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\text { or } \quad \left(\frac {2}{3}\right) \left(\frac {1 3}{9}\right) \frac {A _ {1} E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {5} \end{array} \right] = \left[ \begin{array}{l} R _ {1} + \frac {2 2}{3 6} R _ {2} + \frac {1}{3} R _ {3} + \frac {5}{3 6} R _ {4} \\ \frac {1 4}{3 6} R _ {2} + \frac {2}{3} R _ {3} + \frac {3 1}{3 6} R _ {4} + R _ {5} \end{array} \right] \tag {d}
U _ {3} = \frac {1}{3} \left[ \frac {9}{1 3} \frac {L}{A _ {1} E} \left(R _ {3} + \frac {7}{1 2} R _ {2} + \frac {5}{1 2} R _ {4}\right) + U _ {1} + 2 U _ {5} \right]
We should note that the stiffness matrix of the second-level substructure in (d) is simply \frac{2}{3} times the stiffness matrix of the first-level substructure in (a). Therefore, we could continue to build up even higher-level substructures in an analogous manner; i.e., the stiffness matrix of the nth-level substructure would simply be a factor times the stiffness matrix given in (a).
In most cases loads are applied only at the boundary degrees of freedom between substructures, such as in this example. Using the stiffness matrix of the second-level substructure to assemble the stiffness matrix of the complete bar and assembling the actual load vector for this example, we obtain
\frac {2}{3} \left(\frac {1 3}{9}\right) \frac {A _ {1} E}{L} \left[ \begin{array}{r r r} 1 & - 1 & 0 \\ - 1 & 5 & - 4 \\ 0 & - 4 & 4 \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {5} \\ U _ {9} \end{array} \right] = \left[ \begin{array}{c} 0 \\ R _ {5} \\ 0 \end{array} \right]
Eliminating U_{5} , we have
\left(\frac {4}{5}\right) \left(\frac {2}{3}\right) \left(\frac {1 3}{9}\right) \frac {A _ {1} E}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {9} \end{array} \right] = \left[ \begin{array}{l} \frac {1}{5} R _ {5} \\ \frac {4}{5} R _ {5} \end{array} \right]
where the stiffness matrix is simply the third-level substructure stiffness matrix corresponding to the algorithm given above. We also have
U _ {5} = \frac {1}{5} \left(\frac {2 7}{3 6} \frac {L}{A _ {1} E} R _ {5} + U _ {1} + 4 U _ {9}\right)
To solve for specific displacements, it is necessary to impose boundary conditions on the bar, hence obtain U_{1} and U_{9} , and then obtain the internal bar displacements using previously derived relations. It should be noted that corresponding relations must also be employed to evaluate U_{6} to U_{8} .
So far, we have not mentioned how to proceed in the solution if the total system matrix cannot be contained in high-speed storage. If substructuring is used, it is effective to keep the size of each uncondensed substructure stiffness matrix small enough so that the static condensation of the internal degrees of freedom can be carried out in high-speed core. Therefore, disk storage would mainly be required to store the required information for the calculation of the displacements of the substructure internal nodes as expressed in (8.29).
However, it may be necessary to use multilevel substructuring (i.e., to define substructures of substructures) in order that the final equations to be solved can be taken into high-speed storage.
In general, it is important to use disk storage effectively since a great deal of reading and writing can be very expensive and indeed may limit the system size that can be solved, because not enough backup storage may be available. In out-of-core solutions the particular scheme used for solving the system equilibrium equations is largely coupled with the specific procedure employed to assemble the element stiffness matrices to the global structure stiffness matrix. In many programs the structure stiffness matrix is assembled prior to
performing the Gauss solution. In the program ADINA the equations are considered in blocks that can be taken into high-speed core. The block sizes (number of columns per block) are automatically established in the program and depend on the high-speed storage available. The solution of the system equations is then obtained in an effective manner by first reducing in blocks the stiffness matrix and the load vectors consecutively and then performing the back-substitution. Similar procedures are presently used in many analysis programs.
Instead of first assembling the complete structure stiffness matrix, we may assemble and reduce the equations at the same time. A specific solution scheme proposed by B. M. Irons [D] called the frontal solution method has been used effectively. In the solution procedure only those equations that are actually required for the elimination of a specific degree of freedom are assembled, the degree of freedom considered is statically condensed out, and so on.
As an example, consider the analysis of the plane stress finite element idealization of the sheet in Fig. 8.8. There are two equations associated with each node of the finite element mesh, namely, the equations corresponding to the U and V displacements, respectively. In the frontal solution scheme the equations are statically condensed out in the order of the elements; i.e., the first equations considered would be those corresponding to nodes 1, 2, . . . . To be able to eliminate the degrees of freedom of node 1 it is only necessary to assemble the final equations that correspond to that node. This means that only the stiffness matrix of element 1 needs to be calculated, after which the degrees of freedom corresponding to node 1 are statically condensed out. Next (for the elimination of the equations corresponding to node 2), the final equations corresponding to the degrees of freedom at node 2 are required, meaning that the stiffness matrix of element 2 must be calculated and added to the previously reduced matrix. Now the degrees of freedom corresponding to node 2 are statically condensed out; and so on.
It may now be realized that the complete procedure, in effect, consists of statically condensing out one degree of freedom after the other and always assembling only those equations (or rather element stiffness matrices) that are actually required during the specific condensation to be performed. The finite elements that must be considered for the
text_image
Element q Element q + 1 Element q + 2 Element q + 3 m m + 1 m + 2 m + 3 Element 1 Element 2 Element 3 Element 4 Node 1 2 3 4 Wave front for node 1 Wave front for node 2 V U
Figure 8.8 Frontal solution of plane stress finite element idealization
static condensation of the equations corresponding to one specific node define the wave front at that time, as shown in Fig. 8.8.
In principle, the frontal solution is Gauss elimination and the important aspect is the specific computer implementation. Since the equations are assembled in the order of the elements, the length of the wave front and therefore the half-bandwidth dealt with are determined by the element numbering. Therefore, an effective ordering of the elements is necessary, and we note that if the element numbering in the frontal solution corresponds to the nodal point numbering in the active column solution (see Section 8.2.3), the same number of basic (i.e., excluding indexing) numerical operations is performed in both solutions. An advantage of the wave front solution is that elements can be added with relative ease because no nodal point renumbering is necessary to preserve a small bandwidth. But a disadvantage is that if the wave front is large, the total high-speed storage required may well exceed the storage that is available, in which case additional out-of-core operations are required that suddenly decrease the effectiveness of the method by a great amount. Also, the active column solution is implemented in a compact stand-alone solver that is independent of the finite elements processed, whereas a frontal solver is intimately coupled to the finite elements and may require more indexing in the solution.
8.2.5 Positive Definiteness, Positive Semidefiniteness, and the Sturm Sequence Property
In the Gauss elimination discussed so far, we implicitly assumed that the stiffness matrix K is positive definite, that is, that the structure considered is properly restrained and stable. As discussed in Sections 2.5 and 2.6, positive definiteness of the stiffness matrix means that for any displacement vector U, we have
\mathbf {U} ^ {T} \mathbf {K} \mathbf {U} > 0 \tag {8.31}
Since \frac{1}{2}U^{T}KU is the strain energy stored in the system for the displacement vector U, (8.31) expresses that for any displacement vector U the strain energy of a system with a positive definite stiffness matrix is positive.
Note that the stiffness matrix of a finite element is not positive definite unless the element has been properly restrained, i.e., the rigid body motions have been suppressed. Instead, the stiffness matrix of an unrestrained finite element is positive semidefinite,
\mathbf {U} ^ {T} \mathbf {K} \mathbf {U} \geq 0 \tag {8.32}
where U^{T}KU = 0 when U corresponds to a rigid body mode. Considering the finite element assemblage process, it should be realized that positive semidefinite element matrices are added to obtain the positive semidefinite stiffness matrix corresponding to the complete structure. The stiffness matrix of the structure is then rendered positive definite by eliminating the rows and columns that correspond to the restrained degrees of freedom, i.e., by eliminating the possibility for the structure to undergo rigid body motions.
It is instructive to consider in more detail the meaning of positive definiteness of the structure stiffness matrix. In Section 2.5, we discussed the representation of a matrix by its eigenvalues and eigenvectors. Following the development given in Section 2.5, the eigenproblem for the stiffness matrix K can be written
\mathbf {K} \phi = \lambda \phi \tag {8.33}
The solutions to (8.33) are the eigenpairs (\lambda_i, \phi_i), i = 1, \ldots, n , and the complete solution can be written
\mathbf {K} \boldsymbol {\Phi} = \boldsymbol {\Phi} \boldsymbol {\Lambda}
where \Phi is a matrix of the orthonormalized eigenvectors, \Phi = [\phi_{1}, \ldots, \phi_{n}] , and \Lambda is a diagonal matrix of the corresponding eigenvalues, \Lambda = \text{diag}(\lambda_{i}) . Since \Phi^{T}\Phi = \Phi\Phi^{T} = I , we also have
\Phi^ {T} \mathbf {K} \Phi = \Lambda \tag {8.34}
and \mathbf{K} = \mathbf{\Phi}\mathbf{\Lambda}\mathbf{\Phi}^T (8.35)
Referring to Section 2.6, we recall that \lambda_{i} of K represents the minimum that can be reached by the Rayleigh quotient when an orthonormality constraint is satisfied on the eigenvectors \phi_{1},\ldots,\phi_{i-1} :
with \left.\begin{array}{l}\lambda_{i} = \min \left\{\frac{\boldsymbol{\Phi}^{T}\mathbf{K}\boldsymbol{\Phi}}{\boldsymbol{\Phi}^{T}\boldsymbol{\Phi}}\right\} \\ \boldsymbol{\Phi}^{T}\boldsymbol{\Phi}_{r} = 0;\qquad \text{for } r = 1,2,\dots ,i - 1 \end{array}\right\} \tag{8.36}
Therefore, \frac{1}{2}\lambda_{1} is the minimum strain energy that can be stored in the element assemblage, and the corresponding displacement vector is \phi_{1} . For a positive definite system stiffness matrix, we therefore have \lambda_{1}>0 . On the other hand, for the stiffness matrix of an unrestrained system, we have \lambda_{1}=\lambda_{2}=\cdots=\lambda_{m}=0 , where m is the number of rigid body modes present, m<n. As the system is restrained, the number of eigenvalues of K is decreased by 1 for each degree of freedom that is eliminated, and a zero eigenvalue is lost if the restraint results in the elimination of a rigid body mode.
EXAMPLE 8.11: Determine whether the deletion of the four degrees of freedom of the plane stress element in Fig. E8.11 results in the elimination of the rigid body modes.
The plane stress element has three rigid body modes: (1) uniform horizontal translation, (2) uniform vertical translation, and (3) in-plane rotation. Consider the sequential deletion of the degrees of freedom, as shown in Fig. E8.11. The deletion of U_{4} results in eliminating the horizontal translation rigid body mode. Similarly the deletion of V_{4} results in the deletion of
Figure E8.11 Deletion of degrees of freedom of plane stress element
the vertical translation rigid body mode. However, deleting V_{1} in addition does not result in the elimination of the last rigid body mode; i.e., the in-plane rotation rigid body mode is eliminated only with the additional deletion of U_{2} . Therefore, the deletion of U_{4} and V_{4} and V_{1} and U_{2} eliminates all rigid body modes of the element, although we should note that, in fact, by the deletion of U_{4} , V_{4} , and U_{2} alone we would achieve the same result.
The transformation performed in (8.34) has important meaning. Considering the relation and referring to Section 2.5, we realize that in (8.34) a change of basis is performed. The new basis vectors are the finite element interpolations corresponding to the eigenvectors of K, and in this basis the operator is represented by a diagonal matrix with the eigenvalues of K on its diagonal. We may therefore look at \Lambda as being the stiffness matrix of the system when the finite element displacement functions used in the principle of virtual work in (4.7) are those corresponding to nodal point displacements \phi_{i}, i = 1, \ldots, n , instead of unit nodal point displacements U_{i}, i = 1, \ldots, n (see Section 4.2.1). The relation in (8.34) is therefore a statement of virtual work resulting in a diagonal stiffness matrix. If the system considered is properly restrained, all stiffness coefficients in \Lambda are positive; i.e., the stiffness matrix \Lambda (and hence K) is positive definite, whereas for an unrestrained system some diagonal elements in \Lambda are zero.
Before studying the solution of nonpositive definite systems of equations, another most important observation should be discussed. In Section 2.6 we introduced the Sturm sequence property of the leading principal minors of a matrix. We should note here the physical meaning of the Sturm sequence. Let \mathbf{K}^{(r)} be the matrix of order n - r obtained by deleting from K the last r rows and columns and consider the eigenproblem
\mathbf {K} ^ {(r)} \boldsymbol {\phi} ^ {(r)} = \lambda^ {(r)} \boldsymbol {\phi} ^ {(r)} \tag {8.37}
where \Phi^{(r)} is a vector of order n - r . We say that (8.37) is the eigenproblem of the r th associated constraint problem of the problem \mathbf{K}\Phi = \lambda \Phi . Then we have shown in Section 2.6 that the eigenvalues of the (r + 1) st constraint problem separate those of the r th constraint problem,
\lambda_ {1} ^ {(r)} \leq \lambda_ {1} ^ {(r + 1)} \leq \lambda_ {2} ^ {(r)} \leq \lambda_ {2} ^ {(r + 1)} \leq \dots \leq \lambda_ {n - r - 1} ^ {(r)} \leq \lambda_ {n - r - 1} ^ {(r + 1)} \leq \lambda_ {n - r} ^ {(r)} \tag {8.38}
As an example, the eigenproblems of the simply supported beam discussed in Section 8.2.1 and of its associated constraint problems can be considered. Figure 8.9 shows the eigenvalues calculated and, in particular, displays their separation property. We should note that as we proceed from the (r + 1) st constraint problem to the rth constraint problem by including the (n - r) th degree of freedom, the new system has an eigenvalue smaller than (or equal to) the smallest eigenvalue of the (r + 1) st constraint problem, and also an eigenvalue larger than (or equal to) the largest eigenvalue of the (r + 1) st constraint problem.
Using the separation property of the eigenvalues and realizing that any rows and columns may be interchanged at convenience to become the last rows and columns in the matrix K, it follows that if the stiffness matrix corresponding to the n degrees of freedom is positive definite (i.e., \lambda_{1} > 0 ), then any stiffness matrix obtained by deleting any rows and corresponding columns is also positive definite. Furthermore, the smallest eigenvalue of the new matrix can only have increased, and the largest eigenvalue can only have decreased. This conclusion applies also if the matrix K is positive semidefinite and would apply if it were indefinite since we showed the eigenvalue separation theorem to be applicable to all symmetric matrices.
\mathbf {K} ^ {(1)} = \left[ \begin{array}{r r r} 5 & - 4 & 1 \\ - 4 & 6 & - 4 \\ 1 & - 4 & 6 \end{array} \right]
Figure 8.9 Eigenvalue solutions of simply supported beam and of associated constraint problems
We shall encounter the use of the Sturm sequence property of the leading principal minors more extensively in the design of eigenvalue solution algorithms (see Chapter 11). However, in the following we use the property to yield more insight into the solution of a set of simultaneous equations with a symmetric positive definite, positive semidefinite, or indefinite coefficient matrix. A symmetric indefinite coefficient matrix will be encountered in the solution of eigenproblems.
We showed in Sections 8.2.1 and 8.2.2 that if K is the stiffness matrix of a properly restrained structure, we can factorize K into the form
\mathbf {K} = \mathbf {L D L} ^ {T} \tag {8.39}
where \mathbf{L} is a lower unit triangular matrix and \mathbf{D} is a diagonal matrix, with d_{ii} > 0 . It follows that
\begin{array}{l} \det \mathbf {K} = \det \mathbf {L} \det \mathbf {D} \det \mathbf {L} ^ {T} \tag {8.40} \\ = \prod_ {i = 1} ^ {n} d _ {i i} > 0 \\ \end{array}
This result can also be obtained by considering the characteristic polynomial of K, defined as
p (\lambda) = \det (\mathbf {K} - \lambda \mathbf {I}) \tag {8.41}
Since \lambda_{1} is the smallest root of p(\lambda) and \lambda_{1} > 0 for K being positive definite, it follows that det K > 0. However, it does not yet follow that d_{ii} > 0 for all i.
In order to formally prove that d_{ii} > 0, i = 1, \ldots, n , when K is positive definite, and to identify what happens during the factorization of K, it is expedient to compare the triangular factors of K with those of \mathbf{K}^{(i)} , where \mathbf{K}^{(i)} is the stiffness matrix of the ith associated constraint problem. Assuming that the factors L and D of K have been calculated, we have, for the associated constraint problems,
\mathbf {K} ^ {(i)} = \mathbf {L} ^ {(i)} \mathbf {D} ^ {(i)} \mathbf {L} ^ {(i) T} \quad i = 1, \dots , n - 1 \tag {8.42}
where \mathbf{L}^{(i)} and \mathbf{D}^{(i)} are analogously the factors of \mathbf{K}^{(i)} . Since L is a lower-unit triangular matrix and D is a diagonal matrix, the factors \mathbf{L}^{(i)} and \mathbf{D}^{(i)} are obtained from L and D, respectively, by striking out the last i rows and columns. Therefore, \mathbf{L}^{(i)} and \mathbf{D}^{(i)} are the leading principal submatrices of L and D, respectively, and they are actually evaluated in the factorization of K. However, because \lambda_{1}^{(i)} > 0 , it now follows that we can use the argument in (8.39) and (8.40) starting with i = n - 1 to show that d_{ii} > 0 for all i. Therefore, the factorization of K into LDL^{T} is indeed possible if K is positive definite. We demonstrate the result in the following example.
EXAMPLE 8.12: Consider the simply supported beam in Fig. 8.9 and the associated constraint problems. The same beam was used in Section 8.2.1. Establish the \mathbf{L}^{(i)} and \mathbf{D}^{(i)} factors of the matrices \mathbf{K}^{(i)} , i = 1, 2, 3, and show that d_{ii} must be greater than zero because \lambda_{1} > 0 . The required triangular factorizations are
[ 5 ] = [ 1 ] [ 5 ] [ 1 ] \tag {a}
\left[ \begin{array}{c c} 5 & - 4 \\ - 4 & 6 \end{array} \right] = \left[ \begin{array}{l l} 1 & 0 \\ - \frac {4}{5} & 1 \end{array} \right] \left[ \begin{array}{l l} 5 & 0 \\ 0 & \frac {1 4}{5} \end{array} \right] \left[ \begin{array}{l l} 1 & - \frac {4}{5} \\ 0 & 1 \end{array} \right] \tag {b}
\left[ \begin{array}{r r r} 5 & - 4 & 1 \\ - 4 & 6 & - 4 \\ 1 & - 4 & 6 \end{array} \right] = \left[ \begin{array}{r r r} 1 & 0 & 0 \\ - \frac {4}{5} & 1 & 0 \\ \frac {1}{5} & - \frac {8}{7} & 1 \end{array} \right] \left[ \begin{array}{r r r} 5 & 0 & 0 \\ 0 & \frac {1 4}{5} & 0 \\ 0 & 0 & \frac {1 5}{7} \end{array} \right] \left[ \begin{array}{r r r} 1 & - \frac {4}{5} & \frac {1}{5} \\ 0 & 1 & - \frac {8}{7} \\ 0 & 0 & 1 \end{array} \right] \tag {c}
where the matrices \mathbf{L}^{(i)} and \mathbf{D}^{(i)} are obtained from the L and D factors given in Example 8.3 by striking out the last i rows and columns. [As a check, we may want to calculate the product of the matrices on the right sides of the relations in (a) to (c) to obtain the matrices on the left sides.]
Considering the elements d_{ii} , we have \lambda_1^{(3)} \geq \lambda_1^{(2)} \geq \lambda_1^{(1)} \geq \lambda_1 > 0 . But using the relation (a), we have \lambda_1^{(3)} = d_{11} ; hence d_{11} > 0 . Next we consider \mathbf{K}^{(2)} . Since \lambda_1^{(2)} > 0 , we have, using (8.39) and (8.40), d_{11}d_{22} > 0 , which means that d_{22} > 0 . Similarly, considering \mathbf{K}^{(1)} , we have \lambda_1^{(1)} > 0 , and hence d_{11}d_{22}d_{33} > 0 , from which it follows that d_{33} > 0 . Finally, considering \mathbf{K} , we have \lambda_1 > 0 , and hence d_{11}d_{22}d_{33}d_{44} > 0 . Therefore, d_{44} > 0 also.
Assume next that the matrix K is the stiffness matrix of a finite element assemblage that is unrestrained. In this case, K is positive semidefinite, \lambda = 0.0 is a root, and det K is zero, which, using (8.40), means that d_{ii} for some i must be zero. Therefore, the factorization of K as shown in the preceding sections is, in general, not possible because a pivot element will be zero. It is again instructive to consider the associated constraint problems. When K is positive semidefinite, we note that the characteristic polynomial corresponding to \mathbf{K}^{(i)} will have a zero eigenvalue, and this zero eigenvalue will be retained in all matrices \mathbf{K}^{(i-1)}, \ldots, \mathbf{K} . This follows because, first, the Sturm sequence property ensures that the smallest eigenvalue of \mathbf{K}^{(i-1)} is smaller or equal to the smallest eigenvalue of \mathbf{K}^{(i)} , and second, K has no negative eigenvalues. For the ith associated constraint problem, we will therefore have
\det \left(\mathbf {L} ^ {(i)} \mathbf {D} ^ {(i)} \mathbf {L} ^ {(i) T}\right) = 0 \tag {8.43}
from which it follows that an element of \mathbf{D}^{(i)} is zero. However, assuming that the zero root occurs only in the ith associated constraint problem [i.e., \det(\mathbf{L}^{(r)}\mathbf{D}^{(r)}\mathbf{L}^{(r)T})>0 for r>i], it follows that d_{n-i,n-i} is zero. In summary therefore, if K is positive semidefinite, the factorization of K into LDL^{T} (i.e., the Gauss elimination process) will break down at the time a zero diagonal element d_{kk} is encountered, which means that the (n-k) th associated constraint problem with a zero eigenvalue prevents the continuation of the factorization process.
In the case of a positive semidefinite matrix, a zero diagonal element must be encountered at some stage of the factorization. However, considering the decomposition of an indefinite matrix (i.e., some of the eigenvalues of the matrix are negative and some are positive), a zero diagonal element is encountered only if one of the associated constraint problems has a zero eigenvalue. Namely, as in the case of a positive semidefinite matrix, d_{kk} is zero if the (n - k) th associated constraint problem has a zero eigenvalue. However, if none of the associated constraint problems has a zero eigenvalue, all elements d_{ii} are nonzero and, in exact arithmetic, no difficulties are encountered in the factorization. We shall discuss the decomposition of indefinite coefficient matrices further in the solution of eigenproblems (see Section 11.4.2). Figure 8.10 shows typical cases that use the simply supported beam in Fig. 8.9 on spring supports for which, in the decompositions, we would and would not encounter a zero diagonal element.
Assume that a zero diagonal element d_{ii} is encountered in the Gauss elimination. To be able to proceed with the solution, it is necessary to interchange the ith row with another row, say the jth row, where j > i. The new diagonal element should not be zero, and to increase solution accuracy, it should be large (see Section 8.2.6). This row interchange corresponds to a rearranging of the equations, where it should be noted that the row interchange results in the coefficient matrix no longer being symmetric. On the other hand, symmetry would be preserved if we were to interchange not only the ith and jth rows but also the corresponding columns to obtain a nonzero diagonal element in row i, which is not always possible (see Example 10.4). In effect, the interchange of columns and rows corresponds to a rearranging of the associated constraint problems in such a way that these have nonzero eigenvalues.
The remedy of row interchanges assumes that it can be arranged for the new diagonal element to be nonzero. In fact, this will always be the case unless the matrix has a zero eigenvalue of multiplicity m and i = n - m + 1 . In this case the matrix is singular and d_{ii}
line
| Parameter | Value |
|---|---|
| p(λ) | 0.146 |
| p(λ) | 1.91 |
| p(λ) | 6.85 |
| p(λ) | 13.1 |
| p(λ) | 11.9 |
| p(λ) | 4.51 |
| p(λ) | 0.557 |
| p(λ) | 1.47 |
| p(λ) | 5.00 |
| p(λ) | 9.53 |
| p(λ) | 13.1 |
| p(λ) | 11.9 |
| p(λ) | 4.51 |
| p(λ) | 0.557 |
| p(λ) | 1.47 |
| p(λ) | 5.00 |
| p(λ) | 9.53 |
Figure 8.10 Simply supported beam on spring supports; negative spring stiffness can result in solution difficulties.
is zero, but all other elements of the last m rows of the upper triangular factor of the coefficient matrix are also zero, and the factorization of the matrix has already been completed. In other words, since the number of rows that are made up of only zero elements is equal to the multiplicity m of the zero eigenvalue, the last m - 1 matrices \mathbf{L}_{n-1}^{-1}, \mathbf{L}_{n-2}^{-1}, \ldots, \mathbf{L}_{n-m+1}^{-1} in (8.10) cannot and need not be calculated. We shall therefore be able to solve for m linearly independent solutions by assuming appropriate values for the m last entries of the solution vector.
Consider the following example.
EXAMPLE 8.13: Consider the beam element in Fig. E8.13(a). The stiffness matrix of the element is
\mathbf {K} = \left[ \begin{array}{c c c c} 1 2 & - 6 & - 1 2 & - 6 \\ - 6 & 4 & 6 & 2 \\ - 1 2 & 6 & 1 2 & 6 \\ - 6 & 2 & 6 & 4 \end{array} \right]
Show that Gauss elimination results in the third and fourth row consisting of only zero elements and evaluate formally the rigid body mode displacements.
Using the procedure in (8.10), we expect to arrive at a matrix S with its last two rows consisting of only zero elements because the beam element has two rigid body modes corre-
text_image
U₂ EI = 1.0 U₁ L = 1.0 U₃ U₄
(a) Initial degree of freedom numbering
text_image
U₃ U₁ U₂ U₄
(b) Degree of freedom numbering that requires column and row interchange
Figure E8.13 Beam element with two rigid body modes
sponding to vertical translation and rotation. We have
\mathbf {L} _ {1} ^ {- 1} = \left[ \begin{array}{c c c c} 1 & & & \\ \frac {1}{2} & 1 & & \\ 1 & 0 & 1 & \\ \frac {1}{2} & 0 & 0 & 1 \end{array} \right]
Hence,
\mathbf {L} _ {i} ^ {- 1} \mathbf {K} = \left[ \begin{array}{c c c c} 1 2 & - 6 & - 1 2 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right]
Then
\mathbf {L} _ {2} ^ {- 1} = \left[ \begin{array}{c c c c} 1 & & & \\ 0 & 1 & & \\ 0 & 0 & 1 & \\ 0 & 1 & 0 & 1 \end{array} \right]
and
\mathbf {S} = \mathbf {L} _ {2} ^ {- 1} \mathbf {L} _ {1} ^ {- 1} \mathbf {K} = \left[ \begin{array}{c c c c} 1 2 & - 6 & - 1 2 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \tag {a}
Therefore, as expected, the last two rows in S consist of zero elements, and L_{3}^{-1} cannot and need not be calculated. We should also note that if the numbering of the degrees of freedom of the beam element were initially as in Fig. E8.13(b), we would need to interchange rows and columns 2 with rows and columns 3 in order to be able to continue with the triangularization. This, however, is equivalent to using the degree of freedom numbering that we were concerned with in the first place, i.e., the numbering in Fig. E8.13(a).
Using the matrix \mathbf{S} in (a), we can now formally evaluate the rigid body mode displacements of the beam, i.e., solve the equations \mathbf{KU} = \mathbf{0} to obtain two linearly independent solutions for \mathbf{U} .
First, assume that U_{4} = 1 and U_{3} = 0 ; then using \mathbf{S} , we obtain
1 2 U _ {1} - 6 U _ {2} = 6
U _ {2} = 1
Hence,
U _ {1} = 1, \qquad U _ {2} = 1, \qquad U _ {3} = 0, \qquad U _ {4} = 1
Then assume that U_{4} = 0 , U_{3} = 1 , to obtain
1 2 U _ {1} - 6 U _ {2} = 1 2
U _ {2} = 0