김경종
397 lines
24 KiB
Markdown
397 lines
24 KiB
Markdown
<!-- source-page: 711 -->
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7.29. Use a computer program to analyze the forced convection steady-state flow between two parallel plates as shown. Verify that an accurate solution has been obtained.
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<details>
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<summary>text_image</summary>
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L
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∂θ/∂z = 0
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θ = 0
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2h
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q = 0
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q = -k ∂θ/∂z = 1
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dp/dy = -2.0
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z
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y
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</details>
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$$
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\begin{array}{l} h = 1. 0 \quad \rho = 1. 0 \\ k = 1. 0 \quad c _ {p} = 1. 0 \\ \mu = 1. 0 \quad L = 1 5. 0 \\ \end{array}
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$$
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7.30. Use a computer program to analyze the steady-state conjugate heat transfer in a pipe. The analysis problem is described in the figure. Verify that accurate results have been obtained (see J. H. Lienhard [A]).
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$$
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q _ {w} = \text { heat flow input on external surface of pipe }
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$$
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$$
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\frac {d p}{d z} = - 4
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$$
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<details>
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<summary>text_image</summary>
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L = 10.0 For solid: For fluid:
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t = 0.1 k_s = 1.0 k_f = 1.0
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R = 1.0 ρ = 1.0
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c_p = 5.0
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θ = 0 y_r r z R Fluid
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t
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q_w = 0 q_w = 1.0
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5.0 L
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5.0
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Pipe
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q_w = 0
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</details>
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<!-- source-page: 712 -->
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# Solution of Equilibrium Equations in Static Analysis
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# 8.1 INTRODUCTION
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So far we have considered the derivation and calculation of the equilibrium equations of a finite element system. This included the selection and calculation of efficient elements and the efficient assemblage of the element matrices into the global finite element system matrices. However, the overall effectiveness of an analysis depends to a large degree on the numerical procedures used for the solution of the system equilibrium equations. As discussed earlier, the accuracy of the analysis can, in general, be improved if a more refined finite element mesh is used. Therefore, in practice, an analyst tends to employ larger and larger finite element systems to approximate the actual structure. However, this means that the cost of an analysis and, in fact, its practical feasibility depend to a considerable degree on the algorithms available for the solution of the resulting systems of equations. Because of the requirement that large systems be solved, much research effort has gone into optimizing the equation solution algorithms. During the early use of the finite element method, equations of the order 10,000 were in many cases considered of large order. Currently, equations of the order 100,000 are solved without much difficulty.
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Depending on the kind and number of elements used in the assemblage and on the topology of the finite element mesh, in a linear static analysis the time required for solution of the equilibrium equations can be a considerable percentage of the total solution time, whereas in dynamic analysis or in nonlinear analysis, this percentage may be still higher. Therefore, if inappropriate techniques for the solution of the equilibrium equations are used, the total cost of analysis is affected a great deal, and indeed the cost may be many times, say 100 times, larger than is necessary.
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In addition to considering the actual computer effort that is spent on the solution of the equilibrium equations, it is important to realize that an analysis may, in fact, not be possible if inappropriate numerical procedures are used. This may be the case because the analysis is simply too costly using the slow solution methods. But, more seriously, the
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<!-- source-page: 713 -->
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analysis may not be possible because the solution procedures are unstable. We will observe that the stability of the solution procedures is particularly important in dynamic analysis.
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In this chapter we are concerned with the solution of the simultaneous equations that arise in the static analysis of structures and solids, and we discuss first at length (see Sections 8.2 to 8.3) the solution of the equations that arise in linear analysis,
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$$
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\mathbf {K U} = \mathbf {R} \tag {8.1}
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$$
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where K is the stiffness matrix, U is the displacement vector, and R is the load vector of the finite element system. Since R and U may be functions of time t, we may also consider (8.1) as the dynamic equilibrium equations of a finite element system in which inertia and velocity-dependent damping forces have been neglected. It should be realized that since velocities and accelerations do not enter (8.1), we can evaluate the displacements at any time t independent of the displacement history, which is not the case in dynamic analysis (see Chapter 9). However, these thoughts suggest that the algorithms used for the evaluation of U in (8.1) may also be employed as part of the solution algorithms used in dynamic analysis. This is indeed the case; we will see in the following chapters that the procedures discussed here will be the basis of the algorithms employed for eigensolutions and direct step-by-step integrations. Furthermore, as noted already in Chapter 6 and further discussed in Section 8.4, the solution of (8.1) also represents a very important basic step of solution in a nonlinear analysis. Therefore, a detailed study of the procedures used to solve (8.1) is very important.
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Although we consider explicitly in this chapter the solution of the equilibrium equations that arise in the analysis of solids and structures, the techniques are quite general and are entirely and directly applicable to all those analyses that lead to symmetric (positive definite) coefficient matrices (see Chapters 3 and 7). The only sets of equations presented earlier whose solution we do not consider in detail are those arising in the analysis of incompressible viscous fluid flow (see Section 7.4)—because in such analysis a nonsymmetric coefficient matrix is obtained—but for that case most basic concepts and procedures given below are still applicable and can be directly extended (see Exercise 8.11).
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Essentially, there are two different classes of methods for the solution of the equations in (8.1): direct solution techniques and iterative solution methods. In a direct solution the equations in (8.1) are solved using a number of steps and operations that are predetermined in an exact manner, whereas iteration is used when an iterative solution method is employed. Either solution scheme will be seen to have certain advantages, and we discuss both approaches in this chapter. At present, direct techniques are employed in most cases, but for large systems iterative methods can be much more effective.
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# 8.2 DIRECT SOLUTIONS USING ALGORITHMS BASED ON GAUSS ELIMINATION
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The most effective direct solution techniques currently used are basically applications of Gauss elimination, which C. F. Gauss proposed over a century ago (see C. F. Gauss [A]). However, although the basic Gauss solution scheme can be applied to almost any set of simultaneous linear equations (see, for example, J. H. Wilkinson [A], B. Noble [A], and R. S. Martin, G. Peters, and J. H. Wilkinson [A]), the effectiveness in finite element analysis
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<!-- source-page: 714 -->
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depends on the specific properties of the finite element stiffness matrix: symmetry, positive definiteness, and bandedness.
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In the following we consider first the Gauss elimination procedure as it is used in the solution of positive definite, symmetric, and banded systems. We briefly consider the solution of symmetric indefinite systems in Section 8.2.5.
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# 8.2.1 Introduction to Gauss Elimination
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We propose to introduce the Gauss solution procedure by studying the solution of the equations $\mathbf{KU} = \mathbf{R}$ derived in Example 3.27 with the parameters $L = 5$ , $EI = 1$ ; i.e.,
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$$
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\left[ \begin{array}{r r r r} 5 & - 4 & 1 & 0 \\ - 4 & 6 & - 4 & 1 \\ 1 & - 4 & 6 & - 4 \\ 0 & 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array} \right] \tag {8.2}
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$$
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In this case the stiffness matrix K corresponds to a simply supported beam with four translational degrees of freedom, as shown in Fig. 8.1. (We should recall that the equilibrium equations have been derived by finite differences; but, in this case, they have the same properties as in finite element analysis.)
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# The Mathematical Operations
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Let us first consider the basic mathematical operations of Gauss elimination. We proceed in the following systematic steps:
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Step 1: Subtract a multiple of the first equation in (8.2) from the second and third equations to obtain zero elements in the first column of K. This means that $-\frac{4}{5}$ times the first row is subtracted from the second row, and $\frac{1}{5}$ times the first row is subtracted from the third row. The resulting equations are
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$$
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\left[ \begin{array}{c c c c} 5 & - 4 & 1 & 0 \\ 0 & \frac {1 4}{5} & - \frac {1 6}{5} & 1 \\ 0 & - \frac {1 6}{5} & \frac {2 9}{5} & - 4 \\ 0 & 1 & - 4 & 5 \end{array} \right] \quad \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array} \right] \tag {8.3}
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$$
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Step 2: Considering next the equations in (8.3), subtract $-\frac{16}{14}$ times the second equation from the third equation and $\frac{5}{14}$ times the second equation from the fourth equation. The resulting equations are
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$$
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\left[ \begin{array}{c c c c} 5 & - 4 & 1 & 0 \\ 0 & \frac {1 4}{5} & - \frac {1 6}{5} & 1 \\ 0 & 0 & \frac {1 5}{7} & - \frac {2 0}{7} \\ 0 & 0 & - \frac {2 0}{7} & \frac {6 5}{1 4} \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{c} 0 \\ 1 \\ \frac {8}{7} \\ - \frac {5}{1 4} \end{array} \right] \tag {8.4}
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$$
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<!-- source-page: 715 -->
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<details>
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<summary>text_image</summary>
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R₂ = 1
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U₁ U₂ U₃ U₄
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</details>
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(a)
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$$
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\left[ \begin{array}{r r r r} 5 & - 4 & 1 & 0 \\ - 4 & 6 & - 4 & 1 \\ 1 & - 4 & 6 & - 4 \\ 0 & 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array} \right]
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$$
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<details>
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<summary>text_image</summary>
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R₂ = 1
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U₂ U₃ U₄
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</details>
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(b)
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$$
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\left[ \begin{array}{c c c} \frac {1 4}{5} & - \frac {1 6}{5} & 1 \\ - \frac {1 6}{5} & \frac {2 9}{5} & - 4 \\ 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{c} U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]
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$$
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<details>
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<summary>text_image</summary>
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R₃ = 8/7
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R₄ = -5/14
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U₃
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U₄
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</details>
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(c)
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$$
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\left[ \begin{array}{c c} \frac {1 5}{7} & - \frac {2 0}{7} \\ - \frac {2 0}{7} & \frac {6 5}{1 4} \end{array} \right] \quad \left[ \begin{array}{c} U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{c} \frac {8}{7} \\ - \frac {5}{1 4} \end{array} \right]
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$$
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<details>
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<summary>text_image</summary>
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R₄ = 7/6
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U₄
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</details>
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(d)
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$$
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\left[ \frac {5}{6} \right] \quad \left[ U _ {4} \right] = \quad \left[ \frac {7}{6} \right]
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$$
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Figure 8.1 Stiffness matrices and load vectors considered in the Gauss elimination solution of the simply supported beam. The stiffness matrices in (b), (c), and (d) are the entries below the dashed lines in (8.3), (8.4), and (8.5).
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Step 3: Subtract $-\frac{20}{15}$ times the third equation from the fourth equation in (8.4). This gives
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$$
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\left[ \begin{array}{c c c c} 5 & - 4 & 1 & 0 \\ 0 & \frac {1 4}{5} & - \frac {1 6}{5} & 1 \\ 0 & 0 & \frac {1 5}{7} & - \frac {2 0}{7} \\ 0 & 0 & 0 & \vdots \frac {5}{6} \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 \\ \frac {8}{7} \\ \frac {7}{6} \end{array} \right] \tag {8.5}
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$$
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<!-- source-page: 716 -->
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Using (8.5), we can now simply solve for the unknowns $U_4, U_3, U_2$ , and $U_1$ :
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$$
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U _ {4} = \frac {\frac {7}{6}}{\frac {5}{6}} = \frac {7}{5}; \quad U _ {3} = \frac {\frac {8}{7} - (- \frac {2 0}{7}) U _ {4}}{\frac {1 5}{7}} = \frac {1 2}{5}
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$$
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$$
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U _ {2} = \frac {1 - (- \frac {1 6}{5}) U _ {3} - (1) U _ {4}}{\frac {1 4}{5}} = \frac {1 3}{5} \tag {8.6}
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$$
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$$
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U _ {1} = \frac {0 - (- 4) \frac {1 3}{5} - (1) \frac {1 2}{5} - (0) \frac {7}{5}}{5} = \frac {8}{5}
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$$
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The procedure in the solution is therefore to subtract in step number $i$ in succession multiples of equation $i$ from equations $i + 1, i + 2, \ldots, n$ , where $i = 1, 2, \ldots, n - 1$ . In this way the coefficient matrix $\mathbf{K}$ of the equations is reduced to upper triangular form, i.e., a form in which all elements below the diagonal elements are zero. Starting with the last equation, it is then possible to solve for all unknowns in the order $U_n, U_{n-1}, \ldots, U_1$ .
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It is important to note that at the end of step i the lower right submatrix of order n - i [indicated by dashed lines in (8.3) to (8.5)] is symmetric. Therefore, the elements above and including the diagonal can give all elements of the coefficient matrix at all times of the solution. We will see in Section (8.2.3) that in the computer implementation we work with only the upper triangular part of the matrix.
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Another important observation is that this solution assumes in step i a nonzero ith diagonal element in the current coefficient matrix. It is the nonzero value of the ith diagonal element in the coefficient matrix that makes it possible to reduce the elements below it to zero. Also, in the back-substitution for the solution of the displacements, we again divide by the diagonal elements of the coefficient matrix. Fortunately, in the analysis of displacement-based finite element systems, all diagonal elements of the coefficient matrix are positive at all times of the solution, which is another property that makes the application of the Gauss elimination procedure very effective. (This property is not necessarily preserved when the stiffness matrices are derived using a mixed formulation or by finite differences; see Sections 3.3.4 and 4.4.2). We will prove in Section 8.2.5 that the diagonal elements must remain larger than zero, but this property is also observed by considering the physical process of Gauss elimination.
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# The Physical Process
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In order to identify the physical process corresponding to the mathematical operations in Gauss elimination, we note first that the operations on the coefficient matrix K are independent of the elements in the load vector R. Therefore, let us consider now only the operations on the coefficient matrix K and for ease of explanation again use the above example and Fig. 8.1. We consider that no loads are applied and hence have
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$$
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\left[ \begin{array}{r r r r} 5 & - 4 & 1 & 0 \\ - 4 & 6 & - 4 & 1 \\ 1 & - 4 & 6 & - 4 \\ 0 & 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \end{array} \right] \tag {8.7}
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$$
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<!-- source-page: 717 -->
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Using the condition given by the first equation, i.e.,
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$$
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5 U _ {1} - 4 U _ {2} + U _ {3} = 0
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$$
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we can write
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$$
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U _ {1} = \frac {4}{5} U _ {2} - \frac {1}{5} U _ {3} \tag {8.8}
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$$
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and eliminate $U_{1}$ from the three equations remaining in (8.7). We thus obtain
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$$
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- 4 \left(\frac {4}{5} U _ {2} - \frac {1}{5} U _ {3}\right) + 6 U _ {2} - 4 U _ {3} + U _ {4} = 0
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$$
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$$
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\left(\frac {4}{5} U _ {2} - \frac {1}{5} U _ {3}\right) - 4 U _ {2} + 6 U _ {3} - 4 U _ {4} = 0
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$$
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$$
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U _ {2} - 4 U _ {3} + 5 U _ {4} = 0
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$$
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or, in matrix form,
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$$
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\left[ \begin{array}{c c c} \frac {1 4}{5} & - \frac {1 6}{5} & 1 \\ - \frac {1 6}{5} & \frac {2 9}{5} & - 4 \\ 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{l} U _ {2} \\ U _ {3} \\ U _ {4} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ 0 \end{array} \right] \tag {8.9}
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$$
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Comparing (8.9) with (8.3), we observe that the coefficient matrix in (8.9) is actually the lower right $3 \times 3$ submatrix of the coefficient matrix in (8.3). However, we obtained the coefficient matrix of (8.9) by using (8.7) and the condition in (8.8), which expresses that no force is applied at the degree of freedom 1 of the beam. It follows that the coefficient matrix in (8.9) is, in fact, the stiffness matrix of the beam that corresponds to the degrees of freedom 2, 3, and 4 when no force is applied at the degree of freedom 1, i.e., when the degree of freedom 1 has been “released” (which we shall also refer to as “statically condensed out”). By the same reasoning, we have obtained in (8.4) the stiffness matrix of the beam when the first two degrees of freedom have been released; and in (8.5), the element (4, 4) of the coefficient matrix represents the stiffness matrix of the beam corresponding to degree of freedom 4 when the degrees of freedom 1, 2, and 3 have all been released. These stiffness matrices are given in Figs. 8.1(b) to (d).
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Let us gain further insight into the Gauss elimination process by considering a (hypothetical) laboratory experiment—that is, a Gedankenexperiment. Suppose that in the laboratory we construct a physical beam corresponding to the model shown in Fig. 8.1. At the locations where the degrees of freedom are measured in Fig. 8.1(a), we fasten clamps to the beam with a force-measuring device, as shown in Fig. 8.2. We now impose the displacements shown in Figs. 8.3(a) to (d) to the beam and measure the required forces. These forces correspond to the columns of the stiffness matrix in (8.2). (Of course, depending on the appropriateness of our mathematical model, the accuracy of our numerical representation,
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<details>
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<summary>text_image</summary>
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Frictionless hinge
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Clamp 1
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Clamp 2
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Clamp 3
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Clamp 4
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Displacement imposing and force measuring device
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</details>
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Figure 8.2 Experimental set-up for measurements on beam
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<!-- source-page: 718 -->
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<details>
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<summary>text_image</summary>
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(a)
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5
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-4
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1
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0
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1.0
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(b)
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-4
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6
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-4
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1
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1.0
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(c)
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1
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-4
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6
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-4
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1.0
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(d)
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0
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1
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-4
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5
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1.0
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</details>
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Figure 8.3 Experimental results of forces (given to one digit) in clamps due to unit displacements. (Note that the zero force in the top and bottom results is unrealistic with the given beam curvature but the value of the force is so small that we neglect it.)
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and the accuracy of the laboratory measurements, the measured forces will be slightly different, but in our Gedankenexperiment we neglect that difference.)
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We now remove clamp 1 and repeat the experiment in the laboratory on the same physical beam. The results are shown in Fig. 8.4. The measured forces now correspond to the columns in the stiffness matrix (8.9) [shown also in Fig. 8.1(b)]. Next, we also remove clamp 2 and continue the experiment to obtain the force measurements shown in Fig. 8.5. These results correspond to the stiffness matrix in Fig. 8.1(c). Finally, we also remove clamp 3, and the force measurement gives the result shown in Fig. 8.6, which corresponds to the stiffness given in Fig. 8.1(d).
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The important point of our Gedankenexperiment is that the mathematical operations of Gauss elimination correspond to releasing degrees of freedom, one degree of freedom at a time, until only one degree of freedom is left (in our example $U_{4}$ ). The process of releasing a degree of freedom corresponds physically to removing the corresponding clamp. Hence, at every stage of the Gauss elimination a new stiffness matrix of the same physical structure is established, but this matrix corresponds to fewer degrees of freedom than were previously used.
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Of course, in a laboratory experiment we are free to establish stiffness matrices of the structure considered corresponding to any arbitrary set of selected degrees of freedom. Indeed, considering the beam in Fig. 8.2, we could by measurements establish first the stiffness in Fig. 8.1(d), then the stiffness matrix in Fig. 8.1(c), then the stiffness matrix in Fig. 8.1(b), and finally the stiffness matrix in Fig. 8.1(a), or use any other order of measurements. Also, we could move the clamps to any other locations and introduce more clamps,
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<!-- source-page: 719 -->
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Figure 8.4 Experimental results of forces in clamps due to unit displacements with clamp 1 not present.
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<details>
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<summary>text_image</summary>
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15/7
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-20/7
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8/7
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1.0
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5/7
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</details>
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<details>
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<summary>text_image</summary>
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-20/7
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65/14
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2/7
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5/14
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1.0
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</details>
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Figure 8.5 Experimental results of forces in clamps due to unit displacements with clamps 1 and 2 not present.
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<details>
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<summary>text_image</summary>
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2/3
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7/6
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4/3
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1.0
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5/6
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</details>
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Figure 8.6 Experimental results of forces in clamps due to unit displacement with clamps 1, 2, and 3 not present.
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and then establish stiffness matrices for the corresponding displacement degrees of freedom. However, in the finite element analysis we need a certain number of degrees of freedom to accurately describe the behavior of the structure (see Section 4.3), which results in a specific finite element model, and then we are able to establish the stiffness matrices of only this finite element model with certain degrees of freedom released. Gauss elimination is the process of releasing degrees of freedom.
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EXAMPLE 8.1: Assume that you know a laboratory technician who knows nothing about finite elements and equation solutions. She/he has performed an experiment using clamps that measure forces on a beam structure in the laboratory; see Fig. E8.1. By moving the clamps to "unit" and "zero" positions, the following forces have been measured.
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<details>
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<summary>text_image</summary>
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Clamp for U₁ Clamp for U₂ Clamp for U₃ Clamp for U₄
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</details>
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Figure E8.1 Beam with clamps
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The result of the first experiment is
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$$
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\begin{array}{l} \text { Forces in clamps } \\ \text { due to } U _ {1} = 1 \text { and } \\ U _ {2} = U _ {3} = U _ {4} = 0 \\ F _ {1} \quad F _ {2} \quad F _ {3} \quad F _ {4} \end{array} \left[ \begin{array}{c c c c} U _ {1} & U _ {2} & U _ {3} & U _ {4} \\ \left( \begin{array}{c} 7 \\ - 4 \\ 1 \\ 0 \end{array} \right) & - 4 & 1 & 0 \\ 6 & - 4 & 1 \\ - 4 & 5 & - 2 \\ 1 & - 2 & 1 \end{array} \right] \tag {a}
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$$
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The technician has in a second experiment removed the clamp for $U_{2}$ and repeated the measurement to obtain the following forces.
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The result of the second experiment is
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$$
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\begin{array}{l} \text { Forces in clamps } \\ \text { due to } U _ {1} = 1 \\ \text { and } U _ {3} = U _ {4} = 0 \end{array} \left[ \begin{array}{c c c} \frac {1 3}{3} & \frac {- 5}{3} & \frac {2}{3} \\ \frac {- 5}{3} & \frac {1 0}{3} & \frac {- 4}{3} \\ \frac {2}{3} & \frac {- 4}{3} & \frac {5}{6} \end{array} \right] \tag {b}
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$$
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You are suspicious of whether she/he measured the forces correctly in this second experiment. Assume that the first experiment was correctly performed. Check whether the second experiment was also correctly performed.
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The stiffness matrix (b) is correct if it is obtained from (a) after releasing the degree of freedom $U_{2}$ . Performing Gauss elimination on $U_{2}$ , we obtain from the matrix in (a),
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$$
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\mathbf {K} = \left[ \begin{array}{c c c} \frac {1 3}{3} & - \frac {5}{3} & \frac {2}{3} \\ - \frac {5}{3} & \left(\frac {7}{3}\right) & - \frac {4}{3} \\ \frac {2}{3} & - \frac {4}{3} & \frac {5}{6} \end{array} \right]
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$$
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