[Solved] Maximum bending moment for simply supported beam with udl ov (2024)

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Detailed Solution

Concept:

Net weight of the UDL = wL = W

R_A+ R_B= wL

Due to symmetry,

\({R_A} = \frac{{wL}}{2};{R_B} = \frac{{wL}}{2}\)

Taking a moment about B

\(M = {R_A}x - \frac{{w{x^2}}}{{2}} = \frac{{wL}}{2}x - \frac{{w{x^2}}}{2}\)

For maximum B.M,

\(\frac{{dM}}{{dx}} = 0\;i.e.\;\frac{{wL}}{2} - \frac{{w.2x}}{2} = 0\)

\(x = \frac{L}{2}\)

\({M_{max}} = \frac{{wL}}{2} \times \frac{L}{2} - \frac{{w{L^2}}}{8} = \frac{{w{L^2}}}{8}\)

if W = wL⇒ \(M_{max}=\frac{WL}{8}\)

Mistake Points

• Please, read the question carefully. W is not the weight of the beam per unit length it is the weight of the complete beam.
• So if you will consider the load of the beam uniformly distributed throughout its length at intensity w per unit length then the maximum deflection of the beam will be (wl^2 /8).
• Now total weight (W) = w .l hence put (w = W/l) in the maximum bending moment formula you will get (Wl/8). This was the trick in question W mentioned here is not load intensity it's total load of the beam. keep focused on the keyword used in any question.

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