[Brown] a Simple Trasnversely Isotropic Hyperelastic Model

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a valuable engineering solution to many design challenges. Thesecomposites are often the key to success in many components suchas v-belts where they are used to withstand the wedging of the beltibbn

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acterize the material.In this effort, a series of tests was used to gain a constitutive

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The desired material properties can be obtained by mixingmall fibers with an isotropic elastomer and milling the material tolign the fibers. The direction of milling, or the machine direction,an be referred to as the with-grain WG direction while nor-al to this direction can be referred to as the cross-grain XG

irection. Enough fibers are oriented precisely in the with-grainirection to substantially increase the stiffness of the material inhat direction. This process creates an elastomeric composite,hich can be best described as a transversely isotropic hyperelas-

ic material.The degree of anisotropy can be quantified by the ratio of the

tress at a given elongation for a principal material direction to theame stress in another principal material direction

Ra =WGXG

at a given elongation 1

This has been referred to as modular anisotropy 1.While extensive work has been done on characterizing isotropic

lastomers, relatively very little work has been done in character-zing anisotropic elastomers. The existing isotropic elastomer

understanding of a particular anisotropic elastomer, a model con-sistent with those test results was developed, the model imple-mented in a commercial finite element code, and the response ofthe material examined through finite element analysis FEA ofthe tests.

2 Constitutive TestsIn 1944, Treloar provided a set of test data that became a cor-

nerstone of efforts to develop adequate hyperelasticity models forisotropic elastomer studies 2,3.

Using data from more than one constitutive test is recom-mended for an isotropic elastomer. The tests should cover thetypes of deformation, the temperature, the strains, and the strainrate expected in service 4. This practice prevents using an elas-tomeric material model that is not well suited for the analysis 5.

Test data to fit material models for use in commercial finiteelement programs can be obtained from a combination of uniaxialtension, uniaxial compression, biaxial tension, pure shear, andsimple shear tests 6,7. There is even a standard, BS 903-5, toguide the selection of suitable tests for an isotropic elastomer. Asimple efficient method suitable for many industrial laboratories isthe combination of uniaxial tension, pure shear, and uniaxial com-pression tests 8.

Very little has been written about recommended tests for char-acterizing transversely isotropic and anisotropic elastomers de-spite the great deal of testing that has been conducted. A lot oftesting has been done on biological materials, which are also an-isotropic hyperelastic materials, but because of the complexity of

Contributed by the Materials Division of ASME for publication in the JOURNAL OFNGINEERING MATERIALS AND TECHNOLOGY. Manuscript received May 26, 2010; finalanuscript received December 31, 2010; published online March 23, 2011. Assoc.ditor: Hussein Zbib.

ournal of Engineering Materials and Technology APRIL 2011, Vol. 133 / 021021-1Leslee W. BrownEngineering Manager

Mem. ASMEGates Corporation,

2975 Waterview Drive,Rochester Hills, MI 48309e-mail: [email protected]

Lorenzo M. SmithAssociate Professor

Department of Mechanical Engineering,Oakland University,

118 Dodge Hall,Rochester, MI 48309

e-mail: [email protected]

A SimplHyperelSuitableAnalysisElastomA transversely isotroa series of constitutconstrained comprebased strain energyefficients are showncreating a user subrthe material tests. Amation is given. Theably well over the sreveals further insigfunction is suitable fstrains. The necessaratory tests. DOI:

Keywords: hyperelanonlinear materials

IntroductionCombining an elastomeric matrix with a reinforcing material isCopyright 20

om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2Transversely Isotropicstic Constitutive Modelfor Finite Elementof Fiber Reinforcedrsfiber reinforced elastomers hyperelasticity is characterized usingtests (uniaxial tension, uniaxial compression, simple shear, andn test). A suitable transversely isotropic hyperelastic invariantction is proposed and methods for determining the material co-is material model is implemented in a finite element analysis by

ine for a commercial finite element code and then used to analyzeeful set of constitutive material data for multiple modes of defor-oposed strain energy function fits the experimental data reason-n region of interest. Finite element analysis of the material testsnto the materials constitutive nature. The proposed strain energynite element use by the practicing engineer for small to moderate

material coefficients can be determined from a few simple labo-115/1.4003517

ity, elastomers, strain energy functions, transversely isotropicbber, nonlinear finite element analysis

models are insufficient to simulate the typical response of an an-isotropic elastomer. What is needed is an efficient stable aniso-tropic elastomer model and appropriate constitutive tests to char-11 by ASME

013 Terms of Use: http://asme.org/terms

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Table 1 Summary of previous measurements of anisotropic elastomers

Date and Authors UT UC SS PS

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Downloaded Frbtaining materials, creating samples, and maintaining proper testonditions this is often limited to uniaxial tensile or uniaxial com-ression 9,10.

A transversely isotropic lamina oriented fiber embedded in aatrix can be characterized by a combination of uniaxial tension

n the fiber direction with-grain, uniaxial tension in a directionormal to the fiber direction cross-grain, and a uniaxial tensionest at 45 deg to the fiber direction 11. This approach was used toharacterize a fiber reinforced elastomer for medium tensiletrains 0150% 12. The behavior of this elastomer in shear,ompression, and combined deformation states was not examinednd it would have to be used with caution based on the experiencef isotropic elastomer models that fit only to uniaxial tensile data.

The desirability of using both uniaxial tensile and biaxial ten-ile data to obtain a suitable material model has been described13.

The tensile properties of typical v-belt neoprene elastomersere measured in both the with-grain and cross-grain directions

1. The elastomers were reinforced using various fibers cotton,olyester, and cellulose and milled. The only tensile data reportedere the stress at 10% and 20% strains in both the with-grain and

ross-grain directions. The reported anisotropy ranged from 2.0 to.8.

Similar tensile data have been reported for various milled shortber reinforced elastomers similar to those used in v-belts, hoses,nd tires 12,14,15. Tensile stress-strain curves for a milled shortelamine fiber reinforced ethylene propylene diene monomer

EPDM rubber were reported in both the cross- and with-grainirections 16. The anisotropy in mechanical properties washown to be more prominent at low strains than at large strains.

The uniaxial tensile and compression stresses in the fiber andross fiber directions were reported for a nylon fiber reinforcedubber compound used in an industrial v-belt 17.

The uniaxial tension results in the cross- and with-grain andiaxial tension data for a calendared natural rubber have beeneported by Diani et al. 13.

A summary of these previous efforts is provided in Table 1.learly these efforts have focused on uniaxial tension tests andeglected the equally important compression, shear, and volumet-ic tests.

A set of constitutive test data, similar to that provided by Tre-oar for isotropic elastomers, is needed for anisotropic elastomers.hese data would allow the selection and development of accuratend stable material models suitable for finite element analysis andther uses. By combining the lessons learned from characterizingsotropic elastomers and linear composite materials, the materialshould be characterized by tests of multiple modes of deformationnd in multiple directions e.g., 0 deg, 90 deg, and 45 deg.

Test samples of a suitable anisotropic elastomer were obtainednd tested using four different tests: uniaxial tension, uniaxialompression, simple shear, and a confined compression test. Forach test, samples with different material orientations were tested.hese orientations were 0 deg, 45 deg, and 90 deg to the with-

rientation 0 deg 90 deg 45 deg 0 deg 90 de

981 Rogers 987 Foldi 988 Lee 994 van der Pol 002 Rajeev et al. 004 Diani et al. 008 Ishikawa

otes: UTuniaxial tension, UCuniaxial compression, SSsimple shear, PSpu

21021-2 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2grain direction of the material for tensile and simple shear speci-mens and 0 deg and 90 deg for the compression and pure shearsamples.

The tests were done at the same strain rate to minimize theeffects of viscoelasticity and at the same temperature. In eachcase, three samples were tested and the mean stress for the groupof samples is reported.

2.1 Elastomer. Two elastomers were mixed, milled, andformed into test specimens. The elastomer formulation was basedon that used in an earlier published study of anisotropic elas-tomers used in v-belts 1. The formula is given in Table 2. Thefirst elastomer is an isotropic master batch LB1, which was thenused to create the second elastomer LB2 by adding fibers. Thissecond elastomer LB2 is the anisotropic elastomer examined inthis study; the first elastomer LB1 is the isotropic elastomermatrix.

2.2 Uniaxial Tensile Tests. The material was tested inuniaxial tension, as shown in Fig. 1. Samples were created withthree orientations: cross-grain, with-grain, and 45 deg to the with-grain direction.

The mean stress results for three samples are shown in Fig. 2. Itcan clearly be seen that the anisotropic elastomer has becomeoriented with a different response in each material orientation. Theanisotropic elastomer has a much stiffer response than that of theisotropic elastomer matrix.

BT VC.0 deg 90 deg 45 deg 0 deg 90 deg

hear, BTbiaxial tension, and VCvolumetric compression.

Table 2 Elastomer formulations

IngredientCompound LB1

phrCompound LB2

phr

Neoprene 100Carbon black 44Process oil 4Antioxidant 3Magnesium oxide 4Stearic acid 2CurativesLB1 100Cotton fiber 0.01 cm 35

Fig. 1 Tensile test

Transactions of the ASME013 Terms of Use: http://asme.org/terms

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Downloaded FrThe ratio of anisotropy was calculated and the results are shownn Fig. 3. Over the 020% strain range the ratio was approxi-

ately 4 with a maximum of 4.5. This compares well with the.04.6 modular anisotropy reported for the similar compound1. The ratio generally decreases at a larger strain as was reportedn earlier measurements.

There is a curious drop and rise in the ratio of anisotropyround 5% strain that results from a decrease in the tangent modu-us of the with-grain response. This might be attributable to varia-ions in fiber curvature.

2.3 Uniaxial Compression Test. For the v-belt compound, aominant mode of deformation is compression between two metalurfaces. Therefore, this test is of particular interest for thisnvestigation.

The uniaxial compression test is a relatively easy test that cane done in an industrial laboratory. The basic test is to compress aample of material between two platens while measuring the de-ection and load, as shown in Fig. 4. Here samples of the aniso-

ropic material were created with two orientations: the with-grainriented vertically V and the with-grain oriented horizontallyH. The horizontally oriented samples were created by stackingayers with varying orientation; this creates a unique sample with

cross-grain orientation through its thickness, but with no defin-ble with-grain orientation.

The compression test remains a controversial choice due to ef-ects of friction and the resultant tendency of the sample to bulge,r barrel, during testing 4. This can be minimized by lubricating

Fig. 2 Tensile test results

Fig. 3 Tensile anisotropy

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2the sample and using samples with a ratio of loaded area to forcefree area of 1 4,7,8. For this study the samples were lubricateddisks with a 0.6 shape factor.

The mean stress results for the LB2 compound and the refer-ence isotropic elastomer LB1 are shown in Fig. 5. The ratio ofanisotropy was calculated and is shown in Fig. 6. The fiber rein-forcement clearly increased the compressive stiffness of the ma-terial, but the degree of anisotropy is not as strong as in tension.

At about 10% strain, the cross-grain H response is stiffer thanfor the with-grain V samples. The ratio of anisotropy falls below1. This is attributed to two factors. First, the change in tangentstiffness of the with-grain sample is attributed to an effect similarto buckling of the fiber reinforcement in a composite lamina 11.Second, the fibers in the cross-grain H response are expected tobe in a tensile state, which resists the compression of the sample.The overall increase in stiffness compared with the isotropic elas-tomer is attributed to an effective particulate reinforcement of thecompound.

2.4 Dual Lap Shear Tests. The response of a hyperelasticmaterial to shear deformation is an important distinguishing char-acteristic. Hyperelastic materials can be described as materialswith very large resistance to volumetric changes but a relativelysoft resistance to shear. Both linear elasticity and simple hyper-elasticity neo-Hookean and Mooney are based on the assump-tion that the shear stress will be proportional to the shear strain

Fig. 4 Compression test

Fig. 5 Compression test results

APRIL 2011, Vol. 133 / 021021-3013 Terms of Use: http://asme.org/terms

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ypically this test is done with samples bonded to rigid fixtures,hich results in an appreciable bending. For this study a modifiedersion of the test was done where samples are gripped betweenigh friction surfaces and the amount of deflection is restrained, ashown in Fig. 7. A small amount of compression is used to preventlippage; this introduces a small compression into the sample thats neglected. Very small deflections of thin test samples yield veryarge shear deformations; a deflection of 1 mm for a 3 mm thickample yields approximately 0.33 shear strain.

Material samples for the anisotropic elastomer were createdith three orientations: with-grain, cross-grain, and with the with-rain direction oriented 45 deg from the test direction. The meanhear stress results are shown in Fig. 8. It is clear that the shearesponse of the anisotropic rubber is stiffer than the referencesotropic elastomer matrix material and that the material orienta-ion does not affect the shear response. The stiffness of all three

aterial orientations is essentially the same.

2.5 Confined Compression Tests. While very few materials

Fig. 6 Compression anisotropy

Fig. 7 Dual lap shear test

21021-4 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2are truly incompressible, many elastomers can be considerednearly incompressible. Does adding a compressible fiber to anearly incompressible elastomer yield a nearly incompressiblematerial?

A nearly incompressible material can be defined as a materialfor which a much larger effort is required to change its volumethan is required to change its shape while preserving its volume19. The compressibility of a material is quantified by the bulkmodulus. For an incompressible material the bulk modulus wouldbe infinite; for a nearly incompressible material the ratio of theshear modulus to the bulk modulus is considered. For linear iso-tropic materials the bulk modulus K, the shear modulus G, andPoissons ratio are related 20 as follows:

=3K 2G6K + 2G

2

The bulk modulus can be measured using a confined compres-sion test where a sample is compressed in one direction and re-strained from expanding in all other directions 4. This methodhas been found to be easier than trying to measure Poissons ratioand relatively unaffected by small gaps between the sample andthe test fixture 20.

The samples for this test are similar to those used in theuniaxial compression test. Samples were created with two orien-tations: with-grain V and cross-grain H. The cross-grainsamples were made as earlier described.

The constrained volume compression test results are shown inFig. 9. The measured bulk modulus, the initial shear modulus, the

Fig. 8 Dual lap shear test results

Fig. 9 Constrained volume compression test results


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ompressible because of the high ratio of the bulk modulus tohear modulus. A rule of thumb for nearly incompressibility isaving Poissons ratio greater than 0.49 21. There does not ap-ear to be a significant difference between samples with a differ-nt material orientation. The addition of the fiber did not changehe compressibility of the material significantly.

Constitutive ModelThe general approach of using a strain energy function as the

asis for a constitutive model of a transversely isotropic materialubjected to a large elastic deformation was pioneered by Ericksennd Rivlin 22, Spencer 23, and Green and Adkins 24.

In this effort, we restrict our attention to constitutive modelsased on the approach of characterizing the strain energy as aunction of the invariants of the deformations. Other approachesave been tried including linear transversely isotropic materialodels 25 and generalized strain energy functions 13,2628.What is needed is a stable efficient constitutive model suitable

or implementation in the finite element method. The nature ofurrent iterative nonlinear finite element methods is best suited toodels that do not violate the Drucker stability criteria. The effi-

iency of the model is determined by the number of parametersnd the required material tests to accurately describe the material.

A suitable strain energy function was constructed, the constitu-ive stress-strain relationships were derived, and the parametersetermined from the material tests.

3.1 Strain Energy Function. The essence of elasticity is inefining how a material resists applied forces and deformation.nder load the material deforms, storing the work being applied

s internal elastic energy strain energy, and creating internaltresses that react any applied loads. The constitutive relationshipetween the deformation strain and the internal stresses allowshe definition of a strain energy function of the form 19,24,29

=C 3here the deformation is described by the right CauchyGreen

ensor C, which is defined as

C = FTF 4n terms of the deformation gradient F as follows:

FX =xXX

5

he deformed geometry is described by x and the initial geometrys described by X. It is convenient to work with the deformationradient and the right CauchyGreen tensor in terms of their ei-envalues that are the principal stretches the ratio of the de-ormed length to the undeformed length in the three orthogonalirections 1, 2, and 3

Material andorientation

Bulk modulusMPa

Initiamodulu

LB2-0 V 900 5LB2-90 H 900 5LB1-0 V 890 3LB1-90 H 870 3

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2F = 1 0 00 2 00 0 3 6C = 1

2 0 00 2

2 00 0 3

2 7The change in volume of the material can be calculated using

the Jacobian J

J = F = C 8For materials that are transversely isotropic the strain energy

function can be defined as a function of the three invariants of theright CauchyGreen tensor and two additional quasi-invariantsbased on the orientation of the material ao 2224

=I1,I2,I3,I4,I5

I1 = trC

I2 =12 trC

2 trC2

I3 = C

I4 = ao Cao = f2

I5 = ao C2a0 = f4 9A subset of all functions that meet the general definition is

those functions that split the strain energy into two components:The first is the energy stored in the deformation of the matrixmaterial and the second is that energy stored in the reinforcement

=isoI1,I2,I3 +fI4,I5 10If the matrix is incompressible, then I3 becomes 1 and iso

becomes a function of I1 and I2.Based on the results of the simple shear tests where there is no

effect of material orientation the uncoupling of the fiber and ma-trix strain energy functions seems appropriate.

For the contribution of the fiber, several models with variousdegrees of success have been proposed. These models are summa-rized in Table 4 where they are ranked by the number of materialconstants required efficiency and the order of the fiber strainenergy function indicated. These models combine existing hyper-elastic models with different reinforcing functions.

The difficulty in using the existing strain energy functions todescribe the with-grain compressive and tensile nature of the sub-ject elastomer LB2 is shown in Fig. 10. Additionally, the func-tion used by Ishikawa et al. shows an undesirable elastic instabil-ity elastic stress decreases with increasing strain. A better strainenergy function is needed.

3.1.1 Strain Energy of the Elastomer Matrix. For the matrixcontribution the MooneyRivlin or the neo-Hookean elastomermodel is well suited for small to moderate strains as expected forthe following material 3,18:

earPa

to shearmodulus

CalculatedPoissons ratio

170 0.497176 0.497278 0.498264 0.498


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Strain energy functions Rank Order

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3.1.2 Strain Energy of the Fiber Reinforcement. Many of thearlier strain energy functions have been used to describe the re-ponse of biological materials, which have a characteristic highlyonlinear response over very large strains 30. They use an ex-onential function for the fiber strain energy contribution. Thiseems very different from the response of a reinforcing fiber underuch smaller strains.The anisotropic elastomer LB2 has been reinforced with small

bers that have an initial curvature. Even after milling, curvatures expected. Consider the finite element of a curved fiber, similaro that used in the elastomer LB2, as shown in Fig. 11. The ex-ernal work required to extend and compress this fiber model ishown in Fig. 12.

Relatively little strain energy is stored in compression of theber as the degree of curvature increases. As the fiber straightens,

Standard reinforcing model 35:

=C1I13+C2I412

Fiber reinforced cardiac muscle 36:

=C1expC2I132+C3I13I41+C4I412

Caterpillar muscle 38:

= C1I1 3 + C21 C3 tanh I4 1C4 expC5I4Fiber reinforced rubber 17:

=C10I13+C01I23+C42I412+C43I413

C42 and C43 have different values for tension and com

Human medial collateral ligament 33:

=C1I13+C2I23+C3expI41I4

More complex function C3, C4, C5, and C6 was use

Cord-rubber composite 37:

=VmC1I13+C2I23+VCC3expI41I4

ig. 10 Comparison of models and measured results for WGniaxial stress

21021-6 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2the strain energy becomes proportional to I4. As shown in Fig. 12,this response differs substantially from the characteristic responseof the types of reinforcing functions used previously. The follow-ing strain energy function fits the fibers response more accu-rately:

fI4,I5 = C4I41/2

12

C4 = C4t ifI4 1C4c ifI4 1 123.1.3 New Strain Energy Function. Combining the two con-

stituent strain energy functions, given in Eqs. 11 and 12, yieldsa suitable strain energy function for small to moderate strains of atransversely isotropic hyperelastic material

= C1I1 3 + C2I2 3 + C4I41/2

12 13This strain energy function meets two essential criteria for a

strain energy function. First, if there is no deformation I1=I2=3and I4=1 there is no stored elastic energy and the strain energyfunction must be zero. Second, the strain energy should always be

2 4

4 E

2 1

5 E

ssion.

6 6

capture fiber under three states.

8 E

I13+C6I1329 E

Fig. 11 Finite element model of short fiber


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3.2 Stress Tensor. In this paper we will consider the secondiolaKirchoff PK stress and the transformation of this stress

nto the first PiolaKirchoff stress related traction force vectorforce measured per undeformed unit surface area. This choicellows the comparisons of the measured engineering stress andtrain to those calculated in this section. The second Piolairchoff stress can be determined from the strain energy function

19

S = 2

C= 2

i=1

5

Ii

IiC

14

The partials of the invariants with respect to the right Cauchyreen tensor are 19

I1C

= I 15

I2C

= I1I C 16

I3C

= I3C1 = C1 for an incompressible material 17

I4C

= ao ao 18

I5C

= ao Cao + aoC ao 19

The partials of the proposed strain energy function with respecto the invariants are

I1=1 = C1 20

I2=2 = C2 21

I3=3 = 0 22

ig. 12 Comparison of reinforcing functions with predicted re-ults for short fiber

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2

I4=4 = C41 I4

1/2 23

I5=5 = 0 24

Combining Eqs. 1424 yields the stress function for the pro-posed strain energy function

S = 2C1 + C2I1I C + C41 I41/2ao ao pC1 25The second PiolaKirchoff stress can be transformed into the

first PiolaKirchoff stress 19P = FS 26

3.3 Uniaxial Tension and Compression

3.3.1 With-Grain Uniaxial Stress. In this case the material isstretched along its oriented direction ao, as illustrated in Fig. 13.In the isotropic plane normal to this orientation, the materialcontracts equally in both directions. Assuming the material is in-compressible yields the following deformation gradient:

FWGU = 0 00 1/2 00 0 1/2 , ao100 27

Using Eqs. 2527, the nominal traction force required tostretch the material will be

TWGU = 2C1 2 + 2C21 3 + 2C4 1 28

3.3.2 Cross-Grain Uniaxial Stress. The material is stretched ina direction perpendicular to the material orientation, as illustratedin Fig. 13. Because of the difference in stiffness in the remainingtwo principal directions, the stretches in each of these directionsare not the same 13. Assuming incompressibility the third stretchcan be expressed in terms of the first two as follows:

FXGU = 1 0 00 2 00 0 1121 , ao = 010 29

Noting that two of the three principal stresses are zero those inthe contracting directions, Eqs. 25 and 29 yield three equa-tions with three unknowns S11, p, and 2. Using Eq. 26 givesthe following equations for the traction force:

TXGU = 2C1 112

2 + 2C2112

2 1

3 + 2C411

112

230

C1 + C412 + C21

424

C4122

3 C21

2 + C1 = 0 31While a closed form solution for 2 was found, it is more prac-

tical to use a numerical method to solve Eq. 31 for a given 1.

Fig. 13 Uniaxial deformation


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3.4.1 With-Grain Simple Shear. The shear motion is parallelo the material orientation. The deformation is intentionally isoch-ric. The resultant deformation gradient is

FWGS = 1 00 1 00 0 1 , ao = 100 32

Using Eqs. 25, 26, and 32, the nominal traction force isTWGS = 2C1 + C2 33

3.4.2 Cross-Grain Simple Shear. The deformation is similar tohe with-grain simple shear. The difference is that the materialrientation is normal to the plane of shearing. In this case theeformation gradient and material orientation are

FXGS = 1 00 1 00 0 1 , ao = 001 34

Using Eqs. 25, 26, and 34, the nominal traction force isTXGS = 2C1 + C2 35

It is interesting to note that the orientation of the material doesot affect the shear response. Both TWGS and TXGS are the same.

3.5 Plane Strain (Pure Shear). In his earlier work, Treloar2 used the pure shear test to test the validity of the emergingheory of what is now known as the neo-Hookean elastomer

odel. The test was done by stretching a short wide sample 575 mm2. This establishes a state of plane strain in the sample

y constraining the width. The third direction, through the thick-ess, contracts. It is also known as a constrained tension state31. The deformation in the pure shear test is considered to bequivalent to that in a simple shear test 7,30. The principaltretch can be related to the simple shear strain 29 asollows:

= 1

36

This state of plane strain is often referred to in the hyperelas-icity literature as pure shear. The case is illustrated in Fig. 15.

3.5.1 With-Grain Plane Strain. In this case the material istretched in the orientation direction. Assuming the material isncompressible results in the contraction in the remaining freeirection to be the inverse of the applied stretch. The deformationradient and the material orientation are

FWGPS = 0 00 1 00 0 1 , ao = 100 37

Using Eqs. 25, 26, and 37, the nominal traction force isTWGPS = 2C1 + C2 3 + 2C4 1 38

Fig. 14 Simple shear deformation

21021-8 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/23.5.2 Cross-Grain Plane Strain. In this case the material isrestrained from contracting in the orientation direction andstretched in a direction perpendicular to the orientation direction.Assuming the material is incompressible results in the contractionin the remaining free direction to be the inverse of the appliedstretch. The deformation gradient and the material orientation are

FXGPS = 0 00 1 00 0 1 , ao = 010 39

Using Eqs. 25, 26, and 39, the nominal traction force is

TXGPS = 2C1 + C2 3 40The material orientation had a much larger effect on the pure

shear test than on the simple shear test. This difference suggeststhat the two modes of shear deformation are not equivalent as theyare for an isotropic elastomer. This is attributed to the differencein the stiffening effect between shear and extension predicted bycomposites micromechanics theory for short fibers in a matrix11 where the reinforcement is not as effective in increasing theshear modulus of the composite.

3.6 Determining Model Coefficients. As was earlier stated,if the material moduli C1, C2, C4c, and C4t are all positive, thenthe strain energy function is positive for all deformations. Thisalso results in stable stress-strain relationships. If the materialmoduli are positive, then stress always increases with increasingstrain tangent stiffness is always positive. This satisfies theDrucker stability criteria for hyperelastic materials.

The material moduli C1, C2, C4c, and C4t can be determined byminimizing the differences between the measured force per unitundeformed area and the calculated nominal traction force for thesame measured strain. This problem is well suited to a least-squares method to minimize the difference.

This was done using the commercial spreadsheet program Mi-crosoft Excel. The relationships in Secs. 3.13.6 were pro-grammed as user functions, and the built in solver was used tominimize the difference between calculated and measured values.

The resulting fit of the model to the test data is shown in Fig.16. The data were fitted over the range 10% because of thelimits of the with-grain tensile data. The fit did not include thecross-grain compression samples because of their uniqueconstruction.

The resulting fit is quite good for all modes of deformation inthe range of the fit and in many cases outside of the fit. The mostsignificant difference would seem to be the calculated shear re-sponse. The difference in the with-grain tensile response is be-lieved to be related to failure of the material. The difference in thewith-grain tensile compression response is interesting as there is amarked change in the response that occurs after 10% compres-sion; this will be examined in more detail in Sec. 4. The actual

Fig. 15 Pure shear plane strain deformation


cbw

ttdtc

w

lwtidus

4 Implementation Within the Finite Element MethodThe new invariant based strain energy hyperelastic model was

J

Downloaded Frross-grain compression response is stiffer than calculated; this iselieved to be the effect of the unique sample construction, whichill also be examined in Sec. 4.The nature of the constitutive relationships suggests an alterna-

ive method of fitting the data than spreading the error over allhe results implicit in the above method. This method focuses onetermining the material constants from specific tests. Similar al-ernative methods have been used 17,32. The following methodsould be used:

1. determine the shear modulus 2C1+2C2 from simple shearTWGS or cross-grain pure shear TXGPS tests

2. determine C4t from the with-grain tensile response TWGUor the with-grain planar shear test TWGPS

3. determine C4c from the with-grain compression responseTWGU

4. determine a value of C2 that improves the larger strain re-sponse of the tensile and compression data; this is not nec-essary if C1 alone results in a good fit

These methods were also used and the comparison of this fitith the experimental data is shown in Fig. 17.While this second fit has a larger error than the first fit, it high-

ights the strengths and weaknesses of the material model. The fitith the simple shear results is improved while the with-grain

ensile and compression fits remain good. A remarkable differences that the cross-grain compression result is softer than the testata this difference is not considered in either fit because of thenique construction of these samples. The cross-grain tensile re-ponse is softer and this will be investigated further in Sec. 4.

Fig. 16 Minimized difference material model fit

Fig. 17 Deterministic material model fit

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2implemented within the finite element method using the user sub-routine hypela2 available in the commercial finite element codeMSC.MARC. Finite element models of the test specimens were cre-ated and used to investigate in more detail the response of thematerial including the samples where the material was oriented at45 deg and for the unique construction of the cross-grain compres-sion samples.

4.1 Nearly Incompressible Form. In the analysis of nearlyincompressible materials the finite element method can be diffi-cult. These difficulties are described by Weiss et al. 33. Amethod that improves this is to decouple the deviatoric and dila-tional responses of the material and consequently to separate thestrain energy function into two components. First the deformationgradient and the right CauchyGreen tensor can be decoupled 19as follows:

F = FvolF, Fvol = J1/31, F = J1/3F

Fvol = J, F = 1 41

C = FTF = J2/3C, C = 1 42Using the modified deformation gradient F and the invariants

of the modified right CauchyGreen tensor C, the strain energyfunction can be split into volumetric and isochoric parts 19

=volJ +isoI1,I2,I4,I5 43The second PiolaKirchoff stress can now be calculated as 19

S = JpC1 + J2/3P:S

P = I 13C1 C

S = 11 + 2C + 4ao ao + 5ao Cao + aoC ao

1 = 2 isoI1

+ I1isoI2

2 = 2

isoI2

4 = 2isoI4

5 = 2isoI5

44

Using the incompressible form of the strain energy function inEq. 13 as iso, and the bulk modulus measured in Sec. 2.4, thesecond PiolaKirchoff stress can then be calculated from Eq. 44using the following:

1 = 2C1 + I1C2 45

2 = 2C2 46

4 = 2C41 I41/2 47

5 = 0 48

p = KJ 1 49

4.2 Elasticity Tensor. The methods to solve nonlinear hyper-elasticity problems depend heavily on iterative incremental tech-niques that are based on linearizing the material stiffness at a


given state. Within the hypela2 subroutine, MSC.MARC requires theoutput of the tangent material stiffness matrix. The rate of conver-gence, or even obtaining convergence, depends on this stiffness

iicssfec

tu

K

fics

sfi

cci

tas

0

Downloaded Fr34.The tangent material stiffness matrix for a hyperelastic material

n 3 dimensions can be visualized as being defined as the changen 6 stresses to an incremental change in the 6 strains and isomprised of 36 separate components relating the change in eachtress to a change in each strain. Given the constitutive relation-hip between stress and strain and the hyperelastic strain energyunction, the change in stress is related to the change in strainnergy resulting from any change in deformation. This is moreoncisely expressed using the fourth order elasticity tensor C19 as follows:

C = 42

C C50

The calculation of the 21 independent components of the elas-icity tensor using a continuum mechanics approach can be donesing the following equations 19:

C = Cvol + Ciso

Cvol = JpC1 C1 2pC1 C1

Ciso = P:C:PT + 23 trJ

2/3SP 23 C1 Siso + Siso C1

C1 C1abcd 12 Cac

1Cbd1 + Cad1Cbc1 51For the strain energy function used here, Eq. 13, and using the

ronecker delta ij the C tensor would be

C = J4/311 1 + 4S + 7ao ao ao ao

S = 12 I + I

Iijkl ikjl

Iijkl = iljk 52The coefficients 1, 4, and 7 in Eq. 52 are 19

1 = 4 2isoI1 I1

+ I12isoI1 I2

+isoI2

+ I12

2iso

I2 I2 = 4C2

53

4 = 4isoI2

= 4C2 54

7 = 42isoI4 I4

= 2C4I43/2 55

4.3 Finite Element Analysis of Uniaxial Tension Tests. Anite element model of the tensile specimen used in Sec. 2.2 wasreated and solutions obtained using the subroutine. The secondet of material coefficients, the deterministic fit, was used.

The finite element model and the strain energy contours arehown in Figs. 18 and 19. The engineering stress was calculatedrom the reaction forces and plotted against the measured resultsn Fig. 20. The analysis was done for several material orientations0 deg, 15 deg, 30 deg, 45 deg, and 90 deg

The calculated stresses for the with-grain samples are verylose to the measured stresses. The calculated stresses for theross-grain and 45 deg orientation were less than those measuredn the test.

Suspecting that this difference could be attributed to a distribu-ion of fiber orientation rather than the single orientation modeled,

distributed fiber orientation finite element model was con-tructed. The orientation was assumed to be a normal distribution

21021-10 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2of orientation between the limits of 90 deg from the with-graindirection. The distribution was divided into three sections and thethickness of each section was proportional to the cumulative dis-tribution for the section. This model is shown in Fig. 21.

The resulting calculated stress and strain are compared with themeasured values in Fig. 22. While there is a considerable im-provement, particularly at strains less than 10%, there are stilldifferences. Using a more accurate definition of the distributedorientations of the actual elastomer combined with more laminasin the model might improve the comparison further.

It is also apparent that there is a point or a strain where the

Fig. 18 Tensile test finite element model

Fig. 19 Tensile FEA second PK stress contours MPa

Fig. 20 Tensile FEA results


mRiioa

csuqlbm

Maep

paIdccrd

J

Downloaded Fraterials with-grain stress separates from the predicted response.ather than stiffening as occurs with biotissues, there is a soften-

ng of the material. This is attributed to a failure in the fiber/matrixnterface and is believed to represent the beginning of an inelastic,r damaged, response. This effect is also expected for cross-grainnd other material orientations.

4.4 Uniaxial Compression. Finite element models of theompression tests were created and solutions obtained using theubroutine. Again, the second set of material coefficients wassed. The cross-grain compression model was created using auasi-isotropic laminate 0 deg/45 deg/45 deg/90 deg to simu-ate the unique construction of the cross-grain H samples createdy randomly orienting layers of elastomer in the compressionold.The compression test finite element model is shown in Fig. 23.odels were created for each of the two orientations tested plus

nother cross-grain sample without the varying orientation. Thengineering stress was calculated from the reaction forces andlotted against the measured results in Fig. 24.

The deformed shape and stress contours of the with-grain com-ression sample V are shown in Fig. 25. The calculated stressesre very close to the measured results through 15% compression.t is interesting to note that at near 25% compression the stiffnessramatically changes and the finite element stiffness matrix be-omes nonpositive definite. Forcing the solution confirms thehange, and is indicative of a buckling of the specimen. Theseesults are very consistent with the change in stiffness measureduring the actual test.

The laminated structure, the stress contours, and the deformed

Fig. 21 Distributed fiber orientation finite element model

Fig. 22 Distributed fiber FEA results

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/2shape of the H compression sample are shown in Fig. 26. Thefinite element results for the cross-grain compression samples arecloser to the measured results than that predicted by Eqs. 30 and31. This is due to the sample being stiffened by the additionallaminate constraints that act to keep the radial deformationuniform.

The deformed shape and stress contours for a cross-grain com-pression sample are shown in Fig. 27. This model demonstratesthe difference in deformation of the with-grain direction from thecross-grain response the ovalization of the model. The calculatedstresses compare well with those calculated using Eqs. 30 and31.

4.5 Simple Shear Finite Element Model. Finite elementmodels of the simple shear test were created and solutions ob-tained for each of the three material orientations tested. The finite

Fig. 23 Compression finite element model

Fig. 24 Compression FEA results

Fig. 25 V Compression FEA second PK stress contours MPa


eTe

pdastR

odtri

Fa

0

Downloaded Frlement model undeformed and deformed is shown in Fig. 28.he calculated engineering stress is compared with the measuredngineering stress in Fig. 29.

As expected the shear response matches very closely the ex-erimental results; the shear response was the foundation of theeterministic fit used. The separation that is evident starting withbout 10% shear stress may be the result of either a nonlinearhear response or the beginning of slipping of the samples withinhe test fixture. A known characteristic of the two term Mooneyivlin model is its inherent linear shear response.The deformed shape and stress contours for each of the three

rientations are shown in Fig. 30. It is interesting to note thatespite the similarity in the overall stress results the surface con-ours vary. However, if we look at the stress contours in the inte-ior of the models, we find very similar stress contours, as shownn Fig. 31.

ig. 26 H Compression FEA second PK stress contours MPand laminated structure

Fig. 27 XG Compression second PK stress contours MPa

Fig. 28 Simple shear finite element model

21021-12 / Vol. 133, APRIL 2011om: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/25 ConclusionsThe combination of uniaxial tension, compression, simple

shear, and volumetric compression tests provides a very thoroughdescription of the response of a transversely isotropic elastomer tovarious modes of deformation. These data fill gaps in previouslypublished data for similar materials. The data should provide areasonable basis to test the accuracy of proposed models to awider range of deformation types.

The strain energy function proposed here, and implemented inthe finite element model, is well suited for use by the practicingengineer in analyzing parts comprised of this type of fiber rein-forced elastomer over a small to moderate strain range 030%. Itis efficient, requiring only three material moduli in its simplestform, and inherently stable as long as positive moduli are used.These material constants can be determined from a few simplelaboratory tests e.g., tension, compression, and simple shear.

Fig. 29 Simple shear FEA results

Fig. 30 Simple shear FEA second PK stress contours MPa

Fig. 31 Simple Shear FEA internal second PK stress contoursMPa


If the material has a significant fiber orientation distribution,then a laminated approach is recommended. This should improvethe accuracy of the model for cross-grain tensile states. Alterna-tively, using a minimized overall error fit will also improve thegeneral accuracy of the model.

As indicated by the successful use of the model within thecontext of a laminated structure to model the unique structure ofthe cross-grain test specimens used here, this simple model can beused effectively to simulate the effects of more complex lami-nates.

A plane strain constitutive test for this type of elastomer wouldbe very useful. This will require a very capable clamping systemto handle the relatively larger samples with this type of substantialrfe

slet

oi

A

tGat

R

Polym. Eng., 834, pp. 257282.13 Diani, J., Brieu, M., Vacherand, J.-M., and Rezgui, A., 2004, Directional

Model for Isotropic and Anisotropic Hyperelastic Rubber-Like Materials,Mech. Mater., 36, pp. 313321.

14 van der Pol, J. F., 1994, Short Para Aramid Fiber Reinforcement, RubberWorld June, pp. 3237.

15 Foldi, A. P., 1987, Reinforcement of Rubber Compounds With Short, Indi-vidual Fibers, Rubber World May, pp. 1926.

16 Rajeev, R. S., Bhowmick, A. K., De, S. K., Kao, G. J. P., and Bandyopadhyay,S., 2002, New Composites Based on Short Melamine Fiber ReinforcedEPDM Rubber, Polym. Compos., 234, pp. 574591.

17 Ishikawa, S., Tokuda, A., and Kotera, H., 2008, Numerical Simulation forFiber Reinforced Rubber, Journal of Computational Science and Technology,24, pp. 587596.

18 Mooney, M., 1940, A Theory of Large Elastic Deformation, J. Appl. Phys.,11, pp. 582592.

J

Downloaded Freinforcement. The force required to stretch a 200 mm wide rein-orced rubber is much greater than expected for a typical isotropiclastomer.

Further work on incorporating the effects of fiber distributionhould improve the material model. This is not an inherent prob-em with the transversely isotropic model but rather requires thextension of this simple model to incorporate layers of a dis-ributed orientation.

This study did not consider strain rate effects viscoelasticityr repeated loading effects Mullins effect on the material behav-or. Future research on incorporating these effects would be useful3840.

cknowledgmentWe would like to thank the Gates Corporation for supporting

his research, the staff of the Advanced Materials Developmentroup at the Gates Material Center in Columbia, MO, for mixing

nd preparing the test samples, and Randy Roller who performedhe tests at the Gates Technical Center in Rochester Hills, MI.

eferences1 Rogers, J. W., 1981, The Use of Fibers in V-Belt Compounds, Rubber World

March, pp. 2731.2 Treloar, L. R. G., 1944, Stress-Strain Data for Vulcanised Rubber Under

Various Types of Deformation, Trans. Faraday Soc., 40, pp. 5970.3 Marckmann, G., and Verron, E., 2006, Comparison of Hyperelastic Models

for Rubber-Like Materials, Rubber Chem. Technol., 79, pp. 835858.4 Brown, R., 2006, Physical Testing of Rubber, Springer ScienceMedia Inc.,

New York.5 Peeters, F. J. H., and Kussner, M., 1999, Material Law Selection in the Finite

Element Simulation of Rubber-Like Materials and Its Practical Application inthe Industrial Design Process, Constitutive Models for Rubber, A. Dorfmannand A. Muhr, eds., Balkema, Rotterdam, pp. 2936.

6 Finney, R. H., and Kumar, A., 1988, Development of Material Constants forNonlinear Finite-Element Analysis, Rubber Chem. Technol., 61, pp. 879891.

7 Charlton, D. J., Yang, J., and The, K. K., 1994, A Review of Methods toCharacterize Rubber Elastic Behavior for Use in Finite Element Analysis,Rubber Chem. Technol., 673, pp. 481503.

8 Kim, W.-D., Kim, W.-S., Woo, C.-S., and Lee, H.-J., 2004, Some Consider-ations on Mechanical Testing Methods of Rubber Materials Using NonlinearFinite Element Analysis, Polym. Int., 53, pp. 850856.

9 Holzapfel, G. A., 2005, Similarities Between Soft Biological Tissues andRubberlike Materials, Constitutive Models for Rubber IV, P.-E. Austrell andL. Kari, eds., Taylor & Francis, London, pp. 607617.

10 Krouskop, T. A., Wheeler, T. M., Kallel, F., Garra, B. S., and Hall, T., 1998,Elastic Moduli of Breast and Prostate Tissues Under Compression, Ultrason.Imaging, 20, pp. 260274.

11 Jones, R. M., 1975, Mechanics of Composite Materials, Scripta Book Com-pany, Washington, D.C.

12 Lee, M. C. H., 1988, The Mechanical Properties and Fractural Morphology ofUnidirectional Short-Fiber Reinforced Polychloroprene Composites, J.

ournal of Engineering Materials and Technologyom: http://materialstechnology.asmedigitalcollection.asme.org/ on 11/25/219 Holzapfel, G. A., 2006, Nonlinear Solid Mechanics: A Continuum Approachfor Engineering, Wiley, Chichester.

20 Peng, S. H., Shimbori, T., and Naderi, A., 1994, Measurement of ElastomersBulk Modulus by Means of a Confined Compression Test, Rubber Chem.Technol., 67, pp. 871879.

21 MARC Analysis Research Corporation, 1996, Nonlinear Finite ElementAnalysis of Elastomers, MARC Analysis Research Corporation, Palo Alto, CA.

22 Ericksen, J. L., and Rivlin, R. S., 1954, Large Elastic Deformations of Ho-mogeneous Anisotropic Materials, Journal of Rational Mechanics and Analy-sis, 3, pp. 281301.

23 Spencer, A. J. M., 1984, Continuum Theory of the Mechanics of Fiber-Reinforced Composites, Springer, New York.

24 Green, A. E., and Adkins, J. E., 1970, Large Elastic Deformations, OxfordUniversity Press, Belfast, UK.

25 Moghe, S. R., 1974, Short Fiber Reinforcement of Elastomers, RubberChem. Technol., 47, pp. 10741081.

26 Itskov, M., and Aksel, N., 2004, A Class of Orthotropic and TransverselyIsotropic Hyperelastic Constitutive Models Based on a Polyconvex Strain En-ergy Function, Int. J. Solids Struct., 41, pp. 38333848.

27 Itskov, M., Aksel, N., and Ehret, A., 2003, Constitutive Modelling of Calen-daring Induced Anisotropy in Rubber Sheets, Constitutive Models for RubberIII, J. J. C. Busfield and A. H. Muhr, eds., Sets & Zeitlinger, Lisse, TheNetherlands, pp. 401404.

28 Shariff, M. H. B. M., 2008, Transversely Isotropic Strain Energy With Physi-cal Invariants, Constitutive Models for Rubber V, Taylor & Francis, London,pp. 6772.

29 Ogden, R. W., 1997, Non-Linear Elastic Deformations, Dover, Mineola, NY.30 Fung, Y. C., 1994, A First Course in Continuum Mechanics for Physical and

Biological Engineers and Scientists, Prentice-Hall, Englewood Cliffs, NJ.31 Gent, A. N., 1992, Elasticity, Engineering With Rubber How to Design

Rubber Components, A. N. Gent, ed., Carl Hanser, Munich, Germany, pp.3366.

32 Peng, X. Q., Guo, Z. Y., and Moran, B., 2006, An Anisotropic HyperelasticConstitutive Model With Fiber Matrix Shear Interaction for the Human Annu-lus Fibrosis, ASME J. Appl. Mech., 73, pp. 815824.

33 Weiss, J. A., Maker, B. N., and Govindjee, S., 1996, Finite Element Imple-mentation of Incompressible, Transversely Isotropic Hyperelasticity, Comput.Methods Appl. Mech. Eng., 135, pp. 107128.

34 MSC Software, 2008, MARC, Vol. D, User Subroutines and Special Routines.35 Qiu, G. Y., and Pence, T. J., 1997, Loss of Ellipticity in Plane Deformation of

a Simple Directionally Reinforced Incompressible Nonlinearly Elastic Solid,J. Elast., 49, pp. 3163.

36 Ishikawa, S., and Kotera, H., 2005, Constitutive Equations for Fiber Rein-forced Hyperelasticity, Constitutive Models for Rubber IV, P.-E. Austrell andL. Kari, eds., Taylor & Francis, London, pp. 619625.

37 Zhang, F., Fan, Z., Du, X., and Kuang, Z., 2004, Study on Constitutive Modeland Failure Criterion of Cord-Rubber Composite, J. Elastomers Plast., 36,pp. 351362.

38 Dorfmann, A. L., Woods, W. A., Jr., and Trimmer, B. A., 2008, MusclePerformance in a Soft-Bodied Terrestrial Crawler: Constitutive Modeling ofStrain-Rate Dependency, J. R. Soc., Interface, 5, pp. 349362.

39 Pea, E., and Doblar, M., 2009, An Anisotropic Pseudo-Elastic Approach forModeling Mullins Effect in Fibrous Biological Materials, Mech. Res. Com-mun., 36, pp. 784790.

40 Horgan, C. O., Ogden, R. W., and Saccomandi, G., 2004, A Theory of StressSoftening of Elastomers Based on Finite Chain Extensibility, Proc. R. Soc.London, Ser. A, 460, pp. 17371754.


[Brown] a Simple Trasnversely Isotropic Hyperelastic Model

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