Derivation of Deflection from EulerBernoulli Beam Equationby bill nye scienceguy! Tags: beam, deflection, derivation, equation, eulerbernoulli 

#1
Feb1610, 11:03 AM

P: 127

1. The problem statement, all variables and given/known data
I want to derive a formula for deflection w(x) from the EulerBernoulli beam equation. It's essentially only four integrations but I'm not sure about my boundary conditions, particularly wrt shear. The beam is a cantilever with a point load at the unsupported end. And apologies in advance for the clumsy latex... 2. Relevant equations P = load w = deflection = 0 when x =0 [tex]\frac{dw}{dx}[/tex]= slope = 0 when x =0 EI[tex]\frac{d^{2}w}{dx^{2}}[/tex] = bending moment = 0 when x = L EI[tex]\frac{d^{3}w}{dx^{}3}[/tex] = shear force = 0 when? I guess my question is: what boundary condition do I need to get rid of the C[tex]_{1}[/tex] after the first integration and I suppose if this is the right way to go about this at all! 3. The attempt at a solution Here's what I've done so far: EI[tex]\frac{d^{4}w}{dx^{4}}[/tex]=P EIEI[tex]\frac{d^{3}w}{dx^{3}}[/tex]=Px + C[tex]_{1}[/tex] I've left C[tex]_{1}[/tex] here and carried it through since I don't have a clue about the shear BC. EI[tex]\frac{d^{2}w}{dx^{2}}[/tex]=P[tex]\frac{x^{2}}{2}[/tex] + C[tex]_{1}[/tex]x +C[tex]_{2}[/tex] EI[tex]\frac{d^{2}w}{dx^{2}}[/tex] = bending moment = 0 when x = L, so C[tex]_{2}[/tex]=[tex]\frac{PL^{2}}{2}[/tex]C[tex]_{1}[/tex]L EI[tex]\frac{d^{2}w}{dx^{2}}[/tex]=P[tex]\frac{x^{2}}{2}[/tex] + C[tex]_{1}[/tex]x  [tex]\frac{PL^{2}}{2}[/tex]C[tex]_{1}[/tex]L EI[tex]\frac{dw}{dx}[/tex]=[tex]\frac{Px^{3}}{6}[/tex]+C[tex]_{1}[/tex][tex]\frac{x^{2}}{2}[/tex][[tex]\frac{PL^{2}}{2}[/tex]C[tex]_{1}[/tex]L]x + C[tex]_{3}[/tex] [tex]\frac{dw}{dx}[/tex] = 0 when x = 0 so C[tex]_{3}[/tex]=0 and finally EIw=[tex]\frac{Px^{4}}{24}[/tex]+[tex]\frac{C_{1}x^{3}}{6}[/tex][[tex]\frac{PL^{2}}{2}[/tex]C[tex]_{1}[/tex]L][tex]\frac{x^{2}}{2}[/tex] + C[tex]_{4}[/tex] w=0 when x=0 so C[tex]_{4}[/tex]=0 so: EIw=[tex]\frac{Px^{4}}{24}[/tex]+[tex]\frac{C_{1}x^{3}}{6}[/tex][[tex]\frac{PL^{2}}{2}[/tex]C[tex]_{1}[/tex]L][tex]\frac{x^{2}}{2}[/tex] 



#2
Feb1610, 11:32 AM

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P: 2,532

Why did you set [itex]EI(d^4 w/dx^4)[/itex] equal to P? The units don't match up, for one. Isn't P a force? (Know what I'm getting at? )




#3
Feb1610, 02:48 PM

P: 692

Why has your first eqn got EI in it? What is your definition of shear force? That should tell you the constant you are looking for. It doesn't have to be zero.




#4
Feb1710, 08:09 AM

P: 127

Derivation of Deflection from EulerBernoulli Beam Equation
I don't have a good text on beam theory to hand so this is all coming from a mixture of wikipedia and efunda. So from the statement on the wiki page (http://en.wikipedia.org/wiki/Euler%E..._beam_equation) that:
[tex]\frac{d^{2}}{dx}[/tex](EI[tex]\frac{d^{2}u}{dx^{2}}[/tex])=P [where I've called deflection u and load P] I was trying to get to the statement of deflection as a function of length, distance from supported end, load and constant EI from efunda (http://www.efunda.com/formulae/solid...ing=cantilever) w(x)=[tex]\frac{Px^{2}}{6EI}[/tex][tex]\left(3Lx\right)[/tex] I've looked at the units in the first statement and I'm not sure what the function [tex]\frac{d^{4}u}{dx^{4}}[/tex] actually is? Apart from being the fourth derivative of displacement, but what is it physically? 



#5
Feb1710, 08:16 AM

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P: 2,532

I see [tex]\frac{d^{2}}{dx^2}\left(EI\frac{d^{2}u}{dx^{2}}\right)=w(x)[/tex] for a distributed load and [tex]\frac{d}{dx}\left(EI\frac{d^{2}u}{dx^{2}}\right)=F\,(\mathrm{or~}P)[/tex] for a point load. 



#6
Feb1710, 10:08 AM

P: 127

1st integration EI([tex]\frac{d^{2}u}{dx^{2}}[/tex])=Px +C1 [tex]\frac{d^{2}u}{dx^{2}}[/tex]=0 when x=L so C1=PL 2nd integration EI[tex]\frac{du}{dx}[/tex]=[tex]\frac{Px^{2}}{2}[/tex]PLx+C2 [tex]\frac{du}{dx}[/tex]=0 when x=0 so C2=0 3rd integration EIu=[tex]\frac{Px^{3}}{6}[/tex][tex]\frac{PLx^{2}}{2}[/tex]+C3 u=0 when x=0 so C3=0 which leaves me with: EIu=[tex]\frac{Px^{3}}{6}[/tex][tex]\frac{PLx^{2}}{2}[/tex] u=[tex]\frac{Px^{2}}{2}[/tex](L[tex]\frac{x}{3}[/tex]) multiplying through by 3 gives me 3u=[tex]\frac{3Px^{2}}{2}[/tex](3Lx) and then u=[tex]\frac{Px^{2}}{2}[/tex](3Lx) which still isn't quite right. Can you spot where I've gone wrong in the maths? 



#7
Feb1710, 12:47 PM

P: 692

For your first integration, how do you get Px for the moment at x? Is x measured from the support or from the point load?




#8
Feb1710, 02:52 PM

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P: 2,532

You multiplied the right side by 3 twice.




#9
Feb1810, 09:52 AM

P: 127




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