Slide 1

Sensitivity of EigenproblemsReview of properties of vibration and buckling modes. What is nice about them?Sensitivities of eigenvalues are really cheap! Sensitivities of eigevectors. Why bother getting them? Think of where you want your car to have the least vibrations

When we design subject to frequency or buckling constraints, we need the derivatives of the vibration frequencies and buckling loads with respect to design variables

Derivatives of static response can be calculated more cheaply than finite difference derivatives. This is even truer of derivatives of frequencies and buckling loads, because they are calculated as solution to eigenvalue problems. Derivatives of eigenvalues are really easy to calculate and implement.

We will also cover the calculation of derivatives of eigenvectors. These are of interest in fewer practical problems. A typical example is in the design of cars. It is impossible to eliminate vibrations altogether, but there is an advantage of having vibration modes that have the smallest amplitude at the seat of the driver or at the steering column.1The eigenproblemCommon notation for vibration and buckling

For vibration M is mass matrix, for buckling it is geometric stiffness matrix.Usually W=Mu is vibration or buckling mode, and is the square of the frequency of buckling loadWhat are the properties of K and M?What do we know about the eigenvalues and eigenvectors?

There is the standard eigenvalue problem Ku=ru, where r is the eigenvalue, but for vibration and buckling problems, we solve instead the generalized eigenvalue problem given by the first equation in the slide. I am using the notation for the vibration problem as M is the standard notation for the mass matrix. For buckling that matrix represents the geometric stiffness matrix.

For vibration problems the eigenvalues are the square of the natural frequencies, so that the eigenvalues are always positive. For buckling problems they are buckling loads, and they can be either positive or negative but real.

The properties that are most relevant for us is that because K and M are symmetric and K is positive definite, there is a full set of eigenvalues (though they can be repeated) each with its own eigenvector, and they are orthogonal with respect to M and K.2Derivatives of eigenvaluesDifferentiate:

Pre multiply by :

What is the physical meaning?Why is it cheap to calculate?

We start by differentiating the two equations with respect to a design variable x. The second equation we will need later for derivatives of eigenvectors. The derivative of the eigenvalue is obtained by multiplying the first equation on the left by the eigenvector. Because of the symmetry of the matrices the left and right eigenvectors are the same, so the first term drops out and we are left with the other two terms that gives us the formula for the eigenvalue derivative.

If we denote the derivative with respect to x by a prime, we will see later that uTKu is the derivative of the elastic energy and uTMu is proportional to the derivative of the kinetic energy. So the prescription for raising a frequency is to increase the stiffness where you have maximum elastic energy and reduce the mass where you have maximum kinetic energy.

Obviously this is cheap to calculate. For static sensitivity the derivative calculation required solution for another load vector. Here we do not need to solve for anything.3Problems eigenvalue sensitivityHow you would apply the physical interpretation of the derivatives of eigenvalues to raising or lowering the frequency of a cantilever beam?Check this by using the beam in the semi-analytical problem, assuming that it has a cross-section of 4.5x2, and is made of steel with density of 0.3 lb/in3. Compare the effect of halving the height of the first and last of the 10 elements. Check the frequency of the original beam against a formula from a textbook or web.4Eigenvector derivativesCollecting equations

Difficult to solve because top-left matrix is singular. Why is it?Textbook explains Nelsons method, which uses intermediate step of setting one components of the eigenvector to 1.

If we just collect the equations from the previous slide together, we get combined equations for the eigenvalue and eigenvector derivatives. It may appear that because we already know what is the derivative of the eigenvalue we can solve the first line without paying any attention to the derivative of the normalizing condition.

However, the matrix K-muM is singular, because mu is an eigenvalue, so we do need the normalizing condition. The textbook describes the Nelson method for solving these equations by first adding a fictitious condition that one component of the eigenvector is equal to 1. Then the derivative is corrected by using the derivative of the normalization condition.

All of that is applicable to the case that the eigenvalues are distinct. When we have repeated eigenvalues the situation is more complicated.5Spring-mass exampleFig. 7.3.1

Stiffness and mass matrices (all springs and masses initially equal to one.

Solution of eigenproblem

We consider a simple spring mass example, where initially there is no jackpot and all the springs and masses have unit values. We will want to calculate the derivative with respect to the stiffness of the left spring, so we denote it by k. However, we solve first the eigenvalue problem for k=1. in Matlabk=[2 -1;-1 2]; [u,mu]=eig(k)u = -0.7071 -0.7071 -0.7071 0.7071mu = 1 0 0 3The frequencies are the square roots of the eigenvalues, and the first column of u is the eigenvector corresponding to the lowest frequency. Matlab normalizes it with respect to the unit matrix. For the derivative calculation it is also useful to look for the normalization where the largest component is one.

6Derivative w.r.t kDerivatives of matrices

Derivative of eigenvalue

See in textbook derivative of eigenvector

Do those pass sanity checks?

When we calculate the derivative of the eigenvalue with respect to k using the euqaution we get 0.5. The textbook shows the calculation of the derivative of the eigenvector.

We need to check whether the results are reasonable. First, increasing stiffness, increases frequency, so it is good that the derivative is positive. Second, the logarithmic derivative is 0.5, and that makes sense because if we expect that the logarithmic derivative with respect to both spring constants will be one.

For the derivative of the eigenvector, the sign is right, because if we stiffen the left spring we expect to have less motion on the left side.7Eigenvectors are not always uniqueWhen can we expect two different vibration modes with the same frequency?Why does optimization with frequency constraints likely to lead to repeated eigenvalues?Vibration modes are orthogonal when eigenvalues are distinct, but any combination of modes corresponding to the same frequency is also a vibration mode!

When we do optimization and we try to increase the first vibration frequency, we often will run into it crossing the second frequency, so that at the optimum both frequencies will be the same. That creates a problem because we have a repeated eigenvector, that is two eigenvectors that have the same eigenvalue. Then the eigenvector is not unique, and we have a problem calculating derivatives.8Example 7.3.2Problem definition and solution

Eigenvectors for x=0

Eigenvectors for y=0

At x=y=0 eigenvalues are the same and eigenvectors are discontinuous

For an example of the problems of repeated eigenvalues we use Example 7.3.2, which is purely algebraic. The two eigenvalues coalesce for x=y=0. However, the eigenvectors for x=0 are totally different from the ones for y=0. So at x=y=0 there is discontinuity in the eigenvectors and the eigenvalues are not differentiable.9Eigenvalues for example 7.3.2 .

Figure showing the eigenvalues.10Deriviatives of repeated eigenvaluesAssume m repeated eigenvectors

To find eigenvalue derivatives need to solve a second eigenvalue problem!

11Calculation of derivatives w.r.t xAt x=y=0 any vector is an eigenvector.

Similarly get

Why are these derivatives of limited valueWhat happens if we try to use them for dy=2dx=2dt?

Problems (optional)Explain in 50 words or less why derivatives of vibration frequencies are relatively cheaper than derivatives of stressesWhen eigenvalues coalesce, they are not differentiable even though we can still use Nelsons method to calculate derivatives. How can you reconcile the two statements?Why is the accuracy of lower frequencies (and their derivatives) better than that of higher frequencies?

Source: Smithsonian InstitutionNumber: 2004-57325