# Does a sine-shaped beam remain sine when pushed?

1. Apr 24, 2013

### 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

2. Apr 24, 2013

### 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.

3. Apr 24, 2013

### 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.

4. Apr 24, 2013

### 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?

Last edited: Apr 24, 2013
5. Apr 24, 2013

### a1call

I would say yes.