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Maximum deflection / slope of beam

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data
    As in the diagram




    2. Relevant equations


    3. The attempt at a solution
    For part a ) , my ans is EI(d2y/dx2) = - Px

    For part b) , my ans is EI(dy/dx) = -P(x^2)/2 +c1 EIy = -P(x^3)/6 +c1x +c2

    at x= 0 , y = 0 , so c2 = 0 , at x = 0 , dy/dx = 0 , so c1 = 0 ,

    so , EIy = -P(x^3)/6 , so , EIy max occur at L=3 , so EIy max = -10(3^3) / 6 = -45 , but the ans is EIy max =-46.67 , which part of my working is wrong ??

    for slope at x = 2 , my ans is EI(dy/dx) = -P(x^2)/2 = EI(dy/dx) = -10(2^2)/2 = -40/EI
     

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  3. Nov 29, 2016 #2
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  4. Nov 29, 2016 #3

    PhanthomJay

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    This is the moment
    you are not reading the question correctly for one thing. But also, x is zero at the free end of the cantilever, such that y is not zero and the slope is not zero when x=0. They are zero when x = ???.
     
  5. Nov 29, 2016 #4
    is there anything wrong with my moment ? In this question , i am taking the right end as x =0 ,
    they are zero when x =L
     
  6. Nov 29, 2016 #5
    EI(d2y/dx2) = - Px

    EI(dy/dx) = -P(x^2)/2 +c1 , EIy = -P(x^3)/6 +c1x +c2

    at x = L , dy/dx = 0 , so c1 = P(L^2)/2 ,
    at x= L , y=0 , c2 = (-1/3)(P)(L^3) ,

    at 2m from A = 1m from B
    So , EIy = -10(1^3)/3 +10(1^2)/2 -(1/3)(10)(3^3) = -86.67

    Here's after making correction , but i still didnt get the ans though
     
  7. Nov 29, 2016 #6

    PhanthomJay

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    Your equations for slope and deflection (part a) look good, but what you have done (I didn't check your math) is to solve for the deflection at 1 m from the free end of the beam when the load is applied at the free end. That is not what the problem is asking in part b, it wants to solve for the deflection at the free end when the load is applied at 1 m from the free end. So you first must calculate the deflection at the load point using x=0 at that point and L = 2 m. Then the 1 m portion of the beam from the free end to the load just goes along for the ride without bending since there is no load on it. The deflection at the free end can then be found by knowing the slope of the curve at the load point.
     
  8. Nov 29, 2016 #7
    well , by using x = 0 , L = 2 , i ended up getting EIy = -0-(1/3)(10)(2^3) = -26.67 , which is still not -46.67 , did i miss out anything ?
     
  9. Nov 29, 2016 #8
    Can you show the shape of deflection of graph ? i have no idea
     
  10. Nov 29, 2016 #9
    do you mean find the y by using x = 0 and y = 2 first , then find the area under the graph of dy/dx from x = 0 to x = 1 from the right end ?
     
  11. Nov 30, 2016 #10

    PhanthomJay

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    Best to find the slope (dy/dx) at the point of application of the load at x=0, then since the shape of the curve from that point to the free end is a straight line (no bending in that region), use basic trig to get the additional deflection to the free end.
     
  12. Nov 30, 2016 #11
    why cant I just use x = 0 , L = 2 , i ended up getting EIy = -0-(1/3)(10)(2^3) = -26.67 to get the max deflection ?
     
  13. Nov 30, 2016 #12

    PhanthomJay

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    that is not the max deflection, that is the deflection at 1 m from the right end. The max deflection occurs at the right end. Solve for the slope of the curve at x= 0 !
     
  14. Nov 30, 2016 #13
    x=
    x = 0 , L = 2 already means the deflection of the beam at x = 0 when the load is applied at 2m from A, right ?
     
  15. Nov 30, 2016 #14

    PhanthomJay

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    Remember we have chosen x=0 to be 2 m from the left end. We want the deflection at 3 m from the left end. Again, solve for dy/dx at x=0 using your equation.
     
  16. Nov 30, 2016 #15
    is it possible to
    find EIymax directly at L = 3 using equation of EIymax only ?
     
  17. Nov 30, 2016 #16
    Huh ?? x = 0 is located at the free end....
     
  18. Nov 30, 2016 #17

    PhanthomJay

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    deflected curve.jpg
    No, we chose x = 0 at the load point and L = 2 because the last 1 m of the beam is stress free and it does not deflect the same way , it just goes along for the ride with a straight line deflection, not a curved cubic deflection. Please solve for dy/dx at x = 0 and what do you get? See attached.
     
  19. Nov 30, 2016 #18
    do you mean if th eload is applied at the stress free end , then , we can detremine the max deflection at the stress free end directly by x = 0 , L= 3 , without involving the dy/dx equation multiply with the length ?
     
  20. Nov 30, 2016 #19

    PhanthomJay

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