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Singularity function

  1. Jul 12, 2016 #1
  2. jcsd
  3. Jul 12, 2016 #2
    or ( <x-0.5L>^4 )= 0 is due to the author taking x less than 0.5L ?
     
  4. Jul 12, 2016 #3
    bump
     
  5. Jul 12, 2016 #4

    SteamKing

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    The boundary condition for the beam is evaluated at x = 0, so the singularity function evaluates to zero.
     
  6. Jul 12, 2016 #5
    if we do not consider x = 0 , then x can by any point along the beam ABC ?
     
  7. Jul 13, 2016 #6

    SteamKing

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    That's not the point.

    The deflection function has been constructed for this beam using singularity functions. Because the deflection function was obtained using double integration, there are some unknown constants of integration which must be determined so that the particular deflection function for this beam can be determined.

    At point A, the length coordinate is x = 0 and the deflection there is y = 0. This is one of the boundary conditions for this beam. By taking the general deflection function and substituting x = 0 and y = 0 into it, one can see that the constant c2 = 0. That's why this exercise was done in the first place. By using a different boundary condition, such as the fixed end at point C, then substituting x = L and y = 0 should determine the value of the constant c1.

    Because this beam is statically indeterminate, the reactions RA, RC, and MC will also require solution. You should study the worked out example from the OP, since there are several pages of calculations altogether.
     
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