 60 6.93 kN m A = ° = ⋅M ............................................................................Ans.

(b) ( )3 4sin 60 10.39 kN m D = ° = ⋅M ..........................................................................Ans.


113

5-11 The crane and boom shown in Fig. P5-11 is lifting a 4000-lb load. Determine the moment of the load about point C.

SOLUTION

( )4000 24cos30 1 1 83,100 lb ft C = °− + = ⋅M .....................................................Ans.

5-12 Due to combustion, a compressive force P is exerted on the connecting rod of an automobile engine as shown in Fig. P5-12. The lengths of the crank throw AB and connecting rod BC are 75 mm and 225 mm, respectively. Determine the moment of the force P about the bearing at A in terms of the crank angle θ.

SOLUTION From the Law of Sines

sin sin75 225φ θ=

sinsin3θφ =

Also

2

2 sincos 1 sin 19θφ φ= − = −

Then

( )( ) ( )( )cos 75sin sin 75cosA P Pφ θ φ θ= +M

( )75 sin cos cos sinP θ φ θ φ= +

2sin sin cos75 sin 1

9 3P θ θ θθ

= − + ..........................................................Ans.

5-13 A 760-lb force acts on a bracket as shown in Fig. P5-13. Determine the moment of the force about point A (a) Using the vector approach. (b) Using the scalar approach. SOLUTION

(a) ( ) ( )10 12 760cos 40 760sin 40A = − × ° + °M i j i j

( ) ( ) ( )10 760sin 40 12 760cos 40 11,870 lb in.= ° + ° = ⋅ k k .....................Ans.

(b) ( )( ) ( )( )760cos 40 12 760sin 40 10A = ° + °M

( )11,870 lb in. 11,870 lb in.= ⋅ = ⋅k ............................................................Ans.

5-14 Determine the moment of the 675-N force shown in Fig. P5-14 about point O (a) Using the vector approach. (b) Using the scalar approach. SOLUTION

(a) ( ) ( )0.230 0.250 675cos 22 675sin 22O = + × ° − °M i j i j

( ) ( ) ( )0.230 675sin 22 0.250 675cos 22 215 N m= − ° − ° = − ⋅ k k .........Ans.

(b) ( )( ) ( )( )675cos 22 0.250 675sin 22 0.230O = ° + °M

( )215 N m 215 N m= ⋅ = − ⋅k ......................................................................Ans.


114

5-15 Two forces F1 and F2 are applied to a gusset plate as shown in Fig. P5-15. Determine the moment (a) Of force F1 about point A. (b) Of force F2 about point B. SOLUTION

(a) ( ) ( )9 4 550sin 45 550cos 45A = + − ° + °M i j × i j

( ) ( ) ( )9 550cos 45 4 550sin 45 5060 lb in.= ° + ° = ⋅ k k ............................Ans.

(b) ( ) ( )7 4 750sin 60 750cos 60B = − + × ° + °M i j i j

( ) ( ) ( )7 750cos 60 4 750sin 60 5220 lb in.= − ° − ° = − ⋅ k k ......................Ans.

5-16 Two forces F1 and F2 are applied to a bracket as shown in Fig. P5-16. Determine the moment (a) Of force F1 about point O. (b) Of force F2 about point A. SOLUTION

(a) ( ) ( )0.300 0.500 5cos 45 5sin 45O = + × ° + °M i j i j

( ) ( ) ( )0.300 5sin 45 0.500 5cos 45 0.707 kN m= ° − ° = − ⋅ k k ................Ans.

(b) ( ) ( )0.265 0.375 3cos 45 3sin 45A = + × ° − °M i j i j

( ) ( ) ( )0.265 3sin 45 0.375 3cos 45 1.358 kN m= − ° − ° = − ⋅ k k .............Ans.

5-17 Determine the moment of the contact force F of the crutch shown in Fig. P5-17 about point A. The length of the crutch is 5 ft and F = 35 lb.

SOLUTION

( ) ( )5cos 70 5sin 70 35cos 45 35sin 45A = ° − ° − ° + °M i j × i j

( )( ) ( )( )5cos 70 35sin 45 5sin 70 35cos 45= ° ° − − ° − ° k

( )74.0 lb ft= − ⋅k ....................................................................................................Ans.

5-18 The moment of the force F shown in Fig. P5-18 about point A is (–2700k) N-mm, and its moment about point C is ( )7500 N mm− ⋅k . Determine the magnitude and orientation (angle θ) of the force F.

SOLUTION

( ) ( )120 75 cos sinA F Fθ θ= + × +M i j i j

( ) ( ) ( )120 sin 75 cos 2700 N mmF Fθ θ= − = − ⋅ k k

( ) ( ) ( ) ( ) ( )75 cos sin 75 cos 7500 N mmC F F Fθ θ θ= × + = − = − ⋅ M j i j k k

Therefore 75 cos 7500 N mmF θ = ⋅

120 sin 7500 2700 4800 N mmF θ = − = ⋅

sin 40tancos 100FF

θθθ

= =

21.80θ = ° .................................................................................................................Ans.

107.7 NF = ..............................................................................................................Ans.


115

5-19 A 970-lb force acts at a point in a body as shown in Fig. P5-19. Determine the moment of the force about point C.

SOLUTION

( )2 2

25 30970 620.98 745.17 lb25 30

− += = − ++

i jF i j

( ) ( )30 20 620.98 745.17C = − + − +M j k × i j

( )14,900 12,420 18,630 lb in.= − − − ⋅i j k .......................................................Ans.

5-20 A 200-N force is applied to a pipe wrench as shown in Fig. P5-20. Determine the moment of the force about point A. Express the result in Cartesian vector form.

SOLUTION

( ) ( )0.175 0.580 0.250 200A = − + + × −M i j k k

( )116.0 35.0 N m= − − ⋅i j ....................................................................................Ans.

5-21 A 650-lb force acts at a point in a body as shown in Fig. P5-21. Determine (a) The moment of the force about point A. (b) The direction angles associated with the moment vector. SOLUTION

(a) ( )2 2 2

8 16 30650 148.876 297.751 558.283 lb8 16 30+ += = + ++ +

i j kF i j k

( ) ( )15 24 148.876 297.751 558.283A = − − + +M i j × i j k

( )13,399 8374 893.2 lb in.= − + − ⋅i j k .............................................................Ans.

(b) ( ) ( ) ( )2 2 213,399 8374 893.2 15,826 lb in.AM = + + = ⋅

1 13,399cos 147.8515,826xθ

− −= = ° ......................................................................................Ans.

1 8374cos 58.0515,826yθ

−= = ° ............................................................................................Ans.

1 893.2cos 93.2415,826zθ

− −= = ° ...........................................................................................Ans.

5-22 A pipe bracket is loaded as shown in Fig. P5-22. Determine the moment of the force F about point O. SOLUTION

( )2 2 2

75 150 140875 300.40 600.80 560.75 N75 150 140

+ += = + ++ +

i j kF i j k

( ) ( )0.2 0.25 0.15 300.40 600.80 560.75O = + + + +M i j k × i j k

( )50.1 67.1 45.1 N m= − + ⋅i j k ..........................................................................Ans.


116

5-23 The magnitude of the tension T in cable CD of Fig. P5-23 is 150 lb. Determine the moment of T about point B

(a) Using a position vector from B to D. (b) Using a position vector from B to C. SOLUTION

( )2 2 2

20 22 18150 86.315 94.947 77.684 lb20 22 18

− += = − ++ +

i j kT i j k

( ) ( )28 18 86.315 94.947 77.684B = + − +M i k × i j k

( )1709 621 2660 lb in.= − − ⋅i j k ........................................................................Ans.

( ) ( )8 22 86.315 94.947 77.684B = + − +M i j × i j k

( )1709 621 2660 lb in.= − − ⋅i j k ........................................................................Ans.

5-24 A 450-N force F acts on a machine component as shown in Fig. P5-24. The direction angles of F are 70xθ = ° , 30yθ = ° , and 69zθ = ° . Determine the moment of the force about point A. Express your

answer in Cartesian vector form. SOLUTION

( ) ( )450 cos 70 cos30 cos 69 153.909 389.711 161.266 N= ° + ° + ° = + +F i j k i j k

( ) ( )0.3 0.6 0.4 153.909 389.711 161.266A = − + + +M i j k × i j k

( )253 13.18 209 N m= − + + ⋅i j k .......................................................................Ans.

5-25 The magnitude of the force F shown in Fig. P5-25 is 100 lb. Determine the moment of F about the bearing at C.

SOLUTION

( )2 2 2

10 7 6100 73.521 51.465 44.113 lb10 7 6

− + += = − + ++ +

i j kF i j k

( ) ( )9 14 73.521 51.465 44.113C = − − − + +M j k × i j k

( )324 1029 662 lb in.= + − ⋅i j k ..........................................................................Ans.

5-26 Determine the moment of the 800-N force shown in Fig. P5-26 about point D (a) Using a position vector from D to B. (b) Using a position vector from D to E. SOLUTION

(a) ( )2 2 2

400 400 600800 388.057 388.057 582.086 N400 400 600

− −= = − −+ +

i j kF i j k

( ) ( )0.4 0.8 0.6 388.057 388.057 582.086D = − + − −M i j k × i j k

( )699 466 155 N m= + + ⋅i j k .............................................................................Ans.

(b) ( ) ( )0.8 1.2 388.057 388.057 582.086D = − − −M i j × i j k

( )699 466 155 N m= + + ⋅i j k .............................................................................Ans.


117

5-27 The magnitudes of the forces F1, F2, and F3 shown in Fig. P5-27 are 550 lb, 300 lb, and 600 lb, respectively. Determine the sum of the moments of the three forces about point B.

SOLUTION

( )1 2 2 2

7 11 3550 287.763 452.198 123.327 lb7 11 3

− − += = − − ++ +

i j kF i j k

( ) ( )1 11 6 287.763 452.198 123.327B = − − − +M j k × i j k

( )1357 1727 3165 lb ft= − + + ⋅i j k

( ) ( ) ( )2 11 6 300 1800 3300 lb ftB = − = − − ⋅M j k × i j k

( ) ( ) ( )3 6 6 600 3600 lb ftB = − − = − ⋅M j k × k i

( )4960 73 135 lb ftBΣ = − − − ⋅M i j k .........................................................................Ans.

5-28 If the magnitude of the moment of the cable force T shown in Fig. P5-28 about the hinge at B is 1150 N mBM = ⋅ , determine the magnitude of the force T.

SOLUTION

500 mmxCD = − 1400cos30 1212.436 mmyCD = − = −

750 1400sin 30 1450 mmzCD = + =

( )2 2 2

500 1212.436 1450500 1212.436 1450

0.25574 0.62013 0.74164 N

T

T T T

− − +=+ +

= − − +

i j kT

i j k

( )0.5 1.1cos30 0.75 1.1sin 30B = − ° + + ° M i j k ×T

( )0.09966 0.70328 0.55369 N mT T T= − − ⋅i j k

( ) ( ) ( )2 2 20.09966 0.70328 0.55369

0.90062 1150 N mBM T T T

T

= + +

= = ⋅

1277 NT = .........................................................................................................................Ans.

5-29 The force F in Fig. P5-29 can be expressed in Cartesian vector form as ( )60 100 120 lb= + +F i j k . Determine the scalar component of the moment at point B about line BC.

SOLUTION

( ) ( ) ( )18 32 60 100 120 3200 240 1800 lb in.C = + + + = − − + ⋅M i k × i j k i j k

( )2 2

9 18 0.44721 0.894439 18

BC− += = − +

+i je i j

( )( ) ( )( )3200 0.44721 240 0.89443BC C BCM = = − − + −M e

1216 lb in.= ⋅ ...........................................................................................................Ans.


118

5-30 A 3000-N force is applied to the pipe shown in Fig. P5-30. Determine the scalar component of the moment of the force about the x-axis.

SOLUTION

( )2 2 2

280 300 5303000 1253.173 1342.686 2372.078 N280 300 530

− −= = − −+ +

i j kF i j k

( ) ( )0.65 0.53 1253.173 1342.686 2372.078O = + − −M j k × i j k

( )830.227 664.182 814.562 N m= − + − ⋅i j k

830 N mOx OM = = − ⋅M i .............................................................................................Ans.

5-31 A force ( )30 50 40 lb= − + −F i j k is applied to the machine component shown in Fig. P5-31. Determine the scalar component of the moment of the force about the z-axis.

SOLUTION

( ) ( )10 24 16 30 50 40O = − + − − + −M i j k × i j k

( )160 80 220 lb in.= − + + ⋅i j k

220 lb in.Oz OM = = ⋅M k .............................................................................................Ans.

5-32 The magnitude of the force F in Fig. P5-32 is 595 N. Determine the scalar component of the moment at point O about line OC.

SOLUTION

( )2 2

220 200595 440.26 400.24 N220 200

− += = − ++

i kF i k

( ) ( )0.22 0.24 440.26 400.24O = + − +M i j × i k

( )96.06 88.05 105.66 N m= − + ⋅i j k

( )2 2

220 200 0.73994 0.67267220 200

OC+= = ++

i ke i k

( )( ) ( )( )96.06 0.73994 105.66 0.67267OC O OCM = = +M e

142.2 N m= ⋅ ..........................................................................................................Ans.

5-33 A 40-lb vertical force F is applied to a lug wrench as shown in Fig. P5-33. Determine the magnitude of the component of the moment that would tighten the lug nut.

SOLUTION

( ) ( ) ( )4 8 3 40 320 160 lb in.O = + − − = − + ⋅M i j k × k i j

320 lb in.Oz OM = = ⋅M i ............................................................................................Ans.

5-34 If the magnitude of the force T shown in Fig. P5-34 is 1000-N, determine the scalar component of the moment of the force about the line CD.

SOLUTION

( )2 2

1.8 3.61000 447.21 894.43 N1.8 3.6

+= = ++

j kT j k

( ) ( )1.2 2.7 447.21 894.43C = − + +M i j × j k


119

( )2414.96 1073.32 536.65 N m= + − ⋅i j k

( )2 2

2.4 1.5 0.84800 0.530002.4 1.5

CD− −= = − −

+i je i j

( )( ) ( )( )2414.96 0.84800 1073.32 0.53000CD C CDM = = − + −M e

2620 N m= − ⋅ .........................................................................................................Ans.

5-35 A 120-lb force F is applied to a lever-shaft assembly as shown in Fig. P5-35. Determine the moment of the force about each coordinate axis.

SOLUTION

( )2 2 2

2 14 16120 11.2390 78.6732 89.9122 lb2 14 16

− + −= = − + −+ +

i j kF i j k

( ) ( )13 16 11.2390 78.6732 89.9122O = + − + −M i k × i j k

( )1258.77 989.03 1022.75 lb in.= − + + ⋅i j k

1259 lb in.OxM = − ⋅ .........................................................................................................Ans.

989 lb in.OyM = ⋅ ..............................................................................................................Ans.

1023 lb in.OzM = ⋅ ............................................................................................................Ans.

5-36 A 650-N force acts on the awning structure shown in Fig. P5-36. Determine the moment of the force about line BC. Express the result in Cartesian vector form.

SOLUTION

( ) ( ) ( )0.6 0.6 650 390.0 N mB = − − = ⋅M i k × k j

( )2 2

1.2 0.9 0.8000 0.60001.2 0.9

BC−= = −+

j ke j k

( )( )390.0 0.8000 312 N mBC B BCM = = = ⋅M e

( ) ( )312 0.8000 0.6000 250 187.2 N mBC BC BCM= = − = − ⋅M e j k j k .............Ans.

5-37 The magnitude of the force F in Fig. P5-37 is 107 lb. Determine the component of the moment of the force that rotates the door about the axis of the hinges.

SOLUTION

34 tan 20 12.375 in.Az = ° = 32 19.625 in.Az− =

( )2 2 2

36 34 19.625107 72.318 68.300 39.423 lb36 34 19.625

− − += = − − ++ +

i j kF i j k

( ) ( )36 32 72.318 68.300 39.423O = − + − − +M i k × i j k

( )2185.60 894.95 2458.80 lb in.= − + ⋅i j k

2190 lb in.OxM = ⋅ ...........................................................................................................Ans.


120

5-38 A bracket is rigidly attached to a wall at O and is subjected to a 384-N force F as shown in Fig. P5-38. Determine the component of the moment of the force

(a) That twists the bracket about the y-axis. (b) That bends the bracket about the x-axis. SOLUTION

( )2 2 2

150 250 250384 149.978 249.963 249.963 N150 250 250

− += = − ++ +

i j kF i j k

( ) ( )0.3 0.5 0.2 149.978 249.963 249.963O = + − − +M i j k × i j k

( )74.989 104.985 149.978 N m= − − ⋅i j k

(a) 105.0 N mOyM = − ⋅ .........................................................................................................Ans.

(b) 75.0 N mOxM = ⋅ ..............................................................................................................Ans.

5-39 A bar is bent in a circular arc of radius R = 6 ft and is subjected to a 660-lb force F as shown in Fig. P5-39. The force tends to twist and bend the member about the coordinate axes. Determine the twisting and bending moments and state the axes about which each occurs.

SOLUTION

( )2 2

4 4660 466.690 466.690 lb4 4−= = −+

i kF i k

( ) ( )6 6 466.690 466.690O = + −M j k × i k

( )2800.14 2800.14 2800.14 lb in.= − + − ⋅i j k

Twist: 2800 lb in.⋅ about the negative z-axis ..........................................................Ans.

Bend: 2800 lb in.⋅ about the negative x-axis ..........................................................Ans.

Bend: 2800 lb in.⋅ about the positive y-axis ...........................................................Ans.

5-40 The magnitude of the force F in Fig. P5-40 is 976 N. Determine (a) The component of the moment at point C parallel to line CE. (b) The component of the moment at point C perpendicular to line CE and the direction angles associated

with this moment vector. SOLUTION

(a) ( )2 2 2

350 300 160976 700.06 600.06 320.03 N350 300 160

− − += = − − ++ +

i j kF i j k

( ) ( )( )0.14 0.4 700.06 600.06 320.03

128.01 44.80 196.02 N mC = + − − +

= − + ⋅

M i j × i j k

i j k

( )2 2

210 400 0.46483 0.88540210 400

CE− += = − +

+i je i j

( )( ) ( )( )128.01 0.46483 44.80 0.88540 99.169 N mCE C CEM = = − + − = − ⋅M e

( )46.10 87.80 N mCE CE CEM= = − ⋅M e i j ................................................................Ans.

(b) ( ) ( )128.01 44.80 196.02 46.10 87.80C CE⊥ = − = − + − −M M M i j k i j

( )81.91 43.00 196.02 N m= + + ⋅i j k ................................................................Ans.


121

( ) ( ) ( )2 2 281.91 43.00 196.02 216.75 N mM⊥ = + + = ⋅

1 81.91cos 67.80216.75xθ

−= = ° ............................................................................................Ans.

1 43.00cos 78.56216.75yθ

−= = ° ...........................................................................................Ans.

1 196.02cos 25.26216.75zθ

−= = ° ............................................................................................Ans.

5-41 The magnitude of the force F in Fig. P5-41 is 781 lb. Determine (a) The component of the moment at point C parallel to line CD. (b) The component of the moment at point C perpendicular to line CD and the direction angles associated

with this moment vector. SOLUTION

(a) ( )2 2

12 10781 599.98 499.98 lb12 10

+= = ++

i kF i k

( ) ( )12 25 599.98 499.98C = − + +M i j × i k

( )12,499.60 5999.81 14,999.52 lb in.= + − ⋅i j k

( )2 2

12 10 0.76822 0.6401812 10

CD− += = − +

+i ke i k

( )( ) ( )( )12,499.60 0.76822 14,999.52 0.64018CD C CDM = = − + −M e

19,204.8 lb in.= − ⋅

( )14,753.5 12,294.6 lb in.CD CD CDM= = − ⋅M e i k .................................................Ans.

(b) ( )2253.9 5999.8 2704.9 lb in.C CD⊥ = − = − + − ⋅M M M i j k ..............................Ans.

( ) ( ) ( )2 2 22253.9 5999.8 2704.9 6956.6 lb in.M⊥ = + + = ⋅

1 2253.9cos 108.906956.6xθ

− −= = ° .......................................................................................Ans.

1 5999.8cos 30.416956.6yθ

−= = ° ...........................................................................................Ans.

1 2704.9cos 112.886956.6zθ

− −= = ° .......................................................................................Ans.

5-42 Determine the moment of the couple shown in Fig. P5-42 and the perpendicular distance between the two forces.

SOLUTION

( )( ) ( )( )760cos35 0.2 760sin 35 0.1 168.103 N m = ° + ° = ⋅M ...................Ans.

168.103 N m 760d⋅ =

0.221 m 221 mmd = = ...................................................................................................Ans.


122

5-43 A lug wrench is being used to tighten a lug nut on an automobile wheel as shown in Fig. P5-43. Two equal magnitude, parallel forces of opposite sense are applied to the wrench. If the magnitude of each force is 25 lb, determine the couple applied to a lug nut and express the result in Cartesian vector form.

SOLUTION

( )( )25 12 300 lb in.= − = − ⋅M i i ..................................................................................Ans.

5-44 To open a valve to a steam line in a power plant requires a couple of magnitude 54 N m⋅ . Determine the magnitude of each force F shown in Fig. P5-44 required to open the valve.

SOLUTION

( )54 N m 0.3M F= ⋅ =

180 NF = ..........................................................................................................................Ans.

5-45 A beam is loaded with a system of three couples as shown in Fig. P5-45. Express the resultant of the couple system in Cartesian vector form.

SOLUTION

( )( ) ( )( ) ( )( )500 30 250 14 300 30= − −M

( )2500 lb in. 2500 lb in.= ⋅ = ⋅k ..................................................................Ans.

5-46 The input and output torques (couples) from a gear box are shown in Fig. P5-46. Determine the magnitude and direction of the resultant torque T.

SOLUTION

( )750 200 125 150 cos 45 sin 45= + + + ° + °T i j k j k

( )750 306 231 N m= + + ⋅i j k

( ) ( ) ( )2 2 2750 306 231 842.36 N mT = + + = ⋅ .......................................................Ans.

1 750cos 27.08842.36xθ

−= = ° ............................................................................................Ans.

1 306cos 68.70842.36yθ

−= = ° ............................................................................................Ans.

1 231cos 74.08842.36zθ

−= = ° ............................................................................................Ans.

5-47 Three couples are applied to a rectangular block as shown in Fig. P5-47. Determine the magnitude of the resultant couple C and the direction angles associated with the resultant couple vector.

SOLUTION

( ) ( ) ( ) ( ) ( ) ( )15 10 18 8 10 20 8 15 15= − − + − + + +M j k × i i k × j i j × k

( )25 60 110 lb in.= + + ⋅i j k

( ) ( ) ( )2 2 225 60 110 127.77 lb in.M = + + = ⋅ ..........................................................Ans.

1 25cos 78.72127.77xθ

−= = ° ............................................................................................Ans.

1 60cos 61.99127.77yθ

−= = ° ............................................................................................Ans.


123

1 110cos 30.58127.77zθ

−= = ° ............................................................................................Ans.

5-48 Three couples are applied to a bent bar as shown in Fig. P5-48. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple

vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION

(a) ( )80 65 95 N m= + + ⋅C i j k

( ) ( ) ( )2 2 280 65 95 140.178 N mC = + + = ⋅ ............................................................Ans.

1 80cos 55.20140.178xθ

−= = ° ..........................................................................................Ans.

1 65cos 62.37140.178yθ

−= = ° .........................................................................................Ans.

1 95cos 47.34140.178zθ

−= = ° ..........................................................................................Ans.

(b) ( )2 2 2

140 120 80 0.69653 0.59702 0.39801140 120 80

OA+ −= = + −+ +

i j ke i j k

( )( ) ( )( ) ( )( )80 0.69653 65 0.59702 95 0.39801OA OAC = = + + −C e

56.7 N m= ⋅ ..............................................................................................................Ans.

5-49 Three couples are applied to a rectangular block as shown in Fig. P5-49. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple

vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION

(a) ( )90 125 75 lb in.= + + ⋅C i j k

( ) ( ) ( )2 2 290 125 75 171.318 lb in.C = + + = ⋅ .........................................................Ans.

1 90cos 58.31171.318xθ

−= = ° ..........................................................................................Ans.

1 125cos 43.14171.318yθ

−= = ° .........................................................................................Ans.

1 75cos 64.04171.318zθ

−= = ° ..........................................................................................Ans.

(b) ( )2 2 2

24 8 32 0.58835 0.19612 0.7844624 8 32

OA+ += = + ++ +

i j ke i j k

( )( ) ( )( ) ( )( )90 0.58835 125 0.19612 75 0.78446OA OAC = = + +C e

136.3 lb in.= ⋅ ............................................................................................................Ans.


124

5-50 Replace the 3-kN force shown in Fig. P5-50 by a force at point A and a couple. SOLUTION

3 kN A = ↓F ......................................................................................................................Ans.

( )( )3000 0.15 =450 N m A = ⋅C ...............................................................................Ans.

5-51 Replace the 50-lb force shown in Fig. P5-51 by a force at point A and a couple. Express your answer in Cartesian vector form.

SOLUTION

( )50 lbA =F j .....................................................................................................................Ans.

( )( )( ) ( )50 6 300 lb in.A = − = − ⋅C k k ........................................................................Ans.

5-52 Replace the 130-N vertical force shown in Fig. P5-52 by the mechanic’s hand to the wrench by an equivalent force-couple system at the lug nut.

SOLUTION

130 N A = ↓F ....................................................................................................................Ans.

( )( )130 0.3 =39.0 N m A = ⋅C ...................................................................................Ans.

5-53 A gusset plate is riveted to a beam by three rivets as shown in Fig. P5-53. Replace the 2500-lb force by a force-couple system at the top rivet.

SOLUTION 2500 lb = →F

( )( )2500 6 15,000 lb in. = = ⋅C

5-54 Four forces are applied to a truss as shown in Fig. P5-54. Determine the magnitude and direction of the resultant of the four forces and the perpendicular distance dR from point A to the line of action of the resultant.

SOLUTION

( )3 12 kN 12.369 kN A = − =R i j 75.96° ............................................................Ans.

( )( ) ( )( ) ( )( )6 2 4 4 2 6 40 kN m 12.369A d= + + = ⋅ =C

3.23 md = .........................................................................................................................Ans.

5-55 Three forces are applied to the locked pulley shown in Fig. P5-55. Determine the magnitude and direction of the resultant of the three forces and the perpendicular distance from the axle of the pulley to the line of action of the resultant.

SOLUTION

( ) ( ) ( )( )

160cos 20 160sin 20 90

150.351 35.277 lbA = − ° + ° −

= − −

R i j j

i j

154.43 lb = 13.20° ..............................................................................................Ans.

( )( ) ( )( ) ( )( )120 1 40 1 90 2 100 lb ft 154.43A d= − − = − ⋅ = −C

0.648 ftd = .......................................................................................................................Ans.

5-56 Determine the resultant of the four forces acting on the bell crank shown in Fig. P5-56, and determine where the resultant intersects the x-axis.

SOLUTION

( ) ( )175 400cos 45 400sin 45 300 200= + ° + ° + −R i i j j j


125

( )457.84 382.84 N 596.82 N = + =i j 39.90° ...................................................Ans.

( )( ) ( )( )400 50 200 80 36,000 N mm 382.84A d= + = ⋅ =C

94.0 mm left of d O= ....................................................................................................Ans.

5-57 Three 75-lb traffic lights are suspended over a roadway as shown in Fig. P5-57. Determine the resultant of the weights of the traffic lights and locate the resultant with respect to point A.

SOLUTION

( )( )3 75 225 lb = = ↓R ..............................................................................................Ans.

( )( ) ( )( ) ( )( )75 8 75 11 75 14 2475 lb ft 225d= + + = ⋅ =C

11 ftd = ..............................................................................................................................Ans.

5-58 Four parallel forces act on a concrete slab as shown in Fig. P5-58. Determine the resultant of the forces and locate the intersection of the line of action of the resultant with the xy-plane.

SOLUTION

( ) ( )4 3 3 2 12 kN= − − − − = −R k k ...........................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 1.5 4 3 3 3 1.5 3 3 4.5 2= + − + + − + + − + −C i j × k i j × k i j × k j × k

( )33 25.5 kN m= − + ⋅i j

( ) ( ) ( )12 12 12x y y x= + − = − +i j × k i j

25.5 12 2.12 mx = = ......................................................................................................Ans.

33 12 2.75 my = = ..........................................................................................................Ans.

5-59 The forces exerted on the wheels of an airplane by the runway are shown in Fig. P5-59. Determine the resultant of the three forces and locate the intersection of the line of action of the resultant with the xy-plane (the runway).

SOLUTION

( ) ( )5000 5000 2500 12,500 lb= + + =R k k ...........................................................Ans.

( ) ( ) ( ) ( ) ( )20 8 5000 20 8 5000 200,000 lb ftO = + + − = − ⋅C i j × k i j × k j

( ) ( ) ( )12,500 12,500 12,500x y y x= + = −i j × k i j

200,000 12,500 16 ftx = = ..........................................................................................Ans.

0 12,500 0 fty = = ..........................................................................................................Ans.

5-60 The magnitude of the force F acting on the casting shown in Fig. P5-60 is 2 kN. Replace the force with a force R through point A and a couple C.

SOLUTION

( )2 2 2

3 4 122000 461.538 615.385 1846.154 N3 4 12+ += = + ++ +

i j kR i j k .................Ans.

( ) ( )0.25 0.075 461.538 615.385 1846.154= + + +C j k × i j k

( )415.4 34.6 115.4 N m= + − ⋅i j k ..........................................................................Ans.


126

5-61 A force ( )50 50 200 lb+ −F = i j k acts on the wall bracket shown in Fig. P5-61. Replace the force by a force-couple system at rivet B. The eye bolt is small and the force F may be considered as acting at point C.

SOLUTION

( )50 50 200 lb= + −F i j k ............................................................................................Ans.

( ) ( )2 4 3 50 50 200= + − + −C i j k × i j k

( )650 250 100 lb in.= − + − ⋅i j k ...............................................................................Ans.

5-62 The homogeneous plate shown in Fig. P5-62 has a mass of 90 kg. The magnitude of the force T in cable BC is 800 N. Replace the weight and cable forces by an equivalent force-couple system at hinge A.

SOLUTION

( )( )( ) ( )90 9.81 882.90 N= − = −W k k

( )2 2 2

350 610 500800 324.486 565.533 463.552 N350 610 500

− − += = − − ++ +

i j kT i j k

( )324 566 419 N= + = − − −R W T i j k ...................................................................Ans.

( ) ( ) ( ) ( )0.65 0.28 324.486 565.533 463.552 0.35 882.90= + − − + + −C i j × i j k i × k

( )129.8 7.71 277 N m= + − ⋅i j k ............................................................................Ans.

5-63 Forces act at A, D, and E of the member shown in Fig. P5-63. If the equivalent force-couple system at B is ( )100 100 lb= − −R i j and a couple C, determine the magnitude of FE, the angle θ, and the required

couple C. SOLUTION

( ) ( ) ( )100 300 100 100 lbEx EyF F= − + + − = − +R i j i j

0 lbExF = 200 lbEyF =

200 lb E = ↑F ...................................................................................................................Ans.

( )( ) ( )( ) ( )( )100 10 300 14 200 20 800 lb in. = − + = ⋅C ..................................Ans.

5-64 Four forces and a couple are applied to a rectangular plate as shown in Fig. P5-64. Determine the magnitude and direction of the resultant of the force-couple system and the distance xR from point O to the intercept of the line of action of the resultant with the x-axis.

SOLUTION

( ) ( ) ( )90 60 75 50 30 25 N= − + − = +R i j i j

39.1 N = 39.81° ....................................................................................................Ans.

( )( ) ( )( ) ( )( ) ( )( )10 90 0.25 50 0.3 60 0.25 75 0.3

65.00 N m 25Rx= − − − −= − ⋅ =

C

65.00 25 2.6 mRx = − = − ..............................................................................................Ans.

5-65 The magnitude of the resultant R of the four forces acting on the legs of the table shown in Fig. P5-65 is 80 lb. Determine the magnitude of F and the location, xR and yR, of the line of action of R.

SOLUTION

( ) ( )30 20 15 80 lbF= + + + =R j j


127

15 lbF = .............................................................................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( )3 30 3 15 3 3 20 105 150 lb ft= + + + = − ⋅C i × k j × k i j × k i j

( ) ( ) ( )80 80 80R R R Rx y y x= + = −i j × j i j

150 80 1.875 ftRx = = ....................................................................................................Ans.

105 80 1.313 ftRy = = ....................................................................................................Ans.

5-66 Figure P5-66 shows a crankshaft-flywheel arrangement of a one-cylinder engine. A 1000-N force P is supplied by the connecting rod, and a couple C of magnitude 250 N-m is delivered to the crankshaft by the flywheel. Replace the force and couple by an equivalent force-couple system at the bearing A.

SOLUTION

( )( ) ( )( )1000sin 20 1000cos 20= ° − + ° −P j k

( )342.02 939.69 N= − −j k .......................................................................................Ans.

( ) ( ) ( )0.225 0.125 342.02 939.69 250A = − − − − + −M i j × j k i

( )132.5 211.4 77.0 N m= − − + ⋅i j k .................................................................Ans.

5-67 A farmer is using the hand winch shown in Fig. P5-67 to raise a 40-lb bucket of water from a well. When the force P and the weight of the bucket are replaced by an equivalent force-couple system at bearing D, the result is 10 60 lb= − −R j k and a couple CD Determine the force P applied to the handle of the winch and the couple CD.

SOLUTION

( ) ( )40 10 60 lbx y zP P P= + + − = − −R i j k k j k

0 lbxP = 10 lbyP = − 20 lbzP = − ..................................Ans.

( ) ( ) ( ) ( )30 2.5 40 61 20cos30 20sin 30 10 20D = + − + − ° + ° − −C i j × k i j k × j k

( )346 2420 610 lb in.= + − ⋅i j k .........................................................................Ans.

5-68 In order to remove a rusty screw from a steel plate, a worker attaches a screwdriver to the bent bar shown in Fig. P5-68. To hold the screwdriver in place, the worker applies a 30-N force at C; a 50-N force is applied to the handle of the bar in an attempt to remove the screw. Forces Fy and Fz are also applied at C. It is desirable that the screwdriver not bend about the y and z axes.

(a) Determine the forces Fy and Fz. (b) Replace the forces at B and C by an equivalent force-couple system at A. SOLUTION

(a) ( ) ( ) ( ) ( )0.45 30 0.25 0.15 50A y zF F= − + + + − −M i × i j k i j × k

( ) ( )7.5 12.5 0.45 0.45 0 0 N mz y xF F M= + − + = + + ⋅i j k i j k

0 NyF = 27.78 NzF = ............................................................Ans.

(b) ( ) ( ) ( ) ( )30 27.78 50 30 22.22 N= − + + − = − −R i k k i k ....................................Ans.

( )7.5 N m= ⋅C i ...............................................................................................................Ans.


128

5-69 Reduce the forces shown in Fig. P5-69 to a wrench and locate the intersection of the wrench with the xy-plane.

SOLUTION

( )2 2 2

3 5 4420 178.191 296.985 237.588 lb3 5 4

B− += = − ++ +

i j kF i j k

( )60 178.191 296.985 237.588 250= + − + +R i i j k j

( )238.191 46.985 237.588 lb= − +i j k ...............................................................Ans.

( ) ( ) ( ) ( ) ( ) ( )3 60 5 178.191 296.985 237.588 4 250O = + − + +M j × i j × i j k k × j

( )187.940 1070.955 lb ft= − ⋅i k

( )2 2 2

238.191 46.985 237.588 0.70120 0.13832 0.69942238.191 46.985 237.588

R− += = − ++ +

i j ke i j k

( )( ) ( )( )187.940 0.70120 1070.955 0.69942 617.264 lb ftR O RM = = + − = − ⋅M e

( )432.826 85.380 431.727 lb ftR R RM= = − + − ⋅M e i j k ....................................Ans.

O R= + ×M M r R

( )620.766 85.380 639.228 lb ftO R− = − − ⋅M M i j k

( ) ( )

( )238.191 46.985 237.588

237.588 237.588 46.985 238.191R R

R R R R

x y

y x x y

= + − +

= − − +

r × R i j × i j k

i j k

85.380 237.588 0.359 ftRx = = ..................................................................................Ans.

620.766 237.588 2.61 ftRy = = ..................................................................................Ans.

5-70 Locate the center of mass for the three particles shown in Fig. P5-70 if mA = 26 kg, mB = 21 kg, and mC = 36 kg.

SOLUTION 26 21 36 83 kgm = + + =

( ) ( ) ( )83 26 200cos 60 21 200cos30 36 200 962.693Gx = − ° + − ° + =

( ) ( )83 26 200sin 60 21 200sin 30 2403.332Gy = ° + − ° =

962.693 83 11.60 mmGx = = .......................................................................................Ans.

2403.332 83 29.0 mmGy = = ......................................................................................Ans.

5-71 Locate the center of gravity for the four particles shown in Fig. P5-71 if WA = 20 lb, WB = 25 lb, WC = 30 lb, and WD = 40 lb.

SOLUTION 20 25 30 40 115 lbW = + + + =

( ) ( ) ( ) ( )115 20 0 25 0 30 8 40 0 240Gx = + + + =

( ) ( ) ( ) ( )115 20 12 25 12 30 12 40 0 900Gy = + + + =

( ) ( ) ( ) ( )115 20 10 25 0 30 0 40 0 200Gz = + + + =


129

240 115 2.09 in.Gx = = ..................................................................................................Ans.

900 115 7.83 in.Gy = = ..................................................................................................Ans.

200 115 1.739 in.Gz = = ................................................................................................Ans.

5-72 Locate the center of mass for the four particles shown in Fig. P5-72 if mA = 16 kg, mB = 24 kg, mC = 14 kg, and mD = 36 kg.

SOLUTION 16 24 14 36 90 kgm = + + + =

( ) ( ) ( ) ( )90 16 300 24 0 14 0 36 0 4800Gx = + + + =

( ) ( ) ( ) ( )90 16 0 24 0 14 500 36 0 7000Gy = + + + =

( ) ( ) ( ) ( )90 16 0 24 0 14 0 36 400 14,400Gz = + + + =

4800 90 53.3 mmGx = = ...............................................................................................Ans.

7000 90 77.8 mmGy = = ..............................................................................................Ans.

14,400 90 160 mmGz = = ............................................................................................Ans.

5-73 Locate the center of gravity for the five particles shown in Fig. P5-73 if WA = 25 lb, WB = 35 lb, WC = 15 lb, WD = 28 lb, and WE = 16 lb.

SOLUTION 25 35 15 28 16 119 lbW = + + + + =

( ) ( ) ( ) ( ) ( )119 25 10 35 10 15 0 28 10 16 0 880Gx = + + + + =

( ) ( ) ( ) ( ) ( )119 25 0 35 11 15 16 28 16 16 0 1073Gy = + + + + =

( ) ( ) ( ) ( ) ( )119 25 0 35 0 15 0 28 11 16 11 484Gz = + + + + =

880 119 7.39 in.Gx = = ..................................................................................................Ans.

1073 119 9.02 in.Gy = = ................................................................................................Ans.

484 119 4.07 in.Gz = = ..................................................................................................Ans.

5-74 Locate the center of mass for the five particles shown in Fig. P5-74 if mA = 2 kg, mB = 3 kg, mC = 4 kg, mD = 3 kg, and mE = 2 kg.

SOLUTION 2 3 4 3 2 14 kgm = + + + + =

( ) ( ) ( ) ( ) ( )14 2 300 3 150 4 300 3 300 2 0 3150Gx = + + + + =

( ) ( ) ( ) ( ) ( )14 2 240 3 400 4 400 3 0 2 200 3680Gy = + + + + =

( ) ( ) ( ) ( ) ( )14 2 0 3 0 4 270 3 270 2 270 2430Gz = + + + + =

3150 14 225 mmGx = = ................................................................................................Ans.

3680 14 263 mmGy = = ................................................................................................Ans.

2430 14 173.6 mmGz = = .............................................................................................Ans.


130

5-75 Three bodies with masses of 2, 4, and 6 slugs are located at points (2, 3, 4), (3, −4, 5), and (−3, 4, 6), respectively. Locate the mass center of the system if the distances are measured in feet.

SOLUTION 2 4 6 12 slugm = + + =

( ) ( ) ( )12 2 2 4 3 6 3 2Gx = + + − = −

( ) ( ) ( )12 2 3 4 4 6 4 14Gy = + − + =

( ) ( ) ( )12 2 4 4 5 6 6 64Gz = + + =

2 12 0.1667 ftGx = − = − ................................................................................................Ans.

14 12 1.167 ftGy = = ......................................................................................................Ans.

64 12 5.33 ftGz = = .........................................................................................................Ans.

5-76 Locate the centroid of the shaded area shown in Fig. P5-76 if b = 200 mm and h = 300 mm. SOLUTION

( )VdA hx b dx=

( )HdA b by h dy= −

2

002 2

bb

Vhx hx bhA dA dxb b

= = = =

∫ ∫

3 2

003 3

bb

y Vhx hx hbM xdA x dxb b

= = = =

∫ ∫

2 3 2 133.3 mm

2 3y

C

M hb bxA hb

= = = = ...........................................................................Ans.

2 3 2

002 3 6

hh

x Hby by by bhM y dA y b dyh h

= = − = − = ∫ ∫

2 6 100.0 mm

2 3x

CM bh hyA bh

= = = = ............................................................................Ans.

5-77 Determine the y-coordinate of the centroid of the shaded area shown in Fig. P5-77. SOLUTION

( )2 2VdA x b dx=

2HdA b by dy = −

2 3 2

002 6 6

bb

Vx x bA dA dxb b

= = = =

∫ ∫

/ 22 5/ 2 3 3 3/ 2

00

2 22 5 2 8 10 40

bb

x Hby y b b bM y dA y b by dy b = = − = − = − =

∫ ∫


131

3

2

40 36 20

xC

M b byA b

= = = ...................................................................................................Ans.

5-78 Determine the x-coordinate of the centroid of the shaded area shown in Fig. P5-78. SOLUTION

2

21VxdA b dxa

= −

2

20

2 2 2 1

0

1

sin2 4

a

V

a

xA dA b dxa

b x abx a x aa a

π−

= = −

= − + =

∫ ∫

( )2 232 220

0

13 3

aa

y Vx b baM xdA x b dx a xa a

− = = − = − = ∫ ∫

2 3 4

4 3y

C

M ba axA abπ π

= = = .................................................................................................Ans.

5-79 Locate the centroid of the shaded area shown in Fig. P5-79. SOLUTION

2

225VxdA x dx

= −

2 2

3 42

1 2 22 25 25

1 442 25 625

x C Vx xdM y dA x x dx

x xx

= = − −

= − +

502 350 2 2

00

2 833.33 in.25 75Vx xA dA x dx x

= = − = − =

∫ ∫

250

0

503 43

0

225

2 20,833.33 in.3 100

y VxM xdA x x dx

x x

= = −

= − =

∫ ∫

20,833.33 25.0 in.

833.33y

C

Mx

A= = = .................................................................................Ans.

503 4 3 4 550 2

00

3

1 4 1 442 25 625 2 3 25 3125

8333.33 in.

x xx x x x xM dM x dx

= = − + = − +

=

∫ ∫


132

8333.33 10.00 in.833.33

xC

MyA

= = = ...................................................................................Ans.


( )2 2HdA ay b dy=

2 3

2 2003 3

bb

Hay ay abA dA dyb b

= = = =

∫ ∫

2 4 2

2 2004 4

bb

x Hay ay abM y dA y dyb b

= = = =

∫ ∫

2 4 3

3 4x

CM ab byA ab

= = = ...................................................................................................Ans.


( )VdA ax x dx= −

( )2HdA y y a dy = −

( )3/ 2 2 2

00

3 2 2 6

aa

Vx x aA dA ax x dx a

= = − = − =

∫ ∫

( )5/ 2 3 3

00

5 2 3 15

aa

y Vx x aM xdA x ax x dx a

= = − = − =

∫ ∫

3

2

15 26 5

yC

M a axA a

= = = ...................................................................................................Ans.

2 3 4 3

003 4 12

aa

x Hy y y aM y dA y y dya a

= = − = − =

∫ ∫

3

2

126 2

xC

M a ayA a

= = = .....................................................................................................Ans.


4 59 9Vx xdA x dx dx = − =

929 2

33

5 5 20 mm9 18Vx xA dA dx

= = = = ∫ ∫

24 1 5 5 65

9 2 9 9 162x C Vx x x xdM y dA dx dx

= = + =


133

92 39 3

33

65 65 93.889 mm162 486x x C Vx xM dM y dA dx

= = = = =

∫ ∫ ∫

93.889 4.69 mm

20x

CMyA

= = = ....................................................................................Ans.

5-83 Determine the x-coordinate of the shaded area shown in Fig. P5-83. SOLUTION

( )3 2VdA x dx=

23 42 2

11

15 in.2 8 8Vx xA dA dx

−−

= = = =

∫ ∫

23 52 3

11

33 in.2 10 10y Vx xM xdA x dx

−−

= = = =

∫ ∫

33 10 44 1.760 in.15 8 25

yC

Mx

A= = = ≅ .............................................................................Ans.

5-84 Locate the centroid of the shaded area shown in Fig. P5-84. SOLUTION For the quarter circle

( ) ( )2 2 2x r y r r− + − =

( )22y r r x r= − − −

( )( )

( )

22

22 22

VdA r r r x r dx

r x r dx rx x dx

= − − − −

= − − = −

( )2

2 2 2 1

00

12 2 sin2 4

rr

Vx r rA dA rx x dx x r rx x rr

π− − = = − = − − + = ∫ ∫

( ) ( )

2

0

3/ 22 3 32 2 1

0

2

22 sin

3 2 4 3

r

y V

r

M x dA x rx x dx

rx x r x r r rx r rx x rr

π−

= = −

− − = − + − − + = −

∫ ∫

( ) ( )3 3

2

4 3 44 3

yC

r rM rx rA r

ππ π

−= = = − ........................................................................Ans.

Since the line x y= is an axis of symmetry,

43C Cry x rπ

= = − ...............................................................................................................Ans.


134

5-85 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-85 about the x-axis.

SOLUTION

222

2 4

28

42 64

xdV r dx dx

x x dx

π π

π

= = −

= − +

2 44

0

43 53

0

42 64

4 26.808 in.6 320

x xV dV dx

x xx

π

π

= = − +

= − + =

∫ ∫

42 4 4 64 2 4

00

4 2 33.510 in.2 64 8 384yzx x x xM xdV x dx xπ π

= = − + = − + =

∫ ∫

33.510 1.250 in.26.808

yzC

Mx

V= = = ....................................................................................Ans.

Since the x-axis is an axis of symmetry,

0C Cy z= = .........................................................................................................................Ans.

5-86 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-86 about the x-axis.

SOLUTION

( )2 50dV r dx x dxπ π= =

2 2

0050 25 25b b

V dV x dx x bπ π π = = = = ∫ ∫

( )3 3

00

50 50503 3

bb

yzx bM xdV x x dx π ππ

= = = =

∫ ∫

3

2

50 3 225 3

yzC

M b bxV b

ππ

= = =

( )2 100

66.7 mm3

= ≅ .......................................Ans.

Since the x-axis is an axis of symmetry,

0C Cy z= = .........................................................................................................................Ans.

5-87 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-87. SOLUTION

2 6x y= 3dx dy y=

( ) ( )2 22 2 211 3 1 93

dL dx dy dx dy dy y dy y dy= + = + = + = +


135

( ) 1212 2 2 2

0 0

1 19 9 9ln 9 27.8807 in.3 6

L dL y dy y y y y = = + = + + + + = ∫ ∫

( ) ( )

212 2

0

12232 2 2 2

0

1 96 3

1 9 99 9 ln 9 304.962 in.18 4 8 8

yyM xdL y dy

y yy y y y

= = +

= + − + − + + =

∫ ∫

304.962 10.94 in.27.8807

yC

Mx

L= = = ...................................................................................Ans.

( )12

12 32 2 2

00

1 19 9 207.278 in.3 3 3xyM y dA y dy y = = + = + = ∫ ∫

207.278 7.43 in.27.8807

xC

MyL

= = = .....................................................................................Ans.

5-88 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-88 if b = 50 mm. SOLUTION

2 2y x b= dy dx x b= 50 mmb =

( ) ( )2 22 2 2 211 1 50 5050

dL dx dy dy dx dx x dx x dx= + = + = + = +

( )

( ) ( )

5050 2 2 2 2 2 2 2

0 0

2 2 2

1 150 50 50 ln 5050 100

1 2 50 50 ln 50 2 50 50 ln 50 57.3897 mm100

L dL x dx x x x x = = + = + + + +

= + + − =

∫ ∫

( )50

50 32 2 2 2 2

00

1 150 50 1523.689 mm50 50 3yxM xdL x dx x = = + = + = ∫ ∫

1523.689 26.5 mm57.3897

yC

Mx

L= = = ................................................................................Ans.

( ) ( )

250 2 2

0

502 432 2 2 2 2 2

02

1 50100 50

1 50 5050 50 ln 505000 4 8 8

525.198 mm

xxM y dL x dx

x x x x x x

= = +

= + − + − + +

=

∫ ∫

525.198 9.15 mm57.3897

xC

MyL

= = = ..................................................................................Ans.

5-89 Locate the centroid of the volume of the portion of a right circular cone shown in Fig. P5-89. SOLUTION

( )ˆ rr h zh

= −


136

( )2 2

22

ˆ4 4r rdV dz h z dz

hπ π= = −

( ) ( )32 2 22

2 200

4 4 3 12

hh h zr r r hV dV h z dzh h

π π π −= = − = − =

∫ ∫

( )2

220

2 2 2 3 4 2 2

20

4

24 2 3 4 48

h

xy

h

rM z dV z h z dzh

r h z hz z r hh

π

π π

= = −

= − + =

∫ ∫

2 2

2

4812 4

xyC

M r h hzV r h

ππ

= = = .............................................................................................Ans.

( )3

33

ˆ43 3yzr rdM dV h z dz

hπ= = −

( ) ( )43 3 33

3 300

3 3 4 12

hh

yz yz

h zr r r hM dM h z dzh h

−= = − = − =

∫ ∫

3

2

1212

yzC

M r h rxV r hπ π

= = = ...............................................................................................Ans.

Since the plane x y= is a plane of symmetry,

C Cy x r π= = ....................................................................................................................Ans.

5-90 Locate the mass center of the hemisphere shown in Fig. P5-90 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P.

SOLUTION

( )22 2r y r z= + −

( )22 2 22y r r z rz z= − − = −

( ) ( )2 2 32dm dV kz y dz k rz z dzρ π π= = = −

( )3 4 4

2 3

00

2 523 4 12

rr rz z krm dm k rz z dz k ππ π

= = − = − =

∫ ∫

( )4 5 5

2 3

00

2 324 5 10

rr

xyrz z krM z dm z k rz z dz k ππ π

= = − = − =

∫ ∫

5

4

3 10 185 12 25

xyG

M kr rzm kr

ππ

= = = .........................................................................................Ans.

By symmetry,

0G Gx y= = ........................................................................................................................Ans.


137

5-91 Locate the mass center of the right circular cone shown in Fig. P5-91 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P.

SOLUTION

r̂ rz h=

( )2 2

22

ˆ r zdm dV kz r dz kz dzh

ρ π π

= = =

2 2 4 2 2

32 20

04 4

hh kr kr z kr hm dm z dzh h

π π π = = = =

∫ ∫

2 2 5 2 3

42 20

05 5

hh

xykr kr z kr hM z dm z dzh h

π π π = = = =

∫ ∫

2 3

2 2

5 44 5

xyG

M kr h hzm kr h

ππ

= = = ...........................................................................................Ans.

By symmetry,

0G Gx y= = ........................................................................................................................Ans.

5-92 Locate the centroid of the volume of the tetrahedron shown in Fig. P5-92. SOLUTION By similar triangles

( )ˆ aa c zc

= −

( )ˆ bb c zc

= −

( )22

1 ˆˆ2 2

abdV abdz c z dzc

= = −

( )2 220

32 2

20

22

2 3 6

c

c

abV dV c cz z dzc

ab z abcc z czc

= = − +

= − + =

∫ ∫

( )2 2 3 4 2

2 2 32 20

0

222 2 2 3 4 24

cc

xyab ab c z cz z abcM z dV c z cz z dzc c

= = − + = − + =

∫ ∫

2 24

6 4xy

C

M abc czV abc

= = = ...............................................................................................Ans.

Similarly

4Cax = .................................................................................................................................Ans.

4Cby = .................................................................................................................................Ans.


143

( )( )2 0.316 3.5343 1.1168 lbW = − = −

( )( )( ) 3

3

9 9 0.520.2500 in.

2V = = ( )( )3 0.100 20.2500 2.0250 lbW = =

348.5243 in.totV = 10.9597 lbtotW =

( )

1 1

4 93.8197 in.

3C Cx yπ

= = =

(a) ( )( ) ( )( ) ( )( )3.8197 31.8086 3.5 3.5343 0.25 20.2500

2.35 in.48.5243Cx

+ − += = ......Ans.

( )( ) ( )( ) ( )( )3.8197 31.8086 3 3.5343 3 20.2500

3.54 in.48.5243Cy

+ − += = ................Ans.

( )( ) ( )( ) ( )( )0.25 31.8086 0.25 3.5343 3.5 20.2500

1.606 in.48.5243Cz

+ − += = .........Ans.

(b) ( )( ) ( )( ) ( )( )3.8197 10.0515 3.5 1.1168 0.25 2.0250

3.19 in.10.9597Gx

+ − += = .........Ans.

( )( ) ( )( ) ( )( )3.8197 10.0515 3 1.1168 3 2.0250

3.75 in.10.9597Gy+ − +

= = ...................Ans.

( )( ) ( )( ) ( )( )0.25 10.0515 0.25 1.1168 3.5 2.0250

0.851 in.10.9597Gz

+ − += = ............Ans.

5-106 A cylinder with a conical cavity and a hemispherical cap is shown in Fig. P5-106. Locate (a) The centroid of the composite volume if R = 200 mm, and h = 250 mm. (b) The center of gravity of the composite volume if the cylinder is made of brass (ρ = 8750 kg/m3) and the

cap is made of aluminum (ρ = 2770 kg/m3). SOLUTION

Cylinder: ( ) ( )2 6 31 200 250 31.416 10 mmV π= = ×

( )( )31 8750 31.416 10 274.89 kgm −= × =

Hemisphere: ( ) ( )3

6 32

4 3 20016.755 10 mm

2V

π= = ×

( )( )32 2770 16.755 10 46.412 kgm −= × =

Cavity: ( ) ( )2

6 33

200 25010.472 10 mm

3V

π= − = − ×

( )( )33 8750 10.472 10 91.630 kgm −= − × = −

Totals: 6 337.699 10 mmtotV = × 229.672 kgtotm =

( )

2 2

3 200250 325 mm

8C Gz z= = + =

3 3250 62.5 mm4C Gz z= = =


144

(a) ( )( ) ( )( ) ( )( )125 31.416 325 16.755 62.5 10.472

231 mm37.699Cz

+ + −= = ...............Ans.

0 mm (by symmetry)C Cx y= = ..................................................................................Ans.

(b) ( )( ) ( )( ) ( )( )125 274.89 325 46.412 62.5 91.630

190.4 mm229.672Gz

+ + −= = ...........Ans.

0 mm (by symmetry)G Gx y= = ..................................................................................Ans.

5-107 A slender rod is made of two materials; the segment along the x-axis is aluminum (γ = 0.100 lb/in3) and the remainder is made of brass (γ = 0.316 lb/in3). Locate the center of gravity of the slender rod.

SOLUTION Note that the curved portion of the wire is a half circle. Then,

1 16 in.L = ( )1 0.100 16 1.600 lbW = = 1 8 in.Gx = 1 0 in.Gy = 1 0 in.Gz =

2 14 in.L = ( )2 0.316 14 4.424 lbW = = 2 0 in.Gx = 2 7 in.Gy = 2 0 in.Gz =

( )3 9.9 31.102 in.2dL π π= = = ( )3 0.316 31.102 9.828 lbW = =

3 0 in.Gx =

( )

3

9.9sin 27 cos 45 11.457 in.

2Gyπ

π= + ° =

( )

3

9.9sin 27 sin 45 11.457 in.

2Gzπ

π= + ° =

15.852 lbtotW =

( )( ) ( )( ) ( )( )8 1.600 0 4.424 0 9.828

0.807 in.15.852Gx

+ += = ...........................................Ans.

( )( ) ( )( ) ( )( )0 1.600 7 4.424 11.457 9.828

9.06 in.15.852Gy

+ += = ..................................Ans.

( )( ) ( )( ) ( )( )0 1.600 0 4.424 11.457 9.828

7.10 in.15.852Gz

+ += = ...................................Ans.

5-108 The loads acting on a beam are distributed in a triangular manner as shown in Fig. P5-108. Determine and locate the resultant with respect to the left end of the beam.

SOLUTION

( )( )

1

2 750750 N

2R = =

( )( )

2

4 7501500 N

2R = =

750 1500 2250 N R = + = ↓ .........................................................................................Ans.

( )( ) ( )2250 750 2 3 2 1500 2 4 3AM d= = + +

2.67 md = .........................................................................................................................Ans.


145

5-109 Determine the resultant of the distributed loads acting on the beam shown in Fig. P5-109, and locate its line of action with respect to the support at A.

SOLUTION

( )( )1 200 3 600 lbR = =

( )( )2 150 3 450 lbR = =

( )( )3 100 3 300 lbR = =

600 450 300 1350 lb R = + + = ↓ ................................................................................Ans.

( ) ( ) ( )1350 600 1.5 450 4.5 300 7.5AM d= = + +

3.83 ftd = ..........................................................................................................................Ans.

5-110 A distributed load acts on the beam shown in Fig. P5-110. Determine the resultant of the distributed load and locate its line of action with respect to the support at A.

SOLUTION

( )( )1 2.5 4 10 kNR = =

( )( )

2

2.5 45 kN

2R = =

10 5 15 kN R = + = ↓ ......................................................................................................Ans.

( ) ( )15 10 4 5 2 8 3AM d= = + +

4.22 md = .........................................................................................................................Ans.

5-111 A distributed load acts on a beam as shown in Fig. P5-111. Determine and locate the resultant of the distributed load with respect to the support at A.

SOLUTION

10 102 3

1 003 1000 lbR wdx x dx x = = = = ∫ ∫

10410 3

1 1 1 00

33 7500 lb ft4AxM R d xwdx x dx

= = = = = ⋅

∫ ∫

( )( )2 300 5 1500 lbR = =

1000 1500 2500 lb R = + = ↓ .......................................................................................Ans.

( )2500 7500 1500 10 2.5AM d= = + +

10.50 ftd = ........................................................................................................................Ans.

5-112 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-112 and locate its line of action with respect to the support.

SOLUTION

232 2

1 00

100100 266.667 N3xR wdx x dx

= = = =

∫ ∫

2 23 4

1 1 1 00100 25 400 N mAM R d xwdx x dx x = = = = = ⋅ ∫ ∫


146

( )( )

2

400 2400 N

2R = =

266.667 400 666.667 667 N R = + = ≅ ↓ ................................................................Ans.

( )666.667 400 400 2 2 3AM d= = + +

2.20 md = .........................................................................................................................Ans.

5-113 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-113 and locate its line of action with respect to the support.

SOLUTION

( )( )

1

200 6600 lb

2R = = ↓ ( )( )2 200 2 400 lb R = = ↓

( )( )3 100 6 600 lb R = = ↑

600 400 600 400 lb 400 lb R = − − + = − = ↓ ............................................................Ans.

( ) ( ) ( )400 600 4 400 7 600 3AM d= = + −

8.50 ftd = ..........................................................................................................................Ans.

5-114 A gate is used to hold water as shown in Fig. P5-114. If the width of the gate is 2 m. determine the resultant of the water pressure acting on the gate, and locate its line of action with respect to the bottom of the gate.

SOLUTION

( )( ) ( )( )9800 9 9 2

793,800 N 794 kN2

R = = = ....................................................Ans.

9 3 3 m (from bottom)d = = ........................................................................................Ans.

5-115 A flexible cable is used to tether the balloon shown in Fig. P5-115. The cable weighs 1.2 lb/ft along its length. Determine the magnitude of the resultant force and its location with respect to A.

SOLUTION

2 10y x= 5dy dx x=

( ) ( ) ( ) ( )2 2 2 2 2 211 1 5 55

dL dx dy dy dx dx x dx x dx= + = + = + = +

15 2 2

0

11.2 55

R wdL x dx = = + ∫ ∫

( ) 152 2 2 2 2

0

10.24 5 5 ln 5 33.916 lb2x x x x = + + + + =

...........................Ans.

( )

15 2 2

015

32 2

0

0.24 5

10.24 5 306.228 lb ft3

RRx xwdL x x dx

x

= = +

= + = ⋅

∫ ∫

306.228 9.029 ft33.916Rx = = .................................................................................................Ans.

2 210 9.029 10 8.15 ftR Ry x= = = ................................................................................Ans.


147

5-116 A loaded rivet causes a force distribution or pressure p on a plate, as shown in Fig. P5-116. The rivet has a diameter of 25 mm, and the plate is 15 mm thick. Determine the magnitude of the resultant force acting on the plate. Let po = 400 N/m2.

SOLUTION

( ) ( ) ( )40.025 0.015 1.875 102

dA r d t d dθ θ θ− = = = ×

( ) ( ) ( )4 2cos 400cos 1.875 10 cos 0.075cosydR p dA d dθ θ θ θ θ θ− = = × =

/ 2 2

/ 2/ 2

/ 2

/ 2/ 2

0.075cos

1 cos 2 sin 20.075 0.0752 2 4

y yR dR d

d

π

ππ

π

ππ

θ θ

θ θ θθ

−

−−

= =

+ = = +

∫ ∫

∫

0.1178 N = ↓ ............................................................................................................Ans.

( ) ( ) ( )4sin 400cos 1.875 10 sin 0.0375sin 2xdR p dA d dθ θ θ θ θ θ− = = × =

/ 2

/ 2

/ 2/ 2

cos 20.0375sin 2 0.03752x xR dR d

ππ

ππ

θθ θ−

−

= = = − ∫ ∫

0 N= ............................................................................................................................Ans.

5-117 The lift on the wing of an airplane due to aerodynamic forces is shown in Fig. P5-117. The lift L may be described by the function 25 lb/ftL x= . Determine the moment of the resultant lift force about point A.

SOLUTION

( )25dM xdL x x dx= =

155/ 215 3/ 2

00

225 25 8710 lb ft5xM dM x dx

= = = = ⋅

∫ ∫ .........................................Ans.

5-118 Determine the moment of the 1650-N force shown in Fig. P5-118 about point O. SOLUTION

( )2 2 2

180 180 801650 1113.055 1113.055 494.691 N180 180 80

− + += = − + ++ +

i j kF i j k

( ) ( )0.36 0.24 0.4 1113.055 1113.055 494.691O = + + − + +M i j k × i j k

( )326 623 668 N m= − − + ⋅i j k ..........................................................................Ans.

5-119 The driving wheel of a truck is subjected to the force-couple system shown in Fig. P5-119. Replace this system by an equivalent single force and determine the point of application of the force along the vertical diameter of the wheel.

SOLUTION

750 lbxR = 1800 1800 0 lbyR = − =

750 lb = →R ...................................................................................................................Ans.

600 lb ft 750OM y= ⋅ =

0.8 ft 9.60 in. up from the groundy = = ..................................................................Ans.


148

5-120 A 200-N force is applied at corner B of a rectangular plate as shown in Fig. P5-120. Determine the moment of the force

(a) About point O. (b) About line OD. SOLUTION

(a) ( )2 2 2

0.46 0.5 0.9200 81.5854 88.6798 159.6237 N0.46 0.5 0.9

+ −= = + −+ +

i j kF i j k

( ) ( )0.8 2 81.5854 88.6798 159.6237O = − + + −M i k × i j k

( )177.360 35.472 70.944 N m= − + − ⋅i j k .....................................................Ans.

(b) ( )2 2

0.8 0.5 0.84800 0.530000.8 0.5

OD− += = − +

+i je i j

( )( ) ( )( )177.360 0.84800 35.472 0.53000OD O ODM = = − − +M e

169.2 N m= ⋅ ..........................................................................................................Ans.

( )169.2 0.84800 0.53000OD OD ODM= = − +M e i j

( )143.5 89.7 N m= − + ⋅i j ..................................................................................Ans.

5-121 A 2500-lb jet engine is suspended from the wing of an airplane as shown in Fig. P5-121. Determine the moment produced by the engine at point A in the wing when the plane is

(a) On the ground with the engine not operating. (b) In flight with the engine developing a thrust T of 15,000 lb. SOLUTION

(a) ( )2500 8sin 35 11,470 lb ft = ° = ⋅M ..................................................................Ans.

(b) ( ) ( )2500 8sin 35 15,000 8cos35= ° − °M

86,800 lb ft 86,800 lb ft = − ⋅ = ⋅ ...............................................................Ans.

5-122 Two forces and a couple act on a beam as shown in Fig. P5-122. Determine the resultant R and its location x.

SOLUTION

1.8 1 2.8 kN = + = ↓R ...................................................................................................Ans.

( )( ) ( )( )1.8 1 1 2 10 13.800 kN m 2.8OM x= + + = ⋅ =

4.92 mx = .........................................................................................................................Ans.

5-123 Replace the force and couple acting on the wall bracket shown in Fig. P5-123 by a force-couple system at point A.

SOLUTION

0 lbxR = 40 lbyR = −

40 lb = ↓R .......................................................................................................................Ans.

( )40 8 30 350 lb in.AM = + = ⋅ ..................................................................................Ans.


149

5-124 Determine the resultant of the parallel force system shown in Fig. P5-124 and locate the intersection of its line of action with the xy-plane.

SOLUTION

( ) ( ) ( ) ( ) ( ) ( )125 80 100 200 75 130 N= + − + + − + − = −R k k k k k k ................Ans.

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )3 1.5 125 1 1.5 80 2 3.5 100

3 5.5 200 1 5.5 75O = + + + − + +

+ + − + + −

M i j × k i j × k i j × k

i j × k i j × k

( )1095 180 N m= − + ⋅i j

( ) ( ) ( )130 130 130x y y x= + − = − +i j × k i j

180 1.385 m130

x = = 1095 8.42 m130

y = = ..........................................Ans.

5-125 The concrete floor of a building supports four wood columns as shown in Fig. P5-125. The resultant of the forces transmitted through the columns to the floor is R = 75 kip at the location shown in the figure. Determine the magnitude of the forces F1 and F2.

SOLUTION

1 215 20 75 kip R F F= + + + = ↓

( ) ( ) ( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )

1 2

1 2 1

20 15 20 30 30

30 300 20

13 16 75 1200 975 kip ft

O F F

F F F

= − + + − + −

= − + + +

= + − = − + ⋅

M i × k i j × k j × k

i j

i j × k i j

1 33.7 kip F = ↓ ................................................................................................................Ans.

2 6.25 kip F = ↓ ................................................................................................................Ans.

5-126 A bent rod supports a 450-kN force F as shown in Fig. P5-126. (a) Replace the 450-N force with a force R through the coordinate origin O and a couple C. (b) Determine the twisting moments produced by force F in the three different segments of the rod. SOLUTION

425cos 45 750 449.48 mmDx = °− = −

300 150 150 mmDy = − = 425sin 45 300.52 mmDz = − ° = −

(a) ( )450 N=R k ..................................................................................................................Ans.

( ) ( )0.44948 0.150 0.30052 450= − + −C i j k × k

( )67.5 202.3 N m= + ⋅i j ..........................................................................................Ans.

(b) 202.3 N mOA OM = = ⋅M j ............................................................................................Ans.

( ) ( ) ( )0.75 0.15 450 67.5 337.5 N mB = − − = − + ⋅M i j × k i j

( ) ( ) ( )425cos 45 425sin 45

0.70711 0.70711425AB

° − °= = −

i ke i k

( )( )67.5 0.70711 47.7 N mAB B ABM = = − = − ⋅M e ................................................Ans.

67.5 N mBC BM = = − ⋅M i ...........................................................................................Ans.


150


( )22 0.5VdA x x dx= −

( )22 0.5HdA y y dy= −

( )2 2

0

23/ 2 32

0

2 0.5

2 4 in.3 2 6 3

VA dA x x dx

x x

= = −

= − =

∫ ∫

( )25/ 2 42 2 3

00

2 62 0.5 in.5 2 8 5y Vx xM xdA x x x dx

= = − = − =

∫ ∫

6 5 18 0.9 in.4 3 20

yC

Mx

A= = = = ......................................................................................Ans.

( )25/ 2 42 2 3

00

2 62 0.5 in.5 2 8 5x Hy yM y dA y y y dy

= = − = − =

∫ ∫

6 5 18 0.9 in.4 3 20

xC

MyA

= = = = ......................................................................................Ans.

5-128 Two channel sections and a plate are used to form the cross section shown in Fig. P5-128. Each of the channels has a cross-sectional area of 2605 mm2. Locate the y-coordinate of the centroid of the composite section with respect to the top surface of the plate.

SOLUTION

737,190 64.3 mm11,460Cy−= = − ........................Ans.

5-129 Locate the centroid of the volume shown in Fig. P5-129 if R = 10 in. and h = 32 in. SOLUTION

22

22

RydV r dy dyh

π π

= =

2 4 2 5

4 400

2

5

5

hh R y R yV dV dyh h

R h

π π

π

= = =

=

∫ ∫


151

22 2 6 2 2

2 4006 6

hh

xzRy R y R hM y dV y dyh h

π ππ

= = = =

∫ ∫

( )2 2

2

5 326 5 26.7 in.5 6 6

xzC

M R h hyV R h

ππ

= = = = = ........................................................Ans.

0 in. (by symmetry)C Cx z= = .....................................................................................Ans.

5-130 Locate the centroid and the mass center of the volume shown in Fig. P5-130, which consists of an aluminum cylinder (ρ= 2770 kg/m3) and a steel (ρ = 7870 kg/m3) cylinder and sphere.

SOLUTION

( ) ( )2 3 31 0.1 0.3 9.4248 10 mV π −= = ×

( )31 2770 9.4248 10 26.1066 kgm −= × =

( ) ( )2 3 32 0.05 0.175 1.3744 10 mV π −= = ×

( )32 7870 1.3744 10 10.8169 kgm −= × =

( )3

3 33

4 0.14.1888 10 m

3V

π −= = ×

( )33 7870 4.1888 10 32.9658 kgm −= × =

3 314.9880 10 mtotV−= × 69.8893 kgtotm =

( )( ) ( )( ) ( )( )0 9.4248 187.5 1.3744 375 4.1888

122.0 mm14.9880Cy

+ += = .................Ans.

( )( ) ( )( ) ( )( )0 26.1066 187.5 10.8169 375 32.9658

205.9 mm69.8893Gy

+ += = .........Ans.

0 mm (by symmetry)C G C Gx x z z= = = = ................................................................Ans.

5-131 Determine the resultant R of the system of distributed loads on the beam of Fig. P5-131, and locate its line of action with respect to the left support of the beam.

SOLUTION

( )( ) ( )( )4

0

400 4200 400 4

2R x dx= + +∫

4

3/ 2

0

2200 1600 800 3466.667 lb 3470 lb 3x = + + = ≅ ↓

....................................Ans.

( ) ( )( ) ( ) ( )( ) ( )4

0

45/ 2

0

400 4200 400 4 6 8 4 3

2

2200 9600 7466.667 19,626.667 lb ft5

OM x x dx

x

= + + +

= + + = ⋅

∫

19,626.667 5.66 ft3466.667

x = = ...............................................................................................Ans.


152

Chapter 6 6-1 Draw a free-body diagram of the angle bracket

shown in Fig. P6-1. SOLUTION

6-2 Draw a free-body diagram of the diving board shown in Fig. P6-2. The surface at B is smooth. Neglect the weight of the diving board.

SOLUTION

6-3 Draw a free-body diagram of the lawn mower shown in Fig. P6-3. The lawn mower has a weight W and is resting on a rough surface.

SOLUTION

6-4 Draw a free-body diagram of the sled shown in Fig. P6-4.

SOLUTION

6-5 Draw a free-body diagram of the bracket shown in Fig. P6-5. The contact surfaces between the cylinders and bracket are smooth.

SOLUTION

6-6 Forces P are applied to the handles of the pipe pliers shown in Fig. P6-6. Draw free-body diagrams of each handle.

SOLUTION


153

6-7 The man shown in Fig. P6-7 has a weight W1; the beam has a weight W2 and a center of gravity G. Draw a free-body diagram of the beam.

SOLUTION

6-8 Forces P are applied to the handles of the bolt cutter shown in Fig. P6-8. Draw a free-body diagram of

(a) The lower handle. (b) The lower cutter jaw. SOLUTION

6-9 Draw a free-body diagram of the door shown in Fig. P6-9. The homogeneous door has weight W.

SOLUTION

6-10 Draw a free-body diagram of the bent bar shown in Fig. P6-10. The support at A is fixed, and the bar has negligible mass.

SOLUTION

6-11 Draw a free-body diagram of the bar shown in Fig. P6-11. The support at A is a journal bearing and the supports at B and C are ball bearings.

SOLUTION

6-12 Draw a free-body diagram of the shaft shown in Fig. P6-12. The bearing at A is a thrust bearing, and the bearing at D is a ball bearing. Neglect the weights of the shaft and the levers.

SOLUTION


154

6-13 A beam is loaded and supported as shown in Fig. P6-13. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xA =

0 :AMΣ =�

( ) ( )( ) ( )

15 3 500 6 800

9 700 12 400 0

B − −

− − =

1160 lbB =

0 :yF↑ Σ = 500 800 700 400 0yA B+ − − − − =

1240 lbyA =

1240 lb = ↑A ............................................................................................................................ Ans.

1160 lb = ↑B ............................................................................................................................ Ans.

6-14 A beam is loaded and supported as shown in Fig. P6-14. Determine the reaction at support A. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment

0 :xF→ Σ = 0xA =

0 :yF↑ Σ = 2 0yA − =

2 kNyA =

0 :AMΣ =� ( )2 4 3 0AM − − =

11 kN mAM = ⋅

2 kN = ↑A ................................................................................................................................Ans.

11 kN m A = ⋅M � ..................................................................................................................... Ans.

6-15 A 30-lb force P is applied to the brake pedal of an automobile as shown in Fig. P6-15. Determine the force Q applied to the brake cylinder and the reaction at support A.

SOLUTION From a free-body diagram of the brake pedal, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( )( ) ( ) ( )5.5 30cos30 11 30sin 30 4 0Q − ° − ° =

62.871 lbQ =

0 :xF→ Σ = 30cos30 0xA Q− + ° =

36.890 lbxA =

0 :yF↑ Σ = 30sin 30 0yA − ° =

15 lbyA =

39.8 lb=A 22.13° ............................................................Ans.

62.9 lb = ←Q ........................................................................Ans.


155

6-16 A beam is loaded and supported as shown in Fig. P6-16. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xA =

0 :AMΣ =�

( )( ) ( )

( )( ) ( )

4.5 300 1.5 0.75

1 400 1.5 3 0.5 02

B −

− + =

308.333 NB =

0 :yF↑ Σ = ( )( ) ( )( )1300 1.5 400 1.5 02yA B − − + =

441.667 NyA =

442 N = ↑A .............................................................................................................................. Ans.

308 N = ↑B .............................................................................................................................. Ans.

6-17 A rope and pulley system is used to support a body W as shown in Fig. P6-17. Each pulley is free to rotate and the rope is continuous over the pulleys. Determine the tension T in the rope required to hold body W in equilibrium if the weight of body W is 400 lb. Assume that all rope segments are vertical.

SOLUTION The pulleys are rigidly connected together and can be treated as a single body for the purpose of drawing a free-body diagram. From a free-body diagram of the pulleys, the equilibrium equations are solved to get the tension

0 :yF↑ Σ = 4 400 0T − =

100 lbT = ................................................................................................ Ans.

6-18 Pulleys 1 and 2 of the rope and pulley system shown in Fig. P6-18 are connected and rotate as a unit. The radii of pulleys 1 and 2 are 100 mm and 300 mm, respectively. Rope A is wrapped around pulley 1 and is fastened to pulley 1 at point A�. Rope B is wrapped around pulley 2 and is fastened to pulley 2 at point B�. Rope C is continuous over pulleys 3 and 4. Determine the tension T in rope C required to hold body W in equilibrium if the mass of body W is 225 kg.

SOLUTION

( )225 9.81 2207.25 NW = =

From a free-body diagram of the pulley 4, the equilibrium equations are solved to get

0 :yF↑ Σ = 2 0BT T W+ − =

2BT W T= −

Next, from a free-body diagram of the pulley 3, the equilibrium equations are solved to get

0 :yF↑ Σ = 2 0AT T− =

2AT T=


156

Finally, from a free-body diagram of the pulleys 1 and 2, the equilibrium equations are solved to get

0 :axleMΣ =� 100 300 0A BT T− =

100 300A BT T=

( ) ( )100 2 300 2T W T= −

8 3 6621.75 NT W= =

828 NT = ................................................................................................ Ans.

6-19 Three pipes are supported in a pipe rack as shown in Fig. P6-19. Each pipe weighs 100 lb. Determine the reactions at supports A and B.

SOLUTION From a free-body diagram of the pipes and the rack, the equilibrium equations are solved to get the forces

0 :AMΣ =�

( ) ( ) ( ) ( )( ) ( ) ( )( )

18 100cos30 9 100cos30 15

100cos30 21 3 100sin 30 4 0

B − ° − °

− ° − ° =

249.840 lbB =

0 :xF→ Σ = sin 30 0xA B+ ° =

124.920 lbxA = −

0 :yF↑ Σ = 300 cos30 0yA B− + ° =

83.632 lbyA =

150.3 lb=A 33.80° ............................................................................................................Ans.

250 lb=B 60° ..................................................................................................................... Ans.

6-20 The man shown in Fig. P6-20 has a mass of 75 kg; the beam has a mass of 40 kg. The beam is in equilibrium with the man standing at the end and pulling on the cable. Determine the force exerted on the cable by the man and the reaction at support C.

SOLUTION From a free-body diagram of the man and the beam, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 0xC =

0 :CMΣ =�

( ) ( ) ( ) ( )75 9.81 3 40 9.81 1.5

3 1.5 0T T

+

− − =

621.300 NT =

0 :yF↑ Σ = ( ) ( )2 75 9.81 40 9.81 0yT C+ − − =

114.450 NyC = −

114.5 N = ↓C ........................................................................................................................... Ans.

621 NT = ....................................................................................................................................Ans.


157

6-21 A 75-lb load is supported by an angle bracket, pulley, and cable as shown in Fig. P6-21. Determine (a) The force exerted on the bracket by the pin at C. (b) The reactions at supports A and B of the bracket. (c) The shearing stress on a cross section of the ¼-in.-diameter pin at C, which is in double shear. SOLUTION (a) First, from a free-body diagram of the pulley, the equilibrium

equations are solved to get the axle forces

0 :xF→ Σ = ( )( )4 5 75 0xC− =

60 lbxC =

0 :yF↑ Σ = ( ) ( )75 3 5 75 0yC − − =

120 lbyC =

134.2 lb=C 63.43 on the pulley°

134.2 lb=C 63.43 on the bracket° ..........................Ans. (b) Next, from a free-body diagram of the bracket, the equilibrium

equations are solved to get the forces

0 :xF→ Σ = 60 0xA + =

60 lbxA = −

0 :AMΣ =� ( )10 18 60 0B − =

108 lbB =

0 :yF↑ Σ = 120 0yA B+ − =

12 lbyA =

61.2 lb=A 11.31° ...........................................................................................................Ans.

108.0 lb = ↑B ....................................................................................................................... Ans. (c) Finally, dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear)

gives the shear stress

( )2

134.2 1367 psi2 2 0.25 4

Cs

CA

τπ

= = =

...........................................................................Ans.

6-22 A pipe strut BC is loaded and supported as shown in Fig. P6-22. Determine (a) The reactions at supports A and C. (b) The shearing stress on a cross section of the 10-mm-diameter pin at C, which is in double shear. (c) The elongation of cable AB if it is made of aluminum alloy (E = 73 GPa) and has a diameter of 6 mm. SOLUTION (a) From a free-body diagram of the pipe strut, the equilibrium equations are solved to get the forces

0 :CMΣ =� ( )1000 800 750 0ABT − =

600 NABT = ...................................................................Ans.

0 :xF→ Σ = 0x ABC T− =

600 NxC =


158

0 :yF↑ Σ = 750 0yC − =

750 NyC =

960.47 N 960 N= ≅C 51.34° .....................................................................................Ans. (b) Dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear) gives the

shear stress

( )

6 22

960.47 6.11 10 N/m 6.11 MPa2 2 0.010 4

Cs

CA

τπ

= = = × =

...................................Ans.

(c) Finally, the stretch of the cable AB is given by

( )

( ) ( )29

600 16000.465 mm

73 10 0.006 4AB

PLEA

δπ

= = = ×

.....................................................Ans.

6-23 The lawn mower shown in Fig. P6-23 weighs 35 lb and has a center of gravity at G. Determine (a) The magnitude of the force P required to push the mower at a constant velocity. (b) The forces exerted on the front and rear wheels by the inclined surface. (c) The shearing stresses in the ½-in.-diameter shoulder bolts of the front and rear wheels. SOLUTION (a) From a free-body diagram of the lawn mower, the equilibrium equations are solved to get the forces

0 :OMΣ =� ( ) ( ) ( ) ( )227 35cos15 13 35sin15 4 0N − ° + ° =

2 14.936 lbN = .................................................Ans.

0 :xFΣ = cos30 35sin15 0P ° − ° =

10.460 lbP = ...................................................Ans.

0 :yFΣ = 1 2 35cos15 sin 30 0N N P+ − ° − ° =

1 24.101 lbN = .................................................Ans.

(c) Finally, dividing the shear forces on the pins by their cross-sectional areas gives the shear stresses

( )

11 2

1

24.101 122.7 psi0.5 4s

NA

τπ

= = = ......................................................................................Ans.

( )

22 2

2

14.936 76.1 psi0.5 4s

NA

τπ

= = = .......................................................................................Ans.

6-24 The coal wagon shown in Fig. P6-24 is used to haul coal from a mine. If the mass of the coal and wagon is 2000 kg (with its center of mass at G), determine

(a) The magnitude of the force P required to move the wagon at a constant velocity. (b) The force exerted on each of the front wheels by the inclined surface. (c) The shearing stress in each of the 25-mm-diameter front axles. SOLUTION

( )2000 9.81 19,620 NW = =

(a) From a free-body diagram of the coal wagon, the equilibrium equations are solved to get the forces

0 :xFΣ = sin 30 0P W− ° =


159

9810 NP = ....................................................................Ans.

1 0 :MΣ =�

( )( ) ( ) ( )( )( ) ( )

22 3 1 cos30 2

sin 30 2 0

N P W

W

− − °

+ ° =

2 4028.81 NN = ............................................................Ans.

0 :yFΣ = 1 22 2 cos30 0N N W+ − ° =

1 4466.90 NN =

(c) Finally, dividing the shear force on the front axle by its cross-sectional area gives the shear stress

( )

6 222 2

2

4028.81 8.21 10 N/m 8.21 MPa0.025 4s

NA

τπ

= = = × = ............................................Ans.

6-25 A lever is loaded and supported as shown in Fig. P6-25. Determine (a) The reactions at A and C. (b) The normal stress in the ½-in.-diameter rod CD. (c) The shearing stress in the ½-in.-diameter pin at A, which is in double shear. (d) The change in length of rod CD, which is made of a material with a modulus of elasticity of 30(106) psi. SOLUTION

( )1tan 6 12 26.565θ −= = ° 2 26 12 13.416 in.CDL = + =

(a) From a free-body diagram of the lever, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )cos 6 40 4 60 8 80 12 0CDT θ − − + =

59.628 lbCDT = −

0 :xF→ Σ = 125 cos 0x CDA T θ+ + =

71.667 lbxA = −

0 :yF↑ Σ = 40 60 80 sin 0y CDA T θ− − + − =

6.666 lbyA = −

72.0 lb=A 5.31° ..................................Ans.

59.6 lb (C)CDT = .........................................Ans.

(b) Dividing the normal force in the rod CD by its cross-sectional area gives the normal stress

( )2

59.628 304 psi 304 psi (C)0.5 4

CDCD

n

TA

σπ

−= = = − = .......................................................Ans.

(c) Dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )2

72.0 183.3 psi2 2 0.5 4A

As

VA

τπ

= = =

............................................................................Ans.

(d) Finally, the change in length of the rod CD is given by


160

( )( )

( ) ( )26

59.628 13.416

30 10 0.5 4CD

PLEA

δπ

−= =

×

4 41.358 10 in. 1.358 10 in. (shrink)− −= − × = × ................................................................Ans.

6-26 The wood plane shown in Fig. P6-26 moves with a constant velocity when subjected to the forces shown. Determine

(a) The shearing force of the wood on the plane. (b) The normal force, and its location, of the wood on the plane. SOLUTION From a free-body diagram of the plane, the equilibrium equations are solved to get the forces

0 :xF→ Σ = 40cos 70 70cos16 0F − ° − ° =

0 :yF↑ Σ = 40sin 70 70sin16 0N − ° − ° =

0 :AMΣ =� ( )( ) ( ) ( )70cos16 75 70sin16 220° − °

( ) ( )40cos 70 60 0Nd+ ° + =

81.0 NF = ............................................................................................................Ans.

56.882 NN = .......................................................................................................Ans.

28.5 mmd = −

31.5 mm (from front of plane) ........................................................................Ans.

6-27 A bracket of negligible weight is used to support the distributed load shown in Fig. P6-27. Determine (a) The reactions at the supports A and B. (b) The shearing stress in the 0.25-in.-diameter pin at A, which is in single shear. (c) The bearing stress between the bracket and the 1-in.×1-in. bearing plate at B. SOLUTION (a) From a free-body diagram of the bracket, the equilibrium equations are solved to get the forces

0 :AMΣ =� ( )( ) ( )18 12 10 2 12 3 0B − =

0 :xF→ Σ = 0xA B+ =

0 :yF↑ Σ = ( )( )12 10 2 0yA − =

13.333 lbB =

13.333 lbxA = − 60.0 lbyA =

61.464 lb 61.5 lb= ≅A 77.47° ............................................Ans.

13.33 lb = →B ...............................................................................Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress

( )2

61.464 1252 psi0.25 4

AA

s

VA

τπ

= = = .....................................................................................Ans.

(c) Finally, the bearing stress is given by

13.333 13.33 psi

1 1bb

BA

σ = = =×

.............................................................................................Ans.


161

6-28 Determine the force P required to push the 135-kg cylinder over the small block shown in Fig. P6-28. SOLUTION

1 220 75cos 48.769220

θ − −= = °

90 20 21.231φ θ= ° − − ° = °

From a free-body diagram of the cylinder, the equilibrium equations are solved to get the forces

0 :centerMΣ =� 220 0F− =

0 NF =

0 :xF→ Σ = cos 0P N φ− =

0 :yF↑ Σ = ( )sin 135 9.81 0N φ − =

3657.2 NN =

3410 N P = → ...................................................................................................................... Ans.

6-29 The electric motor shown in Fig. P6-29 weighs 25 lb. Due to friction between the belt and pulley, the belt forces have magnitudes of T1 = 21 lb and T2 = 1 lb. Determine

(a) The support reactions at A and B. (b) The shearing stress in the ¼-in.-diameter pin at A, which is in single shear. SOLUTION (a) From a free-body diagram of the motor and support,

the equilibrium equations are solved to get the forces

0 :xF→ Σ = 21 1 0xA − − =

22 lbxA =

0 :AMΣ =� ( ) ( ) ( )12 8 25 5.5 1 10.5 21 0B − + + =

2.167 lbB = −

0 :yF↑ Σ = 25 0yA B+ − =

27.167 lbyA =

34.958 lb 35.0 lb= ≅A 51.00° ...................................................................................Ans.

2.17 lb = ↓B ......................................................................................................................... Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress

( )2

34.958 712 psi0.25 4

AA

s

VA

τπ

= = = ......................................................................................Ans.

6-30 Bar AB of Fig. P6-30 has a uniform cross section, a mass of 25 kg, and a length of 1 m. Determine the angle θ for equilibrium.

SOLUTION From a free-body diagram of the bar, the equilibrium equations are

0 :xF→ Σ = sin 30 sin 45 0A B° − ° =

0 :yF↑ Σ = ( )cos30 cos 45 25 9.81 0A B° + ° − =

0 :AMΣ =� ( )( ) ( )( ) ( ) ( )cos 45 1cos sin 45 1sin 25 9.81 0.5cos 0B Bθ θ θ° + ° − =


162

Rearranging the first equation gives

1.41421A B= Substituting this into the second equation gives

126.951 NB =

179.535 NA = Finally, the third equation gives

tan 0.366θ =

20.10θ = ° ................................................................................................................................Ans.

6-31 The wrecker truck of Fig. P6-31 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Determine the maximum pull P that the wrecker can exert when θ = 30° if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).

SOLUTION From a free-body diagram of the truck, the equilibrium equations are

0 :xF→ Σ = sin 30 0xP B° − =

0 :yF↑ Σ = 15,000 cos30 0y yA B P+ − − ° =

0 :AMΣ =� ( ) ( ) ( )( ) ( ) ( )15,000 8 14.5 cos30 5 sin 30 10 0yA P P− − ° − ° =

The fourth equation needed to solve for the four unknowns is either 0yA = (the front wheels are on the verge of

lifting off the ground) or 0.8x yB B= (the rear wheels are on the verge of slipping). Guessing that the front wheels are on the verge of lifting off the ground gives the solution

0 lbyA =

12,861.56 lbP =

26,138.44 lbyB =

6430.78 lbxB =

Since

6430.78 lb 0.8 20,911 lbx yB B= < =

the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and

max 12,860 lbP = ..................................................................................................................... Ans.

6-32 Pulleys A and B of the chain hoist shown in Fig. P6-32 are connected and rotate as a unit. The chain is continuous, and each of the pulleys contains slots that prevent the chain from slipping. Determine the force F required to hold a 450-kg block W in equilibrium if the radii of pulleys A and B are 90 mm and 100 mm, respectively.

SOLUTION From a free-body diagram of the lower pulley, vertical equilibrium equation gives the tension

0 :yF↑ Σ = ( )2 450 9.81 0T − =

2207.25 NT = Then, from a free-body diagram of the lower pulley, moment equilibrium equation gives the force F


163

0 :axleMΣ =� 100 90 100 0T T F− − =

221 NF = ................................................................................................................................Ans.

6-33 The crane and boom shown in Fig. P6-33 weigh 12,000 lb and 600 lb, respectively. When the boom is in the position shown, determine

(a) The maximum load that can be lifted by the crane. (b) The tension in the cable used to raise and lower the boom when the load being lifted is 3600 lb. (c) The pin reaction at boom support A when the load being lifted is 3600 lb. (d) The required size of pin A ( which is in double shear) if the shearing stress must not exceed 12 ksi. SOLUTION (a) From a free-body diagram of the entire crane,

the equilibrium equations are

0 :yF↑ Σ = 12,000 600 0N W− − − =

0 :CMΣ =� ( ) ( ) ( )( )12,000 9 600 12cos30 1Nd− − ° −

( )24cos30 1 1 0W− ° − + =

The maximum load that can be lifted corresponds to impending tip; at which point the normal force acts at the front corner C, and 0d = . At this point

max 4930 lbW W= = ...............................................................................................................Ans.

(b) From a free-body diagram of the pulley, the equilibrium equations give the axle forces

0 :xF→ Σ = 3600cos10 0xB − ° =

3545.31 lbxB =

0 :yF↑ Σ = 3600 3600sin10 0yB − − ° =

4225.13 lbyB =

Then, from the free-body diagram of the boom, the equilibrium equations give the forces

0 :AMΣ =�

( ) ( ) ( )( ) ( ) ( )( )

3545.31 24sin 30 4225.13 24cos30 600 12cos30

cos10 24sin 30 sin10 24cos30 0T T

° − ° − °

+ ° ° − ° ° =

6275.13 lb 6280 lbT = ≅ ....................................... Ans.

0 :xF→ Σ = cos10 3545.31 0xA T− ° − =

9725.11 lbxA =

0 :yF↑ Σ = sin10 4225.13 600 0yA T− ° − − =

5914.80 lbyA =

11,382.55 lb 11.38 kip= ≅A 31.31° ............ Ans.

(d) Finally, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress (which must not exceed 12 ksi)

( )2

11,382.55 12,000 psi2 2 4A

s A

VA d

τπ

= = =

0.777 in.Ad = ......................................................................................................................... Ans.


164

6-34 A bar AB of negligible mass is supported in a horizontal position by two cables as shown in Fig. P6-34. Determine

(a) The magnitude of force P and the force in each cable. (b) The change in length of the 15-mm-diamerter cable BD, if it is initially 1 m long and is made of steel

with a modulus of elasticity of 200 GPa. SOLUTION (a) From a free-body diagram of the bar, the equilibrium equations give the forces

0 :BMΣ =� ( )( )1300 sin 60 0AL T L+ ° =

1501.11 N 1501 NAT = ≅ ........................................ Ans.

0 :xF→ Σ = cos 68 cos60 0B AT T° − ° =

2003.58 N 2004 NBT = ≅ ...................................... Ans.

0 :yF↑ Σ = sin 60 sin 68 1300 0A BT T P° + ° − − =

1857.69 N 1858 NP = ≅ .....................................................................................................Ans. (b) The stretch of wire BD is given by

( )

( ) ( )29

2003.58 10000.0567 mm

200 10 0.015 4PLEA

δπ

= = = ×

...................................................Ans.

6-35 The garage door ABCD shown in Fig. P6-35 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown. Determine

(a) The force in the cable and the reactions at the rollers when d = 75 in. (b) The shearing stress (single shear) in the 3/8-in.-diameter rods that connect the roller to the door. SOLUTION

( )1cos 25 48 58.612θ −= = °

1 12 6sintan 5.46675 6cos

θφθ

− −= = °−

(a) From a free-body diagram of the door, the equilibrium equations are

0 :xF→ Σ = cos 2 0DT Bφ − =

0 :yF↑ Σ = sin 225 2 0DT Cφ − − =

0 :DMΣ =� ( ) ( ) ( ) ( ) ( ) ( )225 36cos 2 48sin sin 6cos cos 6sin 0D DB T Tθ θ φ θ φ θ− + − =

The first equation gives

0.49773 DB T=

Substituting this into the moment equation gives

92.5356 lb 92.5 lbDT = ≅ ....................................................................................................Ans.

46.0574 lb 46.1 lbB = ≅ .....................................................................................................Ans.

108.0926 lb 108.1 lb C = − ≅ ↑ .........................................................................................Ans. (b) Dividing the shear force on the rods by their cross-sectional areas gives the shear stresses

( )2

46.0574 417 psi3 8 4

BB

s

VA

τπ

= = = .........................................................................................Ans.


165

( )2

108.0926 979 psi3 8 4

CC

s

VA

τπ

= = = .........................................................................................Ans.

6-36 The lever shown in Fig. P6-36 is formed in a quarter circular arc of radius 450 mm. Determine the angle θ if neither of the reactions at A or B can exceed 200 N.

SOLUTION From a free-body diagram of the lever, the equilibrium equations are

0 :xF→ Σ = sin 0xA B θ− =

0 :yF↑ Σ = 125 cos 0yA B θ− + =

0 :DMΣ =� ( )450 0yA− =

The fourth needed equation is either 2 2 200 Nx yA A A= + = or

200 NB = . In either case, the moment equation gives

0 NyA =

Now, guessing that 200 NB = gives

51.32θ = ° ................................................................................................................................Ans.

156.1 NxA =

2 2 156.1 N 200 Nx y xA A A A= + = = <

Therefore, the guess was correct.

6-37 A man is slowly raising a 20-ft-long homogeneous pole weighing 150 lb as shown in Fig. P6-37. The lower end of the pole is kept in place by smooth surfaces. Determine the force exerted by the man to hold the pole in the position shown.

SOLUTION From a free-body diagram of the pole, the equations of equilibrium give the tension

0 :AMΣ =� ( )( ) ( )sin 30 20 150 10cos60 0T ° − ° =

75.0 lbT = ............................................................................Ans.

6-38 The lever shown in Fig. P6-36 is a quarter circular arc of radius 450 mm. The 25-mm-diameter pin at A is smooth and frictionless and is in single shear.

(a) Plot A, the magnitude of the pin force at A, and B, the force on the smooth support B, as functions of θ (10° ≤ θ ≤ 80°), the angle at which the support is located.

(b) Plot τ, the average shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the lever, the equilibrium equations are


166

0 :DMΣ =� ( )450 0yA− =

0 NyA =

0 :yF↑ Σ = 125 cos 0yA B θ− + =

125 cos NB θ=

0 :xF→ Σ = sin 0xA B θ− =

sin 125 tan NxA B θ θ= =

Therefore

2 2 125 tan Nx y xA A A A θ= + = = ......................................................................................Ans.

( )125 cos NB θ= .................................................................................................................Ans.

(b) Dividing the force in the pin at A by its cross-sectional area gives the shear stress

( ) ( )

22 2

125 tan 500 tan N/m0.025 4 0.025

A

s

VA

θ θτπ π

= = = ....................................................................Ans.

6-39 The crane and boom shown in Fig. P6-39 weigh 12,000 lb and 600 lb, respectively. The pulleys at D and E are small and the cables attached to them remain essentially parallel. The 3-in.-diameter pin at A is smooth and frictionless and is in double shear. The distributed force exerted on the treads by the ground is equivalent to a single resultant force N acting at some distance d behind point C.

(a) Plot d, the location of the equivalent force N relative to point C, as a function of the boom angle θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load.

(b) Plot τ, the average shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load.

(c) It is desired that the resultant force on the tread always be at least 1 ft behind C to ensure that the crane is never in danger of tipping over. Plot Wmax, the maximum load that may be lifted, as a function of θ (0° ≤ θ ≤ 80°).

SOLUTION (a) From a free-body diagram of the entire crane,


0 :yF↑ Σ = 12,000 600 0N W− − − =

0 :CMΣ =� ( )( ) ( )( )12,000 9 600 12cos 1Nd θ− − −

( )24cos 1 1 0W θ− − + =


167

If 3600 lbW = , then

16,200 lbN =

108,600 93,600cos ft

16,200d θ−= ...........................................................................................Ans.

(c) If the normal force must be no closer than 1 ft to the front of the tread, then

max12,600 lbN W= +

( ) ( )max max108,600 12,600 1 7200cos 24 cos 0W Wθ θ− + − − =

max96,000 7200cos lb

1 24cosW θ

θ−=

+...........................................................................................Ans.

(b) As the boom is raised, the cable angle φ (assumed to be the

same for both cables) and the boom angle θ are related by

24cosb θ=

24sinh θ=

6 24sin 6tan9 24cos 9

hb

θφθ

− −= =+ +

Then, from the free-body diagram of the pulley, the equilibrium equations give the forces

0 :xF→ Σ = 3600cos 0xB φ− =

3600cos lbxB φ=

0 :yF↑ Σ = 3600 3600sin 0yB φ− − =

( )3600 1 sin lbyB φ= +

Finally, from the free-body diagram of the boom, the equilibrium equations

0 :xF→ Σ = cos 0x x BDA B T φ− − =

0 :yF↑ Σ = 600 sin 0y y BDA B T φ− − − =

0 :AMΣ =� ( )( ) ( )cos 24sin 600 12cosx BDB T φ θ θ+ −

( )( )sin 24cos 0y BDB T φ θ− + =


168

are solved to get the forces

( )24 7200 cos 24 sin

lb24cos sin 24sin cos

y xBD

B BT

θ θφ θ φ θ

+ −=

−............Ans.

cos lbx x BDA B T φ= +

600 sin lby y BDA B T φ= + +

2 2x yA A A= +

Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 2

2psi

14.13722 3 4x yAA AVτ

π

+= =

..................................................................................... Ans.

Although these graphs appear to be okay, a check of the data reveals that for angles greater than about 66°, the cable tension TBD is negative when the crane is lifting 3600 lb. But the cable cannot push on the boom. Therefore, for angles greater than 66°, the boom would fall back on the cab of the crane and the solution for angles greater than 66° are meaningless. Further analysis would be required to find the critical angle for other weights being lifted.

6-40 An extension ladder arrangement is being raised into position by a hydraulic cylinder as shown in Fig. P6-40. The center of gravity of the 500-kg ladder is at G. The 25-mm-diameter pin at A is smooth, frictionless, and in double shear.

(a) Plot σn, the normal stress in the 40-mm-diameter piston rod of the hydraulic cylinder, as a function of the angle θ (10° ≤ θ ≤ 90°).

(b) Plot τ, the shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 90°). SOLUTION

( )500 9.81 4905 NW = =

(a) From a free-body diagram of the ladder, the equilibrium equations are

0 :xF→ Σ = sin 0x BA F φ+ =

0 :yF↑ Σ = cos 4905 0y BA F φ+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )cos 3cos sin 3sinB BF Fφ θ φ θ−

( )( ) ( ) ( )4905cos 8 4905sin 1 0θ θ− + =

in which

1 3cos 2tan3sin 1.5

θφθ

− − = +

Solving the equilibrium equations gives


169

( )( ) ( ) ( )4905cos 8 4905sin 1

N3cos cos 3sin sinBF

θ θφ θ φ θ

−=

−

sin Nx BA F φ= −

4905 cos Ny BA F φ= −

Dividing the piston force by its cross-sectional area gives the normal stress

( )20.04 4

B Bn

n

F FA

σπ

= =

( )

22

4 N/m0.04

BFπ

= ...................... Ans.

(b) Dividing the force in the pin at A by twice its cross-sectional area (since it is in double shear) gives the shear stress

( ) ( )

2 2 2 22

22

2N/m

2 0.0252 0.025 4x y x yA

s

A A A AVA

τππ

+ += = =

.........................................................Ans.

6-41 The hydraulic cylinder BC is used to tip the box of the dump truck shown in Fig. P6-41. The pins at A (one on each side of the box) each have a diameter of 1.5 in. and are in double shear. If the combined weight of the box and the load is 22,000 lb and acts through the center of gravity G,

(a) Plot C, the force in the hydraulic cylinder, as a function of the angle θ (0° ≤ θ ≤ 80°). (b) Plot τ, the shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the truck box,


0 :xF→ Σ = 2 cos 0x BCA F φ− =

0 :yF↑ Σ = 2 sin 22,000 0y BCA F φ+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )22,000cos 8.5 22,000sin 2θ θ−

( ) ( )sin 12.5 0BCF φ θ− − =

in which

1 12.5sin 0.5tan12.5cos 2.0

θφθ

− +=−

Therefore

( )187,000cos 44,000sin lb

12.5sinBCFθ θ

φ θ−=

−..........................Ans.

( )cos 2 lbx BCA F φ= ( )22,000 sin 2 lby BCA F φ= −

(b) Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 2

2psi

3.534292 1.5 4x yAA AVτ

π

+= =

..................................................................................Ans.


170

6-42 The wrecker truck of Fig. P6-42 has a mass of 6800 kg and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Plot P, the maximum pull that the wrecker can exert, as a function of θ (0° ≤ θ ≤ 90°) if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground).

SOLUTION

( )6800 9.81 66,708 NW = =

From a free-body diagram of the truck, the equilibrium equations are

0 :xF→ Σ = sin 0xP Bθ − =

0 :yF↑ Σ = 66,708 cos 0y yA B P θ+ − − =

0 :BMΣ =� ( )( ) ( ) ( )66,708 2.4 4.4 cos 1.5yA P θ− −

( )( )sin 3 0P θ− =

The fourth equation needed to solve for the four unknowns is either 0yA = (the front wheels are on the verge of

lifting off the ground) or 0.8x yB B= (the rear wheels are on the verge of slipping). Guessing that the front wheels are on the verge of lifting off the ground gives the solution

0 NyA =

( )66,708 2.4

N1.5cos 3sin

Pθ θ

=+

......................................................................................................Ans.

sin NxB P θ=

66,708 cos NyB P θ= +

The forces xB and 0.8 yB are plotted on the same

graph as the force P . Since xB is always less than

0.8 yB , the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and the solution is valid for all values of θ .


171

6-43 The garage door ABCD shown in Fig. P6-43 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown.

(a) Plot T, the tension in the cable, as a function of d (0 ≤ d ≤ 100 in.). (b) Plot B and C, the forces on the frictionless rollers, as a function of d (0 ≤ d ≤ 100 in.). SOLUTION

(a) From a free-body diagram of the door with both rollers on the horizontal track ( )0 40 in.d< < , the equilibrium equations give

0 :xF→ Σ = cos 0T φ =

0 lbT =

0 :CMΣ =� ( ) ( ) ( ) ( )36 2 48 sin 6 0W B T φ− + =

3 8 675 8 84.4 lbB W= = =

0 :yF↑ Σ = sin 2 2 0T B C Wφ + + − =

8 225 8 28.1 lbC W= = =

From a free-body diagram of the door with roller B on the curve and roller C on the horizontal track ( )40 in. 53.5 in., 0 90d γ< < ° < < ° , the equilibrium equations are

0 :xF→ Σ = cos 2 sin 0T Bφ γ− =

0 :yF↑ Σ = sin 2 cos 2 0T B C Wφ γ+ + − =

0 :CMΣ =� ( ) ( ) ( )36cos 2 cos 48W Bθ γ θ− −

( )( ) ( )( )sin 6cos cos 6sin 0T Tφ θ φ θ+ − =

in which

12 12cos 1 cossin

48 4γ γθ − −= =

( )100 48cos 12 1 sind θ γ= − − −

12 6sintan

6cosdθφθ

−=−

Therefore

( )

6 cos lb8cos cos sin sin cos cos sin

WT θφ γ θ γ θ φ θ φ

=− + −

cos lb

2sinTB φ

γ= ( )2 cos sin 2 lbC W B Tγ φ= − −

From a free-body diagram of the door with roller B on the vertical track and roller C on the horizontal track ( )53.5 in. 88 in.d< < , the equilibrium equations are

0 :xF→ Σ = cos 2 0T Bφ − =

0 :yF↑ Σ = sin 2 0T C Wφ + − =

0 :CMΣ =� ( ) ( )( ) ( ) ( ) ( )( )36cos 2 48sin sin 6cos cos 6sin 0W B T Tθ θ φ θ φ θ− + − =


172

in which

100cos

48dθ −=

12 6sintan

6cosdθφθ

−=−

Therefore

6 cos lb

9sin cos cos sinWT θ

θ φ θ φ=

−

cos 2 lbB T φ=

( )sin 2 lbC W T φ= −

Finally, from a free-body diagram of the door with roller B on the vertical track and roller C on the curve ( )88 in. 100 in., 0 90d γ< < ° < < ° , the equilibrium equations are

0 :xF→ Σ = cos 2 2 sin 0T B Cφ γ− − =

0 :yF↑ Σ = sin 2 cos 0T C Wφ γ+ − =

0 :CMΣ =� ( ) ( )36cos 2 48sinW Bθ θ−

( )( ) ( )( )sin 6cos cos 6sin 0T Tφ θ φ θ+ − =

in which

12 12sin 1 sincos

48 4γ γθ − −= =

( )100 12 1 sind γ= − −

12 6sintan

6cosdθφθ

−=−

Therefore

6 cos 8 sin tan lb

9cos sin sin cos 8sin sin tanW WT θ θ γ

φ θ φ θ φ θ γ+=

− +

cos sin tan tan lb

2T T WB φ φ γ γ+ −=

sin lb

2cosW TC φ

γ−=


173

6-44 Determine all forces acting on member ABE of the frame of Fig. P6-44. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support forces

0 :AMΣ =� ( ) ( )300 150 300 0D − =

150 ND =

0 :xF→ Σ = 150 0xA + =

150 NxA = −

0 :yF↑ Σ = 0yA D+ =

150 NyA = −

Next, from a free-body diagram of member BCD, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( )150 300 100 0yC+ =

450 NyC = −

0 :CMΣ =� ( ) ( )150 200 100 0yB− =

300 NyB =

Finally, from a free-body diagram of member ABE, the equilibrium equations give the pin forces

0 :EMΣ =� ( ) ( ) ( ) ( )300 100 150 100 150 200 100 0xB+ − − =

150 NxB =

0 :xF→ Σ = 150 150 0xE − − =

300 NxE =

0 :yF↑ Σ = 300 150 0yE − − =

450 NyE =

Therefore, the forces acting on member ABE are:

212 N=A 45° ..........................................................................................................Ans.

335 N=B 63.43° .....................................................................................................Ans.

541 N=E 56.31° .....................................................................................................Ans.

6-45 In the linkage of Fig. P6-45, a = 2.0 ft, b = 1.5 ft, θ = 30°, and P = 40 lb. Determine all forces acting on member BCD.

SOLUTION Note that member AC is a two-force member. Then, from a free-body diagram of the member BCD, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( ) ( ) ( )cos 45 2 40cos30 3.5 0ACT ° − ° =

85.732 lbACT =

0 :xF→ Σ = 40cos30 cos 45 0x ACB T+ ° − ° =

25.981 lbxB =


174

0 :yF↑ Σ = sin 45 40sin 30 0y ACB T− ° + ° =

40.622 lbyB =

Therefore, the forces acting on member BCD are:

48.2 lb=B 57.40° ...................................................................................................Ans.

85.7 lbAC =T 45° ......................................................................................................Ans.

6-46 Forces of 5 N are applied to the handles of the paper punch of Fig. P6-46. Determine the force exerted on the paper at D and the force exerted on the pin at B by handle ABC.

SOLUTION From a free-body diagram of handle EBD, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( )40 5 70 0DF − =

8.75 NDF =

0 :xF→ Σ = 0 NxB =

0 :yF↑ Σ = 5 0y DB F− + =

13.75 NyB =

13.75 N = ↓B .................................................................................................................Ans. The force exerted on the paper is equal and opposite to the force exerted on the handle

8.75 N D = ↓F .................................................................................................................Ans.

6-47 Forces of 25 lb are applied to the handles of the pipe pliers shown in Fig. P6-47. Determine the force exerted on the pipe at D and the force exerted on handle DAB by the pin at A.

SOLUTION From a free-body diagram of handle DAB, the equilibrium equations give the forces

0 :AMΣ =� ( ) ( )1.25 25 9 0DF − =

180 lbDF =

0 :xF→ Σ = sin 38 0D xF A° − =

110.819 lbxA =

0 :yF↑ Σ = 25 cos38 0y DA F− − ° =

166.842 lbyA =

200 lb=A 56.41° ....................................................................................................Ans. The force exerted on the pipe at D is equal and opposite to the force exerted on the handle

180 lbD =F 52° .........................................................................................................Ans.

6-48 The jaws and bolts of the wood clamp in Fig. P6-48 are parallel. The bolts pass through swivel mounts so that no moments act on them. The clamp exerts forces of 300 N on each side of the board. Treat the forces on the boards as uniformly distributed over the contact areas and determine the forces in each of the bolts. Show on a sketch all forces acting on the upper jaw of the clamp.

SOLUTION The resultant of the distributed force exerted on the upper jaw of the clamp is a force of 300 N at the center of the uniformly distributed load. From a free-body diagram of the upper jaw, the equilibrium equations give the forces


175

0 :DMΣ =� ( ) ( )300 225 100 0CF− =

675 N (T)CF = .............................................................Ans.

0 :CMΣ =� ( ) ( )300 125 100 0DF− =

375 N (C)DF = .............................................................Ans.

6-49 Determine all forces acting on member ABCD of the frame of Fig. P6-49. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )12 100 24 0F − =

200 lbF =

0 :xF→ Σ = 100 0xA − =

100 lbxA =

0 :yF↑ Σ = 0yA F+ =

200 lbyA = −

Next, note that member BE is a two-force member. Then, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( ) ( ) ( )cos 45 6 100 18 100 6 0BEF ° − − =

565.685 lbBEF =

0 :xF→ Σ = 100 100 cos 45 0BE xF C− − ° + =

400 lbxC =

0 :yF↑ Σ = 200 sin 45 0y BEC F− + ° =

200 lbyC = −

Therefore, the forces acting on member ABCD are:

224 lb=A 63.43° ..................................................Ans.

566 lb=B 45° .........................................................Ans.

447 lb=C 26.57° ..................................................Ans.

6-50 In Fig. P6-50, a cable is attached to the structure at E, passes around the 0.8-m-diameter, frictionless pulley at A, and then is attached to a 1000-N weight W. Determine

(a) The support reaction at G. (b) All forces acting on member ABCD. SOLUTION (a) From a free-body diagram of the complete structure,

the equilibrium equations give the support reactions

0 :DMΣ =� ( ) ( )1000 3.4 2 0G− =

1700 NG =

0 :xF→ Σ = 0xD G− =


176

1700 NxD =

0 :yF↑ Σ = 1000 0yD − =

1000 NyD =

1700 N = ←G ...............................................................................................................Ans. (b) Next, from a free-body diagram of the pulley, the equilibrium equations give the pin forces

0 :xF→ Σ = 1000 0xA− =

1000 NxA =

0 :yF↑ Σ = 1000 0yA − =

1000 NyA =

Then, from a free-body diagram of member ECF, the equilibrium equations give the pin forces

0 :FMΣ =� ( ) ( )1 1000 1.4 0xC + =

1400 NxC = −

Finally, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( ) ( )1 1000 2 1000 1 0yC + + =

3000 NyC = −

0 :CMΣ =� ( ) ( ) ( )1000 2 1 1000 1 0yB− + =

3000 NyB =

0 :xF→ Σ = 1000 1700 0x xB C+ + + =

1300 NxB = −

Therefore the forces on member ABCD are

1414 N=A 45° ........................................................................................................Ans.

3270 N=B 66.57° ..................................................................................................Ans.

3310 N=C 64.98° ..................................................................................................Ans.

1972 N=D 30.47° ...................................................................................................Ans.

6-51 A pin-connected system of levers and bars is used as a toggle for a press as shown in Fig. P6-51. Three members are joined by pin D, as shown in the insert. Determine

(a) The force exerted on the can at A when a force of 1000 lb is applied to the lever at G. (b) All forces that act on member CD. SOLUTION (a) From a free-body diagram of the lever, the moment equilibrium

equation gives the force in the link DE

0 :FMΣ =� ( ) ( )1000 30 8 0DEF− =

3750 lbDEF =

Note that all three links connected to the pin D are two-force members. From the free-body diagram of the pin, the equilibrium equations give the forces in the links CD and BD


177

0 :xF→ Σ = cos78 cos 67 3750 0CD BDF F° + ° − =

0 :yF↑ Σ = sin 67 sin 78 0BD CDF F° − ° =

6395.05 lbBDF = 6018.19 lbCDF =

Finally, from a free-body diagram of the plunger, the vertical equilibrium equation gives the crushing force

0 :yF↑ Σ = 6395.05sin 67 0AF − ° =

5890 lb on the plungerA = ↑F

5890 lb on the canA = ↓F .......................................Ans.

(b) Member CD is a two force member in compression; therefore,

6020 lbCD =F 78 on at CD C° .............................................................................Ans.

6020 lbCD =F 78 on at CD D° .............................................................................Ans.

6-52 The front-wheel suspension of an automobile is shown in Fig. P6-52. The pavement exerts a vertical force of 2700 N on the tire. Determine the force in the spring and the forces at A, B, and D.

SOLUTION From a free-body diagram of the wheel, the equilibrium equations give the pin forces

0 :BMΣ =� ( ) ( )325 2700 150 0CDF − =

1246.154 NCDF =

0 :xF→ Σ = 0x CDB F− =

1246.154 NxB =

0 :yF↑ Σ = 2700 0yB− =

2700 NyB =

Next, from a free-body diagram of the control arm, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( ) ( )250 500 50 0s y xF B B− − =

5649.23 NsF =

0 :xF→ Σ = 0x xA B− =

1246.154 NxA =

0 :yF↑ Σ = 5649.23 0y yA B+ − =

2949.23 NyA =

Therefore the force on the spring is

5650 N s = ↓F ................................................................................................................Ans.

and the forces at A, B, and D are 3200 N=A 67.09° ..................................................................................................Ans.

2970 N=B 65.22° ..................................................................................................Ans.

1246 N (C)CDF = ...........................................................................................................Ans.


178

6-53 The fold-down chair of Fig. P6-53 weighs 30 lb and has its center of gravity at G. Determine (a) All forces acting on member ABC. (b) The shearing stress on a cross section of the 3/8-in.-diameter pin at B, which is in single shear. SOLUTION (a) From a free-body diagram of the chair, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )30 3 21 0E− =

4.28571 lbE =

0 :xF→ Σ = 0xA E− =

4.28571 lbxA =

0 :yF↑ Σ = 30 0yA − =

30 lbyA =

Note that member BD is a two force member which makes an angle of

( )1tan 10 12 39.81θ −= = °

relative to the horizontal. Then, from the free-body diagram of the seat, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( )( )30 3 sin 12 0BDT θ− =

11.7154 lbBDT =

0 :xF→ Σ = cos 4.28571 0BD xT Cθ − − =

4.7429 lbxC =

0 :yF↑ Σ = sin 30 0BD yT Cθ + − =

22.500 lbyC =

The forces acting on ABC are equal and opposite to those acting on the seat

30.3 lb=A 81.87° ...................................................................................................Ans.

11.72 lbB =F 39.81° ................................................................................................Ans.

22.99 lb=C 78.10° .................................................................................................Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress

( )2

11.7154 106.1 psi3 8 4

B

s

FA

τπ

= = = .................................................................................Ans.

6-54 A scissors jack for an automobile is shown in Fig. P6-54. The screw threads exert a force F on the blocks at joints A and B. Determine

(a) The force P exerted on the automobile if F = 800 N and θ = 15°, θ = 30°, and θ = 45°. (b) The shearing stress on a cross section of the 10-mm diameter pin at C, which is in single shear. Solve for

each angle in part (a). SOLUTION (a) Note that all three members attached to pin A are two-force members. From a free-body diagram of pin A, the

equilibrium equations give the forces in members AC and AE

0 :yF↑ Σ = sin sin 0AE ACF Fθ θ− =


179

AE ACF F=

0 :xF→ Σ = 800 cos cos 0AE ACF Fθ θ− − =

( )800 2cos NAE ACF F θ= =

Similarly

( )800 2cos NBD ACF F θ= =

Then, from the free-body diagram of the platform, the equilibrium equations give the force P

0 :yF↑ Σ = sin sin 0AC BDF F Pθ θ+ − =

2 sin 800 tan NACP F θ θ= =

If 15θ = ° 214 NP = ...................................Ans.

If 30θ = ° 462 NP = ...................................Ans.

If 45θ = ° 800 NP = ...................................Ans. (b) Finally, dividing the force on the pin at C by its cross-sectional area gives the shear stress

( )

( )6

22

800 2cos 5.09296 10 N/mcos0.010 4

AC

s

FA

θτ

θπ×= = =

If 15θ = ° 6 25.27 10 N/m 5.27 MPaτ = × = ............................................Ans.

If 30θ = ° 6 25.88 10 N/m 5.88 MPaτ = × = ............................................Ans.

If 45θ = ° 6 27.20 10 N/m 7.20 MPaτ = × = ............................................Ans.

6-55 A force of 20 lb is required to pull the stopper DE in Fig. P6-55. Determine (a) All forces acting on member BCD. (b) The shearing stress on the cross section of the 1/8-in.-diameter pin at B, which is in single shear. (c) The deformation of the 1/8×3/8-in. member AB, which is made of steel with a modulus of elasticity of

30(106) psi. SOLUTION (a) From a free-body diagram of the entire cork puller,

the equilibrium equations give the pulling forces

0 :GMΣ =� ( ) ( )1.5 1.5 0y yE D− =

y yE D=

0 :yF↑ Σ = 20 0y yE D− − =

10 lby yD E= =

Note that both links attached to the pin at A are two-force members. Then, from the free-body diagram of pin A, the equilibrium equations give the forces in the links

0 :xF→ Σ = cos 45 cos 45 0AF ABT T° − ° =

AF ABT T=

0 :yF↑ Σ = 20 sin 45 sin 45 0AB AFT T− ° − ° =

14.14214 lbAB AFT T= =

Next, from the free-body diagram of member BCD, the equilibrium equations give the pin forces


180

0 :CMΣ =� ( ) ( ) ( ) ( ) ( )( )2 1.5 cos 45 2 sin 45 1 0x y AB ABD D T T− − ° − ° =

22.500 lbxD =

0 :xF→ Σ = cos 45 0AB x xT C D° + + =

0 :yF↑ Σ = sin 45 0AB y yT C D° + − =

32.500 lbxC = − 0 lbyC =

Therefore, the forces acting on member BCD are

14.14 lbB =F 45° .................................................................. Ans.

32.5 lb = ←C ............................................................................ Ans.

24.6 lb=D 23.96° ............................................................... Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress

( )2

14.14214 1152 psi1 8 4

B

s

FA

τπ

= = = ..................................................................................Ans.

(c) The change in length of the link AB is given by

( )( )

( ) ( ) ( )5

6

14.14214 21.42 10 in.

30 10 1 8 3 8ABPLEA

δ −= = = ×× ×

..........................................Ans.

6-56 Member BD of the frame shown in Fig. P6-56 is made of structural steel (E = 200 GPa) and has a rectangular cross section 50 mm wide by 15 mm thick. All pins have 15 mm diameters. Determine

(a) The axial stress in member BD. (b) The shearing stress on a cross section of pin C if it is loaded in double shear. (c) The change in length of member BD. SOLUTION (a) From a free-body diagram of the entire structure, the equilibrium equations give the support reactions

0 :AMΣ =� ( ) ( )( ) ( ) ( )1.2 2000 1.2 0.6 2000 0.7 0E − − =

2366.67 NE =

0 :xF→ Σ = 2000 0xA− =

2000 NxA =

0 :yF↑ Σ = ( )2000 1.2 0yA E+ − =

33.333 NyA =

Next, from the free-body diagram of member CDE (in which sin 3 5θ = and cos 4 5θ = ), the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( ) ( ) ( ) ( )( )2366.667 1.2 2000 1.2 0.6 sin 0.8 0BDT θ− − =

2916.67 NBDT =

0 :xF→ Σ = 2916.67 cos 0xC θ− =

2333.33 NxC =


181

0 :yF↑ Σ = ( )2366.67 2000 1.2 2916.67sin 0yC θ+ − − =

1783.33 NyC =

2 2 2936.78 Nx yC C C= + =

Then, dividing the force in the member BD by its cross-sectional area gives the axial (normal) stress

( )

( )6 22916.67

3.89 10 N/m 3.89 MPa0.05 0.015

BDBD

n

TA

σ = = = × =×

...............................Ans.

(b) Dividing the force on the pin at C by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )( )

6 22

2936.788.31 10 N/m 8.31 MPa

2 2 0.015 4C

s

FA

τπ

= = = × =

...............................Ans.

(c) Finally, the change in length of the two-force member BD is given by

( )( )

( )( )9

2916.67 10000.01944 mm

200 10 0.05 0.015BDPLEA

δ = = =× ×

........................................Ans.

6-57 The hoist pulley structure of Fig. P6-57 is rigidly attached to the wall at C. A load of sand hangs from the cable that passes around the 1-ft-diameter, frictionless pulley at D. The weight of the sand can be treated as a triangular distributed load with a maximum intensity of 70 lb/ft. Determine

(a) All forces acting on member ABC. (b) The shearing stress on the cross section of the ½-in.-diameter pin D, which is in double shear. (c) The change in length of the ¼ × 1-in. member BE [E = 29(106) psi]. SOLUTION

( )( )1 70 2 70 lb2

T W= = =

(a) From a free-body diagram of the entire structure, the equilibrium equations give the reaction forces

0 :xF→ Σ = 0xT C− =

0 :yF↑ Σ = 0yC T− =

0 :DMΣ =� ( ) ( )5 1.5 0y x CTr Tr C C M− + − − =

245 lb ftCM = ⋅

70 lbxC = 70 lbyC =

Then, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )70 7 4 245 0BF− − =

61.25 lbBF =

0 :xF→ Σ = 70 0xA − =

0 :yF↑ Σ = 70 0y BA F− − =

70 lbxA =

8.75 lbyA =


182

Therefore, the forces acting on member ABC are

70.5 lb=A 7.13° ......................................................................................................Ans.

61.3 lb = ↓B .................................................................................................................. Ans.

99.0 lb=C 45° .........................................................................................................Ans.

245 lb ft C = ⋅M � ..........................................................................................................Ans.

(b) Next, from the free-body diagram of the pulley, the equilibrium equations give the pin forces

0 :xF→ Σ = 70 0xD− =

0 :yF↑ Σ = 70 0yD − =

70 lbxD =

70 lbyD =

2 2 98.995 lbx yD D D= + =

Dividing the force on the pin at D by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )2

98.995 252 psi2 2 0.5 4s

DA

τπ

= = = ...............................................................................Ans.

(c) The change in length of the two-force member BE is given by

( )( )

( ) ( )4

6

61.25 3 123.04 10 in.

29 10 1 4 1BEPLEA

δ −×= = = ×

× × .................................................Ans.

6-58 A pair of vice grip pliers is shown in Fig. P6-58. Determine the force exerted on the bolt by the jaws of the pliers when a force of magnitude 100 N is applied to the handle. Let d = 30 mm.

SOLUTION

( )1tan 30 40 36.870θ −= = ° sin 3 5θ = cos 4 5θ =

From a free-body diagram of the upper handle, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( )( ) ( )( )100 93 sin 38 cos 5 0A AF Fθ θ− + =

494.681 NAF =

0 :xF→ Σ = cos 0A xF Bθ − =

0 :yF↑ Σ = sin 100 0A yF Bθ − − =

395.745 NxB =

196.809 NyB =

Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is

0 :CMΣ =� ( ) ( )35 12 0x yDa B B− − =

in which 2 235 15 38.079 mma = + = Therefore

426 ND = ........................................................................................................................ Ans.


183

6-59 Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P6-59. Determine (a) All forces acting on the handle ABC. (b) The force exerted on the bolt at E. (c) The axial stress in the links at D (one on each side) if each has a 1/8 × 3/4-in. rectangular cross section. (d) The change in length of the links at D if they are 4 in. long and made of SAE 4340 heat-treated steel. SOLUTION (a) From a free-body diagram of the jaw CDE, horizontal equilibrium gives

0 lbxC =

Then, from a free-body diagram of the handle ABC, the equilibrium equations give the pin forces

0 :xF→ Σ = 0x xC B− =

0 :yF↑ Σ = 50 0y yB C− + =

0 :BMΣ =� ( ) ( )1 50 20 0yC − =

1000 lbyC =

0 lbxB = 1050 lbyB =

Therefore, the forces acting on the handle ABC are

1050 lb = ↓B .................................................................................................................Ans.

1000 lb = ↑C .................................................................................................................Ans. (b) Then, returning to the free-body diagram of the jaw CDE, the equilibrium equations give the force on the bolt

0 :DMΣ =� ( ) ( )3 2 0yC E− =

1500 lb on the jaw= ↓E

1500 lb on the bolt= ↑E ..........................................................................................Ans.

0 :yF↑ Σ = 1000 1500 0DT − − = 2500 lbDT =

(c) Dividing the link force by twice its cross-sectional area (since there are two links) gives the normal stress

( )( )

31500 13.33 10 psi 13.33 ksi2 2 1 8 3 4D

n

TA

σ = = = × =

.......................................Ans.

(d) Finally, the change in length of the two-force link is given by

( ) ( )

( ) ( ) ( )3

6

2500 2 41.839 10 in.

29 10 1 8 3 4DPLEA

δ −= = = ××

............................................Ans.

6-60 The tower crane of Fig. P6-60 is rigidly attached to the building at F. A cable is attached at D and passes over small frictionless pulleys at A and E. The object suspended from C has a mass of 1530 kg. Determine

(a) All forces acting on member ABCD. (b) The support reactions at F. (c) The shearing stress on a cross section of pin A if it has a diameter of 15 mm and is loaded in double

shear. SOLUTION

( )1530 9.81 15,009.30 NW = = 1tan 6 14.4 22.620φ −= = °

sin 3 5θ = cos 4 5θ =


184

(a) From a free-body diagram of member ABCD and pulley A, the equilibrium equations give the forces

0 :BMΣ =� ( )( ) ( )( ) ( ) ( )8 sin 8 sin 14.4 15,009.30 10 0T T Tθ φ− + − =

17,176.136 NT =

0 :xF→ Σ = cos cos 0xT B Tθ φ+ − =

2113.97 NxB =

0 :yF↑ Σ = sin 15,009.30 sin 0yT T B Tθ φ− + − + =

15, 273.51 NyB =

Therefore, the forces acting on the member ABCD are

15,363 N 15,360 N= ≅A 26.565° ......................................................................Ans.

15, 420 N=B 82.120° ............................................................................................Ans.

15,010 N = ↓C .............................................................................................................Ans.

17,180 N=D 22.620° ............................................................................................Ans.

(b) Next, from the free-body diagram of the entire structure, the equilibrium equations give the support reactions

0 :xF→ Σ = 0 NxF =

0 :yF↑ Σ = 17,176.136 15,009.30 0yF − − =

0 :FMΣ =� ( )( )17,176.136 8 FM+

( ) ( )15,009.30 10 0− =

32,185.4 N 32.2 kNyF = ≅

12,684 N mFM = ⋅

Therefore, the support reactions are

32.2 kN ≅ ↑F ................................. Ans.

12.68 kN m F ≅ ⋅M � ..................... Ans.

(c) Finally, dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

6 22

15,363 43.5 10 N/m 43.5 MPa2 2 0.015 4A

s

VA

τπ

= = = × =

..............................Ans.

6-61 Three bars are connected with smooth pins to form the frame shown in Fig. P6-61. The weights of the bars are negligible. Determine

(a) The reactions at supports A and E. (b) The resultant forces at pins B, C, and D. SOLUTION

( )1tan 1 2 26.565θ −= = °

From a free-body diagram of the entire structure, the equilibrium equations give

0 :AMΣ =� ( )26 675 29.5 0yE − =


185

765.865 lbyE =

0 :yF↑ Σ = 675 0y yE A− − =

0 :xF→ Σ = 0x xA E− =

x xA E= 90.865 lbyA =

Next, from the free-body diagram of member BD, the equilibrium equations give the pin forces

0 :DMΣ =� ( )( ) ( )( )sin 13 675 10 0BF θ − =

1161.035 lbBF =

0 :xF→ Σ = cos 0B xF Dθ − =

1038.462 lbxD =

0 :yF↑ Σ = sin 675 0y BD F θ− − =

1194.230 lbyD =

Finally, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :CMΣ =� ( ) ( )2 13 13 6.5 5 0x y BA A F+ − =

603.606 lbx xA E= =

0 :xF→ Σ = cos 0x B xA F Cθ− + =

0 :yF↑ Σ = sin 0y B yA F Cθ− + − =

434.856 lbxC = 428.365 lbyC =

(a) Therefore, the reaction forces are

610 lb=A 8.56° ................................................. Ans.

975 lb=E 51.76° ............................................... Ans. (b) The forces acting on pins B, C, and D are

1161 lb=B 26.57 (on )ABC° ......................... Ans.

610 lb=C 44.57 (on )ABC° .................................................................................Ans.

1583 lb=D 48.99 (on )CDE° ...............................................................................Ans.

6-62 Figure P6-62 is a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity at B; arm DEFG has a weight of 1 kN and a center of gravity at E. The weight of the hydraulic cylinders can be ignored.

(a) Calculate the force in the horizontal cylinders CJ and EI and all forces acting on arm DEFG for the position shown.

(b) Determine the required diameter of the pin at E if the shearing stress cannot exceed 120 MPa. The pin is in double shear.

SOLUTION (a) From a free-body diagram of the bucket, the equilibrium equations give the force in cylinder EI

0 :GMΣ =� ( ) ( )( )1.2cos30 10 0.3 0EIT ° − =


186

2.8868 kN 2.89 kNEIT = ≅ .............................................. Ans.

and the pin forces at G

0 :xF→ Σ = 0x EIG T− =

2.8868 kNxG =

0 :yF↑ Σ = 10 0yG − =

10 kNyG =

Next, from a free-body diagram of member DEFG , the equilibrium equations are

0 :DMΣ =� ( ) ( ) ( ) ( ) ( )0.6cos30 1 0.6sin 30 cos 1.2cos30EI BFT F φ° − ° + °

( ) ( ) ( ) ( )sin 1.2sin 30 1.8cos30 1.8sin 30 0BF x yF G Gφ+ ° − ° − ° =

0 :xF→ Σ = cos 0BF x EI xF D T Gφ − + − =

0 :yF↑ Σ = sin 1 0BF y yF D Gφ + − − =

in which

1 1.8cos30 1.2cos30tan 19.1071.8sin 30 1.2sin 30

φ − ° − ° = = ° ° + °

These equations are solved to give

10.438 kNBFF =

9.863 kNxD = 7.583 kNyD =

Therefore, the remaining forces acting on the member DEFG are

12.44 kN=D 37.55° ...............................................................................................Ans.

10.44 kN=F 19.11° ................................................................................................Ans.

10.41 kN=G 73.90° ...............................................................................................Ans. Then, from the free-body diagram of the entire bucket assembly,

the moment equilibrium equation gives the force in the cylinder CJ

0 :AMΣ =� ( ) ( )( )1.8cos30 2 0.9cos30CJT ° − °

( )( )1 2.4cos30 0.6cos30− ° + °

( ) ( )10 2.4cos30 1.8cos30 0.3 0− ° + ° + =

27.9 kNCJT = ..................................................Ans.

(b) Finally, dividing the force on pin E by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )3

22

2.8868 10 N/m2 2 4E

s

VA d

τπ

×= = .....................................................................................Ans.

Since the shear stress must be no greater than 120 MPa, the minimum diameter of the pin is

( )

( )3

36

2 2.8868 103.91 10 m 3.91 mm

120 10d

π−

×= = × =

×.................................................Ans.


187

6-63 The mechanism of Fig. P6-63 is designed to keep its load level while raising it. A pin on the rim of the 4-ft-diameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 feet long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley. Determine the force P applied to the rope and all forces acting on the arm ABC when the package has been lifted 4 feet, as shown.

SOLUTION

When 4 fth = ( )1sin 2 4 30θ −= = °

From a free-body diagram of the platform, the equilibrium equations give

0 :CMΣ =� ( )( ) ( )cos30 3 80 2 0DEF ° − =

61.584 lbDEF =

0 :xF→ Σ = cos30 0DE xF C° − =

0 :yF↑ Σ = sin 30 80 0y DEC F+ ° − =

53.333 lbxC = 49.208 lbyC =

Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )2 4sin 30 4cos30 0x yB C C− ° − ° =

138.564 lbB =

0 :xF→ Σ = sin 30 0x xC A B− − ° =

0 :yF↑ Σ = cos30 0y yB C A° − − =

15.949 lbxA = − 70.792 lbyA =

Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force

0 :AMΣ =� 2 2 0P B− =

138.564 lb 138.6 lbP B= = ≅ ....................................................................................Ans. Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley. The forces acting on arm ABC are then

72.6 lb=A 77.30° ...................................................................................................Ans.

138.6 lb=B 60° .......................................................................................................Ans.

72.6 lb=C 42.70° ...................................................................................................Ans.

6-64 A drum of oil with a mass of 200 kg is supported by a frame (of which there are two) as shown in Fig. P6-64. Determine

(a) All forces acting on member ACE. (b) The elongation of the 20-mm diameter wire if it is made of a material with a modulus of elasticity of 200

GPa. SOLUTION (a) From a free-body diagram of the drum, the equilibrium equations give the forces

0 :xF→ Σ = 2 cos 45 2 cos 45 0D E° − ° =

0 :yF↑ Σ = ( )2 sin 45 2 sin 45 200 9.81 0D E° + ° − =

693.672 ND E= =


188

Next, from a free-body diagram of member BCD , the moment equilibrium equation is

0 :BMΣ =� ( ) ( ) ( )1 1 0.8 2 0y xC C D− + + =

Then, from a free-body diagram of member ACE , the equilibrium equations are

0 :AMΣ =� ( ) ( ) ( )1 1 0.8 2 0x yC C E+ − + =

0 :xF→ Σ = cos 45 0xT C E− + ° =

0 :yF↑ Σ = sin 45 0yA C E+ − ° =

Solving these last four equations gives

1535.937 NxC =

0 NyC =

1045.437 NT =

490.500 NA = Therefore, the forces acting on the member ACE are

491 N = ↑A ................................................................................................................... Ans.

1045 N = →T ................................................................................................................Ans.

1536 N = ←C ................................................................................................................Ans.

694 N=E 45° ...........................................................................................................Ans. (b) Finally, the stretch of the wire is given by

( )( )

( ) ( )29

1045.437 20000.0333 mm

200 10 0.02 4PLEA

δπ

= = = ×

...............................................Ans.

6-65 The mechanism shown in Fig. P6-65 is designed to keep its load level while raising it. A pin on the rim of the 4-ft-diameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 ft long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley.

(a) Plot P, the force required to hold the platform as a function of the platform height h (0 ≤ h ≤ 5.75 ft). (b) Plot A, C, and E, the magnitudes of the pin reaction forces at A, C, and E as a function of h (0 ≤ h ≤ 5.75

ft). SOLUTION

1 2sin4hθ − − =

(a) From a free-body diagram of the platform, the equilibrium equations give

0 :CMΣ =� ( )( ) ( )cos 3 80 2 0DEF θ − =

160 lb

3cosDEF θ=

0 :xF→ Σ = cos 0DE xF Cθ − =

0 :yF↑ Σ = sin 80 0y DEC F θ+ − =

53.333 lbxC = 160 tan80 lb

3yCθ = −


189

Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces

0 :AMΣ =� ( ) ( )2 4sin 4cos 0x yB C Cθ θ− − =

( )2 sin 2 cos lbx yB C Cθ θ= +

0 :xF→ Σ = sin 0x xC A B θ− − =

0 :yF↑ Σ = cos 0y yB C Aθ − − =

( )sin lbx xA C B θ= −

( )cos lby yA B Cθ= −

Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force

0 :AMΣ =� 2 2 0P B− =

( )2 sin 2 cos lbx yP B C Cθ θ= = + ............................................................................Ans.

Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley.

(b) The magnitudes of the pin forces A, C, and E are

2 2 lbx yA A A= + ....................... Ans.

2 2 lbx yC C C= + ....................... Ans.

160 lb

3cosDEF θ= ........................ Ans.

6-66 Forces of P = 100 N are being applied to the handles of the vise grip pliers shown in Fig. P6-66. Plot the force applied on the bolt by the jaws as a function of the distance d (20 mm ≤ d ≤ 30 mm).

SOLUTION

( )1tan 40dφ −= 2 235 15 38.079 mma = + =

From a free-body diagram of the upper handle, the equilibrium equations give the forces

0 :BMΣ =� ( ) ( ) ( ) ( )( )100 93 sin 38 cos 35 0A AF F dφ φ− + − =

( )9300 N

38sin 35 cosAF dφ φ=

− −

0 :xF→ Σ = cos 0A xF Bφ − =

0 :yF↑ Σ = sin 100 0A yF Bφ − − =

cos Nx AB F φ= ( )sin 100 Ny AB F φ= −

Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is

0 :CMΣ =� ( ) ( )35 12 0D x yF a B B− − =


190

35 12

N38.079x y

D

B BF

+= ...................................................Ans.

6-67 A load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy (E = 10,600 ksi) bars B, and a stainless steel (E = 28,000 ksi) bar C, as shown in Fig. P6-67. Each bar has a cross sectional area of 2.00 in.2 If the bars are unstressed before the load P is applied, determine the normal stresses in the bars after a 40-kip load is applied.

SOLUTION From a free-body diagram of bar A, the equilibrium equations give

0 :CMΣ =� 2 1 0B BT a T a− =

1 2B B BT T T= =

0 :yF↑ Σ = 31 2 40 10 0B B CT T T+ + − × =

4 2 40,000 lbB Cσ σ+ =

Since the force in each of the aluminum bars is the same, they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, B Cδ δ= , which in terms of the stresses can be written

( ) ( )

6 6

36 7210.6 10 28 10

B Cσ σ=

× × 1.32075C Bσ σ=


6020 psiBσ = .................................................................................................................. Ans.

7950 psiCσ = .................................................................................................................. Ans.

6-68 The rigid bar CDE, shown in Fig. P6-68, is horizontal before the load P is applied. Tie rod A is a hot-rolled steel (E = 210 GPa) bar with a length of 450 mm and a cross-sectional area of 300 mm2. Post B is an oak timber (E = 12 GPa) with a length of 375 mm and a cross-sectional area of 4500 mm2. After the 225-kN load P is applied, determine

(a) The normal stresses in bar A and post B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CDE, the moment

equilibrium equation gives

0 :CMΣ =� ( )3500 1350 1350 225 10 0AD BET F+ − × =

32.70 607.5 10 NAD BET F+ = ×


191

From similar triangles, the stretch of bar AD and the shrink of bar BE are related by

500 1350AD BEδ δ=

which gives

( )

( ) ( )( )

( )( )9 6 9 6

450 3751350 500

210 10 300 10 12 10 4500 10AD BET F

− −=

× × × ×

2.77714BE ADF T=


( )671,485.0 N (T) 300 10AD ADT σ −= = ×

( )6198,523.9 N (C) 4500 10BE BEF σ −= = ×

6 2238 10 N/m 238 MPa (T)ADσ = × = ......................................................................Ans.

6 244.1 10 N/m 44.1 MPa (C)BEσ = × = ....................................................................Ans.

(b) The vertical displacement of point D is the same as the stretch of the tie rod

( ) ( )

( )( )9 6

71,485.0 4500.511 mm

210 10 300 10ADδ−

= = ↓× ×

........................................................Ans.

6-69 A pin-connected structure is loaded and supported as shown in Fig. P6-69. Member CD is rigid and is horizontal before the load P is applied. Member A is an aluminum alloy bar with a modulus of elasticity of 10,600 ksi and a cross-sectional area of 2.25 in2. Member B is a stainless steel bar with a modulus of elasticity of 28,000 ksi and a cross-sectional area of 1.75 in2. After the load is applied to the structure, determine

(a) The normal stresses in bars A and B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CD,

the moment equilibrium equation gives

0 :CMΣ =� ( )35.5 10 10 8 3 0A BF F× − − =

38 3 55 10 lbA BF F+ = ×

From similar triangles, the shrink of the two bars and the movement of point D are related by

8 3 10A B dδ δ= =

which gives

( )

( ) ( )( )

( )( )6 6

48 363 8

10.6 10 2.25 28 10 1.75A BF F

=× ×

1.02725B AF F=


( )4963.11 lb (C) 2.25A AF σ= =


192

( )5098.35 lb (C) 1.75B BF σ= =

32.21 10 psi 2.21 ksi (C)Aσ = × = ..............................................................................Ans.

32.91 10 psi 2.91 ksi (C)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of point D is then

( ) ( )

( )( )6

10 4963.11 4810 0.01249 in. 8 8 10.6 10 2.25Ad δ= = = ↓

×.................................................Ans.

6-70 Bar B of the pin-connected system of Fig. P6-70 is made of an aluminum alloy [Ea = 70 GPa, Aa = 300 mm2, and αa = 22.5(10-6)/°C] and bar A is made of a hardened carbon steel [Es = 210 GPa, As = 1200 mm2, and αs = 11.9(10-6)/°C]. Bar CDE is to be considered rigid. When the system is unloaded at 40°C, bars A and B are unstressed. After the load P is applied, the temperature of both bars decreases to 15°C. Determine

(a) The normal stresses in bars A and B. (b) The vertical displacement (deflection) of pin E. SOLUTION (a) From a free-body diagram of the bar CDE,


0 :CMΣ =� ( )3150 450 100 10 0AD BEF F+ − × =

33 300 10 NAD BEF F+ = ×

From similar triangles, the stretch of the two bars are related by

150 450AD BEδ δ=

which gives

( )( )( ) ( )( )( )

( )( ) ( ) ( )( ) ( )

69 6

69 6

250450 11.9 10 25 250

210 10 1200 10

500150 22.5 10 25 500

70 10 300 10

AD

BE

F

F

−−

−−

+ × −

× ×

= + × −

× ×

8 19,530 NAD BEF F= −


( )6212,885 N (T) 1200 10AD ADF σ −= = ×

( )629,048.2 N (T) 300 10BE BEF σ −= = ×

6 2177.4 10 N/m 177.4 MPa (T)ADσ = × = ...............................................................Ans.

6 296.8 10 N/m 96.8 MPa (T)BEσ = × = ....................................................................Ans.

(b) The vertical displacement of pin E is the same as the stretch of bar B

( ) ( )

( ) ( ) ( )( )( )69 6

29,048.2 50022.5 10 25 500

70 10 300 10BEδ −−

= + × −

× ×

0.410 mm BEδ = ↓ .........................................................................................................Ans.


193

6-71 A rigid bar CD is loaded and supported as shown in Fig. P6-71. Bars A and B are unstressed before the 30-kip load P is applied. Bar A is made of steel (E = 30,000 ksi) and has a cross-sectional area of 2 in2. Bar B is made of brass (E = 15,000 ksi) and has a cross-sectional area of 1.5 in2. Determine

(a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin C. (c) The shearing stress on a cross section of the ¾-in.-diameter pin at D, which is in double shear. SOLUTION (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives

0 :DMΣ =� ( )310 30 10 2 6 0A BF F× − − =

33 150 10 lbA BF F+ = ×

From similar triangles, the stretch of the two bars and the movement of point C are related by

2 6 10A B cδ δ= =

which gives

( )

( )( )( )

( )( )6 6

8 156 2

30 10 2 15 10 1.5A BF F

=× ×

1.66667A BF F=


( )53,571 lb (T) 2A AF σ= =

( )32,143 lb (T) 1.5B BF σ= =

326.8 10 psi 26.8 ksi (T)Aσ = × = ..............................................................................Ans.

321.4 10 psi 21.4 ksi (T)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of point C is then

( )( )

( ) ( )6

10 53,571 810 0.0357 in. 2 2 30 10 2Ac δ= = = ↓

×.............................................................Ans.

(c) Returning to the free-body diagram of the bar CD, the force equilibrium equations give

0 :xF→Σ = 0xD =

0 :yF↑ Σ = 330 10 0A B yF F D+ − − × =

3 330 10 55.714 10 lby A BD F F= + − × = ×

2 2 355.714 10 lbx yD D D= + = ×

Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

33

2

55.714 10 63.1 10 psi 63.1 ksi2 2 0.75 4D

Ds

VA

τπ

×= = = × =

.......................................Ans.


194

6-72 A rigid bar CD is loaded and supported as shown in Fig. P6-72. Bars A and B are unstressed before the 150-kN load P is applied. Bar A is made of stainless steel (E = 190 GPa) and has a cross-sectional area of 750 mm2. Bar B is made of an aluminum alloy (E = 73 GPa) and has a cross-sectional area of 1250 mm2. Determine

(a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin D. SOLUTION (a) From a free-body diagram of the bar CD,


0 :CMΣ =� ( )0.2 0.5 4 5B AF F+

( )30.6 150 10 0− × =

32 450 10 NB AF F+ = ×

From similar triangles, the stretch of bar B and the vertical movement of pins E and D are related by

0.2 0.5 0.6B e dδ = =

When pin E moves down a distance e, bar A stretches ( )4 5A eδ = . Therefore, 2A Bδ δ= and

( )

( )( )( )

( ) ( )9 6 9 6

1000 5002

190 10 750 10 73 10 1250 10A BF F

− −=

× × × ×

1.56164A BF F=


( )3 6170.432 10 N (T) 750 10A AF σ −= × = ×

( )3 6109.136 10 N (T) 1250 10B BF σ −= × = ×

6 2227 10 N/m 227 MPa (T)Aσ = × = .......................................................................Ans.

6 287.3 10 N/m 87.3 MPa (T)Bσ = × = ......................................................................Ans.

(b) The vertical displacement of pin D is then

( )( )

( )( )3

9 6

109.136 10 5003 3 1.794 mm

73 10 1250 10Bd δ−

×= = = ↓

× ×..............................................Ans.

6-73 A 40-kip load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy [Ea = 10,600 ksi and αa = 12.5(10-6)/°F] bars B, and a stainless steel [Es = 28,000 ksi and αs = 9.6(10-6)/°F] bar C, as shown in Fig. P6-73. The bars are unstressed when the structure is assembled at 72°F. Each bar has a cross-sectional area of 2.00 in2. Determine the normal stresses in the bars after the 40-kip load is applied and the temperature is increased to 250°F.

SOLUTION From a free-body diagram of bar A, the equilibrium equations give

0 :CMΣ =� 2 1 0B BT a T a− =

1 2B B BT T T= =


195

0 :yF↑ Σ = 31 2 40 10 0B B CT T T+ + − × =

2 40,000 lbB CT T+ =

Since stresses are asked for, the equilibrium equation will be written in terms of stresses

4 2 40,000 psiB Cσ σ+ =

Since the force in each of the aluminum bars is the same (and the temperature change is the same for both bars), they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, B Cδ δ= , which in terms of the stresses can be written

( )

( ) ( )( ) ( ) ( )( ) ( )( )( )6 6

6 6

36 7212.5 10 168 36 9.6 10 168 72

10.6 10 28 10B Cσ σ− −+ × = + ×

× ×

31.32075 15.7584 10 psiC Bσ σ= − ×


310.77 10 psi 10.77 ksi (T)Bσ = × = ..........................................................................Ans.

31.536 10 psi 1.536 ksi (C)Cσ = − × = ......................................................................Ans.

6-74 A pin-connected structure is loaded and supported as shown in Fig. P6-74. Member CD is rigid and is horizontal before the 75-kN load P is applied. Bar A is made of structural steel (E = 200 GPa), and bar B is made of an aluminum alloy (E = 73 GPa). The cross-sectional areas of members A and B are 625 mm2 and 2570 mm2, respectively. Determine

(a) The vertical displacement of the pin used to apply the load. (b) The shearing stress on a cross section of the 25-mm diameter pin at C, which is in double shear. SOLUTION

( )1tan 3 4 36.870Aθ −= = ° 2 23 4 5 mAL = + =

( )1tan 3 2 56.310Bθ −= = ° 2 23 2 13 mBL = + =

(a) From a free-body diagram of the bar CD, the moment equilibrium equation gives

0 :CMΣ =� ( ) ( )4 sin 2 sinA A B BF Fθ θ+

( )35 75 10 0− × =

32.4 1.66410 375 10 NA BF F+ = ×

From similar triangles, the vertical movement of pins A, B, and D are related by

2 4 5b a d= =

When pin B moves down a distance b, bar B stretches sinB Bbδ θ= . Then, when pin A moves down a distance

2a b= , bar A will stretch

sin 2 sin

2 sin sin 1.44222A A A

B A B B

a bδ θ θδ θ θ δ

= == =


196

( )

( )( )( )

( )( )9 6 9 6

1351.44222

200 10 625 10 73 10 2570 10BAFF

− −=

× × × ×

1.44315B AF F=


378.0998 10 NAF = × 3112.7097 10 NBF = ×

The vertical displacement of pin D is then

( )( )( )( )

3

9 6

112.7097 10 132.52.5 2.5sin sin 56.310 73 10 2570 10

B

B

d b δθ −

× = = = ° × ×

0.00651 m 6.51 mm d = = ↓ ......................................................................................Ans. (b) Returning to the free-body diagram of the bar CD, the force equilibrium equations give

0 :xF→Σ = cos cos 0x A A B BC F Fθ θ− − =

0 :yF↑ Σ = 3sin sin 75 10 0A A B B yF F Cθ θ+ − − × =

3124.9997 10 NxC = × 365.6402 10 NyC = ×

2 2 3141.186 10 Nx yC C C= + = ×

Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

36 2

2

141.186 10 143.8 10 N/m 143.8 MPa2 0.025 4

CC

s

VA

τπ

×= = = × =

..........................Ans.

6-75 Before the 20 kip load P is applied, the arms of the crank C shown in Fig. P6-75 are horizontal and vertical; there is a 0.009-in. gap between the horizontal arm and the brass (Eb = 15,000 ksi and Ab = 12 in2) post B; and the aluminum (Ea = 10,000 ksi and Aa = 2 in2) rod A is horizontal. If the crank C is rigid, determine

(a) The stresses in members A and B. (b) The change in length of members A and B. SOLUTION (a) From a free-body diagram of the crank C,


0 :CMΣ =� ( )35 20 10 10 6 0A BT F× − − =

35 3 50 10 lbA BT F+ = ×

From similar triangles,

10 6a b=

in which Aa δ= is the stretch of rod A, 0.009 in.Bb δ= + ,

and Bδ is the shrink of the post B . Therefore

( )5 3 5 3 0.015 in.A Bbδ δ= = +

( )

( )( )( )

( )( )6 6

50 155 0.015 in.310 10 2 15 10 12

A BT F = + × ×


197

( )18 108,000 lbB AF T= −


( )36.3390 10 lb 2A AT σ= × =

( )36.1017 10 lb 12B BF σ= × =

3170 psi (T)Aσ = ...........................................................................................................Ans.

508 psi (C)Bσ = .............................................................................................................Ans.

(b) The change in length of the two members is then

( )( )

( )( )

33

6

6.3390 10 5015.85 10 in. (stretch)

10 10 2Aδ −×

= = ××

.............................................Ans.

( )( )

( )( )

33

6

6.1017 10 150.508 10 in. (shrink)

15 10 12Bδ −×

= = ××

..............................................Ans.

6-76 The mechanism of Fig. P6-76 consists of a structural steel (E = 200 GPa) rod A with a cross-sectional area of 350 mm2, a cold-rolled brass (E = 100 GPa) rod B with a cross sectional area of 750 mm2, and a rigid bar C. The nuts at the top ends of rods A and B are initially tightened to the point where all slack is removed from the mechanism but the bars remain free of stress. If a nut advances 2.5 mm with each full turn (360°), determine

(a) The stresses in the rods when the nut at the top end of rod B rotates 180°. (b) The vertical displacement at the top end of rod A for part (a). SOLUTION (a) From a free-body diagram of the bar C, the moment equilibrium equation gives

0 :CMΣ =� 200 125 0A BT T− =

5 8B AT T=

From similar triangles, the vertical movement of points A and B are related by

200 125a b=

in which Aa δ= is the stretch of rod A,

1.25 mmBb δ= − , and Bδ is the stretch of rod B. Therefore

( )5 8 1.25 10 8A B Bδ δ δ= − = −

( )

( )( )( )

( )( )9 6 9 6

1500 15005 10 8

200 10 350 10 100 10 750 10A BT T

− −= −

× × × ×

( )393.33333 10 1.49333 NA BT T= × −


( )3 627.537 10 N (T) 350 10A AT σ −= × = ×

( )3 644.060 10 N (T) 750 10B BT σ −= × = ×


198

6 278.7 10 N/m 78.7 MPaAσ = × = ............................................................................Ans.

6 258.7 10 N/m 58.7 MPaBσ = × = ............................................................................Ans.

(b) The vertical displacement of the top of rod A is the same as the stretch of rod A

( ) ( )

( )( )3

9 6

27.537 10 15000.590 mm

200 10 350 10Aδ−

×= = ↑

× ×.........................................................Ans.

6-77 The pin connected structure shown in Fig. P6-77 consists of a cold-rolled bronze [Eb = 15,000 ksi and αb = 9.4(10-6)/°F] bar A which has a cross-sectional area of 3.00 in2 and two 0.2% C hardened steel [Es = 30,000 ksi and αs = 6.6(10-6)/°F] bars B which have cross-sectional areas of 2.50 in2. If the temperature of bar A decreases 50°F and the temperature of bars B increases 30°F after the 200-kip load is applied, determine

(a) The normal stresses in the bars. (b) The displacement of pin C. SOLUTION (a) From a free-body diagram of the pin, the vertical equilibrium equation gives

0 :yF↑Σ = ( ) 32 4 5 200 10 0B AT T+ − × =

31.6 200 10 lbB AT T+ = ×

The stretch of the bronze and the steel bars are related by cos 4 5B A Aδ δ θ δ= = . Therefore

( )

( )( ) ( )( )( ) ( )( )( ) ( )( ) ( )6 6

6 6

5 46.6 10 30 5 0.8 9.4 10 50 4

30 10 2.50 15 10 3B AT T− −

+ × = + × −

× ×

( )1.06667 37,410 lbB AT T= −


( )396.006 10 lb 3A AT σ= × =

( )364.996 10 lb 2.5B BT σ= × =

332.0 10 psi 32.0 ksi (T)Aσ = × = ..............................................................................Ans.

326.0 10 psi 26.0 ksi (T)Bσ = × = ..............................................................................Ans.

(b) The vertical displacement of pin C is the same as the stretch of bar A

( )( )

( )( ) ( )( )( )3

66

96.006 10 489.4 10 50 48 0.0798 in.

15 10 3Aδ −×

= + × − = ↓×

....................Ans.

6-78 Three bars are connected by smooth frictionless pins as shown in Fig. P6-78. Bar BCD is rigid, bar AB is aluminum (E = 73 GPa; L = 750 mm; d = 40 mm; σmax = 240 MPa), and bar DE is steel (E = 200 GPa; L = 500 mm; d = 30 mm; σmax = 400 MPa). The 50-mm-diameter pivot pin C is aluminum (τmax = 180 MPa) and is in double shear. After the unit is assembled, the nut D is slowly tightened.

(a) Plot σAB, σDE, and τC as functions of the distance that the nut advances (0 ≤ δnut ≤ 2 mm). (b) Plot δAB and δDE as function of δnut (0 ≤ δnut ≤ 2 mm). (c) What is the maximum amount δnut that the nut can be tightened? SOLUTION (a) From a free-body diagram of the bar BCD,

the equilibrium equations give


199

0 :xF→Σ = 0xC =

0 :yF↑Σ = 0y AB DEC T T− − =

y AB DEC T T= +

0 :CMΣ =� 200 125 0AB DET T− =

5 8DE ABT T=

From similar triangles, the vertical movement of points B and D are related by

200 125b d=

in which ABb δ= is the stretch of rod AB, mmnut DEd δ δ= − , DEδ is the stretch of rod DE, and nutδ is the movement of the nut along the rod. Therefore

( )5 8AB nut DEδ δ δ= −

( )

( ) ( )( )

( ) ( )2 29 9

750 5005 8 8

73 10 0.04 4 200 10 0.03 4AB DE

nut

T Tδ

π π= −

× ×

1.44478 282,743.34 NAB DE nutT T δ+ =


282,743.34 92,861.66 N

3.04478nut

AB nutT δ δ= =

1.6DE ABT T=

Dividing the axial forces by the cross-sectional areas gives the normal stresses

( )

22 N/m (T)

0.04 4AB

ABTσ

π= ......................................................................................Ans.

( )

22 N/m (T)

0.03 4DE

DETσ

π= ......................................................................................Ans.

Dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress

( )

2 22

2N/m

2 0.05 4x yC

Cs

C CVA

τπ

+= =

...............................................................................Ans.

(b) The stretches of the two rods are

( )

( ) ( )29

750mm (stretch)

73 10 0.04 4AB

AB

Tδ

π=

×

.........................................................Ans.

( )

( ) ( )29

500mm (stretch)

200 10 0.03 4DE

DE

Tδ

π=

×

......................................................Ans.

(c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for 1.90 mmnutδ ≅ . The other stresses remain below their limiting values for the entire range graphed. Therefore,


200

( )max1.90 mmnutδ = .......................................................................................................Ans.

6-79 Three bars are connected by smooth frictionless pins as shown in Fig. P6-79. Bar ABCD is rigid, bar AE is aluminum (E = 10,600 ksi; L = 24 in.; d = ½ in.; σmax = 40 ksi), and bar CF is steel (E = 30,000 ksi; L = 18 in.; d = ¾ in.; σmax = 50 ksi). The ¾-in.-diameter pivot pin B is steel (τmax = 25 ksi) and is in single shear. The holes in bar CF are slightly overdrilled so that pin C moves down 0.06 in. before contact is made with bar CF.

(a) Plot σAE, σCF, and τB as functions of P (0 ≤ P ≤ 10 kip). (b) Plot δAE and δCF as functions of P (0 ≤ P ≤ 10 kip). (c) What is the maximum force P that the system can withstand? SOLUTION (a) From a free-body diagram of the bar ABCD,


0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AE CFB T F P− + − =

y AE CFB P T F= + −

0 :BMΣ =� 12 9 24 0AE CFT F P+ − =

4 3 8AE CFT F P+ =

From similar triangles, the vertical movement of points A, C, and D are related by

12 9 24a c d= =

in which AEa δ= is the stretch of rod AE, 0.06 in.CFd δ= + , and CFδ is the shrink of rod CF. Therefore

3 4 0.24 in.AE CFδ δ= +

( )

( ) ( )( )

( ) ( )2 26 6

24 183 4 0.24 in.

10.6 10 0.5 4 30 10 0.75 4AE CFT F

π π= +

× ×

6.36792 44,178.65 lbAE CFT F− =


132,536 8 lb

23.10377AEPT +=


201

8 4

3AE

CFP TF −=


( )2 psi (T)0.5 4AE

AETσ

π= .............................................................................................Ans.

( )2 psi (C)0.75 4

CFCF

Fσπ

= ..........................................................................................Ans.

Dividing the force on the pin by its cross-sectional area gives the shear stress

( )

2 2

2 psi0.75 4x yB

Bs

B BVA

τπ

+= = ...........................................................................................Ans.


( )

( ) ( )26

24in. (stretch)

10.6 10 0.5 4AE

AE

Tδ

π=

×

...........................................................Ans.

( )

( ) ( )26

18in. (shrink)

30 10 0.75 4CF

CF

Fδ

π=

×

............................................................Ans.

(c) Checking the graph reveals that the normal stress in rod AE reaches its limiting value for 6.2 kipP ≅ . The other stresses remain below their limiting values for the entire range graphed. Therefore,

max 6.2 kipP = ................................................................................................................... Ans.

6-80 Member ABCD of the pin-connected structure shown in Fig. P6-80 is rigid; bar BF is made of steel [EBF = 210 GPa; ABF = 1200 mm2; αBF = 11.9(10-6)/°C]; and bar CE is made of an aluminum alloy [ECE = 73 GPa; ACE = 900 mm2; αCE = 22.5(10-6)/°C]. As a result of a misalignment of the pin holes at A, B, and C, bar CE must be heated 80°C (after pins A and B are in place) to permit insertion of pin C. Compute and plot

(a) The axial stresses σBF (in the steel bar) and σCE (in the aluminum bar) as functions of the temperature decrease (as bar CE cools back down to room temperature) ∆T (0°C ≥ ∆T ≥ −80°C).

(b) The elongations δBF (of the steel bar) and δCE (of the aluminum bar) as functions of the temperature decrease ∆T (0°C ≥ ∆T ≥ −80°C).

SOLUTION When bar CE is heated 80°C, it stretches


202

( )( ) ( )622.5 10 80 600 1.08000 mminitδ −= × =

(a) From a free-body diagram of the bar ABCD, the moment equilibrium equation gives

0 :AMΣ =� 240 80 0CE BFT T− =

3BF CET T=

From similar triangles, the stretch of bar BF and the movement of pin C are related by

80 240BF cδ =

in which ( )1.08000 mmCEc δ= − and CEδ is the stretch of bar CE. Therefore

( )3 1.0800 mmBF CEδ δ= −

( )( ) ( )

( )( )( ) ( )( )( )

9 6

69 6

10003 1.08000

210 10 1200 10

600 22.5 10 80 600

73 10 900 10

BF

CE

T

TT

−

−−

=

× ×

− + × + ∆

× ×

1.30357 1478.250 NCE BFT T T+ = − ∆

in which 80 C 0 CT− ° ≤ ∆ ≤ ° . Combining the equilibrium equation and the deformation equation gives

1478.250 301.026 N

4.91071CETT T− ∆= = − ∆

3BF CET T=


26 N/m MPa (T)

1200 10 1200BF BF

BFT Tσ −= =

×..............................................................Ans.

26 N/m MPa (T)

900 10 900CE CE

CET Tσ −= =×

..................................................................Ans.


( )

( )( )9 6

1000mm (stretch)

210 10 1200 10BF

BF

Tδ

−=

× ×.........................................................Ans.

( )

( )( )9 6

60073 10 900 10

CECE

Tδ

−

=

× ×

( )( ) ( )622.5 10 80 600 mm (stretch)T− + × + ∆ ............................................Ans.


203

6-81 Three bars are connected by smooth frictionless pins as shown in Fig. P6-81. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; σmax = 20 ksi; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; σmax = 10 ksi; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel (τmax = 5 ksi) and is in single shear. If the temperature drops after the unit is assembled,

(a) Plot σAC, σDE, and τB as functions of the temperature drop ∆T (0 ≥ ∆T ≥ −60°F). (b) Plot δAC and δDE as functions of ∆T (0 ≥ ∆T ≥ −60°F). (c) What is the maximum temperature drop that the system can withstand? SOLUTION (a) From a free-body diagram of the bar BCD,


0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AC DEB T T+ − =

y DE ACB T T= −

0 :BMΣ =� 20 30 0AC DET T− =

1.5AC DET T=

From similar triangles, the shrink of bar AC and the stretch of bar DE are related by

20 30AC DEδ δ=

3 2AC DEδ δ=

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )( ) ( )

626

626

363 12.5 10 36

12 10 1 4

30 2 6.6 10 30

30 10 0.5 4

AC

DE

TT

TT

π

π

−

−

− + × ∆ ×

= + × ∆ ×

1.12500 171.41315 lbDE ACT T T+ = − ∆

in which 60 F 0 FT− ° ≤ ∆ ≤ ° . Combining the equilibrium equation and the deformation equation gives

63.78164 lbDET T= − ∆ 1.5AC DET T=


204


( )2 psi (T)1 4AC

ACTσ

π= .................................................................................................Ans.

( )2 psi (T)0.5 4DE

DETσ

π= .............................................................................................Ans.

Dividing the force on the pin by its cross-sectional area gives the shear stress

( )

2 2

2 psi5 8 4x yB

Bs

B BVA

τπ

+= = .............................................................................................Ans.

(b) The shrink of bar AC and the stretch of bar DE are

( )

( ) ( )( )( ) ( )6

26

3612.5 10 36 in. (shrink)

12 10 1 4AC

ACT

Tδπ

− = − + × ∆ ×

.............Ans.

( )

( ) ( )( ) ( )( )6

26

306.6 10 30 in. (stretch)

30 10 0.5 4DE

DET

Tδπ

− = + × ∆ ×

.............Ans.

(c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for 31 FT∆ ≅ − ° . The other stresses remain below their limiting values for the entire range graphed. Therefore,

max 31 FT∆ = − ° ................................................................................................................Ans.

6-82 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-82, if a = 2 m and P = 5 kN.

SOLUTION 2 tan 30 1.15470 mb = ° =

tan 30 0.66667 md b= ° = From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :AMΣ =� ( )2.66667 2 5 0C − =

0 :yF↑Σ = 5 0yA C+ − =


205

3.7500 kNC =

1.2500 kNyA =

From a free-body diagram for joint A, the equilibrium equations give

0 :yF↑Σ = 1.25 sin 30 0ABT+ ° =

0 :xF→Σ = cos30 0AD ABT T+ ° =

2.500 kN 2.50 kN (C)ABT = − = ...............................Ans.

2.1651 kN 2.17 kN (T)ADT = ≅ ...............................Ans.

From a free-body diagram for joint D, the equilibrium equations give

0 :xF→Σ = 0CD ADT T− =

0 :yF↑Σ = 5 0BDT − =

2.1651 kN 2.17 kN (T)CDT = ≅ .............................................. Ans.

5.000 kN 5.00 kN (T)BDT = = ................................................ Ans.

Finally, from a free-body diagram for joint C, the equilibrium equations give

0 :yF↑Σ = 3.75 sin 60 0BCT+ ° =

4.3301 kN 4.33 kN (C)BCT = − = ........................................... Ans.

6-83 A 4000-lb crate is attached by light, inextensible cables to the truss of Fig. P6-83. Determine the force in each member of the truss using the method of joints.

SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :AMΣ =� ( ) ( )6 4 2000 8 2000 0E − − =

0 :xF→Σ = 0xE A− =

0 :yF↑Σ = 4000 0yA − =

4000 lbE =

4000 lbxA =

4000 lbyA =


0 :xF→Σ = ( )4 5 4000 0ABT − =

0 :yF↑Σ = ( )4000 3 5 0AE ABT T− − =

5000 lb (T)ABT = ................................................................. Ans.

1000 lb (T)AET = ................................................................. Ans.

From a free-body diagram for joint E, the equilibrium equations give

0 :xF→Σ = ( )4 5 4000 0DE BET T+ + =

0 :yF↑Σ = ( )3 5 0AE BET T+ =

1666.67 lb 1667 lb (C)BET = − ≅ .........................................Ans.


206

2666.67 lb 2670 lb (C)DET = − ≅ ........................................Ans.


0 :xF→Σ = 0CD DET T− =

0 :yF↑Σ = 2000 0BDT − =

2666.67 lb 2670 lb (C)CDT = − ≅ ........................................... Ans.

2000 lb (T)BDT = ....................................................................... Ans.


0 :yF↑Σ = ( )3 5 2000 0BCT − =

3333.33 lb 3330 lb (T)BCT = = .............................................. Ans.

6-84 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-84. All members are 3 m long.


0 :xF→Σ = 0xA =

0 :FMΣ =� ( )3 3 3 0yA − = 3 kNyA =

0 :yF↑Σ = 5 3 0yF A− − − = 11 kNF =


0 :yF↑Σ = 3 sin 60 0DET− − ° =

0 :xF→Σ = cos 60 0CD DET T− − ° =

3.46410 kN 3.46 kN (C)DET = − ≅ ........................................ Ans.

1.73205 kN 1.732 kN (T)CDT = ≅ .......................................... Ans.

From a free-body diagram for joint C, the equilibrium equations give

0 :xF→Σ = cos 60 cos 60 0CD CE BCT T T+ ° − ° =

0 :yF↑Σ = sin 60 sin 60 5 0CE BCT T− ° − ° − =

1.15470 kN 1.155 kN (C)BCT = − ≅ ...................................... Ans.

4.61880 kN 4.62 kN (C)CET = − = ........................................ Ans.


0 :yF↑Σ = sin 60 3 0ABT ° − =

0 :xF→Σ = cos60 0AF ABT T+ ° =

3.46410 kN 3.46 kN (T)ABT = ≅ ........................................... Ans.

1.73205 kN 1.732 kN (C)AFT = − ≅ ...................................... Ans.

From a free-body diagram for joint B, the equilibrium equations give

0 :yF↑Σ = sin 60 sin 60 sin 60 0BC BF ABT T T° − ° − ° =


207

4.61880 kN 4.62 kN (C)BFT = − ≅ ........................................ Ans.

0 :xF→Σ = cos 60 cos 60 cos 60 0BE BC BF ABT T T T+ ° + ° − ° =

4.61880 kN 4.62 kN (T)BET = ≅ .................................... Ans.

Finally, from a free-body diagram for joint F, the equilibrium equations give

0 :yF↑Σ = 11 sin 60 sin 60 0BF EFT T+ ° + ° =

8.08291 kN 8.08 kN (C)EFT = − = .................................. Ans.

6-85 Each truss member in Fig. P6-85 is 5 ft long. Find the forces in members CD and EF using the method of sections.


0 :xF→Σ = 0xA =

0 :BMΣ =� ( ) ( ) ( )800 5 600 10 800 15+ +

( )1000 20 25 0yA+ − =

1680 lbyA =

Cut a section through CD, DE, and EF, and draw a free-body diagram of the left-hand side of the truss. The height of the truss is

5sin 60 4.33013 fth = ° = and the equilibrium equations give

0 :EMΣ =� ( ) ( ) ( )1000 5 1680 10 4.33013 0CDT− − =

2725.09 lb 2730 lb (C)CDT = − ≅ ......................Ans.

0 :DMΣ =� ( ) ( ) ( ) ( )800 2.5 1000 7.5 1680 12.5 4.33013 0EFT+ − + =

2655.81 lb 2660 lb (T)EFT = ≅ ..................................................................................Ans.

6-86 Find the forces in members CJ and KJ of the roof truss shown in Fig. P6-86 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( ) ( ) ( ) ( )4 4 3 8 5 12 4 16 3 20 24 0yA+ + + + − = 9.33333 kNyA =

Cut a section through CD, CJ, and JK, and draw a free-body diagram of the left-hand side of the truss.


208

( )1tan 10 12 39.806CDθ −= = °

( )1tan 4 59.036CJ hθ −= = °

( )( )2 3 10 6.66667 mh = =

Then, the equilibrium equations give

0 :AMΣ =� ( ) ( ) ( )( )3 4 4 8 sin 12 0CJ CJT θ− − − =

4.28 kN 4.28 kN (C)CJT = − = ....................................................................................Ans.

0 :CMΣ =� ( ) ( ) ( )3 4 9.33333 8 6.66667 0JKT− + =

9.40 kN (T)JKT = ...........................................................................................................Ans.

6-87 Find the forces in members BC, BF, and AF, of the stairs truss of Fig. P6-87 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :DMΣ =� ( ) ( ) ( )200 4 400 8 600 12 12 0yA+ + − =

933.333 lbyA =

Cut a section through AF, BF, and BC, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give

0 :BMΣ =� ( ) ( ) ( ) ( )600 4 933.333 4 3 5 4 0AFT− − =

555.556 lb 556 lb (T)AFT = ≅ .......................................Ans.

0 :FMΣ =� ( ) ( ) ( ) ( )600 4 933.333 4 4 5 3 0BCT− − =

555.556 lb 556 lb (C)BCT = − ≅ ....................................Ans.

0 :yF↑Σ = ( ) ( )933.333 600 3 5 3 5 0AF BC BFT T T− − − − =

333.333 lb 333 lb (T)BFT = ≅ .......................................Ans.

6-88 The Gambrel truss shown in Fig. P6-88 supports one side of a bridge; an identical truss supports the other side. Floor beams carry vehicle loads to the truss joints. Calculate the forces in members BC, BG, and DE when a truck having a mass of 3500 kg is stopped in the middle of the bridge as shown. The center of gravity of the truck is 1 m in front of the rear wheels.

SOLUTION

( )3500 9.81 34,335 NW = =

From a free-body diagram of the entire bridge (which is consists of two trusses), the equilibrium equations give

0 :xF→Σ = 2 0xA =

0 :EMΣ =� ( ) ( ) ( )34,335 4 2 10 0yA− =

0 :AMΣ =� ( ) ( ) ( )( )2 10 34,335 6 0E − =

6867.0 NyA = 10,300.5 kNE =


209

Half of the truck�s weight is supported by each of the two trusses. From a free-body diagram of the truck, the equilibrium equations give

0 :RMΣ =� ( ) ( ) ( )34,335 1 2 4 0FN− =

0 :FMΣ =� ( ) ( ) ( )2 4 34,335 3 0RN − =

4291.88 NFN =

12,875.6 NRN =

From a free-body diagram of the floor panel between joints H and G, the equilibrium equations give

0 :GMΣ =� ( )2 3 0FN H− =

2861.25 NH = Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss.

( )1tan 2 3 33.690θ −= = °

( )1tan 3 3 45φ −= = °


0 :GMΣ =� ( ) ( ) ( ) ( )2861.25 3 6867 5 cos 5 0BCT θ− − =

6190 N 6.19 kN (C)BCT = − = ......................................Ans.

0 :yF↑Σ = 6867 2861.25 sin sin 0BC BGT Tθ φ− + − =

809.30 N 0.809 kN (T)BGT = ≅ ....................................... Ans.

Finally, from a free-body diagram for joint E, the equilibrium equations give

0 :yF↑Σ = 10,300.5 cos 0DET θ+ =

12,380 N 12.38 kN (C)DET = − = .......................................Ans.

6-89 A truss is loaded and supported as shown in Fig. P6-89. Determine (a) The normal stress in member CD if it has a diameter of ½ in. (b) The change in length of member CF if it has a diameter of ½ in. and a modulus of elasticity of 29(106)

psi. (c) The change in length of member EF if it has the same diameter and modulus of elasticity as member CF. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give

0 :AMΣ =� ( ) ( ) ( )18 1000 6 2000 12F − −

( ) ( )2000 18 1000 24 0− − =

5000 lbF = Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give


210

0 :FMΣ =� ( ) ( )8 1000 6 0CDT − =

( )2750 lb 0.5 4CD CDT σ π= =

(a) 3820 psi (T)CDσ = ...........................................................Ans.

0 :yF↑Σ = ( )4 5 5000 3000 0CFT + − =

2500 lbCFT = −

(b) ( ) ( )

( ) ( )26

2500 1200.0527 in.

29 10 0.5 4CF

PLEA

δπ

−= = = −

×

...............................................Ans.

Finally, from a free-body diagram for joint E, the equilibrium equations give

0 :yF↑Σ = ( )1000 4 5 0EFT− − =

1250 lbEFT = −

(c) ( )( )

( ) ( )26

1250 1200.0263 in.

29 10 0.5 4EF

PLEA

δπ

−= = = −

×

...............................................Ans.

6-90 A transmission line truss supports a 5-kN load, as shown in Fig. P6-90. All members of the truss are made of structural steel (see Appendix A for properties). Determine the change in length of members FG and CD if they have cross-sectional areas of 300 mm2 each.

SOLUTION Cut a section through CD, DG, and FG, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give

0 :GMΣ =� ( )( ) ( )( ) ( )5cos30 9 5sin 30 4 3 0CDT° − ° + =

9.65705 kNCDT = −

0 :DMΣ =� ( )( ) ( )5cos30 6 3 0FGT° − =

8.66025 kNFGT =

( ) ( )

( ) ( )9 6

9657.05 40000.644 mm

200 10 300 10CDPLEA

δ−

−= = = −

× ×..............................................Ans.

( ) ( )

( )( )9 6

8660.25 40000.577 mm

200 10 300 10FGPLEA

δ−

= = = +× ×

..............................................Ans.

6-91 The truss shown in Fig. P6-91 supports a sign that weighs 3000 lb. The sign is connected to the truss at joints E, G, and H, and the connecting links are adjusted so that each joint carries one-third of the load. All members of the truss are made of structural steel, and each has a cross-sectional area of 0.564 in2. Determine the axial stresses and strains in members CD, CF, CG, and FG of the truss.


0 :AMΣ =� ( ) ( ) ( )48 1000 8 1000 24BT − −

( )1000 40 0− =


211

1500 lbBT =

Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give

0 :FMΣ =� ( ) ( ) ( )6 1500 16 1000 8 0CDT + − =

2666.667 lbCDT = −

0 :CMΣ =� ( ) ( ) ( )1500 24 1000 16 6 0FGT− − =

3333.333 lbFGT =

0 :yF↑Σ = ( )1500 1000 3 5 0CFT− + = 833.333 lbCFT = −

Next, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = 1000 0CGT − = 1000 lbCGT =

Finally, calculating the stresses and strains

2666.667 4728 psi 4.73 ksi (C)0.564

CDCD

TA

σ −= = = − ≅ ............................................Ans.

66

4728 163.0 10 163.0 in./in.29 10CD E

σε µ−−= = = − × = −×

..........................................Ans.

833.333 1478 psi 1.478 ksi (C)0.564CFσ −= = − ≅ ........................................................Ans.

66

1478 50.9 10 50.9 in./in.29 10CFε µ−−= = − × = −

×........................................................Ans.

1000 1773 psi 1.773 ksi (T)

0.564CGσ += = + ≅ ...............................................................Ans.

66

1773 61.1 10 61.1 in./in.29 10CFε µ−+= = + × = +

×........................................................Ans.

3333.333 5910 psi 5.91 ksi (T)0.564FGσ += = + ≅ ........................................................Ans.

66

5910 204 10 204 in./in.29 10CFε µ−+= = + × = +

×..........................................................Ans.

6-92 All members of the truss shown in Fig. P6-92 are made of structural steel (E = 200 GPa) and are 25-mm in diameter. Determine the normal stresses in and the change in length of members CD, DI, and HI.


0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( ) ( )10 6 18 12 12 0yA+ − =

23 kNyA =

Cut a section through CD, DI, and HI, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give


212

0 :IMΣ =� ( ) ( ) ( )18 3 23 3 3 0CDT− − =

5 kNCDT = −

0 :DMΣ =� ( ) ( ) ( )18 6 3 23 6 0HIT+ − =

10 kNHIT =

0 :yF↑Σ = 23 18 sin 45 0DIT− + ° =

7.07107 kNDIT = −

Finally, calculating the stresses and changes in length

( )

6 22

5000 10.19 10 N/m 10.19 MPa (C)0.025 4

CDσπ

−= = − × = ..............................Ans.

( )

6 22

7071.07 14.41 10 N/m 14.41 MPa (C)0.025 4

DIσπ

−= = − × = ................................Ans.

( )

6 22

10,000 20.4 10 N/m 20.4 MPa (T)0.025 4

HIσπ

+= = + × = ...................................Ans.

( ) ( )

( ) ( )29

5000 60000.306 mm

200 10 0.025 4CD

PLEA

δπ

−= = = −

×

........................................Ans.

( )( )

( ) ( )29

7071.07 3000 20.306 mm

200 10 0.025 4DIδ

π

−= = −

×

....................................................Ans.

( ) ( )

( ) ( )29

10,000 60000.611 mm

200 10 0.025 4HIδ

π= = +

×

....................................................Ans.

6-93 Determine the force in member BD of the truss shown in Fig. P6-93. SOLUTION

( )130 tan 12 16 66.870θ −= ° + = °

10 11.54701 ft

cos30b = =

° 1 12 sinsin 6.870b

bθφ − −= = °

Cut a section through members BD and CD, and draw a free-body diagram of the left-hand side of the truss. The moment equilibrium equation gives


213

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )sin cos 2000 cos cos sin 0BD BDT b b T bφ θ θ φ θ− − =

907 lb 907 lb (C)BDT = − = ..........................................................................................Ans.

6-94 For the simple truss of Fig. P6-94 show that overall equilibrium of the truss is a consequence of equilibrium of all of the pins; hence the equations of overall equilibrium give no new information. (Hint: Write equations of equilibrium for each of the pins and eliminate the unknown member forces from these equations.)


0 :xF→Σ = 0xP A− = (a)

0 :yF↑Σ = 0yC A− = (b)

0 :AMΣ =� 0Ca Pb− = (c)

xA P= yA C Pb a= =

From a free-body diagram of joint A, the equilibrium equations give

0 :xF→Σ = cos60 0AC AB xT T A+ ° − = (d)

0 :yF↑Σ = sin 60 0AB yT A° − = (e)

From a free-body diagram of joint B, the equilibrium equations give

0 :xF→Σ = cos30 cos60 0BC ABP T T+ ° − ° = (f)

0 :yF↑Σ = sin 30 sin 60 0BC ABT T− ° − ° = (g)

From a free-body diagram of joint B, the equilibrium equations give

0 :xF→Σ = cos30 0BC ACT T− ° − = (h)

0 :yF↑Σ = sin 30 0BCC T+ ° = (i)

Adding Eqs. (d), (f), and (h) gives Eq. (a)

0xP A− = ..................................................................................................................(a)

Adding Eqs. (e), (g), and (i) gives Eq. (b)

0yC A− = ..................................................................................................................(b)

Then, Eqs. (e) and (i) can be rewritten

0sin 60 sin 60

yAB

A CT = = =° °

sin 30BCCT = −

°

Equation (f) can be rewritten

cos30 cos60cos30 cos60sin 30 sin 60 sin 30 sin 60

sin 60 cos30 sin 30 cos 60 sin 90sin 30 sin 60 sin 30 sin 60

C CP C

C C

− ° ° = − ° + ° = + ° ° ° °

° ° + ° ° °= =° ° ° °

But sin 90 1° = , sin 60 b d° = , sin 30 d a° = , sin 30 sin 60 b a° ° = , and therefore, Eq. (f) can be rewritten


214

CPb a

= ................................................................................................................... (c)

which is equivalent to Eq. (c).

6-95 Show that the overall equilibrium of a truss is a consequence of the equilibrium of the two separate parts generated by the method of sections. That is, section the bridge truss of Fig. P6-95 as indicated, and write the equilibrium equations for each piece. Eliminate the member forces from the resulting six equations and show that the result is equivalent to the equilibrium of the whole truss.


0 :xF→Σ = 0xA = (a)

0 :yF↑Σ = 0yA P D− + = (b)

0 :AMΣ =� ( ) ( )3 2 0D w P w− = (c)

Cut a section through BC, CF, and EF, and draw a free-body diagram of the left-hand side of the truss. The equilibrium equations give

0 :xF→Σ = cos 0x BC CF EFA T T Tθ+ + + = (d)

0 :yF↑Σ = sin 0CF yT Aθ + = (e)

0 :CMΣ =� ( ) ( ) ( )2 0EF x yT h A h A w+ − = (f)

0 :FMΣ =� ( ) ( ) 0BC yT h A w− − = (g)

Cut a section through BC, CF, and EF, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give

0 :xF→Σ = cos 0BC CF EFT T Tθ− − − = (h)

0 :yF↑Σ = sin 0CFT P Dθ− − + = (i)

0 :CMΣ =� ( ) ( ) 0EFD w T h− = (j)

0 :FMΣ =� ( ) ( ) ( )2 0BCD w T h P w+ − = (k)

Adding Eqs. (d) and (h) gives Eq. (a)

0xA = ................................................................................................................................ (a)

Adding Eqs. (e) and (i) gives Eq. (b)

0yA P D− + = ................................................................................................................. (b)

Adding two times Eq. (g) to two times Eq. (k) gives

4 2 2 0yDw Pw A w− − =

Adding Eq. (f) to Eq. (j) gives

2 0x yA h A w Dw− + =

Subtracting this last equation from the previous (and using 0xA = ) gives Eq. (c)

3 2 0Dw Pw− = ................................................................................................................ (c)


215

6-96 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-96). Calculate the forces in all the members of a typical truss when water collects to a depth of 0.2 m as shown. The density of water is 1000 kg/m3.

SOLUTION From a free-body diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end of the roof panel as

( ) ( ) ( )1 2 2 1000 9.81 2 1.2 0.2 2

2354.4 NF F W= = = × ×

=

Then, from a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :yF↑Σ = 16 0y yA F F+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )( )1 13.6 2 1.2 2 2.4yF F F− −

( ) ( )1 3.6 0F− =

13 7063.20 Ny yA F F= = =


0 :xF→Σ = 0BCT =

0 :yF↑Σ = 1 0ABT F− − =

2354.4 N 2.35 kN (C)ABT = − ≅ ............................................. Ans.

0 kNBCT = .................................................................................... Ans.


0 :xF→Σ = ( )3 5 0AH ACT T+ =

0 :yF↑Σ = ( )7063.2 4 5 0AB ACT T+ + =

5886.0 N 5.89 kN (C)ACT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)AHT = ≅ ................................................ Ans.

From a free-body diagram for joint H, the equilibrium equations give

0 :xF→Σ = 0GH AHT T− =

0 :yF↑Σ = 0CHT =

0 kNCHT = ................................................................................... Ans.

3531.6 N 3.53 kN (T)GHT = ≅ ................................................ Ans.


0 :xF→Σ = 0DET =

0 :yF↑Σ = 1 0EFT F− − =

0 kNDET = .................................................................................... Ans.

2354.4 N 2.35 kN (C)EFT = − ≅ ............................................. Ans.

From a free-body diagram for joint F, the equilibrium equations give


216

0 :xF→Σ = ( )3 5 0FG DFT T− − =

0 :yF↑Σ = ( )7063.2 4 5 0EF DFT T+ + =

5886.0 N 5.89 kN (C)DFT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)FGT = ≅ ................................................ Ans.


0 :xF→Σ = ( )3 5 0DE CD DFT T T− + =

0 :yF↑Σ = ( )12 4 5 0DG DFF T T− − − =

3531.6 N 3.53 kN (C)CDT = − ≅ ...................................... Ans.

0 kNDGT = ............................................................................ Ans.

Finally, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = ( )4 5 0DG CGT T+ =

0 kNCGT = ............................................................................. Ans.

6-97 Snow on a roof supported by the Howe truss of Fig. P6-97 can be approximated as a distributed load of 20 lb/ft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of the member. Determine the forces in members BC, BG, CG, and GH.


( )( )1 2 2 20 80 2 89.4427 lbF F W= = = =


0 :xF→Σ = 0xA =

0 :yF↑Σ = 18 0yA E F+ − =

0 :AMΣ =� ( ) ( )( ) ( )( )1 132 2 8 2 16E F F− −

( )( ) ( ) ( )1 12 24 32 0F F− − =

13 357.771 lbyA E F= = =

Next, cut a section through members BC, BG, and GH, and draw a free-body diagram of the left-hand portion of the truss.

1tan 8 16 26.565φ −= = °

Then equilibrium equations give

0 :AMΣ =� ( )( ) ( )( )12 8 sin 16 0BGF T φ− − =

200 lb 200 lb (C)BGT = − = ..........................................................................................Ans.

0 :BMΣ =� ( ) ( ) ( )( )14 8 357.771 8 0GHT F+ − =

536.656 lb 537 lb (T)GHT = ≅ ....................................................................................Ans.


217

0 :GMΣ =� ( )( ) ( ) ( ) ( ) ( ) ( )1 12 8 16 357.771 16 sin 16 0BCF F T φ+ − − =

400 lb 400 lb (C)BCT = − = ..........................................................................................Ans.


0 :xF→Σ = cos cos 0CD BCT Tφ φ− =

0 :yF↑Σ = 12 sin sin 0CG CD BCF T T Tφ φ− − − − =

400 lbCDT = −

178.9 lb (T)CGT = ................................................................ Ans.

6-98 Determine the forces in members CD, DE, and DF of the roof truss shown in Fig. P6-98. Triangle CDF is an equilateral triangle and joints E and G are at the midpoints of their respective sides.

SOLUTION

2 23 4 5 mAFL = + = 2.5 m2AF

AELL = = 3 2sin 60 1.26795 mb = − ° =

sin 3 5φ = cos 4 5φ =


0 :xF→Σ = ( )5 8 10 sin 0xAφ+ + − =

0 :yF↑Σ = ( )5 8 10 cos 0yA B φ+ − + + =

0 :AMΣ =� ( ) ( ) ( )8 2.5 8 5 10 0B − − =

8.75 kNB =

13.800 kNxA = 9.650 kNyA =

Next, cut a section through members CD, DF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then the equilibrium equations give

0 :xF→Σ = 10sin cos cos 60 0EF DF CDT T Tφ φ− − ° − =

0 :yF↑Σ = 8.75 10cos sin sin 60 0EF DFT Tφ φ− − − ° =

0 :FMΣ =� ( ) ( )8.75 4 2sin 60 0CDT− ° =

32.2765 kN 32.3 kN (C)EFT = − ≅

20.2073 kN 20.2 kN (T)CDT = ≅ ...............................................................................Ans.

23.2279 kN 23.2 kN (T)DFT = ≅ ..............................................................................Ans.

Finally, from a free-body diagram for joint E, the component of the equilibrium equations normal to members AE and EF gives

1 1.5tan 13.0643 2

bθ − −= = °−

0 :nFΣ = ( )8 sin 0DET φ θ+ + =

10.4533 kN 10.45 kN (C)DET = − ≅ ..........................................................................Ans.


218

6-99 Find the forces in members DE, DJ, and JK of the truss of Fig. P6-99. SOLUTION Then, from a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 0xA =

0 :GMΣ =� ( ) ( )8000 20 8000 40+

( ) ( )6000 60 80 0yA+ − =

10,500 lbyA =

Next, cut a section through members CD, JK, and JL, and draw a free-body diagram of the left-hand portion of the truss. Then, the equilibrium equations give

0 :JMΣ =� ( ) ( )( ) ( )6000 20 10,500 40 30 0CDT− − =

10,000 lb 10,000 lb (C)CDT = − =

0 :CMΣ =� ( ) ( )( )30 10,500 20 0JKT − =

7000 lb 7000 lb (T)JKT = = ........................... Ans.

Finally, from a free-body diagram for joint D, the equilibrium equations give

0 :xF→Σ = 0DE CDT T− =

0 :yF↑Σ = 8000 0DJT− − =

10,000 lb 10,000 lb (C)DET = − = ..............................................................................Ans.

8000 lb 8000 lb (C)DJT = − = ......................................................................................Ans.

6-100 The Gambrel truss shown in Fig. P6-100 supports one side of a bridge; an identical truss supports the other side. A 3400-kg truck is stopped on the bridge at the location shown and floor beams carry the vehicle load to the truss joints. If the center of gravity of the truck is located 1.5 m in front of the rear wheels, plot the force in members BC, BG, and GH as a function of the truck�s location d (0 ≤ d ≤ 20 m)

SOLUTION

( )3400 9.81 33,354 NW = =

From a free-body diagram of the entire truss (one side of the bridge), the equilibrium equations give

0 :xF→Σ = 0xA =

0 :EMΣ =� 4 8 12 16 16 0F G H A yP P P P A+ + + − =

2 3 4

4F G H A

yP P P PA + + +=

To determine the joint forces PA, PE, PF, PG, and PH, we need to first consider a free-body diagram of the truck. The equilibrium equations give the forces exerted on the front and rear wheels by the bridge deck panels

0 :RMΣ =� ( ) ( )33,354 1.5 4 0FN− =

0 :FMΣ =� ( ) ( )4 33,354 2.5 0RN − =

12,507.75 NFN = 20,846.25 NRN =


219

Next, consider a bridge deck panel. Each of the two trusses supports the left end of the panel with a force F1 and the right end of the panel with a force F2. At the instant shown, one of the truck�s wheels exerts a force N on the panel at a distance b from the right end of the panel. Then the forces transmitted to the trusses are

0 :yF↑Σ = 1 22 2 0F F N+ − =

2 0 :MΣ =� ( ) ( )( )12 4 0N b F− =

1 8F Nb= ( )

2

42 8 8

N bN NbF−

= − =

As the truck moves across the bridge, there are five different cases to consider:

Case 1: 0 4 md≤ ≤ ; FN N= , b d=

0A H GP P P= = =

1 8 8F F

FN b N dP F= = =

( )

2

48

FE

N dP F

−= =

Case 2: 4 m 8 md≤ ≤ ; ( )4 mb d= −

0A HP P= =

( )

1

48 8

FFG

N dN bP F−

= = =

( ) ( )

1 2

4 88 8

R FF

N d N dP F F

− −= + = +

( )2

88

RE

N dP F

−= =

Case 3: 8 m 12 md≤ ≤ ; ( )8 mb d= −

0A EP P= =

( )

1

88 8

FFH

N dN bP F−

= = =

( ) ( )

1 2

8 128 8

R FG

N d N dP F F

− −= + = +

( )

2

128

RF

N dP F

−= =

Case 4: 12 m 16 md≤ ≤ ; ( )12 mb d= −

0E FP P= =

( )

1

128 8

FFA

N dN bP F−

= = =

( ) ( )

1 2

12 168 8

R FH

N d N dP F F

− −= + = +


220

( )

2

168

RG

N dP F

−= =

Case 5: 16 m 20 md≤ ≤ ; ( )16 mb d= −

0E F GP P P= = =

( )

1

168 8

RRA

N dN bP F−

= = =

( )

2

208

RH

N dP F

−= =

Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss.

( )1tan 3 4 36.870θ −= = ° ( )1tan 1 4 14.036φ −= = °


0 :GMΣ =� ( ) ( )4 8 4 cos 0H A y BCP P A T φ+ − − =

0 :BMΣ =� ( )3 4 0GH y AT A P− − =

0 :yF↑Σ = sin sin 0y A BC BGA P T Tφ θ− + − =

2 2cos

H A yBC

P P AT

φ+ −

= ..................... Ans.

( )4

3y A

GH

A PT

−= ............................. Ans.

sin

siny A BC

BG

A P TT

φθ

− += ................ Ans.

6-101 An overhead crane consists of an I-beam supported by a simple truss as shown in Fig. P6-101. If the uniform I-beam weighs 400 lb and is supporting a load of 1000 lb at a distance d from its left end, plot the force in members AB, BC, EF, and FG as a function of the position d (0 ≤ d ≤ 8 ft).


0 :xF→Σ = 0xG A− =

0 :yF↑Σ = 0y F DA P P− − =

0 :AMΣ =� 3 3 9 0F DG P P− − =

3F DG P P= +

3x F DA G P P= = + 1400 lby F DA P P= + =

in which the joint forces PF and PD are determined from equilibrium of the I-beam

0 :DMΣ =� ( ) ( ) ( )400 3 1000 7 6 0Fd P+ − − =

0 :FMΣ =� ( ) ( )6 400 3 1000 1 0DP d− − − =


221

( )500 7

200 lb3F

dP

−= +

( )500 1

200 lb3D

dP

−= +

Then, from a free-body diagram for joint A, the equilibrium equations give

0 :xF→Σ = cos 0AB xT Aφ − =

0 :yF↑Σ = 1400 sin 0AG ABT T φ− − =

1400 sinAG ABT T φ= −

cos

xAB

ATφ

= ....................................................................................................................... Ans.

in which ( )1tan 3 9 18.435φ −= = ° .

Then, from a free-body diagram for joint G, the equilibrium equations give

0 :xF→Σ = cos 0FG BGT T Gθ+ + =

0 :yF↑Σ = sin 0AG BGT T θ+ =

sin

AGBG

TTθ

−=

cosFG BGT G T θ= − − ......................................................................................................Ans.

in which ( )1tan 2 3 33.690θ −= = ° .

Finally, cut a section through members BC, CF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then, the equilibrium equations give

0 :FMΣ =� ( )6 sin 6 0BC DT Pφ − =

0 :CMΣ =� 1 3 0EF DT P− − =

sin

DBC

PTφ

= ......................................... Ans.

3EF DT P= − ......................................... Ans.

6-102 The masses of carton 1, 2, and 3, which rest on the platform shown in Fig. P6-102, are 300 kg, 100 kg, and 200 kg, respectively. The mass of the platform is 500 kg. Determine the tensions in the three cables A, B, and C that support the platform.

SOLUTION From a free-body diagram of the platform, the equilibrium equations give

:OΣ =M 0 ( ) ( ) ( ) ( )300 2 100g g+ × − + + × −i j k i j k


222

( ) ( ) ( )3 1.5 2 500BT g+ + × + + × −i j k i j k

( ) ( ) ( )3 200 4 Cg T+ + × − + + × =i j k i j k 0

:x 300 100 1000 600 4 0B Cg g T g g T− − + − − + =

:y 300 200 3 750 200 0B Cg g T g g T+ + − + + − =

0 :zFΣ = 1100 0A B CT T T g+ + − =

340.91 3340 NAT g= = ............................. Ans.

345.45 3390 NBT g= = ................................................................................................Ans.

413.64 4060 NCT g= = ................................................................................................Ans.

6-103 Determine the reaction at support A of the pipe system shown in Fig. P6-103 when the force applied to the pipe wrench is 50 lb.

SOLUTION From a free-body diagram of the platform, the equilibrium equations give

:AΣ =M 0 ( ) ( )7 23 10 50A + − + + × − =M i j k k 0

:x 1150 0AxM − =

:y 350 0AyM − =

:z 0AzM =

0 :xFΣ = 0xA =

0 :yFΣ = 0yA =

0 :zFΣ = 50 0zA − =

( )50 lb=A k ............................. ( )1150 350 lb ftA = + ⋅M i j ........................................Ans.

6-104 The rectangular plate of uniform thickness shown in Fig. P6-104 has a mass of 500 kg. Determine (a) The tensions in the three cables supporting the plate. (b) The deformation of the cable at A, which has a diameter of 10-mm, a length of 1.5 m and a modulus of

elasticity of 200 GPa. SOLUTION

( )500 500 9.81 4905 NW g= = =

(a) From a free-body diagram of the plate, the equilibrium equations give

0 :zFΣ = 4905 0A B CT T T+ + − =

:OΣ =M 0 ( ) ( ) ( ) ( )1.5 0.9 0.5 2.5B CT T− + × + − + ×i j k i j k

( ) ( )0.625 1.25 4905+ − + × − =i j k 0

:x 0.9 2.5 6131.250 0B CT T+ − =

:y 1.5 0.5 3065.625 0B CT T+ − =

1560.682 N 1561 NAT = ≅ ...........................................................................................Ans.

1393.466 N 1393 NBT = ≅ ..........................................................................................Ans.


223

1950.852 N 1951 NCT = ≅ ..........................................................................................Ans.

(b) The stretch of wire A is given by

( ) ( )

( ) ( )29

1560.862 15000.1490 mm

200 10 0.010 4A

PLEA

δπ

= = = ×

..........................................Ans.

6-105 A shaft is loaded through a pulley and a lever (Fig. P6-105) that are fixed to the shaft. Friction between the belt and pulley prevents slipping of the belt. Determine the force P required for equilibrium and the reactions at supports A and B. The support at A is a ball bearing and the support at B is a thrust bearing. The bearings exert only force reactions on the shaft.

SOLUTION From a free-body diagram of the shaft, the moment equilibrium equation gives

:BΣ =M 0 ( ) ( )14 14 200 P+ × −i j j k

( ) ( )18 x zA A+ − × +j i k

( ) ( )30 6 500+ − + × −j k i

( ) ( )30 6 150+ − − × − =j k i 0

:x 14 18 0zP A− − = 116.667 lbzA = −

:y 14 2100 0P − = 150 lbP =

:z 18 16,700 0xA − = 927.778 lbxA =

Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = 650 0x xA B+ − = 277.778 lbxB = −

0 :yFΣ = 200 0yB + = 200 lbyB = −

0 :zFΣ = 0z zA B P+ − = 266.667 lbzB =

( )928 116.7 lb= −A i k ................................................................................................Ans.

( )278 200 267 lb= − − +B i j k ...................................................................................Ans.

150 lbP = ......................................................................................................................... Ans.

6-106 The homogeneous door shown in Fig. P6-106 has a mass of 60 kg, and is held in the position shown by the rod AB. The rod is held in place by smooth horizontal pins at A and B. The hinges at C and D are smooth, and the hinge at C can support thrust along its axis. Determine all forces that act on the door.

SOLUTION

( )60 60 9.81 588.60 NW g= = =

Note that the rod AB is a two-force member.

( )( )

cos 60 sin 60

0.5000 0.86603 NAB AB AB

AB AB

F F

F F

= − ° + °

= − +

F j k

j k

Then, from a free-body diagram of the door, the moment equilibrium equation gives

:CΣ =M 0 ( ) ( )950 y zD D− × +i j k


224

( ) ( )475 475cos 60 475sin 60 588.60+ − + ° + ° × −i j k k

( ) ( )950cos 60 950sin 60 0.5000 0.86603AB ABF F+ ° + ° × − + =j k j k 0

:x 139,792.5 822.724 0ABF− + = 169.914 NABF =

:y 950 279,585 0zD − = 294.300 NzD =

:z 950 0yD− = 0 NyD =


0 :xFΣ = 0xC = 0 NxC =

0 :yFΣ = ( )0.5000 169.914 0y yC D+ − = 84.957 NyC =

0 :zFΣ = ( )0.86603 169.914 588.60 0z zC D+ + − = 147.15 NzC =

( )85.0 147.2 N= +C j k ...............................................................................................Ans.

( )294 N=D k .................................................................................................................Ans.

( )85.0 147.2 NAB = − +F j k .........................................................................................Ans.

6-107 A beam is supported by a ball-and-socket joint and two cables as shown in Fig. P6-107. Determine (a) The tensions in the two cables. (b) The reaction at support A (the ball and socket joint). (c) The axial stress and deformation in each of the cables if they are made of steel (E = 30,000 ksi) and have

¼- in. diameters. SOLUTION

2 211 6 12.530 ftBDL = + = 2 2 27 11 3 13.379 ftCDL = + + =

( )2 2

11 6 0.87790 0.4788511 6

BD BD BD BDT T T− += = − ++

j kT j k

( )2 2 2

7 11 3 0.52320 0.82218 0.224237 11 3

CD CD CD CD CDT T T T− − += = − − ++ +

i j kT i j k

(a) From a free-body diagram of the beam, the moment equilibrium equation gives

:AΣ =M 0 ( ) ( ) ( ) ( )11 300 6 800BD CD× + + + × − =j T T i j k 0

:x 5.26735 2.46653 4800 0BD CDT T+ − =

:y 0 0=

:z 5.75520 3300 0CDT − =

642.77 lb 643 lbBDT = ≅ ..............................................................................................Ans.

573.39 lb 573 lbCDT = ≅ ..............................................................................................Ans.

(b) Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = ( )0.52320 573.39 300 0xA − + = 0 lbxA =

0 :yFΣ = ( ) ( )0.87790 642.77 0.82218 573.39 0yA − − = 1035.72 lbyA =


225

0 :zFΣ = ( ) ( )0.47885 642.77 0.22423 573.39 800 0zA + + − = 363.64 lbzA =

( )1036 364 lb= +A j k .................................................................................................Ans.

(c) Finally, the stresses and deformation of the two cables is

( )

32

642.77 13.09 10 psi 13.09 ksi0.25 4

BDTA

σπ

= = = × = ...........................................Ans.

( )

32

573.39 11.68 10 psi 11.68 ksi0.25 4

CDσπ

= = × = ....................................................Ans.

( )( )

( ) ( )26

642.77 12.530 120.0656 in.

30 10 0.25 4BD

TLEA

δπ

×= = =

×

................................................Ans.

( )( )

( ) ( )26

573.39 13.379 120.0625 in.

30 10 0.25 4CDδ

π

×= =

×

...........................................................Ans.

6-108 The crankshaft-flywheel arrangement of a one-cylinder engine is shown in Fig. P6-108. A 250-N-m couple C is delivered to the crankshaft by the flywheel, which is rotating at a constant angular velocity. The support at A is a ball bearing and the support at B is a thrust bearing, both of which support only force reactions on the shaft. Determine the magnitudes of the force P and of the resultant bearing reactions at A and B.

SOLUTION sin 20 cos 20 0.34202 0.93969P P P P= − ° − ° = − −P j k j k

From a free-body diagram of the crankshaft, the moment equilibrium equation gives

:BΣ =M 0 ( ) ( )250 0.45 y zA A− + × +i i j k

( )0.225 0.125+ − × =i j P 0

:x 250 0.11746 0P− + =

:y 0.45 0.21149 0zA P− + =

:z 0.45 0.07695 0yA P− =

2128.38 NP =

363.954 NyA =

1000.293 NzA =


0 :xFΣ = 0xB = 0 NxB =

0 :yFΣ = ( )0.34202 2128.38 0y yA B+ − = 363.996 NyB =

0 :zFΣ = ( )0.93969 2128.38 0z zA B+ − = 999.728 NzB =

2 2364 1000 1064 NA = + = ......................................................................................Ans.

2 2364 1000 1064 NB = + = ......................................................................................Ans.

2130 NP = ...................................................................................................................... Ans.


226

6-109 A farmer is using the hand winch shown in Fig. P6-109 to slowly raise a 40-lb bucket of water from a well. In the position shown, the force P is vertical. The bearings at C and D exert only force reactions on the shaft. Bearing C can support thrust loading; bearing D cannot. Determine the magnitude of force P and the components of the bearing reactions.

SOLUTION From a free-body diagram of the winch drum and axle, the moment equilibrium equation gives

:CΣ =M 0 ( ) ( )20cos30 20sin 30 11 P− ° + ° + × −i j k j

( ) ( )2.5 20 40+ − × −i k j

( ) ( )50 x yD D+ − × + =k i j 0

:x 11 800 50 0yP D− + =

:y 50 0xD− =

:z 20 cos30 100 0P ° − =

5.7735 lb 5.77 lbP = ≅ ................................................................................................Ans.

0 lbxD = ............................................ 14.7298 lbyD = ..............................................Ans.

(b) Then, the force equilibrium equation Σ =F 0 gives

0 :xFΣ = 0x xC D+ = 0 lbxC = .............................................Ans.

0 :yFΣ = 40 0y yC D P+ − − = 31.0437 lbyC = ................................Ans.

0 :zFΣ = 0zC = 0 lbzC = .............................................Ans.

6-110 The block W shown in Fig. P6-110 has a mass of 250 kg. Bar AB rests against a smooth vertical wall at end B and is supported at end A with a ball-and-socket joint. The two cables are attached to a point on the bar midway between the ends. Determine

(a) The reactions at supports A and B and the tension in cable CD. (b) The axial stress and deformation in cable CD if it is made of steel (E = 210 GPa) and has an 8-mm

diameter. SOLUTION

( )250 250 9.81 2452.50 NW g= = = 2 2 2200 200 500 574.456 mmCDL = + + =

( )2 2 2

200 200 500200 200 500

0.34816 0.34816 0.87039

CD CD

CD CD CD

T

T T T

− + +=+ +

= − + +

i j kT

i j k

(a) From a free-body diagram of the beam, the moment equilibrium equation gives

:AΣ =M 0 ( ) ( )200 500 300 2452.50CD− + × −i j k T k

( ) ( )400 1000 600 yB+ − + × =i j k j 0

:x 539.643 1, 226, 250 600 0CD yT B− + − =

:y 278.526 490,500 0CDT− + = 1761.056 NCDT =

:z 104.448 400 0CD yT B− + = 459.847 NyB =

1761.056 N 1761 NCDT = ≅ .........................................................................................Ans.


227

( )460 N=B j .................................................................................................................. Ans.


0 :xFΣ = ( )0.34816 1761.056 0xA − = 613 NxA =

0 :yFΣ = ( )0.34816 1761.056 0y yA B+ + = 1073 NyA = −

0 :zFΣ = ( )0.87039 1761.056 2452.50 0zA + − = 920 NzA =

( )613 1073 920 N= − +A i j k ....................................................................................Ans.

(c) Finally, the stresses and deformation of the cable is

( )

6 22

1761.056 35.0 10 N/m 35.0 MPa0.008 4

CDσπ

= = × = .............................................Ans.

( )( )

( ) ( )29

1761.056 594.4560.0958 mm

210 10 0.008 4CDδ

π= =

×

...................................................Ans.

6-111 A 20-lb piece of electronic equipment is placed on a wooden skid that weighs 10 lb and that rests on a concrete floor (Fig. P6-111). The coefficient of static friction between the skid and the floor is 0.45.

(a) Determine the minimum pushing force along the handle necessary to cause the skid to start sliding across the floor (Fig. P6-111a).

(b) Determine the minimum force necessary to start the skid in motion if a pulling force instead of a pushing force is applied to the handle (Fig. P6-111b).

SOLUTION (a) Impending motion is to the right and friction acts to the left

to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give

0 :xF→Σ = cos30 0.45 0P N° − =

0 :yF↑Σ = 30 sin 30 0N P− − ° =

40.5 lbN =

21.1 lbP = ....................................................................................................................... Ans. (b) Now, impending motion is to the left and friction acts to the

right to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give

0 :xF→Σ = 0.45 cos30 0N P− ° =

0 :yF↑Σ = 30 sin 30 0N P− + ° =

23.8 lbN =

12.37 lbP = ..................................................................................................................... Ans.

6-112 A skier is at the point of experiencing impending motion on a ski slope as illustrated in Fig. P6-112. Determine the coefficient of static friction between a ski and the slope if the angle θ (the angle of repose) is 5°.

SOLUTION Impending motion is down the hill and friction acts up the hill to oppose the impending motion. From a free-body diagram of the skier, the equilibrium equations give


228

0 :nFΣ = cos5 0N mg− ° = cos5N mg= °

0 :tFΣ = sin 5 0sN mgµ − ° = sin 5sN mgµ = °

sin 5 tan 5 0.0875cos5s

mgmg

µ °= = ° =°

...............................................................................Ans.

6-113 The block in Fig. P6-113 weighs 500 lb and the coefficient of friction between the block and the inclined plane is 0.2. Determine

(a) Whether the system would be in equilibrium for P = 400 lb. (b) The minimum force P to prevent motion. (c) The maximum force P for which the system is in equilibrium. SOLUTION If the impending motion is down the hill, then the friction force would be up the hill. From a free-body diagram of the block, the equilibrium equations are

0 :nFΣ = 500cos30 sin 20 0N P− ° + ° =

0 :tFΣ = 500sin 30 cos 20 0F P− ° + ° =

(a) If 400 lbP = , then the equilibrium equations give

296.205 lbN =

125.877 lbF = −

max 0.2 59.24 lbF N= =

The negative sign on the friction indicates that the direction of impending motion is actually up the hill and the friction is acting down the hill to oppose the motion. However, since the amount of friction required for equilibrium is greater than the maximum available friction,

Not in equilibrium. ........................................................................................................Ans.

(b) If 0.2F N= , then the equilibrium equations give

368.872 lbN =

0.2 73.7744 lbF N= =

min187.5 lbP P= = ..........................................................................................................Ans.

(c) If the impending motion is up the hill, then the friction force would be down the hill. From a free-body diagram of the block, the equilibrium equations are

0 :nFΣ = 500cos30 sin 20 0N P− ° + ° =

0 :tFΣ = 500sin 30 cos 20 0F P− − ° + ° =

Now, if 0.2F N= , the equilibrium equations give

318.812 lbN =

63.7624 lbF =

max334 lbP P= = .............................................................................................................Ans.

6-114 A 75-kg man starts climbing a 5-m long ladder leaning against a wall (Fig. P6-114). The coefficient of friction is 0.25 at both surfaces. Neglect the weight of the ladder and determine how far up the ladder the man can climb before the ladder starts to slip.

SOLUTION Impending motion of the ladder is down and to the left, and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the man and the ladder, the equilibrium equations give


229

0 :xF→Σ = 1 20.25 0N N− =

0 :yF↑Σ = ( )1 275 9.81 0.25 0N N− + =

1 0 :MΣ =� ( ) ( ) ( )2 25sin 60 0.25 5cos 60N N° + °

( ) ( )75 9.81 cos60 0d− ° =

1 692.471 NN =

2 173.118 NN =

2.33 md = .....................................................................Ans.

6-115 A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-115. Determine the minimum coefficient of static friction between the contacting surfaces required to make the device work.

SOLUTION Impending motion is for the block to slide downward. Friction must act upward on the block to oppose this impending motion. The friction force on the member AB must be equal and opposite to that on the block � it must act downward on the member. From a free-body diagram for member AB, the moment equilibrium equation gives

0 :BMΣ =� 4 12 0sN Nµ− =

4 12 0.333sµ = = ...........................................Ans.

6-116 A boy is pulling a sled with a box (at constant velocity) up an inclined surface as shown in Fig. P6-116. The mass of the box and sled is 50 kg; the mass of the boy is 40 kg. If the coefficient of kinetic friction between the sled runners and the icy surface is 0.05, determine the minimum coefficient of static friction needed between the boy�s shoes and the icy surface.

SOLUTION Impending motion is up the hill and friction acts down the hill to oppose the motion. From a free-body diagram of the sled, the equilibrium equations give

0 :nFΣ = ( )1 50 9.81 cos15 sin 25 0N P− ° + ° =

0 :tFΣ = ( ) 1cos 25 50 9.81 sin15 0.05 0P N− ° − =

1 405.142 NN =

162.426 NP = Impending motion of the boy�s foot is down the hill and friction acts up the hill to oppose the motion. From a free-body diagram of the boy, the equilibrium equations give

0 :nFΣ = ( )2 40 9.81 cos15 sin 25 0N P− ° − ° =

0 :tFΣ = ( )2 40 9.81 sin15 cos 25 0sN Pµ − ° − =

2 447.673 NN =

2 2248.769 NsN Fµ = =

0.556sµ = ......................................................................Ans.


230

6-117 A 120-lb girl is walking up a 48-lb uniform beam as shown in Fig P6-117. Determine how far up the beam the girl can walk before the beam starts to slip if

(a) The coefficient of friction is 0.20 at all surfaces. (b) The coefficient of friction at the bottom end of the beam is increased to 0.40 by placing a piece of rubber

between the beam and the floor. SOLUTION Impending motion of the beam is down and to the right and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the girl and the beam, the equilibrium equations are

0 :xF→Σ = ( ) ( )1 1 23 5 4 5 0N F F− − =

0 :yF↑Σ = ( ) ( )1 1 24 5 3 5 168 0N F N+ − + =

1 0 :MΣ =� ( )( ) ( ) 148 4 5 6 120 4 5 10 0x N+ − =

(a) If 1 10.2F N= and 2 20.2F N= , then 0 :xF→Σ = gives

2 111 5N N=

and the other two equations give

1 53.846 lbN = 2 118.462 lbN =

3.21 ftx = ......................................................................................................................... Ans.

(b) If 1 10.2F N= and 2 20.4F N= , then 0 :xF→Σ = gives

2 111 10N N=


1 83.168 lbN = 2 91.485 lbN =

6.26 ftx = ........................................................................................................................ Ans.

6-118 The broom shown in Fig. P6-118 weighs 8 N and is held up by the two cylinders, which are wedged between the broom handle and the side rails. The coefficient of friction between the broom and cylinders and between the cylinders and side rails is 0.30. The side rails are at an angle of θ = 30° to the vertical. The weight of the cylinders may be neglected. Determine whether or not this system is in equilibrium. If the system is in equilibrium, determine the force exerted on the broom handle by the rollers.

SOLUTION Impending motion of the broom handle is downward and friction acts upward to oppose the motion. From a free-body diagram of the broom handle, the equilibrium equations give

0 :xF→Σ = 2 1 0N N− =

0 :yF↑Σ = 1 2 8 0F F+ − =

0 :centerMΣ =� 1 2 0Fb F b− =

1 2 4 NF F= =

1 2N N=

From a free-body diagram of the left-hand roller, the equilibrium equations give

0 :xF→Σ = 3 3 2cos60 cos30 0F N N° + ° − =

0 :yF↑Σ = 3 3 2sin 60 sin 30 0F N F− ° + ° − =


231

0 :centerMΣ =� 3 2 0F r F r− =

3 4 NF =

1 2 3 14.92820 NN N N= = =

max 0.3 4.478 NF N= =

Since the amount of friction required for equilibrium is less than the maximum amount of friction available at all surfaces, the broom handle

Is in equilibrium .............................................................................................................Ans.

14.93 NN = ..................................................................................................................... Ans.

4 NF = ............................................................................................................................. Ans.

6-119 The device shown in Fig. P6-119 is used to raise boxes and crates between floors of a factory. The frame, which slides on the 4-in.-diameter vertical post, weighs 50 lb. If the coefficient of friction between the post and the frame is 0.10, determine the force P required to raise a 150-lb box.

SOLUTION Impending motion of the platform is upward and friction with the post acts to oppose the impending motion. Guessing that the platform rotates slightly clockwise, contact with the post (and the location of the normal and friction forces) is on the top-left and bottom-right sides of the post. Then, from a free-body diagram of the platform and box, the equilibrium equations give

0 :xF→Σ = 0A BN N− =

0 :yF↑Σ = ( )0.1 200 0A BP N N− + − =

0 :AMΣ =� ( ) ( ) ( )26 4 0.1 8 4 50 20 150 0B BN N P+ + − − =

57.143 lbA BN N= =

211 lbP = ........................................................................................................................ Ans. (Note that the two normal forces are both positive indicating that the guess about how the platform rubs against the post was correct.)

6-120 The automobile shown in Fig. P6-120 has a mass of 1500 kg. The coefficient of friction between the rubber tires and the pavement is 0.70. Determine the maximum incline θ that the automobile can drive up if the automobile has

(a) A rear-wheel drive. (b) A front-wheel drive. SOLUTION

( )1500 9.81 14,715 NW = =

Impending slip between the tires and the ground is for the bottom of the tires to move backwards and the friction force acts to oppose the slip. From a free-body diagram of the car, the equilibrium equations are

0 :nFΣ = 1 2 14,715cos 0N N θ+ − =

0 :tFΣ = 1 2 14,715sin 0F F θ+ − =

0 :GMΣ =� ( )2 1 1 21.2 1.7 0.85 0N N F F− + + =

(a) If the vehicle has rear wheel drive, then 1 10.7F N= and 2 0F = and the moment equilibrium equation gives

2 10.92083N N=


232


1

1

14,715sin 0.7tan 0.3644314,715cos 1.92083

NN

θθθ

= = =

20.02θ = ° ........................................................................................................................ Ans.

(b) If the vehicle has front wheel drive, then 1 0F = and 2 20.7F N= and the moment equilibrium equation gives

1 21.05588N N=


2

2

14,715sin 0.7tan 0.3404914,715cos 2.05588

NN

θθθ

= = =

18.80θ = ° ......................................................................................................................... Ans.

6-121 When a drawer is pulled by only one of the handles, it tends to twist and rub as shown (highly exaggerated) in Fig. P6-121, which is a top view of the drawer. The weight of the drawer and its contents is 2 lb and is uniformly distributed. The coefficient of friction between the sides of the drawer and the sides of the dresser is 0.6; between the bottom of the drawer and the side rails the drawer rides on, µs = 0.1. Determine the minimum amount of force P necessary to pull the drawer out.

SOLUTION

( )3 4 0.1 2 0.1 lbF F W= = =

Impending slip of the drawer is to the front and friction acts to oppose the impending motion. From a free-body diagram of the drawer, the equilibrium equations give

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 3 4 0F F F F P+ + + − =

2 0 :MΣ =� 3 427 30.375 0.375P F F− −

1 130.75 18 0F N− − =

1 2 0.57407 lbN N= =

1 2 10.6 0.34444 lbF F N= = =

0.889 lbP = ..................................................................................................................... Ans.

6-122 An ill-fitting window is about 10 mm narrower than its frame (see Fig. P6-122). The window weighs 40 N, and the coefficient of friction between the window and the frame is 0.20. Determine the amount of force P that must be applied at the lower corner to keep the window from lowering.

SOLUTION Impending slip of the window is downward and friction acts to oppose the motion. From a free-body diagram of the window, the equilibrium equations give

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 40 0F F P+ − + =

2 0 :MΣ =� ( ) 1 1405 40 600 810 0N F− − =

1 2 21.260 NN N= =

1 2 10.2 4.252 NF F N= = =

31.5 NP = ....................................................................................................................... Ans.


233

6-123 Determine the minimum coefficient of static friction necessary for the pliers shown in Fig. P6-123 to grip the bolt.

SOLUTION Impending motion is for the bolt to slip out of the pliers and friction acts to oppose the motion. From a free-body diagram for bolt, the equilibrium equations give

0 :centerMΣ =� 1 2 0F r F r− =

1 2 1 2s sF F N Nµ µ= = =

1 2N N=

0 :xF→Σ = ( ) ( )1 2 1 2sin17.5 cos17.5 0N N F F+ ° − + ° =

( ) ( )1 2 1 2sin17.5 cos17.5sN N N Nµ+ ° = + °

tan17.5 0.315sµ = ° = ...................................................................................................Ans.

(Note that even though we did not use the third equation of equilibrium 0 :yF↑Σ = , it is also satisfied by the values solved above.)

6-124 A homogeneous book of mass 0.46 kg rests in a bookshelf as shown in Fig. P6-104. The thickness of the book is small compared to the other dimensions shown. The coefficient of friction at all surfaces is 0.4. Determine the minimum angle θ for which the book is in equilibrium.

SOLUTION

( )0.46 9.81 4.5126 NW = =

Impending motion of the book is down and to the right. Friction acts to oppose this motion. From a free-body diagram of the book, the equilibrium equations give

0 :xF→Σ = 1 20.4 0N N− =

0 :yF↑Σ = 2 10.4 4.5126 0N N+ − =

1 1.55607 NN = 2 3.89017 NN =

2 0 :MΣ =� ( ) ( ) ( )1 14.5126 130cos 260sin 0.4 260cos 0N Nθ θ θ− − =

tan 1.0500θ = 46.40θ = ° ......................................................................Ans.

6-125 A wedge is used to raise a 350-lb refrigerator onto a platform (Fig. P6-125). The coefficient of friction is 0.2 at all surfaces.

(a) Determine the minimum force P needed to insert the wedge. (b) Determine if the system would still be in equilibrium if P = 0. (c) If the system is not in equilibrium when P = 0, determine the force necessary to keep the wedge in place,

or if the system is in equilibrium when P = 0, determine the force necessary to remove the wedge. SOLUTION When the wedge is being pushed to the right, friction acts down and to the left to oppose the motion. From a free-body diagram of the wedge and refrigerator, the equilibrium equations are

0 :xF→Σ = sin15 cos15 0P N F− ° − ° =

0 :yF↑Σ = cos15 sin15 350 0N F° − ° − =

(a) If 0.2F N= , then the equilibrium equations give

382.864 lbN =


234

0.2 76.573 lbF N= =

173.1 lbP = ...................................................................................................................... Ans.

(b) If 0P = , then the equilibrium equations give

390.374 lbN = 104.600 lbF = However, this solution is not valid since the amount of friction required for equilibrium is greater than the

maximum amount of friction available max 0.2 78.075 lbF N= = . Therefore, the wedge will not stay in place if the force P is removed.

When the force P is reduced, the wedge wants to slip to the left and friction must act up and to the right to oppose the impending motion. Now from a free-body diagram of the wedge and refrigerator, the equilibrium equations give (using 0.2F N= )

0 :xF→Σ = sin15 cos15 0P N F− ° + ° =

0 :yF↑Σ = cos15 sin15 350 0N F° + ° − =

343.916 lbN =

0.2 68.783 lbF N= =

22.6 lbP = ....................................................................................................................... Ans.

6-126 A 50-kg uniform plank rests on rough supports at A and B (Fig. P6-126). The coefficient of friction is 0.60 at both surfaces. If a man weighing 800 N pulls on the rope with a force of P = 400 N, determine

(a) The minimum and maximum angles θmin and θmax for which the system will be in equilibrium. (b) The minimum coefficient of friction that must exist between the man�s shoes and the ground for each of

the cases in part (a). SOLUTION

( )50 9.81 490.5 NW = =

(a) Impending motion of the plank is to the left and friction acts to the right to oppose the motion. From a free-body diagram of the plank, the equilibrium equations are

0 :xF→Σ = 400cos 0A BF F θ+ − =

0 :yF↑Σ = 490.5 400sin 0A BN N θ+ − − =

0 :BMΣ =� ( ) ( ) ( ) ( )490.5 1 2 400sin 2 0AN θ− − =

If 0.6A AF N= and 0.6B BF N= (impending slip), then the equilibrium equations give

( )0.6 400cosA BN N θ+ = 490.5 400sinA BN N θ+ = +

666.667 cos 490.5 400sinA BN N θ θ+ = = +

19.920θ = °

If 0A AF N= = (impending tip), then the equilibrium equations give

( )1 490.5sin 37.816

2 400θ −= = °

316.0 NBF = 735.75 NBN = ( )max 0.6 441.45 NB BF N= =

Since the amount of friction required is less than the maximum amount of friction available, the solution is okay and the allowed range of angles is

min 19.92θ = ° ................. max 37.82θ = ° .........................................................................Ans.


235

(b) Impending motion of the man�s feet is to the right and friction acts to the left to oppose the motion. From a free-body diagram of the plank, the equilibrium equations are

0 :xF→Σ = 400cos 0Fθ − =

0 :yF↑Σ = 400sin 800 0N θ+ − =

If 19.920θ = ° , the equilibrium equations give

376.07 NF = 663.72 NN =

( )min

376.07 0.567663.72s

FN

µ = = = ....................................................................................Ans.

If 37.816θ = ° , the equilibrium equations give

315.99 NF = 554.75 NN =

( )min

315.99 0.570554.75sµ = = .............................................................................................Ans.

6-127 The plunger of a door latch is held in place by a spring as shown in Fig. P6-127. Friction on the sides of the plunger may be ignored. If a force of 2 lb is required to just start closing the door and the coefficient of friction between the plunger and the striker plate is 0.25, determine the force exerted on the plunger by the spring.

SOLUTION When the door is pushed in the direction of the 2-lb force, friction acts upward and to the left to oppose the motion. From a free-body diagram of the plunger, the equilibrium equations give

0 :xF→Σ = sin 50 0.25 cos50 0N N P° − ° − =

0 :yF↑Σ = cos50 0.25 sin 50 2 0N N° + ° − =

2.39722 lbN =

1.451 lbP = ...................................................................................................................... Ans.

6-128 The 25-kg block in Fig. P6-128 is held against the wall by the brake arm. The coefficient of friction between the wall and the block is 0.20; between the block and the brake arm, 0.50. Neglect the weight of the brake arm. Determine

(a) Whether the system would be in equilibrium for P = 230 N. (b) The minimum force P for which the system would be in equilibrium. (c) The maximum force P for which the system would be in equilibrium. (d) The shearing stress on a cross section of the pin at A when impending motion of the block is downward if

the pin has a 6-mm diameter and is in double shear. SOLUTION

( )25 9.81 245.25 NW = =

If the impending motion of the block is downward, friction would act upward on the block (and downward on the brake handle) to oppose the motion. From a free-body diagram of the brake handle, the equilibrium equations are

0 :xF→Σ = 2 140 0xN A− − =

0 :yF↑Σ = 2 0yA F− =

0 :AMΣ =� ( )( ) ( )2 2140 50 40 0N F− + =


236

From a free-body diagram of the block, the equilibrium equations are

0 :xF→Σ = 1 2 0N N− =

0 :yF↑Σ = 1 2 245.25 0P F F+ + − =

(a) If 230 NP = , the equilibrium equations give

1 2N N= 1 2 15.25 NF F+ = 2 2140 0.8N F= +

( )1 2 1 2 2 2max 0.2 0.5 0.7 98 0.56 98 NF F N N N F+ = + = = + >

Since the amount of friction required for equilibrium ( )15.25 N is less than the maximum amount of friction

available ( )at least 98 N , the system

Is in equilibrium .............................................................................................................Ans.

(b) If 1 10.2F N= and 2 20.5F N= (impending motion downward), the equilibrium equations give

( )( ) ( ) ( )2 2140 50 0.5 40 0N N− + = 1 2 233.333 NN N= =

1 116.667 NF = 2 46.667 NF =

min81.9 NP P= = ............................................................................................................Ans.

93.333 NxA = 116.667 NyA = 2 2 149.406 Nx yA A A= + =

(c) If impending motion of the block is upward, the friction forces ( 1 10.2F N= and 2 20.5F N= ) change direction and the equilibrium equations give

( )( ) ( )( )2 2140 50 0.5 40 0N N− − = 1 2 100.00 NN N= =

1 20.00 NF = 2 50.00 NF =

max315 NP P= = .............................................................................................................Ans.

(d) When the impending motion of the block is downward, the force on the pin is 149.406 NA = . Dividing the pin force by twice its cross sectional area (since it is in double shear) gives the shear stress

( )

6 22

149.406 2.64 10 N/m 2.64 MPa2 2 0.006 4A

As

VA

τπ

= = = × =

...........................Ans.

6-129 A 250-lb weight is suspended from a lightweight rope wrapped around the inner cylinder of a drum (Fig. P6-129). A brake arm is pressed against the outer cylinder of the drum by a hydraulic cylinder. The coefficient of friction between the brake arm and the drum is 0.40. Determine

(a) The smallest force in the hydraulic cylinder necessary to prevent motion. (b) The shearing stress on a cross section of the pin at A when motion is impending if the drum weighs 75 lb

and the pin has a ½-in. diameter and is in double shear. SOLUTION (a) Impending motion of the drum is a counterclockwise rotation and friction

acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 75 250 0yA N− − − =

0 :AMΣ =� ( )( ) ( )250 6 12 0F− =


237

125 lb 0.4F N= = 312.5 lbN =

125 lbxA = 637.5 lbyA =

2 2 649.639 lbx yA A A= + =

From a free-body diagram of the brake arm, the moment equilibrium equation gives

0 :CMΣ =� 9 9 27 0P F N− − =

1063 lbP = ...................................................................................................................... Ans. (b) Dividing the force on the pin by twice its cross sectional area (since it is in double shear) gives the shear stress

( )2

649.639 1654 psi2 2 0.5 4A

As

VA

τπ

= = =

......................................................................Ans.

6-130 The brake shown in Fig. P6-130 is used to control the motion of block B. If the mass of block B is 25 kg and the kinetic coefficient of friction between the brake drum and brake pad is 0.30, determine

(a) The force P required for a constant-velocity descent. (b) The shearing stress on cross sections of both pins. Each pin has a diameter of 10 mm and is in double

shear. SOLUTION

( )25 25 9.81 245.25 NW g= = =

(a) Motion of the drum is a clockwise rotation and friction acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give

0 :xF→Σ = 0xB F− =

0 :yF↑Σ = 245.25 0yB N− − =

0 :BMΣ =� ( ) ( )180 245.25 90 0F − =

122.625 N 0.3F N= = 408.750 NN =

122.625 NxB = 654.000 NyB =

2 2 665.397 Nx yB B B= + =

From a free-body diagram of the brake arm, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 0yN A P− − =

0 :AMΣ =� 200 40 600 0N F P+ − =

122.625 NxA = 264.325 NyA =

144.4 NP = ..................................................................................................................... Ans.

2 2 291.384 Nx yA A A= + =

(b) Dividing the force on the pins by twice their cross sectional areas (since they are in double shear) gives the shear stresses


238

( )

6 22

291.384 1.855 10 N/m 1.855 MPa2 2 0.01 4A

As

VA

τπ

= = = × =

..........................Ans.

( )

6 22

665.397 4.24 10 N/m 4.24 MPa2 2 0.01 4B

Bs

VA

τπ

= = = × =

..............................Ans.

6-131 Three identical cylinders are stacked as shown in Fig. P6-131. The cylinders each weigh 22 lb and are 8 in. in diameter. The coefficient of friction is µs = 0.40 at all surfaces. Determine the maximum force P that the cylinders can support without moving.

SOLUTION From a free-body diagram for the upper cylinder, the equilibrium equations give

0 :xF→Σ = cos30 cos30 cos60 cos 60 0AC AB AB ACF F N N° − ° + ° − ° =

0 :yF↑Σ = ( ) ( )sin 60 sin 30 22 0AB AC AB ACN N F F P+ ° + + ° − − =

0 :centerMΣ =� 0AC ABF r F r− =

AC ABF F= AC ABN N=

2 sin 60 2 sin 30 22AB ABP N F= ° + ° +

From a free-body diagram for the lower-left cylinder, the equilibrium equations give

0 :xF→Σ = cos30 cos60 0B AB ABF F N+ ° − ° =

0 :yF↑Σ = 22 sin 30 sin 60 0B AB ABN F N− − ° − ° =

0 :centerMΣ =� 0B ABF r F r− =

0.26795B AB ABF F N= =

Since the amount of friction required for equilibrium ( )0.26795AB ABF N= is less than the maximum amount of

friction available ( )max 0.4AB ABF N= , the cylinders will not slip at the surface between the cylinders regardless of the value of P. Then,

( )22 sin 30 3.73205 sin 60 0B AB ABN F F− − ° − ° =

0.26795 5.89488B AB BF F N= = −

and again the amount of friction required for equilibrium ( )0.26795 5.89488B BF N= − is less than the

maximum amount of friction available ( )max 0.4B BF N= so the cylinders will not slip at the lower surface regardless of the value of P. Therefore, the cylinders will not slip for any value of P and

maxP = ∞ ............................................................................................................................ Ans.

6-132 A 10-kg drum rests on a thin lightweight piece of cardboard as shown in Fig. P6-132. The coefficient of friction is the same at all surfaces.

(a) Plot P, the maximum force that may be applied to the cardboard without moving it, as a function of the coefficient of friction µs (0.05 ≤ µs ≤ 0.8).

(b) On the same graph, plot (Af)actual and (Bf)actual, the actual amounts of friction force that act at points A and B, and (Af)avail and (Bf)avail, the maximum amounts of friction force available for equilibrium at points A and B.

(c) What happens to the system at µs ≅ 0.364? (d) What does the solution for µs > 0.364 mean?


239

SOLUTION

( )10 9.81 98.1 NW = =

Impending motion of the cardboard is downward and friction acts to oppose the motion. From a free-body diagram of the cardboard, the equilibrium equations are

0 :nFΣ = 0C BN N− =

0 :tFΣ = 0B CF F P+ − =

C s CF Nµ=

From a free-body diagram of the cylinder, the equilibrium equations are

0 :nFΣ = 98.1cos 20 sin 40 0B AN F− ° − ° =

0 :tFΣ = 98.1sin 20 cos 40 cos50 0B A AF F N− − ° − ° + ° =

0 :centerMΣ =� 0A BF r F r− =

Therefore

A BF F= B CN N=

If impending slip at A, then A s A BF N Fµ= = and

98.1sin 20 cos 40 cos50 0s A s A AN N Nµ µ− − ° − ° + ° =

( )98.1sin 20

cos50 1 cos 40As

Nµ

°=° − + °

98.1cos 20 sin 40 sin 50B s A A CN N N Nµ= ° + ° + ° =

( ) ( )B C A C s A C s A BP F F F F N N N Nµ µ= + = + = + = + .......................................Ans.

( )maxA B A s AF F F Nµ= = = ............................................................................................Ans.

( )maxB s BF Nµ= ...............................................................................................................Ans.

These five functions are all plotted on the graph below. Note that the force ( )maxB BF F< for the entire range graphed. This verifies the guess that impending slip occurs first at A. (c) When the coefficient of friction is 0.364, the normal

force AN goes to infinity. That is, the system locks

at 0.364sµ = , and no amount of pulling force will be able to move the cardboard.

(d) For coefficients of friction greater than 0.364, the

friction and normal forces become negative. The best interpretation of the solution is that a pushing force will be necessary to move the cardboard � therefore, the friction force must be in the opposite direction.


240

6-133 A 150-lb uniformly-loaded file cabinet sits on an inclined surface as shown in Fig. P6-133. A horizontal force P acts on the cabinet, and the coefficient of friction between the cabinet and the inclined surface is 0.4.

(a) Determine Pmin and Pmax, the minimum and maximum horizontal forces for which the file cabinet will be in equilibrium.

(b) Calculate and plot N and F, the normal and friction forces acting on the bottom of the file cabinet, as functions of P (Pmin ≤ P ≤ Pmax).

(c) Calculate and plot d, the distance between the line of action of the normal force and the corner C, as a function of P (Pmin ≤ P ≤ Pmax).

SOLUTION Assuming that impending motion is up the incline, friction will act down the incline to oppose the motion. From a free-body diagram of the cabinet, the equilibrium equations are

0 :xF→Σ = sin 25 cos 25 0P N F− ° − ° =

0 :yF↑Σ = cos 25 sin 25 150 0N F° − ° − =

0 :CMΣ =� ( ) ( ) ( )( )150cos 25 9 150sin 25 24° + °

( ) ( ) ( )( )sin 25 18 cos 25 32 0Nd P P− + ° − ° =

(a) If 0.4F N= (impending slip up the incline), then

203.456 lbN = 81.382 lb F =

159.742 lbP = 3.306 in.d = − But this solution is not possible since the normal force acts at a point

outside the body. If 0d = (impending tip), then

190.168 lbN = 52.887 lb F =

max128.300 lbP P= = .....................................................................................................Ans.

If impending slip is down the incline, then the friction force acts up the incline (changes sign in the equilibrium equations above), 0.4F N= , and the equations of equilibrium give

139.489 lbN = 55.796 lb F =

8.383 lbP = 18.393 in.d = But this solution is also not possible since the normal force again acts at a point outside the body. If

18 in.d = (impending tip), then

140.287 lbN = 54.083 lb F =

min10.272 lbP P= = .......................................................................................................Ans.

(b) For 10.272 lb 128.300 lbP< < , the normal and friction forces are

cos 25 150sin 25F P= ° − ° ..........................................................................................Ans.

sin 25 150cos 25N P= ° + ° ..........................................................................................Ans.

(c) For 10.272 lb 128.300 lbP< < , the location of the normal force is

2744.9413 21.39472Pd

N−= .......................................................................................Ans.


241

6-134 A 35-kg child is sitting on a swing suspended by a rope that passes over a tree branch (Fig. P6-134). The coefficient of friction between the rope and the branch (which can be modeled as a flat belt over a drum) is 0.5 and the weight of the rope can be ignored. Determine the minimum force the child must exert on the rope to keep suspended.

SOLUTION

( )35 9.81 343.35 NW = =

From a free-body diagram of the child, the vertical equilibrium equation gives

0 :yF↑Σ = 1 2 343.35 0T T+ − =

which combined with the belt friction equation

0.52 1 14.81048T T e Tπ= =

gives

1 59.1 NT = ....................................................................................................................... Ans.

6-135 A rope attached to a 500-lb block passes over a frictionless pulley and is wrapped for one full turn around a fixed post as shown in Fig. P6-135. If the coefficient of friction between the rope and the post is 0.25, determine

(a) The minimum force P that must be used to keep the block from falling. (b) The minimum force P that must be used to begin to raise the block. SOLUTION

( )0.25 2500 Pe π= min103.9 lbP P= = ........................................................Ans.

( )0.25 2500P e π= max2410 lbP P= = ........................................................Ans.

6-136 A rope attached to a 220-kg block passes over a fixed drum (Fig. P6-136) If the coefficient of friction between the rope and the drum is 0.30, determine the minimum force P that must be used to

(a) Keep the block from falling. (b) Begin to raise the block. SOLUTION

( ) ( )0.3 2220 9.81 Pe π= min1347 NP P= = .........................................................Ans.

( ) ( )0.3 2220 9.81P e π= max3460 NP P= = ........................................................Ans.

6-137 A rotating shaft with a belt-type brake is shown in Fig. P6-137. The static coefficient of friction between the brake drum and the brake belt is 0.20. When a 75-lb force P is applied to the brake arm, rotation of the shaft is prevented. Determine the maximum torque that can be resisted by the brake if the shaft is tending to rotate

(a) Counterclockwise. (b) Clockwise.


242

SOLUTION From a free-body diagram for the pulley, the moment equilibrium equation gives

0 :BMΣ =� 1 26 6 0T T C− + =

( )2 16C T T= −

From a free-body diagram for the brake arm, the moment equilibrium equation gives

0 :AMΣ =� ( )2 19 3 27 75 0T T− − =

2 13 675T T− =

(a) If the couple C is counterclockwise, then 2 1T T> and the belt friction equation gives

0.22 1 11.87446T T e Tπ= =

1 145.997 lbT = 2 273.666 lbT =

766 lb in. = ⋅C � ............................................................................................................Ans.

(a) If 1 2T T> , then the belt friction equation gives

0.21 2 21.87446T T e Tπ= =

1 1124.130 lbT = 2 599.710 lbT =

3150 lb in. 3150 lb in. = − ⋅ = ⋅C � .............................................................................Ans.

6-138 The scaffolding of Fig. P6-138 is raised using an electric motor that sits on the scaffolding. Frictionless wheels at the ends of the scaffold restrict horizontal motion. The 250-mm-diameter pulley at the top is jammed and will not rotate. The coefficient of friction between the rope and the pulley is 0.25 and the weight of the motor, scaffold, and supplies is 2500 N. Determine the minimum torque that must be supplied by the motor

(a) To raise the scaffold at a constant rate. (b) To lower the scaffold at a constant rate. SOLUTION From a free-body diagram for the scaffolding, the vertical equilibrium equation gives

0 :yF↑Σ = 2500 0A BT T+ − =

(a) If A BT T> , then the belt friction equation gives

0.25 2.19328A B BT T e Tπ= =

1717.106 NAT = 782.894 NBT =

( )( )Torque 1717.106 0.075AT r= =

128.8 N m (to raise the scaffolding)= ⋅ ......................................................Ans.

(b) If B AT T> , then the belt friction equation gives

0.25 2.19328B A AT T e Tπ= =

782.894 NAT = 1717.106 NBT =

( )( )Torque 782.894 0.075AT r= =

58.7 N m (to lower the scaffolding)= ⋅ ......................................................Ans.


243

6-139 A rope is wrapped one full turn around each of two posts as shown in Fig. P6-139. If the coefficient of friction between the rope and the posts is 0.5, determine

(a) The ratio of TA to TB. (b) The ratio of TA to TB if only one post is used. SOLUTION

(a) ( )0.5 2 23.14069B C CT T e Tπ= =

( )0.5 223.14069 535.492A AT e Tπ = =

535B AT T = ...................................................................................................................... Ans.

(b) ( )0.5 2 23.14069B A AT T e Tπ= =

23.1B AT T = ..................................................................................................................... Ans.

6-140 A conveyor belt is driven with the 200-mm-diameter multiple-pulley drive shown in Fig. P6-140. Couples CA and CB are applied to the system at pulleys A and B, respectively. The angle of contact between the belt and a pulley is 225° for each 6.5 kg pulley, and the coefficient of friction is 0.30. Determine

(a) The maximum force T that can be developed by the drive and the magnitudes of the input couples CA and CB.

(b) The shearing stress on a cross section of each 25-mm diameter pin, if each pin is in double shear. SOLUTION

( )6.5 9.81 63.765 lbW = = ( ) ( )225 360 2 3.92699 radβ π= =

(a) ( )0.3 3.926992 6.49638 kNCT e= =

( )0.3 3.92699 21.101 kNCT T e= = .......................................................................................Ans.

From a free-body diagram for pulley A, the equilibrium equations give

0 :xF→Σ = 21,101 6496.38cos 45 0xA+ ° − =

0 :yF↑Σ = 63.765 6496.38sin 45 0yA − + ° =

0 :AMΣ =� ( ) ( )21,101 0.1 6496.38 0.1 0AC − + =

25,695 NxA = 4530 NyA = −

2 2 26,091 Nx yA A A= + =

1460 N mAC = ⋅ ..............................................................................................................Ans.

From a free-body diagram for pulley B, the equilibrium equations give

0 :xF→Σ = 2000 6496.38cos 45 0xB − − ° =

0 :yF↑Σ = 63.765 6496.38sin 45 0yB − − ° =

0 :BMΣ =� ( ) ( )6496.38 0.1 2000 0.1 0BC− − =

6594 NxB = 4657 NyB =

2 2 8073 Nx yB B B= + =

450 N mBC = ⋅ ................................................................................................................Ans.


244

(b) The shearing stress on the pins is obtained by dividing the forces on the pins by twice their cross sectional areas (since they are in double shear),

( )

6 22

26,091 26.6 10 N/m 26.6 MPa2 2 0.025 4A

As

VA

τπ

= = = × =

............................Ans.

( )

6 22

8073 8.22 10 N/m 8.22 MPa2 2 0.025 4B

Bs

VA

τπ

= = = × =

............................Ans.

6-141 The electric motor shown in Fig. P6-141 weighs 30 lb and delivers 50 lb-in. of torque to pulley A of a furnace blower by means of a V-belt. The effective diameters of the 36° pulleys are 5 in. The coefficient of friction is 0.30. Determine the minimum distance a to prevent slipping of the belt if the rotation of the motor is clockwise.

SOLUTION

( )enh0.3 0.9708

sin 36 2sµ = =

From a free-body diagram for the pulley A, the moment equilibrium equation

0 :AMΣ =� ( )( )2 150 2.5 0T T− − =

is combined with the belt friction equation

0.97082 1 121.1137T T e Tπ= =

to get

1 0.99435 lbT =

2 20.9944 lbT =

Then, from a free-body diagram for the motor, the moment equilibrium equation gives

0 :CMΣ =� ( ) ( )20.9944 10.5 0.99435 5.5 50 0a+ − =

4.52 in.a = ....................................................................................................................... Ans.

6-142 Complete the derivation of Eq. (6-20). SOLUTION Starting from

( )cos 2 2T Fθ∆ ∆ = ∆

( ) ( ) ( )2 sin 2 2 sin 2 sin 2P T Tα θ θ∆ = ∆ + ∆ ∆

in the limit as 0θ∆ →

2T F∆ = ∆

( ) ( )2 sin 2 2P T Tα θ θ∆ = ∆ + ∆ ∆

and therefore as 0θ∆ →

2 0P→ 0F∆ → 0T∆ →

Then, at the point of impending slip, sF Pµ∆ = ∆ . Divide through by θ∆ and let 0θ∆ → to get

( )( )0

22limsin 2ss T TPT

θ

µµθ θ α∆ →

+ ∆∆∆ = =∆ ∆


245

( )sin 2sTdT

dµ

θ α=

( )2

1 0sin 2T sT

dT dT

βµ θα

=∫ ∫

( )2

2 11

ln ln lnsin 2

sTT TT

µ βα

− = =

( )sin 22

1

sT eT

µ β α= .................................................................................................................Ans.

6-143 The band wrench of Fig. P6-143 is used to unscrew an oil filter from a car. (The filter acts as if it were a wheel with a resisting torque of 40 lb-ft.) Neglect the weight of the handle and friction between the end of the metal handle and the metal filter case. Determine

(a) The minimum coefficient of friction between the band and the filter that will prevent slippage. (b) The shearing stress on a cross section of the pin at B when slipping is impending if the ¼-in.-diameter pin

is loaded in double shear. SOLUTION

( )1tan 2.5 4.5 29.055φ −= = °

270 299.055 5.2195 radβ φ= ° + = ° =

(a) From a free-body diagram for the filter, the moment equilibrium equation gives

0 :CMΣ =� ( )( )2 140 2.5 12 0T T− − =

2 1 192.00 lbT T− =

From a free-body diagram for the wrench, the moment equilibrium equation gives

0 :PMΣ =� ( ) ( )1 29.5 sin 7.5 0T T φ− =

2 12.60819T T=

1 119.3886 lbT =

2 311.3886 lbT =

Then, the belt friction equation applied to the tensions on the filter gives

5.2195311.3886 119.3886 se µ=

0.1837sµ = ...................................................................................................................... Ans.

(b) Dividing the force on the pin at B by twice its cross sectional area (since it is in double shear) gives the shear stress

( )

322

311.3886 3.17 10 psi 3.17 ksi2 2 0.25 4

Bs

TA

τπ

= = = × =

.......................................Ans.

6-144 A uniform belt 2 m long with a mass of 2 kg hangs over a small fixed peg (Fig. P6-144). If the coefficient of friction between the belt and the peg is 0.40, determine the maximum distance d between the two ends for which the belt will not slip off the peg.

SOLUTION

1 1T W= 2 2T W=


246

( )1 2 2 9.81 19.62 NW W W+ = = =

1 2 2 mL L L+ = =

0.42 1 13.51359W We Wπ= =

1 4.34688 NW = 2 15.27312 NW =

The weight of each portion of the belt is directly proportional to its length. Therefore

1

1

19.622

WL

=

1 0.44311 mL = 2 2 0.44311 1.55689 mL = − =

2 1 1.114 md L L= − = ....................................................................................................Ans.

6-145 The hand brake of Fig. P6-145 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is µs = 0.35, and the weight of the handle may be neglected. If a clockwise torque of 350 lb-ft is applied to the drum, determine the minimum force P that must be applied to the handle to prevent motion when a = 4 in., 8 in., and 12 in.

SOLUTION From a free-body diagram for the drum, the moment equilibrium equation

0 :EMΣ =� ( ) ( )2 1 6 12 350 0T T− − =

is combined with the belt friction equation

0.352 1 13.00284T T e Tπ= =

to get

1 349.5043 lbT =

2 1049.504 lbT =

From a free-body diagram for the brake, the moment equilibrium equation gives

0 :BMΣ =� ( ) 230 0P a T a− − =

4 in.a = 161.5 lbP = .........................................................................................Ans.

8 in.a = 382 lbP = ............................................................................................Ans.

12 in.a = 700 lbP = ............................................................................................Ans.

6-146 The structure shown in Fig. P6-146 consists of a circular tie rod AB and a rigid member BC. If the structure is to support a load P = 40 kN, determine the required diameters of the pins at A, B, and C, and the required diameter of the tie rod. The tie rod is made of structural steel and the pins are made of 0.2% C hardened steel. All pins are in double shear. The tie rod is adequately reinforced around the pins so that tensile failure does not occur at the pins. Failure is by yielding, and the factor of safety is 1.3. Take the yield strength in shear to be one-half the yield strength in tension.

SOLUTION

sin 3 5θ = cos 4 5θ =

structural steel: 250 MPayσ = 200 GPaE =

0.2% hardened steel: 250 MPayσ =

Note that the tie rod is a two-force member. Then from a free-body diagram of the rigid member BC, the equilibrium equations give


247

0 :xF→Σ = cos 0x ABC T θ− =

0 :yF↑Σ = 40 sin 0y ABC T θ− + =

0 :AMΣ =� ( )( ) ( )sin 4 40 3.5 0ABT θ − =

58.333 kNABT =

46.667 kNxC = 5.000 kNyC =

2 2 46.934 kNx yC C C= + =

Dividing the force in the tie rod by its cross sectional area gives the normal stress

3 6

2

58.333 10 250 104 1.3

yABAB

AB

TA d FS

σσ

π× ×= = ≤ =

0.01965 m 19.65 mmABd ≥ = ......................................................................Ans.

The force on the pins are divided by twice their cross sectional area since they are in double shear

( )63

2

430 10 2258.333 102 1.32 4

y yAA

s A

VA FS FSd

τ στ

π××= = ≤ = =

0.01498 m 14.98 mmAd ≥ = ........................................................................Ans.

( )63

2

430 10 258.333 101.32 4B

Bdτ

π××= ≤

14.98 mmBd ≥ ...................Ans.

( )63

2

430 10 246.934 101.32 4C

Cdτ

π××= ≤

13.44 mmCd ≥ ...................Ans.

6-147 A rigid handle is used to twist a 2.5-in.-diameter valve, as shown in Fig. P6-147. A pair of oppositely directed parallel forces P = 100 lb is needed to twist the valve. Force is transmitted from the handle to the valve shaft by means of a 1.0-in.-long (into the page) key with a square cross section. The key is to be made of 0.8% C hot-rolled steel. If failure is by fracture, and the factor of safety is to be 2, determine the cross-sectional dimensions of the square key to the nearest 1/64 in. The yield and ultimate strengths in shear are one-half the corresponding values in tension.

SOLUTION

0.8% hot-rolled steel: 122 ksifσ =

From a free-body diagram of the handle, the moment equilibrium equation gives

0 :centerMΣ =� ( ) ( )100 24 1.25 0V− =

( )1920 lb 1V bτ= = ×

( )3122 10 221920

2f y

b FS FSτ σ ×

≤ = =

0.06295 in.b ≥

( ) ( )4 64 0.06295 5 64< <

min 5 64 in.b = .................................................................................................................Ans.


248

6-148 Each member of the truss of Fig. P6-148 has a circular cross section and is made of structural steel. If failure is by yielding, and the factor of safety is 2.5, determine the minimum permissible diameter of member EJ.

SOLUTION

( )1tan 3 2 56.310θ −= = ° 2 23 2 3.60555 mEJL = + =

structural steel: 250 MPayσ =

From a free-body diagram of the entire truss, the moment equilibrium equation gives

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )8 30 8 30 6 30 4 30 2 0G − − − − =

75 kNG = Next, divide the truss with a section through members DE, EJ, and HJ, and draw a free-body diagram of the right-hand side. From a free-body diagram of the section, the vertical equilibrium equation gives

0 :yF↑Σ = 75 30 30 sin 0EJT θ− − − =

18.0277 kNEJT =

Dividing the force in the member by its cross-sectional area gives the normal stress

3 6

2

18.0277 10 250 104 2.5

yEJEJ

TA d FS

σσ

π× ×= = ≤ =

0.01515 m 15.15 mmd ≥ = Ans.

6-149 A pin-connected system of levers and bars is used as a toggle for a press to crush cans, as shown in Fig. P6-149. The system is designed to provide a crushing force of 550 lb. The handle may be considered rigid. The pin at D is to be made of 2024-T4 wrought aluminum. If failure is by yielding, and the factor of safety is to be 1.5, determine the required diameter of pin D to the nearest 1/32 in.

SOLUTION

2024-T4 wrought aluminum: 48 ksiyσ =

From a free-body diagram of the plunger, the vertical equilibrium equation gives

0 :yF↑Σ = 550 sin 67 0BDF− ° =

597.498 lbBDF =

Note that all three members attached to the pin D are two-force members. Therefore, from a free-body diagram of the pin, the equilibrium equations give

0 :xF→Σ = cos67 cos 78 0BD CD DEF F F° + ° − =

0 :yF↑Σ = sin 67 sin 78 0BD CDF F° − ° =

562.287 lbCDF = 350.367 lbDEF =

Therefore, the maximum shear force on pin D is 597.498 lbBDF = on the shear surface between BD and CD.

( )3

max 2

48 10 22597.4984 1.5

y y

d FS FSτ σ

τπ

×= ≤ = =

0.2181 in.d ≥

6 32 0.2181 7 32< < min 7 32 in.d = .................................................Ans.


249

6-150 A pair of vise grip pliers is shown in Fig. P6-150. All members of the pliers, except for the pins and spring, may be treated as rigid. It is anticipated that the largest applied force on the handles will be P = 200 N. If each of the three pins is in double shear and made of 0.4% C hardened steel, determine the required diameters for the pins to the nearest mm. In the analysis, neglect the spring force. Failure is by yielding, and the factor of safety is 1.25. The yield strength in shear is one-half the tensile yield strength.

SOLUTION

( )1tan 30 50 30.964θ −= = ° ( )1tan 15 35 23.199φ −= = °

0.4% C hardened steel: 360 MPayσ =

From a free-body diagram of the handle, the equilibrium equations give

0 :xF→Σ = cos 0AB xF Cθ − =

0 :yF↑Σ = sin 200 0AB yF Cθ − − =

0 :CMΣ =� ( ) ( )( ) ( )( )200 93 sin 28 cos 5 0AB ABF Fθ θ− + =

1838.212 NABF =

1576.250 NxC = 745.759 NyC =

2 2 1743.766 Nx yC C C= + =

Next, from a free-body diagram of the jaw, the equilibrium equations give

0 :xF→Σ = sin 0x xC D N φ− − =

0 :yF↑Σ = cos 0y yC D N φ− + =

0 :DMΣ =� ( ) ( ) ( )2 235 15 35 12 0x yN C C+ − − =

1683.817 NN =

912.951 NxD = 2293.426 NyD =

2 2 2468.457 Nx yD D D= + =

Dividing the force on the pins by twice their cross-sectional areas (since they are in double shear) gives the shear stresses

( )6

2

360 10 221838.2122 1.252 4

y yABB

s B

FA FS FSd

τ στ

π×

= = ≤ = =

0.00285 mBd ≥ min 3 mmBd = ............................................Ans.

( )6

2

360 10 21743.7662 1.252 4C

s C

CA d

τπ

×= = ≤

0.00222 mCd ≥ min 3 mmCd = ............................................Ans.

( )6

2

360 10 22468.4572 1.252 4D

s D

DA d

τπ

×= = ≤

0.00264 mDd ≥ min 3 mmDd = ............................................Ans.


250

6-151 A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-151. The coefficients of friction at all vertical contact surfaces are µs = 0.4 and µk = 0.3. The device is to lift two blocks, each weighing 15 lb. The pin at B is to be made of structural steel with a strength in shear equal to one-half the strength in tension. For failure by yielding and a factor of safety of 3, determine the minimum permissible diameter of the pin at B, which is in double shear.

SOLUTION

structural steel: 36 ksiyσ =

From a free-body diagram of the blocks, the equilibrium equations give

0 :yF↑Σ = 30 0A CF F+ − =

0 :GMΣ =� 8 8 0A CF F− =

15 lbA CF F= =

Next, from a free-body diagram of member AB, the equilibrium equations give

0 :xF→Σ = 0A xN B− =

0 :yF↑Σ = 0y AB F− =

0 :BMΣ =� 4 12 0A AN F− =

3 45 lbA AN F= =

45 lbxB = 15 lbyB =

2 2 47.434 lbx yB B B= + =

Note that 3A AN F= independent of the weight being lifted. Therefore, as long as the coefficient of friction is

greater than 1 3A AF N = , the device will lift the weight without slipping.


( )3

2

36 10 2247.4342 32 4

y yB

s B

BA FS FSd

τ στ

π×

= = ≤ = =

0.0709 in.Bd ≥ .................................................................................................Ans.

6-152 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-152). Water of density 1000 kg/m3 may collect on the roof to a depth of 0.2 m. All members of the truss are to be the same size. If the truss members are made of structural steel, failure is by yielding, and the factor of safety is 3.0, determine the smallest diameter solid circular rod, to the nearest mm, that can be used. The members are braced so that buckling does not occur.


( ) ( ) ( )1 2 2 1000 9.81 2 1.2 0.2 2

2354.4 NF F W= = = × ×

=


0 :xF→Σ = 0xA =

0 :yF↑Σ = 16 0y yA F F+ − =


251

0 :AMΣ =� ( ) ( ) ( ) ( )( ) ( ) ( )1 1 13.6 2 1.2 2 2.4 3.6 0yF F F F− − − =

13 7063.20 Ny yA F F= = =


0 :xF→Σ = 0BCT =

0 :yF↑Σ = 1 0ABT F− − =

2354.4 N 2.35 kN (C)ABT = − ≅ ............................................. Ans.

0 kNBCT = .................................................................................... Ans.


0 :xF→Σ = ( )3 5 0AH ACT T+ =

0 :yF↑Σ = ( )7063.2 4 5 0AB ACT T+ + =

5886.0 N 5.89 kN (C)ACT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)AHT = ≅ ................................................ Ans.

From a free-body diagram for joint H, the equilibrium equations give

0 :xF→Σ = 0GH AHT T− =

0 :yF↑Σ = 0CHT =

0 kNCHT = ................................................................................... Ans.

3531.6 N 3.53 kN (T)GHT = ≅ ................................................ Ans.


0 :xF→Σ = 0DET− =

0 :yF↑Σ = 1 0EFT F− − =

0 kNDET = .................................................................................... Ans.

2354.4 N 2.35 kN (C)EFT = − ≅ ............................................. Ans.

From a free-body diagram for joint F, the equilibrium equations give

0 :xF→Σ = ( )3 5 0FG DFT T− − =

0 :yF↑Σ = ( )7063.2 4 5 0EF DFT T+ + =

5886.0 N 5.89 kN (C)DFT = − ≅ ............................................. Ans.

3531.6 N 3.53 kN (T)FET = ≅ ................................................ Ans.


0 :xF→Σ = ( )3 5 0DE CD DFT T T− + =

0 :yF↑Σ = ( )12 4 5 0DG DFF T T− − − =

3531.6 N 3.53 kN (C)CDT = − ≅ ...................................... Ans.

0 kNDGT = ............................................................................ Ans.


252

Finally, from a free-body diagram for joint G, the equilibrium equations give

0 :yF↑Σ = ( )4 5 0DG CGT T+ =

0 kNCGT = ............................................................................. Ans.

Dividing the largest force ( )5886.0 N by the cross sectional area of the members gives the normal stress in the member. For structural steel

250 MPayσ = . Therefore

6

max 2

5886.00 250 104 3.0

yPA d FS

σσ

π×= = ≤ =

0.00948 m 9.48 mmd ≥ =

min 10 mmd = ................................................................................................................... Ans.

6-153 The brake shown in Fig. P6-153 is used to control the motion of block B. The weight of the block is 2200 lb, the coefficient of static friction between the brake arm and drum is 0.35, and the coefficient of kinetic friction is 0.30. A force P is applied to the rigid brake arm such that the block moves downward at a constant rate. Determine the minimum permissible diameter for the pin at the end of the brake arm (to the nearest 1/8-in.) if it is in double shear and is made of 0.2% C hardened steel. Failure is by yielding, and the factor of safety is 2.5. The strength in shear is one-half the strength in tension.

SOLUTION

0.2% C hardened steel: 62 ksiyσ =

From a free-body diagram of the drum, the moment equilibrium equation gives

0 :CMΣ =� ( )7 3.5 2200 0F − =

1100 lb 0.35F N= =

3142.857 lbN = Next, from a free-body diagram of the brake handle, the equilibrium equations give

0 :xF→Σ = 0xF A− =

0 :yF↑Σ = 0yN A P− − =

0 :AMΣ =� 8 1.6 24 0N F P+ − =

1120.952 lbP =

1100 lbxA =

4263.809 lbyA =

2 2 4403.416 lbx yA A A= + =


( )3

2

62 10 224403.4162 2.52 4

y yAA

s

VA FS FSd

τ στ

π×

= = ≤ = =

0.47547 in.d ≥

3 8 0.47547 4 8< < min 4 8 1 2 in.d = = .........................................Ans.


253

6-154 Solid circular bars AB and BC of Fig. P6-154 are pinned at the ends, and the structure is subjected to a load P = 26 kN. The angle θ may vary, but pin A is always directly above pin C. Both bars are made of 6061-T6 wrought aluminum alloy. Failure is by yielding, and the factor of safety is 4. In the force analysis assume that the weights of the bars are negligible with respect to the applied loads and that the pins are adequately designed. Determine

(a) The angle θ for a minimum weight structure. (b) The diameters of the bars. (c) The weights of the bars. SOLUTION

6061-T6 wrought aluminum: 270 MPayσ = 32710 kg/mρ =

Note that both AB and BC are two-force members. Then, from a free-body diagram of the pin, the equilibrium equations give

0 :xF→Σ = cos 0BC ABF T θ− =

0 :yF↑Σ = sin 26 0ABT θ − =

( )26 sin kNABT θ= ( )cos 26 tan kNBC ABF T θ θ= =

Dividing the force in the members by their cross sectional areas gives the normal stresses

3 6

2

26 10 sin 270 104 4

yABAB

AB

TA d FS

σθσπ× ×= = ≤ =

322.1457 10 msinABd θ

−×≥

3 6

2

26 10 tan 270 104 4BC

BCdθσ

π× ×= ≤

322.1457 10 mtanBCd θ

−×≥

(a) To minimize the weight of the structure, we need to minimize the volume of the bars making up the structure.

( )1.8 cos mABL θ= 1.8 mBCL =

( )

( )

2 2

23

1.81.84 4 cos

1.8 22.1457 10 cos 14 sin sin cos

BC ABd dV π πθ

π θθ θ θ

−

= +

× = +

To minimize the volume, we set 0dV dθ =

( )23 2

2 2 2

1.8 22.1457 10 cos 1 11 04 sin sin cos

dVd

π θθ θ θ θ

−× = − − − + =

If neither sin 0θ = nor cos 0θ = , we can multiply through by ( )2 2sin cosθ θ− to get

2 2 4 2 2sin cos cos cos sin 0θ θ θ θ θ+ + − =

( ) ( )2 2 2 2cos sin cos 1 sin 0θ θ θ θ+ + − =

2 22cos sin 0θ θ− =

( )2 22 1 sin sin 0θ θ− − =

22 3sin 0θ− =

54.736θ = ° ...................................................................................................................... Ans.


254

324.5 10 m 24.5 mmABd−≥ × = ..................................................................................Ans.

318.62 10 m 18.62 mmBCd−≥ × = ..............................................................................Ans.

( ) ( )20.0245 1.82710 9.81 39.1 N4 cos54.736ABw

π = = ° ..................................Ans.

( ) ( ) ( )20.01862

2710 9.81 1.8 13.03 N4BCw

π = =

...............................................Ans.

6-155 Determine the force P required to pull the 250-lb roller over the step shown in Fig. P6-155. SOLUTION

( )1cos 9 12 41.410θ −= = °

When the roller is about to be pulled over the step, it no longer touches the floor and the force exerted on the roller by the floor is zero. From a free-body diagram of the roller, the equilibrium equations give

0 :xF→Σ = cos30 sin cos 0P N Fθ θ° − − =

0 :yF↑Σ = cos sin sin 30 250 0N F Pθ θ− + ° − =

0 :centerMΣ =� 0Fr =

0 lbF = 1.30931N P=

168.7 lbP = ..................................................................................................................... Ans.

6-156 A machine is activated by force T when a pedal is depressed with a 40-N force, as shown in Fig. P6-156. Determine the magnitude of T and the resultant bearing reaction at B.

SOLUTION From a free-body diagram of the pedal, the equilibrium equations give

0 :xF→Σ = 0xB =

0 :yF↑Σ = 40 0yB T− − =

0 :BMΣ =� ( ) ( ) ( )200 125cos 45 40 0T − ° =

17.68 N T = ↓ ........................................................... Ans.

0 NxB = 57.7 NyB =

57.7 N = ↑B .................................................................................................................. Ans.

6-157 The 500-lb block shown in Fig. P6-157 is supported by a ball-and-socket joint at A, by a smooth pin at B, and by a cable at C. Determine the components of the reactions at supports A and B and the force in the cable at C.

SOLUTION

:AΣ =M 0 ( ) ( )14 8 y zB B− + × +i k j k

( ) ( )7 9 4 500+ − + + × −i j k k

( ) ( )7 18 8 CT+ − + + ×i j k k

( ) ( )14 18 8 200+ − + + × − =i j k k 0


255

:x 8 18 8100 0y CB T− + − = 450 lbCT = ...............................................Ans.

:y 14 7 6300 0z CB T+ − = 225 lbzB = ...............................................Ans.

:z 14 0yB− = 0 lbyB = ....................................................Ans.

0 :xFΣ = 0xA = 0 lbxA = ....................................................Ans.

0 :yFΣ = 350 0y yA B+ + = 350 lbyA = − ............................................Ans.

0 :zFΣ = 700 0z z CA B T+ + − = 25 lbzA = ..................................................Ans.

6-158 The plate shown in Fig. P6-158 has a mass of 80 kg. The brackets at supports A and B exert only force reactions on the plate. Each of the brackets can resist a force along the axis of the pins in one direction only. Determine

(a) The reactions at supports A and B and the tension in the cable. (b) The change in length of the cable. (Use E = 200 GPa and d = 2.5 mm.) SOLUTION

(a) ( )80 9.81 784.800 NW = =

1.4cos30 1.21244 my∆ = ° =

1.2 1.4sin 30 1.9000 mz∆ = + ° =

2 2 21 1.21244 1.9000 2.46577 mCDL = + + =

( )2 2 2

1 1.21244 1.90001 1.21244 1.9000

0.40555 0.49171 0.77055 N

C C

C C C

T

T T T

− − +=+ +

= − − +

i j kT

i j k

From a free-body diagram of the plate, the equilibrium equations give

:BΣ =M 0 ( ) ( ) ( )1 1.2 2C y zA A+ × + × +i k T i j k

( ) ( )1 0.60622 0.35 784.8+ + − × − =i j k k 0

:x 0.59005 475.761 0CT − = 806.307 NCT =

:y 1.25721 2 784.800 0C zT A− − + = 114.449 NzA = −

:z 0.49171 2 0C yT A− + = 198.235 NyA =

0 :xFΣ = ( )0.40555 806.307 0xB − = 326.998 NxB =

0 :yFΣ = ( )0.49171 806.307 0y yA B+ − = 198.234 NyB =

0 :zFΣ = ( )0.77055 806.307 784.800 0z zA B+ + − = 277.949 NzB =

( )198.2 114.4 N= −A j k ...................... 806 NCT = ................................................Ans.

( )327 198.2 278 N= + +B i j k ...................................................................................Ans.

(b) The stretch of the cable CD is given by

( ) ( )

( ) ( )29

806.307 2465.772.03 mm

200 10 0.0025 4CD

PLEA

δπ

= = = ×

...........................................Ans.


256

6-159 The door to an airplane hangar consists of two uniform sections that are hinged at the middle as shown in Fig. P6-159. The door is raised by means of a cable attached to a bar along the bottom edge of the door. Smooth rollers at the ends of the bar C run in a smooth vertical channel. If the door is 30 ft wide, 15 ft tall, and weighs 1620 lb, determine

(a) The force P when the door opening height h = 8 ft. (b) The resultant hinge forces at A and B when h = 8 ft. SOLUTION

( )1sin 3.5 7.5 27.818θ −= = °

(a) From a free-body diagram of the door AB, the equilibrium equations are

0 :xF→Σ = 0x xA B− =

0 :yF↑Σ = 810 0y yA B− − =

0 :AMΣ =� ( ) ( ) ( )810 3.75cos 7.5cos 7.5sin 0y xB Bθ θ θ+ − =

From a free-body diagram of the door AB, the equilibrium equations are

0 :xF→Σ = 0x CB N− =

0 :yF↑Σ = 810 0yB P+ − =

0 :CMΣ =� ( ) ( ) ( )810 3.75cos 7.5cos 7.5sin 0y xB Bθ θ θ− − =

Combining the two moment equations gives

767.562 lbxB = 0 lbyB =

Then the force equations give

767.562 lbxA = 810 lbyA =

767.562 lbCN =

810 lbP = ........................................................................................................................ Ans.

(b) 2 2 1116 lbx yA A A= + = ...............................................................................................Ans.

2 2 768 lbx yB B B= + = .................................................................................................Ans.

6-160 Three bars are connected by smooth frictionless pins as shown in Fig. P6-160. Bar BCDE is rigid, bar AB is aluminum (E = 73 GPa; L = 1.5 m; d = 30 mm) and bar EF is steel (E = 200 GPa; L = 1.2 m; d = 20 mm). The 25-mm-diameter pivot pin D is aluminum and is in double shear. When P = 100 kN determine

(a) The deformation of bars AB and EF. (b) The shearing stress on a cross section of the pin at D. SOLUTION (a) From a free-body diagram of member BCDE,


0 :xF→Σ = 100 0EF x ABT D F− + − =

0 :yF↑Σ = 0yD =

0 :DMΣ =� ( )600 300 300 100 0AB EFF T+ − =

in which EFT is a tensile force and ABF is a compressive force. The stretch of EF and the shrink of AB are related by


257

600 300AB EFδ δ= 2AB EFδ δ=

( )

( ) ( )( )

( ) ( )2 29 9

1.5 1.22

73 10 0.03 4 200 10 0.02 4AB EFF T

π π=

× ×

1.31400AB EFF T=

Combining the equilibrium equations and the deformation equation gives

36.2183 kNABF = 27.5634 kNEFT =

91.3451 kNxD = 0 kNyD =

The deformation of the two rods are

( )( )

( ) ( )

3

29

36.2183 10 15001.053 mm (shrink)

73 10 0.03 4ABδ

π

×= =

×

............................................Ans.

( )( )

( ) ( )

3

29

27.5634 10 12000.526 mm (stretch)

200 10 0.02 4EFδ

π

×= =

×

........................................Ans.

(b) Finally, dividing the force on the pin by twice its cross sectional are (since it is in double shear) gives the shear stress

( )

36 2

2

91.3451 10 93.0 10 N/m 93.0 MPa2 2 0.025 4

Ds

DA

τπ

×= = = × =

............................Ans.

6-161 A person is holding a 20-lb object as shown in Fig. P6-161. Determine the force T in the biceps muscle and the force F of the humerus against the ulna, in terms of the weight W of the forearm which acts through G. For the position shown, both T and F act vertically.

SOLUTION From a free-body diagram of the forearm, the equilibrium equations give

0 :yF↑Σ = 20 0T F W− − − =

0 :FMΣ =� ( )1.5 5.5 11.5 20 0T W− − =

3.667 153.33 lbT W= + ................................Ans.

2.667 133.33 lbF W= + ...............................Ans.

6-162 Bodies A and B shown in Fig. P6-162 have masses of 2200 kg and 450 kg, respectively. The coefficient of friction between B and the horizontal surface is 0.2, between A and B it is 0.2, and between A and the fixed surface is 0.5. Determine the force P required to cause impending motion of body B to the left.

SOLUTION

( )2200 9.81 21,582.0 NAW = =

( )450 9.81 4414.5 NBW = =

For impending motion of block B to the left, the friction forces must act to the right to oppose the motion and

0.2B s B BF N Nµ= =

From a free-body diagram for block B, the equilibrium equations are


258

0 :xF→Σ = 0A BF F P+ − =

0 :yF↑Σ = 4414.5 0B AN N− − =

From a free-body diagram for block A, the equilibrium equations are

0 :xF→Σ = 0C AF F− =

0 :yF↑Σ = 21,582 0C AN N+ − =

0 :AMΣ =� ( ) ( )21,582 300 1500 0CN− =

Therefore

4316.4 NCN = 17, 265.6 NAN = A CF F=

21,680.1 NBN = ( )0.2 21,680.1 4336.02 NBF = =

0.2A s A AF N Nµ≤ = 0.5C s C CF N Nµ≤ =

Guessing that impending slip occurs at C before A gives

( )0.5 4316.4 2158.2 NC AF F= = =

6490 NA BP F F= + = ...................................................................................................Ans.

Since the maximum amount of friction available at A ( )max 0.2 17, 265.6 3453.12 NAF = = is greater than

the amount of friction required ( )2158.2 NAF = , the guess that slip occurs at C first is correct.

6-163 Three bars are connected by smooth frictionless pins as shown in Fig. P6-163. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel and is in single shear. If the temperature drops 20°F after the unit is assembled, determine

(a) The normal stresses in bars AC and DE. (b) The change in length of bars AC and DE. (c) The shearing stress on the cross section of the pin at B. SOLUTION

(a) From a free-body diagram of bar BCD, the equilibrium equations give

0 :xF→Σ = 0xB =

0 :yF↑Σ = 0y AC DEB T T+ − =

0 :AMΣ =� 20 30 0AC DET T− =

1.5AC DET T=

0xB = y DE ACB T T= −

in which ACT and DET are both tensile forces. The stretch of bar DE and the shrink of bar AC are related by

30 20

ACDE δδ =

1.5DE ACδ δ=


259

( )

( ) ( )( ) ( )( )6

26

306.6 10 20 30

30 10 .05 4DET

π−+ × −

×

( )

( ) ( )( )( )( )6

26

361.5 12.5 10 20 36

12 10 1 4ACT

π−

− = − × − ×

5.09296 5.72958 9540.0 lbDE ACT T+ =

Combining the equilibrium equations and the deformation equation gives

1045.493 lb (T)ACT = 696.995 lb (T)DET = 348.498 lbyB = −

( )2

1045.493 1331 psi (T)1 4

ACAC

TA

σπ

= = = ......................................................................Ans.

( )2

696.995 3550 psi (T)0.5 4

DEσπ

= = ...............................................................................Ans.

(b) ( )( )

( ) ( )( )( )( )6

26

1045.493 3612.5 10 20 36

12 10 1 4AC

TL T LEA

δ απ

−−−= − ∆ = − × − ×

0.00501 in. (shrink) ..............................................................................................Ans.

( ) ( )

( ) ( )( ) ( )( )6

26

696.995 306.6 10 20 30

30 10 0.5 4DE

TL T LEA

δ απ

−= + ∆ = + × − ×

0.00751 in. (stretch) .............................................................................................Ans.

(c) ( )

2 2

2348.498 1136 psi

5 8 4x y

Bs

B BA

τπ

+= = = ..................................................................Ans.

6-164 An adjustable bracket is held in place by a force P of magnitude 175 N as shown in Fig. P6-164. Determine the minimum coefficient of friction between the bracket and vertical square member if the bracket is to stay in place.

SOLUTION

From a free-body diagram of the bracket, the equilibrium equations are

0 :xF→Σ = 0B AN N− =

0 :yF↑Σ = 175 0A BF F+ − =

0 :AMΣ =� ( )75 300 417.5 175 0B BF N+ − =

and for impending slip at both A and B

A s AF Nµ= B s BF Nµ=

Therefore

A BN N= 87.5 NA BF F= = 221.667 NA BN N= =

87.5 0.395

221.667A

sA

FN

µ = = = ........................................................................................Ans.


260

6-165 Determine the force in members CD, DE, and FG of the truss shown in Fig. P6-165. SOLUTION

sin 3 5θ = cos 4 5θ =

( )1tan 5 20 14.036φ −= = °

From a free-body diagram of the entire truss, the equilibrium equations give

0 :xF→Σ = 800sin 600sin 0xG A θ θ− − − =

0 :yF↑Σ = 800cos 600cos 500 0yA θ θ− − − =

0 :AMΣ =� ( ) ( ) ( ) ( )10 800 10 600 17.5 500 20 0G − − − =

2850 lbG = 2010 lbxA = 1620 lbyA =

2 447.673 NN =

From a free-body diagram of the joint A, the equilibrium equations give

0 :xF→Σ = cos 2010 0ABT θ − =

0 :yF↑Σ = 1620 sin 0AG ABT T θ− − =

2512.50 lb (T)ABT =

112.50 lb (T)AGT =

From a free-body diagram of the joint G, the equilibrium equations give

0 :xF→Σ = 2850 cos cos 0BG FGT Tβ φ+ + =

0 :yF↑Σ = sin sin 0AG BG FGT T Tβ φ+ − =

( )1tan 4 8 26.565β −= = °

1229.83 lb 1230 lb (C)BGT = − ≅

1803.86 lb 1804 lb (C)FGT = − ≅ ................................................... Ans.

Finally, from a free-body diagram of the joint D, the equilibrium equations give

0 :xF→Σ = cos cos 0CD DET Tθ φ− − =

0 :yF↑Σ = sin sin 500 0CD DET Tθ φ+ − =

1250 lb (T)CDT = ............................................................................... Ans.

1031 lb 1031 lb (C)DET = − = ......................................................................................Ans.

6-166 Determine the change in length of members CF and FG of the truss shown in Fig. P6-166. Each member of the truss is made of structural steel, and is circular with a diameter of 25 mm.

SOLUTION structural steel: 200 GPaE =

4 mCF EF FGL L L= = =

4cos30 3.46410 mDEL = ° =


261

From a free-body diagram of the upper portion of the truss, the equilibrium equations give

0 :EMΣ =� ( ) ( )1200 3.46410 3.46410 0CFT− =

1200 NCFT =

0 :CMΣ =� ( ) ( )1200 3.46410 300 6.92820− −

( )( )sin 30 6.92820 0FGT− ° =

1800 NFGT = −

( )( )

( ) ( )29

1200 40000.0489 mm

200 10 0.025 4CF

TLEA

δπ

= = = ×

........................................Ans.

( )( )

( ) ( )29

1800 40000.0733 mm

200 10 0.025 4FGδ

π

−= = −

×

.................................................Ans.

6-167 The band brake of Fig. P6-167 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is 0.25, and the weight of the handle is negligible. Determine the force P that must be applied to the end of the handle to resist a maximum torque of 200 lb-in. if the torque is applied

(a) Clockwise. (b) Counterclockwise. SOLUTION From a free-body diagram of the drum, the moment equilibrium equation gives

0 :CMΣ =� 4 4 0A BT T M− − =

0.25A BT T M− =

From a free-body diagram of the brake handle, the moment equilibrium equation gives

0 :CMΣ =� ( )3 cos30 15 0AT P° − =

0.2 cos30AP T= °

270 3 2 radβ π= ° =

(a) If 200 lb in.M = + ⋅ (the drum rotates clockwise), then the tension AT must be bigger than BT and the belt friction equation gives

( )0.25 3 2 3.24819A B BT T e Tπ= =

72.240 lbAT = 22.240 lbBT =

12.51 lbP = ...................................................................................................................... Ans.

(b) If 200 lb in.M = − ⋅ (the drum rotates counterclockwise), then the tension BT must be bigger than AT and the belt friction equation gives

( )0.25 3 2 3.24819B A AT T e Tπ= = 22.240 lbAT = 72.240 lbBT =

3.85 lbP = ....................................................................................................................... Ans.


262

Chapter 7 7-1 For the steel shaft shown in Fig. P7-1 (a) Determine the torques transmitted by cross-sections in intervals AB, BC, CD, and DE of the shaft. (b) Draw a torque diagram for the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft

are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives

0 :axisMΣ = 80 0ABT − =

80 kip ftABT = ⋅ ........................................ Ans.

0 :axisMΣ = 80 100 0BCT − + =

20 kip ftBCT = − ⋅ ..................................... Ans.

0 :axisMΣ = 80 100 40 0CDT − + − =

20 kip ftCDT = ⋅ ........................................ Ans.

0 :axisMΣ = 80 100 40 25 0DET − + − − =

45 kip ftDET = ⋅ ........................................ Ans.

(b) A torque diagram for the shaft is shown below the free-body diagrams............................................. Ans.

7-2 For the steel shaft shown in Fig. P7-2 (a) Determine the maximum torque transmitted by any transverse cross section of the shaft. (b) Draw a torque diagram for the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft

are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives

0 :axisMΣ = 30 0ABT − =

30 kN mABT = ⋅

0 :axisMΣ = 30 40 0BCT − + =

10 kN mBCT = − ⋅

0 :axisMΣ = 30 40 15 0CDT − + − =

5 kN mCDT = ⋅

0 :axisMΣ = 30 40 15 5 0DET − + − − =

10 kN mDET = ⋅

max 30 kN mABT T= = ⋅ ..........................................................................................................Ans.



263

7-3 For the steel shaft shown in Fig. P7-3 (a) Determine the maximum torque transmitted by any transverse cross section of the shaft. (b) Draw a torque diagram for the shaft. SOLUTION

0 :axisMΣ = 10 15 15 20 0T− − + + − = 30 kip ftT = ⋅

(a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives

0 :axisMΣ = 10 0ABT − =

10 kip ftABT = ⋅

0 :axisMΣ = 10 15 0BCT − − =

25 kip ftBCT = ⋅

0 :axisMΣ = 10 15 30 0CDT − − + =

5 kip ftCDT = − ⋅

0 :axisMΣ = 10 15 30 15 0DET − − + + =

20 kip ftDET = − ⋅

max 25 kip ftBCT T= = ⋅ ...........................................................................................................Ans.


7-4 The motor shown in Fig. P7-4 supplies a torque of 500 N-m to the 25-mm-diameter shaft BCDE. The torques removed at gears C, D, and E are 100 N-m, 150 N-m, and 250 N-m, respectively.

(a) Determine the torques transmitted by cross sections in intervals BC, CD, and DE of the shaft. (b) Draw a torque diagram for the shaft. (c) Determine the maximum shearing stress in the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of

sections in the intervals BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives

0 :axisMΣ = 500 0BCT − =

500 N mBCT = ⋅ ......................................................Ans.

0 :axisMΣ = 500 100 0CDT − + =

400 N mCDT = ⋅ ......................................................Ans.

0 :axisMΣ = 500 100 150 0DET − + + =

250 N mDET = ⋅ ......................................................Ans.


(c) ( )( )

( )6 2

max 4

500 0.0125163.0 10 N/m 163.0 MPa

0.0125 2TcJ

τπ

= = = × = ..................................Ans.


264

7-5 A solid circular steel shaft 1-in. in diameter is subjected to a torque of 9000 lb-in. The modulus of rigidity (shear modulus) for the steel is 12,000 ksi. Determine

(a) The maximum shearing stress in the shaft. (b) The magnitude of the angle of twist in a 4-ft length. SOLUTION

(a) ( ) ( )

( )4

9000 0.545,837 psi 45.8 ksi

1 32TcJ

τπ

= = = ≅ .............................................................Ans.

(b) ( )( )

( ) ( )46

9000 4 120.367 rad

12 10 1 32TLGJ

θπ

×= = =

×

.................................................................Ans.

7-6 A hollow steel shaft has an outside diameter of 150 mm and an inside diameter of 100 mm. The shaft is subjected to a torque of 35 kN-m. The modulus of rigidity (shear modulus) for the steel is 80 GPa. Determine

(a) The shearing stress on a transverse cross section at the outside surface of the shaft. (b) The shearing stress on a transverse cross section at the inside surface of the shaft. (c) The magnitude of the angle of twist in a 2.5-m length. (d) The magnitude of the angle of twist in a 2.5-m long solid shaft that has the same weight as the hollow

shaft. SOLUTION

( ) ( )4 4 4 4 6 4 6 432 150 100 32 39.88 10 mm 39.88 10 mo iJ d dπ π −= − = − = × = ×

(a) ( ) ( )3

6 26

35 10 0.07565.82 10 N/m 65.8 MPa

39.88 10TcJ

τ −

×= = = × ≅

×....................................Ans.

(b) ( ) ( )3

6 26

35 10 0.05043.88 10 N/m 43.9 MPa

39.88 10TJρτ −

×= = = × ≅

×..................................Ans.

(c) ( )( )

( )( )3

9 6

35 10 2.50.0274 rad

80 10 39.88 10TLGJ

θ−

×= = =

× ×..........................................................Ans.

(d) ( )2 2 2150 100 4 4A dπ π= − = 111.80 mmd =

( )44 6 4 6 432 111.80 32 15.338 10 mm 15.338 10 mJ dπ π −= = = × = ×

( )( )

( )( )3

9 6

35 10 2.50.0713 rad

80 10 15.338 10TLGJ

θ−

×= = =

× ×.........................................................Ans.

7-7 A torque of 30,000 lb-in. is supplied to the 3-in. diameter factory drive shaft of Fig. P7-7 by a belt that drives pulley A. A torque of 18,000 lb-in. is taken off by pulley B and the remainder by pulley C. Shafts AB and BC are 5 ft and 3.75 ft long, respectively. Both shafts are made of steel (G = 12,000 ksi). Determine

(a) The maximum shearing stress in each of the shafts. (b) The magnitude of the angle of twist of pulley B with respect to pulley A. (c) The magnitude of the angle of twist of pulley C with respect to pulley A. SOLUTION

( )44 432 3 32 7.952 inJ dπ π= = =


265

30,000 lb in.ABT = ⋅

30,000 18,000 12,000 lb in.BCT = − = ⋅

(a) ( )( )30,000 1.5

7.952ABTcJ

τ = =

5659 psi 5.66 ksi= ≅ ....................................... Ans.

( ) ( )12,000 1.5

2264 psi 2.26 ksi7.952BCτ = = ≅ ....................................................................Ans.

(b) ( ) ( )( )( )/ 6

30,000 5 120.018863 rad 0.01886 rad

12 10 7.952B ATLGJ

θ×

= = = =×

................................Ans.

(c) ( )( )( )( )/ 6

12,000 3.75 120.005659 rad

12 10 7.952C Bθ×

= =×

/ / / 0.018863 0.005659 0.0245 radC A B A C Bθ θ θ= + = + = .............................................Ans.

7-8 A torque of 10 kN-m is supplied to the steel (G = 80 GPa) factory drive shaft of Fig. P7-8 by a belt that drives pulley A. A torque of 6 kN-m is taken off by pulley B and the remainder by pulley C. Shafts AB and AC are 2.25 m and 1.60 m long, respectively. If the diameter of shaft AB is 80 mm and the diameter of shaft AC is 65 mm, determine

(a) The maximum shearing stress in each of the shafts. (b) The angle of twist of pulley B with respect to pulley A. (c) The angle of twist of pulley C with respect to pulley B. SOLUTION

( )44 6 432 80 32 4.021 10 mmABJ dπ π= = = ×

( )4 6 465 32 1.7525 10 mmACJ π= = ×

6 kN mABT = ⋅

6 10 4 kN mACT = − = − ⋅

(a) ( )( )3

6 26

6 10 0.04059.69 10 N/m 59.7 MPa

4.021 10ABTcJ

τ −

×= = = × ≅

×..................................Ans.

( )( )3

6 26

4 10 0.032574.18 10 N/m 74.2 MPa

1.7525 10ACτ −

×= = × ≅

×..........................................Ans.

(b) ( )( )

( )( )3

/ 9 6

6 10 2.250.04197 rad 0.0420 rad

80 10 4.021 10A BTLGJ

θ−

×= = = ≅

× ×.........................Ans.

(c) ( )( )

( )( )3

/ 9 6

4 10 1.600.04565 rad 0.0456 rad

80 10 1.7525 10C ATLGJ

θ−

− ×= = = − ≅ −

× ×

( ) ( )/ / / 0.04565 0.04197 0.00368 radC B C A A Bθ θ θ= + = − + = − ..................................Ans.


266

7-9 A solid circular aluminum alloy (G = 4000 ksi) shaft with diameters of 2.5 in. and 1.75 in. is subjected to a torque T, as shown in Fig. P7-9. The shearing stress is limited to 8000 psi, and the angle of twist in the 7-ft length cannot exceed 0.04 rad. Determine the maximum permissible value of T.

SOLUTION

( )44 432 2.5 32 3.835 inABJ dπ π= = = ( )4 41.75 32 0.9208 inBCJ π= =

( )

max

0.8758000 psi

0.9208BC

TTcJ

τ τ= = = ≤ 8419 lb in.T ≤ ⋅

( )

( )( )( )

( )( )6 6

3 12 4 120.04 rad

4 10 3.835 4 10 0.9208T TTL

GJθ

× ×= = + ≤

× ×∑

2601 lb in.T ≤ ⋅

max 2601 lb in. 2.60 kip in.T = ⋅ ≅ ⋅ .......................................................................................Ans.

7-10 The solid circular steel (G = 80 GPa) shaft of Fig. P7-10 has a diameter of 80 mm. If the gears are spaced at 1.25-m intervals, determine

(a) The maximum shearing stress in the shaft. (b) The rotation of a section at D with respect to a section at B. (c) The rotation of a section at E with respect to a section at A. SOLUTION

0 :axisMΣ = 8 15 6 14 0T− + − − + = 15 kN mT = ⋅

( )

4

4 6 4

32

80 32 4.021 10 mm

J dπ

π

=

= = ×

(a) max 14 kN mDET T= = ⋅

( )( )3

6 26

14 10 0.040139.27 10 N/m 139.3 MPa

4.021 10DETcJ

τ −

×= = = × ≅

×............................Ans.

(b) / / /D B D C C BCD BC

TL TLGJ GJ

θ θ θ = + = +

( ) ( )

( ) ( )( )( )

( )( )3 3

9 6 9 6

8 10 1.25 7 10 1.25

80 10 4.021 10 80 10 4.021 10− −

× − ×= +

× × × ×

0.00389 rad= ................................................................................................................Ans.

(c) /E ATLGJ

θ θ= =∑ ∑

( ) ( )

( ) ( )( )( )

( )( )3 3

9 6 9 6

8 10 1.25 7 10 1.25

80 10 4.021 10 80 10 4.021 10− −

× − ×= +

× × × ×

( )( )

( )( )( )( )

( ) ( )3 3

9 6 9 6

8 10 1.25 14 10 1.25

80 10 4.021 10 80 10 4.021 10− −

× ×+ +

× × × ×

0.0894 rad= .................................................................................................................. Ans.


267

7-11 The shaft shown in Fig. P7-11 consists of a brass (G = 5600 ksi) tube AB that is securely connected to a solid stainless steel (G = 12,500 ksi) bar BC. Tube AB has an outside diameter of 5 in. and an inside diameter of 2.5 in. Bar BC has a diameter of 3.5 in. Torques T1 and T2 are 100 kip-in. and 40 kip-in., respectively, in the directions shown. Determine

(a) The maximum shearing stress in the shaft. (b) The rotation of a section at C with respect to its no-load position. SOLUTION

( )4 4 45 2.5 32 57.52 inABJ π= − =

( )4 43.5 32 14.732 inBCJ π= =

100 40 60 kip in.ABT = − = ⋅

40 kip in.BCT = − ⋅

(a) ( )( )60 2.5

2.608 ksi57.52AB

TcJ

τ = = =

( )( )

max

40 1.754.752 ksi

14.732BCτ τ= = = .................................................................................Ans.

(b) /C ATLGJ

θ θ= =∑ ∑

( )( )

( ) ( )( ) ( )

( )( )60 2 12 40 3 12

0.00335 rad5600 57.52 12,500 14.732

× − ×= + = − ......................................Ans.

7-12 A motor supplies a torque of 5.5 kN-m to the constant diameter steel (G = 80 GPa) line shaft shown in Fig. P7-12. Three machines are driven by gears B, C, and D on the shaft and they require torques of 3 kN-m, 1.5 kN-m, and 1 kN-m, respectively. Determine

(a) The minimum diameter required if the maximum shearing stress in the shaft is limited to 100 MPa. (b) The rotation of gear D with respect to the coupling at A if the coupling and gears are spaced at 2-m

intervals and the shaft diameter is 75 mm. SOLUTION

5.5 kN mABT = ⋅

5.5 3 2.5 kN mBCT = − = ⋅

5.5 3 1.5 1 kN mCDT = − − = ⋅

(a) ( )( )3

6 2maxmax 4

5.5 10 2100 10 N/m

32dT c

J dτ

π×

= = = ×

( )

( )3

33min 6

16 5.5 1065.43 10 m 65.4 mm

100 10d

π−

×= = × =

×.....................................................Ans.

(b) ( )4 6 4 6 475 32 3.106 10 mm 3.106 10 mJ π −= = × = ×

/D ATLGJ

θ θ= =∑ ∑


268

( )( ) ( )( ) ( )( )

( )( )3 3 3

9 6

5.5 10 2 2.5 10 2 1 10 20.0724 rad

80 10 3.106 10−

× + × + ×= =

× ×...........................Ans.

7-13 The hollow circular steel (G = 12,000 ksi) shaft of Fig. P7-13 is in equilibrium under the torques indicated. Determine

(a) The maximum shearing stress in the shaft. (b) The rotation of a section at D with respect to a section at A. SOLUTION

( )4 4 44 2 32 23.56 inJ π= − =

0 :axisMΣ = 9 21 10 0Q− + − + =

20 kip ftQ = ⋅

9 kip ftABT = ⋅ 9 20 11 kip ftBCT = − = − ⋅ 9 20 21 10 kip ftCDT = − + = ⋅

(a) max 11 kip ftBCT T= = ⋅

( ) ( )

max

11 12 211.21 ksi

23.56BCTcJ

τ τ×

= = = = ......................................................................Ans.

(b) /D ATLGJ

θ θ= =∑ ∑

( ) ( )( )( )

( ) ( )( )( )

( )( )( ) ( )

10 12 4 12 11 12 5 12 9 12 3 1212,000 23.56 12,000 23.56 12,000 23.56

× × − × × × ×= + +

0.00611 rad= ................................................................................................................Ans.

7-14 The solid circular shaft and the hollow tube shown in Fig. P7-14 are both attached to a rigid circular plate at their left ends. A torque TA = 2 kN-m applied to the right end of the shaft is resisted by a torque TB at the right end of the tube. The shaft is made of steel (G = 80 GPa) and the tube is made of an aluminum alloy (G = 28 GPa). If the shaft has a diameter of 50 mm and the tube has an outside diameter of 80 mm, determine

(a) The maximum inside diameter that can be used for the tube if the maximum shearing stress in the tube must be limited to 50 MPa.

(b) The maximum inside diameter that can be used for the tube if the rotation of the right end of the shaft with respect to the right end of the tube must be limited to 0.25 rad.

SOLUTION

(a) ( )( )3

6 22 10 0.040

50 10 N/mB

TcJ J

τ×

= = = ×

( )6 4 6 4 4 41.600 10 m 1.600 10 mm 80 32B iJ dπ−= × = × = −

70.47 mm 70.5 mmid = ≅ .................................................................................................Ans.

(b) ( )4 6 4 6 450 32 0.6136 10 mm 0.6136 10 mAJ π −= = × = ×

( )( )

( )( )( )( )( )

3 3

/ 9 6 9

2 10 3.5 2 10 2.50.25 rad

80 10 0.6136 10 28 10A BB

TLGJ J

θ θ−

× ×= = = + =

× × ×∑ ∑

( )6 4 6 4 4 41.6627 10 m 1.6627 10 mm 80 32B iJ dπ−= × = × = −

70.01 mm 70.0 mmid = ≅ ..................................................................................................Ans.


269

7-15 The 4-in. diameter shaft shown in Fig. P7-15 is composed of brass (G = 5000 ksi) and steel (G = 12,000 ksi) sections rigidly connected. Determine the maximum allowable torque, applied as shown, if the maximum shearing stresses in the brass and the steel are not to exceed 7500 psi and 10,000 psi, respectively, and the distance AC, through which the end of the 10-in. pointer AB moves, is not to exceed 0.60 in.

SOLUTION

( )4 44 32 25.13 inJ π= =

( )2

10,000 psi25.13S

TTcJ

τ = = ≤ 125,650 lb in.T ≤ ⋅

( )2

7500 psi25.13B

TTcJ

τ = = ≤ 94, 238 lb in.T ≤ ⋅

( )

( )( )( )

( )( )6 6

6 12 4 12 0.60 0.060 rad105 10 25.13 12 10 25.13B S

T T sr

θ θ θ× ×

= + = + = ≤ =× ×

81,946 lb in.T ≤ ⋅

max 81,946 lb in. 81.9 kip in.T = ⋅ ≅ ⋅ ...................................................................................Ans.

7-16 A stepped steel (G = 80 GPa) shaft has the dimensions and is subjected to the torques shown in Fig. P7-16. Determine

(a) The maximum shearing stress on a section 3 m from the left end of the shaft. (b) The rotation of the section at the right end of the shaft with respect to its no-load position. SOLUTION

( )4 6 4160 32 64.34 10 mmACJ π= = ×

( )4 6 4100 32 9.817 10 mmCDJ π= = ×

( )4 6 450 32 0.6136 10 mmDEJ π= = ×

0 :axisMΣ = 35 100 20 5 0AT − + − − = 40 kN mAT = − ⋅

40 kN mABT = ⋅

40 35 75 kN mBCT = + = ⋅

40 35 100 25 kN mCDT = + − = − ⋅

40 35 100 20 5 kN mDET = + − + = − ⋅

(a) ( )( )3

6 23 6

25 10 0.050127.33 10 N/m 127.3 MPa

9.817 10TcJ

τ −

×= = = × ≅

×..............................Ans.

(b) ( )( )

( )( )( ) ( )

( )( )3 3

/ 9 6 9 6

40 10 1 75 10 1.5

80 10 64.34 10 80 10 64.34 10E ATLGJ

θ− −

× ×= = +

× × × ×∑

( )( )

( )( )( )( )

( ) ( )3 3

9 6 9 6

25 10 2 5 10 0.5

80 10 9.817 10 80 10 0.6136 10− −

− × − ×+ +

× × × ×

0.0850 rad= − ............................................................................................Ans.


270

7-17 A torque T is applied to the right end of shaft AB of Fig. P7-17. The mean diameter of bevel gear C is twice that of bevel gear B. Both shafts are made of steel (G = 12,000 ksi). Shaft AB has a diameter of 1.5 in., and shaft CD has a diameter of 2.0 in. If the maximum shearing stress in either shaft must not exceed 15 ksi, determine

(a) The maximum permissible torque T. (b) The rotation of a section at A relative to its no-load position. SOLUTION

( )4 41.5 32 0.4970 inABJ π= = ( )4 42.0 32 1.5708 inCDJ π= =

(a) ( )0.75

15,000 psi0.4970AB

AB

TTcJ

τ = = ≤ 9940 lb in.ABT ≤ ⋅

( )1

15,000 psi1.5708CD

CD

TTcJ

τ = = ≤ 2 23,562 lb in.CD ABT T= ≤ ⋅

11,781 lb in.ABT ≤ ⋅

max 9940 lb in. 9.94 kip in.ABT T= = ⋅ ≅ ⋅ ...........................................................................Ans.

(b) ( )2 2 9940 19,880 lb in.CD ABT T= = = ⋅

( ) ( )

( )( )( )( )

( )( )/ / 6 6

9940 4 12 2 19,880 3 122

12 10 0.4970 12 10 1.5708A A B C Dθ θ θ× ×

= + = +× ×

0.1559 rad= ...................................................................................................................... Ans.

7-18 Torque is applied to the steel (G = 80 GPa) shaft shown in Fig. P7-18 through gear C and is removed through gears A and B. If the torque applied to gear C by the motor is 20 kN-m and the torque removed through gear B is 12 kN-m, determine

(a) The minimum permissible diameter for each section of the shaft if the maximum shearing stresses must not exceed 125 MPa.

(b) The minimum permissible uniform diameter for a shaft with L1 = 1.5 m and L2 = 1.25 m if the rotation of gear A relative to gear C must be less than 0.15 rad.

SOLUTION

20 kN mBCT = ⋅ 20 12 8 kN mABT = − = ⋅

(a) ( )

4 3

2 1632

T dTc TJ d d

τπ π

= = = 316Tdπτ

=

( )

( )3 6

16 80000.0688 m 68.8 mm

125 10ABd π= = =

×...............................................................Ans.

( )

( )3 6

16 20,0000.0934 m 93.4 mm

125 10BCd π= = =

×...............................................................Ans.

(b) ( ) 44

3232

TL TL TLGJ G dG d

θππ

= = =

( ) ( )

( )( )( )( )

( )( )3 3

/ / / 9 4 9 4

32 8 10 1.5 32 20 10 1.250.15 rad

80 10 80 10A C A B B C d dθ θ θ

π π× ×

= + = + =× ×

0.0749 m 74.9 mmd = = .............................................................................................Ans.


271

7-19 A torque of 40 lb-ft is applied through gear A to the left end of the gear train shown in Fig. P7-19. The diameters of gears B and C are 5 in. and 2 in., respectively. If the maximum shearing stresses in the aluminum alloy (G = 3800 ksi) shafts AB and CD are limited to 15 ksi, determine

(a) The minimum permissible diameter for shaft AB. (b) The minimum permissible diameter for shaft CD. (c) The maximum length for shaft CD if the rotation of a section at D with respect to a section at C must not

exceed 0.5 rad. SOLUTION

40 lb ftABT = ⋅ ( )2 5 16 lb ftCD ABT T= = ⋅

(a) ( )

4 3

2 1632

T dTc TJ d d

τπ π

= = = 316Tdπτ

=

( )( )

316 40 12

0.5462 in. 0.546 in.15,000ABd π

×= = ≅ .................................................................Ans.

(b) ( )( )

316 16 12

0.40246 in. 0.402 in.15,000CDd π

×= = ≅ ...............................................................Ans.

(c) ( ) ( )

( )( )/ 6

15,000 120.5 rad

3.8 10 0.40246 2D C

LLGcτθ

×= = =

×

2.12 ftL = ........................................................................................................................ Ans.

7-20 The motor shown in Fig. P7-20 supplies a torque of 45 kN-m to shaft AB. Two machines are powered by gears D and E. The torque delivered by gear E to the machine is 8 kN-m. Shafts AB and DCE are made of steel (G = 80 GPa) and have 150-mm and 80-mm diameters, respectively. If the diameters of gears B and C are 450 mm and 150 mm, respectively, determine

(a) The maximum shearing stress in shaft AB. (b) The maximum shearing stress in shaft DCE. (c) The rotation of gear E relative to gear D. SOLUTION

( )4 6 4150 32 49.70 10 mmABJ π= = × ( )4 6 480 32 4.021 10 mmDCEJ π= = ×

( ) ( )( )1 3 1 3 45 15 kN mC BT T= = = ⋅ 8 kN mCET = ⋅

15 8 7 kN mCDT = − = ⋅

(a) ( )( )3

6 26

45 10 0.07567.91 10 N/m 67.9 MPa

49.70 10ABTcJ

τ −

×= = = × ≅

×................................Ans.

(b) ( )( )3

6 26

8 10 0.04079.58 10 N/m 79.6 MPa

4.021 10CEτ −

×= = × ≅

×.............................................Ans.

(c) ( )( )

( ) ( )( )( )

( ) ( )3 3

/ / / 9 6 9 6

8 10 2.5 7 10 1.5

80 10 4.021 10 80 10 4.021 10E D E C D Cθ θ θ− −

× ×= − = −

× × × ×

0.0295 rad= .................................................................................................................. Ans.


272

7-21 The hollow circular tapered shaft of Fig. P7-21 is subjected to a constant torque T. Determine the angle of twist in terms of T, L, G, and r.

SOLUTION

With the left end of the shaft at x L= : 1rxL

ρ = 2 2rxL

ρ =

( )4 4 4 4

1 24

152 32

r xJL

π ρ ρ π−= =

4

4 4

3215

T dx TL dxdGJ Gr x

θπ

= =

42 2

4 4 4

32 2815 45

L L

L L

TL dx TLdGr x Gr

θ θπ π

= = =∫ ∫ ...........................................................................Ans.

7-22 The hollow tapered shaft of Fig. P7-22 has a constant wall thickness t. Determine the angle of twist for a constant torque T in terms of T, L, G, t, and r. Not that when t is small, the approximate expression for the polar second moment of area (J = r2A, where A is the cross-sectional area of the shaft) may be used.

SOLUTION

With the left end of the shaft at x L= : rxL

ρ =

( )3 3 3

2 23

22 2 rx tr xJ A t tL L

πρ ρ πρ π = = = =

3

3 32T dx TL dxdGJ tGr x

θπ

= =

32 2

3 3 3

32 16

L L

L L

TL dx TLdtGr x Gtr

θ θπ π

= = =∫ ∫ ...........................................................................Ans.

7-23 The solid cylindrical shaft of Fig. P7-23 is subjected to a uniformly distributed torque of q lb-in. per inch of length. Determine, in terms of q, L, G, and c, the rotation of the left end caused by the applied torque.

SOLUTION

T qx= 4

2cJ π=

4

2T dx qx dxdGJ Gc

θπ

= =

2

4 40 0

2L Lq qLd x dxGc Gc

θ θπ π

= = =∫ ∫ ....................................................................................Ans.

7-24 The solid cylindrical shaft of Fig. P7-24 is subjected to a distributed torque that varies linearly from zero at the left end to q N-m per meter of length at the right end. Determine, in terms of q, L, G, and c, the rotation of the left end caused by the applied torque.

SOLUTION

2

2qxTL

= 4

2cJ π=

2

4

T dx qx dxdGJ GLc

θπ

= =

2

24 40 0 3

L Lq qLd x dxGLc Gc

θ θπ π

= = =∫ ∫ ..............................................................................Ans.


275

( ) 6 2

max 6

0.04580 10 N/m

5.828 10T

τ −= ≤ ××

10,361 N mT ≤ ⋅

max 9713 N m 9.71 kN mT = ⋅ ≅ ⋅ .........................................................................................Ans.

7-29 A cylindrical tube is fabricated by butt-welding 0.075-in. plate with a spiral seam, as shown in Fig. P7-29. A torque T is applied to the tube through a rigid plate. If the outside diameter of the tube is 1.50 in. and T = 1000 lb-in., determine

(a) The normal stress perpendicular to the weld and the shear stress parallel to the weld. (b) The maximum tensile and compressive stresses in the tube. SOLUTION

(a) ( )4 4 41.50 1.35 32 0.17092 inJ π= − =

( )( )1000 0.75

4388 psi0.17092xy

TcJ

τ−

= = = −

( ) ( ) ( )2 sin cos 2 4388 sin 55 cos 55n xyσ τ θ θ= = − − ° °

4123 psi 4.12 ksi (T)= ≅ ..............................................................................................Ans.

( ) ( ) ( ) ( )2 2 2 2cos sin 4388 cos 55 sin 55nt xyτ τ θ θ = − = − − ° − − °

1500.8 psi 1.501 ksi= ≅ ............................................................................................Ans.

(b) max max 4388 psi (T & C)TcJ

σ τ= = = ...............................................................................Ans.

7-30 The hollow circular steel (G = 80 GPa) shaft shown in Fig. P7-30 has an outside diameter of 120 mm and an inside diameter of 60 mm. Determine

(a) The maximum compressive stress in the shaft. (b) The maximum compressive stress in the shaft after the inside diameter is increased to 100 mm. SOLUTION

( )( )5000 1.5 7500 N mT = = ⋅

(a) ( )4 4 6 4 6 4120 60 32 19.085 10 mm 19.085 10 mJ π −= − = × = ×

( ) 6 2

max max 6

7500 0.06023.58 10 N/m 23.6 MPa (C)

19.085 10TcJ

σ τ −= = = = × ≅×

..................Ans.

(b) ( )4 4 6 4 6 4120 100 32 10.540 10 mm 10.540 10 mJ π −= − = × = ×

( ) 6 2

max max 6

7500 0.06042.69 10 N/m 42.7 MPa (C)

10.540 10TcJ

σ τ −= = = = × ≅×

..................Ans.

7-31 The motor shown in Fig. P7-31 develops a torque T = 1500 lb-ft; the output torques from gears C and D are equal. The mean diameters of gears A and B are 12 in. and 4 in., respectively. If the diameter of the motor shaft is 2 in. and the diameter of the power shaft is 1.25 in., determine

(a) The torque in the power shaft between gears B and C. (b) The torque in the power shaft between gears C and D. (c) The maximum tensile and compressive stresses in each shaft. SOLUTION

( ) ( ) ( )1 3 1 3 1500 500 lb ftB AT T= = = ⋅


276

( ) ( )( )1 2 1 2 500 250 lb ftC D BT T T= = = = ⋅

(a) 500 lb ftBC BT T= = ⋅ .......................................Ans.

(b) 500 250 250 lb ftCD B CT T T= − = − = ⋅ .......Ans.

(c) ( )4 41.25 32 0.2397 inPJ π= =

( )4 42 32 1.5708 inMJ π= =

( ) ( ) ( ) ( )max max

1500 12 1motor motor

1.5708TcJ

σ τ×

= = =

11,459 psi 11.46 ksi (T & C)= ≅ .............................................................Ans.

( ) ( ) ( ) ( )max max

500 12 0.625power power

0.2397TcJ

σ τ×

= = =

15,644 psi 15.64 ksi (T & C)= ≅ .............................................................Ans.

7-32 Five 600-mm diameter pulleys are keyed to a 40-mm diameter solid steel (G = 76 GPa) shaft, as shown in Fig. P7-32. The pulleys carry belts that are used to drive machinery in a factory. Belt tensions for normal operating conditions are indicated on the figure. Each segment of the shaft is 1.5 m long. Determine

(a) The maximum shearing stress in each segment of the shaft. (b) The maximum tensile and compressive stresses in the shaft. (c) The rotation of end E with respect to end A. SOLUTION

0 :axisMΣ = ( )( )2500 500 0.300 0ABT − − =

0 :axisMΣ = ( )( )600 2000 400 0.300 0BCT − + − =

0 :axisMΣ = ( ) ( )600 480 3000 600 0.300 0CDT − + − − =

0 :axisMΣ = ( ) ( )600 480 720 3000 600 0.300 0DET − + − + − =

600 N mABT = ⋅

120 N mBCT = ⋅

840 N mCDT = ⋅

120 N mDET = ⋅

( )4 6 440 32 0.2513 10 mmJ π= = ×

(a) ( ) 6 2

6

600 0.02047.8 10 N/m 47.8 MPa

0.2513 10ABTcJ

τ −= = = × =×

...........................................Ans.

( ) 6 2

6

120 0.0209.55 10 N/m 9.55 MPa

0.2513 10BCτ −= = × =×

.....................................................Ans.

( ) 6 2

6

840 0.02066.9 10 N/m 66.9 MPa

0.2513 10CDτ −= = × =×

.....................................................Ans.

( ) 6 2

6

120 0.0209.55 10 N/m 9.55 MPa

0.2513 10DEτ −= = × =×

.....................................................Ans.


277

(b) max max 66.9 MPa (T & C)CDσ τ τ= = =

(c) ( )

( )( )( )

( )( )/ 9 6 9 6

120 1.5 840 1.576 10 0.2513 10 76 10 0.2513 10E A

TLGJ

θ− −

= = +× × × ×∑

( )

( )( )( )

( ) ( )9 6 9 6

120 1.5 600 1.576 10 0.2513 10 76 10 0.2513 10− −

+ +× × × ×

0.1319 rad= ................................................................................................Ans.

7-33 When the two torques are applied to the steel (G = 12,000 ksi) shaft of Fig. P7-33, point A moves 0.172 in. in the direction indicated by torque T1. Determine

(a) The torque T1. (b) The maximum tensile stress in section BC of the shaft. (c) The maximum compressive stress in section CD of the shaft. SOLUTION

( )4 42 32 1.5708 inBCJ π= = ( )4 44 32 25.13 inCDJ π= =

(a) /0.172 0.02867 rad

6B Dθ = =

0 :axisMΣ = 19000 0LT T− + = 19000 lb ftLT T= − ⋅

( ) ( )( )( ) ( )

( ) ( )( )( )

1 1/ 6 6

9000 12 48 12 240.02867 rad

12 10 25.13 12 10 1.5708B D

T TTLGJ

θ− − ×

= = + =× ×∑

1 2668 lb ft 32,016 lb in. 32.0 kip in.T = ⋅ = ⋅ ≅ ⋅ .........................................................Ans.

(b) 1 32,016 lb in.BCT T= = ⋅

( )( )

max

32,016 120,382 psi 20.38 ksi (T)

1.5708TcJ

σ = = = ≅ ..............................................Ans.

(c) ( )9000 12 32,016 75,984 lb in.CD LT T= = × − = ⋅

( )( )

max

75,984 26047 psi 6.05 ksi (C)

25.13TcJ

σ = = = ≅ ...................................................Ans.

7-34 A solid circular stepped steel (G = 80 GPa) shaft has the dimensions and is subjected to the torques shown in Fig. P7-34. Determine

(a) The maximum tensile stress in section AB of the shaft. (b) The maximum compressive stress in section BC of the shaft. (c) The rotation of a section at C with respect to its no-load position. SOLUTION

(a) ( )4 6 4 6 4150 32 49.70 10 mm 49.70 10 mABJ π −= = × = ×

( )( )3

6 2max max 6

55 10 0.07583.00 10 N/m 83.0 MPa (T)

49.70 10TcJ

σ τ −

×= = = = × =

×...........Ans.

(b) ( )4 6 4 6 4100 32 9.817 10 mm 9.817 10 mBCJ π −= = × = ×


278

( )( )3

6 2max max 6

15 10 0.05076.40 10 N/m 76.4 MPa (C)

9.817 10TcJ

σ τ −

×= = = = × =

×...........Ans.

(c) ( )( )

( )( )( )( )

( ) ( )3 3

/ 9 6 9 6

55 10 0.300 15 10 0.400

80 10 49.70 10 80 10 9.817 10C ATLGJ

θ− −

× ×= = +

× × × ×∑

0.01179 rad= ..............................................................................................Ans.

7-35 A 175-lb man climbs a flight of stairs 12 ft high. Determine (a) The work done by the man. (b) The work done on the man by gravity. SOLUTION

(a) ( ) ( )175 12 2100 lb ftMU F h= ∆ = = ⋅ ..............................................................................Ans.

(b) ( ) ( )175 12 2100 lb ftGU W h= − ∆ = − = − ⋅ ......................................................................Ans.

7-36 A box with a mass of 600 kg is dragged up an incline 12 m long and 4 m high by using a cable that is parallel to the incline. The force in the cable is 2500 N. Determine the work done on the box

(a) By the cable. (b) By gravity. SOLUTION

(a) ( )2500 12 30,000 N m 30.0 kN m 30.0 kJCU Fd= = = ⋅ = ⋅ = ..................................Ans.

(b) ( ) ( )( )600 9.81 4 23,544 N m 23.5 kJGU W h= − ∆ = − × = − ⋅ ≅ − ..............................Ans.

7-37 A 100-lb block is pushed at constant speed for a distance of 20 ft along a level floor by a force F that makes an angle of 35° with the horizontal, as shown in Fig. P7-37. The coefficient of friction between the block and the floor is 0.35. Determine the work done on the block

(a) By the force P. (b) By gravity. (c) By the floor. SOLUTION

From a free-body diagram of the block (with 0.35F k N NA A Aµ= = ), the equilibrium equations give

0 :xF→Σ = cos35 0FP A° − =

0 :yF↑Σ = sin 35 100 0NA P− ° − =

56.60 lbP =

132.47 lbNA = ( )0.35 132.47 46.36 lbFA = =

(a) ( ) ( )( )cos 56.60cos35 20 927 lb ftPU P dα= = ° = ⋅ ....................................................Ans.

(b) ( ) ( )100 0 0 lb ftGU W h= ∆ = = ⋅ ........................................................................................Ans.

(c) ( ) ( ) ( )2 1 46.36 20 132.47 0 927 lb ftF F NU A d A y y= − + − = − + = − ⋅ ........................Ans.

7-38 A block with a mass of 100 kg slides down an inclined surface that is 5 m long and makes an angle of 25° with the horizontal. A man pushes horizontally on the block so that it slides down the incline at a constant speed. The coefficient of friction between the block and the inclined surface is 0.15. Determine the work done on the block

(a) By the man.


279

(b) By gravity. (c) By the inclined surface. SOLUTION

From a free-body diagram of the block (with 0.15F k N NA A Aµ= = ), the equilibrium equations give

0 :xF→Σ = sin 25 cos 25 0N FA A P° − ° − =

0 :yF↑Σ = ( )cos 25 sin 25 100 9.81 0N FA A° + ° − =

290.0 NP =

1011.7 NNA =

( )0.15 1011.7 151.76 NFA = =

(a) ( ) ( ) ( )cos 290.0cos 25 5 1314 N m 1314 JMU P dα= − = − ° = − ⋅ = − .....................Ans.

(b) ( ) ( )( )100 9.81 5sin 25 2073 N m 2073 JGU W h= ∆ = × ° = ⋅ = ..................................Ans.

(c) ( )151.76 5 758.8 N m 759 JS FU A d= − = − = − ⋅ ≅ − .....................................................Ans.

7-39 A constant force acting on a particle can be expressed in Cartesian vector form as

( )8 6 2 lb= − +F i j k . Determine the work done by the force on the particle if the displacement of the

particle can be expressed in Cartesian vector form as ( )5 4 6 ft= + +s i j k .

SOLUTION

( ) ( )8 6 2 5 4 6 28 lb ftU = ⋅ = − + ⋅ + + = ⋅F s i j k i j k ....................................................Ans.

7-40 A constant couple ( )200 300 350 N m= + + ⋅C i j k acts on a rigid body. The unit vector associated

with the fixed axis of rotation of the body for an infinitesimal angular displacement dθ is

0.250 0.350 0.903θ = + −e i j k . Determine the work done on the body by the couple during an angular displacement of 1.5 radians.

SOLUTION

( ) ( )200 300 350 0.250 0.350 0.903 161.05dU d d dθ θ= ⋅ = + + ⋅ + − = −C θ i j k i j k

[ ]1.5 1.5

00161.05 161.05 242 N m 242 JU dθ θ= − = − = − ⋅ = −∫ .....................................Ans.

7-41 Determine the horsepower that a 6-in. diameter shaft can transmit at 200 rpm if the maximum shearing stress in the shaft must be limited to 12,000 psi.

SOLUTION

( )4 46 32 127.2345 inJ π= =

( )

max

312,000 psi

127.2345TTc

Jτ = = =

3 3508.938 10 lb in. 42.4115 10 lb ftT = × ⋅ = × ⋅

( )( )32 200 42.4115 102Power 1615 hp

33,000 33,000NTT

ππω×

= = = = ..................................Ans.


280

7-42 A solid circular steel (G = 80 GPa) shaft transmits 200 kW at 180 rpm. Determine the minimum diameter required if the angle of twist in a 2-m length must not exceed 0.025 rad.

SOLUTION

( ) 3Power 2 2 180 60 200 10 N m/sT NT Tω π π= = = = × ⋅

310.61033 10 N mT = × ⋅

( ) ( )( )( )

3

9 4

10.61033 10 20.025 rad

80 10 32TLGJ d

θπ×

= = =×

0.1020 m 102.0 mmd = = ..................................................................................................Ans.

7-43 A steel (G = 12,000 ksi) shaft with a 4-in. diameter must not twist more than 0.06 rad in a 20-ft length. Determine the maximum power that the shaft can transmit at 270 rpm.

SOLUTION

( )4 44 32 25.13 inJ π= =

( )

( ) ( )6

20 120.06 rad

12 10 25.13TTL

GJθ

×= = =

×

75,390 lb in. 6283 lb ftT = ⋅ = ⋅

( )( )2 270 62832Power 323 hp

33,000 33,000NTT

ππω= = = = ....................................................Ans.

7-44 A 3-m-long hollow steel (G = 80 GPa) shaft has an outside diameter of 100 mm and an inside diameter of 60 mm. The maximum shearing stress in the shaft is 80 MPa, and the angular velocity is 200 rpm. Determine

(a) The power being transmitted by the shaft. (b) The magnitude of the angle of twist in the shaft. SOLUTION

( )4 4 6 4100 60 32 8.545 10 mmJ π= − = ×

( ) 6 2

6

0.05080 10 N/m

8.545 10TTc

Jτ −= = = ×

× 13,672 N mT = ⋅

(a) ( )( )Power 2 2 200 60 13,672T NTω π π= = =

3286.3 10 N m/s 286 kw= × ⋅ ≅ ............................................................................Ans.

(b) ( )( )

( )( )

6

9

80 10 30.0600 rad

80 10 0.050LGcτθ

×= = =

×......................................................................Ans.

7-45 The hydraulic turbines in a water-power plant rotate at 60 rpm and are rated at 20,000 hp with an overload capacity of 25,000 hp. The 30-in.-diameter shaft between the turbine and the generator is made of steel (G = 12,000 ksi) and is 20 ft long. Determine

(a) The maximum shearing stress in the shaft at the rated load. (b) The maximum shearing stress in the shaft at the maximum overload. SOLUTION

( )4 3 430 32 79.52 10 inJ π= = ×


281

(a) ( )2 602Power 20,000 hp

33,000 33,000RTNTT

ππω= = = =

6 61.7507 10 lb ft 21.01 10 lb in.RT = × ⋅ = × ⋅

( )( )6

3

21.01 10 153963 psi 3.96 ksi

79.52 10TcJ

τ×

= = = ≅×

........................................................Ans.

(b) ( )2 60

Power 25,000 hp33,000

OTπ= = 6 62.188 10 lb ft 26.26 10 lb in.OT = × ⋅ = × ⋅

( )( )6

3

26.26 10 154953 psi 4.95 ksi

79.52 10TcJ

τ×

= = = ≅×

........................................................Ans.

7-46 A solid circular steel (G = 80 GPa) shaft 1.5 m long transmits 200 kW at a speed of 400 rpm. If the allowable shearing stress is 70 MPa and the allowable angle of twist is 0.045 rad, determine

(a) The minimum permissible diameter for the shaft. (b) The speed at which this power can be delivered if the shearing stress is not to exceed 50 MPa in a shaft

with a diameter of 75 mm. SOLUTION


4775 N mT = ⋅

(a) ( )

max 4 3

2 1632

T dTc TJ d d

τπ π

= = = 3 16d T πτ=

( )

( )3 6

16 47750.0703 m 70.3 mm

70 10d

π= = =

×

( ) 44

3232

TL TL TLGJ dG d

θππ

= = = 4 32d TL Gπ θ=

( ) ( )

( )( )4 9

32 4775 1.50.0671 m 67.1 mm

80 10 0.045d

π= = =

×

min 70.3 mmd = ...................................................................................................................... Ans.

(b) ( )4 6 475 32 3.106 10 mmJ π= = ×

( ) 6 2

6

0.037550 10 N/m

3.106 10TTc

Jτ −= = = ×

× 4141 N mT = ⋅

( ) 3Power 4141 200 10 N m/sTω ω= = = × ⋅

48.2975 rad/s 461 rpmω = = ..............................................................................Ans.

7-47 The motor shown in Fig. P7-47 develops 100 hp at a speed of 360 rpm. Gears A and B deliver 40 hp and 60 hp, respectively, to operating units in a factory. If the maximum shearing stress in the shafts must be limited to 12 ksi, determine

(a) The minimum satisfactory diameter for the motor shaft. (b) The minimum satisfactory diameter for the power shaft.


282

SOLUTION (a) For the motor shaft:

( )2 3602Power 100 hp

33,000 33,000RTNTT

ππω= = = =

1458.9 lb ft 17,507 lb in.T = ⋅ = ⋅

( )

max 4 3

2 1632

T dTc TJ d d

τπ π

= = = 3 16d T πτ=

( )( )

316 17,507

1.951 in.12,000

dπ

= = ...............................................................................Ans.

(b) For the power shaft:

( ) ( )16 96 17,507 2918 lb in.T = = ⋅

( )

( )3

16 29181.074 in.

12,000d

π= = ................................................................................Ans.

7-48 A motor delivers 149 kW at 250 rpm to gear B of the factory drive shaft shown in Fig. P7-48. Gears A and C transfer 89 kW and 60 kW, respectively, to operating machinery in the factory. Determine

(a) The maximum shearing stress in shaft AB. (b) The maximum shearing stress in shaft BC. (c) The rotation of gear C with respect to gear A if the shafts are both made of steel (G = 80 GPa). SOLUTION

Power 2T NTω π= = 2T P Nπ=

(a) ( )389 10 3399.55 N m

2 250 60ABT π×= = ⋅

( )44 7 432 0.050 32 6.13592 10 mABJ dπ π −= = = ×

( )( ) 6 2

max 7

3399.55 0.025138.51 10 N/m 138.5 MPa

6.13592 10TcJ

τ −= = = × ≅×

..........................Ans.

(b) ( )360 10 2291.83 N m

2 250 60BCT π×= = ⋅

( )4 6 40.075 32 3.10631 10 mBCJ π −= = ×

( )( ) 6 2

max 6

2291.83 0.037527.67 10 N/m 27.7 MPa

3.10631 10TcJ

τ −= = = × =×

...........................Ans.

(c) ( )( )

( )( )( )( )

( )( )/ / / 9 6 9 7

2291.83 1.520 3399.55 0.76080 10 3.10631 10 80 10 6.13592 10C A C B A Bθ θ θ

− −= − = −

× × × ×

0.0386 rad= − ................................................................................................................Ans.


283

7-49 A motor supplies 270 hp at 250 rpm to gear A of the factory drive shaft shown in Fig. P7-49. Gears B and C transfer 170 hp and 100 hp, respectively, to operating machinery in the factory. For an allowable shearing stress of 11 ksi, determine

(a) The minimum permissible diameter d1 for shaft AB. (b) The minimum permissible diameter d2 for shaft BC. (c) The rotation of gear C with respect to gear A if both shafts are made of steel (G = 11,600 ksi) and have

diameters of 3 in. SOLUTION

(a) ( )2 2502Power 270 hp

33,000 33,000ABTNTT

ππω= = = =

5672.28 lb ft 68,067 lb in.ABT = ⋅ = ⋅

( ) 3

max 4

68,067 211 10 psi

32dTc

J dτ

π= = = ×

1 3.16 in.d = ..............................................................................................................Ans.

(b) ( )2 250

Power 100 hp33,000

BCTπ= = 2100.84 lb ft 25, 210 lb in.BCT = ⋅ = ⋅

( ) 3

max 4

25,210 211 10 psi

32d

dτ

π= = ×

2 2.27 in.d = ..............................................................................................................Ans.

(c) ( ) ( )

( ) ( )( ) ( )

( ) ( )/ / / 4 46 6

25,210 72 68,067 36

11.6 10 3 32 11.6 10 3 32C A C B B Aθ θ θ

π π= + = +

× ×

0.0462 rad= .................................................................................................................. Ans.

7-50 A motor provides 180 kW of power at 400 rpm to the drive shafts shown in Fig. P7-50. The maximum shearing stress in the three solid steel (G = 80 GPa) shafts must not exceed 70 MPa. Gears A, B, and C supply 40 kW, 60 kW, and 80 kW, respectively, to operating units in the plant. Determine

(a) The minimum satisfactory diameter for shaft D. (b) The minimum satisfactory diameter for shaft E. (c) The minimum satisfactory diameter for shaft F. SOLUTION

Power 2T NTω π= = 2T P Nπ=

( )

max 4 3

2 1632

T dTc TJ d d

τπ π

= = = 3 16d T πτ=

(a) ( )340 10 477.5 N m

2 800 60DT π×= = ⋅

( )

( )3 6

16 477.50.03263 m 32.6 mm

70 10Dd π= = ≅

×.................................................................Ans.

(b) ( )3180 10 4297 N m

2 400 60ET π×= = ⋅


284

( )

( )3 6

16 42970.06787 m 67.9 mm

70 10Ed π= = ≅

×.................................................................Ans.

(c) ( )380 10 954.9 N m

2 800 60FT π×= = ⋅

( )

( )3 6

16 954.90.04111 m 41.1 mm

70 10Dd π= = ≅

×.................................................................Ans.

7-51 A 3-in. diameter steel drive shaft connects a motor to a machine. The power output of the motor varies with angular speed according to the following data

ωωωω P ωωωω P (rpm) (hp) (rpm) (hp) 60 14.3 700 103.3 360 71.3 940 104.7 450 83.5 1120 90.6 600 96.5 1480 22.5 Calculate and plot the shear stress in the shaft τ as a function of the angular speed ω. SOLUTION

2Power33,000NTT πω= =

( )33,000 hp

2T

Nπ=

( )

( )max 4

33,000 hp 1.52 3 32

TcJ N

τπ π

= =

( )990.696 hp

N= ........................................ Ans.

7-52 A 75-mm diameter steel drive shaft connects a motor to a machine. The power output of the motor varies with angular speed according to the following data

ωωωω P ωωωω P (rpm) (kW) (rpm) (kW) 50 15.5 750 77.2 100 36.0 1000 78.0 250 61.7 1250 67.5 500 73.3 1500 15.0 Calculate and plot the shear stress in the shaft τ as a function of the angular speed ω. SOLUTION

( ) ( )Power 2 60 in revolutions per minuteT N T Nω π= =

( )60

2P

TNπ

=

( ) ( )

( )( ) ( )3

max 4

115.2810 1060 2 0.0375

0.075 32

PP NTcJ N

πτ

π

× = = = ....................................Ans.


286

7.964 kN mmT = ⋅ 2.036 kN msT = ⋅

(a) ( ) ( ) 6 2

6

2036 0.06258.99 10 N/m 8.99 MPa

14.151 10sTcJ

τ −= = = × =×

........................................Ans.

( ) ( ) 6 2

6

7964 0.087510.23 10 N/m 10.23 MPa

68.11 10mτ −= = × =×

.............................................Ans.

(b) ( ) ( )

( )( )3

9 6

7964 23.60 10 rad

65 10 68.11 10θ −

−= = ×

× ×...............................................................Ans.

7-55 A 3-in. diameter cold-rolled steel (G = 11,600 ksi) shaft, for which the maximum allowable shearing stress is 15 ksi, exhibited severe corrosion in a certain installation. It is proposed to replace the shaft with one in which an aluminum alloy (G = 4000 ksi) tube ¼-in. thick is bonded to the outer surface of the cold-rolled steel shaft to produce a composite shaft. If the shearing stress in the aluminum alloy shell is limited to 12 ksi, determine

(a) The maximum torque that the original shaft can transmit. (b) The maximum torque that the replacement shaft can transmit. SOLUTION

( )4 43 32 7.952 insJ π= = ( )4 4 43.5 3 32 6.780 inaJ π= − =

(a) ( )1.5

15 ksi7.952s

TTcJ

τ = = =

79.52 kip in. 79.5 kip in.sT = ⋅ ≅ ⋅ ........................................................................Ans.

(b) Equilibrium:

0 :axisMΣ = 0s a RT T T+ − = R s aT T T= +

Deformations:

a sLGcτθ θ= = ( )( ) ( )( )6 64 10 1.75 11.6 10 1.5

a sL Lτ τ+× ×

0.4023a sτ τ=

max 15 ksisτ τ= = ( ) max0.4023 15 6.035 ksi 12 ksiaτ τ= = < =

( )1.5

15 ksi7.952s

TTcJ

τ = = = 79.52 kip in.sT = ⋅

( )1.75

6.035 ksi6.780a

TTcJ

τ = = = 23.38 kip in.aT = ⋅

79.52 23.38 102.9 kip in.RT = + = ⋅ ....................................................................................Ans.

7-56 The 50-mm diameter steel (G = 80 GPa) shaft shown in Fig. P7-56 is fixed to rigid walls at both ends. When a torque of 4.75 kN-m is applied as shown, determine

(a) The maximum shearing stress in the shaft. (b) The angle of rotation of the section where the torque is applied with respect to its no load position. SOLUTION Equilibrium:

0 :axisMΣ = 4.75 0A CT T+ − =

4.75 kN mA CT T+ = ⋅


287

Deformations:

AB BCTLGJ

θ θ= = ( ) ( )0.6 0.9A CT TGJ GJ

= 1.5A CT T=

2.8500 kN mAT = ⋅ 1.9000 kN mCT = ⋅

(a) ( )4 7 40.050 32 6.1359 10 mJ π −= = ×

( )( )3

6 2max 7

2.8500 10 0.050232 10 N/m 232 MPa

6.1359 10TcJ

τ −

×= = = × =

×...........................Ans.

(b) ( )( )

( )( )3

9 7

2.85 10 0.60.0348 rad

80 10 6.1359 10Bθ−

×= =

× ×.................................................................Ans.

7-57 Two 3-in. diameter solid circular steel (G = 11,600 ksi) and bronze (G = 6500 ksi) shafts are rigidly connected and supported as shown in Fig. P7-57. A torque T is applied at the junction of the two shafts as indicated. The shearing stresses are limited to 18 ksi for the steel and 6 ksi for the bronze. Determine

(a) The maximum torque T that can be applied. (b) The angle of rotation of the section where the torque is applied with respect to its no-load position. SOLUTION Equilibrium:

0 :axisMΣ = 0A CT T T+ − = A CT T T= +

Deformations:

AB BCLGcτθ θ= =

( )( )

( )( )6 6

2 16.5 10 11.6 10

b s

c cτ τ

=× ×

3.569s bτ τ=

max 18 ksisτ τ= = max3.569 5.0431 ksi 6 ksib sτ τ τ= = < =

( )4 43 32 7.9522 inJ π= =

(a) ( )1.5

18 ksi7.9522A

s

TTcJ

τ = = = 95.43 kip in. 7.952 kip ftAT = ⋅ = ⋅

( )1.5

5.0431 ksi7.9522C

b

TTcJ

τ = = = 26.74 kip in. 2.228 kip ftCT = ⋅ = ⋅

7.952 2.228 10.18 kip ftA CT T T= + = + = ⋅ ....................................................................Ans.

(b) ( )( )( )( )

3

6

18 10 1 120.01241 rad

11.6 10 1.5BLGcτθ

× ×= = =

×....................................................................Ans.

7-58 A composite shaft consists of a bronze (G = 45 GPa) shell that has an outside diameter of 100 mm bonded to a solid steel (G = 80 GPa) core. Determine the diameter of the steel core when the torque resisted by the steel core is equal to the torque resisted by the bronze shell.

SOLUTION

( )4 4100 32bJ dπ= − 4 32sJ dπ= b sT T T= =

b sTLGJ

θ θ= = ( ) ( ) ( )6 46 4 4 80 10 3245 10 100 32

TL TLdd ππ

= ×× −

77.5 mmd = ........................................................................................................................... Ans.


288

7-59 The solid steel (G = 12,000 ksi) shaft shown in Fig. P7-59 is fixed to the wall at C. The bolt holes in the flange at A have an angular misalignment of 0.0010 rad with respect to the holes in the wall. Determine

(a) The torque, applied at B, required to align the bolt holes. (b) The maximum shearing stress in the shaft after the bolts are inserted and tightened and the torque at B is

removed. (c) The maximum torque that can be applied at section B after the bolts are tightened if the maximum

shearing stress in the shaft is not to exceed 10 ksi. SOLUTION

(a) ( )4 46 32 127.23 inJ π= =

( )

( )( )/ / / 6

4 120.0010 rad

12 10 127.23A C B C B C

Tθ θ θ

×+ = = =

×

331.8 10 lb in. 31.8 kip in.AT = × ⋅ = ⋅ ........................... Ans.

(b) ( )

( )( )/ 6

6 120.0010 rad

12 10 127.23A C

Tθ

×= =

×

321.205 10 lb in. 21.205 kip in.AT = × ⋅ = ⋅

( )( )

max

21,205 3500 psi

127.23TcJ

τ = = = ...................................................................................Ans.

(c) / / 0.0010 radB C B Aθ θ= +

( )

( )( )( )

( ) ( )6 6

4 12 2 120.0010 rad

12 10 3 12 10 3BC ABτ τ× ×

= +× ×

2 1.5000 ksiBC ABτ τ= +

max 10 ksiABτ τ= = ( ) max10 1.5000 2 5.750 ksi 10 ksiBCτ τ= + = < =

( )3

10 ksi127.23AB

AB

TTcJ

τ = = = 424.10 kip in.ABT = ⋅

( )3

5.750 ksi127.23BC

BC

TTcJ

τ = = = 243.86 kip in.BCT = ⋅

424.10 243.86 668 kip in.AB BCT T T= + = + = ⋅ ..............................................................Ans.

7-60 A composite shaft consists of a solid steel (G = 80 GPa) core with an outside diameter of 40 mm covered by a brass (G = 39 GPa) tube with an inside diameter of 40 mm and a wall thickness of 20 mm which is in turn covered by an aluminum alloy (G = 28 GPa) sleeve with an inside diameter of 80 mm and a wall thickness of 10 mm, as shown in Fig. P7-60. The three materials are bonded so that they act as a unit. Determine

(a) The maximum shearing stress in each material when the assembly is transmitting a torque of 15 kN-m. (b) The angle of twist in a 3-m length when the assembly is transmitting a torque of 10 kN-m. SOLUTION

( )4 6 40.040 32 0.2513 10 msJ π −= = ×

( )4 4 6 40.080 0.040 32 3.770 10 mbJ π −= − = ×

( )4 4 6 40.100 0.080 32 5.796 10 maJ π −= − = ×


289

Equilibrium:

0 :axisMΣ = 15 0s b aT T T+ + − = 15 kN ms b aT T T+ + = ⋅

Deformations: s b aTLGJ

θ θ θ= = =

( )( ) ( )( )9 6 9 680 10 0.2513 10 28 10 5.796 10s aT L T L

− −=

× × × × 0.12388s aT T=

( )( ) ( ) ( )9 6 9 639 10 3.770 10 28 10 5.796 10b aT L T L

− −=

× × × × 0.90598b aT T=

7.390 kN maT = ⋅ 0.915 kN msT = ⋅ 6.695 kN mbT = ⋅

(a) ( )( )3

6 26

0.915 10 0.02072.8 10 N/m 72.8 MPa

0.2513 10sτ −

×= = × =

×.........................................Ans.

( ) ( )3

6 26

6.695 10 0.04071.0 10 N/m 71.0 MPa

3.770 10bτ −

×= = × =

×.........................................Ans.

( )( )3

6 26

7.390 10 0.05063.8 10 N/m 63.8 MPa

5.796 10aTcJ

τ −

×= = = × =

×..............................Ans.

(b) ( )( ) ( )( )( )

3

9 6

10 15 6.695 10 30.0911 rad

39 10 3.770 10bTLGJ

θ−

× = = =× ×

.....................................................Ans.

7-61 The composite shaft shown in Fig. P7-61 is used as a torsional spring. The solid circular polymer (G = 150 ksi) portion of the shaft is encased in and firmly attached to a steel (G = 12,000 ksi) sleeve for part of its length. If a torque T of 1.00 kip-in. is being transmitted by the composite shaft, determine

(a) The rotation of a cross section at C. (b) The rotation of a cross section at C if the steel sleeve is assumed to be rigid. (c) The percent error introduced by assuming the steel sleeve to be rigid. SOLUTION

( )4 42 32 1.5708 inpJ π= =

( )4 4 42.25 2 32 0.9453 insJ π= − =

Equilibrium:

0 :axisMΣ = 1 0p sT T− − = 1 kip in. 1000 lb in.s pT T+ = ⋅ = ⋅

Deformations:

s pTLGJ

θ θ= = ( )( ) ( )( )12,000 0.9453 150 1.5708ps T LT L = 48.14s pT T=

20.35 lb in.pT = ⋅ 979.65 lb in.sT = ⋅

(a) ( )

( )( )( )

( )( )/ / / 3 3

1000 4 20.35 12150 10 1.5708 150 10 1.5708C A C B B Aθ θ θ= + = +

× ×

0.018013 rad 0.01801 rad= ≅ ...................................................................................Ans.


290

(b) If the steel shell is assumed to be rigid, then

( )

( )( )/ / 3

1000 40.016976 rad 0.01698 rad

150 10 1.5708C A C Bθ θ= = = ≅×

..........................Ans.

(c) ( )0.018013 0.016976Error 100 5.76 %0.018013

−= = ...............................................................Ans.

7-62 A disk and two circular shafts are connected and supported between rigid walls, as shown in Fig. P7-62. Shaft AB is made of brass (G = 39 GPa) and has a diameter of 100 mm and a length of 400 mm. Shaft BC is made of Monel (G = 65 GPa) and has a diameter of 80 mm and a length of 600 mm. If a torque of 20 kN-m is applied to the disk, determine

(a) The maximum shearing stress in each of the shafts. (b) The angle of rotation of the disk with respect to its no-load position. (c) The maximum tensile and compressive stresses in the shaft. SOLUTION

( )4 6 4 6 4100 32 9.817 10 mm 9.817 10 mABJ π −= = × = ×

( )4 6 4 6 480 32 4.021 10 mm 4.021 10 mBCJ π −= = × = ×

Equilibrium:

0 :axisMΣ = 20 0AB BCT T+ − = 20 kN mAB BCT T+ = ⋅

Deformations: AB BCTLGJ

θ θ= =

( )

( )( )( )

( ) ( )9 6 9 6

0.400 0.60039 10 9.817 10 65 10 4.021 10

AB BCT T− −

=× × × ×

2.197AB BCT T=

13.744 kN mABT = ⋅ 6.256 kN mBCT = ⋅

(a) ( )( )3

6 26

13.744 10 0.05070.0 10 N/m 70.0 MPa

9.817 10ABτ −

×= = × =

×....................................Ans.

( )( )3

6 26

6.256 10 0.04062.2 10 N/m 62.2 MPa

4.021 10BCτ −

×= = × =

×......................................Ans.

(b) ( )( )

( )( )3

/ 9 6

13.744 10 0.4000.01437 rad

39 10 9.817 10B ATLGJ

θ−

×= = =

× ×..................................................Ans.

(c) max max 70.0 MPa (T & C)ABσ τ τ= = = ............................................................................Ans.

7-63 The steel (G = 12,000 ksi) shaft shown in Fig. P7-63 will be used to transmit a torque of 1000 lb-in. The hollow portion AE of the shaft is connected to the solid portion BF with two pins, at C and D as shown. If the average shearing stress in the pins must be limited to 25 ksi, determine the minimum satisfactory diameters for each of the pins.

SOLUTION

( )4 4 42 1 32 1.4726 inhJ π= − = ( )4 41 32 0.09817 insJ π= =

In the interval CD equilibrium gives

0 :axisMΣ = 1000 0h sT T+ − = 1000 lb in.h sT T+ = ⋅

Deformations:


291

h sTLGJ

θ θ= = ( ) ( )1.4726 0.09817h sT L T L

G G= 15h sT T=

62.5 lb in.sT = ⋅ 937.5 lb in.hT = ⋅

Pin transmits 62.5 lb in.C ⋅ Pin transmits 937.5 lb in.D ⋅

The pins are in double shear and the moment generated by the shear force ( )Aτ is

( ) ( )( )( )225,000 4 1sT A d dτ π= =

( )

( )4 62.5

0.0564 in.25,000Cd π

= = .........................................................................................Ans.

( )

( )4 937.5

0.219 in.25,000Dd π

= = ...........................................................................................Ans.

7-64 A stainless steel (G = 86 GPa) shaft 2.5 m long extends through and is attached to a hollow brass (G = 39 GPa) shaft 1.5 m long, as shown in Fig. P7-64. Both shafts are fixed at the wall. When the two torques shown are applied to the shaft, determine

(a) The maximum shearing stress in the steel. (b) The maximum shearing stress in the brass. (c) The maximum compressive stress in the brass. (d) The rotation of the right end of the shaft. SOLUTION

( )4 6 4 6 480 32 4.021 10 mm 4.021 10 msJ π −= = × = ×

( )4 4 6 4 6 4120 80 32 16.336 10 mm 16.336 10 mbJ π −= − = × = ×

Equilibrium:

0 :axisMΣ = ( ) 36 9 0b sT T+ + − =

27 kN mb sT T+ = − ⋅

Deformation in the interval AB:

s bTLGJ

θ θ= =

( )( ) ( )( )9 6 9 686 10 4.021 10 39 10 16.336 10s bT L T L

− −=

× × × × 0.5428s bT T=

9.499 kN msT = ⋅ 17.501 kN mbT = ⋅

(a) ( )( )3

6 26

9.499 10 0.04094.5 10 N/m 94.5 MPa

4.021 10sTcJ

τ −

×= = = × =

×...............................Ans.

(b) ( )( )3

6 26

17.501 10 0.06078.7 10 N/m 78.7 MPa

13.336 10bτ −

×= = × =

×.......................................Ans.

(c) max 78.7 MPa (C)bσ τ= = ...................................................................................................Ans.


292

(d) ( )( )

( )( )( )( )

( ) ( )3 3

/ / / 9 6 9 6

9 10 1 9.499 10 1.5

86 10 4.021 10 86 10 4.021 10C A C B B Aθ θ θ− −

− × ×= + = +

× × × ×

0.01518 rad= ................................................................................................................Ans.

7-65 The shaft shown in Fig. P7-65 consists of a 6-ft hollow steel (G = 12,000 ksi) section AB and a 4-ft solid aluminum alloy (G = 4000 ksi) section CD. The torque T of 40 kip-ft is applied initially only to the steel section AB. Section CD is then connected and the torque T is released. When the torque is released, the connection slips 0.010 rad before the aluminum section takes any load. Determine

(a) The maximum shearing stress in the aluminum alloy. (b) The maximum shearing stress in the steel after the torque T is released. (c) The final rotation of the collar at B with respect to its no-load position. SOLUTION

( )4 4 46 4 32 102.10 inABJ π= − =

( )4 44 32 25.13 inCDJ π= =

( )( )

( )( )40 12 6 12

0.02821 rad12,000 102.10i

TLGJ

θ× ×

= = =

Equilibrium: After the connection is made and the torque is released,

AB CD fT T T= =

Deformation: / /slipB A i C Dθ θ θ= − −

( )

( )( )( )

( )( )6 12 4 12

0.02821 0.01012,000 102.10 4000 25.13

f fT T× ×= − − 33.96 kip in.fT = ⋅

(a) ( )( )33.96 2

2.70 ksi25.13a CD

TcJ

τ τ= = = = ............................................................................Ans.

(b) ( ) ( )33.96 3

0.998 ksi102.10s ABτ τ= = = ....................................................................................Ans.

(c) ( ) ( )

( )( )/

33.96 6 120.01996 rad

12,000 102.10B ATLGJ

θ×

= = = .............................................................Ans.

7-66 A torque T of 10 kN-m is applied to the steel (G = 80 GPa) shaft shown in Fig. P7-66 without the brass (G = 40 GPa) shell. The brass shell is then slipped into place and attached to the steel. After the original torque is released, determine

(a) The maximum shearing stress in the brass shell. (b) The maximum shearing stress in the steel shaft. (c) The final rotation of the right end of the steel shaft with respect to the left end. SOLUTION

( )4 6 480 32 4.021 10 mmsJ π= = ×

( )4 4 6 4160 140 32 26.62 10 mmbJ π= − = ×

( )( )

( )( )3

9 6

10 10 0.8000.02487 rad

80 10 4.021 10iTLGJ

θ−

×= = =

× ×


293

Equilibrium: After the connection is made and the torque is released,

s b fT T T= =

Deformation: s i bθ θ θ= −

( )

( ) ( )( )

( ) ( )9 6 9 6

0.800 0.8000.02487

80 10 4.021 10 40 10 26.62 10f fT T

− −= −

× × × ×

7680 N mfT = ⋅

(a) ( )( ) 6 2

6

7680 0.08023.1 10 N/m 23.1 MPa

26.62 10bTcJ

τ −= = = × =×

..........................................Ans.

(b) ( ) ( ) 6 2

6

7680 0.04076.4 10 N/m 76.4 MPa

4.021 10sτ −= = × =×

....................................................Ans.

(c) ( ) ( )

( ) ( )/ 9 6

7680 0.8000.01910 rad

80 10 4.021 10R LTLGJ

θ−

= = =× ×

...................................................Ans.

7-67 A hollow steel (G = 11,600 ksi) tube with an inside diameter of 2 in., an outside diameter of 2.5 in., and a length of 12 in. is encased with a brass (G = 5600 ksi) tube that has an inside diameter of 2.5 in. and an outside diameter of 3.25 in. The brass and steel tubes, while unstressed, are brazed together at one end. A torque is then applied to the other end of the brass tube, which twists one degree with respect to the steel tube. The tubes are then brazed together in that position. Determine the maximum shearing stresses in the steel tube and in the brass tube after the torque is removed from the brass tube.

SOLUTION

( )4 4 42.5 2 32 2.264 insJ π= − =

( )4 4 43.25 2.5 32 7.118 inbJ π= − =

( )1

0.017453 rad180i

πθ = =

Equilibrium: After the torque is removed,

b sT T T= =

Deformation: b i sθ θ θ= −

( )

( ) ( )( )

( )( )12 12

0.0174535600 7.118 11,600 2.264T T

= − 23.03 kip in.T = ⋅

( ) ( )23.03 1.25

12.72 ksi2.264s

TcJ

τ = = = ................................................................................Ans.

( )( )23.03 1.625

5.26 ksi7.118bτ = = ..........................................................................................Ans.

7-68 A composite shaft consists of a 2-m long solid circular steel (Gs = 80 GPa) section securely fastened to a 2-m long solid circular bronze (Gb = 40 GPa) section, as shown in Fig. P7-68. Both ends of the composite shaft are attached to rigid supports. The maximum shearing stress τmax in the shaft must not exceed 60 MPa, and the rotation θ of any cross section in the shaft must not exceed 0.04 rad. The ratio of the diameters of the two sections can vary (1/2 ≤ db/ds ≤ 2), but the average diameter (db + ds)/2 must be 100 mm. Compute and plot

(a) The maximum allowable torque T as a function of the diameter ratio db/ds (1/2 ≤ db/ds ≤ 2).


296

When 31 17.4533 10 radBCθ −> ° = ×

Equilibrium: A CT T T+ =

Deformation: 317.4533 10 radBC ABθ θ −= + ×

( )

( )( )9 6

1.580 10 20.36 10

CT−× ×

( )

( )( )3

9 6

0.7517.4533 10

80 10 20.36 10AT −

−= + ×

× ×

0.5 18,951.96 kN mC AT T= + ⋅


0.66667 12,634.64 kN mAT T= − ⋅ 0.33333 12,634.64 kN mCT T= + ⋅

In either case

( ) 2

6

0.060 N/m

20.36 10A

AB

Tτ −=

×...................................................Ans.

( ) 2

6

0.060 N/m

20.36 10C

BC

Tτ −=

×...................................................Ans.

( )

( )( )9 6

1.580 10 20.36 10

CB BC

Tθ θ

−= =

× ×................................Ans.

while for the bolts

26

0.3 N/m2036 10

b A b Ab

b b

V T d TA A

τ −= = =×

..............................................................................Ans.

7-71 A 2-in. diameter solid steel (Gs = 12,000 ksi, τmax = 30 ksi) shaft and a 3-in. diameter hollow aluminum alloy (Ga = 4000 ksi, τmax = 24 ksi) shaft are fastened together with a ½-in. diameter brass (τmax = 36 ksi) pin, as shown in Fig. P7-71. Calculate and plot

(a) The maximum shearing stresses, τa in the aluminum shaft and τs in the steel shaft, as a function of the torque T applied to the end of the shaft (0 ≤ T ≤ 60 kip-in.).

(b) The average shearing stresses τb on the cross-sectional area of the brass pin at the interface between the shafts as a function of T (0 ≤ T ≤ 60 kip-in.).

(c) The rotation of end B with respect to its no-load position θB/D as a function of T (0 ≤ T ≤ 60 kip-in.).


297

(d) What is the maximum torque that can be applied to the shaft without exceeding the maximum shearing stresses in either the shaft or the pin?

SOLUTION

( )4 4 43 2 32 6.3814 inaJ π= − = ( )4 42 32 1.5708 insJ π= =

Equilibrium: s aT T T+ =

Deformation: a sTLGJ

θ θ= = ( )( )

4 6.38141.35417

12 1.5708a s sT T T= =

Combining the deformation equation with the equilibrium equation gives

1.35417s sT T T+ = 0.42478sT T= 0.57522aT T=

(a) ( )1

1.5708s

TTcJ

τ = = .................................................................................................................... Ans.

( )1.5

6.3814a

a

TTcJ

τ = = .................................................................................................................. Ans.

(c) ( ) ( )

( )( )/ 6

12 18rad

12 10 1.5708s

B D

T TTLGJ

θ+

= =×∑ ..........................................................................Ans.

(b) ( )2

2 psi0.5 4

b a s ab

b b

V T d TA A

τπ

= = = .....................................................................................Ans.

(d) The stress in the brass pin reaches its limit first when 24.6 kip in.T = ⋅ .......................................Ans.

7-72 A motor is to transmit 150 kW to a piece of mechanical equipment. The power is transmitted through a solid structural steel shaft. Failure is by yielding and the factor of safety is 1.25. The designer has the freedom to operate the motor at 60 rpm or 6000 rpm. For each case, determine the minimum shaft diameter. Shafts are available with diameters in increments of 5 mm. If weight is important, which speed would be used?

SOLUTION

Structural steel: 250 MPayσ = 125 MPa2y

y

στ = = all

125 100 MPa1.25

y

FSτ

τ = = =

Power 2T NTω π= = 2T P Nπ= ( )

max 4 3

2 1632

T dTc TJ d d

τπ π

= = =

For 60 rpm:


298

( )3

3150 10 23.87 10 N m2 60 60

Tπ

×= = × ⋅ ( )( )

33 3 3

6

16 23.87 101.21569 10 m

100 10d

π−

×= = ×

×

0.1067 m 106.7 mmd = =

Use 110 mmd = .................................................................................................................... Ans. For 6000 rpm:

( )3150 10 238.7 N m

2 6000 60T

π×= = ⋅

( )( )

3 6 36

16 238.712.1569 10 m

100 10d

π−= = ×

×

0.0230 m 23.0 mmd = =

Use 25 mmd = ...................................................................................................................... Ans. If weight is critical,

Use 6000 rpmN = ................................................................................................................Ans.

7-73 A 3-ft long steel pipe is subjected to a torque of 1200 lb-ft at each end. The pipe is made of 0.2% C hardened steel, failure is by yielding, and the factor of safety is 1.5. Determine the nominal diameter of the lightest standard weight steel pipe that can be used for the shaft.

SOLUTION

0.2% C steel: 36 ksiyσ = 18 ksi2y

y

στ = = all

18 12 ksi1.5

y

FSτ

τ = = =

all 2Tc TcJ I

τ = = ( )3

3all

1200 12 0.600 in2 2 12 10

I Tc τ

×= = =×

2.0-in. pipe: 30.561 inS I c= =

2.5-in. pipe: 31.064 inS I c= =

Use 2.5 in.d = ....................................................................................................................... Ans.

7-74 A standard weight structural steel pipe must transmit 150 kW at 60 rpm. The failure mode is yielding and the factor of safety is 1.5.

(a) Select the lightest standard weight steel pipe that can be used. (b) If solid structural steel shafts are available with diameters in increments of 10 mm, determine the

minimum diameter that can be used. (c) Compare the weights of the two shafts. SOLUTION

Structural steel: 250 MPayσ = 37870 kg/mρ =

125 MPa2y

y

στ = = all

125 83.33 MPa1.5

y

FSτ

τ = = =

Power 2T NTω π= = ( )3

3150 10 23.87 10 N m2 2 60 60PTNπ π

×= = = × ⋅

(a) all 2Tc TcJ I

τ = = ( )3

6 36

all

23.87 10 143.22 10 m2 2 83.33 10

I Tc τ

−×= = = ××

152-mm pipe: 6 3139.16 10 mS I c −= = × 28.21 kg/mm =


299

203-mm pipe: 6 3275.67 10 mS I c −= = × 42.46 kg/mm =

Use 203 mmd = ................................................................................................................... Ans.

(b) ( ) ( )3

6 2max 4

23.87 10 283.33 10 N/m

32dTc

J dτ

π×

= = ≤ ×

0.1134 m 113.4 mmd ≥ =

Use 120 mmd = .................................................................................................................... Ans.

(c) Solid shaft ( )27870 0.120 4 89.0 kg/mm Vρ π = = =

............................................Ans.

Hollow shaft 42.5 kg/mm = ..................................................................................................Ans.

7-75 A shaft is to transmit 100 hp at 200 rpm. The designer has a variety of solid bars and standard steel pipes to select from. Both the bars and the pipes are made of structural steel, the failure mode is yielding, and the factor of safety is 2.

(a) Select the lightest standard weight steel pipe that can be used. (b) Select a suitable solid shaft, if they are available with diameters in increments of 1/8 in. (c) If weight is critical, which shaft should be used? SOLUTION

Structural steel: 36 ksiyσ = 30.284 lb/inγ =

18 ksi2y

y

στ = = all

18 9 ksi2

y

FSτ

τ = = =

( )2 2002Power 100 hp

33,000 33,000TNTT

ππω= = = =

2626 lb ft 31,513 lb in.T = ⋅ = ⋅

(a) For a hollow pipe:

all 2Tc TcJ I

τ = = ( )3

3all

31,513 1.75072 in2 2 9 10

I Tc τ

= = =×

3.0-in. pipe: 31.724 inS I c= = 7.58 lb/ftw =

3.5-in. pipe: 32.39 inS I c= = 9.11 lb/ftw =

Use 3.5 in.d = ........................................................................................................................ Ans.

(b) ( ) 3

max 4

31,513 29 10 psi

32dTc

J dτ

π= = ≤ × 2.613 in.d ≥

548 82 2.613 2 in.< <

58Use 2 in.d = ...................................................................................................................... Ans.

(c) Solid shaft ( )20.284 2.625 4 1.537 lb/in.w Vγ π = = =

Hollow pipe 9.11 lb/ft 0.759 lb/in.w = = If weight is critical, Use hollow pipe ..................................................................................................................... Ans.


300

7-76 The motor shown in Fig. P7-76 supplies a torque of 1000 N-m to shaft ABCDE. The torques removed at C, D, and E are 500 N-m, 300 N-m, and 200 N-m, respectively. The shaft is the same diameter throughout and is made of 0.4% C hot-rolled steel. For a factor of safety of 3 and failure by yielding, select a suitable diameter for the shaft, if shafts are available with diameters in increments of 10 mm.

SOLUTION

0.4% C hot-rolled steel: 360 MPayσ = 2 180 MPay yτ σ= =

all 180 3 60 MPay FSτ τ= = =

The internal resisting torques are:

0 :axisMΣ = 1000 0BCT − = 1000 N mBCT = ⋅

0 :axisMΣ = 1000 500 0CDT − + = 500 N mCDT = ⋅

0 :axisMΣ = 1000 500 300 0DET − + + = 200 N mCDT = ⋅

max 1000 N mBCT T= = ⋅

( ) ( ) 6

max 4

1000 260 10 MPa

32dTc

J dτ

π= = ≤ × 0.04395 m 43.95 mmd ≥ =

Use 50 mmd = ...................................................................................................................... Ans.

7-77 A torque of 30,000 lb-in. is supplied to the factory drive shaft of Fig. P7-77 by a belt that drives pulley A. A torque of 10,000 lb-in. is removed by pulley B and 20,000 lb-in by pulley C. The shaft is made of structural steel and has a constant diameter over its length. Segment AB of the shaft is 3 ft long, and segment BC is 4 ft long. Failure is by yielding and the factor of safety is 2.25. Select a suitable diameter for the shaft, if shafts are available with diameters in increments of 1/8 in.

SOLUTION

Structural steel: 36 ksiyσ = 18 ksi2y

y

στ = = all

18 8 ksi2.25

y

FSτ

τ = = =


0 :axisMΣ = 20,000 0BCT − = 20,000 lb in.BCT = ⋅

0 :axisMΣ = 20,000 10,000 0ABT − − = 30,000 lb in.ABT = ⋅

max 30,000 lb in.ABT T= = ⋅

( ) 3

max 4

30,000 28 10 psi

32dTc

J dτ

π= = ≤ × 2.67 in.d ≥

5 68 82 2.67 2 in.< <

6 38 4Use 2 2 in.d = = ...........................................................................................................Ans.

7-78 The band brake shown in Fig. P7-78 is part of a hoisting machine. The coefficient of friction between the 500-mm diameter drum and the flat belt is 0.20. The maximum actuating force P that can be applied to the brake arm is 490 N. Rotation of the drum is clockwise. What minimum size shaft should be used to transmit the resisting torque developed by the brake to the machine if the shaft is to be made of 0.4% C hot-rolled steel. The factor of safety is 3 for failure by yielding. Circular steel bars are available with diameters in increments of 5 mm.

SOLUTION

0.4% C hot-rolled steel: 360 MPayσ = 2 180 MPay yτ σ= =

all 180 3 60 MPay FSτ τ= = =


301

0 :axisMΣ = ( ) ( )50 490 650 0BT − = 6370 NBT =

( )0.20 1.56370 16,348 NsA BT T e e πµ β= = =

( ) ( ) ( )16,348 6370 0.250 2495 N mA BT T T R= − = − = ⋅

( ) ( ) 6

max 4

2495 260 10 MPa

32dTc

J dτ

π= = ≤ × 0.0596 m 59.6 mmd ≥ =

Use 60 mmd = ...................................................................................................................... Ans.

7-79 The motor shown in Fig. P7-79 supplies a torque of 380 lb-ft to shaft BCD. The torques removed at gears C and D are 220 lb-ft and 160 lb-ft, respectively. The shaft BCD has a constant diameter and is made of 0.4% C hot-rolled steel, failure is by yielding, and the factor of safety is 2. Determine

(a) The minimum allowable diameter of the shaft, if shafts are available with diameters in increments of 1/8 in.

(b) The minimum allowable diameter of the bolts used in the coupling, if eight bolts are used, the material is structural steel, the mode of failure is yielding, and the factor of safety is 1.5. The diameter of the bolt circle is d1 = 3.5 in., and the bolts are available with diameters in increments of 1/16 in.

SOLUTION

0.4% C hot-rolled steel: 53 ksiyσ = 26.5 ksi2y

y

στ = = all

18 13.25 ksi2

y

FSτ

τ = = =


0 :axisMΣ = 380 0BCT − = max 380 lb ftBCT T= = ⋅

0 :axisMΣ = 380 220 0CDT − + = 160 lb ftCDT = ⋅

(a) ( ) ( ) 3

max 4

380 12 213.25 10 psi

32dTc

J dτ

π×

= = ≤ × 1.206 in.d ≥

1 28 81 1.206 1 in.< <

2 18 4Use 1 1 in.d = = .............................................................................................................Ans.

Structural steel: 36 ksiyσ = 18 ksi2y

y

στ = = all

18 12 ksi1.5

y

FSτ

τ = = =

0 :axisMΣ = ( )8 3.5 2 0s BCF T− = 325.7 lbs sF Aτ= =

( )( )3 212 10 4 325.7 lbs bA dτ π= × = 0.1859 in.d =

3216 160.1859 in.< <

316Use in.d = ........................................................................................................................ Ans.

7-80 A shaft used to transmit power is constructed by joining two solid segments of shaft with a collar, as shown in Fig. P7-80. The collar has an inside diameter equal to the diameter of the shaft, and both the collar and the shaft are made of the same material. The collar is securely bonded to the shaft segments. Determine the ratio of the diameters of the collar and shaft such that the splice can transmit the same power as the shaft and at the same maximum shearing stress level. Is the solution dependent on the material selected?

SOLUTION For the same power, the values of torque are the same in the shaft and the collar.

For the shaft: ( )

4 4

2 1632

s ss

s s

T d TdTcJ d d

τπ π

= = =


302

For the collar: ( )

( ) ( )4 4 4 4

2 1632

c cc

c s c s

T d TdTcJ d d d d

τπ π

= = =− −

For the same shear stress in the shaft and the collar

4 4 4s c

s c s

d dd d d

=−

( ) ( )4 1 0c s c sd d d d− − =

( ) 1.221 (independent of the type of material)c sd d = ...............................................Ans.

7-81 A solid circular steel (G = 12,000 ksi) shaft is loaded and supported as shown in Fig. P7-81. Determine (a) The maximum shearing stress in the shaft. (b) The rotation of a section at B with respect to its no-load position. (c) The rotation of end C with respect to its no-load position. SOLUTION

( )4 43 32 7.952 inABJ π= =

( )4 42 32 1.5708 inBCJ π= =

(a) ( ) ( )4000 12 1.5

9054 psi7.952AB

TcJ

τ×

= = =

( )( )1000 12 1

7639 psi1.5708BCτ

×= =

max 9054 psi 9.05 ksiτ = ≅ ...................................................................................................Ans.

(b) ( )( )

( )( )/ 6

4000 12 3 120.01811 rad

12 10 7.952B ATLGJ

θ− × ×

= = = −×

.......................................................Ans.

(c) ( )( )

( ) ( )( )( )( )( )/ 6 6

4000 12 3 12 1000 12 4 1212 10 7.952 12 10 1.5708C A

TLGJ

θ− × × − × ×

= = +× ×∑

0.0487 rad= − ................................................................................................................Ans.

7-82 A torque of 12.0 kN-m is supplied to the driving gear B of Fig. P7-82 by a motor. Gear A takes off 4.0 kN-m of torque, and the remainder is taken off by gear C. Determine the angle of twist of gear A with respect to gear C if both shafts are made of steel (G = 80 GPa) and have diameters of 75 mm.

SOLUTION

4 kN mABT = ⋅

4 12 8 kN mBCT = − = − ⋅

( )4 6 475 32 3.106 10 mmBCJ π= = ×

(c) ( )( )

( )( )( )( )

( ) ( )3 3

/ 9 6 9 6

4 10 1 8 10 2

80 10 3.106 10 80 10 3.106 10A CTLGJ

θ− −

× − ×= = +

× × × ×∑

0.0483 rad= − ................................................................................................................Ans.

7-83 An aluminum alloy (G = 3800 ksi) tube will be used to transmit a torque in a control mechanism. The tube has an outside diameter of 1.25 in. and a wall thickness of 0.065 in. Because of the tendency of thin sections to buckle, the maximum compressive stress in the tube must be limited to 8000 psi. Determine

(a) The maximum torque that can be applied.


303

(b) The angle of twist in a 3-ft length when a torque of 1000 lb-in. is applied. SOLUTION

( )4 4 41.25 1.12 32 0.08520 inJ π= − =

(a) ( )

max max

1.25 28000 psi

0.08520TTc

Jσ τ= = = ≤

1091 lb in.T ≤ ⋅ ................................................................................................................Ans.

(b) ( ) ( )

( )( )6

1000 3 120.1112 rad

3.8 10 0.08520TLGJ

θ×

= = =×

................................................................Ans.

7-84 A 2-m long hollow brass (G = 39 GPa) shaft has an outside diameter of 50 mm. The maximum shearing stress in the shaft must be limited to 50 MPa, and the angle of twist in the 2-m length must be limited to 2°. If the shaft will rotate at 1500 rpm and transmit 50 kW, determine the maximum inside diameter that can be used for the shaft.

SOLUTION


318.3 N mT = ⋅ To satisfy the stress specification:

( ) 6 2

max

318.3 0.02550 10 N/mTc

J Jτ = = ≤ × 6 40.15915 10 mJ −≥ ×

To satisfy the deformation specification:

( ) ( )( )9

318.3 22 0.03491 rad

39 10TLGJ J

θ = = ≤ ° =×

6 40.46758 10 mJ −≥ ×

( )4 4 6 4min 50 32 0.46758 10 mmiJ dπ= − = ×

34.9 mmid = .......................................................................................................................... Ans.

7-85 The inner surface of the aluminum alloy (G = 4000 ksi) sleeve A and the outer surface of the steel (G = 12,000 ksi) shaft B of Fig. P7-85 are smooth. Both the sleeve and the shaft are rigidly fixed to the wall at D. The 0.500-in. diameter pin C fills a hole drilled completely through a diameter of the sleeve and shaft. If the average shearing stress on the cross-sectional area of the pin at the interface between the shaft and the sleeve must not exceed 5000 psi, determine

(a) The maximum torque T that can be applied to the right end of the steel shaft B. (b) The maximum shearing stress in the aluminum alloy sleeve A when the torque of part (a) is applied. (c) The rotation of the right end of the shaft when the maximum torque T is applied. SOLUTION

( )4 4 43 2 32 6.3814 inaJ π= − = ( )4 42 32 1.5708 insJ π= =

( )2

2 5000 psi0.5 4

p ap

p

V TA

τπ

= = = 1963.5 lb in.aT = ⋅

Equilibrium: s aT T T+ =

Deformation: a sTLGJ

θ θ= = ( )

( )12 1.5708

0.73846 1450.0 lb in.4 6.3814s a aT T T= = = ⋅


304

(a) 3413.5 lb in. 3410 lb in.s aT T T= + = ⋅ ≅ ⋅ ........................................................................Ans.

(b) ( )1963.5 1.5

462 psi6.3814a

TcJ

τ = = = .......................................................................................Ans.

(c) ( ) ( )

( )( )/ 6

1450.0 12 3413.5 180.00418 rad

12 10 1.5708B DTLGJ

θ+

= = =×∑ ..........................................Ans.

7-86 The 100-mm diameter segment ABC of the shaft shown in Fig. P7-86 is initially not connected to the 60-mm diameter segment CD. Torque TB = 15 kN-m is applied at section B, and a secure connection between the two segments is then made at C, after which the torque TB is removed. Determine the resulting maximum shearing stress in segment CD after torque TB is removed. The moduli of rigidity are 40 GPa for ABC and 80 GPa for CD.

SOLUTION

( )4 6 4100 32 9.817 10 mmAB BCJ J π= = = ×

( )4 6 460 32 1.2723 10 mmCDJ π= = ×

With the torque 15 kN mBT = ⋅ applied:

( )( )( )( )

/ /

3

9 6

15 10 1.20.04584 rad

40 10 9.817 10

Bi B A C ATLGJ

θ θ θ

−

= = =

×= =

× ×

After the connection is made and BT is removed:

AC CDT T T= =

From the deformation of the shaft:

/ /C A C D Biθ θ θ+ =

( )

( )( )( )

( )( )9 6 9 6

2.2 1.60.04584 rad

40 10 9.817 10 80 10 1.2723 10T T

− −+ =

× × × ×

2149.9 N mAC CDT T T= = = ⋅

( ) 6 2

6

2149.9 0.03050.7 10 N/m 50.7 MPa

1.2723 10CDTcJ

τ −= = = × =×

.........................................Ans.

7-87 The motor shown in Fig. P7-87 delivers 200 hp at 300 rpm to a piece of equipment at B. The shaft is made from 0.4% C hot-rolled steel, which is available with diameters in increments of 1/8 in. Determine the shaft diameter required if failure is by yielding and the factor of safety is 1.5.

SOLUTION

0.4% C hot-rolled steel: 53 ksiyσ = 26.5 ksi2y

y

στ = = all

26.5 17.667 ksi1.5

y

FSτ

τ = = =

( )2 3002Power 200 hp

33,000 33,000TNTT

ππω= = = =

3501 lb ft 42.01 kip in.T = ⋅ = ⋅


305

( )( )3

3max 4

42.01 10 217.667 10 psi

32dTc

J dτ

π×

= = ≤ × 2.296 in.d ≥

328 82 2.296 2 in.< <

38Use 2 in.d = ...................................................................................................................... Ans.

7-88 The 160-mm diameter steel (G = 80 GPa) shaft shown in Fig. P7-88 has a 100-mm diameter bronze (G = 40 GPa) core inserted in 3 m of the right end and securely bonded to the steel. Determine

(a) The maximum shearing stress in each of the materials. (b) The rotation of the free end of the shaft. SOLUTION

( )4 6 4 6 4160 32 64.34 10 mm 64.34 10 mABsJ π −= = × = ×

( )4 4 6 4 6 4160 100 32 54.52 10 mm 54.52 10 mBCsJ π −= − = × = ×

( )4 6 4 6 4100 32 9.817 10 mm 9.817 10 mBCbJ π −= = × = ×

(a) In the interval AB: 160 75 85 kN mAT = − = ⋅

( )( )3

6 26

85 10 0.080105.69 10 N/m 105.7 MPa

64.34 10sTcJ

τ −

×= = = × ≅

×

In the interval BC: 75 kN mb sT T+ = ⋅

b sTLGJ

θ θ= = ( )( ) ( )( )9 6 9 640 10 9.817 10 80 10 54.52 10b sT L T L

− −=

× × × ×

11.107s bT T=

6.195 kN mbT = ⋅ 68.805 kN msT = ⋅

( )( )3

6 26

68.805 10 0.080100.96 10 N/m 101.0 MPa

54.52 10sTcJ

τ −

×= = = × ≅

×

( )( )3

6 26

6.195 10 0.05031.55 10 N/m 31.6 MPa

9.817 10bTcJ

τ −

×= = = × ≅

×

For the steel: max 105.7 MPaτ = .............................................................................................Ans.

For the bronze: max 31.6 MPaτ = ...............................................................................................Ans.

(b) ( )( )

( )( )( ) ( )

( )( )3 3

/ 9 6 9 6

85 10 2 68.805 10 1.5

80 10 64.34 10 80 10 54.52 10D A sTLGJ

θ θ− −

× ×= = = −

× × × ×∑

0.00936 rad= ......................................................................................................................... Ans.


306

Chapter 8 8-1 Determine the second moment of the shaded area shown in Fig. P8-1 with respect to (a) The x-axis. (b) The y-axis. SOLUTION

(a) ( ) ( )3 3

46 8 4 6736 in

3 3xI = − = .............................................................................................Ans.

(b) ( ) ( ) ( )3 3 3

42 6 6 4 6 2256 in

3 3 3yI = + − = .............................................................................Ans.

8-2 Determine the second moments of the shaded area shown in Fig. P8-2 with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area.

SOLUTION

( ) ( ) ( ) 2200 60 60 160 200 60 33,600 mmA = + + =

( ) ( ) ( )100 200 60 30 60 160 100 200 60

80 mm33,600Cy

+ + = =

( ) ( ) ( )3 3 3

6 460 200 160 60 60 200331.52 10 mm

3 3 3xI = + + = ×

( )22 6 6 4331.52 10 33,600 80 116.5 10 mmxC x CI I Ay= − = × − = × ..............................Ans.

( ) ( )3 3

6 4200 280 140 160318 10 mm

12 12yCI = − = × ...........................................................Ans.

8-3 Determine the second moments of the shaded area shown in Fig. P8-3 with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area.

SOLUTION

( ) ( )3 3

44 6 3 456.0 in

12 12xCI = − = ..........................................................................................Ans.

( ) ( ) ( )3 3 3

41 4 4 1 1 411.00 in

12 12 12yCI = + + = ..........................................................................Ans.

8-4 Four C305×45 channels are welded together to form the cross section shown in Fig. P8-4. Determine the second moments of the area with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area.

SOLUTION From Table A-6 for a C305 45× channel:

304.8 mmd = 6 467.4 10 mmxI = × 13.0 mmwt =

25690 mmA = 6 42.14 10 mmyI = × 17.1 mmCx =

( ) ( ) ( ) ( )26 6 6 42 67.4 10 2 2.14 10 148.4 5690 390 10 mmxCI = × + × + = ×

.............Ans.

( ) ( ) ( ) ( )26 6 6 42 67.4 10 2 2.14 10 17.1 5690 142.4 10 mmyCI = × + × + = ×

.............Ans.


307

8-5 Two 10×1-in. steel plates are welded to the flanges of an S18×70 I-beam as shown in Fig. P8-5. Determine the second moments of the area with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area.

SOLUTION From Table A-3 for an S18 70× beam:

4926 inxI = 424.1 inyI = 18.00 in.d =

( ) ( ) ( )

32 4 410 1

926 2 9.5 10 1 2733 in 2730 in12xCI

= + + × = ≅

..................................Ans.

( )3

41 1024.1 2 190.8 in

12yCI

= + =

...................................................................................Ans.

8-6 Determine the second moment of the shaded area shown in Fig. P8-6 with respect to (a) The x-axis. (b) The y-axis. SOLUTION

(a) ( ) ( ) ( )4 3 3

6 4120 120 30 60 6037.5 10 mm

16 3 3xIπ

= + − = × ..............................................Ans.

(b) ( ) ( ) ( )4 3 3

6 4120 30 120 60 6053.7 10 mm

16 3 3yIπ

= + − = × ..............................................Ans.

8-7 Determine the second moments of the shaded area shown in Fig. P8-7 with respect to (a) The x- and y-axes shown on the figure. (b) The x- and y-axes through the centroid of the area. SOLUTION

( )

3

4 3420 20 18.727 in.3 3C

Rxπ π

= − = − =

(a) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 4 4 2

2 2220 10 3 3 35 3 5

3 4 8 2xIπ π π

π

= − + − +

4 46667 770 385 5512 in 5510 in= − − = ≅ ................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )3 4 4 4 2

2 2 210 20 3 3 8 3 35 3 18.727

3 4 8 9 2yIπ π π

ππ

= − + − − +

4 426,667 770 4967 20,930 in 20,900 in= − − = ≅ .................................................Ans.

(b) ( ) ( ) ( )22 23

20 10 3 157.59 in2

Aπ

π= − − =

( ) ( ) ( ) ( )2 210 20 10 5 3 18.727 3 2

10.114 in.157.59Cx

π π − − = =

( ) ( )22 45512 5 157.59 1572 inxC x CI I y A= − = − = .........................................................Ans.

( ) ( )22 420,930 10.114 157.59 4810 inyC y CI I x A= − = − = ..........................................Ans.


308

8-8 Determine the second moments of the shaded area shown in Fig. P8-8 with respect to (a) The x- and y-axes shown on the figure. (b) The x- and y-axes through the centroid of the area. SOLUTION

( )

2

4 604160 160 185.46 mm3 3C

Rxπ π

= + = + =

(a) ( ) ( ) ( ) ( ) ( ) ( ) ( )

3 4 2 42 2 2160 120 60 60 30

60 60 3012 8 2 4xI

π π ππ

= + + − +

6 4 6 437.68 10 mm 37.7 10 mm= × ≅ × ..........................................................................Ans.

( ) ( ) ( )( )3

2120 160 160 120320 3

36 2yI

= +

( ) ( ) ( ) ( ) ( ) ( ) ( )

4 4 2 42 2 260 8 60 60 30

185.46 160 308 9 2 4

π π ππ

π

− − + − +

6 4 6 4245.78 10 mm 246 10 mm= × ≅ × ........................................................................Ans.

(b) ( ) ( ) ( )

22 2160 120 60

30 12,427 mm2 2

Aπ

π= − − =

( ) ( ) ( ) ( ) ( ) ( )2 2320 3 160 120 2 185.46 60 2 160 30

130.39 mm12,427Cx

π π + − = =

( ) ( ) ( ) ( ) ( ) ( )2 240 160 120 2 60 60 2 60 30

44.55 mm12, 427Cy

π π + − = =

( ) ( )22 6 6 437.68 10 44.55 12, 427 13.02 10 mmxC x CI I y A= − = × − = × ......................Ans.

( ) ( )22 6 6 4245.78 10 130.39 12,427 34.5 10 mmyC y CI I x A= − = × − = × ....................Ans.

8-9 Determine the second moments of the shaded area shown in Fig. P8-9 with respect to the x- and y-axes. SOLUTION

( )

2

4 4410 10 8.302 in.3 3C

Ryπ π

= − = − =

( )22 22 3 2 4 2 25.13 inA A Rπ π= = = =

( ) ( ) ( ) ( ) ( ) ( )3 4 4 4

220 10 4 8 4 48.302 25.13

3 8 9 8xIπ π

π

= − − + −

4 46667 1760 101 4806 in 4810 in= − − = ≅ ...............................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( )

3 4 42 210 20 4 4

5 25.13 15 25.133 8 8yI

π π = − + − +

4 426,667 729 5755 20,183 in 20, 200 in= − − = ≅ ..................................................Ans.


309

8-10 The maximum flexural stress on a transverse plane in a beam with a rectangular cross section 150 mm wide × 300 mm deep is 15 MPa. Determine the resisting moment Mr transmitted by the plane.

SOLUTION

( )3

6 4 6 4150 300337.5 10 mm 337.5 10 m

12I −= = × = ×

( ) 6 2

max 6

0.15015 10 N/m

337.5 10rr MM c

Iσ −= = = ×

×

333.8 10 N m 33.8 kN mrM = × ⋅ = ⋅ .................................................................................Ans.

8-11 The maximum flexural stress on a transverse plane in a beam with a rectangular cross section 4 in. wide × 8 in. deep is 1500 psi. Determine the resisting moment Mr transmitted by the plane.

SOLUTION

( )3

44 8170.67 in

12I = =

( )max

41500 psi

170.67rr MM c

Iσ = = =

364.0 10 lb in. 64.0 kip in.rM = × ⋅ = ⋅ ...............................................................................Ans.

8-12 Determine the maximum flexural stress required to produce a resisting moment Mr of �15 kN-m if the beam has the cross section shown in Fig. P8-12.

SOLUTION

( ) ( ) ( )

( ) ( ) ( )100 50 200 25 50 200 100 50 200

75 mm50 200 50 200 50 200Cy

+ + = =+ +

( ) ( ) ( )3 3 3100 125 300 75 200 253 3 3

I = + −

6 4 6 4106.25 10 mm 106.25 10 m−= × = ×

( ) ( )3

6 26

15 10 0.12517.65 10 N/m 17.65 MPa (T)

106.25 10rM yI

σ −

− ×= − = − = × =

×.............Ans.

8-13 Determine the maximum flexural stress required to produce a resisting moment Mr of +5000 lb-ft if the beam has the cross section shown in Fig. P8-13.

SOLUTION

( ) ( )( ) ( )

1 6 2 5 6 23 in.

6 2 6 2Cy+ = =+

( ) ( ) ( )3 3 3

42 5 6 3 4 1136 in

3 3 3I = + − =

( ) ( ) 25000 12 5

2206 lb/in. 2210 psi (C)136

rM cI

σ×

= − = − = − ≅ ................................Ans.

8-14 A timber beam consists of three 50×200-mm planks fastened together to form an I-beam 200 mm wide × 300 mm deep, as shown in Fig. P8-14. If the flexural stress at point A of the cross section is 7.5 MPa (C), determine the flexural stress

(a) At point B of the cross section. (b) At point C of the cross section (c) At point D of the cross section.


310

SOLUTION

( ) ( )3 3

6 4 6 4200 300 150 200350 10 mm 350 10 m

12 12I −= − = × = ×

( ) 6 2

6

0.1507.5 10 N/m

350 10rr A

A

MM yI

σ −= − = − = − ××

317.5 10 N mrM = × ⋅

(a) ( ) ( )3

6 26

17.5 10 0.1005.00 10 N/m 5.00 MPa (T)

350 10r B

BM y

Iσ −

× −= − = − = × =

×.........Ans.

(b) ( )( )3

6 26

17.5 10 0.0502.50 10 N/m 2.50 MPa (C)

350 10r C

CM y

Iσ −

×= − = − = − × =

×.........Ans.

(c) ( )( )3

6 26

17.5 10 0.1256.25 10 N/m 6.25 MPa (T)

350 10r D

DM y

Iσ −

× −= − = − = × =

×.........Ans.

8-15 A timber beam is made of three 2×6-in. planks fastened together to form an I-beam 6 in. wide × 10 in. deep. If the maximum flexural stress must not exceed 1200 psi, determine the maximum resisting moment Mr that the beam can support.

SOLUTION

( ) ( )3 3

46 10 4 6428 in

12 12I = − =

( )max

51200 psi

428rr MM c

Iσ = = =

3102.7 10 lb in. 102.7 kip in.rM = × ⋅ = ⋅ ...........................................................................Ans.

8-16 The beam of Fig. P8-16 is made of material that has a yield strength of 250 MPa. Determine the maximum resisting moment that the beam can support if yielding must be avoided.

SOLUTION

( ) ( )

( ) ( )25 100 50 150 200 37.5

100 mm100 50 200 37.5Cy

+ = =+

( ) ( ) ( )3 3 3

6 4100 100 62.5 50 37.5 15072.92 10 mm

3 3 3I = − + = ×

( ) 6 2

max 6

0.150250 10 N/m

72.92 10rr MM c

Iσ −= = = ×

×

3121.5 10 N m 121.5 kN mrM = × ⋅ = ⋅ .............................................................................Ans.

8-17 The maximum flexural stress on a transverse cross section of a beam with a 2×4-in. rectangular cross section must not exceed 10 ksi. Determine the maximum resisting moment Mr that the beam can support if the neutral axis is

(a) Parallel to the 4-in. side. (b) Parallel to the 2-in. side. SOLUTION

(a) ( )3

44 22.667 in

12I = =

( )max

110 ksi

2.667rr MM c

Iσ = = =

26.7 kip in.rM = ⋅ .................................................................................................................. Ans.


311

(b) ( )3

42 410.667 in

12I = =

( )

max

210 ksi

10.667rr MM c

Iσ = = =

53.3 kip in.rM = ⋅ .................................................................................................................. Ans.

8-18 A hardened steel (E = 210 GPa) bar with a 50-mm square cross section is subjected to a flexural form of loading that produces a flexural strain of +1200 µm/m at a point on the top surface of the beam. Determine

(a) The maximum flexural stress at the point. (b) The resisting moment Mr developed in the beam on a transverse cross section through the point. SOLUTION

(a) ( )( )9 6210 10 1200 10Eσ ε −= = × ×

6 2252.0 10 N/m 252 MPa= × = ............................................................ Ans.

(b) ( )3

6 4 6 450 500.5208 10 mm 0.5208 10 m

12I −= = × = ×

( ) 6 2

6

0.025252.0 10 N/m

0.5208 10rr MM y

Iσ −= − = − = ×

×

35.25 10 N m 5.25 kN mrM = − × ⋅ = − ⋅ ............................................................................Ans.

8-19 A steel (E = 29,000 ksi) bar with a rectangular cross section is bent over a rigid mandrel (R = 10 in.) as shown in Fig. P8-19. If the maximum flexural stress in the bar is not to exceed the yield strength (σy = 36 ksi) of the steel, determine the maximum allowable thickness h for the bar.

SOLUTION

10 in.2 2h hRρ = + = +

( )( )

210 2x x

E hcE Eh

σ ερ

= = =+

( )20 3620 0.0249 in.

29,000 36x

x

hE

σσ

= = =− −

............................................................................Ans.

8-20 An aluminum alloy (E = 73 GPa) bar with a rectangular cross section is bent over a rigid mandrel as shown in Fig. P8-19. The thickness h of the bar is 25 mm. If the maximum flexural stress in the bar must be limited to 200 MPa, determine the minimum allowable radius R for the mandrel.

SOLUTION

25 12.5 mm

2 2hR R Rρ = + = + = +

( )( )12.512.5x x

EcE ER

σ ερ

= = =+

( ) ( )12.5 73,000 12.5

12.5 12.5 4550 mm 4.55 m200x

ER

σ= − = − = = ............................Ans.

8-21 The load carrying capacity of an S24×80 American Standard beam (see Appendix A for dimensions) is to be increased by fastening two 8 × ¾-in. plates to the flanges of the beam, as shown in Fig. P8-21. The maximum flexural stress in both the original and modified beams must be limited to 15 ksi. Determine

(a) The maximum resisting moment that the original beam can support. (b) The maximum resisting moment that the modified beam can support. SOLUTION


312

(a) From Table A-3 for an S24 80× section: 24.00 in.d = 42100 inI =

( )

max

12.0015 ksi

2100rr MM c

Iσ = = =

2625 kip in. 2630 kip in.rM = ⋅ ≅ ⋅ ....................................................................................Ans.

(b) ( ) ( ) ( )

32 48 0.75

2100 2 12.375 8 0.75 3938 in12

I

= + + × =

( )

max

12.7515 ksi

3938rr MM c

Iσ = = =

4633 kip in. 4630 kip in.rM = ⋅ ≅ ⋅ ....................................................................................Ans.

8-22 Two L102×102×12.7-mm structural steel angles (see Appendix A for dimensions) are attached back to back to form a T-section, as shown in Fig. P8-22. Determine the maximum resisting moment Mr that can be supported by the beam if the maximum flexural stress must be limited to 125 MPa.

SOLUTION

From Table A-8 for an L102 102 12.7× × angle: 3 3 6 332.3 10 mm 32.3 10 mS −= × = ×

( )6 2

max 6125 10 N/m

2 32.3 10r rM M

Sσ

−= = = ×

×

38.08 10 N m 8.08 kN mrM = × ⋅ = ⋅ ..................................................................................Ans.

8-23 An I-beam is fabricated by welding two 16×2 in. flange plates to a 24×1-in. web plate. The beam is loaded in the plane of symmetry parallel to the web. On a section where the resisting moment Mr = 1400 kip-ft, determine the maximum flexural stress.

SOLUTION

( ) ( )3 3

416 28 15 2411,989 in

12 12I = − =

( )( ) 2

top

1400 12 1419.62 kip/in. 19.62 ksi (C)

11,989rM c

Iσ

×= − = − = − =

( ) ( ) 2

bottom

1400 12 1419.62 kip/in. 19.62 ksi (T)

11,989rM c

Iσ

× −= − = − = + =

max 19.62 ksi (T on bottom & C on top)σ = ...................................................................Ans.

8-24 Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment Mr = −4 kN-m if the beam has the cross section shown in Fig. P8-24. The beam is loaded in the plane of symmetry parallel to the web.

SOLUTION

( ) ( ) ( )( ) ( ) ( )

237.5 100 25 125 200 25 12.5 200 25102.50 mm

100 25 200 25 200 25Cy+ + = =

+ +

( ) ( ) ( ) ( )3 3 3 3

6 4 6 4

100 147.5 75 122.5 200 102.5 175 77.53 3 3 3

105.65 10 mm 105.65 10 m

I

−

= − + −

= × = ×


313

( )( )3

6 2top 6

4 10 0.14755.58 10 N/m 5.58 MPa (T)

105.65 10rM yI

σ −

− ×= − = − = + × =

×..........Ans.

( )( )3

6 2bot 6

4 10 0.10253.88 10 N/m 3.88 MPa (C)

105.65 10rM yI

σ −

− × −= − = − = − × =

×.......Ans.

8-25 Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment Mr = +30,000 lb-in. if the beam has the cross section shown in Fig. P8-25. The beam is loaded in the vertical plane of symmetry.

SOLUTION

( )4 24 0.8488 in.

3 3CRyπ π

= = =

( ) ( )4 44 4

42 8 28 1.7561 in8 9 8 9R RI

πππ π

= − = − =

( )( ) 2

top

30 1.151219.67 kip/in. 19.67 ksi (C)

1.7561rM c

Iσ = − = − = − = ..........................Ans.

( )( ) 2

bottom

30 0.848814.50 kip/in. 14.50 ksi (T)

1.7564rM c

Iσ

−= − = − = + = ...................Ans.

8-26 Determine the maximum resisting moment Mr that can be supported by a beam having the cross section shown in Fig. P8-26 if the maximum flexural stress must be limited to 120 MPa.

SOLUTION

( ) ( )3 4

6 4 6 450 50 100.5130 10 mm 0.5130 10 m

12 4I

π −= − = × = ×

( ) 6 2

max 6

0.025120 10 N/m

0.5130 10rr MM c

Iσ −= = = ×

×

32.46 10 N m 2.46 kN mrM = × ⋅ = ⋅ .................................................................................Ans.

8-27 A beam has the cross section shown in Fig. P8-27. On a section where the resisting moment is �30 kip-ft, determine

(a) The maximum tensile flexural stress. (b) The maximum compressive flexural stress. SOLUTION

( ) ( ) ( )

( ) ( ) ( )2 2

2 2

0.5 8 1 3 4 1 7 2 1.53.113 in.

8 1 4 1 2 1.5Cyπ

π

+ + − = =+ + −

( ) ( ) ( ) ( ) ( )3 3 3

28 1 1 2.113 1 1.8872.613 8 1

12 3 3I

= + × + +

( ) ( ) ( )

4 42 2 2 4

2 1.53.887 2 1.5 152.33 in

4π

π − + + − =

( ) ( ) 2

top

30 12 5.88713.91 kip/in. 13.91 ksi (T)

152.33rM c

Iσ

− ×= − = − = + = ...................Ans.


314

( )( ) 2

bottom

30 12 3.1137.36 kip/in. 7.36 ksi (C)

152.33rM c

Iσ

− × −= − = − = − = ...............Ans.

8-28 Determine the maximum flexural stress on a section where the resisting moment Mr = +100 kN-m if the beam has the cross section shown in Fig. P8-28. The beam is loaded in the vertical plane of symmetry.

SOLUTION

( ) ( ) ( )2 2250 25 150 25 50 17,854 mmA π= + + =

( )( ) ( ) ( )212.5 250 25 100 150 25 225 50

124.36 mm17,854Cy

π + + = =

( ) ( ) ( ) ( ) ( )3 3 3

2250 25 25 99.36 25 50.64111.86 250 25

12 3 3I

= + × + +

( ) ( ) ( )

42 2 6 4 6 450

100.64 50 172.24 10 mm 172.24 10 m4

ππ −

+ + = × = ×

( )( )3

6 2bot 6

100 10 0.1506487.5 10 N/m 87.5 MPa (T)

172.24 10σ −

− × −= − = + × =

×..................Ans.

8-29 The cantilever beam shown in Fig. P8-29 is subjected to a moment M = 20,000 lb-in. at its free end. The beam has a 2×2-in. cross section. Determine the maximum tensile flexural stress in the beam.

SOLUTION

( )3

42 21.3333 in

12I = =

( ) ( ) 2

bottom

20 115.00 kip/in. 15.00 ksi (T)

1.3333rM c

Iσ

−= − = − = + = ..............................Ans.

8-30 The cantilever beam shown in Fig. P8-30a is subjected to a moment M at its free end. The cross section of the beam is shown in Fig. P8-30b. If the allowable stresses are 110 MPa (T) and 170 MPa (C), determine the maximum moment that can be applied to the beam.

SOLUTION

( ) ( )( ) ( )

140 120 40 60 120 40100 mm

120 40 120 40Cy+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2

6 4 6 4

120 60 40 12040 120 60 40 60 120

12 12

21.76 10 mm 21.76 10 m

I

−

= + × + + ×

= × = ×

At the top of the beam, 110 MPa (T)σ =

( ) 6 2

top 6

0.060110 10 N/m

21.76 10rr MM y

Iσ −= − = − ≤ ×

×

339.89 10 N m 39.9 kN mrM ≤ × ⋅ ≅ ⋅

At the bottom of the beam, 170 MPa (C)σ =


315

( ) 6 2

bot 6

0.100170 10 N/m

21.76 10rr MM y

Iσ −

−= − = − ≥ − ×

×

336.99 10 N m 37.0 kN mrM ≤ × ⋅ ≅ ⋅

max 37.0 kN mM = ⋅ ...............................................................................................................Ans.

8-31 A steel pipe with an outside diameter of 4 in. and an inside diameter of 3 in. is simply supported at its ends and carries two concentrated loads, as shown in Fig. P8-31. On a section 5 ft from the right support, determine the flexural stress

(a) At point A (top, outside) on the cross section. (b) At point B (bottom, inside) on the cross section. SOLUTION

0 :CMΣ =� ( ) ( ) ( )20 2000 18 1000 3 0DR − − =

1950 lbDR =

0 :cutMΣ =� ( ) ( )1950 5 2000 3 0M− − =

3750 lb ftM = ⋅

( )4 4

42 1.5

8.590 in4

Iπ −

= =

(a) ( ) ( )3750 12 2

8.590r

AM y

Iσ

×= − = −

210, 477 lb/in. 10.48 ksi (C)= − ≅ ..............................................................................Ans.

(b) ( )( ) 23750 12 1.5

7858 lb/in. 7.86 ksi (T)8.590

rB

M yI

σ× −

= − = − = + ≅ .........................Ans.

8-32 A beam has the cross section shown in Fig. P8-32 and is loaded in the vertical plane of symmetry. On a section where the resisting moment Mr = +50 kN-m, determine

(a) The maximum tensile flexural stress. (b) The maximum compressive flexural stress. SOLUTION

( )2 2 250 40 2827 mmOA π= − =

( )4 4

6 450 40

2.898 10 mm4OI

π −= = ×

( ) ( )

( ) ( ) ( )82.5 30 135 200 2827

52.82 mm3 2827 30 135 30 150Cy

+ = =+ +

( ) ( ) ( ) ( ) ( ) ( )3

2 26 150 302 2.898 10 52.82 2827 52.82 30 150

12I

= × + + + ×

( ) ( ) ( ) ( ) ( ) ( )

32 2630 135

29.68 30 135 2.898 10 147.18 282712

+ + × + × +


319

8-40 A beam is loaded and supported as shown in Fig. P8-40. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < 4 m.

SOLUTION From overall equilibrium:

0 :BMΣ =� ( ) ( ) ( ) ( )20 2 15 4 6 8 0AR+ − = 50 kNAR =

For the initial section of the beam, 0 m 4 mx≤ ≤ :

0 :yF↑Σ = 15 0AR x V− − =

( )15 50 kNV x= − + .............................................................................................................Ans.

cut 0 :MΣ =� ( )( )15 2 0AM x x R x+ − =

( )27.5 50 kN mM x x= − + ⋅ ................................................................................................Ans.

8-41 A beam is loaded and supported as shown in Fig. P8-41. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < 10 ft.


0 :BMΣ =� ( ) ( ) ( ) ( )3000 200 10 5 500 2 10 0AR+ − − = 1200 lbAR =

For the middle section of the beam, 0 ft 10 ftx≤ ≤ :

0 :yF↑Σ = 200 0AR x V− − =

( )200 1200 lbV x= − + .........................................................................................................Ans.

cut 0 :MΣ =� ( )3000 200 2 0AM x x R x+ + − =

( )2 21200 3000 100 100 1200 3000 lb ftM x x x x= − − = − + − ⋅ ...................................Ans.

8-42 A beam is loaded and supported as shown in Fig. P8-42. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < L.


0 :yF↑Σ = 2 0AR wL wL+ − =

0 :AMΣ =� ( ) ( ) ( ) ( )( )22 3 2 2 3 2 0RM wL L wL wL L+ + − =

AR wL= 2RM wL= �

For the initial section of the beam, 0 x L≤ ≤ :

0 :yF↑Σ = 0AR wx V+ − =

( ) NV wL wx w L x= + = + .................................................................................................Ans.

cut 0 :MΣ =� ( ) ( )2 0A RM wx x R x M− − + =

( ) ( )( )2 2 2 22 2 2 2 N mM wx wLx wL w x Lx L= + − = + − ⋅ ........................................Ans.

8-43 The beam shown in Fig. P8-39 has a solid rectangular cross section that is 3 in. wide and 8 in. deep. Determine the maximum tensile flexural stress on a section at x = 10ft.

SOLUTION


320

From overall equilibrium:

0 :AMΣ =� ( ) ( ) ( )12 2000 4 5000 8 0BR − − = 4000 lbBR =

At 10 ftx = (using the right-hand segment of the beam),

cut 0 :MΣ =� ( ) 102 0BR M− = 10 8000 lb ftM = ⋅

( )3 43 8 12 128 inI = =

On the bottom of the beam, 4 in.y = −

( )( ) 28000 12 4

3000 lb/in. 3000 psi (T)128

rM yI

σ× −

= − = − = + = ..............................Ans.

8-44 The beam shown in Fig. P8-40 is an S254×52 steel section. On a section at x = 5 m, determine the maximum tensile and compressive flexural stresses.


0 :BMΣ =� ( ) ( ) ( ) ( )20 2 15 4 6 8 0AR+ − = 50 kNAR =

At 5 mx = (using the left-hand segment of the beam),

cut 0 :MΣ =� ( ) ( ) ( ) ( )5 15 4 3 5 0AM R+ − =

5 70 kN mM = ⋅

From Table A-4 for an S254 52× section: 254.00 mmd = 6 461.2 10 mmI = ×

At the bottom of the beam, 127.0 mmy = −

( )( )3

6 2bot 6

70 10 0.1270145.26 10 N/m 145.3 MPa (T)

61.2 10σ −

× −= − = + × ≅

×..................Ans.

At the top of the beam, 127.0 mmy = +

( )( )3

6 2top 6

70 10 0.1270145.26 10 N/m 145.3 MPa (C)

61.2 10σ −

× += − = − × ≅

×..................Ans.

8-45 A beam is loaded and supported as shown in Fig. P8-45. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam

(a) In the interval �3 ft < x < 0. (b) In the interval 0 < x < 2 ft. (c) In the interval 2 ft < x < 8 ft. (d) In the interval 8 ft < x < 10 ft. SOLUTION From overall equilibrium:

0 :AMΣ =� ( ) ( ) ( ) ( ) ( )2000 3 10 1000 6 5 2000 8 0BR+ − − = 4000 lbBR =

0 :BMΣ =� ( ) ( ) ( ) ( ) ( )2000 13 10 1000 6 5 2000 2 0AR− + + = 6000 lbAR =

(a) 2000 lbV = − .......................................................................................................................... Ans.

( ) ( )2000 3 2000 6000 lb ftM x x= − + = − − ⋅ .................................................................Ans.

(b) 2000 6000 2000 4000 lbAV R= − = − = .........................................................................Ans.


321

cut 0 :MΣ =� ( )2000 3 0AM x R x+ + − =

( ) ( )2000 3 4000 6000 lb ftAM R x x x= − + = − ⋅ ...........................................................Ans.

(c) ( ) ( )2000 1000 2 1000 6000 lbAV R x x= − − − = − + ....................................................Ans.

cut 0 :MΣ =� ( ) ( ) 22000 3 1000 2 02A

xM x R x x − + + − + − =

( ) ( ) 22000 3 1000 22A

xM R x x x − = − + − −

( )2500 6000 8000 lb ftx x= − + − ⋅ ................................................................................Ans.

(d) 4000 lbBV R= − = − ..............................................................................................................Ans.

( ) ( )10 4000 40,000 lb ftBM R x x= − = − + ⋅ ..................................................................Ans.

8-46 A beam is loaded and supported as shown in Fig. P8-46. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam

(a) In the interval �2 m < x < 0. (b) In the interval 0 < x < 4 m. (c) In the interval 4 m < x < 6 m. (d) In the interval 6 m < x < 10 m. SOLUTION From overall equilibrium:

0 :AMΣ =� ( ) ( ) ( ) ( )6 10 12 6 1 24 6 0BR+ − − = 21 kNBR =

0 :BMΣ =� ( ) ( ) ( ) ( )6 12 6 9 24 4 10 0AR+ + − = 75 kNAR =

(a) ( ) ( )12 2 12 24 kNV x x= − + = − − ....................................................................................Ans.

( ) ( )226 12 2 6 24 30 kN m2

xM x x x+ = − − + = − − − ⋅

........................................Ans.

(b) ( ) ( )12 2 12 51 kNAV x R x= − + + = − + ...........................................................................Ans.

( ) ( )226 12 2 6 51 30 kN m2 A

xM x R x x x+ = − − + + = − + − ⋅

............................Ans.

(c) 24 3 kNBV R= − = ...............................................................................................................Ans.

( ) ( ) ( )10 24 6 3 66 kN mBM R x x x= − − − = + ⋅ ............................................................Ans.

(d) 21 kNBV R= − = − .................................................................................................................Ans.

( ) ( )10 21 210 kN mBM R x x= − = − + ⋅ ..........................................................................Ans.

8-47 A beam is loaded and supported as shown in Fig. P8-47. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment

in the beam. SOLUTION From overall equilibrium:


322

0 :BMΣ =� ( ) ( ) ( )1500 12 2 12 3 12 0AR− = 3000 lbAR =

(a) ( ) ( )21500 12

62.5 3000 lb2A

x xV R x= − = − + ..............................................................Ans.

( ) ( )31500 12

20.833 3000 lb ft2 3A

x x xM R x x x = − = − + ⋅

.................................Ans.

(b) The shear force is monotonically decreasing on the interval ( )125dV dx x= − . Therefore, the maximum will occur either at the beginning or the end of the interval.

0 3000 lbxV = =

( )212 max62.5 12 3000 6000 lbxV V= = − + = − = .................................................................Ans.

The maximum bending moment in the interval will occur when

262.5 3000 0dM dx x= − + = 3000 62.5 6.928 ftx = =

6.928 13,858 lb ftxM = = ⋅

Checking the ends of the interval

0 0 lb ftxM = = ⋅ 12 0 lb ftxM = = ⋅

Therefore max 6.928 13,858 lb ft 13.86 kip ftxM M == = ⋅ ≅ ⋅ .................................................Ans.



0 :AMΣ =� ( ) ( ) ( )25 5 2 5 3 5 0BR− = 20.833 kNBR =

(a) ( ) ( ) ( )225 51 5 2.5 25 41.667 kN

2 5 B

xV x R x x

− = − − = − +

...................................Ans.

( ) ( ) ( )225 5 55

10 3B

x xM R x

− −= − −

( )3 20.8333 12.5 41.667 kN mx x x= − + ⋅ ...................................................................Ans.

(b) The shear force is monotonically decreasing on the interval ( )5 25 0dV dx x= − < . Therefore, the maximum will occur either at the beginning or the end of the interval.

0 max41.667 kN 41.7 kNxV V= = ≅ = ..................................................................................Ans.

5 20.833 kNxV = = −


22.5 25 41.667 0dM dx x x= − + = 2.113 m, 7.89 mx =

2.113 40.1 kN mxM = = ⋅

(The second root, 7.89 m is not in the interval and is not a valid solution.) Checking the ends of the interval


323

0 0 kN mxM = = ⋅ 5 0 kN mxM = = ⋅

Therefore 2.113 max40.1 kN mxM M= = ⋅ = ..............................................................................Ans.

8-49 An S8×23 steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-49. (a) Using the coordinate axes shown, write equations for the shear force V and the bending moment M for

any section of the beam in the interval 6 ft < x < 8 ft. (b) Determine the flexural stress at a point 1 in. below the top of the beam on a section at x = 3 ft. (c) Determine the maximum flexural stress on a section at x = 3 ft. SOLUTION From overall equilibrium:

0 :BMΣ =� ( ) ( ) ( ) ( ) ( ) ( )500 6 7 3000 2 10 800 5 2.5 0AR+ − − = 1700 lbAR =

(a) ( )500 6 1300 lbAV R= − = − ...............................................................................................Ans.

( ) ( ) ( )500 6 3 1300 9000 lb ftAM R x x x= − − = − + ⋅ ................................................Ans.

(b) From Table A-3 for an S8 23× section: 8.00 in.d = 464.9 inI =

216.2 inS =

( ) ( ) ( )3 3 500 3 1.5 2850 lb ftAM R= − = ⋅

( ) ( ) 22850 12 3

1581 lb/in. 1581 psi (C)64.9

MyI

σ×

= − = − = − = ....................................Ans.

(c) ( ) 22850 12

2111 lb/in. 2110 psi (C top & T bottom)16.2

MS

σ×

= = − = ≅ .................Ans.

8-50 An S178×30 steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-50. (a) Using the coordinate axes shown, write equations for the shear force V and the bending moment M for

any section of the beam in the interval 0.5 m < x < 2.5 m. (b) Determine the flexural stress at a point 15 mm above the bottom of the beam on a section at x = 1.5 m. (c) Determine the maximum flexural stress on a section at x = 1.5 m. SOLUTION From overall equilibrium:

0 :BMΣ =� ( ) ( ) ( ) ( )9 4 15 2 1.5 3 0AR+ − = 27 kNAR =

(a) ( ) ( )9 15 0.5 15 25.5 kNAV R x x= − + − − = − + ..............................................................Ans.

( ) ( ) 0.59 1 15 0.52A

xM R x x x − = − + − −

( )27.5 25.5 10.88 kN mx x= − + − ⋅ ...............................................................................Ans.

(b) From Table A-4 for an S178 30× section: 177.8 mmd = 6 417.6 10 mmI = ×

3 3198 10 mmS = ×

( ) ( )21.5 7.5 1.5 25.5 1.5 10.875 10.500 kN mM = − + − = ⋅

( )177.8 2 15 73.9 mmy = − + = −


324

( )( )3

6 26

10.5 10 0.073944.1 10 N/m 44.1 MPa (T)

17.6 10MyI

σ −

× −= − = − = + × =

×............Ans.

(c) ( )3

6 26

10.5 1053.0 10 N/m 53.0 MPa (C top & T bottom)

198 10MS

σ −

×= = = × =

×...........Ans.

8-51 A beam is loaded and supported as shown in Fig. P8-51. On a section 6 ft to the right of support A, determine the maximum tensile and compressive flexural stresses if the load P = 3000 lb.


0 :BMΣ =� ( ) ( )3000 6 10 0AR− = 1800 lbAR =

( ) ( )( ) ( )

0.5 4 1 5 1 83.500 in.

4 1 1 8Cy+ = =

+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 44 1 1 8

3 4 1 1.5 1 8 97.00 in12 12

I

= + × + + × =

( ) ( )6 6 3000 2 4800 lb ftAM R= − = ⋅

At the top of the section, 5.5 in.y = +

( ) ( ) 24800 12 5.5

3266 lb/in. 3.27 ksi (C)97.00

MyI

σ×

= − = − = − ≅ ................................Ans.

At the bottom of the section, 3.5 in.y = −

( )( ) 24800 12 3.5

2078 lb/in. 2.08 ksi (T)97.00

MyI

σ× −

= − = − = + ≅ .............................Ans.

8-52 A beam is loaded and supported as shown in Fig. P8-52. On a section 3 m to the right of support A, determine the maximum tensile and compressive flexural stresses if w = 3.5 kN/m.


0 :AMΣ =� ( ) ( ) ( )4 3500 2 1 0BR − = 1750 NBR =

( ) ( )( ) ( )

140 120 40 60 40 120100 mm

120 40 40 120Cy+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2

6 4 6 4

120 40 40 12040 120 40 40 40 120

12 12

21.76 10 mm 21.76 10 m

I

−

= + × + + ×

= × = ×

( )3 1 1750 N mBM R= = ⋅

At the top of the section, 60 mmy = + :

( ) ( ) 6 2

6

1750 0.0604.83 10 N/m 4.83 MPa (C)

21.76 10MyI

σ −= − = − = − × =×

........................Ans.

At the bottom of the section, 100 mmy = −


325

( )( ) 6 2

6

1750 0.1008.04 10 N/m 8.04 MPa (T)

21.76 10MyI

σ −

−= − = − = + × =

×......................Ans.


in the beam. SOLUTION

( )10 10

0 0

10

0

0.5 0.5 1000sin 10

5000 10,000cos lb10

A BR R wds s ds

s

π

ππ π

= = =

= − =

∫ ∫

(a) 0 0

0

10,000 10,000 10,0001000sin cos10 10

xx x

As sV R wds dsπ π

π π π = − = − = +

∫ ∫

10,000 cos lb 3.18cos kip

10 10x xπ π

π= ≅ .........................................................................Ans.

( ) ( )0 0

20 0 0

10,000 1000 sin10

10,000 10,000 100,000 10,000cos sin cos10 10 10

x x

A

x x x

x sM R x w x s ds x s ds

x x s s ss

ππ

π π ππ π π π

= − − = − −

= + + −

∫ ∫

2

100,000 sin lb ft 10.13sin kip ft10 10

x xπ ππ

= ⋅ ≅ ⋅ .......................................................Ans.

(b) max 0 10 10,000 lb 3.18 kipx xV V V π= == = = ≅ ..................................................................Ans.

2max 5 100,000 lb ft 10.13 kip ftxM M π== = ⋅ ≅ ⋅ .........................................................Ans.



0 :BMΣ =� ( ) ( )4

04 4 0AR w s ds− + − =∫

( ) ( )

( ) ( ) ( )

4

0

4 4 4

2 20 0 0

0.25 4 25cos 8

200 400 50 400sin 8 cos 8 sin 8 kN

AR s s ds

ss s s

π

π π ππ π π π

= −

= − − =

∫

(a) 2 20 00

400 400 20025cos sin8 8

xx x

As sV R wds dsπ π

π π π = − = − = −

∫ ∫

( )2

400 200 sin 8 kNxππ π

= − .............................................................................................Ans.


326

( ) ( )20 0

2 20 0 0

400 25 cos8

400 200 1600 200sin cos sin8 8 8

x x

A

x x x

x sM R x w x s ds x s ds

x x s s ss

ππ

π π ππ π π π

= − − = − −

= − + +

∫ ∫

( ) ( )2 2

400 16004 cos 8 kN mx xππ π

= − + ⋅ ...................................................................Ans.

(b) 2max 0 400 kN 40.5 kNxV V π== = ≅ .................................................................................Ans.


( )2

400 200 sin 8 0dM xdx

ππ π

= − = 0.6901 rad8xπ = 1.7573 mx =

max 1.7573 34.1 kN mxM M == = ⋅ ............................................................................................Ans.

8-55 A beam is loaded and supported as shown in Fig. P8-55. Using the coordinate axes shown, (a) Write equations for the shear force V and bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment


0 :BMΣ =� ( ) ( )10

010 10 0AR w s ds− + − =∫

( )( )103 410 2

00

1 1010 10 833.3 lb10 3 4A

s sR s s ds

= − = − =

∫

(a) ( )2 3

0 0833.3 10 3.333 833.3 lb

x x

AV R wds s ds x= − = − ≅ − +∫ ∫ ..................................Ans.

( ) ( )2

0 0833.3 10

x x

AM R x w x s ds x s x s ds= − − = − −∫ ∫

( )40.8333 833 lb ftx x≅ − + ⋅ ........................................................................................Ans.

(b) ( )3max 10 833.3 3.333 10 2500 lbxV V == = − = − .................................................................Ans.


33.333 833.3 0dM dx x= − + = 6.300 ftx =

max 6.300 3937 lb ft 3.94 kip ftxM M == = ⋅ ≅ ⋅ ....................................................................Ans.



0 :BMΣ =� ( ) ( )8

08 8 0AR w s ds− + − =∫


327

( )( ) ( )8 82 3 2

0 0

84 32

0

1 164 8 8 64 5128 8

1 8 32 512 213.3 kN8 4 3

AR s s ds s s s ds

s s s s

= − − = − − +

= − − + =

∫ ∫

(a) ( )3

2

0 0213.3 64 213.3 64

3x x

AxV R wds s ds x= − = − − = − +∫ ∫

( )30.333 64.0 213 kNx x≅ − + ........................................................................................Ans.

( ) ( )( )2

0 0

4 42 2

213.3 64

213.3 32 644 3

x x

AM R x w x s ds x s x s ds

x xx x x

= − − = − − −

= − + + −

∫ ∫

( )4 20.0833 32.0 213 kN mx x x≅ − + ⋅ .......................................................................Ans.

(b) The maximum shear force in the interval will occur when

2 64.0 0dV dx x= − = 8 mx = 8 128.0 kNxV = = −

Checking the beginning of the interval

0 max213.3 kN 213 kNxV V= = ≅ = ......................................................................................Ans.


30.3333 64.0 213.3 0dM dx x x= − + = 3.570 mx =

max 3.570 367 kN mxM M == = ⋅ ..............................................................................................Ans.

8-57 Two C10×15.3 steel channels (see Appendix A) are placed back to back to form a 10-in. deep beam, as shown in Fig. P8-57. The beam is 10 ft long and is simply supported at its ends. The beam carries a uniformly distributed load of 1000 lb/ft over its entire length and a concentrated load P at the center of the span, as shown in Fig. P8-57. If the maximum permissible flexural stress at the center of the span is 16,000 psi, determine the maximum permissible value of the concentrated load P.


0 :BMΣ =� ( ) ( ) ( ) ( )1000 10 5 5 10 0AP R+ − =

5000 0.5 lbAR P= +

( ) ( ) ( ) ( )5 1000 5 2.5 12,500 2.5 lb ftP AM R P= − = + ⋅

From Table A-5 for a C10 15.3× section: 313.5 inS =

( )( ) 3

max

12,500 2.5 1216 10 psi

2 13.5PM

Sσ

+= = ≤ ×

×

9400 lbP ≤ ............................................................................................................................. Ans.

8-58 Two 50×200-mm structural timbers are used to fabricate a beam with an inverted T cross section, as shown in Fig. P8-58. The beam is simply supported at the ends and is 4 m long. If the maximum flexural stress must be limited to 30 MPa, determine

(a) The maximum moment that can be resisted by the beam. (b) The largest concentrated load that can be supported at the center of the span. (c) The largest uniformly-distributed load (over the entire span) that can be supported by the beam.


328

SOLUTION

( ) ( )

( ) ( )25 200 50 150 50 200

87.5 mm200 50 50 200Cy

+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2

6 4 6 4

200 50 50 20062.5 200 50 62.5 50 200

12 12

113.54 10 mm 113.54 10 m

I

−

= + × + + ×

= × = ×

(a) ( ) 6 2

max 6

0.162530 10 N/m

113.54 10MMc

Iσ −= = − = ×

×

320.96 10 N m 21.0 kN mM = × ⋅ ≅ ⋅ ................................................................................Ans. For a concentrated load P:

2A BR R P= =

( )max 2 2AM M R P= = =

max 21.0 kNP M= = .............................................................................................................Ans.

For a distributed load w:

4 2 2A BR R w w= = =

( ) ( )( )max 2 2 2 1 2AM M R w w= = − =

max 2 10.48 kN/mw M= = ..................................................................................................Ans.

8-59 The supports for the beam shown in Fig. P8-59 are symmetrically located. If the distance d from the supports to the ends of the beam is adjustable, compute and plot

(a) The bending moment M(x) in the beam for d = 2 ft, 3 ft, and 4 ft as a function of x (0 ft ≤ x ≤ 15 ft). (b) The maximum bending moments (Mmax)AB in segment AB and (Mmax)BC in segment BC as functions of d

(0 ft ≤ d ≤ 6 ft). (c) What value of d gives the smallest Mmax in the beam? SOLUTION

( )2 1200 15 2 9000 lbB CR R wL= = = =

(a) For the initial section of the beam, 0 ft ftx d≤ ≤ :

( ) 21200 2 600 lb ftM x x x= − = − ⋅ ...................................................................................Ans.

For the middle section of the beam, ( ) ft 15 ftd x d≤ ≤ − :

( ) ( ) ( )21200 2 600 9000 9000 lb ftAM R x d x x x x d= − − = − + − ⋅ ..........................Ans.

For the last section of the beam, ( )15 ft 15 ftd x− ≤ ≤ :

( ) ( ) ( )21200 15 15 2 600 18,000 135,000 lb ftM x x x x= − − − = − + − ⋅ ..............Ans.

(b) 2600 lb ftx dM d= = − ⋅ ( )7.5 33,750 9000 lb ftxM d= = − ⋅

( )27.5 600 33,750 9000x d xM M d d= == = − = − 3.107 ftd =

When 3.107 ftd ≤ :


329

( ) 2max 600 lb ftx dABM M d== = − ⋅ .....................................................................................Ans.

( ) ( )max 7.5 33,750 9000 lb ftxBCM M d== = − ⋅ ................................................................Ans.

When 3.107 ft 6 ftd≤ ≤ :

( ) ( ) 2max max 600 lb ftx dAB BCM M M d== = = − ⋅ ...............................................................Ans.

The smallest maximum moment in the beam occurs when

3.107 ft 3.11 ftd = = ............................................................................................................Ans.

8-60 An overhead crane consists of a carriage which moves along a beam as shown in Fig. P8-60. If the carriage moves slowly along the beam, compute and plot

(a) The bending moment M(x) in the beam for b = 1.2 m, 2 m, and 2.8 as a function of x (0 m ≤ x ≤ 5 m). (b) The bending moments MB under the left wheel and MC under the right wheel as functions of b (0.3 m ≤ b

≤ 4.7 m). (c) What is the largest bending moment in the beam? When and where does it occur? SOLUTION From overall equilibrium:

0 :AMΣ =� ( )5 2500 0DR b− = 500 NDR b=

0 :DMΣ =� ( ) ( )2500 5 5 0Ab R− − = ( )500 5 NAR b= −

(a) For the initial section of the beam, ( )0 m 0.3 mx b≤ ≤ − :

( ) ( )500 5 2500 500 N mAM R x b x x bx= = − = − ⋅

For the middle section of the beam, ( ) ( )0.3 m 0.3 mb x b− ≤ ≤ + :

( )( ) ( )

( )

1250 0.3

500 5 1250 0.3

1250 500 1250 375 N m

AM R x x b

b x x b

x bx b

= − − +

= − − − +

= − + − ⋅

For the last section of the beam, ( )0.3 m 5 mb x+ ≤ ≤ :

( ) ( )

( )5 500 5

2500 500 N mDM R x b x

b bx

= − = −

= − ⋅


330

(b) ( ) ( ) ( )2500 5 0.3 500 2650 750 N mBM b b b b= − − = − + − ⋅

( ) ( )2500 4.7 2350 500 N mCM b b b b= − = − ⋅

(c) The maximum bending moment occurs under the right wheel when 2.35 mb ≅ and under the left wheel when 2.65 mb ≅ :

max 2.76 kN mM ≅ ⋅ ...............................................................................................................Ans.

8-61 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-61. SOLUTION From overall equilibrium:

0 :EMΣ =� ( ) ( )2000 12 4000 8+

( ) ( )6000 4 16 0AR+ − =

5000 lb AR = ↑

0 :AMΣ =� ( ) ( )16 2000 4ER −

( ) ( )4000 8 6000 12 0− − =

7000 lbER = ↑


0 :AMΣ =�

( ) ( ) ( )15 7.5 20 2 5 0CR+ − =

30.5 kNCR = ↑

0 :CMΣ =�

( ) ( ) ( )5 15 2.5 20 3 0AR + − =

4.5 kNAR = ↑


331


0 :CMΣ =�

( ) ( ) ( ) ( )2000 12 10 8000 8 16 0AR− − =

11,000 lbAR = ↑

0 :AMΣ =�

( ) ( ) ( ) ( )16 2000 12 6 8000 24 0CR − − =

21,000 lbCR = ↑


0 :DMΣ =�

( ) ( ) ( ) ( )10 7 20 3 15 5 10 0AR+ − − =

5.5 kNAR = ↑

0 :AMΣ =�

( ) ( ) ( ) ( )10 10 3 20 7 15 15 0DR − − − =

39.5 kNDR = ↑


0 :EMΣ =�

( ) ( ) ( ) ( )300 5 13.5 2000 7 3000 4+ +

( ) ( ) ( )400 5 2.5 11 0BR− − =

3750 lbBR = ↑

0 :BMΣ =�

( ) ( ) ( ) ( )11 300 5 2.5 2000 4ER + −

( ) ( ) ( )3000 7 400 5 13.5 0− − =

4750 lbER = ↑


332


0 :DMΣ =�

( ) ( ) ( ) ( )20 6 30 4 2 8 0AR+ − =

45 kNAR = ↑

0 :AMΣ =�

( ) ( ) ( ) ( )8 20 2 30 4 6 0DR − − =

95 kNDR = ↑


0 :DMΣ =�

( ) ( ) ( )1500 15 250 10 5+

( )1000 20 0AR+ − =

1800 lbAR = ↑

0 :AMΣ =�

( ) ( )20 1000 1500 5DR + −

( ) ( )250 10 15 0− =

2200 lbDR = ↑


0 :EMΣ =�

( ) ( ) ( ) ( )9 18 6 9 36 4 10 0BR+ + − =

112.5 kNBR = ↑

0 :BMΣ =�

( ) ( ) ( ) ( )10 9 18 6 1 36 6 0ER + − − =

31.5 kNER = ↑


333

8-69 The beam shown in Fig. P8-69a has the cross section shown in Fig. P8-69b. Determine the maximum tensile and compressive flexural stresses in the beam.


0 :CMΣ =�

( ) ( ) ( ) ( )6000 400 10 5 1000 2 10 0BR+ − − =

2400 lbBR = ↑

0 :BMΣ =�

( ) ( ) ( ) ( )6000 400 10 5 10 1000 12 0CR− + − =

2600 lbCR = ↑

( ) ( )( ) ( )

7 6 2 3 2 65 in.

6 2 2 6Cy+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 46 2 2 6

2 6 2 2 2 6 136.0 in12 12

I

= + × + + × =

From the moment diagram,

max 6000 lb ftM = − ⋅ max 1200 lb ftM = + ⋅

At the top of the beam, 3.00 in.y = +

( ) ( ) 2

top

6000 12 3.001588 lb/in. 1588 psi (T)

136.0MyI

σ− ×

= − = − = + = ......................Ans.

At the bottom of the beam, 5.00 in.y = −

( )( ) 2

bot

6000 12 5.002647 lb/in. 2650 psi (C)

136.0MyI

σ− × −

= − = − = − ≅ ..................Ans.

At the section where 1200 lb ftM = + ⋅ , the tensile and compressive stresses are both less than these values.

8-70 A W102 × 19 wide-flange beam is loaded and supported as shown in Fig. P8-70. Determine the maximum tensile and compressive flexural stresses in the beam.


0 :EMΣ =� ( ) ( ) ( ) ( )3 2 7 6 5 4 2+ +

( ) ( )3 2 2 6 0BR+ − − =

15.333 kNBR = ↑

0 :BMΣ =� ( ) ( ) ( )6 2 3 2 1 6ER − + +

( ) ( ) ( )5 4 4 3 4 0− − =

13.667 kNER = ↑

From the moment diagram


334

max 17.38 kN mM = + ⋅

For a W102 19× section:

3 489.5 10 mmS = ×

3

6 26

17.38 10 194.19 10 N/m 194.2 MPa (T bottom & C top)89.5 10

MS

σ −

×= = = × ≅×

.............Ans.

At the section where 4 kN mM = − ⋅ , the tensile and compressive stresses are both less than these values.

8-71 A beam is loaded and supported as shown in Fig. P8-71a. Two 10 × 1-in. steel plates are welded to the flanges of an S18 × 70 American standard I-beam to form the cross section shown in Fig. P8-71b. Determine the maximum tensile and compressive flexural stresses in the beam.


0 :DMΣ =�

( ) ( ) ( ) ( )3 17.5 10 7.5 5 7.5 27.5 0AR+ − − =

3.273 kipAR = ↑

0 :AMΣ =�

( ) ( ) ( ) ( )27.5 3 10 10 20 5 35 0DR − − − =

14.727 kipDR = ↑


max 37.5 kip ftM = − ⋅

For an S18 70× section: 18.00 in.d = 4926 inXXI =

( ) ( ) ( )

32 410 1

926 2 9.5 10 1 2733 in12C PI I I

= + = + + × =

At the top of the beam, ( )18 2 1.00 10.00 in.y = + =

( )( ) 2

top

37.5 12 10.001.647 kip/in. 1.647 ksi (T)

2733MyI

σ− ×

= − = − = + = ................Ans.

At the bottom of the beam, 10.00 in.y = −

( )( ) 2

bot

37.5 12 10.001.647 kip/in. 1.647 ksi (C)

2733MyI

σ− × −

= − = − = − = .............Ans.

8-72 A beam is loaded and supported as shown in Fig. P8-72a. Two 250 × 25-mm steel plates and two C254 × 45 channels are welded together to form the cross section shown in Fig. P8-72b. Determine the maximum tensile and compressive flexural stresses in the beam.

SOLUTION

For a C254 45× channel: 254.0 mmd = 6 442.9 10 mmXXI = ×

( ) ( ) ( ) ( )3

26 250 252 42.9 10 2 139.5 250 25

12C PI I I

= + = × + + ×


335

6 4 6 4329.4 10 mm 329.4 10 m−= × = × From overall equilibrium:

0 :DMΣ =� ( ) ( ) ( ) ( )40 2 30 4 6 8 0AR+ − =

100 kNAR = ↑

0 :AMΣ =� ( ) ( ) ( ) ( )8 30 4 2 40 6 0DR − − =

60 kNDR = ↑

From the moment diagram

max 166.667 kN mM = + ⋅

At the top of the beam, ( )254 2 25 152.0 mmy = + = +

( )( )3

2top 6

166.667 10 0.15276.91 N/m 76.9 MPa (C)

329.4 10MyI

σ −

×= − = − = − ≅

×.............Ans.

At the bottom of the beam, 152.0 mmy = −

( )( )3

2bot 6

166.667 10 0.15276.91 N/m 76.9 MPa (T)

329.4 10MyI

σ −

× −= − = − = + ≅

×..........Ans.

8-73 A WT8 × 25 structural steel T-section is loaded and supported as shown in Fig. P8-73. Determine the maximum tensile and compressive flexural stresses in the beam.


0 :BMΣ =� ( ) ( ) ( ) ( )6000 6 1000 12 6 12 0AR+ − =

9000 9.00 kipAR = = ↑


max 36.0 kip ftM = ⋅

For a WT8 25× section: 8.130 in.d =

442.3 inXXI = 1.89 in.top cy y= =

( ) ( )8.130 1.89 6.24 in.bot cy d y= − − = − − = −

( )( ) 2

top

36.0 12 1.8919.302 kip/in. 19.30 ksi (C)

42.3MyI

σ×

= − = − = − ≅ ..................Ans.

( )( ) 2

bot

36.0 12 6.2463.73 kip/in. 63.7 ksi (T)

42.3MyI

σ× −

= − = − = + ≅ ....................Ans.

8-74 A 203 × 203-mm nominal size structural timber (see Appendix A) is supported by two brick columns, as shown in Fig. P8-74. Assume that the brick columns transmit only vertical forces to the timber beam. The beam supports the roof of a building through three timber columns. Columns A and C each transmit forces of 8 kN to the beam; column B transmits a force of 10 kN.

(a) Draw complete shear force and bending moment diagrams for the beam. (b) Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION (a) From overall equilibrium:


336

0 :CMΣ =�

( ) ( ) ( )10 3 8 6 6 0AR+ − =

13 kNAR = ↑

(b) From the moment diagram

max 15.00 kN mM = + ⋅

For a 203 203 mm× − timber:

3 3 6 31150 10 mm 1150 10 mS −= × = ×

3

6 26

15.00 10 13.04 10 N/m1150 10

MS

σ −

×= = = ××

13.04 MPa (T bottom & C top)= .................................................................................Ans.


0 :DMΣ =� ( ) ( )1000 8 2000 6BR− −

( ) ( ) ( )500 4 4 2000 2+ +

( )4000 1000 2 0+ − =

3333 lbBR = ↑

0 :BMΣ =� ( ) ( )6 2000 1000 2DR − +

( ) ( ) ( )500 4 2 2000 4− −

( )4000 1000 8 0+ − =

2667 lbDR = ↑

8-76 An S457 × 81 American standard steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-76. The two segments of the beam are connected with a smooth pin at D.

(a) Draw complete shear force and bending moment diagrams for the beam. (b) Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION (a) From equilibrium for the right-hand segment of the beam:

0 :DMΣ =� ( ) ( ) ( )1.5 40 1.5 2 2.5 0ER − =

50 kNER = ↑

0 :yF↑Σ = ( )50 40 1.5 2 0DR + − =

20 kN 20 kNDR = − = ↓

For the left-hand segment of the beam:

0 :CMΣ =� ( ) ( ) ( ) ( )80 1.5 2.75 40 2 1+

( ) ( )2 20 1 0BR− + =


337

215 kNBR = ↑

0 :BMΣ =� ( ) ( ) ( ) ( ) ( ) ( )2 80 1.5 0.75 40 2 1 20 3 0CR + − + =

35 kN 35 kNCR = − = ↓

(b) From the moment diagram max 90 kN mM = − ⋅

For an S457 81× section: 3 3 6 31465 10 mm 1465 10 mS −= × = ×

3

6 26

90.0 10 61.4 10 N/m 61.4 MPa (T top & C bottom)1465 10

MS

σ −

×= = = × ≅×

...........Ans.

8-77 An S15 × 50 American standard steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-77. The total length of the beam is 15 ft. If the flexural stress is limited to 15 ksi, determine the maximum permissible value for the distributed load w.


0 :CMΣ =� ( ) ( ) ( )3 2 2 0Bw L L R L− =

( )3 4 lbBR wL= ↑

0 :BMΣ =� ( ) ( ) ( )2 3 2 0CR L w L L− =

( )3 4 lbCR wL= ↑

The maximum bending moment between the supports occurs where the shear force is zero,

( )33 0

4 2wx L xwLV = − =

3 2 2.121x L L= ≅

( ) ( ) ( )22

max 2.121

3 2.1213 2.1211.121 0.3107 lb ft4 2 3L

w L LwL LM M L wL = = − ≅ ⋅

For an S15 50× section: 364.8 inS =

( )22

20.3107 5 120.3107 15,000 lb/in.64.8wM wL

S Sσ

×= = = =

869 lb/in. 10.43 kip/ftw = ≅ ...............................................................................................Ans.

8-78 Draw complete shear force and bending moment diagrams for segments AB and CD of the structure shown in Fig. P8-78.

SOLUTION For the complete structure:

0 :DMΣ =� ( ) ( ) ( ) ( )3 3 1.5 3 1.5 1.5 0CR − − = 3.75 kNCR = ↑

0 :yF↑Σ = ( )3 3 1.5 3.75 0DR − − + = 3.75 kNDR = ↑


338

For member AB:

0 :yF↑Σ = 3 0BV− − =

3 kN 3 kNBV = − = ↑

0 :BMΣ =� ( )3 1.5 0BM + =

4.5 kN mBM = − ⋅

4.5 kN m = ⋅ � For member CD:

0 :yF↑Σ = ( )3.75 1.5 3 0CV− − =

0.75 kN 0.75 kNCV = − = ↑

0 :CMΣ =� ( ) ( ) ( )1.5 3 1.5 3.75 3 0CM + − =

4.5 kN m 4.5 kN m CM = ⋅ = ⋅ �

8-79 Three members are connected with smooth pins to form the frame shown in Fig. P8-79. The weights of the members are negligible. Member CD has a 3 × 3-in. cross section, and member BE has a 1 × 1-in. cross section. The pin at A has a ½-in diameter and is in double shear.

(a) Draw complete shear force and bending moment diagrams for member CD. (b) Determine the maximum tensile and compressive flexural stresses in the member CD. (c) Determine the normal stress in member BE. (d) Determine the shearing stress in the pin at A. SOLUTION (a) For the complete frame:

0 :xF→Σ = 500 0xA + =

0 :yF↑Σ = ( )400 12 0y DA R− + =

0 :AMΣ =� ( ) ( ) ( ) ( )12 500 7 400 12 6 0DR − − =

2692 lbDR = ↑

500 lb 500 lbxA = − = ←

2108 lbyA = ↑

For member CD:

0 :yF↑Σ = ( )400 12 2692 0y yC E+ − + =

0 :CMΣ =� ( ) ( ) ( ) ( )8 2692 12 400 12 6 0yE + − =

438 lb 438 lbyE = − = ↓

2546 lbyC = ↑


339

(b) From the moment diagram, max 8103 lb ftM = ⋅ and ( )3 43 3 12 6.75 inI = =

( )( ) 2

top

8103 12 1.5021,608 lb/in. 21.6 ksi (C)

6.75MyI

σ×

= − = − = − ≅ .....................Ans.

( ) ( ) 2

bot

8103 12 1.5021,608 lb/in. 21.6 ksi (T)

6.75MyI

σ× −

= − = = + ≅ .......................Ans.

(c) For member BE: ( )3 5y BEE F= 438 730 lb3 5BEF = =

730 730 psi (T)1 1

BEBE

FA

σ = = =×

..........................................................................................Ans.

(d) For pin A: 2 2 2 2500 2108 2166 lbA x yF A A= + = + =

( )2

2166 5516 psi 5.52 ksi2 2 0.5 4

AA

s

VA

τπ

= = = ≅

.........................................................Ans.

8-80 A body with a mass of 1500 kg is supported by a roller on an I-beam, as shown in Fig. P8-80. The roller moves slowly along the beam, thereby causing the shear force V and the bending moment M to be functions of the position b.

(a) Draw complete shear force and bending moment diagrams when the roller is at position b. (b) Determine the position of the roller when the bending moment is maximum. SOLUTION

(a) ( )1500 9.81 14,715 N 14.715 kNP mg= = = =

0 :BMΣ =� ( ) ( )10 10 0P x A− − =

( )0.1 kNA P Px= −

0 :AMΣ =� ( )10 0B Px− =

( )0.1 kNB Px=

(b) ( )20.1 kN mPM Ax Px Px= = − ⋅

( )1 0.2 0PdM P xdx

= − =

5.00 mx = ........................................................................................................................ Ans.

( ) ( ) ( )2max 14.715 5 0.1 14.715 5 36.8 kN mM = − = ⋅

8-81 Member AB supports a 55-lb sign, as shown in Fig. P8-81. Determine (a) The maximum tensile flexural stress in the ½-in. nominal diameter standard steel pipe AB (see Appendix

A). (b) The normal stress in the 3/16-in. diameter wire BC. (c) The shearing stress in the ¼-in. diameter pin at A, which is in double shear. SOLUTION (a) For pipe AB:

0 :yF↑Σ = ( )2 27.5 0y yA B+ − =


340

0 :AMΣ =� ( ) ( ) ( )64 27.5 16 27.5 48 0yB − − =

27.5 lby yA B= = ↑

From the moment diagram: max 440 lb in.M = ⋅

For a ½-in. diameter pipe: 30.041 inS = Therefore,

2440 10,732 lb/in.0.041

MS

σ = = =

10.73 ksi (T bottom & C top)≅ .................Ans.

(b) ( )1tan 35 64 28.67θ −= = °

27.5 57.32 lb 57.32 lb (T)

sin sin 28.67y

BC

BF

θ= = = =

°

( )2

23 160.02761 in

4BCAπ

= =

57.32 2076 psi 2.08 ksi (T)

0.02761BC

BCFA

σ = = = ≅ ...........................................................Ans.

(c) For the pin A:

0 :xF→Σ = 57.32cos 28.67 0xA − ° = 50.29 lbxA = →

( )

2 2 2 2

2

50.29 27.5 584 psi2 2 0.25 4x y

As

A AA

τπ

+ += = =� ��

.................................................................Ans.

8-82 Two beams AD and EH are spliced, as shown in Fig. P8-82. Draw complete shear force and bending moment diagrams for the beams AD, CF, and EH.

SOLUTION For the system of beams:

0 :BMΣ =� ( ) ( )44 6.5 11 0AR− = 26.0 kNAR = ↑

0 :AMΣ =� ( ) ( )11 44 4.5 0HR − = 18.00 kNHR = ↑

For beam ABD:

0 :yF↑Σ = 26.0 44 0y yB D− − + = 96.0 kNyD = ↑

0 :DMΣ =� ( ) ( ) ( )26.0 4.5 1.5 0yB− + = 78.0 kNyB = ↓

For beam EGH:


341

0 :yF↑Σ = 18.00 0yG E− + = 36.0 kNyE = ↑

0 :EMΣ =� ( ) ( )18.00 4.5 1.5 0yG− = 54.0 kNyG = ↓

8-83 A tractor is moving slowly over a bridge, as shown in Fig. P8-83. The forces exerted on one beam of the bridge by the tractor are 4050 lb by the rear wheels and 1010 lb by the front wheels. Determine the position b of the tractor for which the bending moment in the beam is maximum.

SOLUTION

0 :BMΣ =� ( )1010 40 9.5b− −

( ) ( )4050 40 40 0b A+ − − =

( )4820 126.5 lbA b= − ↑

0 :AMΣ =� ( ) ( )40 4050 1010 9.5 0B b b− − + =

( )239.9 126.5 lbB b= + ↑

The moment under the left wheel

( )2126.5 4820 lb ftCM Ax b b= = − + ⋅

is a maximum when

253 4820 0CdM dx b= − + = 4820 253 19.05 ftb = =

( ) ( ) ( )2

max 126.5 19.05 4820 19.05 45,914 lb ftCM = − + = ⋅

The moment under the right wheel

( ) ( )( )

( )2

30.5 239.9 126.5 30.5

7316.95 3618.35 126.5 lb ftDM B b b b

b b

= − = + −

= + − ⋅

is a maximum when

3618.35 253 0DdM dx b= − = 3618.35 253 14.30 ftb = =

( ) ( ) ( )2

max7316.95 3618.35 14.30 126.5 14.30

33,191 lb ftDM = + −

= ⋅


342

Therefore, the maximum moment occurs when

19.05 ftb = .............................................................................................................................. Ans.

8-84 A timber beam is loaded and supported as shown in Fig. P8-84. At section A-A of the beam, determine the shearing stresses on horizontal planes that pass through points a, b, and c of the cross section.

SOLUTION From overall equilibrium

0 :CMΣ =� ( ) ( )10 2 4 0BR− =

5.00 kNBR = ↑

For the left-hand portion of the beam

0 :yF↑Σ = 5 0AV− =

5.00 kN 5.00 kNAV = + = ↓

( )3

6 4 6 4150 200100.0 10 mm 100.0 10 m

12I −= = × = ×

For point a: 30 mmaQ =

0 MPaaτ = .......................................................................................... Ans.

For point b: ( ) 3 375 150 50 562.5 10 mmb CQ y A= = × = ×

( )( )( )( )

3 63 2

6

5 10 562.5 10187.5 10 N/m 187.5 kPa

100.0 10 0.150bVQIt

τ−

−

× ×= = = × =

×........................Ans.

For point c: ( ) 3 350 150 100 750.0 10 mmc CQ y A= = × = ×

( )( )( ) ( )

3 63 2

6

5 10 750.0 10250 10 N/m 250 kPa

100.0 10 0.150cVQIt

τ−

−

× ×= = = × =

×..............................Ans.

8-85 The transverse shear V at a certain section of a timber beam is 7500 lb. If the beam has the cross section shown in Fig. P8-85, determine

(a) The horizontal shearing stress in the glued joint 2 in. below the top of the beam. (b) The transverse shearing stress at a point 3 in. below the top of the beam. (c) The magnitude and location of the maximum transverse shearing stress on the cross section. SOLUTION

( ) ( )3 3

48 12 4 8981.3 in

12 12I = − =

(a) For the glue joint, 4 in.y = ,

( ) 34 5 8 2 80 inCQ y A= = × =

( )

( )( )2

4

7500 80152.9 lb/in 152.9 psi

981.3 4VQIt

τ = = = = .........................................................Ans.

(b) For a point 3-in. below the top of the beam, 3 in.y = ,

( ) ( ) 33 5 8 2 3.5 4 1 94 inCQ y A= = × + × =


343

( )

( )( )2

3

7500 94179.6 lb/in 179.6 psi

981.3 4VQIt

τ = = = = .........................................................Ans.

(c) The maximum shear stress occurs at the neutral axis, 0 in.y = ,

( ) ( ) 35 8 2 2 4 4 112 inNA CQ y A= = × + × =

( )

( ) ( )2

max

7500 112214 lb/in 214 psi (at NA)

981.3 4VQIt

τ = = = = ............................................Ans.

8-86 A W254 × 89 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-86. Determine the maximum transverse shearing stress at section A-A of the beam

(a) Using Eq. (8-15). (b) Using Eq. (8-17). SOLUTION From overall equilibrium

0 :CMΣ =� ( ) ( ) ( )25 1 20 3 5 0BR+ − =

17.00 kNBR = ↑


0 :yF↑Σ = 17 20 0AV− − =

3.00 kN 3.00 kNAV = − = ↑

For a W 254 89× section, 260 mmd = 10.7 mmwt =

6 4142 10 mmI = × 256 mmfw = 17.3 mmft =

(a) ( ) ( ) 3 3121.35 256 17.3 56.35 10.7 112.7 605.4 10 mmNA CQ y A= = × + × = ×

( )( )

( )( )

3 6

max 6

3.00 10 605.4 10

142 10 0.0107NAVQIt

τ τ−

−

× ×= = =

×

6 21.195 10 N/m 1.195 MPa= × = ................................................ Ans.

(b) ( ) ( ) 2260 2 17.3 10.7 2412 mmwA = − =

3

6 26

3.00 10 1.244 10 N/m 1.244 MPa2412 10avg

w

VA

τ −

×= = = × =×

.........................................Ans.

8-87 A WT7 × 34 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-87. Determine the maximum transverse shearing stress at section A-A of the beam.


0 :CMΣ =� ( ) ( ) ( )500 10 5 10 0BR− =

2500 lbBR = ↑


0 :yF↑Σ = ( )2500 500 3 0AV− − =


344

1000 lb 1000 lbAV = + = ↓

For a WT 7 34× section, 7.020 in.d = 0.415 in.st =

432.6 inI = 1.29 in.Cy =

( ) ( ) 35.73 2 5.73 0.415 6.813 inNA CQ y A= = × =

( )

( )( )2

max

1000 6.813504 lb/in 504 psi (at NA)

32.6 0.415NAVQIt

τ τ= = = = = ...........................Ans.

8-88 A timber beam 4 m long is simply supported at its ends and carries a uniformly distributed load w of 8 kN/m over its entire length. If the beam has the cross section shown in Fig. P8-88, determine

(a) The maximum horizontal shearing stress in the glued joints between the web and flanges of the beam. (b) The maximum horizontal shearing stress in the beam. SOLUTION

( ) ( )3 3

6 4 6 4

180 200 140 12012 12

99.84 10 mm 99.84 10 m

NAI

−

= −

= × = ×

( )

3 3 6 3

80 180 40

576 10 mm 576 10 mJ CQ y A

−

= = ×

= × = ×

( ) ( )

3 3 6 3

80 180 40 30 40 60

648 10 mm 648 10 mNAQ

−

= × + ×

= × = ×

( ) ( )max 2 8 4 2 16.00 kNA BV R R wL= = = = =

(a) ( )( )( )( )

3 6

6

16 10 576 10

99.84 10 0.040JVQIt

τ−

−

× ×= =

×

6 22.31 10 N/m 2.31 MPa= × = ................................................. Ans.

(b) ( ) ( )( ) ( )

3 6

max 6

16 10 648 10

99.84 10 0.040NAVQIt

τ τ−

−

× ×= = =

×

6 22.60 10 N/m 2.60 MPa= × = .................................................................................Ans.

8-89 A W10 × 30 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-89. At section A-A of the beam, determine

(a) The maximum transverse shearing stress due to the 5000-lb load. (b) The maximum transverse shearing stress due to the 5000-lb load plus the weight of the beam. SOLUTION From overall equilibrium

0 :CMΣ =� ( ) ( )10 5000 4 0BR− − =

2000 lb 2000 lbBR = − = ↓


0 :yF↑Σ = 2000 0AV− − =


345

2000 lb 2000 lbAV = − = ↑

For a W10 30× section, 10.47 in.d = 0.300 in.wt =

4170 inI = 5.810 in.fw = 0.510 in.ft =

(a) ( ) ( ) 34.980 5.810 0.510 2.3625 0.300 4.725 18.105 inNA CQ y A= = × + × =

( )

( ) ( )2

max

2000 18.105710 lb/in 710 psi (at NA)

170 0.300NAVQIt

τ τ= = = = = ...........................Ans.

(b) With the weight of the beam included, 30 lb/ftw =

0 :CMΣ =� ( ) ( ) ( )3 30 14 10 5000 4 0BR− − =

1874 lb 1874 lbBR = − = ↓

0 :yF↑Σ = ( )1874 30 5 0AV− − − =

2024 lb 2024 lbAV = − = ↑

( )

( ) ( )2

max

2024 18.105719 lb/in 719 psi (at NA)

170 0.300τ = = = ..................................................Ans.

8-90 A W203 × 60 structural steel wide-flange section (see Appendix A) is used for the cantilever beam shown in Fig. P8-90. Determine the maximum flexural and transverse shearing stresses in the beam and state where they occur.


0 :AMΣ =� ( )30 2.5 0AM− − =

75.0 kN m 75.0 kN m AM = − ⋅ = ⋅ �

0 :yF↑Σ = 30 0AV − =

30 kN 30 kNAV = + = ↑

For a W 203 60× section,

6 460.8 10 mmI = × 210 mmd = 9.1 mmwt =

3 3582 10 mmS = × 205 mmfw = 14.2 mmft =

From the shear force and bending moment diagrams,

max 30.0 kN (over the full length of the beam)V =

max 75.0 kN m (at the wall)M = − ⋅

3

6 2max 6

75 10 128.9 10 N/m 128.9 MPa (T top & C bottom)582 10

MS

σ −

×= = = × =×

...........Ans.

( ) ( ) 3 397.9 205 14.2 45.4 9.1 90.8 322.5 10 mmNA CQ y A= = × + × = ×

( ) ( )

( ) ( )

3 66 2

max 6

30.0 10 322.5 1017.49 10 N/m 17.49 MPa

60.8 10 0.0091NAVQIt

τ τ−

−

× ×= = = = × =

×.........Ans.


346

8-91 The beam shown in Fig. P8-91a is composed of two 1 × 6-in. and two 1 × 3-in. hard maple boards that are glued together as shown in Fig. P8-91b. Determine the magnitude and location of

(a) The maximum tensile flexural stress in the beam. (b) The maximum horizontal shearing stress in the beam. SOLUTION From shear force and bending moment diagrams for the beam

max 1000 lbV =

max 1800 lb ftM = ⋅

( ) ( )3 3

42 6 2 340.5 in

12 12NAI = + =

(a) At the bottom of the beam, 2 ft from the left support,

( ) ( )

max

1800 12 340.5

McI

σ× −

= − = −

21600 lb/in 1600 psi (T)= + = ...................................................................................Ans.

(b) ( ) ( ) 30.75 2 1.5 1.5 2 3 11.25 inNA CQ y A= = × + × =

At the left support and from the end of the distributed load to the right support,

( )

( ) ( )21000 11.25

69.4 lb/in 69.4 psi40.5 4NA

VQIt

τ = = = =

( ) 31.5 2.25 2 1.5 6.75 inQ = × =

( )

( )( )2

1.5

1000 6.7583.3 lb/in 83.3 psi

40.5 4τ = = =

max 1.5 83.3 psi (1.5 in. above and below the NA)τ τ= = .............................................Ans.

8-92 The beam shown in Fig. P8-92a is composed of three pieces of timber that are glued together as shown in Fig. P8-92b. Determine

(a) The maximum horizontal shearing stress in the glued joints. (b) The maximum horizontal shearing stress in the wood. (c) The maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium

0 :BMΣ =� ( ) ( ) ( )9 6 0.5 6 2.5 3 0AR+ + − =

9.00 kNAR = ↑

0 :yF↑Σ = 6 6 0A BR R+ − − =

3.00 kNBR = ↑


max 9.00 kNV =


347

max 10.5 kN mM = ⋅

( )3

6 4 6 4100 240115.2 10 mm 115.2 10 m

12I −= = × = ×

(a) ( ) 3 3 6 395 100 50 475 10 mm 475 10 mJ CQ y A −= = × = × = ×

( )( )( ) ( )

3 63 2

6

9.00 10 475 10371 10 N/m 371 kPa

115.2 10 0.100JVQIt

τ−

−

× ×= = = × =

×....................................Ans.

(b) ( ) 3 3 6 360 100 120 720 10 mm 720 10 mNA CQ y A −= = × = × = ×

( )( )( )( )

3 63 2

6

9.00 10 720 10563 10 N/m 563 kPa

115.2 10 0.100NAVQIt

τ−

−

× ×= = = × =

×.................................Ans.

(c) ( )( )3

max 6

10.5 10 0.120115.2 10

McI

σ −

×= =

×

6 210.94 10 N/m 10.94 MPa (T bottom & C top)= × = .....................................Ans.

8-93 The lintel beam AB shown in Fig. P8-93a has a rectangular cross section as shown in Fig. P8-93b, and is used to support a brick wall over a door opening. The brick wall is assumed to produce a triangular load distribution. The total load carried by the beam is 500 lb. If the beam is simply supported at the ends, determine the maximum flexural and transverse shearing stresses in the beam and state where they occur.


0 :BMΣ =� ( ) ( ) ( )250 4 2 2 4 0AR− =

250 lbA BR R= ↑=


max 250 lb (at ends of the beam)V =

max 333.3 lb ft (at midspan)M = ⋅

( )3

48 0.50.08333 in

12I = =

( ) ( )

max

333.3 12 0.250.08333

McI

σ×

= =

211,999 lb/in 12.00 ksi (T bottom & C top)= ≅ ..................................................Ans.

( ) 30.125 8 0.25 0.25 inNA CQ y A= = × =

( )

( ) ( )2

max

250 0.2593.8 lb/in 93.8 psi (at NA)

0.08333 8VQIt

τ = = = = .....................................Ans.

The results for maxτ are worthless since 16w d = and Eq. 8-15 gives useful results only when 1w d < .

8-94 A timber beam is simply supported and carries a uniformly distributed load of 4 kN/m over the full length of the beam. If the beam has the cross section shown in Fig. P8-94 and a span of 6 m, determine

(a) The horizontal shearing stress in the glued joint 50 mm below the top of the beam and 1 m from the left support.


348

(b) The horizontal shearing stress in the glued joint 50 mm above the bottom of the beam and ½ m from the left support.

(c) The maximum horizontal shearing stress in the beam. (d) The maximum tensile flexural stress in the beam. SOLUTION From overall equilibrium

0 :BMΣ =� ( ) ( ) ( )4 6 3 6 0AR− =� ��

12.00 kNAR = ↑

0 :yF↑Σ = ( )4 6 0A BR R+ − =

12.00 kNBR = ↑


max 12.00 kNV = ( )1.0 12.00 4 1 8.00 kNV = − =

max 18.00 kN mM = + ⋅ ( )0.5 12.00 4 0.5 10.00 kNV = − =

( ) ( ) ( )

( ) ( ) ( )25 150 50 125 50 150 225 250 50

143.18 mm150 50 50 150 250 50Cy

+ +� � � � � �� = =+ +

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 32

3 32

6 4 6 4

150 50 50 93.18118.18 150 50

12 3

50 56.82 250 5081.82 250 50

3 12

209.14 10 mm 209.14 10 m

I

−

� �= + × +� ��

� �+ + + ×� �

� ��

= × = ×

(a) ( ) 3 3 6 381.82 250 50 1022.8 10 mm 1022.8 10 mTJ CQ y A −= = × = × = ×

( )( )

( ) ( )

3 63 2

6

8.00 10 1022.8 10783 10 N/m 783 kPa

209.1 10 0.050TJVQIt

τ−

−

× ×= = = × =

×.....................Ans.

(b) ( ) 3 3 6 3118.18 150 50 886.4 10 mm 886.4 10 mBJQ −= × = × = ×

( ) ( )

( )( )

3 63 2

6

10.00 10 886.4 10848 10 N/m 848 kPa

209.1 10 0.050BJτ−

−

× ×= = × =

×................................Ans.

(c) ( ) ( ) 3 3 6 381.82 250 50 28.41 50 56.82 1103.5 10 mm 1103.5 10 mNAQ −= × + × = × = ×

( )( )

( )( )

3 63 2

6

12.00 10 1103.5 101267 10 N/m 1267 kPa

209.1 10 0.050NAτ−

−

× ×= = × =

×..........................Ans.

(d) ( )( )3

max 6

18.00 10 0.14318209.1 10

MyI

σ −

× −= − = −

×

6 212.33 10 N/m 12.33 MPa (T)= + × = ..................................................................Ans.


349

8-95 The timber beam shown in Fig. P8-95a is fabricated by gluing two 1 × 5-in. and two 1 × 4-in. boards together as shown in Fig. P8-95b. Determine

(a) The maximum horizontal shearing stress in the glued joints. (b) The maximum horizontal shearing stress in the wood. (c) The maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium

0 :BMΣ =� ( ) ( ) ( )200 6 6 600 3+� ��

( ) ( )600 3 9 0AR− − =

800 lbAR = ↑

0 :yF↑Σ =

( )200 6 600 600 0A BR R+ − − − =

1600 lbBR = ↑


max 1000 lbV = −

max 1800 lb ftM = − ⋅

( ) ( )3 3

45 6 3 474.0 in

12 12I = − =

(a) ( ) 32.5 5 1 12.5 inJ CQ y A= = × =

( )

( )( )21000 12.5

84.5 lb/in 84.5 psi74.0 2J

VQIt

τ = = = = ................. Ans.

(b) ( ) ( ) 32.5 5 1 2 1 1 2 16.5 inNAQ = × + × =� ��

( )

( ) ( )21000 16.5

111.5 lb/in 111.5 psi74.0 2NAτ = = = ..................................................................Ans.

(c) ( )( ) 2

max

1800 12 3876 lb/in 876 psi (T top & C bottom)

74.0McI

σ×

= = − = ≅ ...........Ans.

8-96 A cantilever beam is used to support a concentrated load of 20 kN at the end of the beam. The beam is fabricated by bolting two C457 × 86 steel channels (see Appendix A) back-to-back to form the H-section shown in Fig. P8-96. If the pairs of bolts are spaced at 300-mm intervals along the beam, determine

(a) The shear force carried by each of the bolts. (b) The bolt diameter required if the shear and bearing stresses for the bolts must be limited to 60 MPa and

125 MPa, respectively. SOLUTION

For a C457 86× channel, 211,030 mmA = 21.9 mmCx =

6 47.41 10 mmI = × 17.8 mmwt =

For the beam


350

( ) ( ) ( )26 6 4 6 42 7.41 10 21.9 11,030 25.40 10 mm 25.40 10 mNAI − = × + = × = ×

(a) ( ) 3 3 6 321.9 11,030 241.6 10 mm 241.6 10 mNA CQ y A −= = = × = ×

20 kN (entire length of beam)V =

( )( )( ) ( )

3 63 2

6

20.0 10 241.6 10416.1 10 N/m 416.1 kPa

25.40 10 0.4572NAVQIt

τ−

−

× ×= = = × =

×

( )( )3

3416.1 10 0.4572 0.300

28.54 10 N 28.5 kN2 2

NA sB

AF τ × ×= = = × ≅ ...............Ans.

(b) 3

6 22

28.54 10 60 10 N/m4

B

B

FA d

τπ

×= = ≤ × 0.02461 md ≥

( )3

6 228.54 10 125 10 N/m0.0178

Bb

w

Fdt d

σ ×= = ≤ × 0.01283 md ≥

min 0.02461 m 24.6 mmd = ≅ .............................................................................................Ans.

8-97 A timber beam is fabricated from one 2 × 8-in. and two 2 × 6-in. pieces of lumber to form the cross section shown in Fig. P8-97. The flanges of the beam are fastened to the web with nails that can safely transmit a shear force of 100 lb. If the beam is simply supported and carries a 1000-lb load at the center of a 12-ft span, determine

(a) The shear force transferred by the nails from the flange to the web in a 12-in. length of the beam. (b) The spacing required for the nails. SOLUTION

(a) 1000 2 500 lbA BR R= = = ↑

500 lb (over entire length of the beam)V =

( ) ( )3 3

46 12 4 8693.3 in

12 12NAI = − =

( ) 35 6 2 60.0 inJ CQ y A= = × =

( )

( )( )2500 60.0

7.212 lb/in 7.212 psi693.3 6J

VQIt

τ = = = =

( )7.212 12 6 519 lbJ JF Aτ= = × = ....................................................................................Ans.

(b) 519.3 100 5.19 nailsJ NN F F= = =

12 5.19 2.31 in.s = =

Use 6 nails spaced 2 in. apart in a 12-in. length ..........................................................Ans.

8-98 A box beam will be fabricated by bolting two 15 × 260-mm steel plates to two C305 × 45 steel channels (see Appendix A), as shown in Fig. P8-98. The beam will be simply supported at the ends and will carry a concentrated load of 125 kN at the center of a 5-m span. Determine the bolt spacing required if the bolts have a diameter of 20 mm and an allowable shearing stress of 150 MPa.

SOLUTION


351

125 2 62.5 kNA BR R= = = ↑

62.5 kN (over entire length of the beam)V =

For a C305 45× channel, 304.8 mmd =

6 467.4 10 mmI = × 80.5 mmfw =

( ) ( ) ( ) ( )3

26 260 152 67.4 10 2 159.9 260 15

12NAI

= × + + ×

6 4334.4 10 mm= ×

( ) 3 3152.4 260 15 594.4 10 mmJ CQ y A= = × = ×

( )( )( ) ( )

3 63 2

6

62.5 10 594.4 10690.0 10 N/m 690.0 kPa

334.4 10 2 0.0805JVQIt

τ−

−

× ×= = = × =

× ×

( )( ) ( )3690.0 10 2 0.0805 111,087 NJ JF A s sτ= = × × × =

( ) ( )262 150 10 0.020 4 94,248 NJ B B BF F Aτ π = = = × =

94, 248 111,087 0.848 m 848 mms = = = .....................................................................Ans.

8-99 A W21 × 101 structural steel wide-flange section (see Appendix A) is simply supported at its ends and carries a concentrated load at the center of a 20-ft span. The concentrated load must be increased to 125 kip, which requires that the beam be strengthened. It has been decided that two 3/4 × 16-in. steel plates will be bolted to the flanges, as shown in Fig. P8-99. Determine the bolt spacing required if the bolts have a diameter of ¾-in. and an allowable shearing stress of 17.5 ksi.

SOLUTION

125 2 62.5 kipA BR R= = = ↑

62.5 kip (over entire length of the beam)V =

For a W 21 101× section, 21.36 in.d =

42420 inI = 12.290 in.fw =

( ) ( ) ( )

3216 0.75

2420 2 11.055 16 0.7512NAI

= + + ×

45354 in=

( ) 310.68 16 0.75 128.160 inJ CQ y A= = × =

( )( )

( )( )

32

62.5 10 128.16121.73 lb/in 121.73 psi

5354 12.290JVQIt

τ××

= = = =

( ) ( )121.73 12.290 1496.1 lbJ JF A s sτ= = × =

( ) ( )232 17.5 10 0.75 4 15,463 lbJ B B BF F Aτ π = = = × =

15,463 1496.1 10.34 in.s = = .............................................................................................Ans.


352

8-100 A W356 × 122 structural steel wide-flange section (see Appendix A) has a C381 × 74 channel bolted to the top flange, as shown in Fig. P8-100. The beam is simply supported at its ends and carries a concentrated load of 96 kN at the center of an 8-m span. If the pairs of bolts are spaced at 500-mm intervals along the beam, determine

(a) The shear force carried by each of the bolts. (b) The bolt diameter required if the shear and bearing stresses for the bolts must be limited to 60 MPa and

125 MPa, respectively. SOLUTION

96 2 48 kNA BR R= = = ↑

48 kN (over entire length of the beam)V =

For a C381 74× channel, 6 44.58 10 mmI = ×

20.3 mmCx = 29485 mmA = 18.2 mmwt =

For a W 356 122× section, 6 4367 10 mmI = ×

363 mmd = 215,550 mmA = 257 mmfw =

For the beam

( ) ( )360.9 9485 181.5 15,550

249.5 mm9485 15,550Cy

+= =

+

( ) ( ) ( )26367 10 68 15,550NAI = × +

( ) ( ) ( )26 6 44.58 10 111.4 9485 561.2 10 mm + × + = ×

( ) 3 3111.4 9485 1056.6 10 mmJ CQ y A= = = ×

( )( )

( )( )

3 63 2

6

48 10 1056.6 10351.6 10 N/m 351.6 kPa

561.2 10 0.257JVQIt

τ−

−

× ×= = = × =

×

( )( )3 32 351.6 10 0.257 0.500 2 22.6 10 N 22.6 kNB JF Aτ= = × × = × = ...............Ans.

8-101 A timber beam is simply supported and carries a uniformly distributed load w of 360 lb/ft over its entire 18-ft span (Fig. P8-101a). If the beam has the cross section shown in Fig. P8-101b, compute and plot the vertical shearing stress τ as a function of distance y from the neutral axis for a cross section 2 ft from the left end of the beam.

SOLUTION

( )( )360 18 2 3240 lbA BR R= = = ↑

( )2 3240 360 2 2520 lbV = − =

( ) ( ) ( )

( ) ( ) ( )1 6 2 2 1 4 2 1 4

1.400 in.6 2 1 4 1 4Cy

+ + = =+ +

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 46 2 1 4

0.4 6 2 2 0.6 1 4 19.467 in12 12NAI

= + × + + × =


353

For 1.4 in. 0.6 in.y− ≤ ≤ 8 in.t =

( )( ) ( )2 31.40 8 1.40 4 1.9600 in2C

yQ y A y y+= = − = −

( )( )( )( ) ( )

22

2520 4 1.960064.72 1.9600 psi

19.467 8yVQ y

Itτ

−= = = −

For 0.6 in. 2.6 in.y≤ ≤ 2 in.t =

( ) ( )2.62 1 2.62C

yQ y A y+ = = −

( )2 36.7600 iny= −

( )

( )( )

22520 6.760019.467 2

yVQIt

τ−

= =

( )264.72 6.7600 psiy= −

8-102 The transverse shear force V at a certain section of a timber beam is 18 kN. If the beam has the cross section shown in Fig. P8-102, compute and plot the vertical shearing stress τ as a function of distance y (−150 mm < y < 150 mm) from the neutral axis.

SOLUTION

( ) ( )3 3

6 4200 300 100 200383.3 10 mm

12 12I = − = ×

For 100 mm 100 mmy− ≤ ≤ 100 mmt =

( ) ( )( )

( )6 2 2 3

100125 200 50 100 1002

1.250 10 50 100 mm

CyQ y A y

y

+= = × + −

= × + −

( ) ( ) ( )( )( )

( )

3 6 2 2 9

6

6 2 2 2

18 10 1.250 10 50 100 10

383.3 10 0.100

0.46997 1.250 10 50 100 N/m

yVQIt

y

τ−

−

× × + − = =×

= × + −

For 150 mm 100 mmy− ≤ ≤ − and 100 mm 150 mmy≤ ≤ 100 mmt =

( )( )150 200 1502C

yQ y A y+= = −

( )2 2 3100 150 mmy = −

( ) ( ) ( )

( ) ( )

3 2 2 9

6

18 10 100 150 10

383.3 10 0.200

yVQIt

τ−

−

× − = =×

( )2 2 223.48 150 N/my= −


354

8-103 The transverse shear force V at a certain section of a timber beam is 2500 lb. If the beam has the cross section shown in Fig. P8-103, compute and plot the vertical shearing stress τ as a function of distance y (−3 in. < y < 3 in.) from the neutral axis.

SOLUTION

( ) ( )3 3

42 6 2 340.500 in

12 12NAI = + =

For 1.5 in. 1.5 in.y− ≤ ≤ 4 in.t =

( ) ( )( ) ( )2 31.502.25 2 1.5 4 1.50 6.75 2 2.2500 in2C

yQ y A y y+ = = × + − = + −

( )

( ) ( ) ( )2

22500 6.75 2 2.2500

104.167 30.8642 2.2500 psi40.5 4

yVQ yIt

τ + − = = = + −

For 3.0 in. 1.5 in.y− ≤ ≤ − and 1.5 in. 3.0 in.y≤ ≤

2 in.t =

( ) ( )3.0 2 3.02C

yQ y A y+ = = −

( )2 39 iny= −

( )

( ) ( )

22500 940.5 2

yVQIt

τ−

= =

( )230.8642 9 psiy= −

8-104 The beam shown in Fig. 8-104a is fabricated by gluing two pieces of timber together to form the cross section shown in Fig. 8-104b. Compute and plot the vertical shearing stress τ as a function of distance y from the neutral axis for a cross section 0.5 m from the left end of the beam.


0 :BMΣ =� ( ) ( ) ( )3 2 3 4 0AR− =

4.50 kNAR = ↑

0 :yF↑Σ = ( )3 2 0A BR R+ − =

1.50 kNBR = ↑

( )0.5 4.50 3 0.5 3.0 kNV = − =

( ) ( )

( ) ( )100 25 200 212.5 100 25

137.5 mm25 200 100 25Cy

+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2

6 4 6 4

25 200 100 2537.5 25 200 75.0 100 25

12 12

37.89 10 mm 37.89 10 m

I

−

= + × + + ×

= × = ×


355

For 137.5 mm 62.5 mmy− ≤ ≤ 25 mmt =

( )( ) ( )2 2 3137.5 50 137.5 12.5 137.5 mm2C

yQ y A y y+ = = − = −

( ) ( ) ( )

( )( ) ( )3 2 2 9

2 2 26

3 10 12.5 137.5 1039.59 137.5 N/m

37.89 10 0.025

yVQ yIt

τ−

−

× − = = = −×

For 62.5 mm 87.5 mmy≤ ≤ 100 mmt =

( ) ( )87.5 100 87.52C

yQ y A y+= = −

( )2 2 350 87.5 mmy = −

( ) ( ) ( )

( )( )

3 2 2 9

6

3 10 50 87.5 10

37.89 10 0.100

yVQIt

τ−

−

× − = =×

( )2 2 239.59 87.5 N/my= −

8-105 A 20-ft long simply supported timber beam is loaded with 1800-lb concentrated loads applied 6 ft from each support. The allowable flexural stress is 1900 psi, and the allowable shearing stress is 90 psi. Select the lightest standard structural timber that can be used to support the loading.


0 :AMΣ =� ( ) ( ) ( )20 1800 6 1800 14 0B − − =

1800 lbB = ↑

0 :BMΣ =� ( ) ( ) ( )1800 6 1800 14 20 0A+ − =

1800 lbA = ↑

2max

10,800 12 1900 lb/inMS S

σ ×= = ≤

310,800 12 68.21 in1900

S ×≥ =

Try an 8 8-in.× timber with

370.3 inS = 4264 inI = 256.3 inA = 15.6 lb/ftw = With the weight of the beam included, the maximum moment is

( )215.6 20

10,800 11,580 lb ft8

M = + = ⋅ 311,580 12 73.14 in1900

S ×≥ =

which is bigger than the section modulus of the timber. Next, try a 4 12-in.× timber with

379.9 inS = 4459 inI = 241.7 inA = 11.6 lb/ftw = With the weight of the beam included, the maximum moment is

( )211.6 20

10,800 11,380 lb ft8

M = + = ⋅ 311,380 12 71.87 in1900

S ×≥ =


356

which is okay. Next, check the shear stress,

( )

max

11.6 201800 1916 lb

2load weightV V V= + = + =

( ) 2max

max

1.5 19161.5 68.92 lb/in

41.7V

Aτ = = =

which is less than the allowable shear stress of 90 psi . Therefore, this design is okay.

Use a 4 12-in. timber× .........................................................................................................Ans.

8-106 A simply supported timber beam 5 m long is loaded with 6700 N concentrated loads applied 2 m from each support. If the allowable flexural stress is 9 MPa and the allowable shearing stress is 0.6 MPa, select the lightest standard structural timber that can be used to support the loading.


0 :AMΣ =� ( ) ( ) ( )5 6700 2 6700 3 0B − − =

6700 NB = ↑

0 :BMΣ =� ( ) ( ) ( )6700 2 6700 3 5 0A+ − =

6700 NA = ↑

6 2max

13,400 9 10 N/mMS S

σ = = ≤ ×

6 36

13,400 1488.9 10 m9 10

S −≥ = ××

3 31488.9 10 mm= ×

Try a 203 254-mm× timber with

3 31850 10 mmS = × 6 4223 10 mmI = × 246,000 mmA = 29.4 kg/mm =

With the weight of the beam included, the maximum moment is

( ) ( )229.4 9.81 5

13, 400 14,301 N m8load weightM M M

×= + = + = ⋅

6 3 3 36

14,301 1589 10 m 1589 10 mm9 10

S −≥ = × = ××


( ) ( )

max

29.4 9.81 56700 7421 N

2load weightV V V×

= + = + =

( ) 3 2max

max 3

1.5 74211.5 242 10 N/m

46 10V

Aτ −= = = ×

×

which is much less than the allowable shear stress of 600 kPa . Therefore, this design is okay.

Use a 203 254-mm timber× ...............................................................................................Ans.

8-107 The lever shown in Fig. P8-107 is used to lift a 600 lb rock. Select a standard steel pipe to perform the task. The allowable flexural stress is 20 ksi. Neglect the effects of shear.

SOLUTION


357

From overall equilibrium

0 :AMΣ =� ( ) ( )5 600 5.8 0B − =

696 lbB = ↑

0 :BMΣ =� ( ) ( )5 600 0.8 0P − =

96 lbP = ↓

( )max 96 5 480 lb ftBM M= = = ⋅

3 2max

480 12 20 10 lb/inMS S

σ ×= = ≤ × 33

480 12 0.288 in20 10

S ×≥ =×

Use a 1.5-in. piped = .............. ( )30.326 inS = .............................................................Ans.

8-108 A structural steel beam is subjected to the loading shown in Fig. P8-108. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa. Select the lightest American Standard beam that can be used to support the loading.


0 :AMΣ =�

( ) ( ) ( ) ( )5.5 22 1.25 40 2.75 22 4.25 0B − − − =

42 kNB = ↑

0 :BMΣ =�

( ) ( ) ( ) ( )22 4.25 40 2.75 22 1.25 5.5 0A+ + − =

42 kNA = ↑

3

6 2max

82.5 10 152 10 N/mMS S

σ ×= = ≤ ×

3

6 3 3 36

82.5 10 542.8 10 m 542.8 10 mm152 10

S −×≥ = × = ××

Try an S305 47× section with

3 3596 10 mmS = × 8.9 mmwt = 304.8 mmd = 47 kg/mm =


( ) ( )2

3 347 9.81 5.582.5 10 84.24 10 N m

8load weightM M M×

= + = × + = × ⋅

3

6 3 3 36

84.24 10 554.2 10 m 554.2 10 mm152 10

S −×≥ = × = ××


( ) ( )3 3

max

47 9.81 5.542 10 43.27 10 N

2load weightV V V×

= + = × + = ×

3

6 2maxmax

43.27 10 15.95 10 N/m0.0089 0.3048w

VA

τ ×= = = ××


358

which is much less than the allowable shear stress of 100 MPa . Therefore, this design is okay.

Use an S305 47 section× ....................................................................................................Ans.

8-109 A 16-ft long simply supported beam is loaded with a uniform load of 4000 lb/ft over its entire length. If the allowable flexural stress is 22 ksi and the allowable shearing stress is 14.5 ksi, select the lightest structural steel wide-flange beam that can be used to support the loading.


0 :AMΣ =� ( ) ( ) ( )16 4 16 8 0B − =

32 kipB = ↑

0 :BMΣ =� ( ) ( ) ( )4 16 8 16 0A− =

32 kipA = ↑

max128 12 22 ksiM

S Sσ ×= = ≤

3128 12 69.82 in22

S ×≥ =

Try a W18 60× section with

3108 inS = 0.415 in.wt = 18.24 in.d = 60 lb/ftw =


( )20.060 16

128 129.92 kip ft8

M = + = ⋅ 3129.92 12 70.87 in22

S ×≥ =

which is still okay. Next, check the shear stress,

( )

max

0.060 1632 32.48 kip


2maxmax

32.48 4.291 kip/in 4.291 ksi0.415 18.24w

VA

τ = = = =×

which is less than the allowable shear stress of 14.5 ksi . Therefore, this design is okay.

Use a W18 60 section× .......................................................................................................Ans.

8-110 Select the lightest wide-flange beam that can be used to support the loading shown in Fig. P8-110. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa.


0 :AMΣ =� ( ) ( ) ( )7.5 15 2.5 15 5 0B − − =

15 kNB = ↑

0 :BMΣ =� ( ) ( ) ( )15 5 15 2.5 7.5 0A+ − =

15 kNA = ↑

3

6 2max

37.5 10 152 10 N/mMS S

σ ×= = ≤ ×


359

3

6 3 3 36

37.5 10 246.7 10 m 246.7 10 mm152 10

S −×≥ = × = ××

Try a W 254 33× section with

3 3380 10 mmS = × 6.1 mmwt = 258 mmd = 33 kg/mm =


( ) ( )2

3 333 9.81 7.537.5 10 39.78 10 N m

8load weightM M M×

= + = × + = × ⋅

3

6 3 3 36

39.78 10 261.7 10 m 261.7 10 mm152 10

S −×≥ = × = ××


( )( )3 3

max

33 9.81 7.515 10 16.21 10 N

2load weightV V V×

= + = × + = ×

3

6 2maxmax

16.21 10 10.30 10 N/m0.0061 0.258w

VA

τ ×= = = ××


Use a W 254 33 section× ....................................................................................................Ans.

8-111 The floor-framing plan for a residential dwelling is shown in Fig. P8-111. The floor decking is to be supported by 2-in. nominal width joists spaced 16 in. apart. Each joist is to span 12 ft and is simply supported at the ends. The floor decking is subjected to a uniform loading of 60 lb/ft2, which includes the live load plus an allowance for the dead load of the flooring system. The joists are made of construction grade Douglas fir with an allowable flexural stress of 1200 psi and an allowable shearing stress of 120 psi. Determine the required nominal depth of the joists.

SOLUTION A joist is a simply supported beam subjected to a uniformly distributed load on a 12-ft span. Since the joists are spaced 16 in. apart, the uniformly distributed load is

( )60 16 12 80 lb/ftw = =


0 :AMΣ =� ( ) ( ) ( )12 80 12 6 0B − =

480 lbB = ↑

0 :BMΣ =� ( ) ( ) ( )80 12 6 12 0A− =

480 lbA = ↑

2max

1440 12 1200 lb/inMS S

σ ×= = ≤

31440 12 14.40 in1200

S ×≥ =

Try a 2 8-in.× timber with

315.3 inS = 457.1 inI = 212.2 inA = 3.39 lb/ftw = With the weight of the beam included, the maximum moment is


360

( )23.39 12

1440 1501 lb ft8

M = + = ⋅ 31501 12 15.01 in1200

S ×≥ =


( )

max

3.39 12480 500.3 lb


( ) 2max

max

1.5 500.31.5 61.51 lb/in

12.2V

Aτ = = =

which is less than the allowable shear stress of 120 psi . Therefore, this design is okay.

Use a 2 8-in. timber section× ............................................................................................Ans.

8-112 A 15-kN load is supported by a roller on an I-beam, as shown in Fig. P8-112. The roller moves slowly along the beam, thereby causing the shear force and bending moment to be functions of b. Select the lightest permissible American Standard beam to support the loading. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa. Neglect the weight of the beam.


0 :AMΣ =� ( )8 15 0B x− =

( )1.875 kNB x= ↑

0 :BMΣ =� ( ) ( )15 8 8 0x A− − =

( )15 1.875 kNA x= − ↑

The moment under the load is 215 1.875CM Ax x x= = −

The maximum moment occurs under the load when

15 3.75 0CdM dx x= − = 4.00 mx = max 30 kN mM = ⋅

3

6 2max

30 10 152 10 N/mMS S

σ ×= = ≤ ×

3

6 3 3 36

30 10 197.4 10 m 197.4 10 mm152 10

S −×≥ = × = ××

Try an S203 27× section with

3 3236 10 mmS = × 6.9 mmwt = 203.2 mmd = 27 kg/mm =

Next, check the shear stress,

3

6 2maxmax

15 10 10.70 10 N/m0.0069 0.2032w

VA

τ ×= = = ××


Use an S203 27 section× ...................................................................................................Ans.

8-113 A carriage moves slowly along a simply supported I-beam, as shown in Fig. P8-113. Select the lightest permissible wide-flange beam to support the loading, if the allowable flexural stress is 24 ksi and the allowable shearing stress is 14.5 ksi. Neglect the weight of the beam. Note that the shear force and bending moment are functions of b, the position of the left hand wheel.

SOLUTION


361


0 :AMΣ =� ( ) ( )20 2000 2000 5 0D x x− − + =

( )200 500 lbD x= + ↑

0 :yF↑Σ = 4000 0A D+ − =

( )200 3500 lbA x= − + ↑

The moment under the left wheel

( )2200 3500 lb ftBM Ax x x= = − + ⋅

is a maximum when

400 3500 0BdM xdx

= − + = 8.75 ftx = max 15,313 lb ftBM = ⋅

The moment under the right wheel

( ) ( )215 200 2500 7500 lb ftCM D x x x= − = − + + ⋅

is a maximum when

400 2500 0CdM xdx

= − + = 6.25 ftx = max 15,313 lb ftCM = ⋅

Therefore, the maximum bending moment occurs under the wheel closest to the center of the beam,

max 15,313 lb ftM = ⋅

3 2max

15,313 12 24 10 lb/inMS S

σ ×= = ≤ × 33

15,313 12 7.657 in24 10

S ×≥ =×

Try a W8 15× section with

311.8 inS = 0.245 in.wt = 8.11 in.d = 15 lb/ftw =

Next, check the shear stress,

2maxmax

3500 1762 lb/in0.245 8.11w

VA

τ = = =×

which is much less than the allowable shear stress of 14.5 ksi . Therefore, this design is okay.

Use a W 8 15 section× .........................................................................................................Ans.

8-114 A beam has the cross section shown in Fig. P8-114. If the flexural stress at point A is 14 MPa (T), determine

(a) The maximum flexural stress on the section. (b) The resisting moment Mr at the section. SOLUTION

( ) ( ) ( )

( ) ( ) ( )25 200 50 100 50 200 100 50 200

75.00 mm200 50 50 200 50 200Cy

+ + = =+ +

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 6 4200 50 50 200

50 200 50 2 25 50 200 106.25 10 mm12 12

I

= + × + + × = ×


362

(a) ( )max125 14 70 MPa 70 MPa (C)

25AA

cy

σ σ= = = − =−

...................................................Ans.

(b) ( ) 6

6

0.02514 10

106.25 10rr A

A

MM yI

σ −

−= − = − = ×

×

359.5 10 N m 59.5 kN mrM = × ⋅ = ⋅ .................................................................................Ans.

8-115 A T-beam has the cross section shown in Fig. P8-115. Determine the maximum tensile and compressive flexural stresses on a cross section of the beam where the resisting moment being transmitted is 74 kip-ft.

SOLUTION

( ) ( )( ) ( )

4 1.5 8 9 4 26.00 in.

1.5 8 4 2Cy+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 41.5 8 4 2

2 1.5 8 3 4 2 186.667 in12 12

I

= + × + + × =

( ) ( ) 274 12 6

28.5 kip/in. 28.5 ksi (T)186.667bot

MyI

σ× −

= − = − = + = ................................Ans.

( ) ( ) 274 12 4

19.03 kip/in. 19.03 ksi (C)186.667topσ

×= − = − = ..............................................Ans.

8-116 Determine the percentage of the resisting moment Mr carried by the flanges of a W838 × 226 wide-flange beam (see Appendix A for dimensions).

SOLUTION

From Table A-2 for a W838 226× section: 6 43395 10 mmI = ×

851 mmd = 26.8 mmft =

maxMcI

σ = ( ) ( )maxmax

max

33957.9788 kN m

425.5IM

cσσ σ= = = ⋅

For the web 16.1 mmwt = ( )851 2 26.8 797.4 mmwd = − =

( )3

6 416.1 797.4680.3 10 mm

12wI = = ×

( ) ( )max max max398.7web 0.9370425.5

σ σ σ= =

( ) ( ) ( ) ( )max max

max

web 0.9370 680.31.5988 kN m

398.7w

ww

IM

cσ σ

σ= = = ⋅

( )7.9788 1.5988 100 80.0 %7.9788fM −= = .............................................................................Ans.

8-117 The maximum flexural stress on the cross section of the beam shown in Fig. P8-117 is 8000 psi (C). Determine

(a) The resisting moment being transmitted by the section. (b) The magnitude of the flexural force carried by the flange. SOLUTION


363

( ) ( )( ) ( )

3 2 6 7 6 25.00 in.

2 6 6 2Cy+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 42 6 6 2

2 2 6 2 6 2 136 in12 12

I

= + × + + × =

(a) ( ) 35

8 10136rr MM y

Iσ

−= − = − = − ×

3217.6 10 lb ft 218 kip in.rM = − × ⋅ ≅ − ⋅ ..........................................................................Ans.

(b) ( ) ( )3

2217.6 10 2

3200 lb/in136

r CfCf

M yI

σ− ×

= − = − =

( )3200 6 2 38, 400 lb 38.4 kip (T)f Cf fF Aσ= = × = = ................................................Ans.

8-118 A beam is loaded and supported as shown in Fig. P8-118. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Using the coordinate axes shown, write equations for the shear force and bending moment for any section

of the beam in the interval 0 < x < 4 m. SOLUTION From overall equilibrium:

0 :EMΣ =� ( ) ( ) ( )20 2 8 6 7+

( ) ( )10 12 8 0BR+ − =

62 kNBR =

For 0 m 4 mx≤ ≤

( )10 8 2 BV x R= − − + +

( )8 36 kNx= − + ................................ Ans.

( ) ( ) 210 4 8 22B

xM x R x x += − + + − +

( )24 36 56 kN mx x= − + − ⋅ ............ Ans.

8-119 A beam is loaded and supported as shown in Fig. P8-119. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Using the coordinate axes shown, write equations for the shear force and bending moment for any section

of the beam in the interval 0 < x < 10 ft. SOLUTION For 0 ft 10 ftx≤ ≤

( )200 6 760 300V x= + −

( )300 1960 lbx= − + ....................................Ans.

( ) ( ) ( )200 6 3 760 300 2M x x x x= + + −

( )2150 1960 3600 lb ftx x= − + + ⋅ ............Ans.


364

8-120 Select the lightest pair of structural-steel angles (see Appendix A) that may be used for the beam of Fig. P8-120 if the maximum flexural stress must be limited to 60 MPa. The angles will be fastened back-to-back to form a T-section.

SOLUTION From the moment diagram,

max 7.5 kN mM = ⋅

3

6max

7.5 10 60 10MS S

σ ×= = = ×

3

6 3 3 36

7.5 10 125 10 m 125 10 mm60 10

S −×= = × = ××

For each angle

3

3 3125 10 62.5 10 mm2

S ×= = ×

Use two L178 102 9.5-mm angles× × ..............................................................................Ans.

8-121 A timber beam is loaded and supported as shown in Fig. P8-121a. The beam has the cross section shown in Fig. P8-121b. Determine

(a) The flexural stresses at points A and B on a transverse cross section 1 ft from the left end of the beam. (b) The maximum tensile and compressive flexural stresses in the beam. SOLUTION

( ) ( )3 3

48 10 6 6558.7 in

12 12NAI = − =

( ) ( )1 5600 200 1 0.5 5500 lb ftM = − + = − ⋅

(a) ( )( )5500 12 3

558.7AMyI

σ− × −

= − = −

2354.4 lb/in 354 psi (C)= − ≅ ............Ans.

( )( )5500 12 5

558.7BMyI

σ− ×

= − = −

2590.7 lb/in 591 psi (T)= + ≅ ............Ans.

(b) From the moment diagram,

max 5600 lb ftM = − ⋅

( )( )5600 12 5

558.7topσ− ×

= −

2601.4 lb/in 601 psi (T)= + ≅ ............Ans.

( ) ( )5500 12 5

558.7botσ− × −

= −

2601.4 lb/in 601 psi (C)= − ≅ ............Ans.


365

8-122 Select the lightest steel wide-flange or American standard beam (see Appendix A) that may be used for the beam of Fig. P8-122 if the maximum flexural stress must be limited to 75 MPa.

SOLUTION From the moment diagram,

max 20 kN mM = ⋅

3

6max

20 10 75 10MS S

σ ×= = = ×

3

6 36

20 10 266.7 10 m75 10

S −×= = ××

3 3266.7 10 mm= ×

Use a W 254 33 section× ....................Ans.

8-123 The beam shown in Fig. P8-123a has the cross section shown in Fig. P8-113b. Determine the maximum tensile and compressive flexural stresses in the beam.

SOLUTION

( ) ( )

( ) ( )4 1 8 8.5 4 1

5.50 in.1 8 4 1Cy

+ = =+

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 41 8 4 1

1.5 1 8 3 4 1 97 in12 12

I

= + × + + × =


max 14,742 lb ftM = ⋅

( ) ( )14,742 12 3.5

97topMyI

σ×

= − = −

26380 lb/in 6.38 ksi (C)= − =

( ) ( )14,742 12 5.5

97botσ× −

= −

210,030 lb/in 10.03 ksi (T)= + =

Therefore

max 10.03 ksi (T)Tσ = ............................................................................................................Ans.

max 6.38 ksi (C)Cσ = .............................................................................................................Ans.

8-124 A WT305 × 70 structural tee (see Appendix A) is loaded and supported as a beam (with the flange on top) as shown in Fig. P8-124. Determine

(a) The maximum tensile flexural stress in the beam. (b) The maximum compressive flexural stress in the beam. (c) The maximum vertical shearing stress in the beam. (d) The vertical shearing stress at a point in the stem just below the flange on the cross section where the

maximum vertical shearing stress occurs. SOLUTION


366

0 :EMΣ =� ( ) ( ) ( ) ( ) ( )10 1 15 2 2 10 5 4 0BR+ + − =

30 kNBR =

0 :BMΣ =� ( ) ( ) ( ) ( ) ( )10 1 15 2 2 10 3 4 0ER− − + =

30 kNER =

For a WT 305 70× section:

6 477.4 10 mmI = × 308.7 mmd =

22.2 mmft = 230.3 mmfw =

13.1 mmwt = 75.9 mmCy =

(a) At the bottom of the beam 2.33 m from the left support:

( ) ( )3

max 6

23.33 10 0.232877.4 10T

MyI

σ −

× −= − = −

×

6 270.17 10 N/m 70.2 MPa (T)= + × ≅ .................................................................Ans.

(b) At the bottom of the beam at the left support:

( ) ( )3

max 6

10.00 10 0.232877.4 10C

MyI

σ −

− × −= − = −

×

6 230.08 10 N/m 30.1 MPa (C)= − × ≅ .................................................................Ans.

(c) ( ) 3 3116.4 13.1 232.8 355.0 10 mmNA CQ y A= = × = ×

( ) ( )( )( )

3 66 2

max 6

20 10 355.0 107.00 10 N/m 7.00 MPa

77.4 10 0.0131VQIt

τ−

−

× ×= = = × =

×....................Ans.

(d) ( ) 3 364.8 22.2 230.3 331.3 10 mmJQ = × = ×

( )( )( )( )

3 66 2

6

20 10 331.3 106.53 10 N/m 6.53 MPa

77.4 10 0.0131Jτ−

−

× ×= = × =

×...................................Ans.

8-125 A WT12 × 52 structural tee (see Appendix A) is loaded and supported as a beam (with the flange on the bottom), as shown in Fig. P8-125. Determine

(a) The maximum tensile flexural stress in the beam. (b) The maximum compressive flexural stress in the beam. (c) The maximum vertical shearing stress in the beam. (d) The vertical shearing stress at a point in the stem just above the flange on the cross section where the

maximum vertical shearing stress occurs. SOLUTION

0 :EMΣ =� ( ) ( ) ( ) ( ) ( ) ( )7 1 8 7 5 16 1 3 2 2 14 0BR+ + − − =

10 kipBR =

0 :BMΣ =� ( ) ( ) ( ) ( ) ( ) ( )7 1 8 7 5 2 1 3 2 16 14 0ER− + − + =

4.50 kipER =


367

For a WT12 52× section:

4189 inI = 12.030 in.d =

0.750 in.ft = 12.75 in.fw =

0.500 in.wt = 2.59 in.Cy =

(a) At the top of the beam at the left support:

( ) ( )

max

17.00 12 9.44189T

MyI

σ− ×

= − = −

210.19 kip/in 10.19 ksi (T)= + = ............Ans.

(b) At the top of the beam 8 ft from the left support:

( ) ( )

max

10.50 12 9.44189Cσ×

= −

26.29 kip/in 6.29 ksi (C)= − = ..............................................................................Ans.

(c) ( ) 34.72 0.5 9.44 22.28 inNA CQ y A= = × =

( )

( )2

max

5000 22.281179 lb/in 1179 psi

189 0.500VQIt

τ = = = = ....................................................Ans.

(d) ( ) 32.215 12.75 0.750 21.18 inJQ = × =

( )

( )25000 21.18

1121 lb/in 1121 psi189 0.500Jτ = = = ...................................................................Ans.

8-126 An extruded aluminum alloy beam has the cross section shown in Fig. P8-126. All parts of the section are 5 mm thick. When a constant shear force V = 4450 N is being supported by the beam, determine

(a) The shearing stresses at points A and B of the cross section. (b) The maximum horizontal shearing stress in the beam. SOLUTION

( ) 25 50 100 50 100 100 2000 mmA = + + + + =

( ) ( )50 5 100 100 100 5

37.5 mm2000Cy+ = =

( ) ( ) ( ) ( )

3 325 50 100 5

2 37.5 2 5 50 100 512 12

I

= + + × × + ×

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 6 45 100 100 5

12.5 5 100 62.5 100 5 3.960 10 mm12 12

+ + × + + × = ×

(a) ( ) 3 362.5 100 5 31.25 10 mmA CQ y A= = × = ×

( )

( )( )

66 2

6

4450 31.25 107.02 10 N/m 7.02 MPa

3.960 10 0.005AVQIt

τ−

−

×= = = × =

×...........................Ans.

( ) ( ) 3 362.5 100 5 12.5 5 100 37.50 10 mmBQ = × + × = ×


368

( )

( )( )

66 2

6

4450 37.50 108.43 10 N/m 8.43 MPa

3.960 10 0.005Bτ−

−

×= = × =

×.......................................Ans.

(b) ( ) ( ) 3 362.5 100 5 31.25 5 62.5 41.02 10 mmCQ = × + × = ×

( )

( )( )

66 2

6

4450 41.02 109.22 10 N/m 9.22 MPa

3.960 10 0.005Cτ−

−

×= = × =

×.......................................Ans.


369

Chapter 9 9-1 A beam is loaded and supported as shown in Fig. P9-1. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection at the left end of the beam. (c) The slope at the left end of the beam. SOLUTION (a) EIy Px′′ = −

2

12PxEIy C′ = − + At ,x L= 0 :y′ =

2

1 2PLC =

3

1 26PxEIy C x C= − + + At ,x L= 0 :y =

3

2 3PLC −=

( )2 2

2Py x LEI

′ = − +

( )3 2 33 26

Py x L x LEI

= − + − .................................Ans.

(b) ( )3 3

30 0 0 2

6 3 3A xP PL PLy LEI EI EI

δ =−= = + − = = ↓ .............................................................Ans.

(c) ( )2 2

20 0


θ =+′= = + = = .....................................................................Ans.

9-2 A beam is loaded and supported as shown in Fig. P9-2. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection at the right end of the beam. (c) The slope at the right end of the beam. SOLUTION

0 :yF↑Σ = 0AV wL− =

AV wL wL= + = ↑

0 :AMΣ =� ( )2 0AM wL L− − =

2 22 2 AM wL wL= − = �

(a) 2 2

2 2wL wxEIy wLx′′ = − −

2 2 3

12 2 6wLx wL x wxEIy C′ = − − + At 0,x = 0 :y′ = 1 0C =

3 2 2 4

1 26 4 24wLx wL x wxEIy C x C= − − + + At 0,x = 0 :y = 2 0C =

( )3 2 23 36wy x Lx L xEI

′ = − + −


370

( )4 3 2 24 624

wy x Lx L xEI

= − + − .........................................................................................Ans.

(b) ( )4 4

4 4 44 624 8 8B x L

w wL wLy L L LEI EI EI

δ =−= = − + − = = ↓ .................................................Ans.

(c) ( )3 3

3 3 33 36 6 6B x Lw wL wLy L L LEI EI EI

θ =−′= = − + − = = ...................................................Ans.

9-3 A beam is loaded and supported as shown in Fig. P9-3. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection midway between the supports. (c) The slope at the left end of the beam. SOLUTION

(a) 2

2 2wLx wxEIy′′ = −

2 3

14 6wLx wxEIy C′ = − + At ,

2Lx = 0 :y′ =

3

1 24wLC −=

3 4

1 212 24wLx wxEIy C x C= − + + At 0,x = 0 :y = 2 0C =

( )3 2 34 624

wy x Lx LEI

′ = − + −

( )4 3 3224

wy x Lx L xEI

= − + − ..............................................................................................Ans.

(b) 4 4 4 4 4

25 5

24 16 4 2 384 384M x Lw L L L wL wLyEI EI EI

δ = −= = − + − = = ↓

........................................Ans.

(c) ( )3 3

30 0 0

24 24 24A xw wL wLy LEI EI EI

θ =−′= = + − = = ..........................................................Ans.

9-4 A beam is loaded and supported as shown in Fig. P9-4. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection at the left end of the beam. (c) The slope at the left end of the beam. SOLUTION

( )w x wx L=

( ) ( ) ( ) 3

2 3 6wx L x x wxM x

L = − = −

(a) 3

6wxEIy

L′′ = −


371

4

124wxEIy C

L′ = − + At ,x L= 0 :y′ =

3

1 24wLC =

5

1 2120wxEIy C x C

L= − + + At ,x L= 0 :y =

4

2 30wLC −=

( )4 4

24wy x LEIL

′ = − +

( )5 4 55 4120

wy x L x LEIL

= − + − ..........................................................................................Ans.

(b) ( )4 4

50 0 0 4

120 30 30A xw wL wLy LEIL EI EI

δ =−= = + − = = ↓ ....................................................Ans.

(c) ( )3 3

40 0

24 24 24A xw wL wLy LEIL EI EI

θ =+′= = + = = ...............................................................Ans.

9-5 A beam is loaded and supported as shown in Fig. P9-5. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection midway between the supports. (c) The slope at the right end of the beam. SOLUTION

0 :BMΣ =� ( ) ( )2 3 0AwL L R L− =

6AR wL=

( )w x wx L=

( ) ( )( ) 3

2 3 6 6A

wx L x x wLx wxM x R xL

= − = −

(a) 3

6 6wLx wxEIy

L′′ = −

2 4

112 24wLx wxEIy C

L′ = − + At 0,x = 0 :y = 2 0C =

3 5

1 236 120wLx wxEIy C x C

L= − + + At ,x L= 0 :y =

3

17360

wLC −=

( )4 2 2 415 30 7360

wy x L x LEIL

′ = − + −

( )5 2 3 43 10 7360

wy x L x L xEIL

= − + − ..................................................................................Ans.

(b) 5 5 5 4 4

23 10 7 5 5

360 32 8 2 768 768M x Lw L L L wL wLyEIL EI EI

δ = −= = − + − = = ↓

..........................Ans.

(c) ( )3 3

4 4 415 30 7360 45 45B x L

w wL wLy L L LEIL EI EI

θ =+′= = − + − = = ...................................Ans.


372

9-6 For the steel beam [E = 200 GPa and I = 32.0(106) mm4] shown in Fig. P9-6, determine the deflection at a section midway between the supports.

SOLUTION

10 kNA BR R= =

( ) ( )10 1 10 kN mAM x R x x= − + = − ⋅

10 kN mEIy′′ = − ⋅

( ) 2110 kN mEIy x C′ = − + ⋅ At 0 m,x = 0 :y = 2 0C =

( )2 31 25 kN mEIy x C x C= − + + ⋅ At 2 m,x = 0 :y = 3

1 10 kN mC = ⋅

( )2 35 10 kN mEIy x x= − + ⋅

( ) ( )

( ) ( )2 3

1 9 6

5 1 10 1 100.000781 m 0.781 mm

200 10 32.0 10M xyδ = −

− + × = = = + = ↑

× ×......................Ans.

9-7 For the steel beam (E = 30,000 ksi and I = 32.1 in.4) shown in Fig. P9-7, determine the deflection at a section midway between the supports.

SOLUTION

7500 lb ftEIy′′ = ⋅

( ) 217500 lb ftEIy x C′ = + ⋅ At 0 ft,x = 0 :y = 2 0C =

( )2 31 23750 lb ftEIy x C x C= + + ⋅ At 12 ft,x = 0 :y = 3

1 45,000 lb ftC = − ⋅

( )2 33750 45,000 lb ftEIy x x= − ⋅

( ) ( ) ( )( )( )

2 3

6 6

3750 6 45,000 6 120.242 in. 0.242 in.

30 10 32.1M xyδ =

− = = = − = ↓

×..................Ans.

9-8 A cantilever beam is fixed at the left end and carries a uniformly distributed load w over the full length of the beam. In addition, the right end is subjected to a moment of +3wL2/8, as shown in Fig. P9-8. Determine the maximum deflection in the beam if I = 2.5(106) mm4, E = 210 GPa, L = 3 m, and w = 1500 N/m.

SOLUTION

0 :yF↑Σ = 0AV wL− = AV wL wL= = ↑

0 :AMΣ =� ( ) ( ) ( )23 8 2 0AwL wL L M− − = 2 28 8 AM wL wL= − = �

( )2 2A A

wx xM x V x M = + −

2 2

8 2wL wxwLx= − −


373

2 2

8 2wL wxEIy wLx′′ = − −

2 2 3

12 8 6wLx wL x wxEIy C′ = − − + At 0,x = 0 :y′ = 1 0C =

3 2 2 4

1 26 16 24wLx wL x wxEIy C x C= − − + + At 0,x = 0 :y = 2 0C =

( )3 2 24 12 324

wy x Lx L xEI

′ = − + −

( )4 3 2 22 8 348

wy x Lx L xEI

= − + −

The maximum deflection occurs either when 0y′ = or at x L=

( )3 2 24 12 3 024

wy x Lx L xEI

′ = − + − = 0, 0.2753 , 2.725x L L=

41 0.2753 0.001379x Ly wL EIδ == = −

4 416 0.06250B x Ly wL EI wL EIδ == = + = +

( )

( )( )44

max 9 6

1500 30.01446 m 14.46 mm

16 16 210 10 2.5 10BwL

EIδ δ

−= = = = + = ↑

× ×.......Ans.

9-9 A beam is loaded and supported as shown in Fig. P9-9. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The slope at the left end of the beam. (c) The deflection midway between the supports. SOLUTION

(a) 2EIy Px′′ =

2

14PxEIy C′ = + At 2,x L= 0 :y′ =

2

1 16PLC −=

3

1 212PxEIy C x C= + + At 0,x = 0 :y = 2 0C =

( )2 2416

Py x LEI

′ = −

( )3 24 348

Py x L xEI

= − ............................................ Ans.

(b) ( )2 2

20 0


θ =−′= = − = = .................................................................Ans.

(c) 3 3 3 3

24 3

48 8 2 48 48M x LP L L PL PLyEI EI EI

δ = −= = − = = ↓

.....................................................Ans.


374

9-10 A 100 × 300-mm timber having a modulus of elasticity of 8 GPa is loaded and supported as shown in Fig. P9-10. Determine

(a) The deflection at the 7-kN load. (b) The deflection at the free end of the beam. SOLUTION

( )3

6 4 6 4100 300225 10 mm 225 10 m

12I −= = × = ×

7 kNAR = ↑ 14 kN m AM = ⋅ �

( ) ( )7 14 kN mA AM x R x M x= − = − ⋅

(a) ( )7 14 kN mEIy x′′ = − ⋅ 0 m 2 mx≤ ≤

2

21

7 14 kN m2xEIy x C

′ = − + ⋅

At 0,x = 0 :y′ = 1 0C =

3

2 31 2

7 7 kN m6xEIy x C x C

= − + + ⋅

At 0,x = 0 :y = 2 0C =

2

27 14 kN m2xEIy x

′ = − ⋅

0 m 2 mx≤ ≤

3

2 37 7 kN m6xEIy x

= − ⋅

0 m 2 mx≤ ≤

( ) ( )

( )( )3 2 3

2 9 6

7 2 42 2 100.01037 m 10.37 mm

6 8 10 225 10B xyδ = −

− × = = = − = ↓

× ×..........................Ans.

(b) ( ) ( )

( )( )2 3

2 9 6

7 2 28 2 100.007778 rad 0.007778 rad

2 8 10 225 10B xyθ = −

− × ′= = = − =

× ×

( ) ( )( )1.5 0.01037 0.007778 1.5C B Bδ δ θ= + = − + −

0.0220 m 22.0 mm= − = ↓ ..........................................................................................Ans.

9-11 A 130-lb boy is standing on a 1.6 × 12-in. wood (E = 1400 ksi) diving board, as shown in Fig. P9-11. If length AB is 2 ft and length BC is 5 ft, determine the maximum deflection in the diving board.

SOLUTION

0 :AMΣ =� ( )3.5 0BR L P L− =

3.5 3.5BR P P= = ↑

0 :yF↑Σ = 0A BR R P+ − =

2.5 2.5AR P P= − = ↓

For the interval 0 x L≤ ≤

1 2.5EIy Px′′ = −


375

2

1 12.5

2PxEIy C′ = − + At 0,x = 1 0 :y = 2 0C =

3

1 1 22.5

6PxEIy C x C= − + + At ,x L= 1 0 :y =

2

12.5

6PLC =

( )2 21

2.5 36

Py x LEI

′ = − + ( )3 21

2.56

Py x L xEI

= − +

The maximum deflection in the interval 0 x L≤ ≤ occurs when

( )2 21

2.5 3 06

Py x LEI

′ = − + = 0.5774x L=

3 3max 0.5774 0.16038 0.16038x Ly PL EI PL EIδ == = + = ↑

For the interval 3.5L x L≤ ≤

( )2 2.5 3.5EIy Px P x L′′ = − + −

( )22

2 3

3.52.52 2

P x LPxEIy C−

′ = − + +

( )33

2 3 4

3.52.56 6

P x LPxEIy C x C−

= − + + +

The two solutions must match at ,x L=

1 2 :EIy EIy′ ′= 2 2 2

32.5 2.5 2.5 0

2 6 2PL PL PL C− + = − + +

2

32.5

6PLC =

1 2 :EIy EIy= 3 3

1 3 42.5 2.5 0

6 6PL PLC L C L C− + = − + + + 4 0C =

( )3

412 1.64.096 in

12I = =

The maximum deflection in the interval 3.5L x L≤ ≤ occurs at C

( ) ( )

( ) ( )

33

max 3.5 6

7.292 130 2 127.2921.4 10 4.096x L

PLyEI

δ =

− ×−= = =×

2.29 in. 2.29 in.= − = ↓ ...............................................................................................Ans.

9-12 A timber beam 150 mm wide × 300 mm deep is loaded and supported as shown in Fig. P9-12. The modulus of elasticity of the timber is 10 GPa. A pointer is attached to the right end of the beam. Determine

(a) The deflection of the right end of the pointer. (b) The maximum deflection of the beam. SOLUTION

For the interval 0 2x L≤ ≤

(a) 2EIy Px′′ =


376

2

14PxEIy C′ = + At 2,x L= 0 :y′ =

2

1 16PLC −=

3

1 212PxEIy C x C= + + At 0,x = 0 :y = 2 0C =

( )2 2416

Py x LEI

′ = − ( )3 2348

Py x L xEI

= −

( )3

6 4 6 4150 300337.5 10 mm 337.5 10 m

12I −= = × = ×

( )2 2

20 0


θ =−′= = − = =

As a result of symmetry, B Aθ θ= −

( )

( ) ( ) ( )2

9 6

8900 5.51.5 0.00748 m 7.48 mm

16 10 10 337.5 10P B PLδ θ−

= = = + = ↑× ×

...........Ans.

(b) 3 3 3

max 24 3

48 8 2 48x LP L L PLyEI EI

δ = −= = − =

( )

( )( )33

9 6

8900 5.50.00914 m 9.14 mm

48 48 10 10 337.5 10PLEI −

−−= = = − = ↓× ×

................Ans.

9-13 A beam is loaded and supported as shown in Fig. P9-13. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The slope at the right end of the beam. SOLUTION

0 :BMΣ =� ( )3 0APL R L PL− − = 2 3 2 3AR P P= + = ↑

( )3 3 3A

PL PL PxM x R x Px= + − = −

(a) 3 3

PL PxEIy′′ = −

2

13 6PLx PxEIy C′ = − + At 0,x = 0 :y = 2 0C =

2 3

1 26 18PLx PxEIy C x C= − + + At ,x L= 0 :y =

2

1 9PLC −=

( )2 23 6 218

Py x Lx LEI

′ = − + −

( )3 2 23 218

Py x Lx L xEI

= − + − ............................................................................................Ans.


377

( )2 2

2 2 23 6 218 18 18B x L

P PL PLy L L LEI EI EI

θ =+′= = − + − = = .............................................Ans.

9-14 A beam is loaded and supported as shown in Fig. P9-14. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection at the left end of the beam. SOLUTION

( ) ( )2 4

3 4 12kx x x kxM x

− − = =

4

12kxEIy′′ = −

5

160kxEIy C′ = − + At ,x L= 0 :y′ =

5

1 60kLC =

6

1 2360kxEIy C x C= − + + At ,x L= 0 :y =

6

25360

kLC −=

( )5 5

60ky x LEI

′ = − + ( )6 5 66 5360

ky x L x LEI

= − + −

(a) At x L= , 2w kL= . Therefore, 2k w L=

( )6 5 62 6 5

360wy x L x LEIL

= − + − ........................................................................................Ans.

(b) 4 4

05

360 72A xwL wLyEI EI

δ =−= = = ↓ .............................................................................................Ans.

9-15 Determine the deflection midway between the supports for beam AB of Fig. P9-15. Segment BC of the beam is rigid.

SOLUTION

0 :BMΣ =� ( ) ( ) ( )( )2 2 2 0AwL L wL L R L+ − =

3 4 3 4AR wL wL= = ↑

23

4 2wLx wxEIy′′ = −

2 3

13

8 6wLx wxEIy C′ = − + At 0,x = 0 :y = 2 0C =

3 4

1 23

24 24wLx wxEIy C x C= − + + At ,x L= 0 :y =

3

1 12wLC −=

( )3 2 34 9 224

wy x Lx LEI

′ = − + − ( )4 3 33 224

wy x Lx L xEI

= − + −

4 4 4 4

4max 2

3 11 1124 16 8 384 384x L

w L L wL wLy LEI EI EI

δ = −= = − + − = = ↓

..................................Ans.


378

9-16 The beam shown in Fig. P9-16 is a W203 × 60 structural steel (E = 200 GPa) wide-flange section (see Appendix A). Determine

(a) The equation of the elastic curve for the region of the beam between the supports. Use the designated axes.

(b) The deflection midway between the supports if w = 3.5 kN/m and L = 5 m. SOLUTION

0 :BMΣ =� 0A B AM M R L− − =

( ) ( )( )2 2 2 4 0AwL wL L R L− − =

3 8AR wL= ↑

(a) 23

8 2wLx wLEIy′′ = −

2 2

13

16 2wLx wL xEIy C′ = − + At 0,x = 0 :y = 2 0C =

3 2 2

1 23

48 4wLx wL xEIy C x C= − + + At ,x L= 0 :y =

3

1316wLC =

( )2 2 33 8 316

wy Lx L x LEI

′ = − + ( )3 2 2 34 316

wy Lx L x L xEI

= − + ...................Ans.

(b) For a W 203 60× wide-flange section: 6 460.8 10 mmI = ×

( )( )

( ) ( )434 4 4 4

max 2 9 6

5 3.5 10 54 3 516 8 4 2 128 128 200 10 60.8 10x L

w L L L wLyEI EI

δ = −

× = = − + = = × ×

37.03 10 m 7.03 mm−= + × = ↑ ..................................................................................Ans.

9-17 A beam AB is loaded and supported as shown in Fig. P9-17. The load P is applied through a collar which can be positioned on the load bar DE at any location in the interval L/4 < a < 3L/4. Determine

(a) The equation of the elastic curve for beam AB. (b) The location of load P for maximum deflection at end B. (c) The location of load P for zero deflection at end B. SOLUTION

0 :yF↑Σ = 0AV P− = AV P=

0 :AMΣ =� 0AM Pa− − = AM Pa= −

(a) EIy Px Pa′′ = −

2

12PxEIy Pax C′ = − + At 0,x = 0 :y′ = 1 0C =

3 2

1 26 2Px PaxEIy C x C= − + + At 0,x = 0 :y = 2 0C =

( )2 22

Py x axEI

′ = − ( )3 236

Py x axEI

= − .......................................Ans.


379

(b) ( )3 236B x L

Py L aLEI

δ == = −

For 4a L= 3 3 3

3 36 4 24 24B

P L PL PLLEI EI EI

δ += − = = ↑

For 3 4a L= 3 3 3

3 9 5 56 4 24 24B

P L PL PLLEI EI EI

δ −= − = = ↓

The maximum deflection at B occurs when 3 4a L= .............................................................Ans.

0Bδ = when 3 23 0L aL− = 3a L= ..............................................................Ans.

9-18 The cantilever beam ABC shown in Fig. P9-18 has a second moment of area IAB = 2I in the interval AB and a second moment of area IBC = I in the interval BC. Determine

(a) The deflection at section B. (b) The deflection at section C. SOLUTION

0 :yF↑Σ = 2 0AV P− = 2AV P=

0 :AMΣ =� ( ) ( )2 0AM P L P L− − − = 3AM PL= −


12 2 3EIy Px PL′′ = −

21 12 3EIy Px PLx C′ = − + At 0,x = 1 0 :y′ = 1 0C =

3 2

1 1 232

3 2Px PLxEIy C x C= − + + At 0,x = 1 0 :y = 2 0C =

( )21 3

2Py x LxEI

′ = − ( )3 21 2 9

12Py x LxEI

= −

For the interval 2L x L≤ ≤

( )2 2 3EIy Px PL P x L′′ = − − −

( )2

22 33

2P x L

EIy Px PLx C−

′ = − − +

( )33 2

2 3 43

3 2 6P x LPx PLxEIy C x C

−= − − + +


1 2 :EIy EIy′ ′= 2 2

2 23

3 3 02 2

PL PL PL PL C− = − + + 23C PL=

1 2 :EIy EIy= 3 3 3 3

3 43 3 0

6 4 3 2PL PL PL PL C L C− = − + + +

3

4512PLC −=

( )33 2 2 32 4 18 2 12 5

12Py x Lx x L L x LEI

= − − − + −


380

(a) ( ) ( )3 3

1 27 7

12 12B x L x L

PL PLy yEI EI

δ= =

−= = = = ↓ ......................................................................Ans.

(b) ( )3 3

2 2

23 2312 12C x L

PL PLyEI EI

δ=

−= = = ↓ ..................................................................................Ans.

9-19 The simply supported beam ABCD shown in Fig. P9-19 has a second moment of area IBC = 2I in the center section BC and a second moment of area IAB = ICD = I in the other two sections near the supports. Determine

(a) The deflection at section B. (b) The maximum deflection in the beam. SOLUTION For the interval 0 x L≤ ≤

1 2EIy Px′′ =

2

1 14PxEIy C′ = +

3

1 1 212PxEIy C x C= + + At 0,x = 1 0 :y = 2 0C =

For the interval 3 2L x L≤ ≤

22 2EIy Px′′ =

2

2 324

PxEIy C′ = + At 3 2,x L= 2 0 :y′ = 2

3916PLC −=

3

2 3 4212PxEIy C x C= + +


1 2 :EIy EIy′ ′= 2 2 2

11 9

4 8 2 16PL PL PLC

−+ = +

2

11332

PLC −=

1 2 :EIy EIy= 3 3 2

1 41 9 1

12 24 2 16 2PL PL PLC L L C

−+ = + +

3

4 6PLC −=

( )3 21 8 39

96Py x L xEI

= − ( )3 2 32 4 27 8

96Py x L x LEI

= − −

(a) ( ) ( )3 3

1 231 3196 96B x L x L

PL PLy yEI EI

δ= =

−= = = = ↓ ..................................................................Ans.

(b) ( )3 3

max 2 3 2

35 3596 96x L

PL PLyEI EI

δ=

−= = = ↓ ............................................................................Ans.




381

0 :yF↑Σ = 2 0AV wL wL− − = 3AV wL=

0 :AMΣ =� ( ) ( )2 2 2 0AM wL L wL L− − − = 23AM wL= −


2 214 3 3EIy wLx wL wx′′ = − −

2 3

21 1

34 32 3

wLx wxEIy wL x C′ = − − + At 0,x = 1 0 :y′ = 1 0C =

3 2 2 4

1 1 234

2 2 12wLx wL x wxEIy C x C= − − + + At 0,x = 1 0 :y = 2 0C =

( )2 2 31 9 18 2

24wy Lx L x xEI

′ = − − ( )3 2 2 41 6 18

48wy Lx L x xEI

= − −


( )22 22 3 3EIy wLx wL wx w x L′′ = − − + −

( )32 3

22 3

3 32 3 3

w x LwLx wxEIy wL x C−

′ = − − + +

( )43 2 2 4

2 3 43

2 2 12 12w x LwLx wL x wxEIy C x C

−= − − + + +


1 2 :EIy EIy′ ′= 3 3 3 3 3

33

9 2 3 3 024 3 12 2 3wL wL wL wL wLwL C− − = − − + +

3

333

24wLC =

1 2 :EIy EIy= 4 4 4 4 4 4

3 43 3 0

8 8 48 2 2 12wL wL wL wL wL wL C L C− − = − − + + +

4

427

48wLC −=

( )44 3 2 2 3 42 4 24 72 4 66 27

48wy x Lx L x x L L x LEI

= − + − + − + −

(a) ( ) ( )4 4

1 213 1348 48B x L x L

wL wLy yEI EI

δ= =

−= = = = ↓ .................................................................Ans.

(b) ( )4 4

2 2

51 5148 48C x L

wL wLyEI EI

δ=

−= = = ↓ ..................................................................................Ans.

9-21 A beam is loaded and supported as shown in Fig. P9-21. Determine the deflection midway between the supports.

SOLUTION

0 :yF↑Σ = 0A BR R wa+ − =

0 :BMΣ =� ( ) ( )3 2 3 0Awa a R a− =

2A BR R wa= =

For the interval 0 x a≤ ≤

1 2EIy wax′′ =


382

2

1 14waxEIy C′ = +

3

1 1 212waxEIy C x C= + + At 0,x = 1 0 :y = 2 0C =

For the interval 2a x a≤ ≤

( )2

2 2 2w x awaxEIy

−′′ = −

( )32

2 34 6w x awaxEIy C

−′ = − + At 3 2,x a= 2 0 :y′ =

3

313

24waC −=

( )43

2 3 412 24w x awaxEIy C x C

−= − + +

The two solutions must match at ,x a=

1 2 :EIy EIy′ ′= 3 3 3

1130 0

4 4 24wa wa waC− + = − −

3

113

24waC −=

1 2 :EIy EIy= 4 4 3

1 4130 0

12 12 24wa wa waC a a C

−− + = − + +

4 0C =

( )43 32 2 13

24wy ax x a a xEI

= − − −

( )4 4 4 4 4

2 3 2

27 39 205 20524 4 16 2 384 384M x a

w a a a wa wayEI EI EI

δ=

−= = − − = = ↓

..................Ans.

9-22 A beam is loaded and supported as shown in Fig. P9-22. Determine the deflection at the left end of the distributed load.

SOLUTION

0 :BMΣ =� ( ) ( )2 2 0AwL L R L− =

4AR wL=

0 :yF↑Σ = 0A BR R wL+ − =

3 4BR wL=

For the interval 0 x L≤ ≤ 2L x L≤ ≤

1 4EIy wLx′′ = ( )2

2 4 2w x LwLxEIy

−′′ = −

2

1 18wLxEIy C′ = +

( )32

2 38 6w x LwLxEIy C

−′ = − +

3

1 1 224wLxEIy C x C= + +

( )43

2 3 424 24w x LwLxEIy C x C

−= − + +

The boundary conditions give:


383

At 0,x = 1 0 :y = 2 0C =

At 2 ,x L= 2 0 :y = 4

3 47224wLLC C −+ =


1 2 :EIy EIy′ ′= 3 3

1 308 8

wL wLC C+ = − +

1 2 :EIy EIy= 4 4

1 3 4024 24

wL wLxC L C L C+ = − + +

Therefore, 31 3 7 48C C wL= = − 2 4 0C C= =

3 31 2 7

48wy Lx L xEI

= − ( )43 32 2 7

48wy Lx x a L xEI

= − − −

( ) ( )4 4

1 25 5

48 48M x L x L

wL wLy yEI EI

δ= =

−= = = = ↓ ....................................................................Ans.

9-23 A beam is loaded and supported as shown in Fig. P9-23. Determine (a) The slope at the left end of the beam. (b) The maximum deflection between the supports. SOLUTION

0 :BMΣ =� ( ) ( ) ( )2 3 0AP L P L R L+ − =

AR P=

0 :yF↑Σ = 0A BR R P P+ − − =

BR P=

For the interval 0 x L≤ ≤ 2L x L≤ ≤

1EIy Px′′ = ( )2EIy Px P x L′′ = − −

2

1 12PxEIy C′ = +

( )22

2 32 2P x LPxEIy C

−′ = − +

3

1 1 26PxEIy C x C= + +

( )33

2 3 46 6P x LPxEIy C x C

−= − + +

The boundary conditions give:

At 0,x = 1 0 :y = 2 0C =

At 3 2,x L= 2 0 :y′ = 23C PL= −


1 2 :EIy EIy′ ′= 2 2

1 302 2

PL PLC C+ = − + 21 3C C PL= = −

1 2 :EIy EIy= 3 3

1 3 406 6

PL PLC L C L C+ = − + + 4 2 0C C= =


384

(a) ( )3 21 6

6Py x L xEI

= − ( )2 21 2

2Py x LEI

′ = −

( )2 2

1 0A x

PL PLyEI EI

θ=

−′= = = ............................................................................................Ans.

(b) ( )33 22 6

6Py x x L L xEI

= − − −

( )3 3

2 3 2

23 2324 24M x L

PL PLyEI EI

δ=

−= = = ↓ ..............................................................................Ans.



0 :yF↑Σ = 2 0AV wL wL− − = 3AV wL=

0 :AMΣ =� ( ) ( )2 2 3 2 0AM wL L wL L− − − = 25 2AM wL= −


2

21

54 32

wLEIy wLx wx′′ = − −

2 2 3

1 13 54

2 2 3wLx wL x wxEIy C′ = − − + At 0,x = 1 0 :y′ = 1 0C =

3 2 2 4

1 1 254

2 4 12wLx wL x wxEIy C x C= − − + + At 0,x = 1 0 :y = 2 0C =

( )2 2 31 9 15 2

24wy Lx L x xEI

′ = − − ( )3 2 2 41 6 15

48wy Lx L x xEI

= − −


( )22

22

532 2

w x LwLEIy wLx wx−

′′ = − − +

( )32 2 3

2 33 5

2 2 3 6w x LwLx wL x wxEIy C

−′ = − − + +

( )43 2 2 4

2 3 45


−= − − + + +


1 2 :EIy EIy′ ′= 3 3 3 3 3 3

33 5 3 5 0

8 8 12 2 2 3wL wL wL wL wL wL C− − = − − + + 3

3C wL=

1 2 :EIy EIy= 4 4 4 4 4 4

3 45 5 0

8 16 48 2 4 12wL wL wL wL wL wL C L C− − = − − + + +

4

438wLC −=


385

( )44 3 2 2 3 42 2 12 30 24 9

24wy x Lx L x x L L x LEI

= − + − + − + −

(a) ( ) ( )4 4

1 25 5

24 24B x L x L

wL wLy yEI EI

δ= =

−= = = = ↓ .....................................................................Ans.

(b) ( )4 4

2 2

3 32 2C x L

wL wLyEI EI

δ=

−= = = ↓ ......................................................................................Ans.

9-25 A 160-lb diver walks slowly onto a diving board. The diving board is a wood (E = 1800 ksi) plank 10 ft long, 18 in. wide, and 2 in. thick, and is modeled as the cantilever beam shown in Fig. P9-25. For the diver at positions, 5a nL= ( )1,2, ,5n = � , compute and plot the deflection curve for the diving board (plot

y as a function of x for 0 ≤ x ≤ 10 ft). SOLUTION

0 :yF↑Σ = 160 0AV − =

160 lbAV = ↑

0 :AMΣ =� 160 0AM a− − =

160 160 lb ft AM a a= − = ⋅ �

For the interval 0 x a≤ ≤

( )1 160 160 lb ftEIy x a′′ = − ⋅

( )2 21 180 160 lb ftEIy x ax C′ = − + ⋅ At 0,x = 1 0 :y′ = 1 0C =

3

2 31 1 2

80 80 lb ft3xEIy ax C x C

= − + + ⋅

At 0,x = 1 0 :y = 2 0C =

For the interval a x L≤ ≤

( )2 160 160 160 lb ftEIy x a x a′′ = − − − ⋅

( )22 22 380 160 80 lb ftEIy x ax x a C ′ = − − − + ⋅

( )33

2 32 3 4

8080 80 lb ft3 3

x axEIy ax C x C −

= − − − + + ⋅

The two solutions must match at ,x a=

1 2 :EIy EIy′ ′= 2 2 2 2380 160 80 160 0a a a a C− = − − + 3 0C =

1 2 :EIy EIy= 3 3

3 33 4

80 8080 80 03 3a aa a C a C− = − − + + 4 0C =

( ) ( )33 6 218 2

1800 10 21.6 10 lb in12

EI = × = × ⋅

Therefore:

( )3 3

21 6

12 80 80 in.21.6 10 3

xy ax

= − × ........................................ 0 x a≤ ≤ ........................Ans.


386

( ) ( )3 33

22 6

12 8080 80 in.21.6 10 3 3

x axy ax −

= − − ×

............... 10 fta x≤ ≤ .................Ans.

9-26 A diving board consists of a wood (E = 12 GPa) plank which is pinned at the left end and rests on a movable support, as shown in Fig. P9-26. The board is 3 m long, 500 mm wide, and 80 mm thick. If a 70-kg diver stands at the end of the board,

(a) Compute and plot the deflection curve for the beam (plot y as a function of x for 0 ≤ x ≤ 3 m) for the right support at positions b = 0.5 m, 1.0 m, and 1.5 m.

(b) If the stiffness of the board is defined as the ratio of the diver�s weight to the deflection at the end of the board ( )k W y= , compute and plot the stiffness of the board as a function of b for 0 ≤ b ≤ 1.5 m. Does the stiffness depend on the weight of the diver?

SOLUTION

( )70 9.81 686.7 NW = =

0 :AMΣ =� 0Bb WL− =

B WL b= ↑

0 :BMΣ =� ( ) 0Ab W L b− − =

( ) ( )A W L b b W L b b= − − = − ↓

(a) For the interval 0 x b≤ ≤

( )

1

W L b xEIy

b−

′′ = −

( ) 2

1 12W L b x

EIy Cb−′ = − + At 0,x = 1 0 :y = 2 0C =

( ) 3

1 1 26W L b x

EIy C x Cb−

= − + + At ,x b= 1 0 :y = ( )

1 6W L b b

C−

=

For the interval b x L≤ ≤

( ) ( )

2

W L b x WL x bEIy

b b− −

′′ = − +

( ) ( )22

2 32 2W L b x WL x b

EIy Cb b− −

′ = − + +


388

1EIy wx C′′′ = − +

2

1 22wxEIy C x C′′ = − + +

3 2

12 36 2

wx C xEIy C x C′ = − + + +

4 3 2

1 23 424 6 2

wx C x C xEIy C x C= − + + + +

Boundary conditions:

At 0,x = 0 :V EIy′′′= = 1 0C = At ,x L= 0 :y′ = 33 6C wL=

At 0,x = 0 :M EIy′′= = 2 0C = At ,x L= 0 :y = 44 8C wL= −

(a) ( )4 3 44 324

wy x L x LEI

= − + − ..............................................................................................Ans.

(b) 4 4

0 8 8A xwL wLyEI EI

δ =−= = = ↓ ..................................................................................................Ans.

9-28 For the beam and loading shown in Fig. P9-28, determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. SOLUTION

EIy wx L′′′′ = −

2

12wxEIy C

L′′′ = − +

3

1 26wxEIy C x C

L′′ = − + +

4 2

12 324 2

wx C xEIy C x CL

′ = − + + +

5 3 2

1 23 4120 6 2

wx C x C xEIy C x CL

= − + + + +


At 0,x = 0 :M EIy′′= = 2 0C = At ,x L= 0 :M EIy′′= = 1 6C wL=

At 0,x = 0 :y = 4 0C = At ,x L= 0 :y = 33 7 360C wL= −

(a) ( )5 2 3 43 10 7360

wy x L x L xEIL

= − + − ..................................................................................Ans.

(b) 5 5 5 4 4

23 10 7 5 5

360 32 8 2 768 768M x Lw L L L wL wLyEIL EI EI

δ = −= = − + − = = ↓

..........................Ans.

9-29 A beam is loaded and supported as shown in Fig. P9-29. Determine (a) The equation of the elastic curve. (b) The deflection at the left end of the beam. (c) The support reactions VB and MB.


389

SOLUTION

3 3EIy wx L′′′′ = −

4

134wxEIy C

L′′′ = − +

5

1 2320wxEIy C x C

L′′ = − + +

6 2

12 33120 2

wx C xEIy C x CL

′ = − + + +

7 3 2

1 23 43840 6 2


= − + + + +


At 0,x = 0 :V EIy′′′= = 1 0C = At ,x L= 0 :y′ = 33 120C wL=

At 0,x = 0 :M EIy′′= = 2 0C = At ,x L= 0 :y = 44 140C wL= −

(a) ( )7 6 73 7 6

840wy x L x LEIL

= − + − ........................................................................................Ans.

(b) 4 4

0 140 140A xwL wLy

EI EIδ =

−= = = ↓ ............................................................................................Ans.

(c) 4 4B x LV EIy wL wL=′′′= = − = ↑ ..........................................................................................Ans.

(d) 2 220 20 B x LM EIy wL wL=′′= = − = � ..............................................................................Ans.

9-30 A beam is loaded and supported as shown in Fig. P9-30. Determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. (c) The maximum deflection of the beam. (d) The support reactions RA and RB. SOLUTION

3 3EIy wx L′′′′ = −

4

134wxEIy C

L′′′ = − +

5

1 2320wxEIy C x C

L′′ = − + +

6 2

12 33120 2

wx C xEIy C x CL

′ = − + + +

7 3 2

1 23 43840 6 2


= − + + + +


At 0,x = 0 :M EIy′′= = 2 0C = At 0,x = 0 :y = 4 0C =

At ,x L= 0 :M EIy′′= = 1 20C wL= At ,x L= 0 :y = 33 140C wL= −


390

(a) ( )7 4 3 63 7 6

840wy x L x L xEIL

= − + − ....................................................................................Ans.

(b) 7 7 4 4

72 3

7 13 133840 128 8 5120 5120M x L

w L L wL wLy LEIL EI EI

δ = −= = − + − = = ↓

.......................Ans.

(c) The maximum deflection occurs where

( )6 4 2 63 7 21 6 0

840wy x L x LEIL

′ = − + − = 0.5424x L=

4 4

max 0.54240.00256 0.00256 1.008x L M

wL wLyEI EI

δ δ=−= = = ↓= .....................................Ans.

(d) 0 20 20A xR EIy wL wL=′′′= = + = ↑ ....................................................................................Ans.

4 20 5B x L

wL wL wLR EIy = ′′′= − = − − + = ↑

..........................................................................Ans.

9-31 A beam is loaded and supported as shown in Fig. P9-31. Determine (a) The equation of the elastic curve. (b) The deflection at the left end of the beam. (c) The support reactions VB and MB. SOLUTION

cos2

xEIy wL

π′′′′ = −

12 sin

2wL xEIy C

Lπ

π′′′ = − +

2

1 22

4 cos2

wL xEIy C x CL

ππ

′′ = + +

3 2

12 33

8 sin2 2

wL x C xEIy C x CL

ππ

′ = + + +

4 3 2

1 23 44

16 cos2 6 2

wL x C x C xEIy C x CL

ππ

= − + + + +


At 0,x = 0 :V EIy′′′= = 1 0C =

At 0,x = 0 :M EIy′′= = 2 22 4C wL π= −

At ,x L= 0 :y′ = ( )3

3 3

4 2wLC ππ

= −

At ,x L= 0 :y = ( )4

4 3

2 4wLC ππ

= −

(a) ( ) ( )4 2 2 2 3 44 32 cos 4 8 2 4 4

2 2w xy L L x L x L

EI Lπ π π π π π

π = − − + − + −

..................Ans.

(b) ( )4 4

4 40 4

0.1089 0.108932 4 42A x

w wL wLy L LEI EI EI

δ π ππ=

− = = − + − = = ↓ .........Ans.


391

(c) 2 2B x LV EIy wL wLπ π=′′′= = − = ↑ ....................................................................................Ans.

(d) 2 2 2 24 4 B x LM EIy wL wLπ π=′′= = − = � ........................................................................Ans.

9-32 A beam is loaded and supported as shown in Fig. P9-32. Determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. (c) The maximum deflection of the beam. (d) The slope at the left end of the beam. (e) The support reactions RA and RB. SOLUTION

sin2

xEIy wL

π′′′′ = −

12 cos

2wL xEIy C

Lπ

π′′′ = +

2

1 22

4 sin2

wL xEIy C x CL

ππ

′′ = + +

3 2

12 33

8 cos2 2

wL x C xEIy C x CL

ππ

′ = − + + +

4 3 2

1 23 44

16 sin2 6 2


ππ

= − + + + +


At 0,x = 0 :M EIy′′= = 2 0C =

At ,x L= 0 :M EIy′′= = 1 2

4wLCπ

−=

At 0,x = 0 :y = 4 0C =

At ,x L= 0 :y = ( )3

23 4

2 243wLC ππ

= +

(a) ( )4 2 3 2 34

2 24 sin 243 2

w xy L Lx L xEI L

π π ππ

= − − + + .....................................................Ans.

(b) 2 4 2

4 42 4

2 2424 sin3 4 8 2M x L

w Ly L LEI

π π πδπ=

+= = − − +

4 40.00869 0.00869wL wL

EI EI−= = ↓ ..............................................................................Ans.

(c) The maximum deflection occurs where

( )3 2 2 2 34

2 12 cos 3 24 03 2

w xy L Lx LEI L

ππ π ππ

′ = − − + + = 0.5154x L=

4

max 0.51540.00870 1.001x L M

wLyEI

δ δ=−= = = ....................................................................Ans.


392

(d) ( )3 3

3 2 30 4

2 0.0262 0.026212 243A x


θ π ππ=

− ′= = − + + = = .......Ans.

(e) ( )

0 2 2

2 22 4A x

wLwL wLR EIyπ

π π π=

−′′′= = − = ↑ ...................................................................Ans.

2 2

4 4B x L

wL wLR EIyπ π=

− ′′′= − = − = ↑

................................................................................Ans.

9-33 A cantilever beam is loaded and supported as shown in Fig. P9-33. Use singularity functions to determine the deflection

(a) At a distance x = L from the support. (b) At the right end of the beam. SOLUTION

0 :yF↑Σ = 0AV P P+ − =

0AV =

0 :AMΣ =� ( )2 0APL P L M− − =

AM PL PL= − = �

1EIy PL P x L′′ = − + −

2

12P x L

EIy PLx C−

′ = − + + At 0,x = 0 :y′ = 1 0C =

32

1 22 6P x LPLxEIy C x C

−= − + + + At 0,x = 0 :y = 2 0C =

323

6Py Lx x LEI

= − + −

(a) 3 32 2B x Ly PL EI PL EIδ == = − = ↓ .................................................................................Ans.

(b) ( ) ( )3 3

2 32

11 113 26 6 6C x L

P PL PLy L L LEI EI EI

δ =− = = − + = = ↓

......................................Ans.


(a) At a distance x = L from the support. (b) At the right end of the beam. SOLUTION

0 :yF↑Σ = 0AV P P− − =

2AV P=

0 :AMΣ =� ( ) ( )2 0APL P L P L M− − − =

2 2 AM PL PL= − = �


393

0 12 2EIy Px PL PL x L P x L′′ = − − − − −

2

1212

2P x L

EIy Px PLx PL x L C−

′ = − − − − +

2 33

21 23 2 6

PL x L P x LPxEIy PLx C x C− −

= − − − + +


At 0,x = 0 :y′ = 1 0C =

At 0,x = 0 :y = 2 0C =

2 33 22 6 3

6Py x Lx L x L x LEI

= − − − − −

(a) 3 3

3 3 2 22 66 3 3B x LP PL PLy L LEI EI EI

δ =−

= = − = = ↓ ..........................................................Ans.

(b) ( ) ( ) ( ) ( )3 3

3 2 2 32

2 22 2 6 2 36C x L

P PL PLy L L L L L LEI EI EI

δ =− = = − − − = = ↓

........Ans.

9-35 A beam is loaded and supported as shown in Fig. P9-35. Use singularity functions to determine the deflection

(a) At a distance x = L from the left support. (b) At the middle of the span. SOLUTION

0 :yF↑Σ = 0A DR R P P+ − − =

0 :DMΣ =� ( ) ( ) ( )2 3 0AP L P L R L+ − =

A BR R P= =

1 12EIy Px P x L P x L′′ = − − − −

2 22

1

22 2 2

P x L P x LPxEIy C− −

′ = − − +

3 33

1 2

26 6 6

P x L P x LPxEIy C x C− −

= − − + +


At 0,x = 0 :y = 2 0C =

At 3 ,x L= 0 :y = 21C PL= −

3 33 22 6

6Py x x L x L L xEI

= − − − − −

(a) 3 3

3 3 5 566 6 6B x LP PL PLy L LEI EI EI

δ =−

= = − = = ↓ .............................................................Ans.


394

(b) 3 3 3 3

33 2

27 23 2396 8 8 24 24M x L

P L L PL PLy LEI EI EI

δ = −= = − − = = ↓

.....................................Ans.


(a) At the right end of the beam. (b) At a section midway between the supports. SOLUTION

0 :BMΣ =� ( ) ( )2 0AR L P L− − =

2 2AR P P= − = ↓

0 :yF↑Σ = 0A BR R P+ − =

3BR P= ↑

1

2 32LEIy Px P x′′ = − + −

2

21

32 2P LEIy Px x C′ = − + − + At 0,x = 0 :y = 2 0C =

33

1 23

3 6 2Px P LEIy x C x C= − + − + + At 2,x L= 0 :y =

2

1 12PLC =

3

3 24 612 2

P Ly x x L xEI

= − + − +

(a) 3 3 3 3

33 2

108 3612 8 2 2 2C x L

P L L PL PLy LEI EI EI

δ = −= = − + + = = ↓

.....................................Ans.

(b) 3 3 3 3

4 012 16 4 64 64M x L

P L L PL PLyEI EI EI

δ = += = − + + = = ↑

................................................Ans.


(a) At the point of application of the concentrated load 3P. (b) At the right end of the beam. SOLUTION

0 :yF↑Σ = 3 0AV P− =

3AV P=

0 :AMΣ =� ( )2 3 0APL P L M− − =

AM PL PL= − = �

13 3EIy Px PL P x L′′ = − − −


395

22

1

332 2

P x LPxEIy PLx C−

′ = − − +

33 2

1 22 2 2P x LPx PLxEIy C x C

−= − − + +


At 0,x = 0 :y′ = 1 0C =

At 0,x = 0 :y = 2 0C =

33 2

2Py x Lx x LEI

= − − −

(a) 3 3 02B x L

Py L LEI

δ = = = − = ............................................................................................Ans.

(b) 3 3 3 3 3

3 227 9

2 8 4 8 2 2C x LP L L L PL PLyEI EI EI

δ = += = − − = = ↑

.............................................Ans.


(a) At the left end of the beam. (b) At a point midway between the supports. SOLUTION

0 :BMΣ =� ( )4 2 03 3AL PLP R L − − =

2 3AR P= ↑

1 02 2 5

3 3 3 6P L PL LEIy Px x x′′ = − + − + −

2 12

12 5

2 3 3 3 6Px P L PL LEIy x x C′ = − + − + − +

3 23

1 25

6 9 3 3 6Px P L PL LEIy x x C x C= − + − + − + +


At 3,x L= 0 :y = 21 7 36C PL=

At 4 3,x L= 0 :y = 32 19 324C PL= −

3 2

3 2 3554 36 108 63 19324 3 6

P L Ly x x L x L x LEI

= − + − + − + −

(a) 3 3

019 19

324 324L xPL PLyEI EI

δ =−= = = ↓ .........................................................................................Ans.

(b) 3 3

5 6 48 48M x LPL PLyEI EI

δ =+= = = ↑ ..........................................................................................Ans.


396

9-39 Use singularity functions to determine the deflection at the left end of the cantilever beam shown in Fig. P9-39.

SOLUTION

24 3

25 2 2wLx w LEIy x′′ = − − −

32

12 3

25 6 2wLx w LEIy x C′ = − − − +

43

1 22 3

75 24 2wLx w LEIy x C x C= − − − + +


At 5 2,x L= 0 :y′ = 31 2 3C wL=

At 5 2,x L= 0 :y = 42 29 24C wL= −

4

3 3 4316 25 400 725600 2

w Ly Lx x L x LEI

= − − − + −

4 4

0725 29600 24A x

wL wLyEI EI

δ =−= = = ↓ ......................................................................................Ans.


(a) At the left end of the distributed load. (b) At a section midway between the supports. SOLUTION

0 :BMΣ =� ( ) ( ) ( )2 3 0Aw L L R L− =

2 3AR wL= ↑

2

23 2

w x LwLxEIy−

′′ = −

32

13 6w x LwLxEIy C

−′ = − + At 0,x = 0 :y = 2 0C =

43

1 29 24w x LwLxEIy C x C

−= − + + At 3 ,x L= 0 :y =

3

179wLC −=

43 38 3 56

72wy Lx x L L xEI

= − − −

(a) 4 4

4 4 48 28 5672 72 3L x L


δ =−

= = − = = ↓ ...................................................Ans.

(b) 4 4 4

4 43 2

3 305 30527 8472 16 384 384M x L

w L wL wLy L LEI EI EI

δ = −= = − − = = ↓

........................Ans.


397

9-41 A beam is loaded and supported as shown in Fig. P9-41. Use singularity functions to determine (a) The deflection midway between the supports (b) The maximum deflection of the beam. SOLUTION

0 :BMΣ =� ( ) ( ) ( ) ( ) ( )2 2 3 2 5 2 0Aw L L wL L R L+ − =

7 5AR wL= ↑

2

17 325 2 2

wLx w LEIy wL x L x′′ = − − − −

32

21

7 310 6 2wLx w LEIy wL x L x C′ = − − − − +

43

31 2

7 330 3 24 2

wLx wL w LEIy x L x C x C= − − − − + +


At 0,x = 0 :y = 2 0C =

At 5 2,x L= 0 :y = 31 119 120C wL= −

4

33 3328 40 5 119120 2

w Ly Lx L x L x L xEI

= − − − − −

(a) 4 4 4 4 4

5 4875 10 595 101 101


δ = −= = − − = = ↓

................Ans.

(b) The maximum deflection occurs when

3

22 3384 120 20 119 0120 2

w Ly Lx L x L x LEI

′ = − − − − − =

1.2186x L=

4 4 4max 1.2186 50.669 0.418 145.013

120x Lwy L L L

EIδ = = = − −

4 40.7897 0.7897 1.001 M

wL wLEI EI

δ−= = ↓= .............................................................Ans.

9-42 A beam is loaded and supported as shown in Fig. P9-42. Use singularity functions to determine (a) The deflection at the right end of the beam. (b) The deflection midway between the supports. SOLUTION

0 :BMΣ =� ( ) ( ) ( ) ( )3 2 2 4 0AwL L w L L R L− − =

11 8AR wL= ↑

0 :yF↑Σ = ( )2 0A BR R wL w L+ − − =

8BR wL= ↑


398

1

1 1 211 02 8 8 2L wL wL wEIy wL x x x L x L′′ = − + + − + − − −

2

2 2 31

11 02 2 16 16 6

wL L wL wL wEIy x x x L x L C′ = − + + − + − − − +

3

3 3 41 2

11 06 2 48 48 24

wL L wL wL wEIy x x x L x L C x C= − + + − + − − − + +


At 0,x = 0 :y = 42 48C wL=

At ,x L= 0 :y = 31 5 16C wL=

3

3 3 4 3 48 11 0 2 1548 2

w Ly L x L x L x L x L L x LEI

= − + + − + − − − + +

(a) 4 4

3 29 9

128 128C x LwL wLyEI EI

δ =−= = = ↓ .......................................................................................Ans.

(b) 4 4 4 4

4 42

11 15 5 5848 8 2 128 128M x L

w L L wL wLy L LEI EI EI

δ = += = − + + + = = ↑

......................Ans.

9-43 Use singularity functions to determine the deflection at the right end of the cantilever beam shown in Fig. P9-43.

SOLUTION

0 :AMΣ =� ( ) ( ) [ ] ( )8 2 2 0AwL L wL L M− − =

2 24 4 AM wL wL= − = �

0 :yF↑Σ = ( )8 0AV wL wL− + =

7 8AV wL= ↑

22 27

8 4 2 2w x LwLx wL wxEIy

−′′ = − − +

32 2 3

17

16 4 6 6w x LwLx wL x wxEIy C

−′ = − − + +

43 2 2 4

1 27


−= − − + + +


At 0,x = 0 :y′ = 1 0C =

At 0,x = 0 :y = 2 0C =

43 4 2 27 2 6 2

48wy Lx x L x x LEI

= − − + −

4 4

4 4 4 42 56 32 24 2

48 24 24C x Lw wL wLy L L L LEI EI EI

δ =+

= = − − + = = ↑ ..........................Ans.


399


(a) At the left end of the distributed load. (b) At a section midway between the supports. SOLUTION

0 :yF↑Σ = 0A BR R wa+ − =

0 :BMΣ =� ( )( ) ( )3 2 3 0Awa a R a− =

2A BR R wa= = ↑

2 22

2 2 2w x a w x awaxEIy

− −′′ = − −

3 32

1

24 6 6

w x a w x awaxEIy C− −

′ = − − + At 0,x = 0 :y = 2 0C =

4 43

1 2

212 24 24

w x a w x awaxEIy C x C− −

= − − + + At 3 ,x a= 0 :y = 3

113

24waC −=

4 43 32 2 13

24wy ax x a x a a xEI

= − − + − −

(a) 4 4

4 4 11 112 0 0 1324 24 24L x a

w wa way a aEI EI EI

δ =−

= = − + − = = ↓ ....................................Ans.

(b) 4 4 4 4 4

3 254 39 205 2050

24 8 16 2 384 384M x aw a a a wa wayEI EI EI

δ = −= = − + − = = ↓

.................Ans.

9-45 A beam is loaded and supported as shown in Fig. P9-45. Use singularity functions to determine (a) The deflection midway between the supports. (b) The maximum deflection of the beam. SOLUTION

0 :CMΣ =� ( )( ) ( ) ( ) ( )2 3 2 0AwL L wL L R L+ − =

7 12AR wL= ↑

3

1712 6

w x LwLxEIy wL x LL−

′′ = − − −

2 42

17

24 2 24wL x L w x LwLxEIy C

L− −

′ = − − +

3 53

1 27

72 6 120wL x L w x LwLxEIy C x C

L− −

= − − + +


At 0,x = 0 :y = 2 0C =

At 2 ,x L= 0 :y = 31 217 720C wL= −


400

3 52 3 2 470 120 6 217

720wy L x L x L x L L xEIL

= − − − − −

(a) 4 4

5 5 147 14770 0 0 217720 720 720M x L

w wL wLy L LEIL EI EI

δ =−

= = − − − = = ↓ ....................Ans.


2 42 2 2 4210 360 30 217

720wy L x L x L x L LEIL

′ = − − − − −

1.0168x L=

4 4

max 1.01680.2043 0.2043 1.0004x L M

wL wLyEI EI

δ δ=−= = = ↓= .......................................Ans.

9-46 A beam is loaded and supported as shown in Fig. P9-46. Use singularity functions to determine (a) The deflection at the middle of the span. (b) The maximum deflection in the beam. SOLUTION

0 :BMΣ =� ( ) ( ) ( )2 4 3 2 0AwL L R L− =

3AR wL= ↑

2 33

3 6 2 6w x L w x LwLx wxEIy

L L− −

′′ = − + +

3 42 4

16 24 6 24w x L w x LwLx wxEIy C

L L− −

′ = − + + +

4 53 5

1 218 120 24 120w x L w x LwLx wxEIy C x C

L L− −

= − + + + +


At 0,x = 0 :y = 2 0C =

At 2 ,x L= 0 :y = 31 41 360C wL= −

4 52 3 5 420 3 15 3 41

360wy L x x L x L x L L xEIL

= − + − + − −

(a) 4 4

5 5 5 2420 3 0 0 41360 360 15M x L

w wL wLy L L LEIL EI EI

δ =−

= = − + + − = = ↓ ....................Ans.


3 42 2 4 460 15 60 15 41 0

360wy L x x L x L x L LEIL

′ = − + − + − − =

0.9352x L=

4 4

max 0.93520.06703 0.0670 1.005x L M

wL wLyEI EI

δ δ=−= = ≅ ↓= ......................................Ans.

9-47 A 160-lb diver walks slowly onto a diving board. The diving board is a wood (E = 1800 ksi) plank 10 ft long, 18 in. wide, and 2 in. thick, and is modeled as the cantilever beam shown in Fig. P9-47. For the diver at positions, 5a nL= ( )1,2, ,5n = � , compute and plot the deflection curve for the diving board (plot

y as a function of x for 0 ≤ x ≤ 10 ft). SOLUTION


401

0 :yF↑Σ = 160 0AV − =

160 lbAV = ↑

0 :AMΣ =� 160 0AM a− − =

160 160 lb ft AM a a= − = ⋅ �

1160 160 160 lb ftEIy x a x a ′′ = − − − ⋅

22 2

180 160 80 lb ftEIy x ax x a C ′ = − − − + ⋅

33

2 31 2

8080 80 lb ft3 3

x axEIy ax C x C −

= − − + + ⋅


At 0,x = 0 :y′ = 1 0C =

At 0,x = 0 :y = 2 0C =

( ) ( )33

6 2

18 21800 10

1221.6 10 lb in

EI = ×

= × ⋅

Therefore:

( ) 33 3

26

8012 80 80 in.21.6 10 3 3

x axy ax −

= − − ×

............................................................Ans.

9-48 A diving board consists of a wood (E = 12 GPa) plank which is pinned at the left end and rests on a movable support, as shown in Fig. P9-48. The board is 3 m long, 500 mm wide, and 80 mm thick. If a 70-kg diver stands at the end of the board,

(a) Compute and plot the deflection curve for the beam (plot y as a function of x for 0 ≤ x ≤ 3 m) for the right support at positions b = 0.5 m, 1.0 m, and 1.5 m.

(b) If the stiffness of the board is defined as the ratio of the diver�s weight to the deflection at the end of the board ( )k W y= , compute and plot the stiffness of the board as a function of b for 0 ≤ b ≤ 1.5 m. Does the stiffness depend on the weight of the diver?

SOLUTION

0 :AMΣ =� 0Bb WL− = ( )70 9.81 686.7 NW = =

0 :BMΣ =� ( ) 0Ab W L b− − − =

B WL b= ↑

( ) ( )A W L b b W L b b= − − = − ↓

( )3

6 4500 8021.33 10 mm

12I = = ×


403

0 :AMΣ =� ( ) ( )15 210 210 4.5D a a− − +

( ) ( )40 15 7.5 0− =

( )363 28 lbD a= + ↑

0 :yF↑Σ = ( )210 210 40 15 0A D+ − − − =

( )657 28 lbA a= − ↑

( ) 1 12657 28 20 210 210 4.5 lb ftEIy a x x x a x a ′′ = − − − − − − − ⋅

( ) 2 3

2 2 21

657 28 20 105 105 4.5 lb ft2 3

a x xEIy x a x a C −

′ = − − − − − − + ⋅

( ) 3 4

3 3 31 2

657 28 5 35 35 4.5 lb ft6 3

a x xEIy x a x a C x C −

= − − − − − − + + ⋅


At 0 ft,x = 0 :y = 2 0C =

At 15 ft,x = 0 :y = 3 3

1

7 15 7 15 4.51050 19,012.5

3 3a a

C a− − −

= − + +

( ) 3 43 3

3 33

657 28 5 35 35 4.56 3

7 15 7 15 4.51050 19,012.5 lb ft

3 3

a x xEIy x a x a

a x a xax x

−= − − − − − −

− − −+ − + + ⋅

(a) The deflection is maximum at the middle of the span when the cart is at the middle of the bridge. Therefore,

7.5 ftx = 5.25 fta = 2 in.y = − 1800 ksiE =

4 325.109 in 4 12I h= =

4.22 in.h = .............................................................................................................................. Ans.

(b) When the depth of the beam is 5 in.h =

( )3

44 541.667 in

12I = =

( )( )3

6 2

1800 10 41.667

75.00 10 lb in

EI = ×

= × ⋅


404

9-50 The gangplank between a fishing boat and a dock consists of a wood (E = 12 GPa) plank 3 m long, 300 mm wide, and 50 mm thick. If the plank is modeled as the simply supported beam shown in Fig. P9-50, compute and plot the deflection curve for the gangplank (plot y as a function of x for 0 ≤ x ≤ 3 m) as a 75-kg man walks across the plank. Plot curves for the man at positions 7a nL= ( )1,2, ,5n = � .

SOLUTION

0 :AMΣ =� 0CL Wa− =

C WL a= ↑

0 :CMΣ =� ( ) 0W L a AL− − =

( )A W L a L= − ↑

( )85 9.81 833.85 NW = =

( )3

6 4300 503.125 10 mm

12I = = ×

( )( )9 6 3 212 10 3.125 10 37.50 10 N mEI −= × × = × ⋅

( ) 1W L a x

EIy W x aL−

′′ = − −

( ) 22

12 2W x aW L a x

EIy CL

−−′ = − +

( ) 33

1 26 6W x aW L a x

EIy C x CL

−−= − + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y = ( ) ( )3

1 6 6W L a L W L a

CL

− −= − +

Therefore:

( )

( )( ) ( ) ( ) ( ) ( )3

3 2 333

833.85 103 3 3 3 mm

6 37.5 10 3y L a x x a a x a x = − − − − − + −

×.....Ans.


405

9-51 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-51. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with P wL= :

( ) ( ) ( ) ( ) ( )4 3 32 2 38 6 3

C w P Bw Bw CP

w L w L wL LL

EI EI EI

δ δ δ δ θ δ= + = + +

= − − −

4 437 37

3 3wL wL

EI EI−= = ↓ ...................................................................................................Ans.

9-52 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-52. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with P wL= and

2 2M wL= :

( ) ( ) ( )( )23 24 28 3 2

B w P M

wL LwL LwLEI EI EI

δ δ δ δ= + +

= − + +

4 411 11

24 24wL wLEI EI

+= = ↑ .....................................................................................................Ans.

9-53 For the cantilever beam shown in Fig. P9-53, determine (a) The slope and deflection at section B. (b) The slope and deflection at section C. SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with M PL= :

(a) ( ) ( ) ( )22

2B P MP L PL LEI EI

θ θ θ= + = − −

2 22 2PL PL

EI EI−= = ........................................................................................................Ans.

( ) ( ) ( )23 3 32 7 7

3 2 6 6B P M

P L PL L PL PLEI EI EI EI

δ δ δ −= + = − − = = ↓ ........................................Ans.

(b) ( )22 2 2

/2 5 5

2 2 2C B C B

P LPL PL PLEI EI EI EI

θ θ θ −= + = − − = = ................................................Ans.

( ) ( )33 2 3 37 2 7 76 3 2 2C B B P

P LPL PL PL PLL LEI EI EI EI EI

δ δ θ δ −= + + = − − − = = ↓ ..................Ans.


406

9-54 Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-54 when P = 13.5 kN, L = 3 m, I = 80(106) mm4, and E = 200 GPa.

SOLUTION Using the solution for case 5 in Table A-19 with 1 2P P= and 2P P= :

( )( ) ( ) ( )

1 2

2 22 3 4 3 9 4 4 3 4

48P P

P L L L

EIδ δ δ

− = + = −

( ) ( ) ( )2 2

3 32 3 9 4 4 2 53 5348 96 96

P L L L PL PLEI EI EI

− − − = = ↓

( )( )

( ) ( )33

9 6

53 13.5 10 312.58 m 12.58 mm

96 200 10 80 10δ

−

− ×= = − = ↓

× ×.................................................Ans.

9-55 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-55. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with P wL= :

[ ] ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

4 3 3 2

2

2 22

8 6 3 2

D w P Cw Cw BP BPL L

w L w L wL L wL LL L

EI EI EI EI

δ δ δ δ θ δ θ= + = + + +

= − − − −

4 414 14

3 3wL wL

EI EI−= = ↓ .............................................................................................................Ans.

9-56 For the beam shown in Fig. P9-56, determine (a) The deflection at a point midway between the supports. (b) The deflection at the right end of the beam. SOLUTION Using the solutions for cases 6 and 7 in Table A-19 with P wL= :

(a) ( )( )345

384 48B Bw BP

wL LwLEI EI

δ δ δ= + = − −

4 413 13

384 384wL wLEI EI

−= = ↓ ...................................................................................................Ans.

(b) ( ) [ ] ( ) ( )( ) ( )23

3 3 324 16D C Cw CP

wL LwLL L LEI EI

δ θ θ θ

= = + = +

4 45 5

144 144wL wLEI EI

+= = ↑ ......................................................................................................Ans.


407

9-57 Member AB of Fig. P9-57 is the flexural member of a scale that is used to weigh food in a microwave oven. Determine the deflection of point C when W = 5 lb, L = 2 in. and EI = 100 lb-in.2

SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with P W= and

2M WL= :

( ) [ ] [ ] ( )( ) ( ) ( ) ( ) ( ) ( )

2 23

3

2 2

2 22

3 2 2

12

C B B BP BM BP BML L

WL L W L WL LWL LEI EI EI EI

WLEI

δ δ θ δ δ θ θ= − = + − +

= − + − − +

−=

With 5 lbW = , 2 in.L = , and 2100 lb inEI = ⋅ :

( ) ( )

( )

35 20.0333 in. 0.0333 in.

12 100Cδ−

= = − = ↓ ................................................................Ans.

9-58 Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-58. SOLUTION Using the solutions for cases 7 and 8 in Table A-19 with 2 3M wL= :

( )( )22 43 5

16 384M w

wL L wLEI EI

δ δ δ= + = − −

4 413 13

384 384wL wLEI EI

−= = ↓ ...................................................................................................Ans.

9-59 Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-59. SOLUTION Using the solutions for cases 7 and 8 in Table A-19 with 2 8BM wL= and

2 4CM wL= :

( )( ) ( )( )2 22 24 8 45384 16 16

B Cw M M

wL L wL LwLEI EI EI

δ δ δ δ= + +

= − + +

4 4

96 96wL wLEI EI

+= = ↑ ...........................................................................................................Ans.


408

9-60 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-60 when w = 7 kN/m, L = 1.8 m, I = 130(10-6) m4, and E = 200 GPa.

SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with 2BV wL= and

( )( ) ( )2 2 2 0BM wL w L L= − = :

( ) ( )( ) ( )( ) ( )

3 24

4

2 2 28 3 2

2312

C Cw B BL

w L wL L wL LL

EI EI EIwLEI

δ δ δ θ= + +

= − − −

−=

( )( )

( ) ( )43

39 6

23 7 10 1.85.42 10 m 5.42 mm

12 200 10 130 10Cδ −−

− ×= = − × = ↓

× ×..............................Ans.

9-61 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-61. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with BV wL= and

2 2BM wL= − :

( ) ( )( )( ) ( ) ( ) ( )( ) ( )( ) ( )

2 3 22 2 42 22 3 2 8

C B B Cw

BM BV BM BV Cw

LL

wL L wL LwL L wL L wLLEI EI EI EI EI

δ δ θ δδ δ θ θ δ

= + += + + + +

= − − + − − −

( )4 3 4 4 47 41 41

12 8 24 24wL wL wL wL wLLEI EI EI EI EI

− −= − − = = ↓ ..................................................Ans.

9-62 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-62. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with 2P wL= :

( ) ( )( ) ( ) ( )2 432 2 23 2 8

C BP BP CwL

wL L wL L w LL

EI EI EI

δ δ θ δ= + +

= − − −

4 429 29

12 12wL wLEI EI

−= = ↓ ...........................................................................................................Ans.


409

9-63 Determine the midspan deflection of beam AC of Fig. P9-63 if both beams have the same flexural rigidity. SOLUTION Using the solutions for cases 4, 5, and 6 in Table A-19 with 2CP P= and

2CM PL= :

( ) ( ) ( )2 2

3 32 3 4 4 11 11248 12 12F

P L L L PL PLEI EI EI

δ − − = − = = ↓

( ) 23 3 3

/

211 2 212 2 3 3C F C F

PL LPL PL PLEI EI EI EI

δ δ δ −= − = − + = = ↓

( )33 3 3

/

22 5 53 48 6 6B C B C

P LPL PL PLEI EI EI EI

δ δ δ −= + = − − = = ↓ ..............................................Ans.

9-64 Determine the deflection at the right end of the beam shown in Fig. P9-64. SOLUTION Using the solution for case 8 in Table A-19 with 22BM wL= :

( ) ( )( )2 4 42 33 46 4 4 4D C

wL L L wL wLLEI EI EI

δ θ − = = − = = ↓

..............................................Ans.

9-65 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-65. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with BV wL= and

2 2BM wL= :

( ) ( ) ( )( )( ) ( )( ) ( )( ) ( ) ( ) ( )

2 3 22 2 42 22 3 2 8

D B B Cw C

BM BV BM BV Cw BM BV Cw

L LL L

wL L wL LwL L wL L wLLEI EI EI EI EI

δ δ θ δ θδ δ θ θ δ θ θ θ

= + + += + + + + + + +

= − − + − − −

( )( ) ( )( ) ( )

22 322 6

wL L wL L wL LEI EI EI

− + +

4 469 23

24 8wL wLEI EI

−= = ↓ ...................................................................................................Ans.


410

9-66 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-66 when w = 7.5 kN/m, L = 3 m, I = 180(106) mm4, and E = 200 GPa.

SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with 2BV wL= and

( ) ( ) ( )2 23 2 2 2 2BM wL w L L wL= − = :

( ) ( )( ) ( )( ) ( )( )2 324 4 22 2

8 8 2 3

C Cw B B

Cw Bw BM BV Bw BM BV

LL

wL Lw L wL LwLEI EI EI EI

δ δ δ θδ δ δ δ θ θ θ

= + += + + − + + −

= − + − −

( )( ) ( )( ) ( )

223 42 2 76 2 8

wL L wL LwL wLLEI EI EI EI

− + + − =

( )( )

( )( )43

39 6

7 7.5 10 314.77 10 m 14.77 mm

8 200 10 180 10Cδ −−

− ×= = − × = ↓

× ×............................Ans.

9-67 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-67. SOLUTION Using the solution for case 3 in Table A-19:

( )( ) ( ) ( )2

4 34

2 2

2 2 22

30 30 24 2

C C w Bw Bw L

w L w L w L LEI EI EI

δ δ δ θ= + +

= − + +

4 4 4 4 411 11

15 240 192 192 192wL wL wL wL wL

EI EI EI EI EI−= − + + = = ↓ .......................................................Ans.

9-68 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-68. SOLUTION Using the solution for case 3 in Table A-19:

( )( )( ) ( ) ( )

2

4 4 32 230 30 24

C C w Bw Bw L

w L w L wL LEI EI EI

δ δ δ θ= + +

= − + +

4 4 4 4 416 119 119

15 30 24 120 120wL wL wL wL wLEI EI EI EI EI

−= − + + = = ↓ ....................................................Ans.


411

9-69 A beam is loaded and supported as shown in Fig. P9-69. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION

(a) 2

2AwxEIy R x′′ = −

2 3

12 6AR x wxEIy C′ = − +

3 4

1 26 24AR x wxEIy C x C= − + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y′ = 2 3

1 02 6AR L wL C− + =

At ,x L= 0 :y = 3 4

1 2 06 24AR L wL C L C− + + =

Solving simultaneously yields: 31 48C wL= − and

3 8 3 8AR wL wL= + = ↑ ......................................................................................................Ans.

Then, overall equilibrium gives:

0 :yF↑Σ = 0A BR wL V− − =

5 8 5 8BV wL wL= − = ↑ ...............................................................................................Ans.

0 :BMΣ =� 2

02B A

wLM R L+ − =

2 28 8 BM wL wL= − = � .............................................................................................Ans.

(b) ( )3 4 33 248

wy Lx x L xEI

= − −

4 4 4 4 4

23


δ = −= = − − = = ↓

..........................................Ans.

9-70 When a moment M is applied to the left end of the beam shown in Fig. P9-70, the slope at the left end of the beam is zero. Determine the magnitude of the moment M.

SOLUTION

2

2AwxEIy M R x′′ = − + −

2 3

12 6AR x wxEIy Mx C′ = − + − +

2 3 4

1 22 6 24AMx R x wxEIy C x C= − + − + +


412


At 0,x = 0 :y = 2 0C =

At 0,x = 0 :y′ = 1 0C =

At ,x L= 0 :y′ = 2 3

02 6AR L wLML− + − =

At ,x L= 0 :y = 2 3 4

02 6 24

AML R L wL− + − =

Solving simultaneously yields:

2 2AR wL wL= + = ↑

2 212 12 M wL wL= + = � .................................................................................................Ans.

9-71 A beam is loaded and supported as shown in Fig. P9-71. Determine the magnitude of the moment M required to make the slope at the left end of the beam zero.

SOLUTION

2

2wxEIy M′′ = −

3

16wxEIy Mx C′ = − +

2 4

1 22 24Mx wxEIy C x C= − + +


At 0,x = 0 :y′ = 1 0C =

At ,x L= 0 :y′ = 3

06

wLML − =

2 26 6 M wL wL= + = � .....................................................................................................Ans.

9-72 When a moment M is applied to the left end of the cantilever beam shown in Fig. P9-72, the slope at the left end of the beam is zero. Determine the magnitude of the moment M.

SOLUTION EIy Px M′′ = −

2

12PxEIy Mx C′ = − +

3 2

1 26 2Px MxEIy C x C= − + +


At 0,x = 0 :y′ = 1 0C =

At ,x L= 0 :y′ = 2

02

PL ML− =

2 2 M PL PL= + = � .........................................................................................................Ans.


413

9-73 When a load P is applied to the right end of the cantilever beam shown in Fig. P9-73, the deflection at the right end of the beam is zero. Determine the magnitude of the load P.

SOLUTION

2

2wxEIy Px′′ = −

2 3

12 6Px wxEIy C′ = − +

3 4

1 26 24Px wxEIy C x C= − + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y′ = 2 3

1 02 6

PL wL C− + =

At ,x L= 0 :y = 3 4

1 06 24

PL wL C L− + =


3 8 3 8P wL wL= + = ↑ ........................................................................................................Ans.

9-74 A beam is loaded and supported as shown in Fig. P9-74. Determine (a) The reactions at supports A and B. (b) The maximum deflection in the beam. SOLUTION

(a) 3

6AwxEIy R x

L′′ = −

2 4

12 24AR x wxEIy C

L′ = − +

3 5

1 26 120AR x wxEIy C x C

L= − + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y′ = 2 3

1 02 24AR L wL C− + =

At ,x L= 0 :y = 3 4

1 06 120AR L wL C L− + =


10 10AR wL wL= + = ↑ ......................................................................................................Ans.


0 :yF↑Σ = ( )2 0A BR wL V− − =


414

2 5 2 5BV wL wL= − = ↑ ...............................................................................................Ans.

0 :BMΣ =� 2

06B A

wLM R L+ − =

2 215 15 BM wL wL= − = � ........................................................................................Ans.

(b) The maximum deflection occurs where

( )2 2 4 46 5 0120

wy L x x LEIL

′ = − − = 0.4472x L=

( )2 3 5 42120

wy L x x L xEIL

= − −

4 4

max 0.44720.00239 0.00239

x LwL wLy

EI EIδ =

−= = = ↓ ........................................................Ans.

9-75 A beam is loaded and supported as shown in Fig. P9-75. Determine (a) The reactions at supports A and B. (b) The maximum deflection in the beam. SOLUTION

( ) 2w x kx= At :x L= ( )w x w= Therefore, ( ) 2 2w x wx=

(a) 4

212AwxEIy R x

L′′ = −

2 5

122 60AR x wxEIy C

L′ = − +

3 6

1 226 360AR x wxEIy C x C

L= − + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y′ = 2 3

1 02 60AR L wL C− + =

At ,x L= 0 :y = 3 4

1 06 360AR L wL C L− + =


24 24AR wL wL= + = ↑ .....................................................................................................Ans.


0 :yF↑Σ = ( )3 0A BR wL V− − =

7 24 7 24BV wL wL= − = ↑ ..........................................................................................Ans.

0 :BMΣ =� 03 4B A

wL LM R L + − =

2 224 24 BM wL wL= − = � .......................................................................................Ans.

(b) The maximum deflection occurs where


415

( )3 2 5 52 5 4 0

240wy L x x LEIL

′ = − − = 0.4666x L=

( )3 3 6 52 5 2 3

720wy L x x L xEIL

= − −

4 4

max 0.46660.001267 0.001267

x LwL wLy

EI EIδ =

−= = = ↓ ...................................................Ans.

9-76 A beam is loaded and supported as shown in Fig. P9-76. Determine the reactions at supports A and B. SOLUTION

( )sin 2EIy w x Lπ′′′′ = −

12 cos

2wL xEIy C

Lπ

π′′′ = +

2

1 22

4 sin2

wL xEIy C x CL

ππ

′′ = + +

3 2

12 33

8 cos2 2

wL x C xEIy C x CL

ππ

′ = − + + +

4 3 2

1 23 44

16 sin2 6 2


ππ

= − + + + +


At 0,x = 0 :y = 4 0C =

At 0,x = 0 :M EIy′′= = 2 0C =

At ,x L= 0 :y′ = 2

13 0

2C L C+ =

At ,x L= 0 :y = 4 3

134

16 06

wL C L C Lπ

− + + =

Solving simultaneously yields: 41 48C wL π= − and 3 4

3 24C wL π=

( )

0 4 4

02 48 2 48cos2A x

wL wL wL wLV EIyL

ππ π π π=

′′′= = − = − ↑

....................................Ans.

4 4 4

2 48 48 48cos2B x L

wL wL wL wLV EIy ππ π π π=

− ′′′= = − = = ↑ ..........................................Ans.

2 2 2 2

2 4 2 4

4 48 4 48sin 2B x L

wL wL wL wLM EIy ππ π π π=

′′= = − = −

� ..................................Ans.

9-77 A beam is loaded and supported as shown in Fig. P9-77. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION

(a) 1

AEIy R x P x L′′ = − −


416

22

12 2A P x LR xEIy C

−′ = − +

33

1 26 6A P x LR xEIy C x C

−= − + +


At 0,x = 0 :y = 2 0C =

At 2 ,x L= 0 :y′ = ( )2 212 2 0AR L PL C− + =

At 2 ,x L= 0 :y = 3 3

14 2 0

3 6AR L PL C L− + =

Solving simultaneously yields: 21 8C PL= − and

5 16 5 16AR P P= + = ↑ .......................................................................................................Ans.


0 :yF↑Σ = 0A BR P V− − =

0 :BMΣ =� ( )2 0B AM PL R L+ − =

11 16 11 16BV P P= − = ↑ .............................................................................................Ans.

3 8 3 8 BM PL PL= − = � ............................................................................................Ans.

(b) ( )33 25 16 1296

Py x x L L xEI

= − − −

( )3 3

3 3 7 75 0 1296 96 96M x L

P PL PLy L LEI EI EI

δ =−= = − − = = ↓ ...............................................Ans.

9-78 A beam is loaded and supported as shown in Fig. P9-78. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION (a) From symmetry:

2 2AV P P= + = ↑ ................................. 2 2BV P P= − = ↑ .........................................Ans.

Then for the interval 0 2x L≤ ≤

2 A

PxEIy M′′ = +

2

14 APxEIy M x C′ = + +

3 2

1 212 2APx M xEIy C x C= + + +


At 0,x = 0 :y′ = 1 0C =


417

At 0,x = 0 :y = 2 0C =

At 2,x L= 0 :y′ = 2

016 2

APL M L+ =

8 8 AM PL PL= − = � .......................................................................................................Ans.

Then, from symmetry, 8 8 BM PL PL= − = � ..............................................Ans.

(b) ( )3 24 348

Py x LxEI

= − 0 2x L≤ ≤

3 3 3 3

23

48 2 4 192 192M x LP L L PL PLyEI EI EI

δ = −= = − = = ↓

...................................................Ans.

9-79 A beam is loaded and supported as shown in Fig. P9-79. Determine the reactions at supports B, C, and D. SOLUTION

1 12D C BEIy R x R x L R x L′′ = + − + −

2 22

1

22 2 2

C BD R x L R x LR xEIy C− −

′ = + + +

3 33

1 2

26 6 6

C BD R x L R x LR xEIy C x C− −

= + + + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y = ( )316 0DR L C L+ =

At 2 ,x L= 0 :y = ( ) ( )3 314 3 6 2 0D CR L R L C L+ + =

Solving simultaneously yields: 21 6DC R L= and 6D CR R= −


0 :yF↑Σ = 0B C DP R R R− + + + =

0 :BMΣ =� ( ) ( ) ( )2 2 0C DP L R L R L+ + =

13 138 8B

P PR += = ↑ ..........3 34 4CP PR −= = ↓ ..............

8 8DP PR += = ↑ .........................Ans.

9-80 A beam is loaded and supported as shown in Fig. P9-80. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the middle support. (c) The deflection at the middle of span BC. SOLUTION

(a) 22

1

2 2A B

w x LwxEIy R x R x L−

′′ = − + − +

2 32 3

12 6 2 6BA R x L w x LR x wxEIy C

− −′ = − + + +


418

3 43 4

1 26 24 6 24BA R x L w x LR x wxEIy C x C

− −= − + + + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y = 3 4

1 06 24AR L wL C L− + =

At 2 ,x L= 0 :y = 3 4 3 4

14 2 2 0

3 3 6 24A BR L wL R L wL C L− + + + =

Overall equilibrium gives:

0 :yF↑Σ = 0A B CR R R wL+ + − =

0 :CMΣ =� ( ) ( ) ( ) ( )3 2 2 0B AwL L R L R L− − =


7 716 16AwL wLR += = ↑ ........

5 58 8BwL wLR += = ↑ ........

16 16CwL wLR −= = ↓ ....................Ans.

(b) ( ) ( )2 272

16 2 16B AwL wL wLM R L wL L L −= − = − = .........................................................Ans.

(c) ( )3 43 4 37 4 10 4 396

wy Lx x L x L x L L xEI

= − + − + − −

4 4 4 4 4

3 2189 81 10 9

96 8 4 8 4 2MBC x Lw L L L L LyEI

δ =

= = − + + −

4 4

256 256wL wL

EI EI+= = ↑ ..................................................................................................Ans.

9-81 A beam is loaded and supported as shown in Fig. P9-81. Determine (a) The reactions at supports A, C and D. (b) The deflection at the middle of span AC. SOLUTION

(a) 1 12A CEIy R x P x L R x L′′ = − − + −

2 22

1

22 2 2

CA P x L R x LR xEIy C− −

′ = − + +

3 33

1 2

26 6 6

CA P x L R x LR xEIy C x C− −

= − + + +


At 0,x = 0 :y = 2 0C =

At 2 ,x L= 0 :y = 3 3

14 2 0

3 6AR L PL C L− + =


419

At 3 ,x L= 0 :y′ = 22 2

19 4 0

2 2 2CA R LR L PL C− + + =

At 3 ,x L= 0 :y = 33 3

19 4 3 0

2 3 6CA R LR L PL C L− + + =

Solving simultaneously yields: 21 7 44C PL= − and

4 411 11A

P PR += = ↑ ............................23 2322 22C

P PR += = ↑ ......................................Ans.


0 :yF↑Σ = 4 23 011 22 DP PP V− + − =

0 :DMΣ =� ( ) ( ) ( )23 42 3 022 11D

P PM P L L L+ − − =

9 922 22D

P PV += = ↓ ............................3 322 22DPL PLM += = � ....................................Ans.

(b) 3 33 223 22 7

33 6 132 44x L x LP x L xy

EI

− −= − + −

3 3 3 32 7 13 130 0

33 44 132 132B x LP L L PL PLyEI EI EI

δ = −= = − + − = = ↓

.........................................Ans.

9-82 A beam is loaded and supported as shown in Fig. P9-82. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the center support. (c) The maximum bending moment in the beam. SOLUTION

(a) 2

1

2A BwxEIy R x R x L′′ = − + −

22 3

12 6 2BA R x LR x wxEIy C

−′ = − + +

33 4

1 26 24 6BA R x LR x wxEIy C x C

−= − + + +


At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y = 3 4

1 06 24AR L wL C L− + =

At 2 ,x L= 0 :y′ = 3 2

21

42 03 2

BA

wL R LR L C− + + =

At 2 ,x L= 0 :y = 3 4 3

14 2 2 0

3 3 6A BR L wL R L C L− + + =


420


11 1128 28A

wL wLR += = ↑ ..................8 87 7BwL wLR += = ↑ .....................................Ans.


0 :yF↑Σ = 2 0A B CR R V wL+ − − =

13 1328 28C

wL wLV −= = ↑ .................................................................................................Ans.

0 :CMΣ =� ( )( ) ( ) ( )2 2 0B Aw L L R L R L− − =

2 2

14 14CwL wLM −= = � ...................................................................................................Ans.

(b) ( )( )2 211 32

28 2 28B AwL wL wLM R L wL L L −= − = − = .....................................................Ans.

(c) The bending moment ( )12 811

28 2 7wL x LwLx wxM x EIy

−′′= = − +

is a maximum where 0811 0

28 7wL x LwLdM dx wx

−= − + =

11 28 :x L= 2 2121 1568 0.0772M wL wL= ≅

43 28 :x L= 2 257 1568 0.0364M wL wL= ≅

:x L= 2 23 28 0.1071M wL wL= − ≅ −

2 :x L= 2 214 0.0714M wL wL= − ≅ −

2max 3 28M wL= − .................................................................................................................. Ans.

9-83 A beam is loaded and supported as shown in Fig. P9-83. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the center support. SOLUTION

(a) 2

1

2A A BwxEIy V x M R x L′′ = + − + −

22 3

12 6 2BA

A

R x LV x wxEIy M x C−

′ = + − + +

33 2 4

1 26 2 24 6BA A R x LV x M x wxEIy C x C

−= + − + + +


At 0,x = 0 :y′ = 1 0C =

At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y′ = 2 3

02 6A

AV L wLM L+ − =


421

At ,x L= 0 :y = 3 2 4

06 2 24A AV L M L wL+ − =

Solving simultaneously yields:

2 2AV wL wL= + = ↑ .................... 2 212 12 AM wL wL= − = � ........................Ans.

From symmetry:

2 2CV wL wL= − = ↑ .................... Ans.

2 212 12 CM wL wL= − = � ......... Ans.


0 :yF↑Σ = 2 0A B CV R V wL+ − − =

BR wL wL= = ↑ ..............................................................................................................Ans.

(b) ( )( )2 2 2

22 12 2 12B A A

wL wL wL wLM V L M wL L L −= + − = − − = .....................................Ans.

9-84 A beam is loaded and supported as shown in Fig. P9-84. Determine (a) The reactions at supports A and D. (b) The deflection at B if E = 200 GPa and I = 350(106) mm4. SOLUTION

(a) 1 236 3 3 6 kN mAEIy R x x x ′′ = − − − − ⋅

2

2 3 2118 3 6 kN m

2AR xEIy x x C

′ = − − − − + ⋅

43

3 31 2

66 3 kN m

6 4A xR xEIy x C x C

−= − − − + + ⋅


At 0 m,x = 0 :y = 2 0C =

At 12 m,x = 0 :y′ = 172 1674 0AR C− + =

At 12 m,x = 0 :y = 1288 4698 12 0AR C− + =

Solving simultaneously yields: 21 249.75 kN mC = − ⋅ and

26.72 kN 26.7 kNAR = + ≅ + ↑ ...................................................................................Ans.


0 :yF↑Σ = ( )36 6 6 0A DR V− − − =

45.3 kN 45.3 kNDV = − = ↑ .........................................................................................Ans.

0 :DMΣ =� ( ) ( ) ( ) ( )6 6 3 36 9 12 0D AM R+ + − =

111.4 kN m 111.4 kN m DM = − ⋅ = ⋅ � ....................................................................Ans.

(b) ( )3 431 53.44 72 3 3 6 2997 kN m12

EIy x x x x= − − − − − ⋅


422

( )( )3

33 9 6

629 629 10 8.99 10 m 8.99 mm200 10 350 10B xy

EIδ −

= −

− − ×= = = = − × = ↓× ×

.........Ans.

9-85 A beam is loaded and supported as shown in Fig. P9-85. Determine (a) The reactions at supports A, B, and D. (b) The deflection at C if E = 30,000 ksi and I = 26 in.4 SOLUTION

(a) 12600 4A BEIy R x x R x′′ = − + −

2 1

1

600 4 3000 6

10 lb ftD

x x

R x

+ − − −

+ − ⋅

22

33 4200 200 4

2 2BA R xR xEIy x x

−′ = − + + −

2

2 21

101500 6 lb ft

2DR x

x C−

− − + + ⋅

33

44 450 50 4

6 6BA R xR xEIy x x

−= − + + −

3

3 31 2

10500 6 lb ft

6DR x

x C x C−

− − + + + ⋅


At 0 ft,x = 0 :y = 2 0C =

At 4 ft,x = 0 :y = ( ) 132 3 12,800 4 0AR C− + =

At 10 ft,x = 0 :y = ( ) 1500 3 467,200 36 10 0A BR R C− + + =


0 :yF↑Σ = ( )1200 4 3000 2000 0A B DR R R+ + − − − =

0 :DMΣ =� ( ) ( ) ( ) ( ) ( ) ( )10 1200 4 8 6 3000 4 2000 2 0A BR R− + − + − =

Solving simultaneously yields: 21 2027 lb ftC = − ⋅ and

1960 lbAR = ↑ ................ 4467 lbBR = ↑ ............... 3373 lbDR = ↑ ..........................Ans.

(b) 3 43 41 1960 300 4467 4 300 4

6EIy x x x x= − + − + −

3 3 33000 6 3373 10 12,162 lb ftx x x− − + − − ⋅

( )

( )( )

3

6 6

354 12354 0.000784 in. 0.000784 in.30 10 26C xy

EIδ == = = = = ↑

×........................Ans.


431

9-92 A beam is loaded and supported as shown in Fig. P9-92. Determine the magnitude of the load P required to make the slope at the left end of the beam zero.

SOLUTION Using the solutions for cases 1 and 2 in Table A-19:

( ) ( )2 3

02 6A P w

P L w LEI EI

θ θ θ= + = − =

3 3

wL wLP = = ↑ ...................................................................................................................... Ans.

9-93 A beam is loaded and supported as shown in Fig. P9-93. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 4 in Table A-19:

AA R M M Lδ δ δ θ= + +

( ) ( )

3 220

3 2AR L ML ML L

EI EI EI= − − =

9 916 16A

M MRL L

+= = ↑ .........................Ans.


0 :yF↑Σ = 0A BR V− = 9 9

16 16BM MVL L

+= = ↓ .........................Ans.

0 :BMΣ =� ( )2 0B AM M R L+ − = 8 8BM MM = = � ............................Ans.

9-94 A beam is loaded and supported as shown in Fig. P9-94. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 2 in Table A-19:

4 3

08 3B

BB w R

wL R LEI EI

δ δ δ= + = + =

3 8 3 8BR wL wL= + = ↑ ...............................................................................................Ans.


0 :yF↑Σ = 0A BV R wL+ − = 5 58 8AwL wLV += = ↑ .......................Ans.

0 :AMΣ =� ( ) ( )2 0B AR L M wL L− − = 2 2

8 8AwL wLM −= = � ....................Ans.

9-95 A beam is loaded and supported as shown in Fig. P9-95. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with 2BM PL= −

and BV P= :

B B BB BV BM BRδ δ δ δ= + +


432

( ) ( )23 32

03 2 3

BPL LPL R LEI EI EI

= − − + = 7 74 4BP PR += = ↑ ............................Ans.


0 :yF↑Σ = 0A BV R P+ − = 3 4 3 4AV P P= − = ↓ ......................Ans.

0 :AMΣ =� ( ) ( )3 2 0B AR L M P L− − = 4 4 AM PL PL= + = � .................Ans.

9-96 A beam is loaded and supported as shown in Fig. P9-96. Determine the reactions at supports A, B, and C. SOLUTION Using the solutions for cases 1, 2, 7, and 8 in Table A-19 with 2 2BwM wL=

and CBR CM R L= :

CC Cw B CRLδ δ θ δ= + +

( ) ( )( ) ( ) ( ) ( )

3 2 34 2 22 20

8 24 3 3 3C C

wL Lw L R L L R LwL LEI EI EI EI EI

= − + − + + =

8 8CwL wLR += = ↑ .................................................................................................................Ans.


0 :AMΣ =� ( ) ( ) ( ) ( )2 3 3 3 2 0B CR L R L wL L+ − =

33 16 33 16BR wL wL= + = ↑ ......................................................................................Ans.

0 :yF↑Σ = 3 0A B CR R R wL+ + − =

13 16 13 16AR wL wL= + = ↑ ......................................................................................Ans.

9-97 A beam is loaded and supported as shown in Fig. P9-97. When the load P is applied, the slope at the right end of the beam is zero. Determine

(a) The magnitude of the load P. (b) The reactions at supports A and B. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with 4BM PL=

and BV P= :

(a) BC CP Cw CRθ θ θ θ= + +

( ) ( )2 3 25 4

02 6 2

BP L w L R LEI EI EI

= − − + =

75 48 16 0BP R wL− + − =

B B BB Bw BV BM BRδ δ δ δ δ= + + +


433

( ) ( )( )3 24 34

08 3 2 3

BP L PL LwL R LEI EI EI EI

= − − − + =

11 8 3 0BP R wL− + − =

Solving simultaneously gives:

49 4972 72B

wL wLR += = ↑ ........................2 29 9wL wLP += = ↓ .......................................Ans.

(b) Then, overall equilibrium gives:

0 :yF↑Σ = 0A BV wL R P− + − =

13 1324 24A

wL wLV += = ↑ .................................................................................................Ans.

0 :AMΣ =� ( ) ( ) ( ) ( )5 4 2 0B AR L P L wL L M− − − =

2 27 7

72 72AwL wLM −= = � ..............................................................................................Ans.

9-98 A beam is loaded and supported as shown in Fig. P9-98. Determine (a) The reactions at supports A and C. (b) The deflection at the right end of the distributed load. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19:

(a) ( )CC Bw Bw RLδ δ θ δ= + +

( ) ( ) ( )3 34 2

08 6 3

Cw L R LwL LEI EI EI

= − − + =

7 764 64CwL wLR += = ↑ .....................................................................................................Ans.


0 :yF↑Σ = 0C AR wL V− + =

57 5764 64A

wL wLV += = ↑ .................................................................................................Ans.

0 :AMΣ =� ( ) ( )( )2 2 0C AR L wL L M− − =

2 29 9

32 32AwL wLM −= = � ..............................................................................................Ans.

(b) B BV BM Bwδ δ δ δ= + +

( )( ) ( ) ( )23 2 47 647 64

3 2 8wL LwL L wL

EI EI EI= + −

4 413 13

384 384wL wLEI EI

−= = ↓ ...................................................................................................Ans.


434

9-99 A beam is loaded and supported as shown in Fig. P9-99. Determine the magnitude of the moment M required to make

(a) The slope at the right end of the beam zero. (b) The deflection at the right end of the beam zero. SOLUTION Using the solutions for cases 4 and 8 in Table A-19 with 3AM PL=

and BM M= :

(a) A BC B CM BM BM CMθ θ θ θ θ θ = + = + +

( )( ) ( )3 3

06 3

PL L M LMLEI EI EI

= − + + =

12 12PL PLM += = � ........................................................................................................Ans.

(b) ( ) ( )3 3A BC B CM BM BM CML Lδ θ δ θ θ δ = + = + +

( ) ( ) ( ) ( )23 3

3 06 3 2

PL L M LML LEI EI EI

= − + + =

9 9PL PLM += = � ........................................................................................................Ans.

9-100 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P9-100. SOLUTION Using the solutions for cases 1 and 4 in Table A-19:

CC BM BM CRLδ δ θ δ= + +

( ) ( ) ( )2 32

02 3

CM L R LML LEI EI EI

= − − + =

9 9

16 16CM MRL L

+= = ↑ ..................................................................................................................... Ans.


0 :yF↑Σ = 0A CV R+ = 9 9

16 16AM MVL L

−= = ↓ ................................Ans.

0 :AMΣ =� ( )2 0C AR L M M− − = 8 8AM MM += = � .................................Ans.

9-101 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P9-101. SOLUTION Using the solutions for cases 1, 3, and 4 in Table A-19:


435

A AM AV Awδ δ δ δ= + +

( ) ( )2 3 4

02 3 30A AM L V L wLEI EI EI

= + − =

215 10A AM V L wL+ =

A AM AV Awθ θ θ θ= + +

( ) ( )2 3

02 24AA V L w LM L

EI EI EI= − − + =

224 12A AM V L wL+ =

Solving simultaneously gives:

3 320 20AwL wLV += = ↑ ..............................

2 2

30 30AwL wLM −= = � ...................................Ans.


0 :yF↑Σ = ( )2 0A BV wL V− − =

7 720 20BwL wLV −= = ↑ .....................................................................................................Ans.

0 :BMΣ =� ( ) 02 3B A A

wL LM M V L + − − =

2 2

20 20BwL wLM −= = � ...................................................................................................Ans.

9-102 Two beams are loaded and supported as shown in Fig. P9-102. Determine the reactions at supports A, B, and D. Both beams have the same flexural rigidity.

SOLUTION Using the solution for case 1 in Table A-19:

B A B Dδ δ=

( )2B BBR CP CP BRLδ δ θ δ= + +

( ) ( ) ( )

3 23 32 22

3 3 2 3B BP L P LR L R LLEI EI EI EI

− = − − +

5 32BR P=

For beam AB : 5 32BR P= ↓ ..................................................Ans.

0 :yF↑Σ = 0A BV R− = 5 32 5 32AV P P= + = ↑ ................................Ans.

0 :AMΣ =� 0A BM R L− − = 5 32 5 32 AM PL PL= − = � ......................Ans.

For beam CD : 5 32BR P= ↑ ..................................................Ans.

0 :yF↑Σ = 0B DR P V− − = 27 32 27 32DV P P= − = ↑ ..........................Ans.

0 :DMΣ =� ( )2 0D BM P L R L+ − = 11 32 11 32 DM PL PL= − = � ..................Ans.


436

9-103 A steel (E = 29,000 ksi and I = 120 in.4) beam is loaded and supported as shown in Fig. P9-103. The post BD is a 6 × 6-in. timber (E = 1500 ksi) that is braced to prevent buckling. Determine the load carried by the post if it is unstressed before the 530-lb/ft distributed load is applied.

SOLUTION

( ) 26 6 36 inPA = × =

The amount that the center of the beam settles is equal to the amount that the post shrinks:

Bbeam Bw BR postδ δ δ δ= + =

Using the solutions for cases 6 and 7 in Table A-19:

4 35

384 48B B P

P P

wL R L R LEI EI A E

− + = −

( )

( )( )( )

( )( )( ) ( )

( ) ( )

3 4

6 6 6

20 12 20 12 5 530 12 20 1248 29 10 120 36 1.5 10 384 29 10 120

B BR R× × ×+ =

× × ×

6287 lb 6.29 kipBR = ≅ ......................................................................................................Ans.

9-104 A uniformly distributed load of 7 kN/m is supported by two 100 × 100-mm timber (E = 8.5 GPa) beams arranged as shown in Fig. P9-104. Beam AB is fixed at the wall and beam CD is simply supported. Before the load is applied, the beams are in contact at B, but the reaction at B is zero. After the 7-kN/m distributed load is applied, determine

(a) The maximum flexural stress in each beam. (b) The maximum longitudinal shearing stress in each beam. SOLUTION Using the solutions for cases 1, 2, and 6 in Table A-19:

B A B Dδ δ=

B BBw BR BRδ δ δ+ =

( )34 3 2

8 3 48BB R LwL R L

EI EI EI− + = −

4BR wL=

For beam AB : 4BR wL= ↑

0 :yF↑Σ = 0A BV wL R− + =

0 :AMΣ =� ( ) ( )2 0B AR L M wL L− − =

3 4 3 4AV wL wL= + = ↑

2 24 4 AM wL wL= − = �

From the shear force and bending moment diagrams:

max 7.875 kNV =

max 3.938 kN mM = − ⋅


437

( )3

6 4100 1008.333 10 mm

12I = = ×

( ) 3 325 100 50 125.00 10 mmCQ y A= = × = ×

(a) ( )( )3

max 6

3.938 10 0.0508.333 10

McI

σ −

×= =

×

6 223.6 10 N/m 23.6 MPa (T bottom & C top)= × = ...........................................Ans.

(b) ( )( )

( ) ( )

3 6

max 6

7.875 10 125.00 10

8.333 10 0.100VQIt

τ−

−

× ×= =

×

6 21.181 10 N/m 1.181 MPa (at neutral surface)= × = .........................................Ans.

For beam CD : 4BR wL= ↓

From symmetry: 8C DR R wL= = ↑


max 1.3125 kNV =

max 1.9688 kN mM = ⋅

(a) ( )( )3

max 6

1.9688 10 0.0508.333 10

McI

σ −

×= =

×

6 211.81 10 N/m 11.81 MPa (T bottom & C top)= × = ........................................Ans.

(b) ( )( )

( )( )

3 6

max 6

1.3125 10 125.00 10

8.333 10 0.100VQIt

τ−

−

× ×= =

×

6 20.1969 10 N/m 0.1969 MPa (at neutral surface)= × = ...................................Ans.

9-105 The 4-in. wide × 6-in. deep timber (E = 1200 ksi) beam shown in Fig. P9-105 is fixed at the left end and supported at the right end with a tie rod that has a cross-sectional area of 0.125 in.2 Determine the tension in the tie rod if it is unstressed before the load is applied to the timber beam and

(a) The tie rod is made of steel (E = 30,000 ksi). (b) The tie rod is made of aluminum alloy (E = 10,000 ksi). SOLUTION

( )3

44 672 in

12NAI = =

The amount that the end of the beam settles is equal to the amount that the rod shrinks:

Bbeam Bw BR rodδ δ δ δ= + =


4 3

8 3B B R

R R

wL R L R LEI EI A E

− =


438

(a) For the steel ( )30,000 ksiE = rod:

( )

( )( )( )

( )( )( )( )

( )( )

3 4

6 6 6

10 12 30 12 120 12 10 123 1.2 10 72 0.125 30 10 8 1.2 10 72

B BR R× × ×+ =

× × ×

444 lbBR = ...............................................................................................................Ans.

(b) For the aluminum alloy ( )10,000 ksiE = rod:

( )

( ) ( )( )

( )( )( ) ( )

( ) ( )

3 4

6 6 6

10 12 30 12 120 12 10 123 1.2 10 72 0.125 10 10 8 1.2 10 72

B BR R× × ×+ =

× × ×

431 lbBR = ...............................................................................................................Ans.

9-106 In Fig. P9-106, the aluminum alloy tie rod passes through a hole in the aluminum alloy cantilever beam and through the coil spring positioned on the end of the tie rod. Before loading, there is a clearance of 2.5 mm between the bottom of the beam and the top of the spring. The cross-sectional area of the tie rod is 100 mm2, the second moment of area of the cross section of the beam with respect to the neutral axis is 40(106) mm4, the modulus of elasticity of the aluminum alloy is 70 GPa, and the spring modulus is 1000 kN/m.. Determine the axial stress in the tie rod when M = 9 kN-m and w = 90 kN/m.

SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19:

clearanceBw BM BR rod springδ δ δ δ δ+ + = + +

4 2 3 3

8 2 48b b b r

b b b b b b r r

wL ML RL RL R CE I E I E I A E k

− − + = − − −

( ) ( )

( ) ( )( )( )

( )( )( )

( )( )4 2 33 3

9 6 9 6 9 6

90 10 1.25 9 10 1.25 1.258 70 10 40 10 2 70 10 40 10 3 70 10 40 10

R− − −

× ×− − +

× × × × × ×

( )

( )( )3

39 6

1.252.5 10

1000 1070 10 100 10R R −

−= − − − ×

×× ×

36.959 10 NR = ×

3

26

6.959 10 69.6 N/m 69.6 MPa (T)100 10r

RA

σ −

×= = = ≅×

....................................................Ans.

9-107 Two steel (E = 29,000 ksi) beams support a 1600-lb concentrated load, as shown in Fig. P9-107. In the unloaded condition, beam AB touches but exerts no force on beam CD. Beam AB is an S4 × 9.5 American standard section, and beam CD is an S5 × 14.75 American standard section (see Appendix A). Determine

(a) The maximum flexural stress in each beam. (b) The maximum transverse shearing stress in each beam. SOLUTION

For an S4 9.5× section: 46.79 inI = 2.796 in.fw = 0.293 in.ft =

( )Beam AB 33.39 inS = 4.00 in.d = 0.326 in.wt =

For an S5 14.75× section: 415.2 inI = 3.284 in.fw = 0.326 in.ft =

( )Beam CD 36.09 inS = 5.00 in.d = 0.494 in.wt =


439


E A E Dδ δ=

EP EM ER ERδ δ δ δ+ + =

( ) ( )323 32 2

3 2 3 48AB AB AB CD

PL L R LPL RLEI EI EI EI

− − + = −

( ) ( )

( ) ( )7 15.2 16007 2289 lb

2 4 2 6.79 4 15.2CD

AB CD

I PRI I

= = =+ +

For beam AB : 2289 lbR = ↑

0 :yF↑Σ = 1600 0BV R− + − =

689 lb 689 lbBV = + = ↓

0 :BMΣ =� ( ) ( )3 1600 4.5 0BM R− + =

333 lb ft 333 lb ft BM = − ⋅ = ⋅ �


max 1600 lbV = max 2400 lb ftM = ⋅

( )1.8535 2.796 0.293NA CQ y A= = ×

( ) 30.8535 0.326 1.707 1.9936 in+ × =

(a) 2max

2400 12 8496 lb/in 8.50 ksi (T top & C bottom)3.39

MS

σ ×= = = ≅ .....................Ans.

(b) ( )( )

2max

1600 1.99361441 lb/in 1441 psi (at neutral surface)

6.79 0.326VQIt

τ = = = = ...............Ans.

For beam AB : 2289 lbR = ↓

From symmetry 1122.5 lbC DR R= = ↑


max 1144.5 lbV = max 3434 lb ftM = ⋅

( )2.337 3.284 0.326NA CQ y A= = ×

( ) 31.087 0.494 2.174 3.669 in+ × =

(a) 2max

3434 12 6767 lb/in 6.77 ksi (T bottom & C top)6.09

MS

σ ×= = = ≅ .....................Ans.

(b) ( )

( )2

max

1144.5 3.669559 lb/in 559 psi (at neutral surface)

15.2 0.494VQIt

τ = = = = ................Ans.


440

9-108 A beam is loaded and supported as shown in Fig. P9-108. (a) Determine the reactions at supports A and C. (b) Draw complete shear force and bending moment diagrams for the beam. SOLUTION (a) Using the solutions for cases 6 and 8

in Table A-19 with 2CM PL=

and CV P= :

( ) ( )2

4

4 20

16 3 6A C

AA A P AM AM

P L PL LM LEI EI EI

θ θ θ θ= + + = − − + =

2 2APL PLM −= = � ......................................................................................................Ans.


0 :AMΣ =� ( ) ( )( )4 2CR L P L−

( )3 2 0AP L M− − =

3 3CR P P= + = ↑ ............................. Ans.

0 :yF↑Σ = 4 0A CV P R P− + − =

2 2AV P P= + = ↑ .............................. Ans.

9-109 A beam is loaded and supported as shown in Fig. P9-109. (a) Determine the reactions at supports A, B, and C. (b) Draw complete shear force and bending moment diagrams for the beam. SOLUTION (a) Using the solutions for cases 6, 7, and 8 in Table A-19:

( ) ( ) ( )3 22 2 20

24 3 16

A BA Aw AM AR

A Bw L M L R LEI EI EI

θ θ θ θ= + +

= − − + =

gives 28 3 4 0A BM R wL− + − =

and

( )( ) ( ) ( )4 2 35 2 2 20

384 16 48

A BB Bw BM BR

A Bw L M L R LEI EI EI

δ δ δ δ= + +

= − − + =

gives 26 4 5 0A BM R wL− + − =

Solving simultaneously gives

8 87 7BwL wLR += = ↑ ............................. Ans.

2 2

14 14AwL wLM −= = � ........................... Ans.


441


0 :AMΣ =� ( ) ( ) ( ) ( )2 2 0C B AR L R L w L L M+ − − =

11 1128 28C

wL wLR += = ↑ ..........................Ans.

0 :yF↑Σ = ( )2 0A B CV w L R R− + + = 13 1328 28A

wL wLV += = ↑ ..........................Ans.

9-110 A 3-m simply supported beam is loaded with a uniformly distributed load of 2.6 kN/m over its entire length. The beam is made of air-dried Douglas fir (E = 13 GPa) with an allowable flexural stress of 8 MPa and an allowable shearing stress of 0.7 MPa. The maximum deflection at the center of the span must not exceed 10 mm. Select the lightest standard structural timber that can be used for the beam.

SOLUTION

0 :AMΣ =� ( )2 0BL wL L− =

0 :BMΣ =� ( )2 0wL L AL− =

( )2 2 kNA B wL wL= = + = ↑


max 3.9 kNV =

max 2.925 kN mM = ⋅

The properties required by the three specifications are:

maxmax :M

Sσ = ( ) ( )

36 3 3 3max

6max

2.925 10 365.6 10 m 365.6 10 mm8 10

MSσ

−×= = = × = ××

maxmax

1.5 :VA

τ = ( ) ( ) ( )

36 2 2max

6max

1.5 3.9 101.5 8357 10 m 8357 mm0.7 10

VAτ

−×

= = = × =×

4

max5 :

384wL

EIδ =

( )( )( ) ( )

434

9max

5 2.6 10 35384 384 13 10 0.010

wLIEδ

×= =

×

( ) ( )6 4 6 421.09 10 m 21.09 10 mm−= × = ×

Try a 51 254 mm× − timber with 6.38 kg/mm = 29880 mmA =

6 448.3 10 mmI = × 3 3400 10 mmS = × With the weight of the beam included:

( ) ( )3 32.6 10 6.38 9.81 2.663 10 N/m 2.663 kN/mw = × + = × =

( )

max

2.663 33.995 kN

2 2wLV = = =

( )22

max

2.663 32.996 kN m

8 8wLM = = = ⋅

Then, the maximum normal stress, shear stress, and deflection are:


442

( )3

6 2maxmax 6

2.996 10 7.490 10 N/m 8 MPa400 10 all

MS

σ σ−

×= = = × < =×

( ) ( )

36 2max

max 6

1.5 3.995 101.5 0.607 10 N/m 0.7 MPa9880 10 all

VA

τ τ−

×= = = × < =

×

( ) ( )

( )( ) ( )434

3max 9 6

5 2.663 10 35 4.47 10 m 10 mm384 384 13 10 48.3 10 all

wLEI

δ δ−−

×= = = × < =

× ×

Therefore: Use a 51 254 mm timber section× − .......................................................................Ans.

9-111 A portion of a pedestrian walkway along the side of a bridge is shown in Fig. P9-111. Cantilever beams support the loading, one of which is shown. The beams are select structural eastern Hemlock (E = 1200 ksi) with allowable flexural and shearing stresses of 1300 psi and 80 psi, respectively. The allowable deflection is 0.2 in. Select the lightest standard structural timber that can be used for the beams.

SOLUTION

0 :yF↑Σ = ( )600 4 500 0AV − − =

2900 lb 2900 lbAV = + = ↑

0 :AMΣ =� ( ) ( ) ( )600 4 2 500 4 0AM − − =

6800 lb ft 6800 lb ft AM = + ⋅ = ⋅ �


maxmax :M

Sσ = 3max

max

6800 12 62.77 in1300

MSσ

×= = =

maxmax

1.5 :VA

τ = ( ) 2max

max

1.5 29001.5 54.38 in80

VAτ

= = =

3 4

max :3 8PL wLEI EI

δ = + ( )( ) ( )( )

( ) ( )

3 43 44

6max

8 500 4 12 3 600 12 4 128 3 215.0 in24 24 1.2 10 0.2

PL wLIEδ

× + ×+= = =×

Try an 8 8 in.× − timber with 15.6 lb/ftw = 256.3 inA =

4264 inI = 370.3 inS = With the weight of the beam included:

600 15.6 615.6 lb/ftw = + =

( )max 500 615.5 4 2962 lbV P wL= + = + =

( ) ( )22

max

615.6 4500 4 6925 lb ft

2 2wLM PL= + = + = ⋅


( )max

max

6925 121182 psi 1300 psi

70.3 allM

Sσ σ= = = < =

( )max

max

1.5 29621.5 78.9 psi 80 psi56.3 all

VA

τ τ= = = < =


443

( )

( )( )( )( )

( )( )

3 43 4

max 6 6

500 4 12 615.6 12 4 120.1656 in. 0.2 in.

3 8 3 1.2 10 264 8 1.2 10 264 allPL wLEI EI

δ δ× ×

= + = + = < =× ×

Therefore: Use an 8 8 in. timber section× − ...............................................................................Ans.

9-112 A solid circular shaft made of ASTM A36 steel (E = 200 GPa) is supported by bearings spaced 1.5 m apart. The shaft is to support a 4-kN load perpendicular to the shaft; the load may be placed at any point between the bearings. The allowable flexural stress is 152 MPa, the allowable shearing stress is 100 MPa, and the allowable deflection is 5 mm. If shafts are available with diameters in increments of 5 mm, determine the smallest diameter shaft that can be used to support the load. Neglect the weight of the shaft.

SOLUTION

0 :AMΣ =� ( )1.5 4 0B x− = ( )2.667 kNB x= ↑

0 :yF↑Σ = 4 0A B+ − = ( )4 2.667 kNA x= − ↑

The maximum bending moment occurs at C.

( )4 2.667CM Ax x x= = −

The biggest bending moment at C occurs when

4 5.333 0CdM dx x= − =

0.750 mx =

max 1.500 kN mM = ⋅

( )max 4 kN when 0 m or when 1.5 mV x x= ≅ ≅

To satisfy the bending stress condition:

max max maxmax 4 3

44

M c M c MI c c

σπ π

= = =

( )( ) ( )

33 6 3max

6max

4 1.50 104 12.565 10 m152 10

Mcπσ π

−×

= = = ××

( )323.25 10 m 23.25 mmc −= × = 2 46.50 mm 50 mmd c= = ≅

Checking the maximum shear stress and the maximum deflection:

( ) ( )22

34 25 254 10,417 mm3 2 3 2C

r rQ y Aππ

π π

= = = =

( ) ( )

443 425

306.8 10 mm4 4rI

ππ= = = ×

( ) ( )( ) ( ) ( )

3 66 2max

max 9

4 10 10.417 102.716 10 N/m 100 MPa

306.8 10 0.050 allV Q

Itτ τ

−

−

× ×= = = × < =

×

( )( )

( )( ) ( )333

3max 9 9

4 10 1.54.584 10 m 5 mm

48 48 200 10 306.8 10 allPL

EIδ δ−

−

×= = = × < =

× ×

Therefore: Use a 50 mm diameter shaft− .................................................................................Ans.


444

9-113 A simply supported structural steel (E = 29,000 ksi) beam has a span of 24 ft and carries a uniformly distributed load of 1200 lb/ft. The beam has an allowable flexural stress of 24 ksi, an allowable shearing stress of 14 ksi, and an allowable deflection of 1/360 of the span. Select the lightest American standard (S-shape) beam that can be used to support the loading.

SOLUTION

0 :AMΣ =� ( ) ( ) ( )24 1200 24 12 0B − =

0 :BMΣ =� ( ) ( ) ( )1200 24 12 24 0A− =

314.400 10 lb 14.40 kipA B= = × =

max 14.40 kipV A= =

( ) ( ) ( ) 3max 12 12 1200 12 6 86.40 10 lb ft 86.40 kip ftxM M A== = − = × ⋅ = ⋅

( )max 24 12 360 0.800 in.δ = × =


maxmax :M

Sσ = 3max

max

86.4 12 43.2 in24

MSσ

×= = =

maxmax :

w

VA

τ = 2max

max

14.40 1.0286 in14w

VAτ

= = =

4

max5 :

384wL

EIδ =

( )( )( )( )

444

6max

5 1200 12 24 125 386.1 in384 384 29 10 0.800

wLIEδ

×= = =

×

Try an S15 42.9× section with 15.00 in.d = 0.411 in.wt =

4447 inI = 359.6 inS = With the weight of the beam included:

1200 42.9 1242.9 lb/ftw = + =

( ) ( )3max 2 1242.9 24 2 14.915 10 lb 14.915 kipV wL= = = × =

( ) ( )22 3max 8 1242.9 24 8 89.49 10 lb ft 89.49 kip ftM wL= = = × ⋅ = ⋅


( )max

max

89.49 1218.02 ksi 24 ksi

59.6 allM

Sσ σ= = = < =

maxmax

14.915 2.42 ksi 14 ksi15.00 0.411 all

w

VA

τ τ= = = < =×

( )( )

( )( )

44

max 6

5 1242.9 12 24 125 0.716 in. 0.800 in.384 384 29 10 447 all

wLEI

δ δ×

= = = < =×

Therefore: Use an S15 42.9 section× ..........................................................................................Ans.


445

9-114 The simply supported structural steel (E = 200 GPa) beam shown in Fig. P9-114 has an allowable flexural stress of 165 MPa, an allowable shearing stress of 100 MPa, and an allowable deflection of 1/360 of the span. Select the lightest wide-flange beam that can be used to support the loading shown in the figure.

SOLUTION

0 :AMΣ =� ( ) ( ) ( ) ( )9 70 1.5 70 4.5 70 7.5 0E − − − =

0 :EMΣ =� ( ) ( ) ( ) ( )70 1.5 70 4.5 70 7.5 9 0A+ + − =

105 kNA E= = ↑ From the shear force and bending moment diagrams:

max 105 kNV =

max 262.5 kN mM = ⋅


maxmax :M

Sσ =

3max

6max

262.5 10165 10

MSσ

×= =×

( )6 3 31591 10 m 1591 mm−= × =

maxmax :

w

VA

τ = 3

max6

max

105 10100 10w

VAτ

×= =×

( )6 2 21050 10 m 1050 mm−= × =

( )2 2 3

max

2 3 4 9 0.02500 m 25.00 mm48 48 360

Pb L b PLEI EI

δ−

= + = = =

( ) ( )( ) ( ) ( ) ( )( )

( ) ( )

2 23 32 2 3

9max

2 70 10 1.5 3 9 4 1.5 70 10 92 3 448 48 200 10 0.025

Pb L b PLI

Eδ

× − + ×− + = =×

( ) ( )6 4 6 4417.4 10 m 417.4 10 mm−= × = ×

Try a W 533 92× section with ( )92 9.81 902.5 N/mw = = 533 mmd =

6 4554 10 mmI = × 3 32080 10 mmS = × 10.2 mmwt =

With the weight of the beam included:

( ) ( )3 3 3

max

902.5 9105 10 105 10 109.06 10 N

2 2wLV = × + = × + = ×

( ) ( )

223 3 3

max

902.5 9262.5 10 262.5 10 271.6 10 N m

8 8wLM = × + = × + = × ⋅


( )3

6 2maxmax 6

271.6 10 130.58 10 N/m 165 MPa2080 10 all

MS

σ σ−

×= = = × < =×

( )3

6 2maxmax

109.06 10 20.06 10 N/m 100 MPa0.533 0.0102 all

w

VA

τ τ×= = = × < =×


446

( )2 2 3 4

max

2 3 4 548 48 384

Pb L b PL wLEI EI EI

δ−

= + +

( )( ) ( ) ( )

( ) ( )( ) ( )

( )( )2 2 33 3

9 6 9 6

2 70 10 1.5 3 9 4 1.5 70 10 9

48 200 10 554 10 48 200 10 554 10− −

× − × = +× × × ×

( ) ( )

( ) ( ) ( )4

39 6

5 902.2 919.53 10 m 25 mm

384 200 10 554 10 allδ−−

+ = × < =× ×

Therefore: Use a W 533 92 section× ...........................................................................................Ans.

9-115 The simply supported beam shown in Fig. P9-115 is made of air-dried Douglas fir (E = 1900 ksi) with an allowable flexural stress of 1900 psi, an allowable shearing stress of 85 psi, and an allowable deflection of 1/360 of the span. Select the lightest standard structural timber that can be used to support the loads shown in the figure.

SOLUTION

0 :AMΣ =� ( ) ( ) ( )24 4200 8 4200 16 0D − − =

0 :DMΣ =� ( ) ( ) ( )4200 8 4200 16 24 0A+ − =

4200 lbA D= = ↑

max 4200 lbV =

max 33,600 lb ft 33.6 kip ftM = ⋅ = ⋅


maxmax :M

Sσ =

( )33max

max

33.6 10 12212.2 in

1900MSσ

× ×= = =

maxmax

1.5 :VA

τ = ( ) 2max

max

1.5 42001.5 74.12 in85

VAτ

= = =

( )2 2

max

2 3 4 24 12 0.800 in.48 360

Pb L bEI

δ− ×= = =

( ) ( )( ) ( ) ( )

( )( )

2 22 24

3max

2 4200 8 12 3 24 12 4 8 122 3 42343 in

48 48 1900 10 0.800

Pb L bI

Eδ

× × − ×− = = =×

Try a 10 16 in.× − timber with 40.9 lb/ftw = 2147 inA =

42948 inI = 3380 inS = With the weight of the beam included:

( )

max

40.9 244200 4200 4691 lb

2 2wLV = + = + =

( ) ( )

223

max

40.9 2433,600 33,600 36.54 10 lb ft 36.54 kip ft

8 8wLM = + = + = × ⋅ = ⋅


447


( )( )3

maxmax

36.54 10 121154 psi 1900 psi

380 allM

Sσ σ

×= = = < =

( )max

max

1.5 46911.5 47.9 psi 85 psi147 all

VA

τ τ= = = < =

( )2 2 4

max

2 3 4 548 384

Pb L b wLEI EI

δ−

= +

( )( ) ( ) ( )

( )( )( )( )

( )( )

2 2 4

3 3

2 4200 8 12 3 24 12 4 8 12 5 40.9 12 24 1248 1900 10 2948 384 1900 10 2948

× × − × × = +× ×

0.690 in. 0.800 in. allδ= < =

Therefore: Use a 10 16 in. timber section× − .............................................................................Ans.

9-116 The boards for a concrete form are to be bent to a circular curve of 5-m radius. What maximum thickness can be used if the stress is not to exceed 15 MPa? The modulus of elasticity for the wood is 10 GPa.

SOLUTION

From Eq. 9-2: 1 M

EIρ= or

EIMρ

=

( ) ( )

max

2EI c E tMc EcI I

ρσ

ρ ρ= = = =

( )( ) ( )

63max

9

2 15 10 52 15.00 10 m 15.00 mm10 10

tE

σ ρ −×

= = = × =×

...................................Ans.

9-117 During fabrication of a laminated timber arch, one of the 10-in wide × 1-in. thick Douglas fir (E = 1900 ksi) planks is bent to a radius of 12 ft. Determine the maximum flexural stress developed in the plank.

SOLUTION

From Eq. 9-2: 1 M

EIρ= or

EIMρ

=

( ) ( )

max

1900 1 26.60 ksi (T & C)

12 12EI cMc Ec

I Iρ

σρ

= = = = =×

..................................Ans.

9-118 A beam is loaded and supported as shown in Fig. P9-118. Determine (a) The equation of the elastic curve, using the designated axes. (b) The slope at the right end of the beam. (c) The deflection at the right end of the beam. SOLUTION

0 :yF↑Σ = 2 0AV wL− =

0 :AMΣ =� ( )( )2 2 3 0AM wL L− − =

2AV wL= 2 3AM wL= −

( ) ( )( )2 2 3A AM x V x M wx L x= + −


448

2 3

2 3 6wLx wL wxEIy

L′′ = − −

2 2 4

14 3 24wLx wL x wxEIy C

L′ = − − + At 0,x = 0 :y′ = 1 0C =

3 2 2 5

1 212 6 120wLx wL x wxEIy C x C

L= − − + + At 0,x = 0 :y = 2 0C =

(a) ( )5 2 3 3 210 20120

wy x L x L xEIL

= − + − ................................................................................Ans.

(b) ( )4 2 2 36 824

wy x L x L xEIL

′ = − + −

( )3 3

4 4 46 824 8 8x L

w wL wLy L L LEIL EI EI=

−′ = − + − = = ........................................................Ans.

(c) ( )4 4

5 5 5 11 1110 20120 120 120x L

w wL wLy L L LEIL EI EI=

−= − + − = = ↓ ..........................................Ans.

9-119 Select the lightest structural steel (E = 29,000 ksi) wide flange or American standard beam (Appendix A) that can be used for the beam shown in Fig. P9-119 if the maximum flexural stress must not exceed 10 ksi and the maximum deflection must not exceed 0.200 in. when L = 8 ft and w = 2000 lb/ft.

SOLUTION

0 :CMΣ =� ( ) ( ) ( )2 2 0AwL L R L− =

4AR wL=

2

4 2w x LwLxEIy

−′′ = −

32

18 6w x LwLxEIy C

−′ = − + At 0,x = 0 :y = 2 0C =

43

1 224 24w x LwLxEIy C x C

−= − + + At 2 ,x L= 0 :y =

3

1748wLC −=

( )43 32 2 748

wy Lx x L L xEI

= − − −

The maximum deflection occurs when

( )32 36 8 7 048

wy Lx x L LEIL

′ = − − − = 1.0804x L=

4 4

max 1.08040.1050 0.1050

x LwL wLy

EI EIδ =

−= = = ↓

( ) ( )

( )( )

444

min 6max

0.1050 2000 12 8 120.1050 256.3 in29 10 0.200

wLIEδ

×= = =

×

( )( )42

3maxmin 3

max max

9 2000 12 8 12 329 32 43.2 in10 10

M wLSσ σ

×= = = =

×


449

The lightest beam satisfying both requirements is a W16 40× wide flange section. .............................Ans.

9-120 A cantilever beam is fabricated by bolting two structural steel (E = 200 GPa) C127 × 10 channel sections together, as shown in Fig. P9-120. Determine

(a) The deflection at the right end of the beam. (b) The maximum tensile flexural stress in the beam. SOLUTION For a C127 10× channel section:

21270 mmA = ( )6 40.199 10 mmYYI = × 44.5 mmfw =

12.3 mmx = 44.5 12.3 32.2 mmBC fc w x= − = − =

( ) ( )262 0.199 10 1270 12.3ABI = × +

( )6 40.782 10 mm= ×

(a) Using the solutions for cases 1 and 4 in Table A-19:

3 2

3 2AB AB

B BP BMAB AB

PL MLEI EI

δ δ δ= + = − −

( )

( )( )( ) ( )

( )( )23

9 6 9 6

500 1 1500 13 200 10 0.782 10 2 200 10 0.782 10− −

= − −× × × ×

( )32.664 10 m 2.664 mm−= − × = ↓

2

2AB AB

B BP BMAB AB

PL MLEI EI

θ θ θ= + = − −

( )

( )( )( ) ( )

( )( )2

9 6 9 6

500 1 1500 12 200 10 0.782 10 200 10 0.782 10− −

= − −× × × ×

( ) ( )3 34.795 10 rad 4.795 10 rad− −= − × = ×

3

/ 3BC

C B B BC C B B B BCBC

PLL LEI

δ δ θ δ δ θ= + + = + −

( )( ) ( )( ) ( )

33 3

9 6

500 12.664 10 4.795 10 1

3 200 10 0.199 10− −

−= − × − × −

× ×

( )311.65 10 m 11.65 mm−= − × = ↓ ...........................................................................Ans.

(b) ( )( ) ( )6 2

6

500 2 0.044556.9 10 N/m

0.782 10A A

AAB

M cI

σ −

− ×= − = − = + ×

×

( )( ) ( )6 2

6

500 1 0.032280.9 10 N/m

0.199 10B B

BBC

M cI

σ −

− ×= − = − = + ×

×

6 2max 80.9 10 N/m 80.9 MPa (T)Bσ σ= = + × = .............................................................Ans.


450

9-121 An aluminum beam (E = 10,000 ksi and I = 48 in.4) is loaded and supported as shown in Fig. P9-121. Determine

(a) The deflection at the left end of the beam. (b) The deflection at the middle of span BD. SOLUTION

0 :DMΣ =� ( ) ( )2200 13 10 15,000 0BR− − = 1360 lbBR = ↑

0 :BMΣ =� ( ) ( )2200 3 10 15,000 0DR− − = 840 lbDR = ↑

With the origin of coordinates at the right end of the beam,

1 0 1840 0 15,000 5 1360 10 lb ftEIy x x x ′′ = − − − + − ⋅

2 1 2 2

1420 0 15,000 5 680 10 lb ftEIy x x x C ′ = − − − + − + ⋅

3 2 3 3

1 2140 0 7500 5 226.67 10 lb ftEIy x x x C x C = − − − + − + + ⋅

Boundary conditions: At 0 ft,x = 0 :y = 2 0C =

At 10 ft,x = 0 :y = 21 4750 lb ftC = ⋅

3 2 3 3140 0 7500 5 226.67 10 4750 lb ftEIy x x x x = − − − + − + ⋅

(a) At A ( )13 ftx =

( )

( )( ) ( ) ( ) ( ) ( )3

3 2 3

6

12140 13 7500 8 226.67 3 4750 13

10 10 48Aδ = − + + ×

0.376 in. 0.376 in.= − = ↓ ............................................................................................Ans.

(b) At C ( )5 ftx =

( )

( )( ) ( ) ( )3

3

6

12140 5 0 0 4750 5

10 10 48Cδ = − + + ×

0.1485 in. 0.1485 in.= + = ↑ ........................................................................................Ans.

9-122 The steel (E = 200 GPa) beam AB of Fig. P9-122 is fixed at ends A and B and supported at the center by the pin connected timber (E = 10 GPa) struts CD and CE. The cross-sectional area of each strut is 6400 mm2, and the second moment of the cross-sectional area of the beam with respect to its neutral axis is 25(106) mm4. Determine the force in each strut after the 6-kN/m distributed load is applied to the beam.

SOLUTION As a result of symmetry, for a beam with fixed ends and a uniformly distributed load w:

2A BV V wL= = A BM M M= =

Using the solutions for cases 6, 7, and 8 in Table A-19:

3

024 3 6A BA w M MwL ML ML

EI EI EIθ θ θ θ= + + = − − − =

2

12wLM −=


451

4 2 2 4 4 4 45 5

384 16 16 384 192 192 384wwL ML ML wL wL wL wL

EI EI EI EI EI EI EIδ = − + + = − + + = −

As a result of symmetry, for a beam with fixed ends and a concentrated load P at the center:

2A BV V P= = A BM M M= =

( )32.664 10 m 2.664 mm−= − × = ↓

2

016 3 6A BA P M MPL ML ML

EI EI EIθ θ θ θ= + + = − − − =

8PLM −=

3 2 2 3 3 3 3

48 16 16 48 128 128 192PPL ML ML PL PL PL PL

EI EI EI EI EI EI EIδ = − + + = − + + = −

For the system:

( ) ( ) ( ) ( )3 5 3 5 6 5 6 5C strut strut strut sR P P P P= + = =

( )5 3Bbeam Cw CR strutδ δ δ δ= + =

34 5

384 192 3C s s

s s

R L P LwLEI EI A E

− + = −

3 45

192 3 384C s s

s s

R L P L wLEI A E EI

+ =

( ) ( )

( )( )( )

( )( )( )( )

( )( )43 3

9 6 3 9 9 6

6 10 46 5 4 2.553192 200 10 25 10 6.4 10 10 10 384 200 10 25 10

s sP P− − −

× + =

× × × × × ×

( )35.51 10 N 5.51 kNsP = × = ............................................................................................Ans.

9-123 In Fig. P9-123, beam AC is made of brass and beam BC is made of steel. The second moment of the cross-sectional area of beam BC with respect to its neutral axis is twice that of beam AC, and the modulus of elasticity of the steel is twice that of the brass. The reaction at C is zero before the load w is applied. Determine

(a) The reaction at C on beam BC. (b) The reactions at supports A and B. SOLUTION

Using the solutions for cases 1 and 2 in Table A-19 with 2BC ACI I= and 2BC ACE E= :

/ /C A C Bδ δ=

C CCR Cw CRδ δ δ= +

( )

( ) ( ) ( )( )

3 3423 8 2 2 3 2 2

C CR L R LwLEI E I E I

− = − +

4 4CwL wLR += = ↑ ..........................................Ans.


452

For beam AC :

0 :yF↑Σ = 0A CV R− =

0 :AMΣ =� ( )2 0A CM R L− − =

4 4AV wL wL= + = ↑ ...................................................................................................Ans.

2 28 8 AM wL wL= − = � .............................................................................................Ans.

For beam BC :

0 :yF↑Σ = 0C BR wL V− − =

0 :BMΣ =� ( ) ( )2 0B CM wL L R L+ − =

3 4 3 4BV wL wL= − = ↑ ...............................................................................................Ans.

2 24 4 BM wL wL= − = � ............................................................................................Ans.

9-124 A beam is loaded and supported as shown in Fig. P9-124. Determine (a) The reactions at supports A and D. (b) The deflection at B if E = 200 GPa and I = 350(106) mm4. SOLUTION

(a) 1 236 3 3 6 kN mAEIy R x x x ′′ = − − − − ⋅

2

2 3 2118 3 6 kN m

2AR xEIy x x C

′ = − − − − + ⋅

43

3 31 2

66 3 kN m

6 4A xR xEIy x C x C

−= − − − + + ⋅

Boundary conditions: At 0 m,x = 0 :y = 2 0C =

At 12 m,x = 0 :y′ = 172 1674 0AR C− + =

At 12 m,x = 0 :y = 1288 4698 12 0AR C− + =

Solving simultaneously gives 21 249.75 N mC = − ⋅ and

26.72 kN 26.7 kNAR = + ≅ ↑ .............................................................................................Ans.


0 :yF↑Σ = ( )36 6 6 0A DR V− − − =

0 :DMΣ =� ( ) ( ) ( ) ( )36 9 6 6 3 12 0A DR M+ − − =

45.28 kN 45.3 kNDV = − ≅ ↑ .............................................................................................Ans.

111.36 kN m 111.4 kN m DM = − ⋅ ≅ ⋅ � .........................................................................Ans.

(b) ( )3 43 31 53.44 72 3 3 6 2997 kN m12

EIy x x x x= − − − − − ⋅

( )( )3

33 9 6

629 629 10 8.99 10 m 8.99 mm200 10 350 10xy

EI−

= −

×= − = − = − × = ↓× ×

...............Ans.


453

9-125 A beam is loaded and supported as shown in Fig. P9-125. Determine (a) The reactions at supports A, B, and D. (b) The deflection at C if E = 30,000 ksi and I = 26 in.4 SOLUTION

(a) 1 2 1 12600 4 600 4 3000 6 10 lb ftA B DEIy R x x R x x x R x ′′ = − + − + − − − + − ⋅

2 22

3 23 21

4 10200 200 4 1500 6 lb ft

2 2 2B DA R x R xR xEIy x x x C

− −′ = − + + − − − + + ⋅

3 33

4 34 31 2

4 1050 50 4 500 6 lb ft

6 6 6B DA R x R xR xEIy x x x C x C

− −= − + + − − − + + + ⋅

Boundary conditions: At 0 ft,x = 0 :y = 2 0C =

At 4 ft,x = 0 :y = ( ) 132 3 12,800 4 0AR C− + =

At 10 ft,x = 0 :y = ( ) 1500 3 467,200 36 10 0A BR R C− + + =

From equilibrium:

0 :yF↑Σ = ( )1200 4 3000 2000 0A B DR R R+ + − − − =

0 :DMΣ =� ( ) ( ) ( ) ( ) ( ) ( )1200 4 8 3000 4 2000 2 10 6 0A BR R+ − − − =

Solving simultaneously gives 21 2027 lb ftC = − ⋅ and

1960 lbAR = ↑ .................... 4467 lbBR = ↑ ................... 3373 lbDR = ↑ ...................Ans.

(b) 3 43 41 1960 300 4467 4 300 4

6EIy x x x x= − + − + −

3 3 33000 6 3373 10 12,162 lb ftx x x− − + − − ⋅

( )

( ) ( )

3

6 6

354 12354 0.000784 in. 0.000784 in.30 10 26xy

EI= = = = + = ↑×

...............................Ans.


454

9-126 A beam is loaded and supported as shown in Fig. P9-126. Determine (a) The reactions at supports A, B, and C. (b) The deflection midway between supports B and C. SOLUTION

(a) 2

1

2A BwxEIy R x R x L′′ = − + −

22 3

12 6 2BA R x LR x wxEIy C

−′ = − + +

33 4

1 26 24 6BA R x LR x wxEIy C x C

−= − + + +

Boundary conditions: At 0,x = 0 :y = 2 0C =

At ,x L= 0 :y = 2 314 24 0AR L wL LC− + =

At 3 ,x L= 0 :y = 2 3 21108 81 32 72 0A BR L wL R L LC− + + =

From equilibrium:

0 :yF↑Σ = ( )3 0A B CR R R w L+ + − =

0 :DMΣ =� ( ) ( ) ( ) ( )3 3 2 3 2 0A Bw L L R L R L− − =

Solving simultaneously gives 31 48C wL= and

8 8AwL wLR += = ↑ .................................................................................................................Ans.

33 3316 16B

wL wLR += = ↑ .......................................................................................................Ans.

13 1316 16C

wL wLR += = ↑ ........................................................................................................Ans.

(b) ( )33 4 32 4 33 296

wy Lx x L x L L xEI

= − + − +

( )4 4 4 42 16 64 33 4

96x Lwy L L L LEI= = − + +

4 411 11

96 96wL wLEI EI

−= = ↓ .................................................................................................Ans.


455

Chapter 10 10-1 At a point in a thin plate, there are normal stresses of 20 ksi (T) on a vertical plane and 10 ksi (C) on a

horizontal plane, as shown in Fig. P10-1. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION

0 :nFΣ = ( )( )10 cos64 cos64dA dAσ + ° °

( )( )20 sin 64 sin 64 0dA− ° ° =

14.23 ksi 14.23 ksi (T)σ = + = .................................Ans.

0 :tFΣ = ( ) ( )10 cos64 sin 64dA dAτ − ° °

( )( )20 sin 64 cos64 0dA− ° ° =

11.82 ksiτ = + .................................................................................................................Ans.

10-2 At a point in a stressed body, there are normal stresses of 95 MPa (T) on a vertical plane and 125 MPa (T) on a horizontal plane, as shown in Fig. P10-2. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION

0 :nFΣ = ( )( )95 sin 20 sin 20dA dAσ − ° °

( )( )125 cos 20 cos 20 0dA− ° ° =

121.5 MPa 121.5 MPa (T)σ = + = ..........................Ans.

0 :tFΣ = ( ) ( )95 sin 20 cos 20dA dAτ + ° °

( )( )125 cos 20 sin 20 0dA− ° ° =

9.64 MPaτ = + ................................................................................................................Ans.

10-3 The stresses shown in Fig. P10-3 act at a point on the surface of a circular shaft which is subjected to a twisting moment M as shown. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION

0 :nFΣ = ( ) ( )15 cos60 sin 60dA dAσ − ° °

( ) ( )15 sin 60 cos 60 0dA− ° ° =

12.99 ksi 12.99 ksi (T)σ = + = .................................Ans.

0 :tFΣ = ( )( )15 cos 60 cos 60dA dAτ + ° °

( )( )15 sin 60 sin 60 0dA− ° ° =

7.50 ksiτ = + ................................................................................................................... Ans.

10-4 The stresses shown in Fig. P10-4 act at a point on the surface of a cantilever beam. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION


456

0 :nFΣ = ( )( )170 sin 55 sin 55dA dAσ + ° °

( ) ( )70 sin 55 cos55dA− ° °

( ) ( )70 cos55 sin 55 0dA− ° ° =

48.3 MPa 48.3 MPa (C)σ = − = ..............................Ans.

0 :tFΣ = ( ) ( )170 sin 55 cos55dA dAτ − ° °

( ) ( ) ( ) ( )70 sin 55 sin 55 70 cos55 cos55 0dA dA− ° ° + ° ° =

103.8 MPaτ = + .............................................................................................................Ans.

10-5 The stresses shown in Fig. P10-5 act at a point on the surface of a thin-walled pressure vessel that is subjected to an internal pressure, an axial load P, and a torque T. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION

0 :nFΣ = ( ) ( )18 sin 65 sin 65dA dAσ − ° °

( ) ( )6 cos 65 cos 65dA− ° °

( ) ( )15 sin 65 cos 65dA+ ° °

( )( )15 cos65 sin 65 0dA+ ° ° =

4.37 ksi 4.37 ksi (T)σ = + = .....................................Ans.

0 :tFΣ = ( )( ) ( )( )18 sin 65 cos65 6 cos 65 sin 65dA dA dAτ − ° ° + ° °

( ) ( ) ( ) ( )15 sin 65 sin 65 15 cos65 cos65 0dA dA− ° ° + ° ° =

14.24 ksiτ = + .................................................................................................................Ans.

10-6 The stresses shown in Fig. P10-6 act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION

0 :nFΣ = ( )( )65 sin 55 sin 55dA dAσ + ° °

( ) ( )125 cos55 cos55dA+ ° °

( ) ( )75 sin 55 cos55dA+ ° °

( ) ( )75 cos55 sin 55 0dA+ ° ° =

155.2 MPa 155.2 MPa (C)σ = − = ..........................Ans.

0 :tFΣ = ( ) ( ) ( )( )65 sin 55 cos55 125 cos55 sin 55dA dA dAτ + ° ° − ° °

( )( ) ( ) ( )75 sin 55 sin 55 75 cos55 cos55 0dA dA− ° ° + ° ° =

53.8 MPaτ = + ................................................................................................................Ans.

10-7 At a point in a machine component, the normal and shear stresses on an inclined plane are 4800 psi (T) and 1500 psi, respectively, as shown in Fig. P10-7. The normal stress on a vertical plane through the point is zero. Determine

(a) The shear stresses on horizontal and vertical planes. (b) The normal stress on a horizontal plane.


457

SOLUTION

sin 3 5θ = cos 4 5θ =

0 :xF→ Σ = ( ) ( )cos 4800 sinxy dA dAτ θ θ+

( )1500 cos 0dA θ+ =

(a) 5100 psi 5.10 ksixyτ = − = − .......................................Ans.

0 :yF↑ Σ = ( ) ( )cos 4800 cosy dA dAσ θ θ−

( ) ( )1500 sin sin 0xydA dAθ τ θ+ − =

(b) 150.0 psi 150.0 psi (C)yσ = − = .................................................................................Ans.

10-8 The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular bar subjected to an axial load P and a torsional load T are shown in Fig. P10-8. The normal stress on the inclined plane AB is 15 MPa (T). Determine

(a) The normal stress σx on the vertical plane. (b) The magnitude and direction of the shear stress on the inclined plane AB. SOLUTION

sin 5 13θ = cos 12 13θ =

0 :nFΣ = ( )( )15 sin sinxdA dAσ θ θ−

( ) ( )25 sin cosdA θ θ+

( ) ( )25 cos sin 0dA θ θ+ =

(a) 221.4 MPa 221.4 MPa (T)xσ = + = ........................Ans.

0 :tFΣ = ( ) ( ) ( ) ( ) ( )( )sin cos 25 sin sin 25 cos cos 0xdA dA dA dAτ σ θ θ θ θ θ θ− − + =

(b) 61.0 MPa (as shown on sketch)τ = + .......................................................................Ans.

10-9 The thin-walled cylindrical pressure vessel shown in Fig. P10-9 was constructed by wrapping a thin steel plate into a helix that forms an angle θ = 35° with respect to a transverse plane through the cylinder and butt-welding the resulting seam. In a thin-walled cylindrical pressure vessel, the normal stress σy on a horizontal plane through a point on the surface of the vessel is twice as large as the normal stress σx on a vertical plane through the point, and the shear stresses on both the horizontal and vertical planes are zero. If the stresses in the weld material on the plane of the weld must be limited to 10 ksi in tension and 7 ksi in shear, determine the maximum normal stress σx permitted in the vessel.

SOLUTION For 10 ksi (T):σ =

0 :nFΣ = ( ) ( )10 cos35 cos35xdA dAσ− ° °

( ) ( )2 sin 35 sin 35 0xdAσ− ° ° =

7.52 ksixσ = +

For 7 ksi:τ = 0 :tFΣ = ( )( )7 cos35 sin 35xdA dAσ+ ° °

( )( )2 sin 35 cos35 0xdAσ− ° ° =

14.90 ksixσ = +

Therefore: ( )max 7.52 ksixσ = ........................................................................................Ans.


458

10-10 The thin-walled cylindrical pressure vessel shown in Fig. P10-9 was constructed by wrapping a thin steel plate into a helix that forms an angle θ with respect to a transverse plane through the cylinder and butt-welding the resulting seam. The normal stresses σx and σy on vertical and horizontal planes through a point on the surface of the vessel are 50 MPa and 100 MPa, respectively, and the shear stresses are zero. Prepare a plot showing the variation of normal and shear stresses on the plane of the weld as the angle θ is varied from 0° to 180°.

SOLUTION

0 :nFΣ = ( )( )50 cos cosndA dAσ θ θ−

( ) ( )100 sin sin 0dA θ θ− =

( )2 250cos 100sin MPanσ θ θ= + ....................................................................................Ans.

0 :tFΣ = ( )( )50 cos sinntdA dAτ θ θ+

( )( )100 sin cos 0dA θ θ− =

( )50sin cos MPantτ θ θ= .................... Ans.

10-11 In a solid circular shaft subjected to a torsional form of loading, the normal stresses σx and σy on vertical and horizontal planes through a point on the surface of the vessel are zero and the shear stresses are 15 ksi, as shown in Fig. P10-11. Prepare a plot showing the variation of normal and shear stresses on plane AA through the point as the angle θ is varied from 0° to 180°.

SOLUTION

0 :nFΣ = ( ) ( ) ( ) ( )15 cos sin 15 sin cos 0ndA dA dAσ θ θ θ θ− − =

( )30sin cos ksinσ θ θ= .................................................................. Ans.

0 :tFΣ = ( )( ) ( )( )15 cos cos 15 sin sin 0ntdA dA dAτ θ θ θ θ− + =

( )2 215 cos sin ksintτ θ θ= − ...............................................................................................Ans.


459

10-12 Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-12.

SOLUTION The given values are:

20 MPaxσ = + 80 MPayσ = + 0 MPaxyτ = ( )1tan 3 4 36.870θ −= = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

20cos 36.870 80sin 36.870 2 0 sin 36.870 cos 36.870n x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

41.6 MPa 41.6 MPa (T)= + = ......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

20 80 sin 36.870 cos 36.870 0 cos 36.870 sin 36.870

nt x y xyτ σ σ θ θ τ θ θ= − − + −

= − − ° ° + ° − °

28.8 MPa= + ..................................................................................................................... Ans.



25 ksixσ = + 7 ksiyσ = + 12 ksixyτ = + 90 60 30θ = ° − ° = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

25cos 30 7sin 30 2 12 sin 30 cos 30n x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

30.9 ksi 30.9 ksi (T)= + = .............................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

25 7 sin 30 cos 30 12 cos 30 sin 30


= − − ° ° + ° − °

1.794 ksi= − ...................................................................................................................... Ans.



65 MPaxσ = − 125 MPayσ = − 75 MPaxyτ = + 55 90 35θ = ° − ° = − °

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

65 cos 35 125 sin 35 2 75 sin 35 cos 35n x y xyσ σ θ σ θ τ θ θ= + +

= − − ° + − − ° + − ° − °

155.2 MPa 155.2 MPa (C)= − = ..................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

65 125 sin 35 cos 35 75 cos 35 sin 35


= − − − − − ° − ° + − ° − − °

53.8 MPa= + ..................................................................................................................... Ans.


460



25 ksixσ = + 12 ksiyσ = + 10 ksixyτ = − 90 52 38θ = ° − ° = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= ° + ° + − ° °

10.37 ksi 10.37 ksi (T)= + = .........................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

25 12 sin 38 cos 38 10 cos 38 sin 38


= − − ° ° + − ° − °

8.73 ksi= − ......................................................................................................................... Ans.



20 MPaxσ = + 120 MPayσ = + 80 MPaxyτ = − 64θ = − °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − ° + − ° + − − ° − °

163.8 MPa 163.8 MPa (T)= + = ..................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

20 120 sin 64 cos 64 80 cos 64 sin 64


= − − − ° − ° + − − ° − − °

9.85 MPa= + ..................................................................................................................... Ans.

10-17 The stresses shown in Fig. P10-17a act at a point on the free surface of a stressed body. Determine (a) The normal and shear stresses at this point on the inclined plane AB shown in the figure. (b) The normal stresses σn and σt and the shear stress τnt at this point if they act on the rotated stress element

shown in Fig. P10-17b. SOLUTION (a) The given values are:

18 ksixσ = + 13 ksiyσ = + 6 ksixyτ = + 71 90 19θ = ° − ° = − °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − ° + − ° + − ° − °

13.78 ksi 13.78 ksi (T)= + = .........................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

18 13 sin 19 cos 19 6 cos 19 sin 19


= − − − ° − ° + − ° − − °

6.27 ksi= + ......................................................................................................................... Ans.


461

(b) The given values are:

18 ksixσ = + 13 ksiyσ = + 6 ksixyτ = + 26nθ = ° 116tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= ° + ° + ° °

21.8 ksi 21.8 ksi (T)= + = .............................................................................................Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

18cos 116 13sin 116 2 6 sin 116 cos 116t x y xyσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

9.23 ksi 9.23 ksi (T)= + = ..............................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

18 13 sin 26 cos 26 6 cos 26 sin 26


= − − ° ° + ° − °

1.724 ksi= + ...................................................................................................................... Ans.

10-18 The stresses shown in Fig. P10-18a act at a point on the free surface of a stressed body. Determine (a) The normal and shear stresses at this point on the inclined plane AB shown in the figure. (b) The normal stresses σn and σt and the shear stress τnt at this point if they act on the rotated stress element

shown in Fig. P10-18b. SOLUTION

The given values are: 10 MPaxσ = − 70 MPayσ = − 40 MPaxyτ = +

(a) 62 90 28θ = ° − ° = − °

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − − ° + − − ° + − ° − °

56.4 MPa 56.4 MPa (C)= − = ......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

10 70 sin 28 cos 28 40 cos 28 sin 28


= − − − − − ° − ° + − ° − − °

47.2 MPa= + ..................................................................................................................... Ans.

(b) 15nθ = ° 105tθ = °

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − ° + − ° + ° °

5.98 MPa 5.98 MPa (T)= + = .......................................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

10 cos 105 70 sin 105 2 40 sin 105 cos 105t x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − ° + ° °

86.0 MPa 86.0 MPa (C)= − = ......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

10 70 sin 15 cos 15 40 cos 15 sin 15


= − − − − ° ° + ° − °

19.64 MPa= + ................................................................................................................... Ans.


462

10-19 The stresses shown in Fig. P10-19 act at a point in a structural member. Determine the normal stresses σn and σt and the shear stress τnt at this point.


18 ksixσ = + 10 ksiyσ = + 9 ksixyτ = + 15nθ = ° 105tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= ° + ° + ° °

22.0 ksi 22.0 ksi (T)= + = .............................................................................................Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= ° + ° + ° °

6.04 ksi 6.04 ksi (T)= + = .............................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

18 10 sin 15 cos 15 9 cos 15 sin 15


= − − ° ° + ° − °

5.79 ksi= + ......................................................................................................................... Ans.

10-20 The stresses shown in Fig. P10-20 act at a point in a structural member. Determine the normal stresses σn and σt and the shear stress τnt at this point.


170 MPaxσ = + 50 MPayσ = + 30 MPaxyτ = + 19nθ = − ° 71tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − ° + − ° + − ° − °

138.8 MPa 138.8 MPa (T)= + = ..................................................................................Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= ° + ° + ° °

81.2 MPa 81.2 MPa (T)= + = ......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

170 50 sin 19 cos 19 30 cos 19 sin 19


= − − − ° − ° + − ° − − °

60.6 MPa= + ..................................................................................................................... Ans.

10-21 The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular shaft subjected to an axial load P and a torque T are shown in Fig. P10-21. The normal stress on plane AA at this point is 8000 psi (T). Determine

(a) The magnitude of the shearing stresses τh and τv. (b) The magnitude and direction of the shearing stress on the inclined plane AA. SOLUTION (a) The given values are:

8 ksixσ = + 0 ksiyσ = 8 ksinσ = + ( )1tan 4 3 53.130θ −= − = − °

2 2cos sin 2 sin cosn x y xyσ σ θ σ θ τ θ θ= + +


463

( ) ( ) ( ) ( ) ( )2 28 8cos 53.130 0 sin 53.130 2 sin 53.130 cos 53.130vτ= − ° + − ° + − ° − °

5.333 ksivτ = −

5.33 ksiv hτ τ= ≅ ................................................................................................................Ans.

(b) ( ) ( )2 2sin cos cos sinnt x y xyτ σ σ θ θ τ θ θ= − − + −

( ) ( ) ( ) ( ) ( )2 28 0 sin 53.130 cos 53.130 cos 53.130 sin 53.130vτ = − − − ° − ° + − ° − − °

5.33 ksi= + ......................................................................................................................... Ans.

10-22 The stresses on horizontal and vertical planes at a point are shown in Fig. P10-22. The normal stress on plane AB is 15 MPa (T). Determine

(a) The normal stress σx on the vertical plane. (b) The magnitude and direction of the shearing stress on the inclined plane AB. SOLUTION The given values are:

0 MPayσ = 15 MPanσ = + 25 MPaxyτ = + ( )1tan 12 5 67.38θ −= − = − °

(a) 2 2cos sin 2 sin cosn x y xyσ σ θ σ θ τ θ θ= + +

( ) ( ) ( ) ( ) ( ) ( )2 215 cos 67.38 0 sin 67.38 2 25 sin 67.38 cos 67.38xσ= − ° + − ° + − ° − °

221.4 MPa 221 MPa (T)xσ = + ≅ ...............................................................................Ans.


( ) ( ) ( ) ( ) ( )2 2221.4 0 sin 67.38 cos 67.38 25 cos 67.38 sin 67.38 = − − − ° − ° + − ° − − °

61.0 MPa= + ..................................................................................................................... Ans.

10-23 At a point in a structural member, there are stresses on horizontal and vertical planes, as shown in Fig. P10-23. The magnitude of the compressive stress σc is three times the magnitude of the tensile stress σt. Specifications require that the shearing stress on plane AB not exceed 4000 psi, and the normal stress on plane AB not exceed 7800 psi. Determine the maximum value of stress σc that will satisfy the specifications.


x tσ σ= 3y tσ σ= − 0 psixyτ = ( )1tan 4 3 53.130θ −= = °

7800 psinσ = ± 4000 psintτ = ±

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

cos 53.130 3 sin 53.130 2 0 sin 53.130 cos 53.130n x y xy

t t

σ σ θ σ θ τ θ θ

σ σ

= + +

= ° + − ° + ° °

( )0.6410 0.6410 7800 5000 psit nσ σ= − = − − = +

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

3 sin 53.13 cos 53.13 0 cos 53.13 sin 53.13

nt x y xy

t t

τ σ σ θ θ τ θ θ

σ σ

= − − + −

= − − − ° ° + ° − °

( )0.5208 0.5208 4000 2083 psit ntσ τ= = + = +

Therefore:

( ) ( ) ( )max 3 max 3 2083 6249 psi 6.25 ksi (C)c tσ σ= − = − = − ≅ .............................Ans.


464

10-24 At a point in a stressed body, the stresses on two perpendicular planes are as shown in Fig. P10-24. Determine

(a) The stresses on plane aa. (b) The stresses on horizontal and vertical planes at the point. SOLUTION (a) The given values are:

200 MPanσ = + 50 MPatσ = + 0 MPantτ = 135θ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos200cos 135 50sin 135 2 0 sin 135 cos 135

aa n t ntσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

125.0 MPa 125.0 MPa (T)= + = ..................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

200 50 sin 135 cos 135 0 cos 135 sin 135

aa n t ntτ σ σ θ θ τ θ θ= − − + −

= − − ° ° + ° − °

75.0 MPa= + ..................................................................................................................... Ans. (b) The given values are:

200 MPanσ = + 50 MPatσ = + 0 MPantτ = 18xθ = − ° 72yθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos200cos 18 50sin 18 2 0 sin 18 cos 18

x n t ntσ σ θ σ θ τ θ θ= + +

= − ° + − ° + − ° − °

185.7 MPa 185.7 MPa (T)= + = ..................................................................................Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

200cos 72 50sin 72 2 0 sin 72 cos 72y n t ntσ σ θ σ θ τ θ θ= + +

= ° + ° + ° °

64.3 MPa 64.3 MPa (T)= + = .......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

200 50 sin 18 cos 18 0 cos 18 sin 18

xy n t ntτ σ σ θ θ τ θ θ= − − + −

= − − − ° − ° + − ° − − °

44.1 MPa= + ..................................................................................................................... Ans.

10-25 At a point in a stressed body, the stresses on two perpendicular planes are as shown in Fig. P10-25. Determine

(a) The stresses on plane aa. (b) The stresses on horizontal and vertical planes at the point. SOLUTION (a) The given values are:

20 ksinσ = − 3 ksitσ = − 0 ksintτ = 135θ = °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos20cos 135 3 sin 135 2 0 sin 135 cos 135

aa n t ntσ σ θ σ θ τ θ θ= + +

= − ° + − ° + ° °

11.50 ksi 11.50 ksi (C)= − = .........................................................................................Ans.


465

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

20 3 sin 135 cos 135 0 cos 135 sin 135

aa n t ntτ σ σ θ θ τ θ θ= − − + −

= − − − − ° ° + ° − °

8.50 ksi= − ......................................................................................................................... Ans. (b) The given values are:

20 ksinσ = − 3 ksitσ = − 0 ksintτ = 28xθ = ° 118yθ = °

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos20 cos 28 3 sin 28 2 0 sin 28 cos 28

x n t ntσ σ θ σ θ τ θ θ= + +

= − ° + − ° + ° °

16.25 MPa 16.25 ksi (C)= − = .....................................................................................Ans.

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

20 cos 118 3 sin 118 2 0 sin 118 cos 118y n t ntσ σ θ σ θ τ θ θ= + +

= − ° + − ° + ° °

6.75 ksi 6.75 ksi (C)= − = .............................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

20 3 sin 28 cos 28 0 cos 28 sin 28

xy n t ntτ σ σ θ θ τ θ θ= − − + −

= − − − − ° ° + ° − °

7.05 ksi= + ......................................................................................................................... Ans.

10-26 The stresses shown in Fig. P10-26 act at a point on the free surface of a stressed body. (a) Calculate and graph the normal stress nσ and the shearing stress ntτ on the inclined plane AB as a

function of the angle ( )0 180θ θ° ≤ ≤ ° .

(b) For what angle θ is the normal stress a maximum? A minimum? What is the value of the shear stress on these planes?

(c) For what angle θ is the shear stress a maximum? A minimum? What is the value of the normal stress on these planes?

SOLUTION


( ) ( )2 240cos 80 sin 2 80 sin cos MPaθ θ θ θ = + − + ............................................Ans.

( ) ( )2 2sin cos cos sinnt x y xyτ σ σ θ θ τ θ θ= − − + −

( ) ( ){ }2 240 80 sin cos 80 cos sin MPaθ θ θ θ= − − − + − .....................................Ans.

(b) At 26.565θ = °

max 80 MPaσ = ........................................ Ans.

0 MPaτ = ................................................. Ans.

At 116.565θ = °

min 120 MPaσ = − .................................... Ans.

0 MPaτ = ................................................. Ans.

(c) At 71.565θ = °

min 100 MPaτ = − .................................... Ans.


466

20 MPaσ = − .......................................................................................................................... Ans.

At 161.565θ = °

max 100 MPaτ = ...................................................................................................................... Ans.

20 MPaσ = − .......................................................................................................................... Ans.

10-27 The stresses shown in Fig. P10-27 act at a point on the free surface of a stressed body. Calculate the normal stress nσ and the shearing stress ntτ on the inclined plane AB as a function of the angle

( )0 180θ θ° ≤ ≤ ° . For each angle θ , graph the negative of the shearing stress ( ntτ− , vertical axis) as a

function of the normal stress ( nσ , horizontal axis). On your graph, clearly identify the points associated

with the angles 0 , 30 , 45 , 60 , 90 ,120 ,135 ,150 , and 180θ = � � � � � � � � � . SOLUTION

2 2cos sinn x yσ σ θ σ θ= +

2 sin cosxyτ θ θ+

( ) 2 20 cos 60sinθ θ= +

( )2 40 sin cosθ θ+ ............. Ans.

( )sin cosnt x yτ σ σ θ θ= − −

( )2 2cos sinxyτ θ θ+ −

( )0 60 sin cosθ θ= − −

( )2 240 cos sinθ θ+ − ......... Ans.

10-28 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point.


50 MPaxσ = + 20 MPayσ = + 40 MPaxyτ = +

( )

22

1, 2

22

2 2

50 20 50 20 402 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − = ± +

1 35 42.72 77.7 MPa 77.7 MPa (T)pσ = + = = ..............................................................Ans.

2 35 42.72 7.72 MPa 7.72 MPa (C)pσ = − = − = ...........................................................Ans.

3 0 MPap zσ σ= = ........................... max 42.7 MPapτ τ= = ...........................................Ans.

( )

( ) ( )1 12 2 401 1tan tan 34.72

2 2 50 20xy

px y

τθ

σ σ− −= = = °

− −....................................................Ans.


467


SOLUTION

The given values are: 12 ksixσ = + 4 ksiyσ = − 6 ksixyτ = −

( ) ( ) ( )

22

1, 2

22

2 2

12 4 12 46

2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − − − = ± + −

1 4 10.00 14.00 ksi 14.00 ksi (T)pσ = + = = ....................................................................Ans.

2 4 10.00 6.00 ksi 6.00 ksi (C)pσ = − = − = .....................................................................Ans.

3 0 ksip zσ σ= = ............................... max 10.00 ksipτ τ= = .............................................Ans.

( )

( ) ( )1 12 2 61 1tan tan 18.43

2 2 12 4xy

px y

τθ

σ σ− − −

= = = − °− − −

.................................................Ans.


SOLUTION

The given values are: 75 MPaxσ = + 25 MPayσ = − 35 MPaxyτ = −

( ) ( ) ( )

22

1, 2

22

2 2

75 25 75 2535

2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − − − = ± + −

1 25 61.03 86.0 MPa 86.0 MPa (T)pσ = + = = ...............................................................Ans.

2 25 61.03 36.0 MPa 36.0 MPa (C)pσ = − = − = ...........................................................Ans.

3 0 MPapσ = ....................... max 61.0 MPapτ τ= = .........................................................Ans.

( )

( ) ( )1 12 2 351 1tan tan 17.50

2 2 75 25xy

px y

τθ

σ σ− − −

= = = − °− − −

...............................................Ans.



15 ksixσ = − 10 ksiyσ = +

8 ksixyτ = +


468

( ) ( ) ( )

22

1, 2

22

2 2

15 10 15 108

2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

− + − − = ± +

1 2.50 14.84 12.34 ksi 12.34 ksi (T)pσ = − + = = ............................................................Ans.

2 2.50 14.84 17.34 ksi 17.34 ksi (C)pσ = − − = − = ........................................................Ans.

3 0 ksip zσ σ= = ................. max 14.84 ksipτ τ= = ...........................................................Ans.

( )

( ) ( )1 12 2 81 1tan tan 16.31

2 2 15 10xy

px y

τθ

σ σ− −= = = − °

− − −...............................................Ans.



30 MPaxσ = − 40 MPayσ = +

10 MPaxyτ = −

( ) ( ) ( )

22

1, 2

22

2 2

30 40 30 4010

2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

− + − − = ± + −

1 5 36.4 41.4 MPa 41.4 MPa (T)pσ = + = = ...................................................................Ans.

2 5 36.4 31.4 MPa 31.4 MPa (C)pσ = − = − = ................................................................Ans.

3 0 MPap zσ σ= = ........................... max 36.4 MPapτ τ= = ...........................................Ans.

( )

( ) ( )1 12 2 101 1tan tan 7.97

2 2 30 40xy

px y

τθ

σ σ− − −

= = = + °− − −

.................................................Ans.



25 ksixσ = + 7 ksiyσ = +

12 ksixyτ = +

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +


469

( )2

225 7 25 7 122 2+ − = ± +

1 16.00 15.00 31.0 ksi 31.0 ksi (T)pσ = + = = .................................................................Ans.

2 16.00 15.00 1.000 ksi 1.000 ksi (T)pσ = − = + = .........................................................Ans.

3 0 ksip zσ σ= = ............................... 15.00 ksipτ = .........................................................Ans.

max minmax

31 0 15.50 ksi2 2

σ στ − −= = = ............................................................................Ans.

( )

( ) ( )1 12 2 121 1tan tan 26.57

2 2 25 7xy

px y

τθ

σ σ− −= = = + °

− −....................................................Ans.



36 MPaxσ = + 26 MPayσ = +

12 MPaxyτ = +

( )

22

1, 2

22

2 2

36 26 36 26 122 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − = ± +

1 31 13.00 44.0 MPa 44.0 MPa (T)pσ = + = = ...............................................................Ans.

2 31 13.00 18.00 MPa 18.00 MPa (T)pσ = − = + = ........................................................Ans.

3 0 MPap zσ σ= = ........................... 13.00 MPapτ = .....................................................Ans.

max minmax

44 0 22.0 MPa2 2

σ στ − −= = = ..........................................................................Ans.

( )

( ) ( )1 12 2 121 1tan tan 33.69

2 2 36 26xy

px y

τθ

σ σ− −= = = + °

− −.................................................Ans.



2 ksixσ = − 14 ksiyσ = −

8 ksixyτ = −

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +


470

( ) ( ) ( ) ( ) ( )

222 14 2 14

82 2

− + − − − − = ± + −

1 8 10.00 2.00 ksi 2.00 ksi (T)pσ = − + = + = ...................................................................Ans.

2 8 10.00 18.00 ksi 18.00 ksi (C)pσ = − − = − = ..............................................................Ans.

3 0 ksipσ = ......................................... max 10.00 ksipτ τ= = .............................................Ans.

( )

( ) ( )1 12 2 81 1tan tan 26.57

2 2 2 14xy

px y

τθ

σ σ− − −

= = = − °− − − −

..............................................Ans.




24 MPaxyτ = −

( )

22

1, 2

22

2 2

72 36 72 36 242 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − = ± + −

1 54 30.00 84.0 MPa 84.0 MPa (T)pσ = + = = ...............................................................Ans.

2 54 30.00 24.0 MPa 24.0 MPa (T)pσ = − = + = ...........................................................Ans.

3 0 MPap zσ σ= = ........................... 30.00 MPapτ = .....................................................Ans.

max minmax

84 0 42.0 MPa2 2

σ στ − −= = = ..........................................................................Ans.

( )

( ) ( )1 12 2 241 1tan tan 26.57

2 2 72 36xy

px y

τθ

σ σ− − −

= = = − °− −

.................................................Ans.




6 ksixyτ = +

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +


471

( )2

213 18 13 18 62 2+ − = ± +

1 15.50 6.50 22.0 ksi 22.0 ksi (T)pσ = + = = ..................................................................Ans.

2 15.50 6.50 9.00 ksi 9.00 ksi (T)pσ = − = + = ...............................................................Ans.

3 0 ksip zσ σ= = ............................... 6.50 ksipτ = ...........................................................Ans.

max minmax

22 0 11.00 ksi2 2

σ στ − −= = = ...........................................................................Ans.

( )

( ) ( )1 12 2 61 1tan tan 33.69

2 2 13 18xy

px y

τθ

σ σ− −= = = − °

− −...................................................Ans.

10-38 The stresses shown in Fig. P10-38 act at a point on the free surface of a stressed body. The magnitude of the maximum shear stress at the point is 125 MPa. Determine the principal stresses, and the magnitude of the unknown shear stress on the vertical plane, and the angle θp between the x-axis and the maximum tensile stress at the point.

SOLUTION

Since xσ and yσ have opposite signs, max 125 MPapτ τ= =

( )1 2 max2 2 125 250 MPap pσ σ τ− = = =

( )1 2 80 120 40 MPap p x yσ σ σ σ+ = + = + − = −

Therefore: 1 105.0 MPa 105.0 MPa (T)pσ = = ....................................................................Ans.

2 145.0 MPa 145.0 MPa (C)pσ = − = .................................................................Ans.

( ) 2

280 120125 MPa

2p xyτ τ− −

= ± + =

75.0 MPaxyτ = ±

Therefore (as shown on Fig. P10-38): 75.0 MPaxyτ = − ...........................................................Ans.

( )

( ) ( )1 12 2 751 1tan tan 18.43

2 2 80 120xy

px y

τθ

σ σ− − −

= = = + °− − −

............................................Ans.

10-39 The principal compressive stress on a vertical plane through a point in a wooden block is equal to four times the principal compressive stress on a horizontal plane, as shown in Fig. P10-39. The plane of the grain is 30° clockwise from the vertical plane. If the normal and shear stresses on the plane of the grain must not exceed 300 psi (C) and 125 psi shear, determine the maximum allowable compressive stress on the horizontal plane.

SOLUTION

( ) ( )

2 2

2 2

cos sin 2 sin cos

4 cos 30 sin 30 0 300 psin x y xy

c c

σ σ θ σ θ τ θ θ

σ σ

= + +

= − ° + − ° + > −

92.3 psi 92.3 psi (C)cσ > − =

( ) ( )( ) ( ) ( )

2 2sin cos cos sin

4 sin 30 cos 30 0 125 psint x y xy

c c

τ σ σ θ θ τ θ θ

σ σ

= − − + −

= − − − ° − ° + > −

96.2 psi 96.2 psi (C)cσ > − =

Therefore ( )max 92.3 psi (C)cσ = .........................................................................................Ans.


472

10-40 At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an unknown positive shear stress exist on a horizontal plane. One principal stress at the point is 8 MPa (C). The maximum shear stress at the point has a magnitude of 95 MPa. Determine the unknown stresses on the horizontal and vertical planes shown in Fig. P10-40 and the unknown principal stress at the point.

SOLUTION

The principal stress 2pσ must be compressive with a magnitude greater than 64.0 MPa and the principal stress

3 0 MPap zσ σ= = (since it is a free surface). Then

2max minmax

095.0 MPa

2 2pσσ στ

−−= = =

2 190.0 MPa 190.0 MPa (C)pσ = − = ...............................................................................Ans.

1 2p p x yσ σ σ σ+ = +

( ) ( ) ( )8 190 64xσ− + − = + −

134.0 MPa 134.0 MPa (C)xσ = − = .................................................................................Ans.

( ) ( )1 2 8 190

91.0 MPa2 2

p pp

σ στ

− − − −= = =

( ) ( ) 2

2134 6491.0 MPa

2p xyτ τ− − −

= ± + =

84.0 MPaxyτ = ....................................................................................................................... Ans.

10-41 At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a horizontal plane, as shown in Fig. P10-41. An unknown negative shear stress acts on the horizontal and vertical planes. The maximum shear stress at the point has a magnitude of 32 ksi. Determine the principal stresses and the shear stress on the vertical plane at the point.

SOLUTION

Since xσ and yσ have opposite signs, max 32 ksipτ τ= =

( ) 2

220 3032 ksi

2p xyτ τ− −

= ± + =

19.97 ksixyτ = − ..................................................................................................................... Ans.

( )1 2 20 30 10 ksip p x yσ σ σ σ+ = + = + − = −

( )1 2 2 2 32 64 ksip p pσ σ τ− = = =

1 27.0 ksi 27.0 ksi (T)pσ = + = ...........................................................................................Ans.

2 37.0 ksi 37.0 ksi (C)pσ = − = ..........................................................................................Ans.

3 0 ksipσ = − ............................................................................................................................ Ans.


473

10-42 At a point on the free surface of a stressed body, a normal stress of 75 MPa (T) and an unknown negative shear stress exist on a horizontal plane, as shown in Fig. P10-42. One principal stress at the point is 200 MPa (T). The maximum in-plane shear stress at the point has a magnitude of 85 MPa. Determine the unknown stresses on the vertical plane, the unknown principal stress, and the maximum shear stress at the point.

SOLUTION

1 2 220085.0 MPa

2 2p p p

p

σ σ στ

− −= = =

2 30.0 MPa 30.0 MPa (T)pσ = + = ....................................................................................Ans.

max minmax

200 0 100.0 MPa2 2

σ στ − −= = = .....................................................................Ans.

1 2p p x yσ σ σ σ+ = + 200 30 75xσ+ = +

155.0 MPa 155.0 MPa (T)xσ = + = .................................................................................Ans.

2

2155 75 85.0 MPa2p xyτ τ− = ± + =

75.0 MPaxyτ = − .................................................................................................................... Ans.

10-43 The stresses shown in Fig. P10-43 act at a point on the free surface of a stressed body. Use Mohr�s circle to determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.

SOLUTION The given values for use in drawing Mohr�s circle are:

1 12 ksix pσ σ= = +

2 16 ksiy pσ σ= = −

3 0 ksiz pσ σ= =

( )12 16

2 ksi2

a+ −

= = −

16 12 14 ksi

2R += =

2 14cos52 6.62 ksi 6.62 ksi (T)nσ = − + ° = + = ............................................................Ans.

14sin 52 11.03 ksi 11.03 ksintτ = ° = = +� ....................................................................Ans.

10-44 The stresses shown in Fig. P10-44 act at a point on the free surface of a stressed body. Use Mohr�s circle to determine the normal and shear stresses at this point on the inclined plane AB shown in the figure.


2 27 MPax pσ σ= = −

1 45 MPay pσ σ= = +

3 0 MPaz pσ σ= =

( )27 45

9 MPa2

a− +

= =

27 45 36 MPa

2R += =


474

9 36cos 66 23.6 MPa 23.6 MPa (T)nσ = + ° = = ..........................................................Ans.

36sin 66 32.9 MPa 32.9 MPantτ = ° = = −� .................................................................Ans.

10-45 At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-45. Use Mohr�s circle to determine the principal stresses and the maximum shear stress at the point.



12 ksixyτ = + 3 0 ksiz pσ σ= =

25 7 16 ksi

2a += =

2 29 12 15 ksiR = + =

1 16 15 31.0 ksi 31.0 ksi (T)pσ = + = + = ..........................................................................Ans.

2 16 15 1.00 ksi 1.00 ksi (T)pσ = − = + = ..........................................................................Ans.

3 0 ksipσ = ........................... 15.00 ksip Rτ = = ...............................................................Ans.

max minmax

31 0 15.50 ksi2 2

σ στ − −= = = ............................................................................Ans.

11

1 12tan 26.572 9pθ −= = ° .....................................................................................................Ans.




40 MPaxyτ = + 3 0 MPaz pσ σ= =

50 20 35 MPa

2a += =

2 215 40 42.72 MPaR = + =

1 35 42.72 77.7 MPa 77.7 MPa (T)pσ = + = + = ............................................................Ans.

2 35 42.72 7.72 MPa 7.72 MPa (C)pσ = − = − = ...........................................................Ans.

3 0 MPapσ = ....................... max 42.7 MPap Rτ τ= = = .................................................Ans.

11

1 40tan 34.722 15pθ −= = ° .....................................................................................................Ans.


475



15 ksixσ = − 10 ksiyσ = +


( )15 10

2.5 ksi2

a− +

= = −

2 212.5 8 14.841 ksiR = + =

1 2.5 14.841 12.34 ksi 12.34 ksi (T)pσ = − + = + = .........................................................Ans.

2 2.5 14.841 17.34 ksi 17.34 ksi (C)pσ = − − = − = .........................................................Ans.

3 0 ksipσ = ........................... max 14.84 ksip Rτ τ= = = ...................................................Ans.

11

1 8tan 16.312 12.5pθ −= − = − °............................................................................................Ans.



65 MPaxσ = − 125 MPayσ = −

75 MPaxyτ = + 3 0 MPaz pσ σ= =

( ) ( )65 125

95 MPa2

a− + −

= = −

2 230 75 80.78 MPaR = + =

1 95 80.78 14.22 MPa 14.22 MPa (C)pσ = − + = − = .....................................................Ans.

2 95 80.78 175.8 MPa 175.8 MPa (C)pσ = − − = − = .....................................................Ans.

3 0 MPapσ = ....................... 80.8 MPap Rτ = = ..............................................................Ans.

( )max min

max

0 175.887.9 MPa

2 2σ στ

− −−= = = ..............................................................Ans.

11

1 75tan 26.872 55pθ −= = ° .....................................................................................................Ans.


476

10-49 At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-49. Use Mohr�s circle to determine

(a) The principal stresses and the maximum shear stress at the point. (b) The normal and shear stresses on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr�s circle are:

12 ksixσ = + 6 ksiyσ = −


( )12 6

3 ksi2

a+ −

= =

2 29 10 13.454 ksiR = + =

(a) 1 3 13.454 16.45 ksi 16.45 ksi (T)pσ = + = + = ...............................................................Ans.

2 3 13.454 10.45 ksi 10.45 ksi (C)pσ = − = − = ...............................................................Ans.

3 0 ksipσ = ........................... max 13.45 ksip Rτ τ= = = ...................................................Ans.

(b) 3 13.454cos71.99 1.160 ksi 1.160 ksi (C)nσ = − ° = − = ............................................Ans.

13.454sin 71.99 12.79 ksi 12.79 ksintτ = ° = = +� ......................................................Ans.



80 MPaxσ = + 100 MPayσ = −

60 MPaxyτ = − 3 0 MPaz pσ σ= =

( )80 100

10 MPa2

a+ −

= = −

2 290 60 108.17 MPaR = + =

(a) 1 10 108.17 98.2 MPa 98.2 MPa (T)pσ = − + = + = .......................................................Ans.

2 10 108.17 118.2 MPa 118.2 MPa (C)pσ = − − = − = ..................................................Ans.

3 0 MPapσ = ....................... max 108.2 MPap Rτ τ= = = ...............................................Ans.

(b) 10 108.17cos 62.31 60.3 MPa 60.3 MPa (C)nσ = − − ° = − = .....................................Ans.

108.17sin 62.31 95.8 MPa 95.8 MPantτ = ° = = −� ...................................................Ans.


477




10 ksixyτ = − 3 0 ksiz pσ σ= =

25 12 18.50 ksi

2a += =

2 26.5 10 11.927 ksiR = + =

(a) 1 18.5 11.927 30.4 ksi 30.4 ksi (T)pσ = + = + = ..............................................................Ans.

2 18.5 11.927 6.57 ksi 6.57 ksi (T)pσ = − = + = .............................................................Ans.

3 0 ksipσ = ........................... 11.93 ksip Rτ = = ...............................................................Ans.

max minmax

30.42 0 15.21 ksi2 2

σ στ − −= = = ........................................................................Ans.

(b) 18.5 11.927cos 47.02 10.37 ksi 10.37 ksi (T)nσ = − ° = + = .......................................Ans.

11.927sin 47.02 8.73 ksi 8.73 ksintτ = ° = = −� ...........................................................Ans.

10-52 The thin rectangular plate shown in Fig. P10-52 is uniformly deformed such that 2000 m/mxε µ= − ,

1500 m/myε µ= − , and 1250 radxyγ µ= + . Determine the normal strain nε in the plate.


2000 m/mxε µ= − 1500 m/myε µ= − 1250 radxyγ µ= + 35nθ = − °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

2000 cos 35 1500 sin 35 1250sin 35 cos 35n x n y n xy n nε ε θ ε θ γ θ θ= + +

= − − ° + − − ° + − ° − °

2413 m/m 2410 m/mµ µ= − ≅ − ...................................................................................Ans.

10-53 The thin rectangular plate shown in Fig. P10-53 is uniformly deformed such that 880 in./in.xε µ= + ,

960 in./in.yε µ= + , and 750 radxyγ µ= − . Determine the normal strain ACε along diagonal AC of the plate.


880 in./in.xε µ= + 960 in./in.yε µ= + 750 radxyγ µ= −

( )1tan 2 4 26.57ACθ −= = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

880cos 26.57 960sin 26.57 750 sin 26.57 cos 26.57AC x n y n xy n nε ε θ ε θ γ θ θ= + +

= ° + ° + − ° °

596 in./in.µ= + ..............................................................................................................Ans.


478

10-54 The thin rectangular plate shown in Fig. P10-54 is uniformly deformed such that 1500 m/mxε µ= + ,

1250 m/myε µ= − , and 1000 radxyγ µ= + . Determine the normal strain BDε along diagonal BD of the plate.


1500 m/mxε µ= + 1250 m/myε µ= − 1000 radxyγ µ= +

( )1tan 150 200 36.87BDθ −= − = − °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

1500cos 36.87 1250 sin 36.87 1000sin 36.87 cos 36.87BD x n y n xy n nε ε θ ε θ γ θ θ= + +

= − ° + − − ° + − ° − °

30.0 m/mµ= + .................................................................................................................. Ans.

10-55 The strain components for a point in a body subjected to plane strain are 800 in./in.xε µ= − ,

640 in./in.yε µ= + , and 960 radxyγ µ= − . Determine the strain components nε , tε , and ntγ . at the

point if the nt-axes are rotated with respect to the xy-axes by 42θ = ° in the direction indicated in Fig. P10-55.


800 in./in.xε µ= − 640 in./in.yε µ= + 960 radxyγ µ= − 42nθ = ° 132tθ = °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

800 cos 42 640sin 42 960 sin 42 cos 42n x n y n xy n nε ε θ ε θ γ θ θ= + +

= − ° + ° + − ° °

633 in./in.µ= − .................................................................................................................. Ans.

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

800 cos 132 640sin 132 960 sin 132 cos 132t x t y t xy t tε ε θ ε θ γ θ θ= + +

= − ° + ° + − ° °

473 in./in.µ= + .................................................................................................................. Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 800 640 sin 42 cos 42 960 cos 42 sin 42

nt x y n n xy n nγ ε ε θ θ γ θ θ= − − + −

= − − − ° ° + − ° − °

1332 radµ= + .................................................................................................................... Ans.

10-56 The strain components for a point in a body subjected to plane strain are 720 m/mxε µ= + ,

480 m/myε µ= − , and 360 radxyγ µ= + . Determine the strain components nε , tε , and ntγ . at the



720 m/mxε µ= + 480 m/myε µ= − 360 radxyγ µ= + 30nθ = − ° 60tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

720cos 30 480 sin 30 360sin 30 cos 30n x n y n xy n nε ε θ ε θ γ θ θ= + +

= − ° + − − ° + − ° − °

264 m/mµ= + ................................................................................................................... Ans.


479

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

720cos 60 480 sin 60 360sin 60 cos 60t x t y t xy t tε ε θ ε θ γ θ θ= + +

= ° + − ° + ° °

24.1 m/mµ= − ................................................................................................................... Ans.

( ) ( )

( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 720 480 sin 30 cos 30 360 cos 30 sin 30


= − − − − ° − ° + − ° − − °

1219 radµ= + .................................................................................................................... Ans.

10-57 The strain components for a point in a body subjected to plane strain are 900 in./in.xε µ= + ,




900 in./in.xε µ= + 650 in./in.yε µ= + 300 radxyγ µ= − 32nθ = ° 122tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

900cos 32 650sin 32 300 sin 32 cos 32n x n y n xy n nε ε θ ε θ γ θ θ= + +

= ° + ° + − ° °

695 in./in.µ= + .................................................................................................................. Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

900cos 122 650sin 122 300 sin 122 cos 122t x t y t xy t tε ε θ ε θ γ θ θ= + +

= ° + ° + − ° °

855 in./in.µ= + .................................................................................................................. Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 900 650 sin 32 cos 32 300 cos 32 sin 32


= − − ° ° + − ° − °

356 radµ= − ...................................................................................................................... Ans.


950 m/myε µ= − , and 800 radxyγ µ= − . Determine the strain components nε , tε , and ntγ . at the



800 m/mxε µ= + 950 m/myε µ= − 800 radxyγ µ= − 42nθ = − ° 48tθ = °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

800cos 42 950 sin 42 800 sin 42 cos 42n x n y n xy n nε ε θ ε θ γ θ θ= + +

= − ° + − − ° + − − ° − °

414 m/mµ= + ................................................................................................................... Ans.

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

800cos 48 950 sin 48 800 sin 48 cos 48t x t y t xy t tε ε θ ε θ γ θ θ= + +

= ° + − ° + − ° °

564 m/mµ= − ................................................................................................................... Ans.


480

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 800 950 sin 42 cos 42 800 cos 42 sin 42


= − − − − ° − ° + − − ° − − °

1657 radµ= + .................................................................................................................... Ans.





750 in./in.xε µ= + 360 in./in.yε µ= + 300 radxyγ µ= − 52nθ = ° 142tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos


= ° + ° + − ° °

362 in./in.µ= + .................................................................................................................. Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos


= ° + ° + − ° °

748 in./in.µ= + .................................................................................................................. Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 750 360 sin 52 cos 52 300 cos 52 sin 52


= − − ° ° + − ° − °

306 radµ= − ...................................................................................................................... Ans.

10-60 The strain components for a point in a body subjected to plane strain are 100 m/mxε µ= − ,

700 m/myε µ= − , and 400 radxyγ µ= + . Determine the strain components nε , tε , and ntγ . at the



100 m/mxε µ= − 700 m/myε µ= − 400 radxyγ µ= + 28nθ = − ° 62tθ = °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

100 cos 28 700 sin 28 400sin 28 cos 28n x n y n xy n nε ε θ ε θ γ θ θ= + +

= − − ° + − − ° + − ° − °

398 m/mµ= − ................................................................................................................... Ans.

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

100 cos 62 700 sin 62 400sin 62 cos 62t x t y t xy t tε ε θ ε θ γ θ θ= + +

= − ° + − ° + ° °

402 m/mµ= − ................................................................................................................... Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 100 700 sin 28 cos 28 400 cos 28 sin 28


= − − − − − ° − ° + − ° − − °

721 radµ= + ....................................................................................................................... Ans.


481


450 in./in.yε µ= + , and 600 radxyγ µ= + . Determine the strain components nε , tε , and ntγ . at the



850 in./in.xε µ= + 450 in./in.yε µ= + 600 radxyγ µ= + 20nθ = ° 110tθ = °

( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

850cos 20 450sin 20 600sin 20 cos 20n x n y n xy n nε ε θ ε θ γ θ θ= + +

= ° + ° + ° °

996 in./in.µ= + .................................................................................................................. Ans.

( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

850cos 110 450sin 110 600sin 110 cos 110t x t y t xy t tε ε θ ε θ γ θ θ= + +

= ° + ° + ° °

304 in./in.µ= + .................................................................................................................. Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 850 450 sin 20 cos 20 600 cos 20 sin 20


= − − ° ° + ° − °

203 radµ= + ...................................................................................................................... Ans.


200 m/myε µ= + , and 1600 radxyγ µ= − . Determine the strain components nε , tε , and ntγ . at the



1200 m/mxε µ= + 200 m/myε µ= + 1600 radxyγ µ= − 64nθ = − ° 26tθ = °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos


= − ° + − ° + − − ° − °

1023 m/mµ= + .................................................................................................................. Ans.

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos


= ° + ° + − ° °

377 m/mµ= + ................................................................................................................... Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

2 sin cos cos sin

2 1200 200 sin 64 cos 64 1600 cos 64 sin 64


= − − − ° − ° + − − ° − − °

1773 radµ= + .................................................................................................................... Ans.


482

10-63 The thin rectangular plate shown in Fig. P10-63 is uniformly deformed such that 1575 in./in.nε µ= + ,

1350 in./in.tε µ= + , and 1250 in./in.xε µ= + . Determine

(a) The shearing strain ntγ in the plate.

(b) The normal strain yε in the plate.


1575 in./in.nε µ= + 1350 in./in.tε µ= + 1250 in./in.xε µ= +

45nθ = ° / 45x nθ = − ° / 45y nθ = °

(a) 2 2/ / / /cos sin sin cosx n x n t x n nt x n x nε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( )2 21250 1575cos 45 1350sin 45 sin 45 cos 45ntγ= − ° + − ° + − ° − °

425 radntγ µ= + ..................................................................................................................... Ans.

(b) 2 2/ / / /cos sin sin cosy n y n t y n nt y n y nε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( )2 21575cos 45 1350sin 45 sin 45 cos 45ntγ= ° + ° + ° °

1675 in./in.µ= + ...............................................................................................................Ans.

10-64 The thin rectangular plate shown in Fig. P10-64 is uniformly deformed such that 1200 m/mABε µ= − ,

750 m/mBDε µ= + , and 600 m/mADε µ= − . Determine the normal strain yε and the shearing strain

xyγ in the plate.


1200 m/mABε µ= − 750 m/mBDε µ= + 600 m/mADε µ= −

( )1tan 240 200 50.19ABθ −= = ° ( )1tan 240 200 50.19BDθ −= − = − °

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

1200 600 cos 50.19 sin 50.19 sin 50.19 cos 50.19AB x AB y AB xy AB AB

y xy

ε ε θ ε θ γ θ θ

ε γ

= + +

− = − ° + ° + ° °

0.5901 0.4918 954.1y xyε γ+ = − (a)

( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin sin cos

750 600 cos 50.19 sin 50.19 sin 50.19 cos 50.19BD x BD y BD xy BD BD

y xy

ε ε θ ε θ γ θ θ

ε γ

= + +

= − − ° + − ° + − ° − °

0.5901 0.4918 995.9y xyε γ− = + (b)

Solving Eqs. (a) and (b) gives:

35.4 m/myε µ= + .............................................................................................................Ans.

1983 radxyγ µ= − .............................................................................................................Ans.


483


200 in./in.yε µ= − , and 500 radxyγ µ= + . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.


600 in./in.xε µ= + 200 in./in.yε µ= −

500 radxyγ µ= +

( ) ( )

2 2

1, 2

2 2

2 2 2

600 200 600 200 5002 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − − − = ± +

1 200 471.7 671.7 in./in. 672 in./in.pε µ µ= + = ≅ ..........................................................Ans.

2 200 471.7 271.7 in./in. 272 in./in.pε µ µ= − = − ≅ − ....................................................Ans.

3 0 in./in.p zε ε µ= = .................. ( )max 2 471.7 943 radpγ γ µ= = = .............................Ans.

( )1 11 1 500tan tan 16.00

2 2 600 200xy

px y

γθ

ε ε− −= = = °

− − −..................................................Ans.


320 m/myε µ= − , and 480 radxyγ µ= − . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.


960 m/mxε µ= + 320 m/myε µ= − 480 radxyγ µ= −

( ) ( )

2 2

1, 2

2 2

2 2 2

960 320 960 320 4802 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − − − − = ± +

1 320 683.5 1003.5 m/m 1004 m/mpε µ µ= + = ≅ ...........................................................Ans.

2 320 683.5 363.5 m/m 364 m/mpε µ µ= − = − ≅ − .........................................................Ans.

3 0 m/mp zε ε µ= = .................... ( )max 2 683.5 1367 radpγ γ µ= = = ...........................Ans.

( )1 11 1 480tan tan 10.28

2 2 960 320xy

px y

γθ

ε ε− − −= = = − °

− − −...............................................Ans.


484


300 in./in.yε µ= − , and 420 radxyγ µ= − . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION

The given values are: 900 in./in.xε µ= + 300 in./in.yε µ= − 420 radxyγ µ= −

2 2

1, 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − = ± +

( ) ( ) 2 2900 300 900 300 4202 2 2

+ − − − − = ± +

1 300 635.7 935.7 in./in. 936 in./in.pε µ µ= + = ≅ ..........................................................Ans.


3 0 in./in.p zε ε µ= = .................. ( )max 2 635.7 1271 radpγ γ µ= = = ...........................Ans.

( )1 11 1 420tan tan 9.65

2 2 900 300xy

px y

γθ

ε ε− − −= = = − °

− − −..................................................Ans.


600 m/myε µ= + , and 480 radxyγ µ= + . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION

The given values are: 900 m/mxε µ= − 600 m/myε µ= + 480 radxyγ µ= +

2 2

1, 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − = ± +

( ) ( ) 2 2900 600 900 600 480

2 2 2− + − − = ± +

1 150 787.4 637.4 m/m 637 m/mpε µ µ= − + = ≅ ...........................................................Ans.

2 150 787.4 937.4 m/m 937 m/mpε µ µ= − − = − ≅ − .....................................................Ans.

3 0 m/mp zε ε µ= = .................................................................................................................Ans.

( )max 2 787.4 1575 radpγ γ µ= = = ....................................................................................Ans.

( )1 11 1 480tan tan 8.87

2 2 900 600xy

px y

γθ

ε ε− −= = = − °

− − −.................................................Ans.


485


1000 in./in.yε µ= + , and 250 radxyγ µ= − . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.


750 in./in.xε µ= − 1000 in./in.yε µ= + 250 radxyγ µ= −

( ) ( )

2 2

1, 2

2 2

2 2 2

750 1000 750 1000 2502 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

− + − − − = ± +

1 125 883.9 1008.9 in./in. 1009 in./in.pε µ µ= + = ≅ .......................................................Ans.

2 125 883.9 758.9 in./in. 759 in./in.pε µ µ= − = − ≅ − .....................................................Ans.

3 0 in./in.p zε ε µ= = .................. ( )max 2 883.9 1768 radpγ γ µ= = = ...........................Ans.

( )1 11 1 250tan tan 4.07

2 2 750 1000xy

px y

γθ

ε ε− − −= = = °

− − −..................................................Ans.


410 m/myε µ= − , and 360 radxyγ µ= + . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.



( ) ( )

2 2

1, 2

2 2

2 2 2

750 410 750 410 3602 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − − − = ± +

1 170 607.3 777.3 m/m 777 m/mpε µ µ= + = ≅ ...............................................................Ans.

2 170 607.3 437.3 m/m 437 m/mpε µ µ= − = − ≅ − .........................................................Ans.

3 0 m/mp zε ε µ= = .................... ( )max 2 607.3 1215 radpγ γ µ= = = ...........................Ans.

( )1 11 1 360tan tan 8.62

2 2 750 410xy

px y

γθ

ε ε− −= = = °

− − −.....................................................Ans.


486


520 in./in.yε µ= + , and 480 radxyγ µ= + . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION

The given values are: 720 in./in.xε µ= +

520 in./in.yε µ= + 480 radxyγ µ= +

2 2

1, 2

2 2

2 2 2

720 520 720 520 4802 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − = ± +

1 620 260.0 880 in./in.pε µ= + = ........................................................................................Ans.

2 620 260.0 360 in./in.pε µ= − = ........................................................................................Ans.

3 0 in./in.p zε ε µ= = .................. ( )2 260.0 520 radpγ µ= = .........................................Ans.

( ) ( )max max min 880.0 0 880 radγ ε ε µ= − = − = .................................................................Ans.

1 11 1 480tan tan 33.692 2 720 520

xyp

x y

γθ

ε ε− −= = = °

− −........................................................Ans.


980 m/myε µ= − , and 560 radxyγ µ= − . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION

The given values are: 540 m/mxε µ= −

980 m/myε µ= − 560 radxyγ µ= −

( ) ( ) ( ) ( )

2 2

1, 2

2 2

2 2 2

540 980 540 980 5602 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

− + − − − − − = ± +

1 760 356.1 403.9 m/m 404 m/mpε µ µ= − + = − ≅ − .......................................................Ans.

2 760 356.1 1116.1 m/m 1116 m/mpε µ µ= − − = − ≅ − .................................................Ans.

3 0 m/mp zε ε µ= = .................... ( )2 356.1 712 radpγ µ= ≅ ..........................................Ans.

( ) ( )max max min 0 1116 1116 radγ ε ε µ= − = − − = .........................................................Ans.

( ) ( )1 11 1 560tan tan 25.92

2 2 540 980xy

px y

γθ

ε ε− − −= = = − °

− − − −.........................................Ans.


487


432 in./in.yε µ= + , and 288 radxyγ µ= − . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION


432 in./in.yε µ= + 288 radxyγ µ= −

2 2

1, 2

2 2

2 2 2

864 432 864 432 2882 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − − = ± +

1 648 259.6 907.6 in./in. 908 in./in.pε µ µ= + = ≅ ...................Ans.

2 648 259.6 388.4 in./in. 388 in./in.pε µ µ= − = ≅ ...................Ans.

3 0 in./in.p zε ε µ= = .................. ( )2 259.6 519 radpγ µ= ≅ .........................................Ans.

( ) ( )max max min 908 0 908 radγ ε ε µ= − = − = ....................................................................Ans.

1 11 1 288tan tan 16.852 2 864 432

xyp

x y

γθ

ε ε− − −= = = − °

− −......................................................Ans.



SOLUTION

The given values are: 650 m/mxε µ= +

900 m/myε µ= + 300 radxyγ µ= +

2 2

1, 2

2 2

2 2 2

650 900 650 900 3002 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − = ± +

1 775 195.3 970.3 m/m 970 m/mpε µ µ= + = ≅ ...............................................................Ans.

2 775 195.3 579.7 m/m 580 m/mpε µ µ= − = ≅ ..............................................................Ans.

3 0 m/mp zε ε µ= = .................... ( )2 195.3 391 radpγ µ= ≅ ..........................................Ans.


1 11 1 300tan tan 25.092 2 650 900

xyp

x y

γθ

ε ε− −= = = − °

− −......................................................Ans.


488


625 in./in.yε µ= − , and 380 radxyγ µ= + . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION

The given values are: 325 in./in.xε µ= −

625 in./in.yε µ= − 380 radxyγ µ= +

( ) ( ) ( ) ( )

2 2

1, 2

2 2

2 2 2

325 625 325 625 3802 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

− + − − − − = ± +

1 475 242.1 232.9 in./in. 233 in./in.pε µ µ= − + = − ≅ − ...................................................Ans.

2 475 242.1 717.1 in./in. 717 in./in.pε µ µ= − − = − ≅ − ..................................................Ans.

3 0 in./in.p zε ε µ= = .................. ( )2 242.1 484 radpγ µ= ≅ ..........................................Ans.

( ) ( )max max min 0 717 717 radγ ε ε µ= − = − − = .............................................................Ans.

( ) ( )1 11 1 380tan tan 25.85

2 2 325 625xy

px y

γθ

ε ε− −= = = °

− − − −............................................Ans.



SOLUTION


650 m/myε µ= + 600 radxyγ µ= +

2 2

1, 2

2 2

2 2 2

900 650 900 650 6002 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − = ± +

1 775 325 1100 m/mpε µ= + = ............................................................................................Ans.

2 775 325 450 m/mpε µ= − = ..............................................................................................Ans.

3 0 m/mp zε ε µ= = .................... ( )2 325 650 radpγ µ= ≅ .............................................Ans.

( ) ( )max max min 1100 0 1100 radγ ε ε µ= − = − = ................................................................Ans.

1 11 1 600tan tan 33.692 2 900 650

xyp

x y

γθ

ε ε− −= = = °

− −........................................................Ans.


489

10-77 Given that 1 600 in./in.pε µ= + , 2 400 in./in.pε µ= − , and 18.43pθ = + ° at a point in a body

subjected to plane strain, use Mohr�s circle to determine xε , yε , xyγ , pγ , and maxγ , and prepare a sketch

showing the angle pθ , the principal strain deformations, and the maximum shearing strain distortions.


1 600 in./in.pε µ= + 2 400 in./in.pε µ= − 3 0 in./in.p zε ε µ= = 18.43pθ = °

( )600 400

100 in./in.2

a µ+ −

= = ( )600 400

500 in./in.2

R µ− −

= =

( )100 500cos 36.86 500 in./in.xε µ= + ° = + ....................................................................Ans.

( )100 500cos 36.86 300 in./in.yε µ= − ° = − ...................................................................Ans.

( ) ( )2 500 sin 36.86 600 rad 600 radxyγ µ µ= ° = = +� .................................................Ans.

( )max 2 500 1000 radpγ γ µ= = = ........................................................................................Ans.

10-78 Given that 1 785 m/mpε µ= + , 2 945 m/mpε µ= − , and 16.85pθ = + ° at a point in a body




1 785 m/mpε µ= +

2 945 m/mpε µ= −

3 0 m/mp zε ε µ= =

16.85pθ = + °

( )785 945

80 m/m2

a µ+ −

= = −

( )785 945

865 m/m2

R µ− −

= =


490

( )80 865cos 33.70 640 m/mxε µ= − + ° = + .......... Ans.

( )80 865cos 33.70 800 m/myε µ= − − ° = − .......... Ans.

( ) ( )2 865 sin 33.70 960 radxyγ µ= ° = + ................. Ans.

( )max 2 865 1730 radpγ γ µ= = = ............................. Ans.

10-79 Given that 1 708 in./in.pε µ= + , 2 104 in./in.pε µ= − , and 34.10pθ = − ° at a point in a body




1 708 in./in.pε µ= + 2 104 in./in.pε µ= −

3 0 in./in.p zε ε µ= = 34.10pθ = − °

( )708 104

302 in./in.2

a µ+ −

= =

( )708 104

406 in./in.2

R µ− −

= =

( )302 406cos 68.20 453 in./in.xε µ= + − ° = + ...............Ans.

( )302 406cos 68.20 151.2 in./in.yε µ= − − ° = + ...........Ans.

( ) ( )2 406 sin 68.20xyγ = °

754 rad 754 radµ µ= = −� ........................................Ans.

( )max 2 406 812 radpγ γ µ= = = ........................................Ans.

10-80 Given that 1 114 m/mpε µ= − , 2 903 m/mpε µ= − , and 19.26pθ = + ° at a point in a body




1 114 m/mpε µ= − 2 903 m/mpε µ= − 3 0 m/mp zε ε µ= = 19.26pθ = + °

( ) ( )114 903

508.5 m/m2

a µ− + −

= = − ( ) ( )114 903

394.5 m/m2

R µ− − −

= =

( )508.5 394.5cos 38.52 199.8 m/mxε µ= − + ° = − ........................................................Ans.

( )508.5 394.5cos 38.52 817 m/myε µ= − − ° = − ...........................................................Ans.


491

( ) ( )2 394.5 sin 38.52 491 rad 491 radxyγ µ µ= ° = = +� ..............................................Ans.

( )2 394.5 789 radpγ µ= = ...................................................................................................Ans.

( ) ( )max max min 0 903 903 radγ ε ε µ= − = − − = ..............................................................Ans.

10-81 Given that 950 in./in.xε µ= + , 225 in./in.yε µ= − , and 275 radxyγ µ= + at a point in a body

subjected to plane strain, use Mohr�s circle to determine 1pε , 2pε , pγ , maxγ , and pθ , and prepare a

sketch showing the angle pθ , the principal strain deformations, and the maximum shearing strain distortions.


950 in./in.xε µ= + 225 in./in.yε µ= − 275 radxyγ µ= +

( )950 225

362.5 in./in.2

a µ+ −

= = ( ) ( )2 2587.5 137.5 603.4 in./in.R µ= + =

1 362.5 603.4 965.9 in./in. 966 in./in.pε µ µ= + = + ≅ + .................................................Ans.

2 362.5 603.4 240.9 in./in. 241 in./in.pε µ µ= − = − ≅ − .................................................Ans.

3 0 in./in.pε µ= ............ ( )max 2 603.4 1207 radpγ γ µ= = ≅ ..........................................Ans.

11 137.5tan 6.592 587.5pθ −= = ° ..................................................................................................Ans.


492

10-82 Given that 900 m/mxε µ= + , 333 m/myε µ= − , and 982 radxyγ µ= + at a point in a body




900 m/mxε µ= +

333 m/myε µ= −

982 radxyγ µ= +

( )900 333

283.5 m/m2

a µ+ −

= =

( ) ( )2 2616.5 491 788.1 m/mR µ= + =

1 283.5 788.1 1071.6 m/m 1072 m/mpε µ µ= + = + ≅ + ................................................Ans.

2 283.5 788.1 504.6 m/m 505 m/mpε µ µ= − = − ≅ − .....................................................Ans.

3 0 m/mpε µ= ............... ( )max 2 788.1 1576 radpγ γ µ= = ≅ ..........................................Ans.

11 491tan 19.272 616.5pθ −= = ° ................................................................................................Ans.

10-83 Given that 750 in./in.xε µ= − , 390 in./in.yε µ= − , and 900 radxyγ µ= + at a point in a body




750 in./in.xε µ= − 390 in./in.yε µ= − 900 radxyγ µ= +

( ) ( )750 390

570 in./in.2

a µ− + −

= = − 2 2180 450 484.7 in./in.R µ= + =

1 570 484.7 85.3 in./in.pε µ= − + = − ..................................................................................Ans.

2 570 484.7 1054.7 in./in. 1055 in./in.pε µ µ= − − = − ≅ − ............................................Ans.

3 0 in./in.pε µ= ............ ( )2 484.7 969 radpγ µ= ≅ ........................................................Ans.


493

( ) ( )max max min 0 1055 1055 radγ ε ε µ= − = − − = .........................................................Ans.

11 450tan 34.102 180pθ − −= = − ° ..............................................................................................Ans.

10-84 Given that 600 m/mxε µ= + , 480 m/myε µ= + , and 480 radxyγ µ= − at a point in a body




600 m/mxε µ= + 480 m/myε µ= + 480 radxyγ µ= −

600 480 540 m/m

2a µ+= =

2 260 240 247.4 m/mR µ= + =

1 540 247.4 787.4 m/m 787 m/mpε µ µ= + = + ≅ + ........................................................Ans.

2 540 247.4 292.6 m/m 293 m/mpε µ µ= − = + ≅ + ........................................................Ans.

3 0 m/mpε µ= ............................. ( )2 247.4 495 radpγ µ= ≅ ..........................................Ans.


11 480tan 37.982 120pθ − −= = − ° ..............................................................................................Ans.


494

10-85 Given that 680 in./in.xε µ= − , 320 in./in.yε µ= + , and 1 414 in./in.pε µ= + at a point in a body

subjected to plane strain, use Mohr�s circle to determine xyγ , 2pε , pγ , maxγ , and pθ , and prepare a



680 in./in.xε µ= − 320 in./in.yε µ= + 1 414 in./in.pε µ= +

( )680 320

180 in./in.2

a µ− +

= = − ( )1 414 180 594 in./in.pR aε µ= − = − − =

( )222xy xR aγ ε= − −

( ) 222 594 320 180 641 in./in.µ= − − − = ± .........................................................Ans.

2 180 594 774 in./in.pε µ= − − = − ......................................................................................Ans.

3 0 in./in.pε µ= .......................... ( )max 2 594 1188 radpγ γ µ= = ≅ ...............................Ans.

11 321tan 16.352 500pθ − ±= = ± ° ..............................................................................................Ans.

10-86 Given that 450 m/mxε µ= + , 150 m/myε µ= + , and 1 780 m/mpε µ= + at a point in a body




450 m/mxε µ= + 150 m/myε µ= + 1 780 m/mpε µ= +

450 150 300 m/m

2a µ+= = 1 780 300 480 m/mpR aε µ= − = − =

( ) ( )2 22 22 2 480 450 300xy xR aγ ε= − − = − −

911.9 m/m 912 m/mµ µ= ± ≅ ± ..................................................................................Ans.


495

2 300 480 180 m/mpε µ= − = − ...........................................................................................Ans.

3 0 m/mpε µ= ............................. ( )max 2 480 960 radpγ γ µ= = ≅ .................................Ans.

11 456tan 35.892 150pθ − ±= = ± ° ..............................................................................................Ans.

10-87 Given that 360 in./in.xε µ= + , 750 in./in.yε µ= + , and 2 197 in./in.pε µ= + at a point in a body




360 in./in.xε µ= + 750 in./in.yε µ= +

2 197 in./in.pε µ= +

360 750 555 in./in.

2a µ+= = +

2 555 197 358 in./in.pR a ε µ= − = − =

( ) ( )2 22 22 2 358 750 555xy yR aγ ε= − − = − −

600 in./in.µ= ± .......................................................Ans.

1 555 358 913 in./in.pε µ= + = + .................................Ans.

3 0 in./in.pε µ= ..............................................................Ans.

( )2 2 358 716 radp Rγ µ= = = ...................................Ans.

( )max max min 913 0 913 radγ ε ε µ= − = − = ...............Ans.

11 300tan 28.492 195pθ − ±= = ± ° .....................................Ans.


496

10-88 Given that 300 m/mxε µ= − , 600 m/myε µ= + , and 2 522 m/mpε µ= − at a point in a body




300 m/mxε µ= − 600 m/myε µ= + 2 522 m/mpε µ= −

( )300 600

150 m/m2

a µ− +

= = ( )2 150 522 672 m/mpR a ε µ= − = − − =

( ) ( )2 22 22 2 672 600 150xy yR aγ ε= − − = − −

998.2 m/m 998 m/mµ µ= ± ≅ ± ..................................................................................Ans.

1 150 672 822 m/mpε µ= + = + ............................................................................................Ans.

3 0 m/mpε µ= ............................. ( )max 2 672 1344 radpγ γ µ= = = ..............................Ans.

11 499tan 23.982 450pθ − ±= = ± ° ..............................................................................................Ans.

10-89 At a point on the surface of an aluminum alloy (E = 10,000 ksi and G = 3800 ksi) machine part subjected to a biaxial state of stress, the measured strains were 900 in./in.xε µ= + , 300 in./in.yε µ= − , and

400 radxyγ µ= − . Determine the stresses xσ , yσ , and xyτ at the point.


900 in./in.xε µ= + 300 in./in.yε µ= − 400 radxyγ µ= −

3800 ksiG = ( )10,000 ksi 2 1E Gν= = + 0.3158ν =

( ) ( ) ( )62 2

10,000 900 0.3158 300 101 1 0.3158x x yEσ ε νεν

−= + = + − − −

8.94 ksi 8.94 ksi (T)= + = .............................................................................................Ans.


497

( ) ( ) ( ) ( )62 2

10,000 300 0.3158 900 101 1 0.3158y y xEσ ε νεν

−= + = − + − −

0.1753 ksi 0.1753 ksi (C)= − = ....................................................................................Ans.

( )63800 400 10 1.520 ksixy xyGτ γ −= = − × = − ................................................................Ans.

10-90 At a point on the surface of an structural steel (E = 200 GPa and G = 76 GPa) machine part subjected to a biaxial state of stress, the measured strains were 750 m/mxε µ= + , 350 m/myε µ= + , and

560 radxyγ µ= − . Determine the stresses xσ , yσ , and xyτ at the point.



76 GPaG = ( )200 GPa 2 1E Gν= = + 0.3158ν =

( ) ( ) ( )9

62 2

200 10 750 0.3158 350 101 1 0.3158x x yEσ ε νεν

−×= + = + − −

2191.2 N/m 191.2 MPa (T)= + = .................................................................................Ans.

( ) ( ) ( )9

62 2

200 10 350 0.3158 750 101 1 0.3158y y xEσ ε νεν

−×= + = + − −

2130.4 N/m 130.4 MPa (T)= + = .................................................................................Ans.

( )( )9 6 276 10 560 10 42.6 N/m 42.6 MPaxy xyGτ γ −= = × − × = − = − ..........................Ans.

10-91 At a point on the surface of a titanium alloy (E = 14,000 ksi and G = 5300 ksi) machine part subjected to a biaxial state of stress, the measured strains were 1250 in./in.xε µ= + , 600 in./in.yε µ= + , and

650 radxyγ µ= + . Determine the stresses xσ , yσ , and xyτ at the point.


1250 in./in.xε µ= + 600 in./in.yε µ= + 650 radxyγ µ= +

5300 ksiG = ( )14,000 ksi 2 1E Gν= = + 0.3208ν =

( ) ( ) ( )62 2

14,000 1250 0.3208 600 101 1 0.3208x x yEσ ε νεν

−= + = + − −

22.5 ksi 22.5 ksi (T)= + = .............................................................................................Ans.

( ) ( ) ( )62 2

14,000 600 0.3208 1250 101 1 0.3208y y xEσ ε νεν

−= + = + − −

15.62 ksi 15.62 ksi (T)= + = .........................................................................................Ans.

( )65300 650 10 3.45 ksixy xyGτ γ −= = × = + .....................................................................Ans.


498

10-92 At a point on the surface of a stainless steel (E = 190 GPa and G = 76 GPa) machine part subjected to a biaxial state of stress, the measured strains were 1175 m/mxε µ= + , 1250 m/myε µ= − , and

850 radxyγ µ= + . Determine the stresses xσ , yσ , and xyτ at the point.



76 GPaG = ( )190 GPa 2 1E Gν= = + 0.250ν =

( ) ( ) ( )9

62 2

190 10 1175 0.250 1250 101 1 0.250x x yEσ ε νεν

−×= + = + − − −

2174.8 N/m 174.8 MPa (T)= + = .................................................................................Ans.

( ) ( ) ( ) ( )9

62 2

190 10 1250 0.250 1175 101 1 0.250y y xEσ ε νεν

−×= + = − + − −

2193.8 N/m 193.8 MPa (C)= − = .................................................................................Ans.

( )( )9 6 276 10 850 10 64.6 N/m 64.6 MPaxy xyGτ γ −= = × × = + = + .............................Ans.

10-93 Determine the state of strain that corresponds to the following state of stress at a point in a steel (E = 30,000 ksi and 0.30ν = ) machine part: 15,000 psixσ = , 5000 psiyσ = , 7500 psizσ = ,

5500 psixyτ = , 4750 psiyzτ = , and 3200 psizxτ = .


15.0 ksixσ = 5.0 ksiyσ = 7.5 ksizσ = 30,000 ksiE =

5.5 ksixyτ = 4.75 ksiyzτ = 3.2 ksizxτ = 0.30ν =

( )1x x y zE

ε σ ν σ σ = − + ( )2 11

xy xy xyG Eν

γ τ τ+

= =

( ) 61 15.0 0.30 5.0 7.5 375 10 375 in./in.30,000xε µ−= − + = + × = + ........................Ans.

( ) 61 5.0 0.30 7.5 15.0 58.3 10 58.3 in./in.30,000yε µ−= − + = − × = − .....................Ans.

( ) 61 7.5 0.30 15.0 5.0 50.0 10 50.0 in./in.30,000zε µ−= − + = + × = + .....................Ans.

( ) ( ) 62 1 0.30

5.5 477 10 477 rad30,000xyγ µ−+

= = + × = + .......................................................Ans.

( ) ( ) 62 1 0.30

4.75 412 10 412 rad30,000yzγ µ−+

= = + × = + ....................................................Ans.

( ) ( ) 62 1 0.30

3.20 277 10 277 rad30,000zxγ µ−+

= = + × ≅ + ....................................................Ans.


499

10-94 Determine the state of strain that corresponds to the following state of stress at a point in an aluminum alloy (E = 73 GPa and 0.33ν = ) machine part: 120 MPaxσ = , 85 MPayσ = − , 45 MPazσ = ,

35 MPaxyτ = , 48 MPayzτ = , and 76 MPazxτ = .


120 MPaxσ = 85 MPayσ = − 45 MPazσ = 73 GPaE =

35 MPaxyτ = 48 MPayzτ = 76 MPazxτ = 0.33ν =

( )1x x y zE

ε σ ν σ σ = − + ( )2 11

xy xy xyG Eν

γ τ τ+

= =

( ) 61 120 0.33 85 45 1825 10 1825 m/m73,000xε µ−= − − + = + × = + ......................Ans.

( ) 61 85 0.33 120 45 1910 10 1910 m/m73,000yε µ−= − − + = − × = − ......................Ans.

( ) 61 45 0.33 120 85 458 10 458 m/m73,000zε µ−= − − = + × = + ..............................Ans.

( ) ( ) 62 1 0.33

35 1275 10 1275 rad73,000xyγ µ−+

= = + × = + ...................................................Ans.

( ) ( ) 62 1 0.33

48 1749 10 1749 rad73,000yzγ µ−+

= = + × = + ...................................................Ans.

( ) ( ) 62 1 0.33

76 2769 10 2770 rad73,000zxγ µ−+

= = + × ≅ + ...................................................Ans.

10-95 Given that 900 in./in.xε µ= + , 300 in./in.yε µ= − , and 400 radxyγ µ= − at a point on the free

surface of a machine component ( 10,000 ksiE = and 0.30ν = ), determine the principal stresses and the maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch.


900 in./in.xε µ= + 300 in./in.yε µ= − 400 radxyγ µ= −

0.30ν = ( )10,000 ksi 2 1E Gν= = + 3846 ksiG =

( ) ( ) ( )62 2

10,000 900 0.30 300 10 8.901 ksi1 1 0.30x x yEσ ε νεν

−= + = + − = + − −

( ) ( ) ( ) ( )62 2

10,000 300 0.30 900 10 0.3297 ksi1 1 0.30y y xEσ ε νεν

−= + = − + = − − −

( )63846 400 10 1.5384 ksixy xyGτ γ −= = − × = −

2 2

2 21, 2

8.901 0.3297 8.901 0.3297 1.53842 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − − + = ± + = ± +


500

1 4.286 4.865 9.151 ksi 9.15 ksi (T)pσ = + = + ≅ ...........................................................Ans.

2 4.286 4.865 0.579 ksi 0.579 ksi (C)pσ = − = − ≅ .......................................................Ans.

3 0 ksipσ = ............................................................................................................................... Ans.

max 4.865 ksi 4.87 ksipτ τ= = ≅ ........................................................................................Ans.

( )

( )1 12 2 1.53841 1tan tan 9.22

2 2 8.901 0.3297xy

px y

τθ

σ σ− − −

= = = − °− − −

.......................................Ans.

10-96 Given that 750 m/mxε µ= + , 350 m/myε µ= + , and 560 radxyγ µ= − at a point on the free

surface of a machine component ( 200 GPaE = and 0.30ν = ), determine the principal stresses and the maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch.



0.30ν = ( )200 GPa 2 1E Gν= = + 76.92 GPaG =

( ) ( ) ( )9

62 2

200 10 750 0.30 350 101 1 0.30x x yEσ ε νεν

−×= + = + − −

6 2187.91 10 N/m 187.91 MPa (T)= + × =

( ) ( ) ( )9

62 2

200 10 350 0.30 750 101 1 0.30y y xEσ ε νεν

−×= + = + − −

6 2126.37 10 N/m 126.37 MPa (T)= + × =

( )( )9 6 6 276.92 10 560 10 43.08 10 N/m 43.08 MPaxy xyGτ γ −= = × − × = − × = −

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

2187.91 126.37 187.91 126.37 43.082 2+ − = ± +

1 157.14 52.94 210.08 MPa 210 MPa (T)pσ = + = + ≅ ................................................Ans.

2 157.14 52.94 104.20 MPa 104.2 MPa (T)pσ = − = + ≅ .............................................Ans.

3 0 ksipσ = ............................................................................................................................... Ans.

52.94 MPa 52.9 MPapτ = ≅ ..............................................................................................Ans.

max minmax

210.08 0 105.0 MPa 105.0 MPa2 2

σ στ − −= = = ≅ ......................................Ans.

( )1 12 2 43.081 1tan tan 27.25

2 2 187.91 126.37xy

px y

τθ

σ σ− − −

= = = − °− −

.........................................Ans.


501

10-97 Given that 1250 in./in.xε µ= + , 600 in./in.yε µ= + , and 650 radxyγ µ= + at a point on the free

surface of a machine component ( 14,000 ksiE = and 0.32ν = ), determine the principal stresses and the maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch.


1250 in./in.xε µ= + 600 in./in.yε µ= + 650 radxyγ µ= +

0.32ν = ( )14,000 ksi 2 1E Gν= = + 5303 ksiG =

( ) ( ) ( )62 2

14,000 1250 0.32 600 10 22.49 ksi1 1 0.32x x yEσ ε νεν

−= + = + = + − −

( ) ( ) ( )62 2

14,000 600 0.32 1250 10 15.597 ksi1 1 0.32y y xEσ ε νεν

−= + = + = + − −

( )65303 650 10 3.447 ksixy xyGτ γ −= = × = +

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

222.49 15.597 22.49 15.597 3.4472 2+ − = ± +

1 19.044 4.874 23.92 ksi 23.9 ksi (T)pσ = + = + ≅ ........................................................Ans.

2 19.044 4.874 14.170 ksi 14.17 ksi (T)pσ = − = + ≅ ...................................................Ans.

3 0 ksipσ = .................................. 4.874 ksi 4.87 ksipτ = ≅ ...........................................Ans.

max minmax

23.92 0 11.96 ksi2 2

σ στ − −= = = .....................................................................Ans.

( )1 12 2 3.4471 1tan tan 22.50

2 2 22.49 15.597xy

px y

τθ

σ σ− −= = = °

− −.............................................Ans.

10-98 At a point on the free surface of an aluminum alloy (E = 73 GPa and 0.33ν = ) machine part, the strain rosette shown in Fig. P10-98 was used to obtain the following normal strain data: 780 m/maε µ= + ,

345 m/mbε µ= + , and 332 m/mcε µ= − . Determine

(a) The strain components xε , yε , and xyγ at the point.

(b) The principal strains and the maximum shearing strain at the point. SOLUTION The given values are:

780 m/ma xε ε µ= = + 60 345 m/mbε ε µ−= = + 60 332 m/mcε ε µ+= = −

(a) 2 2cos sin sin cosn x n y n xy n nε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( )2 2780cos 60 sin 60 sin 60 cos 60 345 m/mb y xyε ε γ µ= − ° + − ° + − ° − ° = +

( ) ( ) ( ) ( )2 2780cos 60 sin 60 sin 60 cos 60 332 m/mc y xyε ε γ µ= ° + ° + ° ° = −


502

Therefore 780 m/mxε µ= + ......... 251 m/myε µ= − ......... 782 radxyγ µ= − ........................Ans.

(b) 2 2

1, 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − = ± +

( ) ( ) 2 2780 251 780 251 7822 2 2

+ − − − − = ± +

1 264.5 647 911.5 m/m 912 m/mpε µ µ= + = + ≅ + .........................................................Ans.

2 264.5 647 382.5 m/m 383 m/mpε µ µ= − = − ≅ − ........................................................Ans.

( ) ( )30.33 780 251 261 m/m

1 1 0.33p z x yνε ε ε ε µ

ν= = − + = − − = −

− −...........................Ans.

( )max 2 647 1294 radpγ γ µ= = = .......................................................................................Ans.

( )

1 11 1 782tan tan 18.592 2 780 251

xyp

x y

γθ

ε ε− − −= = = − °

− − −................................................Ans.

10-99 At a point on the free surface of a steel (E = 30,000 ksi and 0.30ν = ) machine part, the strain rosette shown in Fig. P10-99 was used to obtain the following normal strain data: 750 in./in.aε µ= + ,

125 in./in.bε µ= − , and 250 in./in.cε µ= − . Determine



750 in./in.a xε ε µ= = + 45 125 in./in.bε ε µ+= = − 250 in./in.c yε ε µ= = −


( ) ( ) ( ) ( ) ( )2 2750cos 45 250 sin 45 sin 45 cos 45 125 in./in.b xyε γ µ= ° + − ° + ° ° = −

Therefore 750 in./in.xε µ= + ....... 250 in./in.yε µ= − ....... 750 radxyγ µ= − ........................Ans.

(b) 2 2 2 2

1, 2750 250 750 250 750

2 2 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − − + − = ± + = ± +

1 250 625 875 in./in.pε µ= + = + ..........................................................................................Ans.

2 250 625 375 in./in.pε µ= − = − .........................................................................................Ans.

( ) ( )30.30 750 250 214 in./in.


ν= = − + = − − = −

− −........................Ans.

( )max 2 625 1250 radpγ γ µ= = = ........................................................................................Ans.

( )

1 11 1 750tan tan 18.432 2 750 250

xyp

x y

γθ

ε ε− − −= = = − °

− − −...............................................Ans.


503

10-100 At a point on the free surface of a steel (E = 200 GPa and 0.30ν = ) machine part, the strain rosette shown in Fig. P10-100 was used to obtain the following normal strain data: 555 m/maε µ= − ,

925 m/mbε µ= + , and 740 m/mcε µ= + . Determine



555 m/ma xε ε µ= = − 60 925 m/mbε ε µ−= = + 60 740 m/mcε ε µ= = +


( ) ( ) ( ) ( )2 2555cos 60 sin 60 sin 60 cos 60 925 m/mb y xyε ε γ µ= − − ° + − ° + − ° − ° = +

( ) ( ) ( ) ( )2 2555cos 60 sin 60 sin 60 cos 60 740 m/mc y xyε ε γ µ= − ° + ° + ° ° = +

Therefore 555 m/mxε µ= − ......... 1295 m/myε µ= + ...... 214 radxyγ µ= − ........................Ans.

(b) 2 2

1, 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − = ± +

( ) ( ) 2 2555 1295 555 1295 214

2 2 2− + − − − = ± +

1 370 931 1301 m/mpε µ= + = + ..........................................................................................Ans.

2 370 931 561 m/mpε µ= − = − ............................................................................................Ans.

( ) ( )30.30 555 1295 317 m/m


ν= = − + = − − + = −

− −......................Ans.

( )max 2 931 1862 radpγ γ µ= = = ........................................................................................Ans.

1 11 1 214tan tan 3.302 2 555 1295

xyp

x y

γθ

ε ε− − −= = = °

− − −......................................................Ans.

10-101 The strain rosette shown in Fig. P10-101 was used to obtain normal strain data at a point on the free surface of a structural steel machine part. The gage readings were 350 in./in.aε µ= − , 1000 in./in.bε µ= + ,

and 550 in./in.cε µ= + . Determine


(b) The stress components xσ , yσ , and xyτ at the point.

(c) The principal stresses and the maximum shearing stress at the point. Express the stresses in U.S. customary units.


350 in./in.a xε ε µ= = − 60 1000 in./in.bε ε µ= = + 60 550 in./in.cε ε µ−= = +



504

( ) ( ) ( ) ( )2 2350cos 60 sin 60 sin 60 cos 60 1000 in./in.b y xyε ε γ µ= − ° + ° + ° ° = +

( ) ( ) ( ) ( )2 2350cos 60 sin 60 sin 60 cos 60 550 in./in.c y xyε ε γ µ= − − ° + − ° + − ° − ° = +

Therefore 350 in./in.xε µ= − ....... 1150 in./in.yε µ= + .... 520 radxyγ µ= + ........................Ans.

(b) For structural steel: 11,000 ksiG = ( )29,000 ksi 2 1E Gν= = + 0.3182ν =

( ) ( ) ( )62 2

29,000 350 0.3182 1150 101 1 0.3182x x yEσ ε νεν

−= + = − + − −

0.514 ksi 0.514 ksi (T)= + = .........................................................................................Ans.

( ) ( ) ( )62 2

29,000 1150 0.3182 350 101 1 0.3182y y xEσ ε νεν

−= + = + − − −

33.514 ksi 33.5 ksi (T)= + ≅ .........................................................................................Ans.

( )611,000 520 10 5.72 ksixy xyGτ γ −= = × = + ..................................................................Ans.

(c) 2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

20.514 33.514 0.514 33.514 5.722 2

+ − = ± +

1 17.012 17.463 34.475 ksi 34.5 ksi (T)pσ = + = + ≅ .....................................................Ans.

2 17.012 17.463 0.451 ksi 0.451 ksi (C)pσ = − = − ≅ ....................................................Ans.

3 0 ksip zσ σ= = ........................ max 17.46 ksipτ τ= = ....................................................Ans.

( )1 12 2 5.721 1tan tan 9.56

2 2 0.514 33.514xy

px y

τθ

σ σ− −= = = − °

− −.............................................Ans.

10-102 The strain rosette shown in Fig. P10-102 was used to obtain normal strain data at a point on the free surface of a 2024-T4 aluminum alloy machine part. The gage readings were 525 m/maε µ= + ,




(c) The principal stresses and the maximum shearing stress at the point. Express the stresses in SI units. SOLUTION The given values are:

525 m/ma xε ε µ= = + 45 450 m/mbε ε µ= = + 45 1425 m/mcε ε µ−= = +


( ) ( ) ( ) ( )2 2525cos 45 sin 45 sin 45 cos 45 450 m/mb y xyε ε γ µ= ° + ° + ° ° = +

( ) ( ) ( ) ( )2 2525cos 45 sin 45 sin 45 cos 45 1425 m/mc y xyε ε γ µ= − ° + − ° + − ° − ° = +

Therefore 525 m/mxε µ= + ......... 1350 m/myε µ= + ...... 975 radxyγ µ= − ........................Ans.


505

(b) For 2024-T4 aluminum alloy: 28 GPaG = ( )73 GPa 2 1E Gν= = + 0.3036ν =

( ) ( ) ( )9

62 2

73 10 525 0.3036 1350 101 1 0.3036x x yEσ ε νεν

−×= + = + − −

6 275.17 10 N/m 75.2 MPa (T)= + × ≅ ........................................................................Ans.

( ) ( ) ( )9

62 2

73 10 1350 0.3036 525 101 1 0.3036y x yEσ ε νεν

−×= + = + − −

6 2121.37 10 N/m 121.37 MPa (T)= + × ≅ .................................................................Ans.

( )( )9 6 6 228 10 975 10 27.30 10 N/m 27.3 MPaxy xyGτ γ −= = × − × = − × ≅ − ..............Ans.

(c) ( )2 2

221, 2

75.17 121.37 75.17 121.37 27.302 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − + − = ± + = ± +

1 98.27 35.76 134.03 MPa 134.0 MPa (T)pσ = + = + ≅ ...............................................Ans.

2 98.27 35.76 62.51 MPa 62.5 MPa (T)pσ = − = + ≅ ...................................................Ans.

3 0 MPap zσ σ= = .................... 35.76 MPa 35.8 MPapτ = ≅ ....................................Ans.

( ) ( )max max min1 1 134.03 0 67.0 MPa2 2

τ σ σ= − = − ≅ .....................................................Ans.

( )1 12 2 27.301 1tan tan 24.88

2 2 75.17 121.37xy

px y

τθ

σ σ− − −

= = = °− −

.............................................Ans.

10-103 The strain rosette shown in Fig. P10-103 was used to obtain normal strain data at a point on the free surface of an 18-8 cold-rolled stainless steel machine part. The gage readings were 665 in./in.aε µ= + ,

390 in./in.bε µ= + , and 870 in./in.cε µ= + . Determine



(c) The principal stresses and the maximum shearing stress at the point. Express the stresses in U.S. customary units.


665 in./in.a xε ε µ= = + 390 in./in.b yε ε µ= = + 60 870 in./in.cε ε µ−= = +


( ) ( ) ( ) ( )2 2665cos 60 390sin 60 sin 60 cos 60 870 in./in.c xyε γ µ= − ° + − ° + − ° − ° = +

Therefore 665 in./in.xε µ= + ....... 390 in./in.yε µ= + ....... 950 radxyγ µ= − ........................Ans.

(b) For 18-8 stainless steel: 12,500 ksiG = ( )28,000 ksi 2 1E Gν= = + 0.12ν =

( ) ( ) ( )62 2

28,000 665 0.12 390 101 1 0.12x x yEσ ε νεν

−= + = + − −

20.22 ksi 20.2 ksi (T)= + ≅ ...........................................................................................Ans.


506

( ) ( ) ( )62 2

28,000 390 0.12 665 101 1 0.12y y xEσ ε νεν

−= + = + − −

13.347 ksi 13.35 ksi (T)= + ≅ .......................................................................................Ans.

( )612,500 950 10 11.875 ksi 11.88 ksixy xyGτ γ −= = − × = − ≅ − .................................Ans.

(c) 2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

220.22 13.347 20.22 13.347 11.8752 2+ − = ± +

1 16.784 12.362 29.146 ksi 29.1 ksi (T)pσ = + = + ≅ ....................................................Ans.

2 16.784 12.362 4.422 ksi 4.42 ksi (T)pσ = − = + ≅ ......................................................Ans.

3 0 ksip zσ σ= = ........................ 12.362 ksi 12.36 ksipτ = ≅ ........................................Ans.

( ) ( )max max min1 1 29.146 0 14.57 ksi2 2

τ σ σ= − = − ≅ ......................................................Ans.

( )1 12 2 11.8751 1tan tan 36.93

2 2 20.22 13.347xy

px y

τθ

σ σ− − −

= = = − °− −

...........................................Ans.

10-104 At a point on the free surface of an aluminum alloy (E = 73 GPa and 0.33ν = ) machine part, the strain rosette shown in Fig. P10-104 was used to obtain the following normal strain data: 875 m/maε µ= + ,

700 m/mbε µ= + , and 650 m/mcε µ= − . Determine

(a) The stress components xσ , yσ , and xyτ at the point.

(b) The principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains.

(c) The principal stresses and the maximum shearing stress at the point. Prepare a sketch showing all of these stresses.


875 m/ma xε ε µ= = + 60 700 m/mbε ε µ−= = + 60 650 m/mcε ε µ= = −


( ) ( ) ( ) ( )2 2875cos 60 sin 60 sin 60 cos 60 700 m/mb y xyε ε γ µ= − ° + − ° + − ° − ° = +

( ) ( ) ( ) ( )2 2875cos 60 sin 60 sin 60 cos 60 650 m/mc y xyε ε γ µ= ° + ° + ° ° = −

Therefore 875 m/mxε µ= + 258 m/myε µ= − 1559 radxyγ µ= −

For aluminum alloy: 0.33ν = ( )73 GPa 2 1E Gν= = + 27.44 GPaG =

( ) ( ) ( )9

62 2

73 10 875 0.33 258 101 1 0.33x x yEσ ε νεν

−×= + = + − − −


507

6 264.71 10 N/m 64.7 MPa (T)= + × ≅ .........................................................................Ans.

( ) ( ) ( )9

62 2

73 10 258 0.33 875 101 1 0.33y x yEσ ε νεν

−×= + = − + − −

6 22.519 10 N/m 2.52 MPa (T)= + × ≅ ........................................................................Ans.

( )( )9 627.44 10 1559 10xy xyGτ γ −= = × − ×

6 242.78 10 N/m 42.8 MPa= − × ≅ − ...........................................................................Ans.

(b) 2 2

1, 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − = ± +

( ) ( ) 2 2875 258 875 258 15592 2 2

+ − − − − = ± +

1 308.5 963.6 1272.1 m/m 1272 m/mpε µ µ= + = + ≅ + .................................................Ans.

2 308.5 963.6 655.1 m/m 655 m/mpε µ µ= − = − ≅ − .....................................................Ans.

( ) ( )30.33 875 258 304 m/m


ν= = − + = − − = −

− −...........................Ans.

( )max 2 963.6 1927.2 rad 1927 radpγ γ µ µ= = = ≅ ........................................................Ans.

( )

1 11 1 1559tan tan 27.002 2 875 258

xyp

x y

γθ

ε ε− − −= = = − °

− − −................................................Ans.

(c) ( ) ( ) ( )9

61 1 22 2

73 10 1272.1 0.33 655.1 101 1 0.33p p pEσ ε νεν

−×= + = + − − −

6 286.50 10 N/m 86.5 MPa (T)= + × ≅ .......................................................................Ans.

( ) ( ) ( )9

62 1 22 2

73 10 655.1 0.33 1272.1 101 1 0.33p p pEσ ε νεν

−×= + = − + − −

6 219.277 10 N/m 19.28 MPa (C)= − × ≅ ..................................................................Ans.

3 0 MPap zσ σ= = .................................................................................................................Ans.

( )( )9 6max max 27.44 10 1927.2 10Gτ γ −= = × ×

6 252.89 10 N/m 52.9 MPa= × ≅ ...............................................................................Ans.

10-105 At a point on the free surface of an aluminum alloy (E = 10,600ksi and 0.33ν = ) machine part, the strain rosette shown in Fig. P10-105 was used to obtain the following normal strain data: 800 in./in.aε µ= + ,

950 in./in.bε µ= + , and 600 in./in.cε µ= + . Determine





508


800 in./in.a xε ε µ= = + 36.870 950 in./in.bε ε µ= = + 600 in./in.c yε ε µ= = +


( ) ( ) ( )( )2 2800 4 5 600 3 5 3 5 4 5 950 in./in.b xyε γ µ= + + = +

Therefore 800 in./in.xε µ= + 600 in./in.yε µ= + 462.5 radxyγ µ= +

For aluminum alloy: 0.33ν = ( )10,600 ksi 2 1E Gν= = + 3985 ksiG =

( ) ( ) ( )62 2

10,600 800 0.33 600 101 1 0.33x x yEσ ε νεν

−= + = + − −

11.872 ksi 11.87 ksi (T)= + ≅ .......................................................................................Ans.

( ) ( ) ( )62 2

10,600 600 0.33 800 101 1 0.33y y xEσ ε νεν

−= + = + − −

10.278 ksi 10.28 ksi (T)= + ≅ .......................................................................................Ans.

( )63985 462.5 10 1.8431 ksi 1.843 ksixy xyGτ γ −= = × = ≅ ..........................................Ans.

(b) 2 2 2 2

1, 2800 600 800 600 462.5

2 2 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − + − = ± + = ± +

1 700 252 952 in./in.pε µ= + = + .........................................................................................Ans.

2 700 252 448 in./in.pε µ= − = + .........................................................................................Ans.

( ) ( )30.33 800 600 690 in./in.


ν= = − + = − + = −

− −.........................Ans.

( )2 252 504 radpγ µ= = ......................................................................................................Ans.

( ) ( )max max min 952 690 1642 radγ ε ε µ= − = − − = ...........................................................Ans.

1 11 1 462.5tan tan 33.312 2 800 600

xyp

x y

γθ

ε ε− −= = = °

− −.........................................................Ans.

(c) ( ) ( ) ( )61 1 22 2

10,600 952 0.33 448 101 1 0.33p p pEσ ε νεν

−= + = + − −

13.083 ksi 13.08 ksi (T)= + ≅ .......................................................................................Ans.

( ) ( ) ( )62 2 12 2

10,600 448 0.33 952 101 1 0.33p p pEσ ε νεν

−= + = + − −

9.066 ksi 9.07 ksi (T)= + ≅ ...........................................................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.

( )63985 504 10 2.008 ksi 2.01 ksip pGτ γ −= = × = ≅ ...................................................Ans.

( )6max max 3985 1642 10 6.543 ksi 6.54 ksiGτ γ −= = × = ≅ ...........................................Ans.


509

10-106 At a point on the free surface of an steel (E = 200 GPa and 0.30ν = ) machine part, the strain rosette shown in Fig. P10-106 was used to obtain the following normal strain data: 875 m/maε µ= + ,






875 m/ma xε ε µ= = + 53.130 700 m/mbε ε µ= = + 350 m/mc yε ε µ= = +


( ) ( ) ( )( )2 2875 3 5 350 4 5 4 5 3 5 700 m/mb xyε γ µ= + + = +

Therefore 875 m/mxε µ= + 350 m/myε µ= + 335.4 radxyγ µ= +

For steel: 0.30ν = ( )200 GPa 2 1E Gν= = + 76.92 GPaG =

( ) ( ) ( )9

62 2

200 10 875 0.30 350 101 1 0.30x x yEσ ε νεν

−×= + = + − −

6 2215.4 10 N/m 215 MPa (T)= + × ≅ ..........................................................................Ans.

( ) ( ) ( )9

62 2

200 10 350 0.30 875 101 1 0.30y x yEσ ε νεν

−×= + = + − −

6 2134.62 10 N/m 134.6 MPa (T)= + × ≅ ....................................................................Ans.

( )( )9 676.92 10 335.4 10xy xyGτ γ −= = × ×

6 225.80 10 N/m 25.8 MPa= + × ≅ + ...........................................................................Ans.

(b) 2 2 2 2

1, 2875 350 875 350 335.4

2 2 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − + − = ± + = ± +

1 612.5 311.5 924 m/mpε µ= + = + ....................................................................................Ans.

2 612.5 311.5 301 m/mpε µ= − = + ....................................................................................Ans.

( ) ( )30.30 875 350 525 m/m


ν= = − + = − + = −

− −...........................Ans.

( )2 311.5 623 radpγ µ= = ...................................................................................................Ans.

( ) ( )max max min 924 525 1449 radγ ε ε µ= − = − − = ...........................................................Ans.

1 11 1 335.4tan tan 16.292 2 875 350

xyp

x y

γθ

ε ε− −= = = °

− −.........................................................Ans.


510

(c) ( ) ( ) ( )9

61 1 22 2

200 10 924 0.30 301 101 1 0.30p p pEσ ε νεν

−×= + = + − −

6 2222.9 10 N/m 223 MPa (T)= + × ≅ ........................................................................Ans.

( ) ( ) ( )9

62 1 22 2

200 10 301 0.30 924 101 1 0.30p p pEσ ε νεν

−×= + = + − −

6 2127.08 10 N/m 127.1 MPa (T)= + × ≅ .....................................................................Ans.

3 0 MPap zσ σ= = .................................................................................................................Ans.

( )( )9 6max max 76.92 10 623 10Gτ γ −= = × ×

6 247.92 10 N/m 47.9 MPa= × ≅ ................................................................................Ans.

( )( )9 6max max 76.92 10 1449 10Gτ γ −= = × ×

6 2111.46 10 N/m 111.5 MPa= × ≅ .............................................................................Ans.

10-107 A spherical gas storage tank, similar to the one shown in Fig. 10-33, has a diameter of 35 ft and a wall thickness of 7/8 in. Determine the maximum normal stress in the tank if the gas pressure is 100 psi.

SOLUTION

( ) ( )

( )100 210 7 8

11,950 psi 11.95 ksi2 2 7 8prt

σ− = = = = .............................................Ans.

10-108 Determine the maximum normal stress in a 300-mm diameter basketball that has a 2-mm wall thickness after it has been inflated to a pressure of 100 kPa.

SOLUTION

( ) ( )

( )

36 2

100 10 0.1483.70 10 N/m 3.70 MPa

2 2 0.002prt

σ×

= = = × = ...................................Ans.

10-109 A steel pipe with an inside diameter of 10 in. will be used to transmit steam under a pressure of 800 psi. If the hoop stress in the pipe must be limited to 10 ksi because of a longitudinal weld in the pipe, determine the minimum satisfactory thickness for the pipe.

SOLUTION

:hprt

σ = ( ) ( )800 5

0.400 in.10,000h

prtσ

= = = ......................................................................Ans.

10-110 A cylindrical propane tank, similar to the ones shown in Fig. 10-35, has an outside diameter of 3.25 m and a wall thickness of 22 mm. If the allowable hoop stress is 100 MPa and the allowable axial stress is 45 MPa, determine the maximum internal pressure that can be applied to the tank.

SOLUTION

:2aprt

σ = ( ) ( )2 0.022 452 1.2352 MPa

1.603h

i

tprσ= = =

:hprt

σ = ( ) ( )0.022 100

1.3724 MPa1.603

h

i

tprσ= = =

Therefore: max 1.2352 MPa 1.235 MPap = ≅ ................................................................Ans.


511

10-111 A cylindrical boiler with hemispherical ends has an outside diameter of 6 ft. If the hoop and axial stresses in the boiler must be limited to 15 ksi and 8 ksi, respectively, determine the thickness of steel plate required if the internal pressure in the boiler will be 200 psi.

SOLUTION

( ) ( )200 36

8000 psi2 2a

tprt t

σ−

= = = 0.4444 in.t =

( ) ( )200 36

15,000 psih

tprt t

σ−

= = = 0.4737 in.t =

Therefore min 0.4737 in. 0.474 in.t = ≅ ..........................................................................Ans.

10-112 The cylindrical tank shown in Fig. P10-112 is 20 m in diameter, is made of structural steel, and will be used

to store stove oil ( )3850 kg/mρ = . Determine the minimum wall thickness required if the allowable

stress is 80 MPa. SOLUTION

( ) ( ) 3 2850 9.81 6 50.03 10 N/m 50.03 kPap ghρ= = = × =

( )( )3

6 250.03 10 10

80 10 N/mh

tprt t

σ× −

= = = ×

0.00625 m 6.25 mmt = = ...................................................................................................Ans.

10-113 A standpipe 12 ft in diameter and 50 ft tall is being constructed for use as a storage tank for water

( )362.4 lb/ftγ = . Determine the minimum thickness of steel plate that can be used if the hoop stress in

the standpipe must be limited to 5000 psi. SOLUTION

( ) 262.4 50 3120 lb/ft 21.67 psip hγ= = = =

( )( )21.67 72

5000 psih

tprt t

σ−

= = =

0.311 in.t = ............................................................................................................................. Ans.

10-114 A cylindrical pressure vessel is fabricated by butt-welding 20-mm plate with a spiral seam, as shown in Fig. P10-114. The pressure in the tank is 2800 kPa. Determine

(a) The normal stress perpendicular to the weld. (b) The shearing stress parallel to the weld. (c) The maximum shearing stress at a point on the outside surface of the vessel. (d) The maximum shearing stress at a point on the inside surface of the vessel. SOLUTION

( )( )

( )

32800 10 0.6002 2 0.020aprt

σ×

= =

6 242.00 10 N/m 42.00 MPa= × =

( ) ( )

( )

32800 10 0.6000.020h

prt

σ×

= =

6 284.00 10 N/m 84.00 MPa= × =


512

( )1tan 3 4 36.870θ −= = °


( ) ( )2 242.00cos 36.870 84.00sin 36.870 0 57.1 MPa= ° + ° + = ..........................Ans.


( ) ( ) ( )42.00 84.00 sin 36.870 cos 36.870 0 20.2 MPa= − − ° ° + = .......................Ans.

(c) max minmax

0 84.00 0 42.0 MPa2 2 2

hσ σ στ − − −= = = = ...................................................Ans.

(d) ( ) ( )max min

max

84.00 2.80043.4 MPa

2 2 2h pσσ στ

− − − −−= = = = .............................Ans.

10-115 A steel boiler 3 ft in diameter is welded using a spiral seam that makes an angle of 30° with respect to a transverse plane of the boiler, as shown in Fig. P10-115. For an internal pressure of 140 psi and a wall thickness of 2 in., determine

(a) The normal stress perpendicular to the weld. (b) The shearing stress parallel to the weld. (c) The maximum shearing stress in the boiler. SOLUTION

( ) ( )

( )140 18

630 psi2 2 2aprt

σ = = =

( ) ( )

( )140 18

1260 psi2h

prt

σ = = =


( ) ( )2 2630cos 30 1260sin 30 0 787 psi= − ° + − ° + = ...............................................Ans.


( ) ( ) ( )630 1260 sin 30 cos 30 0 273 psi= − − − ° − ° + = − ..........................................Ans.

(c) ( ) ( )max min

max

1260 140700 psi

2 2 2h pσσ στ

− − − −−= = = = .......................................Ans.

10-116 A spherical pressure vessel 3 m in diameter is being designed to withstand a maximum internal pressure of 500 kPa. The material being used in its construction has an allowable stress of 105 MPa, a modulus of elasticity of 210 GPa, and a Poisson�s ratio of 0.20. Determine

(a) The minimum satisfactory wall thickness. (b) The circumferential normal strain at maximum pressure when the wall thickness of part (a) is used. (c) The change in diameter of the pressure vessel at maximum pressure when the wall thickness of part (a) is

used. SOLUTION

(a) ( ) ( )0.500 1.5

105 MPa2 2

tprt t

σ−

= = =


513

33.571 10 m 3.57 mmt −= × ≅ .............................................................................................Ans.

(b) ( ) 6

3

105 0.30 105350 10 m/m 350 m/m

210 10Eσ νσε µ−−−= = = × =

×.................................Ans.

(c) ( )( )( )6 6350 10 3 3300 10 mC C Dε επ π− −∆ = = = × = ×

( )C C D D C Dπ π+ ∆ = + ∆ = + ∆

6

63300 10 1050.4 10 m 1.050 mmCDπ π

−−∆ ×∆ = = = × ≅ .............................................Ans.

10-117 A cylindrical tank, similar to the one shown in Fig. 10-35, has an outside diameter of 8 ft and a wall thickness of 3/4 in. The tank is subjected to an internal pressure of 150 psi.

(a) Determine the axial and hoop stresses in the tank. (b) Prepare a plot, similar to Fig. 4-19, showing the variation of normal and shearing stresses on planes

through a point in the wall of the tank as the angle θ, measured counterclockwise from a transverse plane, varies from 0° to 90°.

SOLUTION

(a) ( ) ( )

( )150 47.25

4725 psi 4.73 ksi2 2 3 4aprt

σ = = = ≅ ............................................................Ans.

( )( )

( )150 47.25

9450 psi 9.45 ksi3 4h

prt

σ = = = ≅ ............................................................Ans.

(b) 2 2cos sin 2 sin cosn x y xyσ σ θ σ θ τ θ θ= + +

( )2 24725cos 9450sin 0 psiθ θ= + +

( ) ( )( )

( )

2 2sin cos cos sin

4725 9450 sin cos 0

4725sin cos psi

nt x y xyτ σ σ θ θ τ θ θ

θ θθ θ

= − − + −

= − − +

=

10-118 The strains measured on the outside surface of the cylindrical pressure vessel shown in Fig. P10-118 are

1 619 m/mε µ= + and 2 330 m/mε µ= + . The angle θ = 30°, the outside diameter of the vessel is 510 mm, and the wall thickness is 3 mm. The vessel is made of 0.4 percent carbon hot-rolled steel (see Appendix A for properties). Determine

(a) The stresses 1σ and 2σ in the vessel.

(b) The internal pressure applied to the vessel. SOLUTION For 0.4 percent carbon hot-rolled steel:

80 GPaG = ( )210 GPa 2 1E Gν= = + 0.3125ν =

(a) ( ) ( )9

61 1 22 2

210 10 619 0.3125 330 101 1 0.3125Eσ ε νεν

−×= + = + × − −

6 2168.1 10 N/m 168.1 MPa= × = ................................................................................Ans.


514

( ) ( )9

62 2 12 2

210 10 330 0.3125 619 101 1 0.3125Eσ ε νεν

−×= + = + × − −

6 2121.82 10 N/m 121.8 MPa= × ≅ .............................................................................Ans.

(b) ( )( ) ( )255

42.5 MPa2 2 3a

ppr pt

σ = = =

( )( ) ( )255

85.0 MPa3h

ppr pt

σ = = =

2 22 cos sin 2 sin cosx y xyσ σ θ σ θ τ θ θ= + +

( ) ( ) ( ) ( )2 242.5 cos 30 85.0 sin 30 0 121.82 MPap p= ° + ° + =

2.29 MPap = ......................................................................................................................... Ans.

10-119 A 4-in. diameter shaft is subjected to both a torque of 30 kip⋅in. and an axial tensile load of 50 kip, as shown in Fig. P10-119. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft.

SOLUTION

( )22

2412.566 in

4 4dA

ππ= = = 50 3.979 ksi

12.566xPA

σ = = =

( )44

4425.13 in

32 32dJ

ππ= = = ( )30 2

2.388 ksi25.13xy

TcJ

τ = = =

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

23.979 0 3.979 0 2.3882 2

+ − = ± +

1 1.9895 3.1082 5.0977 ksi 5.10 ksi (T)pσ = + = + ≅ ....................................................Ans.

2 1.9895 3.1082 1.1187 ksi 1.119 ksi (C)pσ = − = − ≅ .................................................Ans.

3 0 ksip zσ σ= = ........................ max 3.1082 ksi 3.11 ksipτ τ= = ≅ ..............................Ans.

( )1 12 2 2.3881 1tan tan 25.10

2 2 3.979 0xy

px y

τθ

σ σ− −= = = °

− −.......................................................Ans.

10-120 A hollow shaft with an outside diameter of 400 mm and an inside diameter of 300 mm is subjected to both a torque of 350 kN⋅m and an axial tensile load of 1500 kN, as shown in Fig. P10-120. Determine the principal stresses and the maximum shearing stress at a point on the outside surface of the shaft.

SOLUTION

( )2 2

3 2 3 2400 300

54.98 10 mm 54.98 10 m4

Aπ

−−

= = × = ×

( )4 4

6 4 6 4400 300

1718.1 10 mm 1718.1 10 m32

Jπ

−−

= = × = ×


515

3

6 23

1500 10 27.28 10 N/m 27.28 MPa54.978 10x

PA

σ −

×= = = × =×

( ) ( )3 3

6 26

350 10 200 1040.74 10 N/m 40.74 MPa

1718.1 10xyTcJ

τ−

−

× ×= = = × =

×

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

227.28 0 27.28 0 40.742 2

+ − = ± +

1 13.64 42.96 56.60 MPa 56.6 MPa (T)pσ = + = + ≅ ....................................................Ans.

2 13.64 42.96 29.32 MPa 29.3 MPa (C)pσ = − = − ≅ ...................................................Ans.

3 0 MPap zσ σ= = .................... max 42.96 ksi 43.0 ksipτ τ= = ≅ ...............................Ans.

( )1 12 2 40.741 1tan tan 35.74

2 2 27.28 0xy

px y

τθ

σ σ− −= = = °

− −........................................................Ans.

10-121 A 2-in. diameter shaft is used in an aircraft engine to transmit 360 hp at 1500 rpm to a propeller that develops a thrust of 2800 lb. Determine the principal stresses and the maximum shearing stress produced at any point on the outside surface of the shaft.

SOLUTION

( ) ( )

( )33,000 hp 33,000 360

1260.5 lb ft2 2 1500

TNπ π

= = = ⋅

( )22

223.142 in

4 4dA

ππ= = = 2800 891.2 psi3.142x

PA

σ = = =

( )44

421.5708 in

32 32dJ

ππ= = = ( )( )1260.5 12 1

9629 psi1.5708xy

TcJ

τ×

= = =

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

2891.2 0 891.2 0 96292 2

+ − = ± +

1 445.6 9639 10,084.6 psi 10.08 ksi (T)pσ = + = + ≅ ...................................................Ans.

2 445.6 9639 9193.4 psi 9.19 ksi (C)pσ = − = − ≅ .........................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.

max 9639 psi 9.64 ksipτ τ= = ≅ ..........................................................................................Ans.

( )1 12 2 96291 1tan tan 43.68

2 2 891.2 0xy

px y

τθ

σ σ− −= = = °

− −........................................................Ans.


516

10-122 A 60-mm diameter shaft must transmit a torque of unknown magnitude while it is supporting an axial tensile load of 150 kN. Determine the maximum allowable value for the torque if the tensile principal stress at a point on the outside surface of the shaft must not exceed 125 MPa.

SOLUTION

( )22

2602827 mm

4 4dA

ππ= = = ( )44

6 4601.2723 10 mm

32 32dJ

ππ= = = ×

3

6 26

150 10 53.06 10 N/m 53.06 MPa (T)2827 10x

PA

σ −

×= = = × =×

2 2

2 21

53.06 0 53.06 0 125 MPa2 2 2 2

x y x yp xy xy

σ σ σ σσ τ τ

+ − + − = + + = + + =

94.83 MPaxyτ = 94.83 MPaxy Tc Jτ = =

( )( )6 694.83 10 1.2723 10

0.030xy JTc

τ −× ×= =

34.022 10 N m 4.02 kN m= × ⋅ ≅ ⋅ ................................................................................Ans.

10-123 The T-section shown in Fig. P10-123 is used as a short post to support a compressive load P = 150 kip. The load is applied on the centerline of the stem at a distance e = 2 in. from the centroid of the cross section. Determine the normal stresses at points C and D on section AB.

SOLUTION

( ) 22 20 6 24 inA = × =

( ) ( )3 2 6 7 2 6

5 in.24Cx

× + ×= =

( ) ( ) ( ) ( ) ( )( )

3 32 2 42 6 6 2

2 6 2 2 6 2 136 in12 12yI

= + × + + × =

( )( )150 2 5150 4.779 ksi 4.78 ksi (T)

24 136CP McA I

σ×

= − + = − + = + ≅ .......................Ans.

( ) ( )150 2 3150 12.868 ksi 12.87 ksi (C)

24 136DP McA I

σ×

= − − = − − = − ≅ ..................Ans.

10-124 The cross section of the straight vertical portion of the coil-loading hook shown in Fig. P10-124a is shown in Fig. P10-124b. The horizontal distance from the line of action of the applied load to the inside face CD of the cross section is 600 mm. Determine the maximum tensile and compressive stresses on section CDEF for a 40-kN load.

SOLUTION

( ) ( ) ( ) 2100 50 30 80 60 20 8600 mmA = × + × + × =

( ) ( ) ( )25 100 50 90 30 80 140 60 20

59.19 mm8600Cx

× + × + ×= =

( ) ( ) ( )

32100 50

100 50 34.1912yI

= + ×


517

( ) ( ) ( ) ( ) ( ) ( )

3 32 230 80 60 20

30 80 30.81 60 20 80.8112 12

+ + × + + ×

6 418.321 10 mm= ×

( )( )3 3 340 10 659.19 10 26.37 10 N m 26.37 kN mM −= × × = × ⋅ = ⋅

( ) ( )33

6 6

26.37 10 0.0591940 108600 10 18.321 10CD

P McA I

σ − −

××= + = +× ×

6 289.845 10 N/m 89.8 MPa (T)= × ≅ .....................................................................Ans.

( )( )33

6 6

26.37 10 0.0908140 108600 10 18.321 10EF

P McA I

σ − −

××= − = −× ×

6 2126.05 10 N/m 126.1 MPa (C)= − × ≅ .................................................................Ans.

10-125 The estimated maximum total force P to be exerted on the C-clamp shown in Fig. P10-125 is 450 lb. If the normal stress on section A�A is not to exceed 16,000 psi, determine the minimum allowable value for the dimension h of the cross section.

SOLUTION

( ) 20.5 2 inA h h= × = ( ) ( )

33 40.5

24 in12h

I h= =

( )450 3 1350 lb inM = = ⋅

( )( )

3

1350 2 2450 16,000 psi2 24

hP McA I h h

σ×

= + = + <

216,000 900 16,200 0h h− − =

1.035 in.h = ............................................................................................................................ Ans.

10-126 A 150-mm diameter shaft will be used to support the axial load and torques shown in Fig. P10-126. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft.

SOLUTION

( )22

3 215017.671 10 mm

4 4dA

ππ= = = × ( )44

6 415049.70 10 mm

32 32dJ

ππ= = = ×

3

6 23

780 10 44.14 10 N/m 44.14 MPa17.671 10x

PA

σ −

×= = = × =×

( )( )3

6 26

19.6 10 0.07529.58 10 N/m 29.58 MPa

49.70 10xyTcJ

τ −

×= = = × =

×

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

244.14 0 44.14 0 29.582 2

+ − = ± +

1 22.07 36.91 58.98 MPa 59.0 MPa (T)pσ = + = + ≅ ....................................................Ans.


518

2 22.07 36.91 14.84 MPa 14.84 MPa (C)pσ = − = − ≅ .................................................Ans.

3 0 MPap zσ σ= = .................... max 36.91 MPa 36.9 MPapτ τ= = ≅ ........................Ans.

( )1 12 2 29.581 1tan tan 26.64

2 2 44.14 0xy

px y

τθ

σ σ− −= = = °

− −.......................................................Ans.

10-127 The thin-walled cylindrical pressure vessel shown in Fig. P10-127 has an inside diameter of 24 in. and a wall thickness of 1/2 in. The vessel is subjected to an internal pressure of 250 psi. In addition, a torque of 150 kip⋅ft is applied to the vessel through rigid plates on the ends of the vessel. Determine the maximum normal and shearing stresses at a point on the outside surface of the vessel.

SOLUTION

( )4 4

425 24

5777.5 in32

Jπ −

= =

( )

( )250 12

3000 psi 3.000 ksi2 2 1 2x aprt

σ σ= = = = =

( )

( )250 12

6000 psi 6.000 ksi1 2y h

prt

σ σ= = = = =

( ) ( )150 12 12.5

3.894 ksi5777.5xy

TcJ

τ×

= = =

2 2

2 21, 2

3 6 3 6 3.8942 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − + − = ± + = ± +

1 4.5 4.173 8.673 ksi 8.67 ksi (T)pσ = + = + ≅ ...............................................................Ans.

2 4.5 4.173 0.327 ksi 0.327 ksi (T)pσ = − = + ≅ ............................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.

max minmax

8.673 0 4.337 ksi 4.34 ksi2 2

σ στ − −= = = ≅ .................................................Ans.

10-128 A steel shaft is loaded and supported as shown in Fig. P10-128. If the maximum shearing stress in the shaft must not exceed 55 MPa and the maximum tensile stress in the shaft must not exceed 83 MPa, determine the maximum torque T that can be applied to the shaft.

SOLUTION

( )22

3 2100

1007.854 10 mm

4 4dA

ππ= = = × ( )22

3 2150

15017.671 10 mm

4 4dA

ππ= = = ×

( )44

6 4100

1009.817 10 mm

32 32dJ

ππ= = = × ( )44

6 4150

15049.70 10 mm

32 32dJ

ππ= = = ×

3

100 3

6 2

550 107.854 10

70.03 10 N/m 70.03 MPa

PA

σ −

×= =×

= × =

3

150 3

6 2

550 1017.671 10

31.12 10 N/m 31.12 MPa

PA

σ −

×= =×

= × =

( ) 2

100 6

0.0505093 N/m

9.817 10TTc T

Jτ −= = =

×

( ) 2150 6

3 0.0754527 N/m

49.70 10TTc T

Jτ −= = =

×


519

Since both 100σ and 100τ are larger than 150σ and 150τ , the maximum stresses will occur in the 100-mm section.

2 2

2 21

70.03 0 70.03 0 83 MPa2 2 2 2

x y x yp xy xy

σ σ σ σσ τ τ

+ − + − = + + = + + ≤

32.67 MPaxyτ ≤

2 2

2 2max

70.03 0 55 MPa2 2

x yp xy xy

σ στ τ τ τ

− − = = + = + ≤

42.41 MPaxyτ ≤

Therefore, ( )max 32.67 MPaxyτ = and

( )( )6 6

100max

100

32.67 10 9.817 106410 N m

0.050xy JTc

τ −× ×= = = ⋅ ..........................................Ans.

10-129 A 30-lb force P is applied to the brake pedal of an automobile as shown in Fig. P10-129. Force Q is applied to the brake cylinder. Determine the maximum tensile and compressive normal stresses on section a�a, which is midway between points A and B. Section a�a may be modeled as a 3/16×1-in. rectangle.

SOLUTION

0 :AMΣ =� ( )( ) ( )( ) ( )30cos30 11 30sin 30 4 5.5 0Q− ° − ° + =

62.87 lb 62.87 lbQ = = ←

0 :xF→Σ = 30cos30 0xA Q+ ° − =

0 :yF↑Σ = 30sin 30 0yA − ° =

37.89 lb 36.89 lbxA = = →

15.00 lb 15.00 lbyA = = ↑

( )1tan 5.5 4 53.97θ −= = °

0 :nFΣ = 36.89cos53.97 15.00sin 53.97 0N° + ° − =

0 :OMΣ =� ( ) ( )36.89 2.75 15.00 2 0OM + − =

33.83 lbN =

71.45 lb in. 71.45 lb in. OM = − ⋅ = ⋅ �

( )( ) 23 16 1 0.1875 inA = =

( )( )3

43 16 10.015625 in

12I = =

( )71.45 0.50033.83 2106 psi 2110 psi (T)

0.1875 0.015625O

topM cN

A Iσ = − + = − + = + ≅ .........Ans.

( )71.45 0.50033.83 2467 psi 2470 psi (C)

0.1875 0.015625O

botM cN

A Iσ = − − = − − = − ≅ .........Ans.


520

10-130 An automobile engine with a mass of 360 kg is supported by an engine hoist, as shown in Fig. P10-130. Determine the maximum tensile and compressive normal stresses on section a�a if member ABC is a hollow square 100×100-mm section with a wall thickness of 20-mm.

SOLUTION

( )360 9.81 3532 NW mg= = =

1430cos16 1374.6 mma = ° =

1430sin16 394.2 mmb = ° =

1 1374.6 890tan 20.53394.2 900

φ − −= = °+

0 :AMΣ =� ( )( ) ( )( ) ( )cos 20.53 1374.6 sin 20.53 394.2 3532 2700cos16 0B B° − ° − ° =

7978 NB =

0 :xF→Σ = 7978sin 20.53 0xA + =

0 :yF↑Σ = 7978cos 20.53 3532 0yA + ° − =

2798 N 2798 NxA = − = ←

3939 N 3939 NyA = − = ↓

0 :nFΣ = 2798cos16 3939sin16 0P − ° − ° =

0 :OMΣ =� ( ) ( )3939 530cos16 2798 530sin16 0OM + ° − ° =

3775 NP =

6 31.5980 10 N mm 1.5980 10 N m OM = − × ⋅ = × ⋅ �

2 2 2100 60 6400 mmA = − =

( ) ( )3 3

6 4100 100 60 607.253 10 mm

12 12I = − = ×

( )( )3

6 6

1.5980 10 0.05037756400 10 7.253 10

Otop

M cPA I

σ − −

×= + = +

× ×

6 211.606 10 N/m 11.61 MPa (T)= + × ≅ .................................................................Ans.

( ) ( )3

6 6

1.5980 10 0.05037756400 10 7.253 10

Obot

M cPA I

σ − −

×= − = −

× ×

6 210.426 10 N/m 10.43 MPa (C)= − × ≅ .................................................................Ans.

10-131 A short post supports a vertical force P = 9600 lb and a horizontal force H = 800 lb, as shown in Fig. P10-131. Determine the vertical normal stresses at corners A, B, C, and D of the post. Neglect stress concentrations.

SOLUTION

( ) 24 6 24 inA = × =

( )( )3

46 432 in

12xI = = ( ) ( )3

44 672 in

12yI = =


521

( )9600 1 9600 lb in.xM = = ⋅

( )800 24 19,200 lb in.yM = = ⋅

9600 400 psi (C)

24PPA

σ = = =

( )9600 2

600 psi (T & C)32x

xM

x

M cI

σ = = =

( )19, 200 3

800 psi (T & C)72y

yM

y

M cI

σ = = =

400 600 800 1000 psi 1000 psi (T)Aσ = − + + = + = .....................................................Ans.

400 600 800 200 psi 200 psi (C)Bσ = − − + = − = .........................................................Ans.

400 600 800 1800 psi 1800 psi (C)Cσ = − − − = − = .....................................................Ans.

400 600 800 600 psi 600 psi (C)Dσ = − + − = − = .........................................................Ans.

10-132 Determine the vertical normal stresses at points A, B, C, and D of the rectangular post shown in Fig. P10-132. Neglect stress concentrations.

SOLUTION

( ) 3 2200 150 30 10 mmA = × = ×

( )( )3

6 4200 15056.25 10 mm

12xI = = ×

( )( )3

6 4150 200100.0 10 mm

12yI = = ×

( )( )375 10 0.075 5625 N mxM = × = ⋅

( )( )375 10 0.050 3750 N myM = × = ⋅

3

6 23

75 10 2.50 10 N/m 2.50 MPa (C)30 10P

PA

σ −

×= = = × =×

( ) 6 2

6

5625 0.0757.50 10 N/m 7.50 MPa (T & C)

56.25 10x

xM

x

M cI

σ −= = = × =×

( ) 6 2

6

3750 0.1003.75 10 N/m 3.75 MPa (T & C)

100.0 10y

yM

y

M cI

σ −= = = × =×

2.50 7.50 3.75 8.75 MPa 8.75 MPa (T)Aσ = − + + = + = .............................................Ans.

2.50 7.50 3.75 1.25 MPa 1.25 MPa (T)Bσ = − + − = + = .............................................Ans.

2.50 7.50 3.75 13.75 MPa 13.75 MPa (C)Cσ = − − − = − = ........................................Ans.

2.50 7.50 3.75 6.25 MPa 6.25 MPa (C)Dσ = − − + = − = ............................................Ans.


522

10-133 Determine the principal stresses and the maximum shearing stress at points on the top and bottom of section a�a of the pipe system shown in Fig. P10-133. The pipe has an outside diameter of 1 in. and a wall thickness of 1/8 in.

SOLUTION

( ) ( )4 4

41 0.750.03356 in

64 64I

π π= − =

( ) ( )4 4

41 0.750.06711 in

32 32J

π π= − =

At Section a-a:

0 :xFΣ = 0xV =

0 :yFΣ = 0N =

0 :zFΣ = 50 0zV − = 50 lbzV =

0 :xMΣ = ( )50 18 0xM − = 900 lb in.xM = ⋅

0 :yMΣ = ( )50 7 0T − = 350 lb in.T = ⋅

0 :zMΣ = 0zM =

( )900 0.5

13, 409 psi (T on top & C on bottom)0.03356

xM cI

σ = = =

( )350 0.5

2608 psi0.06711

TcJ

τ = = =

On the top of the pipe:

2

21, 2

13, 409 0 13,409 0 2608 6705 7194 psi2 2p pσ + − = ± + = ±

1 6705 7194 13,899 psi 13.90 ksi (T)pσ = + = + ≅ ........................................................Ans.

2 6705 7194 489 psi 0.489 ksi (C)pσ = − = − ≅ .............................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.

Since 1pσ and 2pσ have opposite signs:


On the bottom of the pipe:

2

21, 2

13,409 0 13,409 0 2608 6705 7194 psi2 2p pσ − + − − = ± + = − ±

1 6705 7194 489 psi 0.489 ksi (T)pσ = − + = + ≅ ...........................................................Ans.

2 6705 7194 13,899 psi 13.90 ksi (C)pσ = − − = − ≅ .....................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.




523

10-134 The output from a strain gage located on the bottom surface of the hat-section shown in Fig. P10-134 will be used to indicate the magnitude of the load P applied to the section. The hat-section is made of aluminum alloy (E = 73 GPa and 1 3ν = ) and is 24 mm wide. When the maximum load P = 500 N is applied to the section, the strain gage should read 1000 m/mε µ= + . Plot a curve showing the combinations of thickness t and height h that will satisfy the specification. Limit the range of h from 0 to 50 mm.

SOLUTION

224 mmA t= 3

3 424 2 mm12xtI t= =

( )( )( ) ( ) ( )

69 6 2

6 3 12

500 2 10500 73 10 1000 10 N/m24 10 2 10P

h tP McEA I t t

σ ε−

−− −

×= = + = + = × ×

× ×

gives 21752 500 3000t t h− = h (mm) t (mm) 0 0.285 10 4.283 20 5.996 30 7.311 40 8.420

10-135 A solid shaft 4-in. in diameter is acted on by forces P and Q, as shown in Fig. P10-135. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft.

SOLUTION

( )2

2412.566 in

4A

π= =

( )444

12.566 in64

Iπ

= =

( )4

4425.133 in

32J

π= =

At Point A:

0 :xFΣ = 2.25 0xV − = 2.25 kipxV =

0 :yFΣ = 18 0N − = 18 kipN =

0 :zFΣ = 0zV =

0 :xMΣ = 0xM =

0 :yMΣ = ( )2.25 24 0T − 54 kip in.T = ⋅

0 :zMΣ = ( )2.25 36 0zM − = 81 kip in.zM = ⋅

( ) ( )81 218 11.459 ksi (T)

12.566 12.566N McA I

σ = − + = − + =

( )54 2

4.297 ksi25.133

TcJ

τ = = =


524

2

21, 2

11.459 0 11.459 0 4.297 5.579 7.042 ksi2 2p pσ + − = ± + = ±

1 5.579 7.042 12.621 ksi 12.62 ksi (T)pσ = + = + ≅ .......................................................Ans.

2 5.579 7.042 1.463 ksi 1.463 ksi (C)pσ = − = − ≅ ........................................................Ans.

3 0 ksip zσ σ= = ..................................................................................................................... Ans.


max 7.042 ksi 7.04 ksipτ τ= = ≅ ........................................................................................Ans.

10-136 A thin-walled cylindrical pressure vessel with an inside diameter of 1.2 m is fabricated by butt-welding 15-mm plate with a spiral seam as shown in Fig. P10-136. The pressure in the tank is 2500 kPa. Additional loads are applied to the cylinder through a rigid end plate as shown in Fig. P10-136. Determine

(a) The normal and shearing stresses on the plane of the weld at a point on the outside surface of the tank. (b) The principal stresses and the maximum shearing stress at a point on the inside surface of the tank. SOLUTION

( )2 2

3 21230 1200

57.256 10 mm4

Aπ −

= = × ( )1tan 4 3 53.130θ −= − = − °

( )34 4

9 41230 1200

21.134 10 mm32

Jπ −

= = ×

( )( )32500 10 0.600

0.015xprt

σ×

= =

6 2100.0 10 N/m 100.0 MPa (T)= × =

( )( )

( )

3 3

3

2500 10 0.600 125 102 2 0.015 57.256 10ypr Pt A

σ −

× ×= − = −×

6 247.817 10 N/m 47.817 MPa (T)= × =

( ) ( )( )36 2

3

750 10 0.615outside 21.825 10 N/m 21.825 MPa

21.134 10xyTcJ

τ −

×= = = × =

×

( ) ( )( )36 2

3

750 10 0.600inside 21.293 10 N/m 21.293 MPa

21.134 10xyTrJ

τ −

×= = = × =

×


( ) ( )2 2100cos 53.130 47.82sin 53.130= − ° + − °

( ) ( ) ( )2 21.83 sin 53.130 cos 53.130+ − − ° − °

87.56 MPa 87.6 MPa (T)= ≅ .....................................................................................Ans.

( ) ( )2 2sin cos cos sinnt x y xyτ σ σ θ θ τ θ θ= − − + −

( ) ( ) ( )100 47.82 sin 53.130 cos 53.130= − − − ° − °

( ) ( ) ( )2 221.83 cos 53.130 sin 53.130 + − − ° − − °

31.16 MPa 31.2 MPa= ≅ ............................................................................................Ans.


525

(b) 2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

2100 47.82 100 47.82 21.292 2

+ − = ± +

1 73.91 33.67 107.58 MPa 107.6 MPa (T)pσ = + = + ≅ ...............................................Ans.

2 73.91 33.67 40.24 MPa 40.2 MPa (T)pσ = − = + ≅ ...................................................Ans.

6 23 2.50 10 N/m 2.50 MPa (C)p pσ = − = − × = ..............................................................Ans.

( )max min

max

107.58 2.555.04 MPa 55.0 MPa

2 2σ στ

− −−= = = ≅ ..............................Ans.

10-137 A 1-in. diameter steel (E = 30,000 ksi and 0.30ν = ) bar is subjected to a tensile load P and a torque T, as shown in Fig. P10-137. Determine the axial load P and the torque T if the strains indicated by gages a and b on the bar are 1084 in./in.aε µ= + and 754 in./in.bε µ= − .

SOLUTION

( ) ( ) 645 2 2

30,000 1084 0.30 754 10 28.279 ksi1 1 0.30a bEσ ε νεν

−= + = + − × = − −

( ) ( ) ( ) 645 2 2

30,000 754 0.30 1084 10 14.136 ksi1 1 0.30b aEσ ε νεν

−− = + = − + × = − − −

( ) ( ) ( ) ( )2 245 cos 45 sin 45 2 sin 45 cos 45x y xyσ σ σ τ= ° + ° + ° °

( ) ( ) ( ) ( )2 245 cos 45 sin 45 2 sin 45 cos 45x y xyσ σ σ τ− = − ° + − ° + − ° − °

For a shaft subjected to an axial load P and a torque T: 0 ksiyσ =

Therefore: 0.50 28.279 ksix xyσ τ+ =

0.50 14.136 ksix xyσ τ− = −

Solving yields: 14.143 ksixσ = 21.2008 ksixyτ =

Then,

( )2

210.7854 in

4A

π= =

( )441

0.09817 in32

Jπ

= =

( )14.143 0.7854 11.11 kipxP Aσ= = = ...........................................................................Ans.

( )21.208 0.09817

4.16 kip in.0.50

xy JTc

τ= = = ⋅ .................................................................Ans.

10-138 A bag of potatoes is resting on a chair, as shown in Fig. P10-138. The force exerted by the p0tatoes on the frame at one side of the chair is equivalent to horizontal and vertical forces of 24 N and 84 N, respectively, at E and a force of 28 N perpendicular to member BH at G. Determine the maximum tensile and compressive normal stresses on a section midway between pins C and F. Member BH has a 10 × 30-mm cross section; the 30-mm dimension lies in the plane of the page.

SOLUTION


526

( )1tan 3 5 30.96θ −= = °

For the entire chair, the equilibrium equations:

0 :yF↑Σ = 84 28sin 0A B θ+ − − =

0 :BMΣ =� ( ) ( )0.2 84 0.5 24 0.4A− −

( ) ( )0.3 0.5 cos 28 0θ+ + =

give 73.82 NA = 24.58 NB = For member DF the equilibrium equations give:

0 :DMΣ =� ( )0.4 84 0.5 0yF− =

67.2 NyF =

0 :xF→Σ = 24 0x xD F− + =

0 :yF↑Σ = 84 0y yD F+ − =

( )24 Nx xD F= − 16.80 NyD =

For member BH the equilibrium equations give:

0 :CMΣ =� ( ) ( ) ( )0.3 0.1667 sin 28 0.1333 24.58θ+ +

( )0.1667 67.2 0.2777 0xF+ − =

115.1 NxF =

0 :xF→Σ = 28cos 0x xF C θ+ − =

0 :yF↑Σ = 24.58 67.2 28sin 0yC θ+ − − =

91.0 NxC = − 57.0 NyC =

On a section midway between pins C and F:

500 mma b+ = 400 500a b

=

Therefore: 222.2 mma = 277.8 mmb = 2 138.9 mmb =

133.3 mmc = 166.7 mmd = 2 83.4 mmd =

0 :nFΣ = 24.6cos30.96 57.0cos30.96 91.0sin 30.96 0P° + ° + ° − =

0 :OMΣ =� ( ) ( ) ( )24.6 216.7 57.0 83.4 91.0 138.9 0M+ − − =

116.79 NP = 2555 N mm 2.555 N m M = − ⋅ = ⋅ �

210 30 300 mmA = × = ( )3

3 410 3022.5 10 mm

12I = = ×

( )

6 9

2.555 0.015116.79300 10 22.5 10top

P McA I

σ − −= − + = − +× ×

6 21.3140 10 N/m 1.314 MPa (T)= + × ≅ .................................................................Ans.


527

( )

6 3

2.555 0.015116.79300 10 22.5 10bot

P McA I

σ − −= − − = − −× ×

6 22.093 10 N/m 2.09 MPa (C)= − × ≅ .....................................................................Ans.

10-139 Three strain gages are mounted on a 1-in. diameter aluminum alloy (E = 10,000 ksi and 1 3ν = ) rod, as shown in Fig. P10-139. When loads P and Q are applied to the rod, they produce longitudinal strains

550 in./in.Aε µ= + , 400 in./in.Bε µ= + , and 300 in./in.Cε µ= − . Determine the magnitudes of loads P and Q and the location x of load P.

SOLUTION

( )2

210.7854 in

4A

π= =

( )4

410.04909 in

64xIπ

= =

( )( )6 610 10 550 10 5500 psiA AEσ ε −= = × × =

( )( )6 610 10 400 10 4000 psiB BEσ ε −= = × × =

( )( )6 610 10 300 10 3000 psiC CEσ ε −= = × − × = −

( ) ( ) ( )3 3 0.50

5500 psi0.7854 0.04909A

P x c P xQ QA I

σ+ + = + = + = (a)

( ) ( )0.50

4000 psi0.7854 0.04909B

Px c PxQ QA I

σ = + = + = (b)

( ) ( )0.50

3000 psi0.7854 0.04909C

Px c PxQ QA I

σ = − = − = − (c)

From Eqs. (b) and (c): 392.7 lb 393 lbQ = ≅ ................................................................Ans.

From Eq. (b): 343.63 lbPx =

From Eq. (a): ( )3 490.90 lbP x+ =

Solving yields: 49.09 lb 49.1 lbP = ≅ .................... 7.00 in.x = .................................Ans.

10-140 A steel shaft 120 mm in diameter is supported in flexible bearings at its ends. Two pulleys, each 500 mm in diameter are keyed to the shaft. The pulleys carry belts that produce the forces shown in Fig. P10-140. Determine the principal stresses and the maximum shearing stress at point A on the top surface of the shaft.

SOLUTION

( )4

6 412010.179 10 mm

64I

π= = ×

( )46 4120

20.358 10 mm32

Jπ

= = ×

( )3

3 3120144 10 mm

12Q = = × ( ) ( )30 0.250 5 0.250 6.25 kN mAT = − = ⋅

For the horizontal plane:

0 :BMΣ = ( ) ( )2800 35 800 0CR − = 10.00 kNCR =

0 :zFΣ = 35 0B CR R− + = 25.00 kNBR =


528

For the vertical plane:

0 :BMΣ = ( ) ( )2800 35 2000 0CR − = 25.00 kNCR =

0 :zFΣ = 35 0B CR R− + = 10.00 kNBR =


( )10 1.200 12 kN mAM = = ⋅ 35 25 10 kNAV = − =

( )( )3

6 26

12 10 0.06070.73 10 N/m 70.73 MPa (C)

10.179 10xMcI

σ −

×= − = − = − × =

×

( )( ) ( ) ( )

( ) ( )

3 3 6

6 6

6.25 10 0.060 10 10 144 1020.358 10 10.179 10 0.120xy

Tc VQJ It

τ−

− −

× × ×= + = +

× ×

6 219.60 10 N/m 19.60 MPa= × =

2

21, 2

70.73 0 73.70 0 19.60 35.365 40.433 MPa2 2p pσ − + − − = ± + = − ±

1 35.365 40.433 5.068 MPa 5.07 MPa (T)pσ = − + = + ≅ ............................................Ans.

2 35.365 40.433 75.798 MPa 75.8 MPa (C)pσ = − − = − ≅ .........................................Ans.

3 0 MPapσ = ........................................................................................................................... Ans.


max 40.433 MPa 40.4 MPapτ τ= = ≅ ...............................................................................Ans.

10-141 A 120-lb girl is walking up a uniform 2 × 12-in. beam of negligible weight, as shown in Fig. P10-141. The coefficient of friction is 0.20 at all surfaces. When the beam begins to slip, determine the maximum tensile and compressive normal stresses on a section at a distance x/2 from the right end of the beam.

SOLUTION

0.2f n nA A Aµ= =

0.2f n nB B Bµ= =


529

0 :hF→ Σ = ( ) ( )3 5 0.2 4 5 0.2 0n n nB B A− − =

0 :vF↑ Σ = ( ) ( )4 5 0.2 3 5 120 0n n nB B A+ + − =

Solving yields: 84.62 lbnA = 38.46 lbnB =

0 :nFΣ = ( ) ( )( )84.62 3 5 0.2 84.62 4 5 0P+ − =

0 :AMΣ =� ( ) ( )120 4 5 38.46 10 0x − =

4.006 ftx = 64.31 lbP =

0 :OMΣ =� ( )( ) ( )( ) ( )84.62 4 5 2.003 0.2 84.62 3 5 2.003 0M− − =

115.25 lb ft M = ⋅ �

22 12 24.0 inA = × = ( )3

412 28.00 in

12I = =

( ) ( )115.25 12 164.31 175.56 psi 175.6 psi (C)

24.0 8.00topP McA I

σ×

= − − = − − = − ≅ .......Ans.

( ) ( )115.25 12 164.31 170.19 psi 170.2 psi (T)

24.0 8.00botP McA I

σ×

= − + = − + = + ≅ .......Ans.

10-142 Four strain gages are mounted at 90° intervals around the circumference of a 100-mm diameter steel (E = 200 GPa and 0.30ν = ) shaft, as shown in Fig. P10-142. All of the gages are oriented at an angle of 45° with respect to a line on the surface of the shaft that is parallel to the axis of the shaft. Determine the torque T, shear force V, and bending moment M at the cross section where the gages are located if

450 m/mAε µ= + , 325 m/mBε µ= + , 550 m/mCε µ= + , and 675 m/mDε µ= + .

SOLUTION

( )2

21007854 mm

4A

π= = ( )

200 76.92 GPa2 1 2.6EG

ν= = =

+

( )4

6 41004.909 10 mm

64I

π= = ×

( )46 4100

9.817 10 mm32

Jπ

= = ×

( ) ( )2

3 34 50 10083.33 10 mm

3 8CQ y Aπ

π

= = = ×

At point A: 45 45xyTc VQJ It

τ σ σ −= − = = −

( )45 45 451 1

ATc VQ

E E J Itνε σ νσ ε−

+ = − = − =

At point C: 45 45xyTc VQJ It

τ σ σ −= + = = −

( )45 45 451 1

CTc VQ

E E J Itνε σ νσ ε−

+ = − = + =

Therefore: 12A C

TcE Jνε ε + + =


530

from which:

( ) ( ) ( )( )( ) ( ) ( )

9 66

200 10 9.817 10450 550 10

2 1 2 1 0.30 0.050A CEJT

cε ε

ν

−−

× ×= + = + ×

+ +

315.103 10 N m 15.10 kN m= × ⋅ ≅ ⋅ .............................................................................Ans.

Similarly: 12C A

VQE Itνε ε + − =

( ) ( ) ( ) ( )( )( )( ) ( )

9 66

6

200 10 4.909 10 0.100550 450 10

2 1 2 1 0.30 83.33 10C AEItV

Qε ε

ν

−−

−

× ×= − = − ×

+ + ×

345.32 10 N 45.3 kN= × ≅ .............................................................................................Ans.

At point B: ( )( )3

6 26

15.103 10 0.05076.92 10 N/m

9.817 10xyTcJ

τ −

×= = = ×

×

( ) ( )3 2

6

0.05010.185 10 N/m

4.909 10x

MMc MI

σ −= − = − = − ××

2 2cos 45 sin 45 sin 45 cos 45xyx xB E E G

τσ νσε = ° − ° + ° °

( )( ) ( ) ( )

3 66

9 9

1 0.30 10.185 10 76.92 100.5 0.50 325 10200 10 76.92 10

M−

− − × ×= + = ×× ×

from which:

39.818 10 N m 9.82 kN mM = × ⋅ ≅ ⋅ ...........................................................................Ans.

10-143 A 4-in. diameter solid circular steel shaft is loaded and supported as shown in Fig. P10-143. Because of discontinuities at points A and B, the normal and shearing stresses at these points must be limited to 13 ksi (T) and 8.5 ksi, respectively. Determine the maximum permissible value for the load P.

SOLUTION

( )2

2412.566 in

4A

π= =

( ) ( )234 2 4

5.333 in3 8B CQ y A

ππ

= = =

( )4

4412.566 in

64xIπ

= = ( )4

4425.13 in

32J

π= =

At point A:

( ) ( )16 28 3.183

12.566 12.566x

PF Mc P PA I

σ = + = + =

( )( )8 2

0.636725.13xy

PTc PJ

τ = = =

At point B:

( )( )64 28 10.823

12.566 12.566x

PF Mc P PA I

σ = + = + =


531

( )( ) ( )

( )8 2 5.333

0.742825.13 12.566 4xy

P PTc VQ PJ It

τ = + = + =

Therefore, maximum stresses will occur at point B.

( )2

21

10.823 0 10.823 0 0.7428 13 ksi2 2pP P Pσ + − = + + <

gives 1.1955 kipP <

( )2

2max

10.823 0 0.7428 8.5 ksi2pP Pτ τ − = = + <

gives 1.5561 kipP <

Therefore: max 1.1955 kip 1.196 kipP = ≅ ................................................................Ans.

10-144 The frame shown in Fig. P10-144 is constructed of 100 × 100-mm timbers. Determine and show on sketches the principal and maximum shearing stresses at points G and H.

SOLUTION

0 :CMΣ =� ( ) ( ) ( )6 1.5 3 1.5 3 0yA+ − =

4.50 kNyA =

At the section containing G and H:

0 :FΣ = ( )4.50 4 5 0P− = 3.60 kNP =

0 :FΣ = ( )4.50 3 5 0V− = 2.70 kNV =

0 :OMΣ =� ( ) ( )4.50 3 5 1 0M − =

2.7 kN mM = ⋅

3 2100 100 10 10 mmA = × = ×

( )3

6 4100 1008.333 10 mm

12I = = ×

( ) 3 337.5 100 25 93.75 10 mmH CQ y A= = × = ×

At point G:

( )( )33

6 23 6

2.7 10 0.0503.6 10 16.561 10 N/m10 10 8.333 10

P McA I

σ − −

××= − − = − − = − ×× ×

Therefore: 1 0 MPapσ = ............. 6 22 16.561 10 N/m 16.56 MPa (C)pσ = − × ≅ ................Ans.

3 0 MPapσ = .................... 1 2max 8.28 MPa

2p pσ σ

τ−

= = ....................................Ans.

At point H:

( )( )33

6 23 6

2.7 10 0.0253.6 10 7.740 10 N/m10 10 8.333 10

P MyA I

σ − −

××= − + = − + = + ×× ×


535

Maximum-Distortion-Energy:

( )( )2 2 2 21 1 3 3 30 30 50 50 70 ksie p p p pσ σ σ σ σ= − + = − − + =

70 ksi 60 ksi (Failure)e fσ σ= > = ..................................................................................Ans.

10-150 The state of stress at a point on the surface of a machine component is shown in Fig. P10-150. If the yield strength of the material is 250 MPa, determine which, if any, of the theories will predict failure by yielding for this state of stress.

SOLUTION

100 MPaxσ = + 80 MPayσ = + 40 MPaxyτ = −

( )

22

1, 2

22

2 2

100 80 100 80 40 90 41.23 MPa2 2

x y x yp p xy

σ σ σ σσ τ

+ + = ± +

+ − = ± + − = ±

1 90 41.23 131.23 MPa 131.2 MPa (T)pσ = + = + ≅

2 90 41.23 48.77 MPa 48.8 MPa (T)pσ = − = + ≅

3 0 MPap zσ σ= = 250 MPafσ =

Maximum-Normal-Stress:

max 1 131.2 MPa 250 MPa (No failure)p fσ σ σ= = < = ..............................................Ans.

Maximum-Shear-Stress:

( ) ( )max 1 31 1 131.2 0 65.6 MPa2 2p pτ σ σ= − = − =

( )1 1 250 125 MPa2 2f fτ σ= = =

max 65.6 MPa 125 MPa (No failure)fτ τ= < = ..............................................................Ans.


( ) ( )2 2 2 21 1 3 3 131.2 131.2 48.8 48.8 114.9 MPae p p p pσ σ σ σ σ= − + = − + =

114.9 MPa 250 MPa (No failure)e fσ σ= < = .............................................................Ans.

10-151 At a point on the free surface of an alloy steel machine component the principal stresses are 45 ksi (T) and 25 ksi (C). What minimum proportional limit is required according to each of the theories if failure by yielding is to be avoided?

SOLUTION

1 45 ksi (T)pσ = 2 0 ksipσ = 3 25 ksi (C)pσ =

Maximum-Normal-Stress: max 1 45.0 ksiPL pσ σ σ> = = ..................................................... Ans.


( ) ( )max 1 31 1 45 25 35 ksi2 2p pτ σ σ= − = + =

( )max2 2 2 35 70.0 ksiPL fσ τ τ> = = = ..............................................................................Ans.


536


( )( )2 2 2 21 1 3 3 45 45 25 25 61.4 ksie p p p pσ σ σ σ σ= − + = − − + =

61.4 ksiPL eσ σ> = ...............................................................................................................Ans.

10-152 At a point on the free surface of an aluminum alloy machine component the principal stresses are 125 MPa (T) and 190 MPa (C). What minimum proportional limit is required according to each of the theories if failure by yielding is to be avoided?

SOLUTION

1 125 MPa (T)pσ = 2 0 ksipσ = 3 190 MPa (C)pσ =


max 3 190.0 MPaPL pσ σ σ> = = ............................................... Ans.


( ) ( )max 1 31 1 125 190 157.5 MPa2 2p pτ σ σ= − = + =

( )max2 2 2 157.5 315 MPaPL fσ τ τ> = = = ......................................................................Ans.


( )( )2 2 2 21 1 3 3 125 125 190 190 275 MPae p p p pσ σ σ σ σ= − + = − − + =

275 MPaPL eσ σ> = .............................................................................................................Ans.

10-153 A material with a proportional limit of 36 ksi in tension and compression is subjected to a biaxial state of stress. The principal stresses are 18 ksi (T), 16 ksi (T), and 0zσ = . Determine the factor of safety with respect to failure by yielding according to each of the theories of failure.

SOLUTION

36 ksifσ = 1 18 ksi (T)pσ = 2 16 ksi (T)pσ = 3 0 ksipσ =


max 1

36 2.0018

f f

p

FSσ σ

σ σ= = = = .............................................................................................Ans.


( ) ( )max 1 31 1 18 0 9 ksi2 2p pτ σ σ= − = − =

( )1 1 36 18 ksi2 2f fτ σ= = =

max

18 2.009

fFSτ

τ= = = ..........................................................................................................Ans.


( )( )2 2 2 21 1 3 3 18 18 16 16 17.09 ksie p p p pσ σ σ σ σ= − + = − + =

36 2.11

17.09f

e

FSσσ

= = = ......................................................................................................Ans.


537

10-154 A material with a proportional limit of 180 MPa in tension and compression is subjected to a biaxial state of stress. The principal stresses are 90 MPa (T), 80 MPa (C), and 0zσ = . Determine the factor of safety with respect to failure by yielding according to each of the theories of failure.

SOLUTION

180 MPafσ = 1 90 MPa (T)pσ = 2 0 MPapσ = 3 80 MPa (C)pσ =


max 1

180 2.0090

f f

p

FSσ σ

σ σ= = = = ...........................................................................................Ans.


( ) ( )max 1 31 1 90 80 85 MPa2 2p pτ σ σ= − = + =

( )1 1 180 90 MPa2 2f fτ σ= = =

max

90 1.05985

fFSτ

τ= = = ........................................................................................................Ans.


( ) ( )2 2 2 21 1 3 3 90 90 80 80 147.31 MPae p p p pσ σ σ σ σ= − + = − − + =

180 1.222

147.31f

e

FSσσ

= = = ..................................................................................................Ans.

10-155 A material with a yield strength in tension and compression of 60 ksi is subjected to the biaxial state of stress shown in Fig. P10-155. Determine the factor of safety with respect to failure by yielding according to each of the theories of failure.

SOLUTION

20 ksixσ = + 25 ksiyσ = − 36 ksixyτ = + 60 ksifσ =

( ) ( )

22

1, 3

22

2 2

20 25 20 25 36 2.5 42.45 ksi2 2

x y x yp p xy

σ σ σ σσ τ

+ + = ± +

− + = ± + = − ±

1 2.5 42.45 39.95 ksi 40.0 ksi (T)pσ = − + = + ≅

2 0 ksip zσ σ= =

3 2.5 42.45 44.95 ksi 45.0 ksi (C)pσ = − − = − ≅


max 1

60 1.33544.95

f f

p

FSσ σ

σ σ= = = = .....................................................................................Ans.


( ) ( )max 1 31 1 39.95 44.95 42.45 ksi2 2p pτ σ σ= − = + =


538

( )1 1 60 30 ksi2 2f fτ σ= = =

max

30 0.7067 1 (Failure)42.45

fFSτ

τ= = = < .......................................................................Ans.


( )( )2 2 2 21 1 3 3 39.95 39.95 44.95 44.95 73.57 ksie p p p pσ σ σ σ σ= − + = − − + =

60 0.8155 1 (Failure)

73.57f

e

FSσσ

= = = < .........................................................................Ans.

10-156 A point on the free surface of a machine component is subjected to the state of stress shown in Fig. P10-156. If the yield strength of the material is 250 MPa, determine the factor of safety with respect to failure by yielding according to each of the theories of failure.

SOLUTION

140 MPaxσ = + 95 MPayσ = + 80 MPaxyτ = − 250 MPafσ =

( ) ( )

22

1, 2

22

2 2

140 95 140 95 80 117.5 83.10 MPa2 2

x y x yp p xy

σ σ σ σσ τ

+ + = ± +

+ − = ± + − = ±

1 117.5 83.10 200.60 MPa 201 MPa (T)pσ = + = + ≅

2 117.5 83.10 34.40 MPa 34.4 MPa (T)pσ = − = + ≅

3 0 ksip zσ σ= =


max 1

250 1.246200.6

f f

p

FSσ σ

σ σ= = = = .....................................................................................Ans.


( ) ( )max 1 31 1 200.6 0 100.3 MPa2 2p pτ σ σ= − = − =

( )1 1 250 125 MPa2 2f fτ σ= = =

max

125 1.246100.3

fFSτ

τ= = = ..................................................................................................Ans.


( )( )2 2 2 21 1 3 3 200.6 200.6 34.4 34.4 185.8 MPae p p p pσ σ σ σ σ= − + = − + =

250 1.346

185.8f

e

FSσσ

= = = ....................................................................................................Ans.


539

10-157 A thin-walled cylindrical pressure vessel is capped at the ends and is subjected to an internal pressure. The inside diameter of the vessel is 5 ft, and the wall thickness is 1.5 in. The vessel is made of steel with tensile and compressive yield strengths of 36 ksi. Determine the internal pressure required to initiate yielding according to

(a) The maximum-shear-stress theory of failure (b) The maximum-distortion-energy theory of failure. SOLUTION

( )

1

3020

1.5h p

ppr pt

σ σ= = = = 3 0 (Outside surface)r pσ σ= =

( )( )2

3010

2 2 1.5a p

ppr pt

σ σ= = = = 3 (Inside surface)r p pσ σ= = −

(a) Maximum-Shear-Stress: ( )1 1 36 18 ksi2 2f fτ σ= = =

Outside surface:

( ) ( )max 1 31 1 20 0 102 2p p p pτ σ σ= − = − =

max 18 1.800 ksi10 10 10

fpττ= = = =

Inside surface:

( ) ( )max 1 31 1 20 10.52 2p p p p pτ σ σ= − = − − =

max 18 1.714 ksi10.5 10.5 10.5

fpττ= = = = ...................................................................................Ans.

(b) Maximum-Distortion-Energy: Outside surface:

2 21 1 2 2f p p p pσ σ σ σ σ= − +

( ) ( ) ( ) ( )2 236 20 20 10 10 17.321p p p p p= − + =

2.078 ksip =

Inside surface:

( ) ( ) ( )2 2 221 2 2 3 3 12 f p p p p p pσ σ σ σ σ σ σ= − + − + −

( ) ( ) ( ) ( )2 2 2 2 22 36 20 10 10 20 662p p p p p p p= − + + + − − =

1.979 ksip = .......................................................................................................................... Ans.

10-158 The solid circular shaft shown in Fig. P10-158 is subjected to a torque T. The yield strength of the material in tension and compression is 400 MPa. Determine the largest permissible value of the torque T according to

(a) The maximum-shear-stress theory of failure. (b) The maximum-distortion-energy theory of failure. SOLUTION


540

( )1 3 max 34 3

2 2 50932 0.050

p p xyTc Tc T T TJ c c

σ σ τ τπ π π

= − = = = = = = =

(a) Maximum-Shear-Stress: ( )1 1 400 200 MPa2 2f fτ σ= = =

6

3200 10 39.27 10 N m 39.3 kN m5093 5093

fTτ ×= = = × ⋅ ≅ ⋅ .............................................Ans.

(b) Maximum-Distortion-Energy: 400 MPafσ =

2 21 1 2 2f p p p pσ σ σ σ σ= − +

( ) ( ) ( ) ( )2 26400 10 5093 5093 5093 5093 8821T T T T T× = − − + − =

6

3400 10 45.34 10 N m 45.3 kN m8821

T ×= = × ⋅ ≅ ⋅ ...........................................................Ans.

10-159 The shaft shown in Fig. P10-159 is made of steel having a proportional limit of 60 ksi in tension or compression. If a factor of safety of 2.5 with respect to failure by yielding is specified, determine the maximum permissible value for the axial load P according to the maximum-shear-stress theory of failure.

SOLUTION

( )22 0.31833

xP P P PA c

σπ π

= − = − = − = −

( )

( )34 3

2 9000 122 8000 psi2 3

xyTc Tc TJ c c

πτ

π π π×

= = = = =

Maximum-Shear-Stress: 1 1 60 12 ksi2 2 2.5

ff FS

στ = = =

2 2

2 2max

0.3183 0 8000 12,000 psi2 2

x yxy

Pσ στ τ

− − − = + = + =

56, 200 lb 56.2 kipP = = .....................................................................................................Ans.

10-160 A shaft, similar to the one shown in Fig. P10-159, has a 150-mm diameter and is made of steel having a yield strength in tension and compression of 360 MPa. The applied loads are P = 2200 kN and T = 38 kN⋅m. If failure is by yielding, determine the factor of safety according to


( )

36 2

22

2200 10 124.49 10 N/m 124.49 MPa (C)0.075

xP PA c

σπ π

×= − = − = − = − × =

( )( )

36 2

34 3

2 38 102 57.34 10 N/m 57.34 MPa2 0.075

xyTc Tc TJ c c

τπ π π

×= = = = = × =


541

2

21, 3 2 2

x y x yp p xy

σ σ σ σσ τ

+ + = ± +

( ) ( )2

2124.49 0 124.9 0 57.34 62.25 84.63 MPa2 2

− + − − = ± + = − ±

1 62.25 84.63 22.36 MPa 22.4 MPa (T)pσ = − + = + ≅

2 0 ksip zσ σ= =

2 62.25 84.63 146.88 MPa 146.9 MPa (C)pσ = − − = − ≅

Maximum-Shear-Stress: ( )1 1 360 180 MPa2 2f fτ σ= = =

( ) ( )max 1 31 1 22.38 146.88 84.63 MPa2 2p pτ σ σ= − = + =

max

180 2.1384.63

fFSτ

τ= = = ....................................................................................................Ans.

Maximum-Distortion-Energy: 360 MPafσ =

( )( )2 2 2 21 1 2 2 22.38 22.38 146.88 146.88 159.25 MPae p p p pσ σ σ σ σ= − + = − − + =

360 2.26

159.25f

e

FSσσ

= = = ...................................................................................................Ans.

10-161 The yield strength yσ of a ductile material may be determined from a tension test (Fig. P10-161a). Using

the three theories of failure discussed in Section 10-15, show that the yield strength yτ of a member determined from a torsion test (Fig. P10-161b) is predicted to be

(a) y yτ σ= using the maximum-normal-stress theory of failure.

(b) 0.5y yτ σ= using the maximum-shear-stress theory of failure.

(c) 0.577y yτ σ= using the maximum-distortion-energy theory of failure.

SOLUTION

For the tension test: max 1p x f yσ σ σ σ σ= = = =

max max1 1 12 2 2f yτ σ σ σ= = =

For the torsion test: max 1 3 maxp p xy f yσ σ σ τ τ τ τ= = − = = = =

(a) Maximum-Normal-Stress: max yσ σ=

Therefore: y yτ σ= ...............................................................Ans.

(b) Maximum-Shear-Stress: max12 yτ σ=

Therefore: 0.50y yτ σ= ......................................................Ans.


542

(c) Maximum-Distortion-Energy:

2 21 1 3 3p p p p yσ σ σ σ σ− + =

( ) ( )( ) ( )2 2

y y y y yτ τ τ τ σ− − + − =

1.7321 y yτ σ=

Therefore: 0.577y yτ σ= .............................................................................................................Ans.

10-162 The hollow steel ( )250 MPayσ = shaft shown in Fig. P10-162 is subjected to a torque T = 40 kN⋅m.

The factor of safety with respect to failure by yielding is 1.5. Determine the maximum permissible inside diameter for the shaft according to the maximum-shear-stress theory of failure.

SOLUTION

( )max1 250 83.33 MPa2 2 1.5

y

FSσ

τ = = =

( )( )3

max

6 2

40 10 0.075 3000

83.33 10 N/m

xyTcJ J J

τ τ×

= = = =

= ×

( )4 46

30000.15032 83.33 10iJ dπ = − =

×

0.10869 m 108.7 mmid = ≅ ...............................................................................................Ans.

10-163 A solid circular steel ( )36 ksiyσ = shaft is subjected to an axial tensile load P = 10 kip and a bending

moment M = 5 kip⋅in. Determine the minimum permissible diameter for the shaft according to the maximum-shear-stress theory, if the factor of safety for failure by yielding is 2.0.

SOLUTION

( )max1 36 9 ksi2 2 2.0

y

FSσ

τ = = =

max max2 4 2 3

10 5 10 20 2 18 ksi4

P Mc cA I c c c c

σ τπ π π π

= + = + = + = =

318 10 20 0c cπ − − = 0.7902 in.c =

( )min 2 2 0.7902 1.5804 in. 1.580 in.d c= = = ≅ .....................................................................Ans.

10-164 A solid circular shaft has a diameter of 100 mm and is made of a material with a yield strength of 360 MPa in tension and compression. The shaft is subjected to a bending moment M = 12 kN⋅m and a torque T = 20 kN⋅m. Determine the factor of safety with respect to failure by yielding according to the maximum-shear-stress theory of failure.

SOLUTION

( )max1 360 180 MPa2 2

y

FS FS FSσ

τ = = =

( )4 6 450 4.909 10 mm4

I π= = × ( )4 6 450 9.817 10 mm2

J π= = ×


543

( )( )3

6 26

12 10 0.050122.22 10 N/m 122.22 MPa

4.909 10xMcI

σ −

×= = = × =

×

( )( )3

6 26

20 10 0.050101.86 10 N/m 101.86 MPa

9.817 10xyTcJ

τ −

×= = = × =

×

( ) ( )2

2 22max

18061.11 101.86 118.79 MPa2

x yxy FS

σ στ τ

− = + = + = =

180 1.515

118.79FS = = .............................................................................................................Ans.

10-165 The solid circular shaft shown in Fig. P10-165 is subjected to an axial tensile force P, a bending moment M, and a torque T. The shaft is made from a ductile material with a yield strength yσ . Show that the minimum diameter d of the shaft may be found according to the maximum-shear-stress theory of failure from the equation

( ) ( )2 23

2 8 82

y Pd M TFS d

σπ

= + +

where FS is the factor of safety. SOLUTION

( )max12 2

y y

FS FSσ σ

τ = =

( )

2 4 2 3

2 4 324 64x

M dP Mc P P MA I d d d d

σπ π π π

= ± = ± = ±

( )

4 3

2 1632xy

T dTc TJ d d

τπ π

= = =

2 2 2

2max 2 3 3

2 16 162 2

x y yxy

P M Td d d FS

σ σ στ τ

π π π− = + = + + =

( ) ( )2 23

2 8 82

y Pd M TFS d

σπ

= + + ...................................................................................Ans.

10-166 A solid circular shaft has a diameter d and is subjected to a bending moment M and a torque T. The shaft is made from a ductile material with a yield strength yσ . Show that the minimum diameter d of the shaft may be found according to the maximum-shear-stress theory of failure from the equation

( ) ( )1 3

2 232 8y

FSd M Tπσ

= +

where FS is the factor of safety. SOLUTION

( )

4 3

2 3264x

M dMc MI d d

σπ π

= = = ( )

4 3

2 1632xy

T dTc TJ d d

τπ π

= = =


544

2 2 2

2max 3 3

16 162 2

x y yxy

M Td d FS

σ σ στ τ

π π− = + = + =

( ) ( )2 23

2 8 82

y M TFS d

σπ

= +

13

2 232

y

FSd M Tπσ

= +

.....................................................................................................Ans.

10-167 The shaft shown in Fig. P10-167 is made of an aluminum alloy that has a proportional limit of 48 ksi in tension or compression. If a factor of safety of 2.0 with respect to failure by yielding is specified, determine the maximum permissible value for the load R according to


( )2 22 4 inA π π= = ( )4 42 4 in4

I π π= =

( )4 42 8 in2

J π π= = ( )max1 48 12 ksi2 2 2.0

y

FSσ

τ = = =

( )( )24 28 4.4563

4 4x

RQ Mc R RA I

σπ π

= + = + =

( ) ( )20 2

1.59158xy

RTc RJ

τπ

= = =

( )

22

1, 3

22

2 2

4.4563 0 4.4563 0 1.59152 2

2.228 2.738

x y x yp p xy

R R R

R R

σ σ σ σσ τ

+ − = ± +

+ − = ± +

= ±

1 2.228 2.738 4.966 ksi 4.966 ksi (T)p R R R Rσ = + = + =

2 0 ksip zσ σ= =

3 2.228 2.738 0.510 ksi 0.510 ksi (C)p R R R Rσ = − = − =

(a) Maximum-Shear-Stress:

( ) ( )max 1 31 1 4.966 0.510 2.738 12 ksi2 2p p R R Rτ σ σ= − = + = =

12 4.38 kip

5.476R = = ............................................................................................................Ans.

(b) Maximum-Distortion-Energy:

2 21 1 2 2

48 24 ksi2.0

yp p p p FS

σσ σ σ σ− + = = =


545

( ) ( ) ( ) ( )2 24.966 4.966 0.510 0.510 5.2396 24 ksiR R R R R− − + − = =

24 4.58 kip

5.2396R = = .........................................................................................................Ans.

10-168 The shaft shown in Fig. P10-168 is made of steel having a proportional limit of 360 MPa in tension or compression. If a factor of safety of 2.0 with respect to failure by yielding is specified, determine the maximum permissible diameter D according to


( )( ) 2

4 3

15,000 0.600 2 91,673 N/m64x

DMcI D D

σπ×

= = =

( ) ( ) 2

4 3

15,000 0.800 2 61,115 N/m32xy

DTcJ D D

τπ×

= = =

22

1, 3

2 22

3 3 3 3 3

2 2

91,673 91,673 61,115 45,837 76,394 N/m2 2

x y x yp p xy

D D D D D

σ σ σ σσ τ

+ − = ± +

= + + = ±

21 3 3 3

45,837 76,394 122, 231 N/mp D D Dσ = + = +

22 0 N/mp zσ σ= =

23 3 3 3

45,837 76,394 30,557 N/mp D D Dσ = − = −


( )max1 360 90 MPa2 2 2.0

y

FSσ

τ = = =

( ) 6 2max 1 3 3 3 3

1 1 122,231 30,557 76,894 90 10 N/m2 2p p D D D

τ σ σ = − = + = = ×

13

6

76,894 0.0949 m 94.9 mm90 10

D = = = × ......................................................................Ans.


2 21 1 2 2

360 180 MPa2.0

yp p p p FS

σσ σ σ σ− + = = =

2 2

6 23 3 3 3

122, 231 122,231 30,557 30,557 180 10 N/mD D D D

− − + − = ×

0.0920 m 92.0 mmD = = ...................................................................................................Ans.


546

10-169 A steel shaft 4-in. in diameter is supported in flexible bearings at its ends. Two pulleys, each 24 in. in diameter, are keyed to the shaft. The pulleys carry belts that are loaded, as shown in Fig. P10-169. The steel has a proportional limit of 40 ksi in tension and compression and 23 ksi in shear. If a factor of safety of 2.5 with respect to failure by yielding is specified, determine the maximum permissible belt tension P according to


( )4 42 4 in4

I π π= =

( )4 42 8 in2

J π π= =

( ) ( ) ( )4 40 2

25.4654x

PMc PI

σπ

= = =

( )( )2 12 2

1.9108xy

PTc PJ

τπ

×= = =

( )

22

1, 3

22

2 2

25.465 0 25.465 0 1.9102 2

12.733 12.875

x y x yp p xy

P P P

P P

σ σ σ σσ τ

+ − = ± +

+ − = ± +

= ±

1 12.733 12.875 25.608 ksi 25.608 ksi (T)p P Pσ = + = + =

2 0 ksip zσ σ= =

3 12.733 12.875 0.142 ksi 0.142 ksi (C)p P P P Pσ = − = − =


max23 9.200 ksi2.5

y

FSτ

τ = = =

( ) ( )max 1 31 1 25.608 0.142 12.875 9.200 ksi2 2p p P P Pτ σ σ= − = + = =

9.200 0.715 kip 715 lb

12.875P = = = .......................................................................................Ans.


2 21 1 2 2

40 16.00 ksi2.5

yp p p p FS

σσ σ σ σ− + = = =

( ) ( ) ( ) ( )2 225.608 25.608 0.142 0.142 25.68 16.00 ksiP P P P P− − + − = =

16.00 0.623 kip 623 lb25.68

P = = = .........................................................................................Ans.


547

10-170 A structural steel pipe used to transmit steam has an inside diameter of 300 mm. If the steam pressure is 5.5 MPa and a factor of safety of 4 with respect to failure by yielding is specified, determine the minimum permissible wall thickness of the pipe according to the maximum-shear-stress theory of failure. The wall of the pipe does not carry any axial load.

SOLUTION

( )( )6

21

5.5 10 0.150 825,000 N/mh pprt t t

σ σ×

= = = = 3p pσ = −

( )max1 250 31.25 MPa2 2 4.0

y

FSσ

τ = = =

( ) 6

max 1 3

6 2

1 1 825,000 5.5 102 231.25 10 N/m

p p tτ σ σ = − = + ×

= ×

6

825,000 0.01447 m 14.47 mm57 10

t = = = × .......................................................................Ans.

10-171 A solid circular shaft acting as a cantilever beam is subjected to the loading shown in Fig. P10-171. The material is 2024-T4 wrought aluminum with a yield strength in tension and compression of 48 ksi. If the factor of safety for failure by yielding is 2.5, determine the minimum permissible diameter of the shaft according to the maximum-shear-stress theory of failure. Neglect the effects of transverse shear.

SOLUTION

( )

2 4 2 3

2 4 324 64x

M dP Mc P P MA I d d d d

σπ π π π

= ± = ± = ±

( )

4 3

2 1632xy

T dTc TJ d d

τπ π

= = =

2 2 2

2max 2 3 3

2 16 162 2

x y yxy

P M Td d d FS

σ σ στ τ

π π π− = + = + + =

( ) ( )2 23

2 8 82

y Pd M TFS d

σπ

= + + ...................................................................................Ans.

( ) ( ) ( )3

2 2

3

48 10 2 1000 8 200 8 8 50002 2.5

ddπ

× = + × +

1.412 in.d = ............................................................................................................................ Ans. 10-172 Two states of stress are shown in Fig. P10-172. The failure strengths for the material are 152 MPa in

tension and 572 MPa in compression. Use the Coulomb-Mohr theory to determine if these states of stress are safe.

SOLUTION

1 3 100 100 0.8327 1152 572

p p

ut uc

σ σσ σ

−− = − = ≤

(Safe) ................................................................ Ans.

1 3 75 200 0.8431 1152 572

p p

ut uc

σ σσ σ

−− = − = ≤

(Safe) ................................................................ Ans.


548

10-173 The state of stress at the critical point in a machine component is shown in Fig. P10-173. The failure strengths for the material are 26 ksi in tension and 97 ksi in compression. Use the Coulomb-Mohr theory to determine if this state of stress is safe.

SOLUTION

10 ksixσ = + 6 ksiyσ = − 4 ksixyτ = −

( )

( )

2 222

1, 310 6 10 6 4

2 2 2 2

2.0 8.944 ksi

x y x yp p xy

σ σ σ σσ τ

+ − − + = ± + = ± + −

= ±

1 2.0 8.944 10.944 ksi 10.944 ksi (T)pσ = + = + =

2 0 ksip zσ σ= =

3 2.0 8.944 6.944 ksi 6.944 ksi (C)pσ = − = − =

1 3 10.944 6.944 0.4925 1 (Safe)26 97

p p

ut uc

σ σσ σ

−− = − = ≤ ......................................................Ans.

10-174 The state of stress at the critical point in a machine component made of cast iron is shown in Fig. P10-174. The failure strengths for this material are 214 MPa in tension and 750 MPa in compression. Use the Coulomb-Mohr theory to determine if this state of stress is safe.

SOLUTION

150 MPaxσ = + 75 MPayσ = + 60 MPaxyτ = +

( )

( )

2 222

1, 2150 75 150 75 60

2 2 2 2

112.50 70.75 MPa

x y x yp p xy

σ σ σ σσ τ

+ − + − = ± + = ± +

= ±

1 112.50 70.75 183.25 MPa 183.25 MPa (T)pσ = + = + =

2 112.50 70.75 41.75 MPa 41.75 MPa (T)pσ = − = + =

3 0 MPap zσ σ= =

1 3 183.25 0 0.8563 1 (Safe)214 750

p p

ut uc

σ σσ σ

− = − = ≤ .............................................................Ans.

10-175 The solid circular cast iron shaft shown in Fig. P10-175 is subjected to a torque T = 240 kip⋅in. and a bending moment M = 110 kip⋅in. The failure strengths for this material are 43 ksi in tension and 140 ksi in compression. Use the Coulomb-Mohr theory to determine the minimum permissible diameter for this shaft.

SOLUTION

4 3

110 140.06 ksi4x

Mc cI c c

σπ

= = = 4 3

240 152.79 ksi2xy

Tc cJ c c

τπ

= = =

2 2 2

21, 3 3 3 3

140.06 0 140.06 0 152.792 2 2 2

x y x yp p xy c c c

σ σ σ σσ τ

+ − + − = ± + = ± +

3 3

70.03 168.07 ksic c

= ±


549

1 3 3 3 3

70.03 168.07 238.10 238.10ksi ksi (T)p c c c cσ = + = + =

2 0 ksip zσ σ= =

3 3 3 3 3

70.03 168.07 98.04 98.04ksi ksi (C)p c c c cσ = − = − =

1 33 3 3

238.10 98.04 6.237 143 140

p p

ut uc c c cσ σσ σ

−− = − = ≤ 1.8407 in.c ≥

( )min min2 2 1.8407 3.68 in.d c= = ≅ ...................................................................................Ans.

10-176 A solid circular cast iron shaft is subjected to the loads shown in Fig. P10-176. The failure strengths for this material are 214 MPa in tension and 750 MPa in compression. Use the Coulomb-Mohr theory to determine the minimum permissible diameter for the shaft. Neglect the effects of transverse shear.

SOLUTION

( )

4 3

1300 0.100 165.524x

cMcI c c

σπ×

= = = 4 3

6000 38202xy

Tc cJ c c

τπ

= = =

22

1, 3

2 2

3 3 3 3 3

2 2

165.52 0 165.52 0 3820 82.76 3821 MPa2 2

x y x yp p xy

c c c c c

σ σ σ σσ τ

+ − = ± +

+ − = ± + = ±

1 3 3 3 3

82.76 3821 3904 3904MPa MPa (T)p c c c cσ = + = + =

2 0 MPap zσ σ= =

3 3 3 3 3

82.76 3821 3738 3738MPa MPa (C)p c c c cσ = − = − =

( ) ( )6

1 336 3 6 3

3904 3738 23.23 10 1214 10 750 10

p p

ut uc cc cσ σσ σ

−− ×− = − = ≤× ×

328.53 10 mc −≥ ×

( )3 3min min2 2 28.53 10 57.06 10 m 57.1 mmd c − −= = × = × ≅ ........................................Ans.

10-177 A thin-walled cylindrical pressure vessel has an inside diameter of 12 in. and a wall thickness of 0.20 in. The vessel is made of a material with 40 ksiutσ = and 50 ksiucσ = . Use the Coulomb-Mohr theory to determine the maximum internal pressure that the vessel can safely support.

SOLUTION

( )

1

630

0.20p h

ppr pt

σ σ= = = = 3p r pσ σ= = −

( )

( )2

615

2 2 0.20p a

ppr pt

σ σ= = = =

1 3 30 140 50

p p

ut uc

p pσ σσ σ

−− = − ≤

1.299 ksip ≤ .......................................................................................................................... Ans.


550

10-178 The C-clamp shown in Fig. P10-178 is made of a gray cast iron that has an ultimate tensile strength of 180 MPa and an ultimate compressive strength of 670 MPa. Use the Coulomb-Mohr theory to determine the maximum load P that can be applied safely to the clamp.

SOLUTION

212 25 300 mmA = × = ( )3

412 2515,625 mm

12I = =

( ) ( ) ( )3 2

6 12

0.075 0.012563.33 10 N/m

300 10 15,625 10iPP Mc P P

A Iσ − −= + = + = + ×

× ×

( )( ) ( )3 2

6 12

0.075 0.012556.67 10 N/m

300 10 15,625 10oPP Mc P P

A Iσ − −= − = − = − ×

× ×

( )3

16

63.33 101

180 10p

ut

Pσσ

×= ≤

× 2842 NP ≤

( )3

36

56.67 101

670 10p

c

Pσσ

×= ≤

× 11,823 NP ≤

max 2842 N 2840 NP = ≅ .....................................................................................................Ans.

10-179 A solid circular gray cast iron shaft is loaded as shown in Fig. P10-179. The failure strengths for this type of cast iron are 36.5 ksi in tension and 124 ksi in compression. Use the Coulomb-Mohr theory to determine the minimum permissible diameter for the shaft. Neglect the effects of transverse shear.

SOLUTION

( )

4 3

1000 10 12,732 psi4x

cMcI c c

σπ

×= = =

( )

4 3

1000 12 7639 psi2xy

cTcJ c c

τπ

×= = =

22

1, 3

2 2

3 3 3 3 3

2 2

12,732 0 12,732 0 7639 6366 9944 psi2 2

x y x yp p xy

c c c c c

σ σ σ σσ τ

+ − = ± +

+ − = ± + = ±

1 3 3 3 3

6366 9944 16,310 16.310psi ksi (T)p c c c cσ = + = + =

2 0 ksip zσ σ= =

3 3 3 3 3

6366 9944 3578 3.578psi ksi (C)p c c c cσ = − = − =

1 33 3 3

16.310 3.578 0.4757 136.5 124

p p

ut uc c c cσ σσ σ

−− = − = ≤ 0.7806 in.c ≥

( )min min2 2 0.7806 1.561 in.d c= = ≅ .................................................................................Ans.


551

10-180 A 100-mm diameter solid circular shaft is loaded as shown in Fig. P10-180. The shaft is made of a gray cast iron that has an ultimate tensile strength of 150 MPa and an ultimate compressive strength of 570 MPa. Use the Coulomb-Mohr theory to determine if the shaft can safely support the loading shown.

SOLUTION

( )2

21007854 mm

4A

π= =

( )46 4100

9.817 10 mm32

Jπ

= = ×

3

6 26

22 10 2.801 10 N/m 2.801 MPa (T)7854 10x

PA

σ −

×= = = × =×

( ) 6 2

6

8000 0.05040.75 10 N/m 40.75 MPa

9.817 10xyTcJ

τ −= = = × =×

( ) ( )

22

1, 3

22

2 2

2.801 0 2.801 0 40.75 1.401 40.774 MPa2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

+ − = ± + = ±

1 1.401 40.774 42.175 MPa 42.175 MPa (T)pσ = + = + =

2 0 MPap zσ σ= =

3 1.401 40.774 39.373 MPa 39.373 MPa (C)pσ = − = − =

1 3 42.175 39.373 0.3502 1 (Safe)150 570

p p

ut uc

σ σσ σ

−− = − = ≤ ...................................................Ans.

10-181 The stresses shown in Fig. P10-181 act at a point on the free surface of a structural member subjected to plane stress. Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB.


12 ksixσ = + 6 ksiyσ = − 10 ksixyτ = + 60 90 30θ = ° − ° = − °

( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos

12cos 30 6 sin 30 2 10 sin 30 cos 30n x y xyσ σ θ σ θ τ θ θ= + +

= − ° + − − ° + − ° − °

1.1603 ksi 1.160 ksi (C)= − ≅ .......................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

12 6 sin 30 cos 30 10 cos 30 sin 30


= − + − ° − ° + − ° − − °

12.794 ksi 12.79 ksi= + ≅ + ..........................................................................................Ans.

10-182 The stresses shown in Fig. P10-182 act at a point on the free surface of a structural member subjected to plane stress. Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB.


120 MPaxσ = − 50 MPayσ = − 50 MPaxyτ = − 53 90 37θ = ° − ° = − °


552

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

cos sin 2 sin cos


= − − ° + − − ° + − − ° − °

46.58 MPa 46.6 MPa (C)= − ≅ ...................................................................................Ans.

( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2

2 2

sin cos cos sin

120 50 sin 37 cos 37 50 cos 37 sin 37


= − − + − ° − ° + − − ° − − °

47.43 MPa 47.4 MPa= − ≅ − ........................................................................................Ans.

10-183 The stresses shown in Fig. P10-183 act at a point on the free surface of a stressed body. The principal stresses at the point are 20 ksi (C) and 12 ksi (T). Determine the unknown normal stresses on the horizontal and vertical planes and the angle pθ between the x-axis and the maximum tensile stress at the point.

SOLUTION

2

21 8 12 ksi

2 2x y x y

p

σ σ σ σσ

+ − = + + =

2

22 8 20 ksi

2 2x y x y

p

σ σ σ σσ

+ − = − + = −

Adding and subtracting yields: 8 ksix yσ σ+ = −

27.713 ksix yσ σ− = +

9.857 ksi 9.86 ksi (T)xσ = + ≅ ..........................................................................................Ans.

17.857 ksi 17.86 ksi (C)yσ = − ≅ .....................................................................................Ans.

( )1 12 2 81 1tan tan 15.00

2 2 9.857 17.857xy

px y

τθ

σ σ− −= = = °

− +..............................................Ans.



20 MPaxσ = + 120 MPayσ = +

80 MPaxyτ = − 3 0 MPaz pσ σ= =

20 120 70 MPa

2a += =

2 250 80 94.34 MPaR = + =

(a) 1 70 94.34 164.34 MPa 164.3 MPa (T)pσ = + = + ≅ .....................................................Ans.

2 70 94.34 24.34 MPa 24.3 MPa (C)pσ = − = − ≅ .........................................................Ans.


553

3 0 MPapσ = .............................. max 94.3 MPap Rτ τ= = ≅ ..........................................Ans.

(b) 70 94.34cos6 163.82 MPa 163.8 MPa (T)nσ = + ° = + ≅ ...........................................Ans.

94.34sin 6 9.86 MPantτ = ° = .............................................................................................Ans.

10-185 The strain components at a point in a body under a state of plane strain are 1000 in./in.xε µ= + ,

800 in./in.yε µ= − , and 800 in./in.xyγ µ= − Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION


800 in./in.yε µ= − 800 in./in.xyγ µ= −

2 2

1, 2

2 2

2 2 2

1000 800 1000 800 8002 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

− + − = ± +

1 100 984.9 1084.9 in./in. 1085 in./in.pε µ µ= + = + ≅ + ................................................Ans.


3 0 in./in.p zε ε µ= = ...............................................................................................................Ans.

( )max 2 984.9 1970 radpγ γ µ= = ≅ ....................................................................................Ans.

1 11 1 800tan tan 11.982 2 1000 800

xyp

x y

γθ

ε ε− − −= = = − °

− +....................................................Ans.

10-186 The strain components at a point in a body under a state of plane strain are 1200 m/mxε µ= + ,

960 m/myε µ= + , and 960 m/mxyγ µ= − Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch.

SOLUTION


960 m/myε µ= + 960 m/mxyγ µ= −

2 2

1, 2

2 2

2 2 2

1200 960 1200 960 9602 2 2

x y x y xyp p

ε ε ε ε γε

+ − = ± +

+ − − = ± +

1 1080 494.8 1574.8 m/m 1575 m/mpε µ µ= + = + ≅ + ..................................................Ans.

2 1080 494.8 585.2 m/m 585 m/mpε µ µ= − = + ≅ + ......................................................Ans.

3 0 m/mp zε ε µ= = ........................... ( )2 494.8 990 radpγ µ= ≅ ..................................Ans.


554

( ) ( )max max min 1574.8 0 1575 radγ ε ε µ= − = − ≅ .............................................................Ans.

1 11 1 960tan tan 37.982 2 1200 960

xyp

x y

γθ

ε ε− − −= = = − °

− −....................................................Ans.

10-187 At a point on the free surface of steel (E = 30,000ksi and 0.30ν = ) machine part, the strain rosette shown in Fig. P10-187 was used to obtain the following normal strain data: 650 in./in.aε µ= + ,

475 in./in.bε µ= + , and 250 in./in.cε µ= − . Determine





650 in./in.a xε ε µ= = + 45 475 in./in.bε ε µ= = + 250 in./in.c yε ε µ= = −

(a) 2 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( ) ( )2 2650cos 45 250 sin 45 sin 45 cos 45 475 in./in.b xyε γ µ= ° + − ° + ° ° = +

Therefore: 650 in./in.xε µ= + 250 in./in.yε µ= − 550 radxyγ µ= +

( ) ( ) 62 2

30,000 650 0.30 250 101 1 0.30x x yEσ ε νεν

−= + = + − × − −

18.956 ksi 18.96 ksi (T)= + ≅ .....................................................................................Ans.

( ) ( ) ( ) 62 2

30,000 250 0.30 650 101 1 0.30y y xEσ ε νεν

−= + = − + × − −

1.8132 ksi 1.813 ksi (C)= − ≅ .....................................................................................Ans.

( ) ( ) ( )630,000 550 102 1 2 1 0.30xy xy xyEGτ γ γ

ν−= = = ×

+ +

6.346 ksi 6.35 ksi= + ≅ + ..............................................................................................Ans.

(b) 2 2 2 2

1, 2650 250 650 250 550

2 2 2 2 2 2x y x y xy

p p

ε ε ε ε γε

+ − − + = ± + = ± +

1 200 527.4 727.4 in./in. 727 in./in.pε µ µ= + = + ≅ + ....................................................Ans.


( ) ( )30.30 650 250 171.4 in./in.


ν= = − + = − − = −

− −....................Ans.

( )max 2 527.4 1055 radpγ γ µ= = ≅ ....................................................................................Ans.

1 11 1 550tan tan 15.712 2 650 250

xyp

x y

γθ

ε ε− −= = = °

− +.........................................................Ans.


555

(c) ( ) ( ) 61 1 22 2

30,000 727.4 0.30 327.4 101 1 0.30p p pEσ ε νεν

−= + = + − × − −

20.742 ksi 20.7 ksi (T)= + ≅ ......................................................................................Ans.

( ) ( ) ( ) 62 2 12 2

30,000 327.4 0.30 727.4 101 1 0.30p p pEσ ε νεν

−= + = − + × − −

3.599 ksi 3.60 ksi (C)= − ≅ .........................................................................................Ans.

( ) ( ) ( )6max max max

30,000 1054.8 102 1 2 1 0.30EGτ γ γ

ν−= = = ×

+ +

12.171 ksi 12.17 ksi= ≅ ...............................................................................................Ans.

10-188 The strain rosette shown in Fig. P10-188 was used to obtain normal strain data at a point on the free surface of an annealed titanium alloy machine part. The gage readings were 306 m/maε µ= − ,

456 m/mbε µ= − , and 906 m/mcε µ= + . Determine



(c) The principal stresses and the maximum shearing stress at the point. Express the stresses in SI units. SOLUTION The given values are:

30 306 m/maε ε µ= = − 75 456 m/mbε ε µ= = − 120 906 m/mcε ε µ= = +

(a) 2 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( )2 2cos 30 sin 30 sin 30 cos 30 306 in./in.a x y xyε ε ε γ µ= ° + ° + ° ° = −

( ) ( ) ( ) ( )2 2cos 75 sin 75 sin 75 cos 75 456 in./in.b x y xyε ε ε γ µ= ° + ° + ° ° = −

( ) ( ) ( ) ( )2 2cos 120 sin 120 sin 120 cos 120 906 in./in.c x y xyε ε ε γ µ= ° + ° + ° ° = +

Solving these three equations simultaneously gives:

652 in./in.xε µ= + 51.7 in./in.yε µ= − 1806 radxyγ µ= − ......................Ans.

(b) For annealed titanium alloy: 36 GPaG = 2(1 ) 96 GPaE Gν= + = 0.3333ν =

( ) ( )9

62 2

96 10 652 0.3333 51.7 101 1 0.3333x x yEσ ε νεν

−×= + = + − × − −

6 268.55 10 N/m 68.6 MPa (T)= + × ≅ .......................................................................Ans.

( ) ( ) ( )9

62 2

96 10 51.7 0.3333 652 101 1 0.3333y y xEσ ε νεν

−×= + = − + × − −

6 217.886 10 N/m 17.89 MPa (T)= + × ≅ ...................................................................Ans.

( )( )9 636 10 1806 10xy xyGτ γ −= = × − ×

6 265.02 10 N/m 65.0 MPa= − × ≅ − ...........................................................................Ans.


556

(c) 2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

( )2

268.55 17.886 68.55 17.886 65.022 2+ − = ± + −

1 43.22 69.78 113.00 MPa 113.0 MPa (T)pσ = + = + ≅ ...............................................Ans.

2 43.22 69.78 26.56 MPa 26.6 MPa (C)pσ = − = − ≅ ...................................................Ans.

3 0 MPap zσ σ= = ........................... max 69.8 MPapτ τ= ≅ ...........................................Ans.

( )1 12 2 65.021 1tan tan 34.36

2 2 68.55 17.886xy

px y

τθ

σ σ− − −

= = = − °− −

...........................................Ans.

10-189 The thin-walled pressure vessel shown in Fig. P10-189a has an inside diameter of 20 in. and a wall thickness of 3/8 in. The vessel is subjected to an internal pressure of p and a torque of T. At a point on the outside surface of the steel (E = 30,000 ksi and 0.30ν = ) vessel, the strain rosette shown in Fig. P10-189b was used to obtain the following normal strain data: 36 in./in.aε µ= + , 310 in./in.bε µ= + , and

150 in./in.cε µ= + . Gages a and c are oriented in the axial and hoop directions of the vessel, respectively. Determine the pressure p and the torque T.


36 in./in.a xε ε µ= = + 45 310 in./in.bε ε µ= = + 150 in./in.c yε ε µ= = +

2 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + +

( ) ( ) ( ) ( )2 236cos 45 150sin 45 sin 45 cos 45 310 in./in.b xyε γ µ= ° + ° + ° ° = +

Therefore: 36 in./in.xε µ= + 150 in./in.yε µ= + 434 radxyγ µ= +

( ) ( ) 62 2

30,000 36 0.30 150 10 2.670 ksi1 1 0.30x a x yEσ σ ε νεν

−= = + = + × = + − −

( ) ( ) 62 2

30,000 150 0.30 36 10 5.301 ksi1 1 0.30y h y xEσ σ ε νεν

−= = + = + × = + − −

( ) ( ) ( )630,000 434 10 5.008 ksi2 1 2 1 0.30xy xy xyEGτ γ γ

ν−= = = × = +

+ +

( )5.301 3 8

0.19879 ksi 198.8 psi10

htpr

σ= = = + ≅ + ....................................................Ans.

( )4 4

420.75 20

2492 in32

Jπ −

= =

( )5.008 2492

1203 kip in.10.375

xy JTc

τ= = = + ⋅ ......................................................................Ans.

10-190 A human femur is modeled as shown in Fig. P10-190. The abductor muscle force is M = 4060 N, and the femoral load is J = 5210 N.

(a) Determine the maximum tensile and compressive stresses at section a�a if the section is modeled as a solid circular section of 27-mm diameter.


557

(b) Determine the maximum tensile and compressive stresses at section a�a if the section is modeled as a hollow cylinder of outside diameter 27 mm and inside diameter 16 mm.

SOLUTION

0 :yF↑ Σ = 4060 5210 0P + − =

0 :MΣ =� ( ) ( )4060 0.016 5210 0.037 0M − − =

1150 NP = 257.73 N mM = ⋅ (a) Solid:

( )2

6 20.027572.6 10 m

4A

π −= = ×

( )4

9 40.02726.09 10 m

64I

π −= = ×

( ) ( ) 6 2

6 9

257.73 0.01351150 2.0084 133.36 10 N/m572.6 10 26.09 10

σ − −

−= ± = − ± ×× ×

6 2max 131.4 10 N/m 131.4 MPa (T)Tσ = + × = .................................................................Ans.

6 2max 135.4 10 N/m 134.4 MPa (C)Cσ = − × = ................................................................Ans.

(b) Hollow:

( )2 2

6 20.027 0.016

371.5 10 m4

Aπ

−−

= = ×

( )4 4

9 40.027 0.016

22.87 10 m64

Iπ

−−

= = ×

( ) ( ) 6 2

6 9

257.73 0.01351150 3.096 152.14 10 N/m371.5 10 22.87 10

σ − −

−= ± = − ± ×× ×

6 2max 149.0 10 N/m 149.0 MPa (T)Tσ = + × = .................................................................Ans.

6 2max 155.2 10 N/m 155.2 MPa (C)Cσ = − × = ................................................................Ans.

10-191 The clamp of Fig. P10-191 is used to hold two boards. If the clamping force is 80 lb, determine the maximum tensile and compressive stresses at section a�a if the clamp has a 1/2 × 3/16-in. cross section.

SOLUTION

( )( ) 21 2 3 16 0.09375 inA = = ( ) ( )3

3 43 16 1 21.9531 10 in

12I −= = ×

0 :yF↑ Σ = 80 0P− = 80 lbP =

0 :MΣ =� ( )80 3 0M − = 240 lb in.M = ⋅

( ) ( )3

240 1 480 853.33 30,720 psi0.09375 1.9531 10

σ −= ± = ±×

max 31,573 psi 31.6 ksi (T)Tσ = + ≅ ....................................... Ans.

max 29,867 psi 29.9 ksi (C)Cσ = − ≅ ....................................... Ans.


558

10-192 A steel shaft with the left portion solid and the right portion hollow is loaded as shown in Fig. P10-192. If the maximum shearing stress in the shaft must not exceed 80 MPa and the maximum tensile stress in the shaft must not exceed 140 MPa, determine the maximum axial load P that can be applied to the shaft.

SOLUTION

( )2

3 21007.854 10 mm

4ABAπ

= = × ( )2 2

3 2160 100

12.252 10 mm4BCA

π −= = ×

( )4

6 41009.817 10 mm

32ABJπ

= = × ( )4 4

6 4160 100

54.52 10 mm32BCJ

π −= = ×

3 127.327.854 10AB

P P PA

σ −= = =×

3 81.6212.252 10BC

P P PA

σ −= = =×

( ) ( )3

6 26

10 10 0.05050.93 10 N/m 50.93 MPa

9.817 10ABTcJ

τ −

×= = = × =

×

( )( )3

6 26

30 10 0.08044.02 10 N/m 44.02 MPa

54.52 10BCTcJ

τ −

×= = = × =

×

Since both ABσ and ABτ are larger than BCσ and BCτ , the maximum stresses will occur in section AB.

2

21, 2 2 2

x y x yp p xy

σ σ σ σσ τ

+ − = ± +

2

21

0 0 50.93 140 MPa2 2x x

pσ σσ + − = + + ≤

121.47 MPaxσ ≤

2

2max 2

x yp xy

σ στ τ τ

− = = +

2

20 50.93 80 MPa2xσ − = + ≤

123.39 MPaxσ ≤

Therefore: 121.47 MPaxσ ≤ and

( )( )6 3max max 121.47 10 7.854 10ABP Aσ −= = × ×

3954 10 N 954 kN= × = ..............................................................................................Ans.

10-193 A hollow circular shaft has an outside diameter of 12 in. and an inside diameter of 4 in., as shown in Fig. P10-193. The shaft is in equilibrium when the indicated loads and torque are applied. If the maximum normal and shearing stresses at point A must be limited to 5000 psi (T) and 3000 psi, respectively, determine the maximum permissible value for the axial load P.

SOLUTION

( )2 2

212 4

100.53 in4

Aπ −

= = ( )4 4

412 4

1005.3 in64

Iπ −

= =

( )4 4

412 4

2010.6 in32

Jπ −

= =


559

At Point A:

( )( )

( )3

10,000 48 6100.53 1005.3

9.947 10 2865 psi

xP Mc PA I

P

σ

−

×= + = +

= × +

( )400,000 6

1193.7 psi2010.6xy

TcJ

τ = = =

2

21, 2

0 0 1193.7 5000 psi2 2x x

p pσ σσ + − = ± + ≤

4715 psixσ ≤

2

2max

0 1193.7 3000 psi2x

pστ τ − = = + ≤

5505 psixσ ≤

Therefore: ( )39.947 10 2865 4715 psix Pσ −= × + =

from which: 3max 185.99 10 lb 186.0 kipP = × ≅ .......................................................................Ans.

10-194 The frame shown in Fig. P10-194 is subjected to a load of 100 kN. The material has a proportional limit of 220 MPa in tension and compression. Determine the factor of safety with respect to failure by yielding according to

(a) The maximum-shearing-stress theory of failure. (b) The maximum-distortion-energy theory of failure. SOLUTION

( ) 3 22 50 150 15.00 10 mmA = × = ×

( ) ( ) 225 50 150 125 150 50

75.00 mm15,000Cx

× + ×= =

( ) ( ) ( ) ( ) ( ) ( )

3 32 2 6 4

50 150 150 5050 150 50 150 50 50 53.125 10 mm

12 12I

= + × + + × = ×

( )( )3

3 6

6 2

100,000 0.425 0.075100 1015 10 53.125 10

66.67 10 N/m 66.67 MPa (T)

AP McA I

σ − −

××= + = +× ×

= + × =

( )( )3

3 6

6 2

100,000 0.425 0.125100 1015 10 53.125 10

93.33 10 N/m 93.33 MPa (C)

BP McA I

σ − −

××= − = −× ×

= − × =

Since the state of stress is uniaxial at surfaces A and B: (a) Maximum-Shear-Stress:

( ) ( )max 1 31 1 0 93.33 46.67 MPa2 2p pτ σ σ= − = + =

max220 110 MPa

2 2fσ

τ = = =


560

max

110 2.3646.67

fFSτ

τ= = = ...................................................................................................Ans.


2 21 1 2 2e p p p pσ σ σ σ σ= − +

( ) ( )( ) ( )2 20 0 93.33 93.33 93.33 MPa− − + − =

max

220 2.3693.33

fFSσ

σ= = = ...................................................................................................Ans.


561

Chapter 11 11-1 A structural steel rod 1 in. in diameter and 50 in. long will be used to support an axial compressive load P. Determine (a) The slenderness ratio. (b) The smallest slenderness ratio for which the Euler buckling load equation is valid. (c) The Euler buckling load. SOLUTION

( )22

210.7854 in

4 4dA

ππ= = = ( )44

410.04909 in

64 64dI

ππ= = =

0.04909 0.250 in.0.7854

IrA

= = =

(a) 50 200

0.25Lr

= = ......................................................................... Ans.

(b) ( )

2

2cr

cr yP EA L r

πσ σ= = =

( )22 29,000

89.17 89.236y

L Er

ππσ

= = = ≅ ................................................................... Ans.

(c) ( )( )

( )

22

22

29,000 0.049095.62 kip

50cr

EIPL

ππ= = = ........................................................... Ans.

11-2 A hollow, circular structural steel column 6 m long has an outside diameter of 125 mm and an inside diameter of 100 mm. Determine

(a) The slenderness ratio. (b) The smallest slenderness ratio for which the Euler buckling load equation is valid. (c) The Euler buckling load. SOLUTION

( )2 2

2125 100

4418 mm4

Aπ −

= = ( )4 4

6 4125 100

7.075 10 mm64

Iπ −

= = ×

67.075 10 40.02 mm

4418IrA

×= = =

(a) 36 10 149.92 149.9

40.02Lr

×= = ≅ ................................................ Ans.

(b) ( )

2

2cr

cr yP EA L r

πσ σ= = =

( )2 92

6

200 1088.9

250 10y

L Er

ππσ

×= = =

×............................................................................... Ans.

(c) ( )( )

( )

2 9 623

22

200 10 7.075 10387.9 10 N 388 kN

6cr

EIPL

ππ −× ×= = = × ≅ .................... Ans.


562

11-3 A wood column of air-dried Douglas fir is 10 ft long and has a 4 × 4-in. rectangular cross section. Determine (a) The slenderness ratio. (b) The smallest slenderness ratio for which the Euler buckling load equation is valid. (c) The Euler buckling load. SOLUTION

( ) 24 4 16 inA = =

( )3

44 421.33 in

12I = =

21.33 1.1546 in.

16IrA

= = =

(a) 10 12 103.93 103.91.1546

Lr

×= = ≅ .............................................................................................. Ans.

(b) ( )

2

2cr

cr yP EA L r

πσ σ= = =

( )22 1900

54.13 54.16.4y

L Er

ππσ

= = = ≅ ........................................................................ Ans.

(c) ( )( )( )

22

22

1900 21.3327.8 kip

10 12cr

EIPL

ππ= = =×

.................................................................... Ans.

11-4 A 5-m long column with the cross section shown in Fig. P11-4 is constructed from four pieces of timber. The timbers are nailed together so that they act as a unit. Determine

(a) The slenderness ratio. (b) The Euler buckling load. Use E = 14 GPa for the timber. (c) The axial stress in the column when the Euler load is applied. SOLUTION

2 2 2150 100 12,500 mmA = − =

4 4

6 4150 100 33.85 10 mm12

I −= = ×

633.85 10 52.04 mm

12,500IrA

×= = =

(a) 35 10 96.08 96.1

52.04Lr

×= = ≅ .................................................................................................... Ans.

(b) ( )( )

( )

2 9 623

22

14 10 33.85 10187.09 10 N 187.1 kN

5cr

EIPL

ππ −× ×= = = × ≅ .................. Ans.

(c) 3

6 23

187.09 10 14.967 10 N/m 14.97 MPa12.5 10

PA

σ −

×= = = × ≅×

........................................... Ans.


563

11-5 Determine the allowable compressive load that a 5-in. × 5-in. air-dried red oak timber can support if it is 18 ft long, and a factor of safety of 3 is specified.

SOLUTION

From Table A-17: 4.6 ksiyσ = 1800 ksiE =

25 5 25 inA bh= = × = ( )33

45 552.08 in

12 12bhI = = =

52.08 1.4433 in.

25IrA

= = = 18 12 149.661.4433

Lr

×= =

( )( )

( )( )( )( )

22

max 2 2

1800 52.086.61 kip

3 18 12crP EIPFS FS L

ππ= = = =×

.............................................. Ans.

11-6 A 3-m long column with the cross section shown in Fig. P11-6 is constructed from two pieces of air-dried Douglas fir timber. The timbers are nailed together so that they act as a unit. Determine


( ) ( ) 3 2150 50 50 150 15.00 10 mmA = + = ×

( ) ( )

3

25 150 50 125 50 15075.00 mm

15.00 10Cy+ = =×

( ) ( )( )

432150 50

150 50 5012xI

= + ×

( ) ( )( )

32 6 450 150

50 150 50 53.13 10 mm12

+ + × = ×

( ) ( )3 3

6 4150 50 50 15015.625 10 mm

12 12yI = + = ×

6

min3

15.625 10 32.27 mm15.00 10

IrA

×= = =×

(a) 33 10 92.97 93.0

32.27Lr

×= = ≅ ................................................................................................... Ans.

(b) ( )

2

2cr

cr yP EA L r

πσ σ= = =

( )2 92

6

13 1054.0

44 10y

L Er

ππσ

×= = =

×.................................................................................. Ans.

(c) ( )( )

( )

2 9 623

22

13 10 15.625 10223 10 N 223 kN

3cr

EIPL

ππ −× ×= = = × ≅ ........................ Ans.


564

11-7 A WT6 × 36 structural steel section (see Appendix A for cross-sectional properties) is used for an 18-ft long column. Determine


For a WT 6 36× section: 210.6 inA = min 1.48 in.r =

(a) 18 12 145.94 145.91.48

Lr

×= = ≅ ............................................................................................... Ans.

(b) ( )

2

2cr

cr yP EA L r

πσ σ= = =

( )22 29,000

89.236y

L Er

ππσ

= = = ................................................................................... Ans.

(c) ( )

( )( )( )

22

2 2

29,000 10.6142.4 kip

145.94cr

EAPL r

ππ= = = .............................................................. Ans.

11-8 Two L127 × 76 × 12.7-mm structural steel angles (see Appendix A for cross-sectional properties) are used for a column that is 4.5 m long. Determine the total compressive load required to buckle the two members if

(a) They act independently of each other. (b) They are riveted together as shown in Fig. P11-8. SOLUTION

From Table A-18: 250 MPayσ = 200 GPaE =

For an L127 76 12.7× × section: 22420 mmA = min 16.5 mmr =

40.4 mmXXr = 21.1 mmYYr = 19.1 mmCx =

(a) 3

min

4.5 10 272.716.5

Lr

×= =

( )

( )( )( )

2 9 62

2 2

200 10 2 2420 10

272.7cr

EAPL r

ππ −× × ×= =

3128.47 10 N 128.5 kN= × ≅ ........................................................................................ Ans.

(b) For the riveted angles: 2 221.1 19.1 28.46 mmyr = + =

3

min

4.5 10 158.1228.46

Lr

×= =

( )

( )( )( )

2 9 62

2 2

200 10 2 2420 10

158.12cr

EAPL r

ππ −× × ×= =

3382.1 10 N 382 kN= × ≅ ........................................................................... Ans.


565

11-9 Determine the maximum allowable compressive load for a 12-ft long aluminum alloy (E = 10,600 ksi) column having the cross section shown in Fig. P11-9 if a factor of safety of 2.50 is specified.

SOLUTION

( ) ( ) ( )2 22

4 38 1 4 1 17.498 in

4A

π −= + + =

( ) ( ) ( )2 20.5 8 1 3 4 1 7 4 3 4

3.114 in.17.498Cy

π + + − = =

( ) ( )( ) ( ) ( )( )

3 32 28 1 1 4

8 1 2.614 1 4 0.11412 12xI

= + × + + ×

( ) ( ) ( )

4 4 2 22 4

4 3 4 33.886 152.33 in

64 4π π − −

+ + =

( ) ( ) ( )3 3 4 4

44 31 8 4 1

51.59 in12 12 64yI

π −= + + =

( )

( )( )( )

22

max 2 2

10,600 51.59104.1 kip

2.50 12 12crP EIPFS FS L

ππ= = = =×

............................................ Ans.

11-10 A 60-kN load is supported by a tie rod AB and a pipe strut BC, as shown in Fig. P11-10. The tie rod has a diameter of 30 mm and is made of steel with a modulus of elasticity of 210 GPa and a yield strength of 360 MPa. The pipe strut has an inside diameter of 50 mm and a wall thickness of 15 mm and is made of an aluminum alloy with a modulus of elasticity of 73 GPa and a yield strength of 280 MPa. Determine the factor of safety with respect to failure by yielding or buckling for the structure.

SOLUTION

0 :CMΣ = ( ) ( )2.0 60 1.5 0ABF − =

45.0 kN (T)ABF =

( )2

230706.9 mm

4ABAπ

= =

( )( )6 6 3360 10 706.9 10 254.5 10 N 254.5 kNAB AB ABP Aσ −= = × × = × =

254.5 5.66 (with respect to yielding)

45AB

ABAB

PFSF

= = =

( )1tan 1.5 2.0 36.870θ −= = °

0 :yF↑ Σ = cos36.870 60 0BCF ° − = 75.0 kN (C)BCF =

( )2 2

280 50

3063 mm4BCA

π −= =

( )( )6 6 3280 10 3063 10 857.6 10 N 857.6 kNBC BC BCP Aσ −= = × × = × =

857.6 11.43 (with respect to yielding)

75BC

BCBC

PFSF

= = =


566

( )4 4

6 480 50

1.7038 10 mm64BCI

π −= = ×

( ) ( )( )( )

2 9 623

22

73 10 1.7038 10196.3 10 N 196.3 kN

2.5BC cr

EIPL

ππ −× ×= = = × =

196.3 2.62 (with respect to buckling)

75BCFS = = ........................................................... Ans.

11-11 The 2-in. diameter solid circular column shown in Fig. P11-11 is made of an aluminum alloy [E = 10,000 ksi and α = 12.5(10-6)/°F]. If the column is initially stress free, determine the temperature increase that will cause the column to buckle.

SOLUTION

( )22

223.142 in

4 4dA

ππ= = = ( )44

420.7854 in

64 64dI

ππ= = =

0.7854 0.500 in.3.142

IrA

= = = 10 12 2400.500

Lr

×= =

Eσ ε= Tε α= ∆ PA

σ = ( )

2

2crEAP

L rπ=

( ) ( )( )

2 2

2 2613.71 F

12.5 10 240crPTAE L r

ε π πα α α −

∆ = = = = = °×

................................... Ans.

11-12 Two C229 × 30 structural steel channels (see Appendix A for cross-sectional properties) are used for a column that is 12 m long. Determine the total compressive load required to buckle the two members if they are laced 150 mm back-to-back, as shown in Fig. P11-12.

SOLUTION For a C 229 30× section:

81.8 mmxr = 14.8 mmCx =

16.3 mmyr = 23795 mmA =

For the latticed channels: ( )2216.3 75 14.8 91.3 mmyr = + + =

min81.8 mmxr r= = 312 10 146.7

81.8Lr

×= =

( )

( )( )( )

2 9 62

2 2

200 10 2 3795 10

146.7cr

EAP PL r

ππ −× × ×= = =

3696.2 10 N 696 kN= × ≅ ...................................................................................... Ans.

11-13 A simple pin-connected truss is loaded and supported as shown in Fig. P11-13. The members of the truss were fabricated by bolting two C10 × 30 channel sections (see Appendix A for cross-sectional properties) back-to-back to form an H-section. The channels are made of structural steel with a modulus of elasticity of 29,000 ksi and a yield strength of 36 ksi. Determine the maximum load P that can be applied to the truss if a factor of safety of 1.75 with respect to failure by yielding and a factor of safety of 4 with respect to failure by buckling is specified.

SOLUTION


567

From a free-body diagram of the entire truss, the equilibrium equations give:

0 :CMΣ = ( ) ( ) ( )18 2 36 9 0EP P R+ − =

10 10ER P P= + = ↑

( )1tan 12 9 53.130θ −= = °

From a free-body diagram of Pin A, the equilibrium equations give:

0 :xF→ Σ = cos53.130 0AB ADT T+ ° =

0 :yF↑ Σ = sin 53.130 2 0ADT P− °− =

1.50 1.50 (T)ABT P P= + =

2.50 2.50 (C)ADT P P= − =

From a free-body diagram of the part of the truss to the left of Pin B, the equilibrium equations give:

0 :yF↑ Σ = sin 53.130 2 0BDT P° − =

2.50 2.50 (T)BDT P P= + =

From a free-body diagram of the part of the truss to the left of Pin E, the equilibrium equations give:

0 :yF↑ Σ = sin 53.130 3 0BET P− °− =

3.75 3.75 (C)BET P P= − =

From a free-body diagram of the part of the truss to the left of Pin C, the equilibrium equations give:

0 :yF↑ Σ = sin 53.130 10 3 0CET P P° + − =

8.75 8.75 (C)CET P P= − =

0 :BMΣ = ( ) ( )12 2 18 0DET P+ = 3.00 3.00 (C)DET P P= − =

0 :EMΣ = ( ) ( ) ( )9 2 27 12 0BCP P T+ − = 5.25 5.25 (T)BCT P P= + =

For a C10 30× section: 28.82 inA =

For member BC: ( )max 36 2 8.82 635.0 kipyieldF Aσ= = × =

( ) ( )max5.25 635.0 1.75BCT P F FS= ≤ = 69.1 kipP ≤

For a C10 30× section: 4103 inxI = 43.94 inyI = 0.649 in.Cx =

For the bolted channels: ( )2 42 3.94 8.82 0.649 15.31 inxI = + =

For member CE: ( )( )( )

22

22

29,000 15.31135.25 kip

15 12cr

EIPL

ππ= = =×

( ) ( )8.75 135.25 4CE crT P P FS= ≤ = 3.86 kipP ≤

Therefore max 3.86 kipP = ............................................................................................ Ans.


568

11-14 An L102 × 76 × 6.4-mm aluminum alloy (E = 70 GPa) angle is used for a fixed-end, pinned-end column having an actual length of 2.5 m. Determine the maximum safe load for the column if a factor of safety of 1.75 with respect to failure by buckling is specified. See Appendix A for cross-sectional properties; they are the same as those for a steel angle of the same size.

SOLUTION

( )0.7 0.7 2.5 1.75 m 1750 mmL L′ = = = =

For a L102 76 6.4× × section: 21090 mmA = min 16.5 mmr =

min

1750 106.0616.5

Lr

′= =

( )( )

( )( )( )

2 9 62

max 2 2

70 10 1090 10

1.75 106.06crP EAPFS FS L r

ππ −× ×= = =

′

338.25 10 N 38.3 kN= × ≅ ........................................................................................... Ans.

11-15 A W8 × 15 structural steel section (see Appendix A for cross-sectional properties) is used for a fixed-end, free-end column having an actual length of 14 ft. Determine the maximum safe load for the column if a factor of safety of 2 with respect to failure by buckling is specified. Use E = 29,000 ksi.

SOLUTION

( )2 2 14 28 ftL L′ = = =

For a W8 15× section: 24.44 inA = min 0.876 in.r =

min

28 12 383.60.876

Lr

′ ×= =

( )( )

( )( )( )

22

max 2 2

29,000 4.444.32 kip

2 383.6crP EAPFS FS L r

ππ= = = =′

...................................... Ans.

11-16 A W254 × 33 structural steel (E = 200 GPa) section is used for a column with an actual length of 6 m. The column can be considered fixed at both ends for bending about the axis of the cross section with the smallest second moment of area and pinned at both ends for bending about the axis with the largest second moment of area. Determine the maximum axial compressive load P that can be supported by the column if a factor of safety of 1.9 with respect to failure by buckling is specified.

SOLUTION

For a W 254 33× section: 24185 mmA = 108 mmxr = 33.8 mmyr =

For axis x-x: 6.00 m 6000 mmL L′ = = = 6000 55.56108

Lr′= =

For axis y-y: ( )0.5 0.5 6.00 m 3000 mmL L′ = = = 3000 88.7633.8

Lr′= =

( )( )

( )( )( )

2 9 62

max 2 2

200 10 4185 10


ππ −× ×= = =

′

3551.9 10 N 552 kN= × ≅ ................................................................................. Ans.


569

11-17 A WT7 × 24 structural steel section (see Appendix A for cross-sectional properties) is used for a column with an actual length of 45 ft. If the modulus of elasticity for the steel is 29,000 ksi and a factor of safety of 2 with respect to failure by buckling is specified, determine the maximum safe load for the column under the following support conditions:

(a) Pinned-pinned (c) Pinned-fixed (b) Fixed-free (d) Fixed-fixed SOLUTION

For a WT 7 24× section: 27.07 inA = 1.87 in.xr =

424.9 inXXI = 425.7 inYYI = 1.91 in.yr =

( )( )

( )( )( ) ( )

22 6

2 2 2

29,000 24.9 3.563 10 kip2

crallow

P EIPFS FS L L L

ππ ×= = = =′ ′ ′

(a) 45 ft 45 12 540 in.L L′ = = = × =

( ) ( )

6 6

2 23.563 10 3.563 10 12.22 kip

540allowP

L× ×= = =′

................................................................. Ans.

(b) ( )2 2 45 ft 90 ft 90 12 1080 in.L L′ = = = = × =

( ) ( )

6 6

2 23.563 10 3.563 10 3.06 kip

1080allowP

L× ×= = =′

................................................................... Ans.

(c) ( )0.7 0.7 45 ft 31.5 ft 31.5 12 378 in.L L′ = = = = × =

( ) ( )

6 6

2 23.563 10 3.563 10 24.9 kip

378allowP

L× ×= = =′

................................................................... Ans.

(c) ( )0.5 0.5 45 ft 22.5 ft 22.5 12 270 in.L L′ = = = = × =

( ) ( )

6 6

2 23.563 10 3.563 10 48.9 kip

270allowP

L× ×= = =′

................................................................... Ans.

11-18 A solid circular rod with diameter D, length L, and modulus of elasticity E will be used to support an axial compressive load P. The support system for the rod is shown in Fig. P11-18. Determine the critical buckling load in terms of D, L, and E if buckling occurs in the plane of the page.

SOLUTION

4

64DI π=

2LL′ =

( )

( )( )

2 42 3 4

2 2 2

64162

cr

E DEI EDPLL L

π ππ π= = =′

............................................................................ Ans.


570

11-19 The structural steel (E = 29,000 ksi) bar shown in Fig. P11-19 has a 1.0-in. diameter and is 6 ft long. The support at the top of the bar permits free vertical movement but no lateral movement or rotation. The bottom of the bar is free to move laterally but cannot rotate. Determine the maximum load P that can be applied if a factor of safety of 2.5 with respect to failure by buckling is specified.

SOLUTION

( )22

21.00.7854 in

4 4dA

ππ= = =

( )44

41.00.04909 in

64 64dI

ππ= = =

6 ft 6 12 72 in.L L′ = = = × =

0.04909 0.7854 0.2500 in.r I A= = =

72 288 (slender)

0.2500Lr′= =

( )( )

( )( )( )

22

2 2

29,000 0.049091.084 kip

2.5 72cr

allowP EIPFS FS L

ππ= = = =′

................................ Ans.

11-20 Two 25-mm diameter structural steel (E = 200 GPa) columns support a 400-kg mass, as shown in Fig. P11-20. The two short struts at the sides of the mass prevent horizontal movement of the mass, but offer no constraint to vertical motion. Determine the factor of safety with respect to failure by buckling.

SOLUTION

( )22

225490.9 mm

4 4dA

ππ= = =

( )44

3 42519.175 10 mm

64 64dI

ππ= = = ×

3 m 3000 mmL L′ = = =

319.175 10 490.9 6.250 mmr I A= = × =

3000 480 (slender)6.250

Lr′= =

( )

( )( )( )

2 9 92

2 2

200 10 19.175 104206 N

3cr

EIPL

ππ −× ×= = =

′

( )( )

2 42062 2 2.14400 9.81

cr crP PFSW mg

= = = = .............................................................................. Ans.

11-21 Verify the effective length for the fixed-end, fixed-end column shown in Fig. 11-4b by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Place the origin of the xy-coordinate system at the lower end of the column with the x-axis along the axis of the undeformed column.

SOLUTION EIy M Py′′ = −

P My yEI EI

′′ + =


571

which has the solution

( )sin cosy A kx B kx M P= + + 2k P EI=

cos siny Ak kx Bk kx′ = −

Boundary conditions: At 0,x = 0 :y′ = 0A =

At 0,x = 0 :y = B M P= −

At 2 ,x L= 0 :y′ = ( )sin 2 0kL =

If ( )sin 2 0kL = , then 2 2kL P L n

EIπ= =

or 2 2

2

4n EIPLπ=

The smallest value of P is obtained with 1n = . Therefore:

( ) ( )

2 2 2

2 22

42

crEI EI EIPL L L

π π π= = =′

and 2L L′ = ..................................................................................................................................... Ans.

11-22 Verify the effective length for the fixed-end, free-end column shown in Fig. 11-4c by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Place the origin of the xy-coordinate system at the lower end of the column with the x-axis along the axis of the undeformed column.

SOLUTION

( )EIy P yδ′′ = −

P Py yEI EI

δ′′ + =


sin cosy A kx B kx δ= + + 2k P EI=



At 0,x = 0 :y = B δ= −

At ,x L= :y δ= ( )cos 0kL =

If ( )cos 0kL = , then ( )2 1PkL L nEI

π= = +

or ( )2 2

2

2 14

n EIP

Lπ+

=

The smallest value of P is obtained with 0n = . Therefore:

( ) ( )

2 2 2

2 224 2cr

EI EI EIPL L L

π π π= = =′

and 2L L′ = ....................................................................................................................................... Ans.


572

11-23 A free-body diagram for the fixed-end, pinned-end column shown in Fig. 11-4d is shown in Fig. P11-23. Use the free-body diagram to develop the differential equation of the elastic curve. Verify that the effective length for the fixed-end, pinned-end column is 0.7L L′ = by solving the differential equation of the elastic curve and applying the appropriate boundary conditions.

SOLUTION

0 :OMΣ = 0FL M− = F M L=

MEIy M Py xL

′′ = − −

P M My y xEI EI EIL

′′ + = −


sin cos M My A kx B kx xP PL

= + + − 2k P EI=

cos sin My Ak kx Bk kxPL

′ = − −

Boundary conditions: At 0,x = 0 :y′ = A M PLk=

At 0,x = 0 :y = B M P= −

At ,x L= 0 :y = ( )tan kL kL=

When ( )tan kL kL= , 1.430PkL LEI

π= = or 2

2

2.045 EIPLπ=

Therefore: ( ) ( )

2 2 2

2 22

2.0450.6993

crEI EI EIP

L L Lπ π π= = =

′

and 0.6993 0.7L L L′ = ≅ .............................................................................................................. Ans. 11-24 A rigid block is supported by two fixed-end, fixed-end columns, as shown in Fig. P11-24a. Determine the effective lengths

of the columns by solving the differential equation of the elastic curve and applying the appropriate boundary conditions. Assume that buckling occurs as shown in Fig. P11-24b.

SOLUTION EIy M Py′′ = −

P My yEI EI

′′ + =

which has the solution ( )sin cosy A kx B kx M P= + + 2k P EI=



At 0,x = 0 :y = B M P= −

At ,x L= 0 :y′ = ( )sin 0kL =

If ( )sin 0kL = , then PkL L nEI

π= = or 2 2

2

n EIPLπ=

The smallest value of P is obtained with 1n = . Therefore: ( )

2 2

22crEI EIPL L

π π= =′

and L L′ = ......................................................................................................................................... Ans.


573

11-25 An air-dried red oak (E = 1800 ksi and Fc = 4.6 ksi) timber column with an effective length of 5 ft has a 3 × 3-in. rectangular cross section. Determine the maximum compressive load permitted by Code 4.

SOLUTION

23 3 9.0 inA bd= = × =

18000.671 0.671 13.2734.6c

EkF

= = =

5 12 20

3L kd′ ×= = >

( )

( )( )2 2

0.30 18000.30 1.350 ksi20.0

allowE

L dσ = = =

( )1.350 9.0 12.15 kipallowP Aσ= = = ................................................................................ Ans.

11-26 A W254 × 89 structural steel (E = 200 GPa and σy = 250 MPa) column is pinned at both ends, is 4.5 m long, and supports an axial compressive load P. Determine the maximum load permitted by Code 1.

SOLUTION

( )22 2 200,0002 125.66250C

y

ECππ

σ= = =

From Table A-2 (for a W 254 89× wide-flange section):

211,355 mmA = 112 mmXXr = 65.3 mmYYr =

min

4500 68.91 125.6665.3 C

L Cr

′= = < =

35 3 68.91 1 68.91 1.852

3 8 125.66 8 125.66FS = + − =

2250 1 68.911 114.69 MPa

1.852 2 125.66allowσ = − =

( )( )6 6max 114.69 10 11,355 10allowP Aσ −= = × ×

31302.3 10 N 1302 kN= × ≅ ..................................................................... Ans.

11-27 A 3-in. diameter solid circular 6061-T6 aluminum alloy bar is to be used as a column with an effective length of 30 in. Determine the maximum axial compressive load P permitted by Code 3.

SOLUTION

( )22 24 3 4 7.069 inA dπ π= = = ( )44 464 3 64 3.976 inI dπ π= = =

3.976 7.069 0.750 in.r I A= = =

30 40.0 66

0.750Lr

= = <

( ) ( )20.2 0.126 20.2 0.126 40.0 15.16 ksiallow L rσ = − = − =

( )1516 7.069 107.2 kipallowP Aσ= = = ............................................................................. Ans.


574

11-28 Three hollow circular structural steel tubes with inside diameters of 50 mm and outside diameters of 80 mm will be used for a 3.5-m long pin-ended column. Determine the maximum compressive load permitted by Code 1 if

(a) The three tubes act as independent axially loaded members. (b) The three tubes are welded together as shown in Fig. P11-28. SOLUTION


( )22 2 200,0002 125.66250C

y

ECππ

σ= = =

( )2 2

280 50

3063 mm4

Aπ −

= = ( )4 4

6 480 50

1.7038 10 mm64

Iπ −

= = ×

1,703,800 23.58 mm

3063IrA

= = = 3.5 m 3500 mmL L′ = = =

(a) 3500 148.43 125.6623.58 C

L Cr

= = > =

( )

( )( )

2 926 2

2 2

200 1046.66 10 N/m 46.66 MPa

1.92 1.92 148.43allow

EL r

ππσ×

= = = × =

( )( )6 6max 46.66 10 3 3063 10allowP Aσ −= = × × ×

3428.8 10 N 429 kN= × ≅ ......................................................................... Ans.

(b) ( ) 23 3063 9189 mmA = × =

( ) ( )( )26 6 43 1.7038 10 2 3063 40 14.913 10 mmI = × + = ×

14,913,000 40.29 mm

9189IrA

= = = 3500 86.87 125.6640.29 C

L Cr

= = < =

35 3 86.87 1 86.87 1.884

3 8 125.66 8 125.66FS = + − =

2250 1 86.871 100.99 MPa

1.884 2 125.66allowσ = − =

( )( )6 6max 100.99 10 9189 10allowP Aσ −= = × ×

3928 10 N 928 kN= × ≅ ............................................................................. Ans.

11-29 Two C10 × 15.3 structural steel channels 12 ft long are used as a fixed-ended, pin-ended column. Determine the maximum load permitted by Code 1 if the channels are welded together to form a 10 × 5.2-in. box section, as shown in Fig. P11-29.

SOLUTION

From Table A-17: 36 ksiyσ = 29,000 ksiE =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

From Table A-5 (for a C10 15.3× channel): 24.49 inA =


575

3.87 in.XXr = 0.713 in.YYr = 2.600 in.fw = 0.634 in.Cx =

( )0.7 0.7 12 8.40 ftL L′ = = =

For the cross section:

3.87 in.xr = ( ) ( )2 20.713 2.600 0.634 2.091 in.yr = + − =

min

8.4 12 48.21 126.102.091 C

L Cr

′ ×= = < =

35 3 48.21 1 48.21 1.803

3 8 126.10 8 126.10FS = + − =

236 1 48.211 18.508 ksi

1.803 2 126.10allowσ = − =

( )( )max 18.508 2 4.49 166.2 kipallowP Aσ= = × = ............................................................. Ans.

11-30 Four L76 × 76 × 12.7-mm structural steel angles 6 m long are used as a fixed-ended column. Determine the maximum load permitted by Code 1 if the angles are fastened together as shown in Fig. P11-30.

SOLUTION


( )22 2 200,0002 125.66250C

y

ECππ

σ= = =

From Table A-8 (for an L76 76 12.7× × angle):

21775 mmA = 22.8 mmYYr = 23.7 mmCx =

( ) ( )2 222.8 23.7 32.89 mmx yr r= = + =

( )0.5 0.5 6 3 m 3000 mmL L′ = = = =

3000 91.21 125.6632.89 C

L Cr′= = < =

35 3 91.21 1 91.21 1.891

3 8 125.66 8 125.66FS = + − =

2250 1 91.211 97.38 MPa

1.891 2 125.66allowσ = − =

( )( )6 6max 97.38 10 4 1775 10allowP Aσ −= = × × ×

3691.4 10 N 691 kN= × ≅ .......................................................................... Ans.


576

11-31 Four L4 × 3 × 3/8-in. structural steel angles 13 ft long are used as a pin-ended column. Determine the maximum load permitted by Code 1 if the angles are welded together to form a 6 × 8-in. box section as shown in Fig. P11-31.

SOLUTION


( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

From Table A-9 (for an L 4 3 3 8-in.× × angle):

22.48 inA = 0.879 in.YYr = 0.782 in.Cx =

( ) ( )2 2min 0.879 3 0.782 2.386 in.r = + − =

min

13 12 65.38 126.102.386 C

L Cr

′ ×= = < =

35 3 65.38 1 65.38 1.844

3 8 126.10 8 126.10FS = + − =

236 1 65.381 16.899 ksi

1.844 2 126.10allowσ = − =

( )( )max 16.899 4 2.48 167.6 kipallowP Aσ= = × = ............................................................. Ans.

11-32 A column of aluminum alloy 2014-T6 is composed of two L127 × 127 × 19.1-mm angles riveted together as shown in Fig. P11-32. The length between end connections is 3 m, and the end connections are such that there is no restraint to bending about the y-axis, but restraint to bending about the x-axis reduces the effective length to 2.1 m. Determine the maximum axial compressive load permitted by Code 2. The cross-sectional properties of aluminum and steel angles are the same (see Appendix A).

SOLUTION

From Table A-8 (for an L127 127 19.1× × angle): 24475 mmA =

38.4 mmXXr = 38.4 mmYYr = 38.6 mmCx =

38.4 mmx XXr r= = ( ) ( )2 238.4 38.6 54.45 mmyr = + =

2100 54.69 5538.4

x

x

Lr′= = <

3000 55.10 5554.45

y

y

Lr′= = >

( ) ( )

3 3

2 2372 10 372 10 122.53 MPa

55.10allow L r

σ × ×= = =

( )( )6 6max 122.53 10 2 4475 10allowP Aσ −= = × × ×

31096.6 10 N 1097 kN= × ≅ ..................................................................... Ans.

11-33 A strut of aluminum alloy 6061-T6 having the cross section shown in Fig. P11-33 is to carry an axial compressive load of 175 kip. The strut is 4 ft long and is fixed at the bottom and pinned at the top. Determine the required dimension d of the cross section by using Code 3.

SOLUTION


577

( )( ) 21 22

A d d d= = ( )( )3 4

min

2 22

12 24d d dI = =

4

min 2

24 0.2041I dr dA d

= = = ( )0.7 0.7 4 2.80 ft 33.60 in.L L′ = = = =

min

33.60 164.630.2041

Lr d d

′= =

Assume that ( )9.5 66L r≤ ≤

2

175 164.6320.2 0.126allowPA d d

σ = = = −

3.501 in. 3.50 in.d = ≅ .......................................................................................................... Ans.

Check on L r : min

164.63 164.63 47.02 (within range)3.501

Lr d

′= = =

11-34 A sand bin is supported by four 200 × 250-mm fixed ended rectangular timber columns 6 m long. Assume that the load is equally divided among the columns and that the column load is axial. If the columns are made of timber with E = 13 GPa and Fc = 9 MPa, determine the maximum load permitted by Code 4.

SOLUTION

( ) 3 24 200 250 200 10 mmA = × = ×

9

6

13 100.671 0.671 25.59 10C

EkF

×= = =×

( )0.5 0.5 6 3 m 3000 mmL L′ = = = =

4 41 1 151 9 1 8.641 MPa

3 3 25.5allow CL dFk

σ = − = − =

( )( )6 3max 8.641 10 200 10allowP Aσ −= = × ×

31728.2 10 N 1728 kN= × ≅ ..................................................................... Ans. 11-35 Three 2 × 4-in. timber studs are nailed together to form an 8-ft long column with the cross section shown in Fig. P11-35. The

column is fixed at the bottom and pinned at the top. If E = 1600 ksi and Fc = 1100 psi, use Code 4 to determine the maximum permissible axial compressive load that may be applied.

SOLUTION

( ) 23 2 4 24.00 inA = × = 31600 100.671 0.671 25.59

1100C

EkF

×= = =

( )0.7 0.7 8 5.6 ft 67.2 in.L L′ = = = =

67.20 16.80

4L kd′= = <

4 41 1 16.801 1100 1 1031.9 ksi

3 3 25.59allow CL dFk

σ = − = − =

( )( )max 1031.9 24.00 24,766 lb 24.8 kipallowP Aσ= = = ≅ ........................................... Ans.


578

11-36 Four C178 × 22 structural steel channels 12 m long are used to fabricate a column with the cross section shown in Fig. P11-36. The column is fixed at the base and pinned at the top. The pin at the top offers no restraint to bending about the y-axis, but for bending about the x-axis, the pin provides restraint sufficient to reduce the effective length to 7.5 m. Determine the maximum axial compressive load permitted by Code 1.

SOLUTION


( )22 2 200,0002 125.66250C

y

ECππ

σ= = =

From Table A-8 (for a C178 22× channel): 177.8 mmd = 22795 mmA =

63.8 mmXXr = 14.3 mmYYr = 10.6 mmwt = 13.5 mmCx =

( )( ) ( )( ) ( )( )2 2 2 6 42 2795 63.8 2 2795 14.3 2 2795 13.5 24.92 10 mmxI = + + = ×

( )( ) ( )( ) ( )( )2 2 2 6 42 2795 63.8 2 2795 14.3 2 2795 86.0 65.24 10 mmyI = + + = ×

( )

624.92 10 47.21 mm4 2795xr

×= = 7500 158.86 125.6647.21

xC

x

L Cr′= = > =

( )

665.24 10 76.39 mm4 2795yr

×= = ( )0.7 12,000

109.96 125.6676.39

yC

y

LC

r′= = < =

( )

( )( )

2 926 2

2 2

200 1040.74 10 N/m 40.74 MPa

1.92 1.92 158.86allow

EL r

ππσ×

= = = × =

( )( )6 6max 40.74 10 4 2795 10allowP Aσ −= = × × ×

3455.4 10 N 455 kN= × ≅ ......................................................................... Ans.

11-37 A 25-ft long plate and angle column consists of four L5 × 3 1/2 × 1/2-in. structural steel angles riveted to a 10 × 1/2-in. structural steel plate, as shown in Fig. P11-37. Determine the maximum safe load permitted by Code 1

(a) If the column is pinned at both ends. (b) If the column is fixed at the base and pinned at the top. SOLUTION


( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

From Table A-9 (for an 1 12 2L5 3× × angle): 24.00 inA =

1.58 in.xr = 1.01 in.yr = 1.66 in.Cy =

( ) ( ) 210 1 2 4 4.00 21.00 intotalA = + =

( ) ( )( )

32 2 4

min

10 1 24 4.00 1.58 1.91 98.42 in

12yI I= = + + =


579

min98.42 2.165 in.21.00y

Ir rA

= = = =

(a) 25 12 300 in.L L′ = = × = min

300 138.57 126.102.165

Lr

′= = >

( )

( )( )

22

2 2

29,0007.764 ksi

1.92 1.92 138.57allow

EL r

ππσ = = =

( )( )max 7.764 21.00 163.0 kipallowP Aσ= = = ................................................................... Ans.

(b) ( )0.7 0.7 25 17.5 ft 210 in.L L′ = = = =

210 97.0 126.10

2.165Lr′= = <

35 3 97.0 1 97.0 1.898

3 8 126.10 8 126.10FS = + − =

236 1 97.01 13.356 ksi

1.898 2 126.10allowσ = − =

( )( )max 13.356 21.00 280 kipallowP Aσ= = = .................................................................... Ans.

11-38 A connecting rod made of SAE 4340 heat-treated steel has the cross section shown in Fig. P11-38. The pins at the end of the rod are parallel to the x-axis and are 1250 mm apart. Assume that the pins offer no restraint to bending about the x-axis but provide essentially complete fixity for bending about the y-axis. Determine the maximum axial compressive load permitted by Code 1.

SOLUTION


( )22 2 200,0002 62.831000C

y

ECππ

σ= = =

( ) ( ) 22 50 20 30 50 3500 mmA = × + × =

( ) ( )3 3

6 4

50 90 20 5012 12

2.829 10 mm

xI = −

= ×

( ) ( )3 3

6 4

40 50 50 3012 12

0.5292 10 mm

yI = +

= ×

62.829 10 28.43 mm

3500xr×= =

60.5292 10 12.296 mm3500yr

×= =

1250 43.97 62.8328.43

xC

x

L Cr′= = < =

625 50.83 62.8312.296

yC

y

LC

r′= = < =

35 3 50.83 1 50.83 1.904

3 8 62.83 8 62.83FS = + − =

21000 1 50.831 353.3 MPa

1.904 2 62.83allowσ = − =


580

( )( )6 6max 353.3 10 3500 10allowP Aσ −= = × ×

31236.6 10 N 1237 kN= × ≅ ..................................................................... Ans.

11-39 The machine part shown in Fig. P11-39 is made of SAE 4340 heat-treated steel and carries an axial compressive load. The pins at the ends off no restraint to bending about the axis of the pin, but restraint about the perpendicular axis reduces the effective length to 36 in. Determine the maximum load permitted by Code 1.

SOLUTION


( )22 2 29,0002 62.83145C

y

ECππ

σ= = =

( ) ( ) ( ) 22 1 1 2 3 1 4.00 inA = × + × =

( ) ( )3 3

42 1 1 20.8333 in

12 12xI = + = ( ) ( )3 3

41 1 1 32.333 in

12 12yI = + =

0.8333 0.4564 in.4.00xr = =

2.333 0.7637 in.4.00yr = =

36 78.88 62.83

0.4564x

Cx

L Cr′= = > =

60 78.56 62.830.7637

yC

y

LC

r′= = > =

( )

( )( )

22

2 2

29,00023.96 ksi

1.92 1.92 78.88allow

EL r

ππσ = = =

( )( )max 23.96 4.00 95.8 kipallowP Aσ= = = ....................................................................... Ans.

11-40 Two L152 × 89 × 12.7-mm angles of aluminum alloy 2014-T6 are welded together as shown in Fig. P11-40 to form a pin ended column. The pins provide no restraint to bending about the x-axis but reduce the effective length to 0.7L for bending about the y-axis. The sectional properties for one angle are given on the figure where C is the centroid of one angle. Compute and plot the maximum compressive load permitted by Code 2 as a function of the column length L (0.2 m ≤ L ≤ 5.0 m).

SOLUTION

( ) ( )2 248.3 23.9 53.89 mmxr = + = 53.89

x

x

L Lr′=

( ) ( )2 224.9 30.0 38.48 mmyr = + = 0.738.48

y

y

L Lr′=

If both ( ) 12x xL r′ ≤ and ( ) 12y yL r′ ≤ , then

193 MPaallowσ =

If the largest ( )12 55L r′≤ ≤ then

( )212 1.585 MPaallow L rσ = −

If the largest ( )55 L r′≤ then

( )

3

2372 10 MPaallow L r

σ ×=


582

11-43 A 2-in. diameter steel strut is subjected to an eccentric compressive load, as shown in Fig. P11-43. The effective length for bending about the x-axis is 75 in.; but for bending about the y-axis, the end conditions reduce the effective length to 50 in. Determine the maximum load the strut can carry. Use the interaction method with E = 29,000 ksi, σy = 36 ksi, and σb = 24 ksi.

SOLUTION

( )22

223.142 in

4 4dA

ππ= = = ( )44

420.7854 in

64 64dI

ππ= = =

0.7854 0.500 in.3.142

IrA

= = = 75 150.0 126.10

0.500x

CL Cr′= = > =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

50 100.0 126.100.500

yC

LC

r′= = < =

( )

( )( )

22

2 2

29,0006.625 ksi

1.92 1.92 150allow a

x

EL r

ππσ σ= = = =′

( )( )0.125 1.0 0.78543.142 1

6.625 24.0a b

PP A Pec I Pσ σ

+ = + ≤

18.291 kip 18.29 kipP ≤ ≅ .................................................................................................. Ans.

11-44 Determine the maximum load P that can be applied to the timber column shown in Fig. P11-44 if E = 12 GPa and the allowable stress for compression parallel to the grain is 9 MPa. Use the allowable stress method.

SOLUTION

( ) 2150 100 15,000 mmA bh= = = ( )2 2 2 4 mL L′ = = =

( )33

6 4100 15028.125 10 mm

12 12bhI = = = ×

9

6

12 100.671 0.671 24.509 10C

EkF

×= = =×

4000 40.0 24.50100

L kd′= = > =

( )

( )( )

96 2

2 2

0.30 12 100.30 2.25 10 N/m 2.25 MPa40.0

allowE

L dσ

×= = = × =

′

( )( ) 6 2

6 6

0.075 0.0752.25 10 N/m

15,000 10 28.125 10 allow

PP Mc P Pec PA I A I

σ− −+ = + = + ≤ = ×× ×

38.438 10 N 8.44 kNP ≤ × ≅ ............................................................................................... Ans.

11-45 Two C8 × 18.75 structural steel sections 25 ft long are laced 3 in. back-to-back, as shown in Fig. P11-45 to form a pin-ended column. Determine the maximum load permitted by the interaction method if the load is applied at point A of the cross section. Let E = 29,000 ksi, σy = 36 ksi, and σb = 24 ksi.

SOLUTION

From Table A-5 (for a C8 18.75× channel): 25.51 inA = 8.00 in.d =

444.0 inXXI = 0.565 in.Cx = 2.82 in.xr = 0.599 in.yr =


583

( )22 2 29,0002 126.1036C

y

ECππ

σ= = = 25 ft 300 in.L L′ = = =

For the latticed section: ( ) ( )2 20.599 1.5 0.565 2.15 in.yr = + + =

300 106.38 126.102.82 C

x

L Cr′= = < =

300 139.53 126.102.15 C

y

L Cr′= = > =

( )

( )( )

22

2 2

29,0007.657 ksi

1.92 139.531.92allow a

y

EL r

ππσ σ= = = =′

( ) ( )( ) ( )2 5.51 1.5 4.00 2 44.0

17.657 24.0a b

P PP A Pec Iσ σ

× ×+ = + ≤

68.06 kip 68.1 kipP ≤ ≅ ...................................................................................................... Ans.

11-46 A W356 × 64 structural steel section is used for a 12-m long fixed-ended column. The load is applied at a point on the centerline of the web 150 mm from the axis of the column. If E = 200 GPa, σy = 250 MPa, and σb = 160 MPa, determine the maximum safe load according to

(a) The allowable stress method. (b) The interaction method. SOLUTION

From Table A-2 (for a W 356 64× section): 28130 mmA = 347 mmd =

6 4178 10 mmXXI = × 6 418.8 10 mmYYI = × 148 mmxr = 48.0 mmyr =

( )2 92

6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

( )0.5 0.5 12 6 mL L′ = = =

6000 40.54 125.66148 C

x

L Cr′= = < =

6000 125.0 125.6648.0 C

y

L Cr′= = < =

35 3 125.00 1 125.00 1.917

3 8 125.66 8 125.66yFS = + − =

2250 1 125.001 65.89 MPa

1.917 2 125.66allow aσ σ = − = =

(a) ( )( ) 6 2

6 6

0.150 0.173565.89 10 N/m

8130 10 178 10 allowx x


σ− −+ = + = + ≤ = ×× ×

3244.8 10 N 245 kNP ≤ × ≅ ................................................................................................ Ans.

(b) ( ) ( )( ) ( )6 6

6 6

8130 10 0.150 0.1735 178 101

65.89 10 160 10a b

P PP A Pec Iσ σ

− −× ×+ = + ≤

× ×

3359.6 10 N 360 kNP ≤ × ≅ ................................................................................................ Ans.


584

11-47 A hollow square steel member with outside and inside dimensions of 4 in. and 2.8 in., (walls are 0.6 in. thick) functions as a pin-ended column 13 ft long. Determine the maximum load that the column can carry if the load is applied with a known eccentricity of 0.6 in. along a diagonal of the square as shown in Fig. P11-47. Use the allowable stress method and let E = 29,000 ksi and σy = 36 ksi.

SOLUTION

( ) ( )2 2 24 2.8 8.160 inA = − = 2 22 2 2.828 in.c = + =

( ) ( )3 3

44 4 2.8 2.816.2112 in (all axes)

12 12I = − =

16.2112 1.4095 in.

8.160IrA

= = = 13 ft 156 in.L L′ = = =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

156 110.67 126.101.4095 C

L Cr′= = < =

35 3 110.67 1 110.67 1.899

3 8 126.10 8 126.10FS = + − =

236 1 110.671 11.656 ksi

1.899 2 126.10allowσ = − =

( )( ) 30.6 2.828

11.656 10 psi8.160 16.2112 allow

x x


σ+ = + = + ≤ = ×

351.3 10 lb 51.3 kipP ≤ × ≅ ................................................................................................. Ans.

11-48 Two 50 × 150-mm structural steel bars 5 m long will be welded together as shown in Fig. P11-48, and used for a pin-ended column. Determine the maximum load permitted if the load is applied at the point on the cross section indicated in Fig. P11-48. Use the allowable stress method and let E = 200 GPa and σy = 250 MPa.

SOLUTION

( ) 22 50 150 15,000 mmA = × = ( ) ( )25 50 150 125 50 150

75 mm15,000Cy

× + ×= =

( ) ( )( ) ( ) ( )( )

3 32 2 6 4150 50 50 150

50 150 50 50 150 50 53.125 10 mm12 12xI

= + × + + × = ×

( ) ( )3 3

6 450 150 150 5015.625 10 mm

12 12yI = + = ×

5 mL L′ = =

( )2 92

6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

653.125 10 59.51 mm

15,000x

xIrA

×= = = 5000 84.02 125.6659.51 C

x

L Cr′= = < =

615.625 10 32.27 mm

15,000y

y

Ir

A×= = =

5000 154.94 125.6632.27 C

y

L Cr′= = > =


585

( )

( )( )

2 926 2

2 2

200 1042.83 10 N/m 42.83 MPa

1.92 154.941.92allow

y

EL r

ππσ×

= = = × =′

( )( ) 6 2

6 6

0.050 0.07542.83 10 N/m

15,000 10 53.125 10 allowx x


σ− −+ = + = + ≤ = ×× ×

3312.1 10 N 312 kNP ≤ × ≅ ................................................................................................ Ans.

11-49 A WT8 × 25 structural steel T-section is to be used as a compression member to transmit an eccentric load of 100 kip. The effective length of the member is 12 ft. Determine the maximum eccentricity e permitted if the point of load application is on the centerline of the stem between the outer edge of the flange and the centroidal axis of the column. Use the interaction method with E = 29,000 ksi, σy = 36 ksi, and σb = 24 ksi.

SOLUTION

From Table A-11 (for a WT8 25× section): 27.37 inA = 1.89 in.c =

442.3 inXXI = 418.6 inYYI = 2.40 in.xr = 1.59 in.yr =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = = 12 ft 144 in.L L′ = = =

144 60.00 126.102.40 C

x

L Cr′= = < =

144 90.57 126.101.59 C

y

L Cr′= = < =

35 3 90.57 1 90.57 1.890

3 8 126.10 8 126.10yFS = + − =

236 1 90.571 14.135 ksi

1.890 2 126.10allow aσ σ = − = =

( )max 14.135 7.37 104.17 kip 100 kipallowP A Pσ= = = > =

( ) ( )100 7.37 100 1.89 42.3

114.135 24.0a b

eP A Pec Iσ σ

+ = + ≤

0.2153 in. 0.215 in.e ≤ ≅ ...................................................................................................... Ans.

11-50 A 2014-T6 aluminum alloy compression member with an effective length of 1.25 m has the T-cross section shown in Fig. P11-50. Determine

(a) The maximum axial compressive load P permitted by Code 2. (b) Use the allowable stress method to determine the maximum bending moment M in the yz-plane that can be applied as

shown when the column is supporting a 175-kN axial load. SOLUTION

For the section: 22050 mmA = 30.5 mmxr = 20.8 mmyr = 1.25 mL′ =

31.25 10 40.98 55

30.5x

Lr′ ×= = <

31.25 10 60.10 5520.8y

Lr′ ×= = >

( ) ( )

3 3

2 2372 10 372 10 102.99 MPa 103.0 MPa

60.10allow

yL rσ × ×= = = ≅

′

(a) ( )( )6 6 3102.99 10 2050 10 211.1 10 N 211 kNallow allowP Aσ −= = × × = × ≅ ................ Ans.


586

(b) ( )

( ) ( )3

6 226 6

0.0282175 10 102.99 10 N/m2050 10 0.0305 2050 10

allowx

MP McA I

σ− −

×+ = + ≤ = ×× ×

36.964 10 N m 6.96 kN mM ≤ × ⋅ ≅ ⋅ ................................................................................. Ans.

11-51 Two L6 × 3 1/2 × 1/2-in. structural steel (E = 29,000 ksi and σy = 36 ksi) angles are welded together, as shown in Fig. P11-51, to form a column with an effective length of 15 ft. Use the allowable stress method to determine

(a) The maximum axial compressive load P that can be supported by the column. (b) The maximum and minimum values for the distance d when a 50-kip load is applied. SOLUTION

From Table A-9 (for an 1 12 2L6 3× × section): 24.50 inA = 0.833 in.Cx =

416.6 inXXI = 1.92 in.xr = 0.972 in.yr = 2.08 in.Cy =

For the welded angles: ( ) ( )2 20.972 0.833 1.280 in.yr = + =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = = 15 ft 180 in.L L′ = = =

180 93.75 126.101.92 C

x

L Cr′= = < =

180 140.63 126.101.28 C

y

L Cr′= = > =

( )

( )( )

22

2 2

29,0007.538 ksi

1.92 140.631.92allow

y

EL r

ππσ = = =′

( )max 7.538 2 4.50 67.84 kip 67.8 kipallowP Aσ= = × = ≅ ............................................. Ans.

(b) For 2.08 in.c = :

( )( )

50 2.0850 7.538 ksi9.00 2 16.6 allow

x x

eP Mc P PecA I A I

σ+ = + = + ≤ =

0.6329 in.e ≤ min 2.08 0.6329 1.447 in.d = − =

For 6 2.08 3.92 in.c = − = :

( )( )

50 3.9250 7.538 ksi9.00 2 16.6 allow

x x

eP Mc P PecA I A I

σ+ = + = + ≤ =

0.3358 in.e ≤ max 2.08 0.3358 2.42 in.d = + =

Therefore: 1.447 in. 2.42 in.d≤ ≤ ........................................................................................... Ans.

11-52 A W457 × 144 wide-flange section is used for the column shown in Fig. P11-52. The column is made of steel (E = 200 GPa, 290 MPayσ = , 190 MPabσ = ). An eccentric load P is applied on the centerline of the web as shown in the figure.

(a) Calculate and plot the maximum safe load as a function of the eccentricity e (0 mm ≤ e ≤ 236 mm) as given by the allowable stress method for an effective length of 6 m.

(b) Calculate and plot (on the same graph) the maximum safe load as a function of the eccentricity e (0 mm ≤ e ≤ 236 mm) as given by the interaction method for an effective length of 6 m.

(c) Repeat parts (a) and (b) for an effective length of 9 m. SOLUTION

For a W 457 144× section: 218,365 mmA = 472 / 2 236 mmc = =


588

11-53 Two C8 × 18.75 structural steel sections are laced 3-in. back-to-back, as shown in Fig. P11-53 to form a pin ended column. An eccentric load P is applied at point A along the y-axis as shown in the figure. If E = 29,000 ksi, 36 ksiyσ = , and

24 ksibσ = , (a) Calculate and plot the maximum safe load as a function of the eccentricity e (0 in. ≤ e ≤ 4 in.) as given by the allowable

stress method for an effective length of 15 ft. (b) Calculate and plot (on the same graph) the maximum safe load as a function of the eccentricity e (0 in. ≤ e ≤ 4 in.) as given

by the interaction method for an effective length of 15 ft. (c) Repeat parts (a) and (b) for an effective length of 25 ft. SOLUTION

From Table A-5 (for a C8 18.75× channel): 25.51 inA = 8.00 in.d =

444.0 inXXI = 0.565 in.Cx = 2.82 in.xr = 0.599 in.yr =

For the latticed section: ( ) ( )2 20.599 1.5 0.565 2.15 in.yr = + + =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = = 15 ft 180 in.L L′ = = =

min

180 83.72 126.102.15

Lr

′= = <

35 3 83.72 1 83.72 1.879

3 8 126.10 8 126.10FS = + − =

236 1 83.721 14.937 ksi

1.879 2 126.10allow aσ σ = − = =

(a) Allowable stress method:

( )4

14.937 ksi11.02 88 allow

x x

PeP Mc P Pec PA I A I

σ+ = + = + ≤ =

614.485 10 lb

88 44.08P

e×≤

+ 0 in. 4 in.e≤ ≤

(b) Interaction method:

( )4 8811.02 1

14,937 24,000a b

PeP A Mc I Pσ σ

+ = + ≤

6358.49 10 lb

2177.86 678.95P

e×≤

+ 0 in. 4 in.e≤ ≤

(c) 25 ftL = min

300 139.53 126.102.15

Lr

′= = >

( )

( )( )

22

2 2

29,0007.657 ksi

1.92 1.92 139.53allow a

EL r

ππσ σ= = = =′

Allowable stress method:


590

This allowable load is close to the design load of 200 kN; therefore, the next larger size pipe should be satisfactory. For a 127-mm diameter standard weight pipe:

22774 mmA = 47.8 mmr = 4000 83.68 125.6647.8 C

L Cr

= = < =

Therefore, the column is in the intermediate range and:

35 3 83.68 1 83.68 1.879

3 8 125.66 8 125.66FS = + − =

2250 1 83.681 103.6 MPa

1.879 2 125.66allowσ = − =

( )( )6 6 3103.6 10 2774 10 287 10 N 287 kNallow allowP Aσ −= = × × = × ≅

Therefore: Use a 127-mm diameter standard weight pipe. ....................................... Ans.

11-55 Select the lightest structural steel wide-flange section that can be used to support an axial compressive load of 200 kip as a 16-ft long column. Use Code 1.

SOLUTION


If L r is small: 5 3 1.667FS ≅ =

36 21.596 ksi

1.667y

allow FSσ

σ = = = 2min

200 9.261 in21.596allow

PAσ

= = =

First try a W8 40× section (see Table A-1) with:

211.7 inA = min 2.04 in.r = 16 12 94.12

2.04Lr

×= =

For structural steel: ( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

Therefore, the column is in the intermediate range for which

3 35 3 1 5 3 94.12 1 94.12 1.8953 8 8 3 8 126.10 8 126.10C C

L r L rFSC C

= + − = + − =

2 21 36 1 94.121 1 13.706 ksi2 1.895 2 126.10

yallow

C

L rFS Cσ

σ = − = − =

( )( )13.706 11.7 160.4 kipallow allowP Aσ= = =

This allowable load is less than the design load of 200 kip; therefore, a section with a larger area and possibly a larger radius of gyration must be used. Next, try a W12 50× section with:

214.7 inA = min 1.96 in.r = 16 12 97.961.96

Lr

×= =

35 3 97.96 1 97.96 1.899

3 8 126.10 8 126.10FS = + − =


591

236 1 97.961 13.237 ksi

1.899 2 126.10allowσ = − =


This allowable load is close to the design load of 200-kip; therefore, the next larger size section should be satisfactory. Next, try a W10 60× section with:

217.6 inA = min 2.57 in.r = 16 12 74.71

2.57Lr

×= =

35 3 74.71 1 74.71 1.863

3 8 126.10 8 126.10FS = + − =

236 1 74.711 15.932 ksi

1.863 2 126.10allowσ = − =


Therefore: Use a W 10 60 section.× ................................................................................ Ans.

11-56 A 7-m long structural steel column will be used to support an axial compressive load of 400 kN. Select the lightest wide-flange section that can be used. Use Code 1.

SOLUTION



250 150 MPa

1.667y

allow FSσ

σ = = = 3

6 2min 6

400 10 2667 10 m150 10allow

PAσ

−×= = = ××

First try a W 203 22× section (see Table A-2) with:

22865 mmA = min 22.3 mmr = 7000 313.922.3

Lr

= =

For structural steel: ( )2 92

6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

Therefore, the column is in the slender range for which:

( )

( )( )

2 32

2 2

200 1010.43 MPa

1.92 1.92 313.9allow

EL r

ππσ×

= = =

( )( )6 6 310.43 10 2865 10 29.88 10 N 29.9 kNallow allowP Aσ −= = × × = × ≅

This allowable load is much less than the design load of 400 kN; therefore, a section with a much larger area and a larger radius of gyration must be used. Next, try a W 305 74× section with:

29485 mmA = min 49.8 mmr = 7000 140.6 125.6649.8 C

L Cr

= = > =


596

2161.3 mmA = 6.6 mmr = 1500 227.36.6

Lr

= =


6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

Therefore, the column is in the slender range for which:

( )

( )( )

2 32

2 2

200 1019.90 MPa

1.92 1.92 227.3allow

EL r

ππσ×

= = =

( )( )6 6 319.90 10 161.3 10 3.21 10 N 3.21 kNallow allowF Aσ −= = × × = × ≅

This allowable load is much less than the design load of 16.878 kNF = ; therefore, a pipe with a much larger area and a larger radius of gyration must be used. Next, try a 25-mm diameter standard weight pipe with:

2318.7 mmA = 10.7 mmr = 1500 140.210.7

Lr

= =

This column is in the slender range for which:

( )

( )( )

2 32

2 2

200 1052.30 MPa

1.92 1.92 140.2allow

EL r

ππσ×

= = =

( )( )6 6 352.30 10 318.7 10 16.67 10 N 16.67 kNallow allowF Aσ −= = × × = × ≅

This allowable load is close to the design load; therefore, the next larger size pipe should be satisfactory. Use a 32-mm diameter standard weight pipe. ............................................................... Ans.

11-63 Select the lightest structural steel wide-flange section that can be used for the compression members of the truss shown in Fig. P11-63. Assume that buckling is limited to the plane of the structure and that the tension members have been adequately designed. Use Code 1.

SOLUTION

0 :AMΣ = ( ) ( )20 90 10 0C − =

45 kipC A= =

For joint C:

1 7.5tan 36.87010

φ −= = °

0 :xF→ Σ = cos36.870 0BC CDT T− − ° =

0 :yF↑ Σ = 45 sin 36.870 0CDT+ ° =

75 kip 75 kip (C)CD ADT T= = − =

60 kip 60 kip (T)BC ABT T= = + =

For joint B:

0 :yF↑ Σ = 90 0BDT − =

90 kip 90 kip (T)BDT = + =

Only members AD and CD are in compression and 75 kip (C)AD CDT T= =


597



36 21.596 ksi

1.667y

allow FSσ

σ = = = 2min

75 3.473 in21.596allow

PAσ

= = =

First try a W 4 13× section (see Table A-1) with:

23.83 inA = min 1.00 in.r = 12.5 12 150.0

1.00Lr

×= =

For structural steel: ( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

Therefore, the column is in the slender range for which

( )

( )( )

22

2 2

29,0006.625 ksi

1.92 1.92 150.0allow

EL r

ππσ = = =


This allowable load is much less than the design load of 75 kip; therefore, a section with a much larger area and a larger radius of gyration must be used. Next, try a W8 24× section with:

27.08 inA = min 1.61 in.r = 12.5 12 93.17

1.61Lr

×= =

This section is in the intermediate range for which:

3 35 3 1 5 3 93.17 1 93.17 1.8933 8 8 3 8 126.10 8 126.10C C

L r L rFSC C

= + − = + − =

2 21 36 1 93.171 1 13.827 ksi2 1.893 2 126.10

yallow

C

L rFS Cσ

σ = − = − =


This allowable load is larger than the design load of 75 kip; therefore, a slightly smaller section can be used if it can be found. Next, try a W10 22× section with:

26.49 inA = min 1.33 in.r = 12.5 12 112.8

1.33Lr

×= =

This section is in the intermediate range for which:

35 3 112.8 1 112.8 1.913

3 8 126.10 8 126.10FS = + − =

236 1 112.81 11.289 ksi

1.913 2 126.10allowσ = − =


This allowable load is close to but lower than the design load of 75 kip. Therefore: Use a W 8 24 section.× ................................................................................... Ans.


598

11-64 Member ABC of the structure shown in Fig. 11-64 supports a uniformly distributed load of 30 kN/m. Select the lightest standard weight structural steel pipe that can be used for member BD. Use Code 1 and consider BD to be a pin-ended member. Neglect the weight of the structure.

SOLUTION

( )1tan 2 3 33.69φ −= = °

0 :AMΣ = ( )3sin 33.69BDF °

( ) ( )30 4.5 2.25 0− =

182.53 kNBDF P= =



250 150 MPa

1.667y

allow FSσ

σ = = = 3

6 2min 6

182.53 10 1217 10 m150 10allow

PAσ

−×= = = ××

First try a 76-mm diameter standard weight pipe (see Table A-14) with:

21437 mmA = 29.5 mmr = 3606 122.2429.5

Lr

= =


6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

Since CL r C< , the column is in the intermediate range and:

3 35 3 1 5 3 122.24 1 122.24 1.9163 8 8 3 8 125.66 8 125.66C C

L r L rFSC C

= + − = + − =

2 21 250 1 122.241 1 68.74 MPa2 1.916 2 125.66

yallow

C

L rFS Cσ

σ = − = − =


This allowable load is much less than the design load of 182.5 kN; therefore, a pipe with a much larger area and a larger radius of gyration must be used. Next, try a 102-mm diameter standard weight pipe with:

22048 mmA = 38.4 mmr = 3606 93.91 125.6638.4 C

L Cr

= = < =

Therefore, the column is in the intermediate range and:

35 3 93.91 1 93.91 1.895

3 8 125.66 8 125.66FS = + − =

2250 1 93.911 95.09 MPa

1.895 2 125.66allowσ = − =


This allowable load is close to the design load of 182.5 kN; any smaller size pipe would not be satisfactory. Therefore: Use a 102-mm diameter standard weight pipe. ....................................... Ans.


599

11-65 A 25-ft long timber (E = 1200 ksi and 2.4 ksieσ = ) column has the cross section shown in Fig. P11-65. The timbers are nailed together so that they act as a unit. Determine

(a) The slenderness ratio. (b) The smallest slenderness ratio for which the Euler buckling load equation is valid. (c) The Euler buckling load. (d) The axial stress in the column when the Euler load is applied. SOLUTION

( ) ( ) 26 10 2 6 48 inA = − =

( ) ( )3 3

4min

10 6 6 2176 in

12 12I = − =

176 1.9149 in.48

IrA

= = =

(a) 25 12 156.67 156.71.9149

Lr

×= = ≅ .............................................................................................. Ans.

(b) cr yσ σ= ( )22 2 1200

70.32.4cr y

L E Er

ππ πσ σ

= = = = .................................. Ans.

(c) ( )( )( )

22

22

1200 17623.16 kip 23.2 kip

25 12cr

EIPL

ππ= = = ≅×

................................................. Ans.

(d) 23.16 0.483 ksi

48PA

σ = = = .................................................................................................. Ans.

11-66 A 3-m long column with the cross section shown in Fig. P11-66 is fabricated from three pieces of timber (E = 13 GPa and 35 MPaeσ = ). The timbers are nailed together so that they act as a unit. Determine

(a) The slenderness ratio. (b) The smallest slenderness ratio for which the Euler buckling load equation is valid. (c) The Euler buckling load. (d) The axial stress in the column when the Euler load is applied. SOLUTION

( ) ( ) 22 125 20 25 50 6250 mmA = × + =

( ) ( )3 3

6 4125 90 100 506.552 10 mm

12 12xI = − = ×

( ) ( )3 3

6 440 125 50 256.576 10 mm

12 12yI = + = ×

6

min 6.552 10 32.38 mm6250

IrA

×= = =

(a) 3000 92.64 92.632.38

Lr

= = ≅ .................................................................................................... Ans.

(b) cr yσ σ= ( )2 92 2

6

13 1060.6

35 10cr y

L E Er

ππ πσ σ

×= = = =

×............................. Ans.


600

(c) ( )( )

( )

2 9 623

22

13 10 6.552 1093.41 10 N 93.4 kN

3cr

EIPL

ππ −× ×= = = × ≅ ...................... Ans.

(d) 3

6 23

93.41 10 14.946 10 N/m 14.95 MPa (C)6.250 10

PA

σ −

×= = = × ≅×

.................................... Ans.

11-67 Determine the maximum allowable axial compressive load for a 12-ft long aluminum alloy (E = 10,600 ksi) column having the cross section shown in Fig. P11-67 if a factor of safety of 2.25 is specified.

SOLUTION

( ) ( ) 22 8 1 4 1 20 inA = × + =

( ) ( ) ( )0.5 8 1 5 8 1 9.5 4 1

4.10 in.20Cy

× + × + ×= =

( ) ( )( ) ( ) ( )( )

3 32 28 1 1 8

8 1 3.6 8 1 0.912 12xI

= + × + + ×

( ) ( )( )

32 44 1

4 1 5.4 270.5 in12

+ + × =

( ) ( ) ( )3 3 3

4min

8 1 1 8 1 448.67 in

12 12 12yI I= + + = =

min 48.67 1.560 in.20

IrA

= = = 12 12 92.31 (slender)1.560

Lr

×= =

( )

( )( )( )

22

max 2 2

10,600 20109.1 kip


ππ= = = = ............................................. Ans.

11-68 A 25-mm diameter tie rod and a pipe strut with an inside diameter of 100 mm and a wall thickness of 25 mm are used to support a 100-kN load as shown in Fig. P11-68. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of 200 GPa and a yield strength of 250 MPa. Determine

(a) The factor of safety with respect to failure by yielding. (b) The factor of safety with respect to failure by buckling. SOLUTION

1 2.5tan 22.626.0

φ −= = °

1 4.5tan 36.876.0

θ −= = °

0 :xF→ Σ = cos 22.62 cos36.87 0AB ACF F° + ° =

0 :yF↑ Σ = sin 22.62 sin 36.87 100 0AB ACF F° − °− =

92.86 kNABF = 107.14 kNACF = −

( )2

225490.9 mm

4ABAπ

= = ( )2 2

2150 100

9817 mm4ACA

π −= =


601

(a) 3

6 26

92.86 10 189.16 10 N/m 189.2 MPa (T)490.9 10

ABAB

AB

FA

σ −

×= = = × ≅×

250 1.322

189.16ABFS = = .......................................................................................................... Ans.

3

6 26

107.14 10 10.914 10 N/m 10.914 MPa (C)9817 10

ACAC

AC

FA

σ −

− ×= = = − × ≅×

250 22.9

10.914ACFS = = ........................................................................................................... Ans.

(b) ( )4 4

6 4150 100

19.942 10 mm64ACI

π −= = ×

619.942 10 45.07 mm

9817IrA

×= = = 7500 166.41 (slender)45.07

Lr

= =

( )

( )( )( )

2 9 623

2 2

200 10 9817 10699.8 10 N 700 kN

166.41cr

EAPL r

ππ −× ×= = = × ≅ .................. Ans.

699.8 6.53

107.14cr

ACAC

PFSA

= = = ............................................................................................... Ans.

11-69 Three S10 × 35 structural steel sections 40 ft long are used to fabricate a column with the cross section shown in Fig. P11-69. The column is fixed at the base and pinned at the top. The pin at the top offers no restraint to bending about the y-axis; but for bending about the x-axis, the pin provides restraint sufficient to reduce the effective length to 24 ft. Determine the maximum axial compressive load permitted by Code 1.

SOLUTION From Table A-17:

36 ksiyσ = 29,000 ksiE =

( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

From Table A-3 (for an S10 35× section): 210.3 inA =

4147 inXXI = 48.36 inYYI = 0.594 in.wt =

( ) 42 147 8.36 302.4 inxI = + = ( ) 23 10.3 30.9 inA = =

( ) ( )( )2 4147 2 8.36 2 10.3 5.491 784.8 inyI = + + =

24 ft 288 in.xL′ = = ( )0.7 40 28 ft 336 in.yL′ = = =

302.4 3.128 in.30.9

xx

IrA

= = = 288 92.07 126.10

3.128x

Cx

L Cr′= = < =

784.8 5.040 in.30.9

yy

Ir

A= = =

336 66.67 126.105.040

yC

y

LC

r′= = < =


602

3 35 3 1 5 3 92.07 1 92.07 1.8923 8 8 3 8 126.10 8 126.10C C

L r L rFSC C

= + − = + − =

2 21 36 1 92.071 1 13.956 ksi2 1.892 2 126.10

yallow

C

L rFS Cσ

σ = − = − =

( )( )max 13.956 30.9 431 kipallowP Aσ= = = ....................................................................... Ans.

11-70 The compression member AB of the truss shown in Fig. P11-70 is a structural steel W254 × 67 wide-flange section with the xx-axis lying in the plane of the truss. The member is continuous from A to B. Consider all connections to be the equivalent of pin ends. If Code 1 applies, determine

(a) The maximum safe load for member AB. (b) The maximum safe load for member AB if the bracing member CD is removed. SOLUTION From Table A-18:

250 MPayσ = 200 GPaE =

( )2 92

6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

From Table A-2 (for a W 254 67× section):

28580 mmA = 110 mmxr = 51.1 mmyr =

(a) 12,000 109.09 125.66

110 Cx

L Cr′= = < =

6000 117.41 125.6651.1 C

y

L Cr′= = < =

3 35 3 1 5 3 117.41 1 117.41 1.9153 8 8 3 8 125.66 8 125.66C C

L r L rFSC C

= + − = + − =

2 21 250 1 117.411 1 73.56 MPa2 1.915 2 125.66

yallow

C

L rFS Cσ

σ = − = − =

( )( )6 6 373.56 10 8580 10 631.1 10 N 631 kNallow allowP Aσ −= = × × = × ≅ ................... Ans.

(b) 12,000 234.8 125.66

51.1y

Lr′= = >

( )

( )( )

2 32

2 2

200 1018.648 MPa

1.92 1.92 234.8allow

EL r

ππσ×

= = =

( )( )6 6 318.648 10 8580 10 160.0 10 N 160.0 kNallow allowP Aσ −= = × × = × ≅ ............. Ans.

11-71 A cold-rolled steel tension bar AB and a structural steel (E = 29,000 ksi and 36 ksiyσ = ) compression strut BC are used to support a load P = 100 kip, as shown in Fig. P11-71. Assume that the pins at B and C offer no restraint to bending about the x-axis but provide end conditions which are essentially fixed at C and free at B for bending about the y-axis. Select the lightest structural steel T-section listed in Appendix A that can be used for strut BC. Use Code 1.

SOLUTION


603

1 4tan 53.1303

φ −= = ° 1 3tan 36.8704

θ −= = °

0 :xF→ Σ = cos53.130 cos36.870 0AB BCF F− °+ ° =

0 :yF↑ Σ = sin 53.130 sin 36.870 100 0AB BCF F° + °− =

80.0 kipABF = 60.0 kipBCF =

( )22 2 29,0002 126.1036c

y

ECππ

σ= = =


36 21.596 ksi

1.667y

allow FSσ

σ = = = 2min

60 2.778 in21.596allow

PAσ

= = =

200 in.xL L′ = = ( )2 2 200 400 in.yL L′ = = =

Therefore, we need a section with 22.778 inA > .

First, try a WT5 30× section (see Table A-11) with: 28.82 inA =

1.21 in.XXr = 2.57 in.YYr =

200 165.29 126.101.21

xC

x

L Cr′= = > =

400 155.64 126.102.57

yC

y

LC

r′= = > =

This strut is in the slender range for which:

( )

( )( )

22

2 2

29,0005.456 ksi

1.92 1.92 165.29allow

EL r

ππσ = = =


This allowable load is much less than the design load of 60 kip; therefore, a section with a larger area and a larger radius of gyration must be used.

Next, try a WT 7 34× section with: 29.99 inA =

1.81 in.XXr = 2.46 in.YYr =

200 110.49 126.101.81

xC

x

L Cr′= = < =

400 162.60 126.102.46

yC

y

LC

r′= = > =

This strut is also in the slender range for which:

( )

( )( )

22

2 2

29,0005.638 ksi

1.92 1.92 162.60allow

EL r

ππσ = = =


The next heavier section is a WT 6 36× section with: 210.6 inA =

1.48 in.XXr = 3.04 in.YYr =

200 135.14 126.101.48

xC

x

L Cr′= = > =

400 131.58 126.103.04

yC

y

LC

r′= = > =


604

Again the strut is in the slender range for which:

( )

( )( )

22

2 2

29,0008.163 ksi

1.92 1.92 135.14allow

EL r

ππσ = = =


Therefore: Use a WT 6 36 section.× ................................................................................ Ans.

11-72 Two C229 × 30 structural steel channels are laced 120 mm back-to-back, as shown in Fig. P11-68, to form a column with an effective length of 10 m. The load P will be applied at a point 50 mm from the axis of the column on the axis of symmetry parallel to the back of the channels. If 200 GPaE = and 250 MPayσ = , use the allowable stress method to determine the maximum load P that can be supported by the column.

SOLUTION

( )2 92

6

2 200 102 125.66250 10C

y

ECππ

σ×

= = =×

From Table A-6 (for a C 229 30× section):

23795 mmA = 14.8 mmCx =

81.8 mmXXr = 16.3 mmYYr =

228.6 mmd = 6 425.3 10 mmXXI = ×

For the section: ( ) ( )2 216.3 60.0 14.8 76.6 mmyr = + + =

10,000 122.25 125.66

81.8 Cx

L Cr′= = < =

10,000 130.55 125.6676.6 C

y

L Cr′= = > =

( )

( )( )

2 32

2 2

200 1060.32 MPa

1.92 1.92 130.55allow

EL r

ππσ×

= = =

( )( )( )( )

6 26 6

0.050 0.114360.32 10 N/m

2 3795 10 2 25.3 10 allow


σ− −

+ = + = + ≤ = ×× ×

3246.5 10 N 247 kNP ≤ × ≅ ................................................................................................ Ans.

11-73 A W8 × 40 structural steel compression member has an effective length of 25 ft. The member is subjected to an axial load P and a bending moment of 10 kip⋅ft about the x-axis of the cross section (see Appendix A). Use the interaction formula with an allowable flexural stress of 0.66 yσ to determine the maximum safe load P.

SOLUTION


( )22 2 29,0002 126.1036C

y

ECππ

σ= = =

From Table A-1 (for a W8 40× section): 211.7 inA =

4146 inXXI = 3.53 in.XXr = 2.04 in.YYr = 8.25 in.d =


605

25 12 84.99 126.103.53 C

x

L Cr′ ×= = < =

25 12 147.06 126.102.04 C

y

L Cr′ ×= = > =

( )

( )( )

22

2 2

29,0006.893 ksi

1.92 1.92 147.06allow a

EL r

ππσ σ= = = =

( )0.66 0.66 36 23.76 ksib yσ σ= = =

( )( )10 12 4.125 14611.7 1

6.893 23.76a b

P A Mc I Pσ σ

+ = + ≤

69.1 kipP ≤ .............................................................................................................................. Ans.

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