# Problem 1003 | Maximum stresses in wood and steel of composite beam

**Problem 1003**

A simply supported beam 4 m long has the cross section shown in Fig. P-1002. It carries a uniformly distributed load of 20 kN/m over the middle half of the span. If n = 15, compute the maximum stresses in the wood and the steel.

**Solution 1003**

$R = \frac{1}{2}(20)(2) = 20 ~ \text{kN}$

Maximum moment will occur at the midspan:

$M_{max} = 2R - 20(1)(0.5) = 2(20) - 10$

$M_{max} = 30 ~ \text{kN}\cdot\text{m}$

Flexural stress:

$f_b = \dfrac{My}{I}$

$M = M_{max} = 30 ~ \text{kN}\cdot\text{m}$

$I = \dfrac{15(120)(300 + 10 + 10)^3}{12} - \dfrac{[ \, 15(120) - 250 \, ](300^3)}{12}$

$I = 1\,427\,700\,000 ~ \text{mm}^4$

Maximum stress in the wood (y = 150 mm)

$f_{bw} = \dfrac{30(150)(1000^2)}{1\,427\,700\,000}$

$f_{bw} = 3.152 ~ \text{MPa}$

Maximum stress in the steel (y = 160 mm)

$\dfrac{f_{bs}}{15} = \dfrac{30(160)(1000^2)}{1\,427\,700\,000}$

$f_{bs} = 47.279 ~ \text{MPa}$