Beam bending, overhang with 2 concentrated loads

[SolMech]

Homework Statement

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

Homework Equations

M(x) = E*I*W"(X)

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)

 

[haruspex]

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...

 

[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]

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).

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!