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Forces / moment in beam

  1. Oct 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Why we need to replace the 10kip load at 2ft from D to D ?

    2. Relevant equations


    3. The attempt at a solution
    IMO it's wrong ... There's no need to do that .
     

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  2. jcsd
  3. Oct 18, 2016 #2

    PhanthomJay

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    Well you don't need to do that if you just wanted to calculate the end reactions, but if you want to look at shear and moment diagrams in the beam, it is a must. Otherwise your shear and monent diagrams would not be correct, for example, the couple at D wouldn't show. Use free body diagram for the beam.
     
  4. Oct 19, 2016 #3
    Why my shear force diagram and bending moment diagram would be incorrect if I want to do so?
     
  5. Oct 19, 2016 #4

    PhanthomJay

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    Notice the sharp 10 K drop in shear at D in the shear diagram, and the sharp 20 ft-kips rise in moment at D in the moment diagram. You won't be able to discover that unless you isolate the beam from the bracket on a free body diagram.
     
  6. Oct 19, 2016 #5
    Why we wouldn't be able to feel the drop in shear force at 2m from D?

    I can understand that we can feel the drop in moment 20 NM at D since moment = force X distance and moment produced is at D.....
     
  7. Oct 19, 2016 #6

    PhanthomJay

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    And the force at D on the beam comes from the force exerted on the bracket at E. If you draw a free body diagram (FBD) of the bracket DE, the 10 kip force at E produces an end reaction at D on the bracket of 10 kips force up and 20 ft-k moment ccw . By newtons 3rd law, the bracket at D exerts a 10 k downward force and 20 ft-k cw moment on the beam. That is what the main beam 'feels' internally , it doesn't much care about the bracket.
     
  8. Oct 19, 2016 #7

    I am confused.... In the beginning, you said that the moment at D is CCW , and shear force is upwards, then you said that according to newtons 3rd law, the shear force is downward at D, and moment is cw at D?
     
  9. Oct 19, 2016 #8

    PhanthomJay

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    What I said was that the beam exerts an upward force and ccw moment on the vertical part of the bracket, and that therefore in accord with Newton 3, the vertical part of the bracket exerts a downward force and cw moment on the beam. When looking at the beam, the force exerted at D is down and the moment at D is cw, as per book example.
     
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