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

ExAmpLE 6.1
A hollow circular steel shaft with an outside diameter of 1.50 in. and a wall
thickness of 0.125 in. is subjected to a pure torque of 140 lb · ft. The shaft is
90 in. long. The shear modulus of the steel is G = 12,000 ksi. Determine
(a) the maximum shear stress in the shaft.
(b) the magnitude of the angle of twist in the shaft.
Plan the Solution
The elastic torsion formula [Equation (6.5)] will be used to compute the maximum shear
stress, and the angle-of-twist equation [Equation (6.12)] will be used to determine the
angle of twist in the hollow shaft.
A
y
90 in.
140 lb.ft
B
x
SOLUTION
The polar moment of inertia J for the hollow shaft will be required for these calculations.
The shaft has an outside diameter D = 1.50 in. and a wall thickness t = 0.125 in. The inside
diameter of the shaft is d = D − 2t = 1.50 in. − 2(0.125 in.) = 1.25 in. The polar moment
of inertia for the hollow shaft is
π
J D d
32 [ ] π
= − = [(1.50 in.) − (1.25 in.) ] = 0.257325 in.
32
4 4 4 4 4
(a) The maximum shear stress is computed from the elastic torsion formula:
Tc
τ = =
J
(140 lb⋅ft)(1.50 in./2)(12in./ft)
0.257325 in.
4
= 4,896.5 psi = 4,900 psi Ans.
(b) The magnitude of the angle of twist in the 90 in. long shaft is
TL
φ = =
JG
(140 lb⋅ft)(90 in.)(12in./ft)
4 2
(0.257325 in. )(12,000,000 lb/in. )
= 0.0490 rad Ans.
ExAmpLE 6.2
A 500 mm long solid steel [G = 80 GPa] shaft is being designed to transmit a
torque T = 20 N · m. The maximum shear stress in the shaft must not exceed
70 MPa, and the angle of twist must not exceed 3° in the 500 mm length. Determine
the minimum diameter d required for the shaft.
T
T
Plan the Solution
The elastic torsion formula [Equation (6.5)] and the angle-of-twist equation [Equation (6.12)]
will be algebraically manipulated to solve for the minimum diameter required to satisfy
each consideration. The larger of the two diameters will dictate the minimum diameter d
that can be used for the shaft.
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