Strain Energy in Torsion
Problem 3.9-1A solid circular bar of steel (G11.4 106psi)with length L30 in. and diameter d1.75 in. is subjected topure torsion by torques Tacting at the ends (see figure).(a)Calculate the amount of strain energy Ustored in the barwhen the maximum shear stress is 4500 psi. (b)From the strain energy, calculate the angle of twist (in degrees).
Solution 3.9-1
Steel bar
236
CHAPTER 3
Torsion
L
d
T
T
L
d
T
T
G
11.4
10
6
psi
L
30 in.
d
1.75 in.
max
4500 psi
(Eq. 1)
I
P
d
4
32
t
max
16
T
d
3
T
d
3
t
max
16
(a) S
TRAIN ENERGY
(Eq. 2)
Substitute numerical values:
(b) A
NGLE OF TWIST
Substitute for
T
and
U
from Eqs. (1) and (2):
(Eq. 3)
Substitute numerical values:
f
0.013534
rad
0.775
—
f
2
L
t
max
Gd
U
T
f
2
f
2
U
T
U
32.0
in.-lb
—
d
2
L
t
max
2
16
G
U
T
2
L
2
GI
P
¢
d
3
t
max
16
≤
2
¢
L
2
G
≤
¢
32
d
4
≤
Problem 3.9-2A solid circular bar of copper (G45 GPa) with lengthL0.75 m and diameter d40 mm is subjected to pure torsion bytorques Tacting at the ends (see figure). (a)Calculate the amount of strain energy Ustored in the bar when themaximum shear stress is 32 MPa. (b)From the strain energy, calculate the angle of twist (in degrees)

Solution 3.9-2
Copper bar
SECTION 3.9
Strain Energy in Torsion
237
L
d
T
T
G
45 GPa
L
0.75 m
d
40 mm
max
32 MPa
(Eq. 1)
I
P
d
4
32
t
max
16
T
d
3
T
d
3
t
max
16
(
A
) S
TRAIN ENERGY
(Eq. 2)
Substitute numerical values:
(b) A
NGLE OF TWIST
Substitute for
T
and
U
from Eqs. (1) and (2):
(Eq. 3)
Substitute numerical values:
f
0.026667 rad
1.53
—
f
2
L
t
max
Gd
U
T
f
2
f
2
U
T
U
5.36
J
—
d
2
L
t
max
2
16
G
U
T
2
L
2
GI
P
¢
d
3
t
max
16
≤
2
¢
L
2
G
≤
¢
32
d
4
≤
Problem 3.9-3
A stepped shaft of solid circular cross sections
(see figure) has length
L
45 in., diameter
d
2
1.2 in., and
diameter
d
1
1.0 in. The material is brass with
G
5.6
10
6
psi.
Determine the strain energy
U
of the shaft if the angle of twist
is 3.0°.
Solution 3.9-3
Stepped shaft
d
2
d
1
T
T
L
—
2
L
—
2
d
2
d
1
T
T
L
—
2
L
—
2
d
1
1.0 in.
d
2
1.2 in.
L
45 in.
G
5.6
10
6
psi (brass)
3.0
0.0523599 rad
S
TRAIN ENERGY
(Eq. 1)
Also,
(Eq. 2)
Equate
U
from Eqs. (1) and (2) and solve for
T
:
S
UBSTITUTE NUMERICAL VALUES
:
U
22.6
in.-lb
—
U
T
f
2
G
f
2
32
L
¢
d
1
4
d
2
4
d
1
4
d
2
4
≤
f
radians
T
Gd
1
4
d
2
4
f
16
L
(
d
1
4
d
2
4
)
U
T
f
2
8
T
2
L
G
¢
1
d
2
4
1
d
1
4
≤
U
a
T
2
L
2
GI
P
16
T
2
(
L
2)
Gd
2
4
16
T
2
(
L
2)
Gd
1
4