316 Permissible Torque That Can Be Applied to a Compound Shaft

Problem 316
A compound shaft consisting of a steel segment and an aluminum segment is acted upon by two torques as shown in Fig. P-316. Determine the maximum permissible value of T subject to the following conditions: τ_st ≤ 83 MPa, τ_al ≤ 55 MPa, and the angle of rotation of the free end is limited to 6°. For steel, G = 83 GPa and for aluminum, G = 28 GPa.

Solution 316

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Based on maximum shearing stress, τ_max = 16T / πd³:

Steel
$\tau_{st} = \dfrac{16(3T)}{\pi(50^3)} = 83$

$T = 679\,042.16 \, \text{N}\cdot\text{mm}$

$T = 679.04 \, \text{N}\cdot\text{m}$

Aluminum
$\tau_{al} = \dfrac{16T}{\pi (40^3)} = 55$

$T = 691\,150.38 \, \text{N}\cdot\text{mm}$

$T = 691.15 \, \text{N}\cdot\text{m}$

Based on maximum angle of twist, θ_max = 6°:

$\theta = \left( \dfrac{TL}{JG} \right)_{st} + \left( \dfrac{TL}{JG} \right)_{al}$

$6^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{3T(900)}{\frac{1}{32}\pi (50^4)(83\,000)} + \dfrac{T(600)}{\frac{1}{32}\pi (40^4)(28\,000)}$

$T = 757\,316.32 \, \text{N}\cdot\text{mm}$

$T = 757.32 \, \text{N}\cdot\text{m}$

Use the least value of T. Thus, T = 679.04 N·m answer

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