A simply supported beam 4 m long has the cross section shown in Fig. . 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.

 

 A simply supported beam 4 m long has the cross section shown in Fig. . 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

R=12(20)(2)=20 kN$R=\frac{1}{2}(20)(2)=20\text{kN}$
 

Maximum moment will occur at the midspan:
Mmax=2R−20(1)(0.5)=2(20)−10$M_{max}=2R-20(1)(0.5)=2(20)-10$

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

Flexural stress:
fb=MyI$f_b={\displaystyle \frac{My}{I}}$
 

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

 

I=15(120)(300+10+10)312−[15(120)−250](3003)12$I={\displaystyle \frac{15(120)(300+10+10{)}^3}{12}}-{\displaystyle \frac{[15(120)-250]({300}^3)}{12}}$

I=1427700000 mm4$I=1427700000{\text{mm}}^4$

 

Maximum stress in the wood (y = 150 mm)
fbw=30(150)(10002)1427700000$f_{bw}={\displaystyle \frac{30(150)({1000}^2)}{1427700000}}$

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

Maximum stress in the steel (y = 160 mm)
fbs15=30(160)(10002)1427700000${\displaystyle \frac{f_{bs}}{15}}={\displaystyle \frac{30(160)({1000}^2)}{1427700000}}$

fbs=47.279 MPa