Hutton - Fundamentals of Finite Element Analysis (523155), страница 22

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(c) Individualelement displacements.U6Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.6 Work Equivalence for Distributed LoadsTable 4.3 Displacement SchemeGlobalFigure 4.10bElement 1Element 2Element 31U1002U23U34U4v(1)1 (1)1v(1)2 (1)25U506U60v(2)1 (2)1v(2)2 (2)27U70000u (3)1u (3)2000Table 4.4 Element-Displacement CorrespondenceGlobal DisplacementElement 1Element 2Element 31234567123400000123400010003is subjected to bending loads, so the assumptions of the bar element do not apply to thismember.

On the other hand, the vertical support member is subjected to only axial loading, since the pin connections cannot transmit moment. Therefore, we use two differentelement types to simplify the solution and modeling. The global coordinate system andglobal variables are shown in Figure 4.10b, where the system is divided into two flexureelements (1 and 2) and one spar element (3).

For purposes of numbering in the globalstiffness matrix, the displacement scheme in Table 4.3 is used.While the notation shown in Figure 4.10b may appear to be inconsistent with previous notation, it is simpler in terms of the global equations to number displacements successively. By proper assignment of element displacements to global displacements, thedistinction between linear and rotational displacements are clear. The individual elementdisplacements are shown in Figure 4.10c, where we show the bar element in its general(3)2-D configuration, even though, in this case, we know that v(3)1 = v 2 = 0 and those displacements are ignored in the solution.The element displacement correspondence is shown in Table 4.4.

For the beamelements, the moment of inertia about the z axis isIz =bh 340(40 3 )== 213333 mm 41212For elements 1 and 2,EI z207(10 3 )(213333 )== 1635 .6 N/mmL3300 3111Hutton: Fundamentals ofFinite Element Analysis1124. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsTable 4.5 Global Stiffness Matrix12345671219,627.22.944 × 106−19,627.22.944 × 1060002.944 × 105.888 × 108−2.944 × 1062.944 × 108000634−19,627.2−2.944 × 10666,350.40−19,627.22.944 × 106−27,0962.944 × 102.944 × 108011.78 × 108−2.944 × 1062.944 × 1080656700−19,627.2−2.944 × 10619,627.2−2.944 × 1060002.944 × 1062.944 × 108−2.944 × 1065.889 × 108000−27,09600027,096so the element stiffness matrices are (per Equation 4.48)121,800−121,800 1,800 360,000 −1,800 180,000  (1) (2) k= k= 1,635.6 −12 −1,80012−1,800 1,800 180,000 −1,800 360,000while for element 3,AE78.54(69)(10 3 )== 27096 N/mmL200so the stiffness matrix for element 3 isk(3)1= 27,096−1−11Assembling the global stiffness matrix per the displacement correspondence table (notingthat we use a “short-cut” for element 3, since the stiffness of the element in the global Xdirection is meaningless), we obtain the results in Table 4.5.

The constraint conditions areU 1 = U 7 = 0 and the applied force vector is F1 R1  M1 0 F2  00M2 = −10,000 F3 0M3 R4F4where we use R to indicate a reaction force. If we apply the constraint conditions andsolve the resulting 5 × 5 system of equations, we obtain the results1 = 9.3638 (10 −4 ) radv2 = −0.73811 mm2 = −0.0092538 radv3 = −5.5523 mm3 = −0.019444 radHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.6 Work Equivalence for Distributed Loads(Note that we intentionally carry more significant decimal digits than necessary to avoid“round-off” inaccuracies in secondary calculations.) To obtain the axial stress in memberBD, we utilize Equation 3.52 with (3) = /2 :BD1= 69(103 ) −200120000 1 0 0−0.7381 = 254.6 MPa0 0 0 1 00The positive result indicates tensile stress.The reaction forces are obtained by substitution of the computed displacements intothe first and seventh equations (the constraint equations):R 1 = 2.944(10 6 )[9.3638 (10 −4 )] − 19,627.2(−0.73811 )+ 2.944(10 6 )(−0.0092538 ) ≈ −10,000 NR 4 = −27,096(−0.73811 ) + 27,096(0) = 20,000 Nand within the numerical accuracy used in this example, the system is in equilibrium.

Thereader is urged to check moment equilibrium about the left-hand node and note that, bystatics alone, the force in element 3 should be 20,000 N and the axial stress computed byF/ A is 254.6 MPa.The bending stresses at nodes 1 and 2 in the flexure elements are computed via Equations 4.33 and 4.34, respectively, noting that for the square cross section ymax /min =20 mm. For element 1,(1)x (x= 0) = ±20(207)(10 3 )62(−0.738 − 0) −(−(2)0.00093 − 0.0092 )2300300≈0at node 1. Note that the computed stress at node 1 should be identically zero, since thisnode is a pin joint and cannot support bending moment.For node 2 of element 1, we find(1)x (x= L ) = ±20(207)(10 3 )62(0 + 0.738) +(−(2)0.0092 − 0.00093 )2300300≈ ±281.3 MPaFor element 2, we similarly compute the stresses at each node as3 (2)x (x = 0) = ±20(207)(10 )62×(−5.548+0.738)−(−(2)0.0092−0.0194)≈ ±281.3 MPa30023003 (2)x (x = L) = ±20(207)(10 )62×(−0.73811+5.5523)+(−(2)0.019444−0.009538)≈ 0 MPa3002300and the latter result is also to be expected, as the right end of the beam is free of bendingmoment.

We need to carefully observe here that the bending stress is the same at the113Hutton: Fundamentals ofFinite Element Analysis1144. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsjuncture of the two flexure elements; that is, at node 2. This is not the usual situation infinite element analysis. The formulation requires displacement and slope continuity but,in general, no continuity of higher-order derivatives.

Since the flexure element developedhere is based on a cubic displacement function, the element does not often exhibit moment (hence, stress) continuity. The convergence of derivative functions is paramount toexamining the accuracy of a finite element solution to a given problem. We must examine the numerical behavior of the derived variables as the finite element “mesh” is refined.4.7 FLEXURE ELEMENT WITH AXIAL LOADINGThe major shortcoming of the flexure element developed so far is that force loading must be transverse to the axis of the element. Effectively, this means that theelement can be used only in end-to-end modeling of linear beam structures.If the element is formulated to also support axial loading, the applicability isgreatly extended.

Such an element is depicted in Figure 4.11, which shows, in addition to the nodal transverse deflections and rotations, axial displacements at thenodes. Thus, the element allows axial as well as transverse loading. It must bepointed out that there are many ramifications to this seemingly simple extension.If the axial load is compressive, the element could buckle. If the axial load is tensile and significantly large, a phenomenon known as stress stiffening can occur.The phenomenon of stress stiffening can be likened to tightening of a guitarstring.

As the tension is increased, the string becomes more resistant to motionperpendicular to the axis of the string.The same effect occurs in structural members in tension. As shown in Figure 4.12, in a beam subjected to both transverse and axial loading, the effect ofthe axial load on bending is directly related to deflection, since the deflection ata specific point becomes the moment arm for the axial load. In cases of smallelastic deflection, the additional bending moment attributable to the axial loadingis negligible. However, in most finite element software packages, buckling andstress stiffening analyses are available as options when such an element is usedin an analysis. (The reader should be aware that buckling and stress stiffening effects are checked only if the software user so specifies.) For the present purpose,we assume the axial loads are such that these secondary effects are not of concernand the axial loading is independent of bending effects.vii␪ivjuij␪jujFigure 4.11 Nodal displacementsof a beam element with axialstiffness.Hutton: Fundamentals ofFinite Element Analysis4.

Flexure ElementsText© The McGraw−HillCompanies, 20044.7 Flexure Element with Axial LoadingMv(x)FMFMFFM(x)M(x) M Fv(x)x(b)(a)Figure 4.12(a) Beam with bending moment and axial load. (b) Section of beam, illustratinghow tensile load reduces bending moment, hence, “stiffening” the beam.This being the case, we can simply add the spar element stiffness matrix tothe flexure element stiffness matrix to obtain the 6 × 6 element stiffness matrixfor a flexure element with axial loading asAE−AE0000  LL −AEAE0000 L L12EI z6EI z−12EI z6EI z  00L3L2L3L2 [ke ] = (4.59)6EI z4EI z−6EI z2EI z  00L2LL2L −12EI−6EI12EI−6EIzzzz  00L3L2L3L2 6EI z2EI z−6EI z4EI z 00L2LL2Lwhich is seen to be simply[k e ] =[k axial ][0][0][k flexure ](4.60)and is a noncoupled superposition of axial and bending stiffnesses.Adding axial capability to the beam element eliminates the restriction thatsuch elements be aligned linearly and enables use of the element in the analysisof planar frame structures in which the joints have bending resistance.

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