Question

A uniform beam of span L carries an uniformly distributed load 'w' per unit length as shown in the figure given below. The supports are at a distance of 'x' from either end. What is the condition for the maximum bending moment in the beam to be as small as possible?

• A. x = 0.107 L
• B. x = 0.207 L
• C. x = 0.237 L
• D. x = 0.25 L

Answer 1

The correct option is B x = 0.207 L

Maximum positive BM occurs at the centre of beam and it is given as
$\frac{W}{8}(L^2-4xL)$

Maximum negative BM occurs at the supports and it is given as

$=\frac{W{x}^2}{2}$

For the maximum bending moment in the beam to be as small as possible, the maximum positive and negative BM should be equal i.e.

$\frac{W}{8}(L^2-4xL)=\frac{W{x}^2}{2}$

$\Rightarrow 4x^2+4xL-L^2=0$

$\Rightarrow x=\frac{\sqrt{2}L-L}{2}and\frac{-\sqrt{2}L-L}{2}$

$\Rightarrow x=0.207Land-1.207L$