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Distributed Load

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    http://postimg.org/image/7vpxry28t/


    Can someone explain how they calculated the force R representing the distributed load?
    Did they even make use of the value "800N/m" from the question?
     
  2. jcsd
  3. Nov 26, 2013 #2

    SteamKing

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    The calculation for R is included in the solution section. You don't need to use the 800 N/m unless you calculated the value of R using the formula for the area of a trapezoid:

    R = (0.6/2)*(400 + 800) = 360 N

    I think they are trying to show how to break up a trapezoidal load into a constant distributed load and a triangular load.
     
    Last edited: Nov 26, 2013
  4. Nov 26, 2013 #3
    Yes, they used the 800, but they obtained the force R result a stupid (IMHO) way. The way I would have done it would have been to note that the average distributed load over the length of the member is 600 N/m. If we multiply that by the length of the member (0.6), we get 360 N. They did something like the following: the minimum distributed load over the length of the member is 400 N/m, so this contributes 400 (0.6) = 240N. Over and above that, the remainder of the load varies from 0 at the left end to 400 at the right end (400 to 800, minus the 400 already accounted for). The average of this excess is (0+400)/2 = 200. The load contribution of this excess is (400/2)(0.6)=120N. The total load R is again 360 N. As I said, their method is kinda stupid.

    Chet
     
  5. Nov 26, 2013 #4
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