Beam that is simply supported in the middle

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The discussion centers on calculating the force required to flatten a simply supported beam that is .005" higher at both ends compared to the middle support. The user initially broke the beam into two cantilevered sections and calculated the load and pressure for a linearly increasing triangular distributed load. However, discrepancies arose when comparing the pressure from the triangular load to that derived from resultant forces, leading to confusion over the calculations. A suggestion was made that the triangular load could be treated as a combination of uniform and inverse triangular loads, providing a clearer path to reconcile the differing pressures. Ultimately, the calculations yielded a pressure of approximately 10.263 psi at the far end, which is close to the user's initial estimate.
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Hey Guys,

I am working on a situation where I have a beam that is simply supported in the middle, and the two ends of the beam are .005" higher than the middle where the support sits. I am trying to figure out with a linearly increasing load what the force is to make the two ends ,which are sitting .005" higher, the same heigh as the middle. Basically I want to flatten the piece.

EI = 731 lbf*in^2
length = 3.522in
width = .404in

So this is what I have:

I broke it down into two beams that are cantilevered

So for each cantilever:

load = (displacement * 81 * EI ) / (7*(Length/2)^4) Length/2 because I broken the simply supported beam into 2 cantilevered beams

load = 4.399 lbf/in

pressure = load/width of beam = 10.9 psi

My problem is assuming the above is correct, why does it not work out the same when I break that linearly increasing load into the resultant force:

resultant force for one of the cantilevered = .5 * L/2 * load = 3.872lbf

So since we need 2 of those forces, one on each end 2/3rds of the way up the linearly increasing triangle, that would equal 7.744 lbf

My problem is now if I say that force (7.744 lbf) / (L*width) = 5.442 psi

5.442 psi doesn't equal the 10.9 psi?

I know it is different but I did the same thing using a uniform pressure, where I broke the uniform pressure into two resultant forces and then divided those forces by the area and got the same pressure as the uniform pressure I had originally calculated.

Any help would be appreciated, thanks.
 
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Are both ends inclined? (Engineers love drawings so make one)
 


Yes both ends are inclined. They are 0.005" higher than where the support it
 

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You are confusing me with what you term a "linear increasing load." It looks like in one case you mean a simple distributed load and the other you are doing a distributed load that is a function of the distance down the length of the beam, i.e. a triangular shaped distributed load. I understand what you are doing and as a rough first guess I would have done what you did. What I don't understand is the second part you are comparing to.
 


Yeah by linearly increasing load I mean a triangular distributed load that is increasing the further away from the support you get.

So in the first part I have the load required for the triangular distributed load

and then I found the resultant force of that triangular distributed load by using the formula
.5*L*load. Load being in the form of (lbf/in)

When I compare the pressure from the triangular distributed load and the pressure from the resultant forces, I get two different pressures.
 


If you have triangular loads which increase towards the free ends on both sides of the middle support, then you have the equivalent of a cantilever with half the length.
(Ref: http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_bc_cantilever.cfm)
Let the Triangular load be 0 at the support, and p at the far end, then it is equivalent to adding a uniform load of p to the cantilever, and subtracting the inverse triangular load, which is p at the support and 0 at the far end.
The deflections at the "free" end of the cantilever for the first and second cases are
w1=pL4/8EI, and
w2=pL4/30EI
Subtracting
w1-w2=pl4/EI*(1/8-1/30)=11pL4/120EI
now
L=1.761
EI=731
w1-w2=0.005
we have
11*p*L4/(120*EI)=0.005
Solving for p,
p=160039/38600=4.1461
pressure
=4.1461/.404
=10.263 psi (at the far end where the load intensity is maximum)
Very close to what you had at the beginning.
 
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