At least three invariants are necessary to model the mechanical response of incompressible, transversely isotropic materials
Michel Destrade, Brian Mac Donald, Jerry Murphy, Giuseppe Saccomandi
AAt least three invariants are necessary to model the mechanical response ofincompressible, transversely isotropic materials
M. Destrade ab , B. Mac Donald c , J.G. Murphy ca , G. Saccomandi da School of Mathematics, Statistics, and Applied Mathematics,National University of Ireland Galway, University Road, Galway, Ireland. b School of Mechanical and Materials Engineering,University College Dublin, Belfield, Dublin 4, Ireland. c Centre for Medical Engineering Research,Dublin City University, Glasnevin, Dublin 9, Ireland. d Dipartimento di Ingegneria Industriale,Universit`a degli Studi di Perugia, 06125 Perugia, Italy.
Abstract
The modelling of off-axis simple tension experiments on transversely isotropicnonlinearly elastic materials is considered. A testing protocol is proposed wherenormal force is applied to one edge of a rectangular specimen with the oppositeedge allowed to move laterally but constrained so that no vertical displacement isallowed. Numerical simulations suggest that this deformation is likely to remainsubstantially homogeneous throughout the specimen for moderate deformations.It is therefore further proposed that such tests can be modelled adequately as ahomogenous deformation consisting of a triaxial stretch accompanied by a simpleshear. Thus the proposed test should be a viable alternative to the standard biaxialtests currently used as material characterisation tests for transversely isotropicmaterials in general and, in particular, for soft, biological tissue. A consequence ofthe analysis is a kinematical universal relation for off-axis testing that results whenthe strain-energy function is assumed to be a function of only one isotropic andone anisotropic invariant, as is typically the case. The universal relation provides asimple test of this assumption, which is usually made for mathematical convenience.Numerical simulations also suggest that this universal relation is unlikely to agreewith experimental data and therefore that at least three invariants are necessaryto fully capture the mechanical response of transversely isotropic materials. a r X i v : . [ c ond - m a t . s o f t ] S e p Introduction
The recent resurgence in interest in the modelling of the mechanical response of incom-pressible, transversely isotropic, nonlinearly elastic materials is primarily because thereare many examples of biological, soft tissue reinforced with bundles of fibres that have anapproximate single preferred direction, most notably skeletal muscles, ligaments and ten-dons. Developed mainly by Rivlin and co-workers [1], the phenomenological constitutivetheory for such materials was originally used to model elastomers reinforced with steelcords and it is a happy coincidence that such an elegant, rational theory can be appliedto some of the fundamental modelling problems in biomechanics.If soft tissue is assumed non-dissipative, as is commonly the case, then its mechanicalresponse is determined completely by the corresponding potential function, called thestrain-energy function in mechanics. For incompressible, transversely isotropic materials,the general strain-energy function is an arbitrary function of four scalar invariants (see,for example, Spencer [2]). These invariants are functionals of a strain tensor, with theleft Cauchy-Green strain being the measure of choice in biomechanics, and the directionof the fibres in the undeformed configuration. Unfortunately, the corresponding stress-strain relation is algebraically complex (see (5)) and, to facilitate analysis, simplifyingassumptions are, and usually must, be made. By far the most common assumption is toassume that the strain-energy is a function only of the two invariants I , I , where I isthe trace of the Cauchy-Green strain tensor and I is the square of the fibre stretch. Asalready mentioned, this choice is usually made on the basis of mathematical conveniencealone, as a cursory examination of (5) reveals the more complicated terms in the stress-strain relation are eliminated as a result. This choice of the ( I , I ) pair as the basisfor the strain-energy function is the central theme here. It will be critically evaluatedby examining the mechanical response of transversely isotropic materials in the simplestmaterial characterisation test: the simple tension test.In contrast to the situation for isotropic materials, the development of protocols andmethods for the simple tensile testing of anisotropic materials is still on-going despiteits long history (see, for example, Pagano and Halpin [3]). Simple tension testing whenthe direction of anisotropy is oblique to the direction of the applied force is referred toas ‘off-axis testing’. It has long being recognised that the standard rigid grips of mosttensile testing machines induce shear forces and bending moments in the test specimensduring off-axis testing, resulting in large stress concentrations and inhomogeneity in thetest samples. This coupling of simple tension with shearing forces in off-axis testing offibre-reinforced materials has, in fact, been exploited to characterise intralaminar shear(see Mar´ın et al. [4]). References to the different methods employed to reduce stressconcentrations and inhomogeneity of the tested sample in off-axis testing can be foundin Mar´ın et al. [4] and Xiao et al. [5].Heretofore the vast majority of off-axis testing has involved epoxy composites and istherefore concerned with the determination of material constants within the context of thelinear theory. Here the focus is on off-axis testing within the nonlinear regime and a newmethod of testing unidirectional composites is considered. Specifically it is proposed herethat the coupling of shear and simple tension in off-axis testing be fully recognised andthat the specimen, classically fully constrained along one of its edges, be allowed to movelaterally there. A combination of tri-axial stretch and simple shear is therefore proposed2o model the resulting deformation. This stretch/shear combination has previously beenstudied for isotropic materials by, amongst others, Moon and Truesdell [6], Rajagopaland Wineman [7] and, more recently, by Mihai and Goriely [8] and by Destrade et al. [9].The main objective here is to consider the validity of assuming pairs of invariantsas the sole arguments of the strain-energy function. To this end, numerical simulationsbased on the Finite Element Method (FEM) were performed. The commercial FEMprogramme ANSYS was used throughout. There is no direct implementation of a trans-versely isotropic material available in ANSYS. Instead fibre-reinforced nonlinearly elasticmatrices were modelled using a structural modelling approach, although this structuralmodel includes a phenomenological component, the neo-Hookean matrix. This matrix isreinforced by much stiffer linearly elastic cords. On clamping one of the edges to restrictmovement in the vertical direction, but allowing displacement in the lateral direction, anormal force is applied to the opposite edge.Two inferences can be drawn from an analysis of the output of the various simulationsperformed by varying the applied force and the fibre orientation. The first is that ahomogenous deformation field consisting of a triaxial deformation accompanied by asimple shear seems a viable model of material behaviour for off-axis testing. This givessome support to the viability of the experimental method proposed here. The second isthat the assumption that pairs of invariants are sufficient to capture the main featuresof the mechanical response of nonlinearly elastic, transversely isotropic materials is notcompatible with our numerical simulations and therefore that at least three invariantsare necessary to fully capture the mechanical response of transversely isotropic materials .This is because if pairs of invariants are assumed, then a universal kinematical relationbetween the kinematical variables results from satisfaction of the boundary conditionsassociated with simple tension. An illustrative comparison is then made between thekinematical relation that results from the ( I , I ) choice and the FEM results for anillustrative fibre orientation over a range of physiological strain. It will be shown thatthere is a fundamental incompatibility between the two sets of results.Interpretation of this incompatibility will depend on which model is more likely to en-capsulate the mechanical response of biological, soft tissue in the laboratory. Ultimately,of course, this question can only be resolved by conducting simple tension tests of thetype proposed here but, faced with the lack of experimental data, it is our contentionthat the physically well-motivated structural model that is the basis of our FEM resultsis a better choice than a phenomenological model based primarily on mathematical con-venience. Using Finite Element Analysis in this way to inform the constitutive modellingprocess seems a novel application of computational mechanics.The paper is organized as follows: after a section outlining the constitutive theory fortransversely isotropic materials, the modelling of off-axis simple tension tests is discussedin Section 3, with a particular emphasis on the modelling of a class of materials often usedin biomechanics. The results of our numerical experiments are then reported in Section4 and the consequences of these results for the modelling of transversely isotropic, softtissue are discussed.Although the assumed homogeneous deformation consisting of a simple shear super-imposed upon a triaxial stretch seems a natural fit with off-axis testing of transverselyisotropic materials, and seems supported by our numerical experiments, the semi-inverseapproach adopted here is not without its limitations, similar to the problems associated3ith modelling simple shear for isotropic materials (Rivlin [10], Gent et al. [11], Horganand Murphy [12]). Specifically, stresses need to be applied to the inclined faces of thedeformed test pieces in order to maintain the assumed combination of tri-axial stretchand simple shear. This is discussed in the final section. Incompressible fibre-reinforced materials are considered from now on. We call x = x i e i the coordinates in the current configuration B of a particle which was at X = X α E α inthe reference configuration B r . Here the orthonormal vectors E α are aligned with theedges of the test sample, which is assumed to be a cuboid of dimensions − A/ ≤ X ≤ A/ , − L/ ≤ X ≤ L/ , − H/ ≤ X ≤ H/ . (1)We take the the orthonormal vectors ( e , e , e ) to be aligned with ( E , E , E ). Thesample is clamped at X = ± L/ X direction.We assume that there exists a single preferred direction (along the unit vector A , say)to which all reinforcing fibres are parallel (transverse isotropy) and that the fibres areconfined to the ( X , X ) plane in the undeformed configuration. Thus A = C E + S E , where C ≡ cos Φ , S ≡ sin Φ , (2)where Φ (0 < Φ < π/
2) is the angle in the undeformed configuration between the fibersand the direction normal to the tensile force.Let F ≡ ∂ x /∂ X denote the deformation gradient tensor and B = F F T , C = F T F the left and right Cauchy-Green strain tensors, respectively. For incompressible materi-als, det F = 1. The general strain-energy function for incompressible, fibre-reinforced,hyperelastic materials has the form W = W ( I , I , I , I ) (see Spencer [2]), where I , I are the first and second isotropic principal invariants: I = tr C , I = tr C − , (3)and I , I are the anisotropic invariants, I = A · CA , I = A · C A . (4)The corresponding Cauchy stress tensor T is given by [2] T = − p I + 2 W B − W B − + 2 W a ⊗ a + 2 W ( a ⊗ Ba + Ba ⊗ a ) , (5)where W k ≡ ∂W/∂I k , p is a Lagrange multiplier introduced by the incompressibilityconstraint, I is the identity tensor and a ≡ F A .Experience has shown that the technical challenges of analysing general transverselyisotropic materials are formidable and indeed further evidence of this will be provided inlater sections. To make progress simplifying assumptions need to be made. For trans-versely isotropic materials, it is usual to ignore the I , I invariants and to adopt theassumption that W = W ( I , I ) only . (6)4any strain-energy density functions used in biomechanics applications have this form(see, for example, Humphrey and Yin [13], Humphrey et al. [14], Horgan and Murphy[15], Wenk et al. [16]) and a much-used example is the so-called standard reinforcingmaterial W ( I , I ) = µ (cid:2) I − γ ( I − (cid:3) , (7)where µ ( >
0) is the shear modulus of the neo-Hookean potential and γ ( >
0) is a non-dimensional material constant that provides a measure of the strength of reinforcementin the fibre direction, with large values of this parameter typical for soft, biological tissue(see, for example, Ning et al. [17], Destrade et al. [18]) . Another popular choice is theGasser-Ogden-Holzapfel model [19] W ( I , I ) = µ I −
3) + k k (cid:104) e k ( I − − (cid:105) , (8)where µ , k and k are positive constants, to be determined from experimental data.Similarly, the extension of this strain-energy density to include dispersive effects for thefibers in [20], the so-called ‘HGO’ model implemented in the finite element softwareABAQUS, also belongs to the family (6). We focus on the general homogeneous field response generated by a tensile test wherethe tensile force occurs at an angle to the fibres. Hence we take the components F iα = ∂x i /∂X α of the deformation gradient tensor to be constants. One clamp is allowed toslide in the direction of E = e and the line elements that were parallel to the clampsin B r remain parallel to the clamps in B . In other words, the deformation takes the form x = F X + F X , x = F X , x = F X . (9)An illustrative example of this type of deformation is given in Figure 1 below. Deforma-tions of this form are a special case of the homogeneous deformations, with deformationgradient tensor [ F ] iα = F F F F
00 0 F (10)considered by Holzapfel and Ogden [22] who wished to clarify the extent to which bi-axial testing can be used for determining the elastic properties of transversely isotropicmaterials.The deformation (9) can be decomposed as a tri-axial stretch accompanied by a simpleshear, as can be seen from the following identifications in the ( e i ⊗ E α ) coordinate system,[ F ] iα = F F F
00 0 F = λ λ κ λ
00 0 λ = κ
00 1 00 0 1 λ λ
00 0 λ , (11)(see [6, 7, 9] for isotropic materials). Here λ , λ , λ are positive constants, with λ =( λ λ ) − as a result of imposing the incompressibility constraint and κ is the amount of5 l l A l =l H Figure 1:
A cuboid of hyperelastic material reinforced with one family of parallel fibers candeform homogeneously when subjected to a tensile force only, as a combination of triaxial stretch(with stretch ratios λ i ) and simple shear (with amount of shear κ ). shear in the E = e direction. The data to be collected during those tensile tests are: λ , λ , κ and T , the tensile Cauchy stress component. The stretches can be measured withtwo orthogonal LASER tracking devices and κ by measuring the transverse displacementof the sliding clamp, see Fig.1. Alternatively, a Digital Correlation Imaging device canbe used. To measure T , the force is measured by a loadcell attached to a clamp anddivided by λ A × λ H = AH/λ , the current cross-sectional area.We now compute the components of the left and right Cauchy-Green deformationtensors as [ B ] ij = λ + λ κ λ κ λ κ λ
00 0 λ , [ C ] αβ = λ λ λ κ λ λ κ λ (1 + κ ) 00 0 λ , (12)in the ( e i ⊗ e j ) and the ( E α ⊗ E β ) coordinate systems, respectively. The isotropic strain-invariants are given by [7] I = λ + λ (cid:0) κ (cid:1) + λ − λ − , I = λ − (cid:0) κ (cid:1) + λ − + λ λ , (13)and the anisotropic invariants by I = ( λ C + λ κS ) + λ S ,I = λ (cid:0) λ + λ κ (cid:1) C + λ (cid:2) λ + (cid:0) λ + 2 λ (cid:1) κ + λ κ (cid:3) S + 2 λ λ (cid:2) λ + λ (cid:0) κ (cid:1)(cid:3) κCS. (14)6e may then compute the corresponding Cauchy stress components. It follows from (5)that T = T = 0. We complete the plane stress assumption by setting T = 0, whichgives us the expression for p and the in-plane stress components are therefore [21, 22] T = 2 W ( B − B ) + 2 W (cid:2) B ( B − B ) − B (cid:3) + 2 W a + 4 W a ( Ba ) ,T = 2 W ( B − B ) + 2 W (cid:2) B ( B − B ) − B (cid:3) + 2 W a + 4 W a ( Ba ) ,T = 2 W B + 2 W B B + 2 W a a + 2 W [ a ( Ba ) + a ( Ba ) ] , (15)where we used det B = 1 (incompressibility) to compute the components of B − . Here, a i and ( Ba ) i denote the appropriate components of the vectors a and Ba , respectively.Explicitly, they read a = λ C + λ κS, a = λ S, ( Ba ) = λ (cid:0) λ + λ κ (cid:1) C + λ (cid:0) λ + λ + κ λ (cid:1) κS, ( Ba ) = λ λ κC + λ (cid:0) κ (cid:1) S. (16)We remark that a = λ sin Φ (cid:54) = 0, and thus deformed fibres are never aligned with thedirection of the applied force.For tensile testing, T = T (cid:54) = 0 , T = T = 0 . (17)Since two of the in-plane stresses are identically zero, it follows from (15) that for theclasses of materials that depend on only two invariants a relationship between the defor-mation parameters λ , λ , κ will be obatined. For a given general fibre-angle Φ ( (cid:54) = 0 , π/ I , I ) pair being chosen because theresulting form of the stress-strain relation (5) is particularly convenient. For strain-energyfunctions of the form (6), the simultaneous satisfaction of (17) , yields the followinglinear, homogeneous system of two equations for W and W :0 = W ( B − B ) + W a , W B + W a a , which, since a (cid:54) = 0, gives non-trivial solutions for W , W if, and only if, the following, purely kinematical , relation holds: λ (cid:0) − λ − λ − (cid:1) S = λ κC. (18)This relationship is valid for all materials for which W = W ( I , I ). It is therefore anecessary test of this constitutive hypothesis; if for any non-zero angle of orientation (18)is violated at any stage during simple tension, then W is not a function of I and I only. It is shown in the next section that this kinematical relation does not fit the data7btained from the simple tension of a composite consisting of a soft non-linear matrixreinforced with stiff linear fibres.Other pairs of invariants could be considered in the same way as ( I , I ). For example,if the strain energy density is chosen to depend on I and I only, then the correspondingsemi-universal relation has the form (cid:0) λ λ − (cid:1) λ S = κ ( λ C + λ κS ) . (19)Similar considerations apply for the pairs ( I , I ) and ( I , I ).Assume for the moment that (18) holds for all members of the popular family of strainenergies (6). Then since W = W = 0, it follows from (12), (15) that simple tensionfor these materials is described by the following two simultaneous equations in the twounknowns λ , λ : T = 2 (cid:0) λ − λ − λ − (cid:1) W + 2 λ S W , (cid:0) − λ − λ − (cid:1) W + (cid:0) − λ − λ − S (cid:1) W , (20)with I , I now given by I = λ C + λ + λ − λ − (cid:18) − S C (cid:19) , I = λ C (cid:0) − λ − λ − S (cid:1) + λ S , (21)where we eliminated κ using (18). For the particular example of the standard reinforcingmodel (7), these two equations are T ≡ T /µ = λ − λ − λ − + 2 γλ S ( I − , (22)0 = 1 − λ − λ − + 2 γ (cid:0) − λ − λ − S (cid:1) ( I − , (23)with I given just above. These equations suggest a protocol to determine whethera given anisotropic soft tissue can be modelled by the standard reinforcing material,once it has been established that its strain-energy density is of the form (6) by firstchecking experimentally that the semi-universal relation (18) is satisfied. First, plot1 − λ − λ − against − (cid:0) − λ − λ − S (cid:1) ( I − γ . Next, plot the T dataagainst λ − λ − λ − + 2 γλ S ( I − µ . Thus the experimental confirmation of the validity of thestandard reinforcing model requires the satisfaction of at three demanding constitutivetests, given by (18), (22) and (23). To illustrate a typical tensile-stress-tensile stretchresponse for this model, we now fix γ at γ = 10 .
0, say, and vary the angle of the fibres Φto produce Figure 2 in two steps: first, solve (23) for a given λ to find the corresponding λ ; second, substitute into (22) to find T . 8 l F=20F=40F=60F=80
Figure 2:
Tensile stress vs tensile stretch for the standard reinforcing material. Here the stressmeasure is normalised with respect to the base shear modulus, T ≡ T /µ . The stretch λ ismeasured in the direction of the tensile force, but is not the largest stretch in the sample becauseof the fibre reinforcement, see Fig.1. Here the fibers were at a angle Φ = 20 ◦ , ◦ , ◦ , ◦ withrespect to the normal to the tensile force in the reference configuration. We see that, as expected intuitively, the more the fibers were oriented to be aligned withthe direction of the tensile force (i.e. as Φ increases towards 90 ◦ ), the stiffer the materialresponse becomes. Changing the value of γ only brings quantitative changes but thetrend remains the same. For a more complex strain-energy function, a multi-objectiveoptimization exercise must take place in order to evaluate the material parameters [23, 24]. A finite element model of a transversely isotropic block was built using ANSYS Version13, which allows reinforcing fibres to be randomly distributed throughout the matrix,as long as they are aligned in the same direction. If these fibres have identical materialproperties and orientation, then we may use a smeared reinforcement strategy to modelthe contribution of the fibres to the mechanical response of the fibre/matrix composite.The material parameters used in our simulations have a biomechanical motivation. Moul-ton et al. [25] found that the Young’s modulus of passive myocardium is of the order of0.02 MPa. The matrix was therefore assumed to be a neo-Hookean, non-linearly elas-tic material, since it is generally accepted that the neo-Hookean material is an excellentmodel of the mechanical response of general, isotropic materials for strains of the orderconsidered here (Yeoh and Fleming [26]), with this value of Young’s modulus. The fibresare modelled as a relatively stiff linearly elastic material with a Young’s modulus of 200MPa, since Yamamoto et al. [27] found that the Young’s modulus of collagen fasciclesis of this order. In order to make the shearing component of the deformation clearlyvisible in our graphics (see Figure 3 below), a volume fraction for the fibres of 0.1 wasused; this fraction, however, is an order of magnitude greater than the volume fractionof interstitial collagen found in the heart (Van Kerckhoven et al. [28]). The block had9riginally a width of 20mm, a height of 20mm and a thickness of 2mm. The nodes onthe bottom surface of the block were constrained only so that no vertical displacementwas allowed, simulating a clamp which is free to move laterally to allow for shearing ofthe specimen. No lateral displacement was allowed on the top surface of the block (againto simulate clamping), with a force acting in the positive vertical direction. The othersurfaces of the block were assumed stress-free.As an illustrative example of the Finite Element simulations conducted, the mid-plane of the thickness in the initial and deformed configurations is shown in Figure 3 forthe fibre composite with Φ = 45 ◦ , subjected to an axial strain of 1 .
2. For comparisonpurposes, these configurations are also given in Figures 8, 9 in the Appendix for bothΦ = 20 ◦ , and Φ = 80 ◦ . We note that the out-of-plane deformations in all our simulationswere essentially homogenous, with inhomogeneity confined to thin boundary layer-likeregions near the clamped ends. The corresponding contour plots of the axial, transverseand shear strains for Φ = 45 ◦ are given in Figure 4, so that the degree of homogeneity canbe assessed. These graphics confirm our physical intuition that deformations of the form(9) are good models of the deformation that results from subjecting transversely isotropicmaterials to simple tension, especially through the central region of the specimen. It alsosupports our contention that an important application of the analysis presented here isthat of modelling the mechanical response of biological, soft tissue given the excellentqualitative agreement between the edge profiles of the numerical simulations and theexperimental results of, for example, Guo et al. [29], who performed finite simple sheartests on porcine skin in order to obtain guidelines for the selection of specimen aspectratio and clamping prestrain when studying the material response of soft tissues undersimple-shear tests. Although the assumed homogeneous deformation (9) isn’t an exact fitwith the numerical results (the differing amounts through which the bottom corners of thespecimen are sheared are testament to that), nonetheless the homogeneous approximationshould be more than adequate for our constitutive modelling purposes.Figure 3: Initial configuration and final configuration with λ = 1 . ◦ initial fibre orien-tation. Contour plots of (a) normal strain; (b) transverse strain; (c) shear strain
Although the focus here is on simple tension testing, and therefore on stress con-trolled tests, for the numerical experiments it was convenient to control the axial stretch λ so that strains consistent with the physiological regime (of the order of 20%) werereproduced. Consequently axial stretches up to λ = 1 . ◦ are taken as representative. The transverse stretches λ and amounts of shear κ for Φ = 45 ◦ were calculated by measuring the displacementof the edge nodes along the centre of the specimen, where, as can be seen from Figures3, 4, the end effects are minimised and the deformation is essentially homogeneous. Thenumerical results are given in the Appendix and are summarised in graphical form inFigure 5 below: 11 "! !" *+,-./",0"123*4"/4*.153413"1/43/62"7.5*47*./" Figure 5:
Typical transverse stretch, shear and I results. As might be expected, the transverse stretch is a monotonically decreasing and theamount of shear a monotonically increasing function of the imposed axial stretch. Asa check of the above predictions, the invariant I was computed using (14) . Recallingthat this invariant is the square of the fibre stretch, one would expect I to increasewith increasing axial stretch. This is reflected in Figure 5, where the fibre stretch isalways greater than one, thus avoiding possible instabilities arising from fibres being incompression.The validity of kinematical relations like (18), obtained by assuming invariant pairsfor the strain-energy function, is now examined by comparing them with the numericalpredictions of the behaviour of fibre-reinforced composites provided by our Finite Elementsimulations. Only the relation (18) is under consideration here, but similar results wereobtained for the other possible kinematical relations. Strain controlled experiments wereperformed in the Finite Element analysis and we therefore consider λ and κ as functionsof the axial stretch λ and therefore let f ( λ ) ≡ λ (cid:0) − λ − λ − (cid:1) S − λ κC, (24)where λ = λ ( λ ), κ = κ ( λ ). The analysis of Section 3 has shown that if W = W ( I , I ),then f ( λ ) should be zero over the range of axial stretch of interest for all fibre angles .For the physiological range of strain 1 . ≤ λ ≤ .
2, the function values for our illustrativefibre angle of Φ = 45 ◦ are plotted in Figure 6(a). The practical difficulty in interpretingthe data of Figure 6(a) is that a natural measure of ‘closeness’ between the function(24) and 0 is not available. One possible solution is to normalise the absolute differencesbetween the f ( λ ) values and 0 using the applied stretch λ and then interpret the resultsas percentage errors. Plots of these errors, defined therefore bypercentage error ≡ (cid:12)(cid:12)(cid:12)(cid:12) f ( λ ) λ (cid:12)(cid:12)(cid:12)(cid:12) × , (25)are given in Figure 6(b). 12a) !" ! " $ % & ’ ( ) * " + , ’ )-.)*’,/0+/$1’ (b) !" ! " $ " % & ’ ( " ) " * ) ’+,’-).& Figure 6: (a) Test of ( I , I ) hypothesis; (b) Percentage error plots. The percentage error plot strongly suggests that the relation (18) is not valid forΦ = 45 ◦ and thus we conclude that the numerical simulations are not supportive of theconstitutive assumption W = W ( I , I ), since if this assumption were true, then (18)would hold for all fibre angles. Consequently, to model the full range of fibre orienta-tions for physiological strains, the Finite Element simulations suggest that at least threeinvariants are required to fully capture the mechanical response of transversely isotropicmaterials . It should be noted, however, that our simulations show a pronounced fibre-effect on the percentage error values: specifically, the percentage errors decrease withdecreasing orientation angle. This is to be expected because in the limiting case ofΦ = 0 ◦ the fibres are perpendicular to the direction of the applied force and consequentlyhave no effect on the composite response to the applied stress distribution.Holzapfel and Ogden [22] considered the extent to which biaxial testing can be used todetermine the elastic properties of transversely isotropic materials (the same problem forstrain energies based on limited structural information and multiaxial stress-strain datawas considered by Humphrey and Yin [13] and Humphrey et al. [14]). In particular, theyconcluded that if the constitutive assumption (6) is valid , then biaxial tests can be usedto determine the functions W , W and hence to determine the form of the correspondingstrain-energy function (6). Our conclusions do not contradict their results; rather ourresults cast doubt on their premise. It is our contention that strain-energy functions ofthe form (6) are not valid. Although physical intuition and the numerical experiments of the Section 4 suggest thatthe deformation (9) is likely to be an excellent approximation to the deformed stateof a rectangular block subjected to the tension field (17), the semi-inverse approachadopted here results in an over-determined system for the unknowns λ , λ , κ , with someboundary conditions in any physical realisation of the proposed experiments not beingsatisfied. Specifically, it is envisaged that the inclined faces of the specimen will be stress-free but, as is well-known for isotropic materials (see, for example, Atkin and Fox [30] fora clear discussion of the issues involved), normal and shear stresses must be applied tothe inclined faces of the block in order to maintain a state of homogeneous deformation.13t is easily shown here that for the tensile test deformation, the outward unit normal n to the inclined faces in the deformed configuration has the following components: n = (cid:18) √ κ , − κ √ κ , (cid:19) , (26)independent of the axial and lateral stretches. Noting the imposed state of stress, (17),the normal stress N and the shear stress S that therefore have to applied to the inclinedfaces in order to maintain a block in the deformed state (9) are given by N = κ κ T, S = − κ κ T. (27)The normalised stresses ˆ N ≡ N/T , ˆ S ≡ S/T are therefore a function only of the amountof shear κ and are plotted in Figure 7 for the moderate range of κ suggested by thesimulations of the last section. !" " !" ! " $ % &’ ( ) * + ( , ) (( + %$"-!,+".+(/)% ()*+,-./01%2%()*+,-./01%3% Figure 7:
Stresses on inclined faces.
It is clear from this figure that compared to the tensile stress, only insignificant normalstresses are required for physiological strains; consequently the absence of normal stressesapplied to the inclined faces is likely to have a negligible effect on the homogeneity ofthe deformation. In contrast, there is essentially a linear relationship between κ and ˆ S over the the range of strain of interest. That relatively large shear stresses are requiredto maintain homogeneity is not surprising given that even the linear theory for isotropicmaterials requires a shear stress on the inclined face of equal magnitude to the shearstress driving the deformation. It is worth emphasising here, however, that for the givenstrain range, the shear stresses are essentially an order of magnitude smaller than theapplied tensile stress. In practice a lack of shear stress on the inclined faces of shearedblocks of biological tissue does not seem to affect the homogeneity of the deformation. InDokos et al. [31], for example, cuboid specimens of myocardium were sheared up to 40%with no reported mention of any inhomogeneity observed in testing. Some protocols tominimise inhomogeneity when shearing biological, soft tissue were proposed by Horganand Murphy [32]. Certainly for the experiments considered here, Figure 4 suggests thathomogeneity is likely to be maintained, at least within the central region of the specimen.14 Conclusions
A method has been proposed for the off-axis simple tension testing of transversely isotropicnonlinearly hyperelastic materials, a method that should be a viable alternative to thedominant biaxial tension test for material characterisation. This method proposes thata shearing deformation accompanies a triaxial stretching regime. It was shown that if,as is commonly the case, a pair of one isotropic and one anisotropic invariant is chosenas the basis for the strain-energy function, then a kinematical universal relation must besatisfied for this new testing regime, one that must hold for all fibre angles and for the fullrange of applied tension. Finite Element simulations suggest that this is too demandinga requirement and that at least three invariants are necessary to model the full range ofmechanical response of transversely isotropic materials.
The following procedure was adopted to calculate the kinematical quantities λ , λ , κ from the numerical results. The x coordinates of the mid-points of the inclined faceswere isolated from the rest of the output data at specified values of the prescribed axialstretch; call them x l , x r , using an obvious notation. Referring to Figure 3, let the origincoincide the bottom left corner of the undeformed block. Since the dimensions of theblock were chosen to be 20mm × × x l = κλ , x r = λ
20 + κλ . Since λ is controlled, κ, λ are therefore obtained from κ = x l λ , λ = x r − x l . These calculated quantities to four decimal places are tabulated in Table 1.15xial stretch amount of shear transverse stretch1 0 11.02 0.0207 0.98851.04 0.0406 0.97711.06 0.0598 0.96591.08 0.0782 0.95491.1 0.0959 0.94411.12 0.1128 0.93361.14 0.1288 0.92321.16 0.1442 0.91321.18 0.1587 0.90331.2 0.1725 0.8938Table 1: Kinematics for numerical experiments with a 45 ◦ fibre angle Numerical results for a fibre angle of 45 ◦ were presented in the main body of the paper.The simulations for this natural choice of fibre angle are supplemented below for an angleclose to the horizontal and another close to the vertical. For the material parametersused here, it is clear from Figures 8, 9 that shear is negligible for a 20 ◦ angle and muchmore pronounced for 80 ◦ . A comparison of these graphics with Figure 3 shows that theamount of shear for 45 ◦ is between these two limiting cases.Figure 8: Initial and final configurations with λ = 1 . ◦ initial fibre orientation. Initial and final configurations with λ = 1 . ◦ initial fibre orientation. References [1] Rivlin, R.S. Collected Papers of R.S. Rivlin, vol. 1, Barenblatt, G.I., Joseph, D.D.(eds.). Springer, New York (1997)[2] Spencer, A.J.M., 1972. Deformations of Fibre-Reinforced Materials. Oxford Univer-sity Press.[3] N.J. Pagano and J.C. Halpin, 1968. Influence of end constraint in the testing ofanisotropic bodies. J. Compos. Mat. 2, 18–31.[4] J.C. Mar´ın, J. Ca˜nas, F. Par´ıs and J. Morton, 2002. Determination of G12