Question

A beam AD of span 9m of constant cross section is shown in figure

$\text{If EI = 10000}kN-m^2\text{, the vertical deflection of point 'D' is (in mm, upto two decimal places)}$

• A. 11.62

Answer 1

The correct option is A 11.62
${\Delta }_D=\frac{8\times 7^4}{8EI}+\frac{8\times 7^3}{6EI}\times 2-\frac{14.69\times 3^3}{3EI}-\frac{14.69\times 3^2}{2EI}\times 6-\frac{46.10\times 7^3}{3EI}-\frac{46.10\times 7^2}{2EI}\times 2+\frac{20\times 9^3}{3EI}$

$\Rightarrow {\Delta }_D=\frac{2401+914.67-132.21-396.63-5270.77-2258.9+4860}{EI}$

${\Delta }_D=\frac{117.16}{10000}=0.0117m$

${\Delta }_D=11.72mm$