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FINITE ELEMENT METHOD ( BDA 4033 ) Lecture #01
SYLLABUS CHAPTER 1- INTRODUCTION ,[object Object],[object Object],[object Object],[object Object],[object Object]
Definition ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Historical Back Ground ,[object Object],[object Object],[object Object],[object Object],[object Object]
Mathematics… ,[object Object],[object Object],[object Object],[object Object]
FE Programming.. ,[object Object],[object Object],[object Object],[object Object],[object Object]
Basic Steps in Finite Element Method ,[object Object],[object Object],[object Object],[object Object]
Basic Steps  (Contd..) ,[object Object],[object Object],[object Object],[object Object],[object Object]
Basic Steps  (Contd..) ,[object Object],[object Object],[object Object],[object Object]
Basic Steps  (Contd..) ,[object Object],[object Object],[object Object]
Basic Steps  (Contd..) ,[object Object],[object Object],[object Object],[object Object]
Direct Formulation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Direct Formulation (Contd..) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],F
Direct Formulation (Contd..) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],F i j F 1 2
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Direct Formulation (Contd..) i j i j
[object Object],[object Object],[object Object],[object Object],[object Object],Direct Formulation (Contd..)
Direct Formulation (Contd..)- Stiffness Matrix Let us write the stiffness matrix for the first element with CSA, A 1  as and  For the second element with CSA, A 2  as where k 1 =A 1 E 1 /L 1  and k 2 =A 2 E 2 /L 2 Then the force equation can be written as for element 1 and for element 2 F 1 2 1 2 3
Direct Formulation (Contd..)- Assembly Physical Assembly:  Step 0: Nodes  Matrix assembly : Global Matrix Order = No. of nodes X No. of displacements= 3 X 3 1  2  3
Direct Formulation (Contd..)- Assembly Step 1: Physical Assembly: Matrix assembly : Step 2:    Matrix assembly : Physical Assembly: 1  2  3 1 1  2  3 1 2
Direct Formulation (Contd..)- Apply BCs In the given beam, u 1 =F 1 =F 2 =0 Substituting the values in the assembled equation Solution: Solving the above equation, the unknown displacements  u 2  and  u 3  can be found. F 1 2 1 2 3
Direct Formulation (Contd..)- Post Processing From the previous solution, the unknown displacements  u 2  and  u3 a re found. To find the element stresses: For the element 1,  the strain  ε 1 = ( u 2 - u 1 )/L 1 Applying Hooke’s law,   1 =E 1  ε 1 =E 1 (u 2 - u 1 )/L 1 Similarly for the element 2,      2 =E 2   ε 2 =E 2 (u 3 - u 2 )/L 2
Problem 1  ,[object Object],10 kN 1m 1m
Problem 2  ,[object Object],300mm 400mm 200 kN
Problem 3  ,[object Object],300mm 300mm 100 kN 150mm 75mm
Questions ?

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Fem l1(a)

  • 1. FINITE ELEMENT METHOD ( BDA 4033 ) Lecture #01
  • 2.
  • 3.
  • 4.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
  • 16.
  • 17. Direct Formulation (Contd..)- Stiffness Matrix Let us write the stiffness matrix for the first element with CSA, A 1 as and For the second element with CSA, A 2 as where k 1 =A 1 E 1 /L 1 and k 2 =A 2 E 2 /L 2 Then the force equation can be written as for element 1 and for element 2 F 1 2 1 2 3
  • 18. Direct Formulation (Contd..)- Assembly Physical Assembly: Step 0: Nodes Matrix assembly : Global Matrix Order = No. of nodes X No. of displacements= 3 X 3 1 2 3
  • 19. Direct Formulation (Contd..)- Assembly Step 1: Physical Assembly: Matrix assembly : Step 2: Matrix assembly : Physical Assembly: 1 2 3 1 1 2 3 1 2
  • 20. Direct Formulation (Contd..)- Apply BCs In the given beam, u 1 =F 1 =F 2 =0 Substituting the values in the assembled equation Solution: Solving the above equation, the unknown displacements u 2 and u 3 can be found. F 1 2 1 2 3
  • 21. Direct Formulation (Contd..)- Post Processing From the previous solution, the unknown displacements u 2 and u3 a re found. To find the element stresses: For the element 1, the strain ε 1 = ( u 2 - u 1 )/L 1 Applying Hooke’s law,  1 =E 1 ε 1 =E 1 (u 2 - u 1 )/L 1 Similarly for the element 2,  2 =E 2 ε 2 =E 2 (u 3 - u 2 )/L 2
  • 22.
  • 23.
  • 24.