Derivation of resonance fixed-fixed beam

1. Sep 7, 2011

DePurpereWolf

I'm trying to derive the resonance frequencies for a simple fixed-fixed beam, as opposed to a simply-supported beam.

I'm working off the following references:
1. http://emweb.unl.edu/Mechanics-Pages/Scott-Whitney/325hweb/Beams.htm
But this is for a fixed-free cantilever beam.

And
2. alrafidain.engineering-coll-mosul.com/files/132.pdf

But this is for a simply-supported (or 'pinned') beam.

Both sources follow the same derivation steps.
If we write the differntial equation as follows:
$$EI\frac{\partial^{4}z(x,t)}{\partial x^{4}} = \rho A \frac{\partial^{2}z(x,t)}{\partial t^{2}}$$
Than for my application I would like to state the following boundary conditions:
$$1, z(0,t) = 0$$
$$2, z'(0,t) = 0$$
$$3, z(L,t) = 0$$
$$4, z'(L,t) = 0$$
For a fixed - fixed beam.

However, I can't seem to derive this to a manageable equation.

The general solution is in the form of:
$$z(x,t) = (A \cos(\omega t)+ B \sin(\omega t)) \cdot ( C_1 \sin(\alpha x) + C_2 \cos(\alpha x) + C_3 sinh(\alpha x) + C_4 cosh(\alpha x))$$

Assuming the time term is not zero.
Boundary Condition 1 gives:
$$C_2 + C_4 = 0$$
BC 2:
$$C_1 + C_3 = 0$$
BC 3:
$$C_1 sin(\alpha L) + C_2 cos(\alpha L) -C_1 sinh(\alpha L) - C_2 cosh(\alpha L) = 0$$
BC 4:
$$C_1 cos(\alpha L) - C_2 sin(\alpha L) -C_1 cosh(\alpha L) - C_2 sinh(\alpha L) = 0$$

I can get an expression for C1 to C4 but it's incredibly long and I'm not sure what to do with it next. It doesn't derive to a nice equation as in the references.

Could someone with expertise have a look and see if I'm stating my equations correctly. And maybe help me along with the derivations?

2. Sep 7, 2011

Studiot

Look again at your boundary conditions.

I agree z(0) = z(L) = 0

but why is

z'(0) = z'(L) = 0

3. Sep 7, 2011

Bill_K

Standard conditions for a beam problem. The beam is clamped at the ends, so both the displacement and the slope must be zero there.