Hutton - Fundamentals of Finite Element Analysis (523155), страница 23

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For suchapplications, orientation of the element in the global coordinate system must beconsidered, as was the case with the spar element in trusses. Figure 4.13a depictsan element oriented at an arbitrary angle from the X axis of a global referenceframe and shows the element nodal displacements. Here, we use to indicatethe orientation angle to avoid confusion with the nodal slope, denoted .

Figure 4.13b shows the assigned global displacements for the element, where againwe have adopted a single symbol for displacement with a numerically increasingsubscript from node to node. Before proceeding, note that it is convenient here toreorder the element stiffness matrix given by Equation 4.59 so that the element115Hutton: Fundamentals ofFinite Element Analysis1164. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsv2U5yx␪2v1u1u2YU6U2U4␺␪1XU1U3(a)(b)Figure 4.13(a) Nodal displacements in the element coordinate system.

(b) Nodal displacementsin the global coordinate system.displacement vector in the element reference frame is given as u1  v1  1{} =u2  v2 2and the element stiffness matrix becomesAE−AE00 LL12EI6EIzz 00L3L26EI z4EI z 002LL[ke ] =  −AEAE00 LL−12EI z −6EI z 00L3L26EI z2EI z00L2L(4.61)0−12EI zL36EI z− 2L012EI zL3−6EI zL200 −6EI z L2 4EI z L6EI zL22EI zL(4.62)Using Figure 4.13, the element displacements are written in terms of theglobal displacements asu1v11u2v22= U1 cos + U2 sin = −U1 sin + U2 cos = U3= U4 cos + U5 sin = −U4 sin + U5 cos = U6(4.63)Hutton: Fundamentals ofFinite Element Analysis4.

Flexure ElementsText© The McGraw−HillCompanies, 20044.7 Flexure Element with Axial Loading117Equations 4.63 can be written in matrix form as  u1 cos v−sin1  1 0=u2  0v02 20sin cos 00000000100 cos 0 −sin00000sin cos 0 0  U1  0  U2  0  U3= [R]{U }U4 0  0  U5 U61(4.64)where [R] is the transformation matrix that relates element displacements toglobal displacements. In a manner exactly analogous to that of Section 3.3, it isreadily shown that the 6 × 6 element stiffness matrix in the global system isgiven by[K e ] = [R] T [k e ][R](4.65)Owing to its algebraic complexity, Equation 4.65 is not expanded here to obtaina general result. Rather, the indicated computations are best suited for specificelement characteristics and performed by computer program.Assembly of the system equations for a finite element model using the beamaxial element is accomplished in an identical fashion to the procedures followedfor trusses as discussed in Chapter 3.

The following simple example illustratesthe procedure.EXAMPLE 4.4The frame of Figure 4.14a is composed of identical beams having a 1-in. square crosssection and a modulus of elasticity of 10 × 106 psi. The supports at O and C are to be considered completely fixed. The horizontal beam is subjected to a uniform load of intensity10 lb/in., as shown. Use two beam-axial elements to compute the displacements androtation at B.Y10 lb/in.U5U8u2U4BC20 in.2U6␪2U7v2U9120 in.u1U2␺ 兾2U1XU3O(a)(b)v1␪1(c)Figure 4.14(a) Frame of Example 4.4. (b) Global coordinate system and displacementnumbering. (c) Transformation of element 1.Hutton: Fundamentals ofFinite Element Analysis1184.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elements■ SolutionUsing the specified data, The cross-sectional area isA = 1(1) = 1 in.2And the area moment of inertia about the z axis isI z = bh 3 /12 = 1/12 = 0.083 in.4The characteristic axial stiffness isAE/L = 1(10 × 10 6 )/20 = (5 × 10 5 ) lb/in.and the characteristic bending stiffness isEI z /L 3 = 10 × 10 6 (0.083)/20 3 = 104.2 lb/in.Denoting member OB as element 1 and member BC as element 2, the stiffnessmatrices in the element coordinate systems are identical and given by00−5(10 5 )5(10 5 )01,250.412,5040 (1) (2) 012,504166,7200k= k=005(10 5 ) −5(10 5 )0−1,250.4 −12,5040012,50483,360000−1,250.4 12,504 −12,50483,360 001,250.4 −12,504−12,504 166,720Choosing the global coordinate system and displacement numbering as in Figure 4.14b,we observe that element 2 requires no transformation, as its element coordinate system isaligned with the global system. However, as shown in Figure 4.14c, element 1 requirestransformation.

Using = /2 , Equations 4.64 and 4.65 are applied to obtain1,250.40 (1)  −12,504K= 1,250.40−12,50405(10 5 )00−5(10 5 )0−12,5040166,72012,504083,3601,250.4012,5041,250.4012,5040−5(10 5 )005(10 5 )0−12,504083,360 12,504 0166,720Note particularly how the stiffness matrix of element 1 changes as a result of the 90°rotation. The values of individual components in the stiffness matrix are unchanged.

Thepositions of the terms in the matrix are changed to reflect, quite simply, the directions ofbending and axial displacements of the element when described in the global (system)coordinate system.The displacement correspondence table is shown in Table 4.6 and the assembled system stiffness matrix, by the direct assembly procedure, is in Table 4.7. Note, as usual, the“overlap” of the element stiffness matrices at the displacements associated with the common node. At these positions in the global stiffness matrix, the stiffness terms from theindividual element stiffness matrices are additive.Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.7 Flexure Element with Axial Loading119Table 4.6 Displacement CorrespondenceGlobalElement 1Element 2123456789123456000000123456Table 4.7 System Stiffness Matrix1,250.40−12,504 1,250.40−12,5040000500,00000−500,0000000 −12,5040166,72012,5040833,360000 1,250.4012,504501,250.4012,504−500,00000[K ] = 0−500,00000501,250.4 12,5040−1,250.4 12,504  −12,504083,36012,50412,504333,4400−12,50483,360 000−500,00000500,000000000−1,250.4 −12,50401,250.4 −12,504 000012,50483,3600−12,504 166,720Using the system stiffness matrix, the assembled system equations are  U1  RX 1U R2Y1MR1  U3 0 U4  −100[K ] U5 = U6 −333.3 RX 3  U7 U8   RY 3 − 100   U9M R 3 + 333.3where we denote the forces at nodes 1 and 3 as reaction components, owing to the displacement constraints U 1 = U 2 = U 3 = U 7 = U 8 = U 9 = 0 .

Taking the constraints intoaccount, the equations to be solved for the active displacements are then  501,250.4012,504  U4   0 0501,250.4 12,504  U5 = −100  −16.7U612,50412,504333,440Hutton: Fundamentals ofFinite Element Analysis1204. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsSimultaneous solution gives the displacement values asU 4 = 2.47974 (10 −5 ) in.U 5 = −1.74704 (10 −4 ) in.U 6 = −9.94058 (10 −4 ) radAs usual, the reaction components can be obtained by substituting the computed displacements into the six constraint equations.For the beam element with axial capability, the stress computation must take intoaccount the superposition of bending stress and direct axial stress.

For element 1, forexample, we use Equation 4.63 with = /2 to compute the element displacement as u1  v1   0 −1 01=u2   0  0 v2 201000000 0 00 0 01 0 00 0 10 −1 00 0 0  0   U1  0U2 000  U30=−40  −1.74704(10 )  U4  −2.47974(10−5 ) U05  U6−9.94058(10−4 )1The bending stress is computed at nodes 1 and 2 via Equations 4.33 and 4.34 asx (x = 0) = ±0.5(10)(10 6 )= ±495.2 psix (x = L ) = ±0.5(10)(10 6 )62(−2.47974 )(10 −5 ) −(−9.94058 (10 −4 )20 22062(2.47974 )(10 −5 ) +(2)(−9.94058 )(10 −4 )22020= ±992.2 psiand the axial stress isaxial = 10(10 6 )−1.74704 (10 −4 )= −87.35 psi20Therefore, the largest stress magnitude occurs at node 2, at which the compressive axialstress adds to the compressive portion of the bending stress distribution to give = 1079 .6 psi(compressive)4.8 A GENERAL THREE-DIMENSIONALBEAM ELEMENTA general three-dimensional beam element is capable of both axial and torsionaldeflections as well as two-plane bending.

To examine the stiffness characteristicsof such an element and obtain the element stiffness matrix, we first extend thebeam-axial element of the previous section to include two-plane bending, thenadd torsional capability.Figure 4.15a shows a beam element with an attached three-dimensional element coordinate system in which the x axis corresponds to the longitudinal axisHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.8 A General Three-Dimensional Beam Elementyzqz(x)w1zw2␪y112x(a)␪y2 x(b)Figure 4.15(a) Three-dimensional beam element. (b) Nodal displacements inelement xz plane.of the beam and is assumed to pass through the centroid of the beam cross section.

The y and z axes are assumed to correspond to the principal axes for areamoments of inertia of the cross section [1]. If this is not the case, treatment ofsimultaneous bending in two planes and superposition of the results as in thefollowing element development will not produce correct results [2].For bending about the z axis (i.e., the plane of bending is the xy plane), theelement stiffness matrix is given by Equation 4.48. For bending about the y axis,the plane of bending is the xz plane, as in Figure 4.15b, which depicts a beamelement defined by nodes 1 and 2 and subjected to a distributed load q z (x ) shownacting in the positive z direction. Nodal displacements in the z direction are denoted w1 and w2 , while nodal rotations are y1 and y2 .

For this case, it is necessary to add the axis subscript to the nodal rotations to specifically identify theaxis about which the rotations are measured. In this context, the rotations corresponding to xy plane bending henceforth are denoted z1 and z2 . It is also important to note that, in Figure 4.15b, the y axis is perpendicular to the plane of thepage with the positive sense into the page. Therefore, the rotations shown arepositive about the y axis per the right-hand rule. Noting the difference in the positive sense of rotation relative to the linear displacements, a development analogous to that used for the flexure element in Sections 4.3 and 4.4 results in theelement stiffness matrix for xz plane bending as12 −6L −12 −6L2EI y 6L2L 2 [ke ]x z = 3  −6L 4L(4.66)−12 6L126LL−6L 2L 26L4L 2The only differences between the xz plane bending stiffness matrix and that forxy plane bending are seen to be sign changes in the off-diagonal terms and thefact that the characteristic stiffness depends on the area moment of inertia Iy.Combining the spar element stiffness matrix, the xy plane flexure stiffnessmatrix, and the xz plane stiffness matrix given by Equation 4.60, the element121Hutton: Fundamentals ofFinite Element Analysis1224.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsequilibrium equations for a two-plane bending element with axial stiffness arewritten in matrix form as f x1  u1   f x2  u2  vf1y1Mz1z1[0][0][kaxial ]vf2y2 [0]=(4.67)[kbending ]x y[0]z2 Mz2 [0][0][kbending ]x z  w   f 1 z1 My1  y1  w2   f z2 y2M y2where the 10 × 10 element stiffness matrix has been written in the shorthand form[kaxial ][0][0][ke ] =  [0](4.68)[kbending ]x y[0][0][0][kbending ]x zThe equivalent nodal loads corresponding to a distributed load are computed onthe basis of work equivalence, as in Section 4.6.

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