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Moment in beam

  1. Jul 2, 2016 #1
    1. The problem statement, all variables and given/known data
    refer to the circled part , i dont understand the author's working .
    0Ms0rpu.jpg
    KI4VWZy.jpg

    2. Relevant equations


    3. The attempt at a solution
    I think it should be 450x - 150(4-x) -30092)(x-1) , correct me if i'm wrong . (taking moment about x ) ....
     
  2. jcsd
  3. Jul 2, 2016 #2
    is the author wrong , coz the distributed load start from R1 to 2m from R1 ...
     
  4. Jul 2, 2016 #3

    SteamKing

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    No, the author is not wrong.

    The bending moment function EIy" is carefully constructed to show that the distributed load starts at x = 0 and runs to x = 2, after which an equal and opposite distributed load is applied to the beam at x > 2, so that the net distributed load is zero.
     
  5. Jul 2, 2016 #4
    why ? i still dont understand , i can only understand the 450x , why should be -0.5(300)(x^2) +0.5(300)[(x-2)^2 ]
     
  6. Jul 2, 2016 #5
    why shouldnt it be 450x - 150(4-x) -300(0.5)(2)(x-1) ??
     
  7. Jul 2, 2016 #6

    SteamKing

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    I'm not going to speculate on that.

    Look, the author has a method of analysis of this beam which he is trying to teach. If you want to develop your own methods, write your own textbook.
     
  8. Jul 2, 2016 #7
    i dont understand why it is -0.5(300)(x^2) +0.5(300)[(x-2)^2 ] , can you explain ?
     
  9. Jul 2, 2016 #8
    why it's 0.5(300)(x^2) ? the 300N/m only 'defined ' from R1 to 2m from R1, am i right ? why shouldnt it be 0.5(300)(2)(x-2) instead? as you said , after 2m , net distributed load is zero
     
  10. Jul 5, 2016 #9
    the location of 300n/m is at R1 and extend to 2m from R1, so ,the EIy" should be = 450x-300(2)(x-1) ???
    why there's -0.5(300)(x^2) +0.5(300)( (x-2)^2 )????
    I'm confused

    Or do you mean the 300(X) force still exist at X >2?
     
    Last edited: Jul 5, 2016
  11. Jul 6, 2016 #10
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