The paper presents the theoretical background to a numerical method for
calculating the meridian of axisymmetric stressed shapes of infinitesimal
hexagonal grids. Two types of stressing are assumed, one with a pure tension
and another when tension is combined with twist. Also some results of an
illustrative example are given

Hegedűs, István

English

Periodica Polytechnica (Budapest University of Technology and Economics)

PERIODICA POLYTECHNICA SER. CIV. ENG. VOL. 49, NO. 1, PP. 3–12 (2005)THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONALNETSIstván HEGEDŰSDepartment of Structural EngineeringAcademic Research Group of StructuresBudapest University of Technology and EconomicsH–1521 Budapest, Hungarye-mail: hegedus@vtb.bme.huReceived: July 19, 2005AbstractThe paper presents the theoretical background to a numerical method for calculating the meridian ofaxisymmetric stressed shapes of infinitesimal hexagonal grids. Two types of stressing are assumed,one with a pure tension and another when tension is combined with twist. Also some results of anillustrative example are given.Keywords: hexagonal nets, underconstrained structures, membrane analogy.1. IntroductionIn a previous paper ([1]) , the author presented an analysis of axisymmetric shapes ofstressed infinitesimal Chebyshev nets. Chebyshev nets consist of two sets of contin-uous, flexible, inextensional fibres which are connected to each other at equidistantpoints. In original state, the net has a square network. It can be deformed like woventissues. Most analyses assume the distance of points of connections infinitesimallysmall (infinitesimal net), and replace the net by a membrane with incomplete rigidi-ties. Both finite and infinitesimal Chebyshev nets are typical underconstrainedstructures [3] which can develop large displacements without any change in lengthin fibre directions.The shape of stressed membranes of incomplete rigidities is determined usingthe consideration that inextensional deformations modify the initial shape until thesurface gets rigid, that is, the missing rigidities are not necessary to develop stressesresisting further deformation. Equilibrium conditions for the possible membraneforces can be used to determine the assumed shape.For axisymmetric stressed shapes the equilibrium of membrane componentsrequiresnanϑ= −rαrϑ. (1)nα, and nθ in Eq (1) are membrane forces, rα , and nθ are radii of curvaturesin meridian, and in annular directions, respectively.4 I. HEGEDŰSIf initial (unstressed) configuration of the net is a cylinder of the radius a withgiven directions of the fibres, then the stressed shape is a surface of revolution withthe meridian r(z). Ratio r/a determines the changes in directions of the intersectingfibres, directions of the fibres determine the ratio of nα , and nθ . On the other hand,ratio of rα , and nθ can be expressed using derivatives of r(z) as−rαrθ= 1 +(r ′)2rr ′′, (2)in this way an equation can be constructed which only contains r(z), r(z)’, andr(z)’ as unknown functions. Since r(z)” can be expressed from this equation,the rearranged equation permits us a numeric integration of r(z) after taking ap-propriate initial values for r(z0), and r(z0)’. Moreover, for specific initial values,analytic function for the meridian of the stretched axisymmetric Chebysev net canbe obtained.Hexagonal nets are fundamental patterns of assemblies in nature as well asin the theory of nets. Unlike Chebyshev nets, they consist of non-continuous fibres(Fig. 1). In initial state, fibres of hexagonal nets form a regular hexagonal network.Finite and infinitesimal hexagonal nets are also underconstrained surfaces, theirdeformability is more or less the same as that of knitted fabrics or crochets.In the next chapter, homogeneous deformations of finite and infinitesimalplane hexagonal nets are analysed. The results are used in Chapter 3 to determinethe meridian of axisymmetric stressed shapes of infinitesimal hexagonal nets.2. Homogeneous deformations of plane hexagonal netsHomogeneous deformation means in our case that all hexagons of the net deformin the same way. That is only possible if parallel fibres of the net stay parallelafter the deformation. Two types of homogeneous deformation will be assumed:symmetric, and asymmetric deformation. Symmetric means that the deformationdoes not upset the symmetry of the net.First-order theory of deformations assumes that powers of small changes inlength can be neglected. In first-order theory, regular hexagonal nets are discretemodels for surfaces that can only perform deformations which preserve the area ofthe surface meanwhile they develop a hydrostatic state of stress against stretching.These properties can be checked by analysing the deformations of the elementarycell of the net shown in Fig. 1 and by checking the equilibrium of nodal forces.Actually, if fibres of the cell are assumed absolutely rigid and powers of smallchanges in length are neglected, then small deformations of the cell preserve theareaA = 3√3l2. (3)The analysis of the neglected terms shows that A = 3√3l2 is the maximumarea of the elementary cell.THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS 5llll 33lyxFig. 1. Elementary cell of the hexagonal net in initial stateAnalogous joints of the net displace as if they were points of a continuoussurface subjected to a homogeneous state of strain with principal strainsε1 = −ε2. (4)This surface will be referred to as the replacement membrane of the net.Forces acting at a common point of application at lines of action in the di-rections α, α + π / 3, and α + 2π / 3 must be the same. That means, in initial stateof the net, forces acting at the fibres are either zero or equal. This force distrib-ution is analogous to hydrostatic states of stress of continuous membranes. If thevalue of the fibre forces is N , then principal membrane stresses of the replacementmembrane aren1 = n2 =√3N3l. (5)Strains developed in stressing the net, cannot be assumed small. Conse-quently, first-order theory has to be replaced by a geometrically exact theory ofdeformation.2.1. Geometrically exact symmetric homogeneous deformation of the netLet us first analyse finite symmetric deformations shown in Fig. 2. The changes inthe geometry and in the equilibrium conditions of nodal forces will be expressedusing α as geometric parameter.6 I. HEGEDŰSFig. 2. Elementary cell of the symmetrically deformed hexagonal netFinite strains in directions x and y can be expressed matching side lengths ofthe elementary cell in initial and deformed states of the net. Ratios of the lengthsare(1 + εx) = 2√3 cos α3, (6)(1 + εy) = 2 (1 + sin α)3. (7)Assuming N the fibre force at the bars in direction y, equilibrium of jointforces can be ensured by forces N/(2sin α) at the inclined bars. Principal directionsof stresses of the replacement membrane are directions x and y, the replacementmembrane forces arenx = N cos α2l sin α (1 + sin α), (8)ny = N2l cos α, (9)nxy = 0. (10)Limiting values of α are 0, and π /2 . If α = 0, that is, the inclined fibres getparallel to the parallel circles,1 + εx = 2√33, 1 + εy = 12,nynx= 0,if α = π /2, that is, if inclined fibres get parallel to the axis of rotation,1 + εx = 0, 1 + εy = 2, nynx= ∞.THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS 72.2. Asymmetric homogeneous deformations of the netAsymmetric homogeneous deformations are analysed using two angles as geometricparameters. One is α, the same as used in the previous section, another is γ whichis the angle of inclination of fibres 1 in Fig. 3.General description of homogeneous deformation needs a third geometricparameter, e.g. the angle of the rigid body-like rotation of the whole net, but ouranalysis does not need that generalization.Fig. 3. Elementary cell, and vector diagram of equilibrium forces in the state of skew finitedeformationFinite deformations in directions x and y are(1 + εx) = 2√3 cos α3, (11)(1 + εy) = 2 (cos γ + sin α)3, (12)tangent of the finite angle of distortion γ0 istan γ0 = sin γsin α + cos γ . (13)Ratios N2/N1, andN3/N1 of bar forces acting at fibres 1, 2, and 3 can beexpressed from the vector diagram of equilibrium forces (Fig.3.) asN2/N1 = cos (α − γ )sin 2α, N3/N1 = cos (α + γ )sin 2α. (14)Replacement membrane forces can be obtained in two steps. First, obliquecomponents membrane forces can be calculated, which are parallel with the sides of8 I. HEGEDŰSthe deformed elementary cell, then components can be expressed in usual orthogonalresolution. Neglecting details, the lengthy procedure results innx = N1(2 cos2 α cos γ + sin α cos2 γ )2l sin 2α (cos γ + sin α) , (15)ny = N1 cos γ2l cos α, (16)nxy = N1 sin γ2l cos α. (17)Limiting values of γ are ± π / 2.3. Stressing cylinders of hexagonal netA rotationally symmetric stressed hexagonal net is produced in steps as follows.First, a regular plane net is bent and welded into a cylinder of the radius andheight a, H , respectively. Then, cylindrical edges are fixed to rigid boundary rings.Finally, the distance of the boundaries is increased until the net gets stressed.Bending will be assumed in two ways. One when generators of the cylinderare parallel with x (Case 1), and another when they are parallel with y (Case 2).Also stretching will be assumed in two ways. One when the distance of theboundaries is increased with parallel moving in direction of the axis of rotation,and another when the parallel moving is combined with a rotation of the boundarycircles. Both types of stressing preserve the axisymmetric nature of the cylinder.3.1. Analysis using first-order infinitesimal theoryIn first-order theory of infinitesimal deformations, invariance of area of the ele-mentary cell can be assumed. On the basis of that invariance, the problem ofdetermining stressed shape of the cylinder can be directly converted into a minimalsurface problem: what is the meridian of the axisymmetric minimal surface of sur-face area 2πHa and of boundary radii a. Cases 1 and 2 are the same, and rotatingof the boundary circles has no effect on the solution.In [2], a detailed analysis of this problem is presented in a different context.It is proven, that the problem can only be solved for limited values of H/a, and alsoa dual solution exists. Meridians are catenary linesr = C cosh zC. (18)THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS 9Parameter C and the changed height h of the net emerge as solutions ofEqs(19a-b):C coshh2C= a, (19.a)2πCh + πC2 sinh hC= 2π Ha. (19.b)3.2. Analysis using geometrically exact infinitesimal theorySince first order theory yields reliable results only in a drastically limited rangeof parameters, a geometrically exact infinitesimal theory is applied. ‘Infinitesi-mal’ means that a replacement membrane is used, ‘geometrically exact’ means thatgeometric relations at infinitesimal neighbourhood of points of the membrane areanalysed as if they were homogeneous plane deformations in the tangent plane, thatis, in the cases of symmetric deformations Eqs. (4)-(10), in the case of asymmetricdeformation Eqs. (11)-(17) can be used.3.2.1. Stressing with pure tensionCase 1In Case 1 meridian of the replacement membrane coincides with direction x(see Fig. 2). Consequently, nα = nx , nθ = ny, εα = εx , εθ =εy ,nanθ= cos2 αsin α (1 + sin α), (20)and α varies within the limits π / 6, and zero.According to the procedure outlined in Section 1, r(z)” can be expressed onthe basis of Eqs. (1)-(2), and Eqs. (6)-(7) asr ′′ = 1 +(r ′)2rsin α (1 + sin α)cos2 α. (21)Connection between r , and α can be obtained usingra= (1 + εθ ) = 2 (1 + sin α)3. (22)Let the length of an infinitesimal section of the generator of the lattice cylinderbe ds0. The change in length of ds0 due to stressing isdsα = (1 + εα) ds0 = 2√3 cos α3ds0. (23)10 I. HEGEDŰSDifferentials dz, and dr belonging to ds0 aredz = dsα√1 + (r ′)2, dr = dsαr′√1 + (r ′)2. (24)Eqs. (21)-(26) enable us to construct a step-by-step method for computingcoordinates z, and r of meridians belonging to initial values of z, r>2a/3, and r′.Case 2In Case 2 directions x and y change, consequently,r ′′ = 1 +(r ′)2rcos2 αsin α (1 + sin α) (25)and the connection between r and α can be calculated fromra= (1 + εθ ) = 2√3 cos α3, (26)where α varies within the limits π / 6, and π / 2. The equation shows that r can takearbitrary small values.dsα = (1 + εα) ds0 = 2 (1 + sin α)3 ds0 (27)On the basis of Eqs (22)-(27) a similar step-by-step method for computingcoordinates z, and r of meridians can be constructed as used in Case 1.3.2.2. Stressing with a normal force and a torqueEquilibrium conditions for axisymmetric membrane forces of a shell of rotation canalso be met by assuming membrane force nαθ provided the equation2r2πnαθ = T (28)applies to each parallel circle of the shell. T has a direct mechanical meaning inEq. (28). It is a torque. A similarly constant value is2rπnα√1 + (r ′)2= F, (29)which is a tensile force acting at the axis of rotation. The ratio of T and Fa is alsoconstant.In Case 1, the only effect of torque T is that forces at fibres inclined to theright and left get different. This change does not modify the stressed equilibriumshape, provided the tensile force is sufficiently large to keep inclined fibres of bothdirections in tension.THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS 11In Case 2, fibres of meridian direction cannot transfer shear. Fibres 1 also haveto get inclined and the deformation becomes asymmetric. The angle of inclinationis determined by the equationTF= r√1 + (r ′)2 tan γ. (30)Eq. (30) shows that γ varies along the meridian.Inclination of fibres 1 changes the equilibrium shape of the stressed net. Eqs.(1)-(2), and Eqs. (12)-(13) yield for the second derivative of rr ′′ = 1 +(r ′)2r2 cos2 α + sin α cos γ2 sin α (cos γ + sin α) . (31)Connection of α and r is the same as in Case 2 in Section 3.2.2.ra= (1 + εθ) = 2√3 cos α3, (32)but the changed length of ds0 is different:dsα = (1 + εα) ds0 = 2 (cos γ + sin α)3 ds0. (33)The step-by-step method of the solution can be the same as used before.4. Illustrative exampleIn Fig. 4, the results of an illustrative example are presented. The half of themeridian of a stressed infinitesimal net is plotted for Case 1, Case 2, and for Case 2assuming also a torque T = 0.4aF. Ratio H/a is 4, the meridian of the unstressednet is drawn with dotted line.Only geometrically exact theory has been used because H/a = 4 does notpermit any mathematical solution of the minimal surface problem.Meridian belonging to Case 1 obeys the limitation r ≥ 2a/3 and its firstsection is close to a straight line indicating the limit value of r . Increasing ratioH/a results in small changes in the last section, only the length of the almost straightsection increases.In Case 2, the meridian looks different, because the limiting value of r is zeroand the meridian gets the closer to zero the higher the ratioH/a is.The third meridian shows that torque reduces both the stressed height and theminimum radius. For increasing values of the torque the difference of twisted andnot twisted nets rises, beyond a limiting value of T/aF the problem has no solution.12 I. HEGEDŰS0.5 1.0 1.5 2.00.00.00.20.40.60.81.0Case 1Case 2Case 2 with torquer/az/aFig. 4. Meridians of a stressed hexagonal net5. ConclusionsThe results have shown that the analysis of axisymmetric shapes of stressed infini-tesimal hexagonal nets requires the use of a geometrically exact infinitesimal theoryof deformation. Meridians of the stressed net essentially depend on the directionof the non-continuous fibres. Also the effect of twist is different if the orientationof the fibres is different.AcknowledgementThe author thanks for the material and technical support received from OTKA grant T046846 awarded the by the Hungarian National Research Fund and from the ResearchGroup of Structures of the Hungarian Academy of Sciences.References[1] HEGEDŰS, I.: Axisymmetric Csebyshev Surfaces. Scientific Notes of the Department of Bridgesand Structures of BUTE. (In Hungarian), 2004. pp. 57–68.[2] PÁLFALVI, H. D. – HEGEDŰS, I.: On the Multiple Solution of Axisymmetric MinimumSurfaces. Journal of Computational and Applied Mechanics, Vol. 3. No. 2. pp. 157–167. 2003.[3] KUZNETSOV, E. N.: Underconstrained Structures. Springer-Verlag, New York, Heidelberg,etc. 1991.

THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS

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THE STRESSED SHAPE OF AXISYMMETRIC HEXAGONAL NETS

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