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Z axis reaction when using symmetry plane

    • pv00170

      I am analysing the reaction of a bridge under self-weight (Y direction). The bridge is halved, and a symmetry plane is applied to reduce computational time.

      When I check the reactions, I get:

      • X axis = correct (0 N)
      • Y axis = correct (1.46 e6 N, which is the same as the self-weight)
      • Z axis = ?? (5.12 e5 N)

      I have some questions regarding the Z axis:

      • Are these forces existent (even though no Z-direction forces are applied) because of the symmetry plane?
      • Does this imply that if I run the model but symmetrically I should get the same Z but with the opposite sign?

    • pv00170

      For the whole bridge

      I have compared some results. I have used Solution > Insert > Probe > Force reaction, and then I select Location method > Boundary condition and then Fixed support.

      I have 16 supports, 8 on the left-hand side and 8 on the right-hand side. The following table shows the results of the reactions:

      Whole bridgeX (N)Y (N)Z (N)
      Left (8 supports)-3.83E+055.44E+061.24E+06
      Right (8 supports)-3.82E+055.44E+06-1.24E+06
      Sum (by hand)-7.65E+051.09E+07100

      But then, if I consider the 16 supports altogether (rather than sepparatly) I get a different value:

       X (N)Y (N)Z (N)
      Both (16 supports)-6.67E+059.41E+06-1


      For the symmetry bridge

      I get the reactions of the 8 supports which multiplied by two, coincides with the previous one. 

      Symmetry bridgeX (N)Y (N)Z (N)
      Left (8 supports)-3.33E+054.70E+061.66E+06
      2 time left (by hand)-6.66E+059.41E+06? 3.32E+06

      So my question also is: how come the addition of the 8 + 8 supports is not the same as the 16 supports?

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