ENGR0135 - Statics and Mechanics of Materials 1 (2161)Homework #2

Solution Set

1. Summing forces in the y-direction allows one to determine the magnitude of F2:∑Fy = 1000 sin 60◦ − 800 sin 37◦ − F2 sin 45◦ = 0 =⇒ F2 = 543.8689 N

Then, summing forces in the x-direction and using the value of F2 allows one to deter-mine the magnitude of F1:∑

Fx = F1 + F2 cos 45◦ − 1000 cos 60◦ − 800 cos 37◦ = 0 =⇒ F1 = 754.3350 N

Thus,F1 = 754 N , F2 = 544 N

2. (a) Noting that the force F exerted on the block by the smooth surface is in thedirection normal to the surface, the free-body diagram for the block is

20◦

30◦

30◦

P

x

y

F

x′

y′

100 lb

The xy- and x′y′-coordinate systems are for the two alternate solution approachesbelow.

(b) Using the xy-coordinate system, the force equilibrium equations are:∑

Fx = P cos 50◦ − F sin 30◦ = 0∑

Fy = P sin 50◦ + F cos 30◦ − 100 = 0

Solving the two simultaneous equations for P and F ,

P = 53.2089 lb , F = 68.4040 lb

An alternative approach, that avoids having to solve simultaneous equations, isto use the x′y′-coordinate system. First summing forces in the x′-direction,

∑Fx′ = P cos 20◦ − 100 sin 30◦ = 0 =⇒ P = 53.2089 lb

Then summing forces in the y′-direction,∑

Fy′ = F + P sin 20◦ − 100 cos 30◦ = 0 =⇒ F = 68.4040 lb

Using either approach, the answer is

P = 53.2 lb , F = 68.4 lb

3. The vectors and distances from D to A, B, and C are

rDA = 10i + 5j− 8k rDA =√

(10)2 + (5)2 + (−8)2 = 13.7477 m

rDB = 5i− 5j− 8k rDA =√

(5)2 + (−5)2 + (−8)2 = 10.6771 m

rDC = −7i− 2j− 8k rDA =√

(−7)2 + (−2)2 + (−8)2 = 10.8167 m

It follows that the unit vectors in the directions of the cable forces are

eDA =rDArDA

= 0.7274i + 0.3637j− 0.5819k

eDB =rDBrDB

= 0.4683i− 0.4683j− 0.7493k

eDC =rDCrDC

= −0.6471i− 0.1849j− 0.7396k

Noting that the net lift of the balloon is 10k (kN), the force equilibrium equation is

∑F = 10k + FDAeDA + FDBeDB + FDCeDC = 0

where FDA, FDB, and FDC are the forces exerted on the balloon by the three cables.This gives three simultaneous equations to be solved for FDA, FDB, and FDC :

0.7274FDA + 0.4683FDB − 0.6471FDC = 0

0.3637FDA − 0.4683FDB − 0.1849FDC = 0

0.5819FDA + 0.7493FDB + 0.7396FDC = 10 kN

Solving these simultaneous equations gives

FDA = 5.73 kN , FDB = 1.48 kN , FDC = 7.51 kN

4. Note that the weight of the 500 kg mass is (500 kg)(9.81 m/s2) = 4905 N, let the tensionsin cables A and B be FA and FB, respectively, and consider the free-body diagrambelow:

45◦x

y

FA

FB

4905N

∑Fy = FB sin 45◦ − 4905 = 0

=⇒ FB = 6936.72N∑

Fx = FB cos 45◦ − FA = 0

=⇒ FA = 4905N

The minimum allowable cable cross-sectional area, Amin, is given in terms of the cabletension, F , and allowable normal stress, σall = 150 MPa, by

σ =F

A=⇒ Amin =

F

σall.

The minimum allowable cross-sectional area is related to the minimum allowable di-ameter, dmin, by

Amin =π

4d 2min .

Cable Aπ

4d 2min = Amin =

4905 N

150× 106 N/m2= 3.27× 10−5 m2

dmin = 6.4525× 10−3 m = 6.45 mm

Cable Bπ

4d 2min = Amin =

6936.72 N

150× 106 N/m2= 4.6245× 10−5 m2

dmin = 7.6734× 10−3 m = 7.67 mm

5. Note: In a problem, like this one, where an allowable normal stress is given with-out mentioning tension or compression, the allowable normal stress is assumed to bethe same in both tension and compression. This may not be entirely clear from thetextbook.

Use appropriate free-body diagrams, as shown below, to determine the axial force, andsubsequently the minimum allowable cross-sectional area, for each pipe section.

Segment AB

BC

D

30 kip 5 kip 15 kipPAB

∑Fx = −PAB − 30 + 5 + 15 = 0 =⇒ PAB = −10 kip

AAB =PAB

σall=

−10 kip

−24 kip/in2= 0.417 in2

Segment BC

CD

5 kip 15 kipPBC

∑Fx = −PBC + 5 + 15 = 0 =⇒ PBC = 20 kip

ABC =PBC

σall=

20 kip

24 kip/in2= 0.833 in2

Segment CD

D

15 kipPCD

∑Fx = −PCD + 15 = 0 =⇒ PCD = 15 kip

ACD =PCD

σall=

15 kip

24 kip/in2= 0.625 in2

6. Using a free-body diagram of the solid aluminum rod to find the shear force V :

P

L

V

∑Fx = P − V = 0 =⇒ V = P = 100 lb

The interface area that the shear force is distributed over is

A = (πd)L = π(0.75 in)(0.75 in) = 1.7671 in2

Thus, the average shear stress is

τ =V

A=

100 lb

1.7671 in2 = 56.6 psi

7. Note first that the cross-sectional area of the bar is A = (0.125 m)t and the angle ofinclination of the weld is θ = 90◦ − 60◦ = 30◦. There will be a minimum thicknessbased on the allowable normal stress and a minimum thickness based on the allowableshear stress. The answer is the larger of the two, so that neither allowable stress isexceeded.

Normal Stress

σn =P

Acos2 θ =⇒ Amin =

P

σallcos2 θ

Amin =750× 103 N

80× 106 N/m2cos2 30◦ = 7.0312× 10−3 m2

tmin =Amin

0.125 m= 5.625× 10−2 m

Shear Stress

τn =P

Asin θ cos θ =⇒ Amin =

P

τallsin θ cos θ

Amin =750× 103 N

50× 106 N/m2sin 30◦ cos 30◦ = 6.4952× 10−3 m2

tmin =Amin

0.125 m= 5.196× 10−2 m

Answer:tmin = 56.3 mm

8. Note first that the angle of inclination of the joint is θ = 90◦ − φ and the range ofinclinations to be considered is 0 ≤ θ ≤ 45◦. The cross-sectional area of the memberis A = (2 in)(3 in) = 6 in2.

Normal Stress: Given the equation for the normal stress on the plane of the joint,

σn =P

Acos2 θ

the maximum safe load based on the allowable tensile stress in the glue is

Pσ =σallA

cos2 θ=

(50 psi)(6 in2)

cos2 θ=

300 lb

cos2 θ

Shear Stress: Given the equation for the shear stress on the plane of the joint,

τn =P

Asin θ cos θ

the maximum safe load based on the allowable shear stress in the glue is

Pτ =τallA

sin θ cos θ=

(35 psi)(6 in2)

sin θ cos θ=

210 lb

sin θ cos θ

At any given inclination level, θ, the maximum safe load is the smaller of Pσ and Pτ .Plotting these results as a function of θ for 0 ≤ θ ≤ 45◦,

θ

Pall (lb)

θopt 45◦

300

420

600

Popt

Pσ

Pτ

(a) The optimum angle is where the two plots cross, Pσ = Pτ ,

300 lb

cos2 θopt=

210 lb

sin θopt cos θopt=⇒ tan θopt =

210

300=⇒ θopt = 35.0◦

Thus,φopt = 90◦ − θopt = 55.0◦

(b) The maximum safe load at this optimum angle is

Popt =300 lb

cos2 θopt=

210 lb

sin θopt cos θopt= 447 lb