Fundamentals of Finite Element Analysis phần 5 docx

... 10-node triangular element shown in Figure P6.13. Note that the nodes are equally spaced on their edges. 1 2 3 4 5 x x 1 ϭ 2 x 2 ϭ 2. 25 x 3 ϭ 2 .5 x 4 ϭ 2. 75 x 5 ϭ 3 Hutton: Fundamentals of Finite Element Analysis 6. ... Solutions Four Elements, Eight Elements, x (inches) T ( ◦ F) T ( ◦ F) 0 180 180 0 .5 158 .08* 155 .31 1.0 136.16 136.48 1 .5 123 .59 * 122.19 2.0 111.0...

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... Basic Concepts of the Finite Element Method Text © The McGraw−Hill Companies, 2004 1 .5 Examples of Finite Element Analysis 15 0.00197Љ Z X 0. 25 0.488Љ Figure 1.9 Finite element model of a thin-walled heat ... solutions. 36 Hutton: Fundamentals of Finite Element Analysis 1. Basic Concepts of the Finite Element Method Text © The McGraw−Hill Companies,...

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... are 10 5         1.3 25 1.3 25 0 0 −1.3 25 −1.3 25 1.3 25 1.3 25 0 0 −1.3 25 −1.3 25 003. 75 0 −3. 75 0 000000 −1.3 25 −1.3 25 −3. 75 0 5. 0 75 1.3 25 −1.3 25 −1.3 25 0 0 1.3 25 1.3 25         ۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙۙ ۘ ۘ ۘ ۘ ۘ ۘ ۘ                0 0 0 0 U 5 U 6                ۙ =                F 1 F 2 F 3 F 4...

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... resulting 5 × 5 system of equations, we obtain the results ␪ 1 = 9.3638(10 −4 ) rad v 2 =−0.73811 mm ␪ 2 =−0.009 253 8 rad v 3 = 5. 552 3 mm ␪ 3 =−0.019444 rad Hutton: Fundamentals of Finite Element Analysis 4. ... shown, determine (b) 6 in. 6 in. 0.8 75 in. t ϭ 0. 25 in. 10 lb 0.6 25 in. (a) 12 in. 1 in. 0 .5 in. t ϭ 0. 25 in. 10 lb Hutton: Fundamentals of Finite...

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... node. EXAMPLE 5. 5 Hutton: Fundamentals of Finite Element Analysis 5. Method of Weighted Residuals Text © The McGraw−Hill Companies, 2004 5. 4 Application of Galerkin’s Method to Structural Elements 151 Galerkin’s ... 04.40            T 1 T 2 T 3 T 4        =        11.31 11.31 11.31 363.31        Hutton: Fundamentals of Finite Elemen...

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... −0. 258 5 0 0 0 0 −0.1003 0.6906 0 −0. 258 5 −0.0713 0 0 0 −0.0713 −0. 258 5 0 1.3812 −0.2006 0 −0.0713 −0. 258 5 0 −0. 258 5 −0.2006 −0. 258 5 −0.2006 2 .53 08 −0.2006 −0. 258 5 −0.2006 −0. 258 5 0 −0. 258 5 −0.0713 ... −0. 258 5 0 −0. 258 5 −0.0713 0 −0.2006 1.3812 0 −0. 258 5 −0.0713 000−0.0713 −0. 258 5 0 0.7484 −0. 258 5 0 000−0. 258 5 −0.2006 −0. 258 5 −0. 258 5 1.3812 −0.0713 0000−0. 2...

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... 99.37 47.109 45. 2 15 22 67.10 64.67 51 .53 5 49.121 23 40 .55 39.36 57 .836 55 .499 24 18.98 18.28 68.142 65. 4 25 45 67.88 65. 89 41.706 40.799 46 103.87 100.74 42. 359 41.018 This solution is actually ... solution; 40 elements. (b) Streamlines ( ␺ = constant) for finite element solution of Example 8.1. (b) 6 7 2 5 4 3 1 24 23 22 27 42 46 35 41 51 48 47 49 53 21 20 17 25...

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... N 3 ∂z = 25 2A  k (2)  = 1 4A    0 25 − 25    [ 0 25 25 ] + 1 4A    − 25 0 25    [ − 25 0 25 ] [k (2) ] = 1 1 250   00 0 0 6 25 −6 25 0 −6 25 6 25   + 1 1 250   6 25 0 −6 25 000 −6 25 0 6 25   =   0 .50 −0 .5 00 .5 −0 .5 −0 .5 ... Solid Mechanics or  k (1)  = 1 1 250   6 25 −6 25 0 −6 25 6 25 0 000   + 1 1 250   00 0 0 6 25 −6 25...

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... = 1 √ m   0.2 956 0. 657 5 0.6930 0.4289 0 .56 18 −0.7124 0.6064 −0. 355 0 0.0782   To verify Equation 10.109, we form the triple product [ A ] T [M][A] = 1 m   0.2 956 0.4289 0.6064 0. 657 5 0 .56 18 −0. 355 0 0.6930 ... = k m   0.2 956 0.4289 0.6064 0. 657 5 0 .56 18 −0. 355 0 0.6990 −0.7124 0.0782     3 −20 −23−1 0 −11   ×   0.2 956 0. 657 5 0.6990 0.4289 0 .56 18 −0.7124 0...

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... 2 85 what is it, 279 Finite element, 2, 12 Finite element analysis (FEA). See Finite element method (FEM) Finite element formulation axisymmetric heat transfer, 273–276 axisymmetric stress analysis, ... loading? ␴ 0 ␴ 0 6 in. 3 in. 0 .5 in. 150 lb 50 lb/in. 15 in. 40 in. 10,000 N 3 m 9 m Hutton: Fundamentals of Finite Element Analysis Back Matter Appendix D:...

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