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 $c_2$

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

$0=-\frac{20(2)^3}{6}-\frac{5(2-1)^4}{12}+41.67\times 2+c_2$

$0=-26.67-0.4167+83.34+c_2$

$c_2=-56.25 \text{ [put in equation (2)]}$

$EIy=-\frac{20x^3}{6}-\frac{5(x-1)^4}{12}+41.67x+(-56.25)$

$EIy=-\frac{20x^3}{6}-\frac{5(x-1)^4}{12}+41.67x-56.25$--------------(B) [G.D.E.]

To find maximum slope $(Q_{max})$ and maximum deflection $(y_{max})$ : Maximum slope and deflection will occur at B

1) To find $Q_B:$

Put x=0 in equation (G.S.E)

$EI\frac{dy}{dx}=-\frac{20(x)^2}{2}-\frac{5(x-1)^3}{3}+41.67$

$EI\frac{dy}{dx}=-\frac{20(0)^2}{2}-\frac{5(0-1)^3}{3}+41.67$

$EIQ_B=0-0+41.67$

$Q_B=\frac{41.67}{EI}=\frac{41.67}{20}=208 rad$

$EI=2\times 10^7kNmm^2$

$=(2\times 10^7)\times(10^{-3}kNm^2$

$=2\times 10=20kNm^2$

2) To find $y_B:$

Put x=0 in equation (G.D.E)

$y_B=\frac{-56.25}{EI}=\frac{-56.25}{20}=-2.8125m$

$y_B=-2.8125m=-2812.5mm$