Subject : Structural Analysis 1

Topic : Deflection of Beam

Difficulty : High

1. Calculation support reaction:

$ \sum M_A=0(\circlearrowright+ve)\\ 160+120\times4-V_B\times8=0\\ \boxed{V_B=80\ kN}$

$ \sum F_Y=(\uparrow+ve)\\ V_A-120+40\\ \boxed{V_A=40\ kN}$

2. Using Macaulay's Method:

Let, Taking A as origin and using Macaulay's Method, the bending moment at any section x at a distance x from A.

$ BM_x=EI\frac{d^2y}{dx^2}=80x+160(x-3)^0-15x\frac{x}{2}$

$ = 80x+160(x-3)^0-15\frac{x^2}{2}$ --- (1)

Integrating the above equation

$EI\frac{dy}{dx}=80\frac{x^2}{2}+160(x-3)^1-15\frac{x^3}{6}+c_1$ ---- (2)

Integrating once again,

$EIy=80\frac{x^3}{6}+160\frac{(x-3)^2}{2}-\frac{15x^4}{24}+c_1x+c_2$ ----- (3)

3. To find $c_1$ & $c_2$

Boundary condition, we know that when x=0, then y=0 putting in equation (3) we get,

$ 0=0+0-0+0+c_2\\ \boxed{c_2=0}$

when x=8, y=0 putting in equation (3),

$ 0=80\frac{8^3}{6}+160\frac{(8-3)^2}{2}-15\frac{8^4}{24}+c_1\times8+0\\ 0=8c_1+6266.67\\ \boxed{c_1=-783.33}$

Substituting the value of $c_1$ & $c_2$ in the equations (2) & (3)

$ EI\frac{dy}{dx}=80\frac{x^2}{2}+160(x-3)-15\frac{x^3}{6}+783.33\dots\dots\dots[G.S.E]$ --- (A)

$ EIy=80\frac{x^3}{6}+160\frac{(x-3)^2}{2}-15\frac{x^4}{24}+783.33x\dots\dots\dots[G.D.E]$ ----- (B)

4. To find $y_C$

Put $x=3m$ in equation (B)

$EIy=80\times\frac{3^3}{6}+\frac{160(3-3)^2}{2}-15\times\frac{3^4}{24}+783.33\times3\\ = 80\times\frac{3^3}{6}-\frac{15(3)^4}{24}+783.33\times3\\ y_c=\frac{2659.365}{EI}=\frac{2659.365}{40\times10^3}$

$ \boxed{y_c=0.06648\ rad}\underline{Ans}$