<p>Chapter 04.08 Gauss-Seidel – More Examples Civil Engineering</p><p>Example 1 To find the maximum stresses in a compound cylinder, the following four simultaneous linear equations need to be solved. 7 5 3 4.2857 10  9.2307 10 0 0 c1   7.887 10   7 5 7 5     4.2857 10  5.4619 10  4.2857 10 5.4619 10 c 0   2       6.5  0.15384 6.5 0.15384 c3   0.007   7 5      0 0 4.2857 10  3.6057 10 c4   0  In the compound cylinder, the inner cylinder has an internal radius of a  5", and an outer radius c  6.5" , while the outer cylinder has an internal radius of c  6.5" and an outer radius of b  8" . Given E  30 106 psi,   0.3, and that the hoop stress in the outer cylinder is given by E  1     2 c3 1  c4  2  , 1   r  find the stress on the inside radius of the outer cylinder.</p><p>Find the values of c1 , c2 , c3 and c4 using the Gauss-Seidel Method. Use</p><p>c1   0.005 c   0.001   2     c3  0.0002     c 0.03  4    as the initial guess and conduct two iterations. Solution Rewriting the equations gives  7.887 103   9.2307 105 c  0c  0c c  2 3 4 1 4.2857 107 0  4.2857 107 c   4.2857 107 c  5.4619105 c c  1 3 4 2  5.4619105 0.007   6.5c   0.15384c  0.15384c c  1 2 4 3 6.5</p><p>04.08.1 04.08.2 Chapter 04.08</p><p>0  0c  0c  4.2857 107 c c  1 2 3 4  3.6057 105 Iteration #1 Given the initial guess of the solution vector as</p><p>c1   0.005 c   0.001   2     c3  0.0002      c4   0.03  we get  7.887 103   9.2307 105  0.001 c    1 4.2857 107  1.6249104 0  4.2857 107  1.6249104   4.2857 107  0.0002  5.4619105  0.03 c  2  5.4619105  1.5569103 0.007   6.5 1.6249104   0.153841.5569103  0.15384 0.03 c  3 6.5  2.4125104 0  4.2857 107  2.4125104 c  4  3.6057 105  2.8675102</p><p>The absolute relative approximate error for each ci then is 1.6249104   0.005   100 a 1 1.6249104  2977.1% 1.5569103  0.001   100 a 2 1.5569103  35.770% 2.4125104  0.002   100 a 3 2.4125104  17.098% 2.8675102  0.03   100 a 4 2.8675102  4.6223% At the end of the first iteration, the estimate of the solution vector is Gauss-Seidel Method – More Examples: Civil Engineering 04.08.3</p><p>4 c1  1.624910     3  c 1.556910   2     4 c3   2.412510     2  c4   2.867510  and the maximum absolute relative approximate error is 2977.1% . Iteration #2 The estimate of the solution vector at the end of Iteration #1 is 4 c1  1.624910     3  c 1.556910   2     4 c3   2.412510     2  c4   2.867510  Now we get  7.887 103   9.2307 105 1.5569 103 c    1 4.2857 107  1.5050104 0  4.2857 107  1.5050104   4.2857 107  2.4125104         5.4619105  2.8675102  c    2  5.4619105 3  2.063910 0.007   6.5 1.5050104   0.15384 2.0639103       0.15384 2.8675102  c    3 6.5 4  1.989210 0  4.2857 107 1.9892104 c4  5  3.6057 10 2  2.364310 The absolute relative approximate error for each ci then is 1.5050104  1.6249104    100 a 1 1.5050104  7.9702%  2.0639103 1.5569103   100 a 2  2.0639103  175.44% 1.9892104  2.4125104   100 a 3 1.9892104  21.281% 04.08.4 Chapter 04.08</p><p>2.3643102  2.8675102   100 a 4 2.3643102  21.281% At the end of the second iteration, the estimate of the solution vector is 4 c1  1.505010     3  c  2.063910  2     4 c3   1.989210     2  c4   2.364310  and the maximum absolute relative approximate error is 175.44% . At the end of the second iteration the stress on the inside radius of the outer cylinder is calculated E  1     2 c3 1  c4  2  1   r  30106  1 0.3   1.9892104 1 0.3  2.3643102   2     2  1 0.3   6.5   21439psi Conducting more iterations gives the following values for the solution vector and the corresponding absolute relative approximate errors.  %  % Iteration c1 a 1 c2 a 2 4 3 1 1.624910 2977.1 1.556910 35.770 4 3 2 1.505010 7.9702  2.063910 175.44 3  2.284810 4 34.132  9.893110 3 79.138 4  3.9711104 42.464  2.8949102 65.826 5  8.0755104 50.825  6.9799102 58.524 6 3 52.142 1 58.978 1.687410 1.701510</p><p> c  %  % Iteration 3 a 3 c4 a 4 4 2 1 2.412510 17.098 2.867510 4.6223 4 2 2 1.989210 21.281 2.364310 21.281 3 5.471610 5 263.55 6.5035103 263.55 4 1.5927 104 134.35 1.8931102 134.35 5  9.3454104 82.957 1.1108101 82.957 6 3 53.472 1 53.472  2.008510  2.387310 After six iterations, the absolute relative approximate errors are not decreasing. In fact, conducting more iterations reveals that the absolute relative approximate error does not approach zero or converge to any other number. Gauss-Seidel Method – More Examples: Civil Engineering 04.08.5</p><p>SIMULTANEOUS LINEAR EQUATIONS Topic Gauss-Seidel Method – More Examples Summary Examples of the Gauss-Seidel method Major Civil Engineering Authors Autar Kaw Date 2018年4月9日 Web Site http://numericalmethods.eng.usf.edu</p>