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Chapter 12

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]

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Verified Solution

The differential element subjected to the loading has the following stresses:
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]