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

SOLUTION
The elastic torsion formula relates shear stress and torque:
τ =
In this instance, the torque and the allowable shear stress are known for the shaft. Putting
the known terms on the right-hand side of the equation gives
Next, express the left-hand side of this equation in terms of the shaft diameter d:
J
c
=
Tc
J
T
τ
4
( π/32)
d π
3
T
= d =
d/2 16 τ
Now solve for the minimum diameter that will satisfy the 80 MPa allowable shear stress limit:
d
3
16 T 16(20 Nm)(1,000 ⋅ mm/m)
≥ =
= 1, 455.1309 mm
2
π τ π (70 N/mm )
∴d
≥ 11.33 mm
The angle of twist in the shaft must not exceed 3° in a 500 mm length. Solving the angleof-twist
equation
φ = TL
JG
for the moment of inertia J yields
TL
J =
Gφ
Finally, express the polar moment of inertia in terms of the diameter d, and solve for the
minimum diameter that will satisfy the 3° limit:
d
4
32TL
32(20 Nm)(500 ⋅ mm)(1, 000 mm/m)
≥ =
= 24,317.084 mm
2
πGφ π(80,000 N/mm )(3)( ° πrad/180 ° )
∴d
≥ 12.49 mm
On the basis of these two calculations, the minimum diameter that is acceptable for the
shaft is d ≥ 12.49 mm.
Ans.
3
4
ExAmpLE 6.3
146
y
T
(1) (2)
A B C
16 in. 25 in.
x
A compound shaft consists of a solid aluminum segment (1) and a
hollow steel segment (2). Segment (1) is a solid 1.625 in. diameter
aluminum shaft with an allowable shear stress of 6,000 psi and a
shear modulus of 4 × 10 6 psi. Segment (2) is a hollow steel shaft
with an outside diameter of 1.25 in., a wall thickness of 0.125 in.,
an allowable shear stress of 9,000 psi, and a shear modulus of 11 ×
10 6 psi. In addition to the allowable shear stresses, specifications
require that the rotation angle at the free end of the shaft must not
exceed 2°. Determine the magnitude of the largest torque T that
may be applied to the compound shaft at C.