A solid circular shaft of diameter d is subjected to combined bending moment M and twisting moment T. If the shaft is made of brittle material where principal normal stress governs failure, show that the maximum normal stress is given by the following equation:

$\sigma_{\max }=\frac{32 M_{ e }}{\pi d^3} \text { where } M_{ e }=\frac{1}{2}\left[M+\sqrt{M^2+T^2}\right]$

Question

A solid circular shaft of diameter d is subjected to combined bending moment M and twisting moment T. If the shaft is made of brittle material where principal normal stress governs failure, show that the maximum normal stress is given by the following equation:

$\sigma_{\max }=\frac{32 M_{ e }}{\pi d^3} \text { where } M_{ e }=\frac{1}{2}\left[M+\sqrt{M^2+T^2}\right]$

[latex] \sigma_{\max }=\frac{32 M_{ e }}{\pi d^3} ext { where } M_{ e }=\frac{1}{2}\left[M+\sqrt{M^2+T^2} ight] [/latex]

Accepted Answer

The differential element subjected to the loading has the following stresses:
Normal stress due to bending is

[latex] \sigma_n=\frac{32 M_{ e }}{\pi d^3} [/latex]

and shear stress due to torsion is

[latex] au=\frac{16 T}{\pi d^3} [/latex]

Maximum principal stress is given by

[latex] \sigma_1=\frac{\sigma_n}{2}+\sqrt{\left\lgroup \frac{\sigma_n}{2} ight group^2+ au^2} [/latex]

[latex] =\frac{16 M}{\pi d^3}+\frac{16}{\pi d^3} \sqrt{M^2+T^2} [/latex]

[latex] =\left\lgroup \frac{16}{\pi d^3} ight group\left[M+\sqrt{M^2+T^2} ight]=\frac{32}{\pi d^3} M_{ e } [/latex]

where

[latex] M_{ e }=\frac{1}{2}\left[M+\sqrt{M^2+T^2} ight] [/latex]

Q. 12.13

Chapter 12

Q. 12.13

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