Hutton - Fundamentals of Finite Element Analysis (523155), страница 21

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Therefore, if the beam were to be dividedinto two finite elements with a connecting node at the midpoint, the net force atthe node is the applied external force and the net moment at the node is the applied external moment.EXAMPLE 4.1Figure 4.7a depicts a statically inderminate beam subjected to a transverse load applied atthe midspan.

Using two flexure elements, obtain a solution for the midspan deflection.■ SolutionSince the flexure element requires loading only at nodes, the elements are taken to be oflength L /2 , as shown in Figure 4.7b. The individual element stiffness matrices are then126L/2−126L/222 (1) (2) EI z  6L/2 4L /4 −6L/2 2L /4 k= k=12−6L/2 (L/2) 3  −12 −6L/26L/2 2L 2 /4 −6L/2 4L 2 /4123L−123L8EI z  3LL2−3L L 2 /2 = 3 −12 −3L12−3L LL23L L 2 /2 −3LNote particularly that the length of each element is L /2 . The appropriate boundary conditions are v1 = 1 = v3 = 0 and the element-to-system displacement correspondencetable is Table 4.1.L2PL2v1v3␪2␪111␪322(b)(a)v1 ␪1 0v2v3 0v2␪2␪3(c)Figure 4.7(a) Loaded beam of Example 4.1.

(b) Element and displacement designations.(c) Displacement solution.3Hutton: Fundamentals ofFinite Element Analysis1044. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsTable 4.1 Element-to-System Displacement CorrespondenceGlobal DisplacementElement 1Element 2123456123400001234Assembling the global stiffness matrix per the displacement correspondence tablewe obtain in order (and using the symmetry property)(1)K 11 = k 11 =(1)K 12 = k 12 =(1)K 13 = k 13 =(1)K 14 = k 14 =(1)K 22 = k 22 =(1)K 23 = k 23 =(1)K 24 = k 24 =96EI zL324EI zL2−96EI zL324EI zL28EI zL−24EI zL24EI zLK 25 = K 26 = 0(1)(2)(1)(2)K 33 = k 33 + k 11 =192EI zL3K 34 = k 34 + k 12 = 0(2)K 35 = k 13 =(2)K 36 = k 14 =(1)−96EI zL324EI zL2(1)K 44 = k 44 + k 22 =(2)K 45 = k 23 =16EI zL−24EI zL2Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.5 Element Load Vector(2)K 46 = k 24 =(2)K 55 = k 33 =(2)K 56 = k 34 =(2)K 66 = k 44 =4EI zL96EI zL3−24EI zL28EI zLUsing the general form[K ]{U } = {F }we obtain the system equations as9624L8L 2 24LEI z  −96 −24LL 3  24L4L 2 0000−96−24L1920−9624L24L4L 2016L 2−24L4L 200−96−24L9624L    v1 0 F1  0 M11  24L  v2 = F2 2 4L 2 M2 24L vF33  28L3M3Invoking the boundary conditions v1 = 1 = v3 = 0 , the reduced equations become192EI z 0L324L016L 24L 2  24L  v2   −P 04L 2  2 =  08L 23Yielding the nodal displacements asv2 =−7PL 3768EI z2 =−PL 2128EI z3 =PL 232EI zThe deformed beam shape is shown in superposition with a plot of the undeformed shapewith the displacements noted in Figure 4.7c.

Substitution of the nodal displacement values into the constraint equations gives the reactions asF1 =EI z11P(−96v2 + 24 L 2 ) =L316F3 =EI z5P(−96v2 − 24 L 2 − 24 L 3 ) =L316M1 =EI z3PL(−24 L v2 + 4L 2 2 ) =3L16Checking the overall equilibrium conditions for the beam, we findFy =11P5P−P+=01616105Hutton: Fundamentals ofFinite Element Analysis1064.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsand summing moments about node 1,M =3PLL5P−P +L =016216Thus, the finite element solution satisfies global equilibrium conditions.The astute reader may wish to compare the results of Example 4.1 with thosegiven in many standard beam deflection tables, in which case it will be found thatthe results are in exact agreement with elementary beam theory.

In general, thefinite element method is an approximate method, but in the case of the flexureelement, the results are exact in certain cases. In this example, the deflectionequation of the neutral surface is a cubic equation and, since the interpolationfunctions are cubic, the results are exact. When distributed loads exist, however,the results are not necessarily exact, as will be discussed next.4.6 WORK EQUIVALENCEFOR DISTRIBUTED LOADSThe restriction that loads be applied only at element nodes for the flexure element must be dealt with if a distributed load is present.

The usual approach is toreplace the distributed load with nodal forces and moments such that the mechanical work done by the nodal load system is equivalent to that done by thedistributed load. Referring to Figure 4.1, the mechanical work performed by thedistributed load can be expressed asLW =q (x )v(x ) dx(4.51)0The objective here is to determine the equivalent nodal loads so that the workexpressed in Equation 4.51 is the same asLW =q (x )v(x ) dx = F1q v1 + M 1q 1 + F2q v2 + M 2q 2(4.52)0where F1q , F2q are the equivalent forces at nodes 1 and 2, respectively, andM 1q and M 2q are the equivalent nodal moments. Substituting the discretized displacement function given by Equation 4.27, the work integral becomesLW =q (x )[N 1 (x )v1 + N 2 (x )1 + N 3 (x )v2 + N 4 (x )2 ] dx0(4.53)Hutton: Fundamentals ofFinite Element Analysis4.

Flexure ElementsText© The McGraw−HillCompanies, 20044.6 Work Equivalence for Distributed LoadsqL2q12LxqL2qL212qL212(a)107(b)Figure 4.8 Work-equivalent nodal forces and moments for a uniformdistributed load.Comparison of Equations 4.52 and 4.53 shows thatLF1q =q (x ) N 1 (x ) dx(4.54)0LM 1q =q (x ) N 2 (x ) dx(4.55)q (x ) N 3 (x ) dx(4.56)q (x ) N 4 (x ) dx(4.57)0LF2q =0LM 2q =0Hence, the nodal force vector representing a distributed load on the basis of workequivalence is given by Equations 4.54–4.57.

For example, for a uniform loadq (x ) = q = constant, integration of these equations yieldsqL  2  2 qL F1q  M1q12=(4.58)F  qL  2q   2 M2q −qL2 12The equivalence of a uniformly distributed load to the corresponding nodal loadson an element is shown in Figure 4.8.EXAMPLE 4.2The simply supported beam shown in Figure 4.9a is subjected to a uniform transverseload, as shown. Using two equal-length elements and work-equivalent nodal loads, obtain a finite element solution for the deflection at midspan and compare it to the solutiongiven by elementary beam theory.Hutton: Fundamentals ofFinite Element Analysis4.

Flexure Elements108CHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsyv1qv2␪2x␪1L121(a)L2␪332(b)q1v3q223L2qL248qL248qL248qL4qL4qL248qL4qL4(d)(c)Figure 4.9(a) Uniformly loaded beam of Example 4.2. (b) Node, element, and displacement notation. (c) Elementloading. (d) Work-equivalent nodal loads.■ SolutionPer Figure 4.9b, we number the nodes and elements as shown and note the boundary conditions v1 = v3 = 0 .

We could also note the symmetry condition that 2 = 0 . However, inthis instance, we let that fact occur as a result of the solution process. The element stiffness matrices are identical, given by126L/2−126L/222 (1) (2) EI z  6L/2 4L /4 −6L/2 2L /4 k= k=3−12 −6L/212−6L/2 (L/2)26L/2 2L /4 −6L/2 4L 2 /4128EI z 3L= 3 −12L3L3L−12L2−3L−3L12L 2 /2 −3L3LL 2 /2 −3L L2(again note that the individual element length L /2 is used to compute the stiffnessterms), and Table 4.2 is the element connectivity table, so the assembled global stiffnessmatrix is12 3L8EI z [K ] = 3  −12L  3L 003L−12−3LL2−3L240L 2 /20−1203L3L0L 2 /200−122L 2 −3L−3L12L 2 /2 −3L00 3L L 2 /2 −3L L2The work-equivalent loads for each element are computed with reference to Figure 4.9cand the resulting loads shown in Figure 4.9d.

Observing that there are reaction forces atboth nodes 1 and 3 in addition to the equivalent forces from the distributed load, theHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.6 Work Equivalence for Distributed LoadsTable 4.2 Element ConnectivityGlobal DisplacementElement 1Element 2123456123400001234global equilibrium equations become−qL+ F1 42−qL  v1  481  v   −qL 2=[K ]22 0v 3 −qL3+F342qL48where the work-equivalent nodal loads have been utilized per Equation 4.58, with eachelement length = L /2 and q (x ) = −q , as shown in Figure 4.9c. Applying the constraintand symmetry conditions, we obtain the system−3LL28EI z 24 −3LL 3  L 2 /2003LL 2 /202L 2L 2 /2−qL2481 0 −qL3L  v2 =22 L 2 /2  03L22qL48which, on simultaneous solution, gives the displacements as1 = −qL 324EI z2 = 05qL 4384EI zqL 33 =24EI zv2 = −As expected, the slope of the beam at midspan is zero, and since the loading and support conditions are symmetric, the deflection solution is also symmetric, as indicated by109Hutton: Fundamentals ofFinite Element Analysis1104.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsthe end slopes. The nodal displacement results from the finite element analysis of thisexample are exactly the results obtained by a strength of materials approach. This is dueto applying the work-equivalent nodal loads. However, the general deflected shape asgiven by the finite element solution is not the same as the strength of materials result. Theequation describing the deflection of the neutral surface is a quartic function of x and,since the interpolation functions used in the finite element model are cubic, the deflectioncurve varies somewhat from the exact solution.EXAMPLE 4.3In Figure 4.10a, beam OC is supported by a smooth pin connection at O and supported atB by an elastic rod BD, also through pin connections.

A concentrated load F = 10 kN isapplied at C. Determine the deflection of point C and the axial stress in member BD. Themodulus of elasticity of the beam is 207 GPa (steel) and the dimensions of the cross section are 40 mm × 40 mm. For elastic rod BD, the modulus of elasticity is 69 GPa (aluminum) and the cross-sectional area is 78.54 mm2.■ SolutionThis is the first example in which we use multiple element types, as the beam is modeledwith flexure elements and the elastic rod as a bar element. Clearly, the horizontal memberU7DF 10 kN200 mmU1U33U5OBC300 mm300 mm1U2U4(a)2(b)u (3)2v2(3)v1(1)␪1(1)v2(1)␪2(1)v1(2)v2(2)␪1(2)␪2(2)u1(3)v1(3)(c)Figure 4.10(a) Supported beam. (b) Global coordinate system and variables.

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