김경종
470 lines
30 KiB
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470 lines
30 KiB
Markdown
<!-- source-page: 881 -->
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Therefore, the solution to the eigenproblem $\mathbf{K}\phi = \lambda \mathbf{M}\phi$ is
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$$
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\lambda_ {1} = \frac {1}{2} - \frac {\sqrt {2}}{4}; \quad \Phi_ {1} = \left[ \begin{array}{c} \frac {1}{4} \\ \frac {1}{2} \\ \frac {1 + \sqrt {2}}{4} \\ \frac {\sqrt {2}}{2} \end{array} \right]
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$$
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$$
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\lambda_ {2} = \frac {1}{2} + \frac {\sqrt {2}}{4}; \quad \Phi_ {2} = \left[ \begin{array}{c} - \frac {1}{4} \\ - \frac {1}{2} \\ \frac {- 1 + \sqrt {2}}{4} \\ \frac {\sqrt {2}}{2} \end{array} \right]
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$$
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$$
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\lambda_ {3} = \infty ; \quad \phi_ {3} = \left[ \begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]
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$$
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$$
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\lambda_ {4} = \infty ; \quad \phi_ {4} = \left[ \begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \end{array} \right]
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$$
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In the above discussion we gave the formal matrix equations for carrying out static condensation. The main computational effort is in calculating $K_{a}$ given in (10.46), where it should be noted that in practice a formal inversion of $K_{cc}$ is not performed. Instead, $K_{a}$ can be obtained conveniently using the Cholesky factor $\tilde{L}_{cc}$ of $K_{cc}$ . If we factorize $K_{cc}$ ,
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$$
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\mathbf {K} _ {c c} = \tilde {\mathbf {L}} _ {c} \tilde {\mathbf {L}} _ {c} ^ {T} \tag {10.48}
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$$
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we can calculate $\mathbf{K}_a$ in the following way:
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$$
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\mathbf {K} _ {a} = \mathbf {K} _ {a a} - \mathbf {Y} ^ {T} \mathbf {Y} \tag {10.49}
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$$
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where Y is solved from
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$$
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\tilde {\mathbf {L}} _ {c} \mathbf {Y} = \mathbf {K} _ {c a} \tag {10.50}
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$$
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As pointed out earlier, this procedure is, in fact, Gauss elimination of the massless degrees of freedom, i.e., elimination of those degrees of freedom at which no external forces (mass effects) are acting. Therefore, an alternative procedure to the one given in (10.42) to (10.50) is to directly use Gauss elimination on the $\phi_{c}$ degrees of freedom without partitioning K into the submatrices $K_{aa}$ , $K_{cc}$ , $K_{ac}$ , and $K_{ca}$ because Gauss elimination can be performed in any order (see Example 8.1, Section 8.2.1). However, the bandwidth of the stiffness matrix will then, in general, increase during the reduction process, and problems of storage must be considered.
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For the solution of the eigenproblem $K_{a}\phi_{a} = \lambda M_{a}\phi_{a}$ , it is important to note that $K_{a}$ is, in general, a full matrix, and the solution is relatively expensive unless the order of the matrices is small.
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Instead of calculating the matrix $K_{a}$ , it may be preferable to evaluate the flexibility matrix $F_{a} = K_{a}^{-1}$ , which is obtained using
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$$
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\left[ \begin{array}{l l} \mathbf {K} _ {a a} & \mathbf {K} _ {a c} \\ \mathbf {K} _ {c a} & \mathbf {K} _ {c c} \end{array} \right] \left[ \begin{array}{l} \mathbf {F} _ {a} \\ \mathbf {F} _ {c} \end{array} \right] = \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {0} \end{array} \right] \tag {10.51}
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$$
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where I is a unit matrix of the same order as $K_{aa}$ . Therefore, in (10.51), we solve for the displacements of the structure when unit loads are applied in turn at the mass degrees of freedom. Although the degrees of freedom have been partitioned in (10.51), there is no need for it in this analysis (see Example 10.13). Having solved for $F_{a}$ , we now consider instead of (10.45) the eigenproblem
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$$
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\left(\frac {1}{\lambda}\right) \boldsymbol {\phi} _ {a} = \mathbf {F} _ {a} \mathbf {M} _ {a} \boldsymbol {\phi} _ {a} \tag {10.52}
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$$
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Although this eigenproblem is of a slightly different form than the generalized problem $K\phi = \lambda M\phi$ , the transformation to a standard problem proceeds in much the same way (see Section 10.2.5). For the transformation we define
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$$
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\tilde {\boldsymbol {\phi}} _ {a} = \mathbf {M} _ {a} ^ {1 / 2} \boldsymbol {\phi} _ {a} \tag {10.53}
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$$
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where $M_{a}^{1/2}$ is a diagonal matrix with its ith diagonal element equal to the root of the ith diagonal element of $M_{a}$ . Premultiplying both sides of (10.52) by $M_{a}^{1/2}$ and substituting the relation in (10.53), we obtain
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$$
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\tilde {\mathbf {F}} _ {a} \tilde {\boldsymbol {\Phi}} _ {a} = \left(\frac {1}{\lambda}\right) \tilde {\boldsymbol {\Phi}} _ {a} \tag {10.54}
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$$
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$$
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\tilde {\mathbf {F}} _ {a} = \mathbf {M} _ {a} ^ {1 / 2} \mathbf {F} _ {a} \mathbf {M} _ {a} ^ {1 / 2} \tag {10.55}
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$$
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Once the displacements $\phi_{a}$ have been calculated, we obtain the complete displacement vector using
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$$
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\left[ \begin{array}{l} \boldsymbol {\phi} _ {a} \\ \boldsymbol {\phi} _ {c} \end{array} \right] = \left[ \begin{array}{l} \mathbf {I} \\ \mathbf {F} _ {c} \mathbf {K} _ {a} \end{array} \right] \boldsymbol {\phi} _ {a} \tag {10.56}
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$$
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where $F_{c}$ was calculated in (10.51). The relation in (10.56) is arrived at by realizing that the forces applied at the mass degrees of freedom to impose $\phi_{a}$ are $K_{a}\phi_{a}$ . Using (10.51), the corresponding displacements at all degrees of freedom are given in (10.56).
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EXAMPLE 10.13: Use the procedure given in (10.51) to (10.56) to calculate the eigenvalues and eigenvectors of the problem $K\phi = \lambda M\phi$ considered in Example 10.12.
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The first step is to solve the equations
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$$
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\left[ \begin{array}{r r r r} 2 & - 1 & 0 & 0 \\ - 1 & 2 & - 1 & 0 \\ 0 & - 1 & 2 & - 1 \\ 0 & 0 & - 1 & 1 \end{array} \right] \left[ \begin{array}{l l} \mathbf {v} _ {1} & \mathbf {v} _ {2} \end{array} \right] = \left[ \begin{array}{l l} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{array} \right] \tag {a}
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$$
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where we did not interchange rows and columns in $\mathbf{K}$ in order to obtain the form in (10.51).
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For the solution of the equations in (a), we use the $LDL^{T}$ decomposition of K, where
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$$
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\mathbf {L} = \left[ \begin{array}{c c c c} 1 & & & \\ - \frac {1}{2} & 1 & & \\ 0 & - \frac {2}{3} & 1 & \\ 0 & 0 & - \frac {3}{4} & 1 \end{array} \right]; \quad \mathbf {D} = \left[ \begin{array}{c c c c} 2 & & & \\ & \frac {3}{2} & & \\ & & \frac {4}{3} & \\ & & & \frac {1}{4} \end{array} \right]
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$$
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Hence, we obtain
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$$
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\mathbf {v} _ {1} ^ {T} = \left[ \begin{array}{l l l l} 1 & 2 & 2 & 2 \end{array} \right]; \quad \mathbf {v} _ {2} ^ {T} = \left[ \begin{array}{l l l l} 1 & 2 & 3 & 4 \end{array} \right]
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$$
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and hence, $\mathbf{F}_a = \begin{bmatrix} 2 & 2 \\ 2 & 4 \end{bmatrix}$ ; $\mathbf{F}_c = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$
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$$
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\tilde {\mathbf {F}} _ {a} = \left[ \begin{array}{c c} \sqrt {2} & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{c c} 2 & 2 \\ 2 & 4 \end{array} \right] \left[ \begin{array}{c c} \sqrt {2} & 0 \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{c c} 4 & 2 \sqrt {2} \\ 2 \sqrt {2} & 4 \end{array} \right]
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$$
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The solution of the eigenproblem
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$$
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\tilde {\mathbf {F}} _ {a} \tilde {\boldsymbol {\Phi}} _ {a} = \mu \tilde {\boldsymbol {\Phi}} _ {a}
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$$
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$$
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\begin{array}{l} \text { gives } \\ \mu_ {1} = 4 - 2 \sqrt {2}; \quad \tilde {\Phi} _ {a _ {1}} = \left[ \begin{array}{c} - \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \end{array} \right] \\ \mu_ {2} = 4 + 2 \sqrt {2}; \quad \tilde {\Phi} _ {a _ {2}} = \left[ \begin{array}{c} \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \end{array} \right] \end{array} \tag {b}
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$$
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Since $\phi_{a} = \mathbf{M}_{a}^{-1 / 2}\tilde{\phi}_{a}$ , we have
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$$
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\boldsymbol {\Phi} _ {a _ {1}} = \left[ \begin{array}{l l} \frac {1}{\sqrt {2}} & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{c} - \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \end{array} \right] = \left[ \begin{array}{l} - \frac {1}{2} \\ \frac {1}{\sqrt {2}} \end{array} \right]; \quad \boldsymbol {\Phi} _ {a _ {2}} = \left[ \begin{array}{l} \frac {1}{2} \\ \frac {1}{\sqrt {2}} \end{array} \right] \tag {c}
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$$
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The vectors $\phi_{c_1}$ , and $\phi_{c_2}$ are calculated using (10.56); hence,
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$$
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\boldsymbol {\Phi} _ {c _ {1}} = \left[ \begin{array}{c} - \frac {1}{4} \\ \frac {- 1 + \sqrt {2}}{4} \end{array} \right]; \quad \boldsymbol {\Phi} _ {c _ {2}} = \left[ \begin{array}{c} \frac {1}{4} \\ \frac {1 + \sqrt {2}}{4} \end{array} \right] \tag {d}
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$$
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Since $\mu = 1 / \lambda$ we realize that in (b) to (d) we have the same solution as obtained in Example 10.12.
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Considering the different procedures of eliminating the massless degrees of freedom, the results of the eigensystem analysis are the same irrespective of the procedure followed, i.e., whether $K_{a}$ or $F_{a}$ is established and whether the eigenproblem in (10.45) or in (10.52) is solved. The basic assumption in the analysis is that resulting from mass lumping. As we discussed in Section 10.2.4, each zero mass corresponds to an infinite frequency in the
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system. Therefore, in approximating the original system equation $K\phi = \lambda M\phi$ by the equation in (10.42), we replace, in fact, some of the frequencies of $K\phi = \lambda M\phi$ by infinite frequencies and assume that the lowest frequencies solved from either equation are not much different. The accuracy with which the lowest frequencies of $K\phi = \lambda M\phi$ are approximated by solving $K_{a}\phi_{a} = \lambda M_{a}\phi_{a}$ depends on the specific mass lumping chosen and may be adequate or crude indeed. In general, more accuracy can be expected if more mass degrees of freedom are included. However, realizing that the static condensation results in $K_{a}$ having a larger bandwidth than K (and $F_{a}$ is certainly full), the computational effort required in the solution of the reduced eigenproblem increases rapidly as the order of $K_{a}$ becomes large (see Section 11.3). On the other hand, if sufficient mass degrees of freedom for accuracy of solution are selected, we may no longer want to calculate the complete eigensystem of $K_{a}\phi_{a} = \lambda M_{a}\phi_{a}$ but only the smallest eigenvalues and corresponding vectors. However, in this case we may just as well consider the problem $K\phi = \lambda M\phi$ without mass lumping and solve directly only for the eigenvalues and vectors of interest using one of the algorithms described in Chapter 11.
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In summary, the main shortcoming of the mass lumping procedure followed by static condensation is that the accuracy of solution depends to a large degree on the experience of the analyst in distributing the mass appropriately and that the solution accuracy is actually not assessed. We consider the following example to show the approximation that can typically result.
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EXAMPLE 10.14: In Example 10.4 we calculated the eigensystem of the problem $K\phi = \lambda M\phi$ , where K and M are given in the example. To evaluate an approximation to the smallest eigenvalue and corresponding eigenvector, consider instead the following eigenproblem, in which the mass is lumped
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$$
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\left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \boldsymbol {\Phi} = \lambda \left[ \begin{array}{c c c} 0 & & \\ & 2 & \\ & & 0 \end{array} \right] \boldsymbol {\Phi} \tag {a}
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$$
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Using the procedure given in (10.51) to (10.56), we obtain
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$$
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\mathbf {F} _ {a} = \left[ \frac {1}{3} \right]; \quad \mathbf {F} _ {c} = \left[ \begin{array}{l} \frac {1}{6} \\ \frac {1}{6} \end{array} \right]
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$$
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Hence, $\lambda_1 = \frac{3}{2}$ , $\phi_{a_1} = [1 / \sqrt{2}]$ , and
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$$
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\boldsymbol {\Phi} _ {c _ {1}} = \left[ \begin{array}{c} \frac {1}{2 \sqrt {2}} \\ \frac {1}{2 \sqrt {2}} \end{array} \right]
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$$
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The solution of the eigenproblem in (a) for the smallest eigenvalue and corresponding eigenvector is hence
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$$
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\lambda_ {1} = \frac {3}{2}; \quad \phi_ {1} = \left[ \begin{array}{l} \frac {1}{2 \sqrt {2}} \\ \frac {1}{\sqrt {2}} \\ \frac {1}{2 \sqrt {2}} \end{array} \right]
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$$
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<!-- source-page: 885 -->
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whereas the solution of the original problem (see Example 10.4) is
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$$
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\lambda_ {1} = 2; \quad \phi_ {1} = \left[ \begin{array}{l} \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \end{array} \right]
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$$
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It should be noted that using the mass lumping procedure, the eigenvalues can be smaller—as in this example—or larger than the eigenvalues of the original system.
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# 10.3.2 Rayleigh-Ritz Analysis
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A most general technique for finding approximations to the lowest eigenvalues and corresponding eigenvectors of the problem $K\Phi = \lambda M\Phi$ is the Rayleigh-Ritz analysis. The static condensation procedure in Section 10.3.1, the component mode synthesis described in the next section, and various other methods can be understood to be Ritz analyses. As we will see, the techniques differ only in the choice of the Ritz basis vectors assumed in the analysis. In the following we first present the Rayleigh-Ritz analysis procedure in general and then show how other techniques relate to it.
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The eigenproblem that we consider is
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$$
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\mathbf {K} \boldsymbol {\phi} = \lambda \mathbf {M} \boldsymbol {\phi} \tag {10.4}
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$$
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where we now first assume for clarity of presentation that K and M are both positive definite, which ensures that the eigenvalues are all positive; i.e., $\lambda_{1} > 0$ . As we pointed out in Section 10.2.3, K can be assumed positive definite because a shift can always be introduced to obtain a shifted stiffness matrix that satisfies this condition. As for the mass matrix, we now assume that M is a consistent mass matrix or a lumped mass matrix with no zero diagonal elements, which is a condition that we shall later relax.
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Consider first the Rayleigh minimum principle, which states that
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$$
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\lambda_ {1} = \min \rho (\phi) \tag {10.57}
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$$
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where the minimum is taken over all possible vectors $\phi$ , and $\rho(\phi)$ is the Rayleigh quotient
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$$
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\rho (\boldsymbol {\phi}) = \frac {\boldsymbol {\phi} ^ {T} \mathbf {K} \boldsymbol {\phi}}{\boldsymbol {\phi} ^ {T} \mathbf {M} \boldsymbol {\phi}} \tag {10.58}
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$$
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This Rayleigh quotient is obtained from the Rayleigh quotient of the standard eigenvalue problem $\tilde{K}\tilde{\phi} = \lambda\tilde{\phi}$ (see Sections 2.6 and 10.2.5). Since both K and M are positive definite, $\rho(\phi)$ has finite values for all $\phi$ . Referring to Section 2.6, the bounds on the Rayleigh quotient are
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$$
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0 < \lambda_ {1} \leq \rho (\phi) \leq \lambda_ {n} < \infty \tag {10.59}
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$$
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In the Ritz analysis we consider a set of vectors $\overline{\Phi}$ , which are linear combinations of the Ritz basis vectors $\psi_i$ , $i = 1, \ldots, q$ ; i.e., a typical vector is given by
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$$
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\overline {{{\phi}}} = \sum_ {i = 1} ^ {q} x _ {i} \psi_ {i} \tag {10.60}
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$$
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<!-- source-page: 886 -->
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where the $x_{i}$ are the Ritz coordinates. Since $\overline{\Phi}$ is a linear combination of the Ritz basis vectors, $\overline{\Phi}$ cannot be any arbitrary vector but instead lies in the subspace spanned by the Ritz basis vectors, which we call $V_{q}$ (see Sections 2.3 and 11.6). It should be noted that the vectors $\psi_{i}, i = 1, \ldots, q$ , must be linearly independent; therefore, the subspace $V_{q}$ has dimension q. Also, denoting the n-dimensional space in which the matrices K and M are defined by $V_{n}$ , we have that $V_{q}$ is contained in $V_{n}$ .
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In the Rayleigh-Ritz analysis we aim to determine the specific vectors $\overline{\Phi}_{i}, i = 1, \ldots, q$ , which, with the constraint of lying in the subspace spanned by the Ritz basis vectors, "best" approximate the required eigenvectors. For this purpose we invoke the Rayleigh minimum principle. The use of this principle determines in what sense the solution "best" approximates the eigenvectors sought, an aspect that we shall point out during the presentation of the solution procedure.
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To invoke the Rayleigh minimum principle on $\overline{\Phi}$ , we first evaluate the Rayleigh quotient,
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$$
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\rho (\overline {{{\Phi}}}) = \frac {\sum_ {j = 1} ^ {q} \sum_ {i = 1} ^ {q} x _ {i} x _ {j} \tilde {k} _ {i j}}{\sum_ {j = 1} ^ {q} \sum_ {i = 1} ^ {q} x _ {i} x _ {j} \tilde {m} _ {i j}} = \frac {\tilde {k}}{\tilde {m}} \tag {10.61}
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$$
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where $\tilde{k}_{ij} = \pmb{\psi}_i^T\mathbf{K}\pmb{\psi}_j$ (10.62)
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$$
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\tilde {m} _ {i j} = \boldsymbol {\psi} _ {i} ^ {T} \mathbf {M} \boldsymbol {\psi} _ {j} \tag {10.63}
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$$
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The necessary condition for a minimum of $\rho(\overline{\Phi})$ given in (10.61) is $\partial \rho(\overline{\Phi}) / \partial x_i = 0$ , $i = 1, \ldots, q$ , because the $x_i$ are the only variables. However,
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$$
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\frac {\partial \rho (\overline {{{\Phi}}})}{\partial x _ {i}} = \frac {2 \tilde {m} \sum_ {j = 1} ^ {q} x _ {j} \tilde {k} _ {i j} - 2 \tilde {k} \sum_ {j = 1} ^ {q} x _ {j} \tilde {m} _ {i j}}{\tilde {m} ^ {2}} \tag {10.64}
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$$
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and using $\rho = \tilde{k} / \tilde{m}$ , the condition for a minimum of $\rho(\overline{\phi})$ is
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$$
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\sum_ {j = 1} ^ {q} \left(\tilde {k} _ {i j} - \rho \tilde {m} _ {i j}\right) x _ {j} = 0 \quad \text { for } i = 1, \dots , q \tag {10.65}
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$$
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In actual analysis we write the $q$ equations in (10.65) in matrix form, thus obtaining the eigenproblem
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$$
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\tilde {\mathbf {K}} \mathbf {x} = \rho \tilde {\mathbf {M}} \mathbf {x} \tag {10.66}
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$$
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where $\tilde{\mathbf{K}}$ and $\tilde{\mathbf{M}}$ are $q \times q$ matrices with typical elements defined in (10.62) and (10.63), respectively, and $\mathbf{x}$ is a vector of the Ritz coordinates sought:
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$$
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\mathbf {x} ^ {T} = \left[ \begin{array}{l l l l} x _ {1} & x _ {2} & \dots & x _ {q} \end{array} \right] \tag {10.67}
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$$
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The solution to (10.66) yields q eigenvalues $\rho_{1}, \ldots, \rho_{q}$ , which are approximations to $\lambda_{1}, \ldots, \lambda_{q}$ , and q eigenvectors,
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$$
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\mathbf {x} _ {1} ^ {T} = \left[ \begin{array}{l l l l} x _ {1} ^ {1} & x _ {2} ^ {1} & \dots & x _ {q} ^ {1} \end{array} \right]
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$$
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$$
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\begin{array}{l l l l} \mathbf {x} _ {2} ^ {T} = \left[ x _ {1} ^ {2} \quad x _ {2} ^ {2} \quad \dots \quad x _ {q} ^ {2} \right] \\ \vdots & \vdots \end{array} \tag {10.68}
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$$
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$$
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\mathbf {x} _ {q} ^ {T} = \left[ \begin{array}{l l l l} x _ {1} ^ {q} & x _ {2} ^ {q} & \dots & x _ {q} ^ {q} \end{array} \right]
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$$
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<!-- source-page: 887 -->
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The eigenvectors $x_{i}$ are used to evaluate the vectors $\overline{\phi}_{1}, \ldots, \overline{\phi}_{q}$ , which are approximations to the eigenvectors $\phi_{1}, \ldots, \phi_{q}$ . Using (10.68) and (10.60), we have
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$$
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\overline {{{\Phi}}} _ {i} = \sum_ {j = 1} ^ {q} x _ {j} ^ {i} \psi_ {j}; \quad i = 1, \dots , q \tag {10.69}
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$$
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An important feature of the eigenvalue approximations calculated in the analysis is that they are upper bound approximations to the eigenvalues of interest; i.e.,
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$$
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\lambda_ {1} \leq \rho_ {1}; \qquad \lambda_ {2} \leq \rho_ {2}; \qquad \lambda_ {3} \leq \rho_ {3}; \qquad \dots ; \qquad \lambda_ {q} \leq \rho_ {q} \leq \lambda_ {n} \tag {10.70}
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$$
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meaning that since K and M are assumed to be positive definite, $\tilde{K}$ and $\tilde{M}$ are also positive definite matrices.
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The proof of the inequality in (10.70) shows the actual mechanism that is used to obtain the eigenvalue approximations $\rho_{i}$ . To calculate $\rho_{1}$ we search for the minimum of $\rho(\Phi)$ that can be reached by linearly combining all available Ritz basis vectors. The inequality $\lambda_{1} \leq \rho_{1}$ follows from the Rayleigh minimum principle in (10.57) and because $V_{q}$ is contained in the n-dimensional space $V_{n}$ , in which K and M are defined.
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The condition that is employed to obtain $\rho_{2}$ is typical of the mechanism used to calculate the approximations to the higher eigenvalues. First, we observe that for the eigenvalue problem $K\phi = \lambda M\phi$ , we have
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$$
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\lambda_ {2} = \min \rho (\phi) \tag {10.71}
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$$
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where the minimum is now taken over all possible vectors $\phi$ in $V_{n}$ that satisfy the orthogonality condition (see Section 2.6)
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$$
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\boldsymbol {\phi} ^ {T} \mathbf {M} \boldsymbol {\phi} _ {1} = 0 \tag {10.72}
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$$
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Considering the approximate eigenvectors $\overline{\Phi}_i$ obtained in the Rayleigh-Ritz analysis, we observe that
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$$
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\overline {{{\boldsymbol {\Phi}}}} _ {i} ^ {T} \mathbf {M} \overline {{{\boldsymbol {\Phi}}}} _ {j} = \delta_ {i j} \tag {10.73}
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$$
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where $\delta_{ij}$ is the Kronecker delta, and that, therefore, in the above Rayleigh-Ritz analysis we obtained $\rho_{2}$ by evaluating
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$$
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\rho_ {2} = \min \rho (\overline {{{\phi}}}) \tag {10.74}
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$$
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where the minimum was taken over all possible vectors $\overline{\Phi}$ in $V_{q}$ that satisfy the orthogonality condition
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$$
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\overline {{{\boldsymbol {\Phi}}}} ^ {T} \mathbf {M} \overline {{{\boldsymbol {\Phi}}}} _ {1} = 0 \tag {10.75}
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$$
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To show that $\lambda_{2} \leq \rho_{2}$ , we consider an auxiliary problem; i.e., assume that we evaluate
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$$
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\tilde {\rho} _ {2} = \min \rho (\overline {{{\Phi}}}) \tag {10.76}
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$$
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where the minimum is taken over all vectors $\overline{\phi}$ that satisfy the condition
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$$
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\overline {{{\boldsymbol {\Phi}}}} ^ {T} \mathbf {M} \boldsymbol {\phi} _ {1} = 0 \tag {10.77}
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$$
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The problem defined in (10.76) and (10.77) is the same as the problem in (10.71) and (10.72), except that in the latter case the minimum is taken over all $\phi$ , whereas in the problem in (10.76) and (10.77) we consider all vectors $\overline{\phi}$ in $V_{q}$ . Then since $V_{q}$ is contained in $V_{n}$ , we have $\lambda_{2} \leq \tilde{\rho}_{2}$ . On the other hand, $\tilde{\rho}_{2} \leq \rho_{2}$ because the most severe constraint on
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$\overline{\Phi}$ in (10.77) is $\overline{\Phi}_1$ . Therefore, we have
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$$
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\lambda_ {2} \leq \tilde {\rho} _ {2} \leq \rho_ {2} \tag {10.78}
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$$
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The basis for the calculation of $\overline{\Phi}_2$ , and hence $\rho_2$ , is that the minimum of $\rho(\overline{\Phi})$ is sought with the orthogonality condition in (10.75) on $\overline{\Phi}_1$ . Similarly, to obtain $\rho_i$ and $\overline{\Phi}_i$ , we in fact minimize $\rho(\overline{\Phi})$ with the orthogonality conditions $\overline{\Phi}^T\mathbf{M}\overline{\Phi}_j = 0$ for $j = 1, \ldots, i - 1$ . Accordingly, the inequality on $\rho_i$ in (10.70) can be proved in an analogous manner to the procedure used above for $\rho_2$ , but all $i - 1$ constraint equations need to be satisfied.
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The observation that i - 1 constraint equations need to be fulfilled in the evaluation of $\rho_{i}$ also indicates that we can expect less accuracy in the approximation of the higher eigenvalues than in the approximation of the lower eigenvalues, for which fewer constraints are imposed. This is generally also observed in actual analysis.
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Considering the procedure in practical dynamic analysis, the Ritz basis functions may be calculated from a static solution in which q load patterns are specified in R; i.e., we consider
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$$
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\mathbf {K} \boldsymbol {\Psi} = \mathbf {R} \tag {10.79}
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$$
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where $\Psi$ is an $n \times q$ matrix storing the Ritz basis vectors; i.e., $\Psi = [\psi_{1}, \ldots, \psi_{q}]$ . The analysis is continued by evaluating the projections of K and M onto the subspace $V_{q}$ spanned by the vectors $\psi_{i}, i = 1, \ldots, q$ ; i.e., we calculate
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$$
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\tilde {\mathbf {K}} = \boldsymbol {\Psi} ^ {T} \mathbf {K} \boldsymbol {\Psi} \tag {10.80}
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$$
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and $\tilde{\mathbf{M}} = \mathbf{\Psi}^T\mathbf{M}\mathbf{\Psi}$ (10.81)
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where because of (10.79) we have
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$$
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\tilde {\mathbf {K}} = \boldsymbol {\Psi} ^ {T} \mathbf {R} \tag {10.82}
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$$
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Next we solve the eigenproblem $\tilde{K}x = \rho\tilde{M}x$ , the solution of which can be written
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$$
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\tilde {\mathbf {K}} \mathbf {X} = \tilde {\mathbf {M}} \mathbf {X} \boldsymbol {\rho} \tag {10.83}
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$$
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where $\rho$ is a diagonal matrix listing the eigenvalue approximations $\rho_{i}, \rho = \text{diag}(\rho_{i})$ , and X is a matrix storing the $\tilde{M}$ -orthonormal eigenvectors $x_{1}, \ldots, x_{q}$ . The approximations to the eigenvectors of the problem $K\phi = \lambda M\phi$ are then
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$$
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\overline {{{\Phi}}} = \Psi \mathbf {X} \tag {10.84}
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$$
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So far we have assumed that the mass matrix of the finite element system is positive definite; i.e., $\mathbf{M}$ is not a diagonal mass matrix with some zero diagonal elements. The reason for this assumption was to avoid the case $\overline{\Phi}^T\mathbf{M}\overline{\Phi}$ equal to zero in the calculation of the Rayleigh quotient, in which case $\rho (\overline{\Phi})$ gives an infinite eigenvalue. However, the Rayleigh-Ritz analysis can be carried out as described above when $\mathbf{M}$ is a diagonal matrix with some zero diagonal elements, provided the Ritz basis vectors are selected to lie in the subspace that corresponds to the finite eigenvalues. In addition, the Ritz basis vectors must be linearly independent when considering only the mass degrees of freedom in order to obtain a positive definite matrix $\tilde{\mathbf{M}}$ . One way of achieving this in practice is to excite different mass degrees of freedom in each of the load vectors in $\mathbf{R}$ in (10.79) (see Section 11.6.3 and Example 10.16).
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Of particular interest are the errors that we may expect in the solution. Although we have shown that an eigenvalue calculated from the Ritz analysis is an upper bound on the
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corresponding exact eigenvalue of the system, we did not establish anything about the actual error in the eigenvalue. This error depends on the Ritz basis vectors used because the vectors $\overline{\Phi}$ are linear combinations of the Ritz basis vectors $\psi_{i}, i = 1, \ldots, q$ . We can obtain good results only if the vectors $\psi_{i}$ span a subspace $V_{q}$ that is close to the least dominant subspace of K and M spanned by $\phi_{1}, \ldots, \phi_{q}$ . It should be noted that this does not mean that the Ritz basis vectors should each be close to an eigenvector sought but rather that linear combinations of the Ritz basis vectors can establish good approximations of the required eigenvectors of $K\Phi = \lambda M\Phi$ . We further discuss the selection of good Ritz basis vectors and the approximations involved in the analysis in Section 11.6 when we present the subspace iteration method, because this method uses the Ritz analysis technique.
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To demonstrate the Rayleigh-Ritz analysis procedure, consider the following examples.
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EXAMPLE 10.15: Obtain approximate solutions to the eigenproblem $K\phi = \lambda M\phi$ considered in Example 10.4, where
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$$
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\mathbf {K} = \left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right]; \quad \mathbf {M} = \left[ \begin{array}{c c c} \frac {1}{2} & & \\ & 1 & \\ & & \frac {1}{2} \end{array} \right]
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$$
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The exact eigenvalues are $\lambda_1 = 2, \lambda_2 = 4, \lambda_3 = 6$ .
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1. Use the following load vectors to generate the Ritz basis vectors
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$$
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\mathbf {R} = \left[ \begin{array}{l l} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{array} \right]
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$$
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2. Then use a different set of load vectors to generate the Ritz basis vectors
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$$
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\mathbf {R} = \left[ \begin{array}{l l} 1 & 0 \\ 1 & 1 \\ 1 & 0 \end{array} \right]
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$$
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In the Ritz analysis we employ the relations in (10.79) to (10.84) and obtain, in case 1,
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$$
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\left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \boldsymbol {\Psi} = \left[ \begin{array}{l l} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{array} \right]
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$$
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Hence, $\Psi = \begin{bmatrix} \frac{7}{12} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{6} \\ \frac{1}{12} & \frac{7}{12} \end{bmatrix}$
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and $\tilde{\mathbf{K}} = \frac{1}{12}\begin{bmatrix} 7 & 1 \\ 1 & 7 \end{bmatrix}$ ; $\tilde{\mathbf{M}} = \frac{1}{144}\begin{bmatrix} 29 & 11 \\ 11 & 29 \end{bmatrix}$
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The solution of the eigenproblem $\tilde{\mathbf{K}}\mathbf{x} = \rho \tilde{\mathbf{M}}\mathbf{x}$ is
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$$
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\left(\rho_ {1}, \mathbf {x} _ {1}\right) = (2. 4 0 0 4, \left[ \begin{array}{l} 1. 3 4 1 8 \\ 1. 3 4 1 8 \end{array} \right]); \quad \left(\rho_ {2}, \mathbf {x} _ {2}\right) = (4. 0 0 3 2, \left[ \begin{array}{l} 2. 0 0 0 8 \\ - 2. 0 0 0 8 \end{array} \right])
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$$
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Hence, we have as eigenvalue approximations $\rho_{1} = 2.40$ , $\rho_{2} = 4.00$ , and evaluating
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$$
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\overline {{{\boldsymbol {\Phi}}}} = \left[ \begin{array}{l l} \frac {7}{1 2} & \frac {1}{1 2} \\ \frac {1}{6} & \frac {1}{6} \\ \frac {1}{1 2} & \frac {7}{1 2} \end{array} \right] \left[ \begin{array}{c c} 1. 3 4 1 8 & 2. 0 0 0 8 \\ 1. 3 4 1 8 & - 2. 0 0 0 8 \end{array} \right] = \left[ \begin{array}{c c} 0. 8 9 5 & 1. 0 0 \\ 0. 4 4 7 & 0 \\ 0. 8 9 5 & - 1. 0 0 \end{array} \right]
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$$
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we have $\overline{\Phi}_1 = \begin{bmatrix} 0.895 \\ 0.447 \\ 0.895 \end{bmatrix}$ ; $\overline{\Phi}_2 = \begin{bmatrix} 1.00 \\ 0.00 \\ -1.00 \end{bmatrix}$
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Next we assume the load vectors in case 2 and solve
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$$
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\left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 2 \end{array} \right] \boldsymbol {\Psi} = \left[ \begin{array}{l l} 1 & 0 \\ 1 & 1 \\ 1 & 0 \end{array} \right]
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$$
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Hence, $\Psi = \begin{bmatrix} \frac{5}{6} & \frac{1}{6} \\ \frac{2}{3} & \frac{1}{3} \\ \frac{5}{6} & \frac{1}{6} \end{bmatrix}$
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and $\tilde{\mathbf{K}} = \begin{bmatrix} \frac{7}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{1}{3} \end{bmatrix}$ ; $\tilde{\mathbf{M}} = \frac{1}{36} \begin{bmatrix} 41 & 13 \\ 13 & 5 \end{bmatrix}$
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The solution of the eigenproblem $\tilde{\mathbf{K}}\mathbf{x} = \rho \tilde{\mathbf{M}}\mathbf{x}$ gives
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$$
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\left(\rho_ {1}, \mathbf {x} _ {1}\right) = (2. 0 0 0, \left[ \begin{array}{l} 0. 7 0 7 1 1 \\ 0. 7 0 7 1 1 \end{array} \right]); \quad \left(\rho_ {2}, \mathbf {x} _ {2}\right) = (6. 0 0 0 0, \left[ \begin{array}{c} - 2. 1 2 1 3 \\ 6. 3 6 4 0 \end{array} \right])
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$$
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Hence, we have as eigenvalue approximations $\rho_{1} = 2.00$ , $\rho_{2} = 6.00$ , and evaluating
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$$
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\overline {{{\Phi}}} = \left[ \begin{array}{l l} \frac {5}{6} & \frac {1}{6} \\ \frac {2}{3} & \frac {1}{3} \\ \frac {5}{6} & \frac {1}{6} \end{array} \right] \left[ \begin{array}{c c} 0. 7 0 7 1 1 & - 2. 1 2 1 3 \\ 0. 7 0 7 1 1 & 6. 3 6 4 0 \end{array} \right] = \left[ \begin{array}{c c} 0. 7 0 7 1 1 & - 0. 7 0 7 0 8 \\ 0. 7 0 7 1 1 & 0. 7 0 7 1 3 \\ 0. 7 0 7 1 1 & - 0. 7 0 7 0 8 \end{array} \right]
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$$
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we have $\overline{\Phi}_1 = \begin{bmatrix} 0.70711 \\ 0.70711 \\ 0.70711 \end{bmatrix}; \quad \overline{\Phi}_2 = \begin{bmatrix} -0.70708 \\ 0.70713 \\ -0.70708 \end{bmatrix}$
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Comparing the results with the exact solution, it is interesting to note that in case 1, $\rho_{1} > \lambda_{1}$ and $\rho_{2} = \lambda_{2}$ , whereas in case 2, $\rho_{1} = \lambda_{1}$ and $\rho_{2} = \lambda_{3}$ . In both cases we did not obtain good approximations to the lowest two eigenvalues, and it is clearly demonstrated that the results depend completely on the initial Ritz basis vectors chosen.
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EXAMPLE 10.16: Use the Rayleigh-Ritz analysis to calculate an approximation to $\lambda_{1}$ and $\phi_{1}$ of the eigenproblem considered in Example 10.12.
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We note that in this case M is positive semidefinite. Therefore, to carry out the Ritz analysis we need to choose a load vector in R that excites at least one mass. Assume that we use
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$$
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\mathbf {R} ^ {T} = \left[ \begin{array}{c c c c} 0 & 1 & 0 & 0 \end{array} \right]
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$$
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Then the solution of (10.79) yields (see Example 10.13)
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$$
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\boldsymbol {\Psi} ^ {T} = \left[ \begin{array}{l l l l} 1 & 2 & 2 & 2 \end{array} \right]
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$$
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