Question

A circular shaft of 12 cm dia. is subjected to combined bending and twisting moments. The bending moment being three times the twisting moment. If the direct tensile yield point of material is 350 MN/m² and factor of safety on yield is 4, find the allowable twisting moment by     

a) Maximum principal stress theory

b) Maximum shear stress theory

Answer 1

Solution;

Stress due to bending moment;

$\sigma_M=\frac{M\bar{y}}{I}=\frac{32M}{πd^3}$

Stress due to the twisting moment;

$\sigma_T=\frac{T\bar{y}}{J}=\frac{16T}{πd^3}$

Average stress;

$\sigma_a=\frac{\sigma_x+\sigma_y}{2}=\frac{\sigma_m+0}{2}=\frac{\sigma_M}{2}$

$R=\sqrt{(\frac{\sigma_x-\sigma_y)^2}{2}+(\tau_{xy})^2}$

$R=\sqrt{(\sigma_M/2)^2+(\sigma_T)^2}$

$\sigma_1=\sigma_a+R$

By substitution;

$\sigma_1=18161.3474T$

$\sigma_2=\sigma_a-R$

By substitution;

$\sigma_2=-477.46T$

(a)

Maximum principal stress;

$max[\sigma_1,\sigma_2]\leq\sigma_{allowable}$

Using $\sigma_1;$

$18161.3474T\leq\frac{350×10^6}{4}$

Hence,allowable twisting moment is;

$T\leq4817.92Nm$

(b)

Maximum shearing stress theory;

$\frac{\sigma_1-\sigma_2}{2}\leq\sigma_{allowable}$

$\frac{18161.3474+477.46T}{2}\leq87.5×10^6$

$T\leq9389.01Nm$