Hutton - Fundamentals of Finite Element Analysis, страница 13

Node numbers are circled while elementnumbers are in boxes. Element numbers are superscripted in the notation.To obtain the equilibrium conditions, free-body diagrams of the three connecting nodes and the two elements are drawn in Figure 3.3. Note that the external forces are numbered via the same convention as the global displacements.For node 1, (Figure 3.3a), we have the following equilibrium equations in theglobal X and Y directions, respectively:F1 − f(1)1 cos ␪1=0(3.1a)F2 − f(1)1 sin ␪1=0(3.1b)Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.2Nodal Equilibrium EquationsF2F6f (2)2␪2F1F3f (2)3␪1f 1(1)f (1)3F4(b)(a)F5(c)f (1)3f (2)3␪1␪2f 1(1)f (2)2(d)(e)Figure 3.3(a)–(c) Nodal free-body diagrams. (d) and (e) Element free-body diagrams.and for node 2,(2)2 cos ␪2(2)f 2 sin ␪2F3 − f=0(3.2a)F4 −=0(3.2b)while for node 3,F5 − fF6 −(1)3 cos ␪1 −(1)f 3 sin ␪1 −(2)3 cos ␪2(2)f 3 sin ␪2f=0(3.3a)=0(3.3b)Equations 3.1–3.3 simply represent the conditions of static equilibrium from arigid body mechanics standpoint.

Assuming external loads F5 and F6 are known,these six nodal equilibrium equations formally contain eight unknowns (forces).Since the example truss is statically determinate, we can invoke the additionalequilibrium conditions applicable to the truss as a whole as well as those for theindividual elements (Figures 3.3d and 3.3e) and eventually solve for all of theforces. However, a more systematic procedure is obtained if the formulation istransformed so that the unknowns are nodal displacements. Once the transformation is accomplished, we find that the number of unknowns is exactly thesame as the number of nodal equilibrium equations.

In addition, static indeterminacy is automatically accommodated. As the reader may recall from study ofmechanics of materials, the solution of statically indeterminate systems requires55Hutton: Fundamentals ofFinite Element Analysis3. Truss Structures: TheDirect Stiffness Method56CHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodU (e)4ju (e)2u (e)2v (e)2jjU (e)3After loading␪␪Originaliv 1(e)u 1(e)i(a)u 1(e)U (e)2i(b)U 1(e)(c)Figure 3.4(a) Bar element at orientation ␪.

(b) General displacements of a bar element. (c) Bar elementglobal displacements.specification of one or more displacement relations; hence, the displacement formulation of the finite element method includes such situations.To illustrate the transformation to displacements, Figure 3.4a depicts a barelement connected at nodes i and j in a general position in a two-dimensional(2-D) truss structure. As a result of external loading on the truss, we assume thatnodes i and j undergo 2-D displacement, as shown in Figure 3.4b.

Since the element must remain connected at the structural joints, the connected element nodesmust undergo the same 2-D displacements. This means that the element is subjected not only to axial motion but rotation as well. To account for the rotation,we added displacements v1 and v2 at element nodes 1 and 2, respectively, in thedirection perpendicular to the element x axis.

Owing to the assumption of smoothpin joint connections, the perpendicular displacements are not associated withelement stiffness; nevertheless, these displacements must exist so that the element remains connected to the structural joint so that the element displacementsare compatible with (i.e., the same as) joint displacements. Although the elementundergoes a rotation in general, for computation purposes, orientation angle ␪ isassumed to be the same as in the undeformed structure. This is a result of theassumption of small, elastic deformations and is used throughout the text.To now relate element nodal displacements referred to the element coordinates to element displacements in global coordinates, Figure 3.4c shows elementnodal displacements in the global system using the notationU(e)1= element node 1 displacement in the global X directionU(e)2= element node 1 displacement in the global Y directionU(e)3= element node 2 displacement in the global X directionU (e)4= element node 2 displacement in the global Y directionHutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.2Nodal Equilibrium EquationsAgain, note the use of capital letters for global quantities and the superscriptnotation to refer to an individual element. As the nodal displacements must bethe same in both coordinate systems, we can equate vector components of globaldisplacements to element system displacements to obtain the relations(e)1(e)v1=U(e)2(e)v1=Uuu==(e)(e)1 cos ␪ + U 2 sin ␪(e)(e)−U 1 sin ␪ + U 2 cos␪(e)(e)3 cos ␪ + U 4 sin ␪(e)(e)−U 3 sin ␪ + U 4 cos␪(3.4a)(3.4b)As noted, the v displacement components are not associated with element stiffness, hence not associated with element forces, so we can express the axial deformation of the element as (e) (e)(e)(e)(e) (e) ␦(e) = u 2 − u 1 = U 3 − U 1 cos ␪ + U 4 − U 2 sin ␪(3.5)The net axial force acting on the element is then (e) (e)(e) (e) f (e) = k (e) ␦(e) = k (e) U 3 − U 1 cos ␪ + U 4 − U 2 sin ␪(3.6)Utilizing Equation 3.6 for element 1 (Figure 3.3d) while noting that the displacements of element 1 are related to the specified global displacements as(1)(1)(1)(1)U 1 = U 1 , U 2 = U 2 , U 3 = U 5 , U 4 = U 6 , we have the force in element 1 asf(1)3=−f(1)1= k (1) [(U 5 − U 1 )cos ␪1 + (U 6 − U 2 )sin ␪1 ](3.7)and similarly for element 2 (Figure 3.3e):f(2)3=−f(2)2= k (2) [(U 5 − U 3 )cos ␪2 + (U 6 − U 4 )sin ␪2 ](3.8)Note that, in writing Equations 3.7 and 3.8, we invoke the condition that the displacements of node 3 (U5 and U6) are the same for each element.

To reiterate, thisassumption is actually a requirement, since on a physical basis, the structuremust remain connected at the joints after deformation. Displacement compatibility at the nodes is a fundamental requirement of the finite element method.Substituting Equations 3.7 and 3.8 into the nodal equilibrium conditions(Equations 3.1–3.3) yields−k (1) [(U 5 − U 1 )cos ␪1 + (U 6 − U 2 )sin ␪1 ]cos ␪1 = F1−k(3.9)[(U 5 − U 1 )cos ␪1 + (U 6 − U 2 )sin ␪1 ]sin ␪1 = F2(3.10)−k (2) [(U 5 − U 3 )cos ␪2 + (U 6 − U 4 )sin ␪2 ]cos ␪2 = F3(3.11)−k(3.12)(1)(2)[(U 5 − U 3 )cos ␪2 + (U 6 − U 4 )sin ␪2 ]sin ␪2 = F4k (2) [(U 5 − U 3 )cos ␪2 + (U 6 − U 4 )sin ␪2 ] cos ␪2+ k (1) [(U 5 − U 3 )cos ␪1 + (U 6 − U 4 )sin ␪1 ]cos ␪1 = F5k(2)(3.13)[(U 5 − U 3 )cos ␪2 + (U 6 − U 4 )sin ␪2 ]sin ␪2+ k (1) [(U 5 − U 1 )cos ␪1 + (U 6 − U 2 )sin ␪1 ]sin ␪1 = F6(3.14)57Hutton: Fundamentals ofFinite Element Analysis583.

Truss Structures: TheDirect Stiffness MethodCHAPTER 3Text© The McGraw−HillCompanies, 2004Truss Structures: The Direct Stiffness MethodEquations 3.9 through 3.14 are equivalent to the matrix formkc ␪1(1) 2 (1) k s␪1 c␪100 (1) 2 −k c ␪12−k1 s␪1 c␪100−k (1) c2 ␪1k (1) s 2 ␪100−k (1) s␪1 c␪10k (2) c2 ␪2k (2) s␪2 c␪2−k (2) c2 ␪20k (2) s␪2 c␪2k (2) s 2 ␪2−k (2) s␪2 c␪2−k1 s␪1 c␪1−k (2) c2 ␪2−k (2) s␪2 c␪2k (1) c2 ␪1 +k (2) c2 ␪2−k (1) s 2 ␪1−k (2) s␪2 c␪2−k (2) s 2 ␪2k (1) s␪1 c␪1 +k (2) s␪2 c␪2k(1)s␪1 c␪1−k (1) s␪1 c␪1   −k (1) s 2 ␪1 U1 F1  U −k (2) s␪2 c␪2 F22(2) 2UF33−k s ␪2 =(3.15)  U4 F4 k (1) s␪1 c␪1 +    U5  F5    k (2) s␪2 c␪2  U   F 66(1) 2k s ␪+1k (2) s 2 ␪2The six algebraic equations represented by matrix Equation 3.15 express thecomplete set of equilibrium conditions for the two-element truss. Equation 3.15is of the form[K ]{U } = {F }(3.16)where [K ] is the global stiffness matrix, {U } is the vector of nodal displacements, and {F } is the vector of applied nodal forces.

We observe that the globalstiffness matrix is a 6 × 6 symmetric matrix corresponding to six possible globaldisplacements. Application of boundary conditions and solution of the equationsare deferred at this time, pending further discussion.3.3 ELEMENT TRANSFORMATIONFormulation of global finite element equations by direct application of equilibrium conditions, as in the previous section, proves to be quite cumbersome except for the very simplest of models.

By writing the nodal equilibrium equationsin the global coordinate system and introducing the displacement formulation,the procedure of the previous section implicitly transformed the individual element characteristics (the stiffness matrix) to the global system. A direct methodfor transforming the stiffness characteristics on an element-by-element basisis now developed in preparation for use in the direct assembly procedure of thefollowing section.Recalling the bar element equations expressed in the element frame as (e) (e) (e) u1u1f 1AE 1 −1ke−k e==(3.17)(e)(e)(e)−11−kkLeeu2u2f 2the present objective is to transform these equilibrium equations into the globalcoordinate system in the form (e)   (e) U1 F 1  (e)  F (e)  (e) U 22K=(3.18)  F (e)  U (e) 33  (e)  (e) U4F4Hutton: Fundamentals ofFinite Element Analysis3.

Truss Structures: TheDirect Stiffness MethodText© The McGraw−HillCompanies, 20043.3Element TransformationIn Equation 3.18, [K (e) ] represents the element stiffness matrix in the global coordinate system, the vector {F (e) } on the right-hand side contains the element(e)nodal force components in the global frame, displacements U (e)1 and U 3 are(e)(e)parallel to the global X axis, while U 2 and U 4 are parallel to the global Y axis.The relation between the element axial displacements in the element coordinatesystem and the element displacements in global coordinates (Equation 3.4) isu(e)1(e)u2=U=(e)1 cos(e)U 3 cos(e)2 sin␪(3.19)(e)U 4 sin␪(3.20)␪+U␪+which can be written in matrix form asu (e)1u (e)2=cos ␪0sin ␪00cos ␪ (e)  (e) U1 U1 (e) U (e) U 022=[R]sin ␪ U (e)U (e)3 3  (e)  (e) U4U4(3.21)where[R] =cos ␪0sin ␪00cos ␪0sin ␪(3.22)is the transformation matrix of element axial displacements to global displacements.