03-01 Chap Gere - 3 Torsion Torsional Deformations Problem...

Torsional Deformations
Problem 3.2-1 A copper rod of length L 18.0 in. is to be twisted
by torques T (see figure) until the angle of rotation between the ends
of the rod is 3.0°.
If the allowable shear strain in the copper is 0.0006 rad, what is
the maximum permissible diameter of the rod?
3
Torsion
Problem 3.2-2 A plastic bar of diameter d 50 mm is to be twisted by
torques T (see figure) until the angle of rotation between the ends of the
bar is 5.0°.
If the allowable shear strain in the plastic is 0.012 rad, what is the
minimum permissible length of the bar?
L
181

Problem 3.2-3 A circular aluminum tube subjected to pure torsion
by torques T (see figure) has an outer radius r
2
equal to twice the inner
radius r
1
(a) If the maximum shear strain in the tube is measured as
400 10 6 rad, what is the shear strain
1
at the inner
surface?
(b) If the maximum allowable rate of twist is 0.15 degrees
per foot and the maximum shear strain is to be kept at
400 10 6 rad by adjusting the torque T , what is the
minimum required outer radius ( r
2
)
min
?
L
T
r
1
r
2
T
r
2
2 r
1
.
182
CHAPTER 3 Torsion
(a) S
HEAR STRAIN AT INNER SURFACE
Problems 3.2-3, 3.2-4, and 3.2-5
Problem 3.2-4 A circular steel tube of length L 0.90 m is loaded in
torsion by torques T (see figure).
(a) If the inner radius of the tube is r
1
40 mm and the measured
angle of twist between the ends is 0.5°, what is the shear strain
1
(in radians) at the inner surface?
(b) If the maximum allowable shear strain is 0.0005 rad and the angle
of twist is to be kept at 0.5° by adjusting the torque T , what is the
maximum permissible outer radius ( r
2
)
max
?

SECTION 3.2 Torsional Deformations
183
Solution 3.2-4 Circular steel tube
L 0.90 m
r
1
40 mm
(a) S
HEAR STRAIN AT INNER SURFACE
From Eq. (3-5b):
g
1
388 10
6
rad —
g
min
g
1
r
1
f
L
(40 mm)(0.008727 rad)
900 mm
g
max
0.0005 rad
0.008727 rad
f 0.5 (0.5 )
¢
180
rad
degree
≤
(b) M
AXIMUM OUTER RADIUS
From Eq. (3-5a):
( r
2
)
max
51.6 mm
—
( r
2
)
max
(0.0005 rad)(900 mm)
0.008727 rad
r
2
g
max
L
f
g
max
g
2
r
2
f
L
;
r
1
r
2
Problem 3.2-5 Solve the preceding problem if the length L 50 in.,
the inner radius r
1
1.5 in., the angle of twist is 0.6°, and the allowable
shear strain is 0.0004 rad.