Nonlinear finite elements/Natural vibration

Special case : natural vibrations

Recall that the finite element system of equations has the form

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

We could also have written this equation as

$𝐌\ddot{𝐮}+𝐊𝐮=𝐟.$

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

$𝐮={𝐮}^0\exp (i\omega t).$

and

$𝐟={𝐟}^0\exp (i\omega t).$

Then, the accelerations take the form

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

Plugging these into the FE system of equations, we get

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

After simplification, we get

$(-{\omega }^2𝐌+𝐊){𝐮}^0={𝐟}^0.$

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

$(-{\omega }^2𝐌+𝐊){𝐮}^0=0.$

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

$𝐀𝐱=\lambda 𝐱\equiv (𝐀-\lambda 𝐈)𝐱=0.$

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

$\det (𝐊-{\omega }^2𝐌)=0.$

For a two noded element,

$𝐊=\left[\begin{array}{cc}{K}_{11}& K_{12}\\ K_{21}& K_{22}\end{array}\right]\text{and}𝐌=\left[\begin{array}{cc}{M}_{11}& M_{12}\\ M_{21}& M_{22}\end{array}\right].$

Therefore,

$𝐊-{\omega }^2𝐌=\left[\begin{array}{cc}{K}_{11}-{\omega }^2{M}_{11}& K_{12}-{\omega }^2{M}_{12}\\ K_{21}-{\omega }^2{M}_{21}& K_{22}-{\omega }^2{M}_{22}\end{array}\right].$

The determinant is

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

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