Find each internal torque at the solid inner shaft and the outer tube. (FBD)
2
BC
BA
T
T
kip i
=
=
⋅
n
max
4
2(0.5)
(
)
10.2
ksi
=
(0.5)
2
BC
BC
T
c
J
τ
π
=
=
max
4
4
2(1)
(
)
(1
0.75 )
2
BA
BA
T
c
J
τ
π
=
=
−
1.86
ksi
=
Angle of Twist:
4
4
3
2(12)
0.002032
(1
0.75 )11(10 )
2
BA
BA
B
T
L
rad
JG
φ
π
=
=
=
−
/
4
3
2(24)
0.044448
(0.5 )11(10 )
2
BC
BC
C
B
T
L
rad
JG
φ
π
=
=
=
/
0.002032
0.044448
0.04648
2.66
rad
C
B
C
B
φ
φ
φ
=
+
=
+
=
=
°

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Express the radius as a function of x
( )
r
rL
r
r x
r
x
x
L
L
+
=
+
=
Then, the polar moment of inertia is
4
4
4
4
( )
(
)
2
2
rL
rx
r
J x
L
x
L
L
π
π
+
⎛
⎞
=
=
⎜
⎟
⎝
⎠
+
Angle of Twist:
4
4
4
4
4
3
0
0
0
2
2
1
( )
(
)
3(
)
12
L
L
L
Tdx
TL
dx
TL
TL
4
7
J x G
r G
L
x
r G
L
x
r G
φ
π
π
⎛
⎞
=
=
=
−
=
⎜
⎟
+
+
⎝
⎠
∫
∫
π