## Advanced Dynamics and Vibrations: Modal analysis of mdof systems with generalized coordinates, semi-definite systems, repeated roots

### Introduction to Normal Mode Analysis

Our analysis of MDOF systems has shown that regardless of the coordinates used that there are specific properties of the system(natural frequencies, mode shapes) that do not depend on our choices. These properties can be used to uncouple the equations of motion into equations that can be easily solved and then transformed back to the original coordinates which are the most useful form to use the results. The process uses the modal shapes, in the form of vectors, to build a modal matrix. Consider the model vectors . The modal matrix is formed by making these vectors the columns in the matrix:

but instead of using we normalize it to using the mass matrix so that:

or

NOTE:

To start, consider the undamped forced 2DOF system:

Therefore:

Normalize and using where . Therefore:

Therefore:

NOTE Orthogonality:

Modal vectors are orthogonal unit or . Thefore the modal matrix is:

Now transform the equations of motion using by setting then .

Now multiply the equations of motion by to get:

Therefore, the transformed equations become

Where

We can now solve this set of uncoupled SDOF equations for the free, transient on steady-state conditions. In the transformed state the stiffness matrix is now the matrix of natural frequencies:

after solving we transform back to the original condition.

### Normal analysis of MDOF systems

The equations of motion for viscously damped MDOF systems are in general

For up to 2 DOF we can easily use a complex analysis (transform methods). However for larger system we often use modal analysis.

In this approach we start with the homogeneous equations and we will also start with the undamped case. (We will return to it later!) Therefore,

We look for so-called asynchronous motions SSHM (Simultaneous Simple Harmonic motion) SSHM by assuming

or

and get

This is true for only special say and special associated with

We can find the & the within a multiplicative constant. Each and its associated is called a mode.

If the are distinct then we can easily show that the are orthogonal to one another in some sense. To see this consider the two modes say & , then

Multiply the first by

and the second by

Since we can transpose the first equation so that the left hand side are the same. It suggests that

Therefore, the modal vector are orthogonal with respect to the mass and stiffness matrices.

These eigenvectors form a linearly independent set i.e. any motion of the system can be thought of as a linear combination of these modes. Physically this implies that any motion can be thought of as a superposition of these normal modes.

Linear independence means that we cannot write any vector as a combination of the others. If they were linearly dependent there should be some way to combine them to produce one from the others. This means there should be:

But if we premultiply this equation by we get:

but all the terms are zero (orthogonality) except for and therefore . The same is value for each term therefore

as a result any motion can be written as a linear combination of the modal vectors. They are a set of basis vectors in the same way as are for positions. Therefore,

Where is the modal participation factor and

This is called the expansion theorem in vibration where the are functions of time.

It turns out that if we have repeated eigenvalue we can still find a set of eigenvectors that span the system.

We can use these properties of the system to solve these equations in general. The process uses the modal vectors put together in a so-called modal matrix. Before starting we will normalize the modal vectors into as they are only known to within a multiplicative constant.

The normalization is:

Therefore,

Now form the modal matrix

Then

Therefore

Now to solve the original system we transform the equations to a set of generalized coordinates. That is we set:

putting this into the original set of equation gives:

now premultiply by

This uncouples the equations so that each looks like:

The general solution of this equation is:

We can then transform the solution back to the original coordinates using:

This expression contains the initial conditions in the transformed coordinates and are related to the original values by:

We can also solve the steady-state forced problem by simply considering the particular integral instead of the total solution.

### Summary (Free Vibration)

We have been considering the solution of

Where we transform the equation using

If we premultiply by and substitute into we get:

but

Therefore

The solution to these equation is

Where & are constants, Therefore

If and are the initial displacements and velocities then

Therefore,

since , therefore

This solves every free vibration problem for any initial conditions.

Therefore

Characteristic equation is:

Therefore,

Therefore,

Similarly:

now find the force vector in the normal coordinates

Since there are initial conditions with no displacement and no velocity we can simply integrate the convolution integrals to give

We can now find the response in terms of the original coordinates using:

Note that the first mode is most pronounced in the overall response.

Check static deflection:

Therefore, ,

O.K

O.K

Consider a second example with certain initial conditions:

Insert diagram.

A static force is suddenly released for the system of 3 masses mounted on a tightly stretched wire with tension . Determine the resulting free vibration using normal analysis.

Solution:

While the mass matrix is easy we must find the stiffness matrix using say

Apply unit def’l to m,

Therefore,

Therefore,

now find the ‘s & , assume to get:

The characteristic equation is

or

Therefore,

now find , choose

For

For

For

Therefore,

now normalize the to get where , therefore as

Similarly,

Therefore,

Also need the initial conditions

Therefore

We can simplify put this into the general solution:

Again the first mode is the dominant one and the second mode unsymmetric mode is not present if we had chosen different initial condition say, then

Therefore

All three modes would now participate in the motion. To find the response simply insert the new and calculate the .

### e.g Steady State

The equations of motion are

Therefore,

Divide by and set

Therefore,

Note: ,

Therefore, from and

From ,

Therefore:

For

therefore

For

Again

Therefore:

Normalized eigenvectors

therefore:

The transformed forces are:

Therefore transformed equations are:

Steady State Solutions of these equations

For the case in which we have repeated eigenvalues we can still have orthogonal eigenvectors using a process called successive orthogonalization.

If the eigenvalues is repeated p times let be an arbitrary set of modal vectors that satisfy the degenerate set of equations. We construct an orthoganal set using

etc.

To prove take

The characteristic equation is

therefore:

For the repeated root

from

Therefore there are an infinite number of solutions so choose one which satisfies the equations. e.g.:

Now we make them orthogonal

Now and are orthonormal for therefore:

and this is our basis.

Now consider the addition of damping to the system

Now transform this to normal coordinates

and premultiply by

therefore:

but there is no a priori reason that is a diagonal matrix and therefore the equations are coupled. If it is, we can use the same approach as before. One condition that is often used is the so-called proportional damping where

so that the transformed equations are necessarily diagonal. We can then set

then each transformed equation is

The general solution of this equation is

where

Now consider that the are related

therefore

Therefore we can only choose the damping independently for 2 modes. If then so that the damping in each mode is proportional to the frequency of that mode. The higher the frequency the higher the damping so that higher modes damp out more rapidly. [Note: Often called relative damping.]

On the other hand, setting implies the damping is proportional to the mass matrix which damps out the lower modes more than the higher ones. [Note: Often called absolute damping.]

Return to a simple problem we have done previously

Normalize

Therefore:

and the transformal equations for free vibration are

Therefore:

We transform these back to the original coordinates

Assume the initial conditions are:

then:

We can find the transformed initial conditions from

Therefore:

Therefore:

as before.

Now add proportional damping

Select relative damping (proportional to stiffness)

The transformed equations will be

but we called

The solution to the initial value problem for the case discussed is

As before

We can now consider the decay of each of these two components using the idea of the so-called logarithmic decrement. Recall from the SDOF case that

where is the number of cycles to decay to the ratio selected for

Once is chosen decreases linearly with .

If we set

and increases linearly with .

Consider a steady-state damped problem

The equations of motion are:

Consider the undamped problem to find the and

therefore:

therefore:

To find

Therefore

In the same manner,

Therefore:

Therefore undamped solution is from

The undamped steady state solution is

As a result the undamped solution is

As was shown previously at

Now lets add damping

Therefore:

Therefore the equations of motion are:

Consider the standard form where the coefficient of is

Therefore