Question

Determine the maximum uniformly distributed load, a simply supported 40cm deep and 20cm wide timber beam can carry over a span of 4m. The maximum permissible bending stress for the timber is 2N/mm

Answer 1

$I_x=\frac{bh^3}{12}=\frac{0.2\cdot0.4^3}{12}=0.00107(m^4)$

$\sigma_{max}=\frac{M_{max}c}{I_x}\to M_{max}=\sigma_{max}I_x/c=$

$=2\cdot10^6\cdot0.00107/0.2=10700(N\cdot m)$

$M=2qx-qx^2/2$

For $x=2(m)$ $M=M_{max}$

$2q\cdot2-q\cdot2^2/2=10700\to q=5350(N/m)$. Answer