Analytical solution of the uniaxial extension problem for the relaxed micromorphic continuum and other generalized continua (including full derivations)

We derive analytical solutions for the uniaxial extension problem for the relaxed micromorphic continuum and other generalized continua. These solutions may help in the identification of material parameters of generalized continua which are able to disclose size effects.


Introduction
In this paper we continue our investigation of analytical solutions for isotropic the relaxed micromorphic model (and other isotropic generalized continuum models).It follows our recent exposition of analytical solutions for the simple shear [27], bending [26], and torsion problem [13,28].Here, we consider the uniaxial extension problem, which, in classical isotropic linear elasticity, allows to determine the size-independent longitudinal modulus M macro = λ macro + 2µ macro .
Here, we show the genealogy tree of the generalized continuum models: The strain gradient theory and second gradient theory are equivalent [1,17], and contain additionally the couple stress theory as a special case.Using the Curl as primary differential operator for the curvature terms allows a neat unification of concepts.
For some of the traditional models, uniaxial extension gives still rise to size-effects in the sense that thinner samples are comparatively stiffer.In that case, the inhomogeneous response is triggered by the boundary conditions for the additional kinematic fields which are applied at the upper and lower surface.We refer the reader to the introduction of [26,27,28,32] concerning the relevance of the scientific question as well as its importance for the determination of material parameters for generalized continua [33].Indeed, the obtained analytical formulas can be used to determine size-dependent and size-independent material parameters.The notation follows that of [26,27,28].We recapitulate shortly.
The paper is now structured as follows.We start with a recapitulation of the uniaxial extension problem in the classical linear elasticity.The solution is homogeneous and uniquely determines the longitudinal modulus M macro = λ macro + 2µ macro .Then, we consider the isotropic relaxed micromorphic continuum.The boundary conditions for the additional non-symmetric micro-distortion field P derives from the so-called consistent coupling conditions where ν is the normal unit vector to the upper and lower surface.It turns out that for zero Poisson modulus on the micro-and meso-scale, ν micro = ν e = 0, respectively, the solution remains homogeneous and no sizeeffects is observed.In the case with arbitrary ν micro , ν e ∈ [−1, 1/2] the solution will be inhomogeneous and size-effects appear.The limiting stiffness as the ratio between the thickness and the characteristic length tends to zero (h/L c → 0) is given by M = Me Mmicro Me+Mmicro which is both smaller then M micro = λ micro + 2µ micro and M e as well greater than M macro = λ macro + 2µ macro .

Notation
We define the scalar product a, b := n i=1 a i b i ∈ R for vectors a, b ∈ R n , the dyadic product a ⊗ b := (a i b j ) i,j=1,...,n ∈ R n×n and the euclidean norm a2 := a, a .We define the scalar product and the Frobenius-norm P 2 := P , P for tensors P , Q ∈ R n×n in the same way.Moreover, P T := (P ji ) i,j=1,...,n denotes the transposition of the matrix P = (P ij ) i,j=1,...,n , which decomposes orthogonally into the skew-symmetric part skew P := 1 2 (P − P T ) and the symmetric part sym P := 1 the orthogonal decomposition possible for a matrix is P = dev sym P + skew P + tr(P ) 3 1.The Lie-Algebra of skew-symmetric matrices is denoted by so(3) := {A ∈ R 3×3 | A T = −A}, while the vector space of symmetric matrices Sym(3) := {S ∈ R 3×3 | S T = S}.The Jacobian matrix Du and the curl for a vector field u are defined as where × denotes the cross product in R 3 .We also introduce the Curl and the Div operators of the The cross product between a second order tensor and a vector is also needed and is defined row-wise as follow where m ∈ R 3×3 , b ∈ R 3 , and is the Levi-Civita tensor.Using the one-to-one map axl : so(3) → R 3 we have The inverse of axl is denoted by Anti: R 3 → so(3).

Uniaxial extension problem for the isotropic Cauchy continuum
The strain energy density for an isotropic Cauchy continuum is while the equilibrium equations without body forces are Div [2 µ macro symDu + λ macro tr (Du) 1] = 0.
Since the uniaxial extensional problem is symmetric with respect to the x 2 -axis, there will be no dependence of the solution on x 1 and x 3 .The boundary conditions for the uniaxial extension problem are (see Fig. 1) The homogeneous displacement field solution u 2 (x 2 ), the gradient of the displacement Du(x 2 ), and the strain energy W (γ) for the uniaxial extension problem are where is the extensional stiffness (or pressure-wave modulus, longitudinal modulus).
Here and in the remainder of this work, the elastic coefficients µ i , λ i are expressed in [MPa], the coefficients a i and the intensity of the displacement γ are dimensionless, the characteristic lengths L c and the height h are expressed in meter [m].
3 Uniaxial extension problem for the isotropic relaxed micromorphic model The general expression of the strain energy for the isotropic relaxed micromorphic continuum is where we have the parameters related to the meso-scale, the parameters related to the micro-scale, the Cosserat couple modulus, the proportionality stiffness parameter, the characteristic length, and the three dimensionless general isotropic curvature parameters, respectively.This energy expression represents the most general isotropic form possible for the relaxed micromorphic model.In the absence of body forces, the equilibrium equations are then Div σ:= [2µ e sym (Du − P ) + λ e tr (Du − P ) 1 + 2µ c skew (Du − P )] = 0, σ − 2µ micro sym P − λ micro tr (P ) 1 The ansatz for the micro-distortion P (x 2 ), the displacement u(x 2 ), and consequently the gradient of the displacement Du(x 2 ) is It is important to underline that, given the subsequent ansatz (14), it holds that tr (Curl P ) = 0.This reduces immediately the number of curvature parameters appearing in the uniaxial extension solution.
The boundary conditions for the uniaxial extension are Here, the constraint on the components of P is given by the consistent coupling boundary condition where ν is the normal unit vector to the upper and lower surface.After substituting the ansatz ( 14) into the equilibrium equation ( 13) we obtain the following four differ-ential equations where M e = λ e + 2µ e and M micro = λ micro + 2µ micro .Being careful of substituting the system of differential equation with one in which eq.( 17) 2 and eq.( 17) 4 are replaced with their sum and their difference, respectively, we have . It is highlighted that eq.( 18) 4 is a homogeneous second order differential equation depending only on f m (x 2 ) with homogeneous boundary conditions eq.( 15).The fact that eq.( 18) 4 is an independent equation has its meaning in the symmetry constraint of the uniaxial extensional problem in the direction along the x 2 -and x 3 -axis, which requires that P 11 (x 2 ) = P 33 (x 2 ).From eq.( 18) it is possible to obtain the following relation between P 22 (x 2 ) and u 2 (x 2 ) which, after substituting it back into eq.(18), allows us to obtain the following system of three second order differential equations in u 2 (x 2 ), P 22 (x 2 ), and f p (x 2 ) where It is highlighted that due to the positive definiteness conditions ( 12), (z 2 , z 3 ) > 0 and z 1 = 0 if and only if λ micro = λ e = 0 (zero Poisson's ratio case which is studied in Sec.3.1) and Mmicro Me = λmicro λe .If z 1 is zero eqs.(20) uncouples completely into three independent differential equations in u 2 , f p , and f m respectively.
After applying the boundary conditions eqs.(15), the solution in terms of u 2 (x 2 ), P 11 (x 2 ), P 22 (x 2 ), and In the above expressions all the quantities are real and well defined due to the positive definiteness conditions eq.( 12).Indeed, since the coefficients z 1 , z 2 , and z 3 may be rewritten in terms of the meso and micro bulk and shear modulus as we can write the expression of f 1 as follows showing that the positive definiteness of the energy (11) implies that f 1 is a strictly positive real number.Moreover, the function g and it is monotone increasing since its first derivative is given by which it is positive for all x ∈ (0, ∞).Hence, it follows that due to the positive definiteness of the elastic energy which implies that which completes our proof that all the quantities from ( 22) are real and well-defined.The strain energy associated with this solution is The plot of the extensional stiffness M w while varying L c is shown in Fig. 2.  The values of M macro and M micro are where M i = κ i + 4 3 µ i and λ i = κ i − 2 3 µ i with i = {macro, micro, e}. 3  It is highlighted that the structure (•)e (•)micro (•)e+(•)micro is applicable to evaluate the macro coefficients only for the shear and bulk modulus because of the orthogonal energy decomposition "sym dev/tr" of which they are related, and especially here it would be a mistake to use this structure for the coefficient M macro since it will give the value at the micro-scale.For more details about lim Lc→∞ M w see Appendix A.

Uniaxial extension problem for the isotropic relaxed micromorphic model
with ν e = ν micro = 0 A vanishing Poisson's ratio at the meso-and micro-scale (ν e = ν micro = 0) corresponds to a vanishing first Lamé parameter (λ e = λ micro = 0).It is easy to see from eq.( 21) and eq.( 22) that these conditions correspond to with M i = λ i + 2µ i = 2µ i with i = {micro, e}.Since the non-linear terms in the solution eq.( 22) vanish, we retrieve which is a homogeneous elastic solution satisfying the equilibrium equation in the case of a constant microdistortion tensor P (see the Appendix D of [28] for further details) 3 For the sake of completeness are reported here also the relations between the Young's modulus Ei and the Poisson's ratio νi in terms of κi and µi: Ei The strain energy associated with this solution is where M macro = 2µ macro + λ macro = 2µ macro = 2µe µmicro µe+µmicro is the macro extensional stiffness, since λ macro = ν macro = 0.

Uniaxial extension problem for the isotropic micro-stretch model in dislocation format
In the micro-stretch model in dislocation format [5,15,20,22,30], the micro-distortion tensor P is devoid from the deviatoric component dev sym P = 0 ⇔ P = A + ω1, A ∈ so(3), ω ∈ R. The expression of the strain energy for this model in dislocation format can be written as [20]: According with the reference system shown in Fig. 1, the ansatz for the displacement and micro-distortion fields is The boundary conditions at the free surface are then Since the ansatz requires A = 0, the micro-stretch model coincides with the micro-void model which will be presented in the Sec. 6.

Uniaxial extension problem for the isotropic Cosserat continuum
The strain energy for the isotropic Cosserat continuum in dislocation tensor format (curvature energy expressed in terms of CurlA) can be written as [3,8,13,14,18,21,26,27,29] According to the reference system shown in Fig. 1 and the ansatz (14), which has to be particularized as A = skew P ∈ so(3), the ansatz for the displacement field and the micro-rotation for the Cosserat model is Since A = 0, the Cosserat model is not able to catch any non-homogeneous response for the uniaxial extension problem and the classical solution ( 9) is retrieved.
The couple stress model [10,11,16,19,23], which appears by constraining A = skew Du ∈ so(3) in the Cosserat model, is also not able to catch a non-homogeneous response for the uniaxial extension problem since, due to the ansatz, we would have skew Du = 0 as it can be seen in eq.( 40).

Uniaxial extension problem for the isotropic micro-void model in dislocation tensor format
The strain energy for the isotropic micro-void continuum in dislocation tensor format can be obtained from the relaxed micromorphic model by formally letting µ micro → ∞ (while keeping κ micro finite) and can be written as [4,27] W (Du, ω, Curl (ω1)) = µ macro dev sym Du Here, ω : R 3 → R is the additional scalar micro-void degree of freedom [4].The equilibrium equations, in the absence of body forces, are According with the reference system shown in Fig. 1, the ansatz for the displacement field and the function ω(x 2 ) have to be The boundary conditions for the uniaxial extension are After substituting the ansatz (44) into the equilibrium equations (42) we obtain the following two differential equations After applying the boundary conditions eqs.(45), the solution in terms of u 2 (x 2 ) and ω(x 2 ) of the system eqs.(46) is .
where f 1 > 0, z 1 > 0, and z 2 > 0 are strictly positive in order to match the positive definiteness conditions eq.( 43), and the same reasoning applied in the relaxed micromorphic model sections still holds.The strain energy associated with this solution is The plot of the extensional stiffness M w while varying L c is shown in Fig. 3.The values of the extensional stiffness M w for L c → 0 and where µ macro = µ e for µ micro → ∞, according to eq.( 29).We note that the extensional stiffness remains bounded as L c → ∞ (h → 0).
7 Uniaxial extension problem for the classical isotropic micromorphic continuum without mixed terms The expression of the strain energy for the classical isotropic micromorphic continuum [7,17] without mixed terms (like symP , sym (Du − P ) , etc.) and simplified curvature expression [26,28] can be written as: a 3 D (tr (P ) 1) = 0 , where (µ e ,κ e = λ e + 2/3 µ e ), (µ micro ,κ micro = λ micro + 2/3 µ micro ), µ c , L c > 0, and ( a 1 , a 2 , a 3 )> 0 in order to guarantee the positive definiteness of the energy.According with the reference system shown in Fig. 1, the ansatz for the displacement field and the classical micromorphic model is The boundary conditions for the uniaxial extension are assumed to be The calculations are deferred to the micro-strain model Sec.8 since the ansatz, the equilibrium equations, and the boundary conditions are the same, therefore the solution will also be the same.

Uniaxial extension problem for the micro-strain model without mixed terms
The micro-strain model [9,12,31] is the classical Mindlin-Eringen [7,17] model particular case in which it is assumed a priori that the micro-distortion remains symmetric, P = S ∈ Sym(3).
The strain energy which we consider is [26,28] W (Du, S, DS) = µ e (sym Du − S) The chosen 2-parameter curvature expression represents a simplified isotropic curvature (the full isotropic curvature for the micro-strain model would still count 8 parameters [2]).
The equilibrium equations, in the absence of body forces, are therefore the following Div σ:= [2µ e (sym Du − S) + λ e tr (Du − S) 1] = 0, 2µ e (sym Du − S) + λ e tr (Du − S) 1 − 2µ micro S − λ micro tr (S) 1 a 3 D (tr (S) 1) = 0 , where (µ e ,κ e = λ e + 2/3 µ e ), (µ micro ,κ micro = λ micro + 2/3 µ micro ), L c > 0, and ( a 1 , a 3 )> 0 in order to guarantee the positive definiteness of th energy.The boundary conditions for the uniaxial extension are assumed to be According with the reference system shown in Fig. 1, the ansatz for the displacement field and the micro-distortion is (which coincides with the classical micromorphic model eq.( 52)) After substituting the ansatz (57) into the equilibrium equations (55) we obtain the following four differential equations Being careful of substituting the system of differential equation with one in which eq.(58) 2 and eq.(58) 4 are replaced with their sum and their difference, respectively, we have where f p (x 2 ) := P 11 (x 2 ) + P 33 (x 2 ) and f m (x 2 ) := P 11 (x 2 ) − P 33 (x 2 ).It is highlighted that eq.(59) 4 is a homogeneous second order differential equation depending only on f m (x 2 ) with homogeneous boundary conditions eq.( 56).Also here, the fact that eq.(59) 4 is an independent equation has its meaning in the symmetry constraint of the uniaxial extensional problem in the direction along the x 2 -and x 3 -axis, which requires that P 11 (x 2 ) = P 33 (x 2 ).
The solution and the measure of the apparent stiffness are too complicated to be reported here, but nevertheless, it is possible to plot how the apparent stiffness behaves while changing L c (see Fig. 4).We note that the extensional stiffness remains bounded as L c → ∞ (h → 0) and converges to M e .The solution obtained for the micro-strain model for the uniaxial extension problem also holds for the classical micromorphic problem presented in Sec.7.

Uniaxial extension problem for the second gradient continuum
The strain energy density for the isotropic second gradient with simplified curvature [1,6,17,26,28] is where (µ macro , κ macro , µ, a 1 , a 3 ) > 0 in order to guarantee the positive definiteness of the energy.Due to the uniaxial extension problem symmetry the following structure of u = (0, u 2 (x 2 ), 0) T has been chosen, which results in having only the component u 2,2 different from zero in the gradient of the displacement Du.The boundary conditions for the uniaxial extension are (see Fig. 1) assumed to be After substituting the expression of the displacement field in eq.( 61), the non-trivial equilibrium equation reduces to After applying the boundary conditions to the solution of eq.( 63), it results that u 2 (x 2 ) is given by [24,25] where f 1 > 0 is strictly positive in order to match the positive definiteness conditions and the same reasoning applied in the relaxed micromorphic model sections still holds.The strain energy (61) becomes then The plot of the extensional stiffness M w while varying L c is shown in Fig. 5.

Conclusions
Only the second gradient formulation produces an unbounded apparent stiffness as L c → ∞ (h → 0).Otherwise, different bounded limit stiffnesses are observed.The relaxed micromorphic model determines M = Me Mmicro Me+Mmicro , which is less than M micro and M e , while the micro-strain model determines M e as limit stiffness.The Cosserat model is not able to catch a non-homogeneous solution and provides no size-effect.The different limit stiffnesses for the relaxed micromorphic model versus the full micromorphic and microstrain model approach respectively, suggest that the meaning of classical experimental tests does not have an unambiguous deformation and micro-deformation solution field anymore, and this is due to the fact that we can have different boundary conditions on the components of the micro-distortion tensor depending on what each model requires to constrain.This allows the existence of different uniaxial extension-like problems and not just one like for a classical Cauchy material.
where a is an arbitrary constant.This solution to eqs.(66) is therefore not unique.Inserting Du and Dζ from eq.(68) in eq.( 65), the following energy expression is recovered which has to be minimized with respect to a in order to remove the non-uniqueness of the equilibrium system eqs.(66),which means that the following relation The solution eq.( 71) satisfy the equilibrium equations, the boundary conditions, and the minimum energy requirement.The expression of the energy now become with M = Me M micro Me + M micro the extensional stiffness for the relaxed micromorphic when Lc → ∞.

Figure 1 :
Figure 1: Sketch of an infinite stripe of thickness h subjected to uniaxial extension boundary conditions.

2 c
Curl a 1 dev sym Curl P + a 2 skew Curl P + a 3 tr (Curl P ) m = 0 .