Hutton - Fundamentals of Finite Element Analysis, страница 7

If these nodal displacements are known, the total elongation or contractionof the spring is known as is the net force in the spring. At this point in our development, we require that forces be applied to the element only at the nodes (distributed forces are accommodated for other element types later), and these aredenoted as f1 and f2 and are also shown in the positive sense.Assuming that both the nodal displacements are zero when the spring is undeformed, the net spring deformation is given by = u2 − u1(2.1)and the resultant axial force in the spring isf = k = k(u 2 − u 1 )(2.2)as is depicted in Figure 2.1b.For equilibrium, f 1 + f 2 = 0 or f 1 = − f 2 , and we can rewrite Equation 2.2in terms of the applied nodal forces asf 1 = −k(u 2 − u 1 )(2.3a)f 2 = k(u 2 − u 1 )(2.3b)which can be expressed in matrix form (see Appendix A for a review of matrixalgebra) as k−ku1f1=(2.4)−kku2f2or[k e ]{u} = { f }where[k e ] =k−k−kk(2.5)(2.6)is defined as the element stiffness matrix in the element coordinate system (orlocal system), {u} is the column matrix (vector) of nodal displacements, and {f}is the column matrix (vector) of element nodal forces.

(In subsequent chapters,21Hutton: Fundamentals ofFinite Element Analysis222. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elementsthe matrix notation is used extensively. A general matrix is designated bybrackets [ ] and a column matrix (vector) by braces { }.)Equation 2.6 shows that the element stiffness matrix for the linear springelement is a 2 × 2 matrix. This corresponds to the fact that the element exhibitstwo nodal displacements (or degrees of freedom) and that the two displacementsare not independent (that is, the body is continuous and elastic). Furthermore, thematrix is symmetric.

A symmetric matrix has off-diagonal terms such that k i j =kji. Symmetry of the stiffness matrix is indicative of the fact that the body is linearly elastic and each displacement is related to the other by the same physicalphenomenon. For example, if a force F (positive, tensile) is applied at node 2with node 1 held fixed, the relative displacement of the two nodes is the same asif the force is applied symmetrically (negative, tensile) at node 1 with node 2fixed.

(Counterexamples to symmetry are seen in heat transfer and fluid flowanalyses in Chapters 7 and 8.) As will be seen as more complicated structuralelements are developed, this is a general result: An element exhibiting N degreesof freedom has a corresponding N × N, symmetric stiffness matrix.Next consider solution of the system of equations represented by Equation 2.4. In general, the nodal forces are prescribed and the objective is to solvefor the unknown nodal displacements. Formally, the solution is represented by u1f1−1= [k e ](2.7)u2f2where [k e ]−1 is the inverse of the element stiffness matrix. However, this inversematrix does not exist, since the determinant of the element stiffness matrix isidentically zero.

Therefore, the element stiffness matrix is singular, and this alsoproves to be a general result in most cases. The physical significance of thesingular nature of the element stiffness matrix is found by reexamination ofFigure 2.1a, which shows that no displacement constraint whatever has been imposed on motion of the spring element; that is, the spring is not connected to anyphysical object that would prevent or limit motion of either node.

With no constraint, it is not possible to solve for the nodal displacements individually.Instead, only the difference in nodal displacements can be determined, as thisdifference represents the elongation or contraction of the spring element owingto elastic effects. As discussed in more detail in the general formulation of interpolation functions (Chapter 6) and structural dynamics (Chapter 10), a properlyformulated finite element must allow for constant value of the field variable.

Inthe example at hand, this means rigid body motion. We can see the rigid bodymotion capability in terms of a single spring (element) and in the context of several connected elements. For a single, unconstrained element, if arbitrary forcesare applied at each node, the spring not only deforms axially but also undergoesacceleration according to Newton’s second law.

Hence, there exists not onlydeformation but overall motion. If, in a connected system of spring elements, theoverall system response is such that nodes 1 and 2 of a particular element displace the same amount, there is no elastic deformation of the spring and thereforeHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.2 Linear Spring as a Finite Elementno elastic force in the spring.

This physical situation must be included in theelement formulation. The capability is indicated mathematically by singularityof the element stiffness matrix. As the stiffness matrix is formulated on the basisof deformation of the element, we cannot expect to compute nodal displacementsif there is no deformation of the element.Equation 2.7 indicates the mathematical operation of inverting the stiffnessmatrix to obtain solutions.

In the context of an individual element, the singularnature of an element stiffness matrix precludes this operation, as the inverse of asingular matrix does not exist. As is illustrated profusely in the remainder of thetext, the general solution of a finite element problem, in a global, as opposed toelement, context, involves the solution of equations of the form of Equation 2.5. Forrealistic finite element models, which are of huge dimension in terms of the matrixorder (N × N) involved, computing the inverse of the stiffness matrix is a very inefficient, time-consuming operation, which should not be undertaken except for thevery simplest of systems.

Other, more-efficient solution techniques are available,and these are discussed subsequently. (Many of the end-of-chapter problemsincluded in this text are of small order and can be efficiently solved via matrix inversion using “spreadsheet” software functions or software such as MATLAB.)2.2.1 System Assembly in Global CoordinatesDerivation of the element stiffness matrix for a spring element was based onequilibrium conditions. The same procedure can be applied to a connected system of spring elements by writing the equilibrium equation for each node. However, rather than drawing free-body diagrams of each node and formally writingthe equilibrium equations, the nodal equilibrium equations can be obtained moreefficiently by considering the effect of each element separately and adding theelement force contribution to each nodal equation.

The process is described as“assembly,” as we take individual stiffness components and “put them together”to obtain the system equations. To illustrate, via a simple example, the assemblyof element characteristics into global (or system) equations, we next consider thesystem of two linear spring elements connected as shown in Figure 2.2.For generality, it is assumed that the springs have different spring constantsk1 and k2.

The nodes are numbered 1, 2, and 3 as shown, with the springs sharingnode 2 as the physical connection. Note that these are global node numbers. Theglobal nodal displacements are identified as U1, U2, and U3, where the upper caseis used to indicate that the quantities represented are global or system displacements as opposed to individual element displacements. Similarly, applied nodalU2U11F11k1U322F2k23F3Figure 2.2 System of two springs with node numbers,element numbers, nodal displacements, and nodal forces.23Hutton: Fundamentals ofFinite Element Analysis242. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elementsu1(1)u(1)2u1(2)1f 1(1)1u(2)222k1f (1)2f (2)223k2(a)f (2)3(b)f 1(1)F1(2)f (1)2 f212f (2)3F2(d)(c)F3(e)Figure 2.3 Free-body diagrams of elements and nodes for thetwo-element system of Figure 2.2.forces are F1, F2, and F3.

Assuming the system of two spring elements to bein equilibrium, we examine free-body diagrams of the springs individually (Figure 2.3a and 2.3b) and express the equilibrium conditions for each spring, usingEquation 2.4, as (1) (1) u1f 1k1−k 1=(2.8a)(1)(1)−k 1k1u2f 2 (2) (2) u1f 2k2−k 2=(2.8b)(2)(2)−k 2k2uf23where the superscript is element number.To begin “assembling” the equilibrium equations describing the behaviorof the system of two springs, the displacement compatibility conditions, whichrelate element displacements to system displacements, are written asu(1)1= U1u(1)2= U2u(2)1= U2u(2)2= U3(2.9)The compatibility conditions state the physical fact that the springs are connected at node 2, remain connected at node 2 after deformation, and hence, musthave the same nodal displacement at node 2. Thus, element-to-element displacement continuity is enforced at nodal connections.

Substituting Equations 2.9 intoEquations 2.8, we obtain (1) f 1k1−k 1U1=(2.10a)(1)−k 1k1U2f 2and (2) f 2k2−k 2U2=(2.10b)(2)−k 2k2U3f 3Here, we use the notation fnode i.( j)ito represent the force exerted on element j atHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.2 Linear Spring as a Finite ElementEquation 2.10 is the equilibrium equations for each spring element expressedin terms of the specified global displacements.

In this form, the equations clearlyshow that the elements are physically connected at node 2 and have the same displacement U2 at that node. These equations are not yet amenable to direct combination, as the displacement vectors are not the same. We expand both matrixequations to 3 × 3 as follows (while formally expressing the facts that element 1is not connected to node 3 and element 2 is not connected to node 1):  f (1) U1k1 −k1 0 1 U2 = f (1)−k1 k1 0(2.11) 2 00000  0 0000(2)0 k2 −k2U2 = f 2(2.12) (2) 0 −k2 k2U3f3The addition of Equations 2.11 and 2.12 yieldsk1−k10−k1k1 + k2−k20−k2k2U1U2U3=f (1)1(2)f (1)2 + f 2f (2)3(2.13)Next, we refer to the free-body diagrams of each of the three nodes depicted inFigure 2.3c, 2.3d, and 2.3e. The equilibrium conditions for nodes 1, 2, and 3show thatf(1)1= F1f(1)2+ f(2)2= F2f(2)3= F3respectively.

Substituting into Equation 2.13, we obtain the final result: −k 10k1U1F1−k 1 k 1 + k 2 −k 2U 2 = F20−k 2k2U3F3(2.14)(2.15)which is of the form [K ]{U} = {F}, similar to Equation 2.5. However, Equation 2.15 represents the equations governing the system composed of two connected spring elements. By direct consideration of the equilibrium conditions,we obtain the system stiffness matrix [K ] (note use of upper case) ask1−k 10[K ] = −k 1 k 1 + k 2 −k 2(2.16)0−k 2k2Note that the system stiffness matrix is (1) symmetric, as is the case with all linear systems referred to orthogonal coordinate systems; (2) singular, since noconstraints are applied to prevent rigid body motion of the system; and (3) thesystem matrix is simply a superposition of the individual element stiffnessmatrices with proper assignment of element nodal displacements and associatedstiffness coefficients to system nodal displacements.