- #1

SolMech

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## Homework Statement

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

## 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+[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)