$Take \ \ E = 210GPa \ \ and I = 15 × 10^{-6} m^4$.

Mumbai university > MECH > SEM 3 > Strength Of Materials

Marks: 10M

Year: June 2014

$ΣM_A = 0$

$ = -(8 × 3) × 4.5 + V_B × 6 - 10 × 8 = 0$

$V_B = 31.33 kN$

And , $ΣF_X = 0$

$V_A - (8 × 3) + V_B -10 = 0$

Thus , $V_A = 2.67 kN$

Since, $ΣF_Y = 0$

Using the Macaulay’s Method,

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

$EI\frac{d^2y}{dx^2} = 2.67x |-8\frac{(x - 3)^2}{2}| + 31.33 (x - 6) + 8\frac{(x - 6)^2}{2}........(1)$

Integrating, we get

$Ei\frac{dy}{dx} = 2.67\frac{x^2}{2} + C_1 |-8 \frac{(x - 3)^3}{6}| + 31.33\frac{(x - 6)^2}{2} + 8 \frac{(x - 6)^3}{6}.....(2)$

Integrating yet again,

$EIy = 2.67\frac{x^3}{6} + C_1x +C_2 |-8\frac{(x - 3)^4}{24}| + 31.33\frac{(x - 6)^3}{2} + 8\frac{(x - 6)^4}{6}.....,(3)$

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

$At x = 6, y= 0$

Thus, $0 = 2.67\frac{6^3}{6} + 6C_1 | -8\frac{(6 - 3)^4}{24}$

$0 = 96.12 + 6C_1 - 27$

$C_1 = 11.52$

Thus,Slope equation is,

$EI\frac{dy}{dx} = 2.67\frac{x^2}{2} + 11.52 |-8\frac{(x - 3)^3}{6}| + 31.33\frac{(x - 6)^2}{2} + 8\frac{(x - 6)^3}{6}$

And the deflection equation is,

$EIy = 2.67\frac{x^3}{6} + 11.52x |-8 \frac{(x - 3)^4}{24}| + 31.33\frac{(x - 6)^3}{6} + 8\frac{(x - 6)^4}{24}$

Now, to find deflection at the free end ‘C’ for the beam

Put x=8 in eq. (3)

$EIy = 2.67\frac{8^3}{6} + 8C_1|-8\frac{(8 - 3)^4}{24}| + 31.33 \frac{(8 - 6)^3}{6} + 8 \frac{(8 - 6)^4}{24}$

$EIy=227.84 + 92.16 - 208.33 + 41.77 + 5.33$

$y = \frac{158.77}{EI} = \frac{158.77}{210 × 10^9 × 15 × 10^-6} = 5.0403 × 10^-5m$

Hence, deflection at the free end C is $0.0403 × 10^-5m , i.e 0.0504 mm(↑)$