An aluminum alloy is to be used for a driveshaft that transmits 40 hp at 800 rpm. The yield strength of the aluminum alloy is $\sigma_{Y}$ = 37 ksi. If a factor of safety of FS = 3.0 with respect to yielding is required, determine the smallest-diameter shaft that can be selected based on:
(a) the maximum-shear-stress theory.
(b) the maximum-distortion-energy theory.

Question

An aluminum alloy is to be used for a driveshaft that transmits 40 hp at 800 rpm. The yield strength of the aluminum alloy is [latex]\sigma_{Y}[/latex] = 37 ksi. If a factor of safety of FS = 3.0 with respect to yielding is required, determine the smallest-diameter shaft that can be selected based on:
(a) the maximum-shear-stress theory.
(b) the maximum-distortion-energy theory.

Accepted Answer

The torque in the driveshaft is:

[latex]T=\frac{P}{\omega}=\frac{(40 hp )\left(\frac{550 lb - ft / s }{1 hp } ight)}{\left(\frac{800 rev }{\min } ight)\left(\frac{2 \pi rad }{1 rev } ight)\left(\frac{1 min }{60 s } ight)}=262.606 lb - ft[/latex]

(a) Maximum-Shear-Stress Theory
For a shaft subjected to pure torsion only, the principal stresses will be equal in magnitude to the torsional shear stress. The principal stress [latex]\sigma_{p 1}[/latex] will be positive and [latex]\sigma_{p 2}[/latex] will be negative. Failure will occur when [latex]\left|\sigma_{p 1}-\sigma_{p 2} ight| \geq \sigma_{Y}[/latex] , therefore:

[latex]\left|\sigma_{p 1}-\sigma_{p 2} ight|=\left| au_{ ext {torsion }}-\left(- au_{ ext {torsion }} ight) ight|=2 au_{ ext {torsion }} \leq \sigma_{ ext {allow }}[/latex]

The allowable stress for the aluminum alloy is:

[latex]\sigma_{ ext {allow }}=\frac{\sigma_{Y}}{ FS }=\frac{37 ksi }{3}=12.33333 ksi[/latex]

Therefore, the torsion shear stress cannot exceed:

[latex]\begin{aligned}&2 au_{ ext {torsion }}=12.33333 ksi \& herefore au_{ ext {allow }}=\frac{12.33333 ksi }{2}=6.16667 ksi\end{aligned}[/latex]

The minimum diameter required for the shaft can be found from:

[latex]\begin{gathered}\frac{\pi}{16} d^{3} \geq \frac{T}{ au_{ ext {allow }}}=\frac{(262.606 lb - ft )(12 in. / ft )}{6,166.667 psi }=0.51102 in .^{3} \ herefore d \geq 1.375524 in .=1.376 in.\end{gathered}[/latex]

(b) Maximum-Distortion-Energy Theory: For pure torsion,

[latex]\sigma_{p 1}= au_{ ext {torsion }} \quad ext { and } \quad \sigma_{p 2}=- au_{ ext {torsion }}[/latex]

For a state of plane stress, the Mises equivalent stress can be expressed as:

[latex]\begin{aligned}\sigma_{M} &=\left[\sigma_{p 1}^{2}-\sigma_{p 1} \sigma_{p 2}+\sigma_{p 2}^{2} ight]^{1 / 2} \&=\left[\left( au_{ ext {torsion }} ight)^{2}-\left( au_{ ext {torsion }} ight)\left(- au_{ ext {torsion }} ight)+\left(- au_{ ext {torsion }} ight)^{2} ight]^{1 / 2} \&=\left[3 au_{ ext {torsion }}^{2} ight]^{1 / 2}\end{aligned}[/latex]

Set the Mises equivalent stress equal to the allowable stress for the aluminum alloy:

[latex]\left[3 au_{ ext {torsion }}^{2} ight]^{1 / 2}=\frac{37 ksi }{3}=12.33333 ksi[/latex]

and solve for the allowable torsional shear stress:

[latex] au_{ ext {torsion }}=\sqrt{\frac{(12.33333 ksi )^{2}}{3}}=7.12065 ksi[/latex]

Thus, the minimum diameter required to satisfy the maximum-distortion-energy theory is:

[latex]\begin{aligned}\frac{\pi}{16} d^{3} & \geq \frac{T}{ au_{ ext {allow }}}=\frac{(262.606 lb - ft )(12 in . / ft )}{7,120.65 psi }=0.44255 in .^{3} \& herefore d \geq 1.311128 in .=1.311 in .\end{aligned}[/latex]

Question 15.80: An aluminum alloy is to be used for a driveshaft that transm...

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