Does a sine-shaped beam remain sine when pushed?

[refind]

I have a linear elastic thin beam y=sin(pi x) from 0<x<1 and the beam is pin supported (no moment applied) at x=0 and x=1 and constrained so that the ends remain on the x axis. Then I push the end from x=1 to x=1-delta (for some small delta, say 0.1). Will the resulting beam shape still be a sine?

Thanks

 

[sophiecentaur]

It will have a finite modulus and non-zero mass per unit length so the speed of propagation of a compression wave will be finite, hence the x/y relationship will not be a sine wave over the length of the beam during the change because the nearer parts will have compressed before the far parts have moved, distorting the shape.

 

[a1call]

If a slight compression maintaines the sine function, then any compression, large or small should maintain the function since a large compression would equate to many small compressions.
However it is easy to visualize a large compressive force deforming the middle portion due to leverage much more than the sides eventually forcing the bottom to arc inwards. This can not be a sine wave, despite the possibility of a sine wave having infinitely small wave length.

 

[refind]

In case anyone is confused I'm asking this as a static load problem, nothing with wave propagation.

I'm asking because I want to calculate the force applied by the beam when I compress it, and I need to know the function in order to calculate it.

a1call: Can I at least assume that the beam remains approximately sine if the deflection is small?

 

[a1call]

I would say yes.