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Theconstraint forces cannot be obtained until the unconstraineddisplacements are computed. So, we effectively strike out thefirst four rows and columns of the global equations to obtain U5 0 U6 −2000 U07 U80=[K aa ] U9   2000 U0 10U400011 U126000as the system of equations governing the “active” displacements.Step 7. Solve the equations corresponding to the unconstraineddisplacements. For the current example, the equations are solvedusing a spreadsheet program, inverting the (relatively small) globalstiffness matrix to obtain U5 0.02133  0.04085  U6   −0.01600  U7   0.04619U8=in.0.04267 U9  0.15014 U10   U11   −0.00533  0.16614U12Step 8.

Back-substitute the displacement data into the constraint equationsto compute reaction forces. Utilizing Equation 3.37, with {U c } = {0} ,we use the four equations previously ignored to compute the forcecomponents at nodes 1 and 2. The constraint equations are of the formK i5 U 5 + K i6 U 6 + · · · + K i,12 U 12 = Fii = 1, 4and, on substitution of the computed displacements, yield  F−12,000  1 F2 = −4,000 lb F3   6,000 F4077Hutton: Fundamentals ofFinite Element Analysis783. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodThe reader is urged to utilize these reaction force components andcheck the equilibrium conditions of the structure.Step 9.

Compute strain and stress in each element. The major computationaltask completed in Step 7 provides the displacement components ofeach node in the global coordinate system. With this information andthe known constrained displacements, the displacements of eachelement in its element coordinate system can be obtained; hence, thestrain and stress in each element can be computed.For element 2, for example, we haveu(2)1(2)u2= U 1 cos ␪2 + U 2 sin ␪2 = 0√= U 7 cos ␪2 + U 8 sin ␪2 = (−0.01600 + 0.04618 ) 2/2= 0.02134The axial strain in element 2 is thenε(2) =u(2)2−uL (2)(2)10.02133√ = 3.771(10 −4 )40 2=and corresponding axial stress is␴ (2) = E ε(2) = 3771 psiThe results for element 2 are presented as an example only. In finiteelement software, the results for each element are available andcan be examined as desired by the user of the software(postprocessing).Results for each of the eight elements are shown in Table 3.5; andper the usual sign convention, positive values indicate tensile stresswhile negative values correspond to compressive stress.

In obtainingthe computed results for this example, we used a spreadsheet programto invert the stiffness matrix, MATLAB to solve via matrix inversion,and a popular finite element software package. The solutions resultingfrom each procedure are identical.Table 3.5 Results for the Eight ElementsElement12345678Strain−45.33(10 )3.77(10−4)−4.0(10−4)1.33(10−4)5.33(10−4)−5.67(10−4)2.67(10−4)4.00(10−4)Stress, psi53333771−400013335333−565726674000Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.8 Three-Dimensional Trusses3.8 THREE-DIMENSIONAL TRUSSESThree-dimensional (3-D) trusses can also be modeled using the bar element,provided the connections between elements are such that only axial load is transmitted. Strictly, this requires that all connections be ball-and-socket joints. Evenwhen the connection restriction is not precisely satisfied, analysis of a 3-D trussusing bar elements is often of value in obtaining preliminary estimates of member stresses, which in context of design, is valuable in determining requiredstructural properties.

Referring to Figure 3.7 which depicts a one-dimensionalbar element connected to nodes i and j in a 3-D global reference frame, the unitvector along the element axis (i.e., the element reference frame) expressed in theglobal system is␭(e) =1[( X j − X i )I + (Y j − Yi )J + ( Z j − Z i )K]L(3.53)␭(e) = cos ␪x I + cos ␪y J + cos ␪z K(3.54)orThus, the element displacements are expressed in components in the 3-D globalsystem asu(e)1= U 1 cos ␪x + U 2 cos ␪y + U 3 cos ␪z(e)(e)(e)(3.55)u(e)2= U 4 cos ␪x + U 5 cos ␪y + U 6 cos ␪z(e)(e)(e)(3.56)Here, we use the notation that element displacements 1 and 4 are in the global Xdirection, displacements 2 and 5 are in the global Y direction, and elementdisplacements 3 and 6 are in the global Z direction.␭(e)U3j⫺1U3jU3i⫺1U3iU3j⫺2␪Y␪XiYjU3i⫺2␪ZXZFigure 3.7 Bar element in a 3-D global coordinatesystem.79Hutton: Fundamentals ofFinite Element Analysis803.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodAnalogous to Equation 3.21, Equations 3.55 and 3.56 can be expressed as (e) U1 (e) U2  (e)  (e) u1U000cos ␪x cos ␪y cos ␪z3=000cos ␪x cos ␪y cos ␪z u (e)U (e)24 (e) U5 (e) U6 (e) = [R] U(3.57)where [R] is the transformation matrix mapping the one-dimensional elementdisplacements into a three-dimensional global coordinate system. Following theidentical procedure used for the 2-D case in Section 3.3, the element stiffnessmatrix in the element coordinate system is transformed into the 3-D global coordinates via (e) ke−k eTK= [R][R](3.58)−k ekeSubstituting for the transformation matrix [R] and performing the multiplicationresults in 2cxcx c yc x cz−cx2−cx c y −cx czc2yc y cz −cx cx−c2y−c y cz  cx c y c x cz (e) c y czcz2−cx cz −c y cz−cz2  (3.59)K= ke −cx cx −cx czcx2cx c yc x cz  −cx2 −cx c y−c2y−c y cz cx c yc2yc y cz −cx cz−c y cz−cz2c x czc y czcz2as the 3-D global stiffness matrix for the one-dimensional bar element wherec x = cos ␪xc y = cos ␪yc z = cos ␪z(3.60)Assembly of the global stiffness matrix (hence, the equilibrium equations),is identical to the procedure discussed for the two-dimensional case with the obvious exception that three displacements are to be accounted for at each node.EXAMPLE 3.3The three-member truss shown in Figure 3.8a is connected by ball-and-socket joints andfixed at nodes 1, 2, and 3.

A 5000-lb force is applied at node 4 in the negative Y direction,as shown. Each of the three members is identical and exhibits a characteristic axial stiffness of 3(105) lb/in. Compute the displacement components of node 4 using a finiteelement model with bar elements.Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.8 Three-Dimensional Trusses81U52(0, 0, ⫺30)2U4U6Y2ZXU21U11114(40, 0, 0)(0, 0, 30)43 (0, ⫺30, 0)U3U1U10U8335000 lbU7U9(b)(a)Figure 3.8(a) A three-element, 3-D truss. (b) Numbering scheme.■ SolutionFirst, note that the 3-D truss with four nodes has 12 possible displacements. However,since nodes 1–3 are fixed, nine of the possible displacements are known to be zero.

Therefore, we need assemble only a portion of the system stiffness matrix to solve for the threeunknown displacements. Utilizing the numbering scheme shown in Figure 3.8b and theelement-to-global displacement correspondence table (Table 3.6), we need consider onlythe equationsK 10, 10 K 11, 10K 12, 10K 10, 11K 11, 11K 12, 11 K 10, 12  U10   0 K 11, 12  U11 = −5000 K 12, 12U120Prior to assembling the terms required in the system stiffness matrix, the individualelement stiffness matrices must be transformed to the global coordinates as follows.Element 1␭ (1) =1[(40 − 0)I + (0 − 0)J + (0 − 30)K] = 0.8I − 0.6K50Hence, c x = 0.8, c y = 0, c z = −0.6 , and Equation 3.59 gives0.640 −0.48 −0.64000 0 (1) 00.360.485  −0.48K= 3(10 )  −0.64 00.480.64 00000.48 −0 −0.36 −0.480 0.4800 0 −0.36  lb/ln.0 −0.48 00 0 0.36U12Hutton: Fundamentals ofFinite Element Analysis823.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodTable 3.6 Element-to-Global Displacement CorrespondenceGlobal DisplacementElement 1Element 2Element 3123456789101112123000000456000123000456000000123456Element 21[(40 − 0)I + (0 − 0)J + (0 − (−30))K] = 0.8I + 0.6K500.64 0 0.48 −0.64 0 −0.4800000  0 (2) = 3(105 )  0.48 0 0.36 −0.48 0 −0.36  lb/in.K −0.64 0 −0.48 0.64 0 0.48  000000 −0.48 0 −0.36 0.48 0 0.36␭ (2) =Element 31[(40 − 0)I + (0 − (−30))J + (0 − 0)K] = 0.8I + 0.6J500.640.48 0 −0.64 −0.48 00.36 0 −0.48 −0.36 0  0.48 (3) 00000= 3(105 )  0K lb/in. −0.64 −0.48 0 0.640.48 0  −0.48 −0.36 0 0.480.36 0 000000␭ (3) =Referring to the last three rows of the displacement correspondence table, the requiredterms of the global stiffness matrix are assembled as follows:(1)(2)(3)K 10, 10 = k44+ k44+ k44= 3(105 )(0.64 + 0.64 + 0.64) = 5.76(105 ) lb/in.(1)(2)(3)+ k45+ k45= 3(105 )(0 + 0 + 0.48) = 1.44(105 ) lb/in.K 10, 11 = K 11, 10 = k45(1)(2)(3)+ k46+ k46= 3(105 )(−0.48 + 0.48 + 0) = 0 lb/in.K 10, 12 = K 12, 10 = k46(1)(2)(3)+ k55+ k55= 3(105 )(0 + 0 + 0.36) = 1.08(105 ) lb/in.K 11, 11 = k55(1)(2)(3)+ k56+ k56= 3(105 )(0 + 0 + 0) = 0 lb/in.K 11, 12 = K 12, 11 = k56(1)(2)(3)+ k66+ k66= 3(105 )(0.36 + 0.36 + 0) = 2.16(105 ) lb/in.K 12, 12 = k66Hutton: Fundamentals ofFinite Element Analysis3.

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