Hutton - Fundamentals of Finite Element Analysis (523155), страница 19
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In this chapter, elementary beam theory isapplied to develop a flexure (beam) element capable of properly exhibiting transverse bending effects. The element is first presented as a line (one-dimensional)element capable of bending in a plane. In the context of developing the discretized equations for this element, we present a general procedure for determining the interpolation functions using an assumed polynomial form for the fieldvariable. The development is then extended to two-plane bending and the effectsof axial loading and torsion are added.4.2 ELEMENTARY BEAM THEORYFigure 4.1a depicts a simply supported beam subjected to a general, distributed,transverse load q (x ) assumed to be expressed in terms of force per unit length.The coordinate system is as shown with x representing the axial coordinate and ythe transverse coordinate.
The usual assumptions of elementary beam theory areapplicable here:1. The beam is loaded only in the y direction.2. Deflections of the beam are small in comparison to the characteristicdimensions of the beam.3. The material of the beam is linearly elastic, isotropic, and homogeneous.4. The beam is prismatic and the cross section has an axis of symmetry in theplane of bending.91Hutton: Fundamentals ofFinite Element Analysis924. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsOdyVMyq(x)xMV(a)(c)(b)Figure 4.1(a) Simply supported beam subjected to arbitrary (negative) distributed load.(b) Deflected beam element.
(c) Sign convention for shear force and bendingmoment.h(a)(b)(c)Figure 4.2 Beam cross sections:(a) and (b) satisfy symmetry conditionsfor the simple bending theory, (c) doesnot satisfy the symmetry requirement.The ramifications of assumption 4 are illustrated in Figure 4.2, which depicts two cross sections that satisfy the assumption and one cross section thatdoes not. Both the rectangular and triangular cross sections are symmetric aboutthe xy plane and bend only in that plane. On the other hand, the L-shaped sectionpossesses no such symmetry and bends out of the xy plane, even under loadingonly in that plane.
With regard to the figure, assumption 2 can be roughly quantified to mean that the maximum deflection of the beam is much less than dimension h. A generally applicable rule is that the maximum deflection is lessthan 0.1h.Considering a differential length dx of a beam after bending as in Figure 4.1b(with the curvature greatly exaggerated), it is intuitive that the top surface has decreased in length while the bottom surface has increased in length. Hence, thereis a “layer” that must be undeformed during bending. Assuming that this layer islocated distance from the center of curvature O and choosing this layer (which,recall, is known as the neutral surface) to correspond to y = 0 , the length afterbending at any position y is expressed asds = ( − y) d(4.1)Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.2 Elementary Beam Theoryand the bending strain is thenεx =ds − dx( − y) d − dy==−dx d(4.2)From basic calculus, the radius of curvature of a planar curve is given by 2 3/2dv1+dx =(4.3)d2 vdx 2where v = v(x) represents the deflection curve of the neutral surface.In keeping with small deflection theory, slopes are also small, so Equation 4.3 is approximated by1 = 2(4.4)d vdx 2such that the normal strain in the direction of the longitudinal axis as a result ofbending isεx = − yd2 vdx 2(4.5)and the corresponding normal stress isx = E ε x = − E yd2 vdx 2(4.6)where E is the modulus of elasticity of the beam material.
Equation 4.6 showsthat, at a given cross section, the normal stress varies linearly with distance fromthe neutral surface.As no net axial force is acting on the beam cross section, the resultant forceof the stress distribution given by Equation 4.6 must be zero. Therefore, at anyaxial position x along the length, we haved2 vFx =x d A = − E y 2 d A = 0(4.7)dxAANoting that at an arbitrary cross section the curvature is constant, Equation 4.7impliesy dA = 0(4.8)Awhich is satisfied if the xz plane ( y = 0 ) passes through the centroid of the area.Thus, we obtain the well-known result that the neutral surface is perpendicular tothe plane of bending and passes through the centroid of the cross-sectional area.93Hutton: Fundamentals ofFinite Element Analysis944.
Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsSimilarly, the internal bending moment at a cross section must be equivalent tothe resultant moment of the normal stress distribution, sod2 vM (x ) = − yx d A = E 2y2 d A(4.9)dxAAThe integral term in Equation 4.9 represents the moment of inertia of the crosssectional area about the z axis, so the bending moment expression becomesM (x ) = EI zd2 vdx 2(4.10)Combining Equations 4.6 and 4.10, we obtain the normal stress equation forbeam bending:x = −M (x ) yd2 v= −yE 2Izdx(4.11)Note that the negative sign in Equation 4.11 ensures that, when the beam is subjected to positive bending moment per the convention depicted in Figure 4.1c,compressive (negative) and tensile (positive) stress values are obtained correctlydepending on the sign of the y location value.4.3 FLEXURE ELEMENTUsing the elementary beam theory, the 2-D beam or flexure element is now developed with the aid of the first theorem of Castigliano.
The assumptions and restrictions underlying the development are the same as those of elementary beamtheory with the addition of1. The element is of length L and has two nodes, one at each end.2. The element is connected to other elements only at the nodes.3. Element loading occurs only at the nodes.Recalling that the basic premise of finite element formulation is to expressthe continuously varying field variable in terms of a finite number of values evaluated at element nodes, we note that, for the flexure element, the field variable ofinterest is the transverse displacement v(x) of the neutral surface away from itsstraight, undeflected position.
As depicted in Figure 4.3a and 4.3b, transverse deflection of a beam is such that the variation of deflection along the length is notadequately described by displacement of the end points only. The end deflectionscan be identical, as illustrated, while the deflected shape of the two cases is quitedifferent. Therefore, the flexure element formulation must take into account theslope (rotation) of the beam as well as end-point displacement. In addition toavoiding the potential ambiguity of displacements, inclusion of beam elementnodal rotations ensures compatibility of rotations at nodal connections betweenHutton: Fundamentals ofFinite Element Analysis4.
Flexure ElementsText© The McGraw−HillCompanies, 20044.3 Flexure Elementv1v2v1v2(a)v0(c)(b)Figure 4.3(a) and (b) Beam elements with identical end deflections but quite differentdeflection characteristics. (c) Physically unacceptable discontinuity at theconnecting node.yv1v21221LxFigure 4.4 Beam element nodaldisplacements shown in a positivesense.elements, thus precluding the physically unacceptable discontinuity depicted inFigure 4.3c.In light of these observations regarding rotations, the nodal variables to beassociated with a flexure element are as depicted in Figure 4.4. Element nodes 1and 2 are located at the ends of the element, and the nodal variables are the transverse displacements v1 and v2 at the nodes and the slopes (rotations) 1 and 2 .The nodal variables as shown are in the positive direction, and it is to be notedthat the slopes are to be specified in radians. For convenience, the superscript (e)indicating element properties is not used at this point, as it is understood in context that the current discussion applies to a single element.
When multiple elements are involved in examples to follow, the superscript notation is restored.The displacement function v(x) is to be discretized such thatv(x ) = f (v1 , v2 , 1 , 2 , x )(4.12)subject to the boundary conditionsv(x = x 1 ) = v1(4.13)v(x = x 2 ) = v2dv = 1dx x=x1dv = 2dx x=x2(4.14)(4.15)(4.16)95Hutton: Fundamentals ofFinite Element Analysis964. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure ElementsBefore proceeding, we assume that the element coordinate system is chosen suchthat x 1 = 0 and x 2 = L to simplify the presentation algebraically.
(This is not atall restrictive, since L = x 2 − x 1 in any case.)Considering the four boundary conditions and the one-dimensional nature ofthe problem in terms of the independent variable, we assume the displacementfunction in the formv(x ) = a0 + a1 x + a2 x 2 + a3 x 3(4.17)The choice of a cubic function to describe the displacement is not arbitrary.While the general requirements of interpolation functions is discussed inChapter 6, we make a few pertinent observations here. Clearly, with the specification of four boundary conditions, we can determine no more than four constants in the assumed displacement function. Second, in view of Equations 4.10and 4.17, the second derivative of the assumed displacement function v(x ) islinear; hence, the bending moment varies linearly, at most, along the length of theelement.
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