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Moments, completely lost!

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the magnitude of the bending moment and shear forces at the centre of the beam shown in the attached figure

    bsforce.jpg


    2. Relevant equations



    3. The attempt at a solution

    For the shear forces I think its the sum of the vertical components,
    Bending moment = not quite sure.

    I would appreciate if someone could give me a few pointers on bending moments and shear forces. Have looked over some textbooks but it really isn't explained well. Appreciate ANY help.
     
  2. jcsd
  3. May 15, 2009 #2
    cd19: unfortunately you are dead in the water until a mentor approves your attachment.

    I would suggest using a free image host like photobucket.com or imageshack.

    Then you will not have to wait for approval, you can just upload your images to photobucket and then copy/paste the image tag in to your post.
     
  4. May 15, 2009 #3
    sorry here it is:

    bsforce.jpg
     
    Last edited: May 16, 2009
  5. May 16, 2009 #4
    Any hints?
     
  6. May 16, 2009 #5

    PhanthomJay

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    You should first determine the vertical end reactions usung the equilibrium equations, or take advantage of the symmetry of the loading to determine those end reactions. Once you have them, the internal shear and the bending moment at the beam's center can be found by applying the equilibrium equations to a free body diagram of the left part of the beam (from its left end to the beam center). The shear at the center is not just the sum of the external vertical forces.
     
  7. May 16, 2009 #6
    Ok so I have found the vertical end reactions to be:

    r1 = 630.25kN
    r2 = 8.9167kN

    So what i gather is I have to draw a free body diagram exposing sections between the left support and the load. I do this in intervals, summing forces and taking moments about a point in the section.

    so what I have attempted is the above from the intervals 0<x<6, 6<x<9, 9<x<12 but how do i solve for x and get a single answer for the shear force and bending moment?
     
  8. May 16, 2009 #7

    PhanthomJay

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    You did something wrong here; you might want to show your work. Since the load is perfectly symmetrical about the center of the beam, you might want to venture an educated guess as to how the applied loads distribute to each end reaction. But first note that the sum of the end reactions must equal the sum of the applied vertical loads. You've got two concentrated loads of 15 kN each, and a distributed load of 5 kN/m over 6 m, that's another 30 kN, so all together, you have 60 kN acting down. So the end reactions have to sum to 60 kN acting up, from Newton 1. Also, when you sum moments about A to determine the external end reaction at D, note that the distributed load of 5 kN/m over the 6 m length can be represented as a 30 kN load acting at the c.g. of the loading (which in this case is the center of the beam).
    The problem asks for the (internal) shear and bending moment at the center. So just expose the section from the left end to the center. At the center, there wiil be an unknown vertical load (the shear) and an unknown moment (the bending moment). You'll need to solve for these unknowns using sum of F_y = 0 , and sum of moments = 0.
     
  9. May 16, 2009 #8
    Ok so I see that as everything is perfectly symmetrical, the end reactions at A and D must be 30 Kn respectively.

     
  10. May 16, 2009 #9

    PhanthomJay

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    yes, correct.
    You already did, but it's good to check your work.
    No, you must sum ALL forces to solve for the unknown shear, V, in your FBD of the left part of the beam. That includes the concentrated load of 15 kN down, the end reaction of 30 kN up, the downward contribution of the distributed load, and V, all summed to 0 to solve for V.
    to solve for the bending moment, sum moments about the right end of your FBD. It will be easier to assist you if you show your work.
     
  11. May 17, 2009 #10
    This Is what I got:

    For ''V'' ; 15+15-30+V=0 therefore v = 0 Kn (shear)

    Summing the moments about the right end of my FBD i got:

    -(30)*(9) + (15)*(6) = 180kNm (Bending Moment)

    Is this looking correct?
     
  12. May 17, 2009 #11

    PhanthomJay

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    Looks Good!
    I can see you don't like that distributed load, because you keep leaving it out of your FBD! Represent it as a 15 kN load acting at its c.g., then sum moments again. Indicate if the bending moment at the center of the beam is clockwise or counterclockwise.
     
  13. May 17, 2009 #12
    How about:

    30*(9) - 15(6) -15(1.5) = bending moment =157.5(Kn) acting clockwise

    I would have never of got that distributed load right i.e the perpendicular distance (1.5), I'm terrible at FBDs

    thanks for your help!
     
  14. May 17, 2009 #13

    PhanthomJay

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    You're good at them now, no? You're bending moment is correct, except it should be acting counterclockwise on the left part of the beam.
     
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