Beam bending, overhang with 2 concentrated loads


by SolMech
Tags: beam, bending, cantilever, four points bending
SolMech
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#1
Jan21-13, 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)
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haruspex
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#2
Jan21-13, 02:49 PM
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Quote Quote by SolMech View Post
M(x) = F(1+[itex]\frac{x}{a+L}[/itex])*(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
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#3
Jan22-13, 02:36 AM
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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
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#4
Jan22-13, 02:56 AM
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Beam bending, overhang with 2 concentrated loads


Quote Quote by SolMech View Post
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
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#5
Jan22-13, 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.
SolMech
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#6
Jan22-13, 05:24 AM
P: 4
Ah never mind! Got it, thank you for your help!


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