Hutton - Fundamentals of Finite Element Analysis (523155), страница 9
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The31Hutton: Fundamentals ofFinite Element Analysis322. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elementsspring element is also often used to represent the elastic nature of supports formore complicated systems. A more generally applicable, yet similar, element isan elastic bar subjected to axial forces only. This element, which we simply calla bar element, is particularly useful in the analysis of both two- and threedimensional frame or truss structures. Formulation of the finite element characteristics of an elastic bar element is based on the following assumptions:1.2.3.4.The bar is geometrically straight.The material obeys Hooke’s law.Forces are applied only at the ends of the bar.The bar supports axial loading only; bending, torsion, and shear are nottransmitted to the element via the nature of its connections to otherelements.The last assumption, while quite restrictive, is not impractical; this condition issatisfied if the bar is connected to other structural members via pins (2-D) or balland-socket joints (3-D).
Assumptions 1 and 4, in combination, show that this isinherently a one-dimensional element, meaning that the elastic displacement ofany point along the bar can be expressed in terms of a single independent variable. As will be seen, however, the bar element can be used in modeling bothtwo- and three-dimensional structures. The reader will recognize this elementas the familiar two-force member of elementary statics, meaning, for equilibrium, the forces exerted on the ends of the element must be colinear, equal inmagnitude, and opposite in sense.Figure 2.6 depicts an elastic bar of length L to which is affixed a uniaxialcoordinate system x with its origin arbitrarily placed at the left end. This is theelement coordinate system or reference frame. Denoting axial displacement atany position along the length of the bar as u(x), we define nodes 1 and 2 at eachend as shown and introduce the nodal displacements u 1 = u(x = 0) andu 2 = u(x = L ) .
Thus, we have the continuous field variable u(x), which is to beexpressed (approximately) in terms of two nodal variables u 1 and u 2 . To accomplish this discretization, we assume the existence of interpolation functionsN 1 (x ) and N 2 (x ) (also known as shape or blending functions) such thatu(x ) = N 1 (x )u 1 + N 2 (x )u 2(2.17)u1u212xu(x)LFigure 2.6 A bar (or truss) element with elementcoordinate system and nodal displacementnotation.xHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.3 Elastic Bar, Spar/Link/Truss Element(It must be emphasized that, although an equality is indicated by Equation 2.17,the relation, for finite elements in general, is an approximation.
For the bar element, the relation, in fact, is exact.) To determine the interpolation functions, werequire that the boundary values of u(x ) (the nodal displacements) be identicallysatisfied by the discretization such thatu(x = 0) = u 1u(x = L ) = u 2(2.18)Equations 2.17 and 2.18 lead to the following boundary (nodal) conditions:N 1 (0) = 1N 2 (0) = 0(2.19)N1( L ) = 0N2( L ) = 1(2.20)which must be satisfied by the interpolation functions. It is required that the displacement expression, Equation 2.17, satisfy the end (nodal) conditions identically, since the nodes will be the connection points between elements and thedisplacement continuity conditions are enforced at those connections.
As wehave two conditions that must be satisfied by each of two one-dimensional functions, the simplest forms for the interpolation functions are polynomial forms:N 1 (x ) = a0 + a1 x(2.21)N 2 (x ) = b0 + b1 x(2.22)where the polynomial coefficients are to be determined via satisfaction of theboundary (nodal) conditions. We note here that any number of mathematicalforms of the interpolation functions could be assumed while satisfying therequired conditions. The reasons for the linear form is explained in detail inChapter 6.Application of conditions represented by Equation 2.19 yields a0 = 1,b0 = 0 while Equation 2.20 results in a1 = −(1/L ) and b1 = x /L . Therefore,the interpolation functions areN 1 (x ) = 1 − x /L(2.23)N 2 (x ) = x /L(2.24)and the continuous displacement function is represented by the discretizationu(x ) = (1 − x /L )u 1 + (x /L )u 2(2.25)As will be found most convenient subsequently, Equation 2.25 can be expressedin matrix form as uu(x ) = [N 1 (x ) N 2 (x )] 1 = [N ] {u}(2.26)u2where [N ] is the row matrix of interpolation functions and {u} is the columnmatrix (vector) of nodal displacements.Having expressed the displacement field in terms of the nodal variables, thetask remains to determine the relation between the nodal displacements andapplied forces to obtain the stiffness matrix for the bar element.
Recall from33Hutton: Fundamentals ofFinite Element Analysis342. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elementselementary strength of materials that the deflection of an elastic bar of length Land uniform cross-sectional area A when subjected to axial load P is given byPL=(2.27)AEwhere E is the modulus of elasticity of the material. Using Equation 2.27, weobtain the equivalent spring constant of an elastic bar asPAEk==(2.28)Land could, by analogy with the linear elastic spring, immediately write the stiffness matrix as Equation 2.6. While the result is exactly correct, we take a moregeneral approach to illustrate the procedures to be used with more complicatedelement formulations.Ultimately, we wish to compute the nodal displacements given some loadingcondition on the element.
To obtain the necessary equilibrium equations relatingthe displacements to applied forces, we proceed from displacement to strain,strain to stress, and stress to loading, as follows. In uniaxial loading, as in the barelement, we need consider only the normal strain component, defined asduεx =(2.29)dxwhich, when applied to Equation 2.25, givesu2 − u1εx =(2.30)Lwhich shows that the spar element is a constant strain element.
This is in accordwith strength of materials theory: The element has constant cross-sectional areaand is subjected to constant forces at the end points, so the strain does not varyalong the length. The axial stress, by Hooke’s law, is thenu2 − u1x = E ε x = E(2.31)Land the associated axial force isAEP = x A =(u 2 − u 1 )(2.32)LTaking care to observe the correct algebraic sign convention, Equation 2.32 isnow used to relate the applied nodal forces f 1 and f 2 to the nodal displacementsu 1 and u 2 .
Observing that, if Equation 2.32 has a positive sign, the element is intension and nodal force f 2 must be in the positive coordinate direction whilenodal force f 1 must be equal and opposite for equilibrium; therefore,AEf1 = −(u 2 − u 1 )(2.33)LAEf2 =(u 2 − u 1 )(2.34)LHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.3 Elastic Bar, Spar/Link/Truss ElementEquations 2.33 and 2.34 are expressed in matrix form as AE 1 −1u1f1=−11uf2L235(2.35)Comparison of Equation 2.35 to Equation 2.4 shows that the stiffness matrix forthe bar element is given byAE 1 −1[k e ] =(2.36)L −1 1As is the case with the linear spring, we observe that the element stiffness matrixfor the bar element is symmetric, singular, and of order 2 × 2 in correspondencewith two nodal displacements or degrees of freedom.
It must be emphasized thatthe stiffness matrix given by Equation 2.36 is expressed in the element coordinate system, which in this case is one-dimensional. Application of this elementformulation to analysis of two- and three-dimensional structures is considered inthe next chapter.EXAMPLE 2.4Figure 2.7a depicts a tapered elastic bar subjected to an applied tensile load P at one endand attached to a fixed support at the other end.
The cross-sectional area varies linearlyfrom A 0 at the fixed support at x = 0 to A 0 /2 at x = L . Calculate the displacement of theend of the bar (a) by modeling the bar as a single element having cross-sectional areaequal to the area of the actual bar at its midpoint along the length, (b) using two barelements of equal length and similarly evaluating the area at the midpoint of each, and(c) using integration to obtain the exact solution.■ Solution(a) For a single element, the cross-sectional area is 3A 0 /4 and the element “springconstant” isk=AE3A 0 E=L4Land the element equations are3A 0 E4L1−1−1−1U1U2=F1PThe element and nodal displacements are as shown in Figure 2.7b. Applying theconstraint condition U 1 = 0 , we findU2 =4PLPL= 1.3333A 0 EA0 Eas the displacement at x = L .(b) Two elements of equal length L /2 with associated nodal displacements aredepicted in Figure 2.7c.
For element 1, A 1 = 7A 0 /8 sok1 =7A 0 EA1E7A 0 E==L18( L /2)4LHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20041x7A8 oL(A(x) Ao 1 x2LA)3A4 o5A8 o2P(b)(a)(c)1.4ExactOne elementTwo elements1.2PAox(x)0.8Y1.00.60.40.2000.10.20.30.4P0.5xL0.60.70.80.91.0(e)(d)2ExactOne elementTwo elements1.8(Po o Ao)1.61.41.210.80.60.40.2000.10.20.30.40.5xL0.60.70.80.91.0(f)Figure 2.7(a) Tapered axial bar, (b) one-element model, (c) two-element model, (d) free-body diagramfor an exact solution, (e) displacement solutions, (f) stress solutions.36Hutton: Fundamentals ofFinite Element Analysis2.
Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.3 Elastic Bar, Spar/Link/Truss Elementwhile for element 2, we have5A 08A1 =A2 E5A 0 E5A 0 E==L28( L /2)4Land k 2 =Since no load is applied at the center of the bar, the equilibrium equations for thesystem of two elements is 0 U1 F1 −k2 U2 =0 Pk2U3−k1k1 + k2−k2k1 −k10Applying the constraint condition U 1 = 0 results ink1 + k2−k 2−k 2k2U2U3=0PAdding the two equations givesP4PL=k17A 0 EU2 =and substituting this result into the first equation results ink1 + k248PLPL== 1.371k235A 0 EA0 EU3 =(c) To obtain the exact solution, we refer to Figure 2.7d, which is a free-body diagram ofa section of the bar between an arbitrary position x and the end x = L .
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