Solution to Problem 324 Torsion

Problem 324
The compound shaft shown in Fig. P-324 is attached to rigid supports. For the bronze segment AB, the maximum shearing stress is limited to 8000 psi and for the steel segment BC, it is limited to 12 ksi. Determine the diameters of each segment so that each material will be simultaneously stressed to its permissible limit when a torque T = 12 kip·ft is applied. For bronze, G = 6 × 10⁶ psi and for steel, G = 12 × 10⁶ psi.

Solution 324

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$\tau_{max} = \dfrac{16T}{\pi D^3}$

For bronze:

$8000 = \dfrac{16T_{br}}{\pi {D_{br}}^3}$

$T_{br} = 500\pi {D_{br}}^3 \, \text{lb}\cdot\text{in}$

For steel:

$12\,000 = \dfrac{16T_{st}}{\pi {D_{st}}^3}$

$T_{st} = 750\pi {D_{st}}^3 \, \text{lb}\cdot\text{in}$

$\Sigma M = 0$

$T_{br} + T_{st} = T$

$T_{br} + T_{st} = 12(1000)(12)$

$T_{br} + T_{st} = 144\,000 \, \text{lb}\cdot\text{in}$

$500\pi {D_{br}}^3 + 750\pi {D_{st}}^3 = 144\,000$

${D_{br}}^3 = 288 / \pi + 1.5 {D_{st}}^3$ → Equation (1)

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

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

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

$\dfrac{T_{br}}{{D_{br}}^4} = \dfrac{T_{st}}{3{D_{st}}^4}$

$\dfrac{500\pi {D_{br}}^3}{{D_{br}}^4} = \dfrac{750\pi {D_{st}}^3}{3{D_{st}}^4}$

$D_{st} = 0.5D_{br}$

From Equation (1)
${D_{br}}^3 = 288 / \pi - 1.5(0.5 D_{br})^3$

$D_{br} = 288 / \pi$

$D_{br} = 4.26 \, \text{in.}$ answer

$D_{st} = 0.5(4.26) = 2.13 \, \text{in.}$ answer

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