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Stiffness Matrix for a Bar Element

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Stiffness Matrix for a Bar Element CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 1/44 Chapter 3b – Development of Truss Equations Learning Objectives • To derive the stiffness matrix for a bar element. • To illustrate how to solve a bar assemblage by the direct stiffness method. • To introduce guidelines for selecting displacement functions. • To describe the concept of transformation of vectors in two different coordinate systems in the plane. • To derive the stiffness matrix for a bar arbitrarily oriented in the plane. • To demonstrate how to compute stress for a bar in the plane. • To show how to solve a plane truss problem. • To develop the transformation matrix in three- dimensional space and show how to use it to derive the stiffness matrix for a bar arbitrarily oriented in space. • To demonstrate the solution of space trusses. Stiffness Matrix for a Bar Element Inclined, or Skewed Supports If a support is inclined, or skewed, at some angle for the global x axis, as shown below, the boundary conditions on the displacements are not in the global x-y directions but in the x’-y’ directions. CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 2/44 Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports We must transform the local boundary condition of v’3 = 0 (in local coordinates) into the global x-y system. Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports Therefore, the relationship between of the components of the displacement in the local and the global coordinate systems at node 3 is: uu'33cos sin vv'33sin cos We can rewrite the above expression as: cos sin t dtd'[]333 3 sin cos We can apply this sort of transformation to the entire displacement vector as: T dTdordTd'[] 11 []' CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 3/44 Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports T Where the matrix [T1] is: []I [0] [0] []TIT [0][][0] 1 [0] [0] [t3 ] Both the identity matrix [I] and the matrix [t3] are 2 x 2 matrices. The force vector can be transformed by using the same transformation. fTf'[] 1 In global coordinates, the force-displacement equations are: fKd [] Stiffness Matrix for a Bar Element Inclined, or Skewed, Supports Applying the skewed support transformation to both sides of the equation gives: fKd [] []Tf11 [][] TKd By using the relationship between the local and the global displacements, the force-displacement equations become: T T dTd []1 ' fTKTd'[][][]' 11 Therefore the global equations become: F1x u1 F v 1y 1 F2x T u2 [][][]TKT11 F 2y v2 F '3x u ' 3 F ' 3y v '3 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 4/44 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem Consider the plane truss shown below. Assume E = 210 GPa, A = 6 x 10-4 m2 for element 1 and 2, and A = 2 (6 x 10-4)m2 for element 3. Determine the stiffness matrix for each element. CC22CS CS 22 AE CSSS CS k LCSCSCC22 22 CSSS CS Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 1 is: CC22CS CS 22 AE CSSS CS k LCSCSCC22 22 CSSS CS cos (1) 0 sin (1) 1 uvuv1122 0000 62 42 (210 10kN / m )(6 10 m ) 0101 k(1) 1 m 0000 0101 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 5/44 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 2 is: CC22CS CS 22 AE CSSS CS k LCSCSCC22 22 CSSS CS cos (2) 1 sin (2) 0 uv22 u 3 v 3 10 10 62 42 (210 10kN / m )(6 10 m ) 00 00 k(2) 1 m 10 10 00 00 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The global elemental stiffness matrix for element 3 is: CC22CS CS 22 AE CSSS CS k LCSCSCC22 22 CSSS CS 2 2 cos (3) sin (3) 2 2 uv11 uv 3 3 1 111 62 42 (210 10kN / m )(6 2 10m ) 1 111 k(3) 22m 1111 1111 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 6/44 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem Using the direct stiffness method, the global stiffness matrix is: 5.0 5.0 0 0 5.0 5.0 5.0 5.1 0 1 5.0 5.0 0 0 1 0 1 0 K 10260,1 5 N m 0 1 0 1 0 0 5.0 5.0 1 0 5.1 5.0 5.0 5.0 0 0 5.0 5.0 We must transform the global displacements into local coordinates. Therefore the transformation [T1] is: 1000 0 0 0100 0 0 []I [0] [0] 00 10 0 0 []TI1 [0][][0] 000 1 0 0 [0] [0] [t ] 3 22 0000 22 22 0000 22 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The first step in the matrix transformation to find the product of [T1][K]. T1 K 1000 0 0 0.5 0.5 0 0 0.5 0.5 0100 0 00.5 1.5 0 1 0.5 0.5 00 10 0 0 00 1010 TK1, 260 105 N 1 m 010100 000 1 0 0 220.707 0.707 0.707 0 1.414 0.707 0000 22 220 0 0.707 0 0.707 0 0000 22 0.5 0.5 0 0 0.5 0.5 0.5 1.5 0 1 0.5 0.5 5 N 00 1010 TK1 1,260 10 m 010100 0.707 0.707 0.707 0 1.414 0.707 0 0 0.707 0 0.707 0 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 7/44 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The next step in the matrix transformation to find the product of T [T1][K][T1] . T TK1 T1 10 0 0 0 0 0.5 0.5 0 0 0.5 0.5 0.5 1.5 0 1 0.5 0.5 0100 0 0 T 00 1010 00 10 0 0 TKT 1, 2 6 0 1 0 5 N 1 1 m 010100 000 1 0 0 0.707 0.707 0.707 0 1.414 0.707 22 0000 22 0 0 0.707 0 0.707 0 22 0000 22 0.5 0.5 0 0 0.707 0 0.5 1.5 0 1 0.707 0 T 5 N 0 0 1 0 0.707 0.707 TK1 T 1,260 10 1 m 010100 0.707 0.707 0.707 0 1.5 0.5 0 0 0.707 0 0.5 0.5 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem The displacement boundary conditions are: uvv112 v'0 3 u F1x 0.5 0.5 0 0 0.707 0 1 v F1y 0.5 1.5 0 1 0.707 0 1 F 0 0 1 0 0.707 0.707 u2 2x 1, 2 6 0 1 0 5 N F m v 2y 010100 2 F ' 0.707 0.707 0.707 0 1.5 0.5 u' 3x 3 F ' 0 0 0.707 0 0.5 0.5 3y v '3 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 8/44 Stiffness Matrix for a Bar Element Example 9 – Space Truss Problem By applying the boundary conditions the global force- displacement equations are: 1 0.707 u FkN 1,000 1,260 105 N 2 2x m 0.707 1.5 u '3 F '03x Solving the equation gives: umm2 11.91 umm'5.613 Stiffness Matrix for a Bar Element F Example 9 – Space Truss Problem 1x u1 F v 1y 1 F2x T u2 [][][]TKT F 11 2y v2 The global nodal forces are calculated as: F ' u ' 3x 3 F ' 3y v '3 F1x F 0.5 0.5 0 0 0.707 0 0 1y 0.5 1.5 0 1 0.707 0 0 F 2x 2 N 0 0 1 0 0.707 0.707 11.91 1, 2 6 0 1 0 mm F 0101000 2y F ' 0.707 0.707 0.707 0 1.5 0.5 5.61 3x 0 0 0.707 0 0.5 0.5 0 F '3y Therefore: FkNFkN11xy500 500 FFk23yy0'707N CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 9/44 Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations Let’s derive the equations for a bar element using the principle of minimum potential energy. The total potential energy, p, is defined as the sum of the internal strain energy U and the potential energy of the external forces : p U The differential internal work (strain energy) dU in a one- dimensional bar element is: dU xx()()() y z x d Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations If we let the volume of the element approach zero, then: dU xx d dV Summing the differential energy over the whole bar gives: x x 1 2 UddV Ed dV EdV xx xx 2 x V 0 V 0 V For a linear-elastic material (Hooke’s law) as shown below: 1 E UdV xx xx V 2 CIVL 7/8117 Chapter 3 - Truss Equations - Part 2 10/44 Stiffness Matrix for a Bar Element Potential Energy Approach to Derive Bar Element Equations The internal strain energy statement becomes 1 UdV xx 2 V The potential energy of the external forces is: M XudV Tu dS fu bxs ixi VSi1 where Xb is the body force (force per unit volume), Tx is the traction (force per unit area), and fix is the nodal concentrated force.
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