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Using thesame node and element numbering as in Figure 3.2, determine the displacement components of node 3, the reaction force components at nodes 1 and 2, and the elementdisplacements, stresses, and forces. The elements have modulus of elasticity E1 = E2 =10 × 106 lb/in.2 and cross-sectional areas A1 = A2 = 1.5 in.2.Hutton: Fundamentals ofFinite Element Analysis703. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Method300 lb(40, 40)(0, 40)223500 lb11(0, 0)Figure 3.5 Two-element truss withexternal loading.■ SolutionThe nodal√ coordinates are such that ␪1 = ␲/4 and ␪2 = 0 and the element lengths areL 1 = 402 + 402 ≈ 56.57 in., L 2 = 40 in. The characteristic element stiffnesses are thenk1 =A1 E 11.5(10)(10 6 )== 2.65(10 5 ) lb/in.L156.57k2 =A2 E 21.5(10)(10 6 )== 3.75(10 5 ) lb/in.L240As the element orientation angles and numbering scheme are the same as in Example 3.1,we use the result of that example to write the global stiffness matrix as1.3251.32500 −1.325 −1.3251.32500 −1.325 −1.325  1.32503.75 0 −3.750 105 lb/in.[K ] =  0 000000  −1.325 −1.325 −3.75 0 5.0751.325 −1.325 −1.32500 1.3251.325Incorporating the displacement constraints U 1 = U 2 = U 3 = U 4 = 0 , the global equilibrium equations are  1.3251.32500哸−1.325 −1.325 F01哸 1.3251.32500 −1.325 −1.325 0F2哸003.750−3.75050F3=10 哸0000哸 00 0 F4  哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹哹 −1.325 −1.325 −3.75 0 5.0751.325 U500哸5  −1.325 −1.32500哸 1.3251.325U6300Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.6Element Strain and Stressand the dashed lines indicate the partitioning technique of Equation 3.45. Hence, theactive displacements are governed by10 55.0751.3251.3251.325U5U6=500300Simultaneous solution gives the displacements asU 5 = 5.333 × 10 −4 in.andU 6 = 1.731 × 10 −3 in.As all the constrained displacement values are zero, the reaction forces are obtained viaEquation 3.47 as F1 F −1.325 −1.325−300  −1.325 −1.325  0.5333−300 2−310== {Fc } = [K ca ]{Ua } = 105 lb −3.75F 0  1.731−200  3F4000and we note that the net force on the structure is zero, as required for equilibrium.

A checkof moments about any of the three nodes also shows that moment equilibrium is satisfied.For element 1, the element displacements in the element coordinate system areu (1)1(1)u2= R (1)0 √ 2 1 1 0 0  0  −30210−3 in.10 == 0.5333 U5 1.62 0 0 1 1  1.731U6 U 1 U Element stress is computed using Equation 3.52:␴ (1)  U1 1 (1)  U2 RL1 U5 U61= E1 −L1Using the element displacements just computed, we have␴ (1) = 10(10 6 ) −156.57156.57010 −3 ≈ 283 lb/in. 21.6and the positive results indicate tensile stress.The element nodal forces via Equation 3.23 areff(1)1(1)2==k1−k 1−424424−k 1k1(1)1(1)u 2u1= 2.65(10 )−15−11010 −31.6lband the algebraic signs of the element nodal forces also indicate tension.71Hutton: Fundamentals ofFinite Element Analysis723.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodFor element 2, the same procedure in sequence gives U 1 U 0 u101 0 0 0  0  −32= R (2)10−4 in.=10 =(2)U0.533300100.53335u2 U61.73101110−3 ≈ 133 lb/in.2␴ (2) = 10(106 ) −0.533340 40 (2) (2) u1f 1k2 −k21 −1−2000== 3.75(105 )10−3 =lb−k2 k2−1 12000.5333f (2)u (2)22(2)also indicating tension.The finite method is intended to be a general purpose procedure for analyzing problems for which the general solution is not known; however, it is informative in the examples of this chapter (since the bar element poses an exact formulation) to check thesolutions in terms of axial stress computed simply as F/ A for an axially loaded member.The reader is encouraged to compute the axial stress by the simple stress formula for eachexample to verify that the solutions via the stiffness-based finite element method arecorrect.3.7 COMPREHENSIVE EXAMPLEAs a comprehensive example of two-dimensional truss analysis, the structure depicted in Figure 3.6a is analyzed to obtain displacements, reaction forces, strains,and stresses.

While we do not include all computational details, the exampleillustrates the required steps, in sequence, for a finite element analysis.Y6000 lb40 in.U8U440 in.34000 lbU7U3240 in.26(a)8U10U612000 lb16 U114U22000 lbU1274U153U55U9X(b)Figure 3.6(a) For each element, A = 1.5 in.2, E = 10 × 106 psi. (b) Node, element, and global displacement notation.Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.7 Comprehensive ExampleStep 1. Specify the global coordinate system, assign node numbers, anddefine element connectivity, as shown in Figure 3.6b.Step 2. Compute individual element stiffness values:k (1) = k (3) = k (4) = k (5) = k (7) = k (8) =k (2) = k (6) =1.5(10 7 )= 3.75(10 5 ) lb/in.401.5(10 7 )= 2.65(10 5 ) lb/in.√40 2Step 3. Transform element stiffness matrices into the global coordinatesystem. Utilizing Equation 3.28 with␪1 = ␪3 = ␪5 = ␪7 = 0␪4 = ␪8 = ␲/2␪2 = ␲/4we obtain1 (1) (3) (5) (7) 5  0K= K= K= K= 3.75(10 ) −100 0 0 0 (4) (8) 0 1 0 −1 K= K= 3.75(105 ) 0 0 0 0 0 −1 012.65(105 ) 1K= −12−11 (6) 2.65(105 )  −1=K −121(2)␪6 = 3␲/40 −1 00 0 00 1 00 0 011 −1 −11 −1 −1 −1 11 −1 11−1 −1 111 −1 11 −1 −1 −1 1Step 4a. Construct the element-to-global displacement correspondence table.With reference to Figure 3.6c, the connectivity and displacementrelations are shown in Table 3.3.Step 4b.

Alternatively and more efficiently, form the element-nodeconnectivity table (Table 3.4), and the corresponding element globaldisplacement location vector for each element isL (1) = [1256]L(2)= [1278]L(3)= [3478]L (4) = [5678]73Hutton: Fundamentals ofFinite Element Analysis743. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodTable 3.3 Connectivity and Displacement RelationsGlobal Elem.

1 Elem. 2 Elem. 3 Elem. 4 Elem. 5 Elem. 6 Elem. 7 Elem. 8123456789101112120034000000120000340000001200340000000012340000000012003400000000341200000000120034000000001234Table 3.4 Element-Node ConnectivityNodeElementij123456781123354534445466L (5) = [56L (6) = [91078]81112]L(7)= [7L (8) = [910910]1112]Step 5. Assemble the global stiffness matrix per either Step 4a or 4b. Theresulting components of the global stiffness matrix are(1)(2)(1)(2)K 11 = k 11 + k 11 = (3.75 + 2.65/2)10 5K 12 = k 12 + k 12 = (0 + 2.65/2)10 5K 13 = K 14 = 0(1)K 15 = k 13 = −3.75(10 5 )(1)K 16 = k 14 = 0(2)K 17 = k 13 = −(2.65/2)10 5Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.7 Comprehensive Example(2)K 18 = k 14 = −(2.65/2)10 5K 19 = K 1,10 = K 1,11 = K 1,12 = 0(1)(2)K 22 = k 22 + k 22 = 0 + (2.65/2)10 5K 23 = K 24 = 0(1)K 25 = k 23 = 0(1)K 26 = k 24 = 0(2)K 27 = k 23 = −(2.65/2)10 5(2)K 28 = k 24 = −(2.65/2)10 5K 29 = K 2,10 = K 2,11 = K 2,12 = 0(3)K 33 = k 11 = 3.75(10 5 )(3)K 34 = k 12 = 0K 35 = K 36 = 0(3)K 37 = k 13 = −3.75(10 5 )(3)K 38 = k 14 = 0K 39 = K 3,10 = K 3,11 = K 3,12 = 0(3)K 44 = k 22 = 0K 45 = K 46 = 0(3)K 47 = k 23 = 0(3)K 48 = k 24 = 0K 49 = K 4,10 = K 4,11 = K 4,12 = 0(1)(4)(5)(1)(4)(5)K 55 = k 33 + k 11 + k 11 = (3.75 + 0 + 3.75)10 5K 56 = k 34 + k 12 + k 12 = 0 + 0 + 0 = 0(4)K 57 = k 13 = 0(4)K 58 = k 14 = 0(5)K 59 = k 13 = −3.75(10 5 )(5)K 5,10 = k 14 = 0K 5,11 = K 5,12 = 0(2)(4)(5)K 66 = k 44 + k 22 + k 22 = (0 + 3.75 + 0)10 5(4)K 67 = k 23 = 0(4)K 68 = k 24 = −3.75(10) 575Hutton: Fundamentals ofFinite Element Analysis763.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Method(5)K 69 = k 23 = 0(5)K 6,10 = k 24 = 0K 6,11 = K 6,12 = 0(2)(3)(4)(6)(7)K 77 = k 33 + k 33 + k 33 + k 33 + k 11= (2.65/2 + 3.75 + 0 + 2.65/2 + 3.75)10 5(2)(3)(4)(6)(7)K 78 = k 34 + k 34 + k 34 + k 34 + k 12= (2.65/2 + 0 + 0 − 2.65/2 + 0)10 5 = 0(6)K 79 = k 13 = −(2.65/2)10 5(6)K 7,10 = k 23 = (2.65/2)10 5(7)K 7,11 = k 13 = −3.75(10 5 )(7)K 7,12 = k 14 = 0(2)(3)(4)(6)(7)K 88 = k 44 + k 44 + k 44 + k 44 + k 22= (2.65/2 + 0 + 3.75 + 2.65/2 + 0)10 5(6)K 89 = k 14 = (2.65/2)10 5(6)K 8,10 = k 24 = −(2.65/2)10 5(7)K 8,11 = k 23 = 0(7)K 8,12 = k 24 = 0(5)(6)(8)(5)(6)(8)K 99 = k 33 + k 11 + k 11 = (3.75 + 2.65/2 + 0)10 5K 9,10 = k 34 + k 12 + k 12 = (0 − 2.65/2 + 0)10 5(8)K 9,11 = k 13 = 0(8)K 9,12 = k 14 = 0(5)(6)(8)K 10,10 = k 44 + k 22 + k 22 = (0 + 2.65/2 + 3.75)10 5(8)K 10,11 = k 23 = 0(8)K 10,12 = k 24 = −3.75(10 5 )(7)(8)(7)(8)(7)(8)K 11,11 = k 33 + k 33 = (3.75 + 0)10 5K 11,12 = k 34 + k 34 = 0 + 0K 12,12 = k 44 + k 44 = (0 + 3.75)10 5Step 6.

Apply the constraints as dictated by the boundary conditions. In thisexample, nodes 1 and 2 are fixed so the displacement constraints areU1 = U2 = U3 = U4 = 0Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.7 Comprehensive ExampleTherefore, the first four equations in the 12 × 12 matrix system[K ] {U } = {F }are constraint equations and can be removed from consideration sincethe applied displacements are all zero (if not zero, the constraints areconsidered as in Equation 3.46, in which case the nonzero constraintsimpose additional forces on the unconstrained displacements).

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