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Finite element method - Chapter 3: Development of truss equations

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  1. Ministry of Industry & Trade Industrial University of HCM City Chapter 3: DEVELOPMENT OF TRUSS EQUATIONS
  2. 3.1 The Stiffness Matrix for a Bar Element We will now consider the bar element for the linear- elastic, constant cross-sectional area A , modulus of elasticity E, and initial length L. The nodal degrees of freedom are local axial displace- ments (longitudinal displacements directed along the length of the bar) represented by u1 and u2 at the ends of the element as shown in figure. 1
  3. 3.1 The Stiffness Matrix for a Bar Element Hooke’s law xx= E du The strain/displacement relationship  = x dx Force equilibrium A x == T const The differential equation governing the linear-elastic bar behavior as: d du AE = 0 dx dx 2
  4. 3.1 The Stiffness Matrix for a Bar Element The following assumptions are used in deriving the bar element stiffness matrix: + The bar cannot sustain shear force or bending moment + Any effect of transverse displacement is ignored + Hooke’s law applies; that is, axial stress and axial strain + No intermediate applied loads 3
  5. 3.1 The Stiffness Matrix for a Bar Element Step 1 Select the Element Type Step 2 Select a Displacement Function Assume a linear displacement variation along the x axis of the bar u = a12 +a x The total number of coefficients equal to the total number of degrees of freedom associated with the element. 4
  6. 3.1 The Stiffness Matrix for a Bar Element uu− u1 21 =u N N uu= 1 +x 12 L u2 with shape functions given by xx NN=1; − = 12LL 5
  7. 3.1 The Stiffness Matrix for a Bar Element Step 3 Define the Strain/Displacement and Stress/Strain Relationships The Strain/Displacement relationships du uu21−  x == (3.1) dx L The Stress/Strain relationships xx= E (3.2) 6
  8. 3.1 The Stiffness Matrix for a Bar Element Step 4 Derive the Element Stiffness Matrix and Equations The axial force TA=  x (3.3) Using Eqs. (3.1) and (3.2) in Eq. (3.3), we obtain uu21− T= AE (3.4) L 7
  9. 3.1 The Stiffness Matrix for a Bar Element Assume equilibrium node of the element bar uu21− f1x = − T = − AE (3.5) L uu21− f2x == T AE (3.6) L The nodal force in matrix form f1x AE 11− u1 = (3.7) f2x L −11 u2 8
  10. 3.1 The Stiffness Matrix for a Bar Element f1x AE 11− u1 = (3.7) f2x L −11 u2 Equation 3.7 is written:  f =  k d The stiffness matrix for a bar element in local coordinates AE 11− k = L −11 EA/L for a bar element is analogous to the spring constant k for a spring element. 9
  11. 3.1 The Stiffness Matrix for a Bar Element Step 5 Assemble Element Equations to Obtain Global or Total Equations Assemble the global stiffness and force matrices and global equations using the direct stiffness method described in Chapter 2 N N ()e ()e Kk =  Ff =   e=1 e=1 10
  12. 3.1 The Stiffness Matrix for a Bar Element Step 6 Solve for the Nodal Displacements Determine the displacements by imposing boundary conditions and simultaneously solving a system of equations: F =  K d Step 7 Solve for the Element Forces Determine the strains and stresses in each element by back-substitution of the displacements 11
  13. Example 1 For the three-bar assemblage shown, determine (a) the global stiffness matrix, (b) the displacements of nodes 2 and 3, and (c) the reactions at nodes 1 and 4. A force of 3000lb is applied in the x direction at node 2. The length of each element is 30in. Let E =30.106psi and A =1in for elements 1 and 2,and let E = 15 106psi and A =2in. Nodes 1 and 4 are fixed 12
  14. Solution a) The global stiffness matrix The element stiffness matrices are 13
  15. Assembling the element stiffness matrices by the direct stiffness method, we obtain the global stiffness matrix as: 14
  16. b) The displacements of nodes 2 and 3 Equation equilibrium of the three-bar assemblage in global coordinate: The boundary conditions u14 =u =0 15
  17. Applying the boundary conditions, we solve equations 2 and 3 of the equation equilibrium: Solving Eq simultaneously for the displacements yields 16
  18. c) The reactions at nodes 1 and 4 Back-substituting u1, u4 into the equation equilibrium , we obtain the global nodal forces, which include the reactions at nodes 1 and 4, as follows: 17
  19. Homeworks Problem 01 For the bar assemblages shown in Figures, determine the nodal displacements, the forces in each element, and the reactions. Use the direct stiffness method for these problems. 18
  20. Problem 02 Solve for the axial displacement and stress in the tapered bar shown in Figure, using one and then two constant-area elements. Evaluate the area at the center of each element length. Use that area for each element. 7 Let A0=2in, L =20 in,E = 10 psi, and P = 1000 lb. Compare your finite element solutions with the exact solution. 23
  21. 3.2 Transformation of vectors in two dimensions Ox’y’: local coordinates Oxy: global coordinates 24
  22. 3.2 Transformation of vectors in two dimensions We can express vector displacement d in both global and local coordinates by: d= ui + vj = u'''' i + v j 25
  23. 3.2 Transformation of vectors in two dimensions Now a is in the i’ direction and b is in the j’ direction 26
  24. 3.2 Transformation of vectors in two dimensions d= ui + vj = u'''' i + v j 27
  25. 3.2 Transformation of vectors in two dimensions Equation relates the global displacement matrix {d} to the local displacement {d’} as Where: The matrix [T] is called the transformation matrix 28
  26. 3.2 Transformation of vectors in two dimensions For the case of v’=0, we have: We then obtain the magnitude of u’ as 29
  27. Example The global nodal displacements at node 2 have been determined to be u2=0,1 in. And v2 =0,2 in. for the bar element shown in figure flowing . Determine the local x displacement at node 2. 30
  28. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane We now consider a bar inclined at an angle  from the global x axis identified by the local axis x’ directed from node 1 to node 2 along the direction of the bar, as shown in figure: 31
  29. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane Equation equilibrium a bar element in the local coordinate 32
  30. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane Equation equilibrium a bar element in the global coordinate 33
  31. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane By using relationships between local and global force components and between local and global displacement components, we will be able to obtain the global stiffness matrix. 34
  32. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane We replace local and global displacements with local and global forces and obtain Equation equilibrium a bar element in the global coordinate 35
  33. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane Equation equilibrium a bar element in the global coordinate We must invert [T]. This is not immediately possible because [T] is not a square matrix. Using equation for each nodal displacement, we thus obtain 36
  34. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane Equation equilibrium a bar element in the local coordinate for each nodal displacement, f’1y=f’2y=0 T where [T] is the transpose of [T] 37
  35. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane Equation equilibrium a bar element in the global coordinate We obtain the global stiffness matrix for an element as 38
  36. 3.3 Global Stiffness Matrix for Bar Arbitrarily Oriented in the Plane The stiffness matrix for each element can be summed by using the direct stiffness method to obtain Similarly each element global nodal force matrix can be summed such that Equation equilibrium for structure in the global coordinate 39
  37. Example For the bar element shown in figure, evaluate the global stiffness matrix with respect to the x-y coordinate system. Let the bar’s cross-sectional area equal 2 in2. length equal 60 in., and modulus of elasticity equal 30.106psi. The angle the bar makes with the x axis is 300. 40
  38. Solution With angle  defined to be positive when measured counterclockwise from x to x’ 41
  39. Homeworks For each of the bar elements shown in figure, evaluate the global x-y stiffness matrix. 42
  40. 3.4 Computation of Stress for a Bar in the x-y Plane Equation equilibrium for bar element in local coordinate The usual definition of axial tensile stress is axial force divided by cross-sectional area. With: 43
  41. 3.4 Computation of Stress for a Bar in the x-y Plane The stress in the bar element in local coordinate The stress in the bar element in global coordinate 44
  42. Example For the bar shown in Figure 3–14, determine the axial stress. Let A = 4.10-4m2 , E= 210 GPa, and L = 2 m, and let the angle between x and x’ be 60. Assume the global displacements have been previously determined to be u1=0.25 mm, v1= 0, u2= 0.50 mm, and v2= 0.75 mm. 45
  43. Solution We fist caculate [C’] With: The global displacements are given by: 46
  44. The bar axial stress 47
  45. Homeworks Determine the axial stress in each of the bar elements shown in figure 48
  46. 3.5 Solution of plane truss “A plane truss is a structure composed of bar elements that all lie in a common plane and are connected by frictionless pins” The plane truss also must have loads acting only in the common plane and all loads must be applied at the nodes or joints. 49
  47. Example 01 For the plane truss composed of the three elements shown in figure subjected to a downward force of 10000 lb applied at node 1, determine the x and y displacements at node 1 and the stresses in each element. Let E = 30 .10 6 psi and A =22 in2. for all elements. The lengths of the elements are shown in the figure. 50
  48. Solution *The global stiffness matrices for structure The angle  between the global x axis and the local x’ axis for each element No. element Conection 1 1 2 2 1 3 3 1 4 51
  49. *The global stiffness matrices for each element Element 1 Element 2 52
  50. Element 3 The global stiffness matrices for structure 53
  51. Equation equilibrium for structure, when apllied force at node 1 and the boundary constraints at nodes 2–4 as follows 54
  52. We then obtain Solution yields the displacements We determine the stresses in each element as follows: 55
  53. Example 02 Determine the displacement in the y direction of node 1 and the axial force in each element. A force of P = 1000 kN is applied at node 1 in the positive y direction while node 1 settles an amount =50mm in the negative x direction. Let E = 210 GPa and A =6.10-4 m2 for each element. The lengths of the elements are shown in the figure. 57
  54. Solution *The global stiffness matrices for each element Element 1 58
  55. Element 2 59
  56. *The global stiffness matrices for structure The boundary conditions 60
  57. The local element forces are obtained Element 1 Element 2 61
  58. Example 03 To illustrate how we can combine spring and bar elements in one structure, we now solve the two-bar truss supported by a spring shown in figure. Both bars have E =210 GPa and A = 5.10-4m2. Bar one has a length of 5 m and bar two a length of 10 m. The spring stiffness is k=2000 kN/m. 62
  59. Solution The global stiffness matrices for each element Element 1 Element 2 63
  60. Element 3 Applying the boundary conditions, we have 64
  61. Solving equation for the global displacements, we obtain We can obtain the stresses in the bar elements Element 1 Element 2 65
  62. Homeworks Assemble the stiffness matrix for the assemblage shown in figure by superimposing the stiffness matrices of the springs. Here k is the stiffness of each spring. Find the x and y components of deflection of node 1 45N 66
  63. For the plane truss structure shown in figure, determine the displacement of node 2 using the stiffness method. Also determine the stress in element 1. Let A=5 in2, E=7000 N/mm2 , L=100 in 45KN 67
  64. Find the horizontal and vertical displacements of node 1 for the truss shown in figure. Assume AE is the same for each element 630N 68
  65. For the truss shown in figure solve for the horizontal and vertical components of displacement at node 1 and 4,5KN determine the stress in each element Also verify force 4,5KN equilibrium at node 1. All elements have A=1 in2, E=70000 N/mm2 , L=100 in 69
  66. For the truss shown in figure, solve for the horizontal and vertical components of displacement at node 1. Also determine the stress in element 1. Let A =1in2 ,E=70000 N/mm2 , L=100 in 53KN 70
  67. Determine the nodal displacements and the element forces for the truss shown in figure. Assume all elements have the same AE 4,5KN 4,5m 6m 4,5KN 71
  68. Determine the displacement components at node 3 and the element forces for the plane truss shown in figure.Let A =3in2 ,E=207 KN/mm2 6m 22KN 44KN 12m 9m 9m 72
  69. For the plane trusses shown in figure, determine the horizontal and vertical displacements of node 1 and the stresses in each element. All elements A =4.10-4m2 ,E=210 GPa 73
  70. 3.6 Stiffness Matrix for a Bar in 3D Bar in three-dimensional space Ox’y’z’: local coordinates Oxyz: global coordinates 74
  71. 3.6 Stiffness Matrix for a Bar in 3D Relation displacement in local and global coordinate We begin the derivation of [T] by considering the vector d expressed in three dimensions as: The dot product of equation with i’ 75
  72. 3.6 Stiffness Matrix for a Bar in 3D By definition of the dot product Where: We have 76
  73. 3.6 Stiffness Matrix for a Bar in 3D We can write the local axial displacement at node 1 and 2 in explicit form as We write equation in matrix form 77
  74. 3.6 Stiffness Matrix for a Bar in 3D The global force matrix in terms of the local force matrix using [T] as: Now in local coordinates, the local forces are related to the local displacements by 78
  75. 3.6 Stiffness Matrix for a Bar in 3D The global forces are related to the global displacements by: [k] the global stiffness matrix 79
  76. 3.6 Stiffness Matrix for a Bar in 3D We obtain the explicit form of [k] as 80
  77. Example We will now determine the stiffness matrices of the three elements in following figure 81
  78. Solution Element 1 82
  79. Element 2 83
  80. Element 3 84
  81. We have the total stiffness matrix for the truss as We obtain the displacements: 85
  82. The stress of three dimension element 86
  83. Excercises For the space trusses shown in figures, determine the nodal displacements and the stresses in each element. E=210Gpa, A=10-3m2 87