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2.092/2.093 Finite Element Analysis of Solids & Fluids I Fall 09

Lecture 23 - Solution of K = M

Prof. K. J. Bathe MIT OpenCourseWare

Reading assignment: Chapters 10, 11

We have the solutions 0 < 1 2 . . . n . Recall that:

1 2 n

K i = i M i (1)

In summary, a necessary and sufficient condition for i is that Eq. (1) is satised. The orthogonality conditions are not sufficient, unless q = n . In other words, vectors exist which are K - and M -orthogonal, but are not eigenvectors of the problem.

= 1 . . . n (2)

1 zeros

T M = I ; T K = = . . .

(3)

zeros n

Assume we have an n q matrix P which gives us

P T MP = I ; P T KP = A diagonal matrix q q q q

Is a ii necessarily equal to i ?

a 11 zeros

a 22

.zeros

..

If q = n , then A = , P = with some need for rearranging. If q < n , then P may contain eigenvectors (but not necessarily), and A may contain eigenvalues.

Rayleigh-Ritz Method

This method is used to calculate approximate eigenvalues and eigenvectors.

v T Kv (v ) =

v T Mv

1 (v ) n

1 is the lowest eigenvalue, and n is the highest eigenvalue of the system. 1 is related to the least strain energy that can be stored with v T Mv = 1:

T 1 K 1 = 1 (if 1 T M 1 = 1)

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Lecture 23 Solution of K = M 2.092/2.093, Fall 09

Note that twice the strain energy is obtained when the system is subjected to 1 . If the second pick for v gives a smaller value of (v ), then the second pick is a better approximation to 1 .

q

Assume = i x i , and the Ritz vectors i are linearly independent. Also, = [ 1 . . . q ]. The x i will i =1

be selected to minimize ( ). Hence, calculate ( ) = 0. (See Chapter 10.) The result is x i

Mx Kx = (4)

K = T K ; M = T M (5)

We solve Eq. (4) to obtain 1 , 2 , . . . , q and x 1 , x 2 , . . . , x q . Then our approximation to 1 , . . . , q is given by 1 , . . . , q .

1 1 ; 2 2 ; q q

1 1 ; 2 2 ; etc.

where 1 . . . q n q

= n q

[ x 1 . . . x q ] q q

.

If the q Ritz vectors span the subspace given by 1 , . . . , q , then we obtain ( 1 . . . q ) and ( 1 . . . q ). Pictorially, an example:

If 1 and 2 are in the x -y plane, then by the Rayleigh-Ritz analysis we get 1 , 2 . Major shortcoming: in general, we do not know the accuracy of ( i , i ).

The Subspace Iteration Method

Pick X 1 , then calculate for k = 1 , 2, 3, . . . n q

KX k +1 = MX k (a)

This is inverse iteration with q vectors. Now perform the Rayleigh-Ritz solution:

K k +1 = X kT

+1 KX k +1 ; M k +1 = X kT

+1 MX k +1 (b)

K k +1 Q k +1 = M k +1 Q k +1 k +1 (c)

K k +1 , M k +1 , and Q k +1 have dimensions q q . Recall that we have K = M from Eq. (1). We then have

Q T k +1 K k +1 Q k +1 = k +1 ; QT k +1 M k +1 Q k +1 = I (d)

Finally, X k +1 = X k +1 Q k +1 (e)

Equations (b), (c), and (e) correspond to the use of the Rayleigh-Ritz method.

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Lecture 23 Solution of K = M 2.092/2.093, Fall 09

Then, provided the vectors in X 1 are not M -orthogonal to the eigenvectors we seek, we have (with good ordering) that

1

k +1 . . .

q

X k +1 1 . . . q

In practice, we use q vectors to calculate the p lowest eigenvalues, with (say) q = 2 p. In fact, the convergence irate of the vectors is given by q +1 .

If p = 2 and we have a multiplicity of 5 (or higher), q = 2 p corresponds to not enough vectors. Ideally, we want q+1 to be signicantly larger than p , so that i is much less than 1 for i = 1 , . . . , p . The quite q +1 conservative way is to use

q = max(2 p, p + 8)

The textbook gives q = min(2 p, p+8), which can also be used (apply the Sturm sequence check, see textbook); it will use less storage, but will generally need more iterations. For modern computers (specically with

parallel processing), the above formula for q is frequently more effective. Notice that X kT

+1 MX k +1 = I because from (e),

Q T k +1 X T k +1 MX k +1 Q k +1 = I

M k +1

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