Topic 5: Finite Element Method

Size: px

Start display at page:

Download "Topic 5: Finite Element Method"

Jack McKenzie
5 years ago
Views:

1 Topic 5: Finite Element Method 1

2 Finite Element Method (1) Main problem of classical variational methods (Ritz method etc.) difficult (op impossible) definition of approximation function ϕ for non-trivial problems. The solution can be a division of structure to n small parts with simple shapes. Approximational functions ϕ j can be defined on them. Becasue Π is a scalar value: Π approx. = n j=1 Π e,j, (1) where Π e,j... potentional energy of j-th part ( finite element ). 2

3 Finite Element Method (2) The rest of solution can be identical to the Ritz method: Π = 0. (2) Note: here we will utilize the Lagrange variational principle. It implies that unknown variables will be deformations. This will be a deformational version of the Finite Element Method. 3

4 Finite Element Method (3) Finite Element Method (FEM) alternatives: deformational unknowns are displacements and rotations (most common more - than 90% software use it), force unknows are forces/moments mixed. 4

5 Deformational variant of FEM Unknowns are related to theory of elasticity: plane problem (shear walls,... ): u, v slabs (bending problems): w, ϕ x, ϕ y volume.spatial problems: u, v, w Approximational functions mostly have form of polynomials. 5

6 for trusses (1) y u 1 u x Unknown deformational parameters: u in every node. Totally 2 unknown parameters on element: {u 1, u 2 } T. 6

7 for trusses (2) Geometrical equations: ε x = u x (3) In matrix form (ε = T u): { εx } = [ x ] { u } (4) 7

8 for trusses (3) Equilibrium equations: In matrix form: ( σ + X = 0): σ x x + X = 0 (5) [ x ] { σx } + { X } = { 0 } (6) 8

9 for trusses (4) Constitutive equations: In matrix form (σ = D ε): σ x = E ε x (7) { σx } = [ E ] { εx } (8) 9

10 for trusses (5) Approximation of unknowns (node displacements): u(x) = a 1 + a 2 x (9) In matrix form (u = U a): { u } = [ 1 x ] a 1 a 2 (10) 10

11 for trusses (6) Approximation of unknowns in nodes 1, 2 (r = S a): u 1 u 2 = 1 x 1 1 x 2 a 1 a 2 (11) 11

12 for trusses (7) By combination of ε = T u and u = U a we can get ε = B a, where B = T U a: { εx } = [ x ] [ 1 x ] a 1 a 2 (12) 12

13 for trusses (8) By combination of ε = T u and u = U a we can get ε = B a, where B = T U a: { εx } = [ 0 1 ] a 1 a 2 (13) 13

14 for trusses (9) From r = S a: a = S 1 r, (14) kde: S = 1 x 1 1 x 2 S 1 = x 2 x 1 x 2 x 1 x 2 x x 2 x 1 x 2 x 1 (15) Then instead of ε = Ba we can write:ε = B S 1 r: { εx } = [ 0 1 ] x 2 x 1 x 2 x 1 x 2 x x 2 x 1 x 2 x 1 u 1 u 2. (16) 14

15 for trusses (11) Potential energy of internal forces: Π i = 1 2 V εt σ d V = 1 2 V εt D ε d V (17) Potential energy of external forces: Π e = V XT r d V S pt r d S. (18) 15

16 for trusses (12) Potential energy of system: Π = 1 2 V εt D ε d V V XT r d V S pt r d S. (19) After replacemetn of ε and extraction of r: Π = 1 2 rt V S 1T B T DB S 1 d V r T V XT d V r S pt d S r. (20) In short form: Π = 1 2 rt K r F T r. (21) 16

17 for trusses (13) By use of Lagrange variational principle: ( Π = min.) na (21): K r = F, (22) where K... stiffness matrix of finite element: K = V S 1T B T D B S 1 d V, (23) F... load vector of finite element: F = V XT d V S pt d S. (24) 17

18 for trusses (14) For this particular finite element: F = X + p. (25) K = V S 1T B T D B S 1 dv = A with matrices members shown: K = A L 0 x 2 x 2 x 1 x 1 x 2 x 1 1 x 2 x 1 1 x 2 x L 0, S 1T B T D B S 1 dx, [ E ] [ 0 1 ] x 2 (26) x 1 x 2 x 1 x 2 x x 2 x 1 x 2 x 1 (27) dx 18

19 for trusses (15) Full formula (simplified): K = A x 2 x 2 x 1 x 1 x 2 x 1 1 x 2 x 1 1 x 2 x [ E ] [ 0 1 ] x 2 x 1 x 2 x 1 x 2 x x 2 x 1 x 2 x 1 After modification (integration L 0 dx = L and multiplication): K = EAL 1 1 (x 2 x 1 ) 2 (x 2 x 1 ) (x 2 x 1 ) 2 (x 2 x 1 ) 2, x 2 x 1 = L K = EA L EA L L 0 dx (28) EA L EA L (29) which is a sfiffness matrix and it is identical to one that can be derived by slope-deflection method/stiffness method. 19,

20 for trusses (16) The unknowns can be computed by solution of linear equation system: K e r e = F e, In full form: EA L EA L EA L EA L u 1 u 2 = F 1 F 2 (30) 20

21 for trusses (17) Expansion for two unknowns u and v in every node: y v 1 v 2 u u 2 x EA L 0 EA L EA L 0 EA L u 1 v 1 u 2 v 2 = F 1 0 F 2 0 (31) 21

22 Interpolation polynomials (1) Best convergence can be reached for full polynomials of n-th grade (Ženíšek et al). Number of constants (a 1, a 2,...) have to be equal to number of unknowns on finite element For this reason it in not always possible fo use full polynomials 22

23 Interpolation polynomials (2) For one unknown x: 1. a 1 + a 2 x 2. a 1 + a 2 x + a 3 x 2 3. a 1 + a 2 x + a 3 x 2 + a 4 x 3 4. a 1 + a 2 x + a 3 x 2 + a 4 x 3 + a 5 x 4 23

24 Interpolation polynomials (3) For two unknowns x and y: 1. a 1 + a 2 x + a 3 y 2. a 1 + a 2 x + a 3 y + a 4 xy + a 5 x 2 + a 6 x 2 3. a 1 + a 2 x + a 3 y + a 4 xy + a 5 x 2 + a 6 x 2 + a 7 x 3 + a 8 y 3 + a 9 x y 2 + a 10 x 2 y 24

Finite Element Method in Geotechnical Engineering

Finite Element Method in Geotechnical Engineering Short Course on + Dynamics Boulder, Colorado January 5-8, 2004 Stein Sture Professor of Civil Engineering University of Colorado at Boulder Contents Steps