How can I determine horizontal force reactions in a fixed on both ends beam like this one? Fixed on both ends beam


2 Answers 2


The reaction at either end is simply equal and opposite to the axial load in the beam adjacent to it. So what you need to work out is the axial force each side of where F is applied.

To work this out you need the plea formula:

d = PL/EA

where d is extension, P is axial force, L is the original length, E is Young's modulus and A is cross-sectional area.

Using subscript 1 for the left hand side and 2 for the right hand side, we then get two equations:

d1 = P1a/EA

d2 = P2b/EA

We also know that for equilibrium:

P2 + P1 = F

d2 = d1

We can then solve all of these simultaneous equations (I'll leave that step to you), and we'll find:

P1 = F * b/(a+b)

P2 = F * a/(a+b)

NB The plea formula works equally well in tension and compression (assuming no buckling). A tensile force leads to elongation, a compressive force leads to shortening. All my workings are on absolute values, if you want you can make P1 and d1 negative; this is technically more correct but it adds a layer of complexity that I don't feel is necessary.


This seems like a hw question so I'm not going to give you the straight up answer, but the following should help.

The key to this problem is to look at elongation and compression of the beam. To the left of where force F is applied , the beam is in tension and "wants" to elongate. To the right of where force F is applied the opposite is true and the beam is in compression and "wants" to shrink.

Since the beam is constrained we know that the total elongation/deformation is 0.

By applying that constraint we know that the elongation of the left side of the beam is equal to the compression of the right side of the beam, and we can solve for our reactionary forces.

To develop intuition ask yourself this:

What would happen if $a=0$? $b=0$? or $a=b$? Does my answer reflect this?

Best of luck -Brian

  • $\begingroup$ +1 and good point. $\endgroup$
    – Solar Mike
    May 11, 2018 at 6:39

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