- #1
DePurpereWolf
- 15
- 0
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:
[tex] EI\frac{\partial^{4}z(x,t)}{\partial x^{4}} = \rho A \frac{\partial^{2}z(x,t)}{\partial t^{2}} [/tex]
Than for my application I would like to state the following boundary conditions:
[tex] 1, z(0,t) = 0 [/tex]
[tex] 2, z'(0,t) = 0 [/tex]
[tex] 3, z(L,t) = 0 [/tex]
[tex] 4, z'(L,t) = 0 [/tex]
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:
[tex] 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)) [/tex]
Assuming the time term is not zero.
Boundary Condition 1 gives:
[tex] C_2 + C_4 = 0 [/tex]
BC 2:
[tex] C_1 + C_3 = 0 [/tex]
BC 3:
[tex] C_1 sin(\alpha L) + C_2 cos(\alpha L) -C_1 sinh(\alpha L) - C_2 cosh(\alpha L) = 0 [/tex]
BC 4:
[tex] C_1 cos(\alpha L) - C_2 sin(\alpha L) -C_1 cosh(\alpha L) - C_2 sinh(\alpha L) = 0 [/tex]
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?
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:
[tex] EI\frac{\partial^{4}z(x,t)}{\partial x^{4}} = \rho A \frac{\partial^{2}z(x,t)}{\partial t^{2}} [/tex]
Than for my application I would like to state the following boundary conditions:
[tex] 1, z(0,t) = 0 [/tex]
[tex] 2, z'(0,t) = 0 [/tex]
[tex] 3, z(L,t) = 0 [/tex]
[tex] 4, z'(L,t) = 0 [/tex]
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:
[tex] 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)) [/tex]
Assuming the time term is not zero.
Boundary Condition 1 gives:
[tex] C_2 + C_4 = 0 [/tex]
BC 2:
[tex] C_1 + C_3 = 0 [/tex]
BC 3:
[tex] C_1 sin(\alpha L) + C_2 cos(\alpha L) -C_1 sinh(\alpha L) - C_2 cosh(\alpha L) = 0 [/tex]
BC 4:
[tex] C_1 cos(\alpha L) - C_2 sin(\alpha L) -C_1 cosh(\alpha L) - C_2 sinh(\alpha L) = 0 [/tex]
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?