# 421 let us illustrate the e ect of this equilibrated

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Unformatted text preview: . p 1=h ( L) ( R) () With Without With Without With Without Balancing Balancing Balancing Balancing Balancing Balancing 1 32 1.135 0.506 0.879 1.429 1.017 1.049 1 64 1.118 0.498 0.888 1.443 1.012 1.044 2 32 1.162 0.578 0.835 1.175 1.008 0.921 Table 7.4.3: Local and global e ectivity indices for Example 7.4.4 using (7.4.21) with and without equilibration. The essential boundary condition u(r ) = r1=2 cos =2 is prescribed on all boundaries except x > 0, y = 0. Thus, the solution of the Galerkin problem will satisfy the natural boundary condition uy = 0 there. These conditions have been chosen so that the exact solution is the speci ed essential boundary condition. This solution is singular since ur r;1=2 near the origin (r = 0). Results for the e ectivity indices in strain energy for the entire region and for the two elements, L and R , adjacent to the singularity are shown in Table 7.4.3. Computations were performed on a square grid with uniform spacing h in each coordinate direction (Figure 7.4.4). Piecewise line...
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