# GT ME 6443 - Introduction to Finite Element Methods (13 pages)

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## Introduction to Finite Element Methods

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## Introduction to Finite Element Methods

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13
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Georgia Tech
Course:
Me 6443 - Variational Methods
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Introduction to Finite Element Methods ME6443 Dr A A Ferri x u1 u2 Within each element we have a particular E A L u1 u2 x Assume a linear displacement field exact for statics Need to relate the 2 generalized coordinates for each element to the displacement field within each element Assume a linear displacement field u x u1N1 x u2 N 2 x 1 x N1 x 1 L x N 2 x L 1 The functions N x are termed shape functions and they are just Ritz functions that are local to each element Element potential energy 2 1 L V EA u dx 2 0 We can also consider a distributed force per unit length on each element u1 u2 f x P1 P2 x Elemental virtual work can be expressed as L dW 0 f x du dx P1 du 0 P2 du L P1 du1 P2 du2 Let s take P1 P2 zero for now it will be easy to put them back later dV dW 0 Variational Principle L du EA u 0 f x du dx 0 L 2 2 2 EA u N d u N f x d u N j j dx 0 i i i i 0 i 1 i 1 j 1 L 2 d u u EA Ni N i j j dx 0 i 1 j 1 2 kij f x N dx i 0 0 L fi EA 1 1 ke 1 1 L Assuming that the ui s are independent and arbitrary we get 2 u j kij j 1 fi 0 i 1 2 ke ue f e What if there are 2 elements element 1 element 2 E1 A1 L1 u1 u11 u12 combine E2 A2 L2 u11 u12 u12 k1 u1 f1 k1 0 u22 k2 u2 f 2 0 u1 k2 u2 f2 uG FG KG f1 Thus the global or unassembled static equations are KG uG FG But the global coordinate vector uG u11 u12 u12 u22 T has 1 too many degrees of freedom if the elements are connected U1 U2 U3 The coordinates of uG and U U1 U2 U3 T are related as follows u11 1 1 u2 0 2 0 u1 2 0 u2 0 1 1 0 0 U1 0 U2 0 U3 1 uG S U 4x3 compatibility matrix Substitue S into global relationship KG S U FG Now premultiply by S T to eliminate equations and preserve symmetry S T K G S U S T FG K U F Any nodes that are fixed have zero displacement Therefore we can just delete the rows and columns of K and the rows of F corresponding to any fixed nodes Nodal forces If a force P is applied to node 1 of element n then the additional contribution to the elemental force vector is P 0 T likewise if the force P is applied to node 2 of element n then the additional contribution to the elemental force vector is 0 P T Axial Deformation of a Tapered Rod Example A x A0 2 x L P Le x Consider N elements each having length Le L N and area Ai where Ai A xi uexact xi Le 2 i 1 Le PL 2 ln EA0 2 x L midpoints Steel beam of length 1 meter and end width 20mm and end thickness 4mm E 2e11 rho 7800 L 1 A0 0 02 0 004 P 10000 N 1 4 4 5 x 10 finite element exact 4 ke 3 5 1 0e 007 2 4000 2 4000 2 5 u 2 4000 2 4000 3 2 1 5 1 0 5 0 0 0 1 0 2 0 3 0 4 0 5 x 0 6 0 7 0 8 0 9 1 N 2 4 4 5 compatibity matrix x 10 finite element exact 4 S 3 5 0 1 1 0 0 0 0 1 3 2 5 u 1 0 0 0 2 1 5 1 0 5 0 0 0 1 0 2 K Reduced 0 3 0 4 0 5 x 0 6 0 7 F 96000000 40000000 40000000 40000000 0 10000 0 8 0 9 1 N 5 4 4 5 x 10 finite element exact 4 3 5 3 u 2 5 2 1 5 1 0 5 0 0 0 1 0 2 0 3 0 4 0 5 x 0 6 0 7 0 8 0 9 1 N 7 cond K 84 6059 K 1 0e 008 4 1600 2 0000 0 0 0 0 0 2 0000 3 8400 1 8400 0 0 0 0 0 1 8400 3 5200 1 6800 0 0 0 0 0 1 6800 3 2000 1 5200 0 0 0 0 0 1 5200 2 8800 1 3600 0 0 0 0 0 1 3600 2 5600 1 2000 0 0 0 0 0 1 2000 1 2000 Note that for N 7 K is 7x7 and symmetric It is also banded We say that the bandwidth of the matrix is 3 This diagonal dominance of the stiffness matrices from finite element procedures is one of the reasons that the condition number stays reasonable even for models having a large number of elements Connection to Global Ritz Function Approach f 1 x f 2 x f 3 x f 4 x f 5 x f 6 x f 7 x 1 2 3 4 5 6 x Le Using the admissible Ritz functions above would yields identical results to those produced using N 7 finite elements provided that areas were staircased according to the same scheme used with the FEM N u x U j f j x j 1

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