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Question: A cantilever beam of length L is subjected to a UDL of w/unit run. Determine the slope & deflection at the free end. Use Double Integration Method
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Subject : Structural Analysis 1

Topic : Deflection of beam

Difficulty : High

sa1(74) • 193 views
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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 220 written 8 months ago by gravatar for Gyanendra pal Gyanendra pal10
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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}} $

$y_B=\left[0+0 -\frac{wL^4}{8}\right]\frac{1}{EI}\ [Putting \ x=0 \ in \ equation(B)] \\ \therefore y_B=\frac{-wL^4}{8EI}\\ \boxed{y_B=\frac{wL^4}{8EI}(\downarrow)}$

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modified 7 months ago by gravatar for Sanket Shingote Sanket Shingote ♦♦ 220 written 8 months ago by gravatar for Gyanendra pal Gyanendra pal10
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