Hutton - Fundamentals of Finite Element Analysis, страница 8

The superposition procedure is formalized in the context of frame structures in the following paragraphs.25Hutton: Fundamentals ofFinite Element Analysis262. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsEXAMPLE 2.1Consider the two element system depicted in Figure 2.2 given thatNode 1 is attached to a fixed support, yielding the displacement constraint U1 = 0.k1 = 50 lb./in., k2 = 75 lb./in., F2 = F3 = 75 lb.for these conditions determine nodal displacements U2 and U3.■ SolutionSubstituting the specified values into Equation 2.15 yields   50 −500 0   F1  −50 125 −75  U2 = 75   75U30−75 75and we note that, owing to the constraint of zero displacement at node 1, nodal force F1becomes an unknown reaction force. Formally, the first algebraic equation represented inthis matrix equation becomes−50U 2 = F1and this is known as a constraint equation, as it represents the equilibrium conditionof a node at which the displacement is constrained.

The second and third equationsbecome125−75−7575U2U3=7575which can be solved to obtain U2 = 3 in. and U3 = 4 in. Note that the matrix equationsgoverning the unknown displacements are obtained by simply striking out the first rowand column of the 3 × 3 matrix system, since the constrained displacement is zero.Hence, the constraint does not affect the values of the active displacements (we use theterm active to refer to displacements that are unknown and must be computed).

Substitution of the calculated values of U2 and U3 into the constraint equation yields the valueF1 = −150 lb., which value is clearly in equilibrium with the applied nodal forces of75 lb. each. We also illustrate element equilibrium by writing the equations for eachelement as50−5075−75 (1) f 1−1500=lb.=(1)1503f 2 (2) f 2−75−753=lb.=(2)75754f 3−5050for element 1for element 2Example 2.1 illustrates the general procedure for solution of finite element models: Formulate the system equilibrium equations, apply the specified constraintconditions, solve the reduced set of equations for the “active” displacements, andsubstitute the computed displacements into the constraint equations to obtain theunknown reactions.

While not directly applicable for the spring element, forHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.2 Linear Spring as a Finite Element27more general finite element formulations, the computed displacements are alsosubstituted into the strain relations to obtain element strains, and the strains are,in turn, substituted into the applicable stress-strain equations to obtain elementstress values.EXAMPLE 2.2Figure 2.4a depicts a system of three linearly elastic springs supporting three equalweights W suspended in a vertical plane.

Treating the springs as finite elements, determine the vertical displacement of each weight.■ SolutionTo treat this as a finite element problem, we assign node and element numbers as shownin Figure 2.4b and ignore, for the moment, that displacement U1 is known to be zero bythe fixed support constraint. Per Equation 2.6, the stiffness matrix of each element is(preprocessing)k(1)=k (2) =k (3) =3k3k−3k−3k3k2k−2k−2k2k−kkk−k1U13k12k2k3W2U22kW3U3kW4U4(a)(b)Figure 2.4 Example 2.2: elasticspring supporting weights.Hutton: Fundamentals ofFinite Element Analysis282. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsThe element-to-global displacement relations areu(1)1= U1u(1)2=u(2)1= U2u(2)2=u(3)1= U3u(3)2= U4Proceeding as in the previous example, we then write the individual element equations as  U3k −3k 0 0  1  −3k 3k 0 0  U2  = 0U  00 0  3 U4000 0  U0000  1  0 2k −2k 0  U2   0 −2k 2k 0  U3  =   U40000  U0 0 00  1 0 0 0 U2  0= 0 0 k −k  U3    U40 0 −k kf (1)1 f (1)20 00 (2) f 1 f (2)2 00 0 f (3)1 (3) f 2(1)(2)(3)Adding Equations 1–3, we obtain   UF3 −3 00   1 1 −3 5 −2 0  U2   W k= 0 −2 3 −1  U3   W    U4W00 −1 1(4)where we utilize the fact that the sum of the element forces at each node must equal theapplied force at that node and, at node 1, the force is an unknown reaction.Applying the displacement constraint U1 = 0 (this is also preprocessing), we obtain−3kU 2 = F1(5)as the constraint equation and the matrix equation   5 −2 0  U2   W k  −2 3 −1  U3 = W   WU40 −1 1(6)for the active displacements.

Again note that Equation 6 is obtained by eliminating theconstraint equation from 4 corresponding to the prescribed zero displacement.Simultaneous solution (the solution step) of the algebraic equations represented byEquation 6 yields the displacements asU2 =WkU3 =2Wkand Equation 5 gives the reaction force asF1 = −3W(This is postprocessing.)U4 =3WkHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.2 Linear Spring as a Finite Element29Note that the solution is exactly that which would be obtained by the usual staticsequations. Also note the general procedure as follows:Formulate the individual element stiffness matrices.Write the element to global displacement relations.Assemble the global equilibrium equation in matrix form.Reduce the matrix equations according to specified constraints.Solve the system of equations for the unknown nodal displacements (primaryvariables).Solve for the reaction forces (secondary variable) by back-substitution.EXAMPLE 2.3Figure 2.5 depicts a system of three linear spring elements connected as shown.

The nodeand element numbers are as indicated. Node 1 is fixed to prevent motion, and node 3 isgiven a specified displacement as shown. Forces F2 = − F and F4 = 2F are applied atnodes 2 and 4. Determine the displacement of each node and the force required at node 3for the specified conditions.■ SolutionThis example includes a nonhomogeneous boundary condition.

In previous examples, theboundary conditions were represented by zero displacements. In this example, we haveboth a zero (homogeneous) and a specified nonzero (nonhomogeneous) displacementcondition. The algebraic treatment must be different as follows. The system equilibriumequations are expressed in matrix form (Problem 2.6) as−k4k−3k0k −k 000−3k5k−2k    0U1 F1 F1     0  U2 = F2 = −F−2k U  F  F  3   3 3 2kU4F42FSubstituting the specified conditions U 1 = 0 and U 3 = results in the system ofequationsk −k 00−k4k−3k0k  00 F1   0  U2 = −F F3 −2k   2k2FU4F2 F110−3k5k−2k3323k24F4 2F2k␦Figure 2.5 Example 2.3: Three-element system with specifiednonzero displacement at node 3.Hutton: Fundamentals ofFinite Element Analysis302.

Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsSince U 1 = 0 , we remove the first row and column to obtain4k −3k0−3k5k−2k  0 U2   −F =F−2k     3 2FU42kas the system of equations governing displacements U2 and U4 and the unknown nodalforce F3. This last set of equations clearly shows that we cannot simply strike out the rowand column corresponding to the nonzero specified displacement because it appears inthe equations governing the active displacements. To illustrate a general procedure, werewrite the last matrix equation as5k −3k−2k−3k4k0  −2k    F3 0  U2 = −F  U42F2kNext, we formally partition the stiffness matrix and write5k −3k−2k−3k4k0  −2k   {}{F }[K ] [K U ]=0  U2 = {FU }[K U ] [K UU ] {U }U42kwith[K ] = [5k][K U ] = [−3k−2k]−3k[K U ] = [K U ] T =−2k4k 0[K UU ] =0 2k{} = {}U2{U } =U4{F } = {F3 }−F{FU } =2FFrom the second “row” of the partitioned matrix equations, we have[K U ]{} + [K UU ]{U } = {FU }and this can be solved for the unknown displacements to obtain{U } = [K UU ]−1 ({F } − [K U ]{})provided that [K UU ]−1 exists.

Since the constraints have been applied correctly, thisinverse does exist and is given by[K UU ]−11 4k=00 1 2kHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.3 Elastic Bar, Spar/Link/Truss ElementSubstituting, we obtain the unknown displacements as3 F0 +− 4k4  −F + 3k=1  2F + 2kF+ 2kk1 U2 4k={U } =U40The required force at node 3 is obtained by substitution of the displacement into the upperpartition to obtain53F3 = − F + k44Finally, the reaction force at node 1 isF3− k44F1 = −kU 2 =As a check on the results, we substitute the computed and prescribed displacements intothe individual element equations to insure that equilibrium is satisfied.Element 1−kkk−k0U2=−kU2kU2= f (1) 1 f (1) 2which shows that the nodal forces on element 1 are equal and opposite as required forequilibrium.Element 23k−3k−3k3kU2U33F−3k  − + 4k43k 33F (2) − k −f 24k4==f (2) 3F + 3 k 34k43k=−3kwhich also verifies equilibrium.Element 32k−2k−2k2kU3U4=2k−2k−2k2kTherefore element 3 is in equilibrium as well.F+k=−2F2F=ff(3)3(3)42.3 ELASTIC BAR, SPAR/LINK/TRUSS ELEMENTWhile the linear elastic spring serves to introduce the concept of the stiffness matrix, the usefulness of such an element in finite element analysis is rather limited.Certainly, springs are used in machinery in many cases and the availability of afinite element representation of a linear spring is quite useful in such cases.