김경종
486 lines
21 KiB
Markdown
486 lines
21 KiB
Markdown
<!-- source-page: 151 -->
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<details>
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<summary>text_image</summary>
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Transverse load
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p per unit area
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Flexural rigidity D
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Mass per unit area m
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L
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w1
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w2
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L
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w3
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For Table 3.1:
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w1 ≡ w1,1
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w2 ≡ w2,1
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w3 ≡ w3,1
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</details>
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Figure E3.28 Simply supported plate
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The governing differential equation of the plate is (see, for example, S. Timoshenko and S. Woinowsky-Krieger [A])
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$$
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\nabla^ {4} w = \frac {p}{D}
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$$
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where $w$ is the transverse displacement. The boundary conditions are that on each edge of the plate the transverse displacement and the moment across the edge are zero.
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We use the finite difference stencil for $\nabla^{4}w$ given in Table 3.1, with the center point of the molecule placed at the center of the plate. The displacements corresponding to the coefficients -8 and +2 are zero, and the displacements corresponding to the coefficients +1 are expressed in terms of the center displacement. For example, the zero moment condition gives (refer to Fig. E3.28)
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$$
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w _ {1} - 2 w _ {2} + w _ {3} = 0
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$$
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and because $w_{2} = 0$ ,
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$$
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w _ {3} = - w _ {1}
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$$
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Therefore, the governing finite difference equation is
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$$
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1 6 w _ {1} = \frac {p}{D} \left(\frac {L}{2}\right) ^ {4}
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$$
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and we obtain $\left[\frac{16D}{(L / 2)^2}\right]w_1 = R;\quad R = p\left(\frac{L}{2}\right)^2$
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Note that with this relation we in essence represent the plate by a single spring of stiffness $k = 64D/L^{2}$ , and the total load acting on the spring is given by R. The deflection $w_{1}$ thus calculated is only about 4 percent different from the analytically calculated “exact” value.
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For the dynamic analysis, we use d'Alembert's principle and subtract from the externally applied load R the inertia load $M\ddot{w}_{1}$ , where M represents a mass in some sense equivalent to the distributed mass of the plate
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$$
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M = m \left(\frac {L}{2}\right) ^ {2}
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$$
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Hence the dynamic equilibrium equation is
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$$
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m \frac {L ^ {2}}{4} \ddot {w} _ {1} + \frac {6 4 D}{L ^ {2}} w _ {1} = R
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$$
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<!-- source-page: 152 -->
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In these two examples and in the analysis of the bar in Fig. 3.2, the differential equations of equilibrium have been approximated by finite differences. When the differential equations of equilibrium are used to solve a mathematical model, it is necessary to approximate by finite differences and impose on the coefficient matrix both the essential and the natural boundary conditions. In the analysis of the beam and the plate considered in Examples 3.27 and 3.28, these boundary conditions could easily be imposed (the zero displacements on the boundaries are the essential boundary conditions and the zero moment conditions across the boundaries are the natural boundary conditions). However, for complex geometries the imposition of the natural boundary conditions can be difficult to achieve since the topology of the finite difference mesh restricts the form of differencing that can be carried out, and it may be difficult to obtain a symmetric coefficient matrix in a rigorous manner (see A. Ghali and K. J. Bathe [A]).
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The difficulties associated with the use of the differential formulations have given impetus to the development of finite difference analysis procedures based on the principle of minimum total potential energy, referred to as the finite difference energy method (see, for example, D. Bushnell, B. O. Almroth, and F. Brogan [A]). In this scheme the displacement derivatives in the total potential energy, $\Pi$ , of the system are approximated by finite differences, and the minimum condition of $\Pi$ is used to calculate the unknown displacements at the finite difference stations. Since the variational formulation of the problem under consideration is employed, only the essential (geometric) boundary conditions must be satisfied in the differencing. Furthermore, a symmetric coefficient matrix is always obtained.
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As might well be expected, the finite difference energy method is very closely related to the Ritz method, and in some cases the same algebraic equations are generated.
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An advantage of the finite difference energy method lies in the effectiveness with which the coefficient matrix of the algebraic equations can be generated. This effectiveness is due to the simple scheme of energy integration employed. However, the Galerkin method implemented in the form of the finite element procedures discussed in the forthcoming chapters is a much more general and powerful technique, and this of course is the reason for the success of the finite element method.
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It is instructive to examine the use of the finite difference energy method in some examples.
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EXAMPLE 3.29: Consider the cantilever beam in Fig. E3.29. Evaluate the tip deflection using the conventional finite difference method and the finite difference energy method.
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<details>
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<summary>text_image</summary>
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Finite difference
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stations
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R
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Flexural
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rigidity EI
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w1
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w2
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w3
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w4
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w5
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w6
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L/4
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L
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</details>
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Figure E3.29 Finite difference stations on cantilever beam
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<!-- source-page: 153 -->
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The finite difference mesh used is shown in the figure. Using the conventional finite difference procedure and central differencing as in Example 3.27, we obtain the equilibrium equations
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$$
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\frac {6 4 E I}{L ^ {3}} \left[ \begin{array}{r r r r} 7 & - 4 & 1 & 0 \\ - 4 & 6 & - 4 & 1 \\ 1 & - 4 & 5 & - 2 \\ 0 & 1 & - 2 & 1 \end{array} \right] \left[ \begin{array}{l} w _ {2} \\ w _ {3} \\ w _ {4} \\ w _ {5} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ R \\ 0 \end{array} \right] \tag {a}
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$$
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It may be noted that in addition to the equations employed in Example 3.27 the conditions $w' = 0$ at the fixed end and $w''' = 0$ at the free end are also used. For $w'$ and $w'''$ equal to zero at station i, we employ, respectively,
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$$
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w _ {i + 1} - w _ {i - 1} = 0
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$$
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$$
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w _ {i + 2} - 2 w _ {i + 1} + 2 w _ {i - 1} - w _ {i - 2} = 0
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$$
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Using the finite difference energy method, the total potential energy $\Pi$ is given as
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$$
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\Pi = \frac {E I}{2} \int_ {0} ^ {L} [ w ^ {\prime \prime} (x) ] ^ {2} d x - R w \Bigg | _ {x = \frac {3}{4} L}
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$$
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To evaluate the integral we need to approximate $w''(x)$ . Using central differencing, we obtain for station i,
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$$
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w _ {i} ^ {\prime \prime} = \frac {1}{(L / 4) ^ {2}} \left(w _ {i + 1} - 2 w _ {i} + w _ {i - 1}\right) \tag {b}
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$$
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An approximate solution can now be obtained by evaluating $\Pi$ at the finite difference stations using (b) and replacing the integral by a summation process; i.e., we use the approximation
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$$
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\Pi = \frac {L}{8} \Pi_ {1} + \frac {L}{4} \left(\Pi_ {2} + \Pi_ {3} + \Pi_ {4}\right) + \frac {L}{8} \Pi_ {5} - R w _ {4} \tag {c}
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$$
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where $\Pi_{i} = \frac{1}{2} [w_{i - 1}\quad w_{i}\quad w_{i + 1}]\left[ \begin{array}{c}1\\ -2\\ 1 \end{array} \right]\frac{EI}{(L / 4)^{4}} [1\quad -2\quad 1]\left[ \begin{array}{c}w_{i - 1}\\ w_{i}\\ w_{i + 1} \end{array} \right]$
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Therefore, we can write, in analogy with the finite element analysis procedures (see Section 4.2),
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$$
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\Pi_ {i} = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {B} _ {i} ^ {T} \mathbf {C} _ {i} \mathbf {B} _ {i} \mathbf {U}
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$$
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where $B_{i}$ is a generalized strain-displacement transformation matrix, $C_{i}$ is the stress-strain matrix, and U is a vector listing all nodal point displacements. Using the direct stiffness method to calculate the total potential energy as given in (c) and employing the condition that the total potential energy is stationary (i.e., $\delta\Pi = 0$ ), we obtain the equilibrium equations
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$$
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\frac {6 4 E I}{L ^ {3}} \left[ \begin{array}{c c c c c} 7 & - 4 & 1 & & \\ - 4 & 6 & - 4 & 1 & \\ 1 & - 4 & 5. 5 & - 3 & 0. 5 \\ & 1 & - 3 & 3 & - 1 \\ & & 0. 5 & - 1 & 0. 5 \end{array} \right] \left[ \begin{array}{l} w _ {2} \\ w _ {3} \\ w _ {4} \\ w _ {5} \\ w _ {6} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \\ R \\ 0 \\ 0 \end{array} \right] \tag {d}
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$$
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where the condition of zero slope at the fixed end has already been used.
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<!-- source-page: 154 -->
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The close similarity between the equilibrium equations in (a) and (d) should be noted. Indeed, if we eliminate $w_{6}$ from the equations in (d), we obtain the equations in (a). Hence, using the finite difference energy method and the conventional finite difference method, we obtain in this case the same equilibrium equations.
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As an example, let $R = 1$ , $EI = 10^3$ , and $L = 10$ . Then we obtain, using the equations in (a) or (d),
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$$
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\mathbf {U} = \left[ \begin{array}{l} 0. 0 2 3 4 3 7 \\ 0. 0 7 8 1 2 5 \\ 0. 1 4 8 4 3 \\ 0. 2 1 8 7 5 \end{array} \right]
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$$
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The exact answer for the tip deflection is $w_{5} = 0.2109375$ . Hence the finite difference analysis gives a good approximate solution.
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EXAMPLE 3.30: The rod shown in Fig. E3.30 is subjected to a heat flux input of $q^{s}$ at its right end and a constant temperature $\theta_{0}$ at its left end and is in steady-state conditions. The variational indicator is
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} k \left(\frac {\partial \theta}{\partial x}\right) ^ {2} A d x - q ^ {s} A _ {L} \theta_ {L} \tag {a}
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$$
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<details>
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<summary>text_image</summary>
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A(x) = A₀ (1 + x/L)²
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Section BB
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Insulated around
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circumference
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θ₀ → x
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A(x)
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q^S
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B
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L
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B
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L/4
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θ₀ θ₁ θ₂ θ₃ θ₄
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q^S = prescribed heat flow
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input per unit area at x = L
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k = conductivity (constant)
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</details>
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Figure E3.30 Rod in heat transfer condition; finite difference stations used
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Use the finite difference method to obtain an approximate solution for the temperature distribution.
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Let us use five equally spaced finite difference stations as shown in Fig. E3.30. The finite difference approximation of the integral in (a) is then
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$$
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\Pi = \frac {L}{4} \left\{\Pi_ {1 / 2} + \Pi_ {3 / 2} + \Pi_ {5 / 2} + \Pi_ {7 / 2} \right\} - q ^ {s} A _ {L} \theta_ {L}
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$$
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where $\Pi_{1/2} = \frac{1}{2} [\theta_1, \theta_0] \begin{bmatrix} 1 \\ -1 \end{bmatrix} \frac{k(\frac{9}{8})^2 A_0}{(L/4)^2} [1 - 1] \begin{bmatrix} \theta_1 \\ \theta_0 \end{bmatrix}$
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<!-- source-page: 155 -->
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and the values $\Pi_{3/2}$ , $\Pi_{5/2}$ , and $\Pi_{7/2}$ are similarly evaluated. Calculating $\Pi$ , invoking $\delta\Pi = 0$ , and imposing the boundary condition that $\theta_0$ is known, we thus obtain
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$$
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\frac {k A _ {0}}{1 6 L} \left[ \begin{array}{c c c c} 2 0 2 & - 1 2 1 & & \\ - 1 2 1 & 2 9 0 & - 1 6 9 & \\ & - 1 6 9 & 3 9 4 & - 2 2 5 \\ & & - 2 2 5 & 2 2 5 \end{array} \right] \left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \\ \theta_ {4} \end{array} \right] = \left[ \begin{array}{c} \frac {8 1}{1 6 L} k A _ {0} \theta_ {0} \\ 0 \\ 0 \\ 4 A _ {0} q ^ {s} \end{array} \right]
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$$
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Now assume that $\theta_0 = 0$ . Then the solution is
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$$
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\left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \\ \theta_ {4} \end{array} \right] = \left[ \begin{array}{l} 0. 7 9 \\ 1. 3 2 \\ 1. 7 0 \\ 1. 9 8 \end{array} \right] \frac {L q ^ {s}}{k}
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$$
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which compares as follows with the analytical solution
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$$
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\left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \\ \theta_ {4} \end{array} \right] _ {\text { analytical }} = \left[ \begin{array}{l} \frac {4}{5} \\ \frac {4}{3} \\ \frac {1 2}{7} \\ 2 \end{array} \right] \frac {L q ^ {s}}{k}
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$$
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# 3.3.6 Exercises
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3.15. Establish the differential equation of equilibrium of the problem shown and the (geometric and force) boundary conditions. Determine whether the operator $L_{2m}$ of the problem is symmetric and positive definite and prove your answer.
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<details>
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<summary>text_image</summary>
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A(x) = A₀(2 - x/L)
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A₀
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k
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L
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R
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</details>
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Young's modulus E
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Rod with varying cross-sectional area
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3.16. Consider the cantilever beam shown, which is subjected to a moment M at its tip. Determine the variational indicator $\Pi$ and state the essential boundary conditions. Invoke the stationarity of $\Pi$ by using (3.7b) and by using the fact that variations and differentiations are performed using the same rules. Then extract the differential equation of equilibrium and the natural boundary conditions. Determine whether the operator $L_{2m}$ is symmetric and positive definite and prove your answer.
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<!-- source-page: 156 -->
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<details>
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<summary>text_image</summary>
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Flexural stiffness EI
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M
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L
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</details>
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3.17. Consider the heat transfer problem in Example 3.30. Invoke the stationarity of the given variational indicator by using (3.7b) and by using the fact that variations and differentiations are performed using the same rules. Establish the governing differential equation of equilibrium and all boundary conditions. Determine whether the operator $L_{2m}$ is symmetric and positive definite and prove your answer.
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3.18. Consider the prestressed cable shown in the figure. The variational indicator is
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} T \left(\frac {d w}{d x}\right) ^ {2} d x + \int_ {0} ^ {L} \frac {1}{2} k (w) ^ {2} d x - P w _ {L}
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$$
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where w is the transverse displacement and $w_{L}$ is the transverse displacement at x = L. Establish the differential equation of equilibrium and state all boundary conditions. Determine whether the operator $L_{2m}$ is symmetric and positive definite and prove your answer.
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<details>
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<summary>text_image</summary>
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Constant tension T
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P
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Frictionless
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roller
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Cable on distributed vertical
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springs of stiffness k/unit length of cable
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</details>
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3.19. Consider the prestressed cable in Exercise 3.18.
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(a) Establish a suitable trial function that can be employed in the analysis of the cable using the classical Galerkin and least squares methods. Try $w(x) = a_0 + a_1x + a_2x^2$ and modify the function as necessary.
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(b) Establish the governing equations of the system for the selected trial function using the classical Galerkin and least squares methods.
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3.20. Consider the prestressed cable in Exercise 3.18. Establish the governing equations using the Ritz method with the trial function $w(x) = a_0 + a_1x + a_2x^2$ (i.e., a suitable modification thereof).
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<!-- source-page: 157 -->
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3.21. Use the Ritz method to calculate the linearized buckling load of the column shown. Assume that $w = cx^2$ , where $c$ is the unknown Ritz parameter.
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<details>
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<summary>text_image</summary>
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P
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k
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L
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w(x)
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x
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EI(x) = EI0(2 - x/L)
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</details>
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3.22. Consider the structure shown.
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(a) Use the Ritz method to establish the governing equations for the bending response. Use the following functions: (i) $w = a_1 x^2$ and (ii) $w = b_1 [1 - \cos(\pi x / 2L)]$ .
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(b) With $EI_0 = 100, k = 2, L = 1$ estimate the critical load of the column using a Ritz analysis.
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<details>
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<summary>text_image</summary>
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P
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W
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k = spring stiffness per unit length of beam
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L
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w(x)
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EI(x) = EI0(1 - x/2L)
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x
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</details>
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3.23. Consider the slab shown for a heat transfer analysis. The variational indicator for this analysis is
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$$
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\Pi = \int_ {0} ^ {L} \frac {1}{2} k \left(\frac {d \theta}{d x}\right) ^ {2} d x - \int_ {0} ^ {L} \theta q ^ {B} d x
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$$
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State the essential and natural boundary conditions. Then perform a Ritz analysis of the problem using two unknown parameters.
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<!-- source-page: 158 -->
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<details>
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<summary>text_image</summary>
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Prescribed temperature
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θ = 20°
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z
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y
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Infinitely long slab
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in y- and z-directions
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Zero heat flux
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Inside
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k₁
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k₂
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Outside
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∞
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k₁ = conductivity of inner part of slab = 20
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k₂ = conductivity of outer part of slab = 40
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q^B = heat generated per unit volume in total slab = 100
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L/2
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L/2
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L = 10
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</details>
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3.24. The prestressed cable shown is to be analyzed. The governing differential equation of equilibrium is
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$$
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T \frac {\partial^ {2} w}{\partial x ^ {2}} = m \frac {\partial^ {2} w}{\partial t ^ {2}} - p (t)
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$$
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with the boundary conditions
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$$
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w \mid_ {x = 0} = w \mid_ {x = L} = 0
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$$
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and the initial conditions
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$$
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w (x, 0) = 0; \quad \frac {\partial w}{\partial t} (x, 0) = 0
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$$
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(a) Use the conventional finite difference method to approximate the governing differential equation of equilibrium and thus establish equations governing the response of the cable.
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(b) Use the finite difference energy method to establish equations governing the response of the cable.
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(c) Use the principle of virtual work to establish equations governing the response of the cable. When using the finite difference methods, employ two internal finite difference stations. To employ the principle of virtual work, use the two basis functions shown.
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<details>
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<summary>text_image</summary>
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Uniformly distributed
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loading p(t)
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w(x)
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L
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x
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Constant tension T
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Mass/unit length m
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</details>
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<!-- source-page: 159 -->
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<details>
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<summary>text_image</summary>
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1
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2
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L/3 L/3 L/3
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</details>
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Finite difference stations
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<details>
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<summary>text_image</summary>
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L/3 L/3 L/3
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</details>
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Basis functions for use of principle of virtual work
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3.25. The disk shown is to be analyzed for the temperature distribution. Determine the variational indicator of the problem and obtain an approximate solution using the Ritz method with the basis functions shown in Fig. 3.4. Use two unknown temperatures. Compare your results with the exact analytical solution.
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<details>
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<summary>text_image</summary>
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q
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r0
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r1
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θ1
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k
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</details>
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$q^{S}=100\ Btu/(hr\cdot in^{2})$ (prescribed heat flux)
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$\theta_{1} = 70^{\circ}\mathrm{F}$ (prescribed temperature)
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$r_0 = 1.0$ in
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$r_1 = 3.0$ in
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$k = 120 \mathrm{Btu} / (\mathrm{hr} \cdot \mathrm{in} \cdot {}^{\circ} \mathrm{F})$
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h = 0.1 in (thickness of disk)
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The top and bottom faces of the disk are insulated
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3.26. Consider the beam analysis problem shown.
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(a) Use four finite difference stations on the beam with the differential formulation to establish equations governing the response of the beam.
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(b) Use four finite difference stations on the beam with the variational formulation to establish equations governing the response of the beam.
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<!-- source-page: 160 -->
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<details>
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<summary>text_image</summary>
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p = load/unit length
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Flexural stiffness EI
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Spring stiffness k
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h
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h = L/3
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h
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</details>
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3.27. Use the finite difference energy method with only two unknown temperature values to solve the problem in Exercise 3.23.
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3.28. Use the finite difference energy method with only two unknown temperature values to solve the problem in Exercise 3.25.
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3.29. The computer program STAP (see Chapter 12) has been written for the analysis of truss structures. However, by using analogies involving variables and equations, the program can also be employed in the analysis of pressure and flow distributions in pipe networks, current distributions in dc networks, and in heat transfer analyses. Use the program STAP to solve the analysis problems in Examples 3.1 to 3.4.
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3.30. Use a computer program to solve the problems in Examples 3.1 to 3.4.
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# 3.4 IMPOSITION OF CONSTRAINTS
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The analysis of an engineering problem frequently requires that a specific constraint be imposed on certain solution variables. These constraints may need to be imposed on some continuous solution parameters or on some discrete variables and may consist of certain continuity requirements, the imposition of specified values for the solution variables, or conditions to be satisfied between certain solution variables. Two widely used procedures are available, namely, the Lagrange multiplier method and the penalty method (see, for example, D. P. Bertsekas [A]). Applications of these techniques are given in Sections 4.2.2, 4.4.2, 4.4.3, 4.5, 5.4, 6.7.2, and 7.4. Both the Lagrange multiplier and the penalty methods operate on the variational or weighted residual formulations of the problem to be solved.
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# 3.4.1 An Introduction to Lagrange Multiplier and Penalty Methods
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As a brief introduction to Lagrange multiplier and penalty methods, consider the variational formulation of a discrete structural model for a steady-state analysis,
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$$
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\boldsymbol {\Pi} = \frac {1}{2} \mathbf {U} ^ {T} \mathbf {K} \mathbf {U} - \mathbf {U} ^ {T} \mathbf {R} \tag {3.52}
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$$
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with the conditions
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$$
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\frac {\partial \Pi}{\partial U _ {i}} = 0 \quad \text { for all } i \tag {3.53}
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$$
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