Topic 5: Finite Element Method 1
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
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
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
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
for trusses (1) y u 1 u 2 1 2 x Unknown deformational parameters: u in every node. Totally 2 unknown parameters on element: {u 1, u 2 } T. 6
for trusses (2) Geometrical equations: ε x = u x (3) In matrix form (ε = T u): { εx } = [ x ] { u } (4) 7
for trusses (3) Equilibrium equations: In matrix form: ( σ + X = 0): σ x x + X = 0 (5) [ x ] { σx } + { X } = { 0 } (6) 8
for trusses (4) Constitutive equations: In matrix form (σ = D ε): σ x = E ε x (7) { σx } = [ E ] { εx } (8) 9
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
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
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
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
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 1 1 1 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 1 1 1 x 2 x 1 x 2 x 1 u 1 u 2. (16) 14
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
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
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
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 1 0 1 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 1 1 1 x 2 x 1 x 2 x 1 (27) dx 18
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 1 0 1 [ E ] [ 0 1 ] x 2 x 1 x 2 x 1 x 2 x 1 1 1 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 ) 2 1 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,
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
for trusses (17) Expansion for two unknowns u and v in every node: y v 1 v 2 u 1 1 2 u 2 x EA L 0 EA L 0 0 0 0 0 EA L 0 EA L 0 0 0 0 0 u 1 v 1 u 2 v 2 = F 1 0 F 2 0 (31) 21
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
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
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