Journal of Elasticity 65: 149–168, 2001. © 2002 Kluwer Academic Publishers. Printed in the Netherlands.
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Incompressible Elastic Bodies with Non-Convex Energy under Dead-Load Surface Tractions DOMENICO DE TOMMASI, PILADE FOTI, SALVATORE MARZANO and MARIO DANIELE PICCIONI Dipartimento di Ingegneria Civile e Ambientale, Sezione di Ingegneria Strutturale, Politecnico di Bari, Via E. Orabona, 4, 70125 Bari, Italy. E-mail:
[email protected] Received 11 February 2002 Abstract. In this paper we study the equilibrium deformations of an incompressible elastic body with a non-convex strain energy function which is subjected to a homogeneous distribution of dead-load tractions. To determine the stable solutions we consider the mixtures of the phases which minimize the total energy density. In the special case of a trilinear material we discuss the stability of the equilibrium phases in detail. Finally, we show that multiphase solutions are possible when the surface loads correspond to a critical simple shear and we sketch their possible forms. Mathematics Subject Classifications (2000): 74B20, 74G65, 74N10. Key words: non-linear elasticity, phase transitions, non-convex energy.
1. Introduction Non-convexity conditions for the elastic strain energy function have a fundamental role in the description of a number of material behaviours within the framework of the non-linear theory of elasticity. As is well known such constitutive assumptions have provided an efficient tool for predicting thermally induced martensitic transformations in crystalline materials. Furthermore, by considering multi-well energy functions, Ball and James [1, 2] have obtained an interpretation of fine phase mixtures in terms of minimizing sequences for the total energy functional. A model of incompressible, isotropic, elastic material, whose energy function is not rank-one convex, has been analyzed by Fosdick and other authors in a number of papers. In this context, on examination of the helical shearing of a circular tube, Fosdick and MacSithigh [3] have found that stable equilibrium deformations must be discontinuous in the deformation gradient when the boundary data are relatively large. Further, the analysis of the structure of the multiphase equilibrium deformations for the anti-plane shear problem [4] has shown a localization of large deformations which is similar to the results established by means of the theory of plasticity. Finally, such a constitutive model [5–7] has allowed the description of certain internal deformations patterns observed in classical experiments concerning the torsional deformations of steel bars (see, e.g., [8]).
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In the present paper we consider such a model of isotropic, incompressible material. More precisely, by assuming for the strain energy function a “non-convex” form similar to the expression proposed in [3–7], we study the plane equilibrium deformations of a body which is subjected to a homogeneous distribution of deadload surface tractions. Such a loading condition has a long history in non-linear elasticity and has been used to explore the features of the equilibrium problem solutions ever since Rivlin [9, 10] studied homogeneous deformations of a cube of neo-Hookean material. Subsequently, other investigations concerning homogeneous dead-load conditions for elastic bodies have been performed by Ball and Schaeffer [11] for a cube of arbitrary isotropic incompressible material and by Kearsley [12] and MacSithigh [13] for a thin sheet of Mooney–Rivlin material. In the same vein, Chen [14] has studied general stability conditions for homogeneous deformations in elastic bodies. Here, we consider a hyperelastic, incompressible, isotropic and homogeneous body which occupies the infinite rectangular cylinder B in the reference configuration. We choose a Cartesian coordinate system {O; X1 , X2 , X3 } having X3 -axis parallel to the axis of the cylinder and denote by {e1 , e2 , e3 } the orthonormal basis associated with such a Cartesian frame. Thus, if is the cross section of B, we have B = × R. We assume that the body B is held in equilibrium under the sole action of forces sˆ (on the lateral surface ∂ B ) corresponding to a biaxial dead-load traction: sˆ (X) = Sn(X),
X ∈ ∂ B = ∂ × R,
(1.1)
where S is a constant tensor of the form S = s1 e1 ⊗ e1 + s2 e2 ⊗ e2 ,
(1.2)
and n(X) is the unit outward normal at X ∈ ∂ B. This type of boundary condition should be useful in describing meaningful physical situations such as the behaviour of specimens in biaxial loading experiments. In this paper, we restrict our attention to states of plane strain; that is, we consider isochoric deformations f: B X → x = f(X) ∈ R3 of the form x1 = f1 (X1 , X2 ), x2 = f2 (X1 , X2 ), (X1 , X2 ) ∈ , (1.3) x3 = X3 , where (x1 , x2 , x3 ) denote the coordinates of x. As a consequence of our assumptions, the deformation gradient F := ∇f satisfies at each point the following conditions: Fe3 = FT e3 = e3 ,
det F = 1.
(1.4)
Since for a plane isochoric deformation the first and second principal invariants of the right Cauchy–Green strain tensor C = FT F are equal and satisfy the inequality IC = IIC = |F|2 3,
(1.5)
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Figure 1. The strain energy density σ as a function of the strain invariant κ.
the strain energy density σ may be reduced to a function σ¯ of the strain invariant (1.6) κ := IC − 3. Following Fosdick and Zhang [5], we suppose σ¯ : [0, ∞[ κ → σ = σ¯ (κ) to be a class C 1 , piecewise class C 2 , non-convex and strictly increasing function. More precisely, we assume the following constitutive hypotheses: (h1 ) σ¯ (0) = σ¯ (0) = 0, σ¯ (κ) 0 ∀κ ∈ [0, ∞[, lim
κ→∞
σ¯ (κ) = ∞; κ
(h2 ) there exist κ3 and κ4 , with 0 < κ3 < κ4 , such that σ¯ (κ) > 0 ∀κ ∈ ] 0, κ3 [ ∪ ] κ4 , ∞[, σ¯ (κ) < 0 ∀κ ∈ ] κ3 , κ4 [; (h3 ) there exist κ1 and κ2 , with 0 < κ1 < κ3 < κ4 < κ2 , such that σ¯ (κ1 ) = σ¯ (κ2 ), σ¯ (κ2 ) − σ¯ (κ1 ) = σ¯ (κ1 )(κ2 − κ1 ). (σ¯ and σ¯ denote the first and the second derivatives of σ¯ with respect to κ.) Notice that the above assumptions require that σ¯ be convex in the domain [0, κ1 ]∪ [κ2 , ∞[ and strictly non-convex in the interval ]κ1 , κ2 [ (see Figure 1). As a consequence of the constitutive assumptions (h1 )–(h3 ), we have that the strain energy σ is not rank-one convex as a function of the deformation gradient F. In the present work we focus our attention on the stable solutions of the equilibrium problem above defined. Precisely, using the standard minimum energy criterion, we determine the (globally) stable solutions by minimizing the total
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potential energy functional. The special form of the surface tractions allows us to construct the stable deformations as a mixture of the phases which minimize the total energy density. We devote Section 2 to characterize these minimizing phases for different values of the load parameters s1 , s2 . In Section 3 we consider a special class of materials which are described by a trilinear response function σ¯ . In this case, we discuss the stability of the equilibrium phases in detail. In particular, we determine a closed curve in the plane of the load parameters which separates the regions of stability of the phases. We conclude the paper by analyzing piecewise homogeneous solutions (Section 4). We show that multiphase solutions are possible when the surface loads correspond to a “critical” simple shear. Finally, we discuss some kinematical properties of these mixtures of phases. 2. The Minimization Problem The assumptions made in the previous section allow to write the total potential energy per unit length of B as σ¯ (κ) dA − Sn · f dL, (2.1) E(f) =
∂
where the two integrals are the elastic energy stored in the body and the surface potential, respectively. In this paper we restrict our attention to isochoric plane deformations of B which are continuous and piecewise differentiable on B; that is, we consider deformations f defined by (1.3) and (1.4), with the functions f1 , f2 continuous on and piecewise class C 1 . We denote by Ꮽ the set of deformations of B so defined. In order to find stable solutions of the equilibrium problem related to E, we consider the following variational problem: min E(f). f∈Ꮽ
Using the divergence theorem, the functional E can be put in the form σ¯ (κ) − S · F dA, E(f) =
(P )
(2.2)
which shows that the total energy functional E depends on f only through its deformation gradient F. Thus, ¯f ∈ Ꮽ is a solution of problem (P ) if and only if the gradient field ∇ ¯f minimizes the integrand of E ε(F) := σ¯ (κ) − S · F
(2.3)
for almost every X ∈ . Therefore, in the remainder of this section, we fix our attention to the following minimization problem: min ε(F), F∈ᏹ
(p)
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where ᏹ denotes the set of tensors satisfying conditions (1.4): ᏹ := F | det F = 1, Fe3 = FT e3 = e3 .
(2.4)
We postpone (Section 4) the kinematical problem of constructing deformations ¯f whose gradient fields ∇ ¯f minimize the function ε for almost every X ∈ . The special expression assumed by the function ε suggests a minimization procedure for ε of the following form: (i) for each fixed value of the strain invariant κ, we first search for the tensors F which are solutions of the following problem: maximize ϕ(F) := S · F over all tensors F such that
F ∈ ᏹ, F ∈ Lκ := {F | F · F = 3 + κ 2 = const};
(2.5)
(ii) then, we completely determine the solutions of (p) by minimizing the function ε which now depends only on κ. (i) Maximization of the Function ϕ In order to solve problem (2.5), we preliminarily make some helpful observations. We denote by Lin3 the set of linear transformations of {e3 }⊥ into itself: (2.6) Lin3 := H ∈ Lin | He3 = HT e3 = 0 . Moreover, we denote by W3 the skew tensor of Lin3 whose axial vector is e3 : WT3 = −W3 ,
W3 u = e3 × u
for any vector u.
(2.7)
It is immediately seen that if A ∈ Lin3 then A ∈ Sym
⇔
A · W3 = 0.
(2.8)
We have the following characterization for the solutions of the problem (2.5). PROPOSITION 1. Let F be a solution of the problem (2.5). Then the tensors S F and S R are symmetric, where R is the rotation in the polar decomposition of F. Proof. We first consider the case κ > 0. We observe that for κ > 0 the set Aκ := ᏹ ∩ Lκ is a differentiable manifold whose tangent space A˙ κ (F) at F is given by ˙ (F) ∩ L˙ κ (F). A˙ κ (F) = ᏹ
(2.9)
˙ (F) and L˙ κ (F) denote the tangent spaces of ᏹ and Lκ at F, respectively.) (Here ᏹ In particular, by (2.4) and (2.5), we have that ˙ (F) = H ∈ Lin3 | F−T · H = 0 , ᏹ (2.10) L˙ κ (F) = H ∈ Lin | F · H = 0 .
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Now, let F be a solution of the constrained extremum problem (2.5). Then the gradient ∇ϕ( F) of the function ϕ evaluated at F is orthogonal to A˙ κ ( F). Since ∇ϕ F = S, (2.11) in view of (2.9) we have
˙ F ∩ L˙ κ F . S · H = 0 ∀H ∈ ᏹ
(2.12)
Now consider the polar decomposition of F: F= R U, where R is a rotation and U a symmetric and positive definite tensor. In the present T case, since Fe3 = F e3 = e3 , it follows that Ue3 = e3
and
R ∈ Rot(e3 ).
(2.13)
RW3 belong to both of the By (2.10) and (2.13), we have that the tensors FW3 and ˙ ˙ tangent spaces ᏹ(F) and Lκ (F). Hence, by (2.12) we obtain S · FW3 = 0
(2.14)
S· RW3 = 0.
(2.15)
and
Finally, if we put (2.14) and (2.15) in the form T
F S · W3 = 0,
T
R S · W3 = 0, T
(2.16) T
and observe (see (1.2) and (2.13)) that F S, R S ∈ Lin3 , by (2.8) we immediately T T conclude that F S, R S ∈ Sym. To complete the proof, it remains to consider the special case κ = 0. Now, from (2.4) and (2.5) it follows ᏹ ∩ L0 = Rot(e3 ),
(2.17)
so that, if F= R U is a solution of the extremum problem (2.5), we have U = I,
F= R.
(2.18)
Thus, in view of (2.17) and (2.18), we deduce that the gradient of ϕ at the extremum R. Hence, since RW3 point R is orthogonal to the tangent space of Rot(e3 ) at belongs to the tangent space of Rot(e3 ), in view of (2.11) we have S· RW3 = 0.
(2.19)
Therefore, for κ = 0 (2.15) is also valid and we conclude that S F = S R ∈ Sym. ✷ We denote by Rot(e ) the set of rotations about e . Further, R(α) is the rotation about e3 of the 3 3
angle α.
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In order to determine the solutions of problem (2.5), we introduce the decomposition of S in terms of P := I − e3 ⊗ e3
and
D := e1 ⊗ e1 − e2 ⊗ e2 .
(2.20)
Clearly, S, defined by (1.2), may now be written in the form S = sP + τ D,
(2.21)
where we have set s :=
s1 + s2 , 2
τ :=
s1 − s2 . 2
(2.22)
In passing we observe that D satisfies the following identity: R D R = D for any R ∈ Rot(e3 ).
(2.23)
Now, let F= R U be a solution of problem (2.5). By Proposition 1 and (2.21) we have sP R + τ D R ∈ Sym,
sP R U + τ D R U ∈ Sym.
(2.24)
By (2.23) it follows that D R ∈ Sym; thus (2.24)1 yields sP R ∈ Sym,
(2.25)
which, in the case s = 0, is satisfied only if R is a rotation of an angle α = 0 or α = π about e3 : R = I or
R = −P + e3 ⊗ e3 .
(2.26)
Substituting (2.26)1 (or (2.26)2 ) into (2.24)2 , we see that τ D U ∈ Sym.
(2.27)
If we assume τ = 0, we have that (2.27) implies D U= UD,
(2.28)
which means that the symmetric tensors D and U commute. As is well known, such a condition is equivalent to the coaxiality of D and U; hence, in view of (2.20)2 and (2.13)1 , U admits the following representation U = µ1 e1 ⊗ e1 + µ2 e2 ⊗ e2 + e3 ⊗ e3 .
(2.29)
Further, in order to satisfy the incompressibility constraint and the condition F ∈ Lκ , the principal stretches µ1 , µ2 take the following values (2.30) µ2 = µ−1 µ1 ∈ λ, λ−1 , 1 ,
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where we have set κ 1 2 1 2 κ λ := + κ + 4, λ−1 = − + κ + 4. (2.31) 2 2 2 2 The preceding analysis has shown that in order to find the solutions of problem (2.5), we have to consider tensors F= R U with R and U given by (2.26) and (2.29), respectively. When we evaluate the function ϕ in correspondence of such deformation gradients, we obtain one of the following values: ϕ(F) = ±s(µ1 + µ2 ) ± τ (µ1 − µ2 ).
(2.32)
In view of (2.30) and (2.31), it is easy to recognize that, for a given κ, the maximum value ϕ(κ) ¯ := max ϕ(F)
(2.33)
F∈ᏹ∩Lκ
is given by
ϕ(κ) ¯ = |s| κ 2 + 4 + |τ |κ;
(2.34)
moreover, such a value is obtained by taking R and U in the following manner:
R(0) = I for s > 0, (2.35) R=
for s < 0, R(π ) = e3 ⊗ e3 − P U=
λe1 ⊗ e1 + λ−1 e2 ⊗ e2 + e3 ⊗ e3 λ−1 e1 ⊗ e1 + λe2 ⊗ e2 + e3 ⊗ e3
for sτ > 0, for sτ < 0,
(2.36)
where λ is defined by (2.31)1 . We recall that expressions (2.34)–(2.36) have been deduced under the assumptions s = 0 and τ = 0. We now consider the case s = 0 and τ = 0. In this situation, equation (2.27) is identically satisfied so that there are no restrictions on the principal axes of U: T R(α) λe1 ⊗ e1 + λ−1 e2 ⊗ e2 + e3 ⊗ e3 R (α), α arbitrary, U= Uα := (2.37) where λ, the greatest principal stretch, is also given by (2.31)1 and α, the angle between the axis corresponding to λ and e1 , is arbitrary. Then, by means of the same arguments used to determine the solutions of (2.5) in the case τ = 0, we find that the maximum value ϕ(κ) ¯ is also given by equation (2.34) (with τ = 0). F is given by (2.35), whereas Further, the rotation R corresponding to the solution U is an arbitrary tensor expressed by (2.37). Finally, we consider the special case s = 0 and τ = 0. Now, equation (2.25) is identically satisfied for any rotation about e3 : R= R(γ ),
γ arbitrary.
(2.38)
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Hence, equation (2.24)2 takes the form D R(γ ) U= U R T (γ )D.
(2.39)
On the other hand, (2.23) yields the following identity
R(−γ /2)D R(γ /2). R T (γ )D =
(2.40)
Thus, R T (γ )D is symmetric and, in view of (2.39), we recognize that such a tensor commutes with U. Hence, U and R T (γ )D are coaxial and, since the principal axes T R(−γ /2)e1 , R(−γ /2)e2 , e3 (see (2.20)2 and (2.40)), the stretch of R (γ )D are tensor U can be put in the form R(γ /2). U= R(−γ /2)[µ1 e1 ⊗ e1 + µ2 e2 ⊗ e2 + e3 ⊗ e3 ]
(2.41)
Of course, since F ∈ ᏹ ∩ Lκ , the principal stretches µ1 , µ2 are also given by (2.30), (2.31). From (2.38) and (2.41), setting α = γ /2, it follows that R(α), F= R U= R(α)[µ1 e1 ⊗ e1 + µ2 e2 ⊗ e2 + e3 ⊗ e3 ]
α arbitrary. (2.42)
Hence, in the present case s = 0 and τ = 0, if we evaluate the function ϕ, we have ϕ( F) = τ (µ1 − µ2 ).
(2.43)
Therefore, the maximum value ϕ(κ) ¯ is also given by (2.34) (with s = 0); moreover, ϕ(κ) ¯ is obtained for the deformation gradients given by
R(α) λe1 ⊗ e1 + λ−1 e2 ⊗ e2 + e3 ⊗ e3 R(α), α arbitrary, for τ > 0, Fα = −1
R(α), α arbitrary, for τ < 0, R(α) λ e1 ⊗ e1 + λe2 ⊗ e2 + e3 ⊗ e3 (2.44) with λ also defined by (2.31)1 . (ii) Minimization of the Function ε We now turn to problem (p). In order to completely determine the solutions of (p) we must find the values of the strain measure κ which minimize the energy function ε(κ) = σ¯ (κ) − ϕ(κ), ¯
(2.45)
where ϕ(κ) ¯ is defined by (2.34). Therefore, in the remainder of this section we shall consider the following problem min ε(κ). κ0
(2.46)
In passing we observe that, under the current hypotheses on the strain energy function σ¯ (see (h1 )–(h3 )), the problem (2.46) admits at least one solution. In fact,
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by (h1 ) and (2.34) we have that the function ε is bounded below on [0, ∞[ and satisfies the coerciveness condition ε(κ) = ∞; κ→∞ κ lim
(2.47)
therefore any minimizing sequence for ε is bounded. Then, by the continuity of ε it follows that ε attains its minimum in [0, ∞[. We now wish to characterize the solutions of the problem (2.46). Suppose κ¯ is a minimizer of the function ε. If we evaluate the first derivative of ε at κ = κ¯ we have ¯ = σ¯ (κ) ¯ − ϕ¯ (κ) ¯ = 0. ε (κ)
(2.48)
Moreover, the requirement for κ¯ to be an absolute minimizer of ε can be put in the form: σ¯ (κ) − σ¯ (κ) ¯ ϕ(κ) ¯ − ϕ( ¯ κ) ¯
for any κ 0.
(2.49)
By (2.34) we have ϕ¯ (κ) > 0 for any κ 0, so that ϕ¯ is a convex function: ¯ − κ) ¯ ϕ(κ) ¯ − ϕ( ¯ κ) ¯ ϕ¯ (κ)(κ
for any κ 0;
(2.50)
then, in view of (2.48) and (2.50), (2.49) yields the following inequality: ¯ − κ) ¯ for any κ 0, σ¯ (κ) − σ¯ (κ) ¯ σ¯ (κ)(κ
(2.51)
which shows that σ¯ is convex at κ = κ. ¯ Therefore, by the constitutive assumptions (h3 ), we have that κ¯ ∈ [0, κ1 ] ∪ [κ2 , ∞[. We can get an interpretation of (2.48) and (2.51) by considering the graphs of the functions σ¯ and ϕ¯ (see Figure 2). It is readily seen that such conditions tell us that κ¯ is the abscissa of a point of intersection between the curves ϕ¯ and σ¯ within the convexity domain [0, κ1 ] ∪ [κ2 , ∞[. In order for κ¯ to be a minimizer of the energy function ε, it must hold κ (2.52) σ¯ (t) − ϕ¯ (t) dt 0 for any κ 0. ε(κ) − ε(κ) ¯ = κ¯
We can picture energy differences by considering the areas of the regions between the graphs of σ¯ and ϕ¯ in a way similar to that shown by Ericksen [15]. For instance, in the case represented in Figure 2, we have three points of intersection κ < κ < κ of the curves σ¯ and ϕ¯ . Of course, only κ and κ satisfy the necessary condition (2.51). Then, if A1 and A2 denote the areas of the regions between the curves σ¯ and ϕ¯ : κ κ A2 = (2.53) σ¯ (t) − ϕ¯ (t) dt, σ¯ (t) − ϕ¯ (t) dt, A1 = κ
κ
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Figure 2. Plots of σ¯ and ϕ¯ as functions of κ.
it is easy to recognize that condition (2.52) yields for A1 > A2 , κ for A1 < A2 , κ¯ = κ for A1 = A2 . κ ,κ
(2.54)
Remarkably, in the case A1 = A2 , we have two values of the strain measure κ which minimize the energy function ε: ε(κ ) = ε(κ ) ε(κ) for any κ 0.
(2.55)
3. Trilinear Materials In order to discuss in detail the solutions of the problem (2.46), in the following of the paper we refer to a trilinear material defined by the response function κ , 0 κ a, a 5 κ a κ a, (3.1) σ¯ (κ) = − + 2, a 4 5 5 3 bκ + − ab, κ a, 4 4 4 where a and b are positive constants (Figure 3). For such a choice of σ¯ , it is easy to see that, when τ = 0, there exist at most three intersections between σ¯ and ϕ¯ . Whereas, for τ = 0 the curves σ¯ and ϕ¯ also meet at the origin. By the discussion in the previous section, it results that, when the “equal areas” condition A1 = A2 is satisfied, two globally stable solutions κ¯ = κ and κ¯ = κ are possible. We denote by (scr , τcr ) the critical values of the load parameters (s,τ ) which satisfy such a condition. In the present case of a
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Figure 3. Response function σ¯ for a trilinear material defined by (3.1) (a = 0.5, b = 0.1).
Figure 4. The regions of stability of the phases for the trilinear material (3.1) (a = 0.5, b = 0.1).
trilinear material, the condition A1 = A2 defines a closed curve ! in the plane s, τ (see Figure 4). Of course, in view of (2.34) it is readily seen that such a curve is symmetric with respect to the axes. The curve ! plays a fundamental role in discussing the solutions of the problem (2.46). Indeed, if we consider a point (s,τ ) which belongs to the region D bounded by !, it results A1 > A2 . Then, by (2.54)1 , we have κ¯ = κ ; that is the value of κ which minimizes the energy function ε lies on the left increasing branch of σ¯ . Whereas, when the point (s,τ ) is external to !, it results A1 < A2 , so that by (2.54)2 we have κ¯ = κ ; that is the minimizer κ¯ belongs to the right increasing
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branch of σ¯ . Therefore, if we stipulate that each branch of σ¯ determines a phase of the material, we have that the curve ! separates the regions of stability of such phases. A brief comment is necessary for the special case τ = 0. We first note that for values of |s| close to zero the curves σ¯ and ϕ¯ intersect only in the origin, so that κ¯ = 0 is the unique solution of the problem (2.46). In view of (2.34), the slope of ϕ¯ at κ = 0 is given by ϕ¯ (0) = |s|/2. Thus, when |s|/2 is greater than a −1 , the slope of σ¯ at κ = 0, the curve ϕ¯ intersects the left branch of σ¯ in two points (κ = 0, κ = κ < a). By (2.52), it follows ε(0) > ε(κ ), so that, for |s|/2 > a −1 , a stable solution κ¯ = κ bifurcates from the trivial solution κ¯ = 0. However, such a bifurcation occurs if 2a −1 < |scr |, where |scr | is the value defined by the condition A1 = A2 . On the contrary, when 2a −1 |scr | the solution “jumps” from κ¯ = 0 to the value κ¯ = κ belonging to the third branch of σ¯ . Notice that such a result is qualitatively similar to those obtained by Kearsley [12] and MacSithigh [13] in the case of an incompressible elastic sheet stretched by two pairs of equal forces. 4. Stable Deformations We now return to the fundamental problem (P ). Our project is to determine the stable deformations of the body B by making use of the solutions of (p). Of course, every homogeneous deformation ¯f of the form: ¯f(X) = F(X − O) + q,
(4.1)
with F a solution of the problem (p) and q ∈ R3 an arbitrary point, is a solution of problem (P ). Some additional remarks are necessary for the special case s = 0. In view of the results obtained in Section 2, for a given value of τ > 0 we have a class of homogeneous solutions of (P ), whose deformation gradients Fα are defined by (2.44)1 : ¯ 1 ⊗ e1 + λ¯ −1 e2 ⊗ e2 + e3 ⊗ e3 Fα = R(α) λe R(α), α arbitrary, (4.2) where λ¯ is the greatest principal stretch corresponding to κ, ¯ the solution of problem (2.46). In order to discuss such a class of solutions, we notice that λ¯ e1 ⊗ e1 + λ¯ −1 e2 ⊗ e2 + e3 ⊗ e3 corresponds to a pure shear deformation, which can be decomposed into a simple shear of amount κ¯ followed by a rotation about e3 (cf., e.g., [16, Section 43]): R(−2α)(I ˜ + κd ¯ 1 ⊗ d2 ), λ¯ e1 ⊗ e1 + λ¯ −1 e2 ⊗ e2 + e3 ⊗ e3 =
(4.3)
where κ¯ 1 α˜ := − arctan , 2 2
(4.4)
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and di = R(α˜ − π/4)ei ,
i = 1, 2.
(4.5)
In particular, if we take α = α˜ and substitute (4.3) into (4.2), we get Fα˜ = I + κ¯ R(−α)d ˜ 1⊗ R(−α)d ˜ 2 = I + κa ¯ 1 ⊗ a2 ,
(4.6)
where ai := R(−π/4)ei ,
i = 1, 2;
(4.7)
that is, for α = α˜ the deformation of B is a simple shear of amount κ¯ and axis a1 . In the general case α = α, ˜ by using (4.6) and the relation R(α) = R(α − α) ˜ R(α), ˜ we can put the deformation gradient Fα in the form: Fα = R(α − α)(I ˜ + κa ¯ 1 ⊗ a2 ) R(α − α); ˜
(4.8)
hence Fα = R(2α − 2α) ˜ I + κ¯ R(α˜ − α)a1 ⊗ R(α˜ − α)a2 .
(4.9)
Equation (4.9) shows that Fα can be regarded as a simple shear of amount κ¯ ˜ according to the and axis R(α˜ − α)a1 followed by a rotation of angle 2α − 2α, well known characterization for plane isochoric deformations (cf., e.g., [16, Section 43]). The analysis in Section 2 has shown that for some values of the load parameters (s, τ ) there exists a multiplicity of tensors F minimizing the total energy function ε. Such a “freedom” in the choice of the deformation gradients opens the possibility to obtain stable solutions which are qualitatively different from (4.1). Here, we focus our attention on non-trivial piecewise homogeneous deformations. More precisely, we consider a partition of in n ( 2) complementary closed regions i (i = 1, 2, . . . , n) and assume that ¯fi (X) = Fi (X − O) + qi
for X ∈ i ,
i = 1, 2, . . . , n,
(4.10)
where Fi (i = 1, 2, . . . , n) are n constant tensors and qi (i = 1, 2, . . . , n) are n points of R3 . It is well known that the occurrence of such deformations is restricted by Hadamard’s jump condition. Indeed, because ¯f is continuous on , across the (flat) interface Si,j between two contiguous subregions i and j , the corresponding deformation gradients Fi , Fj must satisfy the condition (cf., e.g., [17]) Fj − Fi = a ⊗ n,
(4.11)
where n denotes the unit normal to Si,j and the vector a is the amplitude of the jump across Si,j . Remarkably, such a condition shows that non-trivial piecewise
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deformations are possible if and only if there exist minimizers of the function ε which are rank-one connected. In order to establish existence of piecewise homogeneous deformations, we first consider the case that the deformations Fi and Fj correspond to the same value of the strain measure κ. In this situation, the principal stretches of Fi and Fj coincide so that, in order to satisfy the compatibility condition (4.11), the pair Fi , Fj must be necessarily a twin (cf., e.g., [17, Section 7]). Since Fi , Fj ∈ ᏹ ∩ Lκ , it is easily seen that there exist two rotations Q and Q about e3 such that Fj = Q Fi Q .
(4.12)
On the other hand, the unit vector e3 is also an eigenvector of the right stretch tensor of Fi . Hence, by a property of twins (see [17, Section 7, p. 24]), we recognize that the pair Fi , Fj cannot be a twin. Therefore, two deformations having the same value of κ cannot connect. Now, the analysis in Sections 2, 3 has shown that for values of the load parameters (s,τ ) ∈ / ! the minimizers F of ε correspond to the unique value κ¯ of κ defined by (2.54)1,2 ; therefore, we have that non-trivial piecewise homogeneous deformations are not possible when (s, τ ) ∈ / !. We pass to consider the case (s, τ ) ∈ !. We first assume s = 0. In this case, the rotations R and the stretch tensors U minimizing the energy ε are given by (2.35) and (2.36) for τ = 0, and by (2.35) and (2.37) for τ = 0. As a consequence, in both situations we have that the deformation gradients F = R U are symmetric. Thus, in the present case the compatibility condition (4.11) takes the form Fj = Fi + η n ⊗ n,
(4.13)
with η ∈ R and n a unit vector perpendicular to e3 . Hence, by recalling that Fi and Fj must satisfy the incompressibility condition, we obtain −1
ηFi n · n = 0,
(4.14)
which is satisfied only if η = 0. Thus, when (s,τ ) ∈ ! with s = 0, non-trivial piecewise homogeneous deformations are not possible. It remains to explore the special case s = 0 and τ = τcr , corresponding to a “critical” simple shear. In the present situation we have two different sets of minimizers of ε which are defined by the values κ and κ of κ¯ (see (2.54)3 ). Since we have seen that in each of these two sets there are no rank-one connections, we can directly consider the case that the connecting deformations Fi and Fj are characterized by different values of κ. ¯ In view of the representation formula (4.8), we assume Fi = R αi − α˜ I + κ a1 ⊗ a2 R αi − α˜ , (4.15) Fj = R αj − α˜ I + κ a1 ⊗ a2 R αj − α˜ ,
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where αi , αj ∈ [0, 2π [, and where we have set κ 1 α˜ := − arctan , 2 2 Now, if we let β = αj − α˜ − αi + α˜ , m= R αi − α˜ n,
κ 1 α˜ := − arctan . 2 2
(4.16)
Q= R(β), c= R α˜ − αi a,
(4.17)
compatibility condition (4.11) takes the form Q I + κ a1 ⊗ a2 Q = I + κ a1 ⊗ a2 + c ⊗ m.
(4.18)
Finally, applying (4.18) to the unit vector p := R(π/2)m = R αi − α˜ + π/2 n,
(4.19)
we obtain Q I + κ a1 ⊗ a2 Qp = I + κ a1 ⊗ a2 p.
(4.20)
Therefore, the problem of finding rank-one connections between the two sets of minimizers (4.15) is equivalent to the problem of finding a rotation Q about e3 and a unit vector p orthogonal to e3 such that equation (4.20) is satisfied. The solutions of (4.20) are given by the following PROPOSITION 2. Equation (4.20) has two distinct solutions: (i) β = 0 and p = ± a1 ; R(−2α˜ )a2 . (ii) β = 2(α˜ − α˜ ) and p = ± Proof. We first note that equation (4.20) may be rewritten as Q − QT Qp + κ (Qp · a2 )Qa1 − κ (p · a2 )a1 = 0.
(4.21) (4.22)
(4.23)
In view of the relation Q − QT = 2 sin βW3 (see (4.17)2 and (2.7)), (4.23) takes the form 2 sin βe3 × Qp + κ (Qp · a2 )Qa1 − κ (p · a2 )a1 = 0. If we multiply (4.24) by a2 and observe that Qa1 · a2 = sin β, we obtain sin β 2(Qp · a1 ) + κ (Qp · a2 ) = 0,
(4.24)
(4.25)
which gives sin β = 0
or κ (Qp · a2 ) = −2Qp · a1 .
(4.26)
Notice that, since R(β + π)(I + κ a1 ⊗ a2 ) R(β + π) = R(β)(I + κ a1 ⊗ a2 ) R(β) for any β,
without loss of generality we assume β ∈ [0, π[.
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(4.26)1 yields β = 0; that is, in view of (4.17)2 , Q = I. In this case, (4.24) takes the form (κ − κ )(p · a2 )a1 = 0,
(4.27)
which yields p = ± a1 . Therefore, (4.21) is a solution of (4.20). We now suppose that (4.26)2 holds. By substituting (4.26)2 into (4.24), we obtain 2 sin βe3 × Qp − 2(Qp · a1 )Qa1 − κ (p · a2 )a1 = 0. Multiplying (4.28) by Qp, we have (Qp · a1 ) 2a1 · p + κ (a2 · p) = 0,
(4.28)
(4.29)
which implies Qp · a1 = 0 or
2a1 · p + κ (a2 · p) = 0.
(4.30)
If we assume that (4.30)1 holds, (4.26)2 reduces to κ Qp · a2 = 0.
(4.31)
Since κ > 0 and |p| = 1, we have that such an equation contradicts (4.30)1 ; therefore, (4.29) is satisfied if and only if (4.30)2 holds. In view of (4.16)1 , it is readily seen that (4.30)2 is equivalent to (4.22)2 . Next, by substituting (4.22)2 into (4.26)2 we have R(−2α˜ )a2 · a1 , (4.32) R(−2α˜ )a2 · a2 = −2 Q κ Q which, by (4.16)2 , yields R(−2α˜ )a2 . Q R(−2α˜ )a2 = ±
(4.33)
Finally, if we recall that β ∈ [0, π [ is the rotation angle of Q, equation (4.33) implies (4.22)1 . We have shown that (4.22)1 and (4.22)2 are equivalent to (4.26)2 and (4.29), respectively. Since (4.26)2 and (4.29) have been obtained by multiplying equation (4.24) by the linear independent vectors a2 and Qp, we may conclude that (4.22) is a solution of (4.24). ✷ The above proposition shows that for a critical simple shear (s = 0, τ = τcr ) stable piecewise deformations are possible. In particular, one can choose a value of αi , so that the phase Fi is fixed (see (4.15)1 and (4.9)): Fi = A := R(2αi − 2α˜ ) I + κ R(α˜ − αi )a2 ; R(α˜ − αi )a1 ⊗ (4.34)
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then, the angle αj determining the phase Fj (see (4.15)2 ) is given by (4.21)1 or (4.22)1 . Therefore, for each phase corresponding to the strain κ there are exactly two phases B and C corresponding to the strain κ : R(α˜ − αi )a2 , R(α˜ − αi )a1 ⊗ B := R(2αi − 2α˜ ) I + κ (4.35) C := R(2αi − 4α˜ + 2α˜ ) I R(2α˜ − α˜ − αi )a2 , + κ R(2α˜ − α˜ − αi )a1 ⊗ which satisfy the compatibility condition (4.11). In order to discuss such piecewise deformations, we note that, in view of (4.11), (4.17)3 and (4.19), the unit vector d := R α˜ − αi p (4.36) is parallel to the interfaces between the phases Fi , Fj . Then, by (4.21)2 and (4.22)2 , we have (neglecting the unessential ambiguity of sign of p) (4.37) d= R α˜ − αi a1 when Fj is the phase B, and d= R(−α˜ − αi )a2
(4.38)
when Fj is the phase C. We first consider the interface between the phases A and B. In view of (4.34) and (4.35)1 , we recognize that A and B define two deformations each of which corR(α˜ − αi )a1 responds to a simple shear deformation of (common) axis sA = sB := followed by a rotation of amount 2αi − 2α˜ . Therefore, by (4.37) we have that the interface A/B is a shearing plane for these two simple shear deformations; thus, the interface A/B is a plane which is transported undeformed in the deformation of B. Figure 5(a) shows a possible piecewise stable deformation of B which consists of a mixture of parallel layers of the phases A and B. We now analyze the interface between the phases A and C. From (4.35)2 we see that the phase C corresponds to a simple shear of amount κ having the axis
(a)
(b)
Figure 5. Phase mixtures for a critical simple shear (s = 0, τ = τcr ).
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R(2α˜ − α˜ − αi )a1 followed by a rotation of angle 2αi − 4α˜ + 2α˜ . sC := Therefore, in the present case the axis sC of C is different from the axis sA of A. By (4.38), we can put the unit vector d (parallel to the interface) in the following alternative forms: R(α˜ − αi )a1 = R −2α˜ + π/2 sA , d= R −2α˜ + π/2 (4.39) R 2α˜ − α˜ − αi a1 d= R −2α˜ + π/2 = R −2α˜ + π/2 sC . In view of (4.39) and (4.16), we see that d and sA subtend the angle π/2 + arctan(κ /2) and that d and sC subtend the angle π/2 + arctan(κ /2). By well known properties of simple shear deformations (see also [16, Section 43]), we readily check that the interface A/C is an undeformed plane for both phases A and C. In other words, the interface between the phases A and C is the unique plane of the simple shear deformations (corresponding to A and C ) which is transported undeformed and does not coincide with the shearing plane. Figure 5(b) is a schematic picture of a stable deformation of B which is a mixture of the phases A, B and C. REMARK. It is interesting to notice that for both connections A/B and A/C the surface of separation between the phases is an undeformed plane. This property is consistent with the result obtained by Abeyaratne and Knowles [18] in the analysis of piecewise plane equilibrium states for incompressible isotropic materials.
Acknowledgements The authors would like to express their gratitude to Professor Roger Fosdick for helpful discussions during his visit in Bari which stimulated the present research. The authors are also indebted to Professor Gianpietro Del Piero for his valuable comments. This work has been supported by COFIN–MURST “Modelli Matematici per la Scienza dei Materiali” and by “Fondi di Ricerca di Ateneo”. References 1. 2. 3. 4. 5.
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