Solution to Problem 317 Torsion

Problem 317
A hollow bronze shaft of 3 in. outer diameter and 2 in. inner diameter is slipped over a solid steel shaft 2 in. in diameter and of the same length as the hollow shaft. The two shafts are then fastened rigidly together at their ends. For bronze, G = 6 × 10⁶ psi, and for steel, G = 12 × 10⁶ psi. What torque can be applied to the composite shaft without exceeding a shearing stress of 8000 psi in the bronze or 12 ksi in the steel?

Solution 317

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$\theta_{st} = \theta_{br}$

$\left( \dfrac{TL}{JG} \right)_{st} = \left( \dfrac{TL}{JG} \right)_{br}$

$\dfrac{T_{st} L}{\frac{1}{32}\pi (2^4)(12 \times 10^6)} = \dfrac{T_{br} L}{\frac{1}{32}\pi (3^4 - 2^4)(6 \times 10^6)}$

$\dfrac{T_{st}}{192 \times 10^6} = \dfrac{T_{br}}{390 \times 10^6}$ → Equation (1)

Applied Torque = Resisting Torque
$T = T_{st} + T_{br}$ → Equation (2)

Equation (1) with T_st in terms of T_br and Equation (2)
$T = \dfrac{192 \times 10^6}{390 \times 10^6} T_{br} + T_{br}$

$T_{br} = 0.6701T$

Equation (1) with T_br in terms of T_st and Equation (2)
$T = T_{st} + \dfrac{390 \times 10^6}{192 \times 10^6} T_{br}$

$T_{st} = 0.3299T$

Based on hollow bronze (T_br = 0.6701T)
$\tau_{max} = \left[ \dfrac{16TD}{\pi(D^4 - d^4)} \right]_{br}$

$8000 = \dfrac{16(0.6701T)(3)}{\pi (3^4 - 2^4)}$

$T = 50 789.32 \, \text{lb}\cdot\text{in}$

$T = 4232.44 \, \text{lb}\cdot\text{ft}$

Based on steel core (T_st = 0.3299T):
$\tau_{max} = \left[ \dfrac{16TD}{\pi D^3} \right]_{st}$

$12\,000 = \dfrac{16(0.3299T)}{\pi (2^3)}$

$T = 57\,137.18 \, \text{lb}\cdot\text{in}$

$T = 4761.43 \, \text{lb}\cdot\text{ft}$

Use T = 4232.44 lb·ft. answer

Submitted by Romel Verterra on January 3, 2009 - 5:06pm

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