Fundamentals of Finite Element Analysis phần 8 pptx

... Node 107 of Example 9.4 ␴ x ␴ y ␶ xy Element 1 2049.3 187 .36 1 18. 4 Element 2 2149.4 315.59 91 .89 Element 12 1 987 .3 322.72 204.13 Element 99 185 3 .8 186 .88 3 78. 36 Average 2009 .8 253.14 1 98. 19 and ... 9.4 ␴ x ␴ y ␶ xy ␴ e Element 1 2553.5 209.71 179 .87 2475 .8 Element 2 1922.7 351.69 43.55 1774 .8 Element 12 182 7.5 264.42 154.44 1731.5 Element 99 2 189 .0 24...

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... 2.53 08 −0.2006 −0.2 585 −0.2006 −0.2 585 0 −0.2 585 −0.0713 0 −0.2006 1. 381 2 0 −0.2 585 −0.0713 000−0.0713 −0.2 585 0 0.7 484 −0.2 585 0 000−0.2 585 −0.2006 −0.2 585 −0.2 585 1. 381 2 −0.0713 0000−0.2 585 −0.0713 ... 0.2(60)(0.523)(40.9 28) = 25 68. 6W The results show that only about 18 percent of input heat is removed, so the cooler is not very efficient. Hutton: Fundamentals of...

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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 0.25Љ 0. 488 Љ Figure 1.9 Finite element model of a thin-walled heat ... the domain. (c) Additional elements showing a partial finite element mesh of the domain. Hutton: Fundamentals of Finite Element Analysis 1. Basic Con...

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... 0 (3.1b) Hutton: Fundamentals of Finite Element Analysis 2. Stiffness Matrices, Spring and Bar Elements Text © The McGraw−Hill Companies, 2004 48 CHAPTER 2 Stiffness Matrices, Spring and Bar Elements 3. ... which form the finite element approximation to the solution of the differential equation(s) gov- erning the response of a structural system. Hutton: Fundamentals of...

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... ×10 8 −2.944 ×10 6 5 .88 9 ×10 8 0 70 0 −27,096 0 0 0 27,096 so the element stiffness matrices are (per Equation 4. 48)  k (1)  =  k (2)  = 1,635.6     12 1 ,80 0 −12 1 ,80 0 1 ,80 0 360,000 −1 ,80 0 ... 2.944 ×10 6 5 .88 8 ×10 8 −2.944 ×10 6 2.944 ×10 8 000 3 −19,627.2 −2.944 ×10 6 66,350.4 0 −19,627.2 2.944 ×10 6 −27,096 4 2.944 ×10 6 2.944 ×10 8 0 11. 78 ×10 8 −2.94...

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... straining of the element occurs. The first Hutton: Fundamentals of Finite Element Analysis 6. Interpolation Functions for General Element Formulation Text © The McGraw−Hill Companies, 2004 1 78 CHAPTER ... Element Application of the Galerkin method to the beam element begins with consid- eration of the equilibrium conditions of a differential section taken along t...

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... include the effect of the specified tem- perature at node 1 on the remaining equations gives     4. 289 6 −1 .85 71 0 0 −1 .87 21 4. 289 6 −1 .87 21 0 0 −1 .87 21 4. 289 6 −1 .87 21 00−1 .87 21 2. 281 2            T 2 T 3 T 4 T 5        =        376.2530 39.2750 39.2750 25.0917        Solving ... quadrilateral elements. Hutton: Fundamentals of Finite...

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... 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 31 32 38 39 34 36 43 40 44 45 37 33 26 50 52 30 29 28 91011121 38 15 16 14 18 19 ␺ ϭ 200 ␺ ϭ 0 (a) 309 Hutton: Fundamentals of Finite Element Analysis 8. Applications in Fluid Mechanics Text © The McGraw−Hill Companies, 2004 3 18 CHAPTER 8 ... representation of the displacement field, the ele...

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... of the calculations as given, the sum of all terms in the rectangular element mass matrix in Example 10.6 is twice the total mass of the element. Why? 10. 18 What are the values of the terms of ... diagonal matrix. Hutton: Fundamentals of Finite Element Analysis 10. Structural Dynamics Text © The McGraw−Hill Companies, 2004 10.5 Beam Elements 409 As in the case of t...

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... 285 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, ... ft 9 ft 18 ft 18 ft 18 ft 18 ft 9 ft12 ft 700 lb 400 lb 3 1 25 8 4 6 7 600 lb 3 m 9 m 3 m 3 m 4 m 4 m 4 m 5 kN 9 kN 9 kN 5 kN 12 4 5 7 9 3 681 0 Hutton: Fundamental...

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