A rectangular beam section 300mm wide and 500mm deep is simply supported over a span of 4m.It carries full span UDL of 10 KN/m.Find max. bending stress induced in the section and draw the distribution

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written 5.6 years ago by

Given

$L = 4\hspace{0.05cm}m,\hspace{0.25cm}UDL = 10 \hspace{0.05cm}KN/m\hspace{0.05cm} = 10\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}N/m$

To Find$\hspace{0.25cm}\sigma_b = ?$

Solution

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$\frac{E}{R} = \frac{M}{I} = \frac{\sigma_b}{y}$

$M_{max} = \frac{WL^2}{8} = \frac{10\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}\times\hspace{0.05cm}4^2}{8}\\ \hspace{02cm} = 20000\hspace{0.05cm}Nm\\ \hspace{02cm} = 20000\hspace{0.05cm}\times\hspace{0.05cm}10^3\hspace{0.05cm}Nmm$

$y = \frac{d}{2} = \frac{500}{2} = 250\hspace{0.05cm}mm$

$I = \frac{bd^3}{12} = \frac{300\hspace{0.05cm}\times\hspace{0.05cm}500^3}{12} = 31.25\hspace{0.05cm}\times\hspace{0.05cm}10^8\hspace{0.05cm}mm^4$

$\frac{M}{I} = \frac{\sigma_b}{y}\\ \frac{20000\hspace{0.05cm}\times\hspace{0.05cm}10^3}{31.25\hspace{0.05cm}\times\hspace{0.05cm}10^8} = …

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