Anisotropic Two-Dimensional, Plane Strain, and Plane Stress Models in Classical Linear Elasticity and Bond-Based Peridynamics
mmanuscript No. (will be inserted by the editor)
Anisotropic Two-Dimensional, Plane Strain, and Plane Stress Models inClassical Linear Elasticity and Bond-Based Peridynamics
Jeremy Trageser · Pablo Seleson
Received: date / Accepted: date
Abstract
This paper concerns anisotropic two-dimensional and planar elasticity models within the frame-works of classical linear elasticity and the bond-based peridynamic theory of solid mechanics. We begin byreviewing corresponding models from the classical theory of linear elasticity. This review includes a newelementary and self-contained proof that there are exactly four material symmetry classes of the elasticitytensor in two dimensions. We also summarize classical plane strain and plane stress linear elastic modelsand explore their connections to the pure two-dimensional linear elastic model, relying on the definitionsof the engineering constants. We then provide a novel formulation for pure two-dimensional anisotropicbond-based linear peridynamic models, which accommodates all four material symmetry classes. We fur-ther present innovative formulations for peridynamic plane strain and plane stress, which are obtainedusing direct analogies of the classical planar elasticity assumptions, and we specialize these formulationsto a variety of material symmetry classes. The presented anisotropic peridynamic models are constrainedby Cauchy’s relations, which are an intrinsic property of bond-based peridynamic models. The uniquenessof the presented peridynamic plane strain and plane stress formulations in this work is that we directlyreduce three-dimensional models to two-dimensional formulations, as opposed to matching two-dimensionalperidynamic models to classical plane strain and plane stress formulations. This results in significant com-putational savings, while retaining the dynamics of the original three-dimensional bond-based peridynamicproblems under suitable assumptions.
Keywords peridynamics · linear elasticity · plane strain · plane stress · anisotropy · two-dimensional · Cauchy’s relations · engineering constants Modeling material failure and damage is an essential consideration in the engineering and materials sciencecommunities. In classical continuum mechanics, the governing equations are based on spatial derivatives,which poses difficulties when discontinuities such as cracks develop. As a remedy, the nonlocal peridynamictheory of solid mechanics was proposed in [32,34]. Rather than utilizing spatial derivatives, peridynamics
Jeremy TrageserComputer Science and Mathematics Division,Oak Ridge National Laboratory,One Bethel Valley Road, P.O. Box 2008, MS-6211, Oak Ridge, TN 37831-6211E-mail: [email protected] SelesonComputer Science and Mathematics Division,Oak Ridge National Laboratory,One Bethel Valley Road, P.O. Box 2008, MS-6211, Oak Ridge, TN 37831-6211E-mail: [email protected] a r X i v : . [ phy s i c s . c l a ss - ph ] A p r Jeremy Trageser, Pablo Seleson employs spatial integrals, which model long-range interactions between material points. The idea of long-range interactions parallels the molecular dynamics theory and, in some sense, peridynamics could beconsidered a continuum version of molecular dynamics [29,30].With the exception of various works on fiber-reinforced composites (e.g., [1,17,19,23,24]) and a few otherworks on materials with some degree of anisotropy (e.g., [2,8,14,41]), most peridynamic models in theliterature describe isotropic materials; however, in practice, many materials are anisotropic [38]. To facilitateanalysis of a more diverse group of materials, a three-dimensional anisotropic peridynamic model wasproposed in [31]. The model was shown to be capable of describing any of the eight material symmetryclasses in three-dimensional classical linear elasticity. In two dimensions, there are exactly four materialsymmetry classes in classical linear elasticity [15]. In this work, we provide an elementary self-containedproof of this fact and present a two-dimensional peridynamic model able to accommodate each of thosefour material symmetry classes.A common characteristic of nonlocal models is a penchant to be computationally expensive, particularly inhigher dimensions. Even in classical (local) continuum mechanics, three-dimensional models are frequentlysimplified to two-dimensional formulations, e.g., plane strain and plane stress [38]. Notably, peridynamicplane strain and plane stress models were considered in, e.g., [13,14,20,26]. However, those works simplyemploy two-dimensional peridynamic models rather than placing assumptions on three-dimensional peri-dynamic models to derive two-dimensional formulations. In this work, we look in depth at two-dimensionalsimplifications of three-dimensional anisotropic bond-based linear peridynamic models. The simplificationsare facilitated by the use of nonlocal analogues of the planar elasticity assumptions commonly appearingin classical linear elasticity.It is well established that bond-based peridynamic models can only describe a limited number of materials.In [32], it was demonstrated that isotropic three-dimensional bond-based peridynamic models can onlydescribe materials with a Poisson’s ratio of . This was accomplished by introducing the concept of peri-dynamic traction. A similar derivation leads to a Poisson’s ratio of in two dimensions ( cf . Appendix A).When anisotropy is considered, additional lesser-known restrictions are imposed by bond-based peridy-namic models. These restrictions are due to the utilization of a pair potential in bond-based peridynamics.In molecular dynamics, it is well known that such a potential imposes Cauchy’s relations on a correspond-ing linear elasticity model [35], and in this work we show that Cauchy’s relations are similarly imposed bybond-based linear peridynamic models as demonstrated in [31]. Moreover, we show these Cauchy’s relationsrestrictions are an inherent property of the bond-based peridynamic model and independent of the defini-tion of peridynamic traction. Furthermore, we show the Poisson’s ratio restrictions of in three dimensionsand in two dimensions are a specific case of Cauchy’s relations for isotropic materials.The organization of this paper is as follows. In Section 2, we review the classical theory of linear elasticityto provide a background for material anisotropy and planar elasticity (specificially plane strain and planestress), in order to make connections with peridynamic models. Specifically, in Section 2.1, we providean elementary proof of the fact that there are exactly four material symmetry classes in two-dimensionalclassical linear elasticity. In Section 2.2, we review the equations of motion for classical pure two-dimensionallinear elasticity. We continue in Section 2.3 by deriving plane strain and plane stress formulations in classicallinear elasticity, following [38], whose derivations we mimic in peridynamics. We finalize the discussion ofclassical linear elasticity by reviewing engineering constants and Cauchy’s relations in Sections 2.4 and 2.5,respectively. In Section 3, we delve into the bond-based peridynamic theory. In particular, in Section 3.1,we develop a two-dimensional linear bond-based peridynamic model capable of describing materials in anyof the four material symmetry classes appearing in two-dimensional classical linear elasticity. We continuein Section 3.2 by deriving anisotropic peridynamic formulations of plane strain and plane stress. Finally,conclusions are presented in Section 4. In classical linear elasticity, the stress tensor σ and strain tensor ε are related via the generalized Hooke’slaw [38]: σ ij = C ijkl ε kl , (1) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 3 where C ijkl are the components of the fourth-order elasticity tensor C and Einstein summation conventionfor repeated indices is employed. Due to the symmetries of the stress and strain tensors, C inherits theminor symmetries: C ijkl = C jikl = C ijlk . (2)Furthermore, the relation between the strain energy density W = σ ij ε ij and C , C ijkl = ∂ W∂ε ij ∂ε kl = ∂ W∂ε kl ∂ε ij = C klij , guarantees the major symmetry of C : C ijkl = C klij . (3) Remark 1
Note that the relation C ijkl = C ikjl does not hold in general. However, employing a moleculardescription of a material where interactions between particles are described by pairwise potentials willapply these restrictions to the elasticity tensor [35]. These relations are commonly referred to as Cauchy’srelations [21] and are discussed in further detail in Section 2.5.Due to the major and minor symmetries, we may utilize Voigt notation to express the fourth-order tensor C as a symmetric second-order tensor C and, similarly, represent the second-order tensors σ and ε as vectors.In this formulation, we express (1) in R as σ σ σ σ σ σ = C C C C C C · C C C C C · · C C C C · · · C C C · · · · C C · · · · · C ε ε ε ε ε ε (4)and in R as σ σ σ = C C C · C C · · C ε ε ε . (5)The inverse relation of (1) can be expressed as ε ij = S ijkl σ ij , (6)where S ijkl are the components of the fourth-order compliance tensor S . Similarly to C , we may employVoigt notation to express the fourth-order tensor S as a symmetric second-order tensor S [38]. The strain-stress relation (6) may be expressed in R as ε ε ε ε ε ε = S S S S S S · S S S S S · · S S S S · · · S S S · · · · S S · · · · · S σ σ σ σ σ σ (7)and in R as ε ε ε = S S S · S S · · S σ σ σ . In classical linear elasticity, the strain tensor ε is related to the displacement field u through the relation ε ij = 12 (cid:18) ∂u i ∂x j + ∂u j ∂x i (cid:19) . (8) Jeremy Trageser, Pablo Seleson
We are now able to provide the equation of motion in classical linear elasticity [37]. Given a body
B ⊂ R d ,where d is the dimension, the equation of motion for a material point x ∈ B at time t (cid:62) ρ ( x )¨ u ( x , t ) = ∇ · σ ( x , t ) + b ( x , t ) , (9)where ρ is the mass density, ¨ u is the second derivative in time of the displacement field u , and b is aprescribed body force density field. In component form, (9) may be written as ( cf. (1) and (8)) ρ ( x )¨ u i ( x , t ) = ∂σ ij ∂x j ( x , t ) + b i ( x , t ) = C ijkl ∂ε kl ∂x j ( x , t ) + b i ( x , t )= C ijkl (cid:18) ∂ u k ∂x j ∂x l ( x , t ) + ∂ u l ∂x j ∂x k ( x , t ) (cid:19) + b i ( x , t )= C ijkl ∂ u k ∂x j ∂x l ( x , t ) + b i ( x , t ) , (10)where we used the minor symmetries C ijlk = C ijkl ( cf. (2)). Remark 2
For the sake of brevity, we often omit the arguments x and t .2.1 Material symmetry classes in two-dimensional classical linear elasticitySuppose we have a fourth-order tensor A and a second-order tensor T with components A ijkl and T pq ,respectively, relative the basis { e i } i =1 ,...,d , where d is the dimension. Then, we define A [ T ] to be thesecond-order tensor with components given by A ijkl T kl (this is sometimes called a double contraction anddenoted by A : T [16]). With this formulation, the components of C relative to the basis { e i } i =1 ,...,d are C ijkl = tr { ( e i ⊗ e j ) C [ e k ⊗ e l ] } , (11)where tr denotes the trace. A transformation between orthonormal bases of R d , { e i } i =1 ,...,d and { e (cid:48) i } i =1 ,...,d may be represented by an orthogonal matrix Q , where the components are given by Q ij = e (cid:48) i · e j . (12)We call Q a symmetry transformation of C when the components of C are invariant under the transforma-tion. In Definition 1 we formalize this concept. Definition 1
An orthogonal transformation Q between orthonormal bases { e i } i =1 ,...,d and { e (cid:48) i } i =1 ,...,d ( cf. (12)) is a symmetry transformation of C iftr { ( e i ⊗ e j ) C [ e k ⊗ e l ] } = tr (cid:8) ( e (cid:48) i ⊗ e (cid:48) j ) C [ e (cid:48) k ⊗ e (cid:48) l ] (cid:9) . (13)Equivalently, one may write C ijkl = Q ip Q jq Q kr Q ls C pqrs . (14)In view of (14), one may show that if Q and Q are symmetry transformations of C then Q − (as well as Q − ) and Q Q are also symmetry transformations of C [38]. Clearly, the identity transformation, I , is alsoa symmetry transformation of C . Therefore, the set of symmetry transformations of C forms a group (seee.g. [9]), which we call the symmetry group of C and denote it by G C . We call the set of symmetry groupsthat are equivalent up to a change in orientation, the symmetry class of C . Given a material described by C , its material symmetry class is the corresponding symmetry class of C .In two dimensions, it is well known that every orthogonal transformation is either a reflection or a rotation.For convenience, we recall the corresponding transformation matrices. For a (counterclockwise) rotation byan angle θ about the origin, the corresponding transformation matrix is given by Rot ( θ ) = (cid:20) cos( θ ) − sin( θ )sin( θ ) cos( θ ) (cid:21) . (15) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 5 For a reflection about the line through the origin making an angle of θ with the x -axis, the correspondingtransformation matrix is given by Ref ( θ ) = (cid:20) cos(2 θ ) sin(2 θ )sin(2 θ ) − cos(2 θ ) (cid:21) . (16)From the periodicity of sine and cosine, it is clear that all rotations may be represented by a rotationof θ ∈ [0 , π ), while all reflections may be represented by a reflection through a line making an angle of θ ∈ [0 , π ) with the x -axis. We recall a useful identity for later use: Ref ( φ ) Ref ( θ ) = Rot (2[ φ − θ ]) . (17)Since reflections are their own inverses, we additionally have Rot (2[ φ − θ ]) Ref ( θ ) = Ref ( φ ) . (18)It is well known that there are exactly eight symmetry classes of C in three-dimensional classical linearelasticity [4,12,38]. In the two-dimensional case, there are exactly four symmetry classes of C [15]. Our firstresult provides an alternate proof that there are exactly four symmetry classes of C in two dimensions. Theproof utilizes elementary methods with minimal machinery from abstract algebra. Theorem 1
Up to a change in orientation, there are exactly four symmetry groups of the elasticity tensor C in two dimensions: oblique, rectangular, square, and isotropic. The corresponding elasticity tensors andgroup generators for each symmetry group are given by: Symmetry Group Elasticity Tensor Group Generators
Oblique C C C · C C · · C {− I } Rectangular C C · C · · C {− I , Ref (0) } Square C C · C · · C (cid:8) − I , Ref (0) , Ref ( π ) (cid:9) Isotropic C C · C · · C − C { Ref ( θ ) : θ ∈ [0 , π ) } Proof
Suppose G C is the symmetry group of C . We first show that, up to a change in orientation, G C isone of the four symmetry groups from Lemma 1 below. By (14), it is clear that − I ∈ G C . Consequently,at least one of the four symmetry groups described in Lemma 1 is a subgroup of G C . Let N be the mostsymmetric (largest quantity of reflection transformations) such subgroup of G C . By definition, N ⊆ G C .We then show that G C ⊆ N , which implies G C = N . By Lemma 1, up to a change in orientation, everyreflection in G C is contained in N . Furthermore, by Lemma 2, we know every rotation in G C is containedin N . Since orthogonal transformations in two dimensions are either rotations or reflections, every elementof G C is contained in N , i.e. G C ⊆ N . (cid:117)(cid:116) Lemma 1
Up to a change in orientation, there are only four symmetry groups of the elasticity tensor C generated by reflections and − I : Jeremy Trageser, Pablo Seleson
ObliqueRectangularSquareIsotropic
Ref (0)
Ref ( π ) Ref ( θ ) ∀ θ Fig. 1: The four symmetry groups (up to a change in orientation) of the elasticity tensor in two dimensions.
Symmetry Group Lines of Reflection Symmetry Group Generators
Oblique No lines of reflection symmetry {− I } Rectangular Two lines of reflection symmetry {− I , Ref (0) } Square Four lines of reflection symmetry (cid:8) − I , Ref (0) , Ref (cid:0) π (cid:1)(cid:9) Isotropic All lines of reflection symmetry { Ref ( θ ) : θ ∈ [0 , π ) } Proof
No lines of reflection symmetry
By (14), the group generated by {− I } (19)is a subgroup of any symmetry group of C . This group, { I , − I } , imposes no restrictions on the elasticitytensor, i.e., (14) holds under any transformation in the group. We call the symmetry group generated by(19) the oblique symmetry group and the corresponding elasticity tensor is given by ( cf. (5)) C = C C C · C C · · C . (20) One line of reflection symmetry
We now consider the implications of adding a reflection transformation to the subgroup { I , − I } . Withoutloss of generality, we choose an orthonormal basis { e i } i =1 , so that the line of reflection symmetry coincideswith the x -axis. In this basis, we may write the corresponding reflection transformation as ( cf. (16)) Ref (0) = (cid:20) − (cid:21) . (21)We are interested in the restrictions the group generated by {− I , Ref (0) } (22) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 7First line of reflectionSecond line of reflection xy e e e (cid:48) e (cid:48) θ Fig. 2: Basis { e i } i =1 , and rotated basis { e (cid:48) i } i =1 , for two non-orthogonal lines of reflection symmetry.imposes on the elasticity tensor C .In order for (14) to be satisfied for Q = Ref (0), any C ijkl where 2 appears an odd number of times in theindices requires C ijkl = − C ijkl and thus C ijkl = 0. Consequently, C = C = 0 . (23)The transformation in (22) impose no additional restrictions on C . We call the symmetry group generatedby (22) the rectangular symmetry group and the corresponding elasticity tensor is given by C = C C · C · · C . (24)Notice Rot ( π ), a rotation by π , is equivalent to − I ( cf. (15)). By (18), we have Rot ( π ) Ref ( θ ) = Ref (cid:16) θ + π (cid:17) . (25)Since − I is in every symmetry group of C , a line of reflection symmetry automatically induces a second lineof reflection symmetry orthogonal to the first line of reflection symmetry by (25). In particular, this impliesimplies Ref (cid:0) π (cid:1) belongs to the rectangular symmetry group since the reflection transformation Ref ( )belongs to the group. Thus, to consider a symmetry group distinct from the rectangular symmetry group,we next consider a group containing two non-orthogonal lines of reflection symmetry. Two non-orthogonal lines of reflection symmetry and isotropy
Consider a group generated by − I and two non-orthogonal lines of reflection symmetry. Suppose the anglebetween the two lines of reflection symmetry is given by θ . We further require θ (cid:54) = kπ for k ∈ Z , so that thelines of reflection symmetry are distinct and not orthogonal to each other. We may choose an orthonormalbasis { e i } i =1 , and the rotated basis { e (cid:48) i } i =1 , such that e (cid:48) = cos( θ ) e + sin( θ ) e , e (cid:48) = − sin( θ ) e + cos( θ ) e , where e and e (cid:48) coincide with the two lines of reflection symmetry, see Figure 2. In the basis { e i } i =1 , , thegroup is generated by {− I , Ref (0) , Ref ( θ ) } . (26)From the argument for one line of reflection, we know (23) holds in both bases ( cf. [4,38] for similararguments in three dimensions). Consequently, by (13) we find Jeremy Trageser, Pablo Seleson C (cid:48) = tr { ( e (cid:48) ⊗ e (cid:48) ) C [ e (cid:48) ⊗ e (cid:48) ] } = tr { ( e ⊗ e ) C [ e ⊗ e ] } = cos( θ ) sin( θ )[cos ( θ )( C − C − C ) + sin ( θ )( C − C + 2 C )] (27a)0 = C (cid:48) = tr { ( e (cid:48) ⊗ e (cid:48) ) C [ e (cid:48) ⊗ e (cid:48) ] } = tr { ( e ⊗ e ) C [ e ⊗ e ] } = cos( θ ) sin( θ )[sin ( θ )( C − C − C ) + cos ( θ )( C − C + 2 C )] . (27b)We now consider all solutions of system (27). First, recall sin( θ ) and cos( θ ) are nonzero since θ (cid:54) = kπ with k ∈ Z . Dividing both equations in (27) by cos( θ ) sin( θ ), we are left with0 = a cos ( θ ) + b sin ( θ ) , (28a)0 = a sin ( θ ) + b cos ( θ ) , (28b)where a := C − C − C and b := C − C + 2 C . Summing (28a) and (28b), and applyingthe Pythagorean identity, sin ( θ ) + cos ( θ ) = 1, we deduce a + b = 0 and therefore a = − b or a = b = 0.If a = − b (cid:54) = 0, then by (28) we have cos ( θ ) = sin ( θ ), which implies θ = π or π . Moreover, by thedefinitions of a and b we also have C = C . Recall from (25) that a line of reflection symmetryinduces a second line of reflection symmetry perpendicular to the first. Thus, the choice of θ = π or π isirrelevant as either transformation generates the other (note Ref (cid:0) π (cid:1) is equivalent to Ref (cid:0) π (cid:1) by (16)).The transformation Ref (cid:0) π (cid:1) imposes no additional restrictions on C . We call the symmetry group generatedby (cid:110) − I , Ref (0) , Ref (cid:16) π (cid:17)(cid:111) (29)the square symmetry group and the corresponding elasticity tensor has the restrictions C = C = 0 and C = C . (30)The elasticity tensor corresponding to the square symmetry group is given by C = C C · C · · C . (31)Alternatively, if a = b = 0, we have from a = 0, C = C − C , and substituting this into b = 0, weobtain C = C . The corresponding elasticity tensor has the restrictions C = C = 0 , C = C , and C = C − C , (32)which produces the elasticity tensor C = C C · C · · C − C . (33)Moreover, given (33) one may show (14) holds for Q = Ref ( θ ) with any choice of θ . Thus, this elasticitytensor remains invariant under any choice of line of reflection. We call this group the isotropic symmetrygroup . The generators are given by { Ref ( θ ) : θ ∈ [0 , π ) } . (34)The last piece to tie up the proof is to consider adding an additional line of reflection to the square symmetrygroup. In this case, we would have two lines of reflection which do not intersect at an angle of π or π , andconsequently (28) would yield only the trivial solution a = b = 0. In this case, we immediately obtain (34)as a set of generators for the group. (cid:117)(cid:116) The next lemma shows that adding rotations to the symmetry groups in Lemma 1 does not produce newsymmetry groups and therefore those symmetry groups actually describe all symmetry groups of C . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 9 Lemma 2
The set of symmetry groups described in Lemma 1 is closed under the introduction of rotationsymmetry transformations of C . More specifically, if C is invariant under a rotation transformation Q andthe members of a symmetry group G from Lemma 1, then the symmetry group generated by { Q } ∪ G is oneof the four symmetry groups from Lemma 1.Proof Let C be invariant under a rotation transformation Q and the members of a symmetry group G fromLemma 1. If Q ∈ G then { Q } ∪ G = G and the conclusion of the lemma is immediate. Consequently, wesuppose Q / ∈ G . We discuss each symmetry group G : isotropic, square, rectangular, and oblique. Isotropic : In this case G is generated by (34) and consequently contains all rotation transformations. Thisis easily seen through the closure property of groups and relation (17). Ergo, Q ∈ G , which contradicts theassumption Q / ∈ G . Square : In this case G is generated by (29). Recall the composition of a reflection and a rotation isa reflection in two dimensions ( cf. (18)). Since Q / ∈ G , the closure property of groups implies the groupgenerated by { Q }∪G contains a reflection transformation not in G . Consequently, by Lemma 1, the symmetrygroup generated by { Q } ∪ G is the isotropic symmetry group. Rectangular : In this case G is generated by (26). As in the square case, since Q / ∈ G , the group generatedby { Q } ∪ G contains a reflection transformation not in G . Therefore, by Lemma 1, the isotropic symmetrygroup or square symmetry group is a subgroup of the group generated by { Q } ∪ G . Furthermore, the squareand isotropic cases above guarantee the group generated by { Q } ∪ G is either the square symmetry groupor the isotropic symmetry group. Oblique : In this case G is generated by (19). We further suppose the rotation transformation Q correspondsto a counterclockwise rotation by θ (cid:54) = kπ for k ∈ Z since rotations by kπ are already in the group generatedby (19) ( cf. (15)). Since C is invariant under the rotation transformation Q , we have by (14) that C = C cos ( θ ) − C sin( θ ) cos ( θ ) + 2 C cos ( θ ) sin ( θ ) + 4 C cos ( θ ) sin ( θ ) − C sin ( θ ) cos( θ ) + C sin ( θ ) , (35a) C = C sin( θ ) cos ( θ ) + C (4 sin ( θ ) − ( θ ) + 1)+ C (sin ( θ ) cos( θ ) − sin( θ ) cos ( θ )) + 2 C (sin ( θ ) cos( θ ) − sin( θ ) cos ( θ ))+ C (3 sin ( θ ) − ( θ )) − C sin ( θ ) , cos( θ ) (35b) C = C cos ( θ ) sin ( θ ) + 2 C (sin( θ ) cos ( θ ) − sin ( θ ) cos( θ ))+ C (sin ( θ ) + cos ( θ )) − C cos ( θ ) sin ( θ )+ 2 C (sin ( θ ) cos( θ ) − sin( θ ) cos ( θ )) + C cos ( θ ) sin ( θ ) , (35c) C = C cos ( θ ) sin ( θ ) + 2 C (sin( θ ) cos ( θ ) − sin ( θ ) cos( θ )) − C cos ( θ ) sin ( θ ) + C (sin ( θ ) − cos ( θ )) + 2 C (sin ( θ ) cos( θ ) − sin( θ ) cos ( θ )) + C cos ( θ ) sin ( θ ) , (35d) C = C sin ( θ ) cos( θ ) + C (3 sin ( θ ) − ( θ ))+ C (sin( θ ) cos ( θ ) − sin ( θ ) cos( θ )) + 2 C (sin( θ ) cos ( θ ) − sin ( θ ) cos( θ ))+ C (4 sin ( θ ) − ( θ ) + 1) − C sin( θ ) cos ( θ ) , (35e) C = C sin ( θ ) + 4 C sin ( θ ) cos( θ ) + 2 C cos ( θ ) sin ( θ ) + 4 C cos ( θ ) sin ( θ )+ 4 C sin( θ ) cos ( θ ) + C cos ( θ ) . (35f)Take the difference of the equations for C and C in (35) and then simplify to find0 = ( C − C ) sin ( θ ) + 2( C + C ) sin( θ ) cos( θ ) . (36)Similarly, sum the equations for C and C and then simplify to find0 = ( C − C ) sin( θ ) cos( θ ) − C + C ) sin ( θ ) . (37) Since θ (cid:54) = kπ , we know sin( θ ) (cid:54) = 0 and we may divide (36) and (37) by sin( θ ) to obtain0 = ( C − C ) sin( θ ) + 2( C + C ) cos( θ ) , (38a)0 = ( C − C ) cos( θ ) − C + C ) sin( θ ) . (38b)Multiplying (38a) by cos( θ ) and (38b) by sin( θ ), and taking the difference of the two equations, yields0 = 2( C + C ) ⇒ C = − C . (39)Taking the difference of (35a) and (35f), imposing (39), and then simplifying produces( C − C ) sin ( θ ) = 0 ⇒ C = C . (40)The implication in (40) follows by recalling θ (cid:54) = kπ for k ∈ Z .Substituting (39) and (40) into (35a) and (35b) and simplifying results in0 = ( C − C − C ) sin (2 θ ) + 4 C cos(2 θ ) sin(2 θ ) , (41a)0 = − C sin (2 θ ) + ( C − C − C ) cos(2 θ ) sin(2 θ ) . (41b)We will consider two cases: sin(2 θ ) (cid:54) = 0 and sin(2 θ ) = 0. Case 1 : Let us consider sin(2 θ ) (cid:54) = 0. Multiply (41a) by cot(2 θ ) and then subtract the result by (41b) tofind ( cf. (40)) 0 = 4 C (cid:0) sin (2 θ ) + cos (2 θ ) (cid:1) ⇒ C = − C . (42)Imposing (42) on (41a) results in 0 = ( C − C − C ) sin (2 θ ) . Since sin(2 θ ) (cid:54) = 0, we are able to conclude C − C − C = 0 ⇒ C = C − C . (43)Imposing (40), (42), and (43) on the elasticity tensor (20), we obtain (33) and thus the group generated by G ∪ { Q } is the isotropic group. Case 2 : We suppose sin(2 θ ) = 0, i.e., θ = (2 k +1) π (recall θ (cid:54) = kπ by assumption) for some k ∈ Z . Wefurther suppose C (cid:54) = 0 as otherwise (20) reduces to the square tensor ( cf. (39)) and we have alreadytreated the case where a rotation was added to the square group. Note that f ( x ) := 2(cot(2 x ) − tan(2 x )) = 2 (cid:18) cos(2 x )sin(2 x ) − sin(2 x )cos(2 x ) (cid:19) is continuous on (cid:0) , π (cid:1) . Moreover,lim x → f ( x ) = −∞ and lim x → π f ( x ) = ∞ , and thus f ( x ) has range ( −∞ , ∞ ) on the domain (cid:0) , π (cid:1) by the Intermediate Value Theorem. Consequently,we may find an α ∈ (cid:0) , π (cid:1) such that C − C − C C = 2(cot(2 α ) − tan(2 α )) . (44)It turns out that C is invariant with respect to reflection transformation Ref ( α ) ( cf. (16)). Thus, the groupgenerated by G ∪ { Q } contains the reflection transformation Ref ( α ). Since Rot (cid:0) π (cid:1) is guaranteed to be in G because Rot (cid:16) (2 k +1) π (cid:17) and Rot ( π ) are, using (17) we see that Ref (cid:16) α + π (cid:17) = Rot (cid:16) π (cid:17) Ref ( α ) (45)is also in the group generated by G ∪ { Q } . Thus, the group generated by K := (cid:8) − I , Ref ( α ) , Ref (cid:0) α + π (cid:1)(cid:9) is a subgroup of the group generated by G ∪ { Q } . On the other hand, by noticing Ref ( α ) is its own inverse,from (45) we deduce Q is in the group generated by K . Consequently, the groups generated by K and G ∪ { Q } are the same. By considering a change in orientation, specifically a clockwise rotation by α , we seethat the group generated by K is in the square symmetry class. (cid:117)(cid:116) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 11 Remark 3
Notice that adding a rotation to the oblique symmetry group produces either the isotropic or thesquare symmetry group, but not the rectangular symmetry group. In Table 1, we summarize the symmetrytransformations for each symmetry group. Unlike in the three-dimensional case, we cannot distinguishbetween the symmetry groups of classical linear elasticity solely based on the rotations contained in thegroup.
Symmetry Group Reflections Rotations C RestrictionsOblique None
Rot ( π ) NoneRectangular Ref (0) , Ref ( π ) Rot ( π ) C = C = 0Square Ref (0) , Ref ( π ) , Ref ( π ) , Ref ( π ) Rot ( π ) , Rot ( π ) , Rot ( π ) C = C = 0 C = C C = C = 0Isotropic Ref ( θ ) , θ ∈ [0 , π ) Rot ( θ ) , θ ∈ [0 , π ) C = C C = C − C Table 1: Symmetry transformations by symmetry class in two dimensions.2.2 Pure two-dimensional classical linear elasticityIn this section, we review pure two-dimensional models in classical linear elasticity. We consider the equationof motion (10) for each of the four symmetry classes. The corresponding elasticity tensors are given inTheorem 1.Oblique: There are no restrictions on C under oblique symmetry and thus the oblique equation of motionis given by (10), which we write out explicitly below: ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + b , (46a) ρ ¨ u = C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b . (46b)Rectangular: Imposing the rectangular symmetry restrictions (23) on (46) produces the rectangular equationof motion: ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + ( C + C ) ∂ u ∂x∂y + b , (47a) ρ ¨ u = ( C + C ) ∂ u ∂x∂y + C ∂ u ∂x + C ∂ u ∂y + b . (47b)Square: Imposing the square symmetry restrictions (30) on (46) produces the square equation of motion: ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + ( C + C ) ∂ u ∂x∂y + b , (48a) ρ ¨ u = ( C + C ) ∂ u ∂x∂y + C ∂ u ∂x + C ∂ u ∂y + b . (48b) Isotropic: Imposing the isotropic symmetry restrictions (32) on (46) produces the isotropic equation ofmotion: ρ ¨ u = C ∂ u ∂x + 12 ( C − C ) ∂ u ∂y + 12 ( C + C ) ∂ u ∂x∂y + b , (49a) ρ ¨ u = 12 ( C + C ) ∂ u ∂x∂y + 12 ( C − C ) ∂ u ∂x + C ∂ u ∂y + b . (49b)2.3 Planar approximations in three-dimensional classical linear elasticityIn some situations, a three-dimensional problem may be simplified to a two-dimensional formulation. Thisoften greatly reduces the computational cost of numerically solving the problem, and sometimes allowssolutions of intractable three-dimensional problems to be successfully approximated. In fact, some of the firstsuccessful applications of the finite element method were performed on two-dimensional elastic problems[5,39]. In this section, we consider two-dimensional simplifications of the three-dimensional equations ofmotion of classical linear elasticity. In particular, we focus on planar approximations of anisotropic models,specifically plane strain and plane stress models, where the in-plane and out-of-plane deformations aredecoupled. In Sections 3.2.1 and 3.2.2 we will derive the peridynamic analogues of plane strain and planestress, and the derivations will mirror those utilized in the classical theory. For this reason, we providederivations of the classical planar elasticity models in this section, following [38]. Before we delve intothe planar elasticity models, we briefly review symmetry classes of C in three-dimensional classical linearelasticity. Recall there are exactly eight symmetry classes in classical linear elasticity [4,12,38]: triclinic, monoclinic,orthotropic, trigonal, tetragonal, transversely isotropic, cubic, and isotropic. Similarly to two-dimensionalclassical linear elasticity, the inversion transformation − I is a member of every symmetry group of C .We now briefly consider a representative group from each symmetry class. For each symmetry group, weconsider a generating set as well as the resulting elasticity tensor and its corresponding restrictions.Let a given plane be defined by the unit normal n = (cid:104) n , n , n (cid:105) . The transformation corresponding toreflection through the plane is given by Ref ( n ) = I − n ⊗ n . It turns out that in classical linear elasticity, a symmetry group is entirely determined by the reflectiontransformations in the group [4]. Let a coordinate system be given by the basis { e i } i =1 ,..., . The followingare our representative symmetry groups of the symmetry classes of C in this coordinate system:Triclinic: The triclinic symmetry group may be generated by {− I } . There are no restrictions on the elasticity tensor. The triclinic elasticity tensor is given by ( cf. (4)) C = C C C C C C · C C C C C · · C C C C · · · C C C · · · · C C · · · · · C . (50)Monoclinic: A monoclinic symmetry group may be generated by {− I , Ref ( e ) } . (51) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 13 The corresponding monoclinic symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 . (52)The resulting monoclinic elasticity tensor is given by C = C C C C · C C C · · C C · · · C C · · · · C · · · · · C . Orthotropic: An orthotropic symmetry group may be generated by {− I , Ref ( e ) , Ref ( e ) , Ref ( e ) } . The corresponding orthotropic symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (53a) C = C = C = C = 0 . (53b)The resulting orthotropic elasticity tensor is given by C = C C C · C C · · C · · · C · · · · C · · · · · C . Trigonal: A trigonal symmetry group may be generated by (cid:26) − I , Ref ( e ) , Ref (cid:18) (cid:68) , √ , ± (cid:69)(cid:19)(cid:27) . The corresponding trigonal symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (54a) C = 0 , C = C , C = C , C = C , (54b) C = C = − C , C = C − C . (54c)The resulting trigonal elasticity tensor is given by C = C C C · C C C · · C − C · · · C − C − C · · · · C · · · · · C . Tetragonal: A tetragonal symmetry group may be generated by (cid:26) − I , Ref ( e ) , Ref ( e ) , Ref ( e ) , Ref (cid:18) (cid:68) √ , ±√ , (cid:69)(cid:19)(cid:27) . Typically, trigonal symmetry is presented with the normals in the plane z = 0, but the present formulation has advantagesthat will become apparent when we discuss planar linear elasticity models in Sections 2.3.2 and 2.3.3.4 Jeremy Trageser, Pablo Seleson The corresponding tetragonal symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (55a) C = C = C = C = 0 , (55b) C = C , C = C , C = C . (55c)The resulting tetragonal elasticity tensor is given by C = C C C · C C · · C · · · C · · · · C · · · · · C . Transversely Isotropic: A transversely isotropic symmetry group may be generated by {− I , Ref ( e ) , Ref ( (cid:104) cos( θ ) , sin( θ ) , (cid:105) ) : θ ∈ [0 , π ) } . The corresponding transversely isotropic symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (56a) C = C = C = C = 0 , (56b) C = C , C = C , C = C , (56c) C = C − C . (56d)The resulting transversely isotropic elasticity tensor is given by C = C C C · C C · · C · · · C · · · · C · · · · · C − C . Cubic: A cubic symmetry group may be generated by (cid:26) − I , Ref ( e ) , Ref ( e ) , Ref ( e ) , Ref (cid:18) (cid:104)√ , ±√ , (cid:105) (cid:19) , Ref (cid:18) (cid:104) , √ , ±√ (cid:105) (cid:19) , Ref (cid:18) (cid:104)√ , , ±√ (cid:105) (cid:19)(cid:27) . The corresponding cubic symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (57a) C = C = C = C = 0 , (57b) C = C = C , C = C = C , (57c) C = C = C . (57d)The resulting cubic elasticity tensor is given by C = C C C · C C · · C · · · C · · · · C · · · · · C . Isotropic: The isotropic symmetry group is O (3), the set of orthogonal transformations in R . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 15 The isotropic symmetry restrictions on the elasticity tensor are given by C = C = C = C = C = C = C = C = 0 , (58a) C = C = C = C = 0 , (58b) C = C = C , C = C = C , (58c) C = C = C = C − C . (58d)The isotropic elasticity tensor is given by C = C C C · C C · · C · · · C − C · · · · C − C · · · · · C − C . Classical plane strain is often associated with thick structures [38]. Due to the thickness of the structure,deformations in the thickness direction are often constrained. Provided certain assumptions are met, eachcross-section of the material perpendicular to the thickness direction is in approximately the same deformedstate. In this situation, a two-dimensional formulation of the dynamics in one cross-section is sufficient toprovide information about the dynamics of the entire structure.The classical plane strain assumptions are given as follows:(C ε
1) The geometric form and mass density of the body, and the external loads exerted on it, do not changealong some axis, which we take as the z -axis. In particular, ρ = ρ ( x, y ) and b = b ( x, y, t ) . (C ε
2) The deformation of any arbitrary cross-section perpendicular to the z -axis is identical, i.e., thedisplacements are function of x , y , and t only: u = u ( x, y, t ) . (C ε
3) The material has at least monoclinic symmetry with a plane of reflection corresponding to the plane z = 0, i.e., (52) holds .A system satisfying Assumptions (C ε
1) and (C ε
2) is said to be in a state of classical generalized planestrain. See Figure 3 for an illustration of a body in a state of plane strain.We start with the generalized Hooke’s law (1). By Assumption (C ε σ ij = C ij ∂u ∂x + C ij (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ij ∂u ∂y + C ij ∂u ∂x + C ij ∂u ∂y . (59)Classical generalized plane strain: Combining (59) with the three-dimensional equation of motion (10) and(C ε
1) results in In [38] it is shown the slightly weaker condition C = C = C = C = C = C = 0is sufficient for the derivation of generalized plane stress.6 Jeremy Trageser, Pablo Seleson z b b Fig. 3: Illustration of plane strain. ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + b , (60a) ρ ¨ u = C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + b , (60b) ρ ¨ u = C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b . (60c)Equation (60) is the classical generalized plane strain equation of motion and reduces the degrees of freedomin the system significantly ( u is a function of only x , y , and t , and thus the system may be solved on atwo-dimensional spatial domain). However, the in-plane displacements, u and u , are coupled with theout-of-plane displacement, u . Nevertheless, provided the material has sufficient symmetry, it is possibleto further simplify the model. In fact, having a single plane of reflection symmetry (Assumption (C ε z = 0, as demonstrated below. A system satisfying (C ε
1) – (C ε
3) is said to be ina state of classical plane strain. wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 17
Classical plane strain: Imposing the monoclinic restrictions (52) on (60), we obtain the classical plane strainmodel. The in-plane equations of motion are given by ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + b , (61a) ρ ¨ u = C ∂ u ∂x + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (61b)and the out-of-plane equation of motion is given by ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b . (62)The in-plane equations of motion (61a) and (61b) are equivalent mathematically to the pure two-dimensionaloblique equations of motion (46a) and (46b), respectively. However, it should be noted the elasticity coef-ficients C ijkl do not have the same physical meaning between the two sets of equations, as we will see inSection 2.4 where we discuss the engineering constants.For monoclinic symmetry, there is only one choice of plane of reflection orientation which decouples thein-plane displacements, u and u , and the out-of-plane displacement, u , i.e., when the plane of reflectionsymmetry coincides with the plane z = 0. However, for the higher symmetry classes listed in Section 2.3.1,there are multiple planes of reflection symmetry. Therefore, there are multiple orientations which decouplethe in-plane and out-of-plane displacements, and consequently it is not possible to propose a unique planestrain model for those symmetry classes. For each symmetry class, we consider the equations of motion forthe representative symmetry group presented in Section 2.3.1.Orthotropic: Imposing (53) on (60), we arrive at the following orthotropic plane strain model: ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + ( C + C ) ∂ u ∂x∂y + b , (63a) ρ ¨ u = ( C + C ) ∂ u ∂x∂y + C ∂ u ∂x + C ∂ u ∂y + b , (63b) ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + b . (63c)Notice the in-plane equations of motion (63a) and (63b) are equivalent mathematically to the two-dimensionalrectangular equations of motion (47a) and (47b), respectively.Trigonal: Imposing (54) on (60) we arrive at the following trigonal plane strain model: ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + b , (64a) ρ ¨ u = ( C + C ) ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (64b) ρ ¨ u = C ∂ u ∂x − C ∂ u ∂x∂y + (cid:18) C − C (cid:19) ∂ u ∂y + b . (64c)The in-plane equations of motion of motion (64a) and (64b) are mathematically equivalent to the two-dimensional oblique equations of motion (46a) and (46b), respectively (with C = 0 imposed).Tetragonal and Cubic: Imposing (55) or (57) on (60), we arrive at the following in-plane equations of motion for both the tetragonaland cubic plane strain models: ρ ¨ u = C ∂ u ∂x + C ∂ u ∂y + ( C + C ) ∂ u ∂x∂y + b , (65a) ρ ¨ u = ( C + C ) ∂ u ∂x∂y + C ∂ u ∂x + C ∂ u ∂y + b . (65b)With (55) imposed, the tetragonal out-of-plane deformation satisfies ρ ¨ u = C (cid:18) ∂ u ∂x + ∂ u ∂y (cid:19) + b , (66)whereas with (57) imposed, the cubic out-of-plane deformation satisfies ρ ¨ u = C (cid:18) ∂ u ∂x + ∂ u ∂y (cid:19) + b . (67)Notice the in-plane equations of motion (65a) and (65b) are mathematically equivalent to the two-dimensionalsquare equations of motion (48a) and (48b), respectively.Transversely Isotropic and Isotropic: Imposing (56) or (58) on (60), we arrive at the following in-planeequations of motion for both the transversely isotropic and isotropic plane strain models: ρ ¨ u = C ∂ u ∂x + 12 ( C − C ) ∂ u ∂y + 12 ( C + C ) ∂ u ∂x∂y + b , (68a) ρ ¨ u = 12 ( C + C ) ∂ u ∂x∂y + 12 ( C − C ) ∂ u ∂x + C ∂ u ∂y + b . (68b)With (56) imposed, the transversely isotropic out-of-plane deformation satisfies ρ ¨ u = C (cid:18) ∂ u ∂x + ∂ u ∂y (cid:19) + b , (69)whereas with (58) imposed, the isotropic out-of-plane deformation satisfies ρ ¨ u = (cid:18) C − C (cid:19)(cid:18) ∂ u ∂x + ∂ u ∂y (cid:19) + b . (70)Notice the in-plane equations of motion (68a) and (68b) are mathematically equivalent to the two-dimensionalisotropic equations of motion (49a) and (49b), respectively. Classical plane stress is often associated with thin plate-like structures subjected to edge forces which pro-duce no normal displacement of the mid-plane of the structure [25,38]. Due to the thinness of the structure,if certain assumptions are met, the stress tensor components σ , σ , and σ are approximately constantover the thickness of the structure. A generalization of this, classical generalized plane stress, permits somevariation in those components over the thickness of the structure. In this case, a two-dimensional formu-lation is obtained by considering averages of the displacement field u over the thickness of the structure.Derivations of classical generalized plane stress appear in a variety of sources. Derivations may be foundin [11,21] for the isotropic case and in [38] for the anisotropic case. We begin this section by consideringclassical generalized plane stress. Classical plane stress will be discussed later in Section 2.4. Since we aredealing with anisotropic material models, the following derivation mirrors the work presented in [38].The classical generalized plane stress assumptions are given as follows:(C σ
1) The body is a thin plate of thickness 2 h occupying the region − h (cid:54) z (cid:54) h . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 19 z h b Fig. 4: Illustration of plane stress.(C σ
2) The density is constant in the third dimension: ρ = ρ ( x, y ).(C σ
3) The body is subjected to a loading parallel and symmetric relative to the plane z = 0: b ( x , t ) = 0 and b ( x, y, z, t ) = b ( x, y, − z, t ) . (71)(C σ
4) The first and second components of the initial and boundary conditions are symmetric while theirthird component is antisymmetric relative to the plane z = 0.(C σ
5) The surfaces of the plate are stress-free, i.e., σ = σ = σ = 0 for z = ± h .(C σ
6) The average stress σ ( cf. (74)) is zero throughout the body.(C σ
7) The material has at least monoclinic symmetry with a plane of reflection corresponding to the plane z = 0, i.e., (52) holds .A system satisfying Assumptions (C σ σ
7) is said to be in a state of classical generalized plane stress.See Figure 4 for an illustration of a body in a state of plane stress. For a system in a state of classicalgeneralized plane stress, the displacement field u has certain symmetries imposed on it, which we summarizein Lemma 3. Lemma 3
Under Assumptions (C σ σ
4) and (C σ u of (10) satisfies u i ( x, y, − z, t ) = (cid:26) u i ( x, y, z, t ) , i = 1 , − u i ( x, y, z, t ) , i = 3 , (73) i.e., the in-plane displacements, u and u , are symmetric while the out-of-plane displacement, u , is anti-symmetric relative to the plane z = 0 .Proof See Appendix B.1. (cid:117)(cid:116) It is possible to show that σ = O ( h ). To see this, perform a Taylor series about z = 0 to obtain (omitting x , y , and t dependence for brevity) σ ( z ) = σ (0) + ∂σ ∂z (0) z + O ( z ) . Since the plate is traction-free on the top and bottom surfaces by (C σ σ ( h ) + σ ( − h ) = 2 σ (0) + O ( h ) ⇒ σ (0) = O ( h ) . Thus, σ = 12 h (cid:90) h − h σ ( z ) dz = 12 h (cid:90) h − h σ (0) + ∂σ ∂z (0) z + O ( z ) dz = O ( h ) . (72)Consequently, if the plate is very thin, so that terms of order O ( h ) may be neglected, we may suppose σ ≈ If one were to consider a triclinic material, the in-plane stresses σ , σ , and σ would generate shear strains ε and ε ( cf. (7)). This, in turn, would cause the mid-plane of the plate, the plane z = 0, to no longer remain planar after thein-plane loading. Much like in the plane strain case, this can be alleviated by assuming monoclinic symmetry with the planeof reflection symmetry coinciding with the plane z = 0.0 Jeremy Trageser, Pablo Seleson Before we derive the classical generalized plane stress equation of motion, let us first introduce a shorthandnotation. Given a function f ( x, y, z, t ), we define f ( x, y, t ) := 12 h (cid:90) h − h f ( x, y, z, t ) dz and [ f ]( x, y, t ) := f ( x, y, h, t ) − f ( x, y, − h, t )2 h , (74)where f is the average of f along the z -direction and [ f ] is the average of the values of f at the top andbottom surfaces of the plate. We observe that (73) implies[ u ] = [ u ] = u = 0 . Consequently, we have12 h (cid:90) h − h ∂u k ∂x s dz = ∂u k ∂x s , k, s (cid:54) = 3[ u ] , k = s = 30 , otherwise and ε ks = (cid:16) ∂u k ∂x s + ∂u s ∂x k (cid:17) , k, s (cid:54) = 3[ u ] , k = s = 30 , otherwise . (75)Recalling the stress-strain relation (1) and imposing Assumption (C σ cf. (52)) σ = C ε + C ε + C ε + 2 C ε , (76a) σ = C ε + C ε + C ε + 2 C ε , (76b) σ = C ε + C ε + C ε + 2 C ε , (76c) σ = 2 C ε + 2 C ε , (76d) σ = 2 C ε + 2 C ε , (76e) σ = C ε + C ε + C ε + 2 C ε . (76f)Taking the average along the z -direction for each of the stress-strain relations in (76), we obtain ( cf. (75)) σ = C ∂u ∂x + C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ∂u ∂y + C [ u ] , (77a) σ = C ∂u ∂x + C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ∂u ∂y + C [ u ] , (77b) σ = C ∂u ∂x + C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ∂u ∂y + C [ u ] , (77c) σ = 0 , (77d) σ = 0 , (77e) σ = C ∂u ∂x + C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ∂u ∂y + C [ u ] . (77f)By (77c) and Assumption (C σ σ = C ∂u ∂x + C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C ∂u ∂y + C [ u ] . Solving for [ u ], we find [ u ] = − (cid:20) C C ∂u ∂x + C C (cid:18) ∂u ∂y + ∂u ∂x (cid:19) + C C ∂u ∂y (cid:21) . (78)Plugging (78) into the equations of (77), we find Note for material stability, C > σ = (cid:18) C − C C (cid:19) ∂u ∂x + (cid:18) C − C C C (cid:19)(cid:18) ∂u ∂y + ∂u ∂x (cid:19) + (cid:18) C − C C C (cid:19) ∂u ∂y , (79a) σ = (cid:18) C − C C C (cid:19) ∂u ∂x + (cid:18) C − C C C (cid:19)(cid:18) ∂u ∂y + ∂u ∂x (cid:19) + (cid:18) C − C C (cid:19) ∂u ∂y , (79b) σ = 0 , (79c) σ = 0 , (79d) σ = 0 , (79e) σ = (cid:18) C − C C C (cid:19) ∂u ∂x + (cid:18) C − C C (cid:19)(cid:18) ∂u ∂y + ∂u ∂x (cid:19) + (cid:18) C − C C C (cid:19) ∂u ∂y . (79f)Classical generalized plane stress: Averaging the equation of motion (10) along the z -direction (recall ρ isconstant in z by Assumption (C σ
2) and [ σ ] = [ σ ] = 0 by Assumption (C σ ρ ¨ u = ∂σ ∂x + ∂σ ∂y + [ σ ] + b = (cid:18) C − C C (cid:19) ∂ u ∂x + 2 (cid:18) C − C C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + (cid:18) C − C C C (cid:19) ∂ u ∂x + (cid:18) C − C C C + C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C C (cid:19) ∂ u ∂y + b , (80a) ρ ¨ u = ∂σ ∂x + ∂σ ∂y + [ σ ] + b = (cid:18) C − C C C (cid:19) ∂ u ∂x + (cid:18) C − C C C + C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C C (cid:19) ∂ u ∂y + (cid:18) C − C C (cid:19) ∂ u ∂x + 2 (cid:18) C − C C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (80b)Equation (80) is the classical generalized plane stress equation of motion. By performing the substitution˜ C ijkl = C ijkl − C ij C kl C (81)in (80), we see that up to a change in coefficients, the pure two-dimensional oblique equation of motion(46) and the generalized plane stress equation of motion (80) are equivalent mathematically (if we replace b and u in (46) by b and u , respectively). Moreover, in Section 2.4 we demonstrate that, with respect toengineering constants, these two equations are equivalent. Remark 4
It is possible to perform the steps for the derivation of the classical generalized plane stressmodel (80) in an alternative order (given below), which facilitates the comparison with the derivation ofthe peridynamic generalized plane stress model in Section 3.2.2:Step 1: Take the average of the equation of motion (10) along the thickness of the plate.Step 2: Utilize Assumption (C σ
5) to eliminate [ σ ] and [ σ ].Step 3: Employ Assumption (C σ
6) to replace [ u ] by expressions in u and u .In the derivation, Lemma 3 is utilized to eliminate various terms. Similarly to the plane strain case, for monoclinic symmetry in generalized plane stress, there is only choiceof plane of reflection orientation which decouples the in-plane and out-of-plane displacements (i.e., whenthe plane of reflection coincides with the plane z = 0). However, for many of the other symmetry classesin Section 2.3.1 there are multiple planes of reflection symmetry. Therefore, there are multiple orientationsthat decouple the in-plane and out-of-plane displacements and, consequently, it is not possible to proposea unique generalized plane stress model for those symmetry classes. We consider the equation of motionfor one possible orientation for each symmetry class. We suppose the elasticity tensors for each symmetrygroup are given by those found in Section 2.3.1.Orthotropic: Imposing (53) on (80), we arrive at the following orthotropic generalized plane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + C ∂ u ∂y + (cid:18) C + C − C C C (cid:19) ∂ u ∂x∂y + b , (82a) ρ ¨ u = (cid:18) C + C − C C C (cid:19) ∂ u ∂x∂y + C ∂ u ∂x + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (82b)By performing the substitution (81) in (82), we see that up to a change in coefficients, the orthotropicgeneralized plane stress equation of motion (82) and the pure two-dimensional rectangular equation ofmotion (47) are equivalent mathematically (if we replace b and u in (47) by b and u , respectively).Trigonal: Imposing (54) on (80), we arrive at the following trigonal generalized plane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + 2 C C C ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + C C C ∂ u ∂x + (cid:18) C − C C C + C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C + C C C (cid:19) ∂ u ∂y + b , (83a) ρ ¨ u = C C C ∂ u ∂x + (cid:18) C − C C C + C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C + C C C (cid:19) ∂ u ∂y + (cid:18) C − C C (cid:19) ∂ u ∂x + 2 (cid:18) C + C C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (83b)By performing the substitution (81) in (83), we see that up to a change in coefficients, the trigonal gener-alized plane stress equation of motion (83) and the pure two-dimensional oblique equation of motion (46)are equivalent mathematically (if we replace b and u in (46) by b and u , respectively).Tetragonal: Imposing (55) on (80), we arrive at the following tetragonal generalized plane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + C ∂ u ∂y + (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + b , (84a) ρ ¨ u = (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + C ∂ u ∂x + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (84b)By performing the substitution (81) in (84), we see that up to a change in coefficients, the tetragonalgeneralized plane stress equation of motion (84) and the pure two-dimensional square equation of motion(48) are equivalent mathematically (if we replace b and u in (48) by b and u , respectively).Transversely Isotropic: Imposing (56) on (80), we arrive at the following transversely isotropic generalizedplane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + 12 ( C − C ) ∂ u ∂y + 12 (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + b , (85a) ρ ¨ u = 12 (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + 12 ( C − C ) ∂ u ∂x + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (85b)By performing the substitution (81) in (85), we see that up to a change in coefficients, the transverselyisotropic generalized plane stress equation of motion (85) and the pure two-dimensional isotropic equationof motion (49) are equivalent mathematically (if we replace b and u in (49) by b and u , respectively). wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 23 Pure Two-Dimensional Classical Model Classical Plane Strain or Stress Model
Oblique Monoclinic and TrigonalRectangular OrthotropicSquare Tetragonal and CubicIsotropic Transversely Isotropic and Isotropic
Table 2: Model equivalence (up to a change in constants) between classical pure two-dimensional modelsand classical planar models (provided the three-dimensional elasticity tensors have the form of those inSection 2.3.1).Cubic: Imposing (57) on (80), we arrive at the following cubic generalized plane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + C ∂ u ∂y + (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + b , (86a) ρ ¨ u = (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + C ∂ u ∂x + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (86b)By performing the substitution (81) in (86), we see that up to a change in coefficients, the cubic generalizedplane stress equation of motion (86) and the pure two-dimensional square equation of motion (48) areequivalent mathematically (if we replace b and u in (48) by b and u , respectively).Isotropic: Imposing (58) on (80), we arrive at the following isotropic generalized plane stress model: ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + 12 ( C − C ) ∂ u ∂y + 12 (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + b , (87a) ρ ¨ u = 12 (cid:18) C + C − C C (cid:19) ∂ u ∂x∂y + 12 ( C − C ) ∂ u ∂x + (cid:18) C − C C (cid:19) ∂ u ∂y + b . (87b)By performing the substitution (81) in (87), we see that up to a change in coefficients, the isotropicgeneralized plane stress equation of motion (84) and the pure two-dimensional isotropic equation of motion(49) are equivalent mathematically (if we replace b and u in (49) by b and u , respectively). Remark 5
In Table 2, we summarize the equivalence, up to a change in coefficients, between the in-planeequations of motion for the classical planar models and the corresponding classical pure two-dimensionalmodels. Note that, unlike in classical plane strain ( cf.
Section 2.3.2), in classical generalized plane stress,the in-plane equations of motion for the tetragonal and cubic models as well as those for the transverselyisotropic and isotropic models are not identical.
Remark 6
In classical plane stress, Assumption (C σ
6) is replaced by the assumption σ = σ = σ = 0throughout the body. Note in classical generalized plane stress, we analogously obtain σ = σ = σ = 0;however, only the assumption σ = 0 is directly imposed. It turns out the classical plane stress model isequivalent, in terms of engineering constants ( cf . Section 2.4), to the classical generalized plane stress model(80) if one replaces the displacement field u with its average u and the body force density field b with itsaverage b . This is shown in more detail in Section 2.4. Consequently, we only consider classical generalizedplane stress in this paper.2.4 Engineering constantsAn important consideration is the abuse of notation in the elasticity tensor between the two- and three-dimensional formulations of classical linear elasticity. In particular, even though the in-plane equationsof motion for classical plane strain, (61a) and (61b), look identical to the classical pure two-dimensionaloblique equations of motion, (46a) and (46b), the coefficients are not equivalent. For instance, the elasticityconstant C in two dimensions does not have the same physical meaning as the one in three dimensions.In fact, it can be shown that the classical pure two-dimensional model (46) is, in some sense, equivalent tothe classical generalized plane stress model (80). To facilitate these comparisons, it is useful to express theelasticity constants in terms of engineering constants [18,22,40], which we now briefly introduce. Young’s moduli : For a given i , under a uniaxial stress σ ii , we define the Young’s modulus E i := σ ii ε ii . Shear moduli : Given i and j , under a shear stress σ ij , we define the shear modulus G ij := σ ij ε ij . Poisson’s ratios : Given i and j , under a uniaxial stress σ ii , we define the Poisson’s ratio ν ij := − ε jj ε ii . Chentsov’s coefficients : Given i, j, k, and l , under a shear stress σ ij , we define the Chentsov’s coefficient µ kl,ij := ε kl ε ij , { ij } (cid:54) = { kl } . Coefficients of mutual influence of the first type : Given i, j, and k , under a shear stress σ ij , we definethe coefficient of mutual influence of the first type η kk,ij := ε kk ε ij . Coefficients of mutual influence of the second type : Given i, j, and k , under a uniaxial stress σ ii , wedefine the coefficient of mutual influence of the second type η jk,ii := 2 ε jk ε ii . Remark 7
Note these coefficients can be divided into two groups, those which are measured under a uniaxialstress: E i , ν ij , and η jk,ii , and those which are measured under a shear stress: G ij , µ kl,ij , η kk,ij .To express the elasticity constants in terms of the engineering constants, one typically relates the compliancetensor S ( cf. (6)) to the engineering constants and then inverts S to obtain C . In order to relate a componentof the compliance tensor S ijkl to the engineering constants, one imposes a single nonzero component of thestress in (7), either a uniaxial stress σ ii for a given i to obtain S iiii = 1 E i , S jjii = − ν ij E i , and 2 S jkii = η jk,ii E i , (88)or a shear stress σ ij for a given i and j to obtain4 S ijij = 1 G ij , S klij = µ kl,ij G ij , and 2 S kkij = η kk,ij G ij . (89)Additionally, the major symmetry ( S ijkl = S klij ) of the compliance tensor (7) implies that not all of theseconstants are independent. In fact, ν ij E i = ν ji E j , η ij,kk E k = η kk,ij G ij , and µ ij,kl G kl = µ kl,ij G ij . (90) As we only have one shear strain component in two dimensions, the Chentsov’s coefficients do not play a role in the classicalpure two-dimensional model. However, we include them here for the comparisons between the classical pure two-dimensionalmodel and the classical plane strain and classical generalized plane stress models.wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 25
In three dimensions, the strain-stress relation in (7) can be expressed in terms of engineering constants as ε ε ε ε ε ε = E − ν E − ν E η , E η , E η , E · E − ν E η , E η , E η , E · · E η , E η , E η , E · · · G µ , G µ , G · · · · G µ , G · · · · · G σ σ σ σ σ σ . (91)In classical plane strain and classical generalized plane stress monoclinic symmetry is assumed. Undermonoclinic symmetry, we may simplify (91) to ε ε ε ε = E − ν E − ν E η , E · E − ν E η , E · · E η , E · · · G σ σ σ σ and (cid:20) ε ε (cid:21) = (cid:20) G µ , G · G (cid:21)(cid:20) σ σ (cid:21) . (92) Remark 8
In classical plane strain and classical generalized plane stress, ε = ε = 0 and σ = σ = 0,respectively. Immediately from the second system in (92), we obtain ε = ε = σ = σ = 0 for classicalplane strain and ε = ε = σ = σ = 0 for classical generalized plane stress.By inverting the systems in (92), we obtain expressions for the elasticity constants C ijkl for monoclinicsymmetry in terms of the engineering constants: C = − E G η , η , ν + E G η , + E G η , + E ν − E E E E G | A | ,C = E G η , η , ν + E G η , η , − (cid:0) E G η , − E E (cid:1) ν + (cid:0) E G η , η , + E ν (cid:1) ν E E E G | A | ,C = E G η , η , ν + E G η , η , − (cid:0) E G η , − E E (cid:1) ν + ( E G η , η , + E E ν ) ν E E E G | A | ,C = E η , ν − E η , ν − E η , ν − E η , − ( E η , ν + E η , ν ) ν E E E | A | ,C = − E G η , η , ν + E G η , + E G η , + E ν − E E E E G | A | ,C = E G η , η , ν + E G η , η , + ( E G η , η , + E E ν ) ν − (cid:0) E G η , − E E (cid:1) ν E E E G | A | ,C = − E η , ν ν − E η , ν + E η , ν + E η , + ( E η , ν + E η , ) ν E E E | A | ,C = − E G η , η , ν + E G η , + E G η , + E ν − E E E E G | A | ,C = E η , ν − E E η , − ( E E η , ν + E E η , ) ν − ( E E η , ν + E E η , ) ν E E E | A | ,C = G G − G µ , ,C = − G G µ G − G µ , ,C = G G G − G µ , ,C = − E E ν ν ν + E ν + E E ν + E E ν − E E E E E | A | , (93) where | A | := det E − ν E − ν E η , E · E − ν E η , E · · E η , E · · · G . In classical plane stress, one typically supposes σ = σ = σ = 0. Imposing this on the systems in (92),we may relate the strains ε , ε , and ε to the stresses σ , σ , and σ as follows: ε ε ε = E − ν E η , E · E η , E · · G σ σ σ . (94)Note that (94) is exactly the strain-stress relationship one obtains for classical pure two-dimensional linearelasticity, i.e., the two-dimensional analogue of (91). Moreover, since σ and σ are null in classical planestress, the resulting in-plane equations of motion are identical in terms of engineering constants to theclassical pure two-dimensional equation of motion (46).By inverting the system (94), we obtain expressions for the elasticity constants C ijkl in terms of theengineering constants for the classical pure two-dimensional model (and the classical plane stress model): C = E − G η , E G | S | ,C = E ν + G η , η , E E G | S | ,C = − η , ν + η , E E | S | ,C = E − G η , E G | S | ,C = − E η , + E η , ν E E | S | ,C = E − E ν E E | S | , (95)where | S | := det E − ν E η , E · E η , E · · G . From this analysis, it is clear that the elasticity constants C ijkl appearing in the classical plane strain andclassical generalized plane stress models are not equivalent to those in classical pure two-dimensional linearelasticity. However, it is interesting to note that substituting (93) into the classical generalized plane stressequation of motion (80) and substituting (95) into the classical pure two-dimensional equation of motion(46) results in the exact same equation in terms of engineering constants (if we replace b and u in (46) by b and u , respectively). Further details on engineering constants in three dimensions can be found in [31].2.5 Cauchy’s RelationsIn the nineteenth century, there was some contention about the quantity of independent constants in theequations of classical linear elasticity. Proponents of the multi-constant theory, such as Green, Stokes,and Thomson, supported a model with twenty-one elasticity constants. Alternatively, proponents of therari-constant theory, such as Navier, Poisson, and Saint-Venant, supported a model with fifteen elasticityconstants. The fifteen elasticity constant formulation, first derived by Cauchy, is obtained by assuming amolecular description of materials based on central forces between pairs of molecules. This fifteen constantformulation places six additional restrictions on the elasticity tensor C : wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 27 C ijkl = C ikjl . (96)In [21], the term Cauchy’s relations was coined to refer to these additional symmetries. A far more detailedand complete historical account of the controversy can be found in [21].Eventually the fifteen constant formulation from the rari-constant theory was shown to be invalid for manymaterials; however, as we will see later, there are materials whose properties satisfy (96). Cauchy’s rela-tions make appearances in a variety of theories of mechanics. For example, it is well known that a moleculartheory of elasticity based on pairwise potentials imposes Cauchy’s relations (96) on the corresponding elas-ticity tensor [35]. Since bond-based peridynamics is similarly based on a pairwise potential formulation, itis perhaps unsurprising that bond-based peridynamics suffers from the same limitations, i.e., it should onlybe utilized to describe materials with properties satisfying Cauchy’s relations. This is most evident whenconsidering isotropic materials where it is well known that Poisson’s ratio is restricted to in three dimen-sions [32] and in two dimensions ( cf. Appendix A). Since this work deals with bond-based peridynamics,a discussion of Cauchy’s relations is essential.In three dimensions, there are six Cauchy’s relations: C = C , C = C , C = C ,C = C , C = C , C = C . (97)With the addition of Cauchy’s relations (97) to the major symmetry (3) and the minor symmetries (2), thenumber of independent elasticity constants reduces from twenty-one to fifteen, and C becomes a completelysymmetric tensor. See [31] for an in-depth discussion of the implications of Cauchy’s relations in threedimensions.In two dimensions, there is a single Cauchy’s relation: C = C . (98)With the addition of Cauchy’s relation (98) to the major symmetry (3) and the minor symmetries (2),the number of independent elasticity constants reduces from six to five ( cf. Table 4), and C becomes acompletely symmetric tensor. In terms of the engineering constants, (98) can be expressed as ( cf. (95)and (90)) G = E ν − ν ν − η , η , . (99)Imposing (98) on the two-dimensional elasticity tensor for each of the symmetry classes ( cf. Theorem 1)results in the following forms for the elasticity tensor:
Symmetry Group Elasticity Tensor(Cauchy’s relation imposed)
Oblique C C C · C C · · C (100a)Rectangular C C · C · · C (100b)Square C C · C · · C (100c)Isotropic C C · C · · C . (100d) The restrictions each symmetry class imposes on the engineering constants as well as the correspondingCauchy’s relation is summarized in Table 3. To obtain the engineering constant symmetry restrictions,simply invert each elasticity tensor in Theorem 1 and equate it to the compliance tensor in (94). Then, toget the corresponding Cauchy’s relation in terms of technical constants, impose the engineering constantsymmetry restrictions on (99). Additionally, due to Cauchy’s relation, we lose one degree of freedom ineach symmetry class. The number of independent constants in the elasticity tensor with and without theCauchy’s relation imposed is summarized for each symmetry class in Table 4.Table 3: Restrictions on engineering constants by symmetry class in two-dimensional classical linear elas-ticity.
Symmetry Class Symmetry Restrictions Cauchy’s RelationOblique G = E ν − ν ν − η , η , Rectangular η , = η , = 0 G = E ν − ν ν Square η , = η , = 0 , ν = ν , E = E G = E ν − ν Isotropic η , = η , = 0 , ν = ν , E = E , G = E ν ) G = 3 E ν = 13 Table 4: Number of independent constants by symmetry class in two-dimensional classical linear elasticity.
Symmetry Class Number of Constants Number of Constants(Cauchy’s relation imposed)Oblique 6 5Rectangular 4 3Square 3 2Isotropic 2 1
A goal of this paper is to develop bond-based peridynamic analogues of the classical two-dimensional andplanar linear elastic anisotropic models. In the bond-based peridynamic theory [32], given a body
B ⊂ R d ,where d is the dimension, the equation of motion for a material point x ∈ B at time t (cid:62) cf .[10]) ρ ( x )¨ u ( x , t ) = (cid:90) H x f ( u ( x (cid:48) , t ) , u ( x , t ) , x (cid:48) , x ) dV x (cid:48) + b ( x , t ) , (101)where ρ is the mass density, ¨ u is the second derivative in time of the displacement field u , f is the pairwiseforce function representing the nonlocal interaction a material point x (cid:48) exerts on the material point x , and b is a prescribed body force density field. The neighborhood of x , H x , represents the domain over which thematerial point x is able to directly interact. A material point x is typically assumed to only directly interactwith material points in the body B within some prescribed distance δ , called the peridynamic horizon. Inthis case , H x = B ∩ B δ ( x ) , (102)where B δ ( x ) := (cid:8) x (cid:48) ∈ R d : (cid:107) x (cid:48) − x (cid:107) < δ (cid:9) is the ball in R d of radius δ centered at x . In particular, whena material point x is in the bulk of the body, i.e., further than δ from the boundary of the body ∂ B , H x = B δ ( x ). In the isotropic case, Cauchy’s relation reduces to C = C − C and thus C = C = C . When interactions are limited to the body, boundary conditions are effectively imposed on the model.wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 29
For convenience, we introduce the usual shorthand notation ξ := x (cid:48) − x and η := u ( x (cid:48) , t ) − u ( x , t ), whichrepresent the relative position in the undeformed configuration and the relative displacement, respectively,of the material points x and x (cid:48) ; we refer to ξ as a peridynamic bond.There are several essential conditions which must be placed upon the pairwise force function f in (101). Toensure invariance under rigid body motion, we require f ( u ( x (cid:48) , t ) , u ( x , t ) , x (cid:48) , x ) = f ( η , x (cid:48) , x ) , ∀ η , x (cid:48) , x ∈ R d . (103)To enforce conservation of linear momentum, we require f ( η , x (cid:48) , x ) = − f ( − η , x , x (cid:48) ) , ∀ η , x (cid:48) , x ∈ R d . (104)To guarantee balance of angular momentum, we require( ξ + η ) × f ( η , x (cid:48) , x ) = , ∀ η , x (cid:48) , x ∈ R d , (105)i.e., the pairwise force function acts along the direction of the deformed bond ξ + η . We further suppose thematerial is microelastic, i.e., the pairwise force function f derives from a scalar-valued pairwise potentialfunction w : f ( η , x (cid:48) , x ) = ∂w∂ η ( η , x (cid:48) , x ) , ∀ η , x (cid:48) , x ∈ R d . (106)In order to obtain a linear bond-based peridynamic model, we suppose a small deformation, in particular (cid:107) η (cid:107) (cid:28) δ [33], and linearize the pairwise force function f ( η , · , · ) while holding x (cid:48) and x fixed to obtain f ( η , x (cid:48) , x ) = f ( , x (cid:48) , x ) + C ( x (cid:48) , x ) η , (107)where C ( x (cid:48) , x ) is the second-order micromodulus tensor given by C ( x (cid:48) , x ) := ∂ f ∂ η ( , x (cid:48) , x ) (108)and terms of order O ( (cid:107) η (cid:107) ) have been omitted. Next, we look at the implications of enforcing conditions(103), (104), (105), and (106) on the linearized pairwise force function (107). Combining (104) and (107)we obtain C ( x , x (cid:48) ) = C ( x (cid:48) , x ) , ∀ x (cid:48) , x ∈ R d . (109)Additionally, by considering (106) and (108), we deduce C T ( x (cid:48) , x ) = C ( x (cid:48) , x ) , ∀ x (cid:48) , x ∈ R d . (110)An immediate consequence of (105), is the existence of a scalar-valued function F ( η , x (cid:48) , x ) such that f ( η , x (cid:48) , x ) = ( ξ + η ) F ( η , x (cid:48) , x ) , ∀ η , x (cid:48) , x ∈ R d . (111)Differentiating (111) with respect to η , we obtain from (108) that C ( x (cid:48) , x ) = ξ ⊗ ∂F∂ η ( , x (cid:48) , x ) + F ( , x (cid:48) , x ) I , ∀ x (cid:48) , x ∈ R d . (112)A necessary and sufficient condition for (110) to be imposed on (112) is the existence of a scalar-valuedfunction λ ( x (cid:48) , x ) such that ξ ⊗ ∂F∂ η ( , x (cid:48) , x ) = λ ( x (cid:48) , x ) ξ ⊗ ξ , ∀ x (cid:48) , x ∈ R d . (114) Sufficiency is trivial. To see necessary, set A ( x (cid:48) , x ) := ∂F∂ η ( , x (cid:48) , x ). Then ξ ⊗ A ( x (cid:48) , x ) is symmetric if and only if ξ i A j ( x (cid:48) , x ) = ξ j A i ( ξ , ξ ) , ∀ i, j. (113)If ξ i = 0 and ξ j (cid:54) = 0 then A i ( x (cid:48) , x ) = 0 by (113). Thus we may suppose there exists a function a i ( x (cid:48) , x ) such that A i ( x (cid:48) , x ) = ξ i a i ( x (cid:48) , x ) (no summation implied by repeated indices). From (113) we find that a i ( x (cid:48) , x ) = a j ( x (cid:48) , x ) for all i, j . Consequentlythere is a scalar-valued function a ( x (cid:48) , x ) such that A ( x (cid:48) , x ) = a ( x (cid:48) , x ) ξ .0 Jeremy Trageser, Pablo Seleson The general form of the linearized pairwise force function (107) satisfying conditions (103), (104), (105),and (106) is therefore given by f ( η , x (cid:48) , x ) = f ( , x (cid:48) , x ) + [ λ ( x (cid:48) , x ) ξ ⊗ ξ + F ( , x (cid:48) , x ) I ] η . (115)Henceforth, we refer to λ ( x (cid:48) , x ) as the micromodulus function. Due to (109), the micromodulus functionhas the symmetry λ ( x (cid:48) , x ) = λ ( x , x (cid:48) ) , ∀ x (cid:48) , x ∈ R d . (116)Commonly in bond-based peridynamics, one supposes the system is pairwise equilibrated in the referenceconfiguration, i.e., f ( , x (cid:48) , x ) = , ∀ x (cid:48) , x ∈ R d . In this case (115) becomes ( cf. (111)) f ( η , x (cid:48) , x ) = λ ( x (cid:48) , x ) ξ ⊗ ξ ( u ( x (cid:48) , t ) − u ( x , t )) . (117)Substituting (117) into the peridynamic equation of motion (101) ( cf. (103)), we obtain the linear bond-based peridynamic equation of motion: ρ ( x )¨ u ( x , t ) = (cid:90) H x λ ( x (cid:48) , x ) ξ ⊗ ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) + b ( x , t ) . (118)Written in component form, (118) is expressed as (for i ∈ { , . . . , d } ) ρ ( x )¨ u i ( x , t ) = (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ k ( u k ( x (cid:48) , t ) − u k ( x , t )) d x (cid:48) + b i ( x , t ) . (119)In order to consider material symmetry classes and facilitate the utilization of (119), we relate the classicalelasticity tensor C from (1) to the micromodulus function λ appearing in (119). For this purpose, we suppose u is smooth, perform a Taylor expansion, and equate the coefficients of the derivatives of u in the resultingperidynamic equation of motion with the coefficients of the corresponding derivatives of u in the classicalequation of motion. More formally, by expanding u ( x (cid:48) , t ) about x , we obtain u k ( x (cid:48) , t ) − u k ( x , t ) = ∂u k ∂x j ( x , t ) ξ j + 12 ∂ u k ∂x j ∂x l ( x , t ) ξ j ξ l + · · · , j, l ∈ { , . . . , d } . (120)Substituting (120) into (119), we obtain ρ ( x )¨ u i ( x , t ) = (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ k (cid:18) ∂u k ∂x j ( x , t ) ξ j + 12 ∂ u k ∂x j ∂x l ( x , t ) ξ j ξ l + · · · (cid:19) d x (cid:48) + b i ( x , t ) . (121)Equating (121) with (10) and comparing coefficients for terms up to second order, we find the followingconditions imposed on λ ( x (cid:48) , x ): 0 = (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ j ξ k d x (cid:48) , (122a) C ijkl = 12 (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ j ξ k ξ l d x (cid:48) . (122b)Notice that (122b) can only hold if C is a completely symmetric tensor, i.e., in addition to the minorsymmetries (2) and major symmetry (3) of C , Cauchy’s relations (96) hold. Consequently, the linear bond-based peridynamic model (118) agrees with the classical linear elasticity model (10) up to second-order termsonly when describing a material satisfying Cauchy’s relations . Throughout the remainder of the paper, weassume Cauchy’s relations are imposed on C .We now define the peridynamic analogue of Definition 1. Definition 2
An orthogonal transformation Q is a symmetry transformation of a micromodulus function λ if λ ( Qx (cid:48) , Qx ) = λ ( x (cid:48) , x ) , ∀ x , x (cid:48) ∈ R d . (123) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 31 Note Definition 2 is a generalization of the symmetry definition introduced in [32] for the case when λ is notstrictly a function of the bond ξ . Up to this point, no assumptions have been made on the homogeneity of thematerial response. With the assumption of homogeneity, further simplifications are possible. In particular,in the bulk of the body, the micromodulus function may be assumed to be a function solely of the bond( cf . [32]): λ ( x (cid:48) , x ) = λ ( x (cid:48) − x ) = λ ( ξ ) , ∀ x , x (cid:48) ∈ R d . (124)If we assume a homogeneous material response, for a given material point x in the bulk of the body( H x = B δ ( x )), the peridynamic equation of motion (118) becomes (after the change of variables x (cid:48) → x + ξ ): ρ ( x )¨ u ( x , t ) = (cid:90) B δ ( ) λ ( ξ ) ξ ⊗ ξ ( u ( x + ξ , t ) − u ( x , t )) d ξ + b ( x , t ) . (125)Written in component form, (125) becomes ρ ( x )¨ u i ( x , t ) = (cid:90) B δ ( ) λ ( ξ ) ξ i ξ k ( u k ( x + ξ , t ) − u k ( x , t )) d ξ + b i ( x , t ) . (126)Moreover, for a material point in the bulk of a body, (122a) holds trivially by antisymmetry since λ ( ξ ) = λ ( − ξ ) by (116). Additionally, (122b) reduces to (see [31] and related expressions in [2,27]) C ijkl = 12 (cid:90) B δ ( ) λ ( ξ ) ξ i ξ j ξ k ξ l d ξ . (127)The following proposition shows that if (127) holds, C inherits the symmetries of λ ( ξ ). Proposition 1 ( cf . [31]) Let Q be an orthogonal transformation. Assume the micromodulus functionsatisfies the symmetry property λ ( Q ξ ) = λ ( ξ ) for all ξ ∈ R d . Then, a fourth-order tensor C defined by (127) is symmetric with respect to Q , i.e., (14) holds. Note the converse of Proposition 1 is not necessarily true for an arbitrarily chosen micromodulus function λ ( ξ ).In order to develop anisotropic peridynamic models, we provide a specific form for λ . In [31], the followingmicromodulus was proposed: λ ( ξ ) = 1 m ( ξ ⊗ ξ ) Λ ( ξ ⊗ ξ ) (cid:107) ξ (cid:107) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) , (128)where Λ is a fully symmetric fourth-order tensor, ω is an influence function, and m is the weighted volume[34] m := (cid:90) B δ ( ) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) d ξ . (129)Influence functions in peridynamics are commonly utilized to control the radial dependence of the interactionbetween material points [28,34]. For convenience, we call Λ the peridynamic tensor.Note we have not selected a dimension for the peridynamic model. For a suitable choice of ω , the theorydefined up to this point is valid in R d for d ∈ N . In particular, in R , the micromodulus (128) is given by λ ( ξ ) = 1 m ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) (cid:20) Λ ξ + 4 Λ ξ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 12 Λ ξ ξ ξ (cid:107) ξ (cid:107) + 6 Λ ξ ξ + 4 Λ ξ ξ + 12 Λ ξ ξ ξ + 12 Λ ξ ξ ξ + 4 Λ ξ ξ (cid:107) ξ (cid:107) + Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 12 Λ ξ ξ + Λ ξ (cid:107) ξ (cid:107) (cid:21) , (130) It is shown in [31] that under assumption (127), the micromodulus (130) in spherical coordinates may equivalently beformulated by replacing the anisotropic portion of (130), with a fourth-order spherical harmonics expansion.2 Jeremy Trageser, Pablo Seleson in R , the micromodulus (128) is given by λ ( ξ ) = 1 m Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ (cid:107) ξ (cid:107) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) , (133)and in R , the micromodulus (128) is given by λ ( ξ ) = Λ m ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) , where Λ ijkl are the components of the fourth-order peridynamic tensor Λ .Analogously to classical linear elasticity, where the fourth-order elasticity tensor C can be representedby a second-order tensor C ( cf. (4) and (5)), we may utilize Voigt notation to express the fourth-orderperidynamic tensor Λ as a second-order symmetric tensor Λ . In this case, the anisotropic part of (130) canbe written as ( ξ ⊗ ξ ) Λ ( ξ ⊗ ξ ) = ξ ξ ξ ξ ξ ξ ξ ξ ξ T Λ Λ Λ Λ Λ Λ · Λ Λ Λ Λ Λ · · Λ Λ Λ Λ · · · Λ Λ Λ · · · · Λ Λ · · · · · Λ ξ ξ ξ ξ ξ ξ ξ ξ ξ (134)and the anisotropic part of (133) can be written as( ξ ⊗ ξ ) Λ ( ξ ⊗ ξ ) = ξ ξ ξ ξ T Λ Λ Λ · Λ Λ · · Λ ξ ξ ξ ξ . (135)The micromodulus function (128) has many desirable properties such as: – It can be informed by the classical elasticity tensor through (127). – Its peridynamic tensor Λ and the elasticity tensor C (with Cauchy’s relations imposed) have the samenumber of degrees of freedom. – When it is related to the classical elasticity tensor C through (127), it has the same symmetries as C with respect to Definitions 1 and 2 ( cf. Proposition 2 for two dimensions and [31] for three dimensions).We now have sufficient background to develop pure two-dimensional, plane strain, and plane stress anisotropiclinear bond-based peridynamic models. One can produce an equivalent formulation by replacing the anisotropic portion of (133) with a fourth-order Fourier seriesformulation. In particular, changing to polar coordinates, ( ξ , ξ ) = ( r cos( θ ) , r sin( θ )), in (133) and setting Λ = a + a + a ,Λ =2 a − a ,Λ =2 b + 4 b ,Λ = a − a + a ,Λ =2 b − b , we immediately obtain λ ( r, θ ) = 1 m (cid:32) (cid:88) n =0 a n cos(2 nθ ) + b n sin(2 nθ ) (cid:33) ω ( r ) r . (131)Equation (131) is equivalent to λ ( r, θ ) = 1 m (cid:32) (cid:88) n =0 a n cos( nθ ) + b n sin( nθ ) (cid:33) ω ( r ) r , (132)when invariance with respect to the inversion symmetry λ ( r, θ + π ) = λ ( r, θ ) is assumed. To see this note that sin( nθ ) andcos( nθ ) do not have inversion symmetry when n is odd and thus we must have a = a = b = b = 0.wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 33 ρ ( x )¨ u ( x , t ) = (cid:90) B δ ( ) λ ( ξ ) ξ [ ξ ( u ( x + ξ , t ) − u ( x , t )) + ξ ( u ( x + ξ , t ) − u ( x , t ))] d ξ + b ( x , t ) , (136a) ρ ( x )¨ u ( x , t ) = (cid:90) B δ ( ) λ ( ξ ) ξ [ ξ ( u ( x + ξ , t ) − u ( x , t )) + ξ ( u ( x + ξ , t ) − u ( x , t ))] d ξ + b ( x , t ) . (136b)We further suppose the micromodulus function λ ( ξ ) is described by (133). We relate Λ ijkl to C ijkl throughrelation (127) to obtain Λ = 10 C − C + 2 C , (137a) Λ = 20 C − C , (137b) Λ = 13 ( − C + 76 C − C ) , (137c) Λ = − C + 20 C , (137d) Λ = 2 C − C + 10 C . (137e)We can relate the peridynamic tensor Λ to the engineering constants by substituting (95) into (137). Remark 9
It is interesting to observe the system (137) can be expressed as two decoupled subsystems ofequations: Λ Λ Λ = −
20 2 −
103 763 − −
20 10 C C C and (cid:20) Λ Λ (cid:21) = (cid:20) − −
12 20 (cid:21)(cid:20) C C (cid:21) . (138)The most general peridynamic tensor Λ , the oblique peridynamic tensor, may be represented as ( cf. (135)): Λ = Λ Λ Λ · Λ Λ · · Λ . (139)In Proposition 2 we prove the converse of Proposition 1 when the micromodulus function λ ( ξ ) is given by(128). For an analogous proof in three dimensions see [31]. Proposition 2
Let λ ( ξ ) be given by (128) and suppose (127) holds. If C is invariant with respect to oneof the four symmetry groups in two-dimensional classical linear elasticity ( cf. Theorem 1), then λ ( ξ ) is alsoinvariant under the symmetry group transformations ( cf. Definition 2).Proof
Let λ ( ξ ) be given by (128) and suppose (127) holds. Recall that orthogonal transformations preservelength. Hence (cid:107) ξ (cid:107) and ω ( (cid:107) ξ (cid:107) ) are always invariant under orthogonal transformations. Therefore, we onlyneed to consider the anisotropic part of λ ( ξ ) ( cf. (135)), when showing λ ( ξ ) is invariant with respect to agiven orthogonal transformation. Oblique : Suppose C is given by (100a). The corresponding symmetry group is generated by {− I } ( cf. Lemma 1),i.e., ξ → − ξ and ξ → − ξ . Since the anisotropic part of λ ( ξ ) is a sum of fourth-order monomials of com-ponents of ξ ( cf. (133)), we clearly have λ ( ξ ) is symmetric with respect to − I . Rectangular : Suppose C is given by (100b). Then, by (137) we have ( cf. (23)) Λ = 10 C − C + 2 C , (140a) Λ = 0 , (140b) Λ = 13 ( − C + 76 C − C ) , (140c) Λ = 0 , (140d) Λ = 2 C − C + 10 C , (140e)and λ ( ξ ) = 1 m Λ ξ + 6 Λ ξ ξ + Λ ξ (cid:107) ξ (cid:107) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) . (141)The rectangular peridynamic tensor Λ may be represented as ( cf. (139)) Λ = Λ Λ · Λ · · Λ . (142)The rectangular symmetry group is generated by {− I , Ref (0) } ( cf. Lemma 1). From the oblique portionof the proof, we already know λ ( ξ ) is symmetric with respect to − I . The transformation Ref (0) implies ξ → − ξ ( cf. (21)). Since the anisotropic part of (141) is even in ξ , the invariance with respect to Ref (0)follows. Note that the functional form of λ ( ξ ) is also invariant under the transformation ξ → − ξ , whichis expected since this transformation is given by Ref ( π ) = − I Ref (0).
Square : Suppose C is given by (100c). Then, by (137) we have ( cf. (30)) Λ = 12 C − C , (143a) Λ = 0 , (143b) Λ = 13 ( − C + 76 C ) , (143c) Λ = 0 , (143d) Λ = Λ , (143e)and λ ( ξ ) = 1 m Λ ( ξ + ξ ) + 6 Λ ξ ξ (cid:107) ξ (cid:107) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) . (144)The square peridynamic tensor Λ may be represented as ( cf. (139)) Λ = Λ Λ · Λ · · Λ . (145)The square symmetry group is generated by (cid:8) − I , Ref (0) , Ref (cid:0) π (cid:1)(cid:9) ( cf. Lemma 1). From the rectangularportion of the proof, we already know λ ( ξ ) is symmetric with respect to {− I , Ref (0) } . The transformation Ref (cid:0) π (cid:1) implies ξ → ξ and ξ → ξ ( cf. (16)). We note (144) is clearly invariant when interchanging ξ and ξ . Isotropic : Suppose that C is given by (100d). Then, by (137) we have ( cf. (32)) Λ = 163 C , (146a) Λ = 0 , (146b) Λ = 13 Λ , (146c) Λ = 0 , (146d) Λ = Λ , (146e) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 35 Fig. 5: Isotropic γ ( θ ) with C = 1.and λ ( ξ ) = 1 m Λ ( ξ + 2 ξ ξ + ξ ) (cid:107) ξ (cid:107) ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) = Λ m ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) . (147)The isotropic peridynamic tensor Λ may be represented as ( cf. (139)) Λ = Λ Λ · Λ · · Λ . (148)To obtain (146), note Cauchy’s relation in the isotropic setting implies C = C ( cf. Footnote 7).Notice (147) is invariant under any orthogonal transformation of ξ as λ ( ξ ) is only dependent on (cid:107) ξ (cid:107) , whichis preserved by orthogonal transformations. (cid:117)(cid:116) Remark 10
It is interesting to note that the peridynamic tensor Λ has the same form as the elasticity tensor C for each of the symmetry classes when Cauchy’s relation C = C is imposed. This can be readilyseen by comparing (139), (142), (145), and (148) with (100a), (100b), (100c), and (100d), respectively. In this section, we study the angular dependence of the two-dimensional micromodulus function in each ofthe four symmetry classes. We do this by plotting the angular portion of λ ( ξ ), which we define asˆ γ ( ξ ) := ( ξ ⊗ ξ ) Λ ( ξ ⊗ ξ ) (cid:107) ξ (cid:107) . (149)If we change to polar coordinates, ξ = r cos( θ ) and ξ = r sin( θ ), notice (149) is independent of the radialcomponent r = (cid:107) ξ (cid:107) , and so one may write γ ( θ ) := ˆ γ ( ξ ), where θ is the angle that ξ makes with the positive x -axis. To provide a visualization of the variation in γ with respect to this angle θ , we present plots in polarcoordinates ( θ, s ) with s = γ ( θ ). Throughout this section, we normalize the elasticity constant C to 1,and we consider multiples of C for the other elasticity constants. Isotropic : As one would expect, for isotropic symmetry we obtain a circle (i.e., γ ≡ C is independentof the bond orientation ( cf. (146))) as can be seen in Figure 5. Square : For square symmetry, we have two independent constants C and C ( cf. (143)). By varyingthe ratio between C and C , one is able to increase or decrease the influence of γ in the θ = kπ directions (where ξ = 0 or ξ = 0 and γ ( θ ) = 12 C − C ) while inversely influencing in the θ = (2 k +1) π directions (where ξ = ξ and γ ( θ ) = − C + 28 C ), where k ∈ Z . In Figure 6 wevary C to observe this dependence. When C = < (in solid blue), we see γ favors the θ = kπ directions over the θ = (2 k +1) π directions, where k ∈ Z . When C = (in dash-dotted red), we have the Fig. 6: Square γ ( θ ) with C = 1 for C = (solid blue), (dash-dotted red), (dotted yellow). C = C = C = Fig. 7: Rectangular γ ( θ ) with C = 1 for C = (left), (middle), (right) and C = (solidblue), 1 (dash-dotted red), (dotted yellow).isotropic case ( C = C ). When C = > (in dotted yellow), we see γ favors the θ = (2 k +1) π directions over the θ = kπ directions, where k ∈ Z . Rectangular : For rectangular symmetry, we have, in addition to C and C , a third independentconstant, C ( cf. (140)). By varying the ratio of C and C we are able to increase or decrease theinfluence of γ in the θ = (2 k +1) π directions (where ξ = 0 and γ ( θ ) = 2 C − C + 10 C ) relativeto the influence of γ in the θ = kπ directions (where ξ = 0 and γ ( θ ) = 10 C − C + 2 C ), where k ∈ Z . In Figure 7, we have three plots corresponding to C = (left plot), (middle plot), and (right plot), which are the three cases we considered in the square case. For each of these cases of C ,we consider C = (in solid blue), 1 (in dash-dotted red), and (in dotted yellow). In each imagein Figure 7, we see that when C = < C , γ favors the θ = kπ directions over the θ = (2 k +1) π directions. When C = 1 = C , there is no preference between the axes of the plot. Lastly, when C = > C , we see the θ = (2 k +1) π directions are favored over the θ = kπ directions. Oblique : For oblique symmetry, in addition to C , C , and C , there are two additional constants, C and C . These two additional constants only contribute to the peridynamic tensor components Λ and Λ ( cf. (138)). The contribution to the angular portion of the micromodulus function from C and C can be entirely described by ( cf. (133))4 ξ ξ (cid:0) Λ ξ + Λ ξ (cid:1) (cid:107) ξ (cid:107) . (150)First, note the magnitude of expression (150) favors ξ away from the main axes as ξ ξ approaches zero inthose cases. Second, the term Λ ξ + Λ ξ favors either the ξ ( x -direction) or ξ ( y -direction) depend- wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 37 C = C = C = C = C = C = C = C = C = C = 1 C = 1 C = 1 C = C = C = C = C = C = Fig. 8: Oblique γ ( θ ) for C = 1; C = (left), (middle), (right); C = (top), 1 (middle), (bottom); and C = − C = − (solid blue), 0 (dash-dotted red), (dotted yellow).ing on the values of C and C . There is a multitude of possible behaviors that may be represented by(133). In an attempt to consider the relative influences of C and C on the micromodulus function,we present a small subset of these behaviors in Figure 8. We consider the cases where C = − C = − (solid blue), 0 (dash-dotted red), and (dotted yellow). cf . Section 2.3.1).The resulting models are two-dimensional, anisotropic, and reduce to the classical planar elasticity modelswhen the displacements are smooth and higher-order terms are negligible. To derive the peridynamic plane strain model, we begin with the three-dimensional bond-based linearperidynamic model (118) and impose peridynamic analogues of the classical plane strain assumptions. Wesuppose a homogeneous material response and only consider material points within the bulk of the body. Inthis case, the three-dimensional bond-based linear peridynamic equation of motion is given in componentform by ( cf . (126)): ρ ( x )¨ u ( x , t ) = (cid:90) B Dδ ( ) λ ( ξ ) (cid:2) ξ ( u ( x + ξ , t ) − u ( x , t )) + ξ ξ ( u ( x + ξ , t ) − u ( x , t ))+ ξ ξ ( u ( x + ξ , t ) − u ( x , t ))] d ξ + b ( x , t ) , (151a) ρ ( x )¨ u ( x , t ) = (cid:90) B Dδ ( ) λ ( ξ ) (cid:2) ξ ξ ( u ( x + ξ , t ) − u ( x , t )) + ξ ( u ( x + ξ , t ) − u ( x , t ))+ ξ ξ ( u ( x + ξ , t ) − u ( x , t ))] d ξ + b ( x , t ) , (151b) ρ ( x )¨ u ( x , t ) = (cid:90) B Dδ ( ) λ ( ξ )[ ξ ξ ( u ( x + ξ , t ) − u ( x , t )) + ξ ξ ( u ( x + ξ , t ) − u ( x , t ))+ ξ ( u ( x + ξ , t ) − u ( x , t )) (cid:3) d ξ + b ( x , t ) , (151c)where B Dδ ( ) is the ball in R of radius δ about the origin. To avoid confusion, the ball in R (i.e. a disk)of radius δ about the origin is denoted by B Dδ ( ). We assume almost identical assumptions to those inclassical plane strain ( cf. Section 2.3.2).The peridynamic plane strain assumptions are given as follows:(P ε
1) The geometric form and mass density of the body, and the external loads exerted on it, do not changealong some axis, which we take as the z -axis. In particular, ρ = ρ ( x, y ) and b = b ( x, y, t ) . (P ε
2) The deformation of any arbitrary cross-section perpendicular to the z -axis is identical, i.e., thedisplacements are function of x , y , and t only: u = u ( x, y, t ) . (P ε
3) The material has at least monoclinic symmetry with a plane of reflection corresponding to the plane z = 0, i.e., the micromodulus function is symmetric in its third component ( cf. [31]): λ ( ξ , ξ , ξ ) = λ ( ξ , ξ , − ξ ) . (152) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 39 A system satisfying Assumptions (P ε
1) and (P ε
2) is said to be in a state of peridynamic generalized planestrain. See Figure 3 for an illustration of a body in a state of plane strain.Peridynamic generalized plane strain: Imposing Assumptions (P ε
1) and (P ε
2) on (151) and then integratingover the third component, ξ , we obtain the peridynamic analogue of (60): ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ ) (cid:2) ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) + ξ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) (cid:3) dξ dξ + (cid:90) B Dδ ( ) λ ( ξ , ξ , t ) ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) dξ dξ + b ( x, y, t ) , (153a) ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ ) (cid:2) ξ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) + ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) (cid:3) dξ dξ + (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) dξ dξ + b ( x, y, t ) , (153b) ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ )[ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) + ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t ))] dξ dξ + (cid:90) B Dδ ( ) λ ( ξ , ξ )( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) dξ dξ + b ( x, y, t ) , (153c)where the micromodulus functions { λ n ( ξ , ξ ) } are defined by λ n ( ξ , ξ ) := (cid:90) √ δ − ξ − ξ − √ δ − ξ − ξ λ ( ξ ) ξ n dξ , n ∈ { , , } . (154)Equation (153) is the peridynamic generalized plane strain equation of motion and reduces the degrees offreedom in the system significantly ( u is a function of only x , y , and t , and thus the simulation may berun on a two-dimensional spatial domain). However, the in-plane displacements, u and u , are coupledwith the out-of-plane displacement, u . Nevertheless, provided the material has sufficient symmetry, it ispossible to further simplify the model. In fact, having a single plane of reflection symmetry (Assumption(P ε z = 0, as demonstrated below. A system satisfying (P ε
1) – (P ε
3) is saidto be in a state of peridynamic plane strain.Peridynamic plane strain: Emulating the classical theory by imposing monoclinic symmetry from Assump-tion (P ε
3) on (153), the in-plane displacements, u and u , and the out-of-plane displacement, u , decouple.A system satisfying Assumptions (P ε
1) – (P ε
3) is said to be in a state of peridynamic plane strain. In thiscase, the micromodulus function λ ( ξ ) is even in its third component and therefore λ ( ξ , ξ ) ≡ ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ ) (cid:2) ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) + ξ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) (cid:3) dξ dξ + b ( x, y, t ) , (155a) ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ ) (cid:2) ξ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) + ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) (cid:3) dξ dξ + b ( x, y, t ) , (155b) and the resulting out-of-plane equation of motion (analogue of (62)) is given by ρ ( x, y )¨ u ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ , ξ )( u ( x + ξ , y + ξ , t ) − u ( x, y, t )) dξ dξ + b ( x, y, t ) . (156) Remark 11
One may express the in-plane equations of motion (155) in vector form by letting x = ( x, y ) , ξ =( ξ , ξ ) , u = ( u , u ) , H = B Dδ ( ), and b = ( b , b ). In this case, the in-plane equations of motion areformulated as ρ ( x )¨ u ( x , t ) = (cid:90) H λ ( ξ ) ξ ⊗ ξ ( u ( x + ξ , t ) − u ( x , t )) d ξ + b ( x , t ) . Peridynamic plane strain micromodulus functions: In the peridynamic generalized plane strain model pre-sented in this work, the requirements placed on the micromodulus function λ ( ξ ) have been fairly minimalup to this point. In order to investigate the differences between the pure two-dimensional peridynamic equa-tion of motion (136) and the in-plane equations of motion (155) for the peridynamic plane strain model,we consider the micromodulus function (128) related to the elasticity tensor through (127). In the puretwo-dimensional peridynamic equation of motion, indices of Λ are in { , } , (128) is given by (133), and Λ is related to C through (137). Alternatively, in the in-plane equations of motion for the peridynamic planestrain model, the indices of Λ are in { , , } , (128) is given by (130), and Λ is related to C through ( cf. [31]): Λ = 30 C − C − C + 154 C + 152 C + 154 C , (157a) Λ = − C + 2554 C − C − C − C + 54 C , (157b) Λ = − C − C + 2554 C + 54 C − C − C , (157c) Λ = 70 C − C − C , (157d) Λ = 1052 C − C − C , (157e) Λ = 1052 C − C − C , (157f) Λ = 154 C − C + 152 C + 30 C − C + 154 C , (157g) Λ = 54 C − C − C − C + 2554 C − C , (157h) Λ = − C + 1052 C − C , (157i) Λ = − C + 70 C − C , (157j) Λ = − C + 1052 C − C , (157k) Λ = 154 C + 152 C − C + 154 C − C + 30 C , (157l) Λ = − C − C + 1052 C , (157m) Λ = − C − C + 1052 C , (157n) Λ = − C − C + 70 C . (157o)In order to obtain expressions for λ n ( ξ , ξ ) in (154), it is convenient to introduce a shorthand notation: wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 41 A ( ξ , ξ ) := Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (158a) A ( ξ , ξ ) := 4 (cid:0) Λ ξ + Λ ξ + 3 Λ ξ ξ + 3 Λ ξ ξ (cid:1) , (158b) A ( ξ , ξ ) := 6 (cid:0) Λ ξ + Λ ξ + 2 Λ ξ ξ (cid:1) , (158c) A ( ξ , ξ ) := 4( Λ ξ + Λ ξ ) , (158d) A ( ξ , ξ ) := Λ . (158e)Then, the micromodulus function (130) can be expressed as λ ( ξ ) = 1 m ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) (cid:88) i =0 A i ( ξ , ξ ) ξ i (cid:107) ξ (cid:107) = 1 m ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) A ( ξ , ξ ) + A ( ξ , ξ ) ξ + A ( ξ , ξ ) ξ + A ( ξ , ξ ) ξ + A ( ξ , ξ ) ξ (cid:107) ξ (cid:107) . (159)Depending on the choice of influence function ω , it may be possible to provide closed-form expressionsfor the integrals in (154). Two commonly utilized influence functions in peridynamics are ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) and ω ( (cid:107) ξ (cid:107) ) = 1. For convenience, we additionally introduce the shorthand notation r = (cid:112) ξ + ξ , whichrepresents the magnitude of the projection of ξ on the xy -plane. This is in contrast to (cid:107) ξ (cid:107) = (cid:112) ξ + ξ + ξ ,which is the magnitude of ξ in R . Given ω ( (cid:107) ξ (cid:107) ) = 1 or ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) , the micromodulus functions (154)are given by λ ( ξ , ξ ) = ω ( r ) mr (cid:20) A ( ξ , ξ ) r M (cid:16) rδ (cid:17) + A ( ξ , ξ ) r M (cid:16) rδ (cid:17) + A ( ξ , ξ ) M (cid:16) rδ (cid:17)(cid:21) , (160a) λ ( ξ , ξ ) = ω ( r ) m (cid:20) A ( ξ , ξ ) r M (cid:16) rδ (cid:17) + A ( ξ , ξ ) r M (cid:16) rδ (cid:17)(cid:21) , (160b) λ ( ξ , ξ ) = rω ( r ) m (cid:20) A ( ξ , ξ ) r M (cid:16) rδ (cid:17) + A ( ξ , ξ ) r M (cid:16) rδ (cid:17) + A ( ξ , ξ ) M (cid:16) rδ (cid:17)(cid:21) , (160c)where for ω ( (cid:107) ξ (cid:107) ) = 1 we have: M ( x ) := 34 arctan (cid:16)(cid:112) x − − (cid:17) + 14 (cid:0) x + 2 x (cid:1)(cid:112) − x ,M ( x ) := 14 arctan (cid:16)(cid:112) x − − (cid:17) + 14 (cid:0) x − x (cid:1)(cid:112) − x ,M ( x ) := 34 arctan (cid:16)(cid:112) x − − (cid:17) − (cid:0) x − x (cid:1)(cid:112) − x ,M ( x ) := −
154 arctan (cid:16)(cid:112) x − − (cid:17) + 14 (cid:0) x − + 9 x − x (cid:1)(cid:112) − x , and for ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) we have : M ( x ) := 215 (cid:112) − x (cid:0) x + 3 x (cid:1) ,M ( x ) := 215 (cid:112) − x (cid:0) x − x (cid:1) ,M ( x ) := 25 (cid:112) − x (cid:0) − x + x (cid:1) ,M ( x ) := − (cid:112) − x (cid:0) − x + 3 x (cid:1) + 2 arsinh (cid:16)(cid:112) x − − (cid:17) . The equations for M ( x ) below is only valid for x = rδ >
0. As the set (cid:8) ξ ∈ R : ξ + ξ = 0 (cid:9) is a set of zero measure in B δ ( ), (160c) remains valid with this formulation for M .2 Jeremy Trageser, Pablo Seleson Remark 12
For i ∈ { , . . . , } , note that A i ( ξ ,ξ ) r − i is radially independent ( cf. (158)) and, consequently, onlycontributes to the angular portion of (160). Moreover, as the other terms in (160) are radial functions, the A i ( ξ ,ξ ) r − i terms make up the entirety of the angular dependence of the micromodulus functions.Next, we take a closer look at the micromodulus functions { λ n ( ξ , ξ ) } ( cf. (154)) of the peridynamic planestrain model (155) and (156) for various symmetry classes. Following Remark 12, the choice of symmetryclass only has an effect on the { A i ( ξ , ξ ) } , while the general form of the micromodulus functions (160)remains unchanged. Therefore, we only present the { A i ( ξ , ξ ) } for each symmetry class. As explainedearlier, for many of the symmetry classes, there are multiple planes of reflection symmetry to choose from inorder to satisfy Assumption (P ε cf. (158)): A ( ξ , ξ ) = Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (161a) A ( ξ , ξ ) = 0 , (161b) A ( ξ , ξ ) = 6 (cid:0) Λ ξ + Λ ξ + 2 Λ ξ ξ (cid:1) , (161c) A ( ξ , ξ ) = 0 , (161d) A ( ξ , ξ ) = Λ , (161e)where Λ = 30 C − C − C + 154 C + 152 C + 154 C ,Λ = − C + 2554 C − C − C − C + 54 C ,Λ = − C − C + 2554 C + 54 C − C − C ,Λ = 1052 C − C − C ,Λ = 154 C − C + 152 C + 30 C − C + 154 C ,Λ = 54 C − C − C − C + 2554 C − C ,Λ = − C + 1052 C − C ,Λ = 154 C + 152 C − C + 154 C − C + 30 C ,Λ = − C − C + 70 C . Since A ( ξ , ξ ) = A ( ξ , ξ ) = 0, we immediately have λ ( ξ , ξ ) = 0 as expected for peridynamic planestrain ( cf. (155) and (156)).Orthotropic: We substitute (53), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ ξ + 6 Λ ξ ξ + Λ ξ , (162a) A ( ξ , ξ ) = 0 , (162b) A ( ξ , ξ ) = 6 (cid:0) Λ ξ + Λ ξ (cid:1) , (162c) A ( ξ , ξ ) = 0 , (162d) A ( ξ , ξ ) = Λ , (162e) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 43 where Λ = 30 C − C − C + 154 C + 152 C + 154 C ,Λ = − C + 2554 C − C − C − C + 54 C ,Λ = − C − C + 2554 C + 54 C − C − C ,Λ = 154 C − C + 152 C + 30 C − C + 154 C ,Λ = 54 C − C − C − C + 2554 C − C ,Λ = 154 C + 152 C − C + 154 C − C + 30 C . Trigonal: We substitute (54), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (163a) A ( ξ , ξ ) = 0 , (163b) A ( ξ , ξ ) = 2 (cid:0) Λ ξ + Λ ξ − Λ ξ ξ (cid:1) , (163c) A ( ξ , ξ ) = 0 , (163d) A ( ξ , ξ ) = Λ , (163e)where Λ = 30 C − C + 10 C ,Λ = − C + 1152 C − C ,Λ = 154 C − C + 754 C ,Λ = 3154 C . Tetragonal: We substitute (55), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ ( ξ + ξ ) + 6 Λ ξ ξ , (164a) A ( ξ , ξ ) = 0 , (164b) A ( ξ , ξ ) = 6 Λ r , (164c) A ( ξ , ξ ) = 0 , (164d) A ( ξ , ξ ) = Λ , (164e)where Λ = 1354 C − C − C + 154 C ,Λ = − C + 2554 C − C + 54 C ,Λ = − C − C + 1152 C − C ,Λ = 152 C + 152 C − C + 30 C . Transversely Isotropic: We substitute (56), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ r , (165a) A ( ξ , ξ ) = 0 , (165b) A ( ξ , ξ ) = 6 Λ r , (165c) A ( ξ , ξ ) = 0 , (165d) A ( ξ , ξ ) = Λ , (165e)where Λ = 754 C − C + 154 C ,Λ = − C + 1152 C − C ,Λ = 10 C − C + 30 C . Cubic: We substitute (57), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ ( ξ + ξ ) + 6 Λ ξ ξ , (166a) A ( ξ , ξ ) = 0 , (166b) A ( ξ , ξ ) = 6 Λ r , (166c) A ( ξ , ξ ) = 0 , (166d) A ( ξ , ξ ) = Λ , (166e)where Λ = 752 C − C ,Λ = − C + 2054 C . Isotropic: We substitute (58), with Cauchy’s relations imposed, into (157) to find ( cf. (158)) A ( ξ , ξ ) = Λ r , (167a) A ( ξ , ξ ) = 0 , (167b) A ( ξ , ξ ) = 2 Λ r , (167c) A ( ξ , ξ ) = 0 , (167d) A ( ξ , ξ ) = Λ , (167e)where Λ = 10 C . Remark 13
In classical linear elasticity, the plane strain model for each symmetry class reduces identicallyto a corresponding two-dimensional model ( cf . Table 2). A similar situation occurs with the peridynamicplane strain model. While the plane strain micromodulus function λ , given by (160a), is not equivalentto the two-dimensional micromodulus function (133), it does possess one of the four symmetries of two-dimensional classical linear elasticity ( cf . Theorem 1) for each symmetry class ( cf . Table 5). This canbe observed by considering the symmetries of { A i ( ξ , ξ ) } for each symmetry class. The reason why theplane strain micromodulus functions are not identical to their two-dimensional counterparts is that theyincorporate out-of-plane information. It is also interesting to note that the same correspondence betweeneach three-dimensional symmetry class and the corresponding two-dimensional symmetry class in Table 2occurs for the plane strain micromodulus function, which is summarized in Table 5. However, while differentplane strain micromodulus functions may possess the same two-dimensional symmetry, e.g., tetragonal wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 45 Pure Two-Dimensional Peridynamic Model Peridynamic Plane Strain Model
Oblique Monoclinic and TrigonalRectangular OrthotropicSquare Tetragonal and CubicIsotropic Transversely Isotropic and Isotropic
Table 5: Symmetry equivalence between the pure two-dimensional peridynamic models and the peridynamicplane strain models (provided the micromodulus functions λ and λ is informed by three-dimensionalelasticity tensors having the form of those in Section 2.3.1).and cubic micromodulus functions both have square symmetry, the resulting plane strain micromodulusfunctions are unique for each three-dimensional symmetry class. Due to this fact, the resulting in-planeplane strain equations of motion are unique for each three-dimensional symmetry class. This is in contrastto classical linear elasticity, where the in-plane plane strain equations of motion may be identical for twothree-dimensional symmetry classes, specifically tetragonal and cubic models as well as transversely isotropicand isotropic models.It is of interest to compare our resulting peridynamic plane strain micromodulus function (160a) with planestrain micromodulus functions commonly utilized in the peridynamic literature. This can be directly donein the case of isotropy. The isotropic micromodulus functions are actually quite simple. When the influencefunction ω ( (cid:107) ξ (cid:107) ) = 1, m = πδ ( cf. (129)) and (160) simplifies to λ = 25 C πδ r arctan (cid:32)(cid:114) δ r − (cid:33) , (168a) λ = 0 , (168b) λ = 25 C πδ (cid:34)(cid:114) − r δ − rδ arctan (cid:32)(cid:114) δ r − (cid:33)(cid:35) . (168c)Alternatively, when the influence function ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) , m = πδ ( cf. (129)) and (160) simplifies to λ = 20 C πδ r (cid:114) − r δ , (169a) λ = 0 , (169b) λ = 20 C πδ (cid:34) arsinh (cid:32)(cid:114) δ r − (cid:33) − (cid:114) − r δ (cid:35) . (169c)In terms of the engineering constants, (168a) and (169a) are given, respectively, by: λ = 30 Eπδ r arctan (cid:32)(cid:114) δ r − (cid:33) (170)and λ = 24 Eπδ r (cid:114) − r δ , (171)where E is the Young’s modulus. Here, we used the fact that for an isotropic material with Cauchy’srelations satisfied, Poisson’s ratio ν = and thus ( cf. (93)) C = (1 − ν ) E (1 + ν )(1 − ν ) = 6 E . A common approach for modeling plane strain in the peridynamic literature simply employs a two-dimensional peridynamic model and matches the model constants to those in the in-plane equations of the corresponding classical plane strain model (see, e.g., [13,14,20,26]). To emulate this approach, we con-sider our two-dimensional micromodulus function (133), where m is the two-dimensional weighted volumeand Λ ijkl are given by (137), except that the C ijkl are the three-dimensional elasticity constants appearingin the classical in-plane equations of motion (61). In the isotropic case, we combine (147) with (146) andrecall for isotropic symmetry that C = E when Cauchy’s relations are imposed. When the influencefunction ω ( r ) = 1, m = πδ and the micromodulus function is given by λ ( r ) = 64 E πδ r . (172)Alternatively, when the influence function ω ( r ) = r , m = πδ and the micromodulus function is given by λ ( r ) = 48 E πδ r . (173)The micromodulus function in (173) coincides with the micromodulus function obtained by linearizingthe peridynamic plane strain model derived in [13]. We therefore refer to the two-dimensional peridynamicmicromodulus function (133) with constants matched to those in the in-plane equations of motion (61) of thecorresponding classical plane strain model as the traditional plane strain micromodulus function . We observethat comparing (170) with (172) and (171) with (173), our plane strain micromodulus functions possessa weaker singularity than the traditional plane strain micromodulus functions. Moreover, our plane strainmicromodulus functions continuously transition to zero when r approaches δ . Next, we further explore thedifferences between our plane strain micromodulus functions and the traditional plane strain micromodulusfunctions for various symmetry classes. Micromodulus function visualization for peridynamic plane strain
In order to more easily compare our plane strain micromodulus functions (160a) and the traditional planestrain micromodulus functions based on (133), we create visualizations of their behavior for various sym-metry classes. We observe that in the peridynamic equation of motion (119), the micromodulus function ismultiplied by ξ i ξ j . Consequently, the factor of (cid:107) ξ (cid:107) appearing in the micromodulus function (133) representsa removable (or “artificial”) singularity of the model. To better visualize the nonremovable singularities,when present, in the micromodulus function, we plot r λ ( ξ , ξ ) for the traditional plane strain micromodu-lus function (recall (cid:107) ξ (cid:107) = r in two dimensions). For a proper comparison to the plane strain micromodulusfunction (160a), we accordingly plot r λ ( ξ , ξ ).Our comparison of the micromodulus functions covers materials in every symmetry class except triclinic,because the plain strain Assumption (P ε
3) requires a plane of reflection symmetry. Since we are deal-ing with bond-based peridynamic models, we must consider materials satisfying (at least approximately)Cauchy’s relations. The elasticity tensors for all anisotropic materials were obtained from [7]. For isotropicsymmetry, the elasticity tensor was obtained from [36]. In Table 6, we present the chosen materials andthe corresponding elasticity tensors for each symmetry class. All of the chosen anisotropic materials ap-proximately satisfy Cauchy’s relations. In order to produce elasticity tensors satisfying Cauchy’s relationsexactly, given an elasticity tensor C , we produce a new elasticity tensor ˜ C in the following manner. Formonoclinic, orthotropic, tetragonal, and cubic symmetries, we take˜ C ijkl := C ijkl + C ikjl . For trigonal symmetry, Cauchy’s relations additionally impose C = C , and thus we take˜ C ijkl = (cid:0) C + C + C (cid:1) , { i, j, k, l } = { , , , } (cid:0) C + C + C (cid:1) , i = j = k = l = 2 C ijkl + C ikjl , else . For transversely isotropic symmetry, Cauchy’s relations additionally impose C = C , and thus wetake ˜ C ijkl = (cid:0) C + C + C (cid:1) , { i, j, k, l } = { , , , } (cid:0) C + C + C (cid:1) , i = j = k = l = 1 C ijkl + C ikjl , else . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 47 SymmetryClass Material Elasticity Tensor C (in GPa)
Elasticity Tensor ˜ C (in GPa) (Cauchy’s relations imposed) Monoclinic CoTeO
135 19 54 0 0 42 ·
13 15 0 0 6 · ·
269 0 0 18 · · ·
14 25 0 · · · ·
66 0 · · · · ·
135 18 . ·
13 14 . · ·
269 0 0 21 . · · · . . · · · ·
60 0 · · · · · . Orthotropic Te W
143 1 37 0 0 0 · · ·
102 0 0 0 · · · · · · ·
46 0 · · · · ·
143 1 41 . · . · ·
102 0 0 0 · · · . · · · · . · · · · · Trigonal Ta C
493 141 141 0 0 0 ·
464 159 0 0 45 · ·
464 0 0 − · · · −
45 0 · · · ·
125 0 · · · · ·
493 133 133 0 0 0 ·
467 156 0 0 45 · ·
467 0 0 − · · · −
45 0 · · · ·
133 0 · · · · ·
Tetragonal Si
212 70 58 0 0 0 ·
212 58 0 0 0 · ·
179 0 0 0 · · ·
58 0 0 · · · ·
58 0 · · · · ·
212 77 . ·
212 58 0 0 0 · ·
179 0 0 0 · · ·
58 0 0 · · · ·
58 0 · · · · · . TransverselyIsotropic MoN
499 177 235 0 0 0 ·
499 235 0 0 0 · ·
714 0 0 0 · · ·
241 0 0 · · · ·
241 0 · · · · ·
504 168 238 0 0 0 ·
504 238 0 0 0 · ·
714 0 0 0 · · ·
238 0 0 · · · ·
238 0 · · · · ·
Cubic MgAl O (Spinel)
252 145 145 0 0 0 ·
252 145 0 0 0 · ·
252 0 0 0 · · ·
142 0 0 · · · ·
142 0 · · · · ·
252 143 . . ·
252 143 . · ·
252 0 0 0 · · · . · · · · . · · · · · . Isotropic Pyroceram9608 . . . · . . · · . · · · . · · · · . · · · · · . . . . · . . · · . · · · . · · · · . · · · · · . Table 6: Materials and their corresponding symmetry classes and elasticity tensors, C and ˜ C .For isotropic symmetry, we selected the material Pyroceram 9608 which already satisfies Cauchy’s rela-tions; in this case, we simply take ˜ C = C . The resulting elasticity tensors ˜ C for each symmetry class aresummarized in the fourth column of Table 6.In Figure 9, we present plots for r λ ( ξ , ξ ) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and r λ ( ξ , ξ ) with ω ( r ) = r (inred). Similarly, in Figure 10, we present plots for r λ ( ξ , ξ ) with ω ( (cid:107) ξ (cid:107) ) = 1 (in blue) and r λ ( ξ , ξ ) with ω ( r ) = 1 (in orange). For each symmetry class, these functions are found by substituting the correspondingelasticity tensor ˜ C from Table 6 into (160a) (with (158) and (157)) and (133) (with (137)), respectively.For all of the plots, we took δ = 1. One of the most obvious differences between the plots of r λ ( ξ , ξ )and r λ ( ξ , ξ ) is their behavior near the origin. As remarked earlier for the isotropic case, our plane strainmicromodulus functions have weaker singularities compared to the traditional plane strain micromodulusfunctions. This generalizes to all the symmetry classes, which is clearly observed in Figure 9. Anothersignificant difference between the two micromodulus functions is that our plane strain micromodulus func-tions effectively incorporate out-of-plane information. This is clearly seen in the isotropic and transverselyisotropic cases in Figure 10. Connections between the peridynamic and classical plane strain equations : ) Orthotropic (Te W)Trigonal (Ta C) Tetragonal (Si)Transversely Isotropic (MoN) Cubic (MgAl O )Isotropic (Pyroceram 9608) Fig. 9: Plots of r λ ( ξ , ξ ) ( cf. (160a)) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and r λ ( ξ , ξ ) ( cf. (133)) with ω ( r ) = r (in red). wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 49Monoclinic (CoTeO ) Orthotropic (Te W)Trigonal (Ta C) Tetragonal (Si)Transversely Isotropic (MoN) Cubic (MgAl O )Isotropic (Pyroceram 9608) Fig. 10: Plots of r λ ( ξ , ξ ) ( cf. (160a)) with ω ( (cid:107) ξ (cid:107) ) = 1 (in blue) and r λ ( ξ , ξ ) ( cf. (133)) with ω ( r ) = 1(in orange). We now show that the peridynamic plane strain model reduces to the classical plane strain model undersuitable assumptions.
Proposition 3
Suppose the micromodulus function λ ( ξ ) is related to the elasticity tensor C through (127) .Given a smooth deformation, under a second-order Taylor expansion of the displacement field, the peridy-namic generalized plane strain model (153) reduces to the classical generalized plane strain model (60) withCauchy’s relations imposed. Moreover, under a similar Taylor expansion, the decoupled peridynamic planestrain in-plane equations of motion (155) and out-of-plane equation of motion (156) reduce to the classicalplane strain in-plane equations of motion (61) and out-of-plane equation of motion (62) , respectively, withCauchy’s relations imposed.Proof Recalling (154), one may write (cid:90) B Dδ ( ) λ n ( ξ , ξ ) ξ l ξ m dξ dξ = (cid:90) B Dδ ( ) λ ( ξ ) ξ l ξ m ξ n d ξ , and thus the peridynamic generalized plane strain equation of motion (153) may be expressed as ρ ( x, y )¨ u i ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ ) ξ i ξ j ( u j ( x + ξ , y + ξ , t ) − u j ( x, y, t )) d ξ + b i ( x, y, t ) . (174)Performing a Taylor expansion of u j ( x + ξ , y + ξ , t ) in (174) about ( x , t ) = ( x, y, t ), we obtain ρ ( x, y )¨ u i ( x, y, t ) = (cid:90) B Dδ ( ) λ ( ξ ) ξ i ξ j (cid:20) ξ ∂u j ∂x ( x, y, t ) + ξ ∂u j ∂y ( x, y, t )+ ξ ∂ u j ∂x ( x, y, t ) + ξ ξ ∂ u j ∂x∂y ( x, y, t ) + ξ ∂ u j ∂y ( x, y, t ) + · · · (cid:21) d ξ + b i ( x, y, t ) . (175)Assuming higher-order terms (beyond second-order) are negligible and utilizing antisymmetry under thetransformation ξ → − ξ (recall λ ( − ξ ) = λ ( ξ ) by (116)) to nullify the first-order terms in (175), we obtain ρ ( x, y )¨ u i ( x, y, t ) = 12 (cid:90) B Dδ ( ) λ ( ξ ) ξ i ξ j (cid:20) ξ ∂ u j ∂x ( x, y, t ) + 2 ξ ξ ∂ u j ∂x∂y ( x, y, t ) + ξ ∂ u j ∂y ( x, y, t ) (cid:21) d ξ + b i ( x, y, t ) . (176)Employing the relationship between λ ( ξ ) and C from (127), we rewrite (176) as ρ ( x, y )¨ u i ( x, y, t ) = C ij ∂ u j ∂x ( x, y, t ) + 2 C ij ∂ u j ∂x∂y ( x, y, t ) + C ij ∂ u j ∂y ( x, y, t ) + b i ( x, y, t ) . (177)Writing (177) out, we obtain ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (178a) ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (178b) ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (178c) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 51 where we have omitted the arguments x, y, and t for brevity and a more direct comparison to the classicalmodel. Noticing that (178) is exactly the classical generalized plane strain equation of motion (60) withCauchy’s relations ( cf. (97)) imposed completes the proof of the first portion of Proposition 3.To prove the peridynamic plane strain model converges to the classical plane strain model, first recallthe in-plane and peridynamic out-of-plane plane strain equations of motion (155) and (156), respectively,are obtained by imposing (P ε
3) on the peridynamic generalized plane strain model (153). We thus followthe same derivation as for the generalized plane strain case until (176). As before, we then employ therelationship between λ ( ξ ) and C from (127). However, the additional assumption (P ε
3) results in themonoclinic symmetry restrictions (52) being imposed on the elasticity tensor. Consequently, the peridynamicin-plane equations of motion reduce to ( cf . (178a) and (178b)) ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (179a) ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b , (179b)while the peridynamic out-of-plane equation of motion reduces to ( cf. (178c)) ρ ¨ u = C ∂ u ∂x + 2 C ∂ u ∂x∂y + C ∂ u ∂y + b . (180)Comparing (179a) and (179b) to the classical in-plane equations of motion (61a) and (61b), with Cauchy’srelations imposed, and comparing (180) to the classical out-of-plane equation of motion (62), with Cauchy’srelations imposed, completes the proof. (cid:117)(cid:116) Remark 14
Notice in the in-plane equations of motion for classical generalized plane strain, (60a) and (60b),only three Cauchy’s relations are relevant: C = C , C = C , and C = C , while in the out-of-plane equation of motion for classical generalized plane strain, (60c), five Cauchy’srelations are relevant: C = C , C = C , C = C , C = C , and C = C . In contrast, in the in-plane equations of motion for classical plane strain, (61a) and (61b), only one Cauchy’srelation is relevant: C = C , while in the out-of-plane equation of motion for classical plane strain, (62), three Cauchy’s relations arerelevant: C = C , C = C , and C = C . The above difference between the classical generalized plane strain equations of motion and classical planestrain equations of motion is entirely due to the elimination of two Cauchy’s relations by the monoclinicsymmetry assumption. It is interesting to note that the in-plane equations of motion for classical planestrain, (61a) and (61b), have exactly the same relevant Cauchy’s relation as the classical two-dimensionalequation of motion, (46). x x x e e Fig. 11: Various neighborhoods (in red) of material points x , x , and x in an xz -cross-section of a thinplate. We now focus on the peridynamic analogue of classical plane stress ( cf.
Section 2.3.3). Similarly to theperidynamic plane strain formulation in Section 3.2.1, we begin with the anisotropic three-dimensionalbond-based linear peridynamic model (118) and impose peridynamic analogues of the classical generalizedplane stress assumptions.One key issue with developing a peridynamic formulation for generalized plane stress is that it is unrealisticto restrict the analysis to material points in the bulk of a body for thin plates. In particular, we mustconsider position-dependent neighborhoods of material points, see, e.g., Figure 11. Thus, we must deal witha position-aware micromodulus function λ ( x (cid:48) , x ), and the three-dimensional bond-based linear peridynamicmodel is given by (118). We suppose a homogeneous material response and only consider material pointsfar from the boundaries of the plate in the first and second dimensions. This allows us to assume onlyposition awareness in the third dimension, i.e., λ ( x (cid:48) , x ) = λ ( x (cid:48) − x, y (cid:48) − y, z (cid:48) , z ) . (181)Furthermore, to avoid unintentionally imposing boundary conditions on (118), we additionally suppose H x = B δ ( x ) ( cf. Footnote 8).The peridynamic generalized plane stress assumptions are given as follows:(P σ
1) The body is a thin plate of thickness 2 h (cid:54) δ occupying the region − h (cid:54) z (cid:54) h .(P σ
2) The density is constant in the third dimension: ρ = ρ ( x, y ).(P σ
3) The body is subjected to a loading symmetric and parallel to the plane z = 0: b ( x , t ) = 0 and b ( x, y, z, t ) = b ( x, y, − z, t ) . (182)(P σ
4) The first and second components of the initial and boundary conditions are symmetric while theirthird component is antisymmetric relative to the plane z = 0.(P σ
5) The micromodulus function λ ( x (cid:48) , x ) is null when x (cid:48) and x are not material points of the plate.(P σ
6) The average peridynamic traction τ ( x , t, e ) ( cf. (190)) is zero throughout the body.(P σ
7) The material has at least monoclinic symmetry with a plane of reflection corresponding to the plane z = 0, i.e., the micromodulus function has the following symmetries ( cf . Definition 2 and (51)) : λ ( ξ , ξ , z (cid:48) , z ) = λ ( ξ , ξ , − z (cid:48) , − z ) = λ ( ξ , ξ , z, z (cid:48) ) . (183) Invariance of λ with respect to Ref ( e ) ensures λ ( ξ , ξ , z (cid:48) , z ) = λ ( ξ , ξ , − z (cid:48) , − z ) . Requiring invariance of λ with respect to − I and then recalling (116), we find λ ( ξ , ξ , − z (cid:48) , − z ) = λ ( − ξ , − ξ , z (cid:48) , z ) = λ ( ξ , ξ , z, z (cid:48) ) . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 53 (P σ
8) The displacement u ( x , t ) is smooth in z and its third component, u ( x , t ), is smooth in x .There are a variety of similarities between the peridynamic generalized plane stress assumptions and theclassical generalized plane stress assumptions. Notice Assumption (P σ
1) is Assumption (C σ
1) with anadditional restriction on the thickness 2 h that depends on the peridynamic horizon δ . This assumptionis reasonable for thin plates, since the computational expense associated with the discretization requiredto accommodate a horizon δ < h is likely impractical. This restriction could likely be removed, but itsimplifies the analysis considerably. Assumptions (P σ σ σ
4) are identical to Assumptions(C σ σ σ σ
5) is a nonlocal analogue of Assumption (C σ
5) inthat interactions through the top and bottom surfaces of the plate are nullified . Assumption (P σ
6) is aperidynamic analogue of Assumption (C σ σ peri so that the areal force density or peridynamic traction τ ( x , t, n ) = σ peri ( x , t ) n , and thus τ ( x , t, e ) = σ peri33 ( x , t ) . Assumptions (P σ
7) and (C σ
7) both imposemonoclinic symmetry on their respective models. Finally, Assumption (P σ
8) imposes regularity conditionson the displacement field. These regularity conditions are, however, significantly weaker than those in theclassical theory, where u ( x , t ) is assumed to be smooth in x .A system satisfying all of (P σ σ
8) is said to be in a state of peridynamic generalized plane stress. SeeFigure 4 for an illustration of a body in a state of plane stress. Similarly to the classical theory, we start bysummarizing in Lemma 4 (peridynamic analogue of Lemma 3) the symmetries imposed on the displacementfield u in a system in a state of peridynamic generalized plane stress. Lemma 4
Under Assumptions (P σ σ
5) and (P σ u of (118) satisfies u i ( x, y, − z, t ) = (cid:26) u i ( x, y, z, t ) , i = 1 , − u i ( x, y, z, t ) , i = 3 , (184) i.e., the in-plane displacements, u and u , are symmetric while the out-of-plane displacement, u , is anti-symmetric relative to the plane z = 0 .Proof See Appendix C.1. (cid:117)(cid:116)
In order to derive the peridynamic generalized plane stress model, we perform the following steps, whichare analogous to those presented in the classical theory ( cf . Remark 4):Step 1: Take the average of the equation of motion (118) over the thickness of the plate.Step 2: Utilize Assumption (P σ
5) to eliminate interactions through the top and bottom surfaces of theplate.Step 3: Employ Assumption (P σ
6) to replace expressions in u by expressions in u and u .Step 4: Replace expressions in u and u with expressions in u and u .Step 5: Integrate in z (cid:48) and z to remove the third dimension dependence from the equation of motion.In the derivation, Lemma 4 is utilized to eliminate various terms.We now follow steps 1–5 in order to derive the peridynamic generalized plane stress model. We utilize theshorthand notation introduced in (74). The following steps involve various lemmas whose proofs have beenmoved to Appendix C in order to provide clarity to the derivation of the peridynamic generalized planestress model. Step 1 : We start by taking the average of the peridynamic equation of motion (118) over the thickness ofthe plate (recall ρ is constant in z by (P σ ρ ( x, y )¨ u i ( x, y, t ) = 12 h (cid:90) h − h (cid:90) B Dδ ( x ) λ ( x (cid:48) , x ) ξ i ξ j ( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) dz + b i ( x, y, t ) . (185) For the problems considered here, i.e., thin plates with free surfaces, it is natural to assume all nonlocal interactionsthrough the top and bottom surfaces of the plate are zero. In a more general setting, one could postulate a nonlocal analogueof (C σ σ = σ = σ = 0 on the top and bottom surfaces of the plate, by employing a peridynamic stress tensor σ peri . Such an approach would involve nullifying the net forces on each material point on the top and bottom surfaces of theplate rather than nullifying each pairwise interaction. The dependence of σ on x is introduced due to possible surface effects for points not in the bulk of the body.4 Jeremy Trageser, Pablo Seleson xx − s e δ B + x ( x − s e ) Fig. 12: Illustration of region B + x ( x − s e ). Step 2 : We impose Assumptions (P σ
1) and (P σ
5) on (185). Since λ ( x , x (cid:48) ) = 0 when x (cid:48) is not a materialpoint of the plate and B Dδ ( x ) intersects the top and bottom surfaces of the plate for any material point x of the plate, we may restrict the region of integration for z (cid:48) to [ − h, h ]: ρ ( x, y )¨ u i ( x, y, t ) = 12 h (cid:90) h − h (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( x (cid:48) , x ) ξ i ξ j ( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) dz + b i ( x, y, t ) , (186)where r = (cid:112) δ − ( z (cid:48) − z ) . Furthermore, due to (102) we may suppose (potentially by multiplying by acharacteristic function) λ ( x (cid:48) , x ) = 0 , (cid:107) x (cid:48) − x (cid:107) (cid:62) δ, (187)and extend the region of integration so that ρ ( x, y )¨ u i ( x, y, t ) = 12 h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ j ( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) dz + b i ( x, y, t ) . (188) Step 3 : The goal of this step is to replace the term12 h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz (189)in the right-hand side of (188) by expressions in u and u . This is facilitated by Assumption (P σ σ τ at a material point x inthe direction of e as τ ( x , t, e ) := (cid:90) δ (cid:90) B + x ( x − s e ) f ( u ( x (cid:48) , t ) − u ( x − s e , t ) , x (cid:48) , ( x − s e )) d x (cid:48) ds, (190)where B + x ( x − s e ) := { x (cid:48) ∈ B δ ( x − s e ) : z (cid:48) > z } (191)and f is the pairwise force function. In this work, we are only concerned with linear bond-based peridynamicmodels, and therefore f is given by (117). In Figure 12, we present an illustration of the region B + x ( x − s e ). Remark 15
As explained above, under a homogeneous deformation, one may introduce a peridynamic stresstensor σ peri such that τ = σ peri n . Then σ peri i = τ i ( x , t, e ). By Assumption (P σ λ ( x (cid:48) , x − s e ) = 0 for z = ± h and consequently σ peri13 = σ peri23 = σ peri33 = 0 for z = ± h , just as in the classicalassumption (C σ wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 55 In order to replace the term in (189), we introduce two lemmas. The first lemma, Lemma 5, expressesthe average peridynamic traction τ ( x , t, e ) in a form more readily comparable to (189), a necessary stepto prove the second lemma, Lemma 6. Lemma 6 provides an approximation which allows us to replace(189) with an expression in u and u , thus decoupling the in-plane displacements from the out-of-planedisplacement in (188). Lemma 5
Under Assumptions (P σ σ
5) and (P σ τ ( x , t, e ) ,with the pairwise force function f given by (117) , over the thickness of the plate satisfies τ ( x , t, e ) = 14 h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ξ j ( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) dz. (192) Proof
See Appendix C.2. (cid:117)(cid:116)
Lemma 6
For i = 1 or , under Assumptions (P σ σ u about ( x, y, , t ) : (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ − (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ A ( x (cid:48) , y (cid:48) , t ) d x (cid:48) dz (193) where A ( x (cid:48) , y (cid:48) , t ) := (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x (cid:48) ,y (cid:48) ) λ ( x (cid:48)(cid:48) , x (cid:48) ) ζ [ ζ ( u ( x (cid:48)(cid:48) , t ) − u ( x (cid:48) , t )) + ζ ( u ( x (cid:48)(cid:48) , t ) − u ( x (cid:48) , t ))] d x (cid:48)(cid:48) dz (cid:48) (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x (cid:48) ,y (cid:48) ) λ ( x (cid:48)(cid:48) , x (cid:48) ) ζ d x (cid:48)(cid:48) dz (cid:48) . (194) Here, we defined ζ := x (cid:48)(cid:48) − x (cid:48) to avoid confusion with the terms in (193) .Proof See Appendix C.3. (cid:117)(cid:116)
Utilizing Lemma 6, we may substitute (193) into (188) to obtain our decoupled model (for i = 1 , ρ ( x, y )¨ u i ( x, y, t ) ≈ h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i [ ξ ( u ( x (cid:48) , t ) − u ( x , t )) + ξ ( u ( x (cid:48) , t ) − u ( x , t )) − ξ A ( x (cid:48) , y (cid:48) , t ) (cid:21) d x (cid:48) dz + b i ( x, y, t ) . (195) Step 4 : In this step, we remove the dependence of u and u in (195) on the third dimension in order toobtain expressions in u and u . In classical generalized plane stress, u and u are immediately replacedby expressions in u and u by taking the average of the equation of motion ( cf . Step 1 from Remark 4).However, in peridynamics we cannot simply integrate over the third dimension on the right-hand side of(195) to directly obtain expressions in u and u . This is due to the presence of the micromodulus function λ and the third component of the bond, ξ , in the integrand in (195), which would result in weightedaverages of u and u instead. To overcome this obstacle, we introduce Lemma 7 below. Lemma 7
Under Assumptions (P σ σ
5) and (P σ σ i = 1 or : u i ( x , t ) = u i ( x, y, t ) + O ( h ) . (196) Proof
See Appendix C.4. (cid:117)(cid:116)
Utilizing Lemma 7 and recalling terms of order O ( h ) are small by Assumption (P σ u i ( x , t ) ≈ u i ( x, y, t ). Substituting this approximation into (194) and (195), results in ρ ( x, y )¨ u i ( x, y, t ) ≈ h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i [ ξ ( u ( x (cid:48) , t ) − u ( x , t )) + ξ ( u ( x (cid:48) , t ) − u ( x , t )) − ξ A ( x (cid:48) , y (cid:48) , t ) (cid:21) d x (cid:48) dz + b i ( x, y, t ) , (197)where A ( x (cid:48) , y (cid:48) , t ) ≈ (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x (cid:48) ,y (cid:48) ) λ ( x (cid:48)(cid:48) , x (cid:48) ) ζ [ ζ ( u ( x (cid:48)(cid:48) , y (cid:48)(cid:48) , t ) − u ( x (cid:48) , y (cid:48) , t )) + ζ ( u ( x (cid:48)(cid:48) , y (cid:48)(cid:48) , t ) − u ( x (cid:48) , y (cid:48) , t ))] d x (cid:48)(cid:48) dz (cid:48) (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x (cid:48) ,y (cid:48) ) λ ( x (cid:48)(cid:48) , x (cid:48) ) ζ d x (cid:48)(cid:48) dz (cid:48) . (198) Step 5 : The final step is to integrate over z (cid:48) and z in order to remove the third dimension dependencefrom (197) and (198). For convenience, we introduce the shorthand notation (recall (181)) λ i ( ξ , ξ ) := (cid:90) h − h (cid:90) h − h λ ( x (cid:48) , x ) ξ i dz (cid:48) dz. (199)Since the limits of integration in (197) and (198) are independent of each other, we may change the order ofintegration without altering the limits so that we may apply (199). In addition, we perform the change ofvariables ( x (cid:48) , y (cid:48) ) → ( x + ξ , y + ξ ) in (197) and ( x (cid:48)(cid:48) , y (cid:48)(cid:48) ) → ( x (cid:48) + ζ , y (cid:48) + ζ ) in (198). Up to the approximationsin Steps 1–5, the peridynamic generalized plane stress equation of motion is given by (analogue of (80a)and (80b)): ρ ( x, y )¨ u ( x, y, t ) = 12 h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ [ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t ))+ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t ))] − λ ( ξ , ξ ) A ( x + ξ , y + ξ , t ) ξ dξ dξ + b ( x, y, t ) , (200a) ρ ( x, y )¨ u ( x, y, t ) = 12 h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ [ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t ))+ ξ ( u ( x + ξ , y + ξ , t ) − u ( x, y, t ))] − λ ( ξ , ξ ) A ( x + ξ , y + ξ , t ) ξ dξ dξ + b ( x, y, t ) , (200b)where A ( x (cid:48) , y (cid:48) , t ) = (cid:82) B Dδ ( ) λ ( ζ , ζ )( ζ ( u ( x (cid:48) + ζ , y (cid:48) + ζ , t ) − u ( x (cid:48) , y (cid:48) , t )) + ζ ( u ( x (cid:48) + ζ , y (cid:48) + ζ , t ) − u ( x (cid:48) , y (cid:48) , t ))) dζ dζ (cid:82) B Dδ ( ) λ ( ζ , ζ ) dζ dζ . (201) Remark 16
Due to the presence of A ( x (cid:48) , y (cid:48) , t ) in the equation of motion (200), the resulting peridynamicgeneralized plane stress model is not a bond-based peridynamic model but rather a state-based peridynamicmodel [34]. Analogously to Remark 11, we can express the corresponding equation of motion for (200) invector form by letting x = ( x, y ), ξ = ( ξ , ξ ), ζ = ( ζ , ζ ), u = ( u , u ), H = B Dδ ( ), and b = ( b , b ). Inthis case, using the fact that λ i ( x (cid:48) , y (cid:48) , x, y ) = λ i ( x, y, x (cid:48) , y (cid:48) ) by (116) and Assumption (P σ ρ ( x )¨ u ( x , t ) = (cid:90) H { T [ x , t ] (cid:104) ξ (cid:105) − T [ x + ξ , t ] (cid:104)− ξ (cid:105)} d ξ + b ( x , t ) , where T [ x , t ] (cid:104) ξ (cid:105) = 14 h [ λ ( ξ ) ξ ⊗ ξ ( u ( x + ξ , t ) − u ( x , t )) − λ ( ξ ) A ( x , t ) ξ ]and A ( x , t ) = 1 (cid:82) H λ ( ζ ) d ζ (cid:90) H λ ( ζ ) ζ · ( u ( x + ζ , t ) − u ( x , t )) d ζ . wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 57 Peridynamic plane stress micromodulus functions:In the peridynamic generalized plane stress model presented in this work, the requirements placed on themicromodulus function λ ( x (cid:48) , x ) have been fairly minimal up to this point. In order to investigate the behav-ior of the peridynamic plane stress micromodulus functions λ ( ξ , ξ ) and λ ( ξ , ξ ) appearing in (200) andinform them with the classical elasticity tensor C , we necessarily must provide a prototype micromodulus λ ( x (cid:48) , x ). Unfortunately, the surface effects endemic in peridynamic formulations of generalized plane stressrequire λ ( x (cid:48) , x ) to be position aware in order to satisfy (122). Rather than introducing a new micromodulusfunction to accommodate surface effects, we posit a plausible alternative strategy: relax the requirements(122) on λ ( x (cid:48) , x ) to hold only in the average over the thickness of the plate . Specifically,0 = (cid:90) h − h (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ j ξ k d x (cid:48) dz (202a) C ijkl = 14 h (cid:90) h − h (cid:90) H x λ ( x (cid:48) , x ) ξ i ξ j ξ k ξ l d x (cid:48) dz. (202b)As can be seen in Proposition 4, if we impose the weaker conditions (202) (in relation to (122)) on λ ( x (cid:48) , x ),the peridynamic generalized plane stress model (200) reduces to the classical generalized plane stress model(80) under a second-order Taylor expansion. Moreover, if we suppose the micromodulus is a function onlyof the bond, such as in the case of (128), then (202a) is trivially satisfied. Consequently, when averagingover the plate, the surface effects negate each other for such a micromodulus.With the relaxed formulation (202), we may consider a far larger class of functions. In particular, one caninform the constants of the peridynamic tensor Λ so that (128) satisfies (202). This is precisely the approachwe take in this section in order to investigate the behavior of the peridynamic plane stress micromodulusfunctions λ ( ξ , ξ ) and λ ( ξ , ξ ) appearing in (200). Since no material point is assumed to be in the bulkof the material in our peridynamic plane stress formulation, we multiply (128) by χ B δ ( ) := (cid:26) , (cid:107) ξ (cid:107) < , else (203)in order to enforce (187). Then, given the micromodulus function described by (128) multiplied with χ B δ ( x ) ,the micromodulus functions (199) are formulated as: λ i ( ξ , ξ ) = (cid:90) h − h (cid:90) h − h λ ( x (cid:48) − x ) ξ i χ B δ ( x ) dz (cid:48) dz = (cid:90) h − h (cid:90) h − z − h − z λ ( ξ ) ξ i χ B δ ( ) dξ dz = (cid:90) h − h (cid:90) min { h − z, √ δ − r } max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz, (204)where r = (cid:112) ξ + ξ . In order to remove the piecewise nature of the limits of integration in (204), weconsider two regions: r < δ − h and δ − h (cid:54) r < δ . Region r < δ − : In this region, it follows that 2 h < √ δ − r . Since | z | (cid:54) h , in this region we have h − z (cid:54) h < √ δ − r and −√ δ − r < − h (cid:54) − h − z . Consequently, (204) simplfies to λ i ( ξ , ξ ) = (cid:90) h − h (cid:90) h − z − h − z λ ( ξ ) ξ i dξ dz. (205) This concept parallels the theory for classical generalized plane stress which is developed for quantities averaged over thethickness of the plate ( cf . Section 2.3.3).8 Jeremy Trageser, Pablo Seleson Region δ − (cid:54) r (cid:54) δ : In this region, it follows that √ δ − r (cid:54) h . We first split the integral in(204) to find λ i ( ξ , ξ ) = (cid:90) h −√ δ − r − h (cid:90) min { h − z, √ δ − r } max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) hh −√ δ − r (cid:90) min { h − z, √ δ − r } max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz = (cid:90) h −√ δ − r − h (cid:90) √ δ − r max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) hh −√ δ − r (cid:90) h − z max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz. (206)For the second equality in (206), we utilized the fact that h − z (cid:62) √ δ − r for z ∈ [ − h, h − √ δ − r ] and h − z (cid:54) √ δ − r for z ∈ [ h − √ δ − r , h ]. We then split the integrals in (206) to find λ i ( ξ , ξ ) = (cid:90) − h + √ δ − r − h (cid:90) √ δ − r max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) h −√ δ − r − h + √ δ − r (cid:90) √ δ − r max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) − h + √ δ − r h −√ δ − r (cid:90) h − z max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) h − h + √ δ − r (cid:90) h − z max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz = (cid:90) − h + √ δ − r − h (cid:90) √ δ − r − h − z λ ( ξ ) ξ i dξ dz + (cid:90) h −√ δ − r − h + √ δ − r (cid:90) √ δ − r max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) − h + √ δ − r h −√ δ − r (cid:90) h − z max { − h − z, −√ δ − r } λ ( ξ ) ξ i dξ dz + (cid:90) h − h + √ δ − r (cid:90) h − z −√ δ − r λ ( ξ ) ξ i dξ dz = (cid:90) − h + √ δ − r − h (cid:90) √ δ − r − h − z λ ( ξ ) ξ i dξ dz + (cid:90) h −√ δ − r − h + √ δ − r (cid:90) √ δ − r h − z λ ( ξ ) ξ i dξ dz + (cid:90) h − h + √ δ − r (cid:90) h − z −√ δ − r λ ( ξ ) ξ i dξ dz. (207)In the second equality of (207), we utilized the fact that − h − z > −√ δ − r for z ∈ [ − h, − h + √ δ − r ]and − h − z < −√ δ − r for z ∈ [ − h + √ δ − r , h ]. In the third equality of (207), we changed the orderof the limits of integration in both integrals in the third term and then combined it with the second term.With the more amenable limits of integration in (205) and (207), closed-form expressions for λ i ( ξ , ξ ) canbe deduced. To accomplish this, we utilize the shorthand notation introduced in (159) and (158) and dropterms not satisfying monoclinic symmetry to get λ ( ξ ) = ω ( (cid:107) ξ (cid:107) ) (cid:107) ξ (cid:107) A ( ξ , ξ ) + A ( ξ , ξ ) ξ + A ( ξ , ξ ) ξ (cid:107) ξ (cid:107) , (208)where A ( ξ , ξ ) = Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (209a) A ( ξ , ξ ) = 6 (cid:0) Λ ξ + Λ ξ + 2 Λ ξ ξ (cid:1) , (209b) A ( ξ , ξ ) = Λ , (209c)and the Λ ijkl are to be determined later.Depending on the choice of influence function ω , it may be possible to produce closed-form expressions forthe integrals in (204). Two commonly utilized influence functions in peridynamics are ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) and ω ( (cid:107) ξ (cid:107) ) = 1. With either choice of influence function, the micromodulus functions (204) are given by λ ( ξ , ξ ) = ω ( r ) (cid:20) A ( ξ , ξ ) r M (cid:18) hr , rδ (cid:19) + A ( ξ , ξ ) r M (cid:18) hr , rδ (cid:19) + A ( ξ , ξ ) M (cid:18) hr , rδ (cid:19)(cid:21) , (210a) λ ( ξ , ξ ) = r ω ( r ) (cid:20) A ( ξ , ξ ) r M (cid:18) hr , rδ (cid:19) + A ( ξ , ξ ) r M (cid:18) hr , rδ (cid:19) + A ( ξ , ξ ) M (cid:18) hr , rδ (cid:19)(cid:21) , (210b) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 59 where for ω ( (cid:107) ξ (cid:107) ) = 1 we have M ( x, y ) := x arctan(2 x ) + x x +1 , h δ − h < x x arctan (cid:16)(cid:112) y − − (cid:17) + (cid:0) y − (cid:1) + x (cid:0) y + 2 y (cid:1)(cid:112) − y , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := x arctan(2 x ) − x x +1 , h δ − h < x x arctan (cid:16)(cid:112) y − − (cid:17) − (cid:0) y − (cid:1) + x (cid:0) y − y (cid:1)(cid:112) − y , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := x arctan(2 x ) + x x +1 + ln (cid:16) x +1 (cid:17) , h δ − h < x x arctan (cid:16)(cid:112) y − − (cid:17) + ( y − y − − x (cid:112) − y (5 y − y ) + 2 ln( y ) , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := − x arctan(2 x ) + x +3 x x +1 − (cid:16) x +1 (cid:17) , h δ − h < x − x arctan (cid:16)(cid:112) y − − (cid:17) − y (cid:0) y − (cid:1)(cid:0) y − y − (cid:1) + x (cid:112) − y (cid:0) y − + 9 y − y (cid:1) − y ) , h δ (cid:54) x (cid:54) h δ − h and for ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) we have M ( x, y ) := − + (cid:0) x + 48 x + 3 (cid:1)(cid:0) x + 1 (cid:1) − , h δ − h < x x (cid:0) y + 4 y + 8 (cid:1)(cid:112) − y + (cid:0) y − (cid:1) , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := − + (cid:0) x + 6 x + 1 (cid:1)(cid:0) x + 1 (cid:1) − , h δ − h < x x (cid:0) y − y (cid:1)(cid:112) − y − (cid:0) y − y + 4 (cid:1) , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := − + (cid:0) x + 6 x + 1 (cid:1)(cid:0) x + 1 (cid:1) − , h δ − h < x x (cid:0) − y (cid:1) + (cid:0) y + 9 y + 8 (cid:1) ( y − , h δ (cid:54) x (cid:54) h δ − h M ( x, y ) := − (cid:0) x + 87 x + 12 (cid:1)(cid:0) x + 1 (cid:1) − + 4 x arsinh(2 x ) , h δ − h < x − x (cid:0) y − y + 23 (cid:1)(cid:112) − y − (cid:0) y − + 4 + y (cid:1) ( y − + 4 x arsinh (cid:16)(cid:112) y − − (cid:17) , h δ (cid:54) x (cid:54) h δ − h . For plane strain, we were able to determine a relationship between Λ and C independent of the influencefunction ( cf . (157)). While this may be possible for peridynamic plane stress, the expressions are ratherimpractical and thus it is far more convenient to provide expressions for our specific influence functions.We inform the peridynamic tensor Λ , and consequently (210) (through (209)), with the elasticity tensor C .This is accomplished by employing (202) to find C C C C C C C C C = π α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α Λ Λ Λ Λ Λ Λ Λ Λ Λ , (211) The regions h δ − h < x and h δ (cid:54) x (cid:54) h δ − h reduce to the regions r < δ − h and δ − h (cid:54) r (cid:54) δ , respectively,when x = hr . These are precisely the regions we considered above in order to remove the piecewise function limits in (204).0 Jeremy Trageser, Pablo Seleson where for ω ( (cid:107) ξ (cid:107) ) = 1, (with p = hδ ) α := δ (cid:18) − p + 327 p + 17675 p − p + 14 p − p ln(2 p ) (cid:19) ,α := δ (cid:18) p − p − p + 13 p + 165 p ln(2 p ) (cid:19) ,α := δ (cid:18) − p + 167 p − p − p ln(2 p ) (cid:19) ,α := δ (cid:18) p − p + 415 p (cid:19) ,α := δ (cid:18) − p + 415 p (cid:19) , and for ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) , (with p = hδ ) α := δ (cid:18) − p + 6421 p − p + 51275 p − p + 13 p (cid:19) ,α := δ (cid:18) p − p + 165 p − p + 23 p (cid:19) ,α := δ (cid:18) − p + 3221 p − p + 3275 p (cid:19) ,α := δ (cid:18) p − p + 875 p (cid:19) ,α := δ (cid:18) − p + 425 p (cid:19) . Remark 17
Similarly to plane strain ( cf.
Remark 12), for i ∈ { , . . . , } , note that A i ( ξ ,ξ ) r − i is radiallyindependent ( cf. (209)) and, consequently, only contributes to the angular portion of (210). Moreover,as the other terms in (210) are radial functions, the A i ( ξ ,ξ ) r − i terms make up the entirety of the angulardependence of the micromodulus functions.Next, we take a closer look at the micromodulus functions { λ i ( ξ , ξ ) } ( cf. (204)) of the peridynamic planestress model (200) for various symmetry classes. Following Remark 17, the choice of symmetry class onlyhas an effect on the { A i ( ξ , ξ ) } , while the general form of the micromodulus functions (210) remainsunchanged. Therefore, we only present the { A i ( ξ , ξ ) } for each symmetry class. As explained earlier, formany of the symmetry classes, there are multiple planes of reflection symmetry to choose from in order tosatisfy Assumption (P σ σ cf. (209)): A ( ξ , ξ ) := Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (212a) A ( ξ , ξ ) := 6 (cid:0) Λ ξ + Λ ξ + 2 Λ ξ ξ (cid:1) , (212b) A ( ξ , ξ ) := Λ , (212c)where Λ Λ Λ Λ Λ Λ Λ Λ Λ = 64 π α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α − C C C C C C C C C . (213) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 61 Orthotropic: We substitute (53), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) = Λ ξ + 6 Λ ξ ξ + Λ ξ , (214a) A ( ξ , ξ ) = 6 (cid:0) Λ ξ + Λ ξ (cid:1) , (214b) A ( ξ , ξ ) = Λ , (214c)where Λ Λ Λ Λ Λ Λ = 64 π α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α − C C C C C C . (215)Trigonal: We substitute (54), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) := Λ ξ + 4 Λ ξ ξ + 6 Λ ξ ξ + 4 Λ ξ ξ + Λ ξ , (216a) A ( ξ , ξ ) := 6 (cid:0) Λ ξ + Λ ξ + 2 Λ ξ ξ (cid:1) , (216b) A ( ξ , ξ ) := Λ , (216c)where Λ Λ Λ Λ Λ Λ Λ Λ Λ = 64 π α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α − C C C C C C C − C . (217)Tetragonal: We substitute (55), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) = Λ ( ξ + ξ ) + 6 Λ ξ ξ , (218a) A ( ξ , ξ ) = 6 Λ r , (218b) A ( ξ , ξ ) = Λ , (218c)where Λ Λ Λ Λ = 64 π α α α α α α α α α α α α α α α α − C C C C . (219)Transversely Isotropic: We substitute (56), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) = Λ r , (220a) A ( ξ , ξ ) = 6 Λ r , (220b) A ( ξ , ξ ) = Λ , (220c) While there appear to be nine independent constants in (216), the constants are not actually independent. Since thetrigonal elasticity tensor C with Cauchy’s relations imposed has only four independent constants, there are actually only fourindependent constants in (216).2 Jeremy Trageser, Pablo Seleson where Λ Λ Λ = 64 π α α α α α α α α α − C C C . (221)Cubic: We substitute (57), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) = Λ ( ξ + ξ ) + 6 Λ ξ ξ , (222a) A ( ξ , ξ ) = 6 Λ r , (222b) A ( ξ , ξ ) = Λ , (222c)where Λ Λ Λ Λ = 64 π α α α α α α α α α α α α α α α α − C C C C . (223)Isotropic: We substitute (58), with Cauchy’s relations imposed, into (211) to find ( cf. (209)) A ( ξ , ξ ) = Λ r , (224a) A ( ξ , ξ ) = 6 Λ r , (224b) A ( ξ , ξ ) = Λ , (224c)where Λ Λ Λ = 64 π α α α α α α α α α − C C C . (225) Remark 18
In classical linear elasticity, the plane stress model for each symmetry class reduces identicallyto a corresponding two-dimensional model ( cf . Table 2). A similar situation occurs with the peridynamicplane stress model. While the plane stress micromodulus functions λ and λ , given by (210), are notequivalent to the two-dimensional micromodulus function (133), they do possess one of the four symmetriesof two-dimensional classical linear elasticity ( cf . Theorem 1) for each symmetry class ( cf . Table 7). Thiscan be observed by considering the symmetries of { A i ( ξ , ξ ) } for each symmetry class. Unlike their two-dimensional counterparts, the plane stress micromodulus functions incorporate out-of-plane information.It is also interesting to note that the same correspondence between each three-dimensional symmetry classand the corresponding two-dimensional symmetry class in Table 2 occurs for the plane stress micromodulusfunctions, which is summarized in Table 7. However, while different plane stress micromodulus functions maypossess the same two-dimensional symmetry, e.g., tetragonal and cubic micromodulus functions both havesquare symmetry, the resulting plane stress micromodulus functions are unique for each three-dimensionalsymmetry class. Due to this fact, the resulting plane stress equations of motion are unique for each three-dimensional symmetry class.A common approach for modeling plane stress in the peridynamic literature simply employs a two-dimensionalperidynamic model and matches the model constants to those in the corresponding classical plane stressmodel (see, e.g., [13,20,26]). In bond-based peridynamics, this produces a two-dimensional bond-basedmodel with a single micromodulus function. Unlike in the case for peridynamic plane strain ( cf . Section3.2.1), the peridynamic plane stress model (200) developed in this work is a state-based model with two While there appear to be four independent constants in (222), the constants are not actually independent. Since thecubic elasticity tensor C with Cauchy’s relations imposed has only four independent constants, there are actually only twoindependent constants in (222). While there appear to be three independent constants in (224), the constants are not actually independent. Since theisotropic elasticity tensor C with Cauchy’s relations imposed has only one independent constant, there is actually only oneindependent constant in (224).wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 63 Pure Two-Dimensional Peridynamic Model Peridynamic Plane Stress Model
Oblique Monoclinic and TrigonalRectangular OrthotropicSquare Tetragonal and CubicIsotropic Transversely Isotropic and Isotropic
Table 7: Symmetry equivalence between the pure two-dimensional peridynamic models and the peridynamicplane stress models (provided the micromodulus functions λ and λ are informed by three-dimensionalelasticity tensors having the form of those in Section 2.3.1).micromodulus functions λ and λ , and thus a direct comparison considering only micromodulus functionsis insufficient. Instead, we explore the behavior of the plane stress micromodulus functions (204) for varioussymmetry classes. Micromodulus function visualization for peridynamic generalized plane stress
In this section, we provide visualizations of the behavior of the plane stress micromodulus functions (204)for various symmetry classes. We observe that in the peridynamic plane stress equation of motion (200),the micromodulus function λ ( ξ , ξ ) is multiplied by ξ i ξ j and the micromodulus function λ ( ξ , ξ ) ismultiplied by ξ i . Thus, a singularity of the form r for λ ( ξ , ξ ) and r for λ ( ξ , ξ ), represent removablesingularities of the model. To better visualize the anisotropic behavior of the micromodulus functions, weremove such singularities by plotting r λ ( ξ , ξ ) and rλ ( ξ , ξ ).Our visualization of the micromodulus functions covers materials in every symmetry class except triclinic,since the plane stress Assumption (P ε
7) excludes this class. Since we are dealing with bond-based peri-dynamic models, we must consider materials satisfying (at least approximately) Cauchy’s relations. Forthe sake of consistency, we consider the same materials as were utilized to illustrate anisotropy in theplane strain micromodulus functions. In particular, we inform the peridynamic plane stress micromodulusfunctions (210) with the material properties summarized in Table 6.In Figure 13 we plot r λ ( ξ , ξ ) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and ω ( (cid:107) ξ (cid:107) ) = 1 (in blue). Similarly, in Figure14 we plot rλ ( ξ , ξ ) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and ω ( (cid:107) ξ (cid:107) ) = 1 (in blue). For each symmetry class,these functions are found by substituting the corresponding elasticity tensor ˜ C from Table 6 into (210)(with (209) informed by inverting the system (211)). For all of the plots, we took 3 h = δ = 1. Connections between the peridynamic and classical generalized plane stress equations :We now show that the peridynamic generalized plane stress model reduces to the classical generalized planestress model under suitable assumptions.
Proposition 4
Suppose the micromodulus function λ ( x (cid:48) , x ) is related to the elasticity tensor C through (202b) . Given a smooth deformation, under a second-order Taylor expansion of the displacement field,the peridynamic generalized plane stress model (200) reduces to the classical generalized plane stress model (80) with Cauchy’s relations imposed.Proof First, we simplify the denominator of A ( x (cid:48) , y (cid:48) , t ) ( cf . (201)) by noticing from (199) and (202b) that(recall (187)) (cid:90) B Dδ ( ) λ ( ζ , ζ ) dζ dζ = (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48)(cid:48) , x (cid:48) ) ζ dx (cid:48)(cid:48) dy (cid:48)(cid:48) dz (cid:48)(cid:48) dz (cid:48) = 4 hC . Note that (202b) is a consequence of (122b) and therefore this result also holds under (122b).4 Jeremy Trageser, Pablo SelesonMonoclinic (CoTeO ) Orthotropic (Te W)Trigonal (Ta C) Tetragonal (Si)Transversely Isotropic (MoN) Cubic (MgAl O )Isotropic (Pyroceram 9608) Fig. 13: Plots of r λ ( ξ , ξ ) ( cf. (210a)) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and ω ( (cid:107) ξ (cid:107) ) = 1 (in blue). wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 65Monoclinic (CoTeO ) Orthotropic (Te W)Trigonal (Ta C) Tetragonal (Si)Transversely Isotropic (MoN) Cubic (MgAl O )Isotropic (Pyroceram 9608) Fig. 14: Plots of rλ ( ξ , ξ ) ( cf. (210a)) with ω ( (cid:107) ξ (cid:107) ) = (cid:107) ξ (cid:107) (in green) and ω ( (cid:107) ξ (cid:107) ) = 1 (in blue). Applying a Taylor expansion of u j ( x (cid:48) + ζ , y (cid:48) + ζ , t ) about ( x (cid:48) , y (cid:48) , t ) in the numerator of A ( x (cid:48) , y (cid:48) , t ), weobtain A ( x (cid:48) , y (cid:48) , t ) = (cid:82) B Dδ ( ) λ ( ζ , ζ ) ζ j [ u j ( x (cid:48) + ζ , y (cid:48) + ζ , t ) − u j ( x (cid:48) , y (cid:48) , t )] dζ dζ (cid:82) B Dδ ( ) λ ( ζ , ζ ) dζ dζ = 14 hC (cid:90) B Dδ ( ) λ ( ζ , ζ ) ζ j (cid:20) ζ ∂u j ∂x ( x (cid:48) , y (cid:48) , t ) + ζ ∂u j ∂y ( x (cid:48) , y (cid:48) , t )+ ζ ∂ u j ∂x ( x (cid:48) , y (cid:48) , t ) + ζ ζ ∂ u j ∂x∂y ( x (cid:48) , y (cid:48) , t ) + ζ ∂ u j ∂y ( x (cid:48) , y (cid:48) , t ) + · · · (cid:21) dζ dζ = 14 hC (cid:90) B Dδ ( ) λ ( ζ , ζ ) ζ j (cid:20) ζ ∂u j ∂x ( x (cid:48) , y (cid:48) , t ) + ζ ∂u j ∂y ( x (cid:48) , y (cid:48) , t ) (cid:21) dζ dζ , (226)where the summation in j is over 1 and 2. In the last line of (226) we nullified terms with even-order deriva-tives due to antisymmetry in the transformation ( ζ , ζ ) → ( − ζ , − ζ ), using the fact that λ ( − ζ , − ζ ) = λ ( ζ , ζ ) by Assumption (P σ A ( x (cid:48) , y (cid:48) , t ) = C j C ∂u j ∂x ( x (cid:48) , y (cid:48) , t ) + C j C ∂u j ∂y ( x (cid:48) , y (cid:48) , t ) . (227)Substituting (227) into (200), we find for i = 1 , ρ ( x, y )¨ u i ( x, y, t ) = 12 h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ i ξ j ( u j ( x + ξ , y + ξ , t ) − u j ( x, y, t )) dξ dξ − h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ i (cid:20) C j C ∂u j ∂x ( x + ξ , y + ξ , t ) + C j C ∂u j ∂y ( x + ξ , y + ξ , t ) (cid:21) dξ dξ + b i ( x, y, t ) , (228)where the summation in j is over 1 and 2. Employing a Taylor expansion about ( x, y, t ) of u j ( x (cid:48) , y (cid:48) , t )and its derivatives in (228), eliminating terms antisymmetric with respect to the transformation ( ξ , ξ ) → ( − ξ , − ξ ), and supposing higher-order terms are negligible, we obtain ρ ( x, y )¨ u i ( x, y, t ) = 12 h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ i ξ j (cid:20) ξ ∂ u j ∂x ( x, y, t ) + ξ ξ ∂ u j ∂x∂y ( x, y, t ) + ξ ∂ u j ∂y ( x, y, t ) (cid:21) dξ dξ − h (cid:90) B Dδ ( ) λ ( ξ , ξ ) ξ i (cid:20) C j C (cid:18) ξ ∂ u j ∂x ( x, y, t ) + ξ ∂ u j ∂x∂y ( x, y, t ) (cid:19) + C j C (cid:18) ξ ∂ u j ∂x∂y ( x, y, t ) + ξ ∂ u j ∂y ( x, y, t ) (cid:19)(cid:21) dξ dξ + b i ( x, y, t ) . (229)Applying (199) and (202b) to (229), we find ρ ( x, y )¨ u i ( x, y, t ) = C ij ∂ u j ∂x ( x, y, t ) + 2 C ij ∂ u j ∂x∂y ( x, y, t ) + C ij ∂ u j ∂y ( x, y, t ) − (cid:20) C i C j C ∂ u j ∂x ( x, y, t ) + C i C j C ∂ u j ∂x∂y ( x, y, t ) (cid:21) − (cid:20) C i C j C ∂ u j ∂x∂y ( x, y, t ) + C i C j C ∂ u j ∂y ( x, y, t ) (cid:21) + b i ( x, y, t ) . (230) wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 67 Writing (230) out, we obtain ρ ¨ u = (cid:18) C − C C (cid:19) ∂ u ∂x + 2 (cid:18) C − C C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + (cid:18) C − C C C (cid:19) ∂ u ∂x + (cid:18) C − C C C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C C (cid:19) ∂ u ∂y + b , (231a) ρ ¨ u = (cid:18) C − C C C (cid:19) ∂ u ∂x + (cid:18) C − C C C − C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C C (cid:19) ∂ u ∂y + (cid:18) C − C C (cid:19) ∂ u ∂x + 2 (cid:18) C − C C C (cid:19) ∂ u ∂x∂y + (cid:18) C − C C (cid:19) ∂ u ∂y + b , (231b)where we have omitted the ( x, y, t ) dependence for brevity and a more direct comparison to the classicalmodel. Comparing (231) to (80) with Cauchy’s relations ( cf. (97)) imposed completes the proof. (cid:117)(cid:116) Remark 19
Notice in the equation for classical generalized plane stress, (80a) and (80b), only four Cauchy’srelations are relevant: C = C , C = C , C = C , and C = C . These four Cauchy’s relations are exactly the same ones that are relevant to the classical plane strainmodel (61) and (62). This is not surprising since the remaining Cauchy’s relations are trivially satisfied bymonoclinic symmetry, an assumption of both classical plane strain and classical generalized plane stress.In the in-plane equations of motion for plane strain, (61a) and (61b), the only relevant Cauchy’s relationis C = C ( cf . Remark 14). The entirety of the discrepancy between which Cauchy’s relations arerelevant in classical generalized plane stress and the in-plane equations for classical plane strain is due to thefact that u may be a function of z in the case of classical generalized plane stress, whereas u is independentof z in classical plane strain. In fact, one can easily see in (77) and (78) that the contribution from thesubstitution for [ u ] in the derivation of the classical generalized plane stress model is the sole reason forthe relevance of the additional Cauchy’s relations, C = C , C = C , and C = C . This paper has a twofold objective. First, it reviews pure two-dimensional, plane strain, and plane stressanisotropic models in classical linear elasticity. Second, it introduces novel formulations analogous to theclassical models within the bond-based peridynamic theory of solid mechanics.Our review of classical linear elasticity began with a new elementary and self-contained proof that there areexactly four material symmetry classes in classical linear elasticity in two dimensions: oblique, rectangular,square, and isotropic. Then, we presented pure anisotropic two-dimensional classical linear elasticity mod-els for each of those symmetry classes. We further discussed planar approximations of three-dimensionalanisotropic models in classical linear elasticity, specifically plane strain and plan stress formulations. Theformer normally concerns thick structures, whereas the latter often applies to thin plates. In three di-mensions, there are eight material symmetry classes: triclinic, monoclinic, trigonal, orthotropic, tetragonal,transversely isotropic, cubic, and isotropic. Under certain assumptions, planar approximations reduce three-dimensional models to two-dimensional formulations. We reviewed plane strain and plane stress formula-tions in classical linear elasticity and specialized those formulations to the various symmetry classes. Thesetwo-dimensional approximations are based on decoupling in-plane and out-of-plane deformations which isachieved by assuming the material has a plane of reflection symmetry, i.e., the material symmetry is atleast monoclinic. To discuss connections between the two-dimensional planar approximation models and thepure two-dimensional models in anisotropic classical linear elasticity, we reviewed engineering constants. It turns out that classical generalized plane stress models are equivalent, in terms of engineering constants,to pure two-dimensional models. In contrast, even though the classical plane strain models look identicalto the pure two-dimensional models when expressed in terms of elasticity constants, they differ from thepure two-dimensional models, because the meaning of the elasticity constants varies upon the dimension.We finalized our review of classical linear elasticity with a discussion of Cauchy’s relations in order toconnect classical models to bond-based peridynamic models. While there is a single Cauchy’s relation intwo dimensions, there are six Cauchy’s relations in three dimensions. Overall, Cauchy’s relations reduce thenumber of independent constants in classical linear elasticity from 6 to 5, for general oblique models, andfrom 21 to 15, for general triclinic models.In the context of the bond-based peridynamic theory, we began by introducing new anisotropic models,which can accommodate all four material symmetry classes in two dimensions and discuss related micro-modulus functions with corresponding visualizations. We then derived novel peridynamic plane strain andplane stress formulations. As opposed to common approaches for planar approximations in peridynamics,which are based on simply matching constants of two-dimensional peridynamic models to correspondingconstants appearing in planar approximations in classical linear elasticity, our models directly apply peri-dynamic analogues of classical planar assumptions to reduce three-dimensional models to two-dimensionalformulations. For this purpose, we resorted to the three-dimensional anisotropic peridynamic models de-veloped in [31]. We discussed the resulting plane strain and plane stress micromodulus functions withcorresponding visualizations, and we proved the convergence of our peridynamic plane strain and planestress models to their counterparts in classical linear elasticity with imposed Cauchy’s relations.It is interesting to observe that, as opposed to peridynamic plane strain, which due to the thickness of thestructure can simply consider points in the bulk of a body, peridynamic plane stress deals with thin platesand thus requires examination of surface effects. Furthermore, the peridynamic plane stress approximationsresult in a state-based peridynamic formulation.The work presented in this paper offers a framework for simulation of two-dimensional problems based on thebond-based peridynamic theory, concerning all material symmetry classes found in classical linear elasticity.The newly introduced peridynamic plane strain and plane stress models provide means of reducing fullyanisotropic three-dimensional bond-based peridynamic problems to two-dimensional formulations, resultingin significant computational savings, while retaining the dynamics of the original three-dimensional problemsunder proper assumptions.
A Poisson’s ratio restriction in two-dimensional bond-based peridynamics
In this section, we utilize a peridynamic traction in order to compute the engineering constants for isotropic homogeneousmaterials in the sense of the two-dimensional classical theory of linear elasticity. We suppose the material undergoes a (static)homogeneous deformation given by u ( x ) = (cid:15) x and u ( x ) ≡
0. Then, the pairwise force function ( cf . (117)) has componentsgiven by f ( u ( x (cid:48) ) − u ( x ) , x (cid:48) − x ) = (cid:15) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) ,f ( u ( x (cid:48) ) − u ( x ) , x (cid:48) − x ) = (cid:15) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) ( y (cid:48) − y ) . (232)Analogously to the three-dimensional formulation in [32], given a body B ⊂ R , we define the two-dimensional peridynamictraction τ at a material point x ∈ B in the direction of the unit vector n as τ ( x , t, n ) := (cid:90) δ (cid:90) B + x ( x − s n ) f ( u ( x (cid:48) , t ) − u ( x − s n , t ) , x (cid:48) − ( x − s n )) d x (cid:48) ds, where B + x ( x − s n ) = (cid:110) x (cid:48) ∈ B δ ( x − s n ) : ( x (cid:48) − x ) · n (cid:62) (cid:111) and B Dδ ( x ) is the two-dimensional ball of radius δ centered at x . Under a (static) homogeneous deformation, we may introducea peridynamic stress tensor σ peri , independent of x , such that τ ( x , n ) = σ peri n . In order to compute the engineering constants,we calculate σ peri ij for a material point x in the bulk of the body and relate it to the corresponding component of the classicalwo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 69 x − s e x δ B + x ( x − s e ) x − s e x δ B + x ( x − s e ) Fig. 15: Illustration of regions of integration for the two-dimensional peridynamic traction. stress tensor, σ ij . In the directions of e = (cid:20) (cid:21) and e = (cid:20) (cid:21) , we find (cid:34) σ peri11 σ peri12 (cid:35) = τ ( x , e ) = (cid:90) δ (cid:90) B + x ( x − s e ) f ( u ( x (cid:48) ) − u ( x − s e ) , x (cid:48) − ( x − s e )) d x (cid:48) ds, (cid:34) σ peri21 σ peri22 (cid:35) = τ ( x , e ) = (cid:90) δ (cid:90) B + x ( x − s e ) f ( u ( x (cid:48) ) − u ( x − s e ) , x (cid:48) − ( x − s e )) d x (cid:48) ds. (233)In Figure 15, we present an illustration of the regions B + x ( x − s e ) and B + x ( x − s e ).Now, we compute the stress components σ peri ij = τ i ( x , e j ) in (233). Combining (232) and (233), we find τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) B + x ( x − s e ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) d x (cid:48) ds = (cid:15) (cid:90) δ (cid:90) ˆ x + δx (cid:90) y + √ δ − ( x (cid:48) − ˆ x ) y − √ δ − ( x (cid:48) − ˆ x ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) dy (cid:48) dx (cid:48) ds,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) B + x ( x − s e ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) d x (cid:48) ds = (cid:15) (cid:90) δ (cid:90) ˆ y + δy (cid:90) x + √ δ − ( y (cid:48) − ˆ y ) x − √ δ − ( y (cid:48) − ˆ y ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) dx (cid:48) dy (cid:48) ds,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) B + x ( x − s e ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) ( y (cid:48) − ˆ y ) d x (cid:48) ds = (cid:15) (cid:90) δ (cid:90) ˆ x + δx (cid:90) y + √ δ − ( x (cid:48) − ˆ x ) y − √ δ − ( x (cid:48) − ˆ x ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) ( y (cid:48) − ˆ y ) dy (cid:48) dx (cid:48) ds,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) B + x ( x − s e ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) ( y (cid:48) − ˆ y ) d x (cid:48) ds = (cid:15) (cid:90) δ (cid:90) ˆ y + δy (cid:90) x + √ δ − ( y (cid:48) − ˆ y ) x − √ δ − ( y (cid:48) − ˆ y ) λ ( (cid:107) x (cid:48) − ˆ x (cid:107) )( x (cid:48) − ˆ x ) ( y (cid:48) − ˆ y ) dx (cid:48) dy (cid:48) ds, (234)where in τ i ( x , e j ), we defined ˆ x = (cid:20) ˆ x ˆ y (cid:21) := x − s e j .To simplify (234), we apply the changes of variables x (cid:48) → x (cid:48) − s to the expressions for τ i ( x , e ) and y (cid:48) → y (cid:48) − s to theexpressions for τ i ( x , e ). We follow this by applying the change of variables x (cid:48) → x + ξ to all the resulting equations. This0 Jeremy Trageser, Pablo Selesonprocess yields the following: τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) x + δx + s (cid:90) y + √ δ − ( x (cid:48) − x ) y − √ δ − ( x (cid:48) − x ) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) dy (cid:48) dx (cid:48) ds = (cid:15) (cid:90) δ (cid:90) δs (cid:90) (cid:113) δ − ξ − (cid:113) δ − ξ λ ( (cid:107) ξ (cid:107) ) ξ dξ dξ ds,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) y + δy + s (cid:90) x + √ δ − ( y (cid:48) − y ) x − √ δ − ( y (cid:48) − y ) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) dx (cid:48) dy (cid:48) ds = (cid:15) (cid:90) δ (cid:90) δs (cid:90) (cid:113) δ − ξ − (cid:113) δ − ξ λ ( (cid:107) ξ (cid:107) ) ξ dξ dξ ds = 0 ,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) x + δx + s (cid:90) y + √ δ − ( x (cid:48) − x ) y − √ δ − ( x (cid:48) − x ) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) ( y (cid:48) − y ) dy (cid:48) dx (cid:48) ds = (cid:15) (cid:90) δ (cid:90) δs (cid:90) (cid:113) δ − ξ − (cid:113) δ − ξ λ ( (cid:107) ξ (cid:107) ) ξ ξ dξ dξ ds = 0 ,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) y + δy + s (cid:90) x + √ δ − ( y (cid:48) − y ) x − √ δ − ( y (cid:48) − y ) λ ( (cid:107) x (cid:48) − x (cid:107) )( x (cid:48) − x ) ( y (cid:48) − y ) dx (cid:48) dy (cid:48) ds = (cid:15) (cid:90) δ (cid:90) δs (cid:90) √ δ − ξ − (cid:113) δ − ξ λ ( (cid:107) ξ (cid:107) ) ξ ξ dξ dξ ds. (235)Above we found τ ( x , e ) = 0 and τ ( x , e ) = 0 by antisymmetry in ξ and ξ , respectively. Applying the polar coordinatechange of variables ( ξ , ξ ) → ( r cos( θ ) , r sin( θ )) to (235), we find τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) δs (cid:90) arccos ( sr ) − arccos ( sr ) λ ( r ) r cos ( θ ) dθdrds,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) δs (cid:90) π +arccos ( sr ) π − arccos ( sr ) λ ( r ) r cos ( θ ) sin( θ ) dθdrds. (236)Changing the order of integration between r and s , and then integrating in θ and s , we find τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) r (cid:90) arccos ( sr ) − arccos ( sr ) λ ( r ) r cos ( θ ) dθdsdr = 3 π(cid:15) (cid:90) δ λ ( r ) r dr,τ ( x , e ) = (cid:15) (cid:90) δ (cid:90) r (cid:90) π +arccos ( sr ) π − arccos ( sr ) λ ( r ) r cos ( θ ) sin( θ ) dθdsdr = π(cid:15) (cid:90) δ λ ( r ) r dr. (237)Thus, the peridynamic stress tensor is given by σ peri = (cid:20) A A (cid:21) , where A = π(cid:15) (cid:90) δ λ ( r ) r dr. Under the identical (static) homogeneous deformation, ε = (cid:15) , ε = 0, and ε = 0, the classical stress tensor ( cf. (5)) foran isotropic material ( cf. (33)) is given by σ = (cid:15) (cid:20) C C (cid:21) . Equating the peridynamic and classical stress tensors, we obtain the restriction C = C , which is exactly Cauchy’srelation for an isotropic elasticity tensor in two dimensions ( cf. (100d)). From Table 3, we know that for two-dimensionalisotropic materials satisfying Cauchy’s relation, the Poisson’s ratio ν = .While our derivation above relies on the introduction of the peridynamic traction, in Section 3 we present a derivation,independent of the definition of peridynamic traction, that Cauchy’s relations are imposed on bond-based peridynamic models.Consequently, ν = is an intrinsic restriction on isotropic two-dimensional bond-based peridynamic models.wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 71 B Proof of lemma for classical plane stress
B.1 Proof of Lemma 3
Proof
Let u ( x , t ) satisfy Assumption (C σ v ( x , t ) := u ( x, y, − z, t ) u ( x, y, − z, t ) − u ( x, y, − z, t ) (238)and u ( x , t ) satisfy the same initial and boundary conditions. We show v satisfies (10) whenever u satisfies (10). Since thesolution to (10) with appropriately defined initial and boundary conditions is unique, we may conclude v ( x , t ) = u ( x , t ), andthe result of the lemma follows. Imposing (C σ
7) on the equation of motion (10), we obtain (for brevity, we drop the dependenceon x, y, and t ) ρ ( z )¨ u ( z ) = C ∂ u ∂x ( z ) + 2 C ∂ u ∂x∂y ( z ) + C ∂ u ∂y ( z ) + C ∂ u ∂z ( z )+ C ∂ u ∂x ( z ) + ( C + C ) ∂ u ∂x∂y ( z ) + C ∂ u ∂y ( z ) + C ∂ u ∂z ( z )+ ( C + C ) ∂ u ∂x∂z ( z ) + ( C + C ) ∂ u ∂y∂z ( z ) + b ( z ) ,ρ ( z )¨ u ( z ) = C ∂ u ∂x ( z ) + ( C + C ) ∂ u ∂x∂y ( z ) + C ∂ u ∂y ( z ) + C ∂ u ∂z ( z )+ C ∂ u ∂x ( z ) + 2 C ∂ u ∂x∂y ( z ) + C ∂ u ∂y ( z ) + C ∂ u ∂z ( z )+ ( C + C ) ∂ u ∂x∂z ( z ) + ( C + C ) ∂ u ∂y∂z ( z ) + b ( z ) ,ρ ( z )¨ u ( z ) = ( C + C ) ∂ u ∂x∂z ( z ) + ( C + C ) ∂ u ∂y∂z ( z )+ ( C + C ) ∂ u ∂x∂z ( z ) + ( C + C ) ∂ u ∂y∂z ( z )+ C ∂ u ∂x ( z ) + 2 C ∂ u ∂x∂y ( z ) + C ∂ u ∂y ( z ) + C ∂ u ∂z ( z ) + b ( z ) . (239)By Assumption (C σ z → − z in (239). In addition, we impose Assumptions (C σ
2) and(C σ
3) on (239), and then we multiply the equation for u by negative one, to obtain ρ ¨ u ( − z ) = C ∂ u ∂x ( − z ) + 2 C ∂ u ∂x∂y ( − z ) + C ∂ u ∂y ( − z ) + C ∂ u ∂z ( − z )+ C ∂ u ∂x ( − z ) + ( C + C ) ∂ u ∂x∂y ( − z ) + C ∂ u ∂y ( − z ) + C ∂ u ∂z ( − z )+ ( C + C ) ∂ u ∂x∂z ( − z ) + ( C + C ) ∂ u ∂y∂z ( − z ) + b ( z ) ,ρ ¨ u ( − z ) = C ∂ u ∂x ( − z ) + ( C + C ) ∂ u ∂x∂y ( − z ) + C ∂ u ∂y ( − z ) + C ∂ u ∂z ( − z )+ C ∂ u ∂x ( − z ) + 2 C ∂ u ∂x∂y ( − z ) + C ∂ u ∂y ( − z ) + C ∂ u ∂z ( − z )+ ( C + C ) ∂ u ∂x∂z ( − z ) + ( C + C ) ∂ u ∂y∂z ( − z ) + b ( z ) , − ρ ¨ u ( − z ) = − ( C + C ) ∂ u ∂x∂z ( − z ) − ( C + C ) ∂ u ∂y∂z ( − z ) − ( C + C ) ∂ u ∂x∂z ( − z ) − ( C + C ) ∂ u ∂y∂z ( − z ) − C ∂ u ∂x ( − z ) − C ∂ u ∂x∂y ( − z ) − C ∂ u ∂y ( − z ) − C ∂ u ∂z ( − z ) . (240)By noticing ∂ ∂x j ∂x l u i ( x, y, − z, t ) = ∂ ∂x j ∂x l u i ( x, y, − z, t ) , j, l (cid:54) = 3 or j = l = 3 − ∂ ∂x j ∂x l u i ( x, y, − z, t ) , else , ρ ¨ v ( z ) = C ∂ v ∂x ( z ) + 2 C ∂ v ∂x∂y ( z ) + C ∂ v ∂y ( z ) + C ∂ v ∂z ( z )+ C ∂ v ∂x ( z ) + ( C + C ) ∂ v ∂x∂y ( z ) + C ∂ v ∂y ( z ) + C ∂ v ∂z ( z )+ ( C + C ) ∂ v ∂x∂z ( z ) + ( C + C ) ∂ v ∂y∂z ( z ) + b ( z ) ,ρ ¨ v ( z ) = C ∂ v ∂x ( z ) + ( C + C ) ∂ v ∂x∂y ( z ) + C ∂ v ∂y ( z ) + C ∂ v ∂z ( z )+ C ∂ v ∂x ( z ) + 2 C ∂ v ∂x∂y ( z ) + C ∂ v ∂y ( z ) + C ∂ v ∂z ( z )+ ( C + C ) ∂ v ∂x∂z ( z ) + ( C + C ) ∂ v ∂y∂z ( z ) + b ( z ) ,ρ ¨ v ( z ) = ( C + C ) ∂ v ∂x∂z ( z ) + ( C + C ) ∂ v ∂y∂z ( z )+ ( C + C ) ∂ v ∂x∂z ( z ) + ( C + C ) ∂ v ∂y∂z ( z )+ C ∂ v ∂x ( z ) + 2 C ∂ v ∂x∂y ( z ) + C ∂ v ∂y ( z ) + C ∂ u ∂z ( z ) . (241)Comparing (239) and (241), we see under the assumptions of Lemma 3, v satisfies (10) whenever u does. (cid:117)(cid:116) C Proofs of lemmas for peridynamic plane stress
C.1 Proof of Lemma 4
Proof
Let u ( x , t ) satisfies Assumption (P σ v ( x , t ) := u ( x, y, − z, t ) u ( x, y, − z, t ) − u ( x, y, − z, t ) (242)and u ( x , t ) satisfy the same boundary and initial conditions. We show v satisfies (118) whenever u satisfies (118). Since thesolution to (118) with appropriately defined boundary and initial conditions is unique, we may conclude v ( x , t ) = u ( x , t ) andthe result of the lemma follows.Imposing Assumptions (P σ
1) and (P σ
5) on (118), we obtain ρ ( x )¨ u i ( x , t ) = (cid:90) H x λ ( x (cid:48) , x )( x (cid:48) i − x i )( x (cid:48) j − x j )( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) + b i ( x , t )= (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( x (cid:48) , x )( x (cid:48) i − x i )( x (cid:48) j − x j )( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) + b i ( x , t ) , (243)where r = (cid:112) δ − ( z (cid:48) − z ) and B Dr ( x ) is the two-dimensional ball of radius r centered at x . For brevity, we omit thefunctional dependence of ρ , u , λ , and b on x, y, x (cid:48) , y (cid:48) , and t . By Assumption (P σ − z (since it holds allwo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 73 z ∈ [ − h, h ]). Substitute z → − z and then apply the change of variables z (cid:48) → − z (cid:48) to (243) to obtain (note r is unchanged): ρ ( − z )¨ u ( − z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( x (cid:48) − x ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( x (cid:48) − x )( y (cid:48) − y )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( x (cid:48) − x )( z (cid:48) − z )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + b ( − z ) ,ρ ( − z )¨ u ( − z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( y (cid:48) − y )( x (cid:48) − x )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( y (cid:48) − y ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( y (cid:48) − y )( z (cid:48) − z )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + b ( − z ) ,ρ ( − z )¨ u ( − z ) = − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( z (cid:48) − z )( x (cid:48) − x )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( z (cid:48) − z )( y (cid:48) − y )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( − z (cid:48) , − z )( z (cid:48) − z ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + b ( − z ) . (244)Next, we apply Assumptions (P σ σ σ
7) to (244) to obtain ρ ¨ u ( − z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x )( y (cid:48) − y )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x )( z (cid:48) − z )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + b ( z ) ,ρ ¨ u ( − z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y )( x (cid:48) − x )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y )( z (cid:48) − z )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + b ( z ) ,ρ ¨ u ( − z ) = − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z )( x (cid:48) − x )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) − (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z )( y (cid:48) − y )( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z ) ( u ( − z (cid:48) ) − u ( − z )) d x (cid:48) . (245)4 Jeremy Trageser, Pablo SelesonRecalling (242), we may rewrite (245) as ρ ¨ v ( z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x ) ( v ( z (cid:48) ) − v ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x )( y (cid:48) − y )( v ( z (cid:48) ) − v ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( x (cid:48) − x )( z (cid:48) − z )( v ( z (cid:48) ) − v ( z )) d x (cid:48) + b ( z ) ,ρ ¨ v ( z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y )( x (cid:48) − x )( v ( z (cid:48) ) − u ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y ) ( v ( z (cid:48) ) − v ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( y (cid:48) − y )( z (cid:48) − z )( v ( z (cid:48) ) − v ( z )) d x (cid:48) + b ( z ) ,ρ ¨ v ( z ) = (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z )( x (cid:48) − x )( v ( z (cid:48) ) − v ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z )( y (cid:48) − y )( v ( z (cid:48) ) − v ( z )) d x (cid:48) + (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( z (cid:48) , z )( z (cid:48) − z ) ( v ( z (cid:48) ) − v ( z )) d x (cid:48) . (246)Comparing (246) with (243), we see under the assumptions of Lemma 4, v satisfies (118) whenever u does. (cid:117)(cid:116) C.2 Proof of Lemma 5
Proof
We begin by writing out the limits of integration for the third component of the peridynamic traction (190): τ ( x , t, e ) = (cid:90) δ (cid:90) z − s + δz (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u ( x − s e , t ) , x (cid:48) , ( x − s e )) d x (cid:48) ds, where ˆ r = (cid:112) δ − ( z (cid:48) − ( z − s )) . Next, we apply the change of variable s → z − ˆ z to obtain τ ( x , t, e ) = (cid:90) zz − δ (cid:90) ˆ z + δz (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) d x (cid:48) d ˆ z, where ˆ x = xy ˆ z and ˆ r = (cid:112) δ − ( z (cid:48) − ˆ z ) . Taking the average of the stress τ ( x , t, e ) over the thickness of the plate, we find τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) zz − δ (cid:90) ˆ z + δz (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) d x (cid:48) d ˆ zdz. (247)Appealing to Assumptions (P σ
1) and (P σ | ˆ z | (cid:54) h and | z | (cid:54) h . Thus, we mayrewrite (247) as (noting h < ˆ z + δ and − h > z − δ ) τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) z − h (cid:90) hz (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) d x (cid:48) d ˆ zdz. (248)Next, we change the order of integration so that the integral in z comes first. For clarity, we do this in steps by first changingthe order of integration between ˆ z and z . Viewing Figure 16, we deduce the new limits of integration after changing the orderof integration between ˆ z and z : τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) h ˆ z (cid:90) hz (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) d x (cid:48) dzd ˆ z. (249)Viewing Figure 17, we deduce the new limits of integration after changing the order of integration between z (cid:48) and z in (249): τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) h ˆ z (cid:90) z (cid:48) ˆ z (cid:90) B r ( x,y ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dx (cid:48) dy (cid:48) dzdz (cid:48) d ˆ z. (250)wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 75 z ˆ z h − h h − h Fig. 16: The z and ˆ z region of integration for τ ( x , t, e ) in (248). zz (cid:48) h − h h − h ˆ z ˆ z Fig. 17: The z and z (cid:48) region of integration for τ ( x , t, e ) in (249). Since B r ( x, y ) is independent of z we may integrate over z first to obtain τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) h ˆ z (cid:90) B r ( x,y ) (cid:90) z (cid:48) ˆ z f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dzdx (cid:48) dy (cid:48) dz (cid:48) d ˆ z = 12 h (cid:90) h − h (cid:90) h ˆ z (cid:90) B r ( x,y ) ( z (cid:48) − ˆ z ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dx (cid:48) dy (cid:48) dz (cid:48) d ˆ z. (251)Note that ( z (cid:48) − ˆ z ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) = λ ( x (cid:48) , ˆ x )( z (cid:48) − ˆ z ) ( x (cid:48) j − ˆ x j )( u j ( x (cid:48) , t ) − u j ( x , t ))6 Jeremy Trageser, Pablo Selesonis invariant under the transformation ( z (cid:48) , ˆ z ) → ( − z (cid:48) , − ˆ z ) by Lemma 4 and (183). Thus, applying the change of variables z (cid:48) → − z (cid:48) and ˆ z → − ˆ z to (251) results in τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) ˆ z − h (cid:90) B r ( x,y ) ( z (cid:48) − ˆ z ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dx (cid:48) dy (cid:48) dz (cid:48) d ˆ z. (252)We sum (251) and (252) to obtain2 τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) h − h (cid:90) B r ( x,y ) ( z (cid:48) − ˆ z ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dx (cid:48) dy (cid:48) dz (cid:48) d ˆ z. (253)Since λ ( x (cid:48) , ˆ x ) = 0 when (cid:107) x (cid:48) − ˆ x (cid:107) (cid:62) δ , we may extend the region of integration so that2 τ ( x , t, e ) = 12 h (cid:90) h − h (cid:90) h − h (cid:90) B δ ( x,y ) ( z (cid:48) − ˆ z ) f ( u ( x (cid:48) , t ) − u (ˆ x , t ) , x (cid:48) , ˆ x ) dx (cid:48) dy (cid:48) dz (cid:48) d ˆ z. (254)Dividing both sides of (254) by 2 and then relabeling ˆ z with z completes the proof. (cid:117)(cid:116) C.3 Proof of Lemma 6
Proof
We start by developing an estimate of ∂u ∂z ( x, y, , t ). By Assumption (P σ τ ( x , t, e ) = 0 and, appealing to Lemma5, we obtain 0 = 14 h (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ξ j ( u j ( x (cid:48) , t ) − u j ( x , t )) d x (cid:48) dz. (255)By solving for the integral containing u in (255), we find (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz = − (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ (cid:2) ξ ( u ( x (cid:48) , t ) − u ( x , t )) + ξ ( u ( x (cid:48) , t ) − u ( x , t )) (cid:3) d x (cid:48) dz. (256)Recalling Assumption (P σ u ( x (cid:48) , t ) and u ( x , t ) about the point ( x, y, , t )on the left-hand side of (256) to get (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ (cid:20) u + ∂u ∂x ξ + ∂u ∂y ξ + ∂u ∂z z (cid:48) + 12 ∂ u ∂x ξ + 12 ∂ u ∂y ξ + 12 ∂ u ∂z z (cid:48) + ∂ u ∂x∂y ξ ξ + ∂ u ∂x∂z ξ z (cid:48) + ∂ u ∂y∂z ξ z (cid:48) − (cid:18) u + ∂u ∂z z + 12 ∂ u ∂z z (cid:19)(cid:21) d x (cid:48) dz, (257)where each occurrence of u or its derivatives is evaluated at ( x, y, , t ). Note that many of the terms in (257) are antisymmetricin ( z, z (cid:48) ), i.e., g ( z, z (cid:48) ) = − g ( − z, − z (cid:48) ), by Assumption (P σ − h, h ] × [ − h, h ], those terms are negated under integration and (257) simplifies to (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ (cid:18) ∂u ∂z ( z (cid:48) − z ) + ∂ u ∂x∂z ξ z (cid:48) + ∂ u ∂y∂z ξ z (cid:48) (cid:19) d x (cid:48) dz. (258)Perform the change of variables ( x (cid:48) , y (cid:48) ) → ( x + ξ , y + ξ ) and recall (181) to obtain (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( ) λ ( ξ , ξ , z (cid:48) , z ) ξ (cid:18) ∂u ∂z ξ + ∂ u ∂x∂z ξ z (cid:48) + ∂ u ∂y∂z ξ z (cid:48) (cid:19) dξ dξ dz (cid:48) dz. (259)From (116) and (183) we have λ ( ξ , ξ , z (cid:48) , z ) = λ ( − ξ , − ξ , z, z (cid:48) ) = λ ( − ξ , − ξ , z (cid:48) , z ) , (260)wo-Dimensional and Planar Classical Linear Elasticity and Peridynamics 77i.e., λ ( ξ , ξ , z (cid:48) , z ) is invariant under the transformation ( ξ , ξ ) → ( − ξ , − ξ ). Since the limits of integration in ( ξ , ξ ) areover B Dδ ( ), we may employ antisymmetry to further reduce (259) to (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( ) λ ( x (cid:48) , x ) ξ dξ dξ dz (cid:48) dz ∂u ∂z ( x, y, , t ) . (261)Solving for ∂u ∂z ( x, y, , t ) in (261) provides us with the estimate ∂u ∂z ( x, y, , t ) ≈ (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz (cid:82) h − h (cid:82) h − h (cid:82) B Dr ( ) λ ( x (cid:48) , x ) ξ dξ dξ dz (cid:48) dz . (262)Recalling (256), we may replace the numerator of (262) with an expression solely in u and u : ∂u ∂z ( x, y, , t ) ≈ − (cid:82) h − h (cid:82) h − h (cid:82) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ [ ξ ( u ( x (cid:48) , t ) − u ( x , t )) + ξ ( u ( x (cid:48) , t ) − u ( x , t ))] d x (cid:48) dz (cid:82) h − h (cid:82) h − h (cid:82) B Dr ( ) λ ( x (cid:48) , x ) ξ dξ dξ dz (cid:48) dz . (263)Now that we have the estimate (263), the derivation of (193) is fairly straightforward. By again employing Assumption (P σ x (cid:48) , y (cid:48) , , t ) for u ( x (cid:48) , t ) and ( x, y, , t ) for u ( x )to find (for i = 1 , (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ (cid:20) u ( x (cid:48) , y (cid:48) , , t ) + z (cid:48) ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) + z (cid:48) ∂ u ∂z (cid:48) ( x (cid:48) , y (cid:48) , , t ) − u ( x, y, , t ) − z ∂u ∂z ( x, y, , t ) − z ∂ u ∂z ( x, y, , t ) (cid:21) d x (cid:48) dz. (264)We again exploit antisymmetry with respect to the transformation ( z (cid:48) , z ) → ( − z (cid:48) , − z ) to reduce (264) to (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ (cid:18) z (cid:48) ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) − z ∂u ∂z ( x, y, , t ) (cid:19) d x (cid:48) dz. (265)Recall from (260) that λ is symmetric with respect to ( ξ , ξ ) → ( − ξ , − ξ ). Thus, the second term in the integrand in (265)is antisymmetric in ( ξ , ξ ) and therefore we may reduce (265) to (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dr ( x,y ) λ ( x (cid:48) , x ) ξ i ξ z (cid:48) ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) d x (cid:48) dz. (266)By interchanging z and z (cid:48) and employing (183), we have (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ z (cid:48) ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) d x (cid:48) dz = − (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ z ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) d x (cid:48) dz. (267)Since the right-hand sides of (266) and (267) are equivalent, we may sum them and divide by 2 in order to reformulate (266)as (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ( u ( x (cid:48) , t ) − u ( x , t )) d x (cid:48) dz ≈ (cid:90) h − h (cid:90) h − h (cid:90) B Dδ ( x,y ) λ ( x (cid:48) , x ) ξ i ξ ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) d x (cid:48) dz. (268)Replacing the term ∂u ∂z ( x (cid:48) , y (cid:48) , , t ) in (268) with the corresponding estimate from (263) completes the proof. (cid:117)(cid:116) C.4 Proof of Lemma 7
Proof
This proof has two steps. First, we relate u ( x , t ) and u ( x , t ) to their corresponding evaluations on the plane z = 0, u ( x, y, , t ) and u ( x, y, , t ). This is followed by approximating u ( x, y, , t ) and u ( x, y, , t ) with their corresponding averages u ( x, y, t ) and u ( x, y, t ). Combining the two results completes the proof. Through the relationship (122b) or the relaxed formulation (202), the denominator in (262) will be 4 hC , which ispositive for stable materials [6].8 Jeremy Trageser, Pablo SelesonTo relate u ( x , t ) and u ( x , t ) to their evaluations on the plane z = 0, we appeal to Assumption (P σ
8) to perform a Taylorexpansion: u i ( x , t ) = u i ( x, y, , t ) + ∞ (cid:88) n =1 n ! ∂ n u i ∂z n ( x, y, , t ) z n = u i ( x, y, , t ) + ∞ (cid:88) k =1 k )! ∂ k u i ∂z k ( x, y, , t ) z k = u i ( x, y, , t ) + O ( h ) . (269)In the second equality of (269), we appealed to Lemma 4 so that ∂ n u i ∂z n ( x, y, z, t ) = ( − n ∂ n u i ∂z n ( x, y, − z, t ) ⇒ ∂ n u i ∂z n ( x, y, , t ) = 0 for n odd . (270)Similarly, by appealing to Assumption (P σ u i ( x, y, t ) = 12 h (cid:90) h − h u i ( x , t ) dz = 12 h (cid:90) h − h u i ( x, y, , t ) + ∞ (cid:88) n =1 n ! ∂ n u i ∂z n ( x, y, , t ) z n dz = 12 h (cid:90) h − h u i ( x, y, , t ) + ∞ (cid:88) k =1 k )! ∂ k u i ∂z k ( x, y, , t ) z k dz = u i ( x, y, , t ) + 1 h ∞ (cid:88) k =1 k )! ∂ k u i ∂z k ( x, y, , t ) h k +1 k + 1= u i ( x, y, , t ) + O ( h ) . (271)In the third line of (271), we utilized (270). Combining (269) and (271) completes the proof. (cid:117)(cid:116) Acknowledgements
Research sponsored by the Laboratory Directed Research and Development Program of Oak RidgeNational Laboratory, managed by UT-Battelle, LLC, for the U. S. Department of Energy.
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