Question

A solid circular shaft subjected to a torque T procduces maximum shear stress $f_s$, which is the maximum principal value in the material. The corresponding diameter of the shaft should be

• A. $3\sqrt{\frac{\pi .f_s}{16.T}}$
• B. $3\sqrt{\frac{32.T}{\pi .f_s}}$
• C. $3\sqrt{\frac{\pi }{16.T.f_s}}$
• D. $3\sqrt{\frac{32.T}{\pi .f_s}}$

Answer 1

The correct option is D $3\sqrt{\frac{32.T}{\pi .f_s}}$

From torsional formula, we have

$\frac{{\tau }_{max}}{R}=\frac{T}{I_p}$

$\Rightarrow \frac{{\tau }_{max}}{D/2}=\frac{T}{\frac{\pi }{32}\times D^4}$

$\Rightarrow {\tau }_{max}=\frac{16T}{\pi D^3}[\because {\tau }_{max}=f_s]$

$\Rightarrow D^3=\frac{16T}{\pi f_s}$

$\Rightarrow D=3\sqrt{\frac{16T}{\pi f_s}}$