Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
Chapter 5: Materials and Stresses
105
Chapter 5
Solutions
108
P5.4: A steel cable of diameter 3/16 in. is attached to an eyebolt and tensioned to 500 lb.
Calculate the stress in the cable and express it in the dimensions psi, ksi, Pa, kPa, and
MPa.
Approach:
Calculate the tensile stress σ = F/A with F = 500 lb, and use conversion factors from Table
5.1.
Solution:
Cross–sectional area:
(
)
2
2
2
in02760
4
in18750
4
.
..d
A===
ππ
Tensile stress:
psi10018
in
lb
11618
in02760
lb500
22
,,
.
σ===
σ = 18,100 psi
1000 psi = 1 ksi
σ = 18.1 ksi
( )
Pa10251
psi
Pa
6895psi10018
8
×=
=.,σ
σ = 1.25×10
8
Pa
1000 Pa = 1 kPa
σ = 1.25×10
5
kPa
1000 kPa = 1 MPa
σ = 125 MPa
109
© 2017 Cengage Learning
®
. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
112
P5.7: (a) By using either the vector algebra or polygon methods for finding a resultant,
determine the magnitude of F that will cause the net effect of the three forces to act
vertically. (b) For that value of F, determine the stress in the bolt's 3/8 in. diameter
straight shank.
F = 568.5 N
4
4
Tensile stress:
psi1610
in11040
lb178
2
=== .A
F
σ
σ = 1610 psi
113
© 2017 Cengage Learning
®
. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
P5.10: The tires of the 4555 lb sedan are 6.5 in. wide. Each tire contacts the ground over a
distance of 4.25 in. as measured along the vehicle's length. Calculate the compressive
stress between each tire and the road. The locations of the vehicle's mass center and the
wheelbase dimensions are shown.
2
in6327
.A
F
σ
F
= 49.5 psi
Rear wheel contact stress:
lb911
R
σ
==
Discussion:
The compressive stress on the front tires is larger than the rear tires because the vehicle’s
center of gravity is closer to the front tires, which means the load on the front tires is greater
118
P5.12: A 1 ft long rod is made from the material of P5.11. By what amount must the rod be
stretched from its original length for it to begin yielding?
Approach:
From the results of P5.11, E = 176 ksi and S
y
= 1.04 ksi. Determine the strain in the rod at
the condition where the stress equals the yield strength. Find the change in length GL = εL.
Upper Platform:
The load in the center includes the 250 lb weight and the 200 lb load from the cable at C.
The combined load is split between the two cables, with each supporting 225 lb. Find the
elongation of using Equation (5.6):
124
© 2017 Cengage Learning
®
. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(
)
2
2
2
in2980
in243044 .
.A
d===
ππ
..d in550
=
Lower Platform, Cable at C: The worker on the bottom platform is standing on the left edge
of the platform.
Balance moments about point C to determine F
D
:
(
)
(
)
(
)
(
)
0in190in95lb400
=
+
−
.F.
D
F
D
= 200 lb
Sum forces in the vertical direction to find F
C
:
