Strenght of Materials(ES-64)

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students ‘04

Solved Problems in

Mechanics of Materials

ES-64 1. 12-7. Determine the equations of the elastic curve using the x1 and x2

coordinates. Specify the slope at A and the maximum deflection. EI is constant. 1. 12-10.The beam is made of two rods and is subjected to the concentrated

load P. Determine the slope at C. The moments of inertia of the rods are IAB and IAC, and the modulus of elasticity is E.

2. 12-13. Determine the elastic curve for the cantilevered beam, which is subjected to the couple moment Mo. Also compute the maximum slope and maximum deflection of the beam. EI is constant.

3. 12-15. Determine the deflection at the center of the beam and the slope at B. EI is constant.

4. 12-16. Determine the elastic curve for the simply supported beam, which is subjected to the couple moments Mo. Also, compute the maximum slope and the maximum deflection of the beam. EI is constant.

5. 12-19. Determine the equations of the elastic curve using the coordinates x1 and x2, and specify the slope at A. EI is constant.

6. 12-22.The floor beam of the airplane is subjected to the loading shown. Assuming that the fuselage exerts only vertical reactions on the ends of the beam, determine the maximum deflection of the beam. EI is constant.

7. 12-27. Determine the elastic curve for the simply supported beam using the x coordinates 0 ≤ x ≤ L/2. Also, determine the slope at A and the maximum deflection of the beam. EI is constant.

8. 12-34.The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.

9. 12-35.The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft, and at C by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Determine the equation of the elastic curve. EI is constant.

10. 12-37.The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant.

11. 12-38.The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant.



14. 12-55. Determine the slope and deflection at B. EI is constant. 15. 12-61. Determine the maximum slope and the maximum deflection of the

beam. EI is constant. 16. 12-40.The beam is subjected to the load shown. Determine the equation of

the elastic curve. EI is constant. 17. 12-45.The beam is subjected to the load shown. Determine the equation of

the elastic curve. EI is constant. 18. 12-50. Determine the equation of the elastic curve. Specify the slope at A. EI

is constant. 19. 12-25.The beam is subjected to the linearly varying distributed load.

Determine the maximum slope of the beam. EI is constant. 20. 12-42.The beam is subjected to the load shown. Determine the equation of

the slope and elastic curve. EI is constant. 21. 12-45. The beam is subjected to the load shown. Determine the equation of

the elastic curve. EI is constant. 22. 12-46.The wooden beam is subjected to the load shown. Determine the

equation of the elastic curve. Specify the deflection at the end C. EW=1.6(103) ksi.

23. 12-63. Determine the deflection and slope at C. EI is constant. 24. 12-66. Determine the deflection at C and the slope of the beam at A, B, and

C. EI is constant. 25. 12-73. Determine the slope at B and deflection at C. EI is constant. 26. 12-86.The beam is subjected to the load shown. Determine the slope at B

and deflection at C. EI is constant.

1. 5-5.The copper pipe has an outer diameter of 2.50 in. and an inner diameter

of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown., determine the shear stress developed at points A and

B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B.

2. 5-9. The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench.

3. 5-10.The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section.

4. 5-2.The solid shaft of radius r is subjected to a torque T. Determine the radius r of the inner cone of the shaft that resists one-half of the applied torque (T/2). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear stress distribution.

5. 5-3.The solid shaft of radius r is subjected to a torque T. Determine the radius r of the inner core of the shaft that resists one-quarter of the applied torque (T/4). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear stress distribution.

6. 5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

7. 5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb.

8. 5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm.

9. 5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d.

10. 5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function

( )2

25 xxt = N⋅m/m, where x is in meters. Determine the minimum torque T0

needed to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft.

11. 5-30. The solid shaft has a linear taper from rA at one end to rB at the other. Derive and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.

12. 5-32. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall.

13. 5-33. The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi.

14. 5-34. The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi.

15. 5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi.

16. 5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall.

17. 5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

18. 5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B.

19. 5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

20. 5-46. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What

is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element.

21. 5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed a its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa.

22. 5-78. The composite shaft consists of a mid-section that includes the 1-in.-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 800 lb⋅ft. The material is A-36 steel.

The column is subjected to an axial force of 8 kN at its top. If the cross-sectional area has the dimensions shown in the figure, determine the average normal stress acting at section a-a. Show this distribution of stress acting over the area’s cross section.

( ) ( )[ ]

( )

( )MPa

m

N

A

P

mA

mmA

mmmmmmmmA

82.1

104.4

8000

104.4

4400

14010150102

23

23

2

=

=

=

=

=

+=

−

−

σ

σ

The anchor shackle supports a cable force of 600 lb. If the pin has a diameter of 0.25 in., determine the average shear stress in the pin.

( )

( )ksi

in

lbA

V

in

rA

inr

ind

11.6

04909.02

6.02

:stressshear doubleFor

04909.0

125.0

125.0

25.0

2

2

2

2

=

==

=

=

=

=

=

τ

π

π

The small block has a thickness of 5 mm. if the stress distribution at the support developed by the load varies as shown, determine the force F applied to the block, and the distance d to where it is applied.

( )( )( )( ) ( )( )[ ]( )( )

( )( )( ) ( ) ( )( )( ) ( ) ( )( )( ) ( )

md

d

dF

dF

dforSolving

kNF

NF

F

110

36000

3960000

000,960,3

601203

2512020

2

160120

2

151204060

3

256040

2

1

:

36

000,36

005.012.01060402

1005.006.01040

2

1 66

=

=

=⋅

+

+

+

+

=⋅

=

=

+

+

=

A 175-lb woman stands on a vinyl floor wearing stiletto high-heel shoes. If the heel has the dimensions shown, determine the average normal stress she exerts on the floor and compare it with the average normal stress developed when a man having the same weight is wearing flat-heel shoes. Assume the load is applied slowly, so that dynamic effect can be ignored. Also assume the entire is supported only by the heel of one shoe.

( )( ) ( )( )

( )( ) ( )( )

psi

in

lb

A

F

inA

inininA

psi

in

lb

A

F

inA

ininA

W

W

WW

W

W

M

M

MM

M

M

869

201.0

175

201.0

3.02

6.01.0

5.50

462.3

175

462.3

2.12

4.25.0

2

2

2

2

2

2

=

==

=

+=

=

==

=

+=

σ

σ

π

σ

σ

π

The 50-lb lamp is supported by three steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take θ=30°. The diameter of each rod is given in the figure.

( )

( )

( )psi

psi

lbF

lbF

F

lbFF

F

FF

FF

F

AC

AD

AC

AD

AD

ADAC

Y

ADAC

ADAC

X

61.745

22

25.0

6.36

21.634

22

3.0

83.44

6.36

83.44

45sin30cos

30sin45cos

50

5045sin30sin

0

30cos

45cos

45cos30cos

0

=

=

=

=

=

=

°+°

°°=

=°+°

=Σ

°

°=

°=°

=Σ

πσ

πσ

∴Rod AC is subjected to greater average normal stress at 745.61 psi.

The 50-lb lamp is supported by three steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take θ=45°. The diameter of each rod is given in the figure. ∴Rod AC is subjected to greater average normal stress at 745.61 psi.

( )

( )psi

psi

lbF

lbF

lbFF

F

FF

FF

F

AC

AD

AC

AD

ADAC

Y

ADAC

ADAC

X

3.720

22

25.0

36.35

24.500

22

3.0

36.35

36.35

36.35

5045sin45sin

0

45cos45cos

0

=

=

=

=

=

=

=°+°

=Σ

=

°=°

=Σ

πσ

πσ

The two steel members are joined together using a 60° scarf weld. Determine the average normal and average shear stress resisted in the plane of the weld.

( )

( )MPa

mm

kNA

V

MPamm

kNA

N

mmA

mmmmA

kNV

kNN

NN

VN

F

NV

VN

F

x

Y

62.4025.866

10004

8025.866

100093.6

025.866

60sin

3025

4

93.6

060cos60sin

30sin30cos8

060cos30cos8

0

60sin

30sin

60sin30sin

0

2

3

2

3

2

===

===

=

°=

=

=

=°°

°+°+−

=°+°+−

=Σ

°

°=

°=°

=Σ

τ

σ

The build-up shaft consist of a pipe AB and solid rod BC. The pipe has an inner diameter of 20mm and outer diameter of 28mm. The rod has a diameter of 12mm. Determine the average normal stress at points D and E and represent the stress on a volume element located at each of this points.

( )

( )

( )

( )

( )TMPa

mm

kN

CMPa

mm

kN

mmA

A

mmA

A

E

E

D

D

E

E

D

D

75.70

0973.113

10008

26.13

5929.301

10004

0973.113

36

5929.301

4

20

4

28

2

3

2

3

2

2

22

=

=

=

=

=

=

=

−

=

σ

σ

σ

σ

π

π

The plastic block is subjected to an axial compressive force of 600N. Assuming that the caps at the top and bottom distribute the load uniformly throughout the block. Determine the average normal and average shear stress acting along section a-a.

( )( )

( )( ) kPa

mmmm

NA

V

kPammmm

NA

N

NV

NN

NN

F

NV

VN

F

aa

aa

y

x

52

30cos10050

1000300

90

30cos10050

100062.519

300

62.519

030cos30sin30cos

30sin600

0

30cos

30sin

30cos30sin

0

2

2

=

°

==

=

°

==

=

=

=°−°°

°−

=Σ

°

°=

°=°

=Σ

−

−

τ

σ

The specimen failed tension test at an angle of 52° when the axial load was 19.80 kip. If the diameter of the specimen is 0.5 in., determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs?

When failure occurs no shear stress exists, since the shear force at the section is 0.

ksi

in

kipA

P

ave

101

196.0

8.19

0

2

=

==

=

σ

σ

τ

( )

ksi

in

kipA

V

ksi

in

kipA

N

inA

AA

inA

A

kipV

kipN

NNkip

VNkip

F

NV

VN

F

ave

ave

x

y

9.48'

2492.0

19.12'

''

6.62'

2492.0

6.15'

'

2492.0'

52sin'

196.0

25.0

19.12

6.15

052cos52sin

38sin38cos8.19

052cos38cos8.19

0

52sin

38sin

52sin38sin

0

2

2

2

2

2

=

==

=

==∴

=

°=

=

=

=

=

=°°

°+°+−

=°+°+−

=Σ

°

°=

°=°

=Σ

τ

τ

σ

σ

π

A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the maximum average shear stress in the specimen and indicate the orientation θ of a section on which it occurs.

( )

( )

( )

A

P

A

P

A

P

PP

A

P

A

P

d

d

AP

A

P

PV

AAwhereA

VS

S

2

45sin45cos

sincos

45

1tan

sincos

2sin2cos

2sin2cos0

2cos2sin

sincos

sin

cos

cos

sin;

=

°°

=

=

°=

=

=

=

−=

+−=

=

=

=

==

τ

τ

θθτ

θ

θ

θθ

θθ

θθ

θθθ

τ

θθ

θ

θτ

θ

θτ

The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 50 mm thick.

( )( )

( )( )

kPa

mmmm

kN

MPa

mmmm

kN

kNF

FkN

F

kNF

FkN

F

BC

BC

AB

AB

BC

BC

y

AB

AB

x

596

5050

100049.1

04.2

4050

100008.4

49.1

30sin08.445sin5

0

08.4

030cos45cos5

0

3

3

=

=

=

=

=

=°−°

=Σ

=

=°+°−

=Σ

σ

σ

σ

σ

The joint is subjected to the axial member force of 6kip. Determine the average normal stress acting on section AB and BC. Assume the member is smooth and is 1.5in. thick. ΣFy = 0 6sin60º = FBCsin70º FBC = 5.53 kip ΣFx = 0 FAB = 6cos60º+FBCcos70º FAB = 4.89 kip For section AB: σAB=FAB/A = 4.89 kip 2(1.5 in2) σAB=1.63 ksi For section BC: σBC=FBC/A = 5.53 kip 4.5 in(1.5 in) σBC=0.819 ksi

Rod AB and BC have diameters of 4 mm and 6 mm, respectively. If the load of 8 kN is applied to the ring at B, determine the average normal stress in each rod if θ = 60°.

ΣFy = 0 -8 kN+FBCsin60º = 0

ΣFx = 0

- FBC = 9.24 kN FAB+FBCcos60º = 0

FAB = 4.62 kN

σAB=4.62 kN(10003) = 367.65 MPa π(4mm2) σBC=9.24 kN(10003) = 326.8 MPa π(9mm2)

Rods AB and BC have diameters of 4mm and 6mm respectively. If the vertical load of 8kN is applied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rod is equivalent. What is this stress?

Given: dAB=4mm dBC=6mm Find: θ & σ ΣFy = 0 -8 kN+FBCsinθ = 0 FBC=8 kN/sinθ ΣFx = 0 -FAB+FBCcosθ = 0 FAB = FBCcosθ = 8 kNcosθ/sinθ σBC= 8 kN (sinθ)π(0.006m/2)2 σAB= 8 kNcosθ πsinθ(0.004m/2)2 σBC=σAB 8 kN = 8 kNcosθ πsinθ(0.006m/2)2 πsinθ(0.004m/2)2 θ=cos-1(0.0042/0.0062) θ=63.61º σ= 8 kN

πsin63.61º(0.006m/2)2 =316 MPa

The bars of the truss each have a cross-sectional area of 1.25 in2. Determine the average normal stress in each member due to the loading P=8kip. State whether the stress is tensile or compressive.

ΣFx=0 Cx-Bx=0 Cx=Bx ΣFy=0 Cy+Dy-0.75-P=0 Cy+Dy=1.75 P At joint A ΣFy=0 AB(4/5)=P AB=(5/3)P (T) ΣFx=0 AB(4/5)=AE AE=(4/3)P (C) At joint B ΣFy=0 -AB(3/5)-BE+BD(3/5)=0 BD=()5/3[.75P+(3/5)(5/3)P] BD=(35/12)P (C) ΣFx=0 -AB(4/5)-BD(4/5)+BC=0 BC=(11/3)P (T) ΣMD=0 -3Cx+4(.75P)+8P=0 Cx=(11/3)P Dx=(11/3)P At joint E ΣFx=0 AE=ED ED=(4/3)P (C) ΣFy=0 BE=.75P (P)

σAB= (5/3)P 1.25 =(4/3)(8)=10.7 ksi (T) σBC=(11/3)P/1.25=(44/5)(8) = 23.5 ksi (T) σAE=(4/3)P/1.25=(16/15)(8) =8.53 ksi (C) σBE=.75P/1.25=(3/5)(8) =4.8 ksi (T) σED=(4/3)P/1.25=(16/15)(8) =8.53 ksi (C) σBD=(35/12)P/1.25=(7/3)(8) =18.7 ksi (C)

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.

ΣMC = 0 3ΤAB = 12 kN.m TAB=4 kN ΣFy = 0 TCD = 6 kN-4 kN = 2 kN σAB= 4000 N 12(1m2/10002) σAB=333 MPa σCD= 2000 N 8(1m2/10002) =250 MPa

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 12 mm2 and 8 mm2, respectively. If d = 1 m, determine the average normal stress in each rod.

ΣMA = 0 6d = TCD(3) TCD = 2d ΣFy = 0 TAB = 6-2d σAB = σCD 6-2d = 2d 12 m2 8 m2 10002 10002 (6-2d)(8)(10002) = (12)(2d)(10002) 48 = 40d :ּd = 1.2m

The railcar dock light is supported by the 1/8 in diameter pin at A. If the lamp weighs 4lb, and the extension arm AB has a weight of 0.5 lb/ft, determine the average shear stress in the pin needed to support the lamp. Hint: The shear force is cost by the couple moment required for equilibrium at A.

0.5 lb.ft M Ay 4 lb 3 ft ΣFy = 0 Ay = 4+0.5(3) Ay = 5.5 lb ΣMA = 0 M = 4(3)+0.5(3)(1.5) M = 14.25 lb.ft[12 in/1ft] M = 171 lb.in T = V/A = M/1.25 πr2 = 11147.47 lb/in2 = 11.1 ksi

The two member frame is subjected to the distributed loading shown. Determine the intensity w of the largest uniform loading that can be applied to the frame w/o causing either the average normal stress or the average shear stress at section b-b to exceed σ = 15MPa and τ = 16MPa, respectively. Member CB has a square cross-section of 35mm on each side.

( )

( ) ( )

( ) ( )

kNW

kNW

NW

W

W

BC

kNW

NW

W

W

BC

WBC

WBC

M

y

bb

xbb

A

8.21

8.21

78.777,21

3

53516875.1

5

4

53

2.27

22.222,27

3

53515875.1

5

3

53

875.1

5.135

43

0

2

2

=∴

=

=

=

=

=

=

=

=

=

=

=Σ

−

−

l

l

τ

σ

Ay

AAx B

BC

W

3m 3

54

1-81. The 60mm × 60mm oak post is supported on the pin block. If the allowable bearing stresses for this material are σ oak =43MPa and σ pine =25MPa, determine the greatest load P that can be supported. If a rigid bearing plate is used between this material, determine its required area so that the maximum load P that can be supported. What is this load?

σOAK=43 MPa σPINE=25 MPa σOAK=P/A [43 N/mm2][602 mm2]=P 154,800 N=P P=154.8 kN σPINE=P/A [25 N/mm2][602 mm2] 90000 N=P P=90 kN :ּ P=90 kN since this is the greatest load the pine block can told. Since there is a bearing plate between the oak block and the pine block, the maximum load is the allowable load by the pine block that is P≈155 kN AALLOW= PMAX/σOAK = 154.8[1000/1 kN] 25N/mm2[(1000mm)2/1 m2

=6.19x10-3 m

1-82. The join is fastened together using two bolts. Determine the required diameter of the bolts if the allowable shear stress for the bolts is τallow = 110 MPa. Assume each bolt supports an equal portion of the load.

Τallow = V/4A 110 MPa = 80 kN(1000) 4π(d2/4) d = √80(1000)/π(110) d = 15.2 mm

1-83. The lever is attached to the shaft A using a key that has a width d and length of 25mm. If the shaft is fixed and a vertical force of 200N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is τ allow = 35MPa. ΣFy=0 Ay=200 N ΣMA =0 M=200(500 mm)[1 m/1000 mm] M=100 N.m V=M/20 mm=100 N.m/.02 m =5000N A=V/TALLOW dL=V/TALLOW d=5000 N/(25x25) N d=5.71 mm

20 mm

500 mm

A

d

200 N

1-86. The tension member is fastened together using two bolts, one on each side of the member as shown. Each bolt has a diameter of 0.3 inch. Determine the maximum P that can be applied to the member if the allowable shear stress for the bolts is τ allow =12ksi and the allowable average normal stress is σ allow = 20ksi.

ΣFy = 0 Vsin60º = Nsin30º V = Nsin30º sin60º ΣFx = 0 -P+Ncos30º+Vcos60º = 0 N = 0.866 P V = 0.5 P σ = 0.866 P 2π(0.3 in/2)2 Τ = 0.5 P/2π(0.3 in/2)2 If σ=20 ksi P = 3.26 kip Τ = 11.55 ksi (safe) If Τ=12 ksi P = 3.39 kip σ=20.8 ksi (fail) :ּP = 3.26 kip

P P

60°

1-88. The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5 kN.

ΣFx = 0 -FABsin60º+FAC(4/5) = 0

FAB = 4FBC/5sin60º

ΣFy = 0 -5+FABcos60º+FAC(3/5) = 0 FAC = 4.71 kN FAB = 4.35 kN σAB=4.35(10004) π(dAB)2 dAB=√4.35(10004)/π(200) dAB = 5.26mm σAC=4.71(10004) π(dAC)2 dAC = √4.71(10004)/π(200) = 5.48 mm

60°

A

B

C

P

4

35

1-89. The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σ allow = 180MPa, and wire AB has a diameter of 6mm and AC has a diameter of 4mm, determine the greatest force P that can be applied to the chain before one of the wires fails.

ΣFx = 0 -FABsin60º+FAC(4/5) = 0 FAB = 4FAC 5sin60º ΣFy = 0 -P+FABcos60º+FAC(3/5) = 0 FAC = 0.942 P FAB = 0.87 P σAB = 0.87 P (1000) π(9m2) σAC=0.942 P(1000) π(4m2) If σAB = 180 MPa

60°

A

B

C

P

4

35

P = 5.85 kN 2-1. An air filled rubber ball that has a diameter of 6in. If the air pressure w/in it is increase until the ball’s diameter becomes 7in., determine the average normal strain in the robber.

P

6 in.

inin

L 61

6

67=

−=∈=

δ

2-2. A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5in., determine the average normal strain in the strip.

inin

L0472.0

15

708.0==∈=

δ

2-3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10mm downward, determine the normal strain developed in wires CE and BD.

D E

B C

P

A

3 m 2 m 2 m

4 m

mmmm

mm

mm

mmmm

mm

mm

L

mm

BD

BD

CE

CCE

B

BC

3

3

3

3

1007143.1

1047

30

105.2

104

10

730

37

−

−

×=∈

×=∈

×=∈

×==∈

=

=

δ

δ

δδ

2-5. The wire AB is unstretched when θ = 45°. If the load is applied to the bar AC, w/c causes θ = 47°, determine the normal strain in the wire.

C

L

A

B

θ

L

( )

343.0

034290324.1

034290324.1

247sin

∈=

−∈=

=

=

L

LL

Lx

xL

2-7. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire.

C

A

B

30°30°

P

300 mm

300 mm

( )

( )

mmmm

x

x

310779.5

30

734.1

734.1

cos

230cos300300

81.29

230cos300

150tan

−×∈=

°∈=

=

+°=+

°=

+°=

φ

φ

φ

2-9. Part of a control linkage for an airplane consists of rigid member CBD and a flexible cable AB. If the force is applied to the end D of the member and causes a normal strain in the cable of 0.0035mm/mm, determine the displacement of D. Originally the cable is unstretched.

PD

B

CA

θ

300 mm

300 mm

400 mm

2-13. The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x along the x axis.

A A'5 mm

800 mm

800 mm

45°

45°

x'

x

y

45°

( )

( )

mmmm31043.4

800

80075.44cos

545cos800

75.44

545cos800

45sin800tan

−×∈=

−°

+°

∈=

°=

+°

°=

φ

φ

2-14 The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x’ along the x’ axis.

A A'5 mm

800 mm

800 mm

45°

45°

x'

x

y

45°

mmmm

L

x

x

31084.8

45cos800

5

'

'

−×=∈

°==∈

δ

2-15. The corners of the square plate are given the displacements indicated. Determine the average normal strains ∋x and ∋y along the x and y axes.

A

AD

C

10 in.

10 in.

10 in.10 in.2 in.

2 in.

3 in. 3 in

x

y

inin

L

inin

L

y

y

y

x

xx

02.0

10

102.10

03.0

10

3.1010

=∈

=−

=∈

−=∈

=−

=∈

δ

δ

2-21. A thin wire, lying along the x axis, is strained such that each point on the wire is displaced x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?

xP

x

kx

kx

kx

dxxk

kx

dx

dx

kx

x

2

2

1

11

2

1

11

2

1

1

2

1

2

1

0

2

2

2

−∈=

−∈=

−=∈

=∈

=∈

∈=

∫−

2-22. The wire is subjected to a normal strain that is defined by ∋ - xe-x2 , where x is millimeters. If the wire has an initial length L, determine the increase in its length.

x

x

L

2xxe∈=

( )

( )

( )2

2

2

2

2

2

2

12

1

2

1

2

1

2

1

2

1

2

1

2

1

2

LET

0

L

L

Lx

x

x

x

x

eL

e

e

u

duL

xdxedu

dxexdu

eu

dxxeL

−

−

−

−

−

−

−

−=∆

+−=

−=

−=

−=∆

=−

−=

=

=∆

∫

∫

2-25. The piece rubber is originally rectangular. Determine the average shear strain γxy if the corners B and D are subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.

D

C

BA

300 mm

400 mm

3 mm

2 mm

y

x

2-26. The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.

D

C

BA

300 mm

400 mm

3 mm

2 mm

y

x

mmmm

mmmm

AB

AB

AD

AD

6

3

1000.8

500

500229.0cos

500

38.0

3002tan

1081.2

400

40043.0cos

400

43.0

4003tan

−

−

×=∈

−°=∈

°=

=

×=∈

−°=∈

°=

=

θ

θ

φ

φ

mmmm

w

w

BD

BD

3

22

2

108.6

500

5006.496

6.496

188.89cos382.0cos

300

43.0cos

4002

382.cos

300

43.0cos

400

−×−=∈

−=∈

=

°

°

°−

°+

°=

2-28. The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.

( ) ( )( ) ( )

−=∈

=

−−+=

2655.143

2655.143

40573990cos157258024.12580222

CF

CF

CF

B

A

80 mm

100 mm

10 mm

2 mm

y

x

25 mm

15 mm

C D

F

50 mm

( ) ( )mmx

mmmmx

4.125

10125

5739.4

125

10tan

222

=

+=

=

=

θ

θ

2-29. The non uniform loading causes a normal strain in the shaft that can be expressed as ∋ = kx2 , where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod?

A B

L

x

∫∫ =2kx

2-30. The non uniform loading causes a normal strain in the shaft that can be expresses as ∋ = k sin ((π/L)x), where k is a constant. Determine the displacement of the center C and the average normal strain in the entire rod.

A B

L/2 L/2

C

( )

( )

( )

( )

( ) ( )

( )

π

π

π

π

π

π

π

π

π

π

k

L

kL

kLx

kx

xL

Lkx

dxxL

kx

dxxL

kx

dx

xx

Lk

xL

k

x

X

C

C

L

C

L

C

L

C

CL

x

2

2

1

:STRAIN NORMAL

0cos2

cos

cos

sin

sin

sin

sin

2

0

2

0

2

0

2

0

=∈

=∈

=∆

+

−=∆

−

=∆

=∆

=∆

∆=

=∈

∫

∫

∫

2-31. The curved pipe has an original radius of 2ft. If it is heated non uniformly, so that the normal strain along its length is ∋ = 0.05cosθ, determine the increase in length of the pipe.

A

2 ft

θ

( )ftx

dx

drx

rd

x

rL

1.0

sin1.0

cos1.0

cos05.0

cos05.0

90

0

90

0

90

0

=∆

=

=∆

=∆

∆=

=

∫

∫

θ

θθ

θθ

θθ

θ

4-27. Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.

P

P

h

t

d2

d1

dx

y

h

2212 dd

−

221dx

−

21d

2

x

dxdd

hdy

dyh

dddx

dyh

ddx

ydydhdxh

h

dd

y

dx

h

dd

y

dx

12

12

112

121

121

121

2222

−=

−=

+

−=

−=−

−=

−

−=

−

[ ]

( )

1

2

12

12

12

12

12

12

12

12

ln

ln

ln 2

1

2

1

2

1

d

d

dd

h

tE

P

dddd

h

tE

P

xdd

h

tE

P

x

dx

tE

dd

hP

x

dx

tE

dd

hP

xtE

dxdd

hP

AE

dxdd

hP

AE

Pdyd

d

d

d

d

d

d

−=

−

−=

−=

−=

−=

−=

−=

=

∫

∫

δ

δ

δ

δ

4-50. The three suspender bars are made of the same material and have equal cross-sectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P.

D F

C E

P

B

A

d/2 d/2 d

L

4-53. The 10mm diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20mm, and its inner diameter is 10mm. If the bolt is subjected to a compressive force of P= 20kN. Determine the average normal stress in the steel and the bronze. Est = 200Gpa, Ebr = 100GPa.

P

10 mm

P

20 mm

( ) ( )( )( )

( )

( )

( )( )

( )( )MPa

MPa

kN

L

E

kNP

kNP

PLLP

PL

LP

Pa

LPb

PbPstkN

B

B

ST

STSTST

ST

B

BB

BST

BST

BB

B

9.50

1010012102441.4

102

102008103662.6

8

12

20103662.6102441.4

20103662.6

102441.4

1010001.002.0

4.01000

20

95

95

55

5

5

922

=

××=

=

××=

=

=

=

−×=×

=

−×=

×=

×−=

+=

−

−

−−

−

−

σ

σ

σ

δσ

δδ

δ

δ

πδ

4-110. A 0.25in diameter steel rivet having a temperature of 1500°F is secured between two plates such that at this temperature it is 2in long and exerts a clamping force of 250lb between the plates. Determine the approximate clamping force between the plates when the rivet cools to 70°F. For the calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 8(10-6)/°F, Est = 29(103)ksi. Is the result a conservative estimate of the actual answer? Why or why not?

2 in.

An air filled rubber ball that has a diameter of 6in. If the air pressure w/in it is increase until the ball’s diameter becomes 7in., determine the average normal strain in the robber.

P

6 in.

inin

L 61

6

67=

−=∈=

δ

A thin strip of rubber has an unstretched length of 15 in. If it is stretched around a pipe having an outer diameter of 5in., determine the average normal strain in the strip.

inin

L0472.0

15

708.0==∈=

δ

The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10mm downward, determine the normal strain developed in wires CE and BD.

D E

B C

P

A

3 m 2 m 2 m

4 m

mmmm

mm

mm

mmmm

mm

mm

L

mm

BD

BD

CE

CCE

B

BC

3

3

3

3

1007143.1

1047

30

105.2

104

10

730

37

−

−

×=∈

×=∈

×=∈

×==∈

=

=

δ

δ

δδ

The wire AB is unstretched when θ = 45°. If the load is applied to the bar AC, w/c causes θ = 47°, determine the normal strain in the wire.

C

L

A

B

θ

L

( )

343.0

034290324.1

034290324.1

247sin

∈=

−∈=

=

=

L

LL

Lx

xL

The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire.

C

A

B

30°30°

P

300 mm

300 mm

( )

( )

mmmm

x

x

310779.5

30

734.1

734.1

cos

230cos300300

81.29

230cos300

150tan

−×∈=

°∈=

=

+°=+

°=

+°=

φ

φ

φ

Part of a control linkage for an airplane consists of rigid member CBD and a flexible cable AB. If the force is applied to the end D of the member and causes a normal strain in the cable of 0.0035mm/mm, determine the displacement of D. Originally the cable is unstretched.

For the wire:

PD

B

CA

θ

300 mm

300 mm

400 mm

mmx

x

mmAB

AB

383.4

600tan

75.1

5000035.0

=

=

=∆

∆=

θ

( ) ( )( ) ( )( )

°=

+−=

+−+=

4185.0

90cos2400000625.1753

90cos300400230040075.501 222

θ

θ

θ

The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x along the x axis.

A A'5 mm

800 mm

800 mm

45°

45°

x'

x

y

45°

( )

( )

mmmm31043.4

800

80075.44cos

545cos800

75.44

545cos800

45sin800tan

−×∈=

−°

+°

∈=

°=

+°

°=

φ

φ

The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5mm. Determine the average normal strain ∋x’ along the x’ axis.

A A'5 mm

800 mm

800 mm

45°

45°

x'

x

y

45°

mmmm

L

x

x

31084.8

45cos800

5

'

'

−×=∈

°==∈

δ

The corners of the square plate are given the displacements indicated. Determine the average normal strains ∋x and ∋y along the x and y axes.

A

AD

C

10 in.

10 in.

10 in.10 in.2 in.

2 in.

3 in. 3 in

x

y

inin

L

inin

L

y

y

y

x

xx

02.0

10

102.10

03.0

10

3.1010

=∈

=−

=∈

−=∈

=−

=∈

δ

δ

A thin wire, lying along the x axis, is strained such that each point on the wire is displaced x = kx2 along the x axis. If k is constant, what is the normal strain at any point P along the wire?

xP

x

kx

kx

kx

dxxk

kx

dx

dx

kx

x

2

2

1

11

2

1

11

2

1

1

2

1

2

1

0

2

2

2

−∈=

−∈=

−=∈

=∈

=∈

∈=

∫−

The wire is subjected to a normal strain that is defined by ∋ - xe-x2 , where x is millimeters. If the wire has an initial length L, determine the increase in its length.

x

x

L

2xxe∈=

( )

( )

( )2

2

2

2

2

2

2

12

1

2

1

2

1

2

1

2

1

2

1

2

1

2

LET

0

L

L

Lx

x

x

x

x

eL

e

e

u

duL

xdxedu

dxexdu

eu

dxxeL

−

−

−

−

−

−

−

−=∆

+−=

−=

−=

−=∆

=−

−=

=

=∆

∫

∫

The piece rubber is originally rectangular. Determine the average shear strain γxy if the corners B and D are subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.

D

C

BA

300 mm

400 mm

3 mm

2 mm

y

x

The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD.

D

C

BA

300 mm

400 mm

3 mm

2 mm

y

x

mmmm

mmmm

AB

AB

AD

AD

6

3

1000.8

500

500229.0cos

500

38.0

3002tan

1081.2

400

40043.0cos

400

43.0

4003tan

−

−

×=∈

−°=∈

°=

=

×=∈

−°=∈

°=

=

θ

θ

φ

φ

mmmm

w

w

BD

BD

3

22

2

108.6

500

5006.496

6.496

188.89cos382.0cos

300

43.0cos

4002

382.cos

300

43.0cos

400

−×−=∈

−=∈

=

°

°

°−

°+

°=

The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF.

( ) ( )( ) ( )

( )( ) ( )

mmmm

mmAD

AD

mmy

mmmm

mmCF

CF

AD

AD

CF

CF

24708.0

15850

15850003.157

003.157

90cos801585021585080

9.12515850

843.6

125

15tan

03465.0

22025

220252655.143

2655.143

40573990cos157258024.12580

22

222

=∈

−=∈

=

+−+=

==

°=

=

−=∈

−=∈

=

−−+=

φ

φ

φ

B

A

80 mm

100 mm

10 mm

2 mm

y

x

25 mm

15 mm

C D

F

50 mm

( ) ( )mmx

mmmmx

4.125

10125

5739.4

125

10tan

222

=

+=

°=

=

θ

θ

The non uniform loading causes a normal strain in the shaft that can be expresses as ∋ = k sin ((π/L)x), where k is a constant. Determine the displacement of the center C and the average normal strain in the entire rod.

A B

L/2 L/2

C

( )

( )

( )

( )

( ) ( )

( )

π

π

π

π

π

π

π

π

π

π

k

L

kL

kLx

kx

xL

Lkx

dxxL

kx

dxxL

kx

dx

xx

Lk

xL

k

x

X

C

C

L

C

L

C

L

C

CL

x

2

2

1

:STRAIN NORMAL

0cos2

cos

cos

sin

sin

sin

sin

2

0

2

0

2

0

2

0

=∈

=∈

=∆

+

−=∆

−

=∆

=∆

=∆

∆=

=∈

∫

∫

∫

The curved pipe has an original radius of 2ft. If it is heated non uniformly, so that the normal strain along its length is ∋ = 0.05cosθ, determine the increase in length of the pipe.

A

2 ft

θ

( )ftx

dx

drx

rd

x

rL

1.0

sin1.0

cos1.0

cos05.0

cos05.0

90

0

90

0

90

0

=∆

=

=∆

=∆

∆=

=

∫

∫

θ

θθ

θθ

θθ

θ

Determine the relative displacement of one end of the tapered plate with respect to the other end when it is subjected to an axial load P.

P

P

h

t

d2

d1

dxdd

hdy

dyh

dddx

dyh

ddx

ydydhdxh

h

dd

y

dx

h

dd

y

dx

12

12

112

121

121

121

2222

−=

−=

+

−=

−=−

−=

−

−=

−

[ ]

( )

1

2

12

12

12

12

12

12

12

12

ln

ln

ln 2

1

2

1

2

1

d

d

dd

h

tE

P

dddd

h

tE

P

xdd

h

tE

P

x

dx

tE

dd

hP

x

dx

tE

dd

hP

xtE

dxdd

hP

AE

dxdd

hP

AE

Pdyd

d

d

d

d

d

d

−=

−

−=

−=

−=

−=

−=

−=

=

∫

∫

δ

δ

δ

δ

dx

y

h

2212 dd

−

221dx

−

21d

2

x

The three suspender bars are made of the same material and have equal cross-sectional areas A. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P.

D F

C E

P

B

A

d/2 d/2 d

L

EFEF

CDCD

ABAB

FPA

P

FPA

P

FPA

P

FPA

P

==

==

==

==

,

,

,

,

σ

σ

σ

σ

( ) ( )

2

32

2

32

:0

0

PFF

dPdFdF

M

PFFF

Fy

CDAB

CDAB

E

EFCDAB

=+

=+

=Σ

=++

=Σ

3.2

32

2.02

1.

02

22

12

2

`2

eqnP

FF

eqnFFF

eqnPFFF

FFF

FFFF

FFFF

dA

F

A

F

dA

F

A

F

dd

CDAB

EFCDAB

EFCDAB

EFCDAB

EFCDEFAB

EFCDEFAB

EFCDEFAB

EFCDEFAB

=+

=+−

=++

=+−

−=−

−=

−

−=

−

−=

− σσσσ

PF

PF

PFF

pFF

guFinatee

eqnPFF

PFF

FFF

guFinatee

eqnp

FF

PFF

PFFF

guFinatee

AB

AB

EFAB

EFAB

EF

EFAB

CDAB

EFCDAB

CD

EFAB

CDAB

EFCDAB

Cd

12

7

2

76

35

2

)5&4(sinlim

5.352

32

02

)3&2(sinlim

4.2

2

32

)3&1(sinlim

=

−=−

=+

−=+−

=+

=+

=+−

−=+−

=+

=++

A

P

A

P

A

P

PF

PP

FP

guFsolve

pF

pFP

guFsolve

EF

CD

AB

CD

Cd

CD

EF

EF

EF

12

3

12

7

3

1212

7

)1(sin

12

212

7

)4(sin

=

=

=

=

=++

=

−=+−

σ

σ

σ

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students

The 10mm diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20mm, and its inner diameter is 10mm. If the bolt is subjected to a compressive force of P= 20kN. Determine the average normal stress in the steel and the bronze. Est = 200Gpa, Ebr = 100GPa.

P

10 mm

P

20 mm

( ) ( )( )( )

( )

( )

( )( )

( )( )MPa

MPa

kN

L

E

kNP

kNP

PLLP

PL

LP

Pa

LPb

PbPstkN

B

B

ST

STSTST

ST

B

BB

BST

BST

BB

B

9.50

1010012102441.4

102

102008103662.6

8

12

20103662.6102441.4

20103662.6

102441.4

1010001.002.0

4.01000

20

95

95

55

5

5

922

=

××=

=

××=

=

=

=

−×=×

=

−×=

×=

×−=

+=

−

−

−−

−

−

σ

σ

σ

δσ

δδ

δ

δ

πδ

A 0.25in diameter steel rivet having a temperature of 1500°F is secured between two plates such that at this temperature it is 2in long and exerts a clamping force of 250lb between the plates. Determine the approximate clamping force between the plates when the rivet cools to 70°F. For the calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 8(10-6)/°F, Est = 29(103)ksi. Is the result a conservative estimate of the actual answer? Why or why not?

2 in.

( )

( )( )( ) ( )

( ) ( )

kipsP

P

kipsP

P

AE

PLTL

T

T

TH

TH

54.16

25.0289.16

29.16

102925.04

21500702108

32

6

=

+=

=

×

=−×

−=∆

−

π

α

The rubber block is subjected to an elongation of 0.03 in along the x axis, and its vertical faces are given a tilt so that °= 3.89θ .Determine the strains x∈ , x∈ , xyγ .

Take 5.0=rv

Finding x∈

Finding y∈

( )

inin

y

y

xy

00375.0

0075.05.0

−=∈

=∈

∈−=∈ υ

Finding xyγ

inin

in

in

x

x

0075.0

4

03.0

=∈

=∈

4 in.

3 in.θ

y

x

( )( )radxy

xy

0122.0

1803.8990

=

°−°=

γ

πγ

The non uniform loading causes a normal strain in the shaft that can be expressed as ∋ = kx2 , where k is a constant. Determine the displacement of the end B. Also, what is the average normal strain in the rod?

A B

L

x

∫∫ =2kx

5- 2. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r r'

T

by ratio and proportion:

7651.32cos'

841.0'

2'

2'

'2

'

2

'

4

4

44

44

44

max

rr

rr

rr

rr

rr

r

T

r

T

rr

=

=

=

=

=

=

=

ππ

ττ

5-3. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4).

45sin'

707.0'

4'

4'

'2

'

2

'

'2

.)

2

''

)2

(2

4

4

44

44

44

max

3

3max

33

rr

rr

rr

rr

rr

r

T

r

T

rr

r

T

b

r

T

r

T

r

T

=

=

=

=

=

=

=

=

=

==

ππ

ττ

πτ

πτ

ππτ

r r'

T

5-5. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B. sol’n:

C B

A

600 lb-ft

350 lb-ft

450 lb-ft

τmax

450 lb⋅ft = 5400 lb⋅in 350 lb⋅ft = 4200 lb⋅in 600 lb⋅ft = 7200 lb⋅in

5400 + 4200 – 7200 = 2400

( )( )( )

( )

ksi

inlb

dD

TD

B

B

B

76.2

6.2758

3.25.2

5.2240016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

7200 - 4200 = 3000

( )( )( )

( )

ksi

inlb

dD

TD

B

A

A

45.3

3.3448

3.25.2

5.2300016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

5-10. The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section.

600 N

600 N

25mm

5mm

75mm

75mm

MPa

x

J

Tr

MPa

mx

Nm

J

Tr

mxJ

dDJ

NmT

T

M x

40.10

)10108(

)0125.0)(90(

45.14

)10108(

)075.0)(90(

10108

)025.0035.0(32

)(32

90

)075.0)(600(2

;0

9

49

2

max

49

44

44

=

=

=

=

=

=

=

−=

−=

=

=

=∑

−

−

−

τ

τ

π

π

5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

2.5 in.

d

Given:

?

/75.min/2700

/.250,193

/000,10

5.22

=

==

==Ρ

=

d

srevrevf

sftlbhp

inld

inD

allowτ

Solution:

inlbT

T

T

T

.97.4084

)75.0(2

19250

)]75.0(2[19250

=

=

=

=Ρ

π

π

ω

rd

r

r

r

r

r

=

=

=

−=

=−

−

=

2

206.1

649.6

10000

)2)(25.1)(4084097()25.1(

10000

)25.1)(97.4084(

2)(

2)25.1(

)25.1(2

)25.1(97.4084

4

44

2

44

44max

π

ππ

ππ

πτ

ind 4.2= ; 5.08

14 =

ind2

12=

5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. sol’n:

8in. 8in.

F

F

12in

( )( )( )

( )

ksi

inlb

dD

TD

AB

AB

AB

8.7

72.7818

68.075.0

75.021016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ ( )

( )( )( )

ksi

inlb

dD

TD

BC

BC

BC

36.2

2361

86.01

121016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

AB D = 0.75 in d = 0.68 in BC D = 1 in d = 0.86 in

( ) ( )inlbT

TM A

⋅=

=+=∑210

815615

5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm.

d

T

T

20mm

To find the T; steel;

NmT

Tx

r

T

s

s

s

07.157

)010.0(

210100

2

3

6

3

=

=

=

π

πτ

To find the max.., τ; brass;

MPa

rR

TR

3.13

)01.002.0(

)02.0)(07.157(2

)(

2

44

44

=

−=

−=

τ

π

πτ

5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. Soln:

R

T

T

r

2

3

23

3

2

2

2

2

42

2;2

1;4

4

4

d

rn

Rnd

T

r

T

r

T

Rnd

T

Rnd

T

PRnT

dP

A

P

=

=

><=

><=

⋅=

=

⋅=

=

ππ

πτ

πτ

τπ

τπ

τ

5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function

( )2

25 xxt = N⋅m/m, where x is in meters. Determine the minimum torque T0 needed

to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft.

( )2

25 xxet =

To 80mm

x

2m

Solution: d=80mm=0.08 m

( )2

25 xxet =

3

16

d

T

πτ =

dxxeTx2

2

025∫=

dxxeT x

∫=2

0

2

25

let u= 2x du=2x dx du/2=x dx x=0, u=0 x=2,u=4

∫=4

0 2

25 dueT

u

dueu

∫=4

05.12

[ ]4

05.12 ue=

[ ]15.12 4 −= e

NmT 670=

MPa66.6=τ

33 )1080(

)670(16−

=xπ

3

16

d

T

πτ =

5-30. The solid shaft has a linear taper from Ar at one end to Br at the other. Derive

and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.

rB

x L

rA

ro

L

x

T

T

Solution:

1rrr B +=

L

rr

xL

r BA −=

−1

L

xLrrr BA ))((1

−−=

L

xLrrrr BA

B

))(( −−+=

L

xLrrLrr BAB ))(( −−+

=

L

xLrrxLLrr BAB )()( −−−+

=

L

xLrxLLrr AB )()]([ −+−−

=

L

xLrxrr AB )()( −+

=

L

xLrxrr AB )( −+

=

3

16

d

T

πτ = ;

2

dr =

33 8

2

rd

rd

=

=

38

16

r

T

πτ =

3

2

r

T

π=

3)(

2

−+=

L

xLrxr

T

ABπ

τ

[ ]3

3

)(

2

xLrxr

TL

AB −+=

πτ

5-32. The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall.

Given: psiksi 80008 ==τ

inD 5.2= slbfthp /000,100200 ==Ρ sec/19min/1140 revrevf ==

?=t Solutions:

f

Tπ2

Ρ=

)/19(2

/000,100

srev

slbft

π=

lbftT 66.837= or lbin89.051,10=

)(

1644 dD

TD

−=

πτ

)5.2(

)5.2)(89.051,10(16000,8

44 dpsi

−=

π

)000,8(

65.402075)5.2( 44

π=− d

4

4

23

998.1539

=

−=

d

d

ind 19.2= Thickness:

2

DR = ;

2

dr =

2

5.2=R

2

19.2=r

25.1=R 1.1=r rRt −=

int

t

15.0

1.125.1

=

−=

5-33. The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi.

Given

?

000,77

5.2

/25min/500,1

/000,75150

min =

==

=

==

==Ρ

t

psiksi

inD

srevrevf

slbfthp

τ

Solution

f

Tπ2

Ρ=

)/25(2

/000,75

srev

slbft

π=

lbinT

lbftT

58.729,5

46.477

=

=

ind

d

d

d

d

dD

Td

29.2

5968.27

42.1039

)000,7(

183,22939

)5.2(

)5.2)(58.5729(16000,7

)(

16

4

4

4

44

44

=

=

−=

=−

−=

−=

π

π

πτ

Thickness:

inR

R

DR

25.1

2

5.2

2

=

=

=

;

inr

r

dr

146.1

2

29.2

2

=

=

=

int

t

rRt

104.0

15.125.1

=

−=

−=

5-34. The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi. Given:

ksi

inD

srevrevf

sftlbhp

7

3

/20min/1200

/.100000200

=

=

==

==Ρ

τ

psi7000= ?max =t

Solution:

lbftT

lbftT

srev

slbftT

fT

3.9549

8.795

)/20(2

/100000

2

=

=

=

Ρ=

π

π

ind

d

d

d

d

dD

TD

785.2

16.60

84.2081

)7000(

4.45836681

)3(

)3)(3.9549(167000

)(

16

4

4

4

44

44

=

=

−=

=−

−=

−=

π

π

πτ

Max Thickness:

2

DR =

5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi.

A B

6 in.

( )( )( )( )

( )ind

d

d

dD

TD

inlbT

T

pT

75.1

11.616

2

2150001625000

16

15000

200

250000

4

44

44

=

−=

−=

−=

⋅=

=

=

π

πτ

ω

lbfthp

slbfthp

slbfthp

inlb

ksi

/250000/500

500

/5001

/25000

252

=

⋅

⋅=

=

=τ

5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall.

A B

6 in.

5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A B

6 in.

5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B.

A B

6 in.

5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A B

5-46. The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element.

5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. sol’n: θAB = θBC

680 N⋅m 1.60 m

0.75 m

C

A

B

BCAB JG

TL

JG

TL=

( )( )( ) ( )( )

( )( )

( )( )94

94

103932

6.1680

107532

030.0

75.0680

dππ=

( )( )( )( )( )3975.0

75030.06.1 44 =d

( )4 46103.3 md −=

md 0427.0=

mmd 7.42=

5-78. The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A-36 steel.

d

T

T

20mm

The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and B allow free rotation of the shaft.

( )( )( )

( )

( )( )( )

( )

ksi

ksi

ftlb

ftlbT

ftlbT

T

DE

DE

BC

BC

BC

DE

BC

AB

62.3

75.032

12375.025

07.5

75.032

12358.035

.35

.25

.35

0

4

4

=

=

=

=

=

=

=

=

τ

πτ

τ

πτ

τ

The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.

( )( )( )

( )

ksi

ftlbT

T

CD

CD

EF

CD

EF

17.2

75.032

12375.015

0

.15

0

4

=

=

=

=

=

τ

πτ

τ

5-49. The splined ends and gears attached to the A-36 steel shaft are subjected to the torque shown. Determine the angle of twist of end B with respect to the end A. The shaft has a diameter of 40 mm.

5-58. The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel. 5-59.The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel. 5-6. the solid 1.25 in diameter shaft is used to transmit the torques applied to the gears. If it is supported by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points. 5-7. The shaft has an outer diameter of 1.25 in and an inner diameter of 1 in. If it is subjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque.

5-50.The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.

6-1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exerts only vertical reactions on the shaft.

( )

( )

( ) ( )[ ]

( )

[ ]0

5.75.7

5.7

5.7

5.3124

24

0

.8.05.3125.024

.6

25.024

5.7

5.3124

0

5.31

05.1248.0

0

=

−=

=

=

+−=

−=

=

+−=

−=

−=

=

=−=

=Σ

=

=

=Σ

B

B

B

A

A

o

B

B

A

LA

B

B

Y

A

A

B

V

kNV

kNV

kNV

kNV

kNV

M

mkNM

mkNM

M

kNR

R

F

kNR

R

M

L

L

L

L

250 mm 800 mm

A B

24 kN

RA

RB

-24 kN

7.5 kN

-6 kN⋅m

6-2. The load binder is used to support a load. If the force applied to the handle is 50lb, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC.

( )

( )

( )0

200200

200

200

25050

50

50

200

50250

0

250

31550

0

2

2

1

1

=

−=

=

=

+−=

−=

−=

=

−=

=Σ

=

=

=Σ

C

C

C

B

B

B

A

Y

c

V

lbV

lbV

lbV

lblbV

lbV

lbV

lbT

lblbT

F

lbT

T

M

L

L

50 lb

T2

T1

C

B

A

12 in. 3 in.

200 lb

-50 lb

-600 lb⋅in

5- 2. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r r'

T

5-3. The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4). 5-5. The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B.

5-10. The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section.

r r'

T

C B

A

600 lb-ft

350 lb-ft

450 lb-ft

600 N

600 N

25mm

5mm

75mm

75mm

5-13. A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

2.5 in.

d

5-17. The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb.

8in. 8in.

F

F

12in

5-21. The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm. 5-22. The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. 5-29. The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function

( )2



d

T

T

20mm

R

T

T

r

( )2

25 xxet =

To 80mm

x

2m

5-30. The solid shaft has a linear taper from Ar at one end to Br at the other. Derive

and equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.

T T

x

A

L

rB

rA

B

5-35. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi. 5-36. The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall.

A B

A B

6 in.

6 in.

5-37. A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi. 5-38. The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B. 5-42. The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A B

A B

6 in.

6 in.

A B

5-74. A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. 5-78. The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A-36 steel.

680 N⋅m 1.60 m

0.75 m

C

A

B

d

T

T

20mm

6-3. Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B and C and E.

6-4. Draw the shear and moment diagrams for the beam.

( ) ( ) ( )

( )( )

( )( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )0

.}1276.142

2711047806124.82{

.24.420

1511035804924.82

.16.1196

20803424.82

.36.1151

.1424.82

0

35

76.107

24.2

24.82

24.82

76.1423511080

0

76.142

613534110148049

0

=

+

−−=Σ

−=

−−=Σ

=

−=Σ

=

=Σ

=

=

−=

=

=

=

−++=

=Σ

=

++=

=Σ

L

L

L

L

L

L

L

L

E

E

D

D

C

C

B

B

E

D

C

B

A

A

A

Y

D

D

A

M

inlb

M

inlbM

M

inlbM

M

inlbM

inlbM

V

lbV

lbV

lbV

lbV

lbR

R

F

lbR

R

M

A

B C D

E

14 in. 20 in. 15 in. 12 in.

80 lb 110 lb 35 lb

82.4 lb

2.24 lb

107.76 lb

35 lb

1151.36 lb.ft

1196 lb.ft

-420.24 lb.ft

( )( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

0

4282122162204

.16

4282122164

.24

4282124

.24

4284

.16

4

2

42

:symmetry By

=

−−−−=Σ

=

−−−=

=

−−=Σ

=

−=Σ

=Σ

==

==

L

L

L

L

L

L

L

L

L

F

F

E

E

D

D

C

C

B

FA

FA

M

M

ftkipM

M

ftkipM

M

ftkipM

M

ftkipM

kipRR

RR

4 ft 4 ft 4 ft 4 ft 4 ft

2 kip 2 kip 2 kip 2 kip

4 kip

2 kip

-2 kip

-4 kip

16 kip.ft

24 kip.ft

16 kip.ft

6-5. Draw the shear and moment diagrams for the rod. It is supported by a pin at A and a smooth plate at B. The plate slides w/in the groove and so it cannot support a vertical force, although it can support a moment.

( ) ( )

mkNM

mkNM

M

kNR

F

LC

B

B

A

Y

.60

.30

215415

5.1

0

=Σ

=

−=Σ

=

=Σ

A

B

4 m 2 m

15 kN

15 kN

30 kN.m

60 kN.m

6-6. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x w/in the region 125mm<x<725mm.

( ) ( )

( )

( ) ( )

( ) ( ) ( )075.02500675.08008.0625.815

.328.111

6.0800725.0625.815

.953.101

125.0625.815

37.1484

625.8151500800

0

625.815

751500675800800

0

−−=Σ

=

−=Σ

=

=Σ

=

−+=

=Σ

=

+=

=Σ

L

L

L

L

B

D

D

C

C

B

B

Y

A

A

B

M

mNM

M

mNM

M

NR

R

F

NR

R

M

125 mm

A B

800 N

815.625 N

1500 N

600 mm 75 mm

x

15.625 N

-1484.375 N

101953 N.m

111328 N.m


( )( ) mNxM

xxM

NV

.100625.15

125.0800625.815

625.16

725mm x 125mm @

+=

−−=

=

<<

( ) ( )

( )

( )

( )

0

.15

3839

.39

21875

0

8

1018

18

.75

1558210

18

0

=Σ

−=

+−=Σ

−=

+−=Σ

=

=

−=

=

−=

−−−=Σ

=

=Σ

B

B

B

C

C

B

C

C

A

A

A

A

Y

M

mkNM

M

mkNM

M

V

kNV

kNV

kNV

mkNM

M

kNR

F

L

L

L

L

L

L

10 kN 8 kN

2 m 3 m

15 kN.m

18 kN

8 kN

-39 kN.m-15 kN.m

-75 kN.m

6-8. Draw the shear and moment for the pipe. The end screw is subjected to a horizontal force of 5kN. Hint: the reaction at the pin C must be replaced by equivalent loadings at point B on the axis of the pipe.

( )

( )( )

( )

0

4.04.0

4.0

4.01

0

0

1

1

1

8.054.0

0

1

8.054.0

0

=

⋅+⋅−=

⋅−=

−=

=

=

−=

−=

=

=

=Σ

=

=

=Σ

=

mkNmkNM

mkN

mkNM

M

kNV

kNV

kNR

R

M

kNR

R

M

B

BL

A

BL

A

A

A

B

B

B

A

5 kN

400 mm

80 mmA

B

C

-1 kN

-0.4 kN.m

6-9. Draw the shear and moment diagrams for the beam. Hint: the 20kip load must be replaced by equivalent loadings at point C on the axis of the beam.

( ) ( )

( )

( )

( ) ( )

( )0

433.333.13

33.13

20415867.11

67.46

467.11

0

33.3

01567.11

33.333.367.11

33.367.11

67.1138.3

20815122041512

00

=

−=

⋅=

−−=Σ

⋅=

=Σ

=

−=

=−=

+−==

−==

==

+=−=

=Σ=Σ

B

DL

CL

A

C

BCL

BLA

AB

AB

BA

M

ftkip

M

ftkip

M

M

kip

kipV

VkipV

kipVkipV

kipRkipR

RR

MM

6-10. The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.

( )

( )

( )0

60006000

6000

2

32000

0

0

12001200

1200

1200

32002000

2000

2000

3200

12002000

0

520005

0

=

⋅−−=Σ

⋅−=

−=Σ

=Σ

=

−=

=

=

+−=

−=

−=

=

+=

=Σ

=

=Σ

ftlbM

ftlb

M

M

lblbV

lbV

lb

lblbV

lbV

lbV

lbR

lblbR

F

ftlbR

M

CL

BL

A

CL

CL

B

AL

A

B

B

Y

A

B

A

B C

4 ft

3 ft 5 ft

1200 lb

-2000

1200

-6000

6-12. Draw the shear and moment diagrams for the compound beam w/c is pin connected at B. It is supported by a pin at A and a fix wall at C.

( ) ( )

0

44

4

4

84

4

4

106

6

6

4

1014

0

10

32108

4818614

0

=

+−=

−=

−=

−=

=

=

+−=

−=

−=

=

−=

=Σ

=

+=

+=

=Σ

kipkipV

kipV

kip

kipkipV

kipV

kip

kipkipV

kipV

kipV

kipR

R

F

kipR

R

M

C

CL

D

DL

A

AL

O

C

C

y

A

A

C

( )( )

0

1616

16

2440

24

46

0

=

⋅−⋅=

⋅=

⋅−⋅=

⋅−=

−=

=

ftkipftkipM

ftkip

ftkipftkipM

ftkip

ftkipM

M

CL

DL

AL

o

A

8 kip6 kip

BC

4 ft 6 ft 4 ft 4 ft

4

-6

-4

16

-24

6-15. Draw the shear and moment diagrams for the beam. Also, determine the shear and moment in the beam as a function of x, where 3ft<x<15ft.

( )( )( )

( )

( )

ftkip

AreaM

ftx

x

SimilarBy

kipR

R

F

kipR

R

M

C

B

B

A

A

B

⋅=

−=

−=

−∆=

=

=

∆

=

−=

=Σ↑

=

+=

=Σ

79.7

50778.67

50778.82

167.13

50

778.8

167.13

12

18

:

833.4

167.13125.1

0

167.13

6125.15012

0

( )

( ) ( )

( )

25.46667.1775.0

75.65.475.05.39167.13

9675.05.39167.13

32

5.13167.13

5.17667.17

35.1167.13

:153@

2

2

2

2

−+−=

−+−−=

+−−−=

−−−=

−=

−−=

≤<

xxM

xxx

xxx

xxM

xV

xV

ftxft

50 kip.ft

1.5 kip/ft

A B

3 ft 12 ft

x

x

C

13.167

-4.833

7.79 kip.ft

-50 kip.ft


( ) ( )

( )( ) ( )( )

( )( )

( )( ) ( )( )0

4880012880051200

25600

4880051200

51200

48800128800

0

88008800

0

=

−−=Σ

⋅=

−=Σ

⋅=

−=Σ

=

−=

=Σ

C

B

A

A

y

M

ftlb

M

ftlb

M

R

F

( )

0

64006400

6400

8800

0

=

+−=

−=

−=

=

CL

BL

A

V

lb

V

V

800 lb/ft

A B

8 ft

800 lb/ft

8 ft

-6400

25600

51200

6-17. The 50lb man sits in the center of the boat, w/c has a uniform width and weight per linear foot of 3lb/ft. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat.

( )

0

2

175.075

0

13

15

15045

0

0

=

⋅=

∆=

=

=

+=

=Σ

C

B

A

o

y

M

ftlb

AreaM

M

ftlbw

w

F

( ) ( )

( )

o

V

lb

lbV

lb

V

V

CL

B

BL

A

=

+−=

−=

−=

=

−=

=

7575

75

15075

75

5.735.713

0

150 lb

7.5 ft 7.5 ft

75 lb

-75 lb

281.25 lb.ft

2°2°

6-18. The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform.

( )

( )

0

7

7

147

7

126

77

7

147

7

66

7

67

2428

141424

0

=

+−=

−=

−=

=

+−=

−=

−=

=

=

=

=

+=

=Σ

kipkipV

kip

kipkipV

kipV

ftft

kipkipV

kip

kipkipV

kip

ftft

kipV

ftkipw

ftkip

kipkipw

F

DL

C

CL

CL

B

BL

o

o

y

14 kip 14 kip

12 ft 6 ft6 ft

2°

21 kip.ft

-7 kip

7 kip

2°

2°

21 kip.ft

-7 kip

7 kip

A B CD

E

6 ft


( )( )

( )

( )

( )( )

( )

( )[ ]

( )

( )0

5.05.0

5.0

5.952

0

0

5.055.2

5.2

305.27

25

5102

1

0

5.0

5.910

1052

0

5.9

305.125210

0

=

+−=

=

−=

=Σ

=

⋅−=

⋅=

⋅+−=

⋅−=

−=

=

−=

+−=

−=−=

=

=

−=

=Σ

B

B

B

B

y

BL

BL

CL

AL

O

A

AL

O

A

A

B

V

kipV

kipR

R

F

M

ftkipM

ftkip

ftkipM

ftkip

M

M

kip

kipV

kipV

V

kipR

R

M

30 kip.ft2 kip/ft

A B

5 ft 5 ft 5 ft

-10 kip

-0.5 kip

2°

1°-25 kip.ft

-27.5 kip.ft

2.5 kip.ft

6-21. Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as a function of x, where 4ft<x<10ft. by symmetry:

( )

( )

( ) ( )

2

2

2

7510503200

2

41504450200

1501050

4150450

450

2

900

6150

0

xx

xxM

x

xV

RlbR

RR

RR

F

BA

BA

BA

y

−+−=

−−−+−=

−=

−−=

==

==

=+

=Σ

150 lb/ft

A B

4 ft 6 ft 4 ft

x

200 lb.ft200 lb.ft

200 lb.ft

-200 lb.ft -200 lb.ft

-450 lb

450 lb

2°2°

6-23. The T-beam is subjected to the loading shown. Draw the shear and moment diagrams for the beam.

( )

( ) ( ) ( )( )

( )

ftkipM

kipV

ftx

x

xRV

F

R

R

M

lbRlbRR

RR

F

MAX

MAX

A

y

B

B

A

ABA

BA

y

⋅−=

−=

=

+−−=

−−−=

=Σ

=

=+

=Σ

==+

+=+

=Σ

12

2

67.21

600100200067.35660

61002000

0

33.233

9181001862000

0

67.3566;3800

181002000

0

6-24. The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 2ft length. Draw the shear and moment diagrams for the beam if it supports a uniform loading of 2kip/ft.

( ) ( )( )

( )

( )

( )( )( )[ ]

( ) ( )( )[ ]

( ) ( )( ) ( )( )0

124682118

8

48298

24

2428

18

0

4

8

8

16

0

88

8

168

8

8

816

16

0

4

582102

0

=

+−=Σ

⋅=

⋅−=Σ

⋅=

⋅+=Σ

⋅=Σ

=

=

=

=

+−=

−=

−=

=

=

−=

−=

=Σ

=

=

=Σ

B

D

D

C

A

B

E

A

A

OA

y

O

O

A

M

ftkip

ftkipM

ftkip

ftkipM

ftkipM

M

ftx

x

kipV

kip

kipV

kipV

kipR

wR

F

ftkipw

w

M


L/2

Mmax

12

6424max

LW

LLWLLWM

o

oo

=

⋅−⋅=Σ

4

4

1

22

1

22

1

0

4

612

26

1

22

1

22

3

22

1

0

22

LWR

LWLWLW

R

F

LWR

LWLWLR

LLWLLWLR

M

oB

ooo

B

Y

oA

ooA

ooA

B

=

⋅−+⋅=

=Σ

=

+=

++

⋅⋅=

=Σ


89

3

4

42

0

4

323

2

0

2LWM

L

W

L

y

LWR

LWLWR

F

LWR

LLWLR

M

oAL

oA

oB

ooB

Y

oA

oA

B

−=Σ

=

=

−=

=Σ

=

⋅=⋅

=Σ

2

24

240

2

1

0

2

2

Lx

LL

x

L

xWLW

xyRV

F

OO

A

y

=

−=

−=

−=

=Σ

2

22

0244.0

11785.00934.0

232222

1

324

LW

LWLW

LL

L

WL

L

WLLLWM

O

OO

OOOCL

−=

−=

⋅

⋅−

−=Σ

RA

M

x

3

L

x

y

L

WO =


6

6

6

3

3

:

6

036

336

332

1

3

663

3

27654

7

33

2

32

1

3333

2

932

1

0

222

Lx

x

LWW

x

LW

L

LW

similarby

LW

VLWLW

V

LWLWRV

LW

LWLWLWV

LWLWVR

LWV

RLW

R

LWLWLW

LLWLLLWLLLWLR

M

OO

OO

O

BOO

D

OOBB

O

OOOC

OOBA

OA

BO

A

OOO

OOOA

B

=

=

=

∆

−=

=−=

−===

−=−=

−−===

==

++=

+

++

+

=

=Σ

( )

0

9

2

63633

2

966

54

5

662

1

216

23

216

23

72108

5

6

126696633

54

5

9633

2

2

2

222

2

=

−

+−

+−=Σ

=

−=Σ

=

−−=

−

+−

+=Σ

=

−

=Σ

LLWLLLWLLLWL

LWM

LW

LLWLWM

LW

LWLWLW

LLWLLLWLLLWM

LW

LLWLLWM

OOOOB

O

OOD

O

OOO

OOOMAX

O

OOC

6-32. The ski supports the 180lb weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski.

( ) ( )

( )

( ) ( )

( ) ( )

( )

( )( ) ( )

( )

( )

0

5.13

3015

15

5.12

3090105

105

4

340

2

3

2

3

2

15.140

2

1

15

3

1

2

3

2

340

2

1

90

018090

302

340

2

118040

2

330

3030

402

390

2

340

2

1

40

9

180

32

3180

35.12

15.1

2

1180

0

2

=

−=Σ

⋅=

+−=Σ

⋅=

+

+=Σ

⋅=

⋅

=Σ

−=

=−=

−

=−+=

−==

+−=

=

=

=

+=

++=

=Σ

D

C

E

B

E

DE

CB

Y

M

ftlb

M

ftlb

M

ftlb

M

lbV

VV

lblb

VV

ftlb

W

WW

WWW

F


( ) ( ) ( ) ( )

( )

( )

( )

( ) 05.43

5.11275.168

75.1685.43

5.112

0

0

5.1125.112

5.112

5.1120

0

5.1125.112

5.112

5.112

:5.112

75.84375.168

5.43

25.45.4

2

50

3

5.450

2

19

02

=−=

⋅==

=

=

+−=

−=

−=

=

−=

=

==

=

+=

++=

=Σ

CL

BL

A

C

CL

BL

A

BA

B

B

A

M

mkNM

M

kNV

kN

kNV

kNV

kNV

kNRR

symmetrybykNR

R

M

50 kN/m50 kN/m

4.5 m 4.5 m

A B

112.5 kN

-112.5 kN

2°

2°

168.75 kN.m

6-35. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin.

( ) ( )[ ]

( )( )

( )[ ]

( )

( )

( )

mkN

M

mkNM

M

kNV

kN

kNV

kN

kNV

V

mkNW

W

F

MAX

BL

A

DL

CL

BL

A

O

O

y

⋅=

+=

⋅==

=

=

+−=

−=

−=

=

=

=

=

=

=Σ

00026.0

03.02

012.000008.0

00008.002.03

012.0

0

0

02.02

5.1012.0

012.0

06.04.0012.0

012.0

02.05.12

1

0

2.1

06.04.02.02

12

0

2.6x10-3 kN⋅m

60 mm 20 mm20 mm

wowo

0.4 kN/m

0.012 kN

0.012 kN30 mm

2°

2°


( )( )

( )

( )

( )( )[ ]

( )

( )

0

63

90001800

18000

1500121625

1500

0

05.063009000

9000

106251625

1625

1625

1625

62

300010625

0

10625

12

127500

12262

30001500

0

=

+−=

−=

+−=

⋅=

=

−=

=

+−=

−=

−=

=

−=

=Σ

=

=

++=

=Σ

C

B

B

A

CL

B

B

A

A

A

A

y

B

B

A

M

M

M

ftlbM

V

lbV

lbV

lbV

lbV

lbR

R

F

lbR

R

M

3 kip/ft

1500 lb.ftA

B

12 ft 6 ft

900

-1625

2°

3°

1500

-18000

6-37. The compound beam consists of two segments that are pinned together at B. Draw the shear and moment diagrams if it support the distributed loading shown.

0

66

54

336

3/@

22

2

=

−=

=

⋅=

=

WLWLM

WL

LWL

M

Lx

CL

MAX

( )

26

3

6

2

x?0;V @

3

62

6

6

0

WxWLV

Lx

WLWx

WLR

WLWLR

WLR

LWLLR

M

C

C

A

A

C

−=

=

=

=

=

−=

=

=

=Σ


( )( ) ( )( )( )

mkNM

x

xM

xx

xx

xM

kNV

x

xV

xxV

mkN

M

kNR

R

F

B

B

B

B

B

y

⋅−=

=

−=

−−=

−−=

−=

=

−=

−−=

⋅=

+=Σ

=

+=

=Σ

54

3@

7

6

32

6

2

12

45

3@

15

2

612

54

1362

15.1312

45

32

1812

0

2

22

2

18 kN/m

12 kN/m

3 m

A B

3°

2°

-45 kN

-54 kN.m

6-39. Draw the shear and moment diagrams for the beam and determine the shear and moment as a function of x.

( )( ) ( )( )( )

( ) ( ) ( )

( )

( )

( ) ( )

( ) ( )

6005009

100

39

1003100200

3

3

3

2003

2

13

2

200200

8.3

15

3100

500

03

100500

36

200600200200

333

2003

2

13200200

700

120032

600

0

200

1200300900

132002

15.132006

0

3

32

22

2

2

2

−+−=

−−−−=

−

−−−−=

=

=

=

=−=

−−+−=

−−−−−=

=

−

=

=Σ

=

=+=

+=

=Σ

xxM

xxx

xxxxM

x

x

x

x

xx

xxxV

NR

R

F

NR

R

M

B

B

Y

A

A

B

0

700700

700

0;6@

3.690;0

;8.3@

600;200

;3@

0;200

0@

=

+−=

−=

==

⋅==

=

⋅==

=

==

=

B

BL

BL

DLDL

CLCL

AA

V

NV

Mmx

mNMV

mx

mNMNV

mx

MNV

x

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students

6-43. A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. a) about z axis

b) about y axis

6 in.

6 in. x

z

y

psi

kiplbinkip

inI

I

z

z

z

z

z

167

)/1000(/167.0

864

)12)(6(2

864

)12)(6)(12

1(

2

4

3

=

=

=

=

=

σ

σ

σ

psi

kiplbinkip

inI

I

y

y

y

y

y

333

)/1000)(/333.0(

216

)12)(3)(2(

216

12

)6)(12)(1(

2

4

3

=

=

=

=

=

σ

σ

σ

6-45. A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.

We used the smaller value

M

2 in. 2 in.

4 in. 4 in.

inkipM

M

bhM

bh

c

I

c

IM

I

Mc

t

.44

)24

)32)(4((22

)24

(22

24

2

2

2

=

=

=

=

=

=

σ

σ

inkipM

M

c

IM c

.30

)24

)32)(4((15

2

=

=

=σ

ftkipM

in

ftinkipM

.5.2

12.30

=

=

6-46. A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

M

2 in. 2 in.

4 in. 4 in.

( )

( )( )

ksi

inlb

inC

C

C

hC

inI

I

bhI

I

Mc

c

c

c

c

8.4

/4800

62.4

31.29600

31.2

3

34

3

322

3

2

62.4

36

)32)(4(

36

2

1

1

1

1

4

3

3

1

=

=

=

=

=

=

=

=

=

=

=

σ

σ

σ

σ

( )

( )( )

ksi

inlb

inC

C

hC

t

t

t

4.2

/2400

62.4

15.19600

15.1

3

32

3

2

2

2

2

=

=

=

=

=

=

σ

σ

σ

6-49. A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. a) about z-axis

b) about y-axis

3 in.

z

y

3 in.

3 in.

3 in.

0.25 in.

0.25 in.

0.25 in.

( )( ) ( )( )( ) ( )( )

( ) ( )

ftkipM

inkipM

M

inI

I

I

Mc

z

z

z

z

z

.8.20

.7.249

25.03/8125.3324

8125.33

12

6.025.0125.325.06.0

12

25.06.02

4

32

3

=

=

+=

=

+

+=

=σ

( )( ) ( )( )

( )

ftkipM

inkipM

M

inI

I

y

y

y

y

y

.0.6

.06.72

3

924

9

12

25.06.0

12

6.025.02

4

33

=

=

=

=

+

=

6-50. The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points.

B A

50 mm 50 mm

50 mm 50 mm

50 mm

50 mm

M = 40 kN⋅m

( )( ) ( )( )

( )( )( )

( )( )( )

MPa

MPa

I

zM

I

yM

mI

III

B

B

A

A

y

y

z

zA

zy

2.66

)10(51.1

1000025.0400

199

)10(5.1

1000075.0400

)10(51.1

12

05.005.02

12

15.005.0

5

5

45

33

=

+=

=

+=

+−

=

=

+

===

−

−

−

σ

σ

σ

σ

σ

6-73. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa.

0.8 1.2 0.6

600 400

A B

( ) ( ) ( )

( ) ( ) ( )

( )( )( )

mmd

mr

r

Mr

r

M

r

Mr

NR

R

M

NR

R

M

B

B

A

A

A

3.31

0156.0

)10(160

4804

4

4

4

200

8.06008.14002.1

0

800

6.040026002.1

0

36

3

3

4

=

=

=

=

=

=

=

−=

=Σ

=

−=

=Σ

π

πσ

πσ

πσ

6-81. The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress?

For symmetrical loading

4

222

2@

4

022

0@

20

22

22

0

2

2

0

max

PLM

LLPM

La

PLM

LPM

a

La

aLP

M

aLP

M

MM

M

PR

PR

RR

PRR

Fy

A

A

A

B

A

BA

c

AY

AY

BYAY

BYAY

=

−=

=

=

−=

=

≤≤

−

=

−

=

=

=Σ

=

=

=

=+

=Σ↑+

( )( )

( )

2max

2max

3max

3

maxmax

2

3

12

8

12

24

2

12

,0

bd

PL

bd

PL

bd

dPL

dc

bdI

I

cM

La

=

=

=

=

=

=

=

σ

σ

σ

σ

P

d

b L/2 L/2

a a

6-89. The steel beam has the cross-sectional area shown. If w = 5 kip⋅ft, determine the absolute maximum bending stress in the beam.

( ) ( )

( )( ) ( )( ) ( )( )

( )( )( )

ksi

inI

I

kipR

RR

Fy

kipR

R

M

A

BA

B

B

A

8.66

3.152

3.512160

3.152

12

103.015.53.08

12

30.082

40

80

0

40

24044024

0

max

max

4

32

3

=

=

=

+

+=

=

=+

=Σ↑+

=

+=

=Σ

σ

σ

8 ft 8 ft 8 ft

8 in.

0.30 in.

10 in.

0.30 in.

0.30 in.

ww

6-90. The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa.

( )( )

( )

kNP

P

mI

I

I

Mc

PR

PRR

Fy

PR

PPR

M

A

BA

B

B

A

4.10

)10(953.1

125.05.1)10(10

)10(953.1

12

25.05.0

2

0

35.15.1

0

4

6

44

3

max

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

6-91. The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.

( ) ( ) ( )

( )( )

( )

( )

MPa

mI

I

kNR

RR

Fy

kNR

R

M

A

BA

B

B

A

5.11

)10(953125.1

125.018000

)10(953125.1

125.018

)10(953125.1

12

25.015.0

12

24

0

12

3125.1125.1

0

max

4max

4max

44

3

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

−

σ

σ

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

6-93. A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in.

h

b

2 ft

8 ft 8 ft

P

6-102. The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 45°.

θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

6-103. The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 30°.

θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

6-105. The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined.

6-185. Determine the bending stress distribution in the beam at section a-a. sketch the distribution in three dimension acting over the cross section.

80 N 80 N

80 N 80 N

400 mm 400 mm300 mm 300 mm

a

a

100 mm

75 mm

15 mm 15 mm

15 mm

6-51. The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points.

B 10 mm

M = 75 N⋅m

20 mm

10 mm

10 mm

10 mm 10 mm

20 mm

40 mm

C

N

6-53. A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B.

20 mm

20 mm

200 mm

15 mm

A

200 mm

M = 40 kN⋅m

B

15 mm

6-47. The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.

M = 600 N⋅m

20 mm

25 mm

200 mm

20 mm

6-106. If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.

z

y

12 mm

M = 800 N⋅m

12 mm

200 mm

200 mm

150 mm

z

12 mm

60°

6-71. The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine that maximum bending stress developed at the center of the axle, where the diameter is 5.5 in.

10 in.

20 kip

A B DC

10 in.60 in.

20 kip

6-74. Determine the absolute maximum bending stress in the 1.5-in.-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

18 in.

1 in.

15 in.

400 lb

300 lb

A

B

6-75. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 22 ksi.

18 in.

1 in.

15 in.

400 lb

300 lb

A

B

6-72. Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

0.8 m 1.2 m 0.6 m

600 N 400 N

A B

6-79. The steel shaft has a circular cross section with a diameter of 2 in. It is supported on smooth journal bearings A and B, which exert only vertical reactions on the shaft. Determine the absolute maximum bending stress if it is subjected to the pulley loadings shown.

20 in. 20 in. 20 in. 20 in.

500 lb 300 lb 500 lb

A B

The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-half of the applied torque (T/2).

r r'

T

by ratio and proportion:

7651.32cos'

841.0'

2'

2'

'2

'

2

'

4

4

44

44

44

max

rr

rr

rr

rr

rr

r

T

r

T

rr

=

=

=

=

=

=

=

ππ

ττ

The solid shaft of radius r is subjected to a torque T. Determine the radius r’ of the inner core of the shaft that resists one-quarter of the applied torque (T/4).

45sin'

707.0'

4'

4'

'2

'

2

'

'2

.)

2

''

)2

(2

4

4

44

44

44

max

3

3max

33

rr

rr

rr

rr

rr

r

T

r

T

rr

r

T

b

r

T

r

T

r

T

=

=

=

=

=

=

=

=

=

==

ππ

ττ

πτ

πτ

ππτ

r r'

T

The copper pipe has an outer diameter of 2.50 in. and an inner diameter of 2.30 in. If it is tightly secured to the wall at C and the three torques are applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe’s outer surface. Sketch the shear stress on the volume elements located at A and B. sol’n:

C B

A

600 lb-ft

350 lb-ft

450 lb-ft

τmax

450 lb⋅ft = 5400 lb⋅in 350 lb⋅ft = 4200 lb⋅in 600 lb⋅ft = 7200 lb⋅in

5400 + 4200 – 7200 = 2400

( )( )( )

( )

ksi

inlb

dD

TD

B

B

B

76.2

6.2758

3.25.2

5.2240016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

7200 - 4200 = 3000

( )( )( )

( )

ksi

inlb

dD

TD

B

A

A

45.3

3.3448

3.25.2

5.2300016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear stress distribution over the cross section.

600 N

600 N

25mm

5mm

75mm

75mm

MPa

x

J

Tr

MPa

mx

Nm

J

Tr

mxJ

dDJ

NmT

T

M x

40.10

)10108(

)0125.0)(90(

45.14

)10108(

)075.0)(90(

10108

)025.0035.0(32

)(32

90

)075.0)(600(2

;0

9

49

2

max

49

44

44

=

=

=

=

=

=

=

−=

−=

=

=

=∑

−

−

−

τ

τ

π

π

A steel tube having an outer diameter of 2.5 in. is used to transmit 35 hp when turning at 2700 rev/min. Determine the inner diameter d of the tube to nearest 1/8 in. if the allowable shear stress is τallow = 10 ksi.

2.5 in.

d

Given:

?

/75.min/2700

/.250,193

/000,10

5.22

=

==

==Ρ

=

d

srevrevf

sftlbhp

inld

inD

allowτ

Solution:

inlbT

T

T

T

.97.4084

)75.0(2

19250

)]75.0(2[19250

=

=

=

=Ρ

π

π

ω

rd

r

r

r

r

r

=

=

=

−=

=−

−

=

2

206.1

649.6

10000

)2)(25.1)(4084097()25.1(

10000

)25.1)(97.4084(

2)(

2)25.1(

)25.1(2

)25.1(97.4084

4

44

2

44

44max

π

ππ

ππ

πτ

ind 4.2= ; 5.08

14 =

ind2

12=

The steel shaft has a diameter of 1 in. and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft if the couple forces have a magnitude of F = 30 lb. sol’n:

8in. 8in.

F

F

12in

( )( )( )

( )

ksi

inlb

dD

TD

AB

AB

AB

8.7

72.7818

68.075.0

75.021016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ ( )

( )( )( )

ksi

inlb

dD

TD

BC

BC

BC

36.2

2361

86.01

121016

16

2

44

44

=

=

−=

−=

τ

τ

π

πτ

AB D = 0.75 in d = 0.68 in BC D = 1 in d = 0.86 in

( ) ( )inlbT

TM A

⋅=

=+=∑210

815615

The 20-mm-diameter steel shafts are connected using a brass coupling. If the yield point for the steel is (τY)st = 100 MPa, determine the applied torque T necessary to cause the steel to yield. If d = 40 mm, determine the maximum shear stress in the brass. The coupling has an inner diameter of 20 mm.

d

T

T

20mm

To find the T; steel;

NmT

Tx

r

T

s

s

s

07.157

)010.0(

210100

2

3

6

3

=

=

=

π

πτ

To find the max.., τ; brass;

MPa

rR

TR

3.13

)01.002.0(

)02.0)(07.157(2

)(

2

44

44

=

−=

−=

τ

π

πτ

The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the number of bolts necessary to make the maximum shear in the shaft equal to the shear stress in the bolts. Each bolt has a diameter d. Soln:

R

T

T

r

2

3

23

3

2

2

2

2

42

2;2

1;4

4

4

d

rn

Rnd

T

r

T

r

T

Rnd

T

Rnd

T

PRnT

dP

A

P

=

=

><=

><=

⋅=

=

⋅=

=

ππ

πτ

πτ

τπ

τπ

τ

The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function

( )2



( )2

25 xxet =

To 80mm

x

2m

Solution: d=80mm=0.08 m

( )2

25 xxet =

3

16

d

T

πτ =

dxxeTx2

2

025∫=

dxxeT x

∫=2

0

2

25

let u= 2x du=2x dx du/2=x dx x=0, u=0 x=2,u=4

∫=4

0 2

25 dueT

u

dueu

∫=4

05.12

[ ]4

05.12 ue=

[ ]15.12 4 −= e

NmT 670=

MPa66.6=τ

33 )1080(

)670(16−

=xπ

3

16

d

T

πτ =

The solid shaft has a linear taper from Ar at one end to Br at the other. Derive and

equation that gives the maximum shear stress in the shaft at a location x along the shaft’s axis.

T T

x

A

L

rB

rA

B

Solution:

1rrr B +=

L

rr

xL

r BA −=

−1

L

xLrrr BA ))((1

−−=

L

xLrrrr BA

B

))(( −−+=

L

xLrrLrr BAB ))(( −−+

=

L

xLrrxLLrr BAB )()( −−−+

=

L

xLrxLLrr AB )()]([ −+−−

=

L

xLrxrr AB )()( −+

=

L

xLrxrr AB )( −+

=

3

16

d

T

πτ = ;

2

dr =

33 8

2

rd

rd

=

=

38

16

r

T

πτ =

3

2

r

T

π=

3)(

2

−+=

L

xLrxr

T

ABπ

τ

[ ]3

3

)(

2

xLrxr

TL

AB −+=

πτ

The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow = 8 ksi. If the outer diameter of the shaft is 2.5 in. and the engine delivers 200 hp to the shaft when it is turning at 1140 rev/min, determine the minimum required thickness of the shaft’s wall.

Given: psiksi 80008 ==τ

inD 5.2= slbfthp /000,100200 ==Ρ sec/19min/1140 revrevf ==

?=t Solutions:

f

Tπ2

Ρ=

)/19(2

/000,100

srev

slbft

π=

lbftT 66.837= or lbin89.051,10=

)(

1644 dD

TD

−=

πτ

)5.2(

)5.2)(89.051,10(16000,8

44 dpsi

−=

π

)000,8(

65.402075)5.2( 44

π=− d

4

4

23

998.1539

=

−=

d

d

ind 19.2= Thickness:

2

DR = ;

2

dr =

2

5.2=R

2

19.2=r

25.1=R 1.1=r rRt −=

int

t

15.0

1.125.1

=

−=

The drive shaft AB of an automobile is to be designed as a thin walled tube. The engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The material has an allowable shear stress τallow = 7 ksi.

Given

?

000,77

5.2

/25min/500,1

/000,75150

min =

==

=

==

==Ρ

t

psiksi

inD

srevrevf

slbfthp

τ

Solution

f

Tπ2

Ρ=

)/25(2

/000,75

srev

slbft

π=

lbinT

lbftT

58.729,5

46.477

=

=

ind

d

d

d

d

dD

Td

29.2

5968.27

42.1039

)000,7(

183,22939

)5.2(

)5.2)(58.5729(16000,7

)(

16

4

4

4

44

44

=

=

−=

=−

−=

−=

π

π

πτ

Thickness:

inR

R

DR

25.1

2

5.2

2

=

=

=

;

inr

r

dr

146.1

2

29.2

2

=

=

=

int

t

rRt

104.0

15.125.1

=

−=

−=

The drive shaft of a tractor is to be designed as a thin-walled tube. The engine delivers 200 hp when the shaft is turning at 1200 rev/min. Determine the minimum thickness of the wall of the shaft if the shaft’s outer diameter is 3 in. The material has an allowable shear stress of τallow = 7 ksi. Given:

ksi

inD

srevrevf

sftlbhp

7

3

/20min/1200

/.100000200

=

=

==

==Ρ

τ

psi7000= ?max =t

Solution:

lbftT

lbftT

srev

slbftT

fT

3.9549

8.795

)/20(2

/100000

2

=

=

=

Ρ=

π

π

ind

d

d

d

d

dD

TD

785.2

16.60

84.2081

)7000(

4.45836681

)3(

)3)(3.9549(167000

)(

16

4

4

4

44

44

=

=

−=

=−

−=

−=

π

π

πτ

Max Thickness:

ind

r

R

DR

3925.12

5.1

2

==

=

=

int

rRt

12.0=

−=

A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. If it is rotating at 200 rad/s. determine its largest inner diameter to the nearest 1/8 in. If the allowable shear stress for the material is τallow = 25 ksi.

A B

6 in.

( )( )( )( )

( )ind

d

d

dD

TD

inlbT

T

pT

75.1

11.616

2

2150001625000

16

15000

200

250000

4

44

44

=

−=

−=

−=

⋅=

=

=

π

πτ

ω

lbfthp

slbfthp

slbfthp

inlb

ksi

/250000/500

500

/5001

/25000

252

=

⋅

⋅=

=

=τ

The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 6 ksi. If the outer diameter is 3 in. and the engine delivers 175 hp to the shaft when it is turning at 1250 rev/min. determine the minimum required thickness of the shaft’s wall.

A B

6 in.

( )

( )( )( )( )

( )

int

rRt

dr

DR

thickness

ind

ind

d

d

dD

TD

inlbT

lbftT

f

PT

167.0

08.12

165.2

2

25.12

5.2

2

;

165.2

98.21

6000

87.32085639

5.2

5.24.8021166000

16

.4.8021

.45.668

833.202

87500

2

4

4

44

44

=

−=

===

===

=

=

=−

−=

−=

=

=

=

=

π

π

πτ

π

π

( )( )( )( )

2

44

44

/17.483

25.4

5.42.132216

16

inlb

dD

TD

=

−=

−=

τ

π

πτ

A motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A B

6 in.

( )( )

( )

( )

srad

T

P

inlbTT

dD

TD

/296

2.11137

103

.2.11137;84.12

21625000

16

6

44

44

=

=

=

=−

=

−=

ω

ω

π

πτ

The 0.75 in. diameter shaft for the electric motor develops 0.5 hp and runs at 1740 rev/min. Determine the torque produced and compute the maximum shear stress in the shaft. The shaft is supported by ball bearings at A and B.

A B

6 in.

( )

( )( )

psi

d

T

inlbT

lbftT

f

PT

219

75.0

12.1816

16

.12.18

.51.1

292

260

2

3

3

=

=

=

=

=

=

=

τ

π

πτ

π

π

The motor delivers 500 hp to the steel shaft AB, which is tubular and has an outer diameter of 2 in. and an inner diameter of 1.84 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is τallow = 25 ksi.

A B

( )( )

( )

( ) ( )( )

( )

srad

D

dDP

PT

D

dDT

dD

TD

/4.296

22.11137

103

32

84.122500103

16

16;

16

6

446

44

44

44

=

=

−=

−=

=

−=

−=

ω

ω

π

ω

τπ

ω

ω

τπ

πτ

The solid shaft of radius c is subjected to a torque T at its ends. Show that the maximum shear strain developed in the shaft is γmax = Tc/JG. What is the shear strain on an element located at point A, c/2 from the center of the shaft? Sketch the strain distortion of this element.

JG

Tc

G

G

J

Tc

crJ

Tr

=

=

=

=

==

max

max

max

max ;

γ

τγ

γτ

τ

JG

Tc

G

G

J

Tc

c

j

T

2

2

2;

=

=

=

=

==

γ

τγ

γτ

ρρ

τ

A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N⋅m. If the steel portion has a diameter of 30mm. determine the required diameter of the brass portion so the reactions at the walls will be the same. Gst = 75 GPa, Gbr = 39 GPa. sol’n: θAB = θBC

680 N⋅m 1.60 m

0.75 m

C

A

B

BCAB JG

TL

JG

TL=

( )( )( ) ( )( )

( )( )

( )( )94

94

103932

6.1680

107532

030.0

75.0680

dππ=

( )( )( )( )( )3975.0

75030.06.1 44 =d

( )4 46103.3 md −=

md 0427.0=

mmd 7.42=

The composite shaft consists of a mid section that includes the 1-in-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 100-lb.ft. The material is A-36 steel.

d

T

T

20mm

5-14. The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions BC and DE of the shaft. The bearings at A and B allow free rotation of the shaft.

( )( )( )

( )( )( )( )

ksi

ksi

ftlbT

ftlbT

T

DE

DE

BC

BC

DE

BC

AB

62.3

32

75.0

12375.040

07.8

32

75.0

12375.035

.25

.35

0

4

4

=

=

=

=

=

=

=

τ

πτ

τ

πτ

5-15. The solid shaft has a diameter of 0.75 in. If it is subjected to a torque shown, determine the maximum shear stress developed in regions CD and EF of the shaft. The bearings at A and F allow free rotation of the shaft.

( )( )( )

ksi

lbT

T

CD

CD

EF

CD

EF

17.2

32

75.0

12375.015

0

15

0

4

=

=

=

=

=

τ

πτ

τ

5-49. The splined ends and gears attached to the A-36 steel shaft are subjected to the torque shown. Determine the angle of twist of end B with respect to the end A. The shaft has a diameter of 40 mm.

( )

( )( )( ) ( )( )

( )( )( ) ( )( )

( )( )( ) ( )( )

( )°=

=

−+=

Σ=

=

=

=

=

=

=

0578.0

3.57010079813.0

107502.0

23.0300

107502.0

24.0200

107502.0

2.05.0400

02.0

04.0

.300

.200

.400

/1075

/

949494/

29

AB

AB

AC

CD

BD

rad

JG

TL

mr

md

mNT

mNT

mNT

mNG

θ

πππθ

θ

5-58. The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the shaft AB if the allowable shear stress is τ allow= 8 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel.

5-59.The engine of the helicopter is delivering 600 hp to he rotor shaft AB when the blade is rotating at 1200 rev/min. Determine the nearest 1/8 in. diameter of the

( )

( )( )

ind

ind

d

d

T

inlbT

ftlbT

f

PT

75.2

75.08

16.0;6.2

8000

2864816

16

.28648

.3.2387

202

300000

2

3

3

=

=

=

=

=

=

=

=

=

π

πτ

π

π

shaft AB if the allowable shear stress is τ allow= 10.5 ksi and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 2 ft. long and made from 1.2 steel.

( )

( )( )

ind

ind

d

d

T

inlbT

ftlbT

f

PT

5.2

5.08

14.0;4.2

10500

2864816

16

.28648

.2387

202

300000

2

3

3

=

=

=

=

=

=

=

=

=

π

πτ

π

π

5-6. the solid 1.25 in diameter shaft is used to transmit the torques applied to the gears. If it is supported by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points.

( )( )( )

( )( )( )

ksi

ksi

J

Tr

inlbT

inlbT

D

D

C

C

D

C

56.1

25.1

32625.0600

91.3

25.1

3216251500

.600

.1500

4

4

=

=

=

=

=

=

=

τ

πτ

τ

πτ

τ

5-7. The shaft has an outer diameter of 1.25 in and an inner diameter of 1 in. If it is subjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque.

( )( )( )( )

ksi

inlbTdD

TD

62.6

125.1

25.1150016

.1500;16

max

44

max44max

=

−=

=−

=

τ

π

πτ

5-50.The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm.

( )( )( ) ( )( )

( )( )( ) ( )( )

°=

+=

Σ=

=

=

=

=

=

243.0

107502.0

24.0200

107502.0

25.0400

84.0

.300

.200

.400

/)10(75

/

9494/

29

DC

DC

AC

CD

BD

JG

TL

md

mNT

mNT

mNT

mNG

θ

ππθ

θ

rB

x L

rA

ro

L

x

T

T

The simply supported shaft has a moment of inertia of 2I for given region BC and a moment of inertia I for regions AB and CD. Determine the maximum deflection

P

of the beam due to the load P.

256

3

2

128128

5

128

1536512192

128

5

1536

5

512

3

384512

5

1536

7

384

5

2

2

:0

3

33

3

/

333

/

3

/

333333

/

PL

L

PL

L

PL

PLT

PLPLPLT

PLT

PLPLPLPLPLPLT

PRR

LPLR

M

AB

AB

AD

AD

DA

A

D

=

+

=

=

++=

=

+++++=

==

=

=∑

δ

δ

A B C

D

L/4 L/4 L/4 L/4

I I 2I

TP/A

TD/A

δ

8

2PL

8

2PL

8

2PL

16

2PL

16

2PL

Determine the equation of the elastic curve for the beam using the x coordinate that is valid for 0 ≤ x ≤ L/2. Specify the slope at A and the beam’s maximum deflection. EI is constant.

P

2

L

2

L

x

A B

2

2

2

0

0

PR

PR

LPLR

M

PRR

Fy

A

B

B

A

BA

=

=

=

=Σ

=+

=Σ

( )

EI

PLy

Lxy

LxxEI

Py

xPL

xP

EIy

PLC

CLP

Lxy

C

xy

CxCxp

EIy

Cxp

EIy

xp

EIy

xp

M

M

48

2@

3448

1612

16

240

2,0'@

0

0,0@

12

4'

2''

2

0

max

max

22

23

2

1

1

2

2

213

12

−=

=

−=

−=

−=

+

=

==

=

==

++=

+=

=

=

=Σ

EI

PL

EI

Cy

x

A16

0

0@

2

1

−=

+==

=

θ

θ

Determine the equations of the elastic curve using the x1 and x2 coordinates.

Specify the slope at A and the maximum deflection. EI is constant.

P

x1

A B

P

x2

L

a a

The beam is made of two rods and is subjected to the concentrated load P. Determine the slope at C. The moments of inertia of the rods are IAB and IAC, and the modulus of elasticity is E. A

B

P

L

C

l

Determine the elastic curve for the cantilevered beam, which is subjected to the couple moment Mo. Also compute the maximum slope and maximum deflection of the beam. EI is constant.

A B

Mo

L x

OO

A

A

MM

M

R

Fy

=

=Σ

=

=Σ

0

0

0

EI

LMy

LMLM

EIy

Lxy

LxMxM

EIy

LMC

Lxy

C

xy

CxCxM

EIy

CxMEIy

MM

M

O

OO

OO

O

O

O

O

2

2

@

2

,0'@

0

0,0@

2

'

0

2

max

22

max

max

2

1

2

212

1

−=

−=

=

−=

−=

==

=

==

++=

+=

=

=Σ

EI

LM

EI

Cy

x

O−=

+==

=

max

1max

max

0'

0@

θ

θ

θ

Determine the deflection at the center of the beam and the slope at B. EI is constant.

A

B M0

L

x

EI

LMy

LMLM

L

LMEIy

Lx

EI

LMy

LMEIy

x

LMC

LCLM

L

LM

yLx

C

yx

CxCxM

L

xMEIy

CxML

xMEIy

MxRM

L

MR

M

L

MR

MLR

M

O

OOO

O

O

O

OO

OO

OO

OA

OB

A

OA

OA

B

48

3

6848

2@

3'

3'

0@

3

026

0

0,@

0

0,0@

26

2'

;0

;0

2

max

223

max

1

1

23

2

21

23

1

2

−=

−+−=

=

−=

−=

=

−=

+++−

=

==

=

==

+++−=

++−

=

+−=

=

=Σ

−=

−=

=Σ

[ ]xLLxxLEI

My

LxMxM

L

xMEIy

O

OOO

23

23

236

326

−+−=

−+−

=

Determine the elastic curve for the simply supported beam, which is subjected to the couple moments Mo. Also, compute the maximum slope and the maximum deflection of the beam. EI is constant.

M0

L

x

M0

Determine the equations of the elastic curve using the coordinates x1 and x2, and specify the slope at A. EI is constant.

M0

A B

L

C

x1

L

x2

L

MR

L

MR

MLR

M

RR

RR

Fy

OA

OB

OB

A

BA

BA

−=

=

=

=Σ

−=

=+

=Σ

;0

0

;0

belowindicatessignnegative

EI

LM

CEI

x

LMC

Lxy

C

xy

CxCxL

MEIy

CxL

MEIy

xL

MM

M

O

A

O

O

O

O

;6

0@

6

,0@

0

0,0@

6

2'

;0

1

1

2

213

12

−=

−=

=

−=

==

=

==

++=

+=

=

=Σ

θ

θ

θ

The floor beam of the airplane is subjected to the loading shown. Assuming that the fuselage exerts only vertical reactions on the ends of the beam, determine the maximum deflection of the beam. EI is constant.

80 lb/ft

8 ft 2 ft 2 ft

( )

( ) ( )( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

3max

43

max

443

31

1

443

2

21

443

1

332

22

.3

56320

63

147204

3

06

3

160

6@

3

147202

3

1010

3

0

3

160

.3

14720

12103

102

3

012

3

1600

0,12@

0

0,0@

23

1010

3

0

3

160

23

4010

3

40160'

2401040320''

2

2280

2

101080

320

320

688012

;0

640880

0

ftlbEI

y

EIy

xy

xxxxEIy

ftlbC

C

yx

C

yx

CxCxxxEIy

CxxxEIy

xxxEIy

xx

xxxRM

lbR

lbR

R

M

RR

Fy

A

B

A

A

B

BA

−=

−1

+=

=

−−−−1

+=

−=

+−1

+=

==

=

==

++−−−1

+=

+−−−+=

−−−+=

−−−

−−+=

=

=

=

=Σ

==+

=Σ

3max

3max

.77.18

.77.18

ftkipEI

ftkipEI

y

=∆

−=

The beam is subjected to the load shown. Determine the equation of the elastic curve. EI is constant.

A B

8 ft 4 ft 4 ft

3 kip/ft

5 kip-ft 5 kip-ft

( ) ( )( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 343432

2

2

43432

1

21

43432

1

3232

.13624128

11224

8

142

2

51

136

0,4@

24128

11224

8

142

2

5

24

0',8@

128

11224

8

142

2

5

126

312

2

124

6

34

2

125'

2

1212312

2

44345

12

24

0

12

04838

0

ftkipxxxxxx

EIy

C

yx

Cxxxxxx

EIy

C

yx

CxCxxxxx

EIy

CxxxxxEIy

xxxR

xxxRM

kipsR

RR

Fy

kipsR

R

M

BA

B

BA

A

A

B

+−−+−+−−−+−=

=

==

+−−+−+−−−+−=

−=

==

++−+−+−−−+−=

+−+−+−−−+−=

−−+−+

−−−−+−=

=

=+

=Σ

=

=−

=Σ

The shaft is supported at A by a journal bearing that exerts only vertical reactions on the shaft, and at C by a thrust bearing that exerts horizontal and vertical reactions on the shaft. Determine the equation of the elastic curve. EI is constant.

A B

P

a

C

x

b

( ) ( )( )

( )( )( )

( )( )

( )( )

( )( )

( )( )

+

−++−=

+−+

+−=

=

+−

=

==

=

==

++−+

+−=

+−+

+−=

−++−=

−+−=

=

−=

=Σ

+=

+=

=Σ

xPba

a

axbaP

a

Pbx

EIy

xPba

a

axbaP

a

PbxEIy

PbaC

aCa

Pba

yax

C

yx

CxCa

axbaP

a

PbxEIy

CaxbaP

a

PbxEIy

a

axbaP

a

PbxEIy

axRxRM

a

PbR

PRR

Fy

a

baPR

baPaR

M

BA

A

BA

B

B

666

1

666

6

60

0,@

0

0,0@

66

22'

''

0

0

33

33

1

1

3

2

21

33

1

22

The shafts support the two pulley loads shown. Determine the equation of the

elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI

is constant.

A B

20 in. 20 in. 20 in.

x

40 lb 60 lb

( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )[ ] 3333

333

1

133

2

21

333

1

222

.40064033.182067.667.11

40064033.182067.667.1

4006

402067.64067.10

0,40@

0

0,0@

4033.182067.667.1

2

40110

2

2040

2

10'

40110204010''

402040

10

4060

0

110

6060204040

0

inlbxxxxEI

y

xxxxEIy

C

C

yx

C

yx

CxCxxxEIy

Cxxx

EIy

xxxEIy

xRxxRM

lbR

RR

Fy

lbR

R

M

BA

A

AB

B

B

A

+−+−−−=

+−+−−−=

=

+−−=

==

=

==

++−+−−−=

+−

+−

−−

=

−+−−−=

−+−−−=

↓=

++=

=Σ

↑=

+=

=Σ

The shafts support the two pulley loads shown. Determine the equation of the elastic curve. The bearing at A and B exerts only vertical reactions on the shaft. EI is constant.

A B

12 in. 24 in. 24 in.

x

50 lb 80 lb

( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )[ ] 3333

333

2

1

2

21

21

333

1

222

.576016803633.13121.1733.81

576016803633.13121.1733.8

5760

1680

2&1

60950400

0,60@

12144000

0,12@

3633.13121.1733.8

2

3680

2

125.102

2

50'

3680125.10250

5.102

8050

0

5.27

2480481250

0

inlbxxxxEI

y

xxxxEIy

C

C

eqnequating

C

yx

CC

yx

CxCxxxEIy

Cxxx

EIy

xxxM

lbR

RR

Fy

lbR

R

M

A

BA

B

B

A

−+−−−+−=

−+−−−+−=

−=

=

+−=

==

++−=

==

++−−−+−=

+−

−−

+−=

−−−+−=

=

+=+

=Σ

=

=+

=Σ


A B

xl

9 ft. 15 ft.

6 kip/ft

( ) ( )( ) ( ) ( )( )

( )( ) ( )

( ) ( ) ( )( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )2.24924180

1924

6

4.7724

180

10

0:24@

1.99180

1_0

0:9@

9180

19

6

4.77

180

936

19

2

4.77

36'

99

194.77

9

93

196

3

2

2

194.77

3

1

3

2

2

1''

93

196

2

19

3

1

2

1

3

2

9

6

6.39

156692

1

0

4.77

05.71561593

196

2

115

0

21

535

215

21

535

1

424

33

eqnCC

yx

eqnCC

yx

CxCxxx

EIy

Cxxx

EIy

xxx

xxxxxxxEIy

xxyxRxxyM

xy

x

y

kipsR

RR

Fy

kipsR

R

M

A

B

BA

A

A

B

++−+−+−

=

==

++=

==

++−+−+−=

+−+−+−=

−+−+−=

−

−

−+−+

−=

−

−−+−+

−=

=

=

=

−−+

=Σ

=

=−

+

−

=Σ

( ) ( )

+−

−+−+−=

=

−=

55.26365.256

9180

19

6

4.77

1801

55.2636

5.256

2&1.

535

2

1

x

xxx

EIy

C

C

eqnequating

Determine the slope and deflection at C. EI is constant.

A

B 15 ft.

C

30 ft.

15 kip

Determine the slope and deflection at B. EI is constant.

A

B

P

L

( )

EI

PLB

PLL

BEI

EI

PL

LPLEI

B

B

3

23

2

2

2

1

3

2

2

=∆

=∆

=

=

θ

θ

Determine the maximum slope and the maximum deflection of the beam. EI is constant.

A B

L

M0 M0

0

0

;0

;0

=

=

=Σ

−=

=Σ

B

B

A

BA

R

LR

M

RR

Fy

EI

LMy

LMLMEIy

Lxy

LxMxMEIy

LMC

Lxy

C

xy

CxCxM

EIy

CxMEIy

MM

M

O

OO

OO

O

O

O

O

8

48

2@

22

2

2,0'@

0

0,0@

2

'

;0

2

max

22

max

max

2

1

2

212

1

−=

−=

=

−=

−=

==

=

==

++=

+=

=

=Σ

EI

LM

CEI

x

O

2

0

0@

1

max

−=

+=

=

θ

θ

θ


A

B

4 kip/fit

8 ft.

4 kip 2 kip

8 ft. 8 ft.

x


A B

20 kN

1.5 m 3 m

20 kN

1.5 m

x

( ) ( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) 3333

22

2

333

2333

21

1

333

2

21333

1222

.905.673

105.4

3

105.1

3

101

.90

5.45.675.43

100

3

103

3

100

0,5.4@

5.673

105.4

3

105.1

3

10

.5.67

663

105.1

3

105.4

3

100

0,6@

0

0,0@

3

105.4

3

105.1

3

10

105.4105.110'

205.4205.120

205.45.1

20

20

5.1205.4203

;0

40

;0

mkNxxxxEI

y

mkNC

C

yx

CxxxxEIy

mkNC

C

yx

C

yx

CxCxxxEIy

CxxxEIy

xxxM

xxRxRM

kNR

kNR

R

M

kNRR

Fy

BA

B

A

A

B

BA

−+−−+−=

−=

++−+=

==

++−−+−=

=

+−+=

==

=

==

++−−+−=

+−−+−=

−−+−=

−−+−=

=

=

−=

=Σ

=+

=Σ

Determine the equation of the elastic curve. Specify the slope at A. EI is constant.

A

B

L

C

x

w

L

The beam is subjected to the linearly varying distributed load. Determine the maximum slope of the beam. EI is constant.

A B

x

L

w

( )

( ) ( )

( ) ( )

( )

( )

( )

( )360

7

24

1

12

1'

360

7

12036

10

0,@

0

0,0@

20

1

36

1

24

1

12

1'

6

1

6

1''

32

1

3

6

6

1

0

2

0

342

3

1

1

44

2

21

43

1

42

3

2

Lw

L

xwLxwEIy

LwC

LCL

LwLw

yLx

C

yx

CxCL

xwLxwEIy

CL

xwLxwEIy

xwLxwEIy

xxyxRM

L

xwy

L

w

x

y

proportionandratioby

LwR

LwR

LwLR

M

LwRR

Fy

OOO

O

OO

OO

OO

OO

A

O

O

oB

oA

OA

B

OBA

−

−=

−=

+−=

==

=

==

++

−=

+

−=

−=

−=

=

=

=

=

=

=Σ

=+

=Σ

( ) ( )

EI

Lw

LwEIy

LwLwLwEIy

Lx

O

O

OOO

45

45'

360

7

24

1

12

1'

@

3

max

3

333

=

=

−−=

=

θ

The beam is subjected to the load shown. Determine the equation of the slope and elastic curve. EI is constant. A

B

3 kN/m

15 kN⋅m

5 m 3 m

x

C

( ) ( )( )

( )( )( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( ) ( )

( ) ( )

−

−+−+−=

−

−+−+−

=

−=

+−=

==

=

==

++−

+−+−=

+−

+−+−=

−−+−+

−=

=

+=

=Σ

=

=−+

=Σ

xx

xx

xEI

y

xx

xx

EIy

C

C

yx

C

yx

CxCx

xx

xEIy

Cx

xx

xEIy

xxxR

xxxRM

kNR

R

M

kNR

R

M

BA

B

B

A

A

A

B

125.38

5575.1

875.0

1

125.32

5525.5

225.21

'

125.3

58

5575.00

0,5@

0

0,0@

8

5575.1

875.0

2

5525.5

225.2'

2

5535

23

5.10

155.253)5(

0

5.4

0155.2535

;0

43

43

32

32

1

1

43

2

21

43

43

1

32

32

The wooden beam is subjected to the load shown. Determine the equation of the elastic curve. Specify the deflection at the end C. EW=1.6(103) ksi.

A

B

C

9 ft

0.8 kip/ft 1.5 kip

9 ft

x 6 in.

12 in.

( ) ( ) ( )( ) ( )

( )( )

( ) ( ) ( ) ( )( )

( ) ( ) ( ) ( )

( )( ) ( ) ( )

( )

( ) ( ) ( )

( ) ( )

( ).765.0

106.1,18@

9.8;091080

98.0905.0

09@

0;0,0@

1080

98.0

12

94.0

6

94.5

1080

8.0

6

3.0

454

98.0

3

94.0

2

94.5

454

8.0

2

3.0'

98.09

8.0

6

194.094.5

54

8.03.0''

9

8.0

9

8.0

98.06

194.094.5

6

13.0

3.0

04.598.02

15.1

0

4.5

93

298.0

2

1185.19

0

3

11

53

2

21

54353

1

4324

222

222

iny

EIx

CC

yx

Cyx

CxCxxxxx

EIy

Cxxxxx

EIy

xx

xxxxEIy

xy

x

y

proportionandratioby

xyxxyxxM

kipsR

R

Fy

kipsR

R

M

A

A

B

B

A

−=

==

==+−−

==

===

++−

+−

+−

+−−

=

+−

+−

+−

+−−

=

−

−+−+−+

−−=

=

=

−−+−+−+

−−=

−=

=+−−

=Σ

=

+=

=Σ

2

Determine the deflection and slope at C. EI is constant.

A

B

L

C

x

L

M0

Determine the deflection at C and the slope of the beam at A, B, and C. EI is constant.

A

B

6 m

C

3 m

8 kN⋅m

( )

( )

( )

( )

( ) ( )

( )

( ) ( )

( ) ( )

EIor

EIEIy

xCslope

EIor

EIEIy

xBslope

EIEIy

xAslope

xxEiy

mkNC

C

yx

C

xy

CxCxxEIy

Cxx

CxxEIy

xxEIy

xRxRM

kNRkNR

R

M

RR

Fy

CC

BB

A

BA

BA

A

B

BA

4040,40'

9;@

1616,16'

6;@

8,8'

0;@

863

2

3

2'

.8

609

26

9

20

0,6@

0

0,0@

69

2

9

2

63

2

3

2

66

4

6

4'

63

4

3

4''

6

3

4;

3

4

86

0

0

0

22

21

13

2

21

33

1

22

1

22

=−=−=

=

=−

=−=

=

==

=

+−+−=

=

++−=

==

=

==

++−+−=

+−+−=

+−+−=

−

+−=

−+=

=−=

−=

=Σ

=+

=Σ

θθ

θθ

θ

( )

3

33

.84

84

84

869

2

9

2

9@;@

mkNEI

C

EIy

EIy

xxxEIy

xCdeflection

C

=∆

−=

−=

+−+−=

=

Determine the slope at B and deflection at C. EI is constant.

A B

P

C

2

P

2

P

a a a a

The beam is subjected to the load shown. Determine the slope at B and deflection at C. EI is constant.

A B

C

a 2

a

a 2

a

w w

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

EI

waC

EI

way

waEIy

awa

aw

aw

awa

EIy

axCdeflection

EI

wa

aw

aw

aw

awa

EI

axBslope

xwaaxwwxaxwwax

EIy

waaxwwxaxwwax

EIy

waC

aCaw

aw

awwa

axy

C

xy

CxCaxwwxaxwwax

EIy

Caxwwxaxwwax

EIy

axwwx

axw

waxEIy

axaxw

wxaxaxwxRM

C

B

A

48

25;

48

25

48

25

2

3

24

7

2

3

242

1

242

3

6

2

3;@

12

7

12

7

63

62

63

2

3;@

12

72

2424246

12

72

6662'

12

7

324

324

2246

270

3,0@

0

0,0@

22424246

26662

'

2222

''

2

22

22

44

4

3443

3

3332

34443

33332

31

1

4444

2

21

4443

1

3332

22

2

2

=∆−=

−=

−

−

+

=

=

=

−−−+=

=

−−−−−

+=

−−−−−

+=

−=

+−−+=

==

=

==

++−−−−

+=

+−−−−

+=

−−−−+=

−−−−

−−+=

θ

θ

( )

( )

waR

waR

wawaaR

wa

aa

waaR

M

waRR

Fy

B

A

A

A

B

BA

=

=

+=

+

+=

=Σ

=+

=Σ

22

53

2

22

3

;0

2

0

22

2

Determine the elastic curve for the simply supported beam the x coordinate 0 ≤x ≤ L/2. Also, determine the slope at A, and the maximum deflection of the beam. EI is constant.

wo

BA

L

x

( )

192

5

282120

0';2

@

0

0;0@

2460

812'

43

043

2

2

1

0

2

2

:PROPORTION AND RATIO BY

4

4

612

23

1

2423

2

4

0

2

22

1

22

1

0

3

1

1

24

2

21

35

1

24

3

2

22

LwC

CLLwL

L

w

yL

x

C

yx

CxCLxw

L

xwEIy

CLxw

L

xwEIy

Lxw

L

xwM

xLwx

L

LxwM

M

L

xwy

L

w

x

y

LwR

LwR

LwLwLR

LRLLLwLLw

M

LwRR

LwLwRR

Fy

o

oo

oo

oo

oo

oo

o

o

oA

oB

ooB

Boo

A

oBA

ooBA

−=

+

+

−=

==

=

==

+++−=

++−=

+−=

=−

+

=Σ

=

=

=

=

+=

=

++

=Σ

=+

+

=+

=Σ

EI

Lwy

LLwLLwL

L

LwEIy

Lx

oMAX

oooMAX

24

2192

5

224260

:2

@y

4

335

max

−=

−

+

−=

=

EI

Lw

CEI

x

oMAX

MAX

MAX

192

00

0@

3

1

=

++=

=

θ

θ

θ

Draw the shear and moment diagrams for the shaft. The bearings at A and B exerts only vertical reactions on the shaft.

( )

( )

( ) ( )[ ]

( )

[ ]0

5.75.7

5.7

5.7

5.3124

24

0

.8.05.3125.024

.6

25.024

5.7

5.3124

0

5.31

05.1248.0

0

=

−=

=

=

+−=

−=

=

+−=

−=

−=

=

=−=

=Σ

=

=

=Σ

B

B

B

A

A

o

B

B

A

LA

B

B

Y

A

A

B

V

kNV

kNV

kNV

kNV

kNV

M

mkNM

mkNM

M

kNR

R

F

kNR

R

M

L

L

L

L

250 mm 800 mm

A B

24 kN

RA

RB

-24 kN

7.5 kN

-6 kN⋅m

The load binder is used to support a load. If the force applied to the handle is 50lb, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC.

( )

( )

( )0

200200

200

200

25050

50

50

200

50250

0

250

31550

0

2

2

1

1

=

−=

=

=

+−=

−=

−=

=

−=

=Σ

=

=

=Σ

C

C

C

B

B

B

A

Y

c

V

lbV

lbV

lbV

lblbV

lbV

lbV

lbT

lblbT

F

lbT

T

M

L

L

50 lb

T2

T1

C

B

A

12 in. 3 in.

200 lb

-50 lb

-600 lb⋅in

Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B and C and E.

Draw the shear and moment diagrams for the beam.

( ) ( ) ( )

( )( )

( )( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )0

.}1276.142

2711047806124.82{

.24.420

1511035804924.82

.16.1196

20803424.82

.36.1151

.1424.82

0

35

76.107

24.2

24.82

24.82

76.1423511080

0

76.142

613534110148049

0

=

+

−−=Σ

−=

−−=Σ

=

−=Σ

=

=Σ

=

=

−=

=

=

=

−++=

=Σ

=

++=

=Σ

L

L

L

L

L

L

L

L

E

E

D

D

C

C

B

B

E

D

C

B

A

A

A

Y

D

D

A

M

inlb

M

inlbM

M

inlbM

M

inlbM

inlbM

V

lbV

lbV

lbV

lbV

lbR

R

F

lbR

R

M

A

B C D

E

14 in. 20 in. 15 in. 12 in.

80 lb 110 lb 35 lb

82.4 lb

2.24 lb

107.76 lb

35 lb

1151.36 lb.ft

1196 lb.ft

-420.24 lb.ft

( )( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

0

4282122162204

.16

4282122164

.24

4282124

.24

4284

.16

4

2

42

:symmetry By

=

−−−−=Σ

=

−−−=

=

−−=Σ

=

−=Σ

=Σ

==

==

L

L

L

L

L

L

L

L

L

F

F

E

E

D

D

C

C

B

FA

FA

M

M

ftkipM

M

ftkipM

M

ftkipM

M

ftkipM

kipRR

RR

4 ft 4 ft 4 ft 4 ft 4 ft

2 kip 2 kip 2 kip 2 kip

4 kip

2 kip

-2 kip

-4 kip

16 kip.ft

24 kip.ft

16 kip.ft

Draw the shear and moment diagrams for the rod. It is supported by a pin at A and a smooth plate at B. The plate slides w/in the groove and so it cannot support a vertical force, although it can support a moment.

( ) ( )

mkNM

mkNM

M

kNR

F

LC

B

B

A

Y

.60

.30

215415

5.1

0

=Σ

=

−=Σ

=

=Σ

A

B

4 m 2 m

15 kN

15 kN

30 kN.m

60 kN.m

Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x w/in the region 125mm<x<725mm.

125 mm

A B

800 N

815.625 N

1500 N

600 mm 75 mm

x

15.625 N

-1484.375 N

101953 N.m

111328 N.m


( ) ( )

( )

( ) ( )

( ) ( ) ( )075.02500675.08008.0625.815

.328.111

6.0800725.0625.815

.953.101

125.0625.815

37.1484

625.8151500800

0

625.815

751500675800800

0

−−=Σ

=

−=Σ

=

=Σ

=

−+=

=Σ

=

+=

=Σ

L

L

L

L

B

D

D

C

C

B

B

Y

A

A

B

M

mNM

M

mNM

M

NR

R

F

NR

R

M

( )( ) mNxM

xxM

NV

.100625.15

125.0800625.815

625.16

725mm x 125mm @

+=

−−=

=

<<

( ) ( )

( )

( )

( )

0

.15

3839

.39

21875

0

8

1018

18

.75

1558210

18

0

=Σ

−=

+−=Σ

−=

+−=Σ

=

=

−=

=

−=

−−−=Σ

=

=Σ

B

B

B

C

C

B

C

C

A

A

A

A

Y

M

mkNM

M

mkNM

M

V

kNV

kNV

kNV

mkNM

M

kNR

F

L

L

L

L

L

L

10 kN 8 kN

2 m 3 m

15 kN.m

18 kN

8 kN

-39 kN.m-15 kN.m

-75 kN.m

Draw the shear and moment for the pipe. The end screw is subjected to a horizontal force of 5kN. Hint: the reaction at the pin C must be replaced by equivalent loadings at point B on the axis of the pipe.

( )

( )( )

( )

0

4.04.0

4.0

4.01

0

0

1

1

1

8.054.0

0

1

8.054.0

0

=

⋅+⋅−=

⋅−=

−=

=

=

−=

−=

=

=

=Σ

=

=

=Σ

=

mkNmkNM

mkN

mkNM

M

kNV

kNV

kNR

R

M

kNR

R

M

B

BL

A

BL

A

A

A

B

B

B

A

5 kN

400 mm

80 mmA

B

C

-1 kN

-0.4 kN.m

The engine crane is used to support the engine, which has a weight of 1200 lb. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown.

( )

( )

( )0

60006000

6000

2

32000

0

0

12001200

1200

1200

32002000

2000

2000

3200

12002000

0

520005

0

=

⋅−−=Σ

⋅−=

−=Σ

=Σ

=

−=

=

=

+−=

−=

−=

=

+=

=Σ

=

=Σ

ftlbM

ftlb

M

M

lblbV

lbV

lb

lblbV

lbV

lbV

lbR

lblbR

F

ftlbR

M

CL

BL

A

CL

CL

B

AL

A

B

B

Y

A

B

A

B C

4 ft

3 ft 5 ft

1200 lb

-2000

1200

-6000

Draw the shear and moment diagrams for the compound beam w/c is pin connected at B. It is supported by a pin at A and a fix wall at C.

( ) ( )

0

44

4

4

84

4

4

106

6

6

4

1014

0

10

32108

4818614

0

=

+−=

−=

−=

−=

=

=

+−=

−=

−=

=

−=

=Σ

=

+=

+=

=Σ

kipkipV

kipV

kip

kipkipV

kipV

kip

kipkipV

kipV

kipV

kipR

R

F

kipR

R

M

C

CL

D

DL

A

AL

O

C

C

y

A

A

C

( )( )

0

1616

16

2440

24

46

0

=

⋅−⋅=

⋅=

⋅−⋅=

⋅−=

−=

=

ftkipftkipM

ftkip

ftkipftkipM

ftkip

ftkipM

M

CL

DL

AL

o

A

8 kip6 kip

BC

4 ft 6 ft 4 ft 4 ft

4

-6

-4

16

-24

Draw the shear and moment diagrams for the beam. Also, determine the shear and moment in the beam as a function of x, where 3ft<x<15ft.

( )( )( )

( )

( )

ftkip

AreaM

ftx

x

SimilarBy

kipR

R

F

kipR

R

M

C

B

B

A

A

B

⋅=

−=

−=

−∆=

=

=

∆

=

−=

=Σ↑

=

+=

=Σ

79.7

50778.67

50778.82

167.13

50

778.8

167.13

12

18

:

833.4

167.13125.1

0

167.13

6125.15012

0

( )

( ) ( )

( )

25.46667.1775.0

75.65.475.05.39167.13

9675.05.39167.13

32

5.13167.13

5.17667.17

35.1167.13

:153@

2

2

2

2

−+−=

−+−−=

+−−−=

−−−=

−=

−−=

≤<

xxM

xxx

xxx

xxM

xV

xV

ftxft

50 kip.ft

1.5 kip/ft

A B

3 ft 12 ft

x

x

C

13.167

-4.833

7.79 kip.ft

-50 kip.ft


( ) ( )

( )( ) ( )( )

( )( )

( )( ) ( )( )0

4880012880051200

25600

4880051200

51200

48800128800

0

88008800

0

=

−−=Σ

⋅=

−=Σ

⋅=

−=Σ

=

−=

=Σ

C

B

A

A

y

M

ftlb

M

ftlb

M

R

F

( )

0

64006400

6400

8800

0

=

+−=

−=

−=

=

CL

BL

A

V

lb

V

V

800 lb/ft

A B

8 ft

800 lb/ft

8 ft

-6400

25600

51200

The 50lb man sits in the center of the boat, w/c has a uniform width and weight per linear foot of 3lb/ft. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat.

( )

0

2

175.075

0

13

15

15045

0

0

=

⋅=

∆=

=

=

+=

=Σ

C

B

A

o

y

M

ftlb

AreaM

M

ftlbw

w

F

( ) ( )

( )

o

V

lb

lbV

lb

V

V

CL

B

BL

A

=

+−=

−=

−=

=

−=

=

7575

75

15075

75

5.735.713

0

150 lb

7.5 ft 7.5 ft

75 lb

-75 lb

281.25 lb.ft

2°2°

The footing supports the load transmitted by the two columns. Draw the shear and moment diagrams for the footing if the reaction of soil pressure on the footing is assumed to be uniform.

( )

( )

0

7

7

147

7

126

77

7

147

7

66

7

67

2428

141424

0

=

+−=

−=

−=

=

+−=

−=

−=

=

=

=

=

+=

=Σ

kipkipV

kip

kipkipV

kipV

ftft

kipkipV

kip

kipkipV

kip

ftft

kipV

ftkipw

ftkip

kipkipw

F

DL

C

CL

CL

B

BL

o

o

y

14 kip 14 kip

12 ft 6 ft6 ft

2°

21 kip.ft

-7 kip

7 kip

2°

2°

21 kip.ft

-7 kip

7 kip

A B CD

E

6 ft


( )( )

( )

( )

( )( )

( )

( )[ ]

( )

( )0

5.05.0

5.0

5.952

0

0

5.055.2

5.2

305.27

25

5102

1

0

5.0

5.910

1052

0

5.9

305.125210

0

=

+−=

=

−=

=Σ

=

⋅−=

⋅=

⋅+−=

⋅−=

−=

=

−=

+−=

−=−=

=

=

−=

=Σ

B

B

B

B

y

BL

BL

CL

AL

O

A

AL

O

A

A

B

V

kipV

kipR

R

F

M

ftkipM

ftkip

ftkipM

ftkip

M

M

kip

kipV

kipV

V

kipR

R

M

30 kip.ft2 kip/ft

A B

5 ft 5 ft 5 ft

-10 kip

-0.5 kip

2°

1°-25 kip.ft

-27.5 kip.ft

2.5 kip.ft

Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as a function of x, where 4ft<x<10ft. by symmetry:

( )

( )

( ) ( )

2

2

2

7510503200

2

41504450200

1501050

4150450

450

2

900

6150

0

xx

xxM

x

xV

RlbR

RR

RR

F

BA

BA

BA

y

−+−=

−−−+−=

−=

−−=

==

==

=+

=Σ

150 lb/ft

A B

4 ft 6 ft 4 ft

x

200 lb.ft200 lb.ft

200 lb.ft

-200 lb.ft -200 lb.ft

-450 lb

450 lb

2°2°


( ) ( ) ( ) ( )

( )

( )

( )

( ) 05.43

5.11275.168

75.1685.43

5.112

0

0

5.1125.112

5.112

5.1120

0

5.1125.112

5.112

5.112

:5.112

75.84375.168

5.43

25.45.4

2

50

3

5.450

2

19

02

=−=

⋅==

=

=

+−=

−=

−=

=

−=

=

==

=

+=

++=

=Σ

CL

BL

A

C

CL

BL

A

BA

B

B

A

M

mkNM

M

kNV

kN

kNV

kNV

kNV

kNRR

symmetrybykNR

R

M

50 kN/m50 kN/m

4.5 m 4.5 m

A B

112.5 kN

-112.5 kN

2°

2°

168.75 kN.m

The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin.

( ) ( )[ ]

( )( )

( )[ ]

( )

( )

( )

mkN

M

mkNM

M

kNV

kN

kNV

kN

kNV

V

mkNW

W

F

MAX

BL

A

DL

CL

BL

A

O

O

y

⋅=

+=

⋅==

=

=

+−=

−=

−=

=

=

=

=

=

=Σ

00026.0

03.02

012.000008.0

00008.002.03

012.0

0

0

02.02

5.1012.0

012.0

06.04.0012.0

012.0

02.05.12

1

0

2.1

06.04.02.02

12

0

2.6x10-3 kN⋅m

60 mm 20 mm20 mm

wowo

0.4 kN/m

0.012 kN

0.012 kN30 mm

2°

2°


( )( )

( )

( )

( )( )[ ]

( )

( )

0

63

90001800

18000

1500121625

1500

0

05.063009000

9000

106251625

1625

1625

1625

62

300010625

0

10625

12

127500

12262

30001500

0

=

+−=

−=

+−=

⋅=

=

−=

=

+−=

−=

−=

=

−=

=Σ

=

=

++=

=Σ

C

B

B

A

CL

B

B

A

A

A

A

y

B

B

A

M

M

M

ftlbM

V

lbV

lbV

lbV

lbV

lbR

R

F

lbR

R

M

3 kip/ft

1500 lb.ftA

B

12 ft 6 ft

900

-1625

2°

3°

1500

-18000

Draw the shear and moment diagrams for the beam and determine the shear and moment as a function of x.

( )( ) ( )( )( )

( ) ( ) ( )

( )

( )

( ) ( )

( ) ( )

6005009

100

39

1003100200

3

3

3

2003

2

13

2

200200

8.3

15

3100

500

03

100500

36

200600200200

333

2003

2

13200200

700

120032

600

0

200

1200300900

132002

15.132006

0

3

32

22

2

2

2

−+−=

−−−−=

−

−−−−=

=

=

=

=−=

−−+−=

−−−−−=

=

−

=

=Σ

=

=+=

+=

=Σ

xxM

xxx

xxxxM

x

x

x

x

xx

xxxV

NR

R

F

NR

R

M

B

B

Y

A

A

B

18 kN/m

12 kN/m

3 m

A B

3°

2°

-45 kN

-54 kN.m

A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. a) about z axis

b) about y axis

6 in.

6 in. x

z

y

psi

kiplbinkip

inI

I

z

z

z

z

z

167

)/1000(/167.0

864

)12)(6(2

864

)12)(6)(12

1(

2

4

3

=

=

=

=

=

σ

σ

σ

psi

kiplbinkip

inI

I

y

y

y

y

y

333

)/1000)(/333.0(

216

)12)(3)(2(

216

12

)6)(12)(1(

2

4

3

=

=

=

=

=

σ

σ

σ

A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.


M

2 in. 2 in.

4 in. 4 in.

inkipM

M

bhM

bh

c

I

c

IM

I

Mc

t

.44

)24

)32)(4((22

)24

(22

24

2

2

2

=

=

=

=

=

=

σ

σ

inkipM

M

c

IM c

.30

)24

)32)(4((15

2

=

=

=σ

ftkipM

in

ftinkipM

.5.2

12.30

=

=

A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

M

2 in. 2 in.

4 in. 4 in.

( )

( )( )

ksi

inlb

inC

C

C

hC

inI

I

bhI

I

Mc

c

c

c

c

8.4

/4800

62.4

31.29600

31.2

3

34

3

322

3

2

62.4

36

)32)(4(

36

2

1

1

1

1

4

3

3

1

=

=

=

=

=

=

=

=

=

=

=

σ

σ

σ

σ

( )

( )( )

ksi

inlb

inC

C

hC

t

t

t

4.2

/2400

62.4

15.19600

15.1

3

32

3

2

2

2

2

=

=

=

=

=

=

σ

σ

σ

A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. a) about z-axis

b) about y-axis

3 in.

z

y

3 in.

3 in.

3 in.

0.25 in.

0.25 in.

0.25 in.

( )( ) ( )( )( ) ( )( )

( ) ( )

ftkipM

inkipM

M

inI

I

I

Mc

z

z

z

z

z

.8.20

.7.249

25.03/8125.3324

8125.33

12

6.025.0125.325.06.0

12

25.06.02

4

32

3

=

=

+=

=

+

+=

=σ

( )( ) ( )( )

( )

ftkipM

inkipM

M

inI

I

y

y

y

y

y

.0.6

.06.72

3

924

9

12

25.06.0

12

6.025.02

4

33

=

=

=

=

+

=

The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points.

B A

( )( ) ( )( )

( )( )( )

( )( )( )

MPa

MPa

I

zM

I

yM

mI

III

B

B

A

A

y

y

z

zA

zy

2.66

)10(51.1

1000025.0400

199

)10(5.1

1000075.0400

)10(51.1

12

05.005.02

12

15.005.0

5

5

45

33

=

+=

=

+=

+−

=

=

+

===

−

−

−

σ

σ

σ

σ

σ

50 mm 50 mm

50 mm 50 mm

50 mm

50 mm

M = 40 kN⋅m

Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa.

0.8 1.2 0.6

600 400

A B

( ) ( ) ( )

( ) ( ) ( )

( )( )( )

mmd

mr

r

Mr

r

M

r

Mr

NR

R

M

NR

R

M

B

B

A

A

A

3.31

0156.0

)10(160

4804

4

4

4

200

8.06008.14002.1

0

800

6.040026002.1

0

36

3

3

4

=

=

=

=

=

=

=

−=

=Σ

=

−=

=Σ

π

πσ

πσ

πσ

The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress?


4

222

2@

4

022

0@

20

22

22

0

2

2

0

max

PLM

LLPM

La

PLM

LPM

a

La

aLP

M

aLP

M

MM

M

PR

PR

RR

PRR

Fy

A

A

A

B

A

BA

c

AY

AY

BYAY

BYAY

=

−=

=

=

−=

=

≤≤

−

=

−

=

=

=Σ

=

=

=

=+

=Σ↑+

( )( )

( )

2max

2max

3max

3

maxmax

2

3

12

8

12

24

2

12

,0

bd

PL

bd

PL

bd

dPL

dc

bdI

I

cM

La

=

=

=

=

=

=

=

σ

σ

σ

σ

P

d

b L/2 L/2

a a

The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa.

( )( )

( )

kNP

P

mI

I

I

Mc

PR

PRR

Fy

PR

PPR

M

A

BA

B

B

A

4.10

)10(953.1

125.05.1)10(10

)10(953.1

12

25.05.0

2

0

35.15.1

0

4

6

44

3

max

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.

( ) ( ) ( )

( )( )

( )

( )

MPa

mI

I

kNR

RR

Fy

kNR

R

M

A

BA

B

B

A

5.11

)10(953125.1

125.018000

)10(953125.1

125.018

)10(953125.1

12

25.015.0

12

24

0

12

3125.1125.1

0

max

4max

4max

44

3

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

−

σ

σ

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in.

h

b

2 ft

8 ft 8 ft

P

( )

( )

( )( )

( )

( )( )

( )

( )

kipP

kip

RP

ftkip

M

kipR

R

RMS

M

in

bhS

PR

PR

PR

M

in

h

A

A

A

A

B

A

A

B

114

5.113

75.562

2

5448

75.5696

75.56

681

968

96,

681

6

6.228

6

2

2

816

0

6.22

824

4

2

2

22

=

=

=

=

⋅=

=

=

=

==

=

=

=

=

=

=

=Σ

=

−=

σ

The member has a square cross section and is subjected to a resultant moment of M = 850 N⋅m as shown. Determine that bending stress at each corner and sketch the stress distribution produced by M. Set θ = 45°.

θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

( )( ) ( )

( )( ) ( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

0

102552.3

125.004.601

102552.3

125.004.601

462

102552.3

125.004.601

102552.3

125.004.601

462

102552.3

04.601125.0

102552.3

125.004.601

0

102552.3

125.004.601

102552.3

125.004.601

102552.325.025.012

1

102552.325.025.012

1

04.60145sin850

04.60145cos850

45

44

44

44

44

443

443

=

+−

=

−=

+−

=

=

+−−

=

=

−+

−−=

+−

=

=⋅=

=⋅=

==

==

°=

−−

−−

−−

−−

−

−

E

D

B

Y

Y

Z

ZA

Z

Y

Z

Y

kPa

kPa

I

zM

I

yM

mI

mI

NmM

NmM

σ

σ

σ

σ

θ


θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

( )( ) ( )

( )( ) ( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

kPa

kPa

kPa

kPa

I

zM

I

yM

mI

mI

NmM

NmM

E

D

B

Y

Y

Z

ZA

Z

Y

Z

Y

119

102552.3

125.012.736

102552.3

125.0425

446

102552.3

125.012.736

102552.3

125.0425

446

102552.3

125.012.736

102552.3

125.0425

119

102552.3

125.012.736

102552.3

125.0425

102552.325.025.012

1

102552.325.025.012

1

42530sin850

12.73630cos850

30

44

44

44

44

443

443

=

+−

=

−=

+−

=

=

+−−

=

−=

−+

−−=

+−

=

=⋅=

=⋅=

==

==

°=

−−

−−

−−

−−

−

−

σ

σ

σ

σ

θ

The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined.

6in. 6in.

2in.

8in.

2in.

60°

150 kip.in

y

y

z

B

A

( )( )( ) ( )( )( )[ ]( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( ) ( )( ) ( )( )

( ) ( )

( ) ( )

ksi

ksi

ksi

in

I

in

I

in

A

Ayy

inkipM

inkipM

B

A

Z

Y

o

Z

Y

33.3

0178.2

333.293

19.129

33.333

775

33.3

333.293

69.129

333.333

375

33.333

7921221212

1472882

12

1

33.293

12212

128

12

1

7

12228

1229284

7560cos150

9038.12960sin150

max

4

2323

4

33

=∴

−=

−+

−−−=

=

+−

−=

=

−++

−+

=

=

+

=

=

+

+=

Σ

Σ=

⋅−=−=

⋅==

σ

σ

σ

( )

( ) ( )

°−=

−−=

=

=

2.63

302.33

30tan33.293

333.333

tantan

α

θαY

Z

I

I

The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points.

( )( ) ( )( )( )( ) ( )( )

( )( ) ( )( ) ( )( ) ( )

( )( )( )

( )

( )( )( )

MPa

MPa

m

I

m

y

c

C

B

B

Z

55.1

103633.0

0075.075

61.3

103633.0

0175.075

103633.0

0175.003.004.001.012

12)005.00175.0(01.008.001.008.0

12

1

0175.0

01.004.008.001.0

03.001.004.0201.008.0005.0

6

6

46

2323

=

−−=

=

−−=

=

−+

+

−+=

=

+

+=

−

−

σ

σ

σ

σ

B

10 mm

M = 75 N⋅m

20 mm

10 mm

10 mm

10 mm 10 mm

20 mm

40 mm

C

N

A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B.

20 mm

20 mm

200 mm

15 mm

A

200 mm

M = 40 kN⋅m

B

15 mm

( )( ) ( )( )( ) ( )( )

( )

( )( )

( )( )( )

kN

F

MPa

F

I

zM

I

yM

m

I

RB

B

RA

A

y

Y

Z

ZA

y

5.1

1000

015.024.034.410334

410.0

10316.1

12.04500

0

0

00

10316.1

24.0015.012

1211.002.02.002.02.0

12

12

4

44

323

=

=

=

+=

=

=

+=

+−

=

=

+

+=

−

−

σ

σ

σ

The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.

M = 600 N⋅m

20 mm

25 mm

200 mm

20 mm

( )( )( ) ( )( )( )( )( ) ( )( )

( )( ) ( )( )( )

( )( ) ( )( )( )

( )

( )( )

MPa

I

Mc

m

I

mC

my

A

Ayy

06.2

1053.24

11875.0600

1053.34

05625.01.015.002.015.002.012

12

0125.005625.0025.024.0025.024.012

1

11875.005625.0175.0

05625.0

02.015.0224.0025.0

02.015.01.0224.0025.00125.0

6

max

46

23

23

=

=

=

=

−+

+

−+=

=−=

=

+

+=

Σ

Σ=

−

−

σ

If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.

z

y

12 mm

M = 800 N⋅m

12 mm

200 mm

200 mm

150 mm

z

12 mm

60°

( )( )( ) ( )( )( )( )( )( ) ( )( )( )

( )( ) ( )( )( )

( )

( )( ) ( )( )( )

( )( )

( )( )( )

( )( )( )

( )( )

( )( )

( )( )

°−=

=

=

−=

−+

−−=

=

−−+

−=

=

+

+=

=

−+=

=−=

=

+

+=

−

−

−−

−−

−

−

1.87

30tan103374.16

1018869.0tan

tantan

13.1

103374.16

0366.082.692

1011869.0

2.0400

38.4

103374.16

1134.082.692

101886.0

2.0400

1018869.0

4.0012.012

1

194.0012.0138.0012.0138.012

12

103374.16

0366.0081.0012.04.0012.04.012

1

1134.00366.015.0

366.0

138.0012.024.0012.0

138.0012.0081.024.0012.0006.0

6

3

63

63

43

3

23

46

23

α

α

θα

σ

σ

Y

Z

B

A

Z

Y

I

I

MPa

MPa

m

I

m

I

mC

m

z

The beam has a rectangular cross section as shown. Determine the largest load P that can be subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

0.8 m 1.2 m 0.6 m

600 N 400 N

A B

( )( )

2

3

3

2

3

3

/181

030.0

48032

32

4

324

2;

mN

d

M

r

Mr

d

d

II

Mc

=

=

=

=

=

==

σ

π

π

π

ππ

σ

A member having the dimensions shown is to be used to resist an internal bending moment of M = 2 kip⋅ft. Determine the maximum stress in the member if the moment is applied (a) about the z axis, (b) about the y axis. Sketch the stress distribution for each case. a) about z axis

b) about y axis

6 in.

6 in. x

z

y

psi

kiplbinkip

inI

I

z

z

z

z

z

167

)/1000(/167.0

864

)12)(6(2

864

)12)(6)(12

1(

2

4

3

=

=

=

=

=

σ

σ

σ

psi

kiplbinkip

inI

I

y

y

y

y

y

333

)/1000)(/333.0(

216

)12)(3)(2(

216

12

)6)(12)(1(

2

4

3

=

=

=

=

=

σ

σ

σ

A member has the triangular cross section shown. Determine the largest internal moment M that can be applied to the cross section without exceeding allowable tensile and compressive stresses of (σallow)t = 22 ksi and (σallow)s = 15 ksi, respectively.


M

2 in. 2 in.

4 in. 4 in.

inkipM

M

bhM

bh

c

I

c

IM

I

Mc

t

.44

)24

)32)(4((22

)24

(22

24

2

2

2

=

=

=

=

=

=

σ

σ

inkipM

M

c

IM c

.30

)24

)32)(4((15

2

=

=

=σ

ftkipM

in

ftinkipM

.5.2

12.30

=

=

A member has the triangular cross section shown. If a moment of M = 800 lb⋅ft is applied to the cross section, determine the maximum tensile and compressive bending stresses in the member. Also, sketch a three dimensional view of the stress distribution acting over the cross section.

M

2 in. 2 in.

4 in. 4 in.

( )

( )( )

ksi

inlb

inC

C

C

hC

inI

I

bhI

I

Mc

c

c

c

c

8.4

/4800

62.4

31.29600

31.2

3

34

3

322

3

2

62.4

36

)32)(4(

36

2

1

1

1

1

4

3

3

1

=

=

=

=

=

=

=

=

=

=

=

σ

σ

σ

σ

( )

( )( )

ksi

inlb

inC

C

hC

t

t

t

4.2

/2400

62.4

15.19600

15.1

3

32

3

2

2

2

2

=

=

=

=

=

=

σ

σ

σ

A beam has the cross section shown. If it is made of steel that has an allowable stress of σallow =24 ksi, determine the largest internal moment the beam can resist if the moment is applied (a) about the z axis, (b) about the y axis. a) about z-axis

b) about y-axis

3 in.

z

y

3 in.

3 in.

3 in.

0.25 in.

0.25 in.

0.25 in.

( )( ) ( )( )( ) ( )( )

( ) ( )

ftkipM

inkipM

M

inI

I

I

Mc

z

z

z

z

z

.8.20

.7.249

25.03/8125.3324

8125.33

12

6.025.0125.325.06.0

12

25.06.02

4

32

3

=

=

+=

=

+

+=

=σ

( )( ) ( )( )

( )

ftkipM

inkipM

M

inI

I

y

y

y

y

y

.0.6

.06.72

3

924

9

12

25.06.0

12

6.025.02

4

33

=

=

=

=

+

=

The beam is subjected to a moment of M = 40 kN⋅m. Determine the bending stress acting at points A and B. sketch the results on a volume element acting at each of these points.

B A

50 mm 50 mm

50 mm 50 mm

50 mm

50 mm

M = 40 kN⋅m

( )( ) ( )( )

( )( )( )

( )( )( )

MPa

MPa

I

zM

I

yM

mI

III

B

B

A

A

y

y

z

zA

zy

2.66

)10(51.1

1000025.0400

199

)10(5.1

1000075.0400

)10(51.1

12

05.005.02

12

15.005.0

5

5

45

33

=

+=

=

+=

+−

=

=

+

===

−

−

−

σ

σ

σ

σ

σ

Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 160 MPa.

0.8 1.2 0.6

600 400

A B

( ) ( ) ( )

( ) ( ) ( )

( )( )( )

mmd

mr

r

Mr

r

M

r

Mr

NR

R

M

NR

R

M

B

B

A

A

A

3.31

0156.0

)10(160

4804

4

4

4

200

8.06008.14002.1

0

800

6.040026002.1

0

36

3

3

4

=

=

=

=

=

=

=

−=

=Σ

=

−=

=Σ

π

πσ

πσ

πσ

The beam is subjected to the load P at its center. Determine the placement a of the supports so that the absolute maximum bending stress in the beam is as large as possible. What is this stress?


4

222

2@

4

022

0@

20

22

22

0

2

2

0

max

PLM

LLPM

La

PLM

LPM

a

La

aLP

M

aLP

M

MM

M

PR

PR

RR

PRR

Fy

A

A

A

B

A

BA

c

AY

AY

BYAY

BYAY

=

−=

=

=

−=

=

≤≤

−

=

−

=

=

=Σ

=

=

=

=+

=Σ↑+

( )( )

( )

2max

2max

3max

3

maxmax

2

3

12

8

12

24

2

12

,0

bd

PL

bd

PL

bd

dPL

dc

bdI

I

cM

La

=

=

=

=

=

=

=

σ

σ

σ

σ

P

d

b L/2 L/2

a a

The steel beam has the cross-sectional area shown. If w = 5 kip⋅ft, determine the absolute maximum bending stress in the beam.

( ) ( )

( )( ) ( )( ) ( )( )

( )( )( )

ksi

inI

I

kipR

RR

Fy

kipR

R

M

A

BA

B

B

A

8.66

3.152

3.512160

3.152

12

103.015.53.08

12

30.082

40

80

0

40

24044024

0

max

max

4

32

3

=

=

=

+

+=

=

=+

=Σ↑+

=

+=

=Σ

σ

σ

wowo

8 ft8 ft8 ft

0.3 in

8 in 0.30 in

10 in

0.30 in

160 kip.ft

-40 kip

40 kip

The beam has a rectangular cross section as shown. Determine the largest load P that can be supported on its overhanging ends so that the bending stress does not exceed σmax = 10 MPa.

( )( )

( )

kNP

P

mI

I

I

Mc

PR

PRR

Fy

PR

PPR

M

A

BA

B

B

A

4.10

)10(953.1

125.05.1)10(10

)10(953.1

12

25.05.0

2

0

35.15.1

0

4

6

44

3

max

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

The beam has the rectangular cross section shown. If P =12 kN, determine that absolute maximum bending stress in the beam. Sketch the stress distribution acting over the cross section.

( ) ( ) ( )

( )( )

( )

( )

MPa

mI

I

kNR

RR

Fy

kNR

R

M

A

BA

B

B

A

5.11

)10(953125.1

125.018000

)10(953125.1

125.018

)10(953125.1

12

25.015.0

12

24

0

12

3125.1125.1

0

max

4max

4max

44

3

=

=

=

=

=

=

=+

=Σ↑+

=

=+

=Σ

−

−

−

σ

σ

σ

P P

1.5 m 1.5 m 1.5 m

250 mm

150 mm

. A log that is 2 ft in diameter is to be cut into a rectangular section for use as a simply supported beam. If the allowable bending stress for the wood is σallow = 8 ksi, determine the largest load P that can be supported if the width of the beam is b = 8 in.

h

b

2 ft

8 ft 8 ft

P

( )

( )

( )( )

( )

( )( )

( )

( )

kipP

kip

RP

ftkip

M

kipR

R

RMS

M

in

bhS

PR

PR

PR

M

in

h

A

A

A

A

B

A

A

B

114

5.113

75.562

2

5448

75.5696

75.56

681

968

96,

681

6

6.228

6

2

2

816

0

6.22

824

4

2

2

22

=

=

=

=

⋅=

=

=

=

==

=

=

=

=

=

=

=Σ

=

−=

σ


θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

( )( ) ( )

( )( ) ( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

0

102552.3

125.004.601

102552.3

125.004.601

462

102552.3

125.004.601

102552.3

125.004.601

462

102552.3

04.601125.0

102552.3

125.004.601

0

102552.3

125.004.601

102552.3

125.004.601

102552.325.025.012

1

102552.325.025.012

1

04.60145sin850

04.60145cos850

45

44

44

44

44

443

443

=

+−

=

−=

+−

=

=

+−−

=

=

−+

−−=

+−

=

=⋅=

=⋅=

==

==

°=

−−

−−

−−

−−

−

−

E

D

B

Y

Y

Z

ZA

Z

Y

Z

Y

kPa

kPa

I

zM

I

yM

mI

mI

NmM

NmM

σ

σ

σ

σ

θ


θ

125 mm

125 mm

D

C A

B E

250 mm

M = 850 N⋅m

y

z

( )( ) ( )

( )( ) ( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

kPa

kPa

kPa

kPa

I

zM

I

yM

mI

mI

NmM

NmM

E

D

B

Y

Y

Z

ZA

Z

Y

Z

Y

119

102552.3

125.012.736

102552.3

125.0425

446

102552.3

125.012.736

102552.3

125.0425

446

102552.3

125.012.736

102552.3

125.0425

119

102552.3

125.012.736

102552.3

125.0425

102552.325.025.012

1

102552.325.025.012

1

42530sin850

12.73630cos850

30

44

44

44

44

443

443

=

+−

=

−=

+−

=

=

+−−

=

−=

−+

−−=

+−

=

=⋅=

=⋅=

==

==

°=

−−

−−

−−

−−

−

−

σ

σ

σ

σ

θ

The T-beam is subjected to a moment of M = 150 kip⋅in. directed as shown. Determine the maximum bending stress in the beam and the orientation of the neutral axis. The location y of the centroid, C, must be determined.

( )( )( ) ( )( )( )[ ]( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( ) ( )( ) ( )( )

( ) ( )

( ) ( )

ksi

ksi

ksi

in

I

in

I

in

A

Ayy

inkipM

inkipM

B

A

Z

Y

o

Z

Y

33.3

0178.2

333.293

19.129

33.333

775

33.3

333.293

69.129

333.333

375

33.333

7921221212

1472882

12

1

33.293

12212

128

12

1

7

12228

1229284

7560cos150

9038.12960sin150

max

4

2323

4

33

=∴

−=

−+

−−−=

=

+−

−=

=

−++

−+

=

=

+

=

=

+

+=

Σ

Σ=

⋅−=−=

⋅==

σ

σ

σ

( )

( ) ( )

°−=

−−=

=

=

2.63

302.33

30tan33.293

333.333

tantan

α

θαY

Z

I

I

6-185. Determine the bending stress distribution in the beam at section a-a. Sketch the distribution in three dimension acting over the cross section.

( )( ) ( )( )

( )

( )( )

kPa

Nm

I

Mc

m

I

6.634

52.2

050.032

1052.2

12

15.075.0

12

1.015.0

2

46

33

=

=

=

=

+

=

−

σ

The aluminum machine part is subjected to a moment of M = 75 N⋅m. Determine the bending stress created at points B and C on the cross section. Sketch the results on a volume element located at each of these points.

B

10 mm

M = 75 N⋅m

20 mm

10 mm

10 mm

10 mm 10 mm

20 mm

40 mm

C

N

( )( ) ( )( )( )( ) ( )( )

( )( ) ( )( ) ( )( ) ( )

( )( )( )

( )

( )( )( )

MPa

MPa

m

I

m

y

c

C

B

B

Z

55.1

103633.0

0075.075

61.3

103633.0

0175.075

103633.0

0175.003.004.001.012

12)005.00175.0(01.008.001.008.0

12

1

0175.0

01.004.008.001.0

03.001.004.0201.008.0005.0

6

6

46

2323

=

−−=

=

−−=

=

−+

+

−+=

=

+

+=

−

−

σ

σ

σ

σ

A beam is constructed from four pieces of wood, glued together as shown. If the moment acting on the cross section is M = 450 N⋅m, determine the resultant force that bending stress produces on the top board A and on the side board B.

20 mm

20 mm

200 mm

15 mm

A

200 mm

M = 40 kN⋅m

B

15 mm

( )( ) ( )( )( ) ( )( )

( )

( )( )

( )( )( )

kN

F

MPa

F

I

zM

I

yM

m

I

RB

B

RA

A

y

Y

Z

ZA

y

5.1

1000

015.024.034.410334

410.0

10316.1

12.04500

0

0

00

10316.1

24.0015.012

1211.002.02.002.02.0

12

12

4

44

323

=

=

=

+=

=

=

+=

+−

=

=

+

+=

−

−

σ

σ

σ

The beam is made from three boards nailed together as shown. If the moment acting on the cross section is M = 600 N⋅m, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section.

M = 600 N⋅m

20 mm

25 mm

200 mm

20 mm

( )( )( ) ( )( )( )( )( ) ( )( )

( )( ) ( )( )( )

( )( ) ( )( )( )

( )

( )( )

MPa

I

Mc

m

I

mC

my

A

Ayy

06.2

1053.24

11875.0600

1053.34

05625.01.015.002.015.002.012

12

0125.005625.0025.024.0025.024.012

1

11875.005625.0175.0

05625.0

02.015.0224.0025.0

02.015.01.0224.0025.00125.0

6

max

46

23

23

=

=

=

=

−+

+

−+=

=−=

=

+

+=

Σ

Σ=

−

−

σ

If the internal moment acting on the cross section of the strut has a magnitude of M = 800 N⋅m and is directed as shown, determine the bending stress at points A and B. The location z of the centroid C of the strut’s cross-sectional area must be determined. Also, specify the orientation of the neutral axis.

z

y

12 mm

M = 800 N⋅m

12 mm

200 mm

200 mm

150 mm

z

12 mm

60°

( )( )( ) ( )( )( )( )( )( ) ( )( )( )

( )( ) ( )( )( )

( )

( )( ) ( )( )( )

( )( )

( )( )( )

( )( )( )

( )( )

( )( )

( )( )

°−=

=

=

−=

−+

−−=

=

−−+

−=

=

+

+=

=

−+=

=−=

=

+

+=

−

−

−−

−−

−

−

1.87

30tan103374.16

1018869.0tan

tantan

13.1

103374.16

0366.082.692

1011869.0

2.0400

38.4

103374.16

1134.082.692

101886.0

2.0400

1018869.0

4.0012.012

1

194.0012.0138.0012.0138.012

12

103374.16

0366.0081.0012.04.0012.04.012

1

1134.00366.015.0

366.0

138.0012.024.0012.0

138.0012.0081.024.0012.0006.0

6

3

63

63

43

3

23

46

23

α

α

θα

σ

σ

Y

Z

B

A

Z

Y

I

I

MPa

MPa

m

I

m

I

mC

m

z

The axle of the freight car is subjected to wheel loadings of 20 kip. If it is supported by two journal bearings at C and D, determine that maximum bending stress developed at the center of the axle, where the diameter is 5.5 in.

( ) ( )

( )( )( )

ksi

r

Mr

I

Mc

kipR

R

Fy

kipR

R

M

D

D

C

C

D

24.12

75.2

2004

4

20

202020

0

20

80207020

0

3

4

=

=

=

=

=

=−+

=Σ

=

=+

=Σ

σ

π

π

σ

6-74. Determine the absolute maximum bending stress in the 1.5-in.-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

( ) ( ) ( )

( )( )( )Ksi

dI

d

Mr

d

II

Mc

lbR

R

F

lbR

R

M

B

B

Y

A

A

B

6.13

5.1

5.7450064

64

64

4

2

610

90300400

0

90

153001840030

0

4

4

4

4

=

=

==

==

=

−+=

=Σ

=

−=

=Σ

π

π

π

π

σ

6-75. Determine the smallest allowable diameter of the shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces, and the allowable bending stress is σallow = 22 ksi.

Given:

?

22

=

=

d

Ksiσ

Solution:

( ) ( ) ( )

lbR

R

F

lbR

R

M

B

B

Y

A

A

B

610

90300400

0

90

153001840030

0

=

−+=

=Σ

=

−=

=Σ

( )

( )

ind

d

d

d

d

d

r

MrI

d

M

I

Mc

28.1

1.2

22000

144000

45003222

324

2

4

;32

3

3

3

3

3

43

=

=

=

=

=

===

=

π

π

ππ

ππσ

σ

Determine the absolute maximum bending stress in the 30-mm-diameter shaft which is subjected to the concentrated forces. The sleeve bearings at A and B support only vertical forces.

0.8 m 1.2 m 0.6 m

600 N 400 N

A B

( )( )

2

3

3

2

3

3

/181

030.0

48032

32

4

324

2;

mN

d

M

r

Mr

d

d

II

Mc

=

=

=

=

=

==

σ

π

π

π

ππ

σ

6-79. The steel shaft has a circular cross section with a diameter of 2 in. It is supported on smooth journal bearings A and B, which exert only vertical reactions on the shaft. Determine the absolute maximum bending stress if it is subjected to the pulley loadings shown.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( )ksi

r

M

r

Mr

I

Mc

lbR

R

M

lbR

R

M

B

B

A

A

A

B

4.20

1

160004

4

4

650

20500403006080080

0

650

20500403006050080

;0

3

34

=

=

===

=

++=

=Σ

=

++=

=Σ

σ

π

ππσ

Determine the absolute maximum bending stress in the 20mm diameter pin. The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4kN/m caused by bar C. Determine the intensity of the distributed load wo of the leaves on the pin and draw the shear and moment diagrams for the pin.

( )

( )( )

kPa

m

I

I

Mc

331

10853.7

01.000026.0

10853.7

4

01.90

max

9max

49

4

=

=

=

=

=

−

−

σ

σ

π

σ

2.6x10-3 kN⋅m

60 mm 20 mm20 mm

wowo

0.4 kN/m

If the beam ABC has square cross-section 6in by 6in, determine the absolute bending stress in the beam. The 20kip load must be replaced by equivalent loadings at point C on the axis of the beam.

( )( )

( )( )

ksi

I

Mc

inI

I

6.15

108

12368.46

108

6612

1

max

max

4

3

=

=

=

=

=

σ

σ

6-67. If the crane boom ABC has a rectangular cross-section with the base of 2in a height of 3in, determine the absolute maximum bending stress in the boom. The engine crane is used to support the engine, which has a weight of 1200 lb.

( )( )

( )

ksi

inkip

ft

in

lb

kipftlbM

in

I

24

5.4

5.172

.72

12

1000.6000

5.4

3212

1

max

4

3

=

=

=

=

=

=

σ

6-86. The simply supported beam is made from four ¾ in diameter rod w/c are bounded as shown determine the maximum bending stress in the beam due to the loading shown.

( ) ( ) ( )

( ) ( )

( )

ksi

d

M

inlbftlb

d

M

r

Mr

in

inind

lbR

M

lbR

R

M

B

A

A

A

B

4.2

2

192032

32

.1920.160

32

4

2

44

14

4

3

80

0

80

28088010

0

3

3

33

=

=

=

=

==

=

−=

=

=Σ

=

+=

=Σ

π

πσ

ππσ

6-48. The beam is made from three boards nailed together as shown. If the moment acting on the cross-section is M=600N.m, determine the resultant force bending stress produces on the top board.

( )( )( )NF

F

kPastressAve

kPay

y

R

R

591.3

1000/240255.598

5.5982

252945.

252

75.93

945

25

=

=

=+

=

=

=

5-59. The beam is subjected to a moment M=30lb.ft. Determine the bending stress at points A and B. Also, sketch a three dimensional view of the stress distribution acting over the entire cross-sectional area.

(Mindanao State University-General Santos City, Philippines) BSEE-(G.R.F) students 2004

( )( ) ( )( )( )( )( )( ) ( )( )( )

( ) ( )( ) ( )( )( )

( )( )

( )( )

psi

psi

in

I

in

y

D

A

977.32

367.4

4.01230

35.214

367.4

4.141230

367.4

4.12312

1

36

329.02

12

12

4.1

135.0212

2135.025.012

4

23

2

2

=

=

=

−=

=

−+

++=

=

+

+=

σ

σ

Documents

Strenght of Materials(ES-64)