Finite Elements Methods. Formulary for Prof. Estor's exam

Transkript

1 Finite Elements Methods Formulary for Prof. Estor's exam

2 Finite Element Method in General One wants to obtain the equilibrium eqautions for the body, discretized by nite elements in the form M Ü + C U + K U = R Displacement of the nodes: U = [ U 1 U 2... U n T n: degrees of freedom Displacement within the element m: u (m) (x, y, z) = H (m) (x, y, z)û H (m) : Displacement interpolation matrix Strain inside the element m: ɛ (m) (x, y, z) = B (m) (x, y, z)û B (m) : Verzerrungs-Verschiebungs-Matrix Play around with those matrices and from the priciple of virtual displacement ɛ T τ dv = Ū T f B dv + V V SŪ ST f S ds + Ū it F i it follows: i System of equations (stationary case KU = R) [ B (m)t C (m) B (m) dv (m) Û = H (m)t f B(m) dv (m) m v (m) m v (m) + H S(m)T f S(m) ds (m) + B (m)t τ I(m) dv (m) + F m S (m) m v (m) 1

3 From that one can obtain a formular to calculate each of the matrices in the equlibrium equasions: Stiness matrix: K = B (m)t C (m) B (m) dv (m) = K (m) m V (m) Volume forces: R B = H (m)t f B(m) dv (m) = m V (m) m Surface forces: R S = H S(m)T f S(m) ds (m) = m S (m) m Initial stresses: R I = B (m)t τ I(m) dv (m) = m v (m) m Single forces: R C = F R (m) B R (m) S R (m) I With d'alembert Principle the matrices for the time dependent case follow: Mass matrix: M = ρ (m) H (m)t H (m) dv (m) = m V (m) m Damping matrix: C = κ (m) H (m)t H (m) dv (m) = m V (m) m M (m) C (m) According to the choosen elements one has to concidere only certain components of stress, strain and displacement: 2

4 (Isoparametric) Truss Elements Displacement of the nodes: Global: Û = [ u 1 v 1 u 2 v 2... v n T Local: Ũ = [ ũ 1 ṽ 1 ũ 2 ṽ 2... ṽ n T n: number of nodes in element Û can be linked with the overall node displacement vector of the system (has the same orientation). Interpolation: Coordinates: x(r) = h i x i Displacements: ũ(r) = h i ũ i In the following, an element with 2 nodes is considered. Displacement interpolation matrix: H = [ h 1 0 h 2 0 = [ 1 2 (1 r) (1 + r) 0 with x(r) = H X u(r) = H Ũ Setting up the strain displacement matrix: ɛ x = du dx ɛ = B Ũ because h ix = hi x = B = [ h 1x 0 h 2x 0 = [ 1 L 0 1 L 0 = hi x = hi Setting up the stiness matrix: K = B T C B Ĺ dv = AE B T B dx V It follows: K = EA L ( x ) 1 = h i ( L ) 1 2 with A: Cross-sectional area of truss Transformed to global directions: K = T T KT = EA L cos 2 α cosα sinα cosα cosβ cosα sinβ cosα sinα sin 2 α sinα cosβ sinα sinβ cosα cosβ sinα cosβ cos 2 β sinβ cosβ cosα sinβ sinα sinβ sinβ cosβ sin 2 β 3

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6 Isoparametric Plate Elements Displacement of the nodes: Û = [ u 1... u n v 1... v n T n: number of nodes in element Û can be linked with the overall node displacement vector of the system. Interpolation: Coordinates: x(r, s) = h i x i ; y(r, s) = h i y i Displacements: u(r, s) = h i u i ; v(r, s) = h i v i In the following formulars, an element with 4 nodes is concidered. Displacement interpolation matrix: [ h1 h H = 2 h 3 h with h 1 h 2 h 3 h 4 X(r, s) = H ˆX U(r, s) = H Û Match H with the vector of displacements of all nodes U to obtain the global stiness matric H (m) for the element. Setting up the strain displacement matrix: ɛ x = du dx ɛ y = dv dy ɛ xy = du dy + dv dx ɛ = B Û = B = h 1x h 2x h 3x h 4x h 1y h 2y h 3y h 4y h 1y h 2y h 3y h 4y h 1x h 2x h 3x h 4x Match B with the vector of displacements of all nodes U to obtain the global stiness matric B (m) for the element. Calculating derivatives of h i : [ hix = = J 1 h iy [ hi x h i y [ hi h i s [ a b Inverse 4x4 matrix: A = c d with J = [ x x s A 1 = 1 det(a) y y s [ d b c a Setting up the stiness matrix: K = B T CB dv = t B T CB dr ds V 1 1 t: thickness of the element Match K with the vector of displacements of all nodes U to obtain the global stiness matric K (m) for the element. 5

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8 Beam Elements Displacement of the nodes: Û = [ w 1 ϕ 1 w 2 ϕ 2 T for an element with 2 nodes Displacement interpolation matrix (Hermite Beam): w(x) = HÛ [ H = 1 3 x2 L 2 = K = EI + 2 x3 L 3 x 2 x2 L + x3 L 2 12 L L 2 L 12 L L 2 L L 2 L 6 4 L 2 L 3 x2 L 2 x3 2 L 3 x2 L + x3 L 2 For dierent H the strain displacement matrix B has to be determined rst to calculate K = B T CB dv. Isoparametric: Displacement of the nodes: V Û = [ w 1... w q ϕ 1... ϕ q T q: number of nodes in element Interpolation: w(r) = h i w i ϕ(r) = h i ϕ i Use the interpolation functions for a 1D element. For displacement interpolation matrix H and strain interpolation matrix B concider displacement and rotation seperatly: H w = [ h 1... h q H ϕ = [ h 1... h q B w = J 1 [ h 1... with the Jacobian J = x h q B ϕ = J 1 [ h 1... h q 7

9 Material Matrices 8

10 Calculating Loads In general R S = H ST q(x) ds for distributed loads q(x). S In the following a load acting between note 1 and 2 of a 4-node-isoparametric element where s = const = 1 is considered. Set up H S matrix Consider only the displacements of the two nodes contained in the surface the load is working on: [ us = H v S Û S mit Û S = [ u 1 u 2 v 1 v 2 S [ 1 H S = 2 (1 + r) 1 2 (1 r) (1 + r) 1 2 (1 r) Integrate Rewrite q(x) in local coordinates to obtain q(r) and integrate over the surface: R = [ T R x1 R x2 R y1 R y2 = t H ST q(r) det(j S ) dr with q(r) = [ qx (r) q y (r) ˆ=f S 1 and t: thickness of the plate Jabobian matrix Since s = const here, the Jacobian degenerates to J S = [ x y. Because J S is not quadratic anymore, one has to use the Gramian determinant: det(j S ) = ( J S J ST = x ) 2 ( ) 2 + y Load in one direction (after the other) For a vertical load with q x (x) = 0 only the vertical displacement and forces have to be concidered: [ v S = H S v1 = H S = [ [ h 1 h 2 = 1 2 (1 + r) 1 2 (1 r) R = [ Ry1 v 2 = t H R ST q y (r) det(j) dr y2 1 Always try set up single forces that representate the distributed load rst! 9

11 Setting up Mass Matrices M (m) = ρ (m) H (m)t H (m) dv (m) V (m) Consistent mass matrix: 1D: M (m) = 2D: M (m) = Ĺ 0 A(x)ρ (m) H (m)t H (m) dx = 1 1 A(r)ρ (m) H (m)t H (m) det(j) dr 1 t(x)ρ (m) H (m)t H (m) det(j) dr ds Lumped mass matrix: m n 0 0 M (m) = with m: Mass of the element; n: Number of m 0 0 n nodes in element Calculating Stresses In general: σ = C BU Truss: σ = Cɛ = E L L with L = ũ 2 ũ 1 Plate (at s = s, r = r ): σ = C B r, s U 10

12 Jacobian Matrix Transformation of coordinates / chain rule /...: J = h x h y h z x x s x t y y s y t z z s z t Used to calculate: = J 1 h h s h t so that V s t = J x y z and... dv =... det(j) dr ds dt Rules for Variational Operator b b δ du dx = d dx δu δ f(x) dx = δf(s) dx Integration by Parts b b u (x)v(x) dx = [u(x)v(x) b a u(x)v (x) dx a Variation of the Total Potential Total potential: Π = U W U : Strain energy a a W : Potential of external loads a Obtain natural boundary conditions and partial dierential equation by solving δπ = 0, where Π is the (hopefully given) potential of the system. Ritz Method Insert an approximation function for displacement with coecients a 1, a 2,a 3,... and solve Π Π a 1, a 2,... Note that b a 1 a b f(a i ) dx = a 1 f(a i ) dx. a 11

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14 Gauss Integration K = F (r, s, t) dr ds dt mit F (r, s, t) = B T CB det(j) V Gauss: b F (r) dr = α 1 F (r 1 ) + α 2 F (r 2 ) α n F (r n ) a Polynomials up to the order (2n 1) can be integrated axactly. Example The integral F (r) dr shall be approximated using 1 just one approximation point: F (r) dr = 2 F (0) 1 two approximation points: F (r) dr = 1 F ( ) + 1 F ( ) 1 13

15 Newton-Cotes Integration K = F (r, s, t) dr ds dt mit F (r, s, t) = B T CB det(j) V Newton-Cotes: b F (r) dr = (b a) n Ci nf i + R n a i=0 with C n i : Newton-Cotes constants; R n: Remainder (error estimation) The approximation points F i are linearly distributed, for the distance between to points h = b a n holds. Example The integral F (r) dr shall be approximated using 1 just one interval: F (r) dr = 2 ( 1 2 F ( 1) F (1) + R ) 1 1 two intervals: F (r) dr = 2 ( 1 6 F ( 1) F (0) F (1) + R )

16 Centrale Dierences Method ( t Ü = 1 t t t U 2 t U + t+ t U ) ( t U = t t t U + t+ t U ) Error is of order ( t) 2. System of equations at time t: M t U + C t U + K t U = t R Because the system is concidered at time t, not at time t+ t the method is called and explicit integration method. Insert approximations for t Ü and t U: ( 1 t M t C) t+ t U = t R ( K 2 t M ) t U ( 1 2 t M t C) t t U Since 0 U, 0 U and 0 Ü are known, one only needs to calculate t U to start the calculation of one time step after another. From the equations for t Ü and t U one obtains: t U = 0 U t 0 U + t2 0 2 Ü Now one can calculate t U and respectivly t+ t U and then t Ü and t U for every time step. 15

17 Newmark Integration Assumptions: t+ t U = t U + [ (1 δ) t Ü + δ t+ t Ü t t+ t U = t U + t U t + [ ( 1 2 α) t Ü + α t+ t Ü t 2 Obtain t+ t Ü from second equation, insert in rst equation. Insert the resulting eqations for t+ t U and t+ t Ü in the euation of motion at time t + t. M t+ t U + C t+ t U + K t+ t U = t+ t R Because the system is concidered at time t+ t, not at time t the method is called and implicit integration method. 16

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20 All gures and tables are taken from the book Finite-Element- Procedures by Klaus-Jürgen Bathe. It is the only source I used... SoSe 11, Lars Radtke