jj77
posted this
11 February 2019

- Last edited 11 February 2019
Well for the other modes (say11), the two halfs (about centre line) are in opposite phase. So this constant load does not have different phase (constant all over plate with same phase and amplitude) hence it can not excite it that much.

Easy to think about this is for a pinned-pinned beam. The first mode is half a sine (n =1), second is a full sine (n=2). If we have a constant load on the beam, then the anti-symmetric modes about the centre point of the beam (even multiples, that is n= 2, 4,..) can not be excited much since that would require a force that has opposite phase about the centre line. Thus here a symmetric load excites symmetric modes.

Same goes if we had a point force at the centre point that would not excite the anti-symmetric modes that have no movement there.

It is all maths, and it is a term in the solution that involves the inner product (integral not algebra), of the eigenmodes and the excitation. That will give how much a mode is excited. In the case of this point force one gets for the second mode (n=2), <sin(2pix/L), Dirac delta(x-L/2)>, that gives sin(2piL/2L) = sin(pi)=0, thus it can not be excited. One can do similar things for a plate, but that would involve Bessel, and other eigen-solutions, than sine.

The same thing can be found if we instead of a point force (Dirac delta), use just a constant force all over the beam. It will lead to that anti-symmetric modes are not excited.