Finite element method - Chapter 2: Introduction to the stiffness method

Nội dung text: Finite element method - Chapter 2: Introduction to the stiffness method

1. Ministry of Industry & Trade Industrial University of HCM City Chapter 2: INTRODUCTION TO THE STIFFNESS METHOD
2. 2.1 Definition of the Stiffness Matrix. Single spring element The stiffness matrix [k] is amatrix such that f =[] k d {d} nodal displacements {f} nodal forces 1
3. 2.1 Definition of the Stiffness Matrix. Three-spring assemblage F =[] K d [K] stiffness matrix relates global coordinate (x,y,z) {d} glodal displacements {F} glodal forces 2
4. 2.2 Derivation of the stiffness matrix for a spring Element One-dimensional linear spring The net spring deformation is given by  = u21 -u The axial force in the spring is f== k k(u21 -u ) (2.1) 3
5. 2.2 Derivation of the stiffness matrix for a spring Element For equilibrium: ff1x+f 2x= 0 1x =-f 2x fk1x=− (u 2 -u 1 ) We can rewrite Equation 2.1 fk2x= (u 2 -u 1 ) 4
6. 2.2 Derivation of the stiffness matrix for a spring Element fk=− (u -u ) We can rewrite Equation 2.1 1x 2 1 fk2x= (u 2 -u 1 ) which can be expressed in matrix form f1x  kk− u 1  =  f2x  −kk u 2  f =[] ke d −1 d =[] ke f  5
7. 2.3 System Assembly in Global Coordinates We next consider the system of two linear spring elements connected as shown: + The springs have different spring constants k1 and k2 + The nodes are numbered 1, 2, and 3 as shown, with the springs sharing node 2 as the physical connection 6
8. 2.3 System Assembly in Global Coordinates Free-body diagrams of elements and nodes for the two-element system 7
9. 2.3 System Assembly in Global Coordinates The equilibrium conditions for the first spring (1) (1) kk11− u11  f  (1) = (1)  (2.2a ) −kk11 u22  f  8
10. 2.3 System Assembly in Global Coordinates The equilibrium conditions for the second spring (2) (2) kk22− u11  f  (2) = (2)  (2.2b ) −kk22 u22  f  9
11. 2.3 System Assembly in Global Coordinates The displacement compatibility conditions (1) (1) uu1==uu 1 2 2 (2) (2) (2.3) uu1==uu 2 2 3 u1, u2, u3: global nodal displacements 10
12. 2.3 System Assembly in Global Coordinates Substituting Equations 2.3 into Equations 2.2, we obtain (1) k1− k 1 u 1  f1 = (1)  (2.4a ) −k1 k 1 u 2  f2 (2) kk22− u2 f1 = (2)  (2.4b ) −kk22 u3 f2 11
13. 2.3 System Assembly in Global Coordinates Equation 2.4 is the equilibrium equations for each spring element which specified in the global displacements. The elements are physically connected at node 2 but the equations are not yet amenable to direct combination,as the displacement vectors are not the same 12
14. 2.3 System Assembly in Global Coordinates It’s 2 elements and 3 nodes so we have matrix equations to 3 × 3 as follows (1) k1− k 10 u 1  f1 (1) −=k k0 u f (2.5) 1 1 2  2  0 0 0 0  0 0 0 0 0 0 (2) 0kk−=u f (2.6) 22 2  1  0 −kk (2) 22 u3 f2 13
15. 2.3 System Assembly in Global Coordinates The addition of Equations 2.5 and 2.6 yields (1) k1− k 10 u 1  f1 −k k + k − k u = ff(1)+ (2) (2.7) 1 1 2 2 2  22  0 −k k u (2) 2 2 3  f3 which is of the form []K d = F [K] the system stiffness matrix 14
16. 2.4 Boundary conditions We must specify boundary (or support) conditions for structure models such as the spring assemblage. []K d = F ➢[K] will be singular → its inverse will not exist → the structural system is unstable ➢Without our specifying adequate kinematic constraints or support conditions 15
17. 2.4 Boundary conditions + Homogenous boundary conditions are such that the displacements are zero at certain nodes. Example 16
18. 2.4 Boundary conditions F1x is the unknown reaction 17
19. 2.4 Boundary conditions + Homogenous boundary conditions are such that one or more of the specified displacements are nonzero Example 18
20. 2.4 Boundary conditions F1x is now a reaction from the support that has moved an amount  19
21. Some properties of the stiffness matrix ➢[K] square, relates the same number of forces and displacements. ➢[K]e is symmetric ➢ [K] singular → no inverse exists → boundary conditions. ➢The main diagonal terms of [K] are always positive. ➢[K] is positive semidefinite. 20
22. Example 01 For the spring assemblage with arbitrarily numbered nodes shown in Figure, obtain (a) the global stiffness matrix, (b) the displacements of nodes 3 and 4, (c) the reaction forces at nodes 1 and 2, and (d) the forces in each spring. A force of 2KN is applied at node 4 in the x direction. The spring constants are given in the figure. Nodes 1 and 2 are fixed. k1=170N/mm k2=340N/mm k3=510N/mm 2KN 21
23. Solution k1=170N/mm k2=340N/mm k1=510N/mm 2KN a. The global stiffness matrix Step 1: Difine number elements, nodes, nodal degrees freedom No. Element Conected + 3 elements 1 1 3 2 3 4 3 4 2 + 4 nodes + Each node is one nodal degrees of freedom 22
24. Step 2: The element stiffness matrix (local) Elemenets linked with two node → size of k (2x2) Column13 Rows (1) 170− 170 1 Element 1 k = −170 170 3 Column34 Rows Column42 Rows (2) 340− 340 3 510− 510 4 k = (3) k = −340 340 4 −510 510 2 Element 2 Element 3 23
25. Step 3: The global stiffness matrix We have got 3 elements, 4 nodes so size of the global stiffness matrix (4x4) Column1 2 3 4 Rows 170 0− 170 0 1 0 510 0− 510 2 K  = −170 0 170 + 340 − 340 3 0− 510 − 340 340 + 510 4 24
26. k1=170N/mm k2=340N/mm k1=510N/mm 2KN b. The displacements of nodes 3 and 4 Relates global forces to global displacements as follows F1x 170 0− 170 0 u 1 F 0 510 0− 510 u 2x = 2 F3x −170 0 170 + 340 − 340 u 3 F4x 0− 510 − 340 340 + 510 u 4 25
27. k1=170N/mm k2=340N/mm k1=510N/mm 2KN Applying the homogeneous boundary conditions u1=0 and u2=0 0 510− 340 u3 = 2000 − 340 850 u4 Solving equation system, we obtain the global nodal displacements 400 u3 187 = ()mm u4 600 187 26
28. k1=170N/mm k2=340N/mm k1=510N/mm 2KN c. The reaction forces at nodes 1 and 2 substitute u3, u4 in the equation system 0 F 170 0− 170 0 -4000/11 1x 0 F 0 510 0− 510 -18000/11 2x 400 == ()N F3x −170 0 170 + 340 − 340 187 0 F4x 0− 510 − 340 340 + 510 600 2000 187 27
29. k1=170N/mm k2=340N/mm k1=510N/mm 2KN d. The forces in each spring We use local element to obtain the forces in each element Element 1 (1) f1x (1) u1 = []k (1) u f3x 3 4000 (1) 0 − f 1x 170− 170 11 == 400 ()N f (1) −170 170 4000 3x 187 11 28
30. (2) Element 2 f3x (2) u3 = []k (2) u f4x 4 400 -4000 (2) f3x 340− 340 187 11 == ()N (2) −340 340 600 4000 f4x 187 11 (3) Element 3 f4x u4 = []k (3) (3) u f2x 2 18000 600 f (3) 4x 510− 510 11 == 187 ()N f (3) −510 510 -18000 2x 0 11 29
31. Example 02 For the spring assemblage shown in Figure, obtain (a) the global stiffness matrix, (b) the displacements of nodes 2–4, (c) the global nodal forces, and (d) the local element forces. Node 1 is fixed while node 5 is given a fixed, known displacement =20 mm. The spring constants are all equal to k =200 kN/m. 30
32. Solution a. The global stiffness matrix Step 1: Difine number elements, nodes, nodal degrees freedom No. Element Conected 1 1 2 + 4 elements 2 2 3 3 3 4 + 5 nodes 4 4 5 + Each node is one nodal degrees of freedom 31
33. Step 2: The element stiffness matrix (local) Elemenets linked with two node → size of k (2x2) The element stiffness matrixs are same as: Column12 Rows (1) 200− 200 1 Element 1 k = −200 200 2 32
34. Element 2 Element 3 Column23 Rows Column34 Rows (2) 200− 200 2 (3) 200− 200 3 k = k = −200 200 3 −200 200 4 Column45 Rows (4) 200− 200 4 Element 4 k = −200 200 5 33
35. Step 3: The global stiffness matrix We have got 3 elements, 4 nodes so size of the global stiffness matrix (5x5). Using superposition, we obtain the global stiffness matrix as: Column1 2 3 4 5 Rows 200− 200 0 0 0 1 −−200 400 200 0 0 2 kN K  = 0−− 200 400 200 0 3 m 0 0−− 200 400 200 4 0 0 0− 200 200 5 34
36. b. The displacements of nodes 2–4 Relates the global forces to the global displacements as follows: F1x 200− 200 0 0 0 u 1 F −−200 400 200 0 0 u 2x 2 F3x = 0−− 200 400 200 0 u 3 F4x 0 0−− 200 400 200 u 4 F5x 0 0 0− 200 200 u 5 35
37. Applying the boundary conditions u1=0, u5=0.02 (m), F =F =F =0 2x 3x 4x 0 0−− 200 400 200 0 0 u 2 0= 0 − 200 400 − 200 0 u3 0 0 0−− 200 400 200 u4 0.02 0 400− 200 0 u22 u 0.005 0 = − 200 400 − 200 u u = 0.01 ()m 33 4 0− 200 400 u44 u 0.015 36
38. c. The global nodal forces Back-substituting the boundary condition displacements 37
39. d. The local element forces We make use of local element to obtain the forces in each element Element 1 Element 2 38
40. Element 3 Element 4 39
41. Homeworks Problem 01 For the spring assemblage shown in following figure, we determine: a)The global stiffness matrix [K ] b) The displacements of nodes 3 and 4 c) The reaction forces at nodes 1 and 2. 40
42. Homeworks Problem 02 For the spring assemblage shown in following figure, we determine: a)The global stiffness matrix [K ] b) The displacements of nodes 2, 3 and 4 c) The reaction forces at nodes 1 and 5. 41
43. Homeworks Problem 03 For the spring assemblage shown in following figure, we determine the global stiffness matrix [K ]. With k1=1N/cm, k2=2N/cm, k3=3N/cm, k4=4N/cm and k5=5N/cm 42
44. Homeworks Problem 04 For the spring assemblage shown in following figures, we determine the nodal displacements, the forces in each element, and the reactions. Use the direct stiffness method for all problems. 43
45. Homeworks 44
46. Homeworks 45
47. Homeworks 46
48. Homeworks Problem 05 For the spring assemblage shown in following figures, we determine the global stiffness matrix using the system assembly procedure. 47
49. Homeworks Problem 06 For the spring assemblage shown in following figure, we determine the displacements at nodes 2 and 3 and the reactions at nodes 1 and 4. With k1=500N/mm, k2= k3=300N/mm, k4=k5=400N/mm 48