Rediscovering GF Becker’s early axiomatic deduction of a multiaxial nonlinear stress–strain relation based on logarithmic strain

Abstract

We discuss a completely forgotten work of the geologist GF Becker on the ideal isotropic nonlinear stress–strain function (Am J Sci 1893; 46: 337–356). Due to the fact that the mathematical modelling of elastic deformations has evolved greatly since the original publication we give a modern reinterpretation of Becker’s work, combining his approach with the current framework of the theory of nonlinear elasticity.

Interestingly, Becker introduces a multiaxial constitutive law incorporating the logarithmic strain tensor, more than 35 years before the quadratic Hencky strain energy was introduced by Heinrich Hencky in 1929. Becker’s deduction is purely axiomatic in nature. He considers the finite strain response to applied shear stresses and spherical stresses, formulated in terms of the principal strains and stresses, and postulates a principle of superposition for principal forces which leads, in a straightforward way, to a unique invertible constitutive relation, which in today’s notation can be written as

$\begin{array}{l} T^{Biot} & = 2 G \cdot {dev}_{3} \log (U) + K \cdot tr [\log (U)] \cdot 11 \\ = 2 G \cdot \log (U) + Λ \cdot tr [\log (U)] \cdot 11, \end{array}$

where T^Biot is the Biot stress tensor, log(U) is the principal matrix logarithm of the right Biot stretch tensor $U = \sqrt{F^{T} F}$ , $X = \sum_{i = 1}^{3} X_{i, i}$ denotes the trace and dev₃ X = X − (1/3) tr (X) · 11 denotes the deviatoric part of a matrix $X \in ℝ^{3 \times 3}$ .

Here, G is the shear modulus and K is the bulk modulus. For Poisson’s number ν = 0 the formulation is hyperelastic and the corresponding strain energy

$W_{Becker}^{ν = 0} (U) = 2 G [< U, \log (U) - 11 > + 3]$

has the form of the maximum entropy function.

Keywords

Hencky strain logarithmic strain nonlinear elasticity matrix logarithm

1. Introduction

1.1. Some reflections on constitutive assumptions in nonlinear elasticity

The question of proper constitutive assumptions in nonlinear elasticity has puzzled many generations of researchers. The problem of finding simple enough constitutive assumptions which are sufficient to characterize physically plausible behaviour of ‘completely elastic’ materials was even called ‘das ungelöste Hauptproblem der endlichen Elastizitätstheorie’ (the unsolved main problem of finite elasticity theory) by Truesdell [1]. While such assumptions can only lead to idealized material behaviour, the merits of such an ideal model were already described by H Hencky in his 1928 article On the form of the law of elasticity for ideally elastic materials [2,3]:

Like so many mathematical and geometric concepts, it is a useful ideal, because once its deducible properties are known it can be used as a comparative rule for assessing the actual elastic behaviour of physical bodies. […] While it is certainly a matter of empirical observation to determine how actual materials compare to the ideally elastic body, the law itself acts as a measuring instrument which is extended into the realm of the intellect, making it possible for the experimental researcher to make systematic observations.

The range of applicability of such an idealized response, however, must necessarily be restricted to minute strains, perhaps in the order of 1%, for otherwise we are to expect interference with non-elastic effects like plastic deformations, microstructural instabilities or bifurcations. Nonetheless, it should be formulated tensorially correct for arbitrarily large strains. It is also clear that the restriction to small elastic strains does not imply that one can use linear elasticity theory, nor that the ideal elastic response for larger stresses or strains is arbitrary. On the contrary, our idealization should work, as an ideal model, for arbitrarily large strains.

In the past, a large number of possible basic assumptions for elastic materials have been suggested in order to respond to the idealization described above. Among the most commonly accepted are:

Hyperelasticity: the existence of a strain energy function W.

Homogeneity: the strain energy W does not depend on the position in the body.

Simple material: the strain energy W = W(F) depends only on the first deformation gradient F.

Objectivity: W(Q · F) = W(F) for all Q ∈ SO(3).

Isotropy: W(F · Q) = W(F) for all Q ∈ SO(3).

Unique (up to rotations) stress-free reference state $U = \sqrt{F^{T} F} = 11$ .

Linearization consistent with linear elasticity theory at the reference state.

Well-posedness of the corresponding linear elasticity model in statics and dynamics.

Correct stress response for extreme strains: σ → ∞ as $V = \sqrt{F F^{T}} \to \infty$ as well as σ → −∞ as det(V) → 0, where σ denotes the Cauchy stress tensor.

Second-order behaviour in agreement with Bell’s experimental observations [4], that is, the instantaneous elastic modulus E decreases for tension and increases in the case of compression (see Figure 1).

Correct energetic behaviour for extreme strains in order to ensure invertibility of the deformation gradient F: W → ∞ for ∥F∥ → ∞ as well as W → ∞ for det(F) → 0.

Polyconvexity [5 –10], quasiconvexity [8, 11].

Legendre–Hadamard ellipticity [12].

Baker–Ericksen inequalities [13].

Figure 1.

Nonlinear behaviour of incompressible elastic materials in a simple tensile stress test, drawing the Biot stress T^Biot versus the principal stretch λ.

Apart from these conditions there are several properties which, while not generally viewed as necessary for an elasticity model, may be considered as constitutive assumptions for an idealized material as well:

Superposition principle: T(V₁ · V₂) = T(V₁) + T(V₂) for all coaxial stretches V₁ and V₂ and some corresponding stress tensor T.

Invertible stress–strain relation: the mapping E ↦ T(E) is invertible for some stress tensor T and a corresponding work-conjugate strain tensor E (if T is the Cauchy stress tensor, then this invertibility condition is satisfied for example for a variant of the compressible neo-Hookean energy [15]; if T is the Biot stress tensor, then this condition is Truesdell’s invertible force stretch (IFS) relation [16, p. 156]).

Tension–compression symmetry: T(V⁻¹) = −T(V) for some stress tensor T (note that the classical hyperelastic tension–compression symmetry W(F) = W(F⁻¹) is equivalent to τ(V⁻¹) = − τ(V) for the Kirchhoff stress τ and the left stretch tensor V).

Plausible behaviour under simple homogeneous finite stresses (similar to linear elasticity):

– Pure shear stresses of the form

T = (\begin{matrix} 0 & s & 0 \\ s & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

should induce stretches of the form (c.f. Figure 2)

V = (\begin{matrix} B_{11} & B_{12} & 0 \\ B_{12} & B_{22} & 0 \\ 0 & 0 & 1 \end{matrix})

with

\det (\begin{matrix} B_{11} & B_{12} \\ B_{12} & B_{22} \end{matrix}) = 1 .

– Spherical stresses of the form

T = (\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix})

should induce volumetric stretches of the form (c.f. Figure 3)

V = (\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}) .

Ordered stresses (‘greater stress corresponds to greater stretch’): (T_i − T_j) · (λ_i − λ_j) > 0 for all λ_i ≠ λ_j and some stress tensor T, where T_k are the principal stresses, that is, the principal values of T, and λ_k are the principal stretches (note that the Baker–Ericksen inequality can be stated as (σ_i − σ_j) · (λ_i − λ_j) > 0 for all λ_i ≠ λ_j where σ is the Cauchy stress tensor).

Simple volumetric–isochoric decoupling to ensure a suitable formulation of the incompressibility constraint.

Minimal number of physically motivated and experimentally identifiable constitutive coefficients, for example only the two isotropic Lamé constants.

Clear physical interpretation of Poisson’s number ν for finite deformations: ν = 1/2 enforces exact incompressibility (det F = 1) and ν = 0 implies no lateral contraction under uniaxial tension (as in linear elasticity).

Greatest possible extent of elastic determinacy [3, p. 19]: the stress response should not depend on a specific reference state or previously applied deformations; a similar condition was proposed by Murnaghan [17, 18], who argued that the dependence of the stress response on a specific ‘position of zero strain’ was tantamount to an ‘action at a distance’ and should therefore be avoided.

Figure 2.

Pure shear stress should induce pure shear stretch, preserving the area A.

Figure 3.

Spherical stress should induce purely volumetric stretch.

Several attempts to propose such an idealized model of elasticity can be found in the literature. Becker’s deduction can be seen as an early example of such an attempt [19].

1.2. A modern interpretation of Becker’s development

Becker, in his development of a nonlinear law of elasticity, rejects many of his contemporaries’ approaches to the problem of finite elasticity. He starts his introduction with a description of Hooke’s law, stating that apart from its original formulation (‘Strain is proportionate to the load, or the stress initially applied to an unstrained mass’ [19, p. 337]) it is often interpreted in a different way (‘Strain is proportional to the final stress required to hold a strained mass in equilibrium’ [19, p. 337]). These two different interpretations of Hooke’s law for finite deformations¹ can be expressed as

\begin{array}{l} T^{Biot} & = & 2 G \cdot {dev}_{3} (U - 11) + K \cdot tr [U - 11] \cdot 11 \\ = & 2 G \cdot (U - 11) + Λ \cdot tr [U - 11] \cdot 11 \end{array}

(1)

and

\begin{array}{l} σ & = & 2 G \cdot {dev}_{3} (V - 11) + K \cdot tr [V - 11] \cdot 11 \\ = & 2 G \cdot (V - 11) + Λ \cdot tr [V - 11] \cdot 11 \end{array}

(2)

respectively, where $U = \sqrt{F^{T} F}$ is the right Biot stretch tensor, $V = \sqrt{F F^{T}}$ is the left Biot stretch tensor, σ is the Cauchy stress tensor, T^Biot is the Biot stress tensor, dev₃ X = X −(1/3) tr(X) · 11 denotes the deviatoric part of $X \in ℝ^{3 \times 3}$ , K is the bulk modulus and G, Λ are the Lamé constants. According to Becker, it was already ‘universally acknowledged that either law [(1), (2)] is applicable only to strains so small that their squares are negligible’ [19, p. 337]. He gives a number of reasons for this rejection of Hooke’s law as a model for finite deformations, including the fact that it allows for infinite distortions (det F = 0) under finite stresses [19, p. 337]. In addition, Becker states that Hooke’s law ‘rests entirely upon experiment’, that is, that there is no underlying framework necessitating the linearity of the stress–strain relation. However, Becker also rejects the idea that the stress response could be discovered by ‘any process of pure reason’ alone,² an approach which he attributes to Barré de Saint-Venant. In fact, in Becker’s time the elasticity models that had been developed through purely geometrical considerations generally implied the so-called Cauchy relations [20 –22], in other words, they determined the lateral contraction independent of the specific material, corresponding to a fixed value ν = 1/4 for Poisson’s number ν. As Becker points out in a footnote later on [19, p. 348], this value for ν should only be regarded as a special case and not as a general law.³

Instead, his approach to describe the deformation of an ideally elastic body can be summarized as follows: motivated by geometric considerations he postulates a connection between shear stresses and shear strains as well as between volumetric stresses and dilational strains. He then shows that every homogeneous finite deformation can be decomposed into two shear stretches and a purely dilational deformation. Finally Becker assumes that a law of superposition holds for all coaxial finite strains, allowing him to reduce the problem of a general stress–stretch relation to shears and dilations only.

Thus Becker makes a number of assumptions about the stress–stretch relation from which he then deduces a law of elasticity. His final result is a stress–stretch relation which in today’s notation can be written as

\begin{array}{l} T^{Biot} & = & 2 G \cdot {dev}_{3} \log (U) + K \cdot tr [\log (U)] \cdot 11 \\ = & 2 G \cdot \log (U) + Λ \cdot tr [\log (U)] \cdot 11, \end{array}

where log(U) is the principal matrix logarithm of the right Biot stretch tensor $U = \sqrt{F^{T} F}$ .

In order to reproduce Becker’s approach in a more modern framework of elasticity theory we will therefore interpret Becker’s implicit assumptions as axioms for a law of ideal elasticity. While Sections 2 and 3 summarize Becker’s motivation for these axioms as well as some of his computations, a generalized deduction of Becker’s law of elasticity will be given in Section 4. Finally, we will investigate some basic properties of the resulting stress–stretch relation in Section 5. The notation employed throughout the article can be found in Appendix A.

While the axioms are in fact sufficient to completely characterize an isotropic stress–stretch relation uniquely up to two material parameters, it turns out that some of them may be weakened considerably without changing the result.

2. Becker’s assumptions

In order to understand Becker’s approach it is important to distinguish his (often implicitly stated) assumptions from his deduced results. Like Becker, we consider a unit cube in the reference configuration Ω₀ the edges of which are aligned with an orthogonal coordinate system e₁, e₂, e₃. Unless indicated otherwise, all matrix representations of linear mappings are given with respect to this coordinate system.

2.1. Basic assumptions

Becker’s most basic assumptions are that the stress–stretch relation is an analytic function [19, p. 342] as well as isotropic⁴ [19, p. 338] and that it is possible to ‘regard strains as functions of load’ [19, p. 341], that is, that the strain-load mapping is invertible. Note that here and throughout we will interpret Becker’s ‘initial stress’ (which, for a unit cube, is equal to the load) as the Biot stress tensor [23]

T^{Biot} = U \cdot S_{2} = J \cdot R^{T} \cdot σ \cdot F^{- T},

where F denotes the deformation gradient, F = RU is the polar decomposition [24] of F with R ∈ SO(3) and $U = \sqrt{F^{T} F} \in PSym (3)$ , J = det F = det U is the Jacobian determinant, σ is the Cauchy stress tensor and S₂ is the symmetric second Piola–Kirchhoff stress tensor. A justification of this interpretation can be found in Appendix B.2. Note that in the isotropic case, T^Biot is a symmetric tensor as well [25].

2.2. Pure finite shear

One of Becker’s main assumptions is that ‘a simple finite shearing strain must result from the action of two equal loads or initial stresses of opposite signs at right angles to one another’ [19, p. 339]. This assumption is mostly motivated by geometric considerations: if the deformation is a homogeneous pure shear of the form

F = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}), α > 1,

(3)

Figure 4.

Pure shear load and the corresponding shear deformation.

then the so-called planes of no distortion have a number of properties connected to the quantity α. Becker then relates these properties of strain to certain properties of stress, thereby establishing that the stress corresponding to the above shear deformation must be a pure shear stress of the form

T^{Biot} = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}), s \in ℝ .

(4)

His arguments are largely based on the assumption that ‘[in] the particular case of a shear (or a pure shear) there are two sets of planes on which the stresses are purely tangential, for otherwise there could be no planes of zero distortion’ [19, p. 338].

If a Biot shear stress of the form (4) corresponds to a pure shear deformation of the form (3), then the corresponding Cauchy stress tensor σ computes to

σ = \frac{1}{\det U} \cdot U^{- 1} \cdot F \cdot T^{Biot} \cdot F^{T} = 1 \cdot F^{- 1} \cdot F \cdot T^{Biot} \cdot F = (\begin{matrix} α s & 0 & 0 \\ 0 & - \frac{s}{α} & 0 \\ 0 & 0 & 0 \end{matrix}) .

(5)

The principal Cauchy stresses are therefore σ₁ = α s, σ₂ = −s/α and σ₃ = 0. Note carefully that Becker’s shear deformation is oriented differently: in his considerations, the contractile axis (along the eigenvector to the smaller eigenvalue 1/α) is aligned with the e₁-axis. This difference is reflected in Becker’s formula −σ₁α = σ₂/α [19, p. 338]. For our choice of axes, the corresponding equality reads

- σ_{2} α = \frac{σ_{1}}{α} .

A more detailed geometric description of this relation will be given in Section 3. More recent discussions of shear stresses and shear strains can be found in [26] and [27]. The so-called planes of no distortion also play an important role in Becker’s treatment of the rupture of rocks [28] as well as his later works on schistosity and slaty cleavage [29, 30], where properties of the planes of no distortion are linked to failure criteria for deformations beyond the range of elastic deformations. A summary of Becker’s work on yield criteria for rocks can be found in [31], while the concept of planes of no distortion and its relation to the tangential shear strain is described in a more detailed form in [32].

2.2.1. The planes of no distortion

Let F ∈ GL⁺(3) denote an invertible linear mapping. We call a plane $ℰ \subset ℝ^{3}$ through the origin an initial plane of no distortion if the restriction of F to $ℰ$ is a rotation, which is the case if and only if 〈Fx, Fy〉 = 〈x, y〉 for all $x, y \in ℰ$ or, equivalently, if ∥Fx∥ = ∥x∥ for all $x \in ℰ$ , where 〈·,·〉 denotes the Euclidean inner product on $ℝ^{3}$ . Furthermore, we call $ℱ \subset ℝ^{3}$ a final plane of no distortion if $ℱ$ is the image under F of an initial plane of no distortion, which is the case if and only if ∥F⁻¹x∥ = ∥x∥ for all $x \in ℱ$ . Since Becker’s considerations of such planes are confined to the deformed (or final) configuration of a homogeneous deformation, we will often refer to $ℱ$ simply as a plane of no distortion.

Figure 5.

An initial plane of no distortion is only rotated by the linear mapping F, preserving the angles between two vectors as well as their lengths.

The following basic existence property can be found in [32].

Proposition 2.1. Let F ∈ GL⁺(3). Then there exists a plane of no distortion for F if and only if λ₂ = 1, where λ₁ ≥ λ₂ ≥ λ₃ denote the singular values of F. If λ₁ >λ₂ = 1 >λ₃, then there exist exactly two such planes.

Remark 2.2. If λ₂ ≠ 1 for the second singular value λ₂ of a linear mapping F ∈ GL⁺(3), then instead of a plane there exists a surface of no distortion in the form of an elliptical cone [32, p. 133] instead of a plane of no distortion. If, however, we generalize the term to mean any plane $ℱ$ such that the restriction of F to $ℱ$ is a dilated rotation (also called a conformal mapping) of the form λ · Q with $λ \in ℝ^{+}$ and Q ∈ SO(3), then Proposition 2.1 shows that such a plane exists for F ∈ GL⁺(3) if and only if the dilational factor λ is the second singular value of F.

Remark 2.3. It was shown by Ball and James [33, Proposition 4] that the equality λ₂ = 1 holds if and only if the right Cauchy-Green tensor C = F^T F corresponding to F is expressible in the form

C = (11 + ξ \otimes η) \cdot (11 + η \otimes ξ) = {(11 + η \otimes ξ)}^{T} \cdot (11 + η \otimes ξ)

with $ξ, η \in ℝ^{3}$ , where $ξ \otimes η \in ℝ^{n \times n}$ denotes the tensor product of ξ and η. Thus there exists a plane of no distortion for F ∈ GL⁺(3) if and only if there exists a rank-one tensor $H \in ℝ^{n \times n}$ with

F^{T} F = C = {(11 + H)}^{T} \cdot (11 + H) .

Since Becker only considers planes of no distortion for pure shear deformations, we will assume from now on that F has the form

F = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})

(6)

with α > 1. Then clearly λ₂ = 1, hence there are two distinct planes of no distortion which can be determined by direct computation: for $x = {(x_{1}, x_{2}, x_{3})}^{T} \in ℝ^{3}$ with ∥x∥ = 1 we can use the equalities

x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = | | x | |^{2} \Rightarrow x_{3}^{2} = | | x | |^{2} - x_{1}^{2} - x_{2}^{2}

to find

| | F x | |^{2} = α^{2} x_{1}^{2} + \frac{1}{α^{2}} x_{2}^{2} + x_{3}^{2} = (α^{2} - 1) x_{1}^{2} + (\frac{1}{α^{2}} - 1) x_{2}^{2} + | | x | |^{2} .

Thus the equality ∥Fx∥ = ∥x∥ is equivalent to

\begin{array}{l} | | x | |^{2} = (α^{2} - 1) x_{1}^{2} + (\frac{1}{α^{2}} - 1) x_{2}^{2} + | | x | |^{2} \\ \Leftrightarrow (1 - \frac{1}{α^{2}}) x_{2}^{2} = (α^{2} - 1) x_{1}^{2} \Leftrightarrow x_{2}^{2} = α^{2} x_{1}^{2}, \end{array}

(7)

hence every $x \in ℝ^{3}$ with ∥Fx∥ = ∥x∥ is of the form x = (s, ±α s, t)^T with $s, t \in ℝ$ . Since the final directions of no distortion are the images of those vectors, they in turn have the form y = Ax = (±α s, s, t)^T. Therefore, for every shear deformation of the form (6) with α > 1, there are two initial planes of no distortion,

ℰ^{+} = {(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) : x_{2} = α \cdot x_{1}} and ℰ^{-} = {(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) : x_{2} = - α \cdot x_{1}},

(8)

as well as two final planes of no distortion

ℱ^{+} = {(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) : x_{1} = α \cdot x_{2}} and ℱ^{-} = {(\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) : x_{1} = - α \cdot x_{2}} .

(9)

To verify Becker’s claim that [19, p. 339] ‘[in] a finite shearing strain of ratio α, it is easy to see that the normal to the planes of no distortion makes an angle with the contractile axis of shear the cotangent of which is α’, we consider the direction of the contractile axis along the eigenvector (0, 1, 0)^T = e₂ corresponding to the smallest eigenvalue 1/α. Then the cotangent of the angle between this axis and a vector $\hat{n} = {(y_{1}, y_{2}, y_{3})}^{T}$ with y₁ ≠ 0 is given by y₂/y₁. Since

n^{+} = {(1, - α, 0)}^{T} and n^{-} = {(1, α, 0)}^{T}

(10)

are normal vectors to the planes of no distortion $ℱ^{+}$ and $ℱ^{-}$ , respectively, the cotangent of the angle between these normals and the contractile axis is indeed given by ±α. Similarly, it is easy to see that the normals to the initial planes of no distortion $ℰ^{+}$ and $ℰ^{-}$ form angles of cotangent ±1/α with the contractile axis.

This definition of the planes of no distortion is also consistent with the definition by means of the shear ellipsoid⁵ given by Leith⁶ in Structural geology [34]:

In a strain ellipsoid with three unequal principal axes there are only two cross-sections which are circular in outline […] These planes […] are called ‘planes of no distortion’ because they preserve a circular cross-section similar to a section of the original sphere…

The strain ellipsoid of the deformation with principal stretches α, α⁻¹ and 1 is defined by the equation

\frac{x_{1}^{2}}{α^{2}} + \frac{x_{2}^{2}}{α^{- 2}} + x_{3}^{2} = 1,

where x₁ and x₂ denote coordinates with respect to the tensile axis and the contractile axis respectively. The planes $ℱ^{+}$ and $ℱ^{-}$ are characterized by the equation x₁ = ±α x₂, thus we can verify that their intersections with the ellipsoid are indeed circles of radius 1 centred at the origin (0, 0, 0)^T:

\begin{array}{l} \frac{x_{1}^{2}}{α^{2}} + \frac{x_{2}^{2}}{α^{- 2}} + x_{3}^{2} & = & 1 \land x_{1} = \pm α x_{2} \\ \Rightarrow ‖ (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) - (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) ‖ & = & x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = x_{1}^{2} + x_{2}^{2} + 1 - \frac{x_{1}^{2}}{α^{2}} - α^{2} x_{2}^{2} \\ = & = x_{1}^{2} + x_{2}^{2} + 1 - x_{2}^{2} - x_{1}^{2} = 1 . \end{array}

2.2.2. Different characterizations of the planes of no distortion

The initial planes of no distortion can also be characterized in a number of different ways, for example by means of the cofactor matrix Cof F: if the restriction of F to a plane is a rotation, then

〈 F x, F y 〉 = 〈 x, y 〉

for all x, y in the plane. Then for all x, y in the plane with 〈x, y〉 = 0 we find

| | F x \times F y | |^{2} = | | F x | |^{2} \cdot | | F y | |^{2} - 2 \cdot 〈 F x, F y 〉 = | | x | |^{2} \cdot | | y | |^{2} - 2 \cdot 〈 x, y 〉 = | | x | |^{2} \cdot | | y | |^{2} .

Now let n denote a unit normal vector to the plane of no distortion. Then n can be represented as n = x × y with unit vectors x, y in the plane and 〈x, y〉 = 0. Since, in general,

(Cof F) (x \times y) = F x \times F y,

we find

| | (Cof F) n | |^{2} = | | (Cof F) (x \times y) | |^{2} = | | F x \times F y | |^{2} = | | x | |^{2} \cdot | | y | |^{2} = 1 .

Therefore the equality ∥(Cof F)n∥ = 1 is a necessary condition for a unit normal vector n to be the normal to an initial plane of no distortion. We compute

Cof F = \det (F) \cdot F^{- T} = α \cdot \frac{1}{α} \cdot {(\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix})

as well as

\begin{array}{l} | | (Cof F) n | |^{2} & = & \frac{n_{1}^{2}}{α^{2}} + α^{2} n_{2}^{2} + n_{3}^{2} = \frac{n_{1}^{2}}{α^{2}} + α^{2} n_{2}^{2} + 1 - n_{1}^{2} - n_{2}^{2} \\ = & 1 + (\frac{1}{α^{2}} - 1) n_{1}^{2} + (α^{2} - 1) n_{2}^{2} = 1 + (\frac{1}{α^{2}} - 1) (n_{1}^{2} - α^{2} n_{2}^{2}), \end{array}

thus

| | (Cof F) n | | = 1 \Leftrightarrow 0 = (\frac{1}{α^{2}} - 1) (n_{1}^{2} - α^{2} n_{2}^{2}) \Leftrightarrow n_{1}^{2} = α^{2} n_{2}^{2}

for α > 1, showing that the cotangent of the angle between n and the contractile axis is ±1/α. Therefore the equality ∥(Cof F)n∥ = ∥n∥ = 1 holds only if n is a unit normal vector to a plane of no distortion. Note that this equality also implies

0 = 〈 (Cof F) n, (Cof F) n 〉 - 〈 n, n 〉 = 〈 {(Cof F)}^{T} (Cof F) n - n, n 〉 = 〈 (Cof B - 11) n, n 〉,

where B = FF^T denotes the left Cauchy-Green deformation tensor.

Becker gives another important characterization of the planes of no distortion: ‘the planes of no distortion […] are also the planes of maximum tangential strain’ [19, p. 339]. In Becker’s Finite homogeneous strain, flow and rupture of rocks [28], the tangential strain of $x \in ℝ^{3}$ is defined as the tangent of the angle between x and Fx. Accordingly, the plane of maximum tangential strain of a pure shear deformation F is defined as the plane containing the undistorted axis as well as the line for which the tangential strain is maximal, c.f. Figure 6.

Figure 6.

The shear ellipsoid of a deformation F, showing an arbitrary tangential strain $\tan \tilde{ϑ}$ (left) and the maximum tangential strain tan ϑ (right), which is realized for the plane of no distortion.

We will show that the planes of maximum tangential strain and the initial planes of no distortion are, in fact, identical. Let x be of the form x = (x₁, x₂, 0)^T with $x_{1}^{2} + x_{2}^{2} = 1$ ; in other words, we assume that x is a unit vector orthogonal to the undistorted e₃-axis. Then the angle ϑ between x and Fx is given by

\cos ϑ = \frac{〈 x, F x 〉}{| | x | | \cdot | | F x | |} = \frac{〈 x, F x 〉}{| | F x | |} .

The direction x of maximum tangential strain is characterized by the angle ϑ for which tan ϑ is maximal. In order to find x it is therefore sufficient to maximize

ϑ = \arccos \frac{〈 x, F x 〉}{| | F x | |} = \arccos \frac{α x_{1}^{2} + \frac{1}{α} x_{2}^{2}}{\sqrt{α^{2} x_{1}^{2} + \frac{1}{α^{2}} x_{2}^{2}}},

Figure 7.

Another visualization of the tangential strain: the shift between two parallel vertical lines under the deformation is maximal along the plane of no distortion.

since ϑ ↦ tan ϑ is monotone. Using the equality $x_{1}^{2} = 1 - x_{2}^{2}$ , we obtain

\arccos \frac{〈 x, F x 〉}{| | F x | |} = \arccos \frac{α (1 - x_{2}^{2}) + \frac{1}{α} x_{2}^{2}}{\sqrt{α^{2} (1 - x_{2}^{2}) + \frac{1}{α^{2}} x_{2}^{2}}} .

In order to find t ∈ [−1, 1] for which the function

f : [- 1, 1] \to ℝ, f (t) = \arccos \frac{α (1 - t^{2}) + \frac{1}{α} t^{2}}{\sqrt{α^{2} (1 - t^{2}) + \frac{1}{α^{2}} t^{2}}}

attains its maximum, we compute the first derivative of f to be

\frac{d}{d t} f (t) = \frac{(α^{2} t^{2} - 1) + t^{2}}{α t (t^{2} - 1)} \cdot \frac{\sqrt{\frac{{(α^{2} - 1)}^{2} t^{2} (t^{2} - 1)}{α^{4} (t^{2} - 1) - t^{2}}}}{\sqrt{α^{2} - \frac{(α^{4} - 1) t^{2}}{α^{2}}}} .

Thus the possible extremal points of f are 0, ±1 and $\pm α / \sqrt{1 + α^{2}}$ . Since

f (\frac{α}{\sqrt{1 + α^{2}}}) = f (- \frac{α}{\sqrt{1 + α^{2}}})

as well as f(−1) = f(0) = f(1) = 0 and f(t) ≥ 0 for all t ∈ [−1, 1], the global maxima of f are given by $t = \pm α / \sqrt{1 + α^{2}}$ . Applying this result to our original problem, we find that the maximum tangential strain is attained for the directions

{\hat{x}}_{\pm} = {(\sqrt{1 - \frac{α^{2}}{1 + α^{2}}}, \pm \frac{α}{\sqrt{1 + α^{2}}}, 0)}^{T} .

Finally, in order to verify Becker’s claim that ‘the planes of no distortion […] are also the planes of maximum tangential strain’ [19, p. 339], we observe that

\begin{array}{l} | | F {\hat{x}}_{\pm} | |^{2} = {| | (\begin{matrix} α \cdot \sqrt{1 - \frac{α^{2}}{1 + α^{2}}} \\ \frac{1}{α} \cdot (\pm \frac{α}{\sqrt{1 + α^{2}}}) \\ 0 \end{matrix}) | |}^{2} & = & α^{2} \cdot (1 - \frac{α^{2}}{1 + α^{2}}) + \frac{1}{1 + α^{2}} \\ = & α^{2} + \frac{1 - α^{4}}{1 + α^{2}} = α^{2} + 1 - α^{2} = 1 = | | {\hat{x}}_{\pm} | | . \end{array}

Thus the plane of maximum tangential strain is indeed the initial plane of no distortion.

Figure 8.

Simple glide deformation with shear angle tan φ = γ/1; horizontal lines slide relative to each other, vertical lines tilt to accommodate; originally right angles are distorted, the shear strain measures the change of angles.

2.2.3. Simple glide deformation

A homogeneous deformation F is called a simple glide if it has the form [35]

F = (\begin{matrix} 1 & γ & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}),

with $γ \in ℝ$ , γ > 0. Since F · (t, 0, s)^T = (t, 0, s)^T for all $t, s \in ℝ$ , the restriction of F to the e₁, e₃-plane is the identity function, showing that it is the initial plane of no distortion as well as the final plane of no distortion.

We examine the right Cauchy-Green deformation tensor of a simple glide, which is given by

C = F^{T} F = (\begin{matrix} 1 & γ & 0 \\ γ & 1 + γ^{2} & 0 \\ 0 & 0 & 1 \end{matrix}) .

The principal stretches λ_i are the singular values of F which, in turn, are the square roots of the eigenvalues of C,

λ_{1} = \frac{1}{2} (γ + \sqrt{γ^{2} + 4}), λ_{2} = \frac{1}{2} (γ + \sqrt{γ^{2} - 4}), λ_{3} = 1,

and the principal axes are given by the corresponding eigenvectors of C:

v_{1} = (\begin{matrix} 2 \\ γ + \sqrt{γ^{2} + 4} \\ 0 \end{matrix}), v_{2} = (\begin{matrix} - 2 \\ γ + \sqrt{γ^{2} - 4} \\ 0 \end{matrix}), v_{1} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .

Since 1 < λ₁ = 1/λ₂ as well as λ₃ = 1, we can interpret the simple glide as a rotated pure shear with shear ratio α = λ₁, where the direction of the tensile axis is v₁ and the contractile axis is given by v₂. The cotangent of the angle ϑ between v₂ and the e₁-axis is

\cot (ϑ) = \frac{- 2}{γ + \sqrt{γ^{2} - 4}} = \frac{- 2}{2 \cdot λ_{2}} = - \frac{1}{λ_{2}} = - α .

Hence the angle between (1, 0, 0)^T = e₁ and the contractile axis is

\cot (- ϑ) = - \cot (ϑ) = α,

yielding another angular characterization of the shear ratio α.

2.3. Dilation

Similarly to the connection between shear stress and shear stretches, Becker further assumes that ‘dilational forces acting positively and equally in all directions’[19, p. 339] are ‘[the loads] effecting dilation’ [19, p. 342], in other words, that purely volumetric initial stresses correspond to purely dilational stretches. More precisely, this assumption can be stated in the following way: if the principal axes remain fixed, every Biot stress of the form T^Biot = a · 11 with $a \in ℝ$ corresponds to a deformation of the form F =λ · 11 with λ > 0. Again, this assumption is stated only implicitly.

2.4. Superposition

Becker’s most important assumption is that of a law of superposition for coaxial deformations: ‘the load sums correspond to the products of the strain ratios’ [19, p. 341]. For homogeneous, coaxial stretch tensors U₁, U₂ this law can be stated as

T^{Biot} (U_{1} \cdot U_{2}) = T^{Biot} (U_{1}) + T^{Biot} (U_{2}),

where T^Biot(U) denotes the Biot stress tensor corresponding to the right Biot stretch tensor U. Here, U₁ and U₂ are called coaxial if their principal axes coincide.

2.4.1. The decomposition of stresses and strains

An important application of the law of superposition involves the additive decomposition of stresses and the corresponding multiplicative decomposition of stretches. Consider an initial load (i.e. a Biot stress tensor) of the form

T_{1}^{Biot} = (\begin{matrix} P & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

In the case of isotropic materials, the corresponding stretches are identical for all directions orthogonal to the x₁-axis for symmetry reasons, which is the case if and only if the stretch tensor corresponding to T^Biot has the form

U_{1} = (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & b \end{matrix})

for some $a, b \in ℝ^{+}$ with respect to the principal axes. Then, for h₁ = (ab²)^1/3 and $p = {(\frac{a}{b})}^{1 / 3}$ , we find

\begin{array}{l} U_{1} & = & (\begin{matrix} {(a b^{2} \cdot \frac{a^{2}}{b^{2}})}^{\frac{1}{3}} & 0 & 0 \\ 0 & {(a b^{2} \cdot \frac{b}{a})}^{\frac{1}{3}} & 0 \\ 0 & 0 & {(a b^{2} \cdot \frac{b}{a})}^{\frac{1}{3}} \end{matrix}) \\ = & h_{1} \cdot (\begin{matrix} p^{2} & 0 & 0 \\ 0 & \frac{1}{p} & 0 \\ 0 & 0 & \frac{1}{p} \end{matrix}) = (\begin{matrix} h_{1} & 0 & 0 \\ 0 & h_{1} & 0 \\ 0 & 0 & h_{1} \end{matrix}) \cdot (\begin{matrix} p & 0 & 0 \\ 0 & \frac{1}{p} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} p & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{p} \end{matrix}) . \end{array}

In a similar way we can obtain the stretch tensors corresponding to the stresses

T_{2}^{Biot} = (\begin{matrix} 0 & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & 0 \end{matrix}) and T_{3}^{Biot} = (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & R \end{matrix})

respectively:

\begin{array}{l} U_{2} = h_{2} \cdot (\begin{matrix} \frac{1}{q} & 0 & 0 \\ 0 & q^{2} & 0 \\ 0 & 0 & \frac{1}{q} \end{matrix}) = (\begin{matrix} h_{2} & 0 & 0 \\ 0 & h_{2} & 0 \\ 0 & 0 & h_{2} \end{matrix}) \cdot (\begin{matrix} \frac{1}{q} & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & \frac{1}{q} \end{matrix}), \\ U_{3} = h_{3} \cdot (\begin{matrix} \frac{1}{r} & 0 & 0 \\ 0 & \frac{1}{r} & 0 \\ 0 & 0 & r^{2} \end{matrix}) = (\begin{matrix} h_{3} & 0 & 0 \\ 0 & h_{3} & 0 \\ 0 & 0 & h_{3} \end{matrix}) \cdot (\begin{matrix} \frac{1}{r} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & r \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{1}{r} & 0 \\ 0 & 0 & r \end{matrix}) \end{array}

for some $h_{2}, h_{3}, q, r \in ℝ^{+}$ . Therefore every homogeneous deformation corresponding to a uniaxial load can be decomposed into a dilation and two shear deformations with perpendicular axes. Finally, for the general case of arbitrary stresses

\begin{array}{l} T^{Biot} = (\begin{matrix} P & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & R \end{matrix}) & = & (\begin{matrix} P & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) + (\begin{matrix} 0 & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & 0 \end{matrix}) + (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & R \end{matrix}) \\ = & T_{1}^{Biot} + T_{2}^{Biot} + T_{3}^{Biot}, \end{array}

the law of superposition yields the general formula

U = U_{1} \cdot U_{2} \cdot U_{3} = h_{1} h_{2} h_{3} \cdot (\begin{matrix} p^{2} & 0 & 0 \\ 0 & \frac{1}{p} & 0 \\ 0 & 0 & \frac{1}{p} \end{matrix}) \cdot (\begin{matrix} \frac{1}{q} & 0 & 0 \\ 0 & q^{2} & 0 \\ 0 & 0 & \frac{1}{q} \end{matrix}) \cdot (\begin{matrix} \frac{1}{r} & 0 & 0 \\ 0 & \frac{1}{r} & 0 \\ 0 & 0 & r^{2} \end{matrix}) = h_{1} h_{2} h_{3} \cdot (\begin{matrix} \frac{p^{2}}{q r} & 0 & 0 \\ 0 & \frac{q^{2}}{p r} & 0 \\ 0 & 0 & \frac{r^{2}}{p q} \end{matrix})

for the stretch tensor U corresponding to T^Biot. Then U can be decomposed into a dilation and two shears as well:

U = (\begin{matrix} h_{1} h_{2} h_{3} & 0 & 0 \\ 0 & h_{1} h_{2} h_{3} & 0 \\ 0 & 0 & h_{1} h_{2} h_{3} \end{matrix}) \cdot (\begin{matrix} \frac{p^{2}}{q r} & 0 & 0 \\ 0 & \frac{q r}{p^{2}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{p q}{r^{2}} & 0 \\ 0 & 0 & \frac{r^{2}}{p q} \end{matrix}) .

Furthermore we can find an additive decomposition

\begin{array}{l} T^{Biot} & = & (\begin{matrix} \frac{P + Q + R}{3} & 0 & 0 \\ 0 & \frac{P + Q + R}{3} & 0 \\ 0 & 0 & \frac{P + Q + R}{3} \end{matrix}) \\ + (\begin{matrix} - \frac{Q + R - 2 P}{3} & 0 & 0 \\ 0 & \frac{Q + R - 2 P}{3} & 0 \\ 0 & 0 & 0 \end{matrix}) + (\begin{matrix} 0 & 0 & 0 \\ 0 & \frac{P + Q - 2 R}{3} & 0 \\ 0 & 0 & - \frac{P + Q - 2 R}{3} \end{matrix}) \end{array}

(11)

of T^Biot into a spherical stress and two pure shear stresses. Note that the decomposition of U can be found in Becker’s second table [19, p. 340] while the decomposition of T^Biot can be found in the third table [19, p. 340].

In decomposing the strains U₁, U₂ and U₃ into a dilation and two shear strains, the planes of shear were chosen arbitrarily for every strain (or, more precisely, such that the two resulting shear ratios were identical). However, we may also choose two fixed planes (or, equivalently, fixed axes of tension and contraction) and decompose all strains into shears along the same axes:

\begin{array}{l} U_{1} = h_{1} \cdot (\begin{matrix} p^{2} & 0 & 0 \\ 0 & \frac{1}{p} & 0 \\ 0 & 0 & \frac{1}{p} \end{matrix}) = h_{1} \cdot (\begin{matrix} p^{2} & 0 & 0 \\ 0 & \frac{1}{p^{2}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & \frac{1}{p} \end{matrix}), \\ U_{2} = h_{2} \cdot (\begin{matrix} \frac{1}{q} & 0 & 0 \\ 0 & q^{2} & 0 \\ 0 & 0 & \frac{1}{q} \end{matrix}) = h_{2} \cdot (\begin{matrix} \frac{1}{q} & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & \frac{1}{q} \end{matrix}), \\ U_{3} = h_{3} \cdot (\begin{matrix} \frac{1}{r} & 0 & 0 \\ 0 & \frac{1}{r} & 0 \\ 0 & 0 & r^{2} \end{matrix}) = h_{3} \cdot (\begin{matrix} \frac{1}{r} & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{1}{r^{2}} & 0 \\ 0 & 0 & r^{2} \end{matrix}) . \end{array}

Note that the axes chosen here are identical to those chosen for the decomposition of U. Using this approach we obtain a modified version of Becker’s first table, see Table 1.

Table 1.

Modified version of Becker’s first table.

Active force	P			Q			R
Axis of strain	x	y	z	x	y	z	x	y	z
Dilation	h ₁	h ₁	h ₁	h ₂	h ₂	h ₂	h ₃	h ₃	h ₃
ear	p ²	$\frac{1}{p^{2}}$	1	$\frac{1}{q}$	q	1	$\frac{1}{r}$	r	1
Shear	1	p	$\frac{1}{p}$	1	q	$\frac{1}{q}$	1	$\frac{1}{r^{2}}$	r ²

Figure 9.

Load and deformation of the uniaxial case can be decomposed into two finite shear and a dilational mode.

2.4.2. The uniaxial case

Consider, again, a uniaxial initial stress of the form

T^{Biot} = (\begin{matrix} 0 & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & 0 \end{matrix}) .

Then T^Biot can be decomposed into

T^{Biot} = \underset{= : T_{1}^{Biot}}{\underset{︸}{\frac{Q}{3} (\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix})}} + \underset{= : T_{2}^{Biot}}{\underset{︸}{\frac{Q}{3} (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix})}} + \underset{= : T_{3}^{Biot}}{\underset{︸}{\frac{Q}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})}} .

By Becker’s assumption the two shear stresses $T_{1}^{Biot}$ and $T_{2}^{Biot}$ correspond to shear strains

U_{1} = (\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix}) and U_{2} = (\begin{matrix} 1 & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})

respectively, while the volumetric stress $T_{3}^{Biot}$ corresponds to a dilational strain

U_{3} = (\begin{matrix} h & 0 & 0 \\ 0 & h & 0 \\ 0 & 0 & h \end{matrix}) .

Then, according to the law of superposition, the strain corresponding to T^Biot is given by

U = U_{1} \cdot U_{2} \cdot U_{3} = (\begin{matrix} \frac{h}{α} & 0 & 0 \\ 0 & h α^{2} & 0 \\ 0 & 0 & \frac{h}{α} \end{matrix}) .

The resulting principal stretch h α² in the direction of the applied load is referred to by Becker as the ‘length of the strained mass’ [19, p. 343] in his further computations for the uniaxial case.

Now we consider another uniaxial load given by

{\hat{T}}^{Biot} = (\begin{matrix} P & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

Then ${\hat{T}}^{Biot}$ can be decomposed in a number of different ways. For example we could choose, for symmetry reasons, a decomposition similar to that of T^Biot, that is,

{\hat{T}}^{Biot} = \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix}) + \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix}) + \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}),

yielding the strain

\hat{U} = (\begin{matrix} \hat{α} & 0 & 0 \\ 0 & \frac{1}{\hat{α}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} \hat{α} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{\hat{α}} \end{matrix}) \cdot (\begin{matrix} \hat{h} & 0 & 0 \\ 0 & \hat{h} & 0 \\ 0 & 0 & \hat{h} \end{matrix}) = (\begin{matrix} \hat{h} {\hat{α}}^{2} & 0 & 0 \\ 0 & \frac{\hat{h}}{\hat{α}} & 0 \\ 0 & 0 & \frac{\hat{h}}{\hat{α}} \end{matrix}) .

It is also possible to decompose ${\hat{T}}^{Biot}$ into shears coaxial to $T_{1}^{Biot}$ and $T_{2}^{Biot}$ , that is,

{\hat{T}}^{Biot} = - \frac{2 P}{3} (\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) + \frac{P}{3} (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) + \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) .

Note that the resulting strain is independent of this choice: since

- \frac{2 P}{3} (\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) = \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix}) + \frac{P}{3} (\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix}),

the law of superposition yields

\hat{U} = (\begin{matrix} \hat{α} & 0 & 0 \\ 0 & \frac{1}{\hat{α}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} \hat{α} & 0 & 0 \\ 0 & \frac{1}{\hat{α}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & \hat{α} & 0 \\ 0 & 0 & \frac{1}{\hat{α}} \end{matrix}) \cdot (\begin{matrix} \hat{h} & 0 & 0 \\ 0 & \hat{h} & 0 \\ 0 & 0 & \hat{h} \end{matrix}) = (\begin{matrix} \hat{h} {\hat{α}}^{2} & 0 & 0 \\ 0 & \frac{\hat{h}}{\hat{α}} & 0 \\ 0 & 0 & \frac{\hat{h}}{\hat{α}} \end{matrix})

in this case as well.

2.4.3. Decomposition along fixed axes

The additive decomposition of the stress tensor into a volumetric stress and two shears along fixed axes can also be expressed in basic algebraic terms. We will identify the set of all diagonal matrices in $ℝ^{n \times n}$ with the Euclidean space $ℝ^{3}$ in the canonical way. Since the set

ℬ = {(\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ - 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})},

which corresponds to two shear stresses and a volumetric stress in diagonal form, is a basis of $ℝ^{3}$ , every

(\begin{matrix} P \\ Q \\ R \end{matrix}) \in ℝ^{3}

can be written as

(\begin{matrix} P \\ Q \\ R \end{matrix}) = A (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}) + B (\begin{matrix} 0 \\ - 1 \end{matrix}) + C (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} - 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & - 1 & - 1 \end{matrix}) \cdot (\begin{matrix} A \\ B \\ C \end{matrix}) .

(12)

For given $P, Q, R \in ℝ$ we can therefore find the coefficients A, B, C of the decomposition

(\begin{matrix} P & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & R \end{matrix}) = A (\begin{matrix} - 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 0 \end{matrix}) + B (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) + C (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

by rewriting equation (12) to read

(\begin{matrix} A \\ B \\ C \end{matrix}) = {(\begin{matrix} - 1 & 0 & 1 \\ 1 & 1 & - 1 \\ 0 & - 1 & 1 \end{matrix})}^{- 1} \cdot (\begin{matrix} P \\ Q \\ R \end{matrix}) = \frac{1}{3} (\begin{matrix} - 2 & 1 & 1 \\ 1 & 1 & - 2 \\ 1 & 1 & 1 \end{matrix}) \cdot (\begin{matrix} P \\ Q \\ R \end{matrix}) = \frac{1}{3} (\begin{matrix} - 2 P + Q + R \\ P + Q - 2 R \\ P + Q + R \end{matrix}) .

The decomposition obtained in this way is the same as given in (11).

3. Geometry and static analysis of finite shear deformation

In this section we investigate the geometric and static motivation of the relation between shear deformations and shear stresses assumed by Becker. The unit cube Ω₀ aligned to the orthogonal coordinate system e₁, e₂, e₃ in the reference configuration is considered again. We distinguish six variants of orthogonal deformation along the edges of the unit cube such that one edge preserves its length; see Figure 10.

Figure 10.

Six variants of plane, orthogonal deformation along the edges of a unit cube.

Specifying the deformation Φ_α:Ω₀ → Ω₁ as pure finite shear with ratio α, the volume in the actual configuration Ω₁ is invariant. Thus, stretching one edge in the plane of deformation by α must result in a contraction 1/α of the corresponding edge: see also Figure 11. The volume invariance holds for any intermediate configuration Ω_i and goes along with the multiplicative and nonlinear behaviour of finite deformation. We consider the intermediate configuration defined by the symmetry condition

Φ_{\sqrt{α}} : Ω_{0} \mapsto Ω_{i}, Φ_{\sqrt{α}} : Ω_{i} \mapsto Ω_{1} .

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Figure 11.

Finite shear of a unit cube.

Let us inscribe a circle and a square into Ω₀ as drawn on the left-hand side of Figure 11. Then $Φ_{\sqrt{α}}$ deforms the circle into an ellipse⁷ and the square into a rhombus. Increasing values for α decrease the angle ψ between the normal vector

\bar{n} = \frac{1}{\sqrt{α + 1 / α}} (\begin{matrix} \sqrt{α} \\ 1 / \sqrt{α} \end{matrix}) = \frac{1}{\sqrt{α^{2} + 1}} (\begin{matrix} α \\ 1 \end{matrix})

(14)

and the horizontal direction e₁; see Figure 12.

Figure 12.

Inclination ψ of the normal $\bar{n}$ in the intermediate configuration Ω_i.

Becker denotes the scalar product 〈n, e_i〉 between the normal n of a plane and the coordinate axis e_i as ‘direction cosines’ n_i of the plane [19, p. 338]. For the normal $\bar{n}$ we find

n_{1} = 〈 \bar{n}, e_{1} 〉 = \sqrt{α^{2} + 1}, n_{2} = 〈 \bar{n}, e_{2} 〉 = \sqrt{1 + 1 / α^{2}} \Rightarrow n_{1} = α n_{2};

(15)

thus the inclination ψ of the normal $\bar{n}$ is given by

\cot ψ = \frac{n_{1}}{n_{2}} = α .

(16)

Since here the direction e₁ is the contractile axis, it follows from the calculations in Section 2.2.1 that $\bar{n}$ is normal to the plane of no distortion. The rhombus in Figure 12 therefore describes the two directions of no distortion. We explain this fact geometrically for the line perpendicular to $\bar{n}$ in Figure 13. Drawing this line onto the deformed body Ω₁ and then releasing the deformation leads to a rotated but undistorted line in Ω₀. Thus, the intermediate configuration Ω_i displays an important geometrical aspect in a pure shear deformation.

Figure 13.

The plane of no distortion preserves its length after release of the finite shear deformation.

These properties, Becker argues, suggest that the normal component of the Cauchy stress in the directions of no distortion has to vanish, that is, $N = 0$ , ‘for otherwise there could be no planes of zero distortion’ (8, p. 338). Combining this assumption with the equality n₁/n₂ =α from (15), it follows that the components of the principal Cauchy stresses need to fulfil

- σ_{1} α = \frac{σ_{2}}{α}, σ_{3} = 0,

(17)

where −σ₁α and σ₂/α are the resultant loads; note that here, e₁ is the contractile axis. From the principal Cauchy stresses in equation (17) he infers that the action of ‘two equal loads […] of opposite signs at right angles to one another’ [19, p. 339] must correspond to a simple finite shear.⁸

Next, we discuss one of the six variants in Figure 10 without loss of generality. The deformation in Figure 14 results in principal stresses of horizontal and vertical direction. Defining the Cauchy stress as load per actual area, the quantity −σ₁α represents a compressive force in the horizontal axis and σ₂/α is a tensional force in the vertical direction. Equation (17) yields that these forces are of the same amount, say Q. Then, for a static analysis, we cut the body in Ω₁ by a plane of no distortion. The lower (respectively, upper) part of the body is in an equilibrium balanced by

T = Q (\begin{matrix} - 1 / α \\ 1 \end{matrix}); respectively T = Q (\begin{matrix} 1 - 1 + 1 / α \\ - 1 \end{matrix}) = Q (\begin{matrix} 1 / α \\ - 1 \end{matrix}) .

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Figure 14.

Static analysis in the actual configuration along the cut of a plane of no distortion.

The scalar product of $T$ with $\bar{n}$ from equation (14) vanishes and thus, the equilibrium is free of normal components in the plane of no distortion as required; see Figure 14.

Becker concludes that the six variants of simple finite shearing in Figure 10 are caused by pairwise forces as sketched in Figure 15. This is in concordance with one of five possible invariance conditions for the definition of pure shear [36], namely that the shear tensor is a planar deviator. Note that in the case of isotropic material it is obvious to claim that P = Q = R.

Figure 15.

Six variants of finite pure shear and the corresponding loads.

3.1. Mohr’s stress circle for finite shear

In this section we discuss the plane of no distortion in the context of Mohr’s stress circle.⁹ The Mohr circle is a graphical method to determine the stress components acting on a plane rotated against the coordinate system. Given any symmetric stress tensor in $ℝ^{2 \times 2}$ , Mohr’s stress circle allows for a graphical solution of the spectral decomposition. Vice versa, knowing Mohr’s stress circle and its spectral decomposition, the tensor components are graphically given for arbitrary but orthogonal coordinate systems, for example for a coordinate system in alignment with the plane of no distortion.

It is important to note that the plane of maximal shear stress ρ_max does not coincide with the plane of no distortion for finite shear. Considering the loading in Figure 14 together with equation (17) results in the principal stresses

σ_{1} = - \frac{Q}{α}, σ_{2} = Q α .

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Therefore, Mohr’s stress circle for the finite pure shear loading is drawn in Figure 16.

Figure 16.

Mohr’s circle for Cauchy stresses in simple finite shear loading.

Becker introduces the quantity 2s = α − 1/α as the ‘amount of the shear’ [28]. Thus, the centre of the circle is shifted from the origin of the axis by

σ_{m} = \frac{σ_{1} + σ_{2}}{2} = \frac{Q}{2} (α - \frac{1}{α}) = Q s .

(20)

For α > 1 the angle ϑ = π/4 pointing to the plane of maximum shear stress ρ_max does not coincide with the angle ψ describing the plane of no distortion. Demanding σ_ξ = 0 in the plane of no distortion determines the rotation angle of the ξ, η coordinate system by

ψ = \frac{1}{2} \arccos (\frac{α^{2} - 1}{α^{2} + 1}) = arccot (α),

(21)

which is in accordance with equation (16). For ψ = arccot(α) the stress components are given by

σ_{ξ} = 0, σ_{η} = Q \frac{α^{2} - 1}{α}, σ_{ξ η} = ρ_{0} = Q .

(22)

Thus, in the plane of no distortion the shear stress ρ₀ is a simple function of the loading Q and independent of α.

3.2. Tangential force and failure criteria in finite shear

To discuss the tangential force in the spirit of continuum mechanics, we analyse the equilibrium of stresses in the vicinity of a point. In case of simple finite shear the surrounding of a point is a circle in Ω₀ (respectively, a shear ellipse in Ω₁); see Figure 11. Considering the diameter of the circle to be 1, the diameter r of the ellipse is given in terms of α and the direction cosines n₁, n₂ from equation (15) by

\frac{1}{r^{2}} = α^{2} n_{2}^{2} + \frac{n_{1}^{2}}{α^{2}} .

(23)

Considering the finite shear loading from Figure 14 with principal Cauchy stresses from equations (19) gives a resultant force $ℛ$ with constant magnitude¹⁰ Q on any line cutting the ellipsis through the mid-point

ℛ^{2} = r^{2} | | σ n | |^{2} = r^{2} {| | (\begin{matrix} - \frac{Q}{α} n_{1} \\ Q α n_{2} \end{matrix}) | |}^{2} = r^{2} Q^{2} (\frac{n_{1}^{2}}{α^{2}} + α^{2} n_{2}^{2}) = Q^{2} .

(24)

The normal n of the line defines the direction cosines n₁ and n₂ as explained in equation (15). Next, we investigate the magnitude of $ℛ$ normal to the line, which is given by

N^{2} = r^{2} {〈 σ n, n 〉}^{2} = r^{2} {〈 (\begin{matrix} - \frac{Q}{α} n_{1} \\ Q α n_{2} \end{matrix}), (\begin{matrix} n_{1} \\ n_{2} \end{matrix}) 〉}^{2} = r^{2} Q^{2} {(\frac{- n_{1}^{2}}{α} + α n_{2}^{2})}^{2} .

(25)

The magnitude of stress $T$ tangential to the plane follows from the Pythagorean theorem by

T^{2} = ℛ^{2} - N^{2} .

(26)

Thus the plane of maximal tangential force is due to $N^{2} = 0$ and we conclude from equation (25)

\frac{- n_{1}^{2}}{α} + α n_{2}^{2} = 0 \Leftrightarrow n_{1}^{2} = n_{2}^{2} α^{2} .

(27)

The planes of no distortion fulfil equation (27) and attend Becker’s discussion on failure criteria:¹¹ ‘Rupture by shearing is determined by maximum tangential load, not [Cauchy] stress’ [19, p. 339]; see Appendix B.4. In the present example the maximum shear stress appears in a cut inclined at ψ = 45° to the horizontal axis which does not, however, maximize the tangential load. The situation is illustrated in Figure 17.

Figure 17.

Equilibrium and resultant $ℛ$ (respectively $\bar{ℛ}$ ) on lines cutting the finite shear ellipsis; note that ${\bar{ℛ}}^{2} > T^{2}$ .

Note that the shear stress ρ₀ = Q/1 = Q in the plane of no distortion is lower than the Tresca and the von Mises stresses for this kind of loading:

σ_{y}^{Tresca} = Q (α + 1 / α),

(28)

σ_{y}^{Mises} = Q \sqrt{α^{2} + 1 + 1 / α^{2}},

(29)

σ_{ξ η} = σ_{y}^{Becker} = Q .

(30)

Thus, the inequality $σ_{y}^{Becker} < σ_{y}^{Mises} < σ_{y}^{Tresca}$ for α > 0 shows that the tangential force in the plane of no distortion represents a more conservative lower bound as failure criterion.

It is worth mentioning that the Tresca and the von Mises stresses would account for both the loading Q and the deformation α. However, decoupling the failure criteria from the deformation suggests a basic model, which is also simple from an experimental point of view.

4. The axiomatic approach

Our aim in this section is to formulate Becker’s stress–stretch relation for ideally elastic, isotropic materials in terms of the Biot stress tensor T^Biot and the right Biot stretch tensor $U = \sqrt{F^{T} F}$ , in other words, to find a stress response function U ↦ T^Biot (U) or an inverse response function T^Biot ↦ U(T^Biot) which fulfils Becker’s assumptions listed in Section 2. In order to deduce such a law of ideal elasticity we will introduce a number of axioms corresponding to Becker’s assumptions to uniquely characterize this stress–stretch relationship. In this we will, at first, closely follow Becker’s approach, utilizing all of the given axioms. However, we will later show that the same results can be deduced with only a subset of those assumptions. Furthermore, in order to obtain a more general result, we will formulate our computations in terms of an arbitrary stress tensor instead of only the Biot stress.

4.1. The basic axioms for an isotropic stress–stretch relation

Let $T : PSym (3) \to Sym (3), ℰ \mapsto T (ℰ)$ denote a matrix function which maps the set of positive definite symmetric matrices PSym (3) to the set of symmetric matrices Sym (3). In accordance with our interpretation of T as a stress response function relation we will refer to the argument $ℰ$ as the stretch and to $T (ℰ)$ as the stress tensor. Furthermore, if the mapping is invertible we will, for given $\hat{T} \in Sym (3)$ , often write $ℰ (\hat{T})$ to denote the unique $ℰ \in PSym (3)$ with $\hat{T} = T (ℰ)$

Of course we also assume that T fulfils the axioms stated in this section

The first two axioms are common postulates for an arbitrary stress–stretch relation

Axiom 0.1: Continuous stress response function

The function $T : PSym (3) \to Sym (3), ℰ \mapsto T (ℰ)$ is continuous.

Note that this is a weakened version of Becker’s assumption that the function relating stretch and stress is analytic.

Axiom 0.2: Unique stress-free reference configuration

The equivalence

T (ℰ) = (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = 0 \Leftrightarrow ℰ = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) = 11

holds.

Axiom 0.2 states that the undeformed (and possibly rotated) reference configuration is the only stress-free configuration. Again, this is a weakened version of one of Becker’s assumptions, namely that the stress response function is invertible.

Another basic assumption is that of isotropy: the response of many materials can be idealized to be independent of the direction of applied stresses.

Axiom 0.3: Isotropy

The equality

T^{Biot} (Q^{T} ℰ Q) = Q^{T} T^{Biot} (ℰ) Q

holds for all Q ∈ O(3) and $ℰ \in PSym (3)$ .

Note that isotropy is sometimes defined for Q ∈ SO(3) only. However, since −Q ∈ SO(3) for Q ∈ O(3) \SO(3), even under this narrower definition we find

T (Q^{T} ℰ Q) = T ((- Q^{T}) ℰ (- Q)) = (- Q^{T}) T (ℰ) (- Q) = Q^{T} T (ℰ) Q

for all Q ∈ O(3) \SO(3) as well.

Since the assumption of isotropy implies that $ℰ$ and $T (ℰ)$ are coaxial [37, Theorem 4.2.4] for all $ℰ \in PSym (3)$ , Axiom 0.3 restricts the possible choices for the considered stretch–stress pair $ℰ, T$ to coaxial pairs of tensors. However, this axiom allows us to formulate the other axioms regarding strains and stresses in terms of the principal stretches and principal stresses only [38]:

For an isotropic nonlinear elastic solid, the principal directions of Cauchy stress σ must coincide with the axes of the Eulerian strain ellipsoid. Also, to fully specify the state of strain in a material element we need only know the three principal stretches λ_i relative to some reference configuration and the principal directions of strain. Thus, the constitutive law is completely determined once the relations between the principal components of Cauchy stress σ_i and principal stretches λ_i are known.

Since any given $ℰ \in PSym (3)$ is unitarily diagonalizable we can write $ℰ$ as

ℰ = Q^{T} (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}) Q

with Q ∈ O(3), where λ_i denote the (positive) eigenvalues of $ℰ$ . Then $T (ℰ)$ is given by

T (ℰ) = Q^{T} \cdot T (Q ℰ Q^{T}) \cdot Q = Q^{T} \cdot T ((\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix})) \cdot Q .

This allows us to focus on stretches given in a diagonal representation, that is, stretches of the form

ℰ = diag (λ_{1}, λ_{2}, λ_{3}) = (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix})

with $λ_{1}, λ_{2}, λ_{3} \in ℝ^{+}$ .

4.2. Becker’s three main axioms

The first two main axioms for our law of ideal isotropic elasticity refer to two special cases of deformation, pure shear and pure volumetric dilation.

Axiom 1: Pure shear stresses correspond to pure shear stretches

For every $α \in ℝ^{+}$ there exists $s \in ℝ$ such that

ℰ = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \Leftrightarrow T (ℰ) = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) .

If T is an isotropic tensor function (Axiom 0.3), then Axiom 1 can also be stated in a more general way. Let

ℰ = (\begin{matrix} B_{11} & B_{12} & 0 \\ B_{12} & B_{22} & 0 \\ 0 & 0 & 1 \end{matrix})

with

\det (\begin{matrix} B_{11} & B_{12} \\ B_{12} & B_{22} \end{matrix}) = 1 .

Then the two eigenvalues of

(\begin{matrix} B_{11} & B_{12} \\ B_{12} & B_{22} \end{matrix})

are mutually reciprocal, and thus $ℰ$ can be diagonalized to

ℰ = {(\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix})}^{T} \cdot (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix})

with

(\begin{matrix} Q_{11} & Q_{12} \\ Q_{12} & Q_{22} \end{matrix}) \in O (2)

and $α \in ℝ^{+}$ . Using Axiom 1 and Axiom 0.3 we compute

\begin{array}{l} T (ℰ) & = & {(\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix})}^{T} \cdot T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) \cdot (\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix}) \\ = & {(\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix})}^{T} \cdot (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) \cdot (\begin{matrix} Q_{11} & Q_{12} & 0 \\ Q_{12} & Q_{22} & 0 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} A_{11} & A_{12} & 0 \\ A_{12} & A_{22} & 0 \\ 0 & 0 & 1 \end{matrix}) \end{array}

with

tr (\begin{matrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{matrix}) = tr (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) = 0 .

In the isotropic case, Axiom 1 may therefore be equivalently stated as follows (see Figure 2 on p. 5): for every

(\begin{matrix} B_{11} & B_{12} \\ B_{12} & B_{22} \end{matrix}) \in SL (2)

there exists

(\begin{matrix} A_{11} & A_{12} \\ A_{12} & A_{22} \end{matrix}) \in s l (2)

such that

ℰ = (\begin{matrix} B_{11} & B_{12} & 0 \\ B_{12} & B_{22} & 0 \\ 0 & 0 & 1 \end{matrix}) \Leftrightarrow T (ℰ) = (\begin{matrix} A_{11} & A_{12} & 0 \\ A_{12} & A_{22} & 0 \\ 0 & 0 & 1 \end{matrix}),

where SL(n) denotes the group of all X ∈ GL(n) with det X = 1, and $s l (n)$ is the corresponding Lie algebra of all trace-free matrices in $ℝ^{n \times n}$ .

To further understand the relation between shear stress and shear stretch constituted by Axiom 1 we consider two examples. First, assume that the stress tensor T corresponding to a stretch tensor $ℰ$ is a trace-free pure shear stress of the form

T (ℰ) = (\begin{matrix} 0 & s & 0 \\ s & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

The eigenvalues of $T (ℰ)$ are the principal stresses T₁ = s, T₂ = −s, T₃ = 0, and thus $T (ℰ)$ can be diagonalized to

T (ℰ) = Q^{T} (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) Q

with Q ∈ SO(3). If $ℰ \mapsto T (ℰ)$ is a isotropic mapping satisfying Axiom 1, then

\begin{array}{l} Q T (ℰ) Q^{T} = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) & \Rightarrow & T (Q ℰ Q^{T}) = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) \\ \Rightarrow & Q ℰ Q^{T} = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \Rightarrow ℰ = Q^{T} (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) Q \end{array}

for some $α \in ℝ^{+}$ . Thus the stretch $ℰ$ corresponding to a pure shear stress is a pure shear of the form

ℰ = Q^{T} (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) Q = (\begin{matrix} B_{11} & B_{12} & 0 \\ B_{12} & B_{22} & 0 \\ 0 & 0 & 1 \end{matrix})

with

\det (\begin{matrix} B_{11} & B_{12} \\ B_{12} & B_{22} \end{matrix}) = 1 .

Now assume that a deformation gradient F is a simple glide of the form

F = (\begin{matrix} 1 & γ & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) .

Then the right Biot stretch tensor U has the form

U = \sqrt{F^{T} F} = {(\begin{matrix} 1 & γ & 0 \\ γ & γ^{2} + 1 & 0 \\ 0 & 0 & 1 \end{matrix})}^{\frac{1}{2}} .

Since 1 is an eigenvalue of U and det U = det F = 1, the remaining eigenvalues of U must be of the form λ₁ = α and λ₂ = 1/α. Thus the principal stretches of the deformation are λ₁ = α, λ₂ = 1/α, λ₃ = 1 for some $α \in ℝ^{+}$ and U can be diagonalized to

U = Q^{T} (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) Q

with Q ∈ O(3). Then, if the function U ↦ T(U) mapping U to a stress tensor T is isotropic and satisfies Axiom 1, we can compute

T (U) = T (Q^{T} (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) Q) = Q^{T} T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) Q = Q^{T} (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) Q

for some $s \in ℝ$ . Therefore the stress tensor T corresponding to a simple glide is a pure shear stress.

The second axiom relates spherical stresses, that is, purely normal stresses with the same magnitude in each direction, to volumetric stretches, that is, uniform stretches in all directions.

Axiom 2: Spherical stresses correspond to volumetric stretches

For every $λ \in ℝ^{+}$ there exists $a \in ℝ$ such that

ℰ = λ \cdot 11 = (\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}) \Leftrightarrow T (ℰ) = (\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}) = a \cdot 11 .

This relation between spherical stresses and volumetric stretches for isotropic elastic materials is highly intuitive: if the initial load is equal in every direction the resulting deformation should be equal in all directions as well and vice versa. However, whether this feature is true for all magnitudes of applied loads depends on the chosen constitutive framework. It is known that Axiom 2 is not satisfied for a number of well-known isotropic nonlinear elastic formulations, such as the Mooney–Rivlin energy and the Ogden type energy [26]. The loss of uniqueness of the symmetric solution is encountered in Rivlin’s cube problem [39].

While the first two axioms refer only to stresses and stretches of a specific diagonal form, our third and final axiom states a law of superposition that holds for all coaxial stress–stretch pairs. Recall that we call symmetric matrices A, B ∈ Sym(3) coaxial if their principal axes coincide, which is the case if and only if A and B are simultaneously diagonalizable as well as if and only if A and B commute.

Axiom 3: Law of superposition

Let $ℰ_{1}, ℰ_{2} \in PSym (3)$ be coaxial. Then

T (ℰ_{1} \cdot ℰ_{2}) = T (ℰ_{1}) + T (ℰ_{2}) .

Note that, for an invertible stress–stretch relation, the third axiom could equivalently be stated as

ℰ (T_{1} + T_{2}) = ℰ (T_{1}) \cdot ℰ (T_{2})

(31)

for all coaxial T₁,T₂ ∈ Sym(3).

This law of superposition can be summarized as follows: the (multiplicative) concatenation of stretch tensors should effect the (additive) superposition of the corresponding stress tensors. This nonlinear connection is closely related to a modern approach [40] involving the theory of Lie groups: the deformation tensors correspond to the (multiplicative) group GL⁺(3) while the stress tensors can be represented by the (linear) Lie algebra $g l (3)$ . By focusing on symmetric stresses T as well as on positive-definite symmetric stretches $ℰ$ we can also relate the stretch to the subset PSym (3) ⊂ GL⁺(3) and the stress to the subspace $Sym (3) \subset g l (3)$ . Since the canonical homomorphism mapping the additive structure of Sym(3) to the multiplicative group structure of PSym(3) in the way described by equation (31) is exponential function

\exp : Sym (3) \to PSym (3),

which is invertible with its inverse given by principal logarithm

\log : PSym (3) \to Sym (3),

it is to be expected that the resulting stress–stretch relation $ℰ \mapsto T (ℰ)$ is, in turn, logarithmic in nature.

This approach is also closely related to the later deduction of the quadratic Hencky strain energy by Heinrich Hencky, who employed a similar law of superposition to obtain a logarithmic stress response function. However, Hencky considered the superposition of stresses in the deformed configuration: the Cauchy stress σ in his 1928 article [2] as well as the Kirchhoff stress τ in his 1929 article [41] (an English translation of both papers can be found in [3]). Becker on the other hand assumes a law of superposition for initial loads: his law refers to the Biot stress tensor T^Biot. A more detailed comparison of Becker’s and Hencky’s work can be found in Appendix B.1.

4.3. Deduction of the general stress–stretch relation from the axioms

We will now show that the general stress–stretch relation is determined by the given axioms up to only two constitutive parameters. Since our law of elasticity is isotropic by assumption we will mostly consider deformations given in the diagonal form

F = (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}) .

Recall from Section 4.1 that the stress–stretch relation is uniquely determined by the stress response to such deformations.

4.3.1. Basic properties

Before we explicitly compute the stress–stretch relation from the axioms, we will first deduce some basic properties. The following properties of symmetry and invertibility follow directly from the law of superposition and the uniqueness of the stress-free reference state.

Lemma 4.1. Let $ℰ \in PSym (3)$ . Then $T (ℰ^{- 1}) = - T (ℰ)$ .

Proof. Since $ℰ$ and $ℰ^{- 1}$ obviously commute, the law of superposition implies

T (ℰ) + T (ℰ^{- 1}) = T (ℰ \cdot ℰ^{- 1}) = T (11) = 0,

where the last equality is due to Axiom 0.2.

Remark 4.2. The symmetry property given in Lemma 4.1 is generally not equivalent to the symmetric tension–compression symmetry in hyperelasticity, which is defined by the equality

W (F^{- 1}) = W (F)

for the energy function W and all F ∈ GL⁺(3). A hyperelastic stress–stretch relation is tension–compression-symmetric if and only if

τ (V^{- 1}) = - τ (V)

for all V ∈ PSym(3), where τ denotes the Kirchhoff stress tensor.

Lemma 4.3. The mapping $ℰ \mapsto T (ℰ)$ is injective, that is, $T : PSym (3) \to range (T), ℰ \mapsto T (ℰ)$ is invertible.

Proof. Let $ℰ_{1}, ℰ_{2} \in PSym (3)$ with $T (ℰ_{1}) = T (ℰ_{2})$ . Then using Lemma 4.1 we find

0 = T (ℰ_{1}) - T (ℰ_{2}) = T (ℰ_{1}) + T (ℰ_{2}^{- 1}) = T (ℰ_{1} \cdot ℰ_{2}^{- 1}),

and thus Axiom 0.2 yields $ℰ_{1} \cdot ℰ_{2}^{- 1} = 11$ and therefore $ℰ_{1} = ℰ_{2}$ .

Remark 4.4. We will denote the inverse of the stress response by writing $ℰ (\hat{T})$ for $\hat{T} \in range (T)$ to denote the unique $ℰ \in PSym (3)$ with $\hat{T} = T (ℰ)$ .

Combined with the continuity of the stress–stretch relation, the law of superposition allows us to compute the stress response to arbitrary powers of stretches. For a further discussion of non-rational powers of matrices as well as primary matrix functions in general we refer to [42].

Lemma 4.5. Let $ℰ \in PSym (3)$ . Then

T (ℰ^{r}) = r \cdot T (ℰ)

for all $r \in ℝ$ .

Proof. Since

T ({(Q \cdot ℰ \cdot Q^{T})}^{r}) = T (Q \cdot ℰ^{r} \cdot Q^{T}) = Q \cdot T (ℰ^{r}) \cdot Q^{T}

we will assume without loss of generality that $ℰ$ is in diagonal form, that is, $ℰ = diag (λ_{1}, λ_{2}, λ_{3})$ with $λ_{1}, λ_{2}, λ_{3} \in ℝ^{+}$ . Then $ℰ^{r} = diag (λ_{1}^{r}, λ_{2}^{r}, λ_{3}^{r})$ for each $r \in ℝ$ .

For $n \in ℕ$ we can use the law of superposition to compute

\begin{array}{l} T (ℰ_{r}) & = & T ((\begin{matrix} λ_{1}^{n} & 0 & 0 \\ 0 & λ_{2}^{n} & 0 \\ 0 & 0 & λ_{3}^{n} \end{matrix})) = T (\prod_{k = 1}^{n} (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix})) \\ = & \sum_{k = 1}^{n} T ((\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix})) = n \cdot T ((\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix})) . \end{array}

Furthermore we find

0 = T (11) = T (ℰ^{n} \cdot ℰ^{- n}) = n \cdot T (ℰ) + T (ℰ^{- n}) \Rightarrow T (ℰ^{- n}) = - n \cdot T (ℰ),

and thus $T (ℰ^{z}) = z \cdot T (ℰ)$ for all $z \in ℤ$ . Similarly we compute

z \cdot T (ℰ) = T (ℰ^{z}) = T ({(ℰ^{\frac{z}{n}})}^{n}) = n \cdot T (ℰ^{\frac{z}{n}}) \Rightarrow T (ℰ^{\frac{z}{n}}) = \frac{z}{n}

for $z \in ℤ$ and $n \in ℕ$ , and therefore $T (ℰ^{q}) = q \cdot T (ℰ)$ for all $q \in ℚ$ . Finally, we simply point to the continuity of T, which follows from Axiom 0.1, to conclude $T (ℰ^{r}) = r \cdot T (ℰ)$ for all $r \in ℝ$ .

Remark 4.6. Using our notation $T \mapsto ℰ (T)$ for the inverse stress–stretch relation this proposition may equivalently be stated as

ℰ (r \cdot T) = ℰ {(T)}^{r} \forall r \in ℝ, T \in range (T) .

It is obvious that in a one-dimensional setting, Lemma 4.5 would already characterize the stress response as the logarithm to a fixed base. However, this is not immediately clear in the general case since not every stretch $ℰ \in PSym (3)$ can be written as the real power of a single fixed matrix.

Note also that the assumption of continuity is in fact necessary for the proof of Lemma 4.5.

4.3.2. Spherical stresses

While Axiom 2 relates dilations to purely spherical stresses, no assumption about the amount of stress is made. By using Lemma 4.5, however, it is easy to give an explicit formula for $T (ℰ)$ for arbitrary pure dilations $ℰ$ .

Lemma 4.7. There exists $d \in ℝ$ such that

T (λ \cdot 11) = d \cdot \log (λ \cdot 11) = d \cdot \log (λ) \cdot 11

for all $λ \in ℝ^{+}$ .

Proof. Choose $λ_{0} \in ℝ^{+}$ with λ₀ ≠ 1. Then, according to Axiom 2, the stress response to λ₀ · 11 is given by

T (λ_{0} \cdot 11) = a_{0} \cdot 11

for some $a_{0} \in ℝ$ , and we define d = a₀/log λ₀. Now let $λ \in ℝ^{+}$ with λ ≠ 1. Then

λ = λ_{0}^{\frac{\log λ}{\log λ_{0}}}

and

\begin{array}{l} T (λ \cdot 11) & = & T ((\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix})) = T ((\begin{matrix} λ_{0}^{\frac{\log λ}{\log λ_{0}}} & 0 & 0 \\ 0 & λ_{0}^{\frac{\log λ}{\log λ_{0}}} & 0 \\ 0 & 0 & λ_{0}^{\frac{\log λ}{\log λ_{0}}} \end{matrix})) = \frac{\log λ}{\log λ_{0}} \cdot T ((\begin{matrix} λ_{0} & 0 & 0 \\ 0 & λ_{0} & 0 \\ 0 & 0 & λ_{0} \end{matrix})) \\ = & \frac{\log λ}{\log λ_{0}} \cdot (\begin{matrix} a_{0} & 0 & 0 \\ 0 & a_{0} & 0 \\ 0 & 0 & a_{0} \end{matrix}) = \frac{a_{0}}{\log λ_{0}} \cdot \log (λ) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) = d \cdot \log (λ) \cdot 11 = d \cdot \log (λ \cdot 11) . \end{array}

Finally, if λ = 1, then Axiom 0.2 implies T(λ·11) =λ (11) = 0 = c · log(11).

Remark 4.8. Note that this proposition can equivalently be stated as

ℰ (a \cdot 11) = \exp (\frac{1}{d} \cdot a \cdot 11) = e^{\frac{a}{d}} \cdot 11 \forall a \in ℝ .

4.3.3. Shear stresses

Let us now consider a pure shear stretch of the form

ℰ = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})

with the ratio of shear $α \in ℝ^{+}$ . Again, while Axiom 1 only provides a general relation between shear stretches and shear stresses, the law of superposition yields a quantitative result for the case of pure shears.

Lemma 4.9. There exists $c \in ℝ$ such that

T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) = c \cdot \log (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) = c \cdot (\begin{matrix} \log (α) & 0 & 0 \\ 0 & - \log (α) & 0 \\ 0 & 0 & 0 \end{matrix})

(32)

for all $α \in ℝ^{+}$ .

Proof. The proof is analogous to that of Lemma 4.7: choose $α_{0} \in ℝ^{+}$ with α₀ ≠ 1. Then there exists s₀ such that

T ((\begin{matrix} α_{0} & 0 & 0 \\ 0 & \frac{1}{α_{0}} & 0 \\ 0 & 0 & 1 \end{matrix})) = (\begin{matrix} s_{0} & 0 & 0 \\ 0 & - s_{0} & 0 \\ 0 & 0 & 0 \end{matrix}),

according to Axiom 1, and we define c = s₀/logα₀.

Now let $α \in ℝ^{+}$ . Again we can use Axiom 0.2 and the equality log(11) = 0 to show that the equality obviously holds for α = 1, hence we can assume α ≠ 1 without loss of generality. Then $α = α_{0}^{(\log α) / (\log α_{0})}$ and

\begin{array}{l} T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) & = & T ((\begin{matrix} α_{0}^{\frac{\log α}{\log α_{0}}} & 0 & 0 \\ 0 & {(\frac{1}{α_{0}})}^{\frac{\log α}{\log α_{0}}} & 0 \\ 0 & 0 & 1^{\frac{\log α}{\log α_{0}}} \end{matrix})) \\ = & \frac{\log α}{\log α_{0}} \cdot T ((\begin{matrix} α_{0} & 0 & 0 \\ 0 & \frac{1}{α_{0}} & 0 \\ 0 & 0 & 1 \end{matrix})) = \frac{\log α}{\log α_{0}} \cdot (\begin{matrix} s_{0} & 0 & 0 \\ 0 & - s_{0} & 0 \\ 0 & 0 & 0 \end{matrix}) \\ = & \frac{s_{0}}{\log λ_{0}} \cdot (\begin{matrix} \log α & 0 & 0 \\ 0 & - \log α & 0 \\ 0 & 0 & 0 \end{matrix}) = c \cdot \log ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) . \end{array}

Remark 4.10. Again, this proposition can also be stated as

ℰ ((\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix})) = \exp (\frac{1}{c} \cdot (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix})) = (\begin{matrix} e^{\frac{1}{c} \cdot s} & 0 & 0 \\ 0 & e^{- \frac{1}{c} \cdot s} & 0 \\ 0 & 0 & 1 \end{matrix}) \forall s \in ℝ .

Although Lemma 4.9 refers only to deformations without strain along the x₃-axis, the following corollary shows that a similar property holds for shear deformations along the other principal axes as well.

Corollary 4.11. Let $α \in ℝ^{+}$ . Then

T ((\begin{matrix} α & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})) = c \cdot \log (\begin{matrix} α & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})

(33)

as well as

T ((\begin{matrix} 1 & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})) = c \cdot \log (\begin{matrix} 1 & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})

(34)

with $c \in ℝ$ as given in Lemma 4.9.

Proof. Let

Q = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}) \in O (3) .

Then

\begin{array}{l} T ((\begin{matrix} α & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})) & = & T (Q^{T} \cdot (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot Q) = Q^{T} \cdot T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) \cdot Q \\ = & Q^{T} \cdot c \cdot \log (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot Q = c \cdot \log (\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix}), \end{array}

proving (33). To show (34) we let

Q = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}) \in SO (3)

and find

\begin{array}{l} T ((\begin{matrix} 1 & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & \frac{1}{α} \end{matrix})) & = & T (Q^{T} \cdot (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot Q) = Q^{T} \cdot T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) \cdot Q \\ = & Q^{T} \cdot c \cdot \log (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot Q = c \cdot \log (\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix}) . \end{array}

4.3.4. The general case

Finally we consider the general case of an arbitrary stretch tensor $ℰ$ .

Proposition 4.12. A stress response function $ℰ \mapsto T^{Biot} (ℰ)$ fulfils Axioms 0.1 to 0.3 and Axioms 1 to 3 if and only if there exist constants $G, Λ \in ℝ$ , G ≠ 0, 3 Λ + 2 G ≠ 0 such that

T (ℰ) = 2 G \cdot \log (ℰ) + Λ \cdot tr [\log ℰ] \cdot 11

(35)

or, equivalently, constants $G, K \in ℝ ∖ {0}$ with

T (ℰ) = 2 G \cdot {dev}_{3} \log (ℰ) + K \cdot tr [\log ℰ] \cdot 11

(36)

for all $ℰ \in PSym (3)$ , where log: PSym(3) → Sym(3) is the principal matrix logarithm and dev₃X = X − (1/3)tr (X) · 11 denotes the deviatoric part of $X \in ℝ^{3 \times 3}$ .

Remark 4.13. A justification for the use of the Lamé constants G, Λ and the bulk modulus K in these formulae will be given by means of linearization in Section 5.1.

Proof. First we consider a stretch tensor $ℰ$ in the diagonal form

ℰ = (\begin{matrix} p & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & r \end{matrix}) .

Then $ℰ$ can be decomposed multiplicatively into three stretches $ℰ_{1}$ , $ℰ_{2}$ and $ℰ_{3}$ :

\begin{array}{l} ℰ & = & (\begin{matrix} {(p q r)}^{\frac{1}{3}} & 0 & 0 \\ 0 & {(p q r)}^{\frac{1}{3}} & 0 \\ 0 & 0 & {(p q r)}^{\frac{1}{3}} \end{matrix}) \cdot (\begin{matrix} {(\frac{p^{2}}{q r})}^{\frac{1}{3}} & 0 & 0 \\ 0 & {(\frac{q^{2}}{p r})}^{\frac{1}{3}} & 0 \\ 0 & 0 & {(\frac{r^{2}}{p q})}^{\frac{1}{3}} \end{matrix}) \\ = & \underset{ℰ_{1}}{\underset{︸}{{(\begin{matrix} p q r & 0 & 0 \\ 0 & p q r & 0 \\ 0 & 0 & p q r \end{matrix})}^{1 / 3}}} \cdot \underset{ℰ_{2}}{\underset{︸}{{(\begin{matrix} \frac{p^{2}}{q r} & 0 & 0 \\ 0 & \frac{q r}{p^{2}} & 0 \\ 0 & 0 & 1 \end{matrix})}^{1 / 3}}} \cdot \underset{ℰ_{3}}{\underset{︸}{{(\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{p q}{r^{2}} & 0 \\ 0 & 0 & \frac{r^{2}}{p q} \end{matrix})}^{1 / 3}}} . \end{array}

Using the law of superposition we find $T (ℰ) = T (ℰ_{1}) + T (ℰ_{2}) + T (ℰ_{3})$ . Lemma 4.7 allows us to compute

\begin{array}{l} T (ℰ_{1}) & = & T ({(p q r)}^{\frac{1}{3}} \cdot 11) = d \cdot \log ({(p q r)}^{\frac{1}{3}}) \cdot 11 \\ = & \frac{d}{3} \cdot \log (p q r) \cdot 11 = \frac{d}{3} \cdot \log (\det (ℰ)) \cdot 11 = \frac{d}{3} \cdot tr [\log (ℰ)] \cdot 11 \end{array}

with a constant $d \in ℝ$ , while Lemma 4.9 and Corollary 4.11 simply yield

T (ℰ_{2}) = c \cdot \log (ℰ_{2}), T (ℰ_{3}) = c \cdot \log (ℰ_{3})

with $c \in ℝ$ . Therefore

\begin{array}{l} T (ℰ_{2}) + T (ℰ_{3}) & = & c \cdot \log (ℰ_{2}) + c \cdot \log (ℰ_{3}) \\ = & c \cdot (\begin{matrix} \log [{(\frac{p^{2}}{q r})}^{\frac{1}{3}}] & 0 & 0 \\ 0 & \log [{(\frac{q r}{p^{2}})}^{\frac{1}{3}}] & 0 \\ 0 & 0 & 0 \end{matrix}) + c \cdot (\begin{matrix} 0 & 0 & 0 \\ 0 & \log [{(\frac{p q}{r^{2}})}^{\frac{1}{3}}] & 0 \\ 0 & 0 & \log [{(\frac{r^{2}}{p q})}^{\frac{1}{3}}] \end{matrix}) \\ = & c \cdot (\begin{matrix} \log [{(\frac{p^{2}}{q r})}^{\frac{1}{3}}] & 0 & 0 \\ 0 & \log [{(\frac{q r}{p^{2}})}^{\frac{1}{3}}] + \log [{(\frac{p q}{r^{2}})}^{\frac{1}{3}}] & 0 \\ 0 & 0 & \log [{(\frac{r^{2}}{p q})}^{\frac{1}{3}}] \end{matrix}) \end{array}

\begin{array}{l} = c \cdot (\begin{matrix} \log [{(\frac{p^{2}}{q r})}^{\frac{1}{3}}] & 0 & 0 \\ 0 & \log [{(\frac{q^{2}}{p r})}^{\frac{1}{3}}] & 0 \\ 0 & 0 & \log [{(\frac{r^{2}}{p q})}^{\frac{1}{3}}] \end{matrix}) \\ = c \cdot (\begin{matrix} \log [{(\frac{1}{p q r})}^{\frac{1}{3}} \cdot p] & 0 & 0 \\ 0 & \log [{(\frac{1}{p q r})}^{\frac{1}{3}} \cdot q] & 0 \\ 0 & 0 & \log [{(\frac{1}{p q r})}^{\frac{1}{3}} \cdot r] \end{matrix}) \\ = c \cdot (\begin{matrix} \log (p) & 0 & 0 \\ 0 & \log (q) & 0 \\ 0 & 0 & \log (r) \end{matrix}) + c \cdot (\begin{matrix} \log [{(\frac{1}{p q r})}^{\frac{1}{3}}] & 0 & 0 \\ 0 & \log [{(\frac{1}{p q r})}^{\frac{1}{3}}] & 0 \\ 0 & 0 & \log [{(\frac{1}{p q r})}^{\frac{1}{3}}] \end{matrix}), \end{array}

hence

\begin{array}{l} T (ℰ_{2}) + T (ℰ_{3}) & = & c \cdot \log (\begin{matrix} p & 0 & 0 \\ 0 & q & 0 \\ 0 & 0 & r \end{matrix}) + \frac{c}{3} \cdot (\begin{matrix} - \log (p q r) & 0 & 0 \\ 0 & - \log (p q r) & 0 \\ 0 & 0 & - \log (p q r) \end{matrix}) \\ = & c \cdot \log (ℰ) - \frac{c}{3} \cdot \log (\det ℰ) \cdot 11 \\ = & c \cdot (\log (ℰ) - \frac{1}{3} \cdot tr [\log ℰ] \cdot 11) = c \cdot {dev}_{3} \log ℰ . \end{array}

Thus $T (ℰ)$ computes to

T (ℰ) = T (ℰ_{2}) + T (ℰ_{3}) + T (ℰ_{1}) = c \cdot {dev}_{3} \log (ℰ) + \frac{d}{3} \cdot tr [\log (ℰ)] \cdot 11 .

Finally, for arbitrary $ℰ \in PSym (3)$ we can choose Q ∈ O(3) and a diagonal matrix D such that $ℰ = Q^{T} D Q$ . Utilizing the isotropy property as well as the above computations we find

\begin{array}{l} T (ℰ) & = & T (Q^{T} D Q) = Q^{T} T^{Biot} (D) Q \\ = & Q^{T} \cdot (c \cdot {dev}_{3} \log (D) + \frac{d}{3} \cdot tr [\log (D)] \cdot 11) \cdot Q \\ = & c \cdot {dev}_{3} \log (Q^{T} D Q) + \frac{d}{3} \cdot tr [\log (ℰ)] \cdot (Q^{T} Q) = c \cdot {dev}_{3} \log (ℰ) + \frac{d}{3} \cdot tr [\log (ℰ)] \cdot 11, \end{array}

and thus we obtain equation (35) with G = c/2 and K = d/3. It is also easy to see that the restrictions G ≠ 0 and K ≠ 0 follow directly from the injectivity of the response function. Furthermore, with Λ = K − 2G/3 we obtain the equivalent representation

\begin{array}{l} T (ℰ) & = & 2 G \cdot [\log ℰ - \frac{1}{3} tr [\log ℰ] \cdot 11] + K \cdot 11 \\ = & 2 G \cdot \log (ℰ) + (K - \frac{2 G}{3}) tr [\log ℰ] \cdot 11 = 2 G \cdot \log (ℰ) + Λ \cdot tr [\log ℰ] \cdot 11 . \end{array}

It remains to show that the stress response function (35) does indeed satisfy all our axioms. Since the matrix logarithm and the trace operator are continuous functions¹² on PSym(3), Axiom 0.1 obviously holds. The isotropy of the matrix logarithm immediately implies

\begin{array}{l} 2 G \cdot \log ((Q^{T} ℰ Q)) + Λ \cdot tr [\log (Q^{T} ℰ Q)] \cdot 11 & = & 2 G \cdot Q^{T} \log (ℰ) Q + Λ \cdot tr [Q^{T} (\log ℰ) Q] \cdot 11 \\ = & 2 G \cdot Q^{T} \log (ℰ) Q + Λ \cdot tr [\log ℰ] \cdot 11 \\ = & Q^{T} \cdot (2 G \cdot \log (ℰ) + Λ \cdot tr [\log ℰ] \cdot 11) \cdot Q, \end{array}

and thus Axiom 0.3 holds as well. To show Axiom 0.2 we employ the equivalent representation formula (36): for G, K ≠ 0 we first note that the mapping

X \mapsto 2 G \cdot {dev}_{3} X + K \cdot tr [X] \cdot 11

is an isomorphism from Sym(3) onto itself. Thus

2 G \cdot {dev}_{3} \log (ℰ) + K \cdot tr [\log ℰ] \cdot 11 = 0

if and only if $\log ℰ = 0$ , which is the case if and only if $ℰ = 11$ .

We will now consider the remaining three axioms.

Axiom 1: For $ℰ = diag (α, 1 / α, 1)$ we directly compute

\begin{array}{l} 2 G \cdot {dev}_{3} \log (ℰ) + K \cdot tr [\log ℰ] \cdot 11 & = & 2 G \cdot {dev}_{3} (\begin{matrix} \log α & 0 & 0 \\ 0 & - \log α & 0 \\ 0 & 0 & 0 \end{matrix}) + K \cdot [\log α + \log \frac{1}{α}] \cdot 11 \\ = & 2 G \cdot (\begin{matrix} \log α & 0 & 0 \\ 0 & - \log α & 0 \\ 0 & 0 & 0 \end{matrix}), \end{array}

and thus Axiom 1 is fulfilled with s = 2 G log α.

Axiom 2: For $ℰ = λ \cdot 11$ we compute

\begin{array}{l} 2 G \cdot {dev}_{3} \log (ℰ) + K \cdot tr [\log ℰ] \cdot 11 & = & 2 G \cdot {dev}_{3} (\begin{matrix} \log λ & 0 & 0 \\ 0 & \log λ & 0 \\ 0 & 0 & \log λ \end{matrix}) + K \cdot tr [3 \cdot \log λ] \cdot 11 \\ = & K \cdot [3 \log λ] \cdot 11, \end{array}

and thus Axiom 2 is fulfilled with a = 3 K log α.

Axiom 3: First assume that $ℰ_{1}, ℰ_{2} \in PSym (3)$ have the diagonal forms

ℰ_{1} = (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}), ℰ_{2} = (\begin{matrix} {\hat{λ}}_{1} & 0 & 0 \\ 0 & {\hat{λ}}_{2} & 0 \\ 0 & 0 & {\hat{λ}}_{3} \end{matrix}) .

Then

\begin{array}{l} \log (ℰ_{1} \cdot ℰ_{2}) & = & \log (\begin{matrix} λ_{1} {\hat{λ}}_{1} & 0 & 0 \\ 0 & λ_{2} {\hat{λ}}_{2} & 0 \\ 0 & 0 & λ_{3} {\hat{λ}}_{3} \end{matrix}) = (\begin{matrix} \log (λ_{1} {\hat{λ}}_{1}) & 0 & 0 \\ 0 & \log (λ_{2} {\hat{λ}}_{2}) & 0 \\ 0 & 0 & \log (λ_{3} {\hat{λ}}_{3}) \end{matrix}) \\ = & (\begin{matrix} \log (λ_{1}) + \log ({\hat{λ}}_{1}) & 0 & 0 \\ 0 & \log (λ_{2}) + \log ({\hat{λ}}_{2}) & 0 \\ 0 & 0 & \log (λ_{3}) + \log ({\hat{λ}}_{3}) \end{matrix}) = \log (ℰ_{1}) + \log (ℰ_{2}) \end{array}

and therefore

\begin{array}{l} T (ℰ_{1} \cdot ℰ_{2}) & = & 2 G \cdot \log (ℰ_{1} \cdot ℰ_{2}) + Λ \cdot tr [\log ℰ_{1} \cdot ℰ_{2}] \cdot 11 \\ = & 2 G \cdot [\log (ℰ_{1}) + \log (ℰ_{2})] + Λ \cdot tr [\log (ℰ_{1}) + \log (ℰ_{2})] \cdot 11 \\ = & 2 G \cdot \log (ℰ_{1}) + 2 G \cdot \log (ℰ_{2}) + Λ \cdot tr [\log (ℰ_{1})] \cdot 11 + Λ \cdot tr [\log (ℰ_{2})] \cdot 11 \\ = & T (ℰ_{1}) + T (ℰ_{2}) . \end{array}

Now let $ℰ_{1}$ and $ℰ_{2}$ denote arbitrary coaxial matrices. Then $ℰ_{1}$ and $ℰ_{2}$ can be simultaneously diagonalized, in other words, there exist diagonal matrices D₁ and D₂ as well as Q ∈ O(3) with

ℰ_{1} = Q^{T} D_{1} Q, ℰ_{2} = Q^{T} D_{2} Q .

Then

\begin{array}{l} T (ℰ_{1} \cdot ℰ_{2}) & = & T (Q^{T} D_{1} D_{2} Q) = Q^{T} \cdot T (D_{1} D_{2}) \cdot Q \\ = & Q^{T} \cdot [T (D_{1}) + T (D_{2})] \cdot Q = Q^{T} T (D_{1}) Q + Q^{T} T (D_{2}) Q \\ = & T (Q^{T} D_{1} Q) + T (Q^{T} D_{2} Q) = T (ℰ_{1}) + T (ℰ_{2}) . \end{array}

While Becker assumed Axioms 1 and 2 to hold, we will now show that they are, in fact, not necessary to characterize Becker’s law of elasticity but can be deduced from Axioms 0.1–0.3 and Axiom 3 alone.

Lemma 4.14. If Axioms 0.1, 0.2, 0.3 and 3 hold, then Axioms 1 and 2 hold as well.

Proof. First, for α = 1 and λ = 1, we find

(\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) = 11

and thus

T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) = T ((\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix})) = T (11) = 0

due to Axiom 0.2. We can therefore assume without loss of generality that λ ≠ 1 ≠ α.

Axiom 1: Let $α \in ℝ^{+}$ with α ≠ 1 and $ℰ = diag (α, 1 / α, 1)$ . Then, because the principal axes of $T (ℰ)$ and $ℰ$ coincide, $T (ℰ)$ is in diagonal form as well:

T (ℰ) = T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) = (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix})

for some $a, b, c \in ℝ$ . With

Q = (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) \in O (3),

the property of isotropy allows us to compute

T ((\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix})) = T (Q^{T} \cdot ℰ \cdot Q) = Q^{T} \cdot T (ℰ) \cdot Q = (\begin{matrix} b & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{matrix})

and using the law of superposition we find

\begin{array}{l} T (11) & = & T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix})) = T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) + T ((\begin{matrix} \frac{1}{α} & 0 & 0 \\ 0 & α & 0 \\ 0 & 0 & 1 \end{matrix})) \\ = & (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}) + (\begin{matrix} b & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{matrix}) = (\begin{matrix} a + b & 0 & 0 \\ 0 & a + b & 0 \\ 0 & 0 & 2 c \end{matrix}) . \end{array}

Since T(11) = 0 we conclude b = −a as well as c = 0, and thus $T (ℰ)$ has the form

T ((\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix})) = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix})

(37)

with s = a. As was shown in the proof of Lemma 4.3, a function satisfying Axioms 3 and 0.2 is injective, hence

T (ℰ) = (\begin{matrix} s & 0 & 0 \\ 0 & - s & 0 \\ 0 & 0 & 0 \end{matrix}) \Leftrightarrow ℰ = (\begin{matrix} α & 0 & 0 \\ 0 & \frac{1}{α} & 0 \\ 0 & 0 & 1 \end{matrix}) .

Axiom 2: Now let $λ \in ℝ^{+}$ with λ ≠ 1 and $ℰ = diag (λ, λ, λ) = λ \cdot 11$ . Then

T (ℰ) = T ((\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix})) = (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix})

for some $a, b, c \in ℝ$ . We let

Q = (\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}) \in SO (3)

to find

\begin{array}{l} T (ℰ) = T (λ \cdot 11) & = & T (Q^{T} \cdot (λ \cdot 11) \cdot Q) \\ = & Q^{T} \cdot T (λ \cdot 11) \cdot Q = Q^{T} \cdot (\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}) \cdot Q = (\begin{matrix} b & 0 & 0 \\ 0 & c & 0 \\ 0 & 0 & a \end{matrix}) . \end{array}

Therefore a = b and b = c, and hence $T (ℰ)$ has the form

T (λ \cdot 11) = (\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}) = a \cdot 11

(38)

with $a \in ℝ$ . Then the injectivity of T yields

T (ℰ) = a \cdot 11 \Leftrightarrow ℰ = λ \cdot 11,

concluding the proof.

From this lemma and Proposition 4.12, it immediately follows that the reduced set of axioms is sufficient to characterize the stress response function. This result is summarized in the following proposition.

Proposition 4.15. Let T: PSym(3) → Sym(3) be a continuous isotropic tensor function with

T (ℰ) = 0 \Leftrightarrow ℰ = 11

and

T (ℰ_{1} \cdot ℰ_{2}) = T (ℰ_{2}) + T (ℰ_{2})

for all $ℰ_{1}, ℰ_{2} \in PSym (3)$ . Then there exist constants $G, Λ \in ℝ$ , G ≠ 0, 3Λ + 2 G ≠ 0 such that

T (ℰ) = 2 G \cdot \log (ℰ) + 3 Λ \cdot tr [\log ℰ] \cdot 11

(39)

or, equivalently, constants $G, K \in ℝ ∖ {0}$ with

T (ℰ) = 2 G \cdot {dev}_{3} \log (ℰ) + K \cdot tr [\log ℰ] \cdot 11

(40)

for all $ℰ \in PSym (3)$ .

If we assume beforehand that the stress response function is invertible, we can also deduce this general law in terms of the inverse stress–stretch relation: let T denote a given stress tensor of the form

T = (\begin{matrix} P & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & R \end{matrix}) .

Then T can be written in form of the additive decomposition

T = \underset{T_{1}}{\underset{︸}{\frac{P + Q + R}{3} \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})}} + \underset{T_{2}}{\underset{︸}{\frac{Q + R - 2 P}{3} \cdot (\begin{matrix} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix})}} + \underset{T_{3}}{\underset{︸}{\frac{P + Q - 2 R}{3} \cdot (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix})}}

into two pure shear stresses and one spherical stress. Using Remarks 4.8 and 4.10 we compute

ℰ (T_{1}) = e^{\frac{P + Q + R}{3 d}} \cdot 11

as well as

ℰ (T_{2}) = (\begin{matrix} e^{- \frac{Q + R - 2 P}{3 c}} & 0 & 0 \\ 0 & e^{\frac{Q + R - 2 P}{3 c}} & 0 \\ 0 & 0 & 1 \end{matrix})

and

ℰ (T_{3}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & e^{\frac{P + Q - 2 R}{3 c}} & 0 \\ 0 & 0 & e^{- \frac{P + Q - 2 R}{3 c}} \end{matrix}) .

Therefore the law of superposition yields

\begin{array}{l} ℰ (T) & = & ℰ (T_{1}) \cdot ℰ (T_{2}) \cdot ℰ (T_{3}) \\ = & e^{\frac{P + Q + R}{3 d}} \cdot (\begin{matrix} e^{- \frac{Q + R - 2 P}{3 c}} & 0 & 0 \\ 0 & e^{\frac{Q + R - 2 P}{3 c}} & 0 \\ 0 & 0 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & e^{\frac{P + Q - 2 R}{3 c}} & 0 \\ 0 & 0 & e^{- \frac{P + Q - 2 R}{3 c}} \end{matrix}) \\ = & e^{\frac{tr T}{3 d}} \cdot (\begin{matrix} e^{\frac{2 P - Q - R}{3 c}} & 0 & 0 \\ 0 & e^{\frac{2 Q - P - R}{3 c}} & 0 \\ 0 & 0 & e^{\frac{2 R - P - Q}{3 c}} \end{matrix}) \\ = & e^{\frac{tr T}{3 d}} \cdot \exp (\begin{matrix} \frac{2 P - Q - R}{3 c} & 0 & 0 \\ 0 & \frac{2 Q - P - R}{3 c} & 0 \\ 0 & 0 & \frac{2 R - P - Q}{3 c} \end{matrix}) = e^{\frac{tr T}{3 d}} \cdot \exp (\frac{1}{c} \cdot {dev}_{3} T) . \end{array}

With constants G = c/2 and K = d/3 our law of ideal elasticity can therefore be stated as

ℰ (T) = \exp (\frac{1}{2 G} {dev}_{3} T + \frac{1}{9 K} tr (T) \cdot 11) = e^{\frac{1}{9 K} \cdot tr T} \cdot \exp (\frac{1}{2 G} \cdot {dev}_{3} T) .

(41)

4.3.5. Becker’s stress response function

Finally, since Becker assumed Axioms 0.1 to 0.3 and 1 to 3 to hold for the Biot stress T = T^Biot and the right Biot stretch tensors $ℰ = U$ , we conclude that Becker’s law of elasticity is given by

T^{Biot} (U) = 2 G \cdot \log (U) + Λ \cdot tr [\log U] \cdot 11

(42)

or, equivalently, by

T^{Biot} (U) = 2 G \cdot {dev}_{3} \log (U) + K \cdot tr [\log U] \cdot 11

(43)

= \frac{E}{1 + ν} \cdot {dev}_{3} \log (U) + \frac{E}{3 (1 - 2 ν)} \cdot tr [\log U] \cdot 11

(44)

with Young’s modulus

E = \frac{9 K G}{3 K + G}

and Poisson’s number

ν = \frac{3 K - 2 G}{2 (3 K + G)} .

4.4. Application to other stresses and stretches

If we apply Proposition 4.15 to other coaxial stress–stretch pairs, the resulting law of elasticity will, in general, differ from that given by Becker. Two examples of such combinations are especially important: the left stretch tensor $ℰ = V = \sqrt{F F^{T}}$ with the Cauchy stress tensor T = σ as well as the left stretch with the Kirchhoff stress tensor T = τ. Those cases were considered by Heinrich Hencky in 1928 and 1929, respectively [2, 41]. His approach was remarkably similar to Becker’s: from the assumption of a law of superposition for these two stresses he deduced two laws of idealized elasticity.

Corollary 4.16. If the Cauchy stress σ is a continuous isotropic function of the left stretch tensor V with

σ (V) = 0 \Leftrightarrow V = 11

and

σ (V_{1} \cdot V_{2}) = σ (V_{2}) + σ (V_{2})

for all V₁, V₂ ∈ PSym(3), then there exist constants $G, K \in ℝ ∖ {0}$ such that

σ (V) = 2 G \cdot {dev}_{3} \log (V) + K \cdot tr [\log V] \cdot 11

(45)

for all V ∈ PSym(3).

If the Kirchhoff stress τ is a continuous isotropic function of the left Biot stretch tensor V with

τ (V) = 0 \Leftrightarrow V = 11

and

τ (V_{1} \cdot V_{2}) = τ (V_{2}) + τ (V_{2})

for all V₁, V₂ ∈ PSym(3), then there exist constants $G, K \in ℝ ∖ {0}$ such that

τ (V) = 2 G \cdot {dev}_{3} \log (V) + K \cdot tr [\log V] \cdot 11

(46)

for all V ∈ PSym(3).

It was shown by Hencky that only the latter of these two stress response functions constitutes a hyperelastic law of elasticity: the stress–stretch relation in equation (46) can be obtained from the quadratic Hencky strain energy

W (V) = G | | {dev}_{3} \log V | |^{2} + \frac{K}{2} {[tr (\log V)]}^{2} .

(47)

This energy function has also been given another rigorous justification, based on purely differential geometric reasoning, as the geodesic distance of the deformation gradient F to the special orthogonal group SO(3) with respect to the canonical left-invariant metric on GL⁺(3) [40, 43, 44]. The continuing application of the Hencky strain energy or modifications thereof is described in [12]; see also [45].

4.5. The axioms in the linear case of infinitesimal elasticity

Some of Becker’s assumptions seem to have been adapted from simple results for the linearized theory of elasticity. To explain his motivation it is insightful to discover some of these parallels. The most general stress–strain relation for isotropic homogeneous materials in the case of linear elasticity is

σ = 2 G ε + Λ tr (ε) \cdot 11 = 2 G {dev}_{3} ε + K tr (ε) \cdot 11 = \frac{E}{1 + ν} {dev}_{3} ε + \frac{E}{3 (1 - 2 ν)} tr (ε) \cdot 11,

where σ is the linearized stress tensor and ε = sym ∇u = (1/2)(∇u + ∇u^T) is the linearized strain tensor of the deformation φ(x) = x + u(x) with the displacement $u : Ω_{0} \subset ℝ^{3} \to ℝ^{3}$ . Note that this linear relation is invertible with

ε = \frac{1}{2 G} {dev}_{3} σ + \frac{1}{9 K} tr (σ) \cdot 11,

similar to the first equality in equation (41).

Since the trace operator is the linear approximation of the determinant at 11, that is, $\det (11 + H) = 1 + tr (H) + O (| | H | |^{2})$ , the first-order approximation

\det (\nabla ϕ) = \det (11 + \nabla u) \approx 1 + tr (\nabla u)

holds for sufficiently small ∥∇u∥. Therefore, the condition

\det (U) = \det (\nabla ϕ) = 1

can be linearized to the equation

tr (\nabla u) = tr ε = 0 .

For Λ ≠ −2G/3, this is the case if and only if

0 = tr σ = (2 G + 3 Λ) \cdot tr ε,

thus (linearized) isochoric deformations always correspond to trace-free stress tensors σ and vice versa, analogously to Axiom 1 for the nonlinear case. Similarly, volumetric stresses occur if and only if the strain is (linearly) volumetric (Axiom 2):

ε = (\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}) \Leftrightarrow σ = (\begin{matrix} (2 G + 3 Λ) a & 0 & 0 \\ 0 & (2 G + 3 Λ) a & 0 \\ 0 & 0 & (2 G + 3 Λ) a \end{matrix}) .

Furthermore it is easy to see that the linearized shear strain

ε = sym [(\begin{matrix} 1 & γ & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) - 11] = (\begin{matrix} 0 & \frac{γ}{2} & 0 \\ \frac{γ}{2} & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

corresponds to the shear stress (Axiom 1)

σ = (\begin{matrix} 0 & G \cdot γ & 0 \\ G \cdot γ & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

Finally, we consider two deformation gradients ∇φ₁ = 11 + ∇u₁ and ∇φ₂ = 11 + ∇u₂ with the corresponding strain tensors ε₁ and ε₂. Then

\nabla ϕ_{1} \cdot \nabla ϕ_{2} = (11 + \nabla u_{1}) \cdot (11 + \nabla u_{2}) = 11 + \nabla u_{1} + \nabla u_{2} + \nabla u_{2} \cdot \nabla u_{2} .

By omitting the higher-order term ∇u₁·∇u₂, we find the linear approximation

\nabla ϕ_{1} \cdot \nabla ϕ_{2} \approx 11 + \nabla u_{1} + \nabla u_{2},

and hence the strain tensor ε corresponding to ∇φ₁·∇φ₂ has the linear approximation

ε \approx sym (\nabla u_{1} + \nabla u_{2}) = ε_{1} + ε_{2} .

Thus, in the linear case, the multiplicative superposition of deformation gradients corresponds to an additive composition of the strain tensors. The law of superposition (Axiom 3) can therefore be linearized to

σ (ε_{1} + ε_{2}) = σ (ε_{1}) + σ (ε_{2}),

which obviously holds for all ε₁, ε₂ ∈ Sym(3).

The linear analogies of the three main axioms can therefore be summarized as follows.

Axiom 1, linear version:

The equivalence

ε = (\begin{matrix} 0 & \frac{γ}{2} & 0 \\ \frac{γ}{2} & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \Leftrightarrow σ (ε) = (\begin{matrix} 0 & G \cdot γ & 0 \\ G \cdot γ & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

holds for all $γ \in ℝ$ .

Axiom 2, linear version:

The equivalence

ε = (\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{matrix}) \Leftrightarrow σ (ε) = (\begin{matrix} (2 G + 3 Λ) a & 0 & 0 \\ 0 & (2 G + 3 Λ) a & 0 \\ 0 & 0 & (2 G + 3 Λ) a \end{matrix})

holds for all $a \in ℝ$ .

Axiom 3, linear version:

The equality

σ (ε_{1} + ε_{2}) = σ (ε_{1}) + σ (ε_{2})

holds for all ε₁, ε₂ ∈ Sym(3).

Note that many of the properties listed in Section 1.1 have linearized counterparts which are satisfied by the general linear model, for example the (linearized) tension–compression symmetry σ(−ε) = −σ(ε).

4.5.1. Linearized shear

Becker’s comments on the finite shear response show similarities to the linear case as well. We consider the linearized shear stress

σ = (\begin{matrix} 0 & s \\ s & 0 \end{matrix})

with the corresponding linear shear strain¹³

ε = (\begin{matrix} 0 & \frac{γ}{2} \\ \frac{γ}{2} & 0 \end{matrix})

in the two-dimensional case. Then for given $n = {(n_{1}, n_{2})}^{T} \in ℝ^{2}$ with ∥n∥ = 1 we can compute

| | σ n | | = | | (\begin{matrix} s n_{2} \\ s n_{1} \end{matrix}) | | = s \cdot | | n | | = s .

In order to find a direction of maximum tangential linearized stress (not the tangential load), we decompose the resultant traction σ n in the direction of a given unit normal vector n into a normal and a tangential part:

| | σ n | |^{2} = \underset{normal}{\underset{︸}{{〈 σ n, n 〉}^{2}}} + \underset{tangential}{\underset{︸}{{〈 σ n, n_{⊥} 〉}^{2}}},

where n_⊥ is a unit vector normal to n. Since for such a stress the resultant ∥σ n∥² = s² is constant,¹⁴ that is, independent of the unit normal n, the amount of tangential stress

{〈 σ n, n_{⊥} 〉}^{2} = s^{2} - {〈 σ n, n 〉}^{2}

assumes its maximum among all $n \in ℝ^{2}$ with ∥n∥ = 1 if and only if 〈σ n, n〉² attains its minimum. We find

{〈 σ n, n 〉}^{2} = 〈 (\begin{matrix} s n_{2} \\ s n_{1} \end{matrix}), (\begin{matrix} n_{1} \\ n_{2} \end{matrix}) 〉 = s^{2} n_{1} n_{2},

which is minimal if and only if n₁ = 0 or n₂ = 0. Since the directions of the principal axes are given by the eigenvectors (1, 1)^T and (1, −1)^T of ε, the vectors n₁ and n₂ cut these axes at angles of 45°.

5. Applications and properties of Becker’s law of elasticity

5.1. Infinitesimal deformations

For small deformations the linear approximation

\begin{array}{l} T (11 + ε) & = & 2 G \cdot \log (11 + ε) + Λ \cdot tr [\log (11 + ε)] \cdot 11 \\ = & 2 G \cdot (ε + O (| | ε | |^{2})) + Λ \cdot (tr [ε + O (| | ε | |^{2})]) \cdot 11 \\ = & 2 G \cdot ε + Λ \cdot tr (ε) \cdot 11 + O (| | ε | |^{2}) \end{array}

shows that the stress–stretch relation is compatible with the model of linear elasticity if and only if G and Λ are the two Lamé constants. In this case the additional constraints

G > 0, K > 0

follow from the uniform positivity of the linear strain energy density

W_{lin} (ε) = G | | {dev}_{3} sym ε | |^{2} + \frac{K}{2} {[tr (ε)]}^{2} = G | | sym ε | |^{2} + \frac{Λ}{2} {[tr (ε)]}^{2} .

5.2. Uniaxial stresses

If the initial load is given by a Biot stress tensor of the form

T^{Biot} = (\begin{matrix} 0 & 0 & 0 \\ 0 & Q & 0 \\ 0 & 0 & 0 \end{matrix}),

we can give an explicit formula for the stretch tensor U corresponding to T^Biot: since tr T^Biot = Q and

{dev}_{3} T^{Biot} = T^{Biot} - \frac{1}{3} tr [T^{Biot}] \cdot 11 = (\begin{matrix} - \frac{Q}{3} & 0 & 0 \\ 0 & \frac{2 Q}{3} & 0 \\ 0 & 0 & - \frac{Q}{3} \end{matrix})

we can use equation (41) to find

U (T^{Biot}) = e^{\frac{Q}{9 K}} \cdot (\begin{matrix} e^{- \frac{Q}{2 G \cdot 3}} & 0 & 0 \\ 0 & e^{\frac{2 Q}{2 G \cdot 3}} & 0 \\ 0 & 0 & e^{- \frac{Q}{2 G \cdot 3}} \end{matrix}) = e^{\frac{Q}{9 K}} \cdot (\begin{matrix} e^{- \frac{Q}{6 G}} & 0 & 0 \\ 0 & e^{\frac{Q}{3 G}} & 0 \\ 0 & 0 & e^{- \frac{Q}{6 G}} \end{matrix}) .

In particular, the deformation along the axis of stress is given by¹⁴

λ_{2} = e^{\frac{Q}{9 K}} \cdot e^{\frac{Q}{3 G}} = e^{Q \cdot (\frac{1}{9 K} + \frac{1}{3 G})} = e^{\frac{Q}{E}},

(48)

while the deformation along the axes orthogonal to the stress axis is

λ_{1} = λ_{3} = e^{\frac{Q}{9 K}} \cdot e^{- \frac{Q}{6 G}} = e^{Q \cdot (\frac{1}{9 K} - \frac{1}{6 G})} = e^{- \frac{ν Q}{E}} .

(49)

The factors e^Q^/9K and e^Q^/3G appearing in equation (48) are the dilational stretch and the shear stretch, respectively, as given by Becker [19, equation (5) on page 345] as his main result. Furthermore, equation (49) shows that in the case 9K = 6G, which corresponds to ν = 0 for Poisson’s ratio ν, the stretch along the unstressed axes is 1. Therefore, as in the linear model, there is no lateral contraction in Becker’s model for ν = 0. A similar result holds for Hencky’s elastic law [46, 47].

5.2.1. Application to incompressible materials

To apply the uniaxial stress response to incompressible materials we will now consider the limit K → ∞, that is, we approximate the incompressible case through the nearly incompressible case with a sufficiently large ratio K/G. From equations (48) and (49) we readily obtain

\lim_{K \to \infty} λ_{2} = e^{\frac{Q}{3 G}}

as well as

\lim_{K \to \infty} λ_{1} = \lim_{K \to \infty} λ_{3} = e^{- \frac{Q}{6 G}},

and thus in our theory uniaxial stress induces deformations of the form

U = (\begin{matrix} \frac{1}{\sqrt{λ}} & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & \frac{1}{\sqrt{λ}} \end{matrix}) = (\begin{matrix} e^{- \frac{Q}{6 G}} & 0 & 0 \\ 0 & e^{\frac{Q}{3 G}} & 0 \\ 0 & 0 & e^{- \frac{Q}{6 G}} \end{matrix})

in the incompressible case. Going to the inverse we obtain the formula

Q = 3 G \cdot \log (λ) = E \cdot \log (λ),

(50)

where E = 3G denotes Young’s modulus for incompressible materials.

Equation (50) is identical to the uniaxial stress–stretch relation given by Imbert in 1880 as a phenomenological model for the deformation of vulcanized rubber bands under tension [48, p. 53]. Similarly, in 1893 Hartig applied the same logarithmic law to describe the uniaxial tension and compression of rubber [50]. A comparison of Becker’s results for very large strain to experimental data by Jones and Treloar for the uniaxial deformation of vulcanized rubber [14] as well as the corresponding stress responses for the quadratic Hencky energy and Ogden’s elasticity model [50] are shown in Figure 18. Another possible way to apply Becker’s law to incompressible materials is described in Section 5.5.1.

Figure 18.

Comparison of stress responses for incompressible materials.

5.3. Becker’s law of elasticity for simple shear

Consider a simple glide deformation of the form

F = (\begin{matrix} 1 & γ & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})

with γ > 0; see Section 2.2.3. Then the polar decomposition of F = R · U into the right Biot stretch tensor $U = \sqrt{F^{T} F}$ of the deformation and the orthogonal polar factor R = FU⁻¹ is given by

U = \frac{1}{\sqrt{γ^{2} + 4}} \cdot (\begin{matrix} 2 & γ & 0 \\ γ & γ^{2} + 2 & 0 \\ 0 & 0 & \sqrt{γ^{2} + 4} \end{matrix}), R = \frac{1}{\sqrt{γ^{2} + 4}} \cdot (\begin{matrix} 2 & γ & 0 \\ - γ & 2 & 0 \\ 0 & 0 & \sqrt{γ^{2} + 4} \end{matrix})

Further, U can be diagonalized to

U = L \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{1}{2} (\sqrt{γ^{2} + 4} + γ) & 0 \\ 0 & 0 & \frac{1}{2} (\sqrt{γ^{2} + 4} - γ) \end{matrix}) \cdot L^{- 1} = L \cdot (\begin{matrix} 1 & 0 & 0 \\ 0 & λ_{1} & 0 \\ 0 & 0 & \frac{1}{λ_{1}} \end{matrix}) \cdot L^{- 1},

where

L = (\begin{matrix} 0 & 2 & - 2 \\ 0 & \sqrt{γ^{2} + 4} + γ & \sqrt{γ^{2} + 4} - γ \\ 1 & 0 & 0 \end{matrix})

and $λ_{1} = (1 / 2) (\sqrt{γ^{2} + 4} + γ)$ denotes the first eigenvalue of U, and the principal logarithm of U is

\begin{array}{l} \log U & = & L \cdot \log (\begin{matrix} 1 & 0 & 0 \\ 0 & λ_{1} & 0 \\ 0 & 0 & \frac{1}{λ_{1}} \end{matrix}) \cdot L^{- 1} = L \cdot (\begin{matrix} 0 & 0 & 0 \\ 0 & \log (λ_{1}) & 0 \\ 0 & 0 & - \log (λ_{1}) \end{matrix}) \cdot L^{- 1} \\ = & \frac{1}{\sqrt{γ^{2} + 4}} \cdot (\begin{matrix} - γ \log (λ_{1}) & 2 \log (λ_{1}) & 0 \\ 2 \log (λ_{1}) & γ \log (λ_{1}) & 0 \\ 0 & 0 & 0 \end{matrix}) . \end{array}

Then according to Becker’s law of elasticity, the first Piola–Kirchhoff stress tensor S₁ corresponding to F computes to

\begin{array}{l} S_{1} (F) & = & R \cdot T^{Biot} (U) = R \cdot (2 G \cdot \log (U) + Λ \cdot \log (\det U) \cdot 11) = 2 G \cdot R \cdot \log U \\ = & \frac{2 G}{γ^{2} + 4} \cdot (\begin{matrix} 2 & γ & 0 \\ - γ & 2 & 0 \\ 0 & 0 & \sqrt{γ^{2} + 4} \end{matrix}) \cdot (\begin{matrix} - γ \log (λ_{1}) & 2 \log (λ_{1}) & 0 \\ 2 \log (λ_{1}) & γ \log (λ_{1}) & 0 \\ 0 & 0 & 0 \end{matrix}) \\ = & \frac{2 G}{γ^{2} + 4} \cdot (\begin{matrix} 0 & (4 + γ^{2}) \log (λ_{1}) & 0 \\ (4 + γ^{2}) \log (λ_{1}) & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ = & 2 G \cdot \log (\frac{1}{2} (\sqrt{γ^{2} + 4} + γ)) \cdot (\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) . \end{array}

Finally, the Cauchy stress σ for the simple glide deformation F is

σ = \frac{1}{\det F} \cdot S_{1} (F) \cdot F^{T} = 2 G \cdot \log (\frac{1}{2} (\sqrt{γ^{2} + 4} + γ)) \cdot (\begin{matrix} γ & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) .

Note that σ is independent of the Lamé constant Λ. In particular, the simple shear stress σ₁₂ corresponding to the amount of shear γ is given by

σ_{12} = \log (\frac{1}{2} (\sqrt{γ^{2} + 4} + γ))

for Becker’s law of elasticity. Figure 19 shows a comparison of the simple shear stress resulting from different constitutive laws with experimental data measured by Jones and Treloar [14] for shear deformations of vulcanized rubber.

Figure 19.

Shear stress in a simple glide deformation.

5.4. A comparison of Becker’s and Hencky’s laws of elasticity

The stress–stretch relation corresponding to the quadratic Hencky strain energy is, in terms of the Kirchhoff stress τ, the left Biot stretch tensor V and the Finger tensor B, given by

τ_{H} = 2 G \log V + Λ tr (\log V) \cdot 11 = G \log B + \frac{Λ}{2} tr (\log B) \cdot 11,

(51)

while Becker’s stress–stretch relation, expressed in terms of the Biot stress T^Biot, the right stretch tensor U and the right Cauchy-Green deformation tensor C, is

T_{B}^{Biot} = 2 G \log U + Λ tr (\log U) \cdot 11 = G \log C + \frac{Λ}{2} tr (\log C) \cdot 11 .

Since, in general,

T^{Biot} = U \cdot S_{2} = \det (U) \cdot U \cdot F^{- 1} \cdot σ \cdot F^{- T}

and

τ = \det (U) \cdot σ,

where S₂ is the symmetric second Piola–Kirchhoff stress and σ is the Cauchy stress tensor, we find

\begin{array}{l} σ = \det {(U)}^{- 1} \cdot F \cdot U^{- 1} \cdot T^{Biot} \cdot F^{T} \\ \Rightarrow τ = \underset{= R}{\underset{︸}{F \cdot U^{- 1}}} \cdot T^{Biot} \cdot F^{T} = \underset{= R}{\underset{︸}{V^{- 1} \cdot F}} \cdot T^{Biot} \cdot F^{T}, \end{array}

where the last equality follows directly from the polar decomposition F = RU = VR of the deformation gradient F. We employ the identity C = F^T F = F⁻¹ FF^T F = F⁻¹BF to compute

\begin{array}{l} τ_{B} = V^{- 1} \cdot F \cdot T_{B}^{Biot} \cdot F^{T} & = & V^{- 1} F \cdot [G \log C + \frac{Λ}{2} tr (\log C) \cdot 11] \cdot F^{T} \\ = & V^{- 1} F \cdot [G \log (F^{- 1} B F) + \frac{Λ}{2} tr (\log (F^{- 1} B F)) \cdot 11] \cdot F^{T} \\ = & V^{- 1} F \cdot [G F^{- 1} (\log B) F + \frac{Λ}{2} tr (F^{- 1} (\log B) F) \cdot 11] \cdot F^{T} \\ = & G V^{- 1} F F^{- 1} (\log B) F F^{T} + \frac{Λ}{2} tr (\log B) \cdot V^{- 1} \underset{= B}{\underset{︸}{F F^{T}}} \\ = & G V^{- 1} (\log B) B + \frac{Λ}{2} tr (\log B) \cdot V^{- 1} B \\ = & V^{- 1} \cdot [G (\log B) + \frac{Λ}{2} tr (\log B) \cdot 11] \cdot B = V^{- 1} τ_{H} B . \end{array}

The symmetric tensors V⁻¹, τ_H and B commute because their principal axes coincide, therefore

τ_{B} = V^{- 1} τ_{H} B = V^{- 1} B \cdot τ_{H} = R F^{- 1} F F^{T} \cdot τ_{H} = R F^{T} \cdot τ_{H} = V \cdot τ_{H} .

(52)

This identity allows us to obtain an upper estimate for the difference between the Kirchhoff stress corresponding to the Hencky energy and the one given by Becker’s stress–stretch relation:

τ_{B} = V \cdot τ_{H} = τ_{H} + (V - 11) \cdot τ_{H} \Rightarrow | | τ_{B} - τ_{H} | | \leq | | V - 11 | | \cdot | | τ_{H} | |,

where ∥·∥ denotes the Frobenius matrix norm on $ℝ^{3 \times 3}$ . Thus, for very small elastic strains ∥V − 11∥ ≪ 1, the corresponding Kirchhoff tensors τ_B and τ_H coincide to lowest order.

5.5. Hyperelasticity

Unlike Hencky’s logarithmic stress–stretch relation, Becker’s idealized response is generally not hyperelastic for arbitrary parameters G and K. Incidentally, this result was also shown by Carroll [51], who gave an explicit example of a cycle of loading and unloading without conservation of energy, showing that the elastic behaviour modelled, in fact, by Becker’s law¹⁶ is not path-independent.

Proposition 5.1. The stress–stretch relation

T^{Biot} (U) = 2 G \cdot \log (U) + Λ \cdot tr [\log U] \cdot 11

is hyperelastic if and only if Λ = 0 or, equivalently, ν = 0 for Poisson’s number ν. In this case¹⁷ the energy is given by

W_{Becker}^{ν = 0} (U) = 2 G [< U, \log (U) - 11 > + 3] = 2 G [< \exp (\log U), \log U - 11 > + 3],

(53)

which is the maximum entropy function.

However, Becker’s law is, of course, Cauchy-elastic for all admissible choices of parameters since the Cauchy stress depends only on the state of deformation.

Note that the elastic energy $W_{Becker}^{ν = 0}$ given by equality (53) does not fulfil some of the constitutive properties listed in Section 1.1. For example, det F → 0 does not generally imply $W_{Becker}^{ν = 0} (F) \to \infty$ ; in fact, $W_{Becker}^{ν = 0} (F)$ remains finite even for F = 0. However, the implication

\det F \to 0 \Rightarrow T^{Biot} (F) \to \infty

holds true for Becker’s elastic law, even in the case ν = 0, c.f. Figure 20.

Figure 20.

Energy () and Biot stress () according to Becker’s law for extreme volumetric stretches λ; the tangent of $W_{Becker}^{ν = 0}$ in 0 is vertical.

5.5.1. Comparison to the Valanis–Landel energy

In terms of the principal stretches λ₁, λ₂, λ₃, that is, the eigenvalues of a stretch tensor U, the energy function $W_{Becker}^{ν = 0}$ can be expressed as

\begin{array}{l} W_{Becker}^{ν = 0} (U) & = & 2 G [< U, \log (U) - 11 > + 3] \\ = & 2 G \cdot [\sum_{i = 1}^{3} λ_{i} \cdot (\log (λ_{i}) - 1)] + 6 G = : {\hat{W}}_{Becker}^{ν = 0} (λ_{1}, λ_{2}, λ_{3}) . \end{array}

This energy function is identical to the Valanis–Landel energy, which was introduced in 1967 by KC Valanis and RF Landel [52]. However, the Valanis–Landel energy is used as a model for incompressible hyperelastic materials exclusively, while $W_{Becker}^{ν = 0}$ is only applicable to Becker’s law of elasticity in the (compressible) case ν = 0 or, equivalently, for Λ = 0. Since Becker only considers compressible materials, it is not clear how to extend his constitutive law to the incompressible case. One possible way, involving the limit K → ∞, was discussed in Section 5.2.1. Another possibility, however, is to directly apply the incompressibility condition det F = 1, F the deformation gradient, to the hyperelastic model induced by the energy $W_{Becker}^{ν = 0}$ . This approach leads to a different result for uniaxial deformations: using the general formula [50]

t = \frac{d}{d λ} \hat{W} (λ, \frac{1}{\sqrt{λ}}, \frac{1}{\sqrt{λ}}),

where t is the (uniaxial) load, λ is the stretch and $\hat{W}$ is an energy function of an incompressible hyperelastic material expressed in the principal stretches, we find

\begin{array}{l} t_{Becker} & = & \frac{d}{d λ} {\hat{W}}_{Becker}^{ν = 0} (λ, \frac{1}{\sqrt{λ}}, \frac{1}{\sqrt{λ}}) \\ = & 2 G \cdot \frac{d}{d λ} [λ \cdot (\log (λ) - 1) + 2 λ^{- 1 / 2} \cdot (\log (λ^{- 1 / 2}) - 1)] = G \cdot \log (λ) \cdot (2 + λ^{- 3 / 2}) . \end{array}

(54)

Figure 21 shows the Biot stress response for uniaxial deformations, computed from the two different applications (50) and (54) of Becker’s law to the incompressible case.

Figure 21.

Becker’s law for the uniaxial deformation of incompressible materials, obtained via the limit case K → ∞ () and by applying the incompressibility restraint det F = 1 to the energy function $W_{Becker}^{ν = 0}$ ().

5.6. Becker’s law of elasticity in terms of other stresses and stretches

In Section 5.4 we already established the relation between the left Biot stretch tensor V and the Kirchhoff stress tensor τ for Becker’s law of elasticity: combining equations (52) and (51) we find

τ (V) = V \cdot (2 G \log V + Λ tr (\log V) \cdot 11) = 2 G \cdot V \cdot \log V + Λ tr (\log V) \cdot V .

(55)

Since τ = det(V)·σ in general, the Cauchy stress tensor σ can be expressed as

σ (V) = \frac{2 G}{\det (V)} \cdot V \cdot \log V + \frac{Λ}{\det (V)} \cdot tr (\log V) \cdot V .

(56)

Furthermore, we can obtain a representation of the symmetric second Piola–Kirchhoff stress tensor S₂ from the general formula T^Biot = U · S₂:

\begin{array}{l} S_{2} (U) & = & U^{- 1} \cdot T^{Biot} (U) = 2 G \cdot U^{- 1} \cdot \log (U) + Λ \cdot tr [\log U] \cdot U^{- 1} \\ = & (2 G \cdot \log (U) + Λ \cdot tr [\log U]) \cdot U^{- 1} . \end{array}

(57)

5.7. Constitutive inequalities

5.7.1. Invertibility of the force.-stretch relation

The invertibility of the force–stretch relation, also known as Truesdell’s IFS condition [16, p. 156], is fulfilled by a stress–stretch relation if and only if the mapping U ↦ T^Biot(U) is invertible. Becker’s law of elasticity satisfies this condition, as was shown in Section 4.

5.7.2. The M-condition

Since the principal matrix logarithm log: PSym(3) → Sym(3) is strictly monotone [53], the Krawietz M-condition [54]

〈 T^{Biot} (U_{1}) - T^{Biot} (U_{2}), U_{1} - U_{2} 〉 > 0 \forall U_{1}, U_{2} \in PSym (3), U_{1} \neq U_{2},

(58)

where 〈X, Y〉 = tr(Y^T X) denotes the canonical inner product on $ℝ^{3 \times 3}$ , is satisfied by the stress–stretch relation

T^{Biot} (U) = 2 G \cdot \log (U),

that is, in the special case Λ = 0. However, it is not satisfied in the general case

T^{Biot} (U) = 2 G \cdot \log (U) + Λ \cdot tr [\log U] \cdot 11

for sufficiently large K > 0: choosing

U_{1} = (\begin{matrix} 2 & 0 & 0 \\ 0 & \frac{1}{4} & 0 \\ 0 & 0 & 1 \end{matrix}) and U_{2} = 11

we find

\begin{array}{l} T (U_{1}) & = & 2 G \cdot \log (U_{1}) + Λ \cdot tr [\log U] \cdot 11 \\ = & 2 G \cdot (\begin{matrix} \log 2 & 0 & 0 \\ 0 & \log \frac{1}{4} & 0 \\ 0 & 0 & 0 \end{matrix}) + Λ \cdot \log (\det U_{2}) \cdot 11 \\ = & 2 G \cdot (\begin{matrix} \log 2 & 0 & 0 \\ 0 & - \log 4 & 0 \\ 0 & 0 & 0 \end{matrix}) + Λ \cdot \log (\frac{1}{2}) \cdot 11 = 2 G \cdot (\begin{matrix} \log 2 & 0 & 0 \\ 0 & - \log 4 & 0 \\ 0 & 0 & 0 \end{matrix}) - Λ \cdot \log (2) \cdot 11 \end{array}

as well as T(U₂) = T(11) = 0 and thus

\begin{array}{l} 〈 T^{Biot} (U_{1}) - T^{Biot} (U_{2}), U_{1} - U_{2} 〉 & = & 〈 2 G \cdot (\begin{matrix} \log 2 & 0 & 0 \\ 0 & - \log 4 & 0 \\ 0 & 0 & 0 \end{matrix}) - Λ \cdot \log (2) \cdot 11, (\begin{matrix} 1 & 0 & 0 \\ 0 & - \frac{3}{4} & 0 \\ 0 & 0 & 0 \end{matrix}) 〉 \\ = & 2 G \cdot [\log (2) + (- \log (4)) \cdot (- \frac{3}{4})] - Λ \cdot \log (2) \cdot [1 - \frac{3}{4}] \\ = & 2 G \cdot [\log (2) + \frac{3 \cdot \log (2^{2})}{4}] - \frac{Λ \cdot \log (2)}{4} \\ = & 2 G \cdot [\frac{4 \cdot \log (2)}{4} + \frac{6 \cdot \log (2)}{4}] - \frac{Λ \cdot \log (2)}{4} \\ = & \frac{\log (2)}{4} \cdot [20 G - Λ] < 0 \end{array}

for Λ > 20 G.

5.7.3. Hill’s inequality

Since the energy function of any hyperelastic law satisfying the M-condition is convex in terms of the right Biot stretch tensor U, it follows from Section 5.7.2 that the mapping

U \mapsto W_{Becker}^{ν = 0} (U) = 2 G [< U, \log (U) - 11 > + 3]

is convex on PSym(3). However, the mapping X ↦ 〈exp(X),X − 11〉 is not convex on Sym(3). Therefore $W_{Becker}^{ν = 0}$ does not satisfy Hill’s inequality [55], which holds for an energy function $W : PSym (3) \to ℝ, U \mapsto W (U)$ if and only if the mapping

\tilde{W} : Sym (3) \to ℝ, \tilde{W} (X) = W (\exp (X)),

is convex. This condition is often restated as the convexity of the mapping $\log U \mapsto \tilde{W} (\log U)$ for log U ∈ Sym(3). This inequality is independent of the rank-one convexity of the energy: for example, while it is easy to see that the quadratic Hencky strain energy fulfils Hill’s inequality, it is not rank-one convex [56, 57].

5.7.4. The Baker–Ericksen inequality

A stress–stretch relation fulfils the Baker–Ericksen inequality if

(σ_{i} - σ_{j}) \cdot (λ_{i} - λ_{j}) > 0 for all λ_{i} \neq λ_{j},

where λ_k denotes the kth principal stretch and σ_k denotes the corresponding principal Cauchy stress, that is, the corresponding eigenvalue of the Cauchy stress tensor σ.

Proposition 5.2. The stress–stretch relation

σ (V) = \frac{2 G}{\det (V)} \cdot V \cdot \log V + \frac{Λ}{\det (V)} \cdot tr (\log V) \cdot V

does not satisfy the Baker–Ericksen inequality for any G > 0, 3 Λ + 2 G > 0.

Proof. We assume without loss of generality that V is in the diagonal form V = diag(λ₁, λ₂, λ₃) with λ₁, λ₂, λ₃ > 0. Then

\log (V) = \log (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}) = (\begin{matrix} \log λ_{1} & 0 & 0 \\ 0 & \log λ_{2} & 0 \\ 0 & 0 & \log λ_{3} \end{matrix})

and

\begin{array}{l} σ (V) & = & \frac{2 G}{λ_{1} λ_{2} λ_{3}} \cdot V \cdot \log V + \frac{Λ}{λ_{1} λ_{2} λ_{3}} \cdot \log (λ_{1} λ_{2} λ_{3}) \cdot V \\ = & \frac{2 G}{λ_{1} λ_{2} λ_{3}} \cdot (\begin{matrix} λ_{1} \cdot \log λ_{1} & 0 & 0 \\ 0 & λ_{2} \cdot \log λ_{2} & 0 \\ 0 & 0 & λ_{3} \cdot \log λ_{3} \end{matrix}) + \frac{Λ \cdot \log (λ_{1} λ_{2} λ_{3})}{λ_{1} λ_{2} λ_{3}} \cdot (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}) . \end{array}

The principal stresses σ_k are the diagonal entries of σ, thus

σ_{k} = \frac{λ_{k}}{λ_{1} λ_{2} λ_{3}} \cdot (2 G \log λ_{k} + Λ \log (λ_{1} λ_{2} λ_{3})) .

(59)

We let λ₁ = 1/e, λ₂ = 1/e² and λ₃ = e³ to find

σ_{1} = \frac{\frac{1}{e}}{\frac{1}{e} \cdot \frac{1}{e^{2}} \cdot e^{3}} \cdot (2 G \log (\frac{1}{e}) + Λ \log (\frac{1}{e} \cdot \frac{1}{e^{2}} \cdot e^{3})) = \frac{2 G}{e} \cdot \log (\frac{1}{e}) = - \frac{2 G}{e}

as well as

σ_{2} = \frac{\frac{1}{e^{2}}}{\frac{1}{e} \cdot \frac{1}{e^{2}} \cdot e^{3}} \cdot (2 G \log (\frac{1}{e^{2}}) + Λ \log (\frac{1}{e} \cdot \frac{1}{e^{2}} \cdot e^{3})) = \frac{2 G}{e^{2}} \cdot \log (\frac{1}{e^{2}}) = - \frac{4 G}{e^{2}} .

Since λ₁ > λ₂ we find

(σ_{1} - σ_{2}) \cdot (λ_{1} - λ_{2}) > 0 \Leftrightarrow σ_{1} > σ_{2} \Leftrightarrow - \frac{2 G}{e} > - \frac{4 G}{e^{2}} \Leftrightarrow 1 < \frac{2}{e},

showing that the Baker–Ericksen inequality does not hold in this case.

Therefore Becker’s law does not satisfy the rank-one convexity condition either, since a rank-one convex stress–stretch relation always fulfils the Baker–Ericksen inequality. In contrast, Hencky’s elastic law (see (46)) does fulfil the Baker–Ericksen inequality [12].

5.7.5. The ordered force inequalities

An isotropic stress–stretch relation satisfies the ordered force inequalities (or ‘OF inequalities’) if

(T_{i} - T_{j}) \cdot (λ_{i} - λ_{j}) \geq 0 for all i, j \in {1, 2, 3}, i \neq j,

(60)

where λ_i, λ_j are the principal stretches of a deformation and T_i, T_j are the corresponding principal forces, that is, the eigenvalues of the Biot stretch tensor T^Biot. To show that Becker’s law fulfils the OF inequalities for all G > 0, 3 Λ + 2 G > 0, we assume without loss of generality that a given stretch tensor U is in the diagonal form U = diag(λ₁, λ₂, λ₃) and compute

\begin{array}{l} T^{Biot} & = & 2 G \cdot \log (U) + Λ \cdot tr [\log (U)] \cdot 11 \\ = & 2 G \cdot (\begin{matrix} \log (λ_{1}) & 0 & 0 \\ 0 & \log (λ_{2}) & 0 \\ 0 & 0 & \log (λ_{3}) \end{matrix}) + Λ \cdot tr (\begin{matrix} \log (λ_{1}) & 0 & 0 \\ 0 & \log (λ_{2}) & 0 \\ 0 & 0 & \log (λ_{3}) \end{matrix}) \cdot 11 \\ = & (\begin{matrix} 2 G \log (λ_{1}) & 0 & 0 \\ 0 & 2 G \log (λ_{2}) & 0 \\ 0 & 0 & 2 G \log (λ_{3}) \end{matrix}) + Λ \log (λ_{1} λ_{2} λ_{3}) \cdot 11 . \end{array}

The eigenvalues of T_i of T^Biot corresponding to the principal stretches λ_i are therefore

λ_{i} = 2 G \log (λ_{i}) + Λ \log (λ_{1} λ_{2} λ_{3}),

and thus (60) can be written as

\begin{array}{l} (2 G \log (λ_{i}) + Λ \log (λ_{1} λ_{2} λ_{3}) - 2 G \log (λ_{j}) + Λ \log (λ_{1} λ_{2} λ_{3})) \cdot (λ_{i} - λ_{j}) \geq 0 \\ \Leftrightarrow 2 G \cdot (\log (λ_{i}) - \log (λ_{j})) \cdot (λ_{i} - λ_{j}) \geq 0 \end{array}

(61)

Due to the monotonicity of the natural logarithm, (61) holds for all $λ_{1}, λ_{2}, λ_{3} \in ℝ^{+}$ and all G > 0.

5.8. Existence results

The following proposition represents a basic existence result by Ciarlet [58, Theorem 6.7-1] for solutions to the so-called pure displacement problem in nonlinear elasticity.

Proposition 5.3. Let $Ω \subset ℝ^{3}$ be a domain with a boundary Γ of class C², and let

E = \frac{1}{2} (C - 11) = \frac{1}{2} ({(11 + \nabla u)}^{T} (11 + \nabla u) - 11)

denote the Green–Lagrange strain tensor of a deformation φ(x) = x + u(x). Moreover, assume that the constitutive law is of the form

S_{2} (E) = Λ \cdot tr (E) \cdot 11 + 2 G E + O (| | E | |^{2})

with Λ, G > 0, where S₂ denotes the second Piola–Kirchhoff stress tensor. Then for each number p > 3 there exists a neighbourhood Z^p of the origin in the space L^p(Ω) and a neighbourhood U^p of the origin in the subspace

V^{p} (Ω) = {v \in W^{2, p} (Ω) | v = 0 o n Γ}

of the Sobolev space W²,^p(Ω) such that for each f ∈ Z^p, the boundary value problem

\begin{array}{r} - div S_{1} (F) = f & i n Ω \\ u = 0 & o n Γ \\ S_{1} (F) = (11 + \nabla u) \cdot S_{2} (E (u)) \end{array}}

has exactly one solution u in U^p.

To show that Becker’s stress–stretch relation fulfils the conditions of Theorem 5.3 we compute

\begin{array}{l} S_{2} = U^{- 1} \cdot T^{Biot} & = & [2 G \cdot \log (U) + Λ \cdot tr [\log U] \cdot 11] \cdot U^{- 1} \\ = & [2 G \cdot \log (\sqrt{C}) + Λ \cdot tr [\log \sqrt{C}] \cdot 11] \cdot {\sqrt{C}}^{- 1} \\ = & [G \cdot \log (C) + \frac{Λ}{2} \cdot tr [\log C] \cdot 11] \cdot {\sqrt{C}}^{- 1} . \end{array}

For small enough ∥C − 11∥, we can employ the series expansion

\log (C) = (C - 11) - \frac{1}{2} \cdot {(C - 11)}^{2} + \dots

of the matrix logarithm to find

\begin{array}{l} S_{2} & = & [G \cdot \log (C) + \frac{Λ}{2} \cdot tr [\log C] \cdot 11] \cdot {\sqrt{C}}^{- 1} \\ = & [G \cdot (C - 11 + O (| | C - 11 | |^{2})) + \frac{Λ}{2} \cdot tr [(C - 11) + O (| | C - 11 | |^{2})] \cdot 11] \cdot {\sqrt{C}}^{- 1} \\ = & [G \cdot (C - 11) + \frac{Λ}{2} \cdot tr [C - 11] \cdot 11 + O (| | C - 11 | |^{2})] \cdot {\sqrt{C}}^{- 1} . \end{array}

(62)

Since

{\sqrt{C}}^{- 1} = 11 - \frac{1}{2} \cdot (C - 11) + O (| | C - 11 | |^{2})

for small ∥C − 11∥, (62) can be expressed as

\begin{array}{l} S_{2} & = & [G \cdot (C - 11) + \frac{Λ}{2} \cdot tr [C - 11] \cdot 11 + O (| | C - 11 | |^{2})] \cdot [11 - \frac{1}{2} \cdot (C - 11) + O (| | C - 11 | |^{2})] \\ = & G \cdot (C - 11) + \frac{Λ}{2} \cdot tr [C - 11] \cdot 11 + O (| | C - 11 | |^{2}) = Λ \cdot tr (E) \cdot 11 + 2 G E + O (| | E | |^{2}) . \end{array}

Proposition 5.3 can therefore be directly applied to Becker’s law of elasticity.

Footnotes

A. Notation

The following notation is employed throughout the article:

$11 = diag (1, 1, 1) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})$ identity matrix

$Ω_{0} \subset ℝ^{3}$ reference configuration

$ϕ : Ω_{0} \to ℝ^{3}$ deformation mapping

$F = \nabla ϕ (x) \in ℝ^{3 \times 3}$ deformation gradient

$U = \sqrt{F^{T} F}$ right Biot stretch tensor

C = F^T F = U² right Cauchy-Green deformation tensor

$V = \sqrt{F F^{T}}$ left Biot stretch tensor

B = FF^T = V² left Cauchy-Green deformation tensor

R = FU⁻¹ = V⁻¹F ∈ SO(3) orthogonal polar factor of the deformation gradient

σ Cauchy stress tensor, ‘true stress’

τ = det(F)·σ Kirchhoff stress tensor

S₁ = det(F) σF^−T first Piola–Kirchhoff stress tensor, ‘nominal stress’

S₂ = det(F) F⁻¹ σF^−T symmetric second Piola–Kirchhoff stress tensor

T^Biot = US₂ = R^TS₁ Biot stress tensor

G, Λ Lamé constants

K bulk modulus

E Young’s modulus

ν Poisson’s ratio

B. Appendix

Acknowledgements

We discovered Becker’s original paper in late August 2013 and carefully transcribed it with the help of Mrs B Sacha until the end of September 2013. In January 2014 we understood the meaning of Becker’s tables on page 340 and finished a preliminary version in March 2014. We are grateful to Professor Kolumban Hutter from the ETH Zürich for his helpful remarks on the transcription.

Funding

The author(s) received no financial support for the research, authorship, and/or publication of this article.

Notes

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