김경종
486 lines
28 KiB
Markdown
486 lines
28 KiB
Markdown
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To obtain the governing differential equation and natural boundary condition we use, in essence, the same argument as in Example 3.18; i.e., since $\delta u_{0}$ is zero but $\delta u$ is arbitrary at all other points, we must have
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$$
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E A \frac {\partial^ {2} u}{\partial x ^ {2}} + f ^ {B} = 0 \tag {c}
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$$
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and $EA \frac{\partial u}{\partial x} \bigg|_{x=L} = R$ (d)
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In this problem we have $f^B = -A\rho \partial^2 u / \partial t^2$ and hence (c) reduces to the problem-governing differential equation
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$$
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\frac {\partial^ {2} u}{\partial x ^ {2}} = \frac {1}{c ^ {2}} \frac {\partial^ {2} u}{\partial t ^ {2}}; \quad c = \sqrt {\frac {E}{\rho}}
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$$
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The natural boundary condition was stated in (d).
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Finally, it may be noted that the problem in (a) and (b) is a $C^0$ variational problem; i.e., $m = 1$ in this case.
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EXAMPLE 3.20: The functional governing static buckling of the column in Fig. E3.20 is
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} E I \left(\frac {d ^ {2} w}{d x ^ {2}}\right) ^ {2} d x - \frac {P}{2} \int_ {0} ^ {L} \left(\frac {d w}{d x}\right) ^ {2} d x + \frac {1}{2} k w _ {L} ^ {2} \tag {a}
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$$
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where $w_{L} = w|_{x = L}$ and the essential boundary conditions are
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$$
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w \mid_ {x = 0} = 0, \quad \frac {d w}{d x} \Bigg | _ {x = 0} = 0 \tag {b}
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$$
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Invoke the stationarity condition $\delta\Pi = 0$ to derive the problem-governing differential equation and the natural boundary conditions.
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<details>
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<summary>text_image</summary>
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Flexural stiffness
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EI
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w
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x
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P
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Spring stiffness
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k
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L
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</details>
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Figure E3.20 Column subjected to a compressive load
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This problem is a $C^1$ variational problem, i.e., $m = 2$ , because the highest derivative in the functional is of order 2.
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The stationarity condition $\delta\Pi = 0$ yields
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$$
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\int_ {0} ^ {L} E I w ^ {\prime \prime} \delta w ^ {\prime \prime} d x - P \int_ {0} ^ {L} w ^ {\prime} \delta w ^ {\prime} d x + k w _ {L} \delta w _ {L} = 0
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$$
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<!-- source-page: 132 -->
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where we use the notation $w' = dw/dx$ , and so on. But $\delta w'' = d(\delta w')/dx$ , and EI is constant; hence, using integration by parts, we obtain
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$$
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\int_ {0} ^ {L} E I w ^ {\prime \prime} \delta w ^ {\prime \prime} d x = E I w ^ {\prime \prime} \delta w ^ {\prime} | _ {0} ^ {L} - E I \int_ {0} ^ {L} w ^ {\prime \prime \prime} \delta w ^ {\prime} d x
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$$
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If we continue to integrate by parts $\int_0^L w''' \delta w' dx$ and also integrate by parts $\int_0^L w' \delta w' dx$ , we obtain
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$$
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\begin{array}{l} \underbrace {\int_ {0} ^ {L} \left(E I w ^ {\mathrm{iv}} + P w ^ {\prime \prime}\right) \delta w d x} _ {①} + \underbrace {(E I w ^ {\prime \prime} \delta w ^ {\prime}) | _ {L}} _ {②} - \underbrace {(E I w ^ {\prime \prime} \delta w ^ {\prime}) | _ {0}} _ {③} \\ - \underbrace {\left[ \left(E I w ^ {\prime \prime \prime} + P w ^ {\prime}\right) \delta w \right] \left\| _ {L} \right.} _ {④} + \underbrace {\left[ \left(E I w ^ {\prime \prime \prime} + P w ^ {\prime}\right) \delta w \right] \left\| _ {0} \right.} _ {⑤} + \underbrace {k w _ {L} \delta w _ {L}} _ {⑥} = 0 \tag {c} \\ \end{array}
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$$
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Since the variations on w and $w'$ must be zero at the essential boundary conditions, we have $\delta w_{0} = 0$ and $\delta w_{0}' = 0$ . It follows that terms ③ and ⑤ are zero. The variations on w and $w'$ are arbitrary at all other points, hence to satisfy (c) we conclude, using the earlier arguments (see Example 3.18), that the following equations must be satisfied:
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$$
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\text { term 1: } \quad E I w ^ {\mathrm{iv}} + P w ^ {\prime \prime} = 0 \tag {d}
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$$
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$$
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\text { term 2: } \quad E I w ^ {\prime \prime} | _ {x = L} = 0 \tag {e}
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$$
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$$
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\text { terms 4 and 6: } \quad (E I w ^ {\prime \prime \prime} + P w ^ {\prime} - k w) | _ {x = L} = 0 \tag {f}
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$$
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The problem-governing differential equation is given in (d), and the natural boundary conditions are the relations in (e) and (f). We should note that these boundary conditions correspond to the physical conditions of moment and shear equilibrium at x = L.
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We have illustrated in the preceding examples how the problem-governing differential equation and the natural boundary conditions can be derived by invoking the stationarity of the functional of the problem. At this point a number of observations should be made.
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First, considering a $C^{m-1}$ variational problem, the order of the highest derivative present in the problem-governing differential equation is 2m. The reason for obtaining a derivative of order 2m in the problem-governing differential equation is that integration by parts is employed m times.
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A second observation is that the effect of the natural boundary conditions was always included as a potential in the expression for $\Pi$ . Hence the natural boundary conditions are implicitly contained in $\Pi$ , whereas the essential boundary conditions have been stated separately.
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Our objective in Examples 3.18 to 3.20 was to derive the governing differential equations and natural boundary conditions by invoking the stationarity of a functional, and for this purpose the appropriate functional was given in each case. However, an important question then arises: How can we establish an appropriate functional corresponding to a given problem? The two previous observations and the mathematical manipulations in Examples 3.18 to 3.20 suggest that to derive a functional for a given problem we could start with the governing differential equation, establish an integral equation, and then proceed backward in the mathematical manipulations. In this derivation it is necessary to use integration by parts, i.e., the divergence theorem, and the final check would be that the
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stationarity condition on the $\Pi$ derived does indeed yield the governing differential equations. This procedure is employed to derive appropriate functionals in many cases (see Section 3.3.4 and Chapters 4 and 7, and for further treatment see, for example, R. Courant and D. Hilbert [A], S. G. Mikhlin [A], K. Washizu [B], and M.L Bucalem and K.J. Bathe [B]). In this context, it should also be noted that in considering a specific problem, there does not generally exist a unique appropriate functional, but a number of functionals are applicable. For instance, in the solution of structural mechanics problems, we can employ the principle of minimum potential energy, other displacement-based variational formulations, the Hu-Washizu or Hellinger-Reissner principles, and so on (see Section 4.4.2).
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Another important observation is that once a functional has been established for a certain class of problems, the functional can be employed to generate the governing equations for all problems in that class and therefore provides a general analysis tool. For example, the principle of minimum potential energy is general and is applicable to all problems in linear elasticity theory.
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Based simply on a utilitarian point of view, the following observations can be made in regard to variational formulations.
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1. The variational method may provide a relatively easy way to construct the system-governing equations. This ease of use of a variational principle depends largely on the fact that in the variational formulation scalar quantities (energies, potentials, and so on) are considered rather than vector quantities (forces, displacements, and so on).
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2. A variational approach may lead more directly to the system-governing equations and boundary conditions. For example, if a complex system is being considered, it is of advantage that some variables that need to be included in a direct formulation are not considered in a variational formulation (such as internal forces that do no net work).
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3. The variational approach provides some additional insight into a problem and gives an independent check on the formulation of the problem.
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4. For approximate solutions, a larger class of trial functions can be employed in many cases if the analyst operates on the variational formulation rather than on the differential formulation of the problem; for example, the trial functions need not satisfy the natural boundary conditions because these boundary conditions are implicitly contained in the functional (see Section 3.3.4).
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This last consideration has most important consequences, and much of the success of the finite element method hinges on the fact that by employing a variational formulation, a larger class of functions can be used. We examine this point in more detail in the next section and in Section 3.3.4.
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# 3.3.3 Weighted Residual Methods; Ritz Method
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In previous sections we have discussed differential and variational formulations of the governing equilibrium equations of continuous systems. In dealing with relatively simple systems, these equations can be solved in closed form using techniques of integration, separation of variables, and so on. For more complex systems, approximate procedures of solution must be employed. The objective in this section is to survey some classical techniques in which a family of trial functions is used to obtain an approximate solution. We
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shall see later that these techniques are very closely related to the finite element method of analysis and that indeed the finite element method can be regarded as an extension of these classical procedures.
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Consider the analysis of a steady-state problem using its differential formulation
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$$
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L _ {2 m} [ \phi ] = r \tag {3.8}
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$$
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in which $L_{2m}$ is a linear differential operator, $\phi$ is the state variable to be calculated, and r is the forcing function. The solution to the problem must also satisfy the boundary conditions
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$$
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B _ {i} [ \phi ] = q _ {i} | _ {\text { at boundary } S _ {i}}; \quad i = 1, 2, \dots \tag {3.9}
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$$
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We shall be concerned, in particular, with symmetric and positive definite operators that satisfy the symmetry condition
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$$
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\int_ {D} (L _ {2 m} [ u ]) v d D = \int_ {D} (L _ {2 m} [ v ]) u d D \tag {3.10}
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$$
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and the condition of positive definiteness
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$$
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\int_ {D} (L _ {2 m} [ u ]) u d D > 0 \tag {3.11}
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$$
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where D is the domain of the operator and u and v are any functions that satisfy homogeneous essential and natural boundary conditions. To clarify the meaning of relations (3.8) to (3.11), we consider the following example.
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EXAMPLE 3.21: The steady-state response of the bar shown in Fig. E3.17 is calculated by solving the differential equation
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$$
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- E A \frac {\partial^ {2} u}{\partial x ^ {2}} = 0 \tag {a}
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$$
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subject to the boundary conditions
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$$
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u \big | _ {x = 0} = 0; \quad E A \frac {\partial u}{\partial x} \Bigg | _ {x = L} = R \tag {b}
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$$
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Identify the operators and functions of (3.8) and (3.9) and check whether the operator $L_{2m}$ is symmetric and positive definite.
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Comparing (3.8) with (a), we see that in this problem
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$$
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L _ {2 m} = - E A \frac {\partial^ {2}}{\partial x ^ {2}}; \quad \phi = u; \quad r = 0
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$$
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Similarly, comparing (3.9) with (b), we obtain
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$$
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B _ {1} = 1; \quad q _ {1} = 0
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$$
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$$
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B _ {2} = E A \frac {\partial}{\partial x}; \qquad q _ {2} = R
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$$
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To identify whether the operator $L_{2m}$ is symmetric and positive definite, we consider the case $R = 0$ . This means physically that we are concerned only with the structure itself and not
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with the loading applied to it. For (3.10) we have
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$$
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\begin{array}{l} \int_ {0} ^ {L} - E A \frac {\partial^ {2} u}{\partial x ^ {2}} v d x = - E A \frac {\partial u}{\partial x} v \left| _ {0} ^ {L} + \int_ {0} ^ {L} E A \frac {\partial u}{\partial x} \frac {\partial v}{\partial x} d x \right. \\ = - E A \frac {\partial u}{\partial x} v \bigg | _ {0} ^ {L} + E A u \frac {\partial v}{\partial x} \bigg | _ {0} ^ {L} - \int_ {0} ^ {L} E A \frac {\partial^ {2} v}{\partial x ^ {2}} u d x \tag {c} \\ \end{array}
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$$
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Since the boundary conditions are $u = v = 0$ at $x = 0$ and $\partial u / \partial x = \partial v / \partial x = 0$ at $x = L$ , we have
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$$
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\int_ {0} ^ {L} - E A \frac {\partial^ {2} u}{\partial x ^ {2}} v d x = \int_ {0} ^ {L} - E A \frac {\partial^ {2} v}{\partial x ^ {2}} u d x
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$$
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and the operator is symmetric. We can also directly conclude that the operator is positive definite because from (c) we obtain
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$$
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\int_ {0} ^ {L} - E A \frac {\partial^ {2} u}{\partial x ^ {2}} u d x = \int_ {0} ^ {L} E A \left(\frac {\partial u}{\partial x}\right) ^ {2} d x
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$$
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In the following we discuss the use of classical weighted residual methods and the Ritz method in the solution of linear steady-state problems as in (3.8) and (3.9), but the same concepts can also be employed in the analysis of propagation problems and eigenproblems and in the analysis of nonlinear response (see Examples 3.23 and 3.24).
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The basic step in the weighted residual and Ritz analyses is to assume a solution of the form
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$$
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\overline {{{\phi}}} = \sum_ {i = 1} ^ {n} a _ {i} f _ {i} \tag {3.12}
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$$
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where the $f_{i}$ are linearly independent trial functions and the $a_{i}$ are multipliers to be determined in the solution.
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Consider first the weighted residual methods. These techniques operate directly on (3.8) and (3.9). Using these methods, we choose the functions $f_{i}$ in (3.12) so as to satisfy all boundary conditions in (3.9), and we then calculate the residual
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$$
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R = r - L _ {2 m} \left[ \sum_ {i = 1} ^ {n} a _ {i} f _ {i} \right] \tag {3.13}
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$$
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For the exact solution this residual is of course zero. A good approximation to the exact solution would imply that R is small at all points of the solution domain. The various weighted residual methods differ in the criteria that they employ to calculate the $a_{i}$ such that R is small. However, in all techniques we determine the $a_{i}$ so as to make a weighted average of R vanish.
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Galerkin method. In this technique the parameters $a_i$ are determined from the $n$ equations
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$$
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\int_ {D} f _ {i} R d D = 0; \quad i = 1, 2, \dots , n \tag {3.14}
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$$
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where $D$ is the solution domain.
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Least squares method. In this technique the integral of the square of the residual is minimized with respect to the parameters $a_i$ ,
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$$
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\frac {\partial}{\partial a _ {i}} \int_ {D} R ^ {2} d D = 0; \quad i = 1, 2, \dots , n \tag {3.15}
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$$
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Substituting from (3.13), we thus obtain the following $n$ simultaneous equations for the parameters $a_i$ ,
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$$
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\int_ {D} R L _ {2 m} [ f _ {i} ] d D = 0; \quad i = 1, 2, \dots , n \tag {3.16}
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$$
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Collocation method. In this method the residual R is set equal to zero at n distinct points in the solution domain to obtain n simultaneous equations for the parameters $a_{i}$ . The location of the n points can be somewhat arbitrary, and a uniform pattern may be appropriate, but usually the analyst should use some judgment to employ appropriate locations.
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Subdomain method. The complete domain of solution is subdivided into n subdomains, and the integral of the residual in (3.13) over each subdomain is set equal to zero to generate n equations for the parameters $a_{i}$ .
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An important step in using a weighted residual method is the solution of the simultaneous equations for the parameters $a_{i}$ . We note that since $L_{2m}$ is a linear operator, in all the procedures mentioned, a linear set of equations in the parameters $a_{i}$ is generated. In the Galerkin method, the coefficient matrix is symmetric (and also positive definite) if $L_{2m}$ is a symmetric (and also positive definite) operator. In the least squares method we always generate a symmetric coefficient matrix irrespective of the properties of the operator $L_{2m}$ . However, in the collocation and subdomain methods, nonsymmetric coefficient matrices may be generated. In practical analysis, therefore, the Galerkin and least squares methods are usually preferable.
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Using weighted residual methods, we operate directly on (3.8) and (3.9) to minimize the error between the trial solution in (3.12) and the actual solution to the problem. Considering next the Ritz analysis method (due to W. Ritz [A]), the fundamental difference from the weighted residual methods is that in the Ritz method we operate on the functional corresponding to the problem in (3.8) and (3.9). Let $\Pi$ be the functional of the $C^{m-1}$ variational problem that is equivalent to the differential formulation given in (3.8) and (3.9); in the Ritz method we substitute the trial functions $\overline{\phi}$ given in (3.12) into $\Pi$ and generate $n$ simultaneous equations for the parameters $a_i$ using the stationarity condition of $\Pi$ , $\delta\Pi = 0$ [see (3.1)], which now gives
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$$
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\frac {\partial \Pi}{\partial a _ {i}} = 0; \quad i = 1, 2, \dots , n \tag {3.17}
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$$
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An important consideration is the selection of the trial (or Ritz) functions $f_{i}$ in (3.12). In the Ritz analysis these functions need only satisfy the essential boundary conditions and not the natural boundary conditions. The reason for this relaxed requirement on the trial functions is that the natural boundary conditions are implicitly contained in the functional $\Pi$ . Assume that the $L_{2m}$ operator corresponding to the variational problem is symmetric and positive definite. In this case the actual extremum of $\Pi$ is its minimum, and by invoking
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(3.17) we minimize (in some sense) the violation of the internal equilibrium requirements and the violation of the natural boundary conditions (see Section 4.3). Therefore, for convergence in a Ritz analysis, the trial functions need only satisfy the essential boundary conditions, which is a fact that may not be anticipated because we know that the exact solution also satisfies the natural boundary conditions. Actually, assuming a given number of trial functions, it can be expected that in most cases the solution will be more accurate if these functions also satisfy the natural boundary conditions. However, it can be very difficult to find such trial functions, and it is generally more effective to use instead a larger number of functions that satisfy only the essential boundary conditions. We demonstrate the use of the Ritz method in the following examples.
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EXAMPLE 3.22: Consider a simple bar fixed at one end $(x = 0)$ and subjected to a concentrated force at the other end $(x = 180)$ as shown in Fig. E3.22. Using the notation given in the figure, the total potential of the structure is
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$$
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\Pi = \int_ {0} ^ {1 8 0} \frac {1}{2} E A \left(\frac {d u}{d x}\right) ^ {2} d x - 1 0 0 u | _ {x = 1 8 0} \tag {a}
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$$
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and the essential boundary condition is $u|_{x=0}=0$ .
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1. Calculate the exact displacement and stress distributions in the bar.
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2. Calculate the displacement and stress distributions using the Ritz method with the following displacement assumptions:
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$$
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u = a _ {1} x + a _ {2} x ^ {2} \tag {b}
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$$
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and $u = \frac{xu_B}{100};\quad 0\leq x\leq 100$ (c)
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$$
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u = \left(1 - \frac {x - 1 0 0}{8 0}\right) u _ {B} + \left(\frac {x - 1 0 0}{8 0}\right) u _ {C}; \quad 1 0 0 \leq x \leq 1 8 0
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$$
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where $u_{B}$ and $u_{C}$ are the displacements at points B and C.
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<details>
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<summary>text_image</summary>
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Cross-sectional area = (1 + y/40)² cm²
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Area = 1 cm²
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R = 100 N
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x, u
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A
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B
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C
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y
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100 cm
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80 cm
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</details>
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Figure E3.22 Bar subjected to a concentrated end force
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In order to calculate the exact displacements in the structure, we use the stationarity condition of $\Pi$ and generate the governing differential equation and the natural boundary condition. We have
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$$
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\delta \Pi = \int_ {0} ^ {1 8 0} \left(E A \frac {d u}{d x}\right) \delta \left(\frac {d u}{d x}\right) d x - 1 0 0 \delta u | _ {x = 1 8 0} \tag {d}
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$$
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<!-- source-page: 138 -->
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Setting $\delta \Pi = 0$ and using integration by parts, we obtain (see Example 3.19)
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$$
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\frac {d}{d x} \left(E A \frac {d u}{d x}\right) = 0 \tag {e}
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$$
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$$
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E A \left. \frac {d u}{d x} \right| _ {x = 1 8 0} = 1 0 0 \tag {f}
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$$
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The solution of (e) subject to the natural boundary condition in (f) and the essential boundary condition $u|_{x=0} = 0$ gives
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$$
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u = \frac {1 0 0}{E} x; \quad 0 \leq x \leq 1 0 0
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$$
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$$
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u = \frac {1 0 0 0 0}{E} + \frac {4 0 0 0}{E} - \frac {4 0 0 0}{E \left(1 + \frac {x - 1 0 0}{4 0}\right)}; \quad 1 0 0 \leq x \leq 1 8 0
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$$
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The exact stresses in the bar are
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$$
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\sigma = 1 0 0; \quad 0 \leq x \leq 1 0 0
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$$
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$$
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\sigma = \frac {1 0 0}{\left(1 + \frac {x - 1 0 0}{4 0}\right) ^ {2}}; \quad 1 0 0 \leq x \leq 1 8 0
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$$
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Next, to perform the Ritz analyses, we note that the displacement assumptions in (b) and (c) satisfy the essential boundary condition but not the natural boundary condition. Substituting from (b) into (a), we obtain
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$$
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\Pi = \frac {E}{2} \int_ {0} ^ {1 0 0} (a _ {1} + 2 a _ {2} x) ^ {2} d x + \frac {E}{2} \int_ {1 0 0} ^ {1 8 0} \left(1 + \frac {x - 1 0 0}{4 0}\right) ^ {2} (a _ {1} + 2 a _ {2} x) ^ {2} d x - 1 0 0 u | _ {x = 1 8 0}
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$$
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Invoking $\delta \Pi = 0$ , we obtain the following equations for $a_1$ and $a_2$ :
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$$
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E \left[ \begin{array}{l l} 0. 4 4 6 7 & 1 1 5. 6 \\ 1 1 5. 6 & 3 4 0 7 5. 7 \end{array} \right] \left[ \begin{array}{l} a _ {1} \\ a _ {2} \end{array} \right] = \left[ \begin{array}{l} 1 8 \\ 3 2 4 0 \end{array} \right] \tag {g}
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$$
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and $a_1 = \frac{129}{E};\quad a_2 = -\frac{0.341}{E}$
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This Ritz analysis therefore yields the approximate solution
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$$
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u = \frac {1 2 9}{E} x - \frac {0 . 3 4 1}{E} x ^ {2} \tag {h}
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$$
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$$
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\sigma = 1 2 9 - 0. 6 8 2 x; \quad 0 \leq x \leq 1 8 0 \tag {i}
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$$
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Using next the Ritz functions in (c), we have
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$$
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\Pi = \frac {E}{2} \int_ {0} ^ {1 0 0} \left(\frac {1}{1 0 0} u _ {B}\right) ^ {2} d x + \frac {E}{2} \int_ {1 0 0} ^ {1 8 0} \left(1 + \frac {x - 1 0 0}{4 0}\right) ^ {2} \left(- \frac {1}{8 0} u _ {B} + \frac {1}{8 0} u _ {C}\right) ^ {2} d x - 1 0 0 u _ {C}
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$$
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Invoking again $\delta \Pi = 0$ , we obtain
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$$
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\frac {E}{2 4 0} \left[ \begin{array}{c c} 1 5. 4 & - 1 3 \\ - 1 3 & 1 3 \end{array} \right] \left[ \begin{array}{l} u _ {B} \\ u _ {C} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 1 0 0 \end{array} \right] \tag {j}
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$$
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<!-- source-page: 139 -->
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Hence we now have
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$$
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u _ {B} = \frac {1 0 , 0 0 0}{E}; \quad u _ {C} = \frac {1 1 , 8 4 6 . 2}{E}
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$$
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and $\sigma = 100;$ $0\leq x\leq 100$
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$$
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\sigma = \frac {1 8 4 6 . 2}{8 0} = 2 3. 0 8; \quad 1 0 0 \leq x \leq 1 8 0
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$$
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We shall see in Chapter 4 (see Example 4.5) that this Ritz analysis can be regarded to be a finite element analysis.
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EXAMPLE 3.23: Consider the slab in Example 3.16. Assume that
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$$
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\theta (t) = \theta_ {1} (t) + \theta_ {2} (t) x + \theta_ {3} (t) x ^ {2} \tag {a}
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$$
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where $\theta_{1}(t)$ , $\theta_{2}(t)$ , and $\theta_{3}(t)$ are the undetermined parameters. Use the Ritz analysis procedure to generate the governing heat transfer equilibrium equations.
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The functional governing the temperature distribution in the slab is (see Example 3.18)
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$$
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\Pi = \int_ {0} ^ {L} \frac {1}{2} k \left(\frac {\partial \theta}{\partial x}\right) ^ {2} d x - \int_ {0} ^ {L} \theta q ^ {B} d x - \theta \mid_ {x = 0} q _ {0} \tag {b}
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$$
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with the essential boundary condition
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$$
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\theta \mid_ {x = L} = \theta_ {i}
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$$
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Substituting the temperature assumption of (a) into (b), we obtain
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$$
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\Pi = \int_ {0} ^ {L} \frac {1}{2} k ((\theta_ {2}) ^ {2} + 4 \theta_ {2} \theta_ {3} x + 4 (\theta_ {3}) ^ {2} x ^ {2}) d x - \int_ {0} ^ {L} (\theta_ {1} + \theta_ {2} x + \theta_ {3} x ^ {2}) q ^ {B} d x - \theta_ {1} q _ {0}
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$$
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Invoking the stationarity condition of $\Pi$ , i.e., $\delta\Pi = 0$ , we use
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$$
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\frac {\partial \Pi}{\partial \theta_ {1}} = 0; \quad \frac {\partial \Pi}{\partial \theta_ {2}} = 0; \quad \frac {\partial \Pi}{\partial \theta_ {3}} = 0
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$$
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and obtain
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$$
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k \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & L & L ^ {2} \\ 0 & L ^ {2} & \frac {4}{3} L ^ {3} \end{array} \right] \left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \end{array} \right] = \left[ \begin{array}{l} \int_ {0} ^ {L} q ^ {B} d x + q _ {0} \\ \int_ {0} ^ {L} x q ^ {B} d x \\ \int_ {0} ^ {L} x ^ {2} q ^ {B} d x \end{array} \right] \tag {c}
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$$
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In this analysis $q_{0}$ varies with time, so that the temperature varies with time, and heat capacity effects can be important. Using
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$$
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q ^ {B} = - \rho c \frac {\partial \theta}{\partial t} \tag {d}
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$$
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because no other heat is generated, substituting for $\theta$ in (d) from (a), and then substituting into (c), we obtain as the equilibrium equations,
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$$
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k \left[ \begin{array}{c c c} 0 & 0 & 0 \\ 0 & L & L ^ {2} \\ 0 & L ^ {2} & \frac {4}{3} L ^ {3} \end{array} \right] \left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \end{array} \right] + \rho c \left[ \begin{array}{c c c} L & \frac {1}{2} L ^ {2} & \frac {1}{3} L ^ {3} \\ \frac {1}{2} L ^ {2} & \frac {1}{3} L ^ {3} & \frac {1}{4} L ^ {4} \\ \frac {1}{3} L ^ {3} & \frac {1}{4} L ^ {4} & \frac {1}{5} L ^ {5} \end{array} \right] \left[ \begin{array}{l} \dot {\theta} _ {1} \\ \dot {\theta} _ {2} \\ \dot {\theta} _ {3} \end{array} \right] = \left[ \begin{array}{l} q _ {0} \\ 0 \\ 0 \end{array} \right] \tag {e}
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$$
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<!-- source-page: 140 -->
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The final equilibrium equations are now obtained by imposing on the equations in (e) the condition that $\theta|_{x=L} = \theta_i$ ; i.e.,
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$$
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\theta_ {1} (t) + \theta_ {2} (t) L + \theta_ {3} (t) L ^ {2} = \theta_ {i}
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$$
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which can be achieved by expressing $\theta_{1}$ in (e) in terms of $\theta_{2}$ , $\theta_{3}$ , and $\theta_{i}$ .
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EXAMPLE 3.24: Consider the static buckling response of the column in Example 3.20. Assume that
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$$
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w = a _ {1} x ^ {2} + a _ {2} x ^ {3} \tag {a}
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$$
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and use the Ritz method to formulate equations from which we can obtain an approximate buckling load.
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The functional governing the problem was given in Example 3.20,
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} E I \left(\frac {d ^ {2} w}{d x ^ {2}}\right) ^ {2} d x - \frac {P}{2} \int_ {0} ^ {L} \left(\frac {d w}{d x}\right) ^ {2} d x + \frac {1}{2} k (w | _ {x = L}) ^ {2} \tag {b}
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$$
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We note that the trial function on w in (a) already satisfies the essential boundary conditions (displacement and slope equal to zero at the fixed end). Substituting for w into (b), we obtain
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} E I (2 a _ {1} + 6 a _ {2} x) ^ {2} d x - \frac {P}{2} \int_ {0} ^ {L} (2 a _ {1} x + 3 a _ {2} x ^ {2}) ^ {2} d x + \frac {1}{2} k (a _ {1} L ^ {2} + a _ {2} L ^ {3}) ^ {2}
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$$
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Invoking the stationarity condition $\delta \Pi = 0$ , i.e.,
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$$
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\frac {\partial \Pi}{\partial a _ {1}} = 0; \quad \frac {\partial \Pi}{\partial a _ {2}} = 0
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$$
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we obtain
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$$
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\left\{2 E I \left[ \begin{array}{l l} 2 L & 3 L ^ {2} \\ 3 L ^ {2} & 6 L ^ {3} \end{array} \right] + k L ^ {4} \left[ \begin{array}{l l} 1 & L \\ L & L ^ {2} \end{array} \right] \right\} \left[ \begin{array}{l} a _ {1} \\ a _ {2} \end{array} \right] - P L ^ {3} \left[ \begin{array}{l l} \frac {4}{3} & \frac {3 L}{2} \\ \frac {3 L}{2} & \frac {9 L ^ {2}}{5} \end{array} \right] \left[ \begin{array}{l} a _ {1} \\ a _ {2} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \end{array} \right]
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$$
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The solution of this eigenproblem gives two values of P for which w in (a) is nonzero. The smaller value of P represents an approximation to the lowest buckling load of the structure.
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The weighted residual methods presented in (3.14) to (3.16) are difficult to use in practice because the trial functions must be 2m-times-differentiable and satisfy all—essential and natural—boundary conditions [see (3.13)]. On the other hand, with the Ritz method, which operates on the functional corresponding to the problem being considered, the trial functions need to be only m-times-differentiable and do not need to satisfy the natural boundary conditions. These considerations are most important for practical analysis, and therefore the Galerkin method is used in practice in a different form, namely, in a form that allows the use of the same functions as used in the Ritz method. In the displacement-based analysis of solids and structures, this form of the Galerkin method is referred to as the principle of virtual displacements. If the appropriate variational indicator II is used, the equations obtained with the Ritz method are then identical to those obtained with the Galerkin method.
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We elaborate upon these issues in the next section with the objective of providing further understanding for the introduction of finite element procedures.
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