Question

A beam of rectangular cross-section is subjected to a vertical shear force V. The shear force carried by upper one-fourth of the cross-section is __________V.

• A. 0.156

Answer 1

The correct option is A 0.156

$\tau =\frac{SA\overline{y}}{Ib}$

$=\frac{V\times (\frac{d}{2}-y)\times b\times \left[\left(\frac{\frac{d}{2}+y}{2}\right)\right]}{Ib}$

$=\frac{V}{2I}\times (\frac{d^2}{4}-y^2)$

$dF=\tau \times bdy=\frac{V\times (\frac{d^2}{4}-y^2)}{2I}\times bdy$


$F=\int dF=\frac{Vb}{2l}\underset{d/4}{\overset{d/2}{\int }}(\frac{d^2}{4}-y^2)dy$

$=\frac{Vb}{2I}{[\frac{d^2}{4}y-\frac{y^3}{3}]}_{d/4}^{d/2}$

$=\frac{Vb}{2I}[\frac{d^2}{4}\left(\frac{d}{4}\right)-\frac{1}{3}(\frac{d^3}{8}-\frac{d^3}{64})]$

$=\frac{Vb}{I}\left(0.0130\right)d^3=\frac{Vb}{\left(\frac{b{d}^3}{12}\right)}\times \left(0.0130\right)d^3$
$=0.156V$