In a simply supported beam of span L carrying UDL 'w' over the entire span, the maximum moment will be:
wL216\rm \frac{w L^{2}}{16}16wL2
wL24\rm \frac{w L^{2}}{4}4wL2
wL22\rm \frac{w L^{2}}{2}2wL2
wL28\rm \frac{w L^{2}}{8}8wL2
(D) wL28\rm \frac{w L^{2}}{8}8wL2
Explanation:
Concept: Net weight of the UDL = wL = W RA + RB = wL Due to symmetry, RA=wL2;RB=wL2{RA} = \frac{{wL}}{2};{RB} = \frac{{wL}}{2}RA=2wL;RB=2wL Taking a moment about B M=RAx−wx22=wL2x−wx22M = {RA}x - \frac{{w{x^2}}}{{2}} = \frac{{wL}}{2}x - \frac{{w{x^2}}}{2}M=RAx−2wx2=2wLx−2wx2 For maximum B.M, dMdx=0 i.e. wL2−w.2x2=0\frac{{dM}}{{dx}} = 0\;i.e.\;\frac{{wL}}{2} - \frac{{w.2x}}{2} = 0dxdM=0i.e.2wL−2w.2x=0 x=L2x = \frac{L}{2}x=2L Mmax=wL2×L2−wL28=wL28{M{max}} = \frac{{wL}}{2} \times \frac{L}{2} - \frac{{w{L^2}}}{8} = \frac{{w{L^2}}}{8}Mmax=2wL×2L−8wL2=8wL2
