A cantilever of 2m long carries a udl of 10 kN/m over 1m portion from fixed end a point load of 20 kN at free end. Calculate the maximum slope and deflection of the cantilever.Take $EI=2\times10^7kNmm^2$

Using double integration method,

$BM_x=EI\frac{d^2y}{dx^2}=-20x-5(x-1)\frac{(x-1)}{2}$ $=-20x-5(x-1)^2$

Integrating,

$EI\frac{dy}{dx}=\frac{-20x^2}{2}-\frac{5(x-1)^3}{3}+c_1$---------(1)

First boundary condition to find $c_1$

$\text{At } x=2; \frac{dy}{dx}=0 \text{ [put in equation (1)]}$

$0=-\frac{20(2)^2}{2}-\frac{5(2-1)^3}{3}+c_1$

$0=-40-1.67+c_1$

$c_1=41.67 \text{ [put in equation (1)]}$

$EI\frac{dy}{dx}=-\frac{20x^2}{2}-\frac{5(x-1)^3}{3}+41.67$----------------(A) [G.S.E]

Integrating again,

$EIy=-\frac{20x^3}{6}-\frac{5(x-1)^4}{12}+41.67x+c_2$---------(2)

Second boundary condition to find $c_2$

$\text{At x=2 & y=0} \text{ [put in equation …