Hutton - Fundamentals of Finite Element Analysis, страница 14

(Again note that the element nodal displacements in the direction perpendicular to the element axis, v1 and v2 , are not considered in the stiffness matrixdevelopment; these displacements come into play in dynamic analyses inChapter 10.) Substituting Equation 3.22 into Equation 3.17 yields (e) U 1   U (e) f (e)ke −ke cos ␪ sin ␪0021=(3.23)(e) (e)00cos ␪ sin ␪ −ke keUf3 2 (e) U4orke−ke (e) U 1    U (e) f (e)−ke21=[R]keU (e)f (e)3 2 (e) U4(3.24)While we have transformed the equilibrium equations from element displacements to global displacements as the unknowns, the equations are still expressedin the element coordinate system.

The first of Equation 3.23 is the equilibriumcondition for element node 1 in the element coordinate system. If we multiply59Hutton: Fundamentals ofFinite Element Analysis603. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Methodthis equation by cos ␪, we obtain the equilibrium equation for the node in theX direction of the global coordinate system. Similarly, multiplying by sin ␪ , theY direction global equilibrium equation is obtained.

Exactly the same procedurewith the second equation expresses equilibrium of element node 2 in the globalcoordinate system. The same desired operations described are obtained if wepremultiply both sides of Equation 3.24 by [R]T , the transpose of the transformation matrix; that is, (e) (e) f 1 cos ␪ U1  (e)  (e)cos ␪0  f sin ␪  U (e) fk−ksin ␪0ee211==[R]T[R]−ke ke0cos ␪U (e)f (e)f (e)3 22 cos ␪  (e) 0sin ␪U4f (e)sin␪2(3.25)Clearly, the right-hand side of Equation 3.25 represents the components of theelement forces in the global coordinate system, so we now have (e)   (e) F1 U 1  (e) U   F (e) ke −ke22T=[R][R](3.26)(e) (e) −ke keUF3  3  (e)   (e) U4F4Matrix Equation 3.26 represents the equilibrium equations for element nodes 1and 2, expressed in the global coordinate system.

Comparing this result withEquation 3.18, the element stiffness matrix in the global coordinate frame is seento be given by (e) ke−k eT[R]K= [R](3.27)−k ekeIntroducing the notation c = cos ␪ , s = sin ␪ and performing the matrix multiplications on the right-hand side of Equation 3.27 results inc2sc −c2 −sc (e) s 2 −sc −s 2 K= ke  sc(3.28) −c2 −sc c2sc −sc−s 2scs2where ke = AE/L is the characteristic axial stiffness of the element.Examination of Equation 3.28 shows that the symmetry of the element stiffness matrix is preserved in the transformation to global coordinates. In addition,although not obvious by inspection, it can be shown that the determinant is zero,indicating that, after transformation, the stiffness matrix remains singular.

This isto be expected, since as previously discussed, rigid body motion of the elementis possible in the absence of specified constraints.Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.4 Direct Assembly of Global Stiffness Matrix3.3.1 Direction CosinesIn practice, a finite element model is constructed by defining nodes at specifiedcoordinate locations followed by definition of elements by specification of thenodes connected by each element. For the case at hand, nodes i and j are definedin global coordinates by (Xi, Yi) and (Xj, Yj). Using the nodal coordinates, elementlength is readily computed asL = [( X j − X i ) 2 + (Y j − Yi ) 2 ]1/2(3.29)and the unit vector directed from node i to node j is␭=1[( X j − X i )I + (Y j − Yi )J] = cos ␪X I + cos ␪Y JL(3.30)where I and J are unit vectors in global coordinate directions X and Y, respectively.

Recalling the definition of the scalar product of two vectors and referringagain to Figure 3.4, the trigonometric values required to construct the elementtransformation matrix are also readily determined from the nodal coordinates asthe direction cosines in Equation 3.30cos ␪ = cos ␪X = ␭ · I =X j − XiL(3.31)sin ␪ = cos ␪Y = ␭ · J =Y j − YiL(3.32)Thus, the element stiffness matrix of a bar element in global coordinates canbe completely determined by specification of the nodal coordinates, the crosssectional area of the element, and the modulus of elasticity of the element material.3.4 DIRECT ASSEMBLY OF GLOBALSTIFFNESS MATRIXHaving addressed the procedure of transforming the element characteristics ofthe one-dimensional bar element into the global coordinate system of a twodimensional structure, we now address a method of obtaining the global equilibrium equations via an element-by-element assembly procedure.

The technique ofdirectly assembling the global stiffness matrix for a finite element model of atruss is discussed in terms of the simple two-element system depicted in Figure 3.2. Assuming the geometry and material properties to be completely specified, the element stiffness matrix in the global frame can be formulated for eachelement using Equation 3.28 to obtain (1)(1)(1)(1)k11 k12k13k14 (1)(1)(1)(1)  (1)  k21k22k23k24K=  (1)(3.33)(1)(1)(1)  k31 k32 k33 k34 (1)(1)(1)(1)k41k42k43k4461Hutton: Fundamentals ofFinite Element Analysis623. Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness Methodfor element 1 andK(2)(2)k11 (2)k=  21 k (2) 31(2)k41(2)k12(2)k13(2)k22(2)k23(2)k32(2)k33(2)k42(2)k43(2)k14(2) k24(2) k34(2)k44(3.34)for element 2.

The stiffness matrices given by Equations 3.33 and 3.34 contain32 terms, which together will form the 6 × 6 system matrix containing 36 terms.To “assemble” the individual element stiffness matrices into the global stiffnessmatrix, it is necessary to observe the correspondence of individual element displacements to global displacements and allocate the associated element stiffnessterms to the correct location in the global matrix. For element 1 of Figure 3.2, theelement displacements correspond to global displacements per U (e)1  U1  (1)  U (e) 2U=⇒ U2(3.35)U5 U (e)3U6U (e)4while for element 2 (e)  U 1 U3 (e) (2) U4 U2U=⇒(3.36)U5 U (e)3  U6 (e) U4Equations 3.35 and 3.36 are the connectivity relations for the truss and explicitlyindicate how each element is connected in the structure.

For example, Equation 3.35 clearly shows that element 1 is not associated with global displacementsU3 and U4 (therefore, not connected to global node 2) and, hence, contributes nostiffness terms affecting those displacements. This means that element 1 has noeffect on the third and fourth rows and columns of the global stiffness matrix.Similarly, element 2 contributes nothing to the first and second rows and columns.Rather that write individual displacement relations, it is convenient to placeall the element to global displacement data in a single table as shown in Table 3.1.Table 3.1 Nodal Displacement Correspondence TableGlobal DisplacementElement 1 DisplacementElement 2 Displacement123456120034001234Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.4 Direct Assembly of Global Stiffness MatrixThe first column contains the entire set of global displacements in numericalorder. Each succeeding column represents an element and contains the number ofthe element displacement corresponding to the global displacement in each row.A zero entry indicates no connection, therefore no stiffness contribution. Theindividual terms in the global stiffness matrix are then obtained by allocating theelement stiffness terms per the table as follows:(1)K 11 = k 11 + 0(1)K 12 = k 12 + 0K 13 = 0 + 0K 14 = 0 + 0(1)K 15 = k 13 + 0(1)K 16 = k 14 + 0(1)K 22 = k 22 + 0K 23 = 0 + 0K 24 = 0 + 0(1)K 25 = k 23 + 0(1)K 26 = k 24 + 0(2)K 33 = 0 + k 11(2)K 34 = 0 + k 12(2)K 35 = 0 + k 13(2)K 36 = 0 + k 14(2)K 44 = 0 + k 22(2)K 45 = 0 + k 23(2)K 46 = 0 + k 24(1)(2)(1)(2)(1)(2)K 55 = k 33 + k 33K 56 = k 34 + k 34K 66 = k 44 + k 44where the known symmetry of the stiffness matrix has been implicitly used toavoid repetition.

It is readily shown that the resulting global stiffness matrix isidentical in every respect to that obtained in Section 3.2 via the equilibriumequations. This is the direct stiffness method; the global stiffness matrix is“assembled” by direct addition of the individual element stiffness terms per thenodal displacement correspondence table that defines element connectivity.63Hutton: Fundamentals ofFinite Element Analysis643.