Chapter 5: Materials and Stresses
Set the elongation of the cable to 0.01 in. and solve for A using Equation (5.6):
(
)
(
)
( )
( )
2
26
in1480
in010inlb1030
in90lb495
G.
..
.
LE
FL
A=
×
==
Solve for the diameter of the cable at B:
(
)
2
2
2
in1890
in148044 .
.A
d===
ππ
..d in430
=
Upper Platform, Cable at A: The worker on the bottom platform is standing on the left edge
of the platform and the worker on the top platform is standing on the left edge of the
platform.
Sum forces in the vertical direction to find F
A
:
Set the elongation of the cable to 0.01 in. and solve for A using Equation (5.6):
(
)
(
)
( )
( )
2
26
in1480
in010inlb1030
in90lb495
G.
..
.
LE
FL
A=
×
==
Solve for the diameter of the cable at A:
(
)
2
2
2
in1890
in148044 .
.A
d===
ππ
..d in430
=
Discussion:
The required diameter of the cable at point D is the largest because it is the longest cable. It
also makes sense that the required diameters of the cables at A and B are equivalent since
the cables are the same length. If the maximum allowable elongation decreased, then the
required diameters in all the cables would increase as the cables would have to be stronger.
The 0.59in.9diameter cables used in P5.15 would all effectively support the worst case
loading conditions in this problem with elongation less than 0.01 in. except for the cable at
126
P5.17: An engineer determines that a 40 cm long rod of 1020 grade steel will be subjected
to a tension of 20 kN. The following two design requirements must be met: the stress
must remain below 145 MPa, and the rod must stretch less than 0.125 mm. Determine
an appropriate value for the rod’s diameter to meet these two requirements. Round up to
the nearest millimeter when reporting your answer.
4
6
Pa10145
×
π
d = 0.0133 m
d
stress
=13.3 mm
Since GL = FL/EA, find the diameter in terms of the applied force and elongation:
2
4
d FL
LE
π
=
3
Select the larger of the two diameters to meet both requirements, and round up to the
nearest millimeter:
Discussion:
The larger of the two diameters is chosen to make sure both design requirements are met. If
the smaller diameter is chosen, then only one of the requirements would be met. As a factor
of safety, an engineer may choose a slightly larger diameter than 20 mm in practice
because of inherent variations in the material composition and loading conditions.
132
P5.24: A plastic pipe carries deionized water in a microelectronics clean room, and one
end of it is capped. The water pressure is p
0
= 50 psi, and the cap is attached to the end
of the pipe by an adhesive. Calculate the shear stress τ present in the adhesive.
Solution:
proportional to the diameter. This would result in an overall decrease in the shear stress.
133
P5.27:
A small stepladder has vertical rails and horizontal steps formed from C9section
aluminum channel. Two rivets, one in front and one in back, secure the ends of each
step. The rivets attach the steps to the left9 and right9hand rails. A 200 lb person stands
in the center of a step. If the rivets are formed of 60619T6 aluminum, what should be
the diameter d of the rivets? Use a factor of safety of 6, and round your answer to the
nearest 1/16 of an in.
Approach:
Each rivet is placed in single shear. Calculate the rivet shear force V when the person stands
on a step. That force is carried evenly by the four rivets attached to each step. From Table
5.3, S
y
= 40 ksi. From Equation (5.11), S
sy
= S
y
/2= 20 ksi. With the factor of safety from
Equation (5.12), τ
max
= S
sy
/6. Select diameter and round up to nearest standard size.
Solution:
Force on one comer of a step (assuming the weight is evenly–distributed):
1
134
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®
. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
135
P5.28:
For the exercise of P5.27, and for the most conservative design, at what location on
the step should you specify in your calculation that the 200 lb person stands?
Determine the rivet diameter for that loading condition.
Solution:
Select diameter:
2
2
2
in07640
lb200
V
d
===
136
P5.29:
A 3/8 in. diameter bolt connects the marine propeller to a boat’s 191/4 in. driveshaft.
To protect the engine and transmission if the propeller accidentally strikes an
underwater obstacle, the bolt is designed to be cut when the shear stress acting on it
reaches 25 ksi. Determine the contact force between the blade and obstacle that will
cause the bolt to be sheared, assuming a 4 in. effective radius between the point of
contact on the blade and the center of the driveshaft.
Approach:
Determine the relationship between the contact force F on the propeller and the shear force
V that develops between the bolt and the drive shaft. The shear stress is τ = V/A, where A is
the cross–sectional area of the bolt and the bolt is loaded in double shear. Calculate the
contact force from the condition that τ
max
= 25 ksi.
Solution:
137
© 2017 Cengage Learning
®
. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.