Fundamentals of Finite Element Analysis phần 1 pdf

... Analysis 1. Basic Concepts of the Finite Element Method Text © The McGraw−Hill Companies, 2004 1. 5 Examples of Finite Element Analysis 15 0.0 019 7Љ Z X 0.25Љ 0.488Љ Figure 1. 9 Finite element model of ... −k 2 k 2  0 U 2 U 3  =      0 f (2) 2 f (2) 3      (2 .12 ) The addition of Equations 2 .11 and 2 .12 yields  k 1 −k 1 0 −k 1 k 1 +k 2 −...

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... 38.735 38.4 16 12 3.63 12 2 .17 40.533 40. 510 20 14 2.48 13 7.40 44.903 42. 914 21 100.03 99.37 47 .10 9 45. 215 22 67 .10 64.67 51. 535 49 .12 1 23 40.55 39.36 57.836 55.499 24 18 .98 18 .28 68 .14 2 65.425 45 ... 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 31 32 38 39 34 36 43 40 44 45 37 33 26 50...

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... 3.75 (10 5 )     0000 010 1 0000 0 10 1      K (2)  = 2.65 (10 5 ) 2     11 1 1 11 1 1 1 11 1 1 11 1      K (6)  = 2.65 (10 5 ) 2     1 1 11 11 1 1 11 1 1 1 1 11     Step ... 5 3 9 7 21 19 17 15 13 11 10 14 18 3 1 5 7 9 11 4 81 12 16 20 2 4 6 8 10 12 (e) 2 (3, 4) (0, 0) 1 (d) 2 (0, 1. 2) (Ϫ0.5, 0) 1 (c) 2 (1, 2)...

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... 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 −2.944 10 6 2.944 10 8 0 50 0 19 ,627.2 −2.944 10 6 19 ,627.2 −2.944 10 6 0 6 0 0 2.944 10 6 2.944 10 8 −2.944 ... Displacement Scheme Global Figure 4 .10 b Element 1 Element 2 Element 3 1 U 1 v (1) 1 00 2 U 2 ␪ (1) 1 00 3 U 3 v (1) 2 v (2) 1 u (3) 1 4 U 4 ␪ (1) 2 ␪ (2) 1...

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... in terms of the natural coordinates and request that the reader (a) x y z 12 6 78 4 5 2a 2b 2c 3 (b) (1, 1, 1) (1, 1, 1) ( 1, 1, 1) (1, 1, 1) ( 1, 1, 1) ( 1, 1, 1) s t r Figure 6 .19 Eight-node ... elements 3 and 4, [ k cu ] = k x A L  1 1 11  = 389(␲/4)(0.06) 2 0.25  1 1 11  = 4.40  1 1 11  W/ ◦ C Applying the end conditions T 5 = 80°C and q 1 = 4000 W/m...

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... 7 .1 Nodal Temperature Solutions Four Elements, Eight Elements, x (inches) T ( ◦ F) T ( ◦ F) 0 18 0 18 0 0.5 15 8.08* 15 5. 31 1.0 13 6 .16 13 6.48 1. 5 12 3.59* 12 2 .19 2.0 11 1.02 11 1. 41 2.5 10 4 .13 * 10 3. 41 3.0 ... computed by back substitution of T 2 into the first equation: 2 .14 48 (18 0) − 1. 87 21( 136 .16 ) = 19 .6375 + Aq 1 Aq 1 = 11 1. 515 6 Btu/hr...

Ngày tải lên: 08/08/2014, 17:20

... area of the element, as all other terms are known constants. For example, 1  1 1  1 N 1 dr ds = 1  1 1  1 1 4 (1 − r ) (1 − s)dr ds = 1 4 (1 − r ) 2 2     1 1 (1 − s) 2 2     1 1 = ... then written as   4.5804 1. 14 51 0 1. 14 51 4.5804 1. 14 51 01. 14 51 4.5804      ˙ T 2 ˙ T 3 ˙ T 4    +   1. 8 816 −0.9408 0 −0.9408 1. 8 816 −...

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... Node 10 7 of Example 9.4 ␴ x ␴ y ␶ xy Element 1 2049.3 18 7.36 11 8.4 Element 2 214 9.4 315 .59 91. 89 Element 12 19 87.3 322.72 204 .13 Element 99 18 53.8 18 6.88 378.36 Average 2009.8 253 .14 19 8 .19 and ... two-dimensional isoparametric element. (b) The parent element in natural coordinates. 4 3 2 1 y x v 4 u 4 v 3 u 3 v 2 u 2 v 1 u 1 (a) r s (1, 1) 3 ( 1,...

Ngày tải lên: 08/08/2014, 17:20

... example, m 11 = 15 0(5)␳ 1  1 1  1 N 2 1 dr ds = 15 0(5) 16 ␳ 1  1 1  1 (1 − r) 2 (1 − s) 2 dr ds = 15 0(5) 16 ␳  (1 − r) 3 3 (1 − s) 3 3  1 1 = 750 16 ␳  64 9  = 4(750) 9 (7.830) (10 ) −6 = ... 2.6 (10 ) −3 kg Similarly, m 12 = 15 0(5)␳ 1  1 1  1 N 1 N 2 dr ds = 15 0(5) 16 ␳ 1  1 1  1 (1 − r 2 ) (1 − s) 2 dr ds = 15 0(5)...

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... are as follows: xy 10 0 218 0 311 14 4 918 52 718 62 914 736 0 6 ft 6 ft 15 00 lb 12 00 lb 2000 lb 10 00 lb 1 2 12 ft 12 ft 8 ft 4 ft 5 7 4 36 Hutton: Fundamentals of Finite Element Analysis Back Matter ... a 11 as the pivot element, we can multiply the first row by a 21 /a 11 and subtract the result from the second row to obtain a (1) 21 = a 21 − a 11 a...

Ngày tải lên: 08/08/2014, 17:20