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Boundary Conditions for a beam with two supports

  1. May 5, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the boundary conditions for the beam shown in the figure.

    2. Relevant equations
    Notation:
    V= Shear force
    M= Bending moment


    3. The attempt at a solution

    at x=0 V=R1, M=0
    at x=9 V=R3, M=0

    In the solution provided at x=9 V=-R2. I don't understand why it's negative, since I took upwards as positive.

    Help appreciated!
     

    Attached Files:

  2. jcsd
  3. May 5, 2013 #2

    SteamKing

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    Ask yourself:
    1. What are the deflections at the left and right edges of the beam?
    2. Can the supports at the left an right edges offer any reactive moments to the loading of the beam?
     
  4. May 5, 2013 #3
    1) I assumed the deflection at both ends is zero.
    But that doesn't mean the shear is zero?

    2) The supports don't seem to provide any bending moment.
     
  5. May 5, 2013 #4

    SteamKing

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    1. You are correct. What happens at the ends with respect to the shear force?

    2. What must the bending moment be at each end?
     
  6. May 5, 2013 #5
    From what I understood in lectures, the shear force at the ends of the beam has to be equal to the reaction force going through the beam given that they are point loads.

    The bending moment at each end is 0.
     
  7. May 5, 2013 #6

    SteamKing

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    Correct.
     
  8. May 6, 2013 #7
    That brings me back to my initial problem,

    at x=9 Shear force=R2

    But in the solutions it's: Shear force= -R2

    I don't see why it's negative R2.
     
  9. May 6, 2013 #8

    SteamKing

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    Write the equations of equilibrium and solve them. It's not clear from your attachment if you are talking about R2 (which is horizontal) or R3 (which is vertical).
     
  10. May 6, 2013 #9
    Sorry my bad!
    I meant:


    at x=9 Shear force=R3

    But in the solutions it's: Shear force= -R3

    I don't see why it's negative R3.

    I already have the values of R2 and R3 from the question. It's just a matter of determining the shear force at the extremes of the beam.
     
  11. May 6, 2013 #10
    I agree with you. The way you have it drawn, R3 should be positive, and the shear force should be positive. Why don't you provide more details on the "correct" solution that has been provided to you?
     
  12. May 7, 2013 #11
    I've attached the part of the solution,

    Macauly's notation is being used to formulate the equation of shear force.
     

    Attached Files:

  13. May 7, 2013 #12

    SteamKing

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    You've still got to write the equations of static equilibrium for the beam and solve them for R1 and R3.
    This step must be done before you can proceed with Macaulay's method or any other technique to solve the beam.
     
  14. May 7, 2013 #13
    "at x=0 V=R1, M=0
    at x=9 V=R3, M=0
    "
    The above quote from your #1 is stated without regard to signs. The sign conventions for shear force are not universal. What is the definition of shear force that you are using, and what is the sign convention you are adopting?
     
  15. May 7, 2013 #14
    I've attached my working out.
    Hope that clarifies things!
    Sorry about the orientation, I can't seem to rotate it.
     

    Attached Files:

  16. May 7, 2013 #15
    So, at a section say 1m from the right hand end, do you think the shear is positive or negative, according to the sign convention you have given?
     
  17. May 7, 2013 #16
    I'd say the shear is positive R1 at x=1m.
     
  18. May 8, 2013 #17
    Yes if x=1 m from left end. What about x=8?
     
  19. May 8, 2013 #18
    Ahh! Negative R3 because the force R3 has to bring the shear to 0 at x=9.
     
  20. May 8, 2013 #19
    Agreed negative but I hadn't thought of your reasoning before now. The sign convention diagram you gave us showed positive shear as "downwards to the right" of the section. But this shear force is "upwards to the right" and therefore negative. (There are other conventions in use for M and V - that's why you should state them when you use them)
     
  21. May 8, 2013 #20
    at x=8 I'd say the shear force is negative R3.

    Edit: replaced R2 instead of R3.
     
    Last edited: May 8, 2013
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