written 6.8 years ago by | • modified 2.9 years ago |
Subject : Structural Analysis 1
Topic : Deflection of beam
Difficulty : High
written 6.8 years ago by | • modified 2.9 years ago |
Subject : Structural Analysis 1
Topic : Deflection of beam
Difficulty : High
written 6.8 years ago by | • modified 6.7 years ago |
Using the Double Integration Method,
$ BM_x=EI\frac{d^2y}{dx^2}=-w\times x\times\frac{x}{2}=-\frac{wx^2}{2}$
Integrating, $EI\frac{dy}{dx}=-\frac{wx^3}{6}+c_1$ ----- (1)
First boundary condition: at $x=L,\ \frac{dy}{dx}=0$
From equation (1)
$ 0=\frac{-wL^3}{6}+c_1\implies c_1=\frac{wL^3}{6}\\ \boxed{c_1=\frac{wL^3}{6}}\ [put\ in\ equation\ (1)]$
$ EI\frac{dy}{dx}=-\frac{wx^3}{6}+\frac{wL^3}{6}\dots\dots\dots[G.S.E]$ ----- (A)
Integrating again,
$EIy=\frac{wx^4}{24}+\frac{wL^3}{6}x+c_2$ ------ (2)
Second boundary condition at x=L, y=0
$\therefore from\ equation\ (2)\\ 0=\frac{-wL^4}{24}+\frac{wl^4}{6}+c_2\\ \therefore c_2=\frac{wL^4}{24}-\frac{wL^4}{6}\\ c_2=\frac{wL^4-4wL^4}{24}=\frac{-3wL^4}{24}\\ \boxed{c_2=\frac{-wL^4}{8}}$
$[put\ in\ equation(2)]$
$ \therefore EIy=\frac{-wL^4}{24}+\frac{wL^3}{6}.x\frac{-wL^4}{8}\dots\dots\dots[G.D.E]$----- (B)
$ Q_B=\left(\frac{dy}{dx}\right)^{}_{B}=\left[ 0+\frac{wL^3}{6}\right]\frac{1}{EI}$
Putting x=0 in equation (A)
$\boxed{Q_B=\frac{wL^3}{6EI}} $
$y_B=\left[0+0 -\frac{wL^4}{8}\right]\frac{1}{EI}\ [Putting \ x=0 \ in \ equation(B)] \\ \therefore y_B=\frac{-wL^4}{8EI}\\ \boxed{y_B=\frac{wL^4}{8EI}(\downarrow)}$