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Bending Moments Diagram

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a problem which involves me drawing the bending moment diagram for a trapezoidal distributed load. I understand the bending moment diagrams for a uniform distribution, and partially for a triangular distribution, however i am struggling to link the two for a trapezoid shape distribution. How would i go about knowing what sort of shape will be required and how to draw this?
    2. Relevant equations
    The beam is 2.35m in total length, with a 0.1 segment at each end, and the distribution in the centre at a length of 2.15m. The 'small' side of the trapezoid is 20N/m and the longer side is 40N/m. This distribution is seen to be 'hanging' downwards from the beam, and a reaction load from another beam is situated at the far right of the beam, in the upwards direction.
    3. The attempt at a solution
    I think i have calculated the max moment to be 15.4084N. How would i draw this on my diagram.
     
  2. jcsd
  3. Mar 23, 2010 #2

    PhanthomJay

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    Firstly, I'd like to welcome you to PF!

    Please show how you arrived at your number, which should have the units of N-m, if that's a moment you are trying to calculate. This problem looks very similar to this one:
    https://www.physicsforums.com/showthread.php?t=385188

    Is it? If not, what are the end support conditions...is this a cantilever with a fixed support at one end? Remember, the slope of the shear diagram curve at a given point is equal to the value of the load intensity at that point; and the slope of the moment diagram curve at a given point is equal to the value of the shear at that point. Is this a homework problem, or a take-home exam???
     
  4. Mar 23, 2010 #3
    Indeed this is the same question as the problem in 'mikex24' post. Turns out to be a bit of a challenge!
    Anyway after abit more work i managed to complete my horizontal beam calculations, however thankyou for getting back to me!
    The angled beam seemed quite simple compared to the horizontal beam, most likely because of the lack of distributed loads. Just to clarify, it is possible to rotate the angled beam so it is horizontal, providing the geometry is the same to draw the bending moment and shear force diagrams?
    Once again thanks for getting back to me so quick, i will go and have a quick look at the induced stresses!
     
  5. Mar 23, 2010 #4

    PhanthomJay

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    Yes, you are quite correct that the shear and moment diagrams for the angled member is a lot easier to develop than the horizontal member, primarily because there is no distributed load on the angled member. And yes, once you determine the forces and moments acting on the angled member at its ends, rotate the member (and the forces and moments with it) so it's horizontal, and draw the shear and moment diagrams. It's easier than tilting your head :wink:
     
  6. Jan 14, 2011 #5
    hi to all, i jst want to share my problem. it is a simple beam with trapezoidal loading on it. i resolve the trapezoid by making it into a uniformly distributed load and a triangular loading.. but still i can't determing how to solve or to draw the share and moment diagram.. using both the Equation Method and the Area Method.. can any body share your genius idea? thanks here's the image of the problem
     

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  7. Jan 19, 2011 #6
    Let x be the distance from the left hand end.
    Then Vx = ...... and Mx=.....? That is, what are the values of shear force and bending moment at section X, as functions of x. That is how to start.
     
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