김경종
27 KiB
Hence, in general, the pivots should be selected as the largest (in absolute value) of the elements in any column. This process is called partial pivoting. Even better results can be obtained by choosing the pivot as the largest element in the whole matrix of the remaining equations and performing appropriate interchanging of rows. This is called complete pivoting. Complete pivoting requires a large amount of testing, so it is not recommended in general.
The finite element equations generally involve coefficients with different orders of magnitude, so Gaussian elimination with partial pivoting is a useful method for solving the equations.
Finally, it has been shown that for n simultaneous equations, the number of arithmetic operations required in Gaussian elimination is n divisions, { \frac { 1 } { 3 } } n ^ { 3 } + n ^ { 2 } multiplications, and \textstyle { \frac { 1 } { 3 } } n ^ { 3 } + n additions. If partial pivoting is included, the number of comparisons needed to select pivots is n ( n + 1 ) / 2 .
Other elimination methods, including the Gauss–Jordan and Cholesky methods, have some advantages over Gaussian elimination and are sometimes used to solve large systems of equations. For descriptions of other methods, see References [1–3].
Gauss–Seidel Iteration
Another general class of methods (other than the elimination methods) used to solve systems of linear algebraic equations is the iterative methods. Iterative methods work well when the system of equations is large and sparse (many zero coefficients). The Gauss–Seidel method starts with the original set of equations \underline { { a } } \underline { { x } } = \underline { { c } } written in the form
\begin{array}{l} x _ {1} = \frac {1}{a _ {1 1}} \left(c _ {1} - a _ {1 2} x _ {2} - a _ {1 3} x _ {3} - \dots - a _ {1 n} x _ {n}\right) \\ x _ {2} = \frac {1}{a _ {2 2}} \left(c _ {2} - a _ {2 1} x _ {1} - a _ {2 3} x _ {3} - \dots - a _ {2 n} x _ {n}\right) \end{array} \tag {B.3.29}
\begin{array}{l} \vdots \\ x _ {n} = \frac {1}{a _ {n n}} \big (c _ {n} - a _ {n 1} x _ {1} - a _ {n 2} x _ {2} - \dots - a _ {n, n - 1} x _ {n - 1} \big) \end{array}
The following steps are then applied.
- Assume a set of initial values for the unknowns
x _ { 1 } , x _ { 2 } , \ldots , x _ { n } ., and substitute them into the right side of the first of Eqs. (B.3.29) to solve for the newx _ { 1 }. - Use the latest value for
x _ { 1 }obtained from step 1 and the initial values forx _ { 3 } , x _ { 4 } , \ldots , x _ { n }in the right side of the second of Eqs. (B.3.29) to solve for the newx _ { 2 }. - Continue using the latest values of the x’s obtained in the left side of Eqs. (B.3.29) as the next trial values in the right side for each succeeding step.
- Iterate until convergence is satisfactory.
A good initial set of values (guesses) is often x _ { i } = c _ { i } / a _ { i i } . An example will serve to illustrate the method.
Example B.3
Consider the set of linear simultaneous equations given by
\begin{array}{l} 4 x _ {1} - x _ {2} = 2 \\ - x _ {1} + 4 x _ {2} - x _ {3} = 5 \tag {B.3.30} \\ - x _ {2} + 4 x _ {3} - x _ {4} = 6 \\ - x _ {3} + 2 x _ {4} = - 2 \\ \end{array}
Using the initial guesses given by x _ { i } = c _ { i } / a _ { i i } , we have
x _ {1} = \frac {2}{4} = \frac {1}{2} \quad x _ {2} = \frac {5}{4} \approx 1 \quad x _ {3} = \frac {6}{4} \approx 1 \quad x _ {4} = - 1
Solving the first of Eqs. (B.3.30) for x _ { 1 } yields
x _ {1} = \frac {1}{4} (2 + x _ {2}) = \frac {1}{4} (2 + 1) = \frac {3}{4}
Solving the second of Eqs. (B.3.30) for x _ { 2 } . , we have
x _ {2} = \frac {1}{4} (5 + x _ {1} + x _ {3}) = \frac {1}{4} (5 + \frac {3}{4} + 1) = 1. 6 8
Solving the third of Eqs. (B.3.30) for x _ { 3 } , we have
x _ {3} = \frac {1}{4} (6 + x _ {2} + x _ {4}) = \frac {1}{4} [ 6 + 1. 6 8 + (- 1) ] = 1. 6 7 2
Solving the fourth of Eqs. (B.3.30) for x _ { 4 } . we obtain
x _ {4} = \frac {1}{2} (- 2 + x _ {3}) = \frac {1}{2} (- 2 + 1. 6 7) = - 0. 1 6
The first iteration has now been completed. The second iteration yields
\begin{array}{l} x _ {1} = \frac {1}{4} (2 + 1. 6 8) = 0. 9 2 2 \\ x _ {2} = \frac {1}{4} (5 + 0. 9 2 2 + 1. 6 7 2) = 1. 8 9 9 \\ x _ {3} = \frac {1}{4} [ 6 + 1. 8 9 9 + (- 0. 1 6) ] = 1. 9 4 4 \\ x _ {4} = \frac {1}{2} (- 2 + 1. 9 4 4) = - 0. 0 2 8 \\ \end{array}
Table B–1 lists the results of four iterations of the Gauss–Seidel method and the exact solution. From Table B–1, we observe that convergence to the exact solution has proceeded rapidly by the fourth iteration, and the accuracy of the solution is dependent on the number of iterations.
In general, iteration methods are self-correcting, such that an error made in calculations at one iteration will be corrected by later iterations. However, there are certain systems of equations for which iterative methods are not convergent.
Table B--1 Results of four iterations of the Gauss–Seidel method for Eqs. (B.3.30)
| Iteration | $x_{1}$ | $x_{2}$ | $x_{3}$ | $x_{4}$ |
| 0 | 0.5 | 1.0 | 1.0 | -1.0 |
| 1 | 0.75 | 1.68 | 1.672 | -0.16 |
| 2 | 0.922 | 1.899 | 1.944 | -0.028 |
| 3 | 0.975 | 1.979 | 1.988 | -0.006 |
| 4 | 0.9985 | 1.9945 | 1.9983 | -0.0008 |
| Exact | 1.0 | 2.0 | 2.00 | 0 |
When the equations can be arranged such that the diagonal terms are greater than the off-diagonal terms, the possibility of convergence is usually enhanced.
Finally, it has been shown that for n simultaneous equations, the number of arithmetic operations required by Gauss–Seidel iteration is n divisions, n ^ { 2 } multiplications, and n ^ { 2 } - n additions for each iteration.
B.4 Banded-Symmetric Matrices, Bandwidth, Skyline, and Wavefront Methods
The coefficient matrix (stiffness matrix) for the linear equations that occur in structural analysis is always symmetric and banded. Because a meaningful analysis generally requires the use of a large number of variables, the implementation of compressed storage of the stiffness matrix is desirable both from the standpoint of fitting into memory (immediate access portion of the computer) and for computational efficiency. We will discuss the banded-symmetric format, which is not necessarily the most efficient format but is relatively simple to implement on the computer.
Another method, based on the concept of the skyline of the stiffness matrix, is often used to improve the efficiency in solving the equations. The skyline is an envelope that begins with the first nonzero coefficient in each column of the stiffness matrix (Figure B–5). In skylining, only the coefficients between the main diagonal and the skyline are stored (normally by successive columns) in a one-dimensional array. In general, this procedure takes even less storage space in the computer and is more efficient in terms of equation solving than the conventional banded format. (For more information on skylining, consult References [10–12].)
A matrix is banded if the nonzero terms of the matrix are gathered about the main diagonal. To illustrate this concept, consider the plane truss of Figure B–4.
From Figure B–4, we see that element 2 connects nodes 1 and 4. Therefore, the 2 \times 2 submatrices at positions 1–1, 1–4, 4–1, and 4–4 of Figure B–5 have nonzero coefficients. Figure B–5 represents the total stiffness matrix of the plane truss. The X ^ { \ast } \mathrm { s } denote nonzero coefficients. From Figure B–5, we observe that the nonzero terms are within the band shown. When we use a banded storage format, only the main diagonal and the nonzero upper codiagonals need be stored as shown in Figure B–6. Note that any codiagonal with a nonzero term requires storage of the whole
text_image
① 1 2 3 ② 4 5 6 7 8 9 10 11 12
Figure B–4 Plane truss for bandwidth illustration
text_image
n_b = 8 Skyline X X 0 X X X X X X 0 X X X X 0 X X X X X X 0 X X X X 0 X X X X X X 0 X X X 0 0 X 0 Symmetry n = 24
Figure B–5 Stiffness matrix for the plane truss of Figure B–4, where X denotes, in general, blocks of 2 \times 2 submatrices with nonzero coefficients
codiagonal and any codiagonals between it and the main diagonal. The use of banded storage is efficient for computational purposes. The Scientific Subroutine Package gives a more detailed explanation of banded compressed storage [4].
We now define the semibandwidth n _ { b } as n _ { b } = n _ { d } ( m + 1 ) , where n _ { d } is the number of degrees of freedom per node and m is the maximum difference in node numbers determined by calculating the difference in node numbers for each element of a finite element model. In the example for the plane truss of Figure B–4, m = 4 - 1 = 3 and n _ { d } = 2 , \mathrm { { s o } } n _ { b } = 2 ( 3 + 1 ) = 8 .
Execution time (primarily equation-solving time) is a function of the number of equations to be solved. It has been shown [5] that when banded storage of global stiffness matrix \underline { { K } } is not used, execution time is proportional to ( 1 / 3 ) n ^ { \bar { 3 } } , where n is the number of equations to be solved, or, equivalently, the size of \underline { { K } } . When banded storage of \underline { { K } } is used, the execution time is proportional to ( n ) n _ { b } ^ { 2 } . The ratio of time of execution without banded storage to that with banded storage is then
text_image
n_b = 8 \left( \begin{array}{cccc} X & X & 0 & X \ X & X & X & X \ X & 0 & X & X \ X & X & 0 & X \ X & X & X & X \ X & 0 & X & X \ X & X & 0 & X \ X & X & X & X \ X & 0 & X & X \ X & 0 & 0 & 0 \ X & 0 & 0 & 0 \ X & 0 & 0 & 0 \end{array} \right)
Figure B–6 Banded storage format of the stiffness matrix of Figure B–5
\left( 1 / 3 \right) \left( n / n _ { b } \right) ^ { 2 } . For the plane truss example, this ratio is ( 1 / 3 ) ( 2 4 / 8 ) ^ { 2 } = 3 . Therefore, it takes about three times as long to execute the solution of the example truss if banded storage is not used.
Hence, to reduce bandwidth we should number systematically and try to have a minimum difference between adjacent nodes. A small bandwidth is usually achieved by consecutive node numbering across the shorter dimension, as shown in Figure B–4. Some computer programs use the banded-symmetric format for storing the global stiffness matrix, K .
Several automatic node-renumbering schemes have been computerized [6]. This option is available in most general-purpose computer programs. Alternatively, the wavefront or frontal method is becoming popular for optimizing equation solution time. In the wavefront method, elements, instead of nodes, are automatically renumbered.
In the wavefront method, the assembly of the equations alternates with their solution by Gauss elimination. The sequence in which the equations are processed is determined by element numbering rather than by node numbering. The first equations eliminated are those associated with element 1 only. Next, the contributions of stiffness coefficients of the adjacent element, element 2, are added to the system of equations. If any additional degrees of freedom are contributed by elements 1 and 2 only—that is, if no other elements contribute stiffness coefficients to specific degrees of freedom—these equations are eliminated (condensed) from the system of equations. As one or more additional elements make their contributions to the system of equations and additional degrees of freedom are contributed only by these elements, those degrees of freedom are eliminated from the solution. This repetitive alternation between assembly and solution was initially seen as a wavefront that sweeps over the structure in a pattern determined by the element numbering. For greater efficiency of this method, consecutive element numbering should be done across the structure in a direction that spans the smallest number of nodes.
The wavefront method, though somewhat more difficult to understand and to program than the banded-symmetric method, is computationally more efficient. A banded solver stores and processes any blocks of zeros created in assembling the stiffness matrix. In the wavefront method, these blocks of zero coefficients are not stored
or processed. Many large-scale computer programs are now using the wavefront method to solve the system of equations. (For additional details of this method, see References [7–9].) Example B.4 illustrates the wavefront method for solution of a truss problem.
Example B.4
For the plane truss shown in Figure B–7, illustrate the wavefront solution procedure.
We will solve this problem in symbolic form. Merging k’s for elements 1, 2, and 3 and enforcing boundary conditions at node 1, we have
\left[ \begin{array}{c c c c c c} d _ {2 x} & d _ {2 y} & d _ {3 x} & d _ {3 y} & d _ {4 x} & d _ {4 y} \\ \hline k _ {3 3} ^ {(1)} + k _ {1 1} ^ {(2)} + k _ {1 1} ^ {(3)} & k _ {3 4} ^ {(1)} + k _ {1 2} ^ {(2)} + k _ {1 2} ^ {(3)} & k _ {1 3} ^ {(3)} & k _ {1 4} ^ {(3)} & k _ {1 3} ^ {(2)} & k _ {1 4} ^ {(2)} \\ k _ {4 3} ^ {(1)} + k _ {2 1} ^ {(2)} + k _ {2 1} ^ {(3)} & k _ {4 4} ^ {(1)} + k _ {2 2} ^ {(2)} + k _ {2 2} ^ {(3)} & k _ {2 3} ^ {(3)} & k _ {2 4} ^ {(3)} & k _ {2 3} ^ {(2)} & k _ {2 4} ^ {(2)} \\ \hline k _ {3 1} ^ {(2)} & k _ {3 2} ^ {(3)} & k _ {3 3} ^ {(3)} & k _ {3 4} ^ {(3)} & k _ {3 3} ^ {(2)} & k _ {3 4} ^ {(2)} \\ k _ {4 1} ^ {(3)} & k _ {4 2} ^ {(3)} & k _ {4 3} ^ {(3)} & k _ {4 4} ^ {(3)} & k _ {4 3} ^ {(2)} & k _ {4 4} ^ {(2)} \\ k _ {3 1} ^ {(2)} & k _ {3 2} ^ {(2)} & 0 & 0 & 0 & 0 \\ k _ {4 1} ^ {(2)} & k _ {4 2} ^ {(2)} & 0 & 0 & 0 & 0 \end{array} \right] \left\{ \begin{array}{l} d _ {2 x} \\ d _ {2 y} \\ \hline d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \\ d _ {4 x} ^ {\prime} \\ d _ {4 y} ^ {\prime} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ 0 \\ - P \\ 0 \\ 0 \end{array} \right\} \tag {B.4.1}
Eliminating d _ { 2 x } and d _ { 2 y } (all stiffness contributions from node 2 degrees of freedom have been included from these elements; these contributions are from elements 1–3) by static condensation or Gauss elimination yields
\left[ k _ {c} ^ {\prime} \right] \left\{ \begin{array}{l} d _ {3 x} ^ {\prime} \\ d _ {3 y} ^ {\prime} \\ d _ {4 x} ^ {\prime} \\ d _ {4 y} ^ {\prime} \end{array} \right\} = \left\{F _ {c} ^ {\prime} \right\} \tag {B.4.2}
text_image
1 4 P 3 5 5 1 x̂ 3 x̂ 6 7 2 2 4
Figure B–7 Truss for wavefront solution
where the condensed stiffness and force matrices are (also see Section 7.5)
[ k _ {c} ^ {\prime} ] = [ K _ {2 2} ^ {\prime} ] - [ K _ {2 1} ^ {\prime} ] [ K _ {1 1} ^ {\prime} ] ^ {- 1} [ K _ {1 2} ^ {\prime} ] \tag {B.4.3}
\{F _ {c} ^ {\prime} \} = \{F _ {2} ^ {\prime} \} - [ K _ {2 1} ^ {\prime} ] [ K _ {1 1} ^ {\prime} ] ^ {- 1} \{F _ {1} ^ {\prime} \} \tag {B.4.4}
where primes on the degrees of freedom, such as d _ { 3 x } ^ { \prime } in Eq. (B.4.1), indicate that all stiffness coefficients associated with that degree of freedom have not yet been included. Now include elements 4–6 for degrees of freedom at node 3. The resulting equations are
d _ {3 x} \quad d _ {3 y} \quad d _ {4 x} \quad d _ {4 y}
\left[ \begin{array}{c c c c} k _ {c 1 1} ^ {\prime} + k _ {3 3} ^ {(4)} + k _ {1 1} ^ {(5)} + k _ {1 1} ^ {(6)} & k _ {3 4} ^ {(4)} + k _ {1 2} ^ {(5)} + k _ {1 2} ^ {(6)} + k _ {c 1 2} ^ {\prime} & k _ {1 3} ^ {(6)} + k _ {c 1 3} ^ {\prime} & k _ {1 4} ^ {(6)} + k _ {c 1 4} ^ {\prime} \\ k _ {c 2 1} ^ {\prime} + k _ {3 4} ^ {(4)} + k _ {2 1} ^ {(5)} + k _ {2 1} ^ {(6)} & k _ {4 4} ^ {(4)} + k _ {2 2} ^ {(5)} + k _ {2 2} ^ {(6)} + k _ {c 2 2} ^ {\prime} & k _ {2 3} ^ {(6)} + k _ {c 2 3} ^ {\prime} & k _ {2 4} ^ {(6)} + k _ {c 2 4} ^ {\prime} \\ \hline k _ {c 3 1} ^ {\prime} + k _ {3 1} ^ {(6)} & k _ {c 3 2} ^ {\prime} + k _ {3 2} ^ {(6)} & k _ {c 3 3} ^ {\prime} + k _ {3 3} ^ {(6)} & k _ {c 3 4} ^ {\prime} + k _ {3 4} ^ {(6)} \\ k _ {c 4 1} ^ {\prime} + k _ {4 1} ^ {(6)} & k _ {c 4 2} ^ {\prime} + k _ {4 2} ^ {(6)} & k _ {c 4 3} ^ {\prime} + k _ {4 3} ^ {(6)} & k _ {c 4 4} ^ {\prime} + k _ {4 4} ^ {(6)} \end{array} \right]
\times \left\{ \begin{array}{l} d _ {3 x} \\ - \frac {d _ {3 y}}{d _ {4 x} ^ {\prime}} \\ d _ {4 y} ^ {\prime} \end{array} \right\} = \left\{ \begin{array}{l} 0 \\ - \frac {- P}{0} \\ 0 \end{array} \right\} \tag {B.4.5}
Using static condensation, we eliminate d _ { 3 x } and d _ { 3 y } (all contributions from node 3 degrees of freedom have been included from each element) to obtain
\left[ k _ {c} ^ {\prime \prime} \right] \left\{ \begin{array}{l} d _ {4 x} ^ {\prime} \\ d _ {4 y} ^ {\prime} \end{array} \right\} = \left\{F _ {c} ^ {\prime \prime} \right\} \tag {B.4.6}
where [ k _ { c } ^ { \prime \prime } ] = [ K _ { 2 2 } ^ { \prime \prime } ] - [ K _ { 2 1 } ^ { \prime \prime } ] [ K _ { 1 1 } ^ { \prime \prime } ] ^ { - 1 } [ K _ { 1 2 } ^ { \prime \prime } ] ðB:4:7Þ
\left\{F _ {c} ^ {\prime \prime} \right\} = \left\{F _ {2} ^ {\prime \prime} \right\} - \left[ K _ {2 1} ^ {\prime \prime} \right] \left[ K _ {1 1} ^ {\prime \prime} \right] ^ {- 1} \left\{F _ {1} ^ {\prime \prime} \right\} \tag {B.4.8}
Next we include element 7 contributions to the stiffness matrix. The condensed set of equations yield
\left[ k _ {c} ^ {\prime \prime \prime} \right] \left\{ \begin{array}{l} d _ {4 x} \\ d _ {4 y} \end{array} \right\} = \left\{F _ {c} ^ {\prime \prime \prime} \right\} \tag {B.4.9}
\left[ k _ {c} ^ {\prime \prime \prime} \right] = \left[ K _ {2 2} ^ {\prime \prime \prime} \right] - \left[ K _ {2 1} ^ {\prime \prime \prime} \right] \left[ K _ {1 1} ^ {\prime \prime \prime} \right] ^ {- 1} \left[ K _ {1 2} ^ {\prime \prime \prime} \right] \tag {B.4.10}
where \{ F _ { c } ^ { \prime \prime \prime } \} = \{ F _ { 2 } ^ { \prime \prime \prime } \} - [ K _ { 2 1 } ^ { \prime \prime \prime } ] [ K _ { 1 1 } ^ { \prime \prime \prime } ] ^ { - 1 } \{ F _ { 1 } ^ { \prime \prime \prime } \} ðB:4:11Þ
The elimination procedure is now complete, and we solve Eq. (B.4.9) for d _ { 4 x } and d _ { 4 y } . Then we back-substitute d _ { 4 x } and d _ { 4 ) } into Eq. (B.4.5) to obtain d _ { 3 x } and d _ { 3 y } . Finally, we back-substitute d _ { 3 x } through d _ { 4 y } into Eq. (B.4.1) to obtain d _ { 2 x } and d _ { 2 y } . Static condensation and Gauss elimination with back-substitution have been used to solve the
set of equations for all the degrees of freedom. The solution procedure has then proceeded as though it were a wave sweeping over the structure, starting at node 2, engulfing node 2 and elements with degrees of freedom at node 2, and then sweeping through node 3 and finally node 4. 9
We now describe a practical computer scheme often used in computer programs for the solution of the resulting system of algebraic equations. The significance of this scheme is that it takes advantage of the fact that the stiffness method produces a banded K matrix in which the nonzero elements occur about the main diagonal in K. While the equations are solved, this banded format is maintained.
Example B.5
We will now use a simple example to illustrate this computer scheme. Consider the three-spring assemblage shown in Figure B–8. The assemblage is subjected to forces at node 2 of 100 lb in the x direction and 200 lb in the y direction. Node 1 is completely constrained from displacement in both the x and y directions, whereas node 3 is completely constrained in the y direction but is displaced a known amount d in the x direction.
Our purpose here is not to obtain the actual K for the assemblage but rather to illustrate the scheme used for solution. The general solution can be shown to be given by
\left[ \begin{array}{c c c c c c} k _ {1 1} & k _ {1 2} & k _ {1 3} & k _ {1 4} & k _ {1 5} & k _ {1 6} \\ & k _ {2 2} & k _ {2 3} & k _ {2 4} & k _ {2 5} & k _ {2 6} \\ & & k _ {3 3} & k _ {3 4} & k _ {3 5} & k _ {3 6} \\ & & & k _ {4 4} & k _ {4 5} & k _ {4 6} \\ & & & & k _ {5 5} & k _ {5 6} \\ \text { Symmetry } & & & & & k _ {6 6} \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} = \left\{ \begin{array}{c} F _ {1 x} \\ F _ {1 y} \\ F _ {2 x} = 1 0 0 \\ F _ {2 y} = 2 0 0 \\ F _ {3 x} \\ F _ {3 y} \end{array} \right\} \tag {B.4.12}
text_image
200 lb 2 100 lb k k 1 k 3 δ y x
Figure B–8 Three-spring assemblage
where \underline { { K } } has been left in general form. Upon our imposing the boundary conditions, the computer program transforms Eq. (B.4.12) to:
\left[ \begin{array}{c c c c c c} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & k _ {3 3} & k _ {3 4} & 0 & 0 \\ 0 & 0 & k _ {4 3} & k _ {4 4} & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] \left\{ \begin{array}{l} d _ {1 x} \\ d _ {1 y} \\ d _ {2 x} \\ d _ {2 y} \\ d _ {3 x} \\ d _ {3 y} \end{array} \right\} = \left\{ \begin{array}{c} 0 \\ 0 \\ 1 0 0 - k _ {3 5} \delta \\ 2 0 0 - k _ {4 5} \delta \\ \delta \\ 0 \end{array} \right\} \tag {B.4.13}
From Eq. (B.4.13), we can see that d _ { 1 x } = 0 , d _ { 1 y } = 0 , d _ { 3 y } = 0 , and d _ { 3 x } = \delta . These displacements are consistent with the imposed boundary conditions. The unknown displacements, d _ { 2 x } and d _ { 2 y } , , can be determined routinely by solving Eq. (B.4.13).
We will now explain the computer scheme that is generally applicable to transform Eq. (B.4.12) to Eq. (B.4.13). First, the terms associated with the known displacement boundary condition(s) within each equation were transformed to the right side of those equations. In the third and fourth equations of Eq. (B.4.12), k _ { 3 5 } \delta and k _ { 4 5 } \delta were transformed to the right side, as shown in Eq. (B.4.13). Then the right-side force term corresponding to the known displacement row was equated to the known displacement. In the fifth equation of Eq. (B.4.12), where d _ { 3 x } = \delta , the right-side, fifth-row force term F _ { 3 x } was equated to the known displacement \delta , as shown in Eq. (B.4.13). For the homogeneous boundary conditions, the affected rows of \underline { { F } } , corresponding to the zero-displacement rows, were replaced with zeros. Again, this is done in the computer scheme only to obtain the nodal displacements and does not imply that these nodal forces are zero. We obtain the unknown nodal forces by determining the nodal displacements and back-substituting these results into the original Eq. (B.4.12). Because d _ { 1 x } = 0 , d _ { 1 y } = 0 , and d _ { 3 y } = 0 in \operatorname { E q } . . (B.4.12), the first, second, and sixth rows of the force matrix of Eq. (B.4.13) were set to zero. Finally, for both nonhomogeneous and homogeneous boundary conditions, the rows and columns of \underline { { K } } corresponding to these prescribed boundary conditions were set to zero except the main diagonal, which was made unity. That is, the first, second, fifth, and sixth rows and columns of \underline { { K } } in Eq. (B.4.12) were set to zero, except for the main diagonal terms, which were made unity. Although doing so is not necessary, setting the main diagonal terms equal to 1 facilitates the simultaneous solution of the six equations in Eq. (B.4.13) by an elimination method used in the computer program. This modification is shown in the \underline { { K } } matrix of Eq. (B.4.13).
d References
[1] Southworth, R. W., and DeLeeuw, S. L., Digital Computation and Numerical Methods, McGraw-Hill, New York, 1965.
[2] James, M. L., Smith, G. M., and Wolford, J. C., Applied Numerical Methods for Digital Computation, 3rd ed., Harper & Row, New York, 1985.
[3] Bathe, K. J., and Wilson, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1976.
[4] SYSTEM/360, Scientific Subroutine Package, IBM.
[5] Kardestuncer, H., Elementary Matrix Analysis of Structures, McGraw-Hill, New York, 1974.
[6] Collins, R. J., ‘‘Bandwidth Reduction by Automatic Renumbering,’’ International Journal For Numerical Methods in Engineering, Vol. 6, pp. 345–356, 1973.
[7] Melosh, R. J., and Bamford, R. M., ‘‘Efficient Solution of Load-Deflection Equations,’’ Journal of the Structural Division, American Society of Civil Engineers, No. ST4, pp. 661– 676, April 1969.
[8] Irons, B. M., ‘‘A Frontal Solution Program for Finite Element Analysis,’’ International Journal for Numerical Methods in Engineering, Vol. 2, No. 1, pp. 5–32, 1970.
[9] Meyer, C., ‘‘Solution of Linear Equations-State-of-the-Art,’’ Journal of the Structural Division, American Society of Civil Engineers, Vol. 99, No. ST7, pp. 1507–1526, 1973.
[10] Jennings, A., Matrix Computation for Engineers and Scientists, Wiley, London, 1977.
[11] Cook, R. D., Malkus, D. S., Plesha, M. E., and Witt, R. J., Concepts and Applications of Finite Element Analysis, 4th ed., Wiley, New York, 2002.
[12] Bathe, K. J., and Wilson, E. L., Numerical Methods in Finite Element Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1976.
Problems
B.1 Determine the solution of the following simultaneous equations by Cramer’s rule.
1 x _ {1} + 3 x _ {2} = 5
4 x _ {1} - 1 x _ {2} = 1 2
B.2 Determine the solution of the following simultaneous equations by the inverse method.
1 x _ {1} + 3 x _ {2} = 5
4 x _ {1} - 1 x _ {2} = 1 2
B.3 Solve the following system of simultaneous equations by Gaussian elimination.
x _ {1} - 4 x _ {2} - 5 x _ {3} = 4
3 x _ {2} + 4 x _ {3} = - 1
- 2 x _ {1} - 1 x _ {2} + 2 x _ {3} = - 3
B.4 Solve the following system of simultaneous equations by Gaussian elimination.
2 x _ {1} + 1 x _ {2} - 3 x _ {3} = 1 1
4 x _ {1} - 2 x _ {2} + 3 x _ {3} = 8
- 2 x _ {1} + 2 x _ {2} - 1 x _ {3} = - 6