김경종
371 lines
27 KiB
Markdown
371 lines
27 KiB
Markdown
<!-- source-page: 161 -->
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and assume that we want to impose the displacement at the degree of freedom $U_{i}$ with
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$$
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U _ {i} = U _ {i} ^ {*} \tag {3.54}
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$$
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In the Lagrange multiplier method we amend the right-hand side of (3.52) to obtain
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$$
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\Pi^ {*} = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \lambda (U _ {i} - U _ {i} ^ {*}) \tag {3.55}
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$$
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where $\lambda$ is an additional variable, and invoke $\delta \Pi_{i}^{*} = 0$ , which gives
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$$
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\delta \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \delta \mathbf {U} ^ {T} \mathbf {R} + \lambda \delta U _ {i} + \delta \lambda (U _ {i} - U _ {i} ^ {*}) = 0 \tag {3.56}
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$$
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Since $\delta \mathbf{U}$ and $\delta \lambda$ are arbitrary, we obtain
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$$
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\left[ \begin{array}{c c} \mathbf {K} & \mathbf {e} _ {i} \\ \mathbf {e} _ {i} ^ {T} & 0 \end{array} \right] \left[ \begin{array}{c} \mathbf {U} \\ \lambda \end{array} \right] = \left[ \begin{array}{c} \mathbf {R} \\ U _ {i} ^ {*} \end{array} \right] \tag {3.57}
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$$
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where $e_{i}$ is a vector with all entries equal to zero except its ith entry, which is equal to one. Hence the equilibrium equations without a constraint are amended with an additional equation that embodies the constraint condition.
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In the penalty method we also amend the right-hand side of (3.52) but without introducing an additional variable. Now we use
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$$
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\Pi^ {* *} = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \frac {\alpha}{2} (U _ {i} - U _ {i} ^ {*}) ^ {2} \tag {3.58}
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$$
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in which $\alpha$ is a constant of relatively large magnitude, $\alpha \gg \max (k_{ii})$ . The condition $\delta \Pi^{**} = 0$ now yields
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$$
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\delta \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \delta \mathbf {U} ^ {T} \mathbf {R} + \alpha (U _ {i} - U _ {i} ^ {*}) \delta U _ {i} = 0 \tag {3.59}
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$$
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and $(\mathbf{K} + \alpha \mathbf{e}_i\mathbf{e}_i^T)\mathbf{U} = \mathbf{R} + \alpha U_i^*\mathbf{e}_i$ (3.60)
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Hence, using this technique, a large value is added to the ith diagonal element of K and a corresponding force is added so that the required displacement $U_{i}$ is approximately equal to $U_{i}^{*}$ . This is a general technique that has been used extensively to impose specified displacements or other variables. The method is effective because no additional equation is required, and the bandwidth of the coefficient matrix is preserved (see Section 4.2.2). We demonstrate the Lagrange multiplier method and penalty procedure in the following example.
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EXAMPLE 3.31: Use the Lagrange multiplier method and penalty procedure to analyze the simple spring system shown in Fig. E3.31 with the imposed displacement $U_{2} = 1/k$ .
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The governing equilibrium equations without the imposed displacement $U_{2}$ are
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$$
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\left[ \begin{array}{c c} 2 k & - k \\ - k & k \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] = \left[ \begin{array}{l} R _ {1} \\ R _ {2} \end{array} \right] \tag {a}
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$$
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<details>
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<summary>text_image</summary>
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k
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k
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U1, R1
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U2, R2
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U2 = 1/k
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</details>
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Figure E3.31 A simple spring system
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<!-- source-page: 162 -->
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The exact solution is obtained by using the relation $U_{2} = 1/k$ and solving from the first equation of (a) for $U_{1}$ ,
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$$
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U _ {1} = \frac {1 + R _ {1}}{2 k} \tag {b}
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$$
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Hence we also have
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$$
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R _ {2} = 1 - \frac {1 + R _ {1}}{2}
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$$
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which is the force required at the $U_{2}$ degree of freedom to impose $U_{2} = 1/k$ .
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Using the Lagrange multiplier method, the governing equations are
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$$
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\left[ \begin{array}{c c c} 2 k & - k & 0 \\ - k & k & 1 \\ 0 & 1 & 0 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ \lambda \end{array} \right] = \left[ \begin{array}{l} R _ {1} \\ 0 \\ \frac {1}{k} \end{array} \right] \tag {c}
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$$
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and we obtain $U_{1} = \frac{1 + R_{1}}{2k}; \quad \lambda = -1 + \frac{1 + R_{1}}{2}$
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Hence the solution in (b) is obtained, and $\lambda$ is equal to minus the force that must be applied at the degree of freedom $U_{2}$ in order to impose the displacement $U_{2} = 1/k$ . We may note that with this value of $\lambda$ the first two equations in (c) reduce to the equations in (a).
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Using the penalty method, we obtain
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$$
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\left[ \begin{array}{c c} 2 k & - k \\ - k & (k + \alpha) \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right] = \left[ \begin{array}{c} R _ {1} \\ \frac {\alpha}{k} \end{array} \right]
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$$
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The solution now depends on $\alpha$ , and we obtain
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$$
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\text { for } \alpha = 1 0 k: \quad U _ {1} = \frac {1 1 R _ {1} + 1 0}{2 1 k}; \quad U _ {2} = \frac {R _ {1} + 2 0}{2 1 k}
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$$
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$$
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\text { for } \alpha = 1 0 0 k: \quad U _ {1} = \frac {1 0 1 R _ {1} + 1 0 0}{2 0 1 k}; \quad U _ {2} = \frac {R _ {1} + 2 0 0}{2 0 1 k}
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$$
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$$
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\text { and for } \alpha = 1 0 0 0 k: \quad U _ {1} = \frac {1 0 0 1 R _ {1} + 1 0 0 0}{2 0 0 1 k}; \quad U _ {2} = \frac {R _ {1} + 2 0 0 0}{2 0 0 1 k}
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$$
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In practice, the accuracy obtained using $\alpha = 1000k$ is usually sufficient.
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This example gives only a very elementary demonstration of the use of the Lagrange multiplier method and the penalty procedure. Let us now briefly state some more general equations. Assume that we want to impose onto the solution the m linearly independent discrete constraints $BU = V$ where B is a matrix of order $m \times n$ . Then in the Lagrange multiplier method we use
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$$
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\Pi^ {*} (\mathbf {U}, \lambda) = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \lambda^ {T} (\mathbf {B} \mathbf {U} - \mathbf {V}) \tag {3.61}
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$$
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<!-- source-page: 163 -->
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where $\lambda$ is a vector of m Lagrange multipliers. Invoking $\delta\Pi^{*}=0$ we now obtain
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$$
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\left[ \begin{array}{l l} \mathbf {K} & \mathbf {B} ^ {T} \\ \mathbf {B} & \mathbf {0} \end{array} \right] \left[ \begin{array}{l} \mathbf {U} \\ \boldsymbol {\lambda} \end{array} \right] = \left[ \begin{array}{l} \mathbf {R} \\ \mathbf {V} \end{array} \right] \tag {3.62}
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$$
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In the penalty method we use
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$$
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\boldsymbol {\Pi} ^ {* *} (\mathbf {U}) = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \frac {\alpha}{2} (\mathbf {B} \mathbf {U} - \mathbf {V}) ^ {T} (\mathbf {B} \mathbf {U} - \mathbf {V}) \tag {3.63}
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$$
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and invoking $\delta \Pi^{**} = 0$ we obtain
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$$
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(\mathbf {K} + \alpha \mathbf {B} ^ {T} \mathbf {B}) \mathbf {U} = \mathbf {R} + \alpha \mathbf {B} ^ {T} \mathbf {V} \tag {3.64}
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$$
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Of course, (3.57) and (3.60) are special cases of (3.62) and (3.64).
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The above relations are written for discrete systems. When a continuous system is considered, the usual variational indicator $\Pi$ (see, for example, Examples 3.18 to 3.20) is amended in the Lagrange multiplier method with integral(s) of the continuous constraint(s) times the Lagrange multiplier(s) and in the penalty method with integral(s) of the penalty factor(s) times the square of the constraint(s). If the continuous variables are then expressed through trial functions or finite difference expressions, relations of the form (3.62) and (3.64) are obtained (see Section 4.4).
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Although the above introduction to the Lagrange multiplier method and penalty procedure is brief, some basic observations can be made that are quite generally applicable. First, we observe that in the Lagrange multiplier method the diagonal elements in the coefficient matrix corresponding to the Lagrange multipliers are zero. Hence for the solution it is effective to arrange the equations as given in (3.62). Considering the equilibrium equations with the Lagrange multipliers, we also find that these multipliers have the same units as the forcing functions; for example, in (3.57) the Lagrange multiplier is a force.
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Using the penalty method, an important consideration is the choice of an appropriate penalty number. In the analysis leading to $(3.64)$ the penalty number $\alpha$ is explicitly specified (such as in Example 3.31), and this is frequently the case (see Section 4.2.2). However, in other analyses, the penalty number is defined by the problem itself using a specific formulation (see Section 5.4.1). The difficulty with the use of a very high penalty number lies in that the coefficient matrix can become ill-conditioned when the off-diagonal elements are multiplied by a large number. If the off-diagonal elements are affected by the penalty number, it is necessary to use enough digits in the computer arithmetical operations to ensure an accurate solution of the problem (see Section 8.2.6).
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Finally, we should note that the penalty and Lagrange multiplier methods are quite closely related (see Exercise 3.35) and that the basic ideas of imposing the constraints can also be combined, see M. Fortin and R. Glowinski [A], J. C. Simo, P. Wriggers, and R. L. Taylor [A], and Exercise 3.36.
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# 3.4.2 Exercises
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3.31. Consider the system of equations
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$$
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\left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 2 \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right] = \left[ \begin{array}{c} 1 0 \\ - 1 \end{array} \right]
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$$
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<!-- source-page: 164 -->
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Use the Lagrange multiplier method and the penalty method to impose the condition $U_{2}=0$ . Solve the equations and interpret the solution.
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3.32. Consider the system of carts in Example 3.1 with $k_{i} = k, R_{1} = 1, R_{2} = 0, R_{3} = 1$ . Develop the governing equilibrium equations, imposing the condition $U_{2} = U_{3}$ .
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(a) Use the Lagrange multiplier method.
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(b) Use the penalty method with an appropriate penalty factor.
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In each case solve for the displacements and the constraining force.
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3.33. Consider the heat transfer problem in Example 3.2 with $k = 1$ and $\theta_0 = 10$ , $\theta_4 = 20$ . Impose the condition that $\theta_3 = 4\theta_2$ and physically interpret the solution. Use the Lagrange multiplier method and then the penalty method with a reasonable penalty parameter.
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3.34. Consider the fluid flow in the hydraulic network in Example 3.3. Develop the governing equations for use of the Lagrange multiplier method to impose the condition $p_C = 2p_D$ . Solve the equations and interpret the solution.
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Repeat the solution using the penalty method with an appropriate penalty factor.
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3.35. Consider the problem $\mathbf{KU} = \mathbf{R}$ with the $m$ linearly independent constraints $\mathbf{BU} = \mathbf{V}$ (see (3.61) and (3.62)). Show that the stationarity of the following variational indicator gives the equations of the penalty method (3.64),
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$$
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\tilde {\Pi} ^ {* *} (\mathbf {U}, \boldsymbol {\lambda}) = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \boldsymbol {\lambda} ^ {T} (\mathbf {B} \mathbf {U} - \mathbf {V}) - \frac {\boldsymbol {\lambda} ^ {T} \boldsymbol {\lambda}}{2 \alpha}
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$$
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where $\pmb{\lambda}$ is a vector of the $m$ Lagrange multipliers and $\alpha$ is the penalty parameter, $\alpha > 0$ . Evaluate the Lagrange multipliers in general to be $\pmb{\lambda} = \alpha(\mathbf{BU} - \mathbf{V})$ , and show that for the specific case considered in (3.60) $\pmb{\lambda} = \alpha(U_i - U_i^*)$ .
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3.36. In the augmented Lagrangian method the following functional is used for the problem stated in Exercise 3.35:
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$$
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\tilde {\Pi} ^ {*} (\mathbf {U}, \boldsymbol {\lambda}) = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} + \frac {\alpha}{2} (\mathbf {B} \mathbf {U} - \mathbf {V}) ^ {T} (\mathbf {B} \mathbf {U} - \mathbf {V}) + \boldsymbol {\lambda} ^ {T} (\mathbf {B} \mathbf {U} - \mathbf {V}); \alpha \geq 0
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$$
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(a) Invoke the stationarity of $\tilde{\Pi}^*$ and obtain the governing equations.
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(b) Use the augmented Lagrangian method to solve the problem posed in Example 3.31 for $\alpha = 0, k$ , and $1000k$ . Show that, actually, for any value of $\alpha$ the constraint is accurately satisfied. (The augmented Lagrangian method is used in iterative solution procedures, in which case using an efficient value for $\alpha$ can be important.)
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<!-- source-page: 165 -->
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# Formulation of the Finite Element Method—Linear Analysis in Solid and Structural Mechanics
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# 4.1 INTRODUCTION
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A very important application area for finite element analysis is the linear analysis of solids and structures. This is where the first practical finite element procedures were applied and where the finite element method has obtained its primary impetus of development.
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Today many types of linear analyses of structures can be performed in a routine manner. Finite element discretization schemes are well established and are used in standard computer programs. However, there are two areas in which effective finite elements have been developed only recently, namely, the analysis of general plate and shell structures and the solution of (almost) incompressible media.
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The standard formulation for the finite element solution of solids is the displacement method, which is widely used and effective except in these two areas of analysis. For the analysis of plate and shell structures and the solution of incompressible solids, mixed formulations are preferable.
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In this chapter we introduce the displacement-based method of analysis in detail. The principle of virtual work is the basic relationship used for the finite element formulation. We first establish the governing finite element equations and then discuss the convergence properties of the method. Since the displacement-based solution is not effective for certain applications, we then introduce the use of mixed formulations in which not only the displacements are employed as unknown variables. The use of a mixed method, however, requires a careful selection of appropriate interpolations, and we address this issue in the last part of the chapter.
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Various displacement-based and mixed formulations have been presented in the literature, and as pointed out before, our aim is not to survey all these formulations. Instead, we
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<!-- source-page: 166 -->
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will concentrate in this chapter on some important useful principles of formulating finite elements. Some efficient applications of the principles discussed in this chapter are then presented in Chapter 5.
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# 4.2 FORMULATION OF THE DISPLACEMENT-BASED FINITE ELEMENT METHOD
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The displacement-based finite element method can be regarded as an extension of the displacement method of analysis of beam and truss structures, and it is therefore valuable to review this analysis process. The basic steps in the analysis of a beam and truss structure using the displacement method are the following.
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1. Idealize the total structure as an assemblage of beam and truss elements that are interconnected at structural joints.
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2. Identify the unknown joint displacements that completely define the displacement response of the structural idealization.
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3. Formulate force balance equations corresponding to the unknown joint displacements and solve these equations.
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4. With the beam and truss element end displacements known, calculate the internal element stress distributions.
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5. Interpret, based on the assumptions used, the displacements and stresses predicted by the solution of the structural idealization.
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In practical analysis and design the most important steps of the complete analysis are the proper idealization of the actual problem, as performed in step 1, and the correct interpretation of the results, as in step 5. Depending on the complexity of the actual system to be analyzed, considerable knowledge of the characteristics of the system and its mechanical behavior may be required in order to establish an appropriate idealization, as briefly discussed in Chapter 1.
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These analysis steps have already been demonstrated to some degree in Chapter 3, but it is instructive to consider another more complex example.
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EXAMPLE 4.1: The piping system shown in Fig. E4.1(a) must be able to carry a large transverse load P applied accidentally to the flange connecting the small- and large-diameter pipes. "Analyze this problem."
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The study of this problem may require a number of analyses in which the local kinematic behavior of the pipe intersection is properly modeled, the nonlinear material and geometric behaviors are taken into account, the characteristics of the applied load are modeled accurately, and so on. In such a study, it is usually most expedient to start with a simple analysis in which gross assumptions are made and then work toward a more refined model as the need arises (see Section 6.8.1).
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<!-- source-page: 167 -->
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<details>
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<summary>text_image</summary>
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L
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P
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2L
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0.5L
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</details>
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(a) Piping system
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<details>
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<summary>text_image</summary>
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Element 1
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Node 1
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EI
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P
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Element 2
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8EI
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Element 3
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Node 3
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0.50L
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Node 2
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AE
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Node 4
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2L
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Element 4
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(b) Elements and nodal points
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</details>
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<details>
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<summary>text_image</summary>
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U₁
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U₂
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U₃
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U₄
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U₅
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U₆
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U₇
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</details>
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(c) Global degrees of freedom of unrestraint structure
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Figure E4.1 Piping system and idealization
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Assume that in a first analysis we primarily want to calculate the transverse displacement at the flange when the transverse load is applied slowly. In this case it is reasonable to model the structure as an assemblage of beam, truss, and spring elements and perform a static analysis.
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The model chosen is shown in Fig. E4.1(b). The structural idealization consists of two beams, one truss, and a spring element. For the analysis of this idealization we first evaluate the element stiffness matrices that correspond to the global structural degrees of freedom shown in Fig. E4.1(c). For the beam, spring, and truss elements, respectively, we have in this case
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<!-- source-page: 168 -->
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$$
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\mathbf {K} _ {1} ^ {\epsilon} = \frac {E I}{L} \left[ \begin{array}{c c c c} \frac {1 2}{L ^ {2}} & - \frac {6}{L} & - \frac {1 2}{L ^ {2}} & - \frac {6}{L} \\ & 4 & \frac {6}{L} & 2 \\ \text { symmetric } & & \frac {1 2}{L ^ {2}} & \frac {6}{L} \\ & & & 4 \end{array} \right]; \quad U _ {1}, U _ {2}, U _ {3}, U _ {4}
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$$
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$$
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\mathbf {K} _ {2} ^ {e} = \frac {E I}{L} \left[ \begin{array}{c c c c} \frac {1 2}{L ^ {2}} & - \frac {1 2}{L} & - \frac {1 2}{L ^ {2}} & - \frac {1 2}{L} \\ & 1 6 & \frac {1 2}{L} & 8 \\ \text { symmetric } & & \frac {1 2}{L ^ {2}} & \frac {1 2}{L} \\ & & & 1 6 \end{array} \right]; \quad U _ {3}, U _ {4}, U _ {5}, U _ {6}
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$$
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$$
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\mathbf {K} _ {3} ^ {\xi} = k _ {s}; \quad U _ {6}
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$$
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$$
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\mathbf {K} _ {4} ^ {e} = \frac {E A}{L} \left[ \begin{array}{c c} 2 & - 2 \\ - 2 & 2 \end{array} \right]; \quad U _ {5}, U _ {7}
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$$
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where the subscript on $K^{e}$ indicates the element number, and the global degrees of freedom that correspond to the element stiffnesses are written next to the matrices. It should be noted that in this example the element matrices are independent of direction cosines since the centerlines of the elements are aligned with the global axes. If the local axis of an element is not in the direction of a global axis, the local element stiffness matrix must be transformed to obtain the required global element stiffness matrix (see Example 4.10).
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The stiffness matrix of the complete element assemblage is effectively obtained from the stiffness matrices of the individual elements using the direct stiffness method (see Examples 3.1 and 4.11). In this procedure the structure stiffness matrix K is calculated by direct addition of the element stiffness matrices; i.e.,
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$$
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\mathbf {K} = \sum_ {i} \mathbf {K} _ {i} ^ {e}
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$$
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where the summation includes all elements. To perform the summation, each element matrix $\mathbf{K}_i^e$ is written as a matrix $\mathbf{K}^{(i)}$ of the same order as the stiffness matrix $\mathbf{K}$ , where all entries in $\mathbf{K}^{(i)}$ are zero except those which correspond to an element degree of freedom. For example, for element 4 we have
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$$
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\mathbf {K} ^ {(4)} = \begin{array}{l} 1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 7 \leftarrow \text { Degree of freedom } \\ 1 \left[ \begin{array}{l l l l l l l} 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac {2 A E}{L} & 0 & - \frac {2 E A}{L} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & - \frac {2 A E}{L} & 0 & \frac {2 E A}{L} \end{array} \right] \end{array}
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$$
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<!-- source-page: 169 -->
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Therefore, the stiffness matrix of the structure is
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$$
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\mathbf {K} = \left[ \begin{array}{c c c c c c c} \frac {1 2 E I}{L ^ {3}} & - \frac {6 E I}{L ^ {2}} & - \frac {1 2 E I}{L ^ {3}} & - \frac {6 E I}{L ^ {2}} & 0 & 0 & 0 \\ & \frac {4 E I}{L} & \frac {6 E I}{L ^ {2}} & \frac {2 E I}{L} & 0 & 0 & 0 \\ & & \frac {2 4 E I}{L ^ {3}} & - \frac {6 E I}{L ^ {2}} & - \frac {1 2 E I}{L ^ {3}} & - \frac {1 2 E I}{L ^ {2}} & 0 \\ & & & \frac {2 0 E I}{L} & \frac {1 2 E I}{L ^ {2}} & \frac {8 E I}{L} & 0 \\ & \text { symmetric } & & & \frac {1 2 E I}{L ^ {3}} + \frac {2 A E}{L} & \frac {1 2 E I}{L ^ {2}} & - \frac {2 A E}{L} \\ & & & & & \frac {1 6 E I}{L} + k _ {s} & 0 \\ & & & & & & \frac {2 A E}{L} \end{array} \right]
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$$
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and the equilibrium equations for the system are
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$$
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\mathbf {K U} = \mathbf {R}
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$$
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where U is a vector of the system global displacements and R is a vector of forces acting in the direction of these displacements:
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$$
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\mathbf {U} ^ {T} = \left[ U _ {1}, \dots , U _ {7} \right]; \quad \mathbf {R} ^ {T} = \left[ R _ {1}, \dots , R _ {7} \right]
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$$
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Before solving for the displacements of the structure, we need to impose the boundary conditions that $U_{1} = 0$ and $U_{7} = 0$ . This means that we may consider only five equations in five unknown displacements; i.e.,
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$$
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\tilde {\mathbf {K}} \tilde {\mathbf {U}} = \tilde {\mathbf {R}} \tag {a}
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$$
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where $\tilde{K}$ is obtained by eliminating from K the first and seventh rows and columns, and
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$$
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\tilde {\mathbf {U}} ^ {T} = \left[ \begin{array}{l l l l l} U _ {2} & U _ {3} & U _ {4} & U _ {5} & U _ {6} \end{array} \right]; \quad \tilde {\mathbf {R}} ^ {T} = \left[ \begin{array}{l l l l l} 0 & - P & 0 & 0 & 0 \end{array} \right]
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$$
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The solution of (a) gives the structure displacements and therefore the element nodal point displacements. The element nodal forces are obtained by multiplying the element stiffness matrices $K_{i}^{f}$ by the element displacements. If the forces at any section of an element are required, we can evaluate them from the element end forces by use of simple statics.
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Considering the analysis results it should be recognized, however, that although the structural idealization in Fig. E4.1(b) was analyzed accurately, the displacements and stresses are only a prediction of the response of the actual physical structure. Surely this prediction will be accurate only if the model used was appropriate, and in practice a specific model is in general adequate for predicting certain quantities but inadequate for predicting others. For instance, in this analysis the required transverse displacement under the applied load is quite likely predicted accurately using the idealization in Fig. E4.1(b) (provided the load is applied slowly enough, the stresses are small enough not to cause yielding, and so on), but the stresses directly under the load are probably predicted very inaccurately. Indeed, a different and more refined finite element model would need to be used in order to accurately calculate the stresses (see Section 1.2).
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This example demonstrates some important aspects of the displacement method of analysis and the finite element method. As summarized previously, the basic process is that
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the complete structure is idealized as an assemblage of individual structural elements. The element stiffness matrices corresponding to the global degrees of freedom of the structural idealization are calculated, and the total stiffness matrix is formed by the addition of the element stiffness matrices. The solution of the equilibrium equations of the assemblage of elements yields the element displacements, which are then used to calculate the element stresses. Finally, the element displacements and stresses must be interpreted as an estimate of the actual structural behavior, taking into account that a truss and beam idealization was solved.
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Considering the analysis of truss and beam assemblages such as in Example 4.1, originally these solutions were not called finite element analyses because there is one major difference in these solutions when compared to a more general finite element analysis of a two- or three-dimensional problem, namely, in the analysis performed in Example 4.1 the exact element stiffness matrices (“exact” within beam theory) could be calculated. The stiffness properties of a beam element are physically the element end forces that correspond to unit element end displacements. These forces can be evaluated by solving the differential equations of equilibrium of the element when it is subjected to the appropriate boundary conditions. Since by virtue of the solution of the differential equations of equilibrium, all three requirements of an exact solution—namely, the stress equilibrium, the compatibility, and the constitutive requirements—throughout each element are fulfilled, the exact element internal displacements and stiffness matrices are calculated. In an alternative approach, these element end forces could also be evaluated by performing a variational solution based on the Ritz method or Galerkin method, as discussed in Section 3.3.4. Such solutions would give the exact element stiffness coefficients if the exact element internal displacements (as calculated in the solution of the differential equations of equilibrium) are used as trial functions (see Examples 3.22 and 4.8). However, approximate stiffness coefficients are obtained if other trial functions (which may be more suitable in practice) are employed.
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When considering more general two- and three-dimensional finite element analyses, we use the variational approach with trial functions that approximate the actual displacements because we do not know the exact displacement functions as in the case of truss and beam elements. The result is that the differential equations of equilibrium are not satisfied in general, but this error is reduced as the finite element idealization of the structure or the continuum is refined.
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The general formulation of the displacement-based finite element method is based on the use of the principle of virtual displacements which, as discussed in Section 3.3.4, is equivalent to the use of the Galerkin method, and also equivalent to the use of the Ritz method to minimize the total potential of the system.
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# 4.2.1 General Derivation of Finite Element Equilibrium Equations
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In this section we first state the general elasticity problem to be solved. We then discuss the principle of virtual displacements, which is used as the basis of our finite element solution, and we derive the finite element equations. Next we elaborate on some important considerations regarding the satisfaction of stress equilibrium, and finally we discuss some details of the process of assemblage of element matrices.
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