# Beam bending, overhang with 2 concentrated loads

 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: $\downarrow$.........................................$\downarrow$ ____________________________ <--->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+$\frac{x}{a+L}$)*(a+L) with x left from the center and M(x) = F(1-$\frac{x}{a+L}$)*(a+L) with x right from the center. w" = EI/M(x) Integrating this twice should yield the result: w'(x) = $\frac{F}{2EI}$x$^{2}$ + $\frac{F}{EI}$(a+L)x + C But w'(0) = 0 so C = 0 integrating once more gives: w(x) = $\frac{F}{6EI}$x$^{3}$ + $\frac{F}{EI}$(a+L)x$^{2}$ + C solving for C with w(L)=0 gives: C = - $\frac{FL^{2}}{EI}$($\frac{a}{2}$+$\frac{2L}{3}$) 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)
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P: 9,852
 Quote by SolMech M(x) = F(1+$\frac{x}{a+L}$)*(a+L) with x left from the center and
Could you explain how you arrive at that, please? I'm not saying it's wrong...
 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= -a-l -> 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.
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