How do rotated supports work

If you rotate a support, it can only take reaction in the direction of the rotation.

So if you rotate for example a support in the Y direction (= left support in the image) over an angle of α  ↻ in the XY plane, it means that

• the resulting reaction forces (R_X and R_Y) in Y and X direction are related.
  R_X = R * sin (α)
  R_Y = R * cos (α)Otherwise, the resulting vector for the reaction can never have the same angle α as the support.
• the rotated Y support can not be seen as a combination of a support in the Y and X-direction (although that is what you would expect). Because in the latter case, there is no correlation between the magnitude of the reactions in Y and X direction (while that is the case in a rotated support).

• Finding equilibrium with rotated supports is not guaranteed.

An example:

1. Consider a beam loaded in the middle with a point load Q. The boundary conditions are:
   • In A: support in Y direction, rotated 45° ↻
   • In B: support in Y direction, not rotated
     
2. The load Q can be split into two components Q_X and Q_Y.
3. The horizontal load Q_X must be beared by support A, resulting in a reaction R_A.X.
4. Since support A can only bear forces in this  direction, also the vertical component of the load Q_Y must be beared by support A.
5. Since R_A.X = Q_X and R_A.Y = Q_Y, it follows that R_A = Q. Translational equilibrium is met. Now let’s write a rotational equilibrium around point A: Q * 0.5* L = 0Conclusions:
   • With the current boundary conditions no rotational equilibrium can be found.
   • You’ll get the message in Diamonds: equilibrium issues in the following direction(s): X; Y.^*
6. It would be a different story when point A would be a support in the Y and X-direction. Then support A would take the horizontal force Q_X and support B the vertical force Q_Y. There would be no vertical reaction in support A (R_A.Y=0). The rotational equilibrium would be met.