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Statics : Internal loadings

  1. Mar 5, 2016 #1
    Determine the resultant internal loadings acting on the cross section
    at C of the cantilevered beam
    http://imgur.com/Q4ZUTOq
    1. The problem statement, all variables and given/known data

    F = 270 N/m

    2. Relevant equations
    ΣV = 0
    ΣM = 0
    ΣN = 0

    3. The attempt at a solution
    section CB

    6JVrME0.png
    k/6 = 270/9
    k = 180 N/m

    resultant force = area under the curve = 1/2 * 180 * 6 = 540 N

    ΣN = 0
    -N = 0
    N = 0

    ΣV = 0
    V - resultant = 0
    v = resultant = 540 N

    How can i find M ?
     
  2. jcsd
  3. Mar 5, 2016 #2

    SteamKing

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    You can calculate the shear force diagram for this beam given the loading as shown.

    What's the relationship between the bending moment and the shear force?
     
  4. Mar 5, 2016 #3

    PhanthomJay

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    You have already determined the resultant force of the load . At what point does it act? Then sum moments and watch directions.
     
  5. Mar 5, 2016 #4
    both occur because of force perpendicular to beam ?
    center of the beam ?
     
  6. Mar 5, 2016 #5

    SteamKing

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    That's not the relationship which is useful in calculating M.

    At the free end of the cantilever, M = 0. What must M be at point C for that segment of the beam to remain in equilibrium?
     
  7. Mar 5, 2016 #6

    PhanthomJay

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    [QUOTE="newbphysic, post: 5400435, member: 557120]


    center of the beam ?[/QUOTE] in the free body diagram you have drawn, the resultant of the triangularity distributed load acts at the cg of that load. Where's that?
     
  8. Mar 5, 2016 #7
    the relationship between M and shear force is shear force causes bending moment ?

    Since M=0 at the end of cantilever that means M must be 0 at C to remain equilibrium

    if the beam is uniform then cg will be length / 2 = 6/2 = 3m from C
     
  9. Mar 5, 2016 #8

    SteamKing

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    The only problem is, M can't be zero at point C, 'cuz of that applied force. What's the moment due to the applied force?
     
  10. Mar 5, 2016 #9

    PhanthomJay

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    in your diagram, the loading on the beam is not uniform; it is triangular. The centroid of a triangle is not at its center.
     
  11. Mar 5, 2016 #10
    The moment is the sum of all forces from B to C = 540 N
    centroid of triangle is 1/3 * 6 = 2
     
  12. Mar 5, 2016 #11

    SteamKing

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    That's not what a moment is.
    How do you use the centroid of the applied load to calculate the moment due to that load?
     
  13. Mar 5, 2016 #12

    PhanthomJay

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    Yes , 2 m from where? Now place the resultant force at that point and sum moments about the left end of your section to find the internal moment at that end.
     
  14. Mar 5, 2016 #13
    moment is force times distance.
    so it's 540 times the distance to the left side of the beam
    Is that what you mean ?

    total force times the distance from centroid to C

    2m from zero reference point means from the left of the beam.
    so moment = [itex]force * distance = 540 *2 = 1080 Nm[/itex]

    [itex]ΣM = 0[/itex]

    [itex]M + 1080 Nm = 0[/itex]

    [itex]M = -1080 Nm[/itex]
     
  15. Mar 5, 2016 #14

    PhanthomJay

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    Yes, but can you explain the meaning of the minus sign in front of your answer?
     
  16. Mar 5, 2016 #15
    M is anti - clockwise

    Thanks a lot phantom
     
  17. Mar 5, 2016 #16

    PhanthomJay

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    OK!
     
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