김경종
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24 KiB
Markdown
442 lines
24 KiB
Markdown
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# 3.3.4 An Overview: The Differential and Galerkin Formulations, the Principle of Virtual Displacements, and an Introduction to the Finite Element Solution
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In the previous sections we reviewed some classical differential and variational formulations, some classical weighted residual methods, and the Ritz method. We now want to reinforce our understanding of these analysis approaches—by summarizing some important concepts—and briefly introduce a mathematical framework for finite element procedures that we will further use and extend in Chapter 4. Let us pursue this objective by focusing on the analysis of a simple example problem.
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Consider the one-dimensional bar in Fig. 3.2. The bar is subjected to a distributed load $f^{B}(x)$ and a concentrated load R at its right end. As discussed in Section 3.3.1, the differential formulation of the bar gives the governing equations
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$$
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\boxed {\text { Differential formulation }} \left\{ \begin{array}{c} E A \frac {d ^ {2} u}{d x ^ {2}} + f ^ {B} = 0 \quad \text { in the bar } \\ u \big | _ {x = 0} = 0 \\ E A \frac {d u}{d x} \bigg | _ {x = L} = R \end{array} \right. \tag {3.18}
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$$
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Since $f^B = ax$ , we obtain the solution
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$$
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u (x) = \frac {- (a x ^ {3} / 6) + (R + \frac {1}{2} a L ^ {2}) x}{E A} \tag {3.21}
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$$
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<details>
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<summary>text_image</summary>
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Constant cross-sectional area A
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Young's modulus E
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f^B(x) = ax
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x
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f^B(x)
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R
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L
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</details>
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Figure 3.2 Uniform bar subjected to body load $f^{B}$ (force/unit length) and tip load R
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We recall that (3.18) is a statement of equilibrium at any point x within the bar, (3.19) is the essential (or geometric) boundary condition (see Section 3.2.2), and (3.20) is the natural (or force) boundary condition. The exact analytical solution (3.21) of course satisfies all three equations (3.18) to (3.20).
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We also note that the solution $u(x)$ is a continuous and twice-differentiable function, as required in (3.18). Indeed, we can say that the solutions to (3.18) satisfying (3.19) and (3.20) for any continuous loading $f^{B}$ lie in the space of continuous and twice-differentiable functions that satisfy (3.19) and (3.20).
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An alternative approach for the solution of the analysis problem is given by the variational formulation (see Section 3.3.2),
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$$
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\boxed {\text {Variational} \quad \text {formulation}} \quad \left\{ \begin{array}{c c} \Pi = \int_ {0} ^ {L} \frac {1}{2} E A \left(\frac {d u}{d x}\right) ^ {2} d x - \int_ {0} ^ {L} u f ^ {B} d x - R u | _ {x = L} & \text {(3.22)} \\ \delta \Pi = 0 & \text {(3.23)} \\ \text {with} u | _ {x = 0} = 0 & \text {(3.24)} \\ \delta u | _ {x = 0} = 0 & \text {(3.25)} \end{array} \right.
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$$
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where $\delta$ means “variation in” and $\delta u$ is an arbitrary variation on u subject to the condition $\delta u|_{x=0}=0$ . We may think of $\delta u(x)$ as any continuous function that satisfies the boundary condition (3.25). $^{4}$
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Let us recall that (3.22) to (3.25) are totally equivalent to (3.18) to (3.20) (see Section 3.3.2). That is, invoking (3.23) and then using integration by parts and the boundary condition (3.25) gives (3.18) and (3.20). Therefore, the solution of (3.22) to (3.25) is also (3.21).
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The variational formulation can be derived as follows.
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Since (3.18) holds for all points within the bar, we also have
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$$
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\left(E A \frac {d ^ {2} u}{d x ^ {2}} + f ^ {B}\right) \delta u = 0 \tag {3.26}
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$$
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where $\delta u(x)$ is an arbitrary variation on $u$ (or an arbitrary continuous function) with $\delta u|_{x=0} = 0$ . Hence, also
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$$
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\int_ {0} ^ {L} \left(E A \frac {d ^ {2} u}{d x ^ {2}} + f ^ {B}\right) \delta u d x = 0 \tag {3.27}
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$$
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Integrating by parts, we obtain
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$$
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\int_ {0} ^ {L} \frac {d \delta u}{d x} E A \frac {d u}{d x} d x = \int_ {0} ^ {L} f ^ {B} \delta u d x + E A \frac {d u}{d x} \delta u \big | _ {0} ^ {L} \tag {3.28}
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$$
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Substituting from (3.20) and (3.25), we therefore have
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$$
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\boxed {\text { Principle of virtual displacements }} \left\{ \begin{array}{l} \int_ {0} ^ {L} \frac {d \delta u}{d x} E A \frac {d u}{d x} d x = \int_ {0} ^ {L} f ^ {B} \delta u d x + R \delta u | _ {x = L} \\ \text { with } u | _ {x = 0} = 0; \quad \delta u | _ {x = 0} = 0 \end{array} \right. \tag {3.29}
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$$
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Of course, (3.29) gives
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$$
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\delta \left\{\int_ {0} ^ {L} \left[ \frac {E A}{2} \left(\frac {d u}{d x}\right) ^ {2} - f ^ {B} u \right] d x - R u \mid_ {x = L} \right\} = 0 \tag {3.31}
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$$
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which with (3.30) is the variational statement of (3.22) to (3.25).
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The relation (3.29) along with the condition (3.30) is the celebrated principle of virtual displacements (or principle of virtual work) in which $\delta u(x)$ is the virtual displace-
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$^{4}$ In the literature, differential and variational formulations are, respectively, also referred to as strong and weak forms. Variational formulations are also referred to as generalized formulations.
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ment. We discuss this principle extensively in Section 4.2 and note that the derivation in (3.26) to (3.30) is a special case of Example 4.2.
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It is important to recognize that the above three formulations of the analysis problem are totally equivalent, that is, the solution (3.21) is the (unique) solution $^{5}$ $u(x)$ of the differential formulation, the variational formulation, and the principle of virtual displacements. However, we note that the variational formulation and the principle of virtual work involve only first-order derivatives of the functions u and $\delta u$ . Hence the space of functions in which we look for a solution is clearly larger than the space of functions used for the solution of (3.18) [we define the space precisely in (3.35)], and there must be a question as to what it means and how important it is that we use a larger space of functions when solving the problem in Fig. 3.2 with the principle of virtual displacements.
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Of course, the space of functions used with the principle of virtual displacements contains the space of functions used with the differential formulation, hence all analysis problems that can be solved with the differential formulation (3.18) to (3.20) can also be solved exactly with the principle of virtual displacements. However, in the analysis of the bar (and the analysis of general bar and beam structures) additional conditions for which the principle of virtual work can be used directly for solution are those where concentrated loads are applied within the bar or discontinuities in the material property or cross-sectional area are present. In these cases the first derivative of $u(x)$ is discontinuous and hence the differential formulation has to be extended to account for such cases (in essence treating separately each section of the bar in which no concentrated loads are applied and in which no discontinuities in the material property and cross-sectional area are present, and connecting the section to the adjoining sections by the boundary conditions; see, for example, S. H. Crandall, N. C. Dahl, and T. J. Lardner [A]). Hence, in these cases the variational formulation and the principle of virtual displacements are somewhat more direct and more powerful for solution.
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For general two- and three-dimensional stress situations, we will only consider mathematical models of finite strain energy (meaning for example that concentrated loads should only be applied as enumerated in Section 1.2, see Fig. 1.4, and further discussed in Section 4.3.4), and then the differential and principle of virtual work formulations are also totally equivalent and give the same solutions (see Chapter 4).
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These considerations point to a powerful general procedure for formulating the numerical solution of the problem in Fig. 3.2. Consider (3.27) in which we now replace $\delta u$ with the test function $v$ ,
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$$
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\int_ {0} ^ {L} \left(E A \frac {d ^ {2} u}{d x ^ {2}} + f ^ {B}\right) v d x = 0 \tag {3.32}
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$$
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with $u = 0$ and $v = 0$ at $x = 0$ . Integrating by parts and using (3.20), we obtain
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$$
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\int_ {0} ^ {L} \frac {d v}{d x} E A \frac {d u}{d x} d x = \int_ {0} ^ {L} f ^ {B} v d x + R v \mid_ {x = L} \tag {3.33}
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$$
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This relation is an application of the Galerkin method or of the principle of virtual displacements and states that “for $u(x)$ to be the solution of the problem, the left-hand side of (3.33) (the internal virtual work) must be equal to the right-hand side (the external virtual work)
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for arbitrary test or virtual displacement functions $v(x)$ that are continuous and that satisfy the condition v = 0 at x = 0.
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In Chapter 4 we write the formulation (3.33) in the following form:
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$$
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\text { Find } u \in V \text { such that } ^ {6} \quad a (u, v) = (f, v) \quad \forall v \in V \tag {3.34}
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$$
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where the space V is defined as
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$$
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V = \left\{v \mid v \in L ^ {2} (L), \frac {d v}{d x} \in L ^ {2} (L), v \mid_ {x = 0} = 0 \right\} \tag {3.35}
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$$
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and $L^2(L)$ is the space of square integrable functions over the length of the bar, $0 \leq x \leq L$ ,
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$$
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L ^ {2} (L) = \left\{w \mid w \text { is defined over } 0 \leq x \leq L \text { and } \int_ {0} ^ {L} (w) ^ {2} d x = \| w \| _ {L ^ {2}} ^ {2} < \infty \right\} \tag {3.36}
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$$
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Using (3.34) and (3.33), we have
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$$
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a (u, v) = \int_ {0} ^ {L} \frac {d u}{d x} E A \frac {d v}{d x} d x \tag {3.37}
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$$
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and $(f,v) = \int_0^L f^B v dx + Rv|_{x = L}$ (3.38)
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where $a(u, v)$ is the bilinear form and $(f, v)$ is the linear form of the problem.
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The definition of the space of functions $V$ in (3.35) says that any element $v$ in $V$ is zero at $x = 0$ and
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$$
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\int_ {0} ^ {L} v ^ {2} d x < \infty ; \quad \int_ {0} ^ {L} \left[ \frac {d v}{d x} \right] ^ {2} d x < \infty
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$$
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Hence, any element v in V corresponds to a finite strain energy. We note that the elements in V comprise all functions that are candidates for solution of the differential formulation (3.18) to (3.20) with any continuous $f^{B}$ and also correspond to possible solutions with discontinuous strains [because of concentrated loads, in this one-dimensional analysis case, or discontinuities in the material behavior or cross-sectional area]. This observation underlines the generality of the problem formulation given in (3.34) and (3.35).
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For the Galerkin (or finite element) solution we define the space $V_{h}$ of trial (or finite element) functions $v_{h}$ ,
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$$
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V _ {h} = \left\{v _ {h} \mid v _ {h} \in L ^ {2} (L), \frac {d v _ {h}}{d x} \in L ^ {2} (L), v _ {h} \mid_ {S _ {u}} = 0 \right\} \tag {3.39}
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$$
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where $S_{u}$ denotes the surface area on which the zero displacement is prescribed. The subscript h denotes that a particular finite element discretization is being considered (and h actually refers to the size of the elements; see Section 4.3). The finite element formulation of the problem is then
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$$
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\text { Find } u _ {h} \in V _ {h} \text { such that } a (u _ {h}, v _ {h}) = (f, v _ {h}) \quad \forall v _ {h} \in V _ {h} \tag {3.40}
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$$
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Of course, (3.40) is the principle of virtual displacements applied with the functions contained in $V_{h}$ and also corresponds to the minimization of the total potential energy within this space of trial functions. Therefore, (3.40) corresponds to the use of the Ritz method
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described in Section 3.3.3. We discuss the finite element formulation extensively in Chapter 4.
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However, let us note here that the same solution approach can also be used directly for any analysis problem for which we have the governing differential equation(s). The procedure would be: weigh the governing differential equation(s) in the domain with suitable test function(s); integrate the resulting equation(s) with a transformation using integration by parts (or more generally the divergence theorem; see Example 4.2); and substitute the natural boundary conditions—as we did to find (3.33).
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We obtain in this way the principle of virtual displacements for the general analysis of solids and structures (see Example 4.2), the “principle of virtual temperatures” for the general heat flow and temperature analysis of solids (see Example 7.1), and the “principle of virtual velocities” for general fluid flow analysis (see Section 7.4.2).
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To demonstrate the use of the above notation, consider the following examples.
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EXAMPLE 3.25: Consider the analysis problem in Example 3.22. Write the problem formulation in the form (3.40) and identify the finite element basis functions used when employing the displacement assumptions (b) and (c) in the example.
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Here the bilinear form $a(.,.)$ is
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$$
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a (u _ {h}, v _ {h}) = \int_ {0} ^ {1 8 0} \frac {d u _ {h}}{d x} E A \frac {d v _ {h}}{d x} d x
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$$
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and we have the linear form
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$$
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(f, v _ {h}) = 1 0 0 v _ {h} \mid_ {x = 1 8 0}
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$$
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With the displacement assumption (b) we use
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$$
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u _ {h} = a _ {1} x + a _ {2} x ^ {2}
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$$
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Hence $V_{h}$ is a two-dimensional space, and the two basis functions are
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$$
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v _ {h} ^ {(1)} = x \quad \text { and } \quad v _ {h} ^ {(2)} = x ^ {2}
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$$
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With the displacement assumption (c) we use
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$$
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\begin{array}{l} u _ {h} = \frac {x}{1 0 0} u _ {B}; \quad 0 \leq x \leq 1 0 0 \\ u _ {h} = \left(1 - \frac {x - 1 0 0}{8 0}\right) u _ {B} + \frac {(x - 1 0 0)}{8 0} u _ {C}; \quad 1 0 0 \leq x \leq 1 8 0 \\ \end{array}
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$$
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and the two basis functions for $V_{h}$ are
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$$
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v _ {h} ^ {(1)} = \left\{ \begin{array}{c l} \frac {x}{1 0 0} & \text { for } 0 \leq x \leq 1 0 0 \\ 1 - \frac {x - 1 0 0}{8 0} & \text { for } 1 0 0 \leq x \leq 1 8 0 \end{array} \right.
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$$
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and $v_{h}^{(2)} = \frac{x - 100}{80}$ for $100 \leq x \leq 180$
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Clearly, all these functions satisfy the conditions in (3.39). If we use (3.40), the equations in (g) and (j) in Example 3.22 are generated.
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EXAMPLE 3.26: Consider the analysis problem in Example 3.23. Write the problem formulation in the form (3.40) and identify the element basis functions used when employing the temperature assumption given in the example.
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Here the problem formulation is
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$$
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\text { Find } \theta_ {h} \in V _ {h} \text { such that } a (\theta_ {h}, \psi_ {h}) = (f, \psi_ {h}) \quad \forall \psi_ {h} \in V _ {h} \tag {a}
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$$
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where $a(\theta_h,\psi_h) = \int_0^L\frac{d\psi_h}{dx} k\frac{d\theta_h}{dx} dx$
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$$
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(f, \psi_ {h}) = \int_ {0} ^ {L} \psi_ {h} q ^ {B} d x + q _ {0} \psi_ {h} | _ {x = 0}
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$$
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Here $\theta_{h}$ and $\psi_{h}$ correspond to temperature distributions in the slab. With the assumption in Example 3.23 we have for $V_{h}$ the three basis functions
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$$
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\theta_ {h} ^ {(1)} = 1; \quad \theta_ {h} ^ {(2)} = x; \quad \theta_ {h} ^ {(3)} = x ^ {2}
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$$
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Using (a) the governing equations given in (c) in Example 3.23 are obtained. Note that in this formulation we have not yet imposed the essential boundary condition (which is achieved later, as in Example 3.23).
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# 3.3.5 Finite Difference Differential and Energy Methods
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A classical approach to finding a numerical solution to the governing equations of a mathematical continuum model is to use finite differences (see, for example, L. Collatz [A]), and it is valuable to be familiar with this approach because such knowledge will reinforce our understanding of the finite element procedures. In a finite difference solution, the derivatives are replaced by finite difference approximations and the differential and variational formulations of mathematical models can be solved.
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As an example, consider the analysis of the uniform bar in Fig. 3.2 with the governing differential equation (see Example 3.17 and Section 3.3.4),
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$$
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u ^ {\prime \prime} + \frac {f ^ {B}}{E A} = 0 \tag {3.41}
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$$
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and the boundary conditions
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$$
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u = 0 \quad \text { at } x = 0 \tag {3.42}
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$$
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$$
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E A \frac {d u}{d x} = R \quad \text { at } x = L \tag {3.43}
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$$
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Using an equal spacing h between finite difference stations, we can write (see Fig. 3.3)
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$$
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u ^ {\prime} | _ {i + 1 / 2} = \frac {u _ {i + 1} - u _ {i}}{h}; \quad u ^ {\prime} | _ {i - 1 / 2} = \frac {u _ {i} - u _ {i - 1}}{h} \tag {3.44}
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$$
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and $u^{\prime \prime}|_{i} = \frac{u^{\prime}|_{i + 1 / 2} - u^{\prime}|_{i - 1 / 2}}{h}$ (3.45)
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so that $u^{\prime \prime}|_{i} = \frac{1}{h^{2}} (u_{i + 1} - 2u_{i} + u_{i - 1})$ (3.46)
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<!-- source-page: 147 -->
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<details>
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<summary>text_image</summary>
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Young's modulus E
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Cross-sectional area A
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x → f^B(x) → R
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(a) Bar to be analyzed,
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f^B(x) = ax
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</details>
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<details>
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<summary>text_image</summary>
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h/2
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h
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i-1
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i
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i+1
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i-½
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i+½
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h
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h
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</details>
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(b) Finite difference stations $i - 1, i, i + 1$ (locations $i - \frac{1}{2}, i + \frac{1}{2}$ are not stations)
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<details>
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<summary>text_image</summary>
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n-1
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n
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n+1
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h
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h
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</details>
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(c) Fictitious finite difference station $n + 1$ outside bar
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Figure 3.3 Finite difference analysis of a bar
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The relation in (3.46) is called the central difference approximation. If we substitute (3.46) into (3.41), we obtain
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$$
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\frac {E A}{h} \left(- u _ {i + 1} + 2 u _ {i} - u _ {i - 1}\right) = f _ {i} ^ {B} h \tag {3.47}
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$$
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where $f_{i}^{B}$ is the load $f^{B}(x)$ at station i and $f_{i}^{B}h$ can be thought of as the total load applied at that finite difference station.
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Assume now that we use a total of $n + 1$ finite difference stations on the bar, with station i = 0 at the fixed end and station i = n at the other end. Then the boundary conditions are
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$$
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u _ {0} = 0 \tag {3.48}
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$$
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and $EA\frac{u_{n + 1} - u_{n - 1}}{2h} = R$ (3.49)
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<!-- source-page: 148 -->
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where we have introduced the fictitious station $n + 1$ outside the bar [see Fig. 3.3(c)], merely to impose the boundary condition (3.43).
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For the finite difference solution we apply (3.47) at all stations $i = 1, \ldots, n$ and use the boundary conditions (3.48) and (3.49) to obtain
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$$
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\frac {E A}{h} \left[ \begin{array}{c c c c c c} 2 & - 1 & & & & \\ - 1 & 2 & - 1 & & & \\ & - 1 & 2 & - 1 & & \\ & & & \ddots & & \\ & & & - 1 & 2 & - 1 \\ & & & & - 1 & 1 \end{array} \right] \left[ \begin{array}{c} u _ {1} \\ u _ {2} \\ u _ {3} \\ \vdots \\ u _ {n - 1} \\ u _ {n} \end{array} \right] = \left[ \begin{array}{c} R _ {1} \\ R _ {2} \\ R _ {3} \\ \vdots \\ R _ {n - 1} \\ R _ {n} \end{array} \right] \tag {3.50}
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$$
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where $R_{i} = f_{i}^{B}h, i = 1, \ldots, n - 1$ , and $R_{n} = f_{n}^{B}h / 2 + R$ .
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We note that the equations in (3.50) are identical to the equations that would be obtained using a series of n spring elements, each of stiffness EA/h. The loads at the nodes corresponding to $f^{B}(x)$ would be obtained by using the distributed load value at node i and multiplying that value by the contributing length (h for the interior nodes and h/2 for the end node.)
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The same coefficient matrix is also obtained if we use the Ritz method with the variational formulation of the mathematical model and specific Ritz functions. The variational indicator is (see Example 3.19)
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$$
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\Pi = \frac {1}{2} \int_ {0} ^ {L} E A (u ^ {\prime}) ^ {2} d x - \int_ {0} ^ {L} u f ^ {B} d x - R u | _ {x = L} \tag {3.51}
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$$
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and the specific Ritz functions are depicted in Fig. 3.4. While the same coefficient matrix is obtained, the load vector is different unless the loading is constant along the length of the bar.
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<details>
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<summary>text_image</summary>
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Typical Ritz "hat" function
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1.0
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ξ
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i-1
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i+1
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h
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h
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</details>
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$$
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u (\xi) = \left\{ \begin{array}{l l} \left(1 - \frac {\xi}{h}\right) u _ {i} & \text { for } 0 \leq \xi \leq h \\ \left(1 + \frac {\xi}{h}\right) u _ {i} & \text { for } - h \leq \xi \leq 0 \end{array} \right.
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$$
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Figure 3.4 Typical Ritz function or Galerkin basis function used in analysis of bar problem
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<!-- source-page: 149 -->
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The same equations as in the Ritz solution are of course also obtained using the Galerkin method given in Section 3.3.4 (i.e., the principle of virtual work) with the basis functions in Fig. 3.4.
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The preceding discussion indicates that the finite difference method can also be used to generate stiffness matrices, and that in some cases the resulting equations obtained in a Ritz analysis and in a finite difference solution are identical or almost identical.
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Table 3.1 summarizes some widely used finite difference approximations, also called finite difference stencils or molecules. Let us demonstrate the use of these stencils in two examples.
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TABLE 3.1 Finite difference approximations for various differentiations
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<table><tr><td>Differentiation</td><td>Finite difference approximation</td><td>Molecules</td></tr><tr><td></td><td></td><td><img src="images/cf8132695da7818eb5c46b6235fe07a5f8ecfe391d52b70621bbcaba8a4a293e.jpg"/></td></tr><tr><td> $\frac{dw}{dx}\bigg|_{i}$ </td><td> $\frac{w_{i+1} - w_{i-1}}{2h}$ </td><td><img src="images/663fc8f8c6587b4090b87df94d009ede344207e3b83e17758039f069492585ab.jpg"/></td></tr><tr><td></td><td></td><td><img src="images/f222691b46850b7d81442a0aec52b0faac2b900325f463b1ba85016948687127.jpg"/></td></tr><tr><td> $\frac{d^{2}w}{dx^{2}}\bigg|_{i}$ </td><td> $\frac{w_{i+1} - 2w_{i} + w_{i-1}}{h^{2}}$ </td><td><img src="images/ef7aca0a03be2359d9efc460f306e4d7d3a5fd570e94a567cf36a893d4812b81.jpg"/></td></tr><tr><td> $\frac{d^{3}w}{dx^{3}}\bigg|_{i}$ </td><td> $\frac{w_{i+2} - 2w_{i+1} + 2w_{i-1} - w_{i-2}}{2h^{3}}$ </td><td><img src="images/03fc6365dcf49a0fc83ae10b686af00d619e06efd6c898ee833a80265d74e99f.jpg"/></td></tr><tr><td> $\frac{d^{4}w}{dx^{4}}\bigg|_{i}$ </td><td> $\frac{w_{i+2} - 4w_{i+1} + 6w_{i} - 4w_{i-1} + w_{i-2}}{h^{4}}$ </td><td><img src="images/e749dcd428bd1d2a9f7137195f51684ea501cfea511fe57e138caa2f5bbdbbef.jpg"/></td></tr><tr><td> $\nabla^{2}w|_{i,j}$ </td><td> $\frac{-4w_{i,j} + w_{i+1,j} + w_{i,j+1} + w_{i-1,j} + w_{i,j-1}}{h^{2}}$ </td><td><img src="images/49fca1883e03f3008cd38421aeea2e77607ccd221a0070cc3f3faf6dc8c8e189.jpg"/></td></tr><tr><td> $\nabla^{4}w|_{i,j}$ </td><td> $[20w_{i,j} - 8(w_{i+1,j} + w_{i-1,j} + w_{i,j+1} + w_{i,j-1}) + 2(w_{i+1,j+1} + w_{i-1,j-1} + w_{i+1,j-1}) + w_{i+2,j} + w_{i-2,j} + w_{i,j+2} w_{i,j-2}]/h^{4}$ </td><td><img src="images/0bcb41f9f7a3176f8719dff2d868ac59059355b0eda0217995ffe27f8d1e7d74.jpg"/></td></tr></table>
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Uniform spacing h; error in each case is $o(h^{2})$ . Point i or $(i,j)$ is being considered; and $i \pm \cdots$ denotes points in the x-direction; $j \pm \cdots$ denotes points in the y-direction.
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EXAMPLE 3.27: Consider the simply supported beam in Fig. E3.27. Use conventional finite differencing to establish the system equilibrium equations.
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The finite difference grid used for the beam analysis is shown in the figure. In the conventional finite difference analysis the differential equation of equilibrium and the geometric and natural boundary conditions are considered; i.e., we approximate by finite differences at each interior station,
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$$
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E I \frac {d ^ {4} w}{d x ^ {4}} = q \tag {a}
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$$
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and use the conditions that $w = 0$ and $w'' = 0$ at $x = 0$ and $x = L$ .
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<details>
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<summary>text_image</summary>
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Flexural
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rigidity EI
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R1 R2 R3 R4
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i = -1 i = 0 w1 w2 w3 w4 i = 5 i = 6
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L/5
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</details>
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Figure E3.27 Finite difference stations for simply supported beam
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Using central differencing, (a) is approximated at station i by
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$$
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\frac {E I}{(L / 5) ^ {3}} \{w _ {i - 2} - 4 w _ {i - 1} + 6 w _ {i} - 4 w _ {i + 1} + w _ {i + 2} \} = R _ {i} \tag {b}
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$$
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where $R_{i} = q_{i}L/5$ and is the concentrated load applied at station i. The condition that $w''$ is zero at station i is approximated using
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$$
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w _ {i - 1} - 2 w _ {i} + w _ {i + 1} = 0 \tag {c}
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$$
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Applying (b) at each finite difference station, i = 1, 2, 3, 4, and using condition (c) at the support points, we obtain the system of equations
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$$
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\frac {1 2 5 E I}{L ^ {3}} \left[ \begin{array}{r r r r} 5 & - 4 & 1 & 0 \\ - 4 & 6 & - 4 & 1 \\ 1 & - 4 & 6 & - 4 \\ 0 & 1 & - 4 & 5 \end{array} \right] \left[ \begin{array}{l} w _ {1} \\ w _ {2} \\ w _ {3} \\ w _ {4} \end{array} \right] = \left[ \begin{array}{l} R _ {1} \\ R _ {2} \\ R _ {3} \\ R _ {4} \end{array} \right]
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$$
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where the coefficient matrix of the displacement vector can be regarded as a stiffness matrix.
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# EXAMPLE 3.28: Consider the plate shown in Fig. E3.28.
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1. Calculate the center point transverse deflection when the plate is uniformly loaded under static conditions with the distributed load p per unit area. Use only one finite difference station in the interior of the plate.
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2. If the load $p$ is applied dynamically, i.e., $p = p(t)$ , establish an equation of motion governing the behavior of the plate.
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