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## Q. 12.13

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:

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

## Verified Solution

Normal stress due to bending is

$\sigma_n=\frac{32 M_{ e }}{\pi d^3}$

and shear stress due to torsion is

$\tau=\frac{16 T}{\pi d^3}$

Maximum principal stress is given by

$\sigma_1=\frac{\sigma_n}{2}+\sqrt{\left\lgroup \frac{\sigma_n}{2} \right\rgroup^2+\tau^2}$

$=\frac{16 M}{\pi d^3}+\frac{16}{\pi d^3} \sqrt{M^2+T^2}$

$=\left\lgroup \frac{16}{\pi d^3} \right\rgroup\left[M+\sqrt{M^2+T^2}\right]=\frac{32}{\pi d^3} M_{ e }$

where

$M_{ e }=\frac{1}{2}\left[M+\sqrt{M^2+T^2}\right]$