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Reduction of a simple distributed loading

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  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    media%2F57f%2F57f120ce-0d0f-4f53-bb7a-8d2148749813%2FphpSl5E37.png
    This is the question, but with me 6 kN/m is 4 kN/m and 2 kN/m is 2.5 kN/m.

    I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.

    2. Relevant equations
    F= 1/2*b*h
    M= 1/2*b*h*d

    3. The attempt at a solution
    So if I call the top one F1 and the bottom loading F2 I got:

    F1+F2=0= -4*b*1/2+2.5*(b+a)*1/2
    solving this i got a=0.6b

    Moment around the of the bar (not A, but opposite side) due to the loadings is (counter clockwise positive):

    M= -F1*center of triangle + F2*center of triangle
    Because you can replace the loading with force F1 and force F2 and its line of action is through the center of the triangle area (1/3* base)
    M= -8= -4*b*1/2*1/3b+2.5*(b+a)*1/2*1/3*(b+a)

    So what am I doing wrong?
    Because M can never be negative with me, plus the answer should be b= 5.625 and a= 1.539.
    But this to me makes no sense, because then F1+F2 is not 0.
    And if I should take the reactive forces into account at A then you can never have still a moment, because then it is not static anymore.
     
  2. jcsd
  3. Feb 25, 2016 #2

    SteamKing

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    It seems that there is something wrong with this problem you have been given.

    (a+b) ≤ 4 m , since that's the length of the beam, but the values for a and b given as your answers don't satisfy this condition.

    Working out a and b for the original problem gives values such that (a+b) ≤ 4 m.

    Did the length of the beam change when the loads changed? If it did not, then the data for your problem is inconsistent.
     
  4. Feb 25, 2016 #3

    Owh sorry i did not notice. In my question the total length is 9 meter not 4
     
  5. Feb 25, 2016 #4

    SteamKing

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    There is a mistake here in calculating the equivalent force for the evenly distributed load.
    You've carried over the mistake in the equivalent force calculation for the distributed load to the moment calculation, and you've compounded this mistake by not using the correct load center for the evenly distributed load.

    Why are you treating the evenly distributed load and the triangular load the same? Their distributions are totally different.
    When the forces for each load didn't equal each other, that should have told you something was wrong with your calculation.

    BTW, I don't get a = 1.539 or b = 5.625. You can plug these values back into the calculations of the equivalent forces for the triangular load and the evenly distributed load and immediately see that the forces are not equal, which they should be to give zero net force.
     
  6. Feb 25, 2016 #5

    I am so sorry, but i have uploaded the wrong image. I have been on it all night and so was frustrated and just uploaded image from internet that looked like it but did not recognize that it was not same. I will post a new question, could you please look at that. (basically the thing different in my question is that the lower loading is also an triangle and now rectangular).

    Thx in advance
     
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