The ratio of Maximum Bending Stress to Maximum Shear Stress is: i) 2 ii) 4 iii) 3 iv) 1.5

The ratio of maximum bending stress to the maximum shear stress is,

Bending Equation is,

$\frac{M}{I} = \frac{ \sigma}{y}$

$y = \frac{d}{2}$

(Moment of Inertia)

$I = \frac{ \pi }{4} \times 4^4 = \frac{ \pi}{4} \times (\frac{d}{2})^4$

$ \frac{\pi}{64} \times d^4$

$\therefore$ Bending Stress

$\sigma = \frac{M}{I} \times y$

$\sigma = \frac{M}{ \frac{\pi}{64} \times d^4} \times (\frac{d}{2})$

$\sigma_{max} = \frac{M}{\frac{\pi}{32} \times d^3}$ $\rightarrow$ [1]

Torsional Equation is,

$\frac{T}{J_p} = \frac{\zeta}{r}$

$r = \frac{d}{2}$ and $T = \frac{M}{2}$

(Polar moment of inertia)

$J_p = \frac{\pi}{32} \times d^4$

$\therefore$ Shear Stress,

$\zeta_{max} = \frac{T \times r}{J_p}$

$= \frac{ \frac{M}{2} \times \frac{d}{2}}{\frac{ \pi}{32} d^4}$

$\zeta_{max} = \frac{8 \ M}{ \pi \ d^3}$ $\rightarrow$ [2]

The ratio of maximum bending stress to the maximum shear stress is,

$\therefore$ $\frac{\sigma_{max}}{\zeta_{max}}$ = $\frac{ \frac{M}{\frac{32 \ d^3 } {32}}}{\frac{8M}{\pi d^3}}$

$\frac{\sigma_{max}}{\zeta_{max}} = 4$