Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: June 2014

Assuming supports at point A and B,

$ΣM_A = 0$

$-2000 × 2 - 1500 + V_B × 7 = 0$

$V_B = 785.714 N$

And , $ΣF_X = 0$

$V_A - 2000 + V_B = 0$

Thus, $V_A = 1214.285 N$

Using the Macaulay’s Method,

At any section distant x from A, the BM is given by,

$EI\frac{dy}{dx} = 1214.285x | -2000(x - 2) | - 1500 (x - 4)^0...(i)$

Integrating, we get

$EI\frac{dy}{dx} = 1214.285\frac{x^2}{2} + C_1 |-2000\frac{(x - 2)^2}{2} | -1500 (x - 4)^1.....(ii)$

Integrating, yet again,

$EIy = 1214.285\frac{x^3}{6} + C_1x + C_2 |-2000 \frac{(x - 2)^3}{6}| - 1500\frac{(x - 4)^2}{2}.....(iii)$

At $x = 0 , y = 0$ hence,$C_2 = 0$

At $x = 7 , y = 0$

Thus, $0 = 1214.285 \frac{7^8}{6} + 7C_1 | -2000\frac{(7 - 2)^3}{6} | -1500\frac{(7 - 4)^2}{2}$

$0 = 69416.66 + 7C_1 - 41666.67 - 6750$

$C_1 = -3000$

Thus, slope equation is,

$EI\frac{dy}{dx} = 1214.285\frac{x^2}{2} - 3000 |-2000 \frac{(x - 2)^2}{2} | -1500 (x - 4)^1$

And the deflection equation is,

$EIy = 1214.285\frac{x^3}{6} - 3000x |-2000\frac{(x - 2)^3}{6} | -1500\frac{(x - 4)^2}{2}$

Since, B is a support; deflection at B will always be 0.