The solid circular shaft shown in Figure P15.84 has an outside diameter of 50 mm and is made of an alloy that has ultimate failure strengths of 260 MPa in tension and 440 MPa in compression. Determine the largest permissible torque T that may be applied to the shaft based on the Mohr failure

Question

The solid circular shaft shown in Figure P15.84 has an outside diameter of 50 mm and is made of an alloy that has ultimate failure strengths of 260 MPa in tension and 440 MPa in compression. Determine the largest permissible torque T that may be applied to the shaft based on the Mohr failure criterion.

Accepted Answer

Section properties:

[latex]J=\frac{\pi}{32}(50 mm )^{4}=613,592.315 mm ^{4}[/latex]

Principal stresses:
For pure torsion,

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

Mohr failure criterion:
Since [latex]\sigma_{p 1}[/latex] is positive and [latex]\sigma_{p 2}[/latex] is negative, the largest permissible torque T will correspond to:

[latex]\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}}=1[/latex].

Substitute the torsional shear stress relations from Eq. (a) into this interaction equation:

[latex]\frac{\sigma_{p 1}}{\sigma_{U T}}-\frac{\sigma_{p 2}}{\sigma_{U C}}=\frac{ au}{\sigma_{U T}}-\frac{- au}{\sigma_{U C}}= au\left[\frac{1}{\sigma_{U T}}+\frac{1}{\sigma_{U C}} ight]=1[/latex]

Therefore, the torsional shear stress must not exceed:

[latex] au_{ ext {allow }}=\frac{1}{\left[\frac{1}{\sigma_{U T}}+\frac{1}{\sigma_{U C}} ight]}=\frac{1}{\left[\frac{1}{260 MPa }+\frac{1}{440 MPa } ight]}=163.429 MPa[/latex]

Allowable torque:

[latex]\begin{aligned}& au_{ ext {allow }} \geq \frac{T_{ ext {allow }} c}{J} \& herefore T_{ ext {allow }} \leq \frac{ au_{ ext {allow }} J}{c} \&\quad=\frac{\left(163.429 N / mm ^{2} ight)\left(613,592.315 mm ^{4} ight)}{50 mm / 2} \&=4.01114 imes 10^{6} N - mm \&=4.01 kN - m\end{aligned}[/latex]

Question 15.84: The solid circular shaft shown in Figure P15.84 has an outsi...

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