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Element equations. Because this system is so simple, its element equations can be written directly without recourse to mathematical approximations. This is an example of the direct approach for deriving elements. Figure 32.8 shows an individual element. The relationship between force F and displacement x can be represented mathematically by Hooke's law: F = kx where k = the spring constant, which can be interpreted as the force required to cause a unit displacement. If a force F, is applied at node 1, the following force balance must hold: F = k(x1 - X2) where x1 = displacement of node 1 from its equilibrium position and x2 = displacement of node 2 from its equilibrium position. Thus, X2 - x1 represents how much the spring is elongated or compressed relative to equilibrium (Fig. 32.8). This equation can also be written as F1 = kx - kx2 For a stationary system, a force balance also necessitates that F1 = -F2 and, therefore, F2 = -kx + kx2 These two simultaneous equations specify the behavior of the element in response to prescribed forces. They can be written in matrix form as or (x) = (F) (32.18) where the matrix is the element property matrix. For this case, it is also referred to as the element stiffness matrix. Notice that Eq. (32.18) has been cast in the format of Eq. (31.9). Thus, we have succeeded in generating a matrix equation that describes the behavior of a typical element in our system. Before proceeding to the next step-the assembly of the total solution-we will introduce some notation. The elements of and { F) are conventionally superscripted and subscripted, as in where the superscript (e) designates that these are the element equations. The k's are also subscripted as k; to denote their location in the ith row and jth column of the matrix. For the present case, they can also be physically interpreted as representing the force required at node i to induce a unit displacement at node j. Assembly. Before the element equations are assembled, all the elements and nodes must be numbered. This global numbering scheme specifies a system configuration or topology (note that the present case uses a scheme identical to Table 31.1). That is, it documents which nodes belong to which element. Once the topology is specified, the equations for each element can be written with reference to the global coordinates. The element equations can then be added one at a time to assemble the total system. The final result can be expressed in matrix form as [recall Eq. (31.10)] (x') = (F') where - k3 [k] = (32.19) - k$3) -KB - 15 and (F'] = and {x'] and {F'] are the expanded displacement and force vectors, respectively. Notice that, as the equations were assembled, the internal forces cancel. Thus, the final result for {F') has zeros for all but the first and last nodes. Before proceeding to the next step, we must comment on the structure of the assem- blage property matrix [Eq. (32.19)]. Notice that the matrix is tridiagonal. This is a direct result of the particular global numbering scheme that was chosen (Table 31.1) prior to assemblage. Although it is not very important in the present context, the attainment of such a banded, sparse system can be a decided advantage for more complicated problem settings. This is due to the efficient schemes that are available for solving such systems. Boundary Conditions. The present system is subject to a single boundary condition, x1 = 0. Introduction of this condition and applying the global renumbering scheme re- duces the system to (k's = 1) 2 The system is now in the form of Eq. (31.11) and is ready to be solved. Although reduction of the equations is certainly a valid approach for incorporating boundary conditions, it is usually preferable to leave the number of equations intact when performing the solution on the computer. Whatever the method, once the boundary con- ditions are incorporated, we can proceed to the next step-the solution. Generating Solution. Using one of the approaches from Part Three, such as the effi- cient tridiagonal solution technique delineated in Chap. 11, the system can be solved for (with all k's = 1 and F = 1) 12 = 1 X3 = 2 X4 = 3 X's = 4 Postprocessing. The results can now be displayed graphically. As in Fig. 32.9, the results are as expected. Each spring is elongated a unit displacement.Chemical/Bio Engineering 1 Perform the same computation as in Sec. 32.1, but use Ax = 1.25. 2 Develop a finite-element solution for the steady-state system of Sec. 32.1. 3 Compute mass fluxes for the steady-state solution of Sec. 32.1 using Fick's first law. 4 Compute the steady-state distribution of concentration for the tank shown in Fig. P32.4. The PDE governing this system is D kc = 0 and the boundary conditions are as shown. Employ a value of 0.5 for D and 0.1 for k. 5 Two plates are 10 cm apart, as shown in Fig. P32.5. Initially, both plates and the fluid are still. At / = 0, the top plate is moved at a constant velocity of 8 cm/s. The equations governing the motions of the fluids are Vail a Vall d water = Hail and ar = water ar and the following relationships hold true at the oil-water interface: Dall = Vener and Poll ax ax What is the velocity of the two fluid layers at / = 0.5, 1, and 1.5 s at distances x = 2, 4, 6, and 8 cm from the bottom plate? Note that Awater and pan = 1 and 3 cp, respectively