Given

d = 120 mm, L = 10 m, UDL = 1000 N/m

To Find $\sigma_b = ?$

Solution

$\frac{E}{R} = \frac{M}{I} = \frac{\sigma_b}{y}........(1)$

$M_{max} = \frac{WL^2}{8} = \frac{1000\hspace{0.05cm}\times\hspace{0.05cm}10^2}{8}\\ \hspace{02cm} = 12500\hspace{0.05cm}Nm\\ \hspace{02cm} = 12500\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nmm$

$y = \frac{d}{2} = \frac{120}{2} = 60\hspace{0.05cm}mm$

Moment of Inertia (I) = $\frac{\pi}{64}D^4 = 10.17\hspace{0.05cm}\times\hspace{0.05cm}10^6\hspace{0.05cm}mm^4$

From equation (1),

$\frac{M}{I} = \frac{\sigma_b}{y}\\ \frac{12500\hspace{0.05cm}\times\hspace{0.05cm}10^3}{10.17\hspace{0.05cm}\times\hspace{0.05cm}10^6} = \frac{\sigma_b}{60}\\ \sigma_b = 73.75\hspace{0.05cm}N/mm^2$