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Note for Finite Element Methods - FEM by Rohit Sriram

  • Finite Element Methods - FEM
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Misrimal Navajee Munoth Jain Engineering College Department of Mechanical Engineering Formulae for ME 6603 - Finite Element Analysis (6th Semester) Unit -1 Introduction Weighted Residual methods: 1. Point collocation method The residual(R) is forced to zero at discrete number of points. The number of points is equal to the number of undetermined parameters in the assumed trial solution. 2. Sub-domain method The average residual in the sub-domain is forced to zero. The number of sub-domains is equal to the number of undetermined parameters in the assumed trial solution. 3. Least-square method The integral of the weighted square of the residual over the domain is minimized. max R I 2 dx min Suppose if there are two undetermined coefficients C1 and C2, then I R  R dx  0 C1 min C1 max I R  R dx  0 C2 min C2 max 4. Galerkin Weighted residual method Galerkin introduced the idea of letting w(x) to be the same as trial functions max  w ( x) R( x)dx  0 i min Rayleigh-Ritz method: 1. Assume a trial solution (generally polynomial) u ( x)  c1  c2 x  c3 x 2  ... 2. Evaluate the total potential   U V Where U-Strain energy and V-Work done by external forces 3. Setup and solve the system of equations   0 , where i=1,2,3,…No. of undetermined parameters in trial solution ci Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering Page 1

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Unit-2 One Dimensional problems 1. The general finite element equation is {F e }  [ K e ]{u e } Where{Fe}-nodal force vector [Ke]-stiffness matrix {ue}-nodal field variable vector 2. For linear bar element AE  1  1 [K e ]  L  1 1  3. The field variable in terms of shape functions (interpolation functions) u( x)  N1 ( x)u1  N 2 ( x)u2  N3 ( x)u3  ... 4. Strain in the element, u1  dN 3    dN 2 du  dN1   u 2  dx  dx dx dx    u3   Bu Where B-Strain displacement matrix 5. The element stiffness matrix L K   B T DBdx , for 1-D case D=E, Young’s modulus e 0 6. Element stresses   D 7. Body force vector { f e }   [ N T qAdx , where q-body force per unit length 8. For Truss element  l2 lm  l 2  lm    m 2  lm  m 2  AE  lm e [K ]  Le   l 2  lm l 2 lm    2 lm m 2   lm  m x x y  y1  Sin  where l  2 1  Cos and m  2 Le Le 9. One-Dimensional heat transfer  k  1  1 PhL 2 1 T1   Ph L 2   Q0         q   L  1 1  6 A 1 2 T   0 A T L    Q     2   c  c  2   l    Where Q0 and Ql represent the heat flux at the ends of the element (nodes) 10. Plane beam element 6 L  12 6 L   12  6 L 4 L2  6 L 2 L2  EI  [K e ]  3  L  12  6 L 12  6 L   2 2   6L 2L  6L 4L  Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering Page 2

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11. Dynamic analysis K    2 m u  F  Where [K]-stiffness matrix [m]-mass matrix ω – natural frequency 12. Longitudinal vibration  2u  2u Equation of motion is AE 2  A 2 x t AL 2 1 Consistent mass matrix is m  6 1 2 AL 1 0 Lumped mass matrix is m  2 0 1 13. Transverse vibration  4v  2v Equation of motion is EI 4  A 2  0 dx t  156  4 L2 AL  22 L Consistent mass matrix is m  13L 420  54  2  13L  3L   1  AL 0 Lumped mass matrix is m  2 0  0 0 0 0 0 0 0 1 0 symmetric 156  22 L      4 L2  0 0 0  0 Unit-3 Two-Dimensional Scalar Variable Problems 3 ( x3,y3) 1. For a three noded triangular element Field variable, u(x,y) is given as u1  v   1 u   N1 0 N 2 0 N 3 0  u 2      1 2  v   0 N1 0 N 2 0 N 3   v 2  (x1,y1) (x2,y2) u3     v3    3 x   3   1 x   1   2 x   2 Where N1  1 , N2  2 and N 3  3 2A 2A 2A 1  x2 y3  x3 y2 ,  2  x3 y1  x1 y3 and  3  x1 y2  x2 y1 1  y2  y3 ,  2  y3  y1 and  3  y1  y2  1  x3  x2 ,  2  x1  x3 and  3  x2  x1 Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering Page 3

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2. Strain Displacement matrix  N1    x   x     y    0     xy  N1   y N 2 x 0 N1 y N1 x 0 N 2 y N 3 x 0 N 2 y N 2 x 0 N 3 y  u1  0   v1    N 3  u 2    y  v2  N 3  u   3 x   v   3 The above equation is of the form    Bu Therefore the [B] matrix is given as  1 1  B   0 2A  1 2 0 3 1 0 1  2 2 2 0 0 3 0  3   3  Where A=Area of triangle 1 x1 2 A  1 x2 1 x3 y1  y 2  y3  3. The stiffness matrix K    BT C BdV  tABT C B 4. To calculate field variable inside the element T ( x, y)  N1T1  N 2T2  N 3T3 ( x, y)  N11  N 2  2  N 3  3 Unit-4 Two-Dimensional Vector Variable Problems 1. Plane strain problems  x   1   E    1   y     ( 1   )( 1  2  )    0 0  xy  0   x    0    y    1 2  2   xy  2. Plane stress problem  x  1 E      y   2    (1   )  0   xy     x     1  y    1  0 2   xy  0 0 Prepared by Dr. B. Janarthanan and Mr. A. Prakash Faculty members in Mechanical Engineering Page 4

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