Hutton - Fundamentals of Finite Element Analysis, страница 15

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodEXAMPLE 3.1For the truss shown in Figure 3.2, ␪1 = ␲/4, ␪2 = 0, and the element properties are suchthat k 1 = A 1 E 1 /L 1 , k 2 = A 2 E 2 /L 2 .

Transform the element stiffness matrix of each element into the global reference frame and assemble the global stiffness matrix.■ Solution√1For element 1, cos ␪1 = sin ␪1 = 2/2 and c 2 ␪1 = s 2 ␪1 = c␪1 s␪1 = , so substitution2into Equation 3.33 gives11 −1 −1 (1) k1  11 −1 −1 K= 2  −1 −1 11 −1 −1 11For element 2, cos ␪2 = 1, sin ␪2 = 0 which gives the transformed stiffness matrix as1 0 (2) K= k2 −100 −1 00 0 00 1 00 0 0Assembling the global stiffness matrix directly using Equations 3.35 and 3.36 givesK 11 = k 1 /2K 12 = k 1 /2K 13 = 0K 14 = 0K 15 = −k 1 /2K 16 = −k 1 /2K 22 = k 1 /2K 23 = 0K 24 = 0K 25 = −k 1 /2K 26 = −k 1 /2K 33 = k 2K 34 = 0K 35 = −k 2K 36 = 0K 44 = 0K 45 = 0K 46 = 0Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.4 Direct Assembly of Global Stiffness MatrixK 55 = k 1 /2 + k 2K 56 = k 1 /2K 66 = k 1 /2The complete global stiffness matrix is thenk1 /2k1 /20k1 /20 k1 /2 00k2[K ] =  000 −k /2 −k /2 −k12 1−k1 /2 −k1 /200−k1 /20−k1 /20−k200−k1 /20 k1 /2 + k2k1 /20−k1 /2 0 0 k1 /2 k1 /2The previously described embodiment of the direct stiffness method isstraightforward but cumbersome and inefficient in practice. The main probleminherent to the method lies in the fact that each term of the global stiffness matrix is computed sequentially and accomplishment of this sequential constructionrequires that each element be considered at each step.

A technique that is muchmore efficient and well-suited to digital computer operations is now described. Inthe second method, the element stiffness matrix for each element is considered insequence, and the element stiffness terms added to the global stiffness matrix perthe nodal connectivity table. Thus, all terms of an individual element stiffnessmatrix are added to the global matrix, after which that element need not be considered further. To illustrate, we rewrite Equations 3.33 and 3.34 as1256(1)(1)(1)(1)k11 k12 k13 k141(1)(1)(1)(1)  (1)  k21 k22 k23 k24  2(3.37)K=(1)(1)(1)(1)k 31 k32 k33 k34  5(1)(1)(1)(1)6k41k42k43k44K (2) =3k (2) 11 k (2) 21 k (2) 31(2)k41456(2)k12(2)k22(2)k32(2)k42(2)k13(2)k23(2)k33(2)k43(2)k14(2)k24(2)k34(2)k443 4 56(3.38)In this depiction of the stiffness matrices for the two individual elements, thenumbers to the right of each row and above each column indicate the globaldisplacement associated with the corresponding row and column of the elementstiffness matrix.

Thus, we combine the nodal displacement correspondence tablewith the individual element stiffness matrices. For the element matrices, each65Hutton: Fundamentals ofFinite Element Analysis663. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Methodindividual component is now labeled as associated with a specific row-columnposition of the global stiffness matrix and can be added directly to that location.(2)component of element 2 is to beFor example, Equation 3.38 shows that the k 24added to global stiffness component K46 (and via symmetry K64). Thus, we cantake each element in turn and add the individual components of the element stiffness matrix to the proper locations in the global stiffness matrix.The form of Equations 3.37 and 3.38 is convenient for illustrative purposesonly. For actual computations, inclusion of the global displacement numberswithin the element stiffness matrix is unwieldy.

A streamlined technique suitablefor computer application is described next. For a 2-D truss modeled by sparelements, the following conventions are adopted:1. The global nodes at which each element is connected are denoted by i and j.2.

The origin of the element coordinate system is located at node i and theelement x axis has a positive sense in the direction from node i to node j.3. The global displacements at element nodes are U2i−1, U2i, U2j−1, and U2jas noted in Section 3.2.Using these conventions, all the information required to define element connectivity and assemble the global stiffness matrix is embodied in an elementnode connectivity table, which lists element numbers in sequence and shows theglobal node numbers i and j to which each element is connected. For the twoelement truss of Figure 3.2, the required data are as shown in Table 3.2.Using the nodal data of Table 3.2, we define, for each element, a 1 × 4 element displacement location vector as (e) L= [2i − 1 2i 2 j − 1 2 j ](3.39)where each value is the global displacement number corresponding to elementstiffness matrix rows and columns 1, 2, 3, 4 respectively.

For the truss of Figure 3.2, the element displacement location vectors are (1) L= [1 2 5 6](3.40) (2) L= [3 4 5 6](3.41)Before proceeding, let us note the quantity of information that can beobtained from simple-looking Table 3.2. With the geometry of the structuredefined, the (X, Y) global coordinates of each node are specified. Using thesedata, the length of each element and the direction cosines of element orientationTable 3.2 Element-Node Connectivity Tablefor Figure 3.2NodeElementij121233Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.5 Boundary Conditions, Constraint Forcesare computed via Equations 3.29 and 3.30, respectively. Specification of thecross-sectional area A and modulus of elasticity E of each element allows computation of the element stiffness matrix in the global frame using Equation 3.28.Finally, the element stiffness matrix terms are added to the global stiffness matrixusing the element displacement location vector.In the context of the current example, the reader is to imagine a 6 × 6 arrayof mailboxes representing the global stiffness matrix, each of which is originallyempty (i.e., the stiffness coefficient is zero).

We then consider the stiffness matrix of an individual element in the (2-D) global reference frame. Per the locationvector (addresses) for the element, the individual values of the element stiffnessmatrix are placed in the appropriate mailbox. In this fashion, each element isprocessed in sequence and its stiffness characteristics added to the global matrix.After all elements are processed, the array of mailboxes contains the global stiffness matrix.3.5 BOUNDARY CONDITIONS,CONSTRAINT FORCESHaving obtained the global stiffness matrix via either the equilibrium equationsor direct assembly, the system displacement equations for the example truss ofFigure 3.2 are of the form   U1   F1  U  F2  2  U3F3=[K ](3.42) U4   F4   U5   F5 U6F6As noted, the global stiffness matrix is a singular matrix; therefore, a unique solution to Equation 3.42 cannot be obtained directly.

However, in developingthese equations, we have not yet taken into account the constraints imposed onsystem displacements by the support conditions that must exist to preclude rigidbody motion. In this example, we observe the displacement boundary conditionsU1 = U2 = U3 = U4 = 0(3.43)leaving only U5 and U6 to be determined. Substituting the boundary conditionvalues and expanding Equation 3.42 we have, formally,K 15 U 5 +K 25 U 5 +K 35 U 5 +K 45 U 5 +K 55 U 5 +K 56 U 5 +K 16 U 6K 26 U 6K 36 U 6K 46 U 6K 56 U 6K 66 U 6======F1F2F3F4F5F6(3.44)67Hutton: Fundamentals ofFinite Element Analysis683.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Methodas the reduced system equations (this is the partitioned set of matrix equations,written explicitly for the active displacements). In this example, F1, F2, F3, andF4 are the components of the reaction forces at constrained nodes 1 and 2, whileF5 and F6 are global components of applied external force at node 3. Given theexternal force components, the last two of Equations 3.44 can be explicitly solvedfor displacements U5 and U6. The values obtained for these two displacementsare then substituted into the constraint equations (the first four of Equations 3.44)and the reaction force components computed.A more general approach to application of boundary conditions and computation of reactions is as follows.

Letting the subscript c denote constraineddisplacements and subscript a denote unconstrained (active) displacements, thesystem equations can be partitioned (Appendix A) to obtain K cc K caUcFc=(3.45)K ac K aaUaFawhere the values of the constrained displacements Uc are known (but not necessarily zero), as are the applied external forces Fa. Thus, the unknown, activedisplacements are obtained via the lower partition as[K ac ]{U c } + [K aa ]{U a } = {Fa }−1{U a } = [K aa ] ({Fa } − [K ac ]{U c })(3.46a)(3.46b)where we have assumed that the specified displacements {Uc} are not necessarily zero, although that is usually the case in a truss structure.

(Again, note that, fornumerical efficiency, methods other than matrix inversion are applied to obtainthe solutions formally represented by Equations 3.46.) Given the displacementsolution of Equations 3.46, the reactions are obtained using the upper partition ofmatrix Equation 3.45 as{Fc } = [K cc ]{U c } + [K ca ]{U a }(3.47)where [Kca] = [Kac]T by the symmetry property of the stiffness matrix.3.6 ELEMENT STRAIN AND STRESSThe final computational step in finite element analysis of a truss structure is toutilize the global displacements obtained in the solution step to determine thestrain and stress in each element of the truss. For an element connecting nodes iand j, the element nodal displacements in the element coordinate system aregiven by Equations 3.19 and 3.20 asu(e)1=U(e)1 cos␪+U(e)2 sin␪u(e)2=U(e)3 cos␪+U(e)4 sin␪(3.48)Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.6Element Strain and Stress69and the element axial strain (utilizing Equation 2.29 and the discretization andinterpolation functions of Equation 2.25) is then(e)du (e) (x )d(e)u1(e)ε ==[N 1 (x ) N 2 (x )](e)dxdxu2(e)(e)(e)u2 −u1u11−1==(3.49)(e)L (e)L (e) L (e)u2where L (e) is element length. The element axial stress is then obtained via application of Hooke’s law as␴ (e) = E ε(e)(3.50)Note, however, that the global solution does not give the element axial displacement directly. Rather, the element displacements are obtained from the globaldisplacements via Equations 3.48. Recalling Equations 3.21 and 3.22, the element strain in terms of global system displacements is (e) U1  (e)(e)Udu(x)d2(e)=[N1 (x) N2 (x)][R]ε =(3.51) U (e)dxdx3  (e) U4where [R] is the element transformation matrix defined by Equation 3.22.

Theelement stresses for the bar element in terms of global displacements are thosegiven by(e) U1(e) (e)(e)Udu(x)d2=E[N1 (x) N2 (x)][R]␴ (e) = Eε(e) = E(3.52)dxdxU (e)3 U (e) 4As the bar element is formulated here, a positive axial stress value indicates thatthe element is in tension and a negative value indicates compression per the usualconvention. Note that the stress calculation indicated in Equation 3.52 must beperformed on an element-by-element basis. If desired, the element forces can beobtained via Equation 3.23.EXAMPLE 3.2The two-element truss in Figure 3.5 is subjected to external loading as shown.