Question

A beam of overall length l with equal overhangs on both sides, carries a uniformly distributed load over the entire length. To have numerically equal bending moments at the centre of the beam and at its supports, the distance between the supports should be

• A. 0.277 l
• B. 0.403 l
• C. 0.586 l
• D. 0.707 l

Answer 1

The correct option is C 0.586 l


$BMatmidspan=\frac{wL}{2}(\frac{L}{2}-a)-\frac{w{L}^2}{8}\left(sagging\right)$

$BMatsupports=-\frac{w{a}^2}{2}\left(hogging\right)$

For numerically equal neglecting the negative sign of hogging BM.

$\therefore \frac{w{a}^2}{2}=\frac{wL}{2}(\frac{L}{2}-a)-\frac{w{L}^2}{8}$

$\Rightarrow a^2=\frac{L^2}{2}-La-\frac{L^2}{2}$

$\Rightarrow a^2+La-\frac{L^2}{4}=0$

$\therefore a=0.2071L$

$\therefore$ Distance between supports = L - 2a
= 0.586 L