n08_fea 1 of 10 Finite Element Analysis – One Dimensional Example straight cantilever beam with constant thickness in pure tension w = 1 in L = 10 in t = 0.5 in aluminum E = 10.4e6 psi P = 1000 lbf 1000 lbf 10 in PL PL in 2 = 0.001923 in = 1.923 mil 6 0.5 in 1 in 10.4x10 lbf AE twE Pk k twE L 1000 lbf = 2.0 ksi P P A t w 0.5 in 1 in n08_fea 2 of 10 tapered cantilever beam with constant thickness in pure tension L = 10 in b = 2 in x t = 0.5 in a = 1 in aluminum E = 10.4e6 psi P = 1000 lbf w b b a x b m x L P P E A tw L L 0 0 dx m b a L 0xL P Etw P P dx Et E t w L L 0 1 PL P 1 b dx ln ln b m x Et m 0 E t b a a bm x 2 in 1000 lbf 10 in in 2 ln 6 0.5 in 2 in 1 in 10.4x10 lbf 1 in = 1.333 mil (NOT 1.923 / 1.5 = 1.282 mil) n08_fea 3 of 10 five elements – six nodes width 2.0 node length element width 1 1.8 1.6 1.4 1.2 1.0 2 3 4 5 6 2.0 2.0 2.0 2.0 2.0 1 2 3 4 5 1.9 1.7 1.5 1.3 1.1 n08_fea 4 of 10 five elements – six nodes u1 u2 u3 u4 u5 u6 F1 F2 F3 F4 F5 F6 element1 element2 element3 element4 element5 ui = displacement of node i Fi = external force applied at node i li = length of element i Ai = cross sectional area of element i ti = thickness of element i wi = width of element i Ei = modulus for element i ki = stiffness of element i fi = internal force on element i ui ui+1 NO LOAD u WITH LOAD u f i li fi li Ai Ei t i w i Ei f i k i u i1 u i fi fi fi t i wi Ei u k u k i u i 1 u i li ki t i wi Ei li n08_fea 5 of 10 nodei Fi fi-1 fi elementi-1 F on node 1 F on node 2 F on node 3 F on node 4 F on node 5 F on node 6 k 1 k 1 0 0 0 0 F1 f1 FWALL 0 F2 f 2 f1 F3 f 3 f 2 F4 f 4 f 3 F5 f 5 f 4 F6 f 5 0 elementi F1 k 1u 1 k 1u 2 FWALL F2 k 1u 1 k 1 k 2 u 2 k 2 u 3 0 0 0 0 F3 k 2 u 2 k 2 k 3 u 3 k 3 u 4 F4 k 3 u 3 k 3 k 4 u 4 k 4 u 5 F5 k 4 u 4 k 4 k 5 u 5 k 5 u 6 F6 k 5 u 5 k 5 u 6 k1 0 0 0 k1 k 2 k2 0 0 k2 k2 k3 k3 0 0 k3 k3 k4 k4 0 0 k4 k4 k5 0 0 0 k5 0 u 1 F1 FWALL 0 u 2 F2 0 u 3 F3 0 u 4 F4 k5 u5 F5 k 5 u 6 F6 impose restraints u1 = 0 and F1 = 0 in first row of equations 1 k 1 0 0 0 0 0 0 0 0 k1 k 2 k2 0 0 k2 k2 k3 k3 0 0 k3 k3 k4 k4 0 0 k4 k4 k5 0 0 0 k5 0 u1 0 0 u 2 F2 0 u 3 F3 0 u 4 F4 k 5 u 5 F5 k 5 u 6 F6 [K] = system stiffness matrix {u} = vector of nodal displacements {F} = vector of external forces on nodes displacement analysis - solve for u K F 1 Ku F n08_fea 6 of 10 stress analysis i = strain in element i i = stress in element i i u i 1 u i i Ei i li 1 E 1 / l 1 0 2 3 0 0 4 5 0 0 E2 / l2 0 0 0 0 0 E3 / l3 0 0 0 0 0 E4 / l4 0 Ei li u i 1 ui 0 u 2 u 1 0 u 3 u 2 0 u 4 u 3 0 u 5 u 4 E 5 / l 5 u 6 u 5 {} = vector of stresses on each element [C] = elasticity matrix {u} = vector of differences in nodal displacements Cu n08_fea 7 of 10 SUMMARY FOR THREE-DIMENSIONAL ANALYSIS 1) Know material properties – E, G, (Poisson’s ratio) 2) Generate mesh – create elements and nodes reduce mesh size by factor of 2, increases number of nodes by factor of 8 3) Compute stiffness matrix – order n = number of degrees of freedom (DOF) n = 3 * number of nodes 4) Define restraints – modify stiffness matrix to force specific nodal displacements to zero 5) Define loading – form right-hand-side (RHS) vector 6) Invert stiffness matrix and pre-multiply loading RHS vector to solve for nodal displacements computation time proportional to n2 7) Compute nodal deformations RHS vector (differences of nodal displacements) 8) Compute elasticity matrix 9) Pre-multiply elasticity matrix times deformation RHS vector to solve for stresses at nodes find von Mises stress, maximum shear stress, normal stress, Dowling stress 10) Know strength of material – SY, SUT, SUC 11) Compute factor of safety at each node n08_fea 8 of 10 % fea_1d.m - one dimensional finite elemnent analysis % HJSIII, 12.09.05 % P L t E axial load, = 1000; = 10.0; = 0.5; = 10.4e6; geometry, material % axial load [lbf] % length [inch] % thickness [inch] % Young's modulus for aluminum [psi] % analytical solution - constant width w = 1.0; % width [inch] delta_const_w_mil = P *L /w /t /E *1000; % [mils] % analytical solution - tapered a = 1.0; % width at tip [inch] b = 2.0; % width at root [inch] delta_taper_mil = P *L /t /E /(b-a) *log(b/a) *1000; % element stiffness - tapered n = 5; li = L/n * ones(n,1); ti = t * ones(n,1); Ei = E * ones(n,1); wi = [ 1.9 1.7 1.5 1.3 1.1 ]'; % % % % % % [mils] number of finite elements length [inch] thickness [inch] Young's modulus for aluminum [psi] width [inch] k = wi .*ti .*Ei ./li; % global stiffness K = [ +1 0 -k(1) +k(1)+k(2) 0 -k(2) 0 0 0 0 0 0 % external forces F = [ 0 0 0 0 0 0 -k(2) +k(2)+k(3) -k(3) 0 0 0 0 -k(3) +k(3)+k(4) -k(4) 0 P ]'; % displacements u = inv(K) * F; delta_fea_mil = u(6) *1000; delta_const_w_mil delta_taper_mil delta_fea_mil % stress C = diag( Ei ./li ); strain = diff( u ); stress = C * strain; stress_ksi = stress /1000 % [mils] 0 0 0 -k(4) +k(4)+k(5) -k(5) 0 0 0 0 -k(5) +k(5) ; ; ; ; ; ]; n08_fea >> fea_1d delta_const_w_mil = 1.9231 delta_taper_mil = 1.3330 delta_fea_mil = 1.3306 stress_ksi = 1.0526 1.1765 1.3333 1.5385 1.8182 9 of 10 n08_fea FEA sensitivity study for 10 of 10 tee_01.sldprt mesh control size parameter [inch] max von Mises stress [ksi] DOF relative computations 0.086 0.06 0.04 0.02 0.01 45.59 50.80 51.55 50.38 50.42 35496 61335 87162 205329 762018 1 2.99 6.03 33.46 460.86