An aluminum alloy is to be used for a driveshaft that transmits 22 kW at 4 Hz. The yield strength of the aluminum alloy is $\sigma_{Y}$ = 255 MPa. 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 22 kW at 4 Hz. The yield strength of the aluminum alloy is [latex]\sigma_{Y}[/latex] = 255 MPa. 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{(22 kW )\left(\frac{1,000 N - m / s }{1 kW } ight)}{\left(\frac{4 rev }{ s } ight)\left(\frac{2 \pi rad }{1 rev } ight)}=875.352 N - m[/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}}{F S}=\frac{255 MPa }{3}=85 MPa[/latex]

Therefore, the torsion shear stress cannot exceed:

[latex]\begin{aligned}&2 au_{ ext {torsion }}=85 MPa \& herefore au_{ ext {allow }}=\frac{85 MPa }{2}=42.5 MPa\end{aligned}[/latex]

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

[latex]\begin{aligned}\frac{\pi}{16} d^{3} & \geq \frac{T}{ au_{ ext {allow }}}=\frac{(875.352 N - m )(1,000 mm / m )}{42.5 N / mm ^{2}}=20,596.518 mm ^{3} \& herefore d \geq 47.161542 mm =47.2 mm\end{aligned}[/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{255 MPa }{3}=85 MPa[/latex]

and solve for the allowable torsional shear stress:

[latex] au_{ ext {torsion }}=\sqrt{\frac{(85 MPa )^{2}}{3}}=49.07477 MPa[/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{(875.352 N - m )(1,000 mm / m )}{49.07477 N / mm ^{2}}=17,837.109 mm ^{3} \& herefore d \geq 44.953641 mm =45.0 mm\end{aligned}[/latex]

Question 15.81: An aluminum alloy is to be used for a driveshaft that transm...

Related Answered Questions