Buckling of chiral rods due to coupled axial and rotational growth
aa r X i v : . [ c ond - m a t . s o f t ] S e p Buckling of chiral rods due to coupled axial androtational growth
Satya Prakash Pradhan and Prashant Saxena ∗ Department of Mechanical & Aerospace Engineering, Indian Institute of TechnologyHyderabad, Sangareddy 502285, Telangana, India. Glasgow Computational Engineering Centre, James Watt School of Engineering,University of Glasgow, Glasgow G12 8LT, UK.
Abstract
We present a growth model for special Cosserat rods that allows for inducedrotation of cross-sections. The growth law considers two controls, one for lengthwisegrowth and other for rotations. This is explored in greater detail for straight rodswith helical and hemitropic material symmetries by introduction of a symmetrypreserving growth to account for the microstructure. The example of a guided-guided rod possessing a chiral microstructure is considered to study its deformationdue to growth. We show the occurrence of growth induced out-of-plane buckling insuch rods.
Keywords:
Cosserat rod, Hemitropy, Helical symmetry, Growth, Bifurcation
MSC 2010: B · G · K Several theoretical models for elastic rods have been around for a while now. Startingfrom the Euler’s elastica to Kirchhoff rods, a very rich literature is available includingthe general model developed by Green, Naghdi and their collaborators. The general rodtheory proposed by Green and Naghdi subsumes classical theories like the Cosserat rodtheory as special cases under appropriate constraints. A comprehensive description ofdifferent rod theories is provided by Antman (2005) and O’Reilly (2017).Rod theories have been employed in many interesting applications in the last fewdecades, for example in DNA biophysics (Manning et al., 1996), marine cables (Goyalet al., 2005), tendril perversion in plants (Goriely and Tabor, 1998; McMillen et al., 2002),surgical filaments (Nuti et al., 2014), slender viscous jets (Arne et al., 2010), hair curlsMiller et al. (2014) and carbon nanotubes (Chandraseker et al., 2009; Kumar et al., 2011).Growing filamentary structures are ubiquitous in nature. Plant organs such as ten-drils, roots and stem tend to twist while growing axially (Wada and Matsumoto, 2018). ∗ Corresponding author email: [email protected] symmetry preserving , so that any imaginary helix associated with themicrostructure remains unaltered as the rod grows. Such a growth problem dependsonly on the microstructural pitch and the constitutive laws, keeping aside the boundaryconditions and other external factors.A rod constrained to grow (or decay) in a guided-guided environment is considered,with a chiral constitutive law that is applicable to both helical symmetry and transversehemitropy. Out-of-plane buckling is observed to occur at certain growth (or atrophy)stages, corresponding to the bifurcation modes We demonstrate that an exact reversal inchirality of these non-planar solutions requires us to mirror the chiral parameters in bothgrowth and constitutive laws simultaneously. Comparisons are made for the end-to-enddistance in the buckled configuration with that in the virtual state to see if the ends havecome closer or moved apart, than what they would have been in the absence of the guides.We also show that total growth induced extension in rod does not depend monotonicallyon the degree of chirality, that is, total extension in an isotropic rod need not lie betweenthe total extension of rods with opposite material chirality.This paper is organised as follows. We begin with a theoretical background of materialsymmetries in the context of special Cosserat rods in Section 2. A twisting growth lawwith two control parameters is systematically derived using certain kinematic assump-tions such as homogeneity in length-wise growth and relative rotation of cross-sections inSection 3. In Section 4, we solve the problem of growth induced out-of-plane bifurcationin a chiral rod with guided-guided boundary conditions to study the interplay betweenchiralities in growth and material laws. We present our conclusions in Section 5.
Throughout this text, the indices i, j, k ∈ { , , } and α, β ∈ { , } , unless mentionedotherwise. We let { e , e , e } to be a right-handed, fixed, orthonormal basis for theEuclidean space E . Boldface symbols are used to denote tensors, lowercase letters forfirst order e.g. v and uppercase letters for second order tensors e.g. T . Underlinedsymbols such as v and T denote matrix representation of tensors with respect to a basis. Consider a straight rod of unit length in its stress-free reference configuration as shownin Figure 1. Assumption of special Cosserat rod behaviour requires the transverse cross-sections to stay rigid during the deformation. Let s ∈ (cid:2) − , (cid:3) denote a signed arc-lengthparameter of the centre-line in the reference configuration. Let r ( s ) define the centre-lineof the deformed rod. Let R ( s ) ∈ SO (3) be the rotation of transverse cross-sections inthe reference configuration of the rod, mapping the fixed basis { e , e , e } to a triad oforthonormal directors given by d i ( s ) = R ( s ) e i . (1)3igure 1: Kinematics of a special Cosserat rod– depicting the deformed centre-curve andthe triad of orthonormal directors.The vector fields ν := r ′ , κ := axial( R ′ R T ) . (2)define convected coordinates ν = ν i d i and κ = κ i d i with respect to the director framefield, along with the ordered triples v := ( ν , ν , ν ) and k := ( κ , κ , κ ). The strains ν α correspond to shear, ν corresponds to stretch, κ α correspond to curvatures, and κ corresponds to twist.We further assume the rod to be hyperelastic with a differentiable energy density (perunit length) function Φ( r ′ , R , R ′ , s ). Material objectivity allows for a simpler version ofenergy function in terms of strains (Healey, 2002), given byΦ = W (v , k , s ) , (3)where W is another differentiable scalar valued function.The internal force and moment on the transverse cross-section are denoted by n ( s ) = n i d i and m ( s ) = m i d i , respectively, along with the corresponding triples n := ( n , n , n )and m := ( m , m , m ). The components n α are essentially the shear forces, n is axialforce, m α are bending moments and m is the torsional moment. These are related tothe strain components as n = ∂W∂ v , m = ∂W∂ k . (4)To prevent self penetration, we require ν = r ′ · d > , (5)and the unshearability constraint is expressed as ν α = r ′ · d α = 0 . (6)4 .1 Material symmetry in Rods In this section, we present a brief overview of certain classes of material symmetry forspecial Cosserat rods (Healey, 2002, 2011).
Consider a straight rod possessing helical material symmetry with a signed pitch
M 6 = 0with M > M ,keeping its magnitude, the same.We introduce a rotating basis { e ∗ ( φ ) , e ∗ ( φ ) , e ∗ ( φ ) } and a corresponding triad of di-rector fields given by e ∗ i ( φ ) = Q φ e i , ≤ φ < π (7) d ∗ α ( s ) = R ( s ) e ∗ α (cid:16) s M (cid:17) , (8)where Q φ is a proper orthogonal tensor with matrix representation Q φ = cos φ − sin φ φ cos φ
00 0 1 , (9)in the fixed basis.Assuming e ∗ ( φ ) to be the rotating flip axis, we denote by H πφ the flip about e ∗ ( φ ).Material properties with respect to the symmetry axis e ∗ (cid:0) s M (cid:1) are assumed not to changeas the cross-section ‘ s ’ moves along the rod. This motivates the definition of a symmetryadapted energy function (Healey, 2002) independent of s , given by W (v , k , s ) = Φ = W ∗ (v ∗ , k ∗ ) , (10)where v ∗ = ( ν ∗ , ν ∗ , ν ) and k ∗ = ( κ ∗ , κ ∗ , κ ) emerge from the change of coordinates κ = κ ∗ α d ∗ α + κ d , ν = ν ∗ α d ∗ α + ν d . (11)Helical symmetry is characterized by the following equation W ∗ ( ν ∗ , ν ∗ , ν , κ ∗ , κ ∗ , κ ) = W ∗ ( − ν ∗ , ν ∗ , ν , − κ ∗ , κ ∗ , κ ) , (12)in terms of the new energy function without ‘ s ’ as an argument. Consider a rod with a symmetry analogous to n ≥ s has n equally spaced flip axes. A 180-degree rotation about eachof these gives a symmetry. Such a rod is said to have a n-fold dihedral helical symmetrywhich is characterized by the condition W ∗ (cid:16) − H π ∗ πn v ∗ , − H π ∗ πn k ∗ (cid:17) = W ∗ (v ∗ , k ∗ ) , (13)5n addition to (12), where H π ∗ πn is the matrix of H π ∗ πn with respect to the rotating basis(7). H π ∗ πn = cos( πn ) sin( πn ) 0sin( πn ) − cos( πn ) 00 0 − . (14) For n ≫
1, a straight rod with n -fold dihedral helical symmetry approaches to whatis called continuous helical symmetry. In this type of symmetry all vectors of the crosssection act as symmetry axis, or equivalently any fixed flip axis, say e acts as a symmetryaxis for all cross sections. Continuous helical symmetry can be characterized by W ( − H πφ v , − H πφ k) = W (v , k) , ∀ φ ∈ [ 0 , π ) . (15) Let E denote a reflection with a matrix with respect to the fixed basis as E = − . (16)A homogeneous hyperelastic straight rod with energy function W (v , k) is transverselyhemitropic if W ( Q φ v , Q φ k) = W (v , k) ∀ φ ∈ [ 0 , π ) , (17)and flip-symmetric if W ( E v , E k) = W (v , k) . (18)Note that flip-symmetry does not belong to the class of transverse symmetry, definedby Healey (2002). A straight rod is transversely isotropic if in addition to (17), it alsosatisfies W (v , k) = W ( E v , − E k) . (19)Flip-symmetric hemitropy is equivalent to continuous helical symmetry (Healey, 2002).Another way to obtain flip-symmetric hemitropy, is to consider a rod with helical sym-metry and take the limit M →
The energy density per unit length of unshearable hemitropic rods can be expressed as(Antman, 2005; Healey, 2002) W = Υ( κ α κ α , ν , κ ) , (20)6here Υ is a scalar valued function. This representation is also valid for flip-symmetry.For calculations in this paper, we adopt a model considered by Papadopoulos (1999);Healey and Papadopoulos (2013) defined asΥ = 12 h Φ( ν ) + 2 A ( ν − κ + Bκ + Cκ α κ α i , (21)where Φ : (0 , ∞ ) → R is a function such that g := Φ ′ obeys g ( ν ) → −∞ as ν → g ( · ) allows us to modify the axial force response of the model, and it mustsatisfy g (1) = 0. The constant C corresponds to bending stiffness, B − A g ′ (1) is equivalentto torsional rigidity and g ′ (1) − A B to axial stiffness, where A is the degree of hemitropy.We assume B > C > Bg ′ ( ν ) > A for all ν to ensure convexity. This inturn implies that g ( · ) should be monotonic and hence invertible. For example, a responsefunction satisfying all our criteria can be chosen as (Papadopoulos, 1999) g ( ν ) = F ln( ν ) + A B ( ν − , (22)where F > n →−∞ whenever an unrealistically extreme strain ν → n -fold helical symmetry ( n ≥
3) and hemitropy.On similar lines, the energy function (21) can be shown to be applicable to n -fold helicalsymmetry. Growth in elastic bodies is typically modelled by introduction of a multiplicative de-composition of the deformation gradient into pure growth and pure elastic deformationparts (Rodriguez et al., 1994; Ambrosi et al., 2011). This decomposition assumes a vir-tual stress-free incompatible configuration. For one-dimensional structures where growthmanifests as increase in overall length, first the stress-free rod isolated from its environ-ment and boundary conditions can be allowed to grow free into a virtual state, and thenthe boundary and environmental factors can be forcibly imposed (Goriely, 2017; O’Keeffeet al., 2013).
Let R o denote the initial stress-free reference configuration of the rod, occupying { S e : − ≤ S ≤ } and denote by S a signed arc length parameter of the centre-line in R o .Let e r ( S ) be the curve taken by the centre-line in the virtual grown configuration e R , suchthat the point S e in R o gets mapped to e r ( S ) in e R (Figure 2). The virtual configurationis assumed to be stress-free. We define a signed arc-length s ( S ) in e R by s ( S ) := Z S || e r ′ ( τ ) || dτ. (23)7igure 2: Kinematics of an initially straight rod growing from origin S o , depicting theconfigurations– reference R o , virtual e R and current R ; along with the multiplicativedecomposition Q = RW .We denote the transverse cross-section at S in R o by Γ o ( S ) and let it get mapped to e Γ( S ) in the virtual configuration e R . Define W ( S ) ∈ SO (3) to be the rotation of e Γ( S )with respect to Γ o ( S ), and let it map the fixed basis { e , e , e } to a virtual director fieldgiven by e e i ( S ) = W ( S ) e i . (24)When the boundary conditions and environmental factors are imposed, let the centre-linetake the curve r ( S ) in the current configuration R , and the cross-section e Γ( S ) in e R bemapped to Γ( S ) in R . Define R ( S ) ∈ SO (3) to be the rotation of Γ( S ) with respect to e Γ( S ) and Q ( S ) ∈ SO (3) to be the rotation of Γ( S ) with respect to Γ o ( S ), so that Q ( S ) = R ( S ) W ( S ) . (25)The virtual director field is transformed into another director field in the current config-uration given by d i ( S ) = R ( S ) e e i ( S ) = Q ( S ) e i . (26)All the maps we have introduced are assumed to be smooth for the sake of convenience.Analogous to r : (cid:2) − , (cid:3) → E , we define another map b r : (cid:2) s ( − ) , s ( ) (cid:3) → E to denotethe same curve via another parametrization, r ( S ) = (cid:0)b r ◦ s (cid:1) ( S ) . (27)This implies r ′ ( S ) = || e r ′ ( S ) || b r ′ (cid:0) s ( S ) (cid:1) . (28)8imilarly, define b R : (cid:2) s ( − ) , s ( ) (cid:3) → SO (3) by R ( S ) = (cid:0) b R ◦ s (cid:1) ( S ) . (29)We assume the transverse cross-sections to remain orthogonal to centre-line in both vir-tual and current configurations, hence the conditions e r ′ · e e α = 0 (30)and r ′ · d α = 0 , (31)where (31) is equivalent to the unshearability constraint (6). The symbols and notationsintroduced in this section are pictorially represented in Figure 2. We consider the growth to be homogeneous throughout the rod. This assumption leadsto the following constraints: • The length-wise growth parameter denoted by γ := || e r ′ ( S ) || is a constant, that is,it is independent of S . • Let h ∈ R be such that 0 < | h | <
1. Consider the relative rotation of cross-section e Γ( S + h ) with respect to e Γ( S ). e e i ( S + h ) = W ( S + h ) W ( S ) − e e i ( S ) . (32)For all permissible h , the relative rotation W ( S + h ) W ( S ) − is assumed to beindependent of S , and hence can be denoted as a function of h only. W ( S + h ) W ( S ) − =: Π ( h ) . (33)This gives us the decomposition W ( S + h ) = Π ( h ) W ( S ) which leads to Π ( h ) T ∂ Π ( h ) ∂h = ∂ W ( S ) ∂S W ( S ) T . (34)Another way to interpret (33) is to set ∂∂S n W ( S + h ) W ( S ) − o = O , (35)which implies W ( S + h ) T ∂∂S W ( S + h ) = W ( S ) T ∂∂S W ( S ) . (36)We define the tensor fields for later use Λ ( S ) := W ( S ) T ∂ W ( S ) ∂S , and Ω ( S ) := ∂ W ( S ) ∂S W ( S ) T . (37)Equations (34) and (36) imply that Λ and Ω are constant skew-symmetric tensors(The proofs are detailed in Appendix A).9 We fix a point on the centre-line which gets mapped to itself under the growthtransformation, along with its corresponding cross-section. Thus, we assume theexistence of a point S o ∈ (cid:2) − , (cid:3) satisfying e r ( S o ) = S o e and W ( S o ) = I . (38)This can also be interpreted as if the rod is allowed to grow while being held at S o (origin of growth). It is held in such a way that no incompatibility or stress iscaused due to growth. Define vectors a := axial (cid:0) Λ (cid:1) and ω := axial (cid:0) Ω (cid:1) , these areactually constant vectors and can be related by ω = W ( S ) a . (39)Since this is also satisfied for the specific point S = S o , we imply a = ω and Λ = Ω .This also means that axis (cid:0) W ( S ) (cid:1) = ω for all S . Thus one can solve (37) for W ( S )as a differential equation to obtain W ( S ) = e ( S − S o ) Ω , (40)where tensor exponential is defined by the usual series definition. The mathematicaldetails for derivations in this section are provided in Appendix A. Consider a general scenario where the initial configuration R o is a special Cosserat rod.Let ¯ r : (cid:2) − , (cid:3) → E be its centre-curve, where ¯ r ( S ) is arc-length parametrized. Let¯ W ( S ) ∈ SO (3) denote the orientation of Γ o ( S ) with respect to the fixed basis, mappingthose to an orthonormal director field ¯ e i ( S ) := ¯ W ( S ) e i associated with initial configu-ration. Homogeneous growth law still requires γ to be constant while equation (32) ismodified as e e i ( S + h ) = W ( S + h ) ¯ W ( S + h ) ¯ W ( S ) − W ( S ) − e e i ( S ) . (41)The tensor W ( S + h ) W ( S ) − is again independent of S . In addition, the rod is assumedto be held at S o ∈ (cid:2) − , (cid:3) while growing, so that we have e r ( S o ) = ¯ r ( S o ) and W ( S o ) = I . (42)This assumption along with the kind of homogeneity used in induced rotations gives sucha W ( S ) that makes all the cross-sections rotate about the particular axis a . Moreover,the solution is given by (40) which in turn implies e e i ( S ) = e ( S − S o ) Λ ¯ W ( S ) e i , (43)In fact, the constant vector ω = a can be treated as the growth parameter controllingrelative rotation of cross-sections while γ controls the length-wise growth as in the formercase. Whenever the centre-curves are normal to the cross-sections, throughout R o and e R , we deduce e r ( S ) = ¯ r ( S o ) + γ Z SS o e ( τ − S o ) Λ ¯ r ′ ( τ ) dτ . (44)We emphasise that (43) and (40) do not assume the respective centre-curves to be normalto the cross-sections neither in R o nor in e R .10 .2 Growth in straight rods Consider a straight rod with flip-symmetric hemitropy in its reference configuration. Astraight virtual configuration condenses to e r ( S ) = { S o + γ ( S − S o ) } e , (45)which with the aid of (30) results in W ( S ) e = e ∀ S ∈ h − , i . (46)This indicates that ω is along e . We introduce another growth parameter ω defined by ω = ω e , (47)so that its corresponding skew tensor is Ω = ω A , with A = e ⊗ e − e ⊗ e . (48)Since rotation tensor can also be expressed as Q φ = e φ A , (49)we get W ( S ) = Q ( S − S o ) ω . (50)The parameters γ and ω capture all the necessary information regarding growth. Itsevident that γ > γ < ω and − ω signify two opposite cross-sectional rotations caused by growth while ω = 0 indicates nogrowth induced rotation. The growth law adopted here considers rotation of cross sections with respect to eachother in the due course of growth. Consider a rotating basis field { e ∗ ( S ) , e ∗ ( S ) , e ∗ ( S ) = e } given by e ∗ i ( S ) = Q S M e i , (51)representing a helix embedded in the initial configuration of a rod. As the rod grows thistransforms into W ( S ) e ∗ i ( S ) in the virtual configuration. Let us denote this by a basisfield { f ∗ ( s ) , f ∗ ( s ) , f ∗ ( s ) } defined on the virtual arc-length parameter by W ( S ) e ∗ i ( S ) =: (cid:0) f ∗ i ◦ s (cid:1) ( S ) . (52)This is equivalent to f ∗ i ( s ) = Q sγ M +( sγ − S o ) ω e i . (53)Let h = 0 be such that e ∗ i ( S + h ) and f ∗ i ( s + h ) are well defined, then we obtain e ∗ i ( S + h ) = Q h M e ∗ i ( S ) , (54) f ∗ i ( s + h ) = Q hγ ( M + ω ) f ∗ i ( s ) . (55)11his shows that our chosen growth map transforms the initial helix with pitch M intoanother helix with pitch, say µ , which can be expressed as µ = γ M ω M . (56)This motivates us to define a symmetry preserving growth law for rods possessing helicalsymmetry. Rods with helical symmetry
Consider a rod which due to its microstructure pos-sesses simple helical symmetry or n -fold helical symmetry. Let M be the pitch associatedwith its microstructure. Once growth parameters γ and ω are known, (56) serves as anevolution law for the pitch of its microstructure.We introduce the idea of symmetry preserving growth – wherein the growth map fixesall helices with pitch same as that of the microstructure ( µ = M ). Thus, for rods witha pitch associated with their microstructure we have the following helical growth law γ = 1 + ω M , (57)where γ is the only growth parameter and M comes from the material symmetry. Forrods having helical symmetry, this assumption of symmetry preserving growth providesa rationale for relative rotation of cross-sections during growth. Hemitropic rods
Although there are different versions (Healey, 2002, 2011) of howhelical symmetry can be used to arrive at hemitropy, there is no pitch directly associatedwith transverse hemitropy (17). So, for hemitropic rods (and even isotropic) one mayuse the same helical growth law (57) without any notion of microstructural pitch, inwhich case both γ and M are independent growth parameters. For such a growth lawall helices with pitch M remain unaltered under the growth map, so we denote it as the characteristic pitch of growth. The grown configuration is obtained by imposing environmental and boundary effects onthe virtual stress-free configuration. Hence the strain energy is a function of b r ′ ( s ), b R ( s )and b R ′ ( s ). We define the vector fields b ν = b r ′ and b κ = axial (cid:0) b R ′ b R T (cid:1) . (58)Let their components be b ν = b ν i d i and b κ = b κ i d i with respect to the director frame incurrent configuration. Consider the derivative ∂ d i ∂S = ∂ Q ∂S Q − d i = (cid:20) ∂ R ∂S W + R ∂ W ∂S (cid:21) W − R − d i (59a)= (cid:16) ∂ R ∂s (cid:17)(cid:16) ∂s∂S (cid:17) R − d i + R ∂ W ∂S W − e e i (59b)= γ b κ × d i + R ( ω × e e i ) (59c)= ( γ b κ + R ω ) × d i . (59d)12ow define the axial vector β := axial (cid:18) ∂ Q ∂S Q − (cid:19) which along with the straight growthassumption implies β = γ b κ + ω d . (60)Given the growth parameters, this relation will be used in retracting the actual strainsfrom the apparent curvature β . Corresponding to b ν and b κ we define ν = r ′ and κ = axial (cid:0) R ′ R T (cid:1) , (61)along with their convected components ν = ν i d i and κ = κ i d i . These speeds andcurvatures can be related to the actual strains by ν i = γ b ν i and κ i = γ b κ i . (62)Upon use of the energy density function (21), the internal force n ( S ) = n i ( S ) d i ( S ) andmoment m ( S ) = m i ( S ) d i ( S ) in the current configuration can be related to the strains asfollows: n = g ( b ν ) + A b κ , (63) m = A ( b ν −
1) + B b κ , (64) m α = C b κ α . (65) The local linear and angular momentum balance equations for static equilibrium (O’Keeffeet al., 2013; Goriely, 2017) are as follows: ∂ n ∂s + f = , (66) ∂ m ∂s + ∂ r ∂s × n + l = . (67)where f and l respectively denote the body force and body moment per unit virtualarc-length. A change of variable to reference coordinates results in n ′ + γ f = , (68) m ′ + r ′ × n + γ l = . (69) In this section we consider the example of a growing rod with guided ends as shown inFigure 3. A guided boundary condition is equivalent to fixing the end of the rod to ablock constrained by a slot to translate only along the rod’s axis. We use the energyfunction (21) and the growth law (57) to model the rod. Even though all the calculationswould be similar, the results can be discussed separately for two different problems –first, a hemitropic rod and second, a rod with n -fold helical symmetry.13igure 3: Schematic of Guided-Guided boundary condition.The linear and angular momentum balance equations are dds (cid:20) n α Qe α + (cid:8) g ( b ν ) + A b κ (cid:9) Qe (cid:21) = , (70) dds (cid:20) C b κ α Qe α + (cid:2) A { b ν − } + B b κ (cid:3) Qe (cid:21) + b r ′ × h n α Qe α + (cid:8) g ( b ν ) + A b κ (cid:9) Qe i = , (71)along with the boundary conditions n (cid:16) ± (cid:17) · e = 0 , (72) r (cid:16) ± (cid:17) · e α = 0 (73)and Q (cid:16) ± (cid:17) = I . (74)The unshearability constraint (31) results in r ′ · Qe α = 0 . (75)Equations (70)-(75) comprise our boundary value problem to be solved for the fields r , R , and n α . Since we have not imposed any sort of axial constraint, with these set ofboundary conditions we will get a family of solutions differing by a scalar multiple of e .The rod is assumed to be of unit length; thus, all the kinematic quantities are di-mensionless by default. The components of internal force, internal moment, materialconstants A, B and the response function g ( · ) can be all non-dimensionalized against C by either dividing the concerned quantities in (63)-(65) by C, or equivalently setting C = 1 in the boundary value problem (85)-(92). We follow the bifurcation analysismethodology presented by Smith and Healey (2008); Healey and Papadopoulos (2013)wherein first a primary solution is determined which is then perturbed and the boundaryvalue problem is rederived in terms of the perturbations to get linearized equations.14 .1 The primary solution Let us consider a simple solution where the rod always remains straight while growing.A straight solution is of the form r ( S ) = λS e , Q ( S ) = I , n α ( S ) = 0 . (76)where S ∈ (cid:2) − , + (cid:3) . This solution has its local force, moment and strain fields asfollows: b ν ( s ) = λγ e , (77) b κ ( s ) = − ωγ e , (78) n ( S ) = h g (cid:16) λγ (cid:17) − A ωγ i e , (79) m ( S ) = h A n λγ − o − B ωγ i e . (80)For such a solution to comply with the force boundary condition (72) we require λ tosatisfy g (cid:16) λγ (cid:17) = A ωγ . (81)
Consider a first order perturbation of the straight solution (with 0 < ε ≪
1) given by r ( S ) = λS e + ε ρ ( S ) , (82) Q ( S ) = e ε Ψ ( S ) , (83) n α ( S ) = εη α ( S ) , (84)where Ψ ( S ) is skew symmetric with axial( Ψ ) =: ψ . We require these perturbed fieldsto satisfy our boundary value problem. Plugging in the perturbations (82)-(84) into ourboundary value problem (70)-(75) results in the following linearized problem. η ′ α e α = , (85)( ψ ′′ + ω e × ψ ′ ) · e α e α + (cid:2) A { λ − γ } − Bω (cid:3) ψ ′ × e + γλ e × η α e α = , (86) n g ′ (cid:16) λγ (cid:17) ρ ′′ + A ψ ′′ o · e = 0 , (87)( A ρ ′′ + B ψ ′′ ) · e = 0 , (88) (cid:0) ρ ′ − λ ψ × e (cid:1) · e α = 0 , (89) ψ (cid:16) ± (cid:17) = , (90) ρ (cid:16) ± (cid:17) · e α = 0 , (91) (cid:20) g ′ (cid:16) λγ (cid:17) ρ ′ (cid:16) ± (cid:17) + A ψ ′ (cid:16) ± (cid:17)(cid:21) · e = 0 , (92)15ith details provided in Appendix B. Since Bg ′ (cid:0) λγ (cid:1) − A is non-zero (assumed to bepositive), equations (87) and (88) imply ρ ′′ · e = 0 and ψ ′′ · e = 0 . (93)Boundary condition (90) forces us to have ψ ( S ) ∈ span { e , e } , which motivates theintroduction of the decomposition ρ ( S ) = ρ t ( S ) + ρ a ( S ) , (94)where ρ t ( S ) ∈ span { e , e } and ρ a ( S ) ∈ span { e } .Equations (85)-(92) can now be reduced to the following (details in Appendix C): ψ ′′ + ζ ψ ′ × e = ψ ′ (cid:16) + 12 (cid:17) − ψ ′ (cid:16) − (cid:17) , (95) ρ ′ t = λ ψ × e , (96) ρ ′′ a = , (97)accompanied by the boundary conditions ρ t (cid:16) ± (cid:17) = , (98) ρ ′ a (cid:16) ± (cid:17) = . (99)The new parameter ζ appearing in (95) is defined as ζ := A ( λ − γ ) − ( B + 1) ω. (100)It is clear that ρ a ( S ) = C o e for all S , where C o is a constant that appears because wehave put no physical constraint in axial direction. As the rod can slide in the axiallywithout causing any strain, we can fix C o = 0.For ζ = 0, the problem admits only trivial solutions (Appendix C). Now assuming ζ = 0, the differential equations (95) and (96) admit general solutions of the form ψ ( S ) = C ζ sin( ζ S ) − cos( ζ S ) + 2 S sin ζ + C ζ − cos( ζ S ) − S sin ζ − sin( ζ S )0 + C C , (101) ρ t ( S ) = C λζ − sin( ζ S ) + ζ S sin ζ cos( ζ S )0 + C λζ cos( ζ S )sin( ζ S ) + ζ S sin ζ + λ C + C SC − C S , (102)where C , C , · · · , C are generic integration constants in R . The representations ψ and ρ t are with respect to the fixed basis. The boundary conditions (90) when invoked intothe solution (101) leads to ( C − C ) sin ζ , (103)16imultaneously giving C = C ζ cos ζ , C = C ζ cos ζ . (104)The values of ζ = 0 for which sin ζ = 0 eventually lead to the trivial solution (AppendixC). Therefore, we assume C = C , which when plugged into the general solution (102)and forced to satisfy (98), leads to the condition1 ζ sin ζ −
12 cos ζ . (105)It simultaneously leads to the constants C = − C ζ (cid:16) ζ ζ ζ (cid:17) = C . (106)Hence we have an out-of-plane solution, ρ t ( S ) = C λζ (cid:26) cos( ζ S ) + (cid:16) S − (cid:17) ζ sin ζ − cos ζ (cid:27) + C λζ n Sζ cos ζ − sin( ζ S ) o − , (107)whose existence is subject to the condition that parameters γ and λ admit sensible solu-tions ( γ > λ > a n ) ∞ n =1 satisfying tan( a n ) = a n can be defined. The values taken by ζ ∈ {± a n : n ∈ N } correspond to the discretebifurcation modes. In view of the equivariance properties of our problem (Appendix D), any rotation of (107)about e is an acceptable solution, hence the solution can be simplified to ρ ( S ) = C λζ cos( ζ S ) + (cid:0) S − (cid:1) ζ sin ζ − cos ζ sin( ζ S ) − Sζ cos ζ , (108) ψ ( S ) = C λζ cos ζ − cos( ζ S )2 S sin ζ − sin( ζ S )0 , (109) η ( S ) = 0 , (110) η ( S ) = C ζγ cos ζ , (111)17here representations (108) and (109) are with respect to the fixed basis. This solution isclearly flip symmetric about e , thus making it clear that (107) was also flip symmetric,but about an axis different from e . The deformed centre-line r ( S ) for this solution isr( S ) = λζ cos( ζ S ) + (cid:0) S − (cid:1) ζ sin ζ − cos ζ sin( ζ S ) − Sζ cos ζ ζ S , (112)represented with respect to the fixed basis, wherein εC = 1 is set for the sake of simplicity.For a particular ζ ∈ {± a n : n ∈ N } , the end-to-end distance λ and growth stage γ canbe found by solving the system g (cid:16) λγ (cid:17) = A M (cid:18) − γ (cid:19) , (113) ζ = A ( λ − γ ) − B + 1 M ( γ − , (114)simultaneously (Table 1, Appendix E). Equations (113)-(114) couple the axial force re-sponse of the rod with the bifurcation mode caused due to growth, via the kinematicconstraint of symmetry preserving growth (57). Whenever this system does not admit asolution γ > λ >
0, the perturbation chosen gives only trivial solutions, indicatingthat out-of-plane buckling is not guaranteed.An inspection of (112) reveals that the sign change ζ
7→ − ζ reverses the chirality ofsolution curve, reflecting it about e − e plane (Figure 4). Moreover, since our solution isflip symmetric, this is equivalent to reflection in e − e plane. These centre-line solutionswith handedness are similar to those obtained by Healey and Papadopoulos (2013) for afixed-fixed rod under axial compression.Internal chirality of the rod is taken care of by the constants M and A . In caseof hemitropic rods, A captures chirality in load response of the rod while M containsinformation regarding the chiral growth law. For rods with n -fold helical symmetry, A denotes the same thing, but with the assumption of symmetry preserving growth in place, M captures chirality in microstructure.Consider two rods with opposite internal chirality with all other material properties assame. Let one of them with chiral constants A , M have a solution with bifurcation mode ζ , end-to-end distance λ and growth stage γ . Naturally the second rod with oppositeinternal chirality is expected to give rise to a reflected solution with bifurcation mode − ζ while end-to-end distance and growth stage are still the same. Thus equations (113)-(114)imply that the chiral constants associated with the second rod are − A and −M . Weinfer that the complete reversal of internal chirality in rods requires the transformations M 7→ −M and A
7→ − A to be taken simultaneously. In addition, the ζ solution of a rodwith internal chirality M , A and the − ζ solution of a rod with opposite internal chirality −M , − A are mirror images with respect to e − e and e − e planes.Assuming that (113)-(114) admit an acceptable solution, the monotonicity of g ( · ) andthe condition g (1) = 0 reveal the following observations: Growth γ > • A and M are of same sign if and only if λ > γ , signifying that the ends incurrent configuration have moved away from each other, as compared to bothinitial and virtual configurations.18igure 4: Out-of-plane bifurcated solution for the case M = − . A = − B = 1 . F = 10 . The two graphs correspond to projection of the rod centre-line on X − X and X − X planes. 19igure 5: Variation of λ and γ with A for the first mode ( ζ = 8 . • A and M are of opposite if and only if λ < γ , signifying that the ends incurrent configuration have come closer as compared to virtual configuration,but no guaranteed comparison can be made with the initial configuration. Atrophy γ < • A and M are of opposite sign if and only if λ > γ , signifying that the ends incurrent configuration have moved apart as compared to virtual configuration,but no guaranteed comparison can be made with the initial configuration. • A and M are of same if and only if λ < γ , signifying that the ends in cur-rent configuration have come closer as compared to both initial and virtualconfigurations.For a rod with n -fold helical symmetry with growth law assumed to be symmetry preserv-ing, these results reveal an interesting interplay between chiralities in microstructure andload response of the rod. But for a hemitropic rod, the growth law allowing cross-sectionto rotate makes the guided-guided problem similar to a non-growing rod subject to aaxial twist at one end while the other end is free to move axially. And the results abovedirectly reflect the twist-extension type Poisson effect expected in hemitropic rods.20 ase of Isotropy A = 0In this case, the solutions have n = 0 with γ = λ = 1 − ζ M B + 1 . (115)A growing isotropic rod has an out-of-plane solution with sign of ζ opposite to that of M . But for a decaying isotropic rod, (115) guarantees an out-of-plane solution only if |M| < B + 12 a , and hence such solutions exist only up to the first few modes (for thechosen perturbation), with sign of ζ same as that of M .For small A = 0, the solution is close (in terms of γ and λ ) to that of the isotropic casewith B , g ( · ), M and ζ kept same. In addition, the chirality of these solutions are sameas that of the corresponding isotropic case. With A = 0, growing rods admit ζ M > ζ M < A + > A − <
0, such that A + + A − =0, everything else being kept same. Then one of these cases gives a solution where endscome closer, while the ends move apart in the other case (comparisons made here arewith respect to the virtual configuration). Let λ + and λ − denote the respective solutionsfor A + and A − , whereas λ o denotes the same for the isotropic case. While λ o may liebetween λ + and λ − , it is also a possibility that both λ + and λ − might lie on the sameside of λ o (Figure 5), thus suggesting that no definitive comment can be made on this. In this work we study the growth of slender elastic rods with chiral material symmetries–transverse hemitropy and multi-fold dihedral helical symmetry. Based on the intuitivenotion that rods with helical symmetry should twist during growth, we propose a homo-geneous growth law that allows for relative rotation of cross-sections. A guided-guidedrod set-up is considered to illustrate the occurrence of out-of-plane buckling at certainstages of growth (or atrophy). These solutions obtained are flip symmetric and chiral innature. A complete mirroring of the rod, including both growth and constitutive proper-ties gives a solution with opposite chirality, under the same deformation. We show thatthe end-to-end distance at bifurcation modes for the isotropic case need not lie betweenthose for rods of opposite material chiralities, with rest of the elastic and growth proper-ties kept same. End-to-end distance for different combinations of growth (atrophy) andmaterial chiralities have also been examined to understand the effect of twisting growthon the constitutive twist-extension coupling.Embedding our biologically active (growth or atrophy) chiral rod set-up in an elas-tomeric matrix and introducing inhomogeneities similar to (Almet et al., 2019), can bean interesting direction to explore. One can also consider a ply of biologically active rods,like growing bi-rods in (Lessinnes et al., 2017), to study the effect of growth and materialchiralities of individual rods on the total deformation.21 cknowledgements
Prashant Saxena acknowledges the support of startup funds provided by the James WattSchool of Engineering at the University of Glasgow.
A The Growth Map
Λ and Ω are skew-symmetric
Since W ( S ) ∈ SO (3), we get WW T = I = W T W (116a)= ⇒ ∂ W ∂S W T + W ∂ ( W T ) ∂S = O = ∂ ( W T ) ∂S W + W T ∂ W ∂S (116b)= ⇒ ∂ W ∂S W T | {z } = Ω + W (cid:18) ∂ W ∂S (cid:19) T | {z } = Ω T = O = (cid:18) ∂ W ∂S (cid:19) T W | {z } = Λ T + W T ∂ W ∂S | {z } = Λ . (116c)Thus Λ T = − Λ and Ω T = − Ω . Relation between a and ω We observe that W being a proper rotation must satisfy W = W ∗ , where W ∗ denotesthe cofactor of W . Now for any v ∈ E , ω × v = Ωv = WΛW T v (117a)= W ( a × W T v ) (117b)= ( W ∗ a ) × ( W ∗ W T v ) = W a × v . (117c)This implies ω = W a . Λ and Ω are constant
We start with the decomposition W ( S + h ) = Π ( h ) W ( S ) (118)Let ε = 0 be such that all the tensor fields appearing in the following calculation makesense. n Π ( h + ε ) − Π ( h ) o W ( S ) = W ( S + h + ε ) − W ( S + h ) (119a)= Π ( h ) n W ( S + ε ) − W ( S ) o . (119b)Dividing by ε and taking the limit ε → ∂ Π ( h ) ∂h W ( S ) = Π ( h ) ∂ W ( S ) ∂S , (120)22hich in turn implies Π ( h ) T ∂ Π ( h ) ∂h = Ω ( S ) . (121)Since h and S can be chosen arbitrarily, independent of each other, we conclude that Ω ( S ) is constant. Now we expand (35) as O = ∂∂S n W ( S + h ) W ( S ) − o (122a)= ∂∂S (cid:8) W ( S + h ) (cid:9) W ( S ) T − W ( S + h ) W ( S ) T ∂ W ( S ) ∂S W ( S ) T (122b)= W ( S + h ) n Λ ( S + h ) − Λ ( S ) o W ( S ) T . (122c)This implies Λ ( S + h ) = Λ ( S ) for all choices of S and h , chosen independent of eachother, which means Λ ( S ) is constant. Solving for W ( S ) This boils down to solve (37) for W ( S ). Define orthogonal tensor fields Φ := e S Λ and U := ΦW − . Then we have the following: ∂ Φ ∂S = ΛΦ = ΦΛ , (123) Φ T ∂ Φ ∂S = W T U T ∂ U ∂S W + Λ . (124)thus implying that U ( S ) is a constant equal to e S o Λ , which results in W ( S ) = e ( S − S o ) Λ . (125) B Derivation of Perturbed Equations
Perturbations ν = r ′ · d = (cid:0) λ e + ε ρ ′ (cid:1) · (cid:0) e ε Ψ e (cid:1) (126a)= (cid:0) λ e + ε ρ ′ (cid:1) · (cid:0) e + ε ψ × e + · · · (cid:1) (126b)= ( λ + ε ρ ′ · e + · · · ) . (126c) ∂ Q ∂S Q T v = ∂ Q ∂S (cid:16) v − ε ψ × v + · · · (cid:17) (127a)= ε ψ ′ × (cid:16) v − ε ψ × v + · · · (cid:17) + · · · (127b)= ε ψ ′ × v + · · · . (127c)23 R ∂S R T = ∂∂S ( QW T )( QW T ) T (128a)= (cid:16) ∂ Q ∂S W T + Q ∂ W T ∂S (cid:17) WQ T (128b)= ∂ Q ∂S Q T − QΛQ T . (128c) QΛQ T v = Q n a × ( v − ε ψ × v + · · · ) o (129a)= ( I + ε Ψ + · · · ) n a × v − ε a × ( ψ × v ) + · · · o (129b)= a × v − ε a × ( ψ × v ) + ε ψ × ( a × v ) + · · · (129c)= a × v + ε ( ψ × a ) × v + · · · . (129d) κ = axial (cid:16) ∂ R ∂S R T (cid:17) = − a + ε ( ψ ′ + a × ψ ) + · · · . (130)The perturbed strain fields are as follows: b ν = 1 γ ( λ + ε ρ ′ · e + · · · ) , (131) b κ α = ε γ ( ψ ′ + ω e × ψ ) · e α + · · · (132)and b κ = − ωγ + ε γ ψ ′ · e + · · · . (133)Thus g ( b ν ) = g (cid:16) λγ (cid:19) + 1 γ (cid:8) ε ρ ′ · e + · · · (cid:9) g ′ (cid:16) λγ (cid:17) + · · · (134)and n = g (cid:0)b ν (cid:1) + A b κ = ε γ (cid:26) g ′ (cid:16) λγ (cid:17) ρ ′ + A ψ ′ (cid:27) · e + · · · . (135) Linearization
Linear momentum d n dS = ddS (cid:20) n α Qe α + (cid:8) g ( b ν ) + A b κ (cid:9) Qe (cid:21) = ε (cid:20) η ′ α e α + 1 γ n g ′ (cid:16) λγ (cid:17) ρ ′′ + A ψ ′′ o · e e (cid:21) + · · · . (136)Equating the ε term to zero gives (85) and (87).24 ngular momentumr ′ × n = (cid:0) λ e + ε ρ ′ (cid:1) × ε (cid:20) η α e α + 1 γ n g ′ (cid:16) λγ (cid:17) ρ ′ + A ψ ′ o · e e (cid:21) (137a)= ελ e × η α e α . (137b)And d m dS = ddS (cid:20) C b κ α Qe α + h A (cid:8)b ν − (cid:9) + B b κ i Qe (cid:21) = ε γ (cid:20) C ( ψ ′′ + ω e × ψ ′ ) · e α e α + (cid:2) A { λ − γ } − Bω (cid:3) ψ ′ × e + ( A ρ ′′ + B ψ ′′ ) · e e (cid:21) . (138)Plugging these into (71) and equating the ε term to zero gives (86) and (88). Unshearability r ′ ( S ) · Q ( S ) e α = (cid:0) λ e + ε ρ ′ (cid:1) · (cid:0) e α + ε ψ × e α + · · · (cid:1) (139a)= ε (cid:0) ρ ′ − λ ψ × e (cid:1) · e α + · · · . (139b) C Solution for Perturbations
Proceeding on similar lines as that of Healey and Papadopoulos (2013), we eliminate η α to obtain a differential equation in ψ alone. Integrating (85) we get η α e α = c , (140)for some constant c ∈ span { e , e } . Having introduced the parameter ζ in (100), equation(86) transforms into ψ ′′ + ζ ψ ′ × e = γλ c × e , (141)which upon integration and application of boundary condition (90) gives γλ c × e = ψ ′ (cid:16) + 12 (cid:17) − ψ ′ (cid:16) − (cid:17) . (142)thus leading to (95). Solution for ψ Denote by y the two-component representation of ψ ′ t with respect to { e , e } and let b denote a similar representation for ψ ′ t (cid:0) + (cid:1) − ψ ′ t (cid:0) − (cid:1) . Define matrix M = (cid:18) −
11 0 (cid:19) so that (95) can be rewritten as y ′ = ζ M y + b . (143)25ssume ζ = 0 for time being. Observe that solving (143) is equivalent to solving x ′ = ζ M x , (144)so that the general solution of (143) would be given by y = x − ζ M − b , (145) b = y (cid:16) + 12 (cid:17) − y (cid:16) − (cid:17) = x (cid:16) + 12 (cid:17) − x (cid:16) − (cid:17) . (146)Thus we have general solutions for x and y given by x ( S ) = C cos( ζ S )sin( ζ S ) ! + C sin( ζ S ) − cos( ζ S ) ! , (147) y ( S ) = C cos( ζ S )sin( ζ S ) + ζ sin ζ ! + C sin( ζ S ) − ζ sin ζ − cos( ζ S ) ! , (148)where C and C are constants in R .This gives the solution for ψ as (101). Finally all trivial and non-trivial solutionsdiscussed in section 4.2 can be summarized as follows: Case-I
Assume ζ = 0. Equation (95) with boundary condition (91) invoked gives ψ ( S ) = 12 (cid:16) S − (cid:17)n ψ ′ (cid:16) + 12 (cid:17) − ψ ′ (cid:16) − (cid:17)o , (149)substituting which into (96) gives the following relation between boundary values ρ t (cid:16) + 12 (cid:17) − ρ t (cid:16) − (cid:17) = − λ n ψ ′ (cid:16) + 12 (cid:17) − ψ ′ (cid:16) − (cid:17)o × e . (150)Invoking the boundary condition (98), we imply ψ ′ (cid:16) + 12 (cid:17) = ψ ′ (cid:16) − (cid:17) , (151)thus resulting in the trivial solution ψ ( S ) = = ρ ( S ). Case-II
Assume ζ = 0. In this case, a general solution (102) is obtained, which subse-quently gives rise to the following sub-cases based on (103). • Let sin ζ = 0 with ζ = 0. This implies ζ = 2 nπ where n ∈ Z \ { } . Each suchvalue of ζ gives a solution ρ t ( S ) = C λζ − sin( ζ S ) + ( − n ζ S cos( ζ S )0 + C λζ cos( ζ S )sin( ζ S ) − ( − n ζ S + λ C C . (152)But for this to agree with (98), we require C = 0 = C and C = 0 = C ,thus leading to a trivial solution. • Let C = C with ζ = 0. This leads to non-trivial out-of-plane solutions (107),which are discussed further in section 4.3.26 Equivariance Properties of Solutions
Let F be the tensor defined by flip action – a 180-degree rotation– about e axis and Q θ denote the rotation tensor about e axis as defined in (9).For any solution (cid:16) r ( S ) , R ( S ) , n α ( S ) (cid:17) of the boundary value problem (70)-(75), thetuple (cid:16) Q θ r ( S ) , Q θ R ( S ) Q Tθ , ( Q θ ) αβ n β ( S ) (cid:17) (153)also solves the system (70)-(75) for all 0 ≤ θ < π and so does (cid:16) Fr ( − S ) , FR ( − S ) F , − F αβ n β ( − S ) (cid:17) (154)Equivalently in terms of perturbations, any solution (cid:16) ρ ( S ) , ψ ( S ) , η α ( S ) (cid:17) of the bound-ary value problem (85)-(92), generates an entire class of solutions comprising of (cid:16) Q θ ρ ( S ) , Q θ ψ ( S ) , ( Q θ ) αβ η β ( S ) (cid:17) (155)for all 0 ≤ θ < π and (cid:16) F ρ ( − S ) , F ψ ( − S ) , − F αβ η β ( − S ) (cid:17) . (156)Our boundary value problem is equivariant with respect to the action of a groupgenerated by rotations about e axis and flip about e axis.A solution is said to be flip symmetric if (cid:16) Fr ( − S ) , FR ( − S ) F , − F αβ n β ( − S ) (cid:17) = (cid:16) r ( S ) , R ( S ) , n α ( S ) (cid:17) , (157)or equivalently if the perturbations satisfy (cid:16) F ρ ( − S ) , F ψ ( − S ) , − F αβ η β ( − S ) (cid:17) = (cid:16) ρ ( S ) , ψ ( S ) , η α ( S ) (cid:17) (158)for all S ∈ (cid:2) − , + (cid:3) .These equivariance properties of solutions are explained in much greater detail in(Papadopoulos, 1999). E Calculation of λ and γ First of all, numerical values of
A, B, F and M are fixed. Inspired by the calibrationcalculations present in (Papadopoulos, 1999), for a rod of length L = 1 with circularcross section and material constant C = 1, radius r of the cross-section can be shown tobe r = 2 √ F , (159)where both r and F are dimensionless. For instance, F = 10 is equivalent to consider a1 metre rod with diameter 4 millimetres. In addition, we have the following values of ζ corresponding to different bifurcation modes ζ ∈ { ± . , ± . , ± . , ± . , ± . , · · · } . (160)27ntroduce variables x = λγ and y = γ . For a particular ζ , equations (113)-(114) requireus to solve F ln( x ) + (cid:18) A B + A ζ M ζ − B − (cid:19) x = (cid:18) A B + A ( A + ζ ) M ζ − B − (cid:19) (161)for x . Define the following solution set. S ( m, c ) := { x : ln( x ) = mx + c , x ∈ (0 , ∞ ) } . (162)We observe that, |S ( m, c ) | = m ≤
00 if m > m ) + c + 1 >
01 if m > m ) + c + 1 = 02 if m > m ) + c + 1 < , (163)where m, c ∈ R and | · | denotes the cardinality of a set. We set m = − A F (cid:18) B + 1 M ζ − B − (cid:19) and c = AF (cid:18) AB + A + ζ M ζ − B − (cid:19) . (164)Clearly m is positive only when 1 < ζ M < B . Thus if ζ and M have opposite sign(161) has a guaranteed solution. Whenever they are of same sign, the choice |M| < a guarantees a solution to (161), although there may be several other scenarios leading toa solution.Once we have a solution x o ∈ S ( m, c ), we have corresponding y o = M A ( x o − − B − M ζ − B − λ o = x o y o , γ o = 1 y o would give the complete solution (Table 1).Table 1: Sample calculation for F = 10 and ζ = 8 . M A B λ − γ − − . − . . . † − . . . . † − × −
24 0 . − . × − . × − Atrophy 10 − −
16 0 . . × − − . × − − −
16 0 . − . − . ‡ −
16 0 . − . − . ‡† , ‡ Values are really close.
Note that we sometimes we may get an absurd solution y o < M = 0 . A = − B = 0 . F = 10 when solved with ζ = 8 .
986 gives x o = 0 . y o = − . m = − . eferences Almet A.A., Byrne H.M., Maini P.K., and Moulton D.E. “Post-buckling behaviour of agrowing elastic rod”.
Journal of mathematical biology , 78(3):777–814 (2019)Ambrosi D., Ateshian G.A., Arruda E.M., Cowin S., Dumais J., Goriely A., HolzapfelG.A., Humphrey J.D., Kemkemer R., Kuhl E. et al. “Perspectives on biological growthand remodeling”.
Journal of the Mechanics and Physics of Solids , 59(4):863–883 (2011)Antman S.
Nonlinear Problems of Elasticity, 2nd Ed.
Springer-Verlag, New York (2005)Arne W., Marheineke N., Meister A., and Wegener R. “Numerical analysis of cosseratrod and string models for viscous jets in rotational spinning processes”.
MathematicalModels and Methods in Applied Sciences , 20(10):1941–1965 (2010)Chandraseker K., Mukherjee S., Paci J.T., and Schatz G.C. “An atomistic-continuumcosserat rod model of carbon nanotubes”.
Journal of the Mechanics and Physics ofSolids , 57(6):932–958 (2009)Goldstein R.E. and Goriely A. “Dynamic buckling of morphoelastic filaments”.
PhysicalReview E , 74(1):010901 (2006)Goriely A.
The mathematics and mechanics of biological growth , volume 45. Springer(2017)Goriely A. and Tabor M. “Spontaneous helix hand reversal and tendril perversion inclimbing plants”.
Physical Review Letters , 80(7):1564 (1998)Goriely A. and Tabor M. “Spontaneous rotational inversion in phycomyces”.
Physicalreview letters , 106(13):138103 (2011)Goyal S., Perkins N.C., and Lee C.L. “Nonlinear dynamics and loop formation in kirchhoffrods with implications to the mechanics of dna and cables”.
Journal of ComputationalPhysics , 209(1):371–389 (2005)Guillon T., Dumont Y., and Fourcaud T. “A new mathematical framework for modellingthe biomechanics of growing trees with rod theory”.
Mathematical and ComputerModelling , 55(9-10):2061–2077 (2012)Healey T. “Material symmetry and chirality in nonlinearly elastic rods”.
Mathematicsand Mechanics of Solids , 7(4):405–420 (2002)Healey T. and Papadopoulos C. “Bifurcation of hemitropic elastic rods under axialthrust”.
Quarterly of Applied Mathematics , 71(4):729–753 (2013)Healey T.J. “A rigorous derivation of hemitropy in nonlinearly elastic rods”.
Discrete &Continuous Dynamical Systems-B , 16(1):265–282 (2011)Hoang T.M. “Influence of chirality on buckling and initial postbuckling of inextensiblerings subject to central loadings”.
International Journal of Solids and Structures ,172:97–109 (2019) 29umar A., Mukherjee S., Paci J.T., Chandraseker K., and Schatz G.C. “A rod modelfor three dimensional deformations of single-walled carbon nanotubes”.
Internationaljournal of solids and structures , 48(20):2849–2858 (2011)Lauderdale T.A. and O’Reilly O.M. “On the restrictions imposed by non-affine materialsymmetry groups for elastic rods: application to helical substructures”.
EuropeanJournal of Mechanics-A/Solids , 26(4):701–711 (2007)Lauderdale T.A. and OReilly O.M. “On transverse and rotational symmetries in elasticrods”.
Journal of Elasticity , 82(1):31 (2006)Lessinnes T., Moulton D.E., and Goriely A. “Morphoelastic rods part ii: growing birods”.
Journal of the Mechanics and Physics of Solids , 100:147–196 (2017)Luo C. and O’Reilly O.M. “On the material symmetry of elastic rods”.
Journal ofelasticity and the physical science of solids , 60(1):35–56 (2000)Manning R.S., Maddocks J.H., and Kahn J.D. “A continuum rod model of sequence-dependent dna structure”.
The Journal of chemical physics , 105(13):5626–5646 (1996)McMillen T., Goriely A. et al. “Tendril perversion in intrinsically curved rods”.
Journalof Nonlinear Science , 12(3):241–281 (2002)Miller J., Lazarus A., Audoly B., and Reis P.M. “Shapes of a suspended curly hair”.
Physical review letters , 112(6):068103 (2014)Moulton D., Lessinnes T., and Goriely A. “Morphoelastic rods. part i: A single growingelastic rod”.
Journal of the Mechanics and Physics of Solids , 61(2):398–427 (2013)Moulton D.E., Lessinnes T., and Goriely A. “Morphoelastic rods iii: Differential growthand curvature generation in elastic filaments”.
Journal of the Mechanics and Physicsof Solids , page 104022 (2020)Nuti S., Ruimi A., and Reddy J. “Modeling the dynamics of filaments for medicalapplications”.
International Journal of Non-Linear Mechanics , 66:139–148 (2014)O’Keeffe S.G., Moulton D.E., Waters S.L., and Goriely A. “Growth-induced axial buck-ling of a slender elastic filament embedded in an isotropic elastic matrix”.
InternationalJournal of Non-Linear Mechanics , 56:94–104 (2013)O’Reilly O.M.
Modeling Nonlinear Problems in the Mechanics of Strings and Rods .Springer (2017)OReilly O. and Tresierras T. “On the evolution of intrinsic curvature in rod-based modelsof growth in long slender plant stems”.
International journal of solids and structures ,48(9):1239–1247 (2011)Papadopoulos C.M.
Nonplanar buckled states of hemitropic rods . Cornell University, May(1999)Rodriguez E.K., Hoger A., and McCulloch A.D. “Stress-dependent finite growth in softelastic tissues”.
Journal of biomechanics , 27(4):455–467 (1994)30ilverberg J.L., Noar R.D., Packer M.S., Harrison M.J., Henley C.L., Cohen I., and Ger-bode S.J. “3d imaging and mechanical modeling of helical buckling in medicago trun-catula plant roots”.
Proceedings of the National Academy of Sciences , 109(42):16794–16799 (2012)Smith M. and Healey T. “Predicting the onset of dna supercoiling using a non-linearhemitropic elastic rod”.
International Journal of Non-linear Mechanics , 43(10):1020–1028 (2008)Su T., Liu J., Terwagne D., Reis P.M., and Bertoldi K. “Buckling of an elastic rod em-bedded on an elastomeric matrix: planar vs. non-planar configurations”.
Soft Matter ,10(33):6294–6302 (2014)Wada H. “Hierarchical helical order in the twisted growth of plant organs”.