Beam bending, overhang with 2 concentrated loadsby SolMech Tags: beam, bending, cantilever, four points bending 

#1
Jan2113, 08:44 AM

P: 4

1. The problem statement, all variables and given/known data
A beam supported at two locations is subjected to two equal loads at the end points Compute the central deflection W. Schematic: [itex]\downarrow[/itex].........................................[itex]\downarrow[/itex] ____________________________ <>Δ<><>Δ<> ...a..........L............L...........a The delta's are the supports the arrows the forces F and a & L distances 2. Relevant equations M(x) = E*I*W"(X) 3. The attempt at a solution At first I try to formulate the momentum along the beam. x = 0 at the center of the beam. I thought it would be zero at the points where the Forces are, and maximum at the center. So I came up with: M(x) = F(1+[itex]\frac{x}{a+L}[/itex])*(a+L) with x left from the center and M(x) = F(1[itex]\frac{x}{a+L}[/itex])*(a+L) with x right from the center. w" = EI/M(x) Integrating this twice should yield the result: w'(x) = [itex]\frac{F}{2EI}[/itex]x[itex]^{2}[/itex] + [itex]\frac{F}{EI}[/itex](a+L)x + C But w'(0) = 0 so C = 0 integrating once more gives: w(x) = [itex]\frac{F}{6EI}[/itex]x[itex]^{3}[/itex] + [itex]\frac{F}{EI}[/itex](a+L)x[itex]^{2}[/itex] + C solving for C with w(L)=0 gives: C =  [itex]\frac{FL^{2}}{EI}[/itex]([itex]\frac{a}{2}[/itex]+[itex]\frac{2L}{3}[/itex]) which therefore should be the max deflection, because at x=0 deflection is max and w(0) = C The problem is I have no way to check my result, and I'm not quite sure of the result. Could someone check me and explain me where I went wrong (if so) 



#2
Jan2113, 02:49 PM

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#3
Jan2213, 02:36 AM

P: 4

With the argument that it should be zero at the force and maximum at the midpoint, I determined them to be linear in between, so that with negative x, M(x) goes from x= al > M((a+L))=0 to x=0 giving M(0)=F(a+L).
Also I see that i did not integrate right, forgot some constants, but the problem lies mainly in the determination of the momentum along the beam, the rest is straightforward I guess. 



#4
Jan2213, 02:56 AM

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Beam bending, overhang with 2 concentrated loads 



#5
Jan2213, 04:18 AM

P: 4

I'm not sure how you determined these moments.. What do you mean by: "I approached it by actually taking moments". Did you divide the problem into halfs and used M = F*x,
and then used that if there is a load F, the support gives a reaction force F? If this is what you did, it is still not obvious for me how you arrived at the moments from this. 



#6
Jan2213, 05:24 AM

P: 4

Ah never mind! Got it, thank you for your help!



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