# Nonlinear finite elements/Natural vibration

## Special case : natural vibrations

Recall that the finite element system of equations has the form

${\displaystyle \mathbf {M} ~{\ddot {\mathbf {u} }}=\mathbf {f} _{\text{ext}}-\mathbf {f} _{\text{int}}~.}$

We could also have written this equation as

${\displaystyle \mathbf {M} ~{\ddot {\mathbf {u} }}+\mathbf {K} ~\mathbf {u} =\mathbf {f} ~.}$

For natural vibrations, the forces and the displacements are assumed to be periodic in time, i.e.,

${\displaystyle \mathbf {u} =\mathbf {u} ^{0}~\exp(i\omega t)~.}$

and

${\displaystyle \mathbf {f} =\mathbf {f} ^{0}~\exp(i\omega t)~.}$

Then, the accelerations take the form

${\displaystyle {\ddot {\mathbf {u} }}=(i\omega )^{2}~\mathbf {u} ^{0}~\exp(i\omega t)=-\omega ^{2}~\mathbf {u} ^{0}~\exp(i\omega t)~.}$

Plugging these into the FE system of equations, we get

${\displaystyle [-\omega ^{2}~\exp(i\omega t)]\mathbf {M} ~\mathbf {u} ^{0}+\exp(i\omega t)~\mathbf {K} ~\mathbf {u} ^{0}=\exp(i\omega t)~\mathbf {f} ^{0}~.}$

After simplification, we get

${\displaystyle {\left(-\omega ^{2}\mathbf {M} +\mathbf {K} \right)~\mathbf {u} ^{0}=\mathbf {f} ^{0}~.}}$

If there is no forcing, the right hand side is zero, and we get the finite element system of equations for free vibrations

${\displaystyle {\left(-\omega ^{2}\mathbf {M} +\mathbf {K} \right)~\mathbf {u} ^{0}=0~.}}$

The above equation is similar to the eigenvalue problem of the form

${\displaystyle \mathbf {A} ~\mathbf {x} =\lambda ~\mathbf {x} \qquad \equiv \qquad \left(\mathbf {A} -\lambda \mathbf {I} \right)\mathbf {x} =0~.}$

Since the right hand side is zero, the finite element system of equations has a solution only if

${\displaystyle {\det(\mathbf {K} -\omega ^{2}\mathbf {M} )=0~.}}$

For a two noded element,

${\displaystyle \mathbf {K} ={\begin{bmatrix}K_{11}&K_{12}\\K_{21}&K_{22}\end{bmatrix}}~{\text{and}}~\mathbf {M} ={\begin{bmatrix}M_{11}&M_{12}\\M_{21}&M_{22}\end{bmatrix}}~.}$

Therefore,

${\displaystyle \mathbf {K} -\omega ^{2}~\mathbf {M} ={\begin{bmatrix}K_{11}-\omega ^{2}~M_{11}&K_{12}-\omega ^{2}~M_{12}\\K_{21}-\omega ^{2}~M_{21}&K_{22}-\omega ^{2}~M_{22}\end{bmatrix}}~.}$

The determinant is

${\displaystyle \det(\mathbf {K} -\omega ^{2}~\mathbf {M} )=(K_{11}-\omega ^{2}~M_{11})(K_{22}-\omega ^{2}~M_{22})-(K_{12}-\omega ^{2}~M_{12})(K_{21}-\omega ^{2}~M_{21})~.}$

This gives us a quadratic equation in ${\displaystyle \omega ^{2}}$  which can be solved to find the natural frequencies of the element.