A simply supported beam of 5 m span carries a triangular load of 30 kN. Find out the maximum bending moment.

This question was previously asked in

PGCIL DT Civil 2018 Official Paper

Option 2 : 25 kN.m

**Concept:**

Since the load is symmetrical, therefore the reactions R_{A} and R_{B} will be equal

\({R_A} = {R_B} = \frac{1}{2} × w × \frac{l}{2} = \frac{{wl}}{4}\)

R_{A} = R_{B} = W/2 (where W = Total load = wl/2)

The shear force at any section X at a distance x from B,

\({F_x} = - {R_B} + \frac{{w{x^2}}}{l} = \frac{{w{x^2}}}{l} - \frac{{wl}}{4} = \frac{{w{x^2}}}{l} - \frac{W}{2}\)

Thus we see that shear force is equal to -W/2 at B, where x = 0 and increases in the form of a parabolic curve to zero at C i.e., midpoint of the span, beyond which it continues to increase to +W/2 at a where x = l.

The bending moment at any section X at a distance x from B,

\({M_x} = {R_B}x - \frac{{wx}}{{\frac{l}{2}}} × \frac{x}{2} × \frac{x}{3} = \frac{{wlx}}{4} - \frac{{w{x^3}}}{{3l}}\)

Thus we see that the bending moment at A and B is zero and increase in the form of a cubic curve, at C, i.e., mid point of the beam, where the bending moment will be maximum because shear forces change sign.

\({M_{\max }} = \frac{{wl}}{4} × \frac{l}{2} - \frac{w}{{3l}} × {\left( {\frac{l}{2}} \right)^3}\)

\({M_{\max }} = \frac{{w{l^2}}}{{12}} = \frac{{Wl}}{6}\)

**Calculation:**

Given,

W = 30 kN, l = 5 m

Maximum bending moment is given by,

M_{max} = Wl/6 = (30 × 5)/6 = **25 kN-m**