# Beam reaction force

1. Oct 20, 2009

### Ry122

1. The problem statement, all variables and given/known data

http://users.on.net/~rohanlal/udl.jpg [Broken]
If the middle restraint has a vertical reaction of 50kn when the structure is first built,
how do i determine what the reaction would be if the middle restraint was to settle by 13mm?

2. Relevant equations

Is this a simply supported beam? If so
delta=(P(L^3))/48EI

3. The attempt at a solution

Would this reduce the reaction force since the bending of the metal is sucking up some of the applied force?

So would it be Final Reaction = (Initial Reaction) - (P from above equation)

Last edited by a moderator: May 4, 2017
2. Oct 20, 2009

### PhanthomJay

At a glance, I think you have the correct line of reasoning, but that UDL of unknown value complicates the situation. If you neglect the UDL, then initially all of the P concentrated load gets supported at the middle (rigid) support. Then due to the settlement of that support, you are correct that some of that load goes to the end supports based on a simple beam between the end supports with the 13mm deflection at centerline, so that what's left goes to the middle support. But the problem does not seem solvable without a value for the UDL, because you'd have to consider it in your deflelction calculation.

3. Oct 20, 2009

### Ry122

I do know the the value for the UDL but the initial reaction takes the UDL into account already, so knowing this is what I said correct?

4. Oct 20, 2009

### PhanthomJay

No. The center support reaction initially includes a contribution from the applied load, P, and the UDL. I assume you are given P, which must be less than 50 kN. The contribution from the UDL is statically indeterminate, but you can find it in handbook tables. Now as the middle support settles, some of the 'P' load transfers to the end supports, and some of the UDL load also transfers to the end supports. When you use the deflection formula, you must include not only the FL^3/48EI term, which is for the unit load only,
but also the UDL deflection at mid point, which, if memory serves me correctly, is wL^4/384EI. Both contribute to the 13 mm deflection.