Question

A solid circular shaft needs to be designed to transmit a torque of $50N.m$. If the allowable shear stress of the material is $140MPa$, assuming a factor of safety of $2$, minimum allowable design diameter in $mm$ is

• A. $8$
• B. $34$
• C. $24$
• D. $16$

Answer 1

The correct option is D $16$

$T=50Nm$
${\tau }_{\text{allowable /permissible}}=140MPa$
$\therefore {\tau }_{per}=\frac{16T}{\pi d^3}$
$d\ge {\left(\frac{16T}{\pi {\tau }_{per}}\right)}^{1/3}\ge {\left(\frac{16\times 50}{\pi \times 140\times {10}^6}\right)}^{1/3}$
$d\ge 0.0122m$
$d\ge 12.2mm$
Hence from option, $d=16mm$