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

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

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:
$f_b = \dfrac{My}{I}$
 

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

 

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

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

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

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