Nonlinear Finite Elements for Continua and Structures Part 1 pps

... dimensions: σσσσ≡           →                   =                   ≡ {} σσσ σσσ σσσ σ σ σ σ σ σ σ σ σ σ σ σ 11 12 13 21 22 23 31 32 33 11 22 33 23 13 12 1 2 3 4 5 6 (A .1. 2) We will call the correspondence between the square matrix form of the tensor and the column matrix form the Voigt rule. For ... Procedures for Contact-Im...

Ngày tải lên: 12/08/2014, 02:20

... equations for the penalty method, we first evaluate P c by Eq. (34): P c = β 1 φ T φH( g) dΓ Γ c ∫ =β 1 H ( γ) +1 1 0         +1 1 0 [ ] = β 1 H( g) +1 1 0 1 +1 0 0 0 0  ... Fig. . The equations for this system can be obtained by just eliminating rows 1 and 3 and columns 1 and 3, giving k 2 1 1 0       d 1 λ 1    ...

Ngày tải lên: 12/08/2014, 02:20

... F T ⋅F = 1. 25 1 1 1. 25       (E3 .11 .2) The eigenvalues and corresponding eigenvectors of C are λ 1 = 0.25 y 1 T = 1 2 1 1 [ ] λ 2 = 2.25 y 2 T = 1 2 1 1 [ ] (E3 .11 .3) 3-62 T. ...    1 2 0 0 3 2       1 2 1 1 1 1       = 1 2 2 1 1 2       (E3 .11 .5) The rotation matrix is obtained by Eq. (3.7.6): R...

Ngày tải lên: 12/08/2014, 02:20

... linear finite element analysis, the integrand in the expression for the stiffness matrix consists of polynomials for rectangular elements and is smooth and nearly a polynomial for isoparametric elements. ... Chapter 3 and are given in both tensor form and indicial form in Box 4 .1. As can be 4-2 T. Belytschko, Lagrangian Meshes, December 16 , 19 98 CHAPTER 4 LAGRANGIAN ME...

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... M iIjJ = δ ij ˜ M IJ and then using the rule of Eq. (1. 4.26), which gives M = ρ 0 A 0 a 0 12 2 0 1 0 1 0 0 2 0 1 0 1 1 0 2 0 1 0 0 1 0 2 0 1 1 0 1 0 2 0 0 1 0 1 0 2         ... ] = v x1 v x2 v x3 v y1 v y2 v y3       1 2A y 23 x 32 y 31 x 13 y 12 x 21           = = 1 2 A y 23 v x1 + y 31 v x2 + y 12...

Ngày tải lên: 12/08/2014, 02:20

... C 11 23 ε 23 +C 11 31 ε 31 + C 11 32 ε 32 +C 11 3 31 ε 33 = C 11 ε 1 + 1 2 C 16 ε 6 + 1 2 C 15 ε 5 + 1 2 C 16 ε 6 + 1 2 C 12 ε 2 + 1 2 C 14 ε 4 + 1 2 C 15 ε 5 + 1 2 C 14 ε 4 + 1 2 C 13 ε 3 = C 11 ε 1 +C 12 ε 2 + ... J 1 λP 11 e 1 ⊗ e 1 (5 .1. 9) For the special case of an incompressible material J =1 and Eq. (5 .1. 9) is equi...

Ngày tải lên: 12/08/2014, 02:20

... midpoint time step. The central difference formula for the velocity is ˙ d n+ 1 2 ≡ v n+ 1 2 = 1 ∆t n+ 1 2 d n +1 − d n ( ) , d n +1 = d n +∆t n+ 1 2 v n+ 1 2 (6.2.2) where the second equation ... December 16 , 19 98 the discrete momentum equation at time step n +1 in a form applicable to both equilibrium and dynamic problems: 0 = r d n +1 , t n +1 ( ) = s D M ˙...

Ngày tải lên: 12/08/2014, 02:20

... by expanding the jth term in Eq. (7 .14 .13 ): P e 1 0 1 1 0 1 1 0 1             M φ j 1 φ j φ j +1 M             + 1 2 1 1 2 1 1 2 1   ... Eq. (7 .14 .15 ), the j+1th and j-1th terms of φ are: φ j +1 = e a( j +1) ∆x = e aj∆x e a∆x = z j +1 (7 .14 .16 a) φ j 1 = e a( j 1) ∆x = e aj∆x e −a∆x = z j 1 (7 .14 .16 b) Sub...

Ngày tải lên: 12/08/2014, 02:20

... ∑ I =1 n N e (1. 2 .17 ) It is recognized from (11 ) that the coefficients of α i on the right hand side of Eq (17 ) correspond to x i so u = α o s I N I x + α i x i∑ i =1 n D e ∑ I =1 n N e (1. 2 .18 ) We ... material and the finite element mesh coincides with the 20 ∂y ∂ξ = 1 4 y I ξ I (1 + η I η ∑ I =1 4 (1. 2.8c) ∂y ∂η = 1 4 y I η I 1 + ξ I ξ ∑ I =1 4 (1. 2.8d...

Ngày tải lên: 12/08/2014, 02:20

... 2.089 1. 010 1. 029 1. 013 1. 038 .995 1. 0 31 3 2.462 1. 005 1. 015 1. 006 1. 016 .997 1. 012 4 2. 717 1. 002 1. 007 1. 003 1. 007 .999 1. 008 Fig. 21 shows the convergence of the error in the energy norm. All elements ... (0 .1) 15 .9 (0.00) 8.39 (3.40) 8 .18 (4.88) 8 .14 (5.04) 7.22 (1. 05) 7.67 (3.82) FB (0.3) 7.68 (0 .12 ) 7.05 (1. 15) 7.59 (3.74) 7.92 (4.67) 5.3...

Ngày tải lên: 12/08/2014, 02:20