Question

A beam of rectangular cross-section is subjected to a vertical shear force of 100 kN the shear force carried by middle $\frac{1}{3}$rd portion of the cross section is_____kN.

• A. 48.14

Answer 1

The correct option is A 48.14

$\tau =\frac{V\left(A\overline{y}\right)}{Ib}$

$=\frac{V}{2I}\left[\right(d/2{)}^2-y^2]$ (For rectangular)

Shear force in middle third portion

$=2{\int }_0^{d/6}\tau \times bdy$

$=2{\int }_0^{d/6}\frac{V}{2I}\left[\right(d/2{)}^2-y^2]bdy$

$=\frac{Vb}{I}{[\frac{d^2}{4}y-\frac{y^3}{3}]}_0^{d/6}$

$=\frac{Vb}{I}[\frac{d^2}{4}\times \frac{d}{6}-\frac{1}{3}\left(\frac{d^3}{216}\right)]$

$=\frac{Vb}{I}\left[\frac{27-1}{648}\right]d^3$

$=12\frac{Vb}{b{d}^3}\times \frac{26}{648}{d}^3$

$=\frac{13}{27}V$

$=\frac{13}{27}\times 100kN$

$=48.14kN$