Beam bending, overhang with 2 concentrated loads

SolMech

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)

haruspex

Could you explain how you arrive at that, please? I'm not saying it's wrong...

SolMech

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.

haruspex

I approached it by actually taking moments. For a point x left of centre, x < L, there's an anticlockwise moment F(x+L) from the support and a clockwise one F(x+L+a) from the load. Doesn't that make the bending moment Fa for all x between the supports? Beyond the supports it's F(L+a-x).

SolMech

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.

SolMech

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