Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Plan the Solution
To determine the largest torque T that can be applied at C, we must
consider the maximum shear stresses and the angles of twist in both
shaft segments.
SOLUTION
The internal torques acting in segments (1) and (2) can be easily
determined from free-body diagrams cut through each segment.
Cut a free-body diagram through segment (2), and include
the free end of the shaft. A positive internal torque T 2 is assumed
to act in segment (2). The following equilibrium equation is
obtained:
Σ M = T − T = 0 ∴ T = T
x 2 2
Repeat the process with a free-body diagram cut through segment
(1) that includes the free end of the shaft. From this free-body diagram,
a similar equilibrium equation is obtained:
Σ M = T − T = 0 ∴ T = T
x 1 1
y
T
(1) (2)
T 2
A B C
16 in. 25 in.
y
T
(1) (2)
T 1
A B
C
16 in. 25 in.
x
x
Therefore, the internal torque in both segments of the shaft is equal to the external torque
applied at C.
Shear Stress
In this compound shaft, the diameters and allowable shear stresses in segments (1) and (2)
are known. Thus, the elastic torsion formula can be solved for the allowable torque that
may be applied to each segment:
T
1
τ1J1
τ J
= T2
=
c
c
1
Segment (1) is a solid 1.625 in. diameter aluminum shaft. The polar moment of inertia for
this segment is
J
1
2 2
π
= (1.625 in.) = 0.684563 in.
32
4 4
Use this value along with the 6,000 psi allowable shear stress to determine the allowable
torque T 1 :
2
T
1
4
τ1J1
(6,000 psi)(0.684563 in. )
≤ = = 5,055.2 lb⋅ in.
(a)
c
(1.625 in./2)
1
Segment (2) is a hollow steel shaft with an outside diameter D = 1.25 in. and a wall thickness
t = 0.125 in. The inside diameter of this segment is d = D − 2t = 1.25 in. − 2(0.125 in.) =
1.00 in. The polar moment of inertia for segment (2) is
J
2
π
= [(1.25 in.) − (1.00 in.) ] = 0.141510 in.
32
4 4 4
Use this value along with the 9,000 psi allowable shear stress to determine the allowable
torque T 2 :
T
2
4
τ 2J2
(9,000 psi)(0.141510 in. )
≤ = = 2,037.7 lb⋅ in.
(b)
c
(1.25 in./2)
2
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