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Finding Moment

  1. Feb 23, 2014 #1
    Hi can anybody tell me if my working for finding the moment of a b c is correct?

    Taking clockwise as +ve

    Ma+1.2k x 3 +0.7 x 7 = 0
    Ma= -8.5Knm

    Mb-8.5+1.9k x 3=0
    Mb=2.8kNm

    Mc=1.2k x 4 +1.9k x 7 -8.5k=0
    Mc=-117.85Nm


    Hence Point B has the biggest moment am I right? However the book says is Point A. Please advice, thank you.
     

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  3. Feb 23, 2014 #2

    adjacent

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    What is the full question?I don't understand what you are trying to do.Find the largest moment?
     
  4. Feb 23, 2014 #3
    Taking clockwise as +ve

    Ma+1.2k x 3 +0.7 x 7 = 0
    Ma= -8.5Knm

    Can you explain to me why my working is wrong for Moment A, thank you.
     
  5. Feb 23, 2014 #4
    This is the real question. For flexture stress to be maximum at the point, the value of "M' must be largest in the beam
     

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  6. Feb 23, 2014 #5

    adjacent

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    Isn't that the total moment??
    It's also wrong.
    M=F*perpendicular distance.
    M=1.2k x 3 + 0.7 x 7
    =8.5KNm
    Why -8.5kNm?
    If clockwise is +,and all the forces are acting on clockwise direction,Why should there be any negative value?

    EDIT:That tensile flexural state thing is beyond my knowledge:eek:.Maybe someone else can help you from here onwards. Sorry :frown:
     
  7. Feb 23, 2014 #6
    Ma+1.2k x 3 +0.7 x 7 = 0
    Ma+8.5=0
    Ma=-8.5
    I put a = 0 behind my equation and I switched it over, hence I got a negative 8.5
     
  8. Feb 23, 2014 #7
    Formula for question

    :cry:
     

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  9. Feb 23, 2014 #8
    All my 3 moments are wrong? :cry::cry::cry:
     
  10. Feb 23, 2014 #9

    PhanthomJay

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    Moment about A is correct...what does the minus sign mean?
    I have no idea what you are doing when summing moments about B and C. The moment at A is an external reaction moment. Elsewhere , moments ate internal in the beam and can be calculated using free body diagrams or shear and moment diagrams.
    EDIT: o I think I see now what you are doing...you are messing up your plus and minus signs. These moments are more easily calculated using right hand sections.
     
    Last edited: Feb 23, 2014
  11. Feb 23, 2014 #10
    For my moment about A my minus means that it is -8.5Nm CCW

    whereas for moment B and C, let's start with B, I am trying to take the FBD LHS.

    I have attached a picture showing the example of what I am trying to do. The question in this picture is different. I am actually "copying" the steps to help me solve my #1 problem.

    I am confused right now :cry::cry:
     

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  12. Feb 23, 2014 #11
    How I work out C post #1 is based on this question.


    Taking clockwise as +ve

    Ma-5k x 1.2 = 0
    Ma=-6kN CCW

    Mb+5 x1.2 -6k =0 (I added -6kN which is Moment A into this equation)
    Mb =0

    So for post #1 working
    Mc=1.2k x 4 +1.9k x 7 -8.5k=0 (I added the -8.5k Moment A which I found earlier into this equation)
    Mc=-117.85Nm
     

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  13. Feb 23, 2014 #12

    SteamKing

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    You are hopelessly confusing yourself by first not drawing a free-body diagram of your beam. Currently, you are just calculating random moments, adding or subtracting them at will, and hoping you get the right answer somehow.

    Draw the FBD of the cantilever and find the reacting force and moment at A. If you can do that, then further help will be forthcoming. Right now, you are just spinning your wheels, and mixing these other problems in with the original problem is not helpful.
     
  14. Feb 23, 2014 #13
    I refer to other example because I tried to solve the problem as I study alone and have nobody to ask :cry:

    This is my working.. All moment I got are the same. So referring to post #4, which point has the largest moment?
     

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  15. Feb 23, 2014 #14

    SteamKing

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    That's just the point. If you follow the hint given in the problem and draw the bending moment diagram of the beam, the location of the maximum moment is immediately obvious.

    The beam is restrained at point A. Write your equations of static equilibrium using point A as the reference for the moments. Calculating moments about points B and C is a waste of time because the beam is not restrained at those points, and this is keeping you from proceeding with a solution.
     
  16. Feb 23, 2014 #15
    I would like to ask can I just calculate the moment for each point A,B,C manually without drawing the BMD? As shown in post #10.

    And what do you mean by "Write your equations of static equilibrium using point A as the reference for the moments."

    Thank you.
     
  17. Feb 23, 2014 #16

    SteamKing

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    In order to determine the reaction force and moment applied at A which keeps the free body of the beam in static equilibrium, you write equations which contain the unknown reactions. The beam remains in equilibrium as long as the following equations are satisfied:

    ∑F = 0

    ∑M about A = 0

    For this particular beam, let's say there is an unknown reaction R[itex]_{A}[/itex] and unknown moment M[itex]_{A}[/itex] at point A. The equations of equilibrium for the beam will be:

    [+ forces are pointing up; + moments are counterclockwise]

    ∑F = R[itex]_{A}[/itex] - 1.2 - 0.7 = 0 [kN]

    ∑M about A = M[itex]_{A}[/itex] -1.2 * 3 - 0.7 * 7 = 0 [kN-m]

    Now, it should be easy to solve these two equations for R[itex]_{A}[/itex] and M[itex]_{A}[/itex].

    Once you have done that, you can then construct the shear force and bending moment diagram for the beam.
     
  18. Feb 23, 2014 #17
    Yes..I already calculated Ra=1900N and based on the SFD and BMD, I have found that
    Ma=-8500Nm
    Mb=-2800Nm
    Mc=0Nm


    I would like to ask without drawing the BMD, can I calculate each Moment individually?
     

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  19. Feb 23, 2014 #18

    SteamKing

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    Your shear force diagram looks OK, but the value of the bending moment at B is not -2800 Nm, and the bending moment does not go to zero between points B and C.

    You use the value of the moment at A and the area under the shear force diagram to calculate the value of the bending moment at points B and C.

    For example, the BM at point B = MA + Area of the shear force diagram from A to B

    You should have obtained MA = 8500 Nm from the equilibrium equations, rather than -8500 Nm.
    The reaction moment at A is CCW, given the direction of the the forces on the beam.

    Can you calculate the bending moment values at B and C now and correct the bending moment diagram?
     
  20. Feb 23, 2014 #19
    I am really confused on why Moment A is not - 8500Nm CW.. Can you please take a look at post #13 again? I set my orientation as +ve CW
     
  21. Feb 23, 2014 #20
    This is the new BMD, but it don't make any sense to me. I thought for BMD you always start at 0 and end at 0. For this BMD, I start at 0 but end at 17240. Whereas for the old BMD, I start and end at 0.
     

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