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# Note for Finite Element Methods - FEM by Rohit Sriram

• Finite Element Methods - FEM
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Rohit Sriram
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Misrimal Navajee Munoth Jain Engineering College Department of Mechanical Engineering ME 6603 – FINITE ELEMENT ANALYSIS Assignment-1 B.E. Mechanical 3rd Year Due Date:02.01.2018 1. Solve the differential equation for the physical problem expressed as with the boundary d2y  500x 2  0, 0  x  1 conditions as dx2 y(0)=0 and y(1)=0 using a two-term trial function by (i) point collocation method, (ii) subdomain collocation method, (iii) Least square method and (iv) Galerkin’s method. d2y  300x 2  0, 0  x  1 with 2. Consider the differential equation for a problem as dx2 the boundary conditions y(0)=0 and y(1)=0. Find the solution by using trial function as y  a1x(1  x3 ) . Use (i) point collocation method, (ii) subdomain collocation method, (iii) Least square method and (iv) Galerkin’s method. 3. Solve the differential equation for the physical problem expressed as with the boundary d2y  100  0, 0  x  10 conditions as dx2 y(0)=0 and y(10)=0 using (i) point collocation method, (ii) subdomain collocation method, (iii) Least square method and (iv) Galerkin’s method 4. A simply supported beam (span L and flexural rigidity EI) subjected to a point load at the centre of the span. Calculate the deflection at the mid-span using Galerkin’s weighted residual method and compare with the exact solution. Assume trial solution  x  as y  aSin  . L 5. A simply supported beam (span L and flexural rigidity EI) subjected to a point load at the centre of the span. Calculate the deflection at the mid-span using Rayleigh-Ritz method and compare with the exact solution. 6. A cantilever beam is subjected to a uniformly distributed load throughout its length. Find the deflection at free end using Rayleigh-Ritz method. 7. A cantilever beam of length L is loaded with a point load at the free end. Find the maximum deflection and maximum bending moment using Rayleigh-Ritz method   x  using the function y  a 1  Cos  . Given EI is constant.  2 L   d2y  400x 2  0, 0  x  1 subject to the 2 dx boundary conditions y(0)=0 and y(1)=0. The functional corresponding to the problem 2 1    dy  2  I   0 . 5  400 x y to be extremised is given by    dx . Find the solution using 0   dx    Rayleigh-Ritz method by considering a two-term trial solution as y( x)  a1x(1  x)  a2 x(1  x3 ) 8. Consider the differential equation

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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