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Restriction of moment at wall

  1. Jul 13, 2016 #1
    1. The problem statement, all variables and given/known data
    in the area-moment diagram , why the moment restricted at the wall is 0.594)(1200)(8/3 )?

    9CfZ0qv.jpg
    wqdFMiY.jpg
    2. Relevant equations


    3. The attempt at a solution
    I think \the 8/3 should be 1+3(2/3) , am i right ? Since we know that , the centorid of triangle is 2/3 from the right end , and the triangle has the length of 3m .. ( it's stated in the question that slope is 3m from the wall) ....
     
  2. jcsd
  3. Jul 14, 2016 #2
    since 1+3(2/3) = 3 ,
    So, M=(0.5)(4)(1200)(3) = 7200Nm?
     
  4. Jul 14, 2016 #3

    SteamKing

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    No, the centroid of the triangular load is 2/3 from the left end in this case, where the beam is fixed and the triangle has its acute angle.

    You really should check these things out more carefully.

    http://www.ele.uri.edu/~daly/106/06/project/centroid/centroid.html
     
  5. Jul 14, 2016 #4
    why it's 4(2/3)=8/3 ?
    why not 1+3(2/3) = 3 ?
    The triangle only start at 1m away from wall. so length of triangle only 4-1=3m
     
  6. Jul 14, 2016 #5

    SteamKing

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    You might want to look more carefully at the diagram in the OP, Figure E4.10. This shows the layout of the entire beam and the load.

    The unsupported span of the beam is 4 meters.

    Now, the diagram below Fig. E4.10 is for calculating the slope and deflection of the beam at 3 m from the wall. However, in order to make this calculation, you must first determine the reactions R and M at the wall due to the triangular load, which is why R = (1/2)(4)(1200) and M = (1/2)(4)(1200)(8/3).
     
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