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Draw the shear force and bending moment diagrams for beams

  1. Apr 9, 2017 #1
    1. The problem statement, all variables and given/known data
    q4.JPG

    2. Relevant equations

    Shear force is calculated at each point on the beam. Downward forces are negative, upward forces are positive.

    Moments about points are calculated as force multiplied by perpendicular distance. Clockwise moments are negative. Anti clockwise moments are positive.

    3. The attempt at a solution

    SF @ A = -6kN
    SF @ B = -4kN
    SF @ C = 12kN
    SF @ D = 12kN (same as C)

    SFD.JPG

    BM@A=0
    BM@B=6kNx1+12kNx0.5 = 12kN CCW
    BM@C=6kNx1.5+4kNx0.5=11kN CCW
    BM@D=6kNx3+4kNx2-12kNx1.5=8kN CCW

    BMD.JPG


    Answers given are:
    Vmax= -10kN (I can see this from the shear force diagram)
    Mmax = -11kN.m (point c?)
    Points of contra-flexure = C (not sure how this is calculated)
     
  2. jcsd
  3. Apr 9, 2017 #2

    haruspex

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    In your shear force diagram, why does the line slope down from -6 to -10? Wouldn't that be for a uniformly applied load of 4kN along AB?
    How do you get that?
    This is not how to calculate bending moments. You should only consider forces on one side of the point. Which side does not matter in principle as long as you are consistent. Switching sides will just flip the sign.
     
  4. Apr 10, 2017 #3
    Is the 4kN not added to the 6kN therefore giving us 10kN?

    The shear force at D I thought would be the same as at C because there are no further forces between C and D?

    Looking at the RHS for bending moments:
    BM@A=-4kNx1+12kNx1.5=14kN CCW
    BM@B=612kNx0.5 = 6kN CCW
    BM@C=0

    I am not sure how the bending moment diagram is supposed to look but I would assume something like this;
    BMD 2.JPG
     
  5. Apr 10, 2017 #4

    haruspex

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    Yes, but not until you reach that point in the beam.
    For point loads only to the left of some point, the shear diagram up to there should look like a step function and the bending moment would consist of straight line slopes.
    With one or more uniform loads to the left, the shear gives straight line slopes while the bending moment has quadratics (parabolas).
    No, don't do that - stick with working from the left. That seems to be standard and so you may lose marks doing something different.
     
  6. Apr 10, 2017 #5
    Like this:

    SFD 2.JPG

    LHS

    BM@A=0
    BM@B=6kNx1=6kN CCW
    BM@C=6kNx1.5+4kNx0.5=11kN CCW
    BM@D=6kNx3+4kNx2-12kNx1.5=8kN CCW

    I really don't know what the bending moment diagram should look like though?
     
  7. Apr 10, 2017 #6

    haruspex

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    You shear diagram and your moments for the individual points are correct.
    The bending moment diagram should be continuous - no steps.
    To do it properly you should consider a point at between A and B, at distance x from A say, and calculate the bending moment there. Then do likewise for a general point between B and C, etc.
    But knowing that for point loads it is all straight lines, you can cheat and just connect up the plotted individual points.
     
  8. Apr 10, 2017 #7
    Something like this:

    BMD 3.JPG
     
  9. Apr 10, 2017 #8

    haruspex

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  10. Apr 10, 2017 #9
    Thanks again for your help Haruspex
     
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