HOMEWROK PROBLEMS
Chapter 1:
Chapter 2:
Chapter 3:
Chapter 4:
4-66
Use Castigliano’s theorem to verify the maximum deflection for the uniformly loaded
cantilever beam 3 of Appendix Table A-9. Neglect Shear.
Chapter 5:
Chapter 6:
Chapter 7:
HINT:
Trial #1: Choose dr = 0.75 in
Trial #2: Choose dr = 0.799 in
Chapter 8: Please submit 8-19, 8-38 and 8-78
Problem 8-34 solution:
Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in.
L ≥ l + H = 1.125 + 7/16 = 1.563 in.
Round up to L = 1.75 in Ans.
Eq. (8-13): LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in
Table 8-7: ld = L  LT = 1.75  1.25 = 0.5 in, lt = l ld = 1.125  0.5 = 0.625 in
Ad =  (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2
Eq. (8-17):
kb 
0.196 3  0.141 9  30
Ad At E

 4.316 Mlbf/in
Ad lt  At ld 0.196 3  0.625  0.141 9  0.5
Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20),
k1 
0.5774  30  0.5
1.155  0.5  0.75  0.5  0.75  0.5 
ln 
1.155  0.5  0.75  0.5  0.75  0.5 
 33.30 Mlbf/in
Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 =
0.5625 in.
Upper frustum, t = 0.5625 0.5 = 0.0625 in, d = 0.5 in,
D = 0.75 + 2(0.5) tan 30 = 1.327 in, E = 14.5 Mpsi (from Table 8-8)
Eq. (8-20)  k2 = 292.7 Mlbf/in
Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi
Eq. (8-20)  k3 = 15.26 Mlbf/in
Eq. (8-18): km = (1/33.30 + 1/292.7 + 1/15.26)1 = 10.10 Mlbf/in
C = kb / (kb + km) = 4.316/(4.316+10.10) = 0.299
Table 8-9:
Sp = 85 kpsi
For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the
effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus,
P = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28):
np 
S p At
CP  Fi

85  0.141 9 
 1.27
0.299 1.443  9.05
Ans.
Overload factor of safety, Eq. (8-29):
nL 
S p At  Fi
CP

85  0.141 9   9.05
 6.98
0.299 1.443
Ans.
Separation factor of safety, Eq. (8-30):
n0 
Fi
9.05

 8.95
P 1  C  1.443 1  0.299 
Ans.
Chapter 9: Please submit 9-45, 9-46 and 9-52
Chapter 10: Please submit 10-7, 10-13 and 10-40
Chapter 11: Please submit 11-19, 11-37 and 11-44
Solution to Problem 3-79:
Ft 
T
900

 180 lbf
c / 2 10 / 2
Fn  180 tan 20  65.5 lbf
TC  Ft  b 2   180  5 2   450 lbf  in
P
TC
450

 150 lbf
 a 2  6 2
  M A  z  0  20 RDy  65.5(14)  150(4)  RDy  75.9 lbf
 RDz  126 lbf
  M A  y  0  20 RDz  180(14)
 Fy  0  RAy  75.9  65.5  150
 Fz  0  RAz  126  180
 RAy  140 lbf
 RAz  54.0 lbf
The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and
bending moment diagrams. We will directly obtain the combined moments from each plane.
M B  AB RA2y  RA2z  4 1402  542  600 lbf  in
M C  CD RD2 y  RD2 z  6 75.92  1262  883 lbf  in
The maximum stresses occur at C.
C 
32M C
C 
16TC
d3
d
3


32(883)

 3460 psi
 1.3753

16(450)
 882 psi

 1.3753
C

Ans.
 max
3460
 
 3460 
2

  C    C2 
 
  882  3670 psi
2
2
 2 
 2 
 max
 
 3460 
2
  C    C2  
  882  1940 psi
2
2




2
2
2
2
Ans.
Ans.
Chapter 12:
Chapter 13:
Chapter 14:
Chapter 15:
Chapter 16:
Chapter 17: