A cantilever beam of spam 5 m carries two concentrated loads one of 1kN at 3m from the fixed end and the other of 3kN at the free end. If the flexural rigidity EI is 8000 $kN.m^2$. Find the slope and deflection under each load.

Using double integration method,

$BM_x=EI\frac{d^2y}{dx^2}=-3x-1(x-2)$

Integrating,

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

First boundary condition to find $c_1$

$\text{At } x=5; \frac{dy}{dx}=0 \text{ [put in equation (1)]}$

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

$0=-37.5-4.5+c_1$

$c_1=42 \text{ [put in equation (1)]}$

$EI\frac{dy}{dx}=-\frac{3x^2}{2}-\frac{1(x-2)^2}{2}+42$----------------(A)

Integrating again,

$EIy=-\frac{3x^3}{6}-\frac{1(x-2)^3}{6}+42x+c_2$---------(2)

Second boundary condition to find $c_2$

$\text{At x=5 & y=0}$

$0=-\frac{3(5)^3}{6}-\frac{1(5-2)^3}{6}+42\times 5+c_2$

$0=-62.5-4.5+210+c_2$

$c_2=-143 \text{ [put in equation (2)]}$

$EIy=-\frac{3x^3}{6}-\frac{1(x-2)^3}{6}+42x+(-143)$

$EIy=-\frac{3x^3}{6}-\frac{1(x-2)^3}{6}+42x-143$--------------(B) [G.D.E.]

1) To find $Q_C:$

Put x=2 in (G.S.E)

$EI\frac{dy}{dx}=-\frac{3(2)^2}{2}-\frac{1(2-2)^2}{2}+42$

$EIQ_C=-6+42$

$Q_C=\frac{36}{EI}=\frac{36}{8000}=4.5\times 10^{-3} rad$

2) To find $Q_B:$

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

$EI\frac{dy}{dx}=-\frac{3(x)^2}{2}-\frac{1(x-2)^2}{2}+42$

$EIQ_B=-\frac{3(0)^2}{2}-\frac{1(0-2)^2}{2}+42$

$Q_B=\frac{42}{EI}=\frac{42}{8000}=5.25\times 10^{-3} rad$

3) To find $y_C:$

Put x=2 in (G.S.E)

$EIy=-\frac{3(x)^3}{6}-\frac{1(x-2)^3}{6}+42x-143$

$EIy_C=-\frac{3(2)^3}{6}-\frac{1(2-2)^3}{6}+42(2)-143$

$EIy_C=-4+84-143$

$EIy_C=-63$

$y_C=\frac{-63}{EI}=\frac{-63}{8000}=-7.875\times 10^{-3} m$

(-) sign indicates the downward deflection at C.

$\therefore y_C=7.875\times 10^{-3}\times 10^3mm=7.875mm$

4) To find $y_B:$

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

$EIy=-\frac{3(x)^3}{6}-\frac{1(x-2)^3}{6}+42x-143$

$EIy_B=-\frac{3(0)^3}{6}-\frac{1(0-2)^3}{6}+42(2)-143$

$y_B=\frac{-143}{EI}=\frac{-143}{8000}=-0.0178m$

$y_B=0.017875m=17.875mm$