A uniformly distributed load of 300 lb/ft (including the weight of the beam) is simply supported on a 20-ft span. The cross section of the beam is described in Problem 1005. If n = 20, determine the maximum stresses produced in the wood and the steel.

 A uniformly distributed load of 300 lb/ft (including the weight of the beam) is simply supported on a 20-ft span. The cross section of the beam is described in Problem 1005. If n = 20, determine the maximum stresses produced in the wood and the steel.

Solution

 

Maximum moment will occur at the midspan
Mmax=woL28=300(202)8$M_{max}={\displaystyle \frac{w_o{L}^2}{8}}={\displaystyle \frac{300({20}^2)}{8}}$

Mmax=15000 lb⋅ft$M_{max}=15000\text{lb}\cdot \text{ft}$
 

From Solution 105:

y¯=7.1 in$\overline{y}=7.1\text{in}$

INA=34873 in4$I_{NA}=\frac{3487}{3}{\text{in}}^4$

 

fb=McI$f_b={\displaystyle \frac{Mc}{I}}$
 

At the extreme wood fiber (c=y¯$c=\overline{y}$)
fbw=Mmaxy¯INA$f_{bw}={\displaystyle \frac{M_{max}\overline{y}}{I_{NA}}}$

fbw=15000(7.1)(12)34873$f_{bw}={\displaystyle \frac{15000(7.1)(12)}{\frac{3487}{3}}}$

fbw=1099.51 psi$f_{bw}=1099.51\text{psi}$           answer
 

At the extreme steel fiber (c=10.5−y¯$c=10.5-\overline{y}$)
fbsn=Mmax(10.5−y¯)INA${\displaystyle \frac{f_{bs}}{n}}={\displaystyle \frac{M_{max}(10.5-\overline{y})}{I_{NA}}}$

fbs20=15000(10.5−7.1)(12)34873${\displaystyle \frac{f_{bs}}{20}}={\displaystyle \frac{15000(10.5-7.1)(12)}{\frac{3487}{3}}}$

fbs=10530.54 psi$f_{bs}=10530.54\text{psi}$           answer