Gummadi_1998_Large Strain Analysis of Beams and Arches Undergoing Large Rotations

7/28/2019 Gummadi_1998_Large Strain Analysis of Beams and Arches Undergoing Large Rotations

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Pergamon Inl. J. Non-Linear Mechanics. Vol. 33, No. 4, pp. 615-645, 1998Published by Elwier Science LfdPrinted in Great Britam

OOZO-7462/98 $17.00 + 0.00

PII: SOO20-7462(97)00033-4

LARGE STRAIN ANALYSIS OF BEAMS AND ARCHESUNDERGOING LARGE ROTATIONSL. N. B. Gummadi and A. N. Palazotto

Department of Aeronautics and Astronautics, Air Force Institute of Technology, WPAFB,OH-454433, U.S.A.(Receiv ed for publ icat ion 23 M ay 1997)

Abstract-A geometrically non-linear finite element formulation based on the total Lagrangianapproach is developed for laminated composite beams and arches considering large strain effects.Explicit expressions for the changes in the constitutive equations relating the Green strain to the 2ndPiola-Kirchhoff stress due to large strains are presented in curvilinear coordinates. Relationshipsbetween the Euler stress-Almansi strain measures and 2nd Piola-Kirchhoff stress-Green strainmeasures are used to determine the new constitutive relations. Large strain effects on load displace-ment characteristics are studied for both isotropic and laminated structures when they are undergo-ing large displacements and rotations. Published by Elsevier Science Ltd.Keywords 2nd Piola-Kirchhoff stress, Green strain finite element, beams, arches, curvilinearcoordinates

INTRODUCTIONThe non-linear finite element analysis of composite beams and arches has been the subjectof research for the past few years. Some of the studies focused on accurately determining thebehavior of structures undergoing large displacements and large rotations. A few relevantresearch efforts are reviewed here. Initially, formulations that used the small strain assump-tion are discussed. By using this assumption, the constitutive relationship between the stress(2nd Piola-Kirchhoff stress) and the strain (Green strain) is assumed to be constantirrespective of the magnitude of the strain. Surana [l] and Oliver and Oiiate [2] presenteda total Lagrangian finite element formulation using degenerated shell elements to solve theproblems of beams and arches. Mondkar and Powell [3] developed equations of motionbased on the total Lagrangian kinematics for isotropic structures. They retained higherorder terms and considered mid-plane extension and transverse shear in their finite elementformulation, Huddleston [4] used a total Lagrangian formulation with a no small angleassumption to analyze deep, isotropic, circular arches. In his analytical development,transverse shear was neglected while mid-plane extension was allowed. Sabir and Lock [5]also neglected the transverse shear strain. Total Lagrangian kinematics, in conjunctionwith a finite element solution method, was used to analyze isotropic, circular arches.DeDeppo and Schmidt [6] presented an analytical approach for solving non-linear circulararch problems with different boundary conditions. Also in their analysis, transversestrain and higher order mid-plane stretching terms were neglected. They used thetotal Lagrangian kinematics with no small angle approximations and a finite differencesolution method. They considered the effect of weight due to gravity of the arch in theiranalysis.Epstein and Murray [7] retained quadratic in-plane strain displacement terms whileneglecting transverse shear strain. They analyzed only isotropic beams using finite elementsand a total Lagrangian approach. Belytschko and Glaum [8] used small angle approxima-tions in an updated Lagrangian approach. They ignored transverse shear and higher orderterms. Antman [9] presented an updated Lagrangian approach to non-linear shell prob-lems. In his analytical formulation, mid-surface extension, transverse shear, bending

Contributed by G. J. Simitses.615


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616 L. N. B. Gummadiand A. N. Palazottorotation and transverse normal strain were included along with geometrically exact kin-ematic relations. Minguet and Dugundji [lo] used the co-rotational method. Euler angleswere considered for exact rigid body kinematics. Transverse shear and mid-plane extensibil-ity were incorporated without higher order terms. These later references [8-lo] consideredmethods which allowed for the strain functions to be represented in linear displacementform. Yet, each included major revisions in the usual kinematic expressions to capture thelarge rigid body displacements and rotations.Palazotto and Dennis [11] presented the Simplified Large Rotation (SLR) theory toanalyze shells. The SLR theory determines the equilibrium path of orthotropic shells usinga total Lagrangian approach and includes a parabolic transverse shear stress distribution.This approach captures the appropriate kinematics through displacement polynomials.Therefore, all of the final displacement functions are carried into the respective strain-displacement relationships without making any attempt at separating the rigid bodymovement. Thus, the features classically presented for including through the thicknessshear (for example) can be exploited [12]. Creaghan and Palazotto [13] reduced theSLR theory to one dimension for the analysis of beams and arches. Their theory producedaccurate solutions through 23 degrees of rotation. Miller and Palazotto [14] improvedthe rotation limits of Creaghan and Palazotto [13] to approximately 45 degreesby replacing the small angle approximation for the bending rotation using a truncatedseries representation of the tangent function. The truncated series representation usedin Miller and Palazotto [14] does not represent the kinematics of a beam accurately.Also, in Miller and Palazotto [14], the total potential energy was minimized by satis-fying the equilibrium equations only approximately. All these theories [ l&13,14] incorpor-ate polynomial approximations for the kinematics, Greens strains and minimization ofthe total potential to obtain the equilibrium equations. Yet, these theories failed to realizethe characteristics of large displacements and large rotations. It is the objective ofthe present paper to develop a large rotation theory using the conventional polynomialapproximations for the kinematics, complete Greens strains and the total Lagrangianapproach.All the theories discussed thus far assume the strain to be small. For an isotropicstructure, a strain of magnitude less than 0.04 is considered small [15] and when the strain issmall, the constitutive relation between the 2nd Piola-Kirchhoff stress and the Green strainis considered equal to the constitutive relation between the Eulerian stress and Almansistrain which is taken to be the true stress-strain relationship. However, such an assumptionis not valid when the strain is large. Effects of the changes in the constitutive relations areconsidered in Schimmels and Palazotto [16] while analyzing composite cylindrical shells.Schimmels and Palazotto used the SLR theory, which is valid for moderate rotations. Also,they introduced various approximations while dealing with changes in constitutive rela-tions. While these approximations are valid when dealing with problems that experiencemoderate rotations, they may introduce additional errors when dealing with problems oflarge displacement and large rotation.

In this paper, the development of a non-linear finite element model using the Greenstrains and the 2nd Piola-Kirchhoff stresses with large strains (yet not plastic) is discussed.Initially, kinematics that can capture the characteristics of large rotation are developed interms of curvilinear coordinates. Trigonometric relations, in the form of sine and cosinefunctions are used. Unlike the series representation of the sine and cosine functions, thetrigonometric terms are used by themselves. Exact constitutive relations are employed todetermine the total potential. The effects of large strains and the transformation between theEulerian stresses and the 2nd Piola-Kirchhoff stresses are taken into consideration. Explicitexpressions for constitutive relations as a function of displacement and rotation areprovided for beam and arch geometries. Even though the development is restricted to archand beam geometries, they can be extended to shell geometries. The equilibrium equationsare obtained through the minimization of the total potential energy which leads to finiteelement stiffness matrices. By incorporating the transformation of the constitutive relationswithin the non-linear load increment algorithm, the load displacement characteristics of thebeam and arch geometries with and without considering the changes in the constitutivemodel are determined.


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Large strain analysis of beams and arches 617eformed contiguration

defom&on@ration(a) Deformed and undeformed configuration of a beam

k

(b) Deformed kinematicsFig. 1. Kinematics and co-ordinate system for the beam.

THEORYThe beam/arch displacement equations are derived in a vector format similar toCreaghan and Palazotto [13] and Miller and Palazotto [14]. Initially, a normal to the mid

surface is chosen and the position vector of this normal is written in terms of the originalcoordinate system. The position vector of the same normal after deformation is determinedin terms of the original coordinate system. The vectorial difference between the two positionvectors form the desired kinematics. Unlike the earlier references, throughout the analysis,the small strain assumption is corrected and hence the constitutive relationship between2nd Piola stresses and Green strains is determined at every load increment. The displace-ments due to pure bending for a flat beam are shown in Fig. l(a). The dotted lines in thefigure correspond to the configuration after deformation. A cross-section of the flat beam,that is undergoing deformation, is shown in Fig. l(b). Pointsj and k are on the outer surfaceof the beam, and point o is on the mid plane. The y-z coordinates are the originalundeformed coordinates or global coordinates with the total Lagrangian system, and $ isthe rigid body rotation angle due to bending. The natural coordinate [ is aligned in thez direction and has a unit magnitude (+ 1) at the bottom of the beam (positive z direction)and - 1 at the top.Kinematics

The total Lagrangian displacement from any initial configuration i to a deformedconfiguration f for any point p isu(T) = pqf - I-ypl' (1)' . ' LZP_l LZPJ \ I

Here superscripts f; i denote the deformed and initial configurations, respectively.This equation can also be written as

Here 6 corresponds to the movement of point p due to midplane displacement, h is thethickness of the beam/arch and the direction cosines m3, II~ are given by(3)


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618 L. N. B. Gummadiand A. N. PalazottoAssuming the bending motion to be without any stretching of the beam normal, the pointp displaces to a new position p [shown as solid lines in Fig. l(b)]. This motion is equivalentto a zero strain in the normal direction.By substituting $ = 0 in the initial configuration, the displacement function at any i canbe written as

This derivation can be adapted to curved beams by considering the additional move-ment due to curvature. For an arch, the movement of point p due to midplane displace-ment is

-( )-v 1-z Ru=W (5)

where u and w are the displacements of point o on the midplane and R is the radius of thearch.The total kinematics can be expressed in global coordinates as

U=Y l-6 +zsin($)( >u3 = w + z(cos(ll/) - 1) (6)

When the rotation is small, these kinematic approximations result in expressions similarto that of Timoshenko [17]. These kinematic expressions can be written in tensorial form as

u = u2e2 + u3e3 (7)Here u2 and u3 are the contravariant tensorial components and e2 and e3 are the basisvectors in the y and z directions, respectively.

Strain-displacement relationsThe in-plane (physical) Green strain s22 (throughout the paper, the directions 2 and 3 arealso used for y and z) can be written in terms of the contravariant components u2, u3 of the

displacement vector asE22 = $2 + ;

[(u,22)2 /p (u;32)2

2 1Here, ufj is defined as the jth covariant derivative of the contravariant tensor ui and can beexpressed asUij = uf j + Uk

where uf s the jth partial derivative of ui and {: j} is the Christoffel symbol of the secondkind, (described in detail in Fung [18]) defined as= $gil (glj,k + gZk,j - gkj,l)

For the beam and arch problems, the range of the indices i, j and I is 2,3. Here g is thecontravariant metric tensor and glj is the covariant metric tensor. These metric tensors aredefined for curvilinear coordinates (assuming orthogonal curvilinear axes) asgij = -$= f, i=jgi j =g' j =O, i #j (11)

where hi are the scale factors which are defined later.


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Large strain analysis of beams and arches 619The use of the Christoffel symbols is required due to the curvilinear coordinate axes. For

example,(12)

By using the definition of the Christoffel symbol,

(13)Here h2 is the scale factor, and for an arch it is equal to 1 - z/R while for a beam, inCartesian system, it is equal to 1. Hence, all the Christoffel symbols will become identicallyequal to zero.By using the above definitions, Palazotto and Dennis [l 11 presented explicit expressionsfor Green strain-displacement relations. To model arches, the shell scale factorshI = h3 = 1 and h2 = 1 - z/R are used. If the derived kinematics [equation (6)] areincorporated in equation (8), the in-plane strain sz2 can be written as

3E22 = &202 + 1 ZP~22(JJ, (14)

p=1where .si2 is the mid plane strain and ~~~~~~re functions independent of z which are given by

&;2 = U,Y WC+ w,Yc - u,Ywc + $(U,, + w,, + u2c2 + WV)K22(1)= - WC2 v,YwcZ+ w2c3 + WYCZ + w,+ - c(cos($) - 1)

- o,,c(cos(ll/) - 1) + wc2(cos(ti) - 1) + $.ycosW) + a.yti,:ycos($)- WC+,Y os(ll/) + v sin($)c2 + cw,),sin($) - u*,, sin(lC/)c- w,,$,,sinW)

X22(2)= - u,c2 - +(cV + CU,,)+ 2V,,WC3+ *WV - 2ucw,, + w,$c2+ (1 - cos(lj))c(c + cu,), - 2wc2 + c(1 - cos($))/2 + $,;cos($)+ Il/,,cos(W + u,yti.ycos(ti)c - 2w$,,cosW)c2+ f(tj, + sin2(@)c2) + vsin($)c3 f 2w,,csin(+) - uw,,sin($)c2- 2w,,~,,csinW) - II/,,csin(+)

?C22(3) = U2Cs + 11,$C3(COS($)- 1)(2U,,c3 + (COS(+) - 1)C3- 2$,,c2cos($)) - ~c~v,,*,~co~(~) + *,,c - 2vc4sin($)+ 2uc3$,,sin($) -t c3sin2($) - 2*,/,,c2sin2($) (15)

In the above expressions, ,y denote the differentiation with respect to y and c = 1/R.Through the thickness shear strain is derived using linear Green strain components.Assuming a constant shear strain distribution across the thickness, shear strain 723 is given by723 = 2823 - w.~ + sin (+) (16)

Constitutive relationsThe simplified one-dimensional constitutive relationship between the stresses andstrains can be developed for a composite laminated beam made up of differently orien-ted plies. Thus, the constitutive relation for the kth lamina at an orientation of 8 can bewritten as

42 = Qk22.522k

023 = @k5&23 (17)


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620where

L. N. B. Gummadi and A. N. Palazotto

Qt2 = Qk,,cos40 + 2Qk,,sin28cos26 + Q:Isin48Qt5 = $(Qk4sin20 + Q~scos20) (18)

Here Qk,;s (the range of m, n include 2,5) represent the ply stiffness expressed in Euleriancoordinates, the O~jare the Eulerian stresses and the strains (Eij) are the Almansi strains.The factor 2 is the shear correction factor [17].The present formulation uses Green strains and 2nd Piola-Kirchhoff stress measures.When the stresses and strains are infinitesimal, both the stress measures will become equalto Cauchy stresses while both the strain measures will become Cauchy strains. Theconstitutive relation that relates the Green strain and 2nd Piola-Kirchhoff stress tensorscan be written as {sij)bX1 CQ16X6{Eij)6X1 (19)

Here Sij correspond to the second Piola-Kirchhoff stress tensor and sij is the Greenstrain tensor. The constitutive constants Q can be related to the Eulerian constitutiveconstants Q by using transformation relationships as shown below.The Eulerian stress tensor and the 2nd Piola-Kirchhoff stress tensor are related throughthe equation

Q _ aYi YjIJ A atxk &Y, k (20)in which dyi/aak = Bik+ ni;k, 6, is the Kroneker delta, A is the determinant of thedisplacement gradient matrix, F, given by

r 1 + $1 u;2 $3F = $1 1 + U;z n;3 (21)

_ $1 $2 1 + u,j3_The stress vectors (Tij and Ski are given by

{Oij} = (0 11, O22> O33> 012, 023, 013 1

&dT = (%I, s22~ s33, s12> s23> &3} (22)A similar relationship was developed by Washizu [ 191 for Cartesian coordinate system.However when dealing with Cartesian system, covariant derivative a:i becomes equal to thepartial derivative Ufi. By substituting the values ui, and the curvilinear nature of the original

coordinate system, the displacement gradient matrix F can be written as [see equations(9-l 3)l

F=

1 00

3 20 aU+Kay R

Thus, cij can be written asaij = _ TlSklA

0

>hi?!!2aZI+!!?aZ

(23)

(24)where the transformation matrix T1 is given by

1 0 0 00 1: h:(n:J2 2hzu%

T- 0 E2 (1 + a2 212(1 + $A

0 00 0I01 - 0 Zilz h&l + u,z) 11(1 + UT=)+ h,u:J, 0 0 (25)

0 0 0 0 1 + u;r 12_o 0 0 0 hzn:= II_


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Large strain analysis of beams and arches 621In addition, II = 1 + hzufy + h,u3/R and i2 = u2/R + u,,. Since h2, uz and u3 are

functions of z, it should be recognized that /I and l2 and hence T1, are functions of z.Similarly, the Almansi strain and Green strain can be related through the transformationrelationship

aal, da,E,j = - - Eklayi ayj (26)One can arrive at

Eij (27)where

T;=

and

A2 0 0 0 0 00 (1 + z4)2 l2 -/2(1 + $A 0 00 ht(&) l2 -h&l 0 00 -2h&(1 + u,,) -2111, 11(1 + u,z)+ h+f& 0 00 0 0 0 Ilb -h&

.O 0 0 0 -12A (1 + &)A

fEijlT = (El13 Ez2, E33, E12, E23, E13)

(EkI}T = {b &22, E3j, &12r &23, &L3) (28)

It should be noted lhat the transformation matrix 7-l is equal to the transpose of theinverse of the matrix Tz.

By using these transformation relations, the relationship between the constitutive ma-trices e and 0 can be obtained asT ;lQTZ8= A (29)

If one expands equation (29), the constitutive constants &, f&5 required for the beam orarch configuration can be written as&22 = (Q22U + u,z, + 2Q2,h:(u;2,)2(1 + (A2 + 4Qssh:(ufJ2(1 + d,

+ h:Q,&J41/A3

+ 2Q,,14(1 + u.z, + 2Q,,h,u:J:)/A3Here A = [11(1 + u,I) - h2ufZ12]

(30)A point to note here is the dependency of the transformation on the through the thicknessdirection since II, 1, and hI are functions af Z, the thickness coordinate. That is, thetransformation of the constitutive constants throughout the cross section is not uniform.

Typically, in a composite laminated beam or arch, consritutive constants of individuallamina can be transformed by using the transformation relations of equation (30). Sinceindividual layers experience different transformations, the overall cross sectional materialproperties can become unsymmetric. This necessitates the development of an un-symmetrical element for the analysis of large displacement, Large rotation and large strainproblems.However, under certain conditions, the constitutive constants remain constant. Forexample, in an isotropic beam or arch, the whole cross sectional properties are assumed tobe constant and are based on the properties at the neutral axis. A zero strain in an isotropicbeam indicate zero magnitude for the neutral axis (mid plane) strain &;2 which in turn isequal to 1 + V,Y= cos($) and w, , = -sin(ti). This relationship between the rotation rl/ andV,Y. W.Y also indicate zero shear strain in the beam. By substituting these reiatians intothe transformation equations [equation (3011, one can obtain Q2, = QZZ and &, =


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622 L. N. B. Gummadi and A. N. Palazotto~5s(~os2(11/)- sin2(JI)) + 2Q22 sin2(+)cos2(rl/). For an isotropic material, Qz2 = E andQS5 = G which are related as G = E/2(1 + v) where v corresponds to Poissons ratio, G isthe shear modulus and E is Youngs modulus. Considering zero Poissons ratio, QS5 = Qs5.Thus, for an isotropic material, constitutive relations based on the above assumptions donot need to be changed even at large displacements and rotations, as long as the midplanestrains are zero. This simplification satisfies the assumption of constant constitutive rela-tions in the cases of small strain problems. Limitations of the above simplifications arediscussed in the results section of this paper.

Total potentialThe internal strain energy for the arch structure consists of energy due to axial normalstrain and through the thickness shear strain which can be written as

U=;bss

e:IBtldzdy+;bSI Q:&dzdy (31)1 h I h

Here 1 is the length of the beam, b is the width of the beam and h is the thickness ofthe beam. Each of the strain components can be divided into linear and non-linearparts [l l]

E& = L;q + +q*HoqX22(i) = LTq + $qHiq

?23 = L : q +&f &q (32)Here, Li are column arrays for linear parts, Hi are the symmetric matrices for the non-linearparts of the normal strain, L, and H, are the linear and non-linear parts for the shear strainwhile q is the displacement gradient vector which is given by

4* = (r4 o,y, w, w Y)W YY, * , $ . Y)For example, L1 and HI corresponding to rc22(1jcan be written as

L1= 00+0001[ 1

HI =

0

001-R2

0 0 1 sin(*)R2 0 7jZ0 1 1 - cos(9)-

R2 o

Q+R1 2 cos($) - 1-- R2 R3 0 0 $R2

0 0 2Z 0 - in($)$R

0 0 0 0 0 0

sin$1- Rcow

COW-- R- sin($)

0sin($) 1 - cos(J/) cos(+) - 1 sin($)1l/R2 $R $R2 +R

o 2(1 - cos(ll/)) cos(~) - 1ti2R *

sir&V cow cos(J/) - 1-- R COW - - - sin(+) 0R * 0

(33)

(34)


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Large strain analysis of beams and archesand L, , I f , corresponding to y23 can be written as

L , = [O 0 0 1 0 1 O]00000 0 00 0 0 0 0 0 00 0 0 0 0 0 0

H,=O 0 0 0 0 0 00 0 0 0 0 0 00 0 0 0 0 sin(ll/) VQIc/1 0 0 0 0 0 0

623

It can be observed that matrices Hi and H, are functions of bending rotation $. Usingthese relations, the total potential can be written as 1 I_ qTP dy (351

Here P is the load vector, I? is constant, fii is a linear function and fi2 is a quadraticfunction of the displacement gradient vector q and can be expressed asR = ~ il: 3Ci+ j (L iL T + L jL T) + ~CS(L,L : + L,L~)

i=O =oRI = i i Ci+ j (L i qTHj + qTLjH i + H jqLT) + CS(L,qTH, + qTL,H, + H ,qLT )i=o j=o

R2 = i i Ci+ j+(H iqqTH j + H jqqTH i + +(qTHjqH i + qTHi qHj ) )i=o j=(J+ CWKwT K + mqT4 + +(qTfbqK + qTG?fu)

Here, the elastic terms, Ci, CS when i = 0, 1,2,3,4,5,6 can be defined as(36)

CC,, Ci, C2, G, Cq, G, Gil = bs 4?22Cl, i, 12, C3, 14> 15, C61 ihCS = b s &d( (37)h

For a laminate with the through the thickness symmetry, the constants Ci, C3 and C5 arezero. As will be pointed out subsequently, these quantities will become non-zero as thetransformation effects become relevant.

Equilibrium equationsIf the variation of the total potential is taken, the equilibrium equations can be obtained as

That is, even the equilibrium equations can be expressed in terms of the same matrices I?,fil and fi2 as that of the total potential. Next, the formation of the matrices R, fli, f12 andfi3 is described. It should be noted here that the development of the matrices k, fll and fi2is detailed in Palazotto and Dennis [i l] while the extra matrix I?3 is present because of thesine and cosine terms in the strain expression.


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624 L. N. B. Gummadi and A. N. PalazottoFormati on of mat ri ces & , fi , B2 and fl SThe variation of the total potential can be taken with respect to the degrees of freedom4 as XI = b ss & w%,,)dz dy + b ss & ~d(Y~Wdy - dq*pddy (39)1 h 1 h s IIn terms of Li and Hi, the variation of strains ~2~ and ~23 can be written as

6~22 = i LfSq + Gq*Hi q + GLTq + q*SHi qi=O

6~23 = LHSq + Gq*H,q + SLTq + q*dH,q (40)In Palazotto and Dennis [ll] and Creaghan and Palazotto [13], all the elements in L i andH i are constants (independent of q) and hence the last two terms in the right hand side of theabove equations are identically equal to zero. In the case of Miller and Palazotto [13] andthe present formulation, all the elements of L i are constants and some of the elements of H iare functions of $. That is, while the variations of L i are equal to zero, variations of H i arenot equal to zero. The variation of strains can be rearranged as

8~22 = i LTdq + Sq*Hiq + Gq*Aiqi=O

6y23 = L:dq + Gq*H,q + dq*& q (41)This rearrangement process can be explained with the help of the following example. 6H1[in equation (40)] corresponding to the variation of HI , can be written as

6H1 =

0 0 0 0 0 h116 h1170 0 0 0 0 h126 h1270 0 0 0 0 h136 h1370 0 0 0 0 h146 h1470 0 0 000 0

h116 h126 h136 h146 0 h166 h167h117 h127 h137 h147 0 h167 0

where

h117 = -coRs(+)G+

h127 = - sin($)6$

h147 = - COS(J@$

(42)

7x7

(43)


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Large strain analysis of beams and arches 625The notation hijk is used to denote (j, k)th term in ith H. That is, h167 represents (6,7) termin H,. Using this notation, qT6HIq can be written from equation (41) as

qTdH,q = v4Qhll6 + v,,$h126 + w$h136 + w&h146 + $h166/2 + $,,1C/h167+ vll/,,h117 + v,,lC/,,h127 + w$,,h137 + w,,$,,h147

qT6Hlq can be rearranged such that all variational terms in $ from equation (43) can beseparated out giving qT6HIq = GqfiIq. Thus, l?, can be written as

A, =

000000000000000000000000000000000000 0 0 0 0 tl 00000000

(44)

7x7

The term t1 can be obtained from the relation (where tl is positioned appropriately asmultiplier of $)

GqT&q = s*t,+ (45)Thus

t 1 - cos($)1 *2

If all the elements in Li, Hi are contstants, it was shown that [11]

+bis

!%(yz3Ls~q + &Hsq)dzdy - dqTfdy1 h s 1

When Hi terms are not constants,

+b &(yzjL:dq + GqTH,q + Gq=fl,q)dzdy -s

6qTpdyI

= 6qT 1 I-F dy

(46)

(47)

(48)where N3 can be written as

I h

& = t, + zt1 + z2t2 + z3t3 (49)Here ti when i = 0, 1,2,3 correspond to Hi(6, 6) when i = 0, 1,2,3 and t, correspond to H,.


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626 L. N. B. Gummadi and A. N. PalazottoThe analysis presented so far is element independent. In the next section, a 12 degree of

freedom beam element is developed based on the current formulation.

FINI TE ELEMENT FORMULATIONThe beam finite element used for the current formulation is shown in Fig. 2. The

element has two nodes with degrees of freedom (DOF) u, v,~, w, w,~, II/ and JI,, at eachnode.The values of the DOF q(q) at any point can be expressed in natural coordinates in termsof the elements DOF using the shape functions

where

TT=

HII (HII),, 0 0Hlz W12),9 0 0

0 0 0 00 0 0 00 0 H,l WI A,0 0 H12 (Hdg

H21 W21),9 0 0H22 W22l.q 0 00 0 0 00 0 0 00 0 Hz1 (H2d.9,

= Td

0 0 00 0 00 HII WII0 H12

WI l),v,q 0W12L,t, 0

0 00 00 Hz10 H22

( Hz I ) , ~,

)V-&2)::

0000

(Hzd,tW22L

00 0 Hz2 W22l.9 W22),4q 0 0 2x7

(51)

and(52)

Here the shape functions H are

Hll = $2 - 39 + q3) Hi2 = ;(l - rj - q2 + q3)

Hz1 = $ (2 + 3~ - q3) Hz2 = ; (- 1 - q + q2 + q3)

and a is half the length of the element.The derivatives in the displacement vector q are transformed from natural coordinates toglobal coordinates using the inverse of Jacobian matix, J, (I = J- ), asq( y ) = - q( q) =- Td Dd ( 53)


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Large strain analysis of beams and arches 621

2a Fiber Orientation

/ q r-1

-

(1)YYww w (2)wl) ;:2,.Y (2)w.Y

Fig. 2. Element description.

where D = IT while the matrix r can be written as1000 000Ol/aO 0 0 0 00010 000

r= 0 0 0 l/a 0 0 00 0 0 0 l/a2 0 00000 0100000 0 0 l/a

(54)

1

By using these shape functions, the equilibrium equations in terms of the nodal degrees offreedom can be written as

[ K+?+F+N, d-F=O=f(d)1whereK = D=I?DdrjJ I

N1 = DTfiT,Ddyls 1

(55)

F = DTfldrjJ (56)I


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628 L. N. B. Gummadi and A. N. PalazottoThese non-linear equilibrium equations at each load or displacement level can be solvedusing the modified Newton-Raphson incremental methods.If the equilibrium equations at d = d,_ 1can be written as f(dn_ i) = 0, the equilibrium at

d, = d,_ 1 + Ad,,_ 1 can be arrived at by expanding f(dn_ 1 + Ad,_ i) in a truncated Taylorseries for a small value of Ad,_ 1 asf(d,,-1 + Ad,-,) xf(dn-1) +;Ad.l = 0 (57)

Alternatively$Ad_ 1 = -f(d,_ 1) (58)

af/ad can also be written as [K + N1 + N2 + aN,/dd]. Thus, the incremental equations tobe solved can be written as

+ DTfldqs (59)Iwhered,=d,el +Ad,_, (60)

The Gauss quadrature is used to numerically evaluate the integrals in the above equation.The integral on the left hand side of equation (59), as an example, can be written as

s= (61)-1where det J = determinant of the Jacobian matrix, Wi = a weighting factor, 4(qi) = (R +10, + fiZ + afl,/ad)det Jdq evaluated at the Gaussian point vi and m is the number ofGaussian points. Seven point quadrature (m = 7) is used for all the example problems.Either the displacement control method or a modified Riks method developed by Tsai andPalazotto [20] can be incorporated to solve these incremental equations for determiningthe global displacements. The results are discussed for various laminate configurations ofbeams and arches.

RESULTS AND DISCUSSIONThe bending of beams and arches can be modeled as axial loading of individual fibersalong the cross section of the beam. Hence, to determine the strain levels at which the

transformation becomes significant, an axial rod subjected to tension is initially solved. Theequations relating the stesses and strains for one-dimensional isotropic axial rod can bewritten as

(1 -I- u2.2)622 = (1 - vu2,2)2 s22

1E22 = (1 + u2,2)2 E22 (62)

The non-dimensionalized Eulerian stress as a function of Almansi strain is shown in theform of the solid line in Fig. 3. By using a Poissons ratio of v = 0.3 in equation (62), theGreen strains and 2nd Piola-Kirchhoff stresses are obtained and are shown as the dashedline in Fig. 3. For both the curves, stress is non-dimensionalized with respect to thelongitudinal modulus of the problem. It can be observed that there is no deviation betweenthe curves until the strain level reaches 0.04. That is, a strain can be defined to be small if thestrain level is less than 0.04 or 4% [15].


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Large strain analysis of beams and arches 6290.25 I I I I I

0.2 -

0.15 - i.-.i _ Euler - Almansi%85 _____

.1 - _A/. _-_-2 nd Piola Kirchhoff - Green

/0 /, I 1 6 t 80 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5Strain

Fig. 3. Comparison of stress-strain curves for a uniaxial case.

With this function of small strain, the developed element is used to solve a number ofbeam and arch problems. Load displacement characteristics involving very large rotationsare determined using an incremental approach. At the zero load level, all the elements usedfor modeling the structure are assumed to have the same constitutive relations between thestress and strain. At the end of an incremental displacement, the constitutive relationsbetween the Green strains and 2nd Piola-Kirchhoff stresses are reevaluated for eachelement and new constitutive relations are used for the next solution increment. Thematerial considered is AS4-3501-6 graphite epoxy and the material properties used in all theproblems are El1 = 142 GPa, Et2 = 9.8 GPa, Gi2 = Gr3 = 6 GPa, GZ3 = 4.8 GPa andvr2 = 0.24. The thickness of the each layer considered is 0.000127 m. Whenever isotropicmaterial is used, material properties are provided in the corresponding figure.

Canti l ever i sotr opic beam under t i p l oadAn isotropic cantilever beam under a tip concentrated load, as shown in Fig. 4, isconsidered as a first example. In this problem, the beam experiences large displacement and

large rotation but only a small strain (at the neutral surface). To incorporate the boundaryconditions, all the degrees of freedom at the fixed end are constrained. The beam is modeledusing 10 elements of equal length. A convergence study indicated that this many elementswere sufficient for accurate results. A displacement control scheme [ 1 ] is used to solve thenon-linear algebraic equations with a displacement increment of 0.01 m and a tolerance of0.001% between two successive iterations. Initially, the problem is solved without consider-ing any transformation for the elastic constituents. The solid line in Fig. 5 denotes thenon-dimensional load deflection curve for this beam. Deflection is non-dimensionalizedwith respect to the total length of the beam while the load is non-dimensionalized as2 = PLz/C2 where P is the load, L is the length of the beam and CZ is the bending stiffness ofthe beam. For an isotropic beam C2 is equal to EZ where E is the Youngs modulus and I isthe bending moment of inertia. To study the effect of the transformed constitutive relations,the beam is again analyzed using the same geometric and analysis parameters (such asnumber of elements, displacement increment and tolerance level) but with the transformedelastic constants, and the resulting load deflection curve is shown as the dotted line in Fig. 5.


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630 L. N. B. Gummadi and A. N. Palazotto

Lengthof the beam 4.01 mWidthofthebeam=l.OmThickness of the beam =O.OOlm

Youngs Modulus of the beam ~1.2 KN/ mm 2Poissons ratio =0

Fig. 4. Geometry and material properties of a cantilever beam with a tip concentrated load,

7o.7/0.6 I-

0.5I

(r 0.4 -lx.i 0.3 -

0.2 -

0.1I

_ Wiihout Transformation--- Wii Neutral axis Transformation,,.Wiih multilayer Transformation

Ow I I I I I I0 0.1 0.2 0.3 % 0.5 0.6 0.7 0.8

Fig. 5. Tip displacement comparison with and without transformation for an isotropic beam.

In this case, at the end of each displacement increment, displacements, rotations and strainsare determined at the neutral axis and based on these displacement variables, the constitut-ive constants & and & are modified according to equations (30). The same transforma-tion is assumed to be valid throughout the cross section, eliminating the thicknessdependency of the transformation. This is because, in an isotropic beam or arch, the wholecross sectional properties are assumed to be constant and are equal to the properties at theneutral axis. But in reality, different fibers of the cross section experience different strains,displacements and rotations. That is, constitutive constants for the outer fibers will bedifferently transformed from those of inner fibers. To capture the effect of the transforma-tion of individual fibers, the beam or arch is assumed to be divided into a number of layers,each layer having the same material properties at the zero load level. Ideally, by increasing


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Fig. 6. Strain as a function of tip deflection.

the number of layers, modeling of the cross section becomes realistic. This layered approachis used for elasto-plastic analysis of isotropic beams by Owen and Hinton [21].Now, using the layered approach, the cross section of the beam is assumed to be made upof 12 isotropic layers of the same material properties. At the end of each displacementincrement, the transformation is carried out for the individual layers. Load deflectioncharacteristics of the beam using this layered approach are shown in dotted form in Fig. 5.In this figure, the non-dimensionalized displacement of 0.7 corresponds to a rotation of 1.2radians. It can be seen that even for a large displacement (as high as 0.7 times the length ofthe beam), all three load displacement curves deviate very slightly from each other.A reason for the slight deviation of the load deflection curves is the presence of moderatestrains in some layers of the beams cross section. To see the dependence on the strain for theload displacement characteristics, the variation of the total axial strain at the top layer atthe tip of the beam (note here that the tip of the beam corresponds to the Gaussian pointclosest to the tip) is plotted in Fig. 6 as a function of the tip displacement. It can be observedfrom this figure that even at very high tip displacement for this beam configuration, strain isless than 0.01 or 1%. From Fig. 3, where the stress-strain curves are compared, this strainlevel shows very little difference in the constitutive relations. However, if the strain atvarious points along the length of the beam is determined as a function of tip displacement,it was observed that the outer fiber strains became greater than 4% close to the fixedsupport (y < L/10). Yet, these high strain levels do not seem to affect the load-deflectioncharacteristics. A judgement was put forth in Schimmels and Palazotto [16] that a shellshould be defined as undergoing large rotation only when at least 20% of the area of theshell experiences large rotations. This is an arbitrary judgement by the authors of Ref. [16].But this argument did put bounds on the validity of their theory. Using a similar argument,the beam or arch can be defined as undergoing large strains when a minimum of 20% of thelength (arc length in the case of arches) experiences strains larger than 4%. Based on thisdefinition, the present beam is experiencing small strains over 90% of the length. Thus, theload-deflection characteristics were not affected significantly.It can be seen from this strain curve that the axial strain is increasing with the increase oftip deflection until W/L = 0.7. After that, the axial strain starts to decrease. This anomalycan be attributed to the limitation of the conservative force nature of the problem. That is,


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0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8WiLFig. 7. Unsymmetric stiffnesses as a function of tip deflection.

at lower displacements, the applied force was behaving like a conservative force. But as thetip of the beam became very close to the vertical tangent, the vertical applied force becamecloser to a follower force indicating a non-conservative force effect.Another observation is the development of the thickness direction unsymmetric proper-

ties. In the layered approach, as each of the layers undergo different transformations, thecross sectional properties become unsymmetric. When the cross section is symmetric, thequantities Cr, CJ, C5 of equation (37) are equal to zero. As the unsymmetry develops acrossthe cross section, these quantities become non-zero. A non-dimensional curve of C1 (definedas J,, oZ2 5 dc) as a function of the tip displacement is shown in Fig. 7. The non-dimensional-ization of C1 is defined as C, = C1 * h/C2 where h is the thickness of the beam and C2 is thebending stiffness of the beam. The larger the value of C1, the stronger the coupling betweenthe axial and bending characteristics of the beam. It can be observed from this figure that asthe tip displacement increases (that is, load increases), the degree of unsymmetry (C,) alsoincreases. That is, an unsymmetric formulation is required to analyze a symmetric structurewhen it undergoes large deformation. Similar results are observed in the remaining exam-ples considered.

Canti l ever lam inat ed beam under t i p loadTo study the effect of the displacement and rotation in a layered composite beam,a cantilever beam was studied using the same geometric properties and boundary condi-

tions as that of the earlier example, but made of AS4-3501-6 graphite epoxy compositematerial of lay up [O/90/90/0]. The elastic beam was modeled using 10 elements of equallength. The displacement control algorithm was used to solve the non-linear equationsiteratively until a convergence accuracy of 0.0005%. The tip displacement in the verticaldirection at different tip load levels is determined (Fig. 8). The solid line in the figurecorresponds to the load deflection curve when the same constitutive equation is usedthroughout the analysis, while the dotted line corresponds to the solution deflection curvewhen the constitutive relation is transformed at the end of each solution increment. Forsmall deflections, it can be observed that there is no difference in the load deflection curveswhile the transformed load deflection curve deviates slightly as the deflection is increased.


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0.6 -

0.5 -

NO.4 -0NI-z! 0.3 -

0.2 -

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Fig. 8. Tip displacement comparison for a laminated beam.

1.6-

1.4 -

1.2 -.EgfJJ l-52; 0.8 -2

0.6 -

0.4 -

Fig. 9. Strain as a function of tip deflection for a laminated beam.

8

When the tip deflection is 0.7 times the length of the beam, there is a 7% difference in theload deflection curves. A plot of tip outer strain is shown as a function of tip deflection inFig. 9. It can be seen that the axial strain is increasing with tip deflection and is around0.02% when the tip deflection is 0.7 times the length of the beam. Even though the axialstrain near the tip of the beam is very low (as seen in Fig. 9), strain greater than 4% occursnear the support. Furthermore, it can be seen, in Fig. 10, the non-dimensionalized C1increases with the increase of tip displacement, as pointed out in the earlier example.


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0 I I0 0.1 0.2 0.3 0.5 0.6 0.7 0.8Fig. 10. Unsymmetric stiffnesses as a function of tip deflection for a laminated beam.

\\ I:>

Length of the beam =O.OlmWldthofthebeam= l.OmThickness of the beam =O.OOlm

YoungsModulusofthebeam=1.2KN,mm2Poissons ratio =o

Fig. 11. Geometry and material properties of a cantilever beam with a tip concentrated moment.

Isotropic cantilever beam under tip momentIn order to study the effect of a very large rotation, the next problem considered isa cantilever beam with a concentrated moment at the free end. Material and geometricproperties of the cantilever beam are shown in Fig. 11. An exact solution for this problemwas presented by Ramm [22], with the assumptions of inextensibility (strain of the neutralaxis is zero or E& = 0), and zero shear strain across the thickness. Under these assumptions,moment was determined to be equal to M = (EZJI)/L where EI is the bending stiffness, JI isthe bending rotation angle and L is the length of the beam. That is, a constant variation ofrotation II/ will lead to the condition of a cantilever beam under the tip moment. In thepresent formulation, none of the above assumptions are used. However, the condition ofvery low neutral axis direct strain and very low through the thickness shear strain can beobtained by considering very thin beams.


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Bending Moment Distribution

Rotation Distribution

Finite Element Modeling - Displacement MethodFig, 12. Application of moment in the form of constant rotation

In the total Lagrangian formulation, moments are based on 2nd Piola-Kirchhoff stresses.I2ere, moment is defined as l_ h,2Sz2zdz where h is the thickness of the beam and SZ2 is thelongitudinal 2nd Piola-Kirchhoff stress. For an isotropic beam, the equation for themoment becomes M = E j!t,2 z21c22(1j z where ~~~~~~s given in equation (15). When theaxial strain is zero or small [the condition of inextensibility where the relations1 + t,? = cos($) and w,~ = -sin($) are valid], the equation for the moment can beobtained as M = C,$, Ywhere CZ = j!,,, 022 z2 dz is the bending stiffness and equal to Elin the case of an isotropic beam. The application of a tip moment results in a constantbending moment along the length of the beam. It can be observed from this mo-ment-change of rotation relationship that a constant bending moment is a by-product ofa constant rate of change of rotation along the length of the beam. This constant rate ofchange of rotation can be achieved by specifying rotations, tj, at all nodes. Thus, a lineardistribution of rotation is applied along the length of the beam (Fig. 12). Another momentquantity, based on the Eulerian stress, can be defined as ME = j!i,262Zzdz. By using thetransformation relations between the Eulerian stress and 2nd Piola-Kirchhoff stress [equa-tion (24)], the expression for the Eulerian stress based moment can be obtained asME = [(l + uFY)M]/A where M is the moment based on 2nd Piola-Kirchhoff stress anddefinition of A is given in equation (30). If the condition of inextensibility is imposed, thismoment becomes equal to ME = cos2(t,b)M.

It was discussed in the previous paragraph that the assumption of inextensibility led tothe conclusion that the constitutive relations between Eulerian stress-Almansi strain and2nd Piola-Kirchhoff stress-Green strain remain the same. Therefore, the presence of theaxial strain also leads to the changes in the constitutive constants. It can be concluded thatthe presence of axial strain leads to (i) failure of inextensibility and (ii) changes in theconstitutive relations. The effects of each are studied by considering beams of differentthicknesses.The thickness of the beam used for this study is equal to L/h = 100, where L is the lengthof the beam. The rotation controlled approach is carried out as shown in Fig. 12. This issimilar to the displacement control algorithm with a rotation increment of 0.02 radians andconvergence tolerance of 0.001%. Five elements are used in this analysis as they are oservedto be sufficient for a converged solution.


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Nondma~ Len#h, YA(a)Momentwhen rotaWn=0.01 Radii

NondimmsiaWLenah. Y/L(b) h4omeniwhen rotaticm= 1.6 Radians

B. 5 __2ndPbkKl rchhon~ E 1. 5 _2ndPlohkXdbftMomeni-.- Euidul M0rn.m z! i -.- Eukkn Momnt8 -1

%. ./ \

j 1e .\0.5. . ' . , 1. ' ' \ , \ .!! ' \ . \0. 5. /' .

d $. / \. \ . ' \. '

P 0 ' Z N. . '0 0. 5 1 P 0 - ' T - . ' ' .0 0. 5 1?&dimnsiONIllL8n~h, Yk NadimensionelLmglh, Y/L(c) Mortleni when romtion- 3.2 Radiins (d) Manent when roWion= 6.4 Radii

Fig. 13. Moment distribution along the length of the beam when L/h = 100

From the results, it was observed that a constant moment along the length of the beamwas not obtained, as can be seen in Fig. 13. Here, the moment distribution along the lengthof the beam, at four different rotations are shown. Non-dimensionalized moments based on2nd Piola-Kirchhoff stress and on Eulerian stress are shown in this figure. The solid linecorresponds to the non-dimensionalized moment based on 2nd Piola-Kirchhoff stress.Both the moments are non-dimensionalized with respect to the moment Ca$,Y. A constantmoment along the length of the beam indicates a non-dimensional moment of 1 through outthe length. When the rotations are small ($ = O.Ol), it can be seen that moment based on2nd Piola-Kirchhoff stress is equal to one while the moment based on Eulerian stress isclose to one (the vertical axis in this case is between 0.98 and 1.01). At higher rotations, it canbe observed that the moments based on Eulerian stress do not satisfy the condition ofconstant moment along the beam while the moments based on the 2nd Piola-Kirchhoffstress slowly start to deviate from unity. When the rotation is 6.4 radians, there is a 15%deviation from the ideal moment (C,$,,). This deviation can be attributed to the failure ofinextensibility and the transformation of the constitutive relations, or in other words, to thepresence of axial strain. Relative contribution of each is studied by comparing the resultsobtained with and without transformation.Vertical tip deflection as a function of the non-dimensionalized tip moment based on 2ndPiola-Kirchhoff stress is shown in Fig. 14. In this figure, the solid line (referred to as curve 1)corresponds to the moment definition M = Cz$,y. As discussed earlier, this momentdefinition assumes the inextensibility, isotropy and zero shear strain across the thickness,and the validity of this definition is very restrictive. All the remaining three curves in thefigure are based on the moment definitions M = j _h,2/ &.zdz with different assumptions.In all three cases, the assumption of inextensibility is not used. Instead, stresses aredetermined and integrated along the thickness of the beam. The dashed line (referred to ascurve 2) represents the moment when the transformation of the constitutive relations is nottaken into consideration. The only difference between curve 1 and curve 2 is the condition ofinextensibility. The third curve (dash-dot curve or curve 3) is the moment when thetransformation is based on the displacement variables of the neutral axis (single layerapproach). That is, the difference between curves 2 and 3 is the transformation of theconstitutive relations. The dotted line (curve 4) in this figure corresponds to the moment byassuming the cross section of the beam to be made up of 4 isotropic layers of equal


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Large strain analysis of beams and arches 637I I I I I_ Moment based on inextensibility - curve 1 I I--- Moment with out transformation - curve 2-.- Moment with neutral axis transformation - curve 3. Moment with multi-layer transformation -curve 4

0.5 -

9.4 -ES= 0.3 -

-I

4

0.3 0.4 0.5Vertical Deflection,W/L 0.6 0.7 0.8

Fig. 14. Comparison of vertical tip deflections when L / h = 100

0.08 -

.E 0.06;m5; IL _ Neutral axis axial strain--- Top layer axial strain, c - - - x , .

i!

-0.02.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8Vertical Deflection, W/LFig. 15. Strain as a function of tip deflection for the beam.

properties (multi-layer approach). This approach allows the transformation of each layerseparately. For the thickness considered, it can be observed that close agreement can beseen between all the four curves, up to a vertical deflection of 0.7 times the length. Tiprotation corresponding to this deflection is around 2 radians. After that, there is a slightvariation between the four curves. To explain the slight variation between these curves,axial strain at the tip of the beam is plotted as a function of tip deflection in Fig. 15. Here thesolid line corresponds to the axial strain of the neutral axis while the dashed line is the axial


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-5-m3.Ee8N

-lO-

_l5l5 0 5y coordinateFig. 16. Deformed configuration of the beam at various tip rotations.

strain at the top most layer of the beam. Since the axial strain at the neutral axis is observedto be non-zero, the condition of inextensibility was not met exactly. Thus, the slightvariation between curves 1 and 2 (Fig. 14) can be attributed to the failure of inextensibility.From Fig. 15, it can be observed that the maximum strain at any cross section is around0.01 (1%) till the non-dimensionalized tip deflection is 0.7. Afterwards, the magnitude ofstrain increases and so is the difference between the load deflection curves. Also, thedifference in the load deflection curves is not significant since the strains are comparativelysmall. As the strain is observed to be less (less than 0.04 or 4%) at the neutral axis, thetransformation of the constitutive relations does not have a significant effect. That is thereason why curves 2 and 3 in Fig. 14 are right on top of each other, indicating that thetransformation based on the neutral axis does not have any effect for this configuration. Thedifference in axial strain at the neutral axis and the strain at the top of the beam is the reasonfor the deviation of the moments based on the single layer approach and multi-layerapproach. A plot of the deformation configuration at different tip rotations is shown inFig. 16. This plot is obtained by joining the nodal points using straight lines. A similar studyis carried out for a thicker beam as a next case.A thicker beam (L/h = 10) is considered. The same analysis parameters such as incrementsize and convergence criteria are used for this beam. The non-dimensionalized momentdistribution along the length of the beam is shown in Fig. 17 at different tip rotations. It canbe observed that there is a significant deviation from the condition of constant moment.Figure 18 shows the plot of the vertical tip deflection versus non-dimensionalized moment.Significant differences can be observed among the four curves at larger moment values.A plot of axial strain at the tip of the beam is shown in Fig. 19. From this figure it can beobserved that there is significant neutral axis strain for this thickness which is the reason forthe difference between the solid and dotted curves of Fig. 18. Since the neutral axis strain ismore than 4%, there is a difference between the dotted curve and the dash-dot curve at thehigher moment values. The difference between the axial strain at the mid surface and thestrain at the top surface is responsible for the deviation between dash-dot curve and thedotted curve.


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: .\0.98 0 0.5 h oc -_ J1 0 0.5 1Nondimensbnai ength,Y/L(a)Moment when rotation= 0.01 Radians

Nondimemiw~~~engh, Y/l_(b) Moment when rotation= 1.6 Radians

\ -. \. . _

Documents

Gummadi_1998_Large Strain Analysis of Beams and Arches Undergoing Large Rotations