Finite Elements for the Treatment of the Inextensibility Constraint

Comput Mech DOI 10.1007/s00466-017-1437-9

ORIGINAL PAPER

Fiber-reinforced materials: ﬁnite elements for the treatment of the inextensibility constraint

Ferdinando Auricchio1 · Giulia Scalet1 · Peter Wriggers2

Received: 5 April 2017 / Accepted: 29 May 2017 © Springer-Verlag GmbH Germany 2017

Abstract The present paper proposes a numerical frame- 1 Introduction work for the analysis of problems involving fiber-reinforced anisotropic materials. Specifically, isotropic linear elastic The use of fiber-reinforcements for the development of novel solids, reinforced by a single family of inextensible fibers, and efficient materials can be traced back to the 1960s are considered. The kinematic constraint equation of inex- and, to date, it is an active research topic. Such materi- tensibility in the fiber direction leads to the presence of als are composed of a matrix, reinforced by one or more an undetermined fiber stress in the constitutive equations. families of fibers which are systematically arranged in the To avoid locking-phenomena in the numerical solution due matrix itself. The interest towards fiber-reinforced mate- to the presence of the constraint, mixed finite elements rials stems from the mechanical properties of the fibers, based on the Lagrange multiplier, perturbed Lagrangian, and which offer a significant increase in structural efficiency. penalty method are proposed. Several boundary-value prob- Particularly, these materials possess high resistance and lems under plane strain conditions are solved and numerical stiffness, despite the low specific weight, together with an results are compared to analytical solutions, whenever the anisotropic mechanical behavior determined by the fiber derivation is possible. The performed simulations allow to distribution. Typical examples are carbon fiber-reinforced assess the performance of the proposed finite elements and epoxy resins, boron fiber-reinforced aluminium, and nylon- to discuss several features of the developed formulations or steel-reinforced rubbers [1]. Thanks to such important concerning the effective approximation for the displacement features, fiber-reinforced materials are largely employed in and fiber stress fields, mesh convergence, and sensitivity to several fields as mechanical components (e.g., soft actuators), penalty parameters. aeronautical and aerospace parts (e.g., fuselage, radomes), civil engineering structures (e.g., pile foundations massif), Keywords Fiber-reinforced material · Inextensible fibers · or home products (e.g., hoses), and in the domain of biolog- Anisotropic material · Finite element analysis · Mixed finite ical materials studied in biomechanics (e.g., bones, vegetal element method tissues) [2]. Given the widespread use of fiber-reinforced materials, a B Giulia Scalet comprehensive understanding of their behavior is fundamen- [email protected] tal for design and production purposes. Extensive theoretical Ferdinando Auricchio surveys focusing on the mechanics of fiber-reinforced solids [email protected] are available from the literature, e.g., [1,3–5], in addition to Peter Wriggers works investigating the wave propagation problem, e.g., [6]. [email protected] A common approach is to consider the fiber-reinforced material as an elastic continuum which is internally con- 1 Department of Civil Engineering and Architecture, University of Pavia, via Ferrata 3, 27100 Pavia, Italy strained. In particular, constraints of incompressibility and/or inextensibility have been treated theoretically in both small 2 Institut für Kontinuumsmechanik, Gottfried Wilhelm Leibniz Universität Hannover, Appelstraße 11, 30167 Hannover, and large strains [1,7–17]. The presence of these constraints Germany in the formulation determines new features which are not 123 Comput Mech encountered in classical elasticity and which require special the penalty parameter is addressed. The simulations pro- attention in the theoretical and computational formulation. vide interesting results to understand the consequences of From the numerical point of view, the presence of con- constraining the extensional deformation in the fiber direc- straints leads to locking-phenomena in the numerical solution tion. and appropriate numerical methods have to be adopted. The paper is organized as follows. Section 2 presents an Specifically, in the finite element framework, standard dis- overview of the continuum problem. Then, Sect.3 presents placement interpolations cannot be used and ad-hoc inter- the proposed constraint formulations, while Sect. 4 derives polations need to be adopted [18]. This has been largely the related finite element formulations. Section 5 presents discussed in solid and fluid mechanics in case of incom- the numerical results. Finally, summary and conclusions are pressibility or contact at both small and finite strains [19]. given in Sect. 6. For incompressible and inextensible fiber-reinforced materials mixed finite elements based on a penalty formulation have been proposed for small deformations and tested on sim- 2 Continuum formulation ple beam problems under plane stress in [13]. Simacek and Kaliakin [20] developed the governing equations for incom- This section is dedicated to a brief overview of the continuum pressible plates exhibiting a direction of inextensibility. The equations for a three-dimensional body made of isotropic paper by Hayes and Horgan [9] presents a formulation of a linear elastic material reinforced by a single family of parallel class of mixed boundary-value problems for isotropic and inextensible fibers. The reader is referred to extensive surveys transversely isotropic materials and establishes a uniqueness as [1,3–5] for further details. theorem which provides necessary and sufficient conditions for uniqueness of solution to the mixed problems for homo- geneous materials. 2.1 Kinematics and equilibrium equations The present work is motivated by the current interest in the formulation of novel modeling and numerical approaches We consider a three-dimensional deformable continuous ⊂ R3 for the analysis of constrained fiber-reinforced materials. The body, consisting of a domain and a boundary ∂ ∂ ∂ aim of this work is to develop a computational framework , subdivided in turn into u and t , representing for the phenomenological theory of isotropic linear elastic respectively the boundary where displacements and surface ∂ ∪ ∂ = ∂ materials reinforced by a single family of parallel inexten- tractions are imposed, such that u t . sible fibers. These materials offer the framework to easily Assuming small strains, the problem is expressed by the understand the physical problem and to analyze it from the following equilibrium equation: theoretical and numerical point of view. In this way we are able to propose a complete approach to the analysis of divσ + b = 0 in (1) these materials, that can be extended to more complex situa- tions. We recall, in fact, that finite strain approaches, finding where σ is the symmetric Cauchy stress tensor and b is the applications in soft tissue biomechanics and in the mechan- body force per unit volume vector. The symmetric infinites- ics of fiber-reinforced rubber-like materials, have been also imal strain tensor ε is defined as follows: largely studied, e.g., [21–25], but are out of the present scope. 1 ε =∇su = [∇u + (∇u)T ] (2) According to our purpose, we carefully investigate differ- 2 ent novel approaches to fulfill the inextensibility constraint, particularly, the method of Lagrange multipliers and a u being the displacement vector and ∇s the symmetric relaxed version, i.e., the perturbed Lagrangian method. To gradient operator. We introduce the following boundary con- develop the mixed finite elements, different ansatz func- ditions: tions for both the displacement and the Lagrange multiplier fields are considered. As a further comparison, the u = u on ∂u penalty method is adopted to fulfill approximately the (3) σ = ∂ constraint. Several boundary-value problems under plane n t on t strain conditions are considered to analyze the performance of the proposed formulations. For each test, dif- where u and t are the assigned displacement and surface force ferent fiber directions and mesh refinements are consid- field, respectively, and n is the outward normal vector. ered. The proposed finite element formulations are dis- The formulation is completed by the constitutive and fiber cussed and compared to analytical solutions, computed constraint equations, as described in the following subsec- whenever possible. Moreover, the sensitivity analysis to tions. 123 Comput Mech

2.2 Constitutive equation Y

We focus on isotropic linear elastic matrix materials for which the strain energy W iso is deﬁned as:

iso W (ε) = W(I1(ε), I2(ε)) (4) where I1 and I2 are the ﬁrst and second invariant of the strain tensor, deﬁned together with their derivatives as:

I = tr(ε), ∂ I /∂ε = I 1 1 1 2 2 I2 = tr(ε) − tr(ε ) ,∂I2/∂ε = I1I − ε 2 Fig. 1 Schematic representation of a two-dimensional continuum body made of isotropic linear elastic material reinforced by a single I being the second order identity tensor. The energy can be family of parallel fibers whose direction is described by the unit vec- defined as follows: tor a 1 W iso(ε) = λ(tr(ε))2 + μtr(ε2) (5) 2 We assume that the material does not extend in the fiber direction a, i.e., the fibers are inextensible. This means that where λ and μ are the Lamé constants. the kinematic constraint is the following: We obtain the following expression for the stress tensor σ : a · εa = M : ε = tr(ε · M) = 0(10)

∂W iso σ = = λI I + 2με (6) ∂ε 1 3 Constraint formulation and for the fourth-order elasticity tensor D: To treat the inextensible constraint (10), we propose to ∂σ adopt the following approaches: (1) the method of Lagrange D = = λ(I ⊗ I) + μII ∂ε 2 (7) multiplier, (2) a relaxed version based on the perturbed Lagrangian, and (3) a penalty method. This implies to deﬁne where ⊗ represents the tensor outer product and the total strain energy W tot in the following general form:

tot iso f I I + I ˆ I W = W + W (11) II = (8) 2 where W iso is the isotropic strain energy defined in Eq. (5), f with and ˆ being defined as (A B)C = ACBT and while W is the strain energy related to the inextensibil- (A ˆ B)C = ACT BT . ity constraint of the fibers, defined according to the adopted approach. 2.3 Inextensible fibers In the following, the quantities identifying the variables for the Lagrange multiplier, perturbed Lagrangian, and penalty We restrict our attention to materials reinforced by a single formulation will be denoted with subscripts L, PL, and P, family of parallel fibers, whose direction is described by the respectively. unit vector a, as schematically shown in Fig.1. The fibers are considered to be long compared to their diameters and the 3.1 Lagrange multiplier formulation fiber spacings as well as to be sparsely distributed, so that they constitute a limited proportion of the composite (typically Following the definition provided in Eq. (11), we define the tot lower than 50% by volume, according to the definition given total strain energy WL ,as: in [1] for quite densely distributed fibers). Starting from a, tot(ε,σ f ) = iso(ε) + f (ε,σ f ) we construct the fabric tensor M, as follows: WL L W WL L (12)

= ⊗ f M a a (9) where WL is the energy term related to the constraint of a material that is not extendable in the direction a, expressed such that M2 = M. as: 123 Comput Mech

f (ε,σ f ) = σ f (ε · ) WL L L tr M (13) We now have the following system of equations: ⎧ σ f ⎨σ = Dε + σ f Here, L is the Lagrange multiplier that physically represents PL M PL (21) the reaction to the constraint. ⎩ : ε − 1 σ f = M PL 0 Therefore, we obtain the following expression for the CC stress tensor σ L : or, in matrix form: ∂W tot ⎡ ⎤ σ = L = λI I + 2με + σ f M (14) L ∂ε 1 L σ D M ε PL = ⎣ 1 ⎦ − σ f (22) 0 M PL We now consider the following system of equations: CC σ = Dε + Mσ f The solution to problem (22) becomes: L L (15) M : ε = 0 σ f = : (D + ⊗ )−1 σ PL CC M CC M M PL −1 (23) or, in matrix form: ε = (D + CC M ⊗ M) σ PL σ D ε f L = M The fourth-order tensor D such that: σ f (16) PL 0 M 0 L σ = D f ε (24) The solution to problem (16) is straightforward after solving PL PL a boundary-value problem imposing proper boundary condi- is: tions: f f − − − D = D + ⊗ σ = (M : D 1M) 1M : D 1σ PL CC M M (25) L L (17) ε = D−1 − D−1M(M : D−1M)−1M : D−1 σ L which is given by the elasticity tensor D increased by a term depending on the penalty paramater C . where it can easily verified that the scalar quantity M : C D−1M = 0. 3.3 Penalty formulation 3.2 Perturbed Lagrangian formulation Following the definition provided in Eq. (11), we define the total strain energy W tot: Following the definition provided in Eq. (11), we define the P total strain energy W tot,as: PL tot iso f W (ε, CC ) = W (ε) + W (ε, CC ) (26) P P tot ε,σ f , = iso(ε) + f ε,σ f , WPL PL CC W WPL PL CC f where WP is the energy related to the inextensibility con- (18) straint, defined as:

f 1 where WPL is the energy term related to the inextensibility f 2 W (ε, CC ) = CC [tr(ε · M)] (27) constraint, expressed as: P 2

2 It can be noted that, the constraint is exactly satisﬁed if the f ε,σ f , = σ f (ε· )− 1 σ f WPL PL CC PL tr M PL (19) penalty parameter C tends to inﬁnity. 2CC C We obtain the following expression for the stress tensor σ f σ P : Here, PL is the Lagrange multiplier that physically represents the reaction to the constraint, while CC is a penalty ∂W tot parameter. σ = P = λI I + με + C tr(ε · M)M P ∂ε 1 2 C (28) We obtain the following expression for the stress tensor σ PL: We now have: ∂W tot σ = PL = λI I + 2με + σ f M (20) σ = Dε + (ε · ) PL ∂ε 1 PL P CC tr M M (29) 123 Comput Mech

D f σ f The fourth-order tensor P such that: Introducing the interpolations for u and reported in Eqs. (32) and (33), the residual equations can be rewritten σ = D f ε P P (30) as: is: (δu)T (Bu)T DBuuˆd + (Bu)T MNσ σˆ f d+ D f = D + ⊗ P CC M M (31) − ( u)T = f N bd 0 (37) It is observed that the fourth-order tensor D is equivalent to P the one obtained for the perturbed Lagrangian approach [see T σ δσˆ f (N )T MT Buuˆd = 0 Eq. (25)].

4 Finite element formulation We can thus express the problem in matrix form as:

This section presents the investigated finite elements for each Kuu Kuσ uˆ fˆ constraint formulation. σ =− (38) K u 0 σˆ f 0 4.1 Weak form and discretization where: We first derive the weak form for each proposed formulation and then we discretize the governing integral equations. The solution parameters are the displacement field u and the Kuu = [(Bu)T DBu]d σ f mixed variable (introduced in the Lagrange multiplier and perturbed Lagrangian formulations). Therefore, we dis- Kuσ = [(Bu)T MNσ ]d cretize the two fields independently by employing two sets (39) of shape functions, as follows: Kσu = [(Nσ )T MT Bu]d u = Nuuˆ (32) ˆ u T σ f = σ σˆ f f =− (N ) bd N (33) where Nu and Nσ are, respectively, the displacement and fiber stress interpolation functions and uˆ and σˆ f are the vector of Note that the stiffness matrix has zero entries on the diagonal. nodal displacements and Lagrange multipliers, respectively. The strain field is then derived as follows: 4.1.2 Perturbed Lagrangian formulation ε = Buuˆ (34) The total potential energy PL reads as: 4.1.1 Lagrange multiplier formulation ( ,σ f , ) = tot ε( ), σ f , The total potential energy L reads as: PL u PL CC WPL u PL CC d + (u) (40) ,σ f = tot ε( ), σ f + ( ) ext L u L WL u L d ext u (35) where ext is the potential energy of external loads. As before, we write the variations of the total potential As classically done [18], we write the variations of the energy in Eq. (40) with respect to the two fields, obtaining total potential energy in Eq. (35) with respect to the two fields, the residual equations in the weak form: obtaining the residual equations in the weak form: δεT Dε + σ f − δ T = δεT Dε + Mσ f d − δuT bd = 0 M L d u bd 0 PL (36) (41) T T 1 δσ f T ε = δσ f T ε − σ f = L M d 0 PL M PL d 0 CC 123 Comput Mech

Introducing the interpolations for u and σ f reported in Introducing the interpolations for u reported in Eq. (32), Eqs. (32) and (33), the residual equations become: the residual equation can be rewritten as: (δu)T (Bu)T DBuuˆd+ T u T u u T σ f (δu) (B ) D(B )uˆd + (B ) MN σˆ d+ u T T u + CC (B ) MM B uˆd + u T − (N ) bd = 0 − (Nu)T bd = 0 (47) (δσˆ f )T (Nσ )T MT Buuˆd+ σ 1 σ We can thus express the problem in matrix form as: − (N )T N σˆ f d = 0 CC uu ˆ (42) K uˆ =− f (48)

where the stiffness matrix is: We can thus express the problem in matrix form as: uu u T u u T T u K = [(B ) D(B ) + CC (B ) MM (B )]d (49) Kuu Kuσ uˆ fˆ σ σσ =− (43) K u K σˆ f 0 We note that, in the penalty formulation, the constraint is imposed, in an approximated way, introducing some extra where: stiffness terms in the global stiffness matrix, according to the fiber orientation. The ill-conditioning of such a matrix increases as CC tends to infinity. Kuu = (Bu)T DBud 4.2 Interpolation functions Kuσ = (Bu)T MNσ d (44) In the following, we focus on plane strain problems, dis- σ σ K u = (N )T MT Bud cretized through quadrilateral isoparameteric elements. For all the formulations we make use of the following continuous 1 Kσσ =− (Nσ )T Nσ d approximations for u: CC nu e = u,e(ξ, η)ˆ e We note that, differently from the Lagrange multiplier for- u Nk uk (50) mulation, we have non-zero entries on the diagonal of the k=1 stiffness matrix, which depend on the penalty parameter C . C where n is the number of nodes. In particular, we consider The ill-conditioning of such a matrix increases as C tends u C four-node and nine-node elements. For the four-node ele- to infinity. ment (Q1), i.e., nu = 4, we adopt bilinear shape functions to approximate u, while for the nine-node element (Q2), i.e., 4.1.3 Penalty formulation nu = 9, we use biquadratic shape functions. For the Lagrange multiplier and perturbed Lagrangian for- σ f The total potential energy P reads as: mulation, we consider the following approximations for for both elements (Q1 and Q2): (ε, ) = tot(ε( ), ) + ( ) P CC WP u CC d ext u (45) – bilinear continuous approximation 4 , 1 , As before, we write the variation of the total potential σ f e = (1 + ξξ )(1 + ηη )σˆ f e 4 i i i energy in Eq. (45), obtaining the residual equations in the i=1 weak form: – linear discontinuous approximation δεT [Dε + (ε · ) ] − δ T = f,e f,e CC tr M M d u bd 0 (46) σ =ˆσ + ax ξ + ayη 0 123 Comput Mech

Lagrange Y FE label FE view Displacement Multiplier Approximation Approximation B C

Q1Q1 Bilinear Continuous q L

Q1P0 Bilinear Discontinuous D X L Q1P1 Bilinear Discontinuous Fig. 4 Pure traction test. Geometry and boundary conditions

Q2Q1 Biquadratic Continuous 5 Numerical tests

To test the performance of the proposed ﬁnite elements, we consider the following boundary-value problems: Q2P0 Biquadratic Discontinuous

– pure traction test – pure bending test Q2P1 Biquadratic Discontinuous – Cook’s membrane test – two-element distortion test

Displacement For each problem we consider three fiber directions, i.e., a = Lagrange Multiplier √ √ [1, 0], a = [0, 1], and a = 2/2, 2/2 , and we provide an analytical solution whenever it can be derived. In this case, Fig. 2 Investigated finite elements for the Lagrange multiplier and perturbed Lagrangian formulations. The interpolations for u and σ f are the relative error is computed as follows: indicated an − num = x x Er an (51) Displacement x FE label FE view Approximation xan and xnum being the analytical and numerical quantities. We also analyze the behavior of the considered finite ele- Q1 Bilinear ments by performing a convergence analysis varying the mesh size and, for the perturbed Lagrangian and penalty formulations, the penalty parameter CC . To implement the Q2 Biquadratic elements and the tests, we use the Mathematica packages AceGen and AceFEM [26].

Displacement 5.1 Pure traction test

Fig. 3 Investigated finite elements for the penalty formulations. The We consider the square plate of Fig. 4, subject to a uniform interpolations for u are indicated tensile load along the edge CD. This configuration generates a state of uniaxial pure traction. The square plate has an – constant discontinuous approximation edge L = 10 and an applied load q = 1. The mechanical properties are E = 1000 and ν = 0.3. σ f,e =ˆσ f,e 0 Due to the fact that we consider three fiber orientations [see Fig. 5], we have three different traction problems, indi- For the linear and constant discontinuous approximations, cated in the following as T1, T2, and T3, respectively. The static condensation is used. In a qualitative way, Figs.2 and 3 analytical solution for the three cases of Fig. 5 are derived in summarize all the investigated elements, together with the Appendix A.1. For the tests T1 and T2, the boundary condi- labels used in the numerical simulations presented in the next tions applied along the edges AB (i.e., u = 0) and AD (i.e., section. v = 0) are represented in Fig.4. For the test T3, to obtain 123 Comput Mech

Y Y Y components u and v of node C (see Fig. 4) are considered. We note that the Lagrange multiplier formulation provides correct results only with the Q2Q1-L element. If the Lagrange multiplier is interpolated with a constant approximation throughout the element (Q2P0-L and Q1P0-L), the imposed constraint is ineffective. All the remaining cases (Q1Q1-L, = , Fig. 5 Pure traction test with√ ﬁber√ direction (left) a [1 0],(middle) Q1P1-L, and Q2P1-L) provide incorrect results. Both the per- a = [0, 1],and(right) a = 2/2, 2/2 . The three tests are denoted, turbed Lagrangian and penalty formulations provide accurate respectively, as T1, T2 and T3 results, except for the Q2P0-PL and Q2P1-PL elements. Table2 reports the results obtained with the three formulations providing the best results for all the numerical tests, a pure traction condition, we apply the displacements com- i.e., the Q2Q1-L, Q2Q1-PL, and Q2-P elements. The relative puted in Appendix A.1 along the edges AB and AD, since errors for the displacement and Lagrange multiplier of node in these cases u = 0 and v = 0. f C (see Fig. 4), respectively, EU and Eσ , are also reported. Table1 reports the performance of all the ﬁnite element r r The Lagrange multiplier is computed from Eq. (28)forthe formulations, obtained with a 1 × 1 mesh. The displacement

Table 1 Pure traction test Q1Q1-L Q1P0-L Q1P1-L Q2Q1-L Q2P0-L Q2P1-L

uC ✗ ✗✗✓ ✗ ✗

vC ✗ ✗✗✓ ✗ ✗ T2

uC ✗ ✗✗✓ ✗ ✗

vC ✗ ✗✗✓ ✗ ✗ T3

uC ✓ ✗ ✗ ✓ ✗ ✗

vC ✓ ✗ ✗ ✓ ✗ ✗

Cc = 1011 Q1Q1-PL Q1P0-PL Q1P1-PL Q2Q1-PL Q2P0-PL Q2P1-PL

uC ✓✓✓✓✓✗

vC ✓✓✓✓✓✗ T2

uC ✓✓✓✓✓✗

vC ✓✓✓✓✓✗ T3

uC ✓✓✓✓✗✗

vC ✓✓✓✓✗✗

Cc = 1011 Q1-P Q2-P

uC ✓✓

vC ✓✓ T2

uC ✓✓

vC ✓✓ T3

uC ✓✓

vC ✓✓ Schematic view of the performance of the analyzed ﬁnite elements. The green symbol ✓ indicates correct results (i.e., ﬁber-reinforced material), the black symbol ✗ indicates ineffective constraint (i.e., isotropic material), the red symbol ✗ indicates incorrect results

123 Comput Mech

Table 2 Pure traction test Analytical Q2Q1-L Q2Q1-PL Q2-P 11 11 (CC = 10 )(CC = 10 )

No ﬁbers

uC 0.0091

vC −0.0039 σ f C – T1

uC 0000

vC 0000 U Er –000 σ f C 1 1 1 0.9999999890 σ f · −8 Er –00110 T2

uC 0.00742857 0.00742857 0.00742857 0.00742857

vC 0000 U Er –000 σ f − − − − C 0.42857143 0.42857143 0.42857142 0.42857142 σ f · −8 · −8 Er – 0 2.33 10 2.33 10 T3

uC 0.0065 0.0065 0.0065 0.0065

vC −0.0065 −0.0065 −0.0065 −0.0065 U Er –000 σ f C 0.285714 0.285714 0.285714 0.285714 σ f Er –000 Comparison between analytical and numerical results for the Q2Q1-L, Q2Q1-PL, and Q2-P ﬁnite elements (1 × 1mesh)

Table 3 Pure traction test T3 uC vC

Analytical 0.0065 −0.0065 1 × 1 0.006500000000000001 −0.0065000000000000014 10 × 10 0.006500006863952316 −0.006499993136047688 100 × 100 0.006500006863952316 −0.006499993136047688 Mesh convergence for the Q2Q1-L finite element penalty formulation. Concerning the perturbed Lagrangian perturbed Lagrangian and penalty formulations and adopting 11 5 and penalty formulations, a penalty parameter CC = 10 is a1× 1 mesh. Convergent values are reached for CC = 10 15 used. It interesting to note how the solution changes with the and the response is stable up to CC = 10 . fiber orientation, compared to the case of a purely isotropic linear elastic material without fibers. We now perform the traction test, meshing with 1 × 1, 11 5.2 Pure bending test 10 ×10 and 100 ×100 elements and setting CC = 10 .The formulations are not effected by the meshing if the fibers are We consider the beam depicted in Fig.6, subject to a load inclined at an angle of 90◦ or 0◦. As an example, we report along the edge CD, generating a pure bending state. The hor- the results obtained with the Q2Q1-L element in Table3 for izontal displacement u is set equal to zero along the edge AB, fibers inclined at an angle of 45◦. It is interesting to note that while node A is constrained along both directions. The beam exact values are obtained using only one single element. has length L = 10 and height H = 2 and the applied load Finally, we investigate the displacement components by has a maximum value f = 15. The mechanical properties varying the values assigned to the penalty parameter C in the C are E = 1500 and ν = 0.3. 123 Comput Mech

Y Due to the fact that we consider three ﬁber orientations

B C (see Fig. 7), we have three different bending problems, indicated in the following as B1, B2, and B3, respectively. The H analytical solution for the three cases of Fig. 7 is derived in D X Appendix A.2. L Table4 reports the performance of all the formulations, obtained with a 80 × 16 mesh. The displacement compo- Fig. 6 Pure bending test. Geometry and boundary conditions nents u and v of node D (see Fig. 6) are considered. The Y Y Y observed overall element behavior is in general very similar to the one already highlighted in the traction test. In case of the Lagrange multiplier formulation, we obtain correct results only with the Q2Q1-L element. Instead, the imposed = , Fig. 7 Pure bending test with√ ﬁber√ direction (left) a [1 0],(middle) constraint is ineffective with the Q1P0-L, Q1P1-L, Q2P0-L, a = [0, 1],and(right) a = 2/2, 2/2 . The three tests are denoted, and Q2P1-L elements, while incorrect results are obtained respectively, as B1, B2 and B3 with the Q1Q1-L element. All ﬁnite elements based on the

Table 4 Pure bending test Q1Q1-L Q1P0-L Q1P1-L Q2Q1-L Q2P0-L Q2P1-L

u D ✓ ✗✗✓ ✗✗

vD ✓ ✗✗✓ ✗✗ B2

u D ✗ ✗✗✓ ✗✗

vD ✗ ✗✗✓ ✗✗ B3

u D ✗ ✗✗✓ ✗✗

vD ✗ ✗✗✓ ✗✗

Cc = 1011 Q1Q1-PL Q1P0-PL Q1P1-PL Q2Q1-PL Q2P0-PL Q2P1-PL

u D ✓ ✓✓✓ ✓✓

vD ✓ ✓✓✓ ✓✓ B2

u D ✓ ✓✓✓ ✓✓

vD ✓ ✓✓✓ ✓✓ B3

u D ✓ ✓✓✓ ✓✓

vD ✓ ✓✓✓ ✓✓

Cc = 1011 Q1-P Q2-P

u D ✓✓

vD ✓✓ B2

u D ✓✓

vD ✓✓ B3

u D ✗ ✓

vD ✗ ✓ Schematic view of the performance of the analyzed ﬁnite elements. The green symbol ✓ indicates correct results (i.e., ﬁber-reinforced material), the black symbol ✗ indicates ineffective constraint (i.e., isotropic material), the red symbol ✗ indicates incorrect results

123 Comput Mech

Table 5 Pure bending test Analytical Q2Q1-L Q2Q1-PL Q2-P 11 11 (CC = 10 )(CC = 10 )

No ﬁbers

u D 0.091

vD 0.455 σ f D – B1 −8 −8 −9 u D 0 −5.6900×10 −5.5400×10 1.50×10 −8 −8 −9 vD 0 −5.8525×10 −5.1018×10 7.50×10 U × −8 × −8 × −9 Er –8.1610 7.53 10 7.65 10 σ f D 15 14.99987 14.99987 14.99999 σ f × −6 × −6 × −7 Er – 8.6667 10 8.6667 10 6.6667 10 B2

u D 0.0743 0.07418 0.07418 0.07429

vD 0.3710 0.37089 0.37089 0.37143 U × −4 × −4 × −3 Er –4.3010 4.30 10 1.10 10 σ f − − − − D 6.42857 6.42810 6.42810 6.42878 σ f × −5 × −5 × −5 Er – 7.31111 10 7.31111 10 2.02220 10 B3

u D 0.0836 0.08448 0.08448 0.08437

vD 0.4180 0.41405 0.41405 0.41492 U × −3 × −3 × −3 Er –9.5010 9.50 10 7.40 10 σ f D 4.28571 4.28603 4.28603 4.28592 σ f × −5 × −5 × −5 Er –7.3710 7.37 10 4.90 10 Comparison between analytical and numerical results for the Q2Q1-L, Q2Q1-PL, and Q2-P ﬁnite elements (80 × 16 mesh) perturbed Lagrangian formulation provide correct results, Y while the penalty formulation fails with the four-node ele- C ment. Table5 reports the values for the three formulations providing the best results for all the numerical tests, i.e., the F Q2Q1-L, Q2Q1-PL, and Q2-P elements. Again, the relative B D errors for the displacements and Lagrange multiplier of node U σ f D (respectively, Er and Er ) are also reported. Concerning

the perturbed Lagrangian and penalty formulations, a param- 44 11 eter CC = 10 is used. The lower relative errors are obtained with the Q2Q1-L and Q2Q1-PL elements. We now perform a convergence study in terms of mesh performance. Meshes of 5 × 1, 10 × 2, 40 × 8 and 80 × 16 A 11 48 X elements are considered and CC = 10 . As previously observed, the Lagrange multiplier formulation is not affected Fig. 8 Cook’s membrane test. Geometry and boundary conditions by the imposed constraint in the fiber direction in case of constant or linear interpolation. The Q2Q1-L element provides the expected results with a fine mesh. We note We finally analyze the results by varying the penalty that all the used finite elements based on the perturbed parameter CC in the perturbed Lagrangian and penalty for- Lagrangian and penalty formulation reach convergence with mulation. We can affirm that, in general, the convergence is 7 a fairly coarse discretization in the B1 test, while reach reached for CC = 10 . We observe that both the formula- the convergence with a finest discretization in the B2 and tions provide correct results, except for the Q1-P element B3 tests. The Q1-P element fails completely in the B3 in the B3 test. Regarding the B1 test, the results coincide test. for all the elements and converge with the value of the 123 Comput Mech

Y Y Y this value, ill-conditioning occurs. The Q2Q1-PL element provides correct results also for high values of the CC parameter.

5.3 Cook’s membrane test = , Fig. 9 Cook’s membrane tests with√ ﬁber direction√ (left) a [1 0], (middle) a = [0, 1],and(right) a = 2/2, 2/2 . The three tests are The Cook’s membrane test [27] consists of a tapered panel denoted, respectively, as C1, C2 and C3 clamped on the edge AB and subject to a shearing load at the free end CD, as shown in Fig. 8. The applied load determines penalty parameter mentioned above. For the B2 test, we a deformation dominated by a bending response. The applied note that the penalty formulation provides satisfactory results load is F = 100. The mechanical properties are E = 250 with values of the penalty parameter up to 1013. Beyond and ν = 0.3.

Table 6 Cook’s membrane test Q1Q1-L Q1P0-L Q1P1-L Q2Q1-L Q2P0-L Q2P1-L

uC ✗ ✗✗✓ ✗✗

vC ✗ ✗✗✓ ✗✗ C2

uC ✗ ✗✗✓ ✗✗

vC ✗ ✗✗✓ ✗✗ C3

uC ✗ ✗✗✓ ✗✗

vC ✗ ✗✗✓ ✗✗

Cc = 105 Q1Q1-PL Q1P0-PL Q1P1-PL Q2Q1-PL Q2P0-PL Q2P1-PL

uC ✓ ✓✓✓ ✓✓

vC ✓ ✓✓✓ ✓✓ C2

uC ✓ ✓✓✓ ✓✓

vC ✓ ✓✓✓ ✓✓ C3

uC ✓ ✓✓✓ ✓✓

vC ✓ ✓✓✓ ✓✓

Cc = 105 Q1-P Q2-P

uC ✓✓

vC ✓✓ C2

uC ✓✓

vC ✓✓ C3

uC ✓✓

123 Comput Mech

Table 7 Cook’s membrane test Y Q2Q1-L Q2Q1-PL Q2-P d 7 7 B C (CC = 10 )(CC = 10 )

C1 H

uC −2.3032 −2.3043 −2.0232 A D X vC 4.3048 4.3061 3.9667 C2 Fig. 10 Two-element distortion test. Geometry and boundary condi- uC −5.8835 −5.8837 −5.8645 tions vC 7.8872 7.8873 7.8776 C3

uC −1.2975 −1.2987 −1.2873 Y Y Y

vC 1.2981 1.2992 1.2873 Comparison between numerical results for the Q2Q1-L, Q2Q1-PL, and Q2-P finite elements (80 × 80 mesh) = Fig. 11 Two-elements distortion test with fiber√ directions√ (left) a [1, 0],(middle) a = [0, 1],and(right) a = 2/2, 2/2 .Thethree Due to the fact that we consider three fiber orientations tests are denoted, respectively, as D1, D2, and D3 (see Fig. 9), we have three different Cook’s problems, indicated in the following as C1, C2, and C3, respectively. The analytical solution for the three cases of Fig. 9 is unknown. horizontal displacement u is set equal to zero, while node A Table6 reports the performance of all the formulations, is constrained along both the directions. The aim is now to obtained with a 80 × 80 mesh. The displacement compo- demonstrate the sensibility against mesh distortion parameter nents u and v of node C (see Fig. 8) are considered. Using d. The beam has length L = 10 and height H = 2 and the the Lagrange multiplier formulation, if we approximate the applied load has a maximum value f = 60. The mechanical Lagrange multiplier with constant or linear interpolations, properties are E = 3000 and ν = 0. the constraint is ineffective. We note that the perturbed The analytical solutions for the three cases of Fig. 11 Lagrangian and penalty formulations provide comparable are clearly the ones computed in the pure bending test of results with all the element used. Table7 reports the values Sect. 5.2. for the three formulations providing the best results for all the To determine the effect that the imposed inextensibility numerical tests, i.e., again the Q2Q1-L, Q2Q1-PL, and Q2-P constraint produces on the developed finite elements in the elements. Concerning the perturbed Lagrangian and penalty 7 case of a material with fibers, we first report the perfor- formulations, a parameter CC = 10 is used. mance of all the formulations obtained without distortion, We now perform a convergence study in terms of mesh i.e., d = 0, in Table8. We note that only the Q2Q1-L, Q2- performance. Meshes of 5 × 5, 10 × 10, 40 × 40 and P, and Q2Q1-PL elements provide correct results for this 80 × 80 are considered, together with a penalty parameter 7 test in terms of displacement components of node D. Table9 CC = 10 . The observed overall element behavior is in gen- reports the values obtained with these elements. Concerning eral very similar to the one already highlighted in the previous the perturbed Lagrangian and penalty formulations, a param- tests. eter C = 1011 is used. We finally analyze the results by varying the penalty C We now analyze the displacement components of point D parameter CC in the perturbed Lagrangian and penalty for- varying the mesh distortion parameter d, from 0 to 5, and mulation. Regarding the C1 test, we note that all the elements 11 5 setting CC = 10 . In all the cases, the quadratic elements provide reliable results only when CC = 10 . As regards the Q2Q1 L, Q2Q1-PL, and Q2-P provide the correct results, C2 and C3 tests, all the elements show more stable results for independently from the value of d. all the CC , except for the Q2-P element that shows numerical 13 We also analyze the sensitivity of the results with respect issues after CC = 10 . to the penalty parameter CC under the maximum possible distortion, i.e., d = 5. We note that we obtain converged 5.4 Two-element distortion test 5 13 results for CC = 10 . For values of CC greater than 10 the penalty formulation shows ill-conditioning. We consider the cantilever beam depicted in Fig. 10, subject to a loading along the edge CD. Along the edge AB,the

123 Comput Mech

Table 8 Two-elements Q1Q1-L Q1P0-L Q1P1-L Q2Q1-L Q2P0-L Q2P1-L distortion test D1

u D ✓ ✗✗✓ ✗✗

vD ✓ ✗✗✓ ✗✗ D2

u D ✗✗✗✓✓✓

vD ✗✗✗✓✓✓ D3

u D ✗✗✗✓ ✗✗

vD ✗✗✗✓ ✗✗

Cc=1011 Q1Q1-PL Q1P0-PL Q1P1-PL Q2Q1-PL Q2P0-PL Q2P1-PL

u D ✓ ✗✗✓ ✗✗

vD ✓ ✗✗✓ ✗✗ D2

u D ✗✗✗✓✓✓

vD ✗✗✗✓✓✓ D3

u D ✗✗✗✓ ✗✗

vD ✗✗✗✓ ✗✗

Cc = 1011 Q1-P Q2-P

u D ✓✓

vD ✓✓ D2

u D ✗ ✓

vD ✗ ✓ D3

u D ✗ ✓

Table 9 Two-elements Analytical Q2Q1-L Q2Q1-PL Q2-P distortion test 11 11 (CC = 10 )(CC = 10 )

No ﬁbers

u D 0.2

vD 1.0 D1 −9 −9 u D 0 0 6.00000×10 6.00000×10 −8 −8 vD 0 0 3.00000×10 3.00000×10 U × −8 × −8 Er – 0 3.06 10 3.06 10 D2

u D 0.2 0.2 0.2 0.2

123 Comput Mech

Table 9 continued Analytical Q2Q1-L Q2Q1-PL Q2-P 11 11 (CC = 10 )(CC = 10 )

vD 1.0 1.0 1.0 1.0 U Er –0 0 0 D3

u D 0.15 0.15168 0.15168 0.14862

vD 0.75 0.73091 0.73091 0.72703 U × −2 × −2 × −2 Er –2.5110 2.51 10 3.01 10 Comparison between analytical and numerical results for the Q2Q1-L, Q2Q1-PL, and Q2-P ﬁnite elements

6 Conclusions A.1 Traction test

This paper has presented the analysis of isotropic mate- The solution of the traction test carried out in Sect.5.1 is rials with inextensible fibers under small strain. Such a computed by considering that the pure uniaxial traction state case is appropriate to model sparsely reinforced compos- implies: ites and it has been considered thanks to its simple physical interpretation. Extensions to the case of densely reinforced σxx = q composites is however possible by considering the classical σyy = 0 (52) energy definition for transversely isotropic materials, see, σ = e.g., [1]. Several numerical approaches have been presented xy 0 and tested on numerous benchmark tests. The analytical solutions have been also derived whenever possible. In particular, It is possible to compute the displacement field as follows: the quadratic elements Q2Q1-L, Q2Q1-PL, and Q2-P have provided the correct results in all the performed simulations. u(x, y) = εxx x + εxy y (53) Within the framework of classical Lagrange multiplier meth- v(x, y) = εxy x + εyy y ods, the inextensibility condition has been exactly satisfied by introducing an additional variable, i.e., the Lagrange mul- We then consider the stress-strain relationship (refer to sys- tiplier. Such an extra variable has added computational effort tem (16)): to the solution process which may require special proce- ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ dures to handle the presence of zero diagonal terms. The εxx S11 S12 S13 S14 σxx penalty method, on the other hand, has allowed to transform ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ εyy ⎥ ⎢ S21 S22 S23 S24 ⎥ ⎢ σyy ⎥ the constrained problem into an unconstrained one with- ⎣ ⎦ = ⎣ ⎦ ⎣ ⎦ (54) 2εxy S31 S32 S33 S34 σxy out introducing additional variables. However, the constraint σ f S41 S42 S43 S44 0 condition has been shown to be satisfied only approximately for finite values of the penalty parameter CC . The main diffi- where: culty associated with the penalty method, however, has been shown to lie in the poor conditioning of the problem as CC a4μ + a2a2(λ + 2μ) was increased. The augmented Lagrangian procedure has S = y x y 11 (a2 + a2)2μ(λ + 2μ) been shown to be a promising way to partially overcome x y these difficulties and to regularize the penalty formulation. a4μ + a2a2(λ + 2μ) S = x x y The considered cases have been limited to the plane strain 22 ( 2 + 2)2μ(λ + μ) ax ay 2 conditions for which closed-form solutions were easier to −2a2a2λ + (a4 + a4)(λ + 2μ) derive; however, the presented formulations are completely S = x y x y 33 ( 2 + 2)2μ(λ + μ) general and can be implemented in a three-dimensional ax ay 2 μ(λ + μ) framework. S =− 4 44 ( 2 + 2)2(λ + μ) ax ay 2 2 2(λ + μ) ax ay S = S =− (55) A Analytical solutions 12 21 ( 2 + 2)2μ(λ + μ) ax ay 2 a a3λ − a3a (λ + 2μ) In the following sections, we report the derivation of the S = S = x y x y 13 31 ( 2 + 2)2μ(λ + μ) analytical solutions of some performed numerical tests. ax ay 2 123 Comput Mech

− 2λ + 2(λ + μ) ∂2φ ay ax 2 S = S = σxx = = 6Ay 14 41 ( 2 + 2)2(λ + μ) ∂ 2 ax ay 2 y 3 λ − 3(λ + μ) ∂2φ ax ay ax ay 2 σ = = 0 (59) S = S = yy ∂ 2 23 32 (a2 + a2)2μ(λ + 2μ) x x y ∂2φ −a2λ + a2(λ + 2μ) σxy =− = 0 S = S = x y ∂x∂y 24 42 ( 2 + 2)2(λ + μ) ax ay 2 a a (λ + μ) Now, we establish a relation between the moment M and S = S = 4 x y 34 43 ( 2 + 2)2(λ + μ) the stress distribution at the beam ends in integral form to ax ay 2 fully define the stress field in terms of problem parameters. Particularly, after setting h = H/2, we obtain: After applying system (52), we can provide the following analytical solution in terms of displacement compo- h h 2 3 M nents: M = σxxydy= 6A y dy = 4Ah ⇒ A = − − 4h3 h h (60) S31 u(x, y) = S11x + y σxx 2 (56) Therefore, the stress component σxx becomes: S31 v(x, y) = x + S21 y σxx 2 3M σxx = y (61) 2h3 and Lagrange multiplier: Now, we apply the strain-stress relationship (54) and we obtain: σ f = S41σxx (57) ∂u 3M εxx = = S11σxx = S11 y where components S11, S21, S31, and S41 are reported in ∂x 2h3 Eq. (55). ∂v 3M εyy = = S21σxx = S21 y (62) ∂y 2h3 ∂u ∂v 3M A.2 Pure bending test 2εxy = + = S σxx = S y ∂y ∂x 31 31 2h3 The solution of the bending test carried out in Sect. 5.2 can S S S be obtained considering the pure bending of a beam under where 11, 21, and 31 are given in Eq. (55). applied moment M (see Fig. 12). The elementary beam the- After integrating relations (62)1 and (62)2, we obtain the following displacement field: ory predicts that stress component σxx varies linearly with y. Therefore, we consider the following Airy stress function φ: 3M u(x, y) = S xy + g(y) 11 3 2h (63) φ = Ay3 (58) 3M v(x, y) = S y2 + h(x) 21 4h3 where A is a constant to be determined. After introducing the displacement components of Eq. (63) The Airy stress function enable us to determine the stress into (62) , we obtain the following equation: components by applying the following relations: 3

3M 3M S x + g (y) + h (x) − S y = 0 11 2h3 31 2h3

which can be separated into two independent relations in x and y: ⎧ 3M ⎨⎪ h (x) + S x = C 11 2h3 1 ⎩⎪ 3M Fig. 12 Pure bending of a beam subject to moment M g (y) − S y =−C 31 2h3 1 123 Comput Mech which leads to: References

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