Question

As per the maximum principal stress theory, when a shaft is subjected to a bending moment M and torque T. And $\sigma$ is the allowable stress in axial tension, then the diameter d of the shaft is given by

• A. $d^3=\frac{16}{\pi \sigma }[M+\sqrt{M^2+T^2}]$
• B. $d^3=\frac{4}{\pi \sigma }[M+\sqrt{M^2+T^2}]$
• C. $d^3=\frac{32}{\pi \sigma }[M+\sqrt{M^2+T^2}]$
• D. $d^3=\frac{8}{\pi \sigma }[M+\sqrt{M^2+T^2}]$

Answer 1

The correct option is A $d^3=\frac{16}{\pi \sigma }[M+\sqrt{M^2+T^2}]$

Maximum stress due to bending moment

${\sigma }_1=\frac{32M}{\pi d^3}$

Maximum shear stress due to torsion T

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

Maximum normal stress due to combined action of bending moment and torque

${\sigma }_{max}=\frac{{\sigma }_1}{2}+\sqrt{{\left(\frac{{\sigma }_1}{2}\right)}^2+{\tau }^2}$

$=\frac{16}{\pi d^3}[M+\sqrt{M^2+{\tau }^2}]$

According to maximum principal stress theory, for no failure

${\sigma }_{max}\le Allowablestress$

$d^3\le \frac{16}{\pi \sigma }[M+\sqrt{(M{)}^2+{\tau }^2}]$