Fundamentals of Finite Element Analysis phần 2 pps

... K 1, 12 = 0 K 22 = k (1) 22 + k (2) 22 = 0 + (2. 65 /2) 10 5 K 23 = K 24 = 0 K 25 = k (1) 23 = 0 K 26 = k (1) 24 = 0 K 27 = k (2) 23 =− (2. 65 /2) 10 5 K 28 = k (2) 24 =− (2. 65 /2) 10 5 K 29 = K 2, 10 = K 2, 11 = ... k 1 ( U 1 − U 2 ) ∂U e ∂U 2 = F 2 = k 1 ( U 2 − U 1 ) + 2k 2 ( U 3 − U 2 )( −1 ) =−k 1 U 1 + ( k 1 + 2k 2 ) U 2 − 2k 2 U 3 ∂U...

Ngày tải lên: 08/08/2014, 17:20

... by  k (1)  =  k (2)  = EI z (L /2) 3     12 6L /2 − 12 6L /2 6L /24 L 2 /4 −6L /22 L 2 /4 − 12 −6L /21 2−6L /2 6L /22 L 2 /4 −6L /24 L 2 /4     = 8EI z L 3     12 3L − 12 3L 3LL 2 −3LL 2 /2 − 12 −3L 12 −3L 3LL 2 /2 −3LL 2     (again ... 24 L −96 24 L 00 24 L 8L 2 24 L 4L 2 00 −96 24 L 1 92 0 −96 24 L 24 L 4L 2 016L 2 24 L 4L...

Ngày tải lên: 08/08/2014, 17:20

... premise 1 x x 2 x 3 y 3 z 3 z 2 xz xz 2 x 2 z xy xyz xy 2 x 2 y yz y 2 z y 2 yz 2 y z Figure 6.6 Pascal “pyramid” for polynomials in three dimensions. Hutton: Fundamentals of Finite Element Analysis 6. ... 5 .26 yields x 2  x 1 dN 1 dx  y 1 dN 1 dx + y 2 dN 2 dx  dx = x 2  x 1 N 1 (x ) f (x)dx + dy (e) dx     x 1 (5 .27 a) x 2  x 1 dN 2 dx  y...

Ngày tải lên: 08/08/2014, 17:20

... AL 420  156 22 L 22 L 4L 2  ¨v 2 ¨ ␪ 2  + EI z L 3  12 −6L −6L 4L 2  v 2 ␪ 2  =  0 0  For computational convenience, the equations are rewritten as  156 22 L 22 L 4L 2  ¨v 2 ¨ ␪ 2  + 420 ... ␻ f t m x 2 (t ) =  0.17 72 ␻ 2 1 − ␻ 2 f + 0.3156 ␻ 2 2 − ␻ 2 f + 0.5075 ␻ 2 3 − ␻ 2 f  F 0 sin ␻ f t m x 3 (t ) =  0 .25 52 ␻ 2 1 − ␻ 2 f +...

Ngày tải lên: 08/08/2014, 17:20

... k 2 −k 2 0 −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 −k 2 0 −k 2 k 2  U 1 U 2 U 3  =      f (1) 1 f (1) 2 + ... 3k     − F 4k + 3 4 ␦ ␦    =        − 3F 4k − 3 4 k␦ 3F 4k + 3 4 k␦        =  f (2) 2 f (2) 3  which also v...

Ngày tải lên: 08/08/2014, 17:20

... 2, the integral becomes 2  A L 1 L 2 r dr dz = 2  A L 1 L 2 ( L 1 r 1 + L 2 r 2 + L 3 r 3 )dA = 2 r 1  A L 2 1 L 2 d A + 2 r 2  A L 1 L 2 2 d A + 2 r 3  A L 1 L 2 L 3 d A Applying the integration ... General Element Formulation or I = 16␲  A  L 2 1 L 3 2 r 1 + L 1 L 4 2 r 2 + L 1 L 3 2 L 3 r 3  d A − 8␲  A  L 2 1 L 2 2 r 1 + L 1 L 3 2 r...

Ngày tải lên: 08/08/2014, 17:20

... −0 .20 06 −0 .25 85 0 0 0 0 −0.1003 0.6906 0 −0 .25 85 −0.0713 0 0 0 −0.0713 −0 .25 85 0 1.38 12 −0 .20 06 0 −0.0713 −0 .25 85 0 −0 .25 85 −0 .20 06 −0 .25 85 −0 .20 06 2. 5308 −0 .20 06 −0 .25 85 −0 .20 06 −0 .25 85 0 −0 .25 85 ... to obtain [C] −1 =   0 .23 39 −0.0 624 0.0156 −0.0 624 0 .24 95 −0.0 624 0.0156 −0.0 624 0 .23 39   Hutton: Fundamentals of Finite Element Analysi...

Ngày tải lên: 08/08/2014, 17:20

... y 1 ) = ␤ 2 2 A ∂ N 3 ∂ x = 1 2 A ( y 1 − y 2 ) = ␤ 3 2 A (9 .29 ) ∂ N 1 ∂ y = 1 2 A (x 3 − x 2 ) = ␥ 1 2 A ∂ N 2 ∂ y = 1 2 A (x 1 − x 3 ) = ␥ 2 2 A ∂ N 3 ∂ y = 1 2 A (x 2 − x 1 ) = ␥ 3 2 A The [B] ... 38.4 16 123 .63 122 .17 40.533 40.510 20 1 42. 48 137.40 44.903 42. 914 21 100.03 99.37 47.109 45 .21 5 22 67.10 64.67 51.535 49. 121 23 40.55 39.36 57.836 55.499...

Ngày tải lên: 08/08/2014, 17:20

... N 1 ∂z =− 25 2A N 2 = 1 2A (25 y) ∂ N 2 ∂y = 25 2A ∂ N 2 ∂z = 0 N 3 = 1 2A (25 z − 25 y) ∂ N 3 ∂y =− 25 2A ∂ N 3 ∂z = 25 2A  k (2)  = 1 4A    0 25 25    [ 025 25 ] + 1 4A    25 0 25    [ 25 ... at Node 107 of Example 9.4 ␴ x ␴ y ␶ xy Element 1 20 49.3 187.36 118.4 Element 2 2149.4 315.59 91.89 Element 12 1987.3 322 . 72 204.13 Element...

Ngày tải lên: 08/08/2014, 17:20

... difference method, 28 0 key parameter, 28 5 time step, 27 9, 28 5 what is it, 27 9 Finite element, 2, 12 Finite element analysis (FEA). See Finite element method (FEM) Finite element formulation axisymmetric ... the 2 × 2 matrix [ A ] =  a 11 a 12 a 21 a 22  (A.14) for which the determinant is defined as | A | =     a 11 a 12 a 21 a 22     ≡ a...

Ngày tải lên: 08/08/2014, 17:20