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Understanding statically indeterminate beams

  1. Mar 1, 2012 #1
    As a physics student, I never came across the following problem:

    A uniform beam of mass M and length L rests on three identical supports: one at each end of the beam and one in the middle. What are the forces of reaction on each of the supports?

    I now realise that that was the case because this system is statically intederminate in the sense that we have too few equations of equlibrium to solve for the three unknowns. But, surely the problem must have a unique solution. So, how would one solve this (in principle, very simple) problem?

    1) One option is symmetry - since the supports are placed symmetrically, each would carry equal weight. That means that the reaction force is Mg/3 for each support.

    2) Or, one can divide the beam into two equal parts by (imaginary) cutting it in the middle. Then the problem is solvable and gives Mg/2 for the middle support and Mg/4 for the each of the outer supports.

    3) I'm guessing that the correct answer is something else, so please, enlighten me :)
     
  2. jcsd
  3. Mar 1, 2012 #2

    OldEngr63

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    The clue is in the name of the problem class - statically indeterminate - which means that the equations of statics are insufficient for the solution. You must look to other information for a solution, in particular, you must look to information about the beam deformation. Presumably this is a flexible beam, not a rigid beam, so it deforms under gravity. That deformation must be taken into account in order to get a solution.

    There are several ways to do this, but one of the easiest ways to think about it is the method of superposition. Imagine that the center span support is removed. Then you could presumably calculate the midspan deflection, the deflection at the point where the support was removed. Now imagine that we wish to replace that support with a force. How much force will be required to lift that point back up to a net zero deflection? Well, that too can be calculated from beam deflection theory. You know how much distance you need to lift that point, and how much force must be applied to do it. Thus you have effectively found the center support load. Now the equations of statics are sufficient for the other two support loads.
     
  4. Mar 2, 2012 #3
    Thanks for the response, it was very helpful!

    I still have further questions:

    1) Does the final answer depend on rigidity of the beam? Can we take Young's modulus to be infinite in the final expression?
    2) Does the answer depend on which support we remove at the beggining? What if we first remove the left support, then calculate the deflection of the left end etc. Would we obtain the same result as with your method?
     
  5. Mar 2, 2012 #4

    OldEngr63

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    1) It most definitely does depend on the rigidity of the beam. If Youngs Modulus is infinite (Unobtainium 304) then the center span support is entirely redundant (carries no load at all). This is not the case for all real materials, however.

    2) No, you will get the same result if you start out by removing the left support to begin. The picture is just a bit more difficult to visualize, but be my guest, work through it for yourself to convince yourself.
     
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