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Journal of the Mechanics and Physics of Solids

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Morphoelastic rods Part II: Growing birods

Thomas Lessinnes⁎,1, Derek E. Moulton, Alain Goriely

Mathematical Institute, University of Oxford, United Kingdom

A B S T R A C T

The general problem of determining the shape and response of two attached growing elasticKirchhoff rods is considered. A description of the kinematics of the individual interacting rods isintroduced. Each rod has a given intrinsic shape and constitutive laws, and a map associatingpoints on the two rods is defined. The resulting filamentary structure, a growing birod, can beseen as a new filamentary structure. This kinematic description is used to derive the generalequilibrium equations for the shape of the rods under loads, or equivalently, for the new birod. Itis shown that, in general, the birod is not simply a Kirchhoff rod but rather, due to the internalconstraints, new effects can appear. The two-dimensional restriction is then considered explicitlyand the limit for small deformation is shown to be equivalent to the classic Timsohenko bi-metallic strip problem. A number of examples and applications are presented. In particular, theproblem of two attached rods with intrinsic helical shape and uniform growth is computed indetail and a host of new interesting solutions and bifurcations are observed.

1. Introduction

Within the wealth of structures and shapes found in the natural world, we encounter a great number of filamentary structures.These structures, such as polymers, roots, stems, axons, octopus' arms, elephant trunks, and tendrils, to name a few, can typically bedescribed by a one dimensional representation. Further, at the global level, the response of these filaments to external loads canoften be accurately modelled by a theory of continuum elastic rods, as has been demonstrated for proteins and DNA (Benham, 1979,1985; Coleman and Swigon, 2001; Hoffman et al., 2003; Manning et al., 1996; Neukirch, 2004; Neukirch et al., 2008; Neukirch et al.2008), plants (Goldstein and Goriely, 2006; Goriely and Tabor, 1998; Goriely and Neukirch, 2006; Alméras et al., 2002) and in manyproblems in physiology (Audoly and Pomeau, 2010; Gross et al., 1983; Levinson and Segev, 2010a).

The key feature of many such biological filaments is their ability to grow and remodel depending on age and external conditions,a common feature in many three-dimensional processes (Ambrosi et al. 2011; Goriely and Moulton, 2010). This continual growthand remodelling process can serve to accomplish a number of tasks, such as matching growth within an organ or an organism,moving, forming self attachments, or providing mechanical support against external forces. At the level of a rod, growth andremodelling is achieved by an increase in size, a change in reference shape, and an evolution of material properties. Whereas ageneral theory of volumetric growth is, in principle, applicable to these structures, the reduction in geometry associated by thedescription of a one-dimensional object evolving in a three-dimensional space suggests that a general framework for growing elasticrods (morphoelastic rods) offers a simple and more fruitful alternative. This effort was initiated in the first installment of this seriesof papers (Moulton et al., 2013), where it was shown that a self-contained theory of growing one-dimensional rods can be developedby integrating classical rod mechanics with the general prescription of multiplicative decomposition first proposed within the context

http://dx.doi.org/10.1016/j.jmps.2015.07.008Received 2 September 2014; Received in revised form 10 July 2015; Accepted 12 July 2015

⁎ Corresponding author.

1 Present address: EPFL, Lausanne, Switzerland.E-mail address: [email protected] (T. Lessinnes).

J. Mech. Phys. Solids 100 (2017) 147–196

Available online 14 July 20150022-5096/ © 2015 Elsevier Ltd. All rights reserved.

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of biology in Rodriguez et al. (1994). The problem is then to understand the interplay between growth, geometry, and mechanicswithin a one-dimensional theory. Since no residual stress is generated through growth in a 1D structure, interesting effects onlyappear when the rod is constrained geometrically. In Moulton et al. (2013), the authors considered constraining a single rod, boththrough elastic attachment to a substrate, and by considering growth in a closed ring geometry. In this second installment, weconsider the natural generalisation of these problems where the growth of a rod is constrained by another (possibly growing) rod.

The problem of two growing rods forms the first step in understanding the behaviour of multiple interacting filaments, a problemof great generality and applicability, as many biological filaments are composed of sub-filaments bundled together. For instance, cellsin roots are often organised in files that wind around each other helically around a main core. It is the delicate mechanical balancebetween geometry and mechanical stress that dictates the shape and functional properties of the roots. At the microscopic level,plant cells are also built from multiple interacting cellulose microfibrils that wind-up along the cells (Lloyd and Chan, 2002). Again,it is the organisation and interaction of these constituents that dictates the overall response (Buschmann et al., 2009; Furutani et al.,2000). Flagella are another example of well-studied filaments whose structure and response derive from internal coupled sub-filaments organised in the near universal 7+2 pattern (seven microtubules on a circle with two internal microtubules) (Calladine,1982; Kamiya et al. 1979; Srigiriraju and Powers, 2005; Gadêlha et al., 2013). Other examples include the microtubules of axons(Stephen and Peter, 2012), the triple helix arrangement of arteries and veins in umbilical cord (Lacro et al. 1987; Goriely, 2004), thecollagen fibers in arteries (Goriely and Vandiver, 2010; Holzapfel and Ogden 2010), and the muscle groups of octopus arms(Levinson and Segev, 2010b) and elephant trunks (Kier and Smith, 1985). Typically in these systems the different sub-filamentsgrow (or equivalently contract by muscle action) and remodel at different rates and therefore develop different unstressedconfigurations that are incompatible with the geometry of the structure. This incompatibility results both in residual stresses in thesub-filaments and in equilibrium configurations of the global structure in which the sub-filaments undergo large displacements.These two correlated phenomena illustrate how macro-filaments of complex shapes and mechanical properties are produced fromcomparatively simpler sub-filaments.

A general theory for the mechanics of interacting rods has been proposed in Moakher and Maddocks (2005), where the word“birod” was coined to refer to ‘two elastically interacting elastic strands’. This structure was motivated by the study of DNA.Similarly, a number of authors have looked at the mechanics of braids and plies by considering the interaction of two rods infrictionless contact (as typically occurs during the formation of plectoneme structures in the folding of a filament on itself)(Thompson et al., 2020; Starostin and van der Heijden, 2014; Neukirch and van der Heijden, 2002; Chamekh et al., 2009; Olsen andBohr, 2012; Prior 2014). The main difference with a birod is that the two strands of the birod interact elastically while in a ply orbraid the two strands are assumed to have a continuous zone of hard contact and are allowed to slide along one another. Here westudy a different version of the birod where the two rods are “glued” together, this is similar to the original birod (Moakher andMaddocks, 2005) as we assume that the strands are not allowed to slide (see also the recent study Dias and Audoly, 2013 of twostrips connected along one edge) and that there might be an elastic response to the change of angle between the tangents to thestrands. However we also assume that like a ply (Starostin and van der Heijden, 2014), the distance between the centrelines of thestrands is constant (a rigid constraint). The present study is therefore well suited to encompass birods for which the attachmentmechanism is much stiffer than the elastic strands. In this sense it can be viewed as a limit case of the original birod theory.Interestingly, these assumptions motivate the introduction of a new geometric representation of the structure together with a novelsystem of balance equations between ensuing generalised stresses.

The paper is organised as follows: In Section 2, we summarise the formalism and findings of part I (Moulton et al., 2013). Section3 develops the kinematic description necessary for the birod and a derivation of the equilibrium equations for the birod. Under ourgeometrical assumptions, the birod has the same number of degrees of freedom as a single unshearable rod. We show that it ispossible to define a single centreline and an orthogonal local frame for the birod. A natural question is whether a birod is equivalentto a Kirchhoff rod in the sense that there exists a definition of stresses and strains satisfying the Kirchhoff equation for a single rodwith appropriate constitutive laws. We show that the answer to this question depends on the dimension of the embedding space. In1D (rod on the line) and 2D (a planar rod), the equilibrium equations are equivalent to Kirchhoff equations for adequately definedglobal stresses. Furthermore, in the limit of small deformations, the planar case is equivalent to the classical 1925 Timoshenko bi-metallic strip problem (Timoshenko, 1925). In 3D, in general the equilibrium equations differ from a Kirchhoff-like rod. In Sections4 and 5 we consider the one- and two-dimensional cases, computing explicitly the unloaded shape and constitutive laws. The 1Dexamples are used to develop intuition, then the 2D case opens the door to modelling a vast range of applications. Finally, the three-dimensional case is studied in Section 6 in full generality. Because the equilibrium equations are now different from Kirchhoffequations, the standard geometric description – in terms of a centreline and orthonormal frame of directors – is no longer adequate,and an alternative geometric approach is proposed. It naturally leads to a different set of global mechanical quantities and to a newset of equilibrium equations. Once again, the unloaded shape of the birod is computed as well as its constitutive laws, and the theoryis illustrated with a few examples. The case where the material properties of both the strands and the glue are homogeneous is ofparticular importance and we study it in detail in Section 7. Extensions and conclusions are discussed in Section 8.

2. A single morphoelastic rod

In this section, we briefly summarise the notations and concepts introduced in Moulton et al. (2013), to describe the kinematicsand geometry of a growing elastic rod.

T. Lessinnes et al. J. Mech. Phys. Solids 100 (2017) 147–196

148

2.1. Kinematics

2.1.1. Geometry of curvesA dynamical space curve R Tr( , ) is a continuously differentiable function of a material parameter R and time T with values in 3:

A B R T R Tr r: [ , ] × → : ( , ) → ( , ),3 (2.1)

where A and B denote the ends of the curve. At any time T, the arc length s is defined as

∫sσ

σ T σr= ∂∂

( , ) d .A

R

(2.2)

To complete the geometrical description of a curve, we attach a local orthonormal basis (or frame). In part I (Moulton et al.,2013), we used the standard Frenet frame. Here we complete this picture by introducing the natural frame, as first described inBishop 1975. To define the natural frame, we parametrise the curve by its arc length s, assume that the function s Tr( , ) is C2 withrespect to its first variable, and set A=0 without loss of generality (assuming finite length). We define c =

sr

3∂∂ as the unit tangent

vector to the curve. At s=0, Tc (0, )3 is completed by Tc (0, )1 and Tc (0, )2 so that T T Tc c c{ (0, ), (0, ), (0, )}1 2 3 forms a right handedorthonormal frame. This frame is defined up to an arbitrary phase and is transported along the curve by solving

sc

p c∂∂

= × ,ii (2.3)

where p is defined by

sp c

c≔ ×

∂∂

.33

(2.4)

The frame s T s T s Tc c c{ ( , ), ( , ), ( , )}1 2 3 is a natural frame of the curve r. As s spans l[0, ] (l being the total length of the curve), rspans the curve and the natural frame rotates about the direction p at a rate p| | per unit s. Note that the definition (2.4) does notdepend on the choice of phase. Hence, p is a property of the curve itself and all natural frames have the same p. Furthermore, p iseverywhere orthogonal to c3 and can be written as

p pp c c= + .1 1 2 2 (2.5)

The pair p p( , )1 2 is called the normal development of the curve r (Bishop 1975). As s spans l[0, ], the normal development draws a

curve in 2. Although this new curve is defined up to a global rotation about the origin of 2 – corresponding to the arbitrary rotationin the definition of the frame at s=0 – its shape provides much information about the geometry of the three-dimensional curve r. Thenormal development is to normal frames what curvature and torsion are to the Frenet frame. Indeed, p1 (resp. p2) specifies thebending of the curve about c1 (resp. c2). After parameterisation of the normal development in polar coordinates p θ( , ), the curvature κand torsion τ of r are recovered as

κ p p p= = + ,12

22

(2.6)

θpp

= arctan ,2

1 (2.7)

τ θs

p p p p

p p= ∂

∂=

∂ − ∂+

.s s1 2 2 1

12

22

(2.8)

2.1.2. Kinematics of unshearable Kirchhoff rodsAn elastic rod is geometrically defined by a dynamic curve called the centreline R Tr( , ), equipped with a local frame of

orthonormal vectors R T R T R Td d d{ ( , ), ( , ), ( , )}1 2 3 called the frame of directors. The role of this material frame is to keep track of theorientation of the cross section at every point along the centreline. By convention, R Td ( , )3 is normal to the cross section at R, while d1and d2 are tangent to it at the point contained in the centreline. A rod is unshearable if its cross-sections remain perpendicular to itscentreline in all configurations so that d c=3 3. This is the case considered in this paper and no further mention of unshearabilityshall be required. The definition of the frame of directors is completed by specifying d1 as a material direction corresponding to theorientation of the cross section (e.g. along the width of the ribbon in Fig. 1). By orthonormality of the frame of directors, d d d= ×2 3 1.The components of any vector a d d d= a + a + a1 1 2 2 3 3 in this particular frame are denoted in the sans-serif fonts by the triplets(a , a , a )1 2 3 .

A configuration of a rod can be characterised (up to global translations and rotations) by two local quantities (four degrees offreedom): the stretch v and the Darboux vector u of the frame of directors. These quantities are defined (implicitly) by the followingrelations:

Rv

Rir d

du d∂

∂= ;

∂∂

= × ( = 1, 2, 3).ii3 (2.9)


149

Hence v = sR

∂∂ characterises the increase or decrease in arc length and u describes the rotation of the frame of directors: the frame

rotates about axis u u/| | by an amount Ru| |d over an infinitesimal displacement Rd . Once the functions v and u are specified, theconfiguration of the rod can be recovered, up to a global rigid body motion, by integrating (2.9). The curvature and torsion of thecentreline can also be recovered from u, giving (see Appendix A)

⎛⎝⎜⎜

⎞⎠⎟⎟κ

vτ

v= 1 u + u and = 1 u +

u ∂ u − u ∂ uu + u

.R R12

22

31 2 2 1

12

22

(2.10)

Furthermore, u1 and u2 encodes the rate of bending of the centreline about d1 and d2 per unit R, while u3 encodes the twist of the rod.Note that the notion of a configuration does not involve motion, it is a purely geometric statement. The motion is described by

taking a derivative of r by time T, and through the spin vector w implicitly defined by

Ti

dw d

∂∂

= × , = 1, 2, 3.ii (2.11)

Introducing the matrix D d d d= ( )1 2 3 , the relations (2.9) and (2.11) can be written in matrix form D DU∂ =R ; D DW∂ =T , with

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟U W=

0 −u uu 0 −u

−u u 0and =

0 −w ww 0 −w

−w w 0.

3 2

3 1

2 1

3 2

3 1

2 1 (2.12)

This form is useful as it allows one to define the Darboux and spin matrices explicitly

R TU D D W D D= ∂

∂and = ∂

∂.T T

(2.13)

It is important to note that when modelling an actual slender structure as a Kirchhoff rod, a number of choices have to be made.First, the centreline has to be chosen. For a free rod, one may choose to define it through the centre of mass of the sections. Withexternal attachments or constraints, however, alternative choices may work better. Similarly, a choice of orientation for the section-wise directors d d{ , }1 2 is required. In most cases, the directors will be chosen to lie along the principal axes of the distribution ofYoung's modulus in the rod (if the sections are constitutively uniform, this amounts to the principal axes of the second moments ofarea).

2.1.3. Kinematics of growing rodsWe now turn our attention to the kinematics of growing rods, for which three special configurations play important roles: the

initial configuration 0, the reference configuration and the current configuration .The reference configuration of an elastic rod is the configuration it assumes when stress free. The stretch v and Darboux vector u

corresponding to this configuration are denoted with a hat to emphasise their special status. It is often convenient to parameteriserods by their arc length in this state and we denote this quantity with an upper case S. Technically, we can designate one tip of therod as the origin. Then S is defined as the (one-to-one) map between material points of the centreline and the arc length from theorigin to the material point in the reference configuration. With this parameterisation, the reference state by definition has theproperty v = 1 and is completely described by the definition of its Darboux vector u.

To proceed, we must clarify the notion of growth and maps evolving in time. We define a morphoelastic rod as an elastic rod

Fig. 1. A rod (green online) with its centreline (black online) and local material frame drawn at a few points along the centreline: d1 (red online), d2 (blue online) and

d3 (black online). (For interpretation of the references to colour in this figure caption, the reader is referred to the web version of this paper.)


150

whose reference configuration evolves on a growth time scale t that is slow in comparison to the time scale associated with themechanical motion of the rod, associated with the time T. Only the latter gives rise to inertial terms in the balance of momenta, whilethe former changes the properties of the rod in quasi-static fashion.

Having defined the slow (growth) time scale t, the initial configuration is the reference configuration at time t=0. All quantities inthis state are denoted by a subscript 0. In particular, we define S0 as the (one-to-one) map between material points of the centrelineand their reference arc-length in the initial configuration. Inverting this mapping and composing it with S leads to the definition of aone-to-one map Γ so that S Γ S t= ( , )0 . With these definitions, the addition (or resorption) of new matter along the centreline iscaptured by the growth factor γ:

γ S t ΓS

SS

( , ) = ∂∂

= ∂∂

,00 0 (2.14)

where the second equality consists of a benign abuse of notations that amounts to identifying material points and theirparameterisation by the associated arc-length in the initial configuration whereby S becomes a function of S0 and t.

At any time t, the current configuration is defined as the shape assumed by the rod under the actual constraints and loads thatare applied to it. Naturally, the material points of the centreline can also be parameterised by the corresponding current arc-length s.We then define the elastic stretch of the rod as α = s

S∂∂ and its total stretch by λ αγ= =s

S∂

∂ 0. While α is a mechanical deformation, γ is

the growth stretch as it encodes the elongation of the reference configuration purely due to growth. The total stretch λ containsinformation about both stretches. While λ is typically the easiest parameter to access experimentally, a mechanical description of theprocess requires separating growth from mechanical stretch.

Note the subtle difference between the geometric stretch v = sR

∂∂ defined in (2.9) for any parameterisation R of r and the elastic

stretch α = sS

∂∂ . Both quantities are equal whenever the morphoelastic rod is parameterised by its reference arc-length. Similarly,

when studying morphoelastic rods, and unless otherwise specified, we always assume that u refers to the Darboux vector of the rodparameterised by its reference arc-length S, i.e. u is defined by

Sd

u d∂∂

= × .ii

In particular, considering the slow growth time t, a change in γ captures the elongation of the reference configuration as the rodgrows, while a change in u captures a change of reference shape.

2.2. Mechanics of morphoelastic rods

To describe the mechanics of rods, the geometric description of Section 2.1 must be completed by information regarding massdistribution in the cross sections. Following Antman (2005), we define the linear density of mass ρA( ) and second moment of massinertia ρI( )1,2 per unit R along the centreline R Tr( , ) as

∫ρA S ρ S x x x x( )( ) = ( , , ) d d ,1 2 1 2 (2.15)

∫ ∫ρI S ρ S x x x x x ρI S ρ S x x x x x( ) ( ) = ( , , ) d d , ( ) ( ) = ( , , ) d d ,1 1 2 22

1 2 2 1 2 12

1 2 (2.16)

where is the section at S with points labelled x x( , )1 2 according to their location at x xr d d+ +1 1 2 2, and ρ is the mass density per unitstress-free volume.

Finally, we define the resultant force R Tn( , ) and resultant moment of forces R Tm( , ) about R Tr( , ) applied by the material atR R′ > on the material at R R″ < . The balance of momenta yields (Antman, 2005)

RρA

T R RρI

TρI

Tn f r m r n l d

dd

d∂∂

+ = ( ) ∂∂

, ∂∂

+ ∂∂

× + = ( ) ×∂∂

+ ( ) ×∂∂

,2

2 2 1

212 1 2

222 (2.17)

where f and l are the external densities of forces and moments about S Tr( , ) applied on the rod per unit of R.

2.3. Constitutive laws

The system is completed with constitutive laws. We refer to part I (Moulton et al., 2013) for a more detailed discussion. Here, werecall the assumption that there exist functions K and k relating the four strains u u− and v v− to the four stresses that deliverwork under their respective virtual displacements:

v vm K u u= ( − , − ), (2.18)

k v vu un = ( − , − ).3 (2.19)

By definition of the reference configuration, it is required that K 0 0( , 0) = and k 0( , 0) = 0 and that 0( , 0) is the only simultaneousminimum of these functions. We assume that both functions derive from a convex and coercive energy-density function W vu( , ) thatrelates the strains u and v to the line density of mechanical energy stored in the rod with a unique minimum at vu( , ).


151

2.3.1. Diagonal constitutive lawsIt is often assumed that the tension depends solely on the stretch and the bending moment depends only on the curvature and

twist, i.e. k v vn = ( − )3 and m K u u= ( − ), in which case the constitutive laws are said to be diagonal. Because rods are slenderbodies and provided that δu u| − |⪡ where δ1/ is the characteristic radius of the cross-section, a first order development of K about uis found to be an excellent approximation for many problems. In this case, we have explicitly

EI EI μJm d d d= ( ) (u − u ) + ( ) (u − u ) + ( ) (u − u ) ,1 1 1 1 2 2 2 2 3 3 3 (2.20)

where the parameter μJ( ) depends both on the shear modulus and the shape of the cross-section (Goriely et al., 2001; Nizette andGoriely, 1999; Goriely and Tabor, 2000; Antman, 2005) and

∫ ∫EI S E S x x x x x EI S E S x x x x x( ) ( ) = ( , , ) d d , ( ) ( ) = ( , , ) d d ,1 1 2 22

1 2 2 1 2 12

1 2 (2.21)

where E is Young's modulus at S x xr d d( ) + +1 1 2 2 and where the centreline has been defined such that

∫ ∫E S x x x x x E S x x x x x( , , ) d d = 0 and ( , , ) d d = 0.1 2 1 1 2 1 2 2 1 2 (2.22)

Furthermore, the constitutive law (2.20) assumes that d d( , )1 2 correspond to the principal directions of the distribution of E in eachsection. Also, note that if E is constant in the section then EI EI( ) =1 1 and EI EI( ) =2 2 where I1 and I2 are the second moment of areawith respect to the axis along d1 and d2 respectively and going through the point r.

If the stretch2 is small, that is v v v| − |⪡ , then a linear constitutive law for the tension is also found to be a suitable approximationin many situations (but higher stretches may require more refined constitutive laws, especially for biological tissues). In this case, theconstitutive laws (2.18) and (2.19) become

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟v v

EIEI

μJE

m u un = −

−where =

( ) 0 0 00 ( ) 0 00 0 ( ) 00 0 0 ( )

3

1

2

(2.23)

and E( ) is defined as

∫E E S x x x x( ) = ( , , ) d d .1 2 1 2 (2.24)

Note that if Young's modulus E is constant over a section, Eq. (2.24) reduces to E E( ) = where is the area of the cross sectionof the rod.

2.3.2. Inextensible rodsIt is sometimes useful to assume that the rod's elongation under tensile forces is negligible. In those cases, the rod is said to be

inextensible: the arc length between any two points R1 and R2 on the centreline is the same in all configurations and by definition:v v= (equal to unity if R is chosen to be the reference arc-length). In such a case, the tensile stress n3 is no longer knownconstitutively and the constitutive laws reduce to (2.18) alone.

3. Morphoelastic birods: kinematics and mechanics

Our aim is to develop a general framework to study the mechanics of morphoelastic birods. In this section, we introduce anumber of conventions and definitions for the rest of the paper. In the next sections we consider the cases of birods embedded in 1D(Section 4) and 2D (Section 5) Euclidian spaces. The reduced dimensionality is instructive for the theory but is also applicable toseveral applications. We study the more general case of 3D birods in Section 6.

3.1. Geometry of two attached morphoelastic rods

As briefly outlined in the Introduction, a birod consists of two morphoelastic rods, called sub-rods, which are attached to one-another along their length. We label the sub-rods by a sign ‘+’ or ‘−’ and all variables relating to either rod acquire the correspondingsuperscript. For instance, the sub-rods are parameterised by their reference arc-length S+ and S−. The attachment between the sub-rods is not trivial and needs to be fully specified.

3.1.1. Single contact pointWe consider first the simplest case of two non-growing sub-rods with circular cross-sections of constant radii a±. The

construction of a birod is shown schematically in Fig. 2. Define a curve on the surface of each sub-rod (Fig. 2(a)), and then imaginetwisting the rods so that the curves are aligned (Fig. 2(b)). The next step is to “glue” the rods together, which requires defining agluing map between material points of each rod (Fig. 2(c)). In the absence of external loads, the glued birod (Fig. 2(d)) relaxes to its

2 Note we use the term stretch to refer to either axial extension or compression.


152

reference state (Fig. 2(e)).Mathematically, this procedure defines unambiguously a contact line (the green line in Fig. 2(e)), this is a material curve c c=+ −

on the surface of both sub-rods that can be written as

L S a φ S φ Sc r d d: [0, ] → : → + [cos ( ) + sin ( ) ].± ± 3 ± ± ± ± ±1± ± ±

2± (3.1)

The functions φ± are the polar angles of the contact point in the local frame (d1±, d2

±) of its cross section. The functions φ± can bedefined in any configuration of the birod and are invariant under any elastic deformations.

The “gluing map” is a bijection between material points of the sub-rods, i.e. a map

G L L S S Sc c: [0, ] → [0, ]: → = ( ) ( ( )).− + − + + −1 − − (3.2)

Under our assumptions, G is invariant under any elastic deformation of the birod. However, as we will see in the next section, it canbe modified when the sub-rods undergo growth.

3.1.2. General geometric definition of a morphoelastic birodFor sub-rods with general cross sections, the definition of a contact line is not always straightforward. Indeed, in many cases such

as polygonal cross sections, the rods actually have a contact surface. Yet in other cases, there is a layer of “glue” or an attachmentmechanism and the sub-rods have actually no contact points at all (see Example 3 in Section 3.1.3). These defining problems can beavoided by specifying the geometry of the attachment between the two sub-rods in terms of their relative position in some preferredconfiguration. Specific attachment mechanisms will then be modelled by providing adequate constitutive laws for the “glue”.

We start with two elastic sub-rods labelled by + and −, with lengths L0± and arclength parameters S0

± in an initial configuration.Each sub-rod is geometrically described by a space curve r± and a director frame d d d{ , , }1

±2±

3± , with d3

± along the tangent of theirrespective curves. A morphoelastic birod is defined by the two sub-rods and an attachment, which we assume can be characterisedby the following:

1. A gluing function G0: that is, a continuous and monotonic map G L L: [0, ] → [0, ]0 0−

0+ such that

G S S( ) = .0 0−

0+ (3.3)

2. The distance constraint: the distance between paired points on the two sub-rods is uniform. That is a S G Sr r= | ( ) − ( ( ))|0−

0− +

0 0− 3 is

independent of S0−.

3. The polar angle φ L: [0, ] →0−

0− of the vector r r−+ − in the local frame d d( , )1

−2−

Fig. 2. (a) The sub-rods in their reference states with the (material) contact line (green online). (b→d) A sequence of configurations showing the steps leading to abirod in a current configuration (e). Depending on the external loads, the birod of (d) could relax to elastically deform to a shape similar to (e). (For interpretation ofthe references to colour in this figure caption, the reader is referred to the web version of this paper.)

3 In most cases, we will try to avoid such cumbersome notations. Quantities such as S G S G Sr r r r( ) − ( ( )) = ( − ○ )( )−0− +

0− − +

0 0− will simply be written as r r−− + where

the composition by G0 at the adequate place is implicit.


153

4. The polar angle φ L: [0, ] →0+

0+ of the vector r r−+ − in the local frame d d( , )1

+2+ .

These three maps and the distance constraint can evolve in the quasi-static evolution of the system (the slow growth time t) butthey are not the result of a mechanical balance. Rather, they are kinematic constraints imposed on the two sub-rods. In the context ofgrowth, t can be thought of as the time and we will talk of growing rods even though the kinematic description is valid for any otherprocess that changes the intrinsic geometry or material parameters in time. The distance constraint is of great use in themathematical treatment; while it reduces the generality of the approach, the system is still sufficiently general to model manysystems with analytical treatment.

The growth of each sub-rod is specified by the growth maps Γ± (cf. Section 2.1.3) so that their reference arc-lengths S± obey

S Γ S t S Γ S t= ( , ). = ( , ).− −0− + +

0+ (3.4)

These growth functions define an evolving gluing function G L L: [0, ] → [0, ]− + such that

G S Γ G Γ S( ) = ( ∘ ∘ ( ) ) = .− +0

− −1 + (3.5)

By analogy with the definition (2.14) of a growth stretch γ for a single morphoelastic rod, we define the relative growth stretch g ofthe “+” rod with respect to the “−” rod as

g GS

= ∂∂

.− (3.6)

There is an interesting and important equivalence in the growth of birods. The gluing function G0 specifies the sections of the twosub-rods that are connected. Each reference configuration can grow independently while remaining attached. An equivalentviewpoint is to consider two rods free to grow up to a time t and attach them by a gluing function at that time. Therefore, theattachment and growth processes commute in the sense that the mechanical equilibrium in the current configuration is independentof the order in which these two basic processes take place: rods can be glued then grown or, equivalently, first grown and then glued.In this context the growth stretch g really represents the differential growth between the two rods.

The polar angle maps φ0± can also evolve under growth

φ S t φ S ϕ S t( , ) = ( ) + ( , ),± −0± − ± −

(3.7)

where the functions ϕ± allow for an active change of the direction of the attachment as growth and remodelling occurs. The polarangle φ S t( , )± − is an important quantity that will require further study when we model the process of mechanical attachment. Notethat for systems where there is no active reorientation of the attachment, ϕ = 0± and the maps φ± are fixed.

We also let the distance constraint evolve in time t, however while remaining uniform. That is, the distance a t r r( ) = | − |− + isindependent of position.

Note that our definition of a birod implies that the point at r+ lies in the plane spanned by d d( , )1−

2− . Similarly, the point r− lies in

the plane spanned by d d( , )1+

2+ .

3.1.3. ExamplesTo elucidate the role and practical use of the growth function, we briefly outline 3 examples of problems whose geometry can be

well modelled within our framework.

Example 1 (Timoshenko's bi-metallic strip). Consider two initially flat thermo-elastic strips bonded together in their unstressed

Fig. 3. Bi-metallic strips can be modelled within the framework of morphoelastic birods by considering the deformation of the centrelines of each strip as they arebonded through welding and deform by thermal expansion.


154

configuration, each susceptible to thermal expansion and only allowed to deform in the plane perpendicular to their width(Fig. 3(a)). This classic problem was first described by Timoshenko (1925). We can model the deformation of these bonded strips asa birod evolving in the plane with trivial polar angle φ = π±

2 and distance a2 taken as the sum of the strip thickness. We identify thetemperature t0 as the temperature at which both plates of initial length L0 are flat and initially bonded. We assume that the stripshave different thermal expansion coefficients c+ and c−. Hence, when the plates are brought to a new temperature t (such thatt t t| − |⪡0 0), their reference lengths become L L c t t= [1 + ( − )]±

0±

0 and we compute

g t t LL

c t tc t t

( − ) = =1 + ( − )1 + ( − )

.0

+

−

+0

−0 (3.8)

We will return to this example in Section 5.3, where we compute the mechanical equilibrium of a bi-metallic strip and recoverTimoshenko's result in the limit of small deflections.

Example 2 (Uniform growth in 3D). A simple example in 3D is to consider initially two rods of circular cross section and diametera but of different lengths such that the gluing function and one of the polar angle functions are linear. Then we have (Fig. 4 )

S G SLL

S φ φ ωS a a= ( ) = , = 0, = , = .0+

0 0− 0

+

0− 0

−0−

0+

0−

0(3.9)

In this case, the mechanical structure obtained by the attachment is equivalent to a structure where the “+” rod is first twisted so thatthe helical curve traced by the polar angle φ0

+ becomes a straight line, then compressed (or extended) uniformly by an amount

g1/ = LL0

0−

0+ , and then glued to the “−” rod. In the case where ω = 0, so that no twisting is required, the structure might be thought of as

an idealised model of a plant with internal filaments growing at different rates. A natural question is whether the resulting structureforms a planar or non-planar (e.g. helical) shape. We consider this explicitly in Section 7.2.

Example 3 (A ladder with rigid spokes). The interesting design proposed in Lachenal et al. (2012) can also be modelled within theframework of (non-growing) birods. In this system, two elastic flanges are connected by a number of rigid spokes (see Fig. 5). Thespokes insure that the flanges stay at a constant distance a from one another. Since the length of the spokes is much larger than thethickness of the flanges, the structure may adopt shapes for which au| | > 1± while still satisfying the condition a u| |⪡1± ± so that eachflange can be treated as a Kirchhoff rod. This particular structure leads to a number of interesting non-trivial mechanical equilibriastudied in detail (Lachenal et al., 2012). We will show how these solutions can also be recovered within our framework in Section 7.3.

3.1.4. The birod as one slender objectWhen attaching two rods to create a birod, the resulting object is also a slender structure. It is therefore a natural question to ask

whether its mechanical response is equivalent to that of a single elastic Kirchhoff rod. Such an upscaling requires two elements: firstthe definition of a single centreline and frame of directors, and second, the mechanical balance of the birod as a single structure. Weconsider the definition of a birod centreline and frame of directors in this section, and defer consideration of forces and moments tothe subsequent section.

Consider the birod in Fig. 6. We define the centreline of the birod as a curve lying between the centrelines of the two sub-rods,that is

β βr r r= (1 − ) + ,− + (3.10)

where β ∈ [0, 1] is an arbitrary constant.A note regarding parameterisation: With two rods that are each independently growing and changing arclength, we are

confronted with numerous choices for how to parameterise any given curve/vector. The key point is that all maps are one-to-one, so

Fig. 4. Uniform growth of a twisted rod. Here one rod is grown uniformly and attached along a helical path to a second rod. This is equivalent to specifying the initiallength of both rods, and linear gluing and polar angle functions.


155

for instance while the curve above could easily be parameterised by S−, since S G S= ( )+ − , it could also be parameterised by S+ usingthe inverse of G, or S0

+ for that matter. Parameterisation is irrelevant when referring to the curve itself. It is only when consideringreference states, and defining quantities such as elastic stretch – which requires taking a derivative with respect to a referencearclength – that care must be taken. For notational simplicity, we try to avoid making a particular choice of parameterisation exceptwhen needed.

Along these lines, it is important to define a reference arc-length of the birod, which we do below, but we remark that the birodcentreline could be written in any of the parameterisations.

Let us assume the existence of a reference configuration for the birod (the exact definition of which depends on the dimension ofthe embedding space and is therefore postponed to Sections 4–6). The arc length along the curve r in this configuration is called thereference arc-length of the birod S. This defines a bijective map G L L: [0, ] → [0, ]− − between S and S− defined by

∫G S Sσ tσ

σr

( ) : → =∂ ( , )

∂d ,

S− −1 −

0

ref−

(3.11)

where rref is the centreline of the birod in its reference state (defined up to a rigid body motion). The mapping G+ from the birod tothe + rod is obtained by

G G G= ∘ .+ − (3.12)

Fig. 5. Left: An interesting structure can be generated by taking two rods with intrinsic curvature (flanges) and constraining their distances (spokes). Right: Theassembled structure is a ladder that can be modelled as a continuous birod as explained in the main text.

Fig. 6. Details of the centrelines and director bases in a birod. The left (blue online) and right (red online) curves are the centrelines of the sub-rods shown with theirlocal frames. Note the line along d1 connecting one point r− to its associated point r+ in the + rod. The middle vertical curve (yellow online) is the centreline of the

birod shown with its own frame. (For interpretation of the references to colour in this figure caption, the reader is referred to the web version of this paper.)


156

It is useful to define the growth stretch of each rod with respect to the centreline as

g GS

= ∂∂

.±±

(3.13)

Any scalar function χ∼ of S− or S+ can be expressed as a function χ of S by the composition

χ χ G= ∘ ,∼ ± (3.14)

and by application of the chain rule to (3.14), we have

χS

g χS

∂∂

= ∂∂

.∼

±± (3.15)

Also, differentiating (3.12), we obtain

g SS

gg

= ∂∂

= .+

−

+

−

It is important to note that while the map G is prescribed geometrically (Section 3.1.2), the maps G± require the definition of areference state. This reference state is the solution of a mechanical problem, namely, determining the shape of the birod in theabsence of loads, a computation treated explicitly in Sections 4–6.

We can now define the frame of directors attached to the birod. As before, we start by defining d3 as the unit tangent to r and wedefine d1 as the unit vector pointing from r− to r+. We will only consider here the case where this vector is uniquely defined.4 Thisimplies the relations

a a β a βr r d r r d r d= + and similarly = + = − (1 − ) .+ −1

−1

+1 (3.16)

Since d1 is perpendicular to d3 (cf. note at the end of (3.1.2)), we now have the orthonormal birod director frame

S

S

d

r

rd r r

r rd d d=

∂∂∂∂

, = −| − |

, = × .3 1

+ −

+ − 2 3 1

(3.17)

As before, any vector Sa( ) can be written in the birod frame as Sa d d d( ) = a + a + a1 1 2 2 3 3.Having defined a birod centreline and local frame, all associated strains can be easily defined. The birod stretch α and birod

Darboux vector u are defined by

αS S Sr d

ud

= ∂∂

;∂∂

= ×∂∂

.i i

(3.18)

As we shall see, one further geometric quantity of importance is the angle θ between the tangent vectors to the centrelines of the sub-rods

θ d d d d d= Sign[ ·( × )] arccos( · ).1 3−

3+

3−

3+ (3.19)

This angle is given explicitly in terms of α and u in Appendix B.

3.2. Mechanics

Starting from the Kirchhoff equations for both sub-rods, we now derive balance equations for the birod. Each sub-rod is subjectto resultant forces n± and moments m± as defined in Section 2.2. These quantities obey the Kirchhoff equations (2.17). We focus hereon static equilibria, i.e. we omit the time-dependent term (related to inertial effects).5 The static Kirchhoff equations for the sub-rodsare

S g gn f F∂

∂+ + = 0,

±

±

±

±

±

± (3.20)

S S gc

g gm r n l d f L∂

∂+ ∂

∂× + ∓ × + = 0,

±

±

±

±±

±

±±

1

±

±

±

± (3.21)

where c a β= (1 − )+ and c aβ=− . Note that in this formulation we have decomposed the body loads into external loads F± and L± andthe internal loads f± and l± due to the interaction between the sub-rods. Specifically, gF( / )± ± and gL( / )± ± are forces and momentsabout the centrelines of either sub-rod per unit reference arc-length S±.

4 In the case where one rod is inside another rod so that their centrelines coincide, the vector d1 is not uniquely defined. In this case d1 can be chosen convenientlyeither along d1

+ or d1−.

5 For static equilibrium it might seem more natural to use non-partial derivatives. However, since a key idea in our formulation is growth – which althoughoccurring on a slow time scale, nevertheless gives quantities a quasi-static time component – we shall continue to use partial derivatives.


157

3.2.1. External loadsThe external forces gF( / )± ± and moments gL( / )± ± result in forces F± and moments L± per unit reference arc-length S of the birod.

It is therefore natural to define the resultant external force F and moment L about the centreline of the birod as

F F F= + ,+ − (3.22)

c cL L L d F d F= + + × − × .+ − +1

+ −1

− (3.23)

The last two terms in (3.23) account for the fact that F± are applied at the centrelines of the sub-rods and therefore producemoments with respect to the centreline of the birod. In general, the difference of external moments can also produce work.Therefore, we define

L L L dˇ = ( − )· .1+ −

1 (3.24)

3.2.2. Internal loadsAs a result of the attachment, internal stresses are local in the sense that the interaction between the two rods is only done

through the paired cross sections. However, a pair of attached cross sections may have multiple points of contact. Hence, in general,they exert on each other a system of forces and moments which can be reduced to a resultant force and moment about an arbitrarypoint which we choose as Sr( ).6 With this choice gf( / )± ± and gl( / )± ± are the linear densities of force and moment about r per unitreference arc-length S± applied by the ‘∓’ rod on the ‘± ’ rod. Rescaling to the unit of reference arc-length of the birod, we definef f= + and l l= + respectively as the force and moment applied by the − rod on the + rod per unit reference length of the birod. Byapplication of the principle of action-reaction, the + rod applies densities of forces f− and moments l− on the − rod.

3.2.3. Mechanics of the birod and global quantitiesThe system (3.20) and (3.21), expressed in terms of the arc-length parameter of the birod, S, is

cn f F m r n l d f L( )′ ± + = 0, ( )′ + ′ × ± − × + = 0,± ± ± ± ± ±1

± (3.25)

where ( )′ =S

∂( )∂. We can now define N and M as the total resultant force and moment about the centreline as

N n n= + ,+ − (3.26)

c cM m m d n d n= + + × − × .+ − +1

+ −1

− (3.27)

The definitions (3.26) and (3.27), together with the individual Kirchhoff equations for the sub-rods (3.25), and the relations (3.22)and (3.23), lead to the following balance equations for the forces and moments in the birod first derived in (Moakher and Maddocks,2005)

N F′ + = 0, (3.28)

M r N L′ + ′ × + = 0. (3.29)

We note that these equations have the form of the Kirchhoff equations. However, in general they are not sufficient to establish themechanical balance. Indeed, as already discussed, a variation of the angle θ between the tangents of the sub-rods may produce work.To account for this effect, we introduce the difference of forces and moments applied at each section

n n n= −+ − (3.30)

c cm m d n m d n= ( + × ) − ( − × ).+ +1

+ − −1

− (3.31)

The specific form of interaction between the sub-rods is such that the balance of forces and moments (3.28) and (3.29) is completedby a single equation expressing the notion that the moment about d1 on the + sub-rod must equal the moment about d1 of the − rod(as derived in Appendix E), that is

Lm d r n d( ′)· + ( ′ × )· + 2l + ˇ = 0.1 1 1 1 (3.32)

This equation introduces an extra mechanical quantity in the theory, l1 the d1 component of the internal moment l, in addition to thetraditional resultant force and moment N and M. An extra constitutive equation for l1 will therefore be needed.

Solving balance equations for these three mechanical quantities is considerably simpler than solving the Kirchhoff equations ofboth sub-rods. In part, the internal constraints f , l2 and l3, which ensure the cohesion of the paired cross sections, do not need to becomputed. Accordingly, instead of solving the 12 first order differential equations (3.20) and (3.21), we reduce the problem to onlysolving (3.28), (3.29) and (3.32), and in a geometry that is natural for the entire macroscopic structure.

Having established the basic equations for the geometry and mechanical balance of a birod, we need to obtain the constitutivelaws of this new structure, based on both the material response of the attached sub-rods and the attachment. The constitutive lawsfor the birod contain both the unstressed state (the state that the birod takes in the absence of loads) and its material response away

6 Note that the choice of location of the point at which the resultant moments and forces are defined does not change the form of the resulting equations.


158

from that state. We develop these ideas progressively by increasing complexity, first restricting the rod in 1D (linear rods), then 2D(planar rods), and finally 3D (general rods). In the first two cases, the geometry is sufficiently simple to allow for explicit expressionsfor the birod and as such we give a full analysis.

4. A birod constrained on the line

In this section we treat the simplest case where both sub-rods are twistless and restricted to a straight line (i.e. they can expandand contract in length but not bend or twist).

4.1. Geometry of linear birods

If the birod and sub-rods are constrained to a line (taken to be the z-axis so that d d e= = z3±

3 ), then u 0= and we can choosewithout loss of generality a director basis with d1

± and d2± along the x- and the y-axis, respectively. For the kinematics, we choose

φ ≡ 0± , d d=i i±, and a=0. The geometry of the 1D birod is then completely specified by α S t( , ) = r

S∂∂ ; in particular Eq. (3.16) reduces

to

r r r= = .+ − (4.1)

Taking a derivative of (4.1) by S leads to two equations relating the configuration of the 1D birod (fully described by the function α)to that of the sub-rods (characterised by α )± :

α α g g gg

= ; = .± ±+

− (4.2)

4.2. Constitutive laws and reference state

The 1D case is particularly well suited to demonstrate the derivation of constitutive laws. This is typically a three step process.First, a general law is formed from the constitutive laws of the sub-rods; however, these involves the functions g±, still unknown atthis stage. Second, the reference state is defined such that the stresses vanish, and this relation is inverted to determine g− andg gg=+ −. Third, the substitution of g± in the constitutive laws obtained at the first step provides a constitutive law relating directlythe stress to the relevant geometrical quantity. We also note that the 1D case could be easily computed from energy considerationsbut that the goal here is to show how we can apply the ideas outlined in the previous sections.

We assume general constitutive laws for the tensions in both sub-rods:

k αn = ( ).3± ± ± (4.3)

where the functions k± derive from convex potentials (possibly different) that have a minimum at 1 and with k (1) = 0± .Step 1: A general constitutive law. By definition of the tension N3 in the birod given by (3.26) and using the fact that d d=3

±3, we

obtain the constitutive law for the birod:

k α k αN = n + n = ( ) + ( ).3 3+

3− + + − − (4.4)

Since α = αgg

+− and α = α

g−

− , and g− is not a priori known, this is only a formal law at this point.7

Step 2: The reference configuration of the birod. In 1D, we define the reference configuration of the birod as the configurationthat satisfies N = 03 . In this configuration, α = 1, the substitution of which in (4.4) leads to

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟k

ggk

g1 + 1 = 0.+

−−

−(4.5)

If there exists a solution of this equation for g−, then it is unique as it is the (unique) root of a monotonically decreasing function.8

Then we obtain g− as a function of all other parameters appearing in the equation.Remembering that g = S

S− ∂

∂

−, the reference arc-length parameter S and total length L are given by

∫ ∫Sg σ t

σ Lg σ t

σ= 1( , )

d , = 1( , )

d .S L

0 − 0 −

− −

(4.6)

Eq. (4.6) can then be inverted to obtain g− as a function of S.Step 3: The constitutive law. Once S− is known as a function of S, all functions appearing in (4.4) are known as a function of S.

This relation then provides a general constitutive law for the birod in terms of the birod variables α S t g S t g S t( , ), ( , ), ( , )− . In the next

7 Recall that g = SS

− ∂∂ − describes the stretch from the reference length of the ‘−’ rod to the reference length of the birod, so is not known until the birod reference

state has been determined. While the quantity g = SS

∂ +

∂ − is known a priori, it cannot be expressed as a function of S without knowing the reference state of the birod.8 The existence of a solution is not guaranteed as the constitutive laws for the sub-rods may have a vertical asymptote, preventing them from reaching a certain

strain. In the case where the functions k± are continuous on the positive axis, a solution is guaranteed.


159

subsections we illustrate the theory with particular choices of constitutive law.

4.2.1. Linear constitutive lawsStep 1: Let the sub-rods have linear constitutive laws

k αn = ( − 1),3± ± ± (4.7)

where k± are constants. The general constitutive law for the birod is then

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟N α k α

ggk α

g( ) = − 1 + − 1 .3

+−

−−

(4.8)

Step 2: At α = 1, N (1) = 03 is easily solved to obtain

gg

k gkk k

= 1 ++

.−+ −

+ − (4.9)

The arc length of the birod is ∫S k k σ= ( + ) dS k g σ t k

g σ t+ − −1

0+ ( , )

( , )

− + −. If g is constant along the rod, i.e. a constant length differential, then

the length of the birod is L g L= k gkk k

− ++

+ −

+ − .Step 3: Substituting (4.9) in (4.8), we obtain a linear constitutive law for the birod

k k αN = ( + ) ( − 1).3+ − (4.10)

4.2.2. A neo-Hookean birodStep 1: In many biological application, tissues operate in large deformation. In these cases, it is necessary to use non-linear

constitutive laws. The simplest of such laws comes from the uniaxial response of a neo-Hookean material, which reads

⎛⎝⎜

⎞⎠⎟μ α

αn = − 1 ,3

± ± ±2± (4.11)

where μ± are constants. The corresponding Young's moduli are E μ= 3 /± ± ± (with ± the areas of the rods sections). The generalconstitutive law is then

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟N α μ α

ggggα

μ αg

gα

( ) =( )

− +( )

− .3+

2

− 2

−−

2

− 2

−

(4.12)

Step 2: At α = 1, N (1) = 03 is solved to obtain

g gμμ

g μμ

=1 + 1

1 +.−

2

+

−

+

−

3

(4.13)

Step 3: The constitutive law for the birod is obtained by substituting this last expression into (4.12), which after simplification, reads

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟μ α

αμ μ

gμμ

g μμ

N = − 1 , = 1 + 1 1 + .32 −

2

+

−

+

−

2/3

3(4.14)

We have obtained the interesting result that a birod made of two neo-Hookean sub-rods is itself neo-Hookean with an elasticstiffness that can be re-written symmetrically as μ μ g μ g= ++ + − −. That is, the stiffness of the birod is a linear combination of theeffective stiffnesses of the sub-rods μ± weighted by the amount of sub-rods material per unit arc length of the birod g±. By contrast,the linear constitutive laws of Section 4.2.1 lead to a direct sum k k++ − in Eq. (4.10) that does not take this correction into account.The effective Young's modulus of the birod is then

E μ g μ g= 3 ++

.+ + − −

+ − (4.15)

4.3. Application to tissue tension and differential growth stiffening

Many biological structures, such as arteries, roots, and stems, are composed of multiple tubular structures growing at differentrates (Vandiver and Goriely, 2009). A natural problem is to understand the change of material properties resulting from axialdifferential growth. For deformation along the axis, the response of a two-layer cylinder is equivalent to that of a 1D birod. Theeffective Young's modulus (4.15) provides the overall axial response of this structure. In Fig. 7, we show that for any g different from


160

1, Young's modulus of the birod increases, indicating that the structure stiffens whenever there is axial differential growth (the resultremains valid even when μ μ/ ≠ 1+ − ).

We can use this idea to revisit the problem of the growth of rhubarb as discussed in Vandiver and Goriely 2008 and Holland et al.2013 When the stalk of a rhubarb of length L is peeled, it can be easily observed that the peeled skin is shorter than the original stalk(L L<+ ) and that the pith is longer than the original stalk L L>− (see Fig. 8(a)). These experimental observations suggest that therhubarb stalk can be mechanically modelled as a composite beam made of a pith in compression surrounded by a skin in tension.Furthermore, in principle, uniaxial tension tests can be performed on both the skin and the pith so that infinitesimal Young's moduliE± of both components are known. We assume that the differential growth between pith and skin is uniform, so that g and g− are

independent of arc length. This assumption implies that g = LL

+− and g = L

L± ±

. For the purpose of this computation, we suppose that

L± and the material parameters are known while L is unknown. We further assume that the pith is a neo-Hookean rod, so that

⎛⎝⎜

⎞⎠⎟μ α

αn = − 1 ,3

− − −2− (4.16)

whereas the skin is a rod with a Fung-like response (Goriely et al., 2006)

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟μ α

αen = − 1 .

ν α α3+ + +2

++2/ −3+2 +

(4.17)

Within our framework, we have, as before:Step 1: The general law is

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟N α μ α

ggggα

e μ αg

gα

( ) =( )

− +( )

− .ν α gg gg α3

+2

− 2

−( /( ) +2 / −3) −

2

− 2

−2 − 2 −

(4.18)

Step 2: At α = 1, N (1) = 03 is solved to obtain g− or equivalently L as a function of all the parameters. This is a transcendentalequation for which a single solution exists.

Step 3: Once L is known, the constitutive law for the birod is obtained as

Fig. 7. A log-linear plot of the elastic stiffness μ in (4.14) as a function of g in the case μ μ/ = 1+ − . We observe that the elastic stiffness of the birod can be increased as

a result of differential growth.

Fig. 8. (a) Peeled rhubarb stalk and skin. The skin contracts whereas the pith elongates, indicative of the state of tension and compression of both parts (dotted lineindicates length before peeling). (b) Effective Young's modulus of a rhubarb stalk as a function of g L L= /+ −, here we chose + = 1+ − , μ μ/ = 0.01+ − . We observe a

dramatic increase of stiffness with increasing values of the strain-stiffening parameter ν.


161

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟N α μ α L

LLαL

e μ α LL

LαL

( ) =( )

− +( )

− .ν α L L L αL3

+2 2

+ 2

+( /( ) +2 / −3) −

2

− 2

−2 2 + 2 +

(4.19)

Young's modulus is obtained as E = ( | )/( + )Nα α

∂∂ =1

+ −3 , where ( + )+ − is the cross-sectional area of the birod. An example of theincrease of effective Young's modulus as a function of axial differential growth is shown in Fig. 8(b).

5. Planar birods

5.1. Geometry of planar birods

We now consider the case of a 2D birod, i.e. when the birod is twistless and constrained so that the centreline r lies in a plane. Wetake d2

± normal to the plane (along ez). In this case, the three frames d d d{ , , }1±

2±

3± and d d d{ , , }1 2 3 are aligned.9 In particular, we have

SSS S

d d∂∂

= ∂∂

∂∂

i i± ±

± (5.1)

and the Darboux vector of the birod, u d= u2 2, is related to the Darboux vectors of the sub-rods u d d= u = u±2±

2±

2±

2 by

gu = u .2 2± ± (5.2)

The derivative of (3.16) by S− and of (3.10) by S together with (5.2) imply

⎧⎨⎪⎪

⎩⎪⎪

⎧⎨⎪⎪

⎩⎪⎪

g α α ga

α g β α βα g

αg

α a β

α gg

α a β

u = −

= [(1 − ) + ]or, equivalently

= 1 ( + u ),

= 1 [ − (1 − )u ].

2−

− +

− − +

−− 2

+− 2

(5.3)

Therefore, once the relative growth function g is defined and the maps g± are known, a configuration of the planar birod isentirely determined by two generalised strains given by the global stretch α and curvature u2. All other geometric variables can beobtained from (5.2) and (5.3).

Note that we still have the arbitrary parameter β ∈ [0, 1] that describes the position of the birod centreline with respect to thesub-rods. However, in any configuration, α−, α+ and g are independent of β while α, u, g− and g+ depend on it. This reflects the factthat the former quantities are intrinsic to the individual sub-rods, while the latter are intrinsic to the birod and thus depend on howits centreline is defined. Hence, the constitutive laws will depend also on β and some judicious choices can be made depending on theapplication.

5.2. Constitutive laws and reference state

In order to find the constitutive laws and reference state of the planar birod, we follow the strategy developed in the previoussection. First, we obtain a general constitutive law by substituting the constitutive laws of the sub-rods in (3.26) and (3.27). Second,the reference state is computed by solving N 0= and M 0= for g− and u2. Third, once the reference state is know, the birod'sconstitutive law can be written as a function of the birod strain parameters α and u2.

Step 1: The general constitutive laws. Assuming both sub-rods are extensible, their constitutive laws are

k αn = ( , u ),3± ± ±

2± (5.4)

K αm d= ( , u ) .± ± ±2±

2 (5.5)

Upon using (5.2) an (5.3) to express α± and u2± as functions of α and u2, and with the definitions (3.26) and (3.27), the general law for

the birod is of the form

k α gK α g

N = ( , u ; u , ),M = ( , u ; u , ),

3 2 2−

2 2 2−

(5.6)

where k k k= ++ − and K K K= ++ −.Step 2: The reference state. In the reference state of the birod, N 0= , M 0= , α = 1 and u = u2 2, where u2 describes the reference

curvature of the birod. Therefore, g− and u2 are found by solving the equations

k g K g(1, u ; u , ) = 0, (1, u ; u , ) = 0.2 2−

2 2− (5.7)

Step 3: The constitutive law and the planar birod equations. For each particular solution found in Step 2, (5.8) provides aconstitutive law for the birod. The full system for the birod is then

9 Indeed, since the birod is in the plane perpendicular to d d= ±2+

2− and we parameterise the sub-rods so that d d=2

+2−. For the birod, d1 is orthogonal to d3

+, d3− and

d3. Since sub-rods and birod have the same orientation, d d d= =3+

3−

3, which in turn implies d d d= =1+

1−

1. Hence the three frames d d d{ , , }1±

2±

3± and d d d{ , , }1 2 3 are

aligned.


162

α

k α g

K α g

r d

N F 0M r N L 0

′ = ,

′ + = ,′ + ′ × + = ,

N = ( , u ; u , ),

M = ( , u ; u , ).

3

3 2 2−

2 2 2− (5.8)

Hence in the planar case, the birod is a Kirchhoff elastic rod, as its configuration under load is fully specified by a planar system ofelastic rod equations with no extra stress variables. As usual, it is often convenient to let x yr = [ , , 0], n nN = [ , , 0]x y , mM = [0, 0, ]and define θ as the angle between the tangent vector θ θd = [cos , sin , 0]3 and the x-axis. Then u = θ

S2∂∂

and the system can be written as

x α θ y α θn f n f

m α θn α θn ln θ n θ k α θm K α θ

′ = cos , ′ = sin′ + = 0, ′ + = 0′ + cos − sin + = 0,cos + sin = ( , ′),= ( , ′),

x x y y

y x

x y

(5.9)

where f fF = [ , , 0]x y and lL = [0, 0, ]. This constitutes 7 equations for the unknowns x y n n α θ m{ , , , , , , }x y , to be complemented with 6boundary conditions.

The properties of the birod are derived from its constitutive laws, i.e. the form of solutions of (5.7). In the remainder of thissection, we perform steps 1 and 2 explicitly for a number of typical linear cases. Interesting behaviour can emerge through theinteraction of linear sub-rods. Of course, many biological applications rely on the interplay between differential growth and non-linear constitutive laws, such as the rhubarb stalk discussed in Section 4.3. The resulting structures tend to have highly non-linearconstitutive laws and the program highlighted above must be followed for each specific system. Yet, it is sometimes enough to knowthe response of the system to small deformations of the composite structure (i.e. to linearise the theory about its non-linearequilibrium). The interested reader can find this calculation in Appendix D for arbitrary constitutive laws.

5.2.1. Extensible sub-rods with linear constitutive lawsHere we apply the general method outlined above to the particular case where both sub-rods obey linear constitutive laws. In this

case, they can be easily written in matrix notation for the generalised stresses and strains

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

α αz x x z=nm

, = u ; = u , =NM .± 3

±

2±

±±

2± 2

3

2 (5.10)

With these notations, the constitutive laws are

⎛⎝⎜

⎞⎠⎟z K x x x= ( − ), =

1u ,± ± ± ± ±

2±

(5.11)

where K± are 2×2 matrices.Step 1: Since M and N are linearly dependent on n3

± and on m± the constitutive laws of the birods given by (5.2) and (5.3) are alsolinear. Rewriting (3.27) and (3.26) in matrix form, we have

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟g

a β aβx M x M M= 1 with = 1 − (1 − )0 1

and = 10 1

.±±

± + −

(5.12)

With these two matrices, we can write the general form of the constitutive law for the birod as

z M z M z= ( ) + ( ) .T T+ + − − (5.13)

Using (5.11) and (5.12) in (5.13) then gives

gz Kx v= 1 − ;− (5.14)

with

⎡⎣⎢

⎤⎦⎥g

K M K M M K M v M K x M K x= ( ) + ( ) , = [( ) + ( ) ].T T T T++

+ − − − + + + − − −

(5.15)

Step 2: In the reference state of the birod, z = 0 (since N 0= and M 0= ) and x x= . In this configuration, (5.14) gives v Kx=g1− ,

which can be written as

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

v Kv K

g KK

−− u = .1 12

2 22

−

2

11

21 (5.16)

This is a linear system for g− and u2. Its solution provides the information on both the length of the reference state of the birod and


163

its reference curvature.Step 3: Once g− and u2 are known, from (5.14) the linear constitutive law for the birod is simply

gz K x x= 1 ( − ).− (5.17)

5.2.2. Extensible sub-rods with linear and diagonal constitutive lawsIt is of interest to further restrict our attention to the most commonly used constitutive laws: those for which there is no coupling

between stretch and curvature. In this case, K K= = 012±

21± and we can easily obtain explicit expression of u2, g− and K

a κ g ρδ g κρ g h κ κ ρδ β κ g ρδ g κ

u = (1 − ) + ( + )( + )(1 + ) + (1 + ) − [ (1 − ) + ( + )]

,2(5.18)

g κ gκ

a β β κ gκ

= 1 + /1 +

+ u − (1 − ) /1 +

,−2 (5.19)

where we have introduced the ratio of elastic stifnesses: κ K K= /11+

11−, the ratio of bending stiffnesses h K K= /22

+22− , the squared ratio of

the persistence length of one of the rod to the thickness of the birod ρ K K a= /( )22−

11− 2 , and δ a h a= ( u + u )2

−2+ measuring the total effect

of the reference curvature of the sub-rods.The constitutive laws of the birod can then be explicitly written as

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

αKNM = − 1

u − u ,3

2 2 2 (5.20)

where

⎛

⎝⎜⎜⎜

⎛⎝⎜

⎞⎠⎟

⎞

⎠⎟⎟⎟

Kg

κ g aβ a β κ g

aβ a β κ g aβ a β ρaK =

1 + / − (1 − ) /

− (1 − ) / + (1 − ) + 1 +.κ

ghg

11−

− 2 2

(5.21)

If the sub-rods are furthermore uniform and uniformly glued so that K± and g are independent of S, the modelling parameter

β = =κg κ

KgK K+ +

11+

11−

11+ can be chosen such that β β κ g− (1 − ) / = 0, in which case the birod has diagonal constitutive laws.

In the simplest case of uniformly glued rods of the same length and with no growth, i.e. g=1, a simple formula arises. Indeed, wefind

k αN = ( − 1),3 (5.22)

KM = (u − u ),2 2 2 (5.23)

where

k k k= ++ − (5.24)

K K K a k kk

= + +− + 2− +

(5.25)

K KK

u =u + u

.2

−2− +

2+

(5.26)

We see that the axial stiffness of the birod is simply the sum of the stiffnesses of the individual rods, as we found in the 1D case,similarly with the bending stiffness but with an extra term that accounts for compression/extension of the individual rods whenbending, and the intrinsic curvature comes as a weighted sum of u2

+ and u2−.

5.2.3. Linear extensible–inextensible sub-rodsThere is a further simplification that is particularly relevant in numerous applications. If the elastic stiffness of one of the sub-

rods is much larger than the other, then it is reasonable to model it as inextensible. We then have the case of a birod with oneextensible and one inextensible sub-rod (note that the case where both are inextensible is possible but amounts then to a purelygeometric exercise as no axial extension is possible). We assume that the − rod is inextensible while the + rod is extensible. Theconstitutive laws of the sub-rods are then

k αn = ( , u ),3+ + +

2+ (5.27)

K αm d= ( , u ) ,+ + +2+

2 (5.28)

Km d= (u ) ,− −2−

2 (5.29)

Since the − rod is inextensible, we can use it as a centreline (that is we choose β = 0). As a consequence n3− does not appear in the


164

definition (3.27) of M. Furthermore, s S= −, so g = 1− , thus α = 1 and the birod itself is inextensible (in particular there is noconstitutive law for N3). With this choice, Eqs. (5.2) and (5.3) become

gα

gau = u ; u =

u; = 1 (1 − u ).2

−2 2

+ 2 +2 (5.30)

Assuming linear constitutive laws of the form

k α K Kn = ( − 1); m = (u − u ); m = (u − u ),3+ + +

2+ +

2+

2+

2− −

2+

2+ (5.31)

and following the same steps as in the previous cases, the constitutive law for the birod is

KM = (u − u ),2 2 2 (5.32)

with

Kg

K gK a kg K K ak g

K gK a k= 1 [ + + ], u =

( u + u ) + (1 − )+ +

.+ − 2 +2

+2+ −

2− +

+ − 2 + (5.33)

5.3. Timoshenko's bimetallic strips

In his seminal paper (Timoshenko, 1925), Timoshenko studied the mechanics of two metallic strips welded along their length. Inparticular, the different coefficients of thermal expansion (t±) of the strips imply that the bilayer bends under changes oftemperature, as shown in Fig. 6. In this section, we show how Timoshenko's result can be recovered in the limit of smalldeformations. Recalling the discussion of Section 3.1.3 we have

gc t tc t t

gc c t t

c t t=

1 + ( − )1 + ( − )

so that 1 − =( − )( − )

1 + ( − ).t

t

t t

t

+0

−0

− +0

−0 (5.34)

Defining

m aa

n EE

= and = ,−

+

−

+ (5.35)

Timoshenko found that the radius of curvature of a bi-metallic strip of thickness a2 (Equation (4) in Timoshenko, 1925) is

⎛⎝⎜

⎛⎝⎜

⎞⎠⎟⎞⎠⎟

c c t t m

a m mn mmn

u =6( − )( − )(1 + )

2 3(1 + ) + (1 + ) + 1t t

Timo

+ −0

2

2 2

(5.36)

With the definitions (5.35), the non-dimensional parameters κ, h, ρ and δ defined in Section 5.2.2 become

κ mn h nm ρm

δ= , = , = 13

1(1 + )

and = 0.32 (5.37)

The last equality in (5.37) expresses that the individual strips are straight in their own reference configuration.Substituting (5.37) in (5.18) leads to

⎛⎝⎜

⎞⎠⎟

a g mg

mnm β g m

u = 3(1 − )( + 1)

+ + 3(1 − (1 − ))(1 + ).2

2

2 2

(5.38)

Assuming small thermal dilatation, c T Tϵ = ( − )⪡1t−

0 , (5.34) can be expanded as g c c O= 1 + (1 − / )ϵ + (ϵ )t t+ − 2 . Accordingly, (5.38)

becomes

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

a

mcc

mmn

mOu = 1

2

6(1 + ) 1 −

3(1 + ) + 1 +ϵ + (ϵ ),

t

t2

2+

−

2 2

2

(5.39)

which matches (5.36) to first order in c t tϵ = ( − )t−

0 . Note that here and in general, we find a non-linear dependence of u2 on g forlarge deformations. This is not relevant for the case of bimetallic strips but crucial to many biological systems.

5.4. Birings

Multi-layered tubular structures abound in biology, often deriving counterintuitive mechanical properties through mismatchedreference states of the constitutive layers. The mechanical behaviour can sometimes be understood through consideration of a cross-


165

section, for instance measuring the opening angle of a cut ring of artery is used to infer residual stress (Fung, 1991). As a simpleexample to demonstrate our framework in such a setting, we consider here birings, that is a birod constructed of planar sub-rods thateach have constant intrinsic curvature u2

±. We work under the assumption that the sub-rods have linear and diagonal constitutivelaws so that the intrinsic curvature and constitutive laws follow Eqs. (5.18)–(5.21) with β = κ

g κ+.

Since each sub-rod is a (portion of) a circular ring, it is intuitive to think in terms of their radii in their reference state, R±. Itfollows that

g SS

RR R

= ∂∂

= , u = 1 .+

−

+

− 2±

± (5.40)

5.4.1. Trivial caseWe first verify the validity of (5.18) in the trivial case in which the two sub-rods perfectly fit together in their reference state. That

is, suppose

R R a= + .− + (5.41)

Here, we have placed the minus rod on the “outside” so that d1 points inward. Then, if we choose the natural orientation, fixed in thex–y plane, so that increasing arclength corresponds to travelling the circle in a counterclockwise direction, d3 points in thecircumferential direction and d e= z2 . Here, the sub-rods reference configurations being compatible with one another, “gluing” therods to create the birod should have no effect, i.e. creating the birod should not deform the sub-rods or create any stress. To see thatthis is indeed the case, note that the radius of curvature of the birod is r = 1/u2, and the radii of curvature of the sub-rods are, from(3.16),

r r aβ r r a β= + Sign(u ) , = − Sign(u ) (1 − ).−2

+2 (5.42)

After some algebraic manipulations (for which it is useful to use the fact that (5.41) can be rewritten g= 1 −aR− ), it is easily found

that r R=− − and r R=+ +, as expected.The more interesting case is when the sub-rods have mismatched curvatures. Formula (5.42) still holds, but now it will not

generally be true that r R=± ±, instead the radii in the birod will be determined by a competition between bending and stretching ineach sub-rod. Before we analyse this, we must clarify an important point regarding boundary conditions. Note that the referencestate of the birod is such that no force or moment is applied at the ends of the rod. Thus, while we might naturally associate biringswith a closed circle, a closed ring implies a closed boundary condition, and this is not assumed in the birod formulation. In otherwords, the reference state of the biring is such that the centreline forms part of a circle, but it is not necessarily a closed circle and theends are not attached.

5.4.2. Inversion of a ringAs an intuitive example, consider the inversion of a ring, a process seen, for example, in the accretionary growth of a seashell

(Waite, 1983). Two closed elastic rings sit in a reference stress-free state such that the outside edge of the smaller ring aligns with theinside edge of the larger ring. Now stretch the smaller ring outside the larger ring and glue them together. What will be the radius ofthis closed biring?

To answer this in terms of the birods formulation, we first remove the closed boundary condition. Imagine a 3 step process (seeFig. 9A) where we first (i) invert and glue to form the biring, then (ii) cut the rings to remove the closed geometry and form the biringreference state, and finally (iii) glue the ends back together to form the closed biring.10

The subrings: To describe the biring, we use the same orientation as in the trivial case, and thus label the outside rod the minusrod. Then we have

R R a= − ,− +

which underscores the inversion: the outside rod has the smaller reference radius.The open biring: Eq. (5.18) gives the curvature of the open, “cut” biring. In Fig. 9B we plot u2 as a function of E+, Young's

modulus of the plus rod. Fig. 9C shows the reference open biring at the marked points. Here we are assuming circular cross-sectionswith equal radii a a a= = /2+ − , so that

k E π a K E π a= ( ) , = ( ) /4.± ± ± 2 ± ± ± 4 (5.43)

As E → 0+ , u → u2 2− and β → 0. This is as expected, and simply reflects the fact that as the stiffness of the plus rod vanishes, the birod

sits at the steady state of the minus rod. Conversely, as E → ∞+ , u → u2 2+ and β → 1. Between these extremes, Eu ( )2

+ has anintriguing shape. There are two points at which u = 02 , and a region where u < 02 . Where u = 02 , the reference state of the birod isactually flat. This means that two rods that naturally curve in the same direction are glued together and the result is a straight rod.Even more surprising, in the region u < 02 the two rods bend the opposite direction of their reference state! To understand this,imagine that the minus rod is a string, i.e. that it only resists stretching, not bending (indeed, bending stiffness scales with radius to

10 Experimentally, of course the cutting step is unnecessary. We include it here as it is natural from the theoretical point of view to first find the reference state,which requires removing the closed geometry restriction.


166

the fourth and can be negligible for small E). By opening the birod, the minus rod (which has been stretched to be glued) reduces itsstretching; at the same time this can reduce the compressive stress of the plus rod. Of course, opening comes at the cost of bendingenergy. That u2 can be zero or negative reflects the fine balance between bending and stretching energy in the system.

Closed biring: To form the inverted ring, we must close the birod by joining the ends. To determine the radius of the closedbiring, the key observation is that the length of the centreline of the birod is the same in the open and closed configurations. This is aconsequence of the following statement, which we prove in Appendix C: If a planar rod with constant intrinsic curvature and linearand diagonal constitutive laws is closed into a ring, there is no change in length, that is the deformation is such that α = 1.

The biring we are considering is mechanically equivalent to a single rod with linear and diagonal constitutive laws. Therefore, theabove result holds, and the radius of the closed biring is simply r L π= /2ring , where L is the length of the open biring. Since g = S

S− ∂

∂

−,

we have

L Lg

πRg

= = 2 ,−

−

−

−

with g− satisfying

gβ g1 = 1 + ( − 1).−

It follows that the reference curvature of the biring, which we denote u2ring, satisfies

Fig. 9. (A) Depiction of three step process to form the inverted closed biring. (B) Reference curvature u2 of the open biring as a function of stiffness E+. Other

parameters are E = 1− , a=0.1, R = 1.0+ , R = 0.9− . (C) Reference state of the biring at the points I–IV of B.


167

gu = u ,2−

2−

ring (5.44)

It is easy to see that u2ringis bounded between u2

+ and u2−. Moreover, it only depends on the stretching stiffnesses and the length

mismatch of the sub-rods. This highlights the notion that when a closed ring deforms to another closed ring, the deformation ispurely one of stretching. This can be understood via a simple calculation: if a ring with reference radius R is deformed to one ofradius r, then Ru = 1/2 , α r R= / and the geometric curvature κ r= 1/ , and thus

ακ rR r R

u = = 1 = 1 = u2 2 (5.45)

is unaffected. Thus, any deformation in which the biring remains closed and circular is purely stretching; on the other hand, if thebiring is cut from its reference state it will open up, a deformation that will be purely bending.

We make one final remark regarding birings. There is an interesting consequence of the fact that closing a rod into a ring requiresno stretching energy. Suppose we take two straight sub-rods of equal length but different thickness and glue them together to form abirod (again choosing β so that the birod is diagonal), which in this case will be straight in its reference state. Now form a ring byattaching the ends. There are two ways this can be done: the thick rod can be on the inside or on the outside of the ring. The point isthat either way, the length of the birod centreline does not change, and thus the biring radius is the same, even though the sub-rodswill have different radii in either case. Also, the energy is exactly equal in both cases. Similar arguments can be made to show that ifthe straight birod is clamped and a compressive force is applied until a buckling occurs, the two directions of buckling, which are notsymmetric in terms of the individual sub-rods, are equivalent in terms of the birod – the energy is the same and the system has nopreference to buckle one direction or the other.

6. 3D birods

In 3D, the tangents d3± to the sub-rods are not aligned in general (cf. Eq. (3.19) and Fig. 6). Hence the definition N n n= ++ −

mixes components of n± that are not known constitutively so that we cannot derive directly a constitutive law for N3. Similarly, thedefinition of a moment M for the birod depends on n± and hence also include components that are not known constitutively. Theequations for the birod are still well-defined, and it is possible to reach consistent (though implicit) constitutive laws. Thiscalculation is included in Appendix E. However, unlike the 1D and 2D cases, where one could derive explicit constitutive laws for thebirod forces and moments, in 3D the equation for the birod cannot be simply written as a set of Kirchhoff equations.

The key to developing the birod theory in 3D is to identify the correct generalised strains and the associated generalised stresses.This can be accomplished through use of a non-orthogonal frame. We first introduce a non-orthogonal basis and its dual in Section6.1, which as we show still allows a natural description of the birod geometry in terms of 4 scalar functions. We then introduce theforce components and generalised moments associated with the kinematic description, and express the force balance equations inthese components (Section 6.2). The value of the new basis is readily apparent in consideration of constitutive laws, as explicit lawsare immediately obtained in Section 6.3. Finally, we show in Section 6.4 how determining the reference configuration of the birodcomes down to solving a single second order equation in the most typical case of linear diagonal laws for the moments of the sub-rods.

6.1. Geometry of 3D birods

We use the definition of the sub-rods as given in Section 3.1.2 and summarised in Fig. 10. In Section 3, we defined a centrelineand a right-handed orthonormal frame of directors for the birod itself and showed how it could be characterised by 4 geometricdegrees of freedom: α and u. Here we propose an alternative geometric description of the structure. We introduce a non-

Fig. 10. A summary of the key geometric quantities defining a birod from Section 3.1.2.


168

orthonormal frame and use a different four geometric parameters, to be defined. The definition of a centreline of the birod is nolonger required. We will assume that sections of the birod are now labelled by a material parameter R (which depends on theparticular problem of interest and can be different from the unloaded arc length S). The definitions of Section 3 remain valid afterreplacing S by R. In particular, there exist one-to-one mappings G± such that S G R t= ( , )± ± which implies that G G G= ○+ − and wehave

g GR

g GR

= ∂∂

, = ∂∂

,++

−−

(6.1)

while g = SS

∂∂

+− as before. Note that while g is a physical quantity that encodes the differential growth between the two sub-rods, the

functions g± actually relate the (arbitrary) parameterisation R of the birod to the reference arc-lengths S± of the sub-rods. Inparticular one of these latter functions (say g−) may be chosen ad libitum as such a choice is equivalent to making a choice ofparameterisation R. The other function is then given by g gg=+ − according to the chain rule.

The unit vector d1 is defined, as before, as the vector (see Fig. 6) pointing from r− to r+ as given by (3.17). By definition of φ± inSection 3.1.2, we have

φ φ φ φd d d d d= cos + sin = − cos − sin ,1−

1− −

2− +

1+ +

2+ (6.2)

φ φd d d d× = − sin + cos ,3−

1−

1− −

2− (6.3)

φ φd d d d× = sin − cos .3+

1+

1+ +

2+ (6.4)

The frames d{ }i± of the sub-rods are recovered from d1, d3

± and φ± by

φ φ πd d d d d d d R d d d d d d d d R d( ) = ( × ) ( , − ), ( ) = ( × ) ( , + ),1−

2−

3−

1 3−

1 3−

3− −

1+

2+

3+

1 3+

1 3+

3+ + (6.5)

where αR d( , ) indicates a right handed rotation around the vector d by an angle α. The frames D d d d= ( )±1±

2±

3± are related by the

following sequence of rotations (see Fig. 10):

φ θ φ πD D R d R d R d= ( , ) ( , ) ( , − − ),+ −3− −

1 3+ + (6.6)

From these definitions, it follows that (see Appendix F)

g g φ θ φu u d d d= + ( )′ + ′ − ( )′ ,+ + − − −3−

1+

3+ (6.7)

where ()′ denotes a derivative w.r.t. R.We now equip the birod with the non-orthonormal frame d d d{ , , }1 3

−3+ . Although this implies that we will also have to keep track of

the dual basis, we will see in the next two sub-sections that this choice is natural and leads to a complete description of the birod.Since we need the dual basis, it is practical to use upper and lower indices. For this reason we introduce the following vectors(Fig. 11):

c d c d c d= , = , = ,1 1 + 3+

− 3−

(6.8)

c d c d d c d d= , = − × , = × .11

+1 3

− −1 3

+ (6.9)

From (6.2), we have

φ φ φ φc d d c d d= − sin + cos , = ∓ (cos + sin ),∓ ±1± ±

2± 1 ±

1± ±

2± (6.10)

so that

Fig. 11. An alternative way to frame a birod: non-orthonormal birod frame superposed on the sub-rods (left) and the same frame projected on a plane perpendicularto c1 (right). The right panel can be used to easily infer the relations (6.13)–(6.19).


169

φ φ

φ φ

d c c

d c c

= − sin ∓ cos ,

= cos ∓ sin .1± ± ∓ ± 1

2± ± ∓ ± 1 (6.11)

We define θ as the oriented angle from c d=− 3− to c d=+ 3

+ such that θd d d× = sin3−

3+

1 (this definition is equivalent to (3.19)) and notethe following identities:

θc c c c· = 0, · = − cos ,1 ± + − (6.12)

θc c c c c c× = ± , × = − sin ,1±

∓+ − 1

(6.13)

θc c c c· = 0, · = − sin ,±∓

±±

(6.14)

θc c d c c d× = ± cos , × = ± .±± 1

±∓ 1 (6.15)

For any vector v, there exist six11 scalars {v , v , v }1 + − and {v , v , v }1 + − defined as

v c v c v dv = − · , v = − · , v = v = · ,± ±± ± 1

1 1 (6.16)

such that whenever θ ≠ 0,

θ θ θ θv c c c c

c c= v + vsin

+ vsin

= v + vsin

+ vsin

.11

+

+

−

−1

1+ + − −

(6.17)

In particular,

⎧⎨⎪⎪

⎩⎪⎪

⎧⎨⎪⎪

⎩⎪⎪

θθ

θθ

θθ

θθ

c c c

c c c

cc c

cc c

= − cos −sin

= − − cossin

and=

− + cossin

,

=− + cos

sin.

−

+ −

+

+ −

− − +

+ + −

(6.18)

By taking the scalar product of (6.17) by c−, c+, c− and c+, we obtain relations between the components in the two bases

⎧⎨⎪⎪

⎩⎪⎪

⎧⎨⎪⎪

⎩⎪⎪

θθ

θθ

θθ

θθ

−v = v cos + vsin

−v = v + v cossin

and−v =

v − v cossin

,

−v =v − v cos

sin.

−

+ −

+

+ −

− − +

+ + −

(6.19)

The construction (6.16) and (6.17) is not quite standard (usually, the sines would be included within the definitions of thecomponents vα). However, it insures that for a vector v, the components vα and vβ are well defined even when θ → 0. In general thepresence of θsin may be a problem as θ → 0. To circumnavigate this problem, we will occasionally mix the basis c c c{ , , }1 − + and itsdual and express vectors as linear combinations of the right-handed orthonormal bases c c c{ , , }1

+− or c c c{ , , }1 +

− . These frames wherealready used in Starostin and van der Heijden (2014) to study the ply. It is interesting to note that they can be gathered into a single(non-orthogonal frame) if we allow that some physical quantities are more naturally expressed in the dual: geometrically there isonly one orthonormal frame plus one degree of freedom. The implications of choosing one geometric description or the other aredeveloped in Appendix I.

Next, we require the evolution of the new bases along the length of the birod. We compute

φ φ g φc d d d u d d′ = ′ = (cos + sin )′ = ( + ( )′ ) × ,11

−1− −

2− − − −

3−

1 (6.20)

φ φ g φc d d d d u d c′ = ( − × )′ = ( − sin + cos )′ = ( + ( )′ ) × ,+1 3

− −1− −

2− − − −

3− + (6.21)

φ φ g φc d d d d u d c′ = ( × )′ = ( − sin + cos )′ = ( + ( )′ ) × .−1 3

+ +1+ +

2+ + + +

3+ − (6.22)

By defining

g φ θu d c= + ( )′ +′

2,− − −

3− 1

(6.23)

and substituting (6.7) in (6.22), we obtain

⎛⎝⎜

⎞⎠⎟σ θc c c( )′ = −

′2

× ,αα

α1

(6.24)

where Greek indices run through {1, + , − } and σ = ±α if α = ± and 0 if α = 1; no summation implied in RHS. Similarly, we have

⎛⎝⎜

⎞⎠⎟σ θc c c( )′ = +

′2

× .α α α1(6.25)

11 Note however that v = v11 since d1 is always perpendicular to d3

±.


170

Eqs. (6.24) and (6.25) show that the vector plays the role of the Darboux vector for the non-orthogonal bases c c c{ , , }1 + − andc c c{ , , }1 + − . Taking a derivative by R across ar r d− =+ −

1, we obtain

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟α g α g a a

θ θa

θ θc c c c c c

c c− = × =sin

+sin

× =sin

−sin

.+ ++

− −−

1+

+

−

−1

−+

+−

(6.26)

Projecting (6.26) on c± leads to

a θ α g= sin ,±∓ ∓

(6.27)

and therefore

ga

α α gc c c= + ( + ),1−

− + + −(6.28)

where c= · 1. We thus see that a birod, through Eq. (6.28), can be completely parameterised12 by the four scalar functions α R( )+ ,α R( )− , θ R( ) and R( ); these are natural and physically relevant quantities of the structure: the stretches in both strands, the anglebetween the strands, and the bending of the structure about the common chord, respectively.

6.2. 3D birod mechanics in non-orthogonal basis

Having characterised the birod geometry in the new local basis, we turn to the mechanics. In the orthogonal basis approach, asdescribed in Section 3.2.3, the birod mechanics was defined by 7 variables: the total force N, the total moment M (about a chosencentreline), and m1, the d1 component of the difference of moments. As stated, the issue in 3D is that we cannot derive explicitconstitutive laws for these quantities. As we shall see, this problem is circumvented by working with analogous quantities in theframework of the non-orthogonal basis. We still consider the total force, now decomposed into the basis c c c{ , , }1 + − , that is we write

θ θN c c c= N + N

sin+ N

sin.1

1+

+

−

−

Along with the quantities {N , N , N }1 + − we define the generalised moments

m m cP = ( + )· ,1 + − 1 (6.29)

a am m d n c m m cP = − ( + − × )· = − ( + )· − n ,+ + −1

− + + − +3− (6.30)

a am m d n c m m cP = − ( + + × )· = − ( + )· − n ,− + −1

+ − + − −3+ (6.31)

m m cQ = ( − )· .1+ −

1 (6.32)

Note that P1 is the component along c1 of the total moment about any point13 on the common chord, while P+ is the component alongc+ of the total moment about r+, and similarly P− is the component along c− of the total moment about r− . These quantities take theplace of the components of M in the theory, while Q1, similar to m1, is the component along the common chord of the difference ofmoments. These four scalars P±, Q1 and P1 are also found to be the moments of force conjugated to virtual displacements of α±, θ and

(respectively).Projecting the force and moment balance equations, (3.28), (3.29) and (3.32), onto the non-orthogonal bases leads, after some

algebra (see Appendix G), to

ga

α α g(N )′ = ( N − N ) − F ,1

−−

−+

+ 1 (6.33)

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

θ θθ

ga

α α g θ(N )′ = −′

2N − N cos

sin− ( − cos )N − F ,−

+ −−

− +1 −

(6.34)

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

θ θθ

ga

α θ α g(N )′ = − +′

2N − N cos

sin+ ( cos − )N − F ,+

− +−

− +1 +

(6.35)

⎛⎝⎜

⎞⎠⎟

ga

α g θ aθ

αθ a

θ(P )′ = P + P cos − N

sin−

P + P cos − Nsin

− H ,1−

++ −

− −− +

+ 1

(6.36)

⎛⎝⎜

⎞⎠⎟

ga

α θ a θ θ θ aθ

(P )′ = (P sin − N cos ) +′

2+ P + P cos − N

sin− H .−

−− 1

1

+ −− −

(6.37)

12 Since (6.24), (6.25), an (6.28) can be integrated to find the local basis c c c{ , , }1 − + . The local frames of sub-rods are then given by (6.5) and their centrelines arefound by integrating α gr c( )′ =± ± ±

±.13 As stated, the birod does not require an explicitly defined centreline in this framework.


171

⎛⎝⎜

⎞⎠⎟

ga

α g a θ θ θ θ aθ

(P )′ = ( N cos − P sin ) +′

2−

P + P cos − Nsin

− H ,+−

+1

1− +

+ +

(6.38)

⎛⎝⎜

⎞⎠⎟

ga

α g θ aθ

αθ a

θga

α g α(Q )′ = P + P cos − Nsin

+P + P cos − N

sin+ 2 (Q + Q ) − 2l − h ,1

−+

+ −− −

− ++

−

++

−−

1 1(6.39)

where

Q Qm c m c= − · , = − · ,+−

+ −+

− (6.40)

and the external loads are defined by

F c F c F c F F FF = · , F = − · , F = − · , with = + ,1 1 − − + ++ −

(6.41)

and

a aL L c L L c F c L L c L L c F cH = ( + )· , H = − ( + )· − · , h = ( − )· , H = − ( + )· − · .1 + − 1 − + − − ++ 1

+ −1

+ + − + −− (6.42)

Note that with the introduction of two additional moments, Q+ and Q−, we now have 6 moments and only 4 equations. However, dueto the geometric constraints, two are determined constitutively. By comparison, Moakher and Maddocks (2005) theory of birods hasno geometric constraints but Starostin and van der Heijden (2014) theory of braids imposes a different set of geometric constraints(frictionless contact) and their formulation requires the explicit computation of internal loads. Finally, the removable singularity inthe vector field appearing as θ → 0 is discussed in Appendix I.

6.3. Constitutive laws

As we have shown, the birod geometry is completely specified by the 4 scalars α+, α−, and θ, whereas Eqs. (6.33)–(6.39)provide 7 equilibrium equations involving 10 mechanical quantities: N, P1, P+, P−, Q1, Q+, Q− and l1. As a result, we need (10+4)− 7=7constitutive laws.

Constitutive law for the attachment: The moment l1 about d1 applied by the sub-rods on one another is transmitted by theparticular fashion in which the two rods are attached, what we have referred to as the glue. Its dependence on the strain is therefore aconstitutive property of the glue itself and must be specified. For simplicity, we assume that it is a function of θ only. In particular,two special cases are of note: freely rotating sud-rods, i.e. the case where l ≡ 01 , and the case where the two sub-rods are rigidlyattached so that θ R( ) is a prescribed function. In the latter case, l1 is no longer prescribed constitutively but we can drop (6.39) sincespecifying θ removes a degree of freedom.

Constitutive law for extensible sub-rods: The value of the generalised moments in the non-orthogonal bases is readily apparentupon inspection of Eqs. (6.29)–(6.32) and (6.40). Namely, the constitutive laws for the sub-rods provide 6 explicit constitutive lawsfor P1, P+, P−, Q1, Q+ and Q−. Indeed, if both sub-rods are extensible, the moments m± and the tensions n3

±14 are known constitutivelyas a function of α± and u±. But since u± are functions of α±, θ and (see Section 6.1), these relations can be re-expressed as afunction of the latter variables. Hence, we have 6 explicit constitutive equations in terms of the 4 geometric variables.

These 6 relationships, together with the law for the attachment, give the 7 required constitutive laws. An explicit computation ofthese constitutive laws in the particular case of sub-rods with linear constitutive laws for the moments m± is given in Appendix H.

Note that if one of the sub-rods is inextensible, say the + rod, then its stretch is given and the birod loses both a degree of freedomα+ and the constitutive law for P− so that both the total count of variables and equations are decreased by one.

We have now a full system of equations for the growing birod. This system is defined by the three components:

1. The growth laws: A set of growth laws for the slow evolution of the attachment and reference shapes of the sub-rods is specified asexplain in Sections 2.1.3 and 3.1.2.

2. The equilibrium equations: The quasi-static equilibrium problem involving 4 kinematic variables and 10 mechanical variables for7 equilibrium equations (6.33)–(6.39).

3. The constitutive laws: A set of 7 constitutive laws as discussed in this section.

Once the solution of this system is known, the positions of the sub-rods' centrelines are obtained by integration of α gr c( )′ =± ± ±± from

(6.5), (6.24), (6.25), and (6.28).

6.4. Unloaded birod and reference configuration

The first question of interest is to establish the shape of the birod when unloaded (that is in the absence of an external line densityof forces f , moments l and loads at the ends). It is important to realise that an unloaded birod can be stressed in order to balance itsinternal geometry. Therefore, the unloaded shape cannot be simply obtained by imposing the vanishing of the constitutive laws(indeed, this would provide a system of 7 equations for 4 unknown strains).

14 The key difference compared to the orthogonal basis choice is that only the d3± component of n± appears in the definition of the moments.


172

More precisely, an unloaded configuration is defined as a configuration that solves (6.33)–(6.39) with F L 0= =± ± and no endloads. The condition of no end loads implies that the quantities N , N , N , P , P , P1 + −

1 + − and Q1 all vanish at the ends R=0 and R=L. Inthat case, Eqs. (6.33)–(6.38) imply15 N 0≡ together with

P = P = P ≡ 0,1 + − (6.43)

while Eq. (6.39) becomes

ga

α g α12

(Q )′ = (Q + Q ) − l ,1

−

++

−−

1 (6.44)

with the boundary condition LQ (0) = Q ( ) = 01 1 .The system of equations (6.43) and 6.44 can also be written in terms of the mechanical quantities of the sub-rods, in which case it

reads

a ga

α α gm m d d m m d m m d d m d m±( + )·( × ) = − n , ( + )· = 0, (( − )· )′ + 2 ( · + · ) + 2l = 0,+ −1 3

±3± + −

1+ −

1

−−

3− + +

3+ −

1 (6.45)

along with the boundary conditions m m c( − )· = 0+ − 1 at R=0 and at R=L. We note that the system (6.45) is a completely generic set offour equations for the four degrees of freedom of the unloaded birod. It is valid for any choice of constitutive laws of the sub-rods.

6.5. Unloaded configuration with linear moments

We now consider the particular but important case of sub-rods with linear diagonal constitutive laws for the moments m± andarbitrary laws for n3

±:

EI b Γm d d d= ( ) [(u − u ) + (u − u ) + (u − u ) ],± ±1±

1±

1± ±

2±

2±

2± ±

3±

3±

3± (6.46)

where EI EI( ) = [( ) ]±1

±, b EI EI= [( ) ] /[( ) ]±2

±1

±, and Γ μJ EI= ( ) /[( ) ]± ±1

±. We define measures of the total stiffness of the birod EI( ) andrelative stiffness of the sub-rods ± as

EI EIg

EI EIg EI

EIEI

( ) = ( ) + ( ) , = ( )( )

, = ( )( )

,+

− ++

−−

(6.47)

and the following reference curvatures and twists about key directions:

⎛⎝⎜

⎞⎠⎟a g a g

ga g a g

gφR

u u d u c u u d= ( − )· , = ( · ), = ( + )· , = u + dd

.0+ −

1 2± ±

−± ∓

1+ −

1 3± ±

− 3±

±

± (6.48)

The parameters 0 and 1 give measures of reference curvatures of the sub-rods about the common chord, while the 2±measure

reference curvatures about c∓, and the 3±measure reference twist, both due to u3

± and pre-gluing twist contained in φ±.The computation of the general constitutive laws for the components is given in Appendix H. We focus on the case of sub-rods

having cross-sections with at least the symmetries of the square (b = 1± ) or such that one of their principal axes is along c1:φ kπ k= ( ∈ )± (see Appendix H for a discussion). In this case, the constitutive laws for the birod are

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

EIg

ga

θ ga

P = ( ) −2

+ ( − )′

2−

2,1

−

−1 + −

−0

(6.49)

⎡⎣⎢

⎤⎦⎥a α EI

ab α α g θ b θ α g α θ Γ θ α θP = − n ( ) − ( ) ( − cos − ) − cos ( − cos − )+ sin ( sin + ) ,+

3− − − − − +

2− + + + −

2+ + + −

3+

(6.50)

⎡⎣⎢

⎤⎦⎥a α EI

ab θ α g α θ b θ α α g θ Γ θ α g θP = − n ( ) − ( ) cos ( − cos − ) − cos ( − cos − )+ sin ( sin + ) ,−

3+ + + + + −

2+ − − − +

2− − − +

3+

(6.51)

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

EIg

θ ga

ga

Q = ( ) ′2

−2

+ ( − ) −2

,1 −

−0 + −

−1

(6.52)

⎡⎣⎢

⎤⎦⎥

EIa

b θ α α g θ Γ θ α g θQ = ( ) sin ( − cos − ) + cos ( sin + ) ,+− − − +

2− − − +

3−

(6.53)

⎡⎣⎢

⎤⎦⎥

EIa

b θ α g α θ Γ θ α θQ = ( ) sin ( − cos − ) + cos ( sin + ) .−+ + + −

2+ + + −

3+

(6.54)

The unloaded configuration is obtained by substituting these relationships in the condition (6.43) and (6.44) to obtain a set of fourequations for α±, θ and . In particular, the condition P = 0± becomes

15 Indeed the boundary conditions imply A B 0= = in Appendix I which in turn imply P = 0α .


173

aEI

α g b b θ Γ θ α θ b b b b θ Γ θ( )

n + ( + cos + sin ) − cos ( + ) = − cos − sin ,2

3+ + + + − − 2 − − 2 − + + − − + +

2+ − −

2− − −

3−

(6.55)

aEI

α b b θ Γ θ α g θ b b b b θ Γ θ( )

n + ( + cos + sin ) − cos ( + ) = − cos − sin ,2

3− − − − + + 2 + + 2 + + + − − − −

2− + +

2+ + +

3+

(6.56)

while the condition P = 01 reads

⎛⎝⎜⎜

⎞⎠⎟⎟g

aθ g

a−

2+ ( − )

′2

−2

= 0.−

1 + −−

0

(6.57)

Finally, substituting (6.52)–(6.54) in (6.44) and eliminating according to (6.57) gives

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎢

⎤⎦⎥⎥

EIg

θ θ ga

EI θ b α α g α θ b α g α α g θ

θ α Γ θα α g Γ α g θ

( ) ( ′ − )′

+ l ( ) = ( ) sin ( ( − cos − ) + ( − cos − ))

+cos ( (sin + ) + ( sin + )) .

−+ −

0 1

−

2+ + − + −

2+ − − + − +

2−

− + + −3+ + − − +

3−

(6.58)

Similarly substituting (6.52) and eliminating according to (6.57) in the boundary conditions LQ (0) = Q ( ) = 01 1 (see Section 6.4)gives

θ θ L L′(0) = (0) and ′( ) = ( ).0 0 (6.59)

The system of equations (6.55)–(6.59) is a closed set of equations whose solutions make up the set of possible unloadedconfigurations of a birod. If the tensions n3

± have linear constitutive laws, then (6.55) and (6.56) form a linear system for α±. The twostretches can be substituted accordingly in (6.58) and the problem of finding the reference state of linear birods is thus expressed asone second order differential equation with the boundary condition (6.59). Note that under our assumptions of b = 1± or φ kπ=± ,decouples from the other variables as it only appears in the linear equation (6.57).

The problem (6.55)–(6.59) describes many different situations of interest and different particular cases may easily be studied.For instance, in the case of a completely rigid attachment, we prescribe the function θ S( ) as a constraint, then Eq. (6.58) provides anexpression for the moment l1. The unloaded states then simply obey the algebraic system of equations (6.55) and (6.57). Note that ifeither sub-rod is inextensible, then the corresponding stretch is fixed and Eq. (6.55) or (6.56), involving the corresponding tension,disappears from the problem. Finally, another particularly interesting class of structures arises when all parameters are constantalong the length of the structure. We study this case in detail in the following section.

7. Homogeneous birods

7.1. A mechanical analogy

A birod is homogeneous when its sub-rods are homogeneous (the constitutive laws are independent of S±), g and theU’s are allindependent of R and l1 does not explicitly depend on R. It is straightforward to show that in this case, EI( ) is also independent of R.After solving (6.55) and (6.56) for α− and α g( )+ , Eq. (6.58) reduces to a second-order equation that can be written formally as a onedegree-of-freedom Hamiltonian system with a potential V of the form

θ Vθ

″ = − 1 ∂∂

,+ − (7.1)

along with boundary conditions

θ θ L′(0) = ′( ) = .0 (7.2)

The constant 0, together with the function V θ( ), defined up to an additive constant, contain all the information on the possiblesolutions of the system. A judicious choice for the parameterisation of the birod, utilised for the remainder of this section, consists intaking g a=− , which implies that R S a= /−

The form (7.1) is fully generic. Particular examples are the equilibrium equation of the ply considered in Thompson et al. (2020)and that of the braid considered in Starostin and van der Heijden (2014). In the next sections, we focus on the novel effects which aredue to differential growth between the strands.

The problem (7.1) is mathematically equivalent to that of finding the trajectory of a ball sliding on a potential such that the ball

starts and ends at ‘time’ L with a ‘velocity’ 0. In particular, the quantity +θ V′2

2

+ − , the total energy of the ball in the analogy, is afirst integral along the length of the birod, i.e.

⎛⎝⎜

⎞⎠⎟

θ V′2

+′

= 0.2

+ −(7.3)

As a consequence, the local extrema of V θ( ) lead to solutions with constant θ. Since θ is the angle between the tangents of the sub-


174

rods, those solutions correspond to birods wrapping around helices. Also, from the boundary conditions, it can be observed thatwhenever ≠ 00 , there are no solutions with constant θ.

Furthermore, the analogy with the sliding ball implies that whenever V has a local minimum, the problem (7.1) admits a set ofdiscrete sets of oscillating solutions provided that the depth of the well is larger than ( ) /2+ −

02 . Similarly, if the potential V has a

maximum whose height is greater than ( ) /2+ −0

2 , and > 00 (resp. < 00 ), the problem (7.1) admits a solution that starts onthe left (resp. right) of the maximum, goes over the local maximum and ends exactly at the same height on the other side of themaximum. We now examine these solutions further in the case of inextensible sub-rods.

7.2. Inextensible sub-rods

When the sub-rods are inextensible, α = 1± and from (6.55)–(6.59) the tension n3± in each sub-rod is known. Then, Eqs. (6.58)

and (6.59) reduce to (7.1) with

V A θ B θ g C θ ψ θ= −4

cos2 + cos [ − ] − sin − ( ),2 3 (7.4)

where a= lψθ

dd 1 and

b b gb b g

Γ g Γ=++

, = + ,2

+ +2+ − −

2−

+ + − − 3+ +

3+ − −

3−

(7.5)

A b Γ g b Γ B b b= ( − ) + ( − ) , = + .+ + + 2 − − − − − + + (7.6)

The parameter 2 is a measure of the reference curvatures of the sub-rods about the respective axes d d×1 3±, and 3 is related to

both the reference twist and pre-gluing twist of the sub-rods. Parameters A and B give the relative stiffnesses of the sub-rods. Notethat when constructing a birod from an experimental point of view, the natural independent control parameters are the referencecurvature of the sub-rods, their principal stiffnesses and g; nevertheless it is mathematically convenient to study the system (7.1) and(7.4) as a function of the parameters g, A, B, 2 and 3, all 5 of which depend on g.

7.3. Identical sub-rods with free glue

In order to show the great wealth of possible solutions for growing birods we conclude this paper by studying a simple case of twohomogeneous inextensible sub-rods that are identical apart from having possibly different intrinsic curvatures. We let one of therods grow homogeneously and differentially and we assemble the two rods as explained in Section 3.1.1. The question is then tocompute possible unloaded equilibrium states. The stability of such solutions will be studied elsewhere.

In the particular case of constitutively identical sub-rods with isotropic cross-sections: EI EI( ) = ( )+ −, b b= = 1+ − , Γ Γ Γ= =+ −

(taken to be 2/3 in all figures), the expression of the material parameters becomes considerably simpler: A Γ g g= (1 − )(1 − + )2 andB=1. If, furthermore, the glue is assumed to oppose no resistance to the relative rotation of the sub-rods about d1

16, we have ψ = 0 forall θ. Under these hypotheses, V defined in (7.4) further simplifies to

V g θ Γ g g θ θ= ( − ) cos − 1 −4

(1 − + ) cos2 − sin ,22

3 (7.7)

Vg

ϕθ ϕ Γ g g θ=

−cos

cos( + ) − 1 −4

(1 − + ) cos(2 ),2 2

(7.8)

where⎛⎝⎜

⎞⎠⎟ϕ = arctan

g −3

2.

We remark that ϕ = 0 whenever = 03 . This phase shift encodes the effect of reference twist in the system (be it reference twistof the sub-rods, u3

±, or the changes of φ± along the length of the structure).The functionV θ( ) in Eq. (7.7) is shown in Fig. 12 for different values of the parameters. As can be seen from Fig. 12a and b there

are two qualitatively different cases: eitherV θ( ) is concave with one local extremum θm on the domain θ− ≤ ≤π π2 2 (hill-shaped) or it

presents three extrema: two maxima and one minimum (‘M’-shaped).This is a general feature of the system. The second form (7.8) shows that there is a competition between two terms: a cos θ term

shifted by the phase ϕ and a θcos2 term. If the former dominates, the potential is ‘hill’-shaped; if not, the potential is ‘M’-shaped. Thistransition can occur under various combination of parameter values. It is characterised by the maximum of the hill-shaped V θ( )becoming a minimum of the M-shaped V θ( ). The bifurcation criterion is therefore

Vθ

dd

= 0.θ

2

2m (7.9)

When = 03 , both the maximum of the hill-shaped and the minimum of the ‘M’-shaped potentials lie at θ = 0m . In that case, the

16 This is the limiting case of freely rotating sub-rods, i.e. the constitutive law l ≡ 01 .


175

bifurcation condition (7.9) simplifies to

Vθ

g Γ g gdd

= 0 = − + (1 − ) (1 − + ).2

20

22

(7.10)

For each value of the parameter Γ , the bifurcation criterion (7.10) defines a parabola that splits the g( , )2 plane into tworegions: one in which the potential is hill-shaped and the other in which it is M-shaped (see Fig. 13).

The distinction between the two general shapes of V is important because it leads to qualitatively different solutions. The hill-shaped potential present a single maximum and as a result the corresponding birods each have a single unloaded equilibrium.Indeed, the functionV θ( ) being concave, the sliding ball analogy shows that the only way to start with a velocity 0 and end with thesame velocity anywhere is to have a symmetric trajectory over the maximum. On the other hand, M-shaped potentials lead to amultitude of equilibria: one for each maximum and a discrete set of solutions of (7.1) and (7.7) oscillating about the minimum andbounded between the two maxima.

Because birods with hill-shaped potentials have a single equilibrium unloaded configuration that configuration must be stable(otherwise, by contradiction, we would have perpetual motion in the absence of load). We observe that this stable equilibriumcorresponds to either staying on the maximum of V if = 00 or going over the maximum if ≠ 00 . Although studying the stabilityof the problem is beyond the scope of this paper, this fact leads us to conjecture that solutions close to maxima of V may be goodcandidates for stable solutions. In that sense, we expect that the solutions of the M-shaped potentials that are on (if = 00 ) or closeto (if ≠ 00 ) the local maxima should be stable.

Note that equilibrium solutions with θ = 0 lead to planar birods (either straight or forming arcs of circle) and that equilibria with

Fig. 12. The function V controlling Eq. (7.1) and defined in (7.7). Parameters as specified with Γ = 2/3: (a) = 02 & ϕ = 0, (b) =212

& ϕ = 0, (c) = 02 &

ϕ = π24.

Fig. 13. Boundary in parameter space between the two shapes of V θ( ) as defined in (7.7) in the twist less case: ϕ = = 03 . The full curve (blue online) shows the

boundary for Γ = 2/3. The two shades of grey (yellow and red online) colour respecting regions split by that boundary. The dashed curves show the shape of theboundary for different values of Γ . (For interpretation of the references to colour in this figure caption, the reader is referred to the web version of this paper.)


176

Fig. 14. A few particular examples of unloaded birods with hill-shaped V θ( ). (a) Example 1: g = 1, = 02 , = 03 , Γ = 2/3, L = 10. (b) Example 2: g = 5/4, = 02 ,

= 03 , Γ = 2/3, L = 10. (c) Example 3: g = 1, = 02 , =315, Γ = 2/3, L = 50.


177

θ < 0 (resp. θ > 0) lead to right (resp. left) handed helical solutions. The functions V θ( ) displayed in Fig. 12a and b correspond toϕ= = 03 . In Fig. 12c, we observe that ϕ shifts the maximum hill-shaped V to values of θ of the opposite sign; so positive values of

ϕ lead to right-handed helical birods and negative ϕ lead to left-handed helical birods.

7.4. Explicit examples

We conclude this section by illustrating the diversity of birod structures through solving numerically (using a shooting method)the second order ODE problem (7.1) and (7.2) for a few particular choices of parameter values. Once the function θ R( ) is known, wefind R( ) using (6.57). Remembering that the sub-rods are inextensible, we have α = 1± . We can then substitute according to(6.28) and then numerically integrate (6.24) and (6.25) to determine the local frames and subsequently the centrelines. Throughout,we assume constitutively identical sub-rods with isotropic cross-sections: EI EI( ) = ( )+ −, b b= = 1+ − , Γ Γ= = 2/3+ − .

In Fig. 14 we display the result of this process for various parameter values that give hill-shaped potentials. We systematicallyshow the reference shapes of the sub-rods together with the unloaded shape of the birod. In Fig. 14a and b, = 03 and the single

equilibrium is either an arc of a (possibly infinite) circle with θ R( ) ≡ 0 when = 00 or nearly planar in the cases with ≠ 00 . InFig. 14 (c), since ≠ 03 adds an intrinsic twist to the system, even with = 00 the constant θ solution is at a non-zero value and theresulting structure is helical.

Next we turn to the case of M-shaped potentials. According to Fig. 13, there are essentially two ways to achieve this: with largevalues of either 2 or g. The two cases are fundamentally different, the former referring to strong differences in intrinsic curvaturesand the latter dealing with large reference length differentials. We consider each possibility separately. In Figs. 15 and 16 we showthe behaviour of large 2 and g=1. Fig. 15 shows the constant θ solutions, two oppositely oriented helices and one planar solution.These solutions qualitatively correspond to the equilibria of the ladder described in Example 3 of Section 3.1.3 and observed inLachenal et al. (2012). Fig. 16 displays several of the oscillatory solutions about the minimum of V. Note that each crossing of the θ-axis corresponds to a perversion, that is a change in handedness. Such structures are quite common in biology, e.g. tendril perversion

Fig. 15. Equilibria with constant θ R( ) of homogeneous birods with inextensible sub-rods and parameters L L= = 18+ − , = = = 00 1 3 , g=1 and =29

10. Left:

the right handed helical solution. Middle: the left handed helical solution. Right: the neutral (probably unstable) planar solution. In each case, the function V θ( ) isshown in the upper plot together with the position at which the system stays (black dot); the shape of the resulting birod is displayed on the right.

Fig. 16. Equilibria with θ R( ) oscillating k times about θ = 0 with from left to right: k=1 , k=2, k=3 and k=5; same parameters as Fig. 15. In the upper plots, the thin

line (blue online) indicates the shape of the functionV θ( )/( )+ − , while the thick (black online) curve shows the (k times covered) trajectory in the θ V( , /( ))+ − plane;

the function θ R( ) is shown in the bottom plot. The shapes of the resulting structures are displayed (+ rod red online and − rod blue online). (For interpretation of the

references to colour in this figure caption, the reader is referred to the web version of this paper.)


178

in plants (Goriely and Tabor, 1998; McMillen and Goriely, 2002).Figs. 17 and 18 illustrate the case of large g, with all = 0i . Compared to the previous case, we encounter a large difference in

the extremum values of V. These solutions are plotted with small radii for the sub-rods and attaching spokes, to make them easier tovisualise. As before, we illustrate both the constant θ solutions (Fig. 17) and several oscillatory solutions (Fig. 18). The oscillatingsolutions have more difficult to visualise perversions, and also systematically involve self-penetration of the attaching spokes.

Fig. 17. Equilibria with constant θ R( ) of homogeneous birods with inextensible sub-rods and parameters L = 6.08− , = = = = 00 1 2 3 and g=5. From left to

right: the left handed helical solution with θ > 0, the right handed helical solution with θ < 0 and the neutral (probably unstable) planar solution (multiply covered).In each case, the function V θ( )/( )+ − is shown (blue online) in the upper plot where the (black online) dot displays the position at which the system stays (dot); the

function θ R( ) is shown in the bottom plot; and the shape of the resulting birod is displayed on the right. The shapes of the resulting structures are displayed assuming

that the thin spokes (black online) are rigid (similarly to the ladder of Example 3 in Section 3.1.3), the + rod is the longest tube (red online) and the − rod is theshorter inner one (blue online). (For interpretation of the references to colour in this figure caption, the reader is referred to the web version of this paper.)

Fig. 18. Equilibria with θ S( ) oscillating k times about θ = 0 with k=1 (top left), k=2 (top right), k=3 (bottom left) and k=5 (bottom right); same parameters as Fig. 17.

In the top plots, the thin line (blue online) indicates the shape of the function V θ( ), while the thick (black online) curve shows the (k times covered) trajectory in the

V θ( )/( )+ − plane; the function θ R( ) is shown in the bottom plots. The shapes of the resulting structures are displayed assuming that the thin spokes (black online) are

rigid (similarly to the ladder of Example 3 in Section 3.1.3), the + rod is the longest tube (red online) and the − rod is the shorter inner one (blue online). (Forinterpretation of the references to colour in this figure caption, the reader is referred to the web version of this paper.)


179

8. Conclusions

In this paper, we have derived a practical framework to study the mechanical equilibrium of structures composed of twomorphoelastic rods. In particular, we have obtained a set of equations describing the structure as a single macroscopic object. Theconstitutive properties of the two sub-rods, together with their growth laws, were translated into effective constitutive laws for themacroscopic structure. Furthermore, we have obtained equations expressing the balance of external and internal forces andmoments so that the behaviour of the structure under general loads can be studied.

We first considered the cases of 1D and 2D birods. When the structure is constrained on a straight line or when it is planar, it ispossible to describe it as a single Kirchhoff rod. A centreline was defined together with a local material frame of directors. Then,global forces and moments of forces applied at every point of the centreline of the macro-structure were defined, and expressingequilibrium of these moments led to the well-known Kirchhoff equations. The properties of this macro-structure, in particular itsconstitutive laws and its shape in the absence of external loads, can be computed from the underlying properties of the sub-rods, andwe outlined a three-step method for these computations. Even with reduced dimensionality, relevant structures may be modelledand interesting behaviour observed.

In 1D, we found that the macro-structure may be considerably stiffer (or softer depending on the non-linearity of the microscopicconstitutive laws) than its parts, as in the case of the rhubarb stalk, a well-known phenomenon that is easily captured within ourframework. It is worth highlighting the fact that differential growth between the two strands may drive them to their non-linearregimes.

In 2D, we recovered in the limit of small deformation the classic result of Timoshenko on bimetallic strips and also obtainedcounterintuitive results in the simple case of birings. The 2D framework may be particularly appealing for modelling applications, abalance of the geometrical freedom of two dimensions combined with computational simplicity. Indeed, as stated, a planar birod ismechanically equivalent to a rod, albeit one with potentially complex intrinsic and/or constitutive properties that can evolve in time.

In the general three-dimensional case, we showed that the structure is no longer equivalent to a Kirchhoff rod. Accordingly, weproposed a new geometric description of the birod, better suited to this structure than the traditional ‘centreline equipped with alocal material frame’ description. This new geometric description required a new set of generalised moments, for which weestablished both the relevant constitutive laws and equilibrium equations. Computing the unloaded equilibria of these structures is anon-trivial task in general but we described the steps necessary for arbitrary growth and constitutive laws of the sub-rods.

Finally, we applied this method to the particular case of a homogeneous structure composed of inextensible sub-rods with linearmoments. The problem of determining unloaded equilibria was reduced to finding the solutions of a single second-order differentialequation with Neumann–Neumann boundary conditions. We showed that the complete set of solutions can be computed, although itmay be countably infinite. The shape of several of these equilibria was then (numerically) computed and displayed, demonstrating awealth of behaviour even in this simplest of cases, including bifurcations in solution type from planar to helical as well as helicalperversions.

The stability of these equilibrium solutions remains to be established. Indeed, the differential equation in space does not containin itself sufficient information to assess stability. While our basic birod framework can be formulated with time as an explicitmechanical variable, for simplicity of presentation and computation we have primarily only dealt with mechanical equilibrium. Thequestion of stability can be treated through linear stability analysis of the full dynamic equations or by analysing the second variationof the energy, following the techniques developed in Lessinnes and Goriely 2016.

Acknowledgements

A.G. is a Wolfson/Royal Society Merit Award Holder and both A.G. and T.L. acknowledge support from EC Framework VIIthrough a Reintegration Grant and a Marie Curie Fellowship.

Appendix A. Normal frames and frame of directors

The motivation for using normal frames compared to, say, the Frenet frame, in the present context is three fold. First, they play aparticularly illuminating role in the geometric description of elastic rods: Darboux vectors of unshearable rods can be viewed asnatural frames that can twist (cf. below). Second it is somewhat easier to infer geometric information about the 3D curve r from itsnormal development than from a Frenet development, see Table A1. Third, a normal development only requires the curve r to be C2

and does not suffer from pathological behaviour if the curve is straight for a finite length while a Frenet frame needs both C3 and

non-vanishingR

r∂∂

2

2 .

While a normal frame is a property of the centreline of the rod, the frame of directors contain also information about theorientation of cross sections. Because we focus on unshearable rods, both c3 and d3 are aligned with the tangent vector to r.Therefore, there exists a function φ such that


180

⎛

⎝⎜⎜

⎞

⎠⎟⎟R R

φ φφ φd c= ( ) with =

cos −sin 0sin cos 0

0 0 1,j i φ ij φ3, 3,

(A.1)

and

⎛⎝⎜

⎞⎠⎟R s

RR

ssR

d cc

∂∂

=∂∂

( ) +∂( )

∂∂∂

j iφ ij i

φ ij3,

3,

(A.2)

⎛⎝⎜

⎞⎠⎟R

v R R φs

dp c c d

∂∂

= × ( ) + ( × ( ) ) ∂∂

ji φ ij i φ ij3, 3 3,

(A.3)

⎛⎝⎜

⎞⎠⎟R

v φR

dp d d

∂∂

= + ∂∂

×jj3

(A.4)

Eqs. (A.4) together with Eq. (2.9) lead to the relation

v φR

u p d= + ∂∂

.3 (A.5)

Because p is orthogonal to c3 and therefore to d3, this last relation shows that

φR

u = ∂∂3 (A.6)

describes the twist of the rod about the centreline which corresponds to the change of angle φ between a normal frame and the frameof directors. Furthermore (A.5) implies

v p pd d c cu + u = ( + )1 1 2 2 1 1 2 2 (A.7)

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

φ φφ φ

vppso that

cos −sinsin cos

uu = .1

2

1

2 (A.8)

Eqs. (2.6), (2.8), (A.6) and (A.8) lead to

⎛⎝⎜⎜

⎞⎠⎟⎟κ

vτ

v= 1 u + u and = 1 u +

u ∂ u − u ∂ uu + u

.R R12

22

31 2 2 1

12

22

(A.9)

These relations (A.9) compared with 2021 of part I (Moulton et al., 2013) prove that the curvature and torsion defined in the contextof the normal development in Eqs. (2.6) and (2.8) are indeed the same quantities as those defined in the usual Frenet Frame.Furthermore, because p1 and p2 describe the bending of the centreline about c1 and c2, Eq. (A.8) shows that u1 and u2 respectivelydescribe the bending of the rod about d1 and d2.

Appendix B. Geometry of the 3D birod as slender structure themselves

The objective of this appendix is to show how the shape of the sub-rods can be recovered from the centreline of the birod r and itslocal frame d d d{ , , }1 2 3 . In other terms we show that under the definitions of Section 3.1.2 and after the functions g± have beencomputed, upon knowing the stretch α and Darboux vectors u of the birod, we can recover the stretches α± of the sub-rods, the angleθ between their tangents d3

± and their respective Darboux vectors u±.We open by recalling the definition and assumptions listed in Section 3.1.2, most of which is summarised in Fig. B1a. Then we

introduce the angles θ and τ shown in Fig. B1b and consider separately the relations concerning α± and θ (Section B.1) and therelations involving u± (Section B.2). The former are concisely gathered in Fig. B2 while the latter lead to (B.45)–(B.47).

In Section 3.1.2, a birod was defined as a composite object made of two sub-rods the centrelines r±, assumed to be at constantdistance a from one another, were equipped with the local material frames d d d{ , , }1 2 3 . Their local stretch α± and Darboux vector u±

were defined as

Table A1Description of special curves by curvature and torsion vs normal development. Note that the relation between shape and normal development is bijective while theshapes only imply the Frenet development (e.g. uniformly vanishing torsion does not insure planarity). The second column contains trivial results. Proofs regardingthe third column can be found in Bishop 1975.

Shape of r Frenet κ τ( , ) Normal development (p , p )1 2

Straight line κ τ= = 0 p = p = 01 2Circle κ constant, τ = 0 p1 & p2 constant

Planar curve τ = 0 Within a straight line through the originCurve on a sphere of radius ρ ? Within a straight line at distance ρ1/ from the origin


181

αS Sr d

u d= ∂∂

,∂∂

= × ,ii

±±

± ±±

(B.1)

where S± are the reference arc-lengthes along r±.Furthermore, it was assumed that there exists a mapping G pairing points along both centreline. Intuitively, G pairs centre-points

belonging to glued cross-sections.Finally, choosing a point on the centreline χ (with χ =+ or − labelling the sub-rods), the azimuth of the paired point (on χ− ) in the

local frame of χ is recorded by the material angle φ χ (cf. Fig. B1a).The vector field d1 was defined as the unit vector pointing from the point r− to its paired point in the + centreline cf. (3.16)

recalled here

ar r d= + .+ −1 (B.2)

Because the centrelines are at constant distance a, each of them lies on a cylinder of radius a centred on the other. It follows that thetangent vectors d3

± are perpendicular to the radial unit vector d1.The centreline of the birod r was defined by (3.10) and implied

aβr r d= + .−1 (B.3)

Fig. B1. A summary of the key geometrical objects defined in Section 3.1.2. (a) The centrelines r± of the sub-rods (red and blue online) and r of the birod (yellowonline), the local frames of the sub-rods (black online), the common chord (magenta online) and the orthonormal local frame of the birod (blue online). (b) A close upon the tangent vectors to the sub-rods in the frame of the birod together with the definition of the angles θ and τ. (For interpretation of the references to colour in thisfigure caption, the reader is referred to the web version of this paper.)

Fig. B2. This geometric construction allows to find α, au2 and au3 as a function of α± and θ once the functions g± are known. The construction, built in the plane

perpendicular to d1, proceeds as follow. Define a unit reference length, an arbitrary reference point O and an arbitrary unit vector d3− through O. Find d3

+ by rotating

the vector d3− by the oriented angle θ. Find points A± so that OA d±

3± and OA α g| | =± ± ±, where AB and AB| | are respectively the vector and the distance from the point A

to the point B in units of our reference length. Find the point B− on the straight line OA− such that β= (1 − )OBOA| −|| −|

. Draw the parallel to OA− through B− on which we

find the point B+ such that β=B BOA

| − +|| −|

. By virtue of Eq. (B.8), α OB= | |+ and d3 is the unit vector on the oriented line OB+. Define the points P and Q as the intersections

of the perpendicular to OB+ through A− and A+ respectively. By virtue of Eq. (B.15), a OP OQu = | | − | |2 , and a A Q PAu = − | | − | |3+ − .


182

The vector field d3 was defined as unit tangent to r. Furthermore, we defined S as the reference arc-length r and we defined g = SS

± ∂∂

±.

Taking the derivative of (B.3) by S proves that d3 is perpendicular to d1 since d d⊥3−

1 and d d∂ ⊥S 1 1 (as d1 is a unit vector). We also definedd d d= ×2 3 1 so that d d d{ , , }1 2 3 forms a right handed orthonormal frame. In any configuration, the stretch α of the birod and itsDarboux vector u were defined as

αS Sr d

u d= ∂∂

;dd

= × .ii

(B.4)

B.1. Stretches α± and angle θ

All four unit vectors d3±, d3 and d2 are coplanar as they are orthogonal to d1. We therefore define θ the angle from d3

− to d3+ and η

the angle from d3− to d3 such that

τ τd d d= cos + sin ,3−

3 2 (B.5)

τ θ τ θd d d= cos( − ) + sin( − ) .3+

3 2 (B.6)

It is possible to compute the angles τ and τ θ− as a function of α± and θ as follows. Eqs. (B.2) and (B.3) imply

β βr r r= (1 − ) + ,− + (B.7)

the derivative by S of which leads to

α α g β α g βd d d= (1 − ) + .3− −

3− + +

3+ (B.8)

Substituting d3± according to (B.5) and (B.6) in (B.8) and projecting on d3 and d2 leads to the following linear system in τcos and τsin :

⎧⎨⎩α α g β α g β θ τ α g β θ τ

α g β θ τ α g β α g β θ τ= [ (1 − ) + cos ]cos + [ sin ]sin ,

0 = − [ sin ]cos + [ (1 − ) + cos ]sin ,

− − + + + +

+ + − − + +(B.9)

the inversion of which gives

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟τ

τ

αα g β α g β θ

α g β θ

α g β α g β α g α g β β θcossin =

(1 − ) + cossin

( (1 − )) + ( ) + 2 (1 − )cos.

− − + +

+ +

− − 2 + + 2 − − + + (B.10)

Taking the norm of (B.8), we obtain

α α g β α g β α g α g β β θ= ( (1 − )) + ( ) + 2 (1 − )cos ,− − 2 + + 2 − − + +(B.11)

and (B.10) simplifies to

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

ττ α

α g β α g β θα g β θ α

α g βα

R α g βcossin = 1 (1 − ) + cos

sin= 1 (1 − )

0+ 1

0,θ

− − + +

+ +

− − + +

(B.12)

where Rχ is the 2D rotation matrix of angle χ. The second formulation in (B.12) is practical as we can directly obtain

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

τ θτ θ

R ττ α

α g β α g β θα g β θ

cos( − )sin( − ) = cos

sin = 1 + (1 − )cos− (1 − )sin

.θ−

+ + − −

− −(B.13)

We are now ready to express the components of u as a function of α± and θ. The derivative of (B.2) by S yields

α g α g ad d d d= + (u − u ).+ +3+ − −

3−

3 2 2 3 (B.14)

Substituting d3± according to (B.5) and (B.6) in (B.8) and projecting on d3 and d2 gives

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟a M

α gα g

Mη η θη η θ

uu = where =

cos −cos( − )−sin sin( − ) .2

3 1

− −

+ + 1(B.15)

Substituting in turn (B.12) and (B.13) in (B.15), we obtain

a α g β α g β α g α g θ β

α g β α g β α g α g β β θu = ( ) (1 − ) − ( ) + ( )( )cos (2 − 1)

( (1 − )) + ( ) + 2 (1 − )cos,2

− − 2 + + 2 − − + +

− − 2 + + 2 − − + +(B.16)

a α g α g θ

α g β α g β α g α g β β θu = − sin

( (1 − )) + ( ) + 2 (1 − )cos.3

− − + +

− − 2 + + 2 − − + +(B.17)

Relations (B.16) and (B.17) together with (B.11) prove that once the functions g± are known, the three variables α±, θ are in one-to-one correspondence with the three variables α, u2 and u3. Indeed, Eqs. (B.11), (B.16) and (B.17) can be inverted (since α > 0± and


183

θcos > 0) to obtain

αg

a β aβ α= 1 u + ( u + ) ,−−

232 2

22

(B.18)

αg

a β α a β= 1 (1 − ) u + ( − (1 − )u ) ,++

2 232

22

(B.19)

θaα

a β α aβ a β α a βsin =

− u

u + ( + u ) (1 − ) u + ( − (1 − )u ).3

2 232

22 2 2

32

22

(B.20)

Although Eqs. (B.16)–(B.20) appear tedious, these results can be brought together in a relatively simple geometric construction(see Fig. B2) which allows to find the variables α, u2 and u3 as a function of the angle θ and of α g± ± (represented by the positions ofpoints A± along d3

±). We readily observe in Fig. B2 that

(a) if θ → 0, then

a α g α gu → − ,2− − + + (B.21)

au → 0,3 (B.22)

α α g β α g β→ (1 − ) + .− − + + (B.23)

so that we recover the geometric relations (5.3) of the 2D case;(b) if β → 0, then B A→+ −, τ → 0, and

α α g→ ,− − (B.24)

a α g θα gu → − cos ,2− − + + (B.25)

a α g θu → − sin ;3+ + (B.26)

(c) similarly, if β → 1, then B A→+ +, τ θ→ and

α α g→ ,+ + (B.27)

a α g θ α gu → cos − ,2− − + + (B.28)

a α g θu → − sin .3− − (B.29)

The one-to-one correspondence between α±, θ and α, u2, u3 implies we can work with either set of variables to solve problems.While the latter gives direct geometrical information on the birod, we shall see that in some practical cases, working with the formercan be easier. In the derivations of Appendix E we extensively use the matrices Rτ, Rτ θ− and M1. It is therefore practical to be able toexpress them as a function of our two sets of variables (no explicit dependance on τ). Following from (B.12) and (B.13):

αR α g βR α g β= + (1 − ) ,τ θ±+ +

±− −

2 (B.30)

αR α g β α g β R= + (1 − ) .τ θ θ±( − )+ +

2− −

∓ (B.31)

Next we observe from the definition of M1 that

M R M R M= = ,τ θ τ1 −( − ) 2 − 3 (B.32)

R M M= ,θ 2 3 (B.33)

where M = ( )θθ2

cos−sin

−10 , M = ( )θ

θ310

−cos−sin .

Substituting (B.30) and (B.31) in (B.32) and using (B.33), we obtain

M β α gα

M β α gα

M= + (1 − ) .1

+ +

2

− −

3 (B.34)

Substitutions according to (B.18)–(B.20) lead to

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟M = .

α βa

α βa βa

α β a

α β a β a

βa

α βa βa

β a

α β a β a

1

+ u

( + u ) + ( u )

− (1 − ) u

( − (1 − ) u ) + ((1 − ) u )

u

( + u ) + ( u )

(1 − ) u

( − (1 − ) u ) + ((1 − ) u )

2

22

32

2

22

32

3

22

32

3

22

32

(B.35)

Similarly, substituting α± and θ in (B.12) and (B.13) according to (B.18)–(B.20), we obtain


184

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

Rα βa βa

α βa βaβa α βa

Rα β a β a

α β a β aβ a α β a

= 1

( + u ) + ( u )

+ u u− u + u , = 1

( − (1 − ) u ) + ((1 − ) u )

−( − (1 − ) u (1 − ) u−(1 − ) u − (1 − ) u .

τ τ θ2

23

22 3

3 2−

22

32

2 3

3 2 (B.36)

Finally, for any 2×2 matrix A, defining⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟A A= 0 1

−1 00 −11 0

∼so that A A= ∼

A

T−1 1det( ) , we note that

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟M τ τ θ

τ τ θM τ θ τ

τ θ τM

θτ θ τ θ

τ τ= cos −cos( − )

−sin sin( − ) ; = sin( − ) sincos( − ) cos ; = −1

sinsin( − ) cos( − )

sin cos.1 1 1

−1

(B.37)

B.2. Deriving the Darboux vectors of the sub-rods

Next we derive the geometric relations between u± and u. By definition of φ− (cf. Fig. B1a) and η (cf. Fig. B1b),

d d d d d d d( ) = ( × )R τ1 2 3 1 3−

1 3−

1, (B.38)

d d d d d d( ) = ( )R R ,φ τ1 2 3 1−

2−

3−

3, 1,− (B.39)

D D⇔ = R R ,φ τ−

3, 1,− (B.40)

D Dand similarly = R R .φ π τ θ+

3, + 1, −+ (B.41)

where

⎛

⎝⎜⎜

⎞

⎠⎟⎟

χ χχ χR =

cos −sin 0sin cos 0

0 0 1χ3, and

⎛

⎝⎜⎜

⎞

⎠⎟⎟χ χ

χ χR =

1 0 00 cos −sin0 sin cos

χ1, .

Using the matrix form of u defined in (2.13), remembering that g∂ = ∂S S− − and invoking (B.40), we obtain

gS

U D D= ∂∂

T− − −−

(B.42)

gS

U D D= (R ·R ) ∂∂

[ (R ·R ) ],φ τT

φ τT− −

3, 1, 3, 1,− −(B.43)

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟g

φφ

φ φτ φU U= (R ·R ) (R ·R ) −

0 0 sin0 0 −cos

−sin cos 0′ −

0 −1 01 0 00 0 0

′,φ τ φ τT− −

3, 1, 3, 1,

−

−

− −

−− −

(B.44)

where ′ =S∂

∂ .After developing (B.44) in components, we obtain

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟g τ φ

uuu

= R Ruuu

− ′R100

− ′001

.φ τ φ−

1−

2−

3−

3, 1,

1

2

33,

−− −

(B.45)

The same procedure can be applied to U+ after which we obtain

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟g τ θ φ

uuu

= R Ruuu

− ( − )′R100

− ′001

.φ π τ θ φ π+

1+

2+

3+

3, + 1, −

1

2

33, +

++ +

(B.46)

Applying D− and D+ respectively to (B.45) and (B.46), we obtain

g τ φ g τ θ φu u d d u u d d= − ′ − ′ , = − ( − )′ − ′ .− −1

−3− + +

1+

3+ (B.47)

Appendix C. The unloaded 2D ring

Consider a planar, extensible rod, with constant reference curvature u2 and linear constitutive laws. Form a closed ring by joiningthe ends of the rod such that the tangents to the centreline are aligned. What stable configuration does the ring assume when noexternal loads are applied?

Suppose the rod has length L and constitutive laws

k α Kn = ( − 1), m = (u − u ).3 2 2 2 (C.1)

Define one point of the ring as S=0 and d30 the tangent to the centreline at that point. Then define θ S( ) the oriented angle from d30


185

to the tangent at that point Sd ( )3 . By definition of θ, θ(0) = 0. Furthermore, imposing θ S( ) to be continuous on S L∈ [0, ] yields

θ L π( ) = 2 on a ring. Finally, note that u = θS2

dd.

The current configuration is completely specified provided we know the two functions α S( ) and θ S( ). Equilibrium configurationscan be found by finding extrema of the energy functional :

∫α S θ SK

α SK

θ S S[ ( ), ( )] =2

( ( ) − 1) +2

( ′( ) − u ) d ,L

0

1 2 22

2(C.2)

under the constraints

∫ ∫α S θ S S α S θ S S( )cos ( ) d = 0, ( )sin ( ) d = 0,L L

0 0 (C.3)

which ensure that the extremities of the rod meet ( α θ= cosS

r dd ·d

30 , α θ= sinS

r d dd ·( × )d2 30 ), the constraint (C.3) ensures that the tangent is

aligned and that the rod makes a full turn.Consider the easier problem of minimising (C.2) with the boundary conditions θ(0) = 0 and θ L π( ) = 2 alone. The Euler–Lagrange

equations derived from for the integrand are

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟α α

K αθ θ

K θ∂∂ ′

′− ∂

∂= ( − 1) = 0, ∂

∂ ′′

− ∂∂

= ″ = 0.1 2(C.4)

From this and the boundary conditions we conclude α = 1 and θ S= πL

2 . This solution also satisfies the constraints (C.3).As this is the only extremum of (C.2) with adequate boundary conditions but no constraints, it is a global minimum of the

constrained problem. Finding all equilibria of the constrained problem is a much harder task that we will not undertake here. In anycase, this simple exercise demonstrates that if a rod with constant intrinsic curvature is closed into a circular ring, the length does notchange during the deformation, and thus the energy minimising ring is the one with radius L

π2.

Appendix D. Linear approximation to the constitutive laws of 2D morphoelastic birods

The constitutive laws (5.8) can be linearised around the reference state of the birod, giving

k αα

αk α

ggk g k α

gga β k agβ k

ggk g k

N =∂ ( , u )

∂( − 1) +

∂ ( , u )∂u

(u − u ) + h. o. t. = 1 [∂ + ∂ ] ( − 1)

+ 1 [ − (1 − )∂ + ∂ ] (u − u )+ 1 [∂ + ∂ ] (u − u ) + h.o.t.,

α αα α

α α u u

32

=1u =u

2

2 =1u =u

2 2 −+ −

−+ −

2 2 −+ −

2 2

2 2 2 2

+ −

0

+ −

0

+ −

0 (D.1)

and

⎡⎣⎢

⎤⎦⎥

K αα

αK α

ggK g K α

ggK g K

gga β k agβ k α

gga β k ga β k

gga β k agβ k

M =∂ ( , u )

∂( − 1) +

∂ ( , u )∂u

(u − u ) + h. o. t = 1 [∂ + ∂ ] ( − 1)

+ 1 [∂ + ∂ ] (u − u ) + 1 [ − (1 − )∂ + ∂ ] ( − 1)

+ 1 [ (1 − ) ∂ + ∂ ] (u − u ) + 1 − (1 − )∂ + ∂ (u − u ) + h.o.t.,

α αα α

u u α α

α α

22

=1u =u

2

2 =1u =u

2 2 −+ −

−+ −

2 2 −+ −

−2 2 + 2 2 −

2 2 − u+

u−

2 2

2 2 2 2

+ −

0

+ −

0

+ −

0

+ −

02+ 2

−

0 (D.2)

where we have defined the reference state of the birod 0 as α a β gg= (1 − (1 − )u )+2

−, ggu = u2+

2−, α aβ g= (1 + u )−

2− and

gu = u2−

2−.

This linearisation is particularly useful if the sub-rods are uniform and uniformly glued so that k±, K± and g are independent of S.If furthermore their constitutive laws are diagonal so that k∂ = 0u

±2± and K∂ = 0α

±± , we choose

β βk

k g k= ≔

∂

∂ + ∂.α

α α

⋆+

+ −+

+ −0 (D.3)

With this choice, the birod also has diagonal constitutive laws at first order,

ggk g k αN = −1 [∂ + ∂ ] ( − 1) + h.o.t.,α α3 −

+ −+ −

0 (D.4)

⎡⎣⎢⎢

⎤⎦⎥⎥gg

K g K ag k k

k g kM = 1 ∂ + ∂ +

∂ ∂

∂ + ∂(u − u ) + h. o. t.α α

α α2 − u

+u

− 2− +

+ − 2 22+ −

−+

+ −0 (D.5)

Finally, note that although β explicitly appears in the R.H.S. of (D.3), the derivatives are to be evaluated in the reference state of the


186

birod, in which the values of α± are independent of β. Explicitly, the reference stretches can be found either by solving (5.7) and using(5.3) or by directly inverting

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟k α k α K K ak α K α gα

agK α gα

aak αN = ( ) + ( ) = 0, M = (u ) + (u ) + ( ) = − + − + ( ) = 0.3

+ + − −2

+2+ −

2− − − +

− +−

− +− −

(D.6)

Appendix E. Constitutive laws for n, M and m for 3D extensible birod

E.1. Constitutive law for n

Assuming that the sub-rods are extensible and unshearable so that n dn = ·3+ +

3+ and n dn = ·3

− −3− are known constitutively as a

function of α± (and possibly u±). Using the result of Appendix B, we can therefore infer constitutive laws for n3± as a function of α and

u:

α αu un = n ( , ), n = n ( , ).3+

3+

3−

3− (E.1)

Because the mechanical balance of the birod (3.28)–(3.32) involve N and n rather than the stresses n± in the sub-rods, we need totranslate the constitutive information contained in (E.1) into constitutive relations for N and n. To this end, we first note that (3.28)and (3.30) can be inverted to

n N n= ( ± ).± 12 (E.2)

Next, we invoke the geometric relations (B.5) and (B.6) so that

τ θ τ θ τ τn d N n d d n d N n d dn = · = ( + )·(cos( − ) + sin( − ) ), n = · = ( − )·(cos + sin ).3+ +

3+ 1

2 3 2 3− −

3− 1

2 3 2 (E.3)

These two equations can be gathered in matricidal form (see Eq. (B.37))

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟M M2 n

n=

NN + 1 0

0 −1nn .T T3

+

3− 1

2

31

2

3 (E.4)

When θ = 0, the tangents are aligned d d=3±

3 such that N = n + n3 3+

3− and n = n − n3 3

+3−. When θ ≠ 0, we obtain

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟θ

M M Mnn = −2

sin1 00 −1

nn

− 1 00 −1

NN .2

3 13+

3− 1 1

−1 2

3 (E.5)

This generic form for the constitutive laws of n2 and n3 is practical since M1 and θsin can either be expressed as a function of α± and θusing (B.34) or as a function of α, u2 and u3 using (B.18) and (B.35).

Although we have not been able to express n2 and n3 as a function of kinematic variable only, the relation (E.5) still plays the roleof a constitutive law in the theory of the three-dimensional birod as it eliminates two mechanical variables from the problem. It isimportant to note that we have no constitutive law for any component of N.

E.2. Constitutive law for M and m

The definitions (3.27) of both these global moments involve components of n± for which no constitutive laws exist. As discussedin Section 3.2.3, we propose here one way around that problem. An alternative approach is discussed in Section 6 where we definedifferent global strains; see also Appendix I. In the definitions (3.27) and (3.31) of M and m, let us substitutive n± according to (E.2):

a c cM m m d n d N= + +2

× + −2

× ,+ −1

+ −

1 (E.6)

a c cm m m d N d n= − +2

× + −2

× .+ −1

+ −

1 (E.7)

We already know from (E.5) that we have constitutive laws for the n2 and n3 components of n as a function of N and the birodsgeometric variables (α and u for instance). Similarly, Eqs. (E.6) and (E.7) are constitutive laws for M and m as a function of the strainof the birod (e.g. α and u) and of N. While this construction allows to make progress regarding the count of variables vs count ofequations discussed in Section 3.2.3 (after all Eqs. (E.5)–(E.7) provide 9 equations at the cost of adding 5 variables to the theory, 2for n2 and n3 and 3 for m), it does so at the cost of making the moments of forces in the structure a direct explicit function of the forceapplied on the structure.

Finally to actually obtain an explicit version of the constitutive laws mentioned in the previous paragraph, we need to be able toexpress the constitutive laws of m± and n3

± as a function of α and u. While the latter usually poses little problem, we would also like tobe able to express the constitutive laws for m in the frame of director of the birod (they are originally known in the frame of directorof the sub-rods). In the next subsection, we show how this can be done in the case of linear constitutive laws for the moments m±.


187

E.3. Constitutive laws for m± as a function of the kinematic variables of the birod

To be able to treat both the ‘+’ and the ‘−’ rod at once, we introduce

(a) the angles τ τ=− and τ τ θ= −+ ;(b) the matrices R = R Rφ π τ θ

+3, + 1, −+ and R = R Rφ τ

−3, 1,− so that according to (B.40) and (B.41), we have D D= R± ±;

(c) M M=+2 and M M=−

3.

According to (2.20) with the definitions listed below (6.46), linear constitutive laws for the sub-rods can be expressed in thefollowing matricidal form:

⎛

⎝⎜⎜

⎞

⎠⎟⎟⎡

⎣

⎢⎢⎢

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎤

⎦

⎥⎥⎥EI bΓ

m D= ( )1 0 00 00 0

uuu

−uuu

.± ± ± ±

±

1±

2±

3±

1±

2±

3±

(E.8)

⎛

⎝⎜⎜

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎞

⎠⎟⎟

⎡

⎣

⎢⎢⎢⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎤

⎦

⎥⎥⎥EIg

b Γτ g

gg φ

m D= ( ) R + ( − 1)0 0 00 1 00 0 0

+ ( − 1)0 0 00 0 00 0 1

Ru − ′

uu

− Ruu

u + ′,B B B B± ( .45, .46),( .40, .41)

−± ±−1 ± ± ± 1

±

2

3

±−1

±1±

±2±

±3± ±

(E.9)

where EI EI EI( ) = ( ) + ( )g

− 1 +, = EIg EI

+ ( )( )

+and = EI

EI− ( )

( )

−. Furthermore,

⎛⎝⎜

⎛⎝⎜

⎞⎠⎟

⎞⎠⎟

⎡⎣⎢⎢⎛⎝⎜

⎞⎠⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

⎡

⎣⎢⎢⎢

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

⎤

⎦

⎥⎥⎥

EIg

τ g φ φ EIg

Γ R R

Rg φ φ

g φb EI

g

τR

gg

g φ

m d d d

D

= ( ) [u − ′ ± (cos u + sin u )] + ( ) ( ) + ( − 1) 0 00 1

uu −

∓ (cos u − sin u )u + ′

+ ( − 1) ( ) R B Ru − ′

uu

−uu

u + ′,

τ τ

τ τ τ φ π π

±−

±1 1

± ± ±1± ±

2±

−±

2 3±

−

2

3 −

± ±2± ±

1±

±3± ±

±−

±1,−

±1,

1±

2

3

3,−( + 2 ± 2 )

±1±

±2±

±3± ±

± ±

± ± ± ±

(E.10)

where the matrices B± are defined as

⎛⎝⎜⎜

⎞⎠⎟⎟B = R

0 0 00 1 00 0 0

R .φ π π φ π π±

3,−( + /2± /2) 3,( + /2± /2)± ±

(E.11)

Next, let us define

⎛⎝⎜

⎞⎠⎟φ φ φ φ φ

Sd d d= ∓ (cos u + sin u ) ∓ (cos u − sin u ) + u + d

d.

±1

±1± ±

2±

2±

2± ±

1±

3 3±

±

±(E.12)

Furthermore, recalling (B.32): R M M= ( )τ±

1−1± , we note that

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

R R M M M MM θ M

Mθ

M

0 00 1 = ( ) 0 0

0 1 =

0 cos0 1 when ± = − ,

1 0cos 0 when ± =+.

τ τ− 1± −1 ±

1−1

1 1−1

1 1−1

± ±

(E.13)

With (E.13) and the notations (E.11) and (E.12), Eq. (E.10) becomes

⎡⎣⎢

⎡⎣⎢

⎛

⎝

⎜⎜⎜⎜⎜

⎛⎝⎜

⎞⎠⎟

⎧

⎨⎪⎪

⎩⎪⎪

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎫

⎬⎪⎪

⎭⎪⎪

⎞

⎠

⎟⎟⎟⎟⎟

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎤

⎦

⎥⎥⎥⎥

⎛

⎝⎜⎜

⎞

⎠⎟⎟⎡

⎣⎢⎢⎢

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟⎤

⎦⎥⎥⎥

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⎤

⎦

⎥⎥⎥⎥

EIg

τ g EIg

M Γθ

θM g M b

EIg

M M M Mτ

g M

m d d d

D

= ( ) (u − ′ − ) + ( ) ( ) 1 00 1 + ( − 1)

1 0cos 0

0 cos0 1

( )uu − ( ) + ( − 1)

( ) 1 0 000

1 0 00 ( )0

B1 0 000

1 0 00 ( )0

u − ′uu

−1 0 00 ( )0

.

±−

±1 1

± ±1±

−±

2 3 1±

1−1 2

3± ± −1 2

±

3±

±

−±

1± −1 ± ±

1−1

1±

2

3

± ± −11±

2±

3±

(E.14)

The two matrices superposed in the second line of (E.14) mean that the top one must be used for m+ and the bottom one for m−.The expression (E.14) gives the constitutive laws of m± in the orthonormal frame of director of the birod and as a function of α

and u. Indeed, see (B.30)–(B.37) for expressions of all the matrices involved as a function of α and u. Similarly, τ′ could besubstituted (although the expression would become very lengthy) for α and u by using Eqs. (B.12) and (B.18)–(B.20).

The complexity of the expression (E.14) and the difficulty to use it to solve practical birods is one of the major motivations behind


188

the development of the alternative geometric description of the birod in Section 6.1. For comparison, we then obtain the fully explicitconstitutive laws (H.6) for m±.

Appendix F. Proof of (6.7)

In this section, for any vector u ∈ 3, we denote by u[ ]× the operator

u w u w[ ] : → : → × .×3 3 (F.1)

Let d d d{ , , }1 2 3 be a right-handed orthonormal bases of 3 and D the matrix, the ith column of which are the components of d{ }i inthe fixed right handed orthonormal lab frame e e e{ , , }1 2 3 . Let R be an orthogonal matrix of determinant 1 (i.e. a rotation matrix). LetU be a skew symmetric matrix defined by u , u1 2 and u3:

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟U =

0 −u uu 0 −u

−u u 0.

3 2

3 1

2 1

Finally, define c c c{ , , }1 2 3 the right-handed orthonormal frame such that c c c d d d( ) = ( )RT1 2 3 1 2 3 .

Lemma 1. The matrix UR RT is the matrix of the operator c[ u ]i i × in the basis d{ }i .

Proof. Applying standard change of basis formula, the operator A, the matrix of which is R URT in the basis d{ }i has the matrixR(R UR)R = UT T in the basis c{ }i . Direct expansion of u w× in components in the basis c{ }i shows that U is indeed the matrix of theoperator c[u ]i i × in that basis. Hence A c= [u ]i i ×.□

Next let us define the short hands Ri ω, for the matrices:

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟ω ω

ω ω

ω ω

ω ω

ω ωω ωR =

1 0 00 cos −sin0 sin cos

, R =cos 0 sin

0 1 0−sin 0 cos

, R =cos −sin 0sin cos 0

0 0 1.ω ω ω1, 2, 3,

(F.2)

Note that (as can be checked by direct expansion)

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟S

δ δδ δδ δ

ωS

dd

R = R0 −

0 −− 0

dd

,i ω S i ω S

i i

i i

i i

, ( ) , ( )

3, 2,

3, 1,

2, 1,

Ui (F.3)

where there is no summation over repeated indices and δ = 1i j, if i=j and 0 otherwise.We are now ready to prove (6.7). Introducing the matrices D d d d= ( )±

1±

2±

3± , we can rewrite (6.6) as

D D= R R R .φ θ φ π+ −

3, 1, 3,− −

Q

R

− +

(F.4)

Taking the derivative of (F.4) by S an arbitrary parameterisation of the sections of the birod:

⎛⎝⎜

⎞⎠⎟

g g φS

θS

φS

g φS

Q θS

φS

D D D D D

D

U = U R R R + dd

R U R R + dd

R R U R − dd

R R R U

= R U R + dd

Q U + dd

R U R + − dd

U ,

φ θ φ π φ θ φ π φ θ φ π φ θ φ π

T Tφ π

Tφ π

+ + + − − −3, 1, 3,− −

−−

3, 3 1, 3,− −−

3, 1, 1 3,− −

+−

3, 1, 3,− − 3

+ − −−

3 3,− − 1 3,− −

+

3

− + − + − + − +

+ +(F.5)

where the first equality comes from applying (F.3) to the derivative of the LHS of (F.4); and the second equality comes fromsubstituting D− according to (F.4).

Next we compute

D d d d d D d d d dR = ( × ), Q = ( × ).φ πT T+3,− − 1 3

+1 3

+ +1 3

−1 3

−+ (F.6)

Applying Lemma 1, the equality (F.5) between matrices can be translated into an equality between operators:

⎡⎣⎢

⎤⎦⎥g g φ

SθS

φS

g g φS

θS

φS

g

g φS

θS

φS

u u d d d u u d d d u

u d d d

[ ] = [ ] + dd

[ ] + dd

[ ] − dd

[ ] , ⇔ [ ] = + dd

+ dd

− dd

. ⇔

= + dd

+ dd

− dd

.

+ +×

− −×

−

3−

× 1 ×

+

3+

×+ +

×− −

−

3−

1

+

3+

×

+ +

− −−

3−

1

+

3+

(F.7)


189

Appendix G. Calculus for Section 6.2

In this section we derive equilibrium equations (6.33)–(6.39). Formally, the idea is to project the force and moment balanceequations (3.28), (3.29), and (3.32) onto the non-orthogonal bases. For convenience, we reproduce (3.28), (3.29), and (3.32) here:

N F′ + = 0, (G.1)

M r N L′ + ′ × + = 0, (G.2)

Lm d r n d( ′)· + ( ′ × )· + 2l + ˇ = 0.1 1 1 1 (G.3)

First we project (G.1) along c−. Since

⎡⎣⎢

⎤⎦⎥

⎡⎣⎢⎛⎝⎜

⎞⎠⎟

⎤⎦⎥θ θ

θN F c N c N c F c c c c c c F c( ′ + )· = ( · )′ − ·( )′ + · = − (N )′ − N + Nsin

+ Nsin

· −′

2× + · ,− − − − − 1

1+

+

−

−

1 − −

we obtain

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

θ θθ

ga

α α g θ F c(N )′ = −′

2N − N cos

sin− ( − cos )N + · ,− 1

+ −−

− +1 −

(G.4)

where we substituted according to (6.28). Similarly, N F c( ′ + )· = 0+ leads to

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

θ θθ

ga

α θ α g F c(N )′ = − +′

2N − N cos

sin− ( cos − )N + · ,+ 1

− +−

− +1 +

(G.5)

and N F c( ′ + )· = 01 gives (6.33).It turns out that (6.36)–(6.38) can be recovered in a similar fashion by projecting Eq. (G.2) on c± and c1. Before we proceed, we

however need to express M as a function of P. First, we project (E.2) on d c=3±

±:

n = − (N + n ) and n = − (N − n ),3+ 1

2 + + 3− 1

2 − − (G.6)

where the vector n was defined in Appendix E as n n n= −+ −. Eqs. (G.6) can be reorganised as

n = − N − 2 n and n = N + 2 n .+ + 3+

− − 3−

(G.7)

Second, we recall (E.6), where we expand N and n in the basis c c c{ , , }1 + − and use the definition of the scalars c a β= (1 − )+ andc aβ=− :

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

aθ θ

a βθ θ

M m m d c c c d c c c= + +2

× n + nsin

+ nsin

+2

(1 − 2 ) × N + Nsin

+ Nsin

.+ −1 1

1+

+

−−

1 11

+

+

−−

(G.8)

Substituting n± according to (G.7) in (G.8) and remembering according to (6.9) that d c=11, we obtain

⎛⎝⎜

⎞⎠⎟

aθ θ

βθ

βθ

M m m c c c c c= + +2

× − 2 nsin

+ 2 nsin

− 2 Nsin

+ 2 (1 − )Nsin

.+ − 13+

+

3−

−

+

+

−

−

(G.9)

Finally, we expand m m++ − in the basis c c c{ , , }1 + − and use the relations (6.13) and (6.15) to obtain the following expression in whichwe recognise the components P1, P+ and P− defined in (6.29)

θ θa

θa

θa β

θa β

θ

a βθ

aβθ

M m m c c m m cc

m m c c c c c c

Pc c

= (( + )· ) − (( + )· )sin

− (( + )· )sin

− nsin

− nsin

− (1 − )Nsin

− Nsin

= − (1 − )Nsin

−Nsin

.

+ − 11

+ − + + + − − −3− +

3+ −

−+

+−

− + + −(G.10)

We are now ready to project (G.2) on c−. To this end, let us compute

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢⎛⎝⎜

⎞⎠⎟

⎤⎦⎥

aβ

θ θa β

θa β

θθ

g β α βα g

M r N c M c M c c r N

cc c c c

c c

c c c N

( ′ + ′ × )· = ( · )′ − ·( )′ + ( × ′)· = ( − P + N )′

− P + Psin

+ Psin

− (1 − )Nsin

−Nsin

· +′

2×

+ ( × ((1 − ) + ))· .

− − − − −+

11

+ + − − − + + −1

−

− − −−

++ (G.11)

Next, in Eq. (G.11), substitute (N )′+ according to (G.5) and use (6.28) to compute the factor in the square bracket:


190

⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

aβ θ θθ

g β α θ α g aβ ga

θ

α θ θθ

θ aβ β θ

θga

a β α θ βα g

M r N c F c( ′ + ′ × )· = − (P )′ − +′

2N − N cos

sin− ( cos − )N + · + P sin

+ +′

2P + P cos

sin− +

′2

(1 − )N + N cossin

− N ((1 − ) cos + ),

− − − + − − +1 + 1

−

−+ −

− +

−

1− +

(G.12)

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

a θθ

ga

α θ a aβ ga

α θ

θ θθ

M r N c F c( ′ + ′ × )· = − (P )′ − +′

2N

sin− cos N ++ · + sin P

+ +′

2P + P cos

sin.

− − −−

−1 +

−−

1

+ −

(G.13)

Finally substituting (G.13) in (G.2) gives

⎛⎝⎜

⎞⎠⎟

ga

α θ a θ θ θ aθ

aβF c L c(P )′ = (P sin − N cos ) +′

2+ P + P cos − N

sin+ · + · ,−

−− 1

1

+ −−

+−

(G.14)

which is the balance equation (6.37).Similarly, projecting (G.2) on c1 and on c+ gives (6.36) and (6.38) and projecting (G.3) on c1 gives (6.39) after a similar sequence

of steps.

Appendix H. Linear constitutive laws for moments m±

As mentioned in Section 6.3, the constitutive laws for m± can be written as a function of α±, θ and . Here we carry out thiscomputation in the most common case of linear constitutive laws for m±. Note that practically, not only do we need to express thelaws as a function of the adequate variable but also to project them on the frames c c c{ , , }1 + − in which we intend to perform furthercomputations.

⎪

⎪

⎪

⎪

⎧⎨⎩

⎛⎝⎜⎡⎣⎢

⎤⎦⎥

⎞⎠⎟⎫⎬⎭

⎡⎣⎢

⎡⎣⎢

⎤⎦⎥

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎛⎝⎜

⎞⎠⎟⎤⎦⎥

⎡⎣⎢

⎤⎦⎥⎤⎦⎥⎥

EI b Γ EI b Γ

EIg

θ φ g Γ θα g φ g b

φ φ θ φ φ φ g EIag

a θ g φ φ

α g α g ag φ φ ag φS

Γ θα g a g φS

b

φ φ φ a θ φ α g θ α g ag

m d d d u u u u d d

u u d d c c u c

c c c c c c c

c c c c c

c c

= ( ) [(u − u ) + (u − u ) + (u − u ) ] = ( ) [ − + ( − 1)[( − )· ] + ( − 1)

[( − )· ] ] = ( ) ±′

2− ( )′ − − ( − 1)[sin + ( )′ + u ] + ( − 1)

(cos ∓ sin ) ±′

2− ( )′ ·(cos ∓ sin ) − u = ( ) ±

′2

± (u cos + u sin )

+ + − (u cos − u sin ) − u + dd

− ( − 1) sin + u + dd

+ ( − 1)

(cos ∓ sin ) ∓ sin ( ±′

2) − cos ( cos − ) − u

± ±1±

1±

1± ±

2±

2±

2± ±

3±

3±

3± ± ± ± ± ± ±

2±

2± ±

± ±3±

3±

±

±1 ±

±± ± ± ∓ ∓ ± ±

3±

±±

± ∓ ± 1 1 ±±

± ∓ ± 1 ±2±

±

±±

1± ±

2± ± 1

− − + + + − ±2± ±

1± ± ∓ ±

3±

±

± ±± ∓ ∓ ±

3±

±

± ±±

± ∓ ± 1 ± ± ∓ ∓ ± ± ±2±

(H.1)

Next let us define a measure EI( ) of the total stiffness of the birod together with the total reference curvatures and twists i aboutkey directions

⎛⎝⎜

⎞⎠⎟

EI EIg

EI a g EIg EI

a g EIEI

a gg

a gg

φS

u u d u u d

u c

( ) = ( ) + ( ) , = ( − )· , = ( )( )

, = ( + )· , = ( )( )

,

= ( · ), = u + dd

,

+−

0+ −

1+

+

1+ −

1−

−

2± ±

−± ∓

3± ±

− 3±

±

± (H.2)

which can be inverted to obtain

gag

φ φ gag

φ φ ga g

φS

u = − (cos ( ± ) + sin ), u = ( − sin ( ± ) + cos ), u = − dd

.1±

−

±±

0 1±

2±

2±

−

±±

0 1±

2±

3±

−

± 3± ±

± (H.3)

In particular,

⎛⎝⎜⎜

⎞⎠⎟⎟φ φ g

agu cos + u sin = −

2±

2,1

± ±2± ±

−

±0 1

(H.4)

φ φ ga g

u cos − u sin = .2± ±

1± ±

−

± 2±

(H.5)

The substitution of (H.3)–(H.5) in (H.1) gives


191

⎧⎨⎪⎩⎪

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

⎛

⎝⎜⎜

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

⎫⎬⎪⎭⎪

EIag

a ga

θ ga

α g α g Γ θα g g g Γ

b φ φ a φ θ ga

ga

φ α g θ α g g

m c c c c c c

c c

= ( ) −2

±′

2−

2+ + − ( − 1)sin − −

− ( − 1)(cos ∓ sin ) sin′

2−

2± −

2+cos ( cos − + )) .

±±

±

−1

−0 1 − − + + + − ± ∓ ∓

±−

2± ∓ − ±

3±

±

± ± ∓ ± 1 ±−

0−

1 ± ∓ ∓ ± ± −2±

(H.6)

The last term in (H.6) is by far the most complicated one. It accounts for the fact that if one of the sub-sections does not have thesymmetries of the square (b = 1χ ), and if the common chord d1 is not aligned with a principal axis of that sub-section, their is acoupling between the birod bending about d1 and the stretches in the sub-rods. For the sake of clarity, we ignore that case (a merealgebraic complication) and focus on either of two cases: b = 1χ or φ k π k= ( ∈ )χ , both for χ =+ and χ = −.

Under these restrictions (H.6) simplifies to

⎪⎪

⎪⎪

⎧⎨⎩

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

⎫⎬⎭

EIag

a ga

θ ga

α g α g Γ θα g g

g Γ b α g θ α g g

m c c c c c

c c

= ( ) −2

±′

2−

2+ + − ( − 1)sin −

− −( − 1)( cos − + ) .

±±

±

−1

−0 1 − − + + + − ± ∓ ∓

±−

2± ∓

− ±3±

±± ∓ ∓ ± ± −

2± ∓

(H.7)

Substituting Eq. (H.7) in the definitions (6.29) allows to find the constitutive laws for P1, P+, P−, Q1, Q+ and Q−:

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

EIg

ga

θ ga

m m cP = ( + )· = ( ) −2

+ ( − )′

2−

2,1 + − 1

−

−1 + −

−0

(H.8)

⎡⎣⎢

⎤⎦⎥

a a α EIa

b α α g θ b θ α g α θ

Γ θ α θ

m m cP = − ( + )· − n = − n ( ) − ( ) ( − cos − ) − cos ( − cos − )

+ sin ( sin + ) ,

+ + − +3−

3− − − − − +

2− + + + −

2+

+ + −3+

(H.9)

⎡⎣⎢

⎤⎦⎥

a a α EIa

b θ α g α θ b θ α α g θ

Γ θ α g θ

m m cP = − ( + )· − n = − n ( ) − ( ) cos ( − cos − ) − cos ( − cos − )

+ sin ( sin + ) ,

− + − −3+

3+ + + + + −

2+ − − − +

2−

− − +3+

(H.10)

⎡⎣⎢⎢⎛⎝⎜⎜

⎞⎠⎟⎟

⎛⎝⎜⎜

⎞⎠⎟⎟⎤⎦⎥⎥

EIg

θ ga

ga

m m cQ = ( − )· = ( ) ′2

−2

+ ( − ) −2

,1+ −

1 −

−0 + −

−1

(H.11)

EIa

b θ α α g θ Γ θ α g θm cQ = − · = ( ) [ sin ( − cos − ) + cos ( sin + )],+−

+− − − +

2− − − +

3−

(H.12)

EIa

b θ α g α θ Γ θ α θm cQ = − · = ( ) [ sin ( − cos − ) + cos ( sin + )].−+

−+ + + −

2+ + + −

3+

(H.13)

Appendix I. A computationally friendly version of the birod equilibrium equations

In Sections 6.1–6.3 we established that the equilibria of birods may be described by the four scalars α±, θ and and the 7equations (6.33)–(6.39) involving 10 momenta 7 of which have constitutive laws. While the previous discussions suggest that this isactually a minimal description of the system, some factors in Eqs. (6.33)–(6.39) involve divisions by θsin . Not only this poses anobvious numerical issue, but also it makes it difficult to recover a 2D theory for which θ = 0 along the whole length of the birod. Thecurrent section addresses those issues by proving that the 7 equilibrium equations (6.33)–(6.39) can be replaced by the 11equilibrium equations (I.23)–(I.33) with 4 extra mechanical variables. The latter system presents the advantage that it no longerinvolves division by θsin .

Before we proceed, a geometrical property must be highlighted. Let R R Rd d d{ ( ), ( ), ( )}1 2 3 be a smooth one-parameter family ofright handed orthonormal frames (like for instance the frame of directors of a single Kirchhoff rod), the Darboux vector of which is u.For any vector field v depending on R, we have

R R R R R R Rv d d d( ) = v ( ) ( ) + v ( ) ( ) + v ( ) ( ),1 1 2 2 3 3

the derivative of which is

v d u d d u v′ = (v)′ + v( × ), = (v)′ + × .i i i i i i


192

Furthermore, the projection of this derivative on one of the basis vector gives

v d d u v′· = v′ + ( × )· .j j j (I.1)

We are now ready to proceed. First the factors involving division by θsin in (6.34) and (6.35) can easily be substituted for N±

according to (6.19). By so doing, we however add two variables, namely N±, to the theory. Accordingly, two more equations must beprovided. This could be achieved by projecting (3.28) on c± in a similar fashion to what we did in Appendix G to obtain (6.34) and(6.35). Let us however outline an alternative approach.

At the end of Section 6.1 we remarked that c c c{ , , }1+

− form a right handed orthonormal basis. Hence for any vector v, we obtainv c c c= v − v − v1 1

+ +− − where the unfamiliar signs are due to the definitions (6.16). The Darboux vector of S S Sc c c{ ( ), ( ), ( )}1

+− is

⎛⎝⎜

⎞⎠⎟

θ ga

α ga

α gq c c c= −′

2+ + .−

1

−− +

−+ −

(I.2)

Making use of (I.1), we can easily project (3.28) on c c c{ , , }1+

− . As a result, we obtain

⎧⎨⎪⎩⎪

c q Nc q N

c q N

N′ + ( × )· + F = 0,(N )′ − ( × )· + F = 0,N′ − ( × )· + F = 0.

1 1−

1+ + − +

− −−

− (I.3)

Substituting (I.2) in (I.3), we obtain

ga

α g αN′ + ( N − N ) + F = 0,1

−+

+−

− 1 (I.4)

⎛⎝⎜

⎞⎠⎟

ga

α g θ θ(N )′ + sin N − −′

2N + F = 0,+

−+

1 −+

(I.5)

⎛⎝⎜

⎞⎠⎟

ga

α α g θ θN′ + ( − cos )N + −′

2N + F = 0.−

−− +

1+

−(I.6)

With (I.4) and (I.6), we recover (6.33) and (6.34) after making use of (6.19). Eq. (I.5) however was not part of (6.33)–(6.39) and isone of the two equations we are looking for. The second one can be obtained by repeating the steps highlighted here but this timeprojecting (3.28) on the orthonormal basis c c c{ , − , }1

−+ , the Darboux vector of which is

⎛⎝⎜

⎞⎠⎟

θ ga

α ga

α gq c c c= +′

2+ + .+

1

−− +

−+ −

(I.7)

As a result, we obtain (6.33) a third time together with

⎛⎝⎜

⎞⎠⎟

θ ga

α θ(N )′ + +′

2N − sin N + F = 0,−

+

−−

1−

(I.8)

⎛⎝⎜

⎞⎠⎟

ga

α g α θ θN′ − ( − cos )N − +′

2N + F = 0,+

−+ −

1−

+(I.9)

where (I.9) is equivalent to (6.35) after making use of (6.19) and Eq. (I.8) is new. The system of five equations (I.4)–(I.6), (I.8) and(I.9) with the five unknown N1, N±, N± has been obtained by projecting the vectorial equation (3.28) on various unit vectors.17 In thatsense, it leads to the same set of solutions (under suitable initial or boundary conditions) as the system (6.33) and (6.34) which wasobtained by projecting the same vectorial equations on a different set of unit vectors17.

Next let us define the total moment A about r− according to (3.27) with c a=+ and c = 0− and the total moment B about r+

according to (3.27) with c = 0+ and c a=− :

a aA m m d n B m m d n= + + × and = + − × .+ −1

+ + −1

− (I.10)

It is not surprising that A and B being the total moments due to the same system of forces but about different points, we have (as canbe proven by substituting n N n= −+ − in the definition of A)

aA B d N= + × .1 (I.11)

We observe that P1 and P± defined in (6.29) can be expressed as some components of A and B respectively:

A c B c B c A cP = · = A = · = B , P = − · = B and P = − · = A .1 1 1 1 1 + + + − − − (I.12)

Furthermore, according to Eqs. (3.28) and (3.29), if the birod is in equilibrium A and B obey (3.29)

A r N L 0′ + ( )′ × + = ,A− (I.13)

17 So that the set of these unit vectors generate 3.


193

B r N L 0′ + ( )′ × + = ,B+ (I.14)

where LA and LB are defined according to (3.23) with c a=+ , c = 0− and with c = 0+ , c a=− respectively.Projecting (I.13) and (I.14) on both the orthonormal bases c c c{ , − , }1

−+ and c c c{ , , }1

+− the Darboux vectors of which are q±

defined in (I.2) and (I.7), we find

ga

α g α α gA′ + ( A − A ) + N + L = 0,1

−+

+−

−− − +

1 (I.15)

⎛⎝⎜

⎞⎠⎟

θ ga

α θ a θ(A )′ + +′

2A + (cos N − sin A ) + (L ) = 0,A

−+

−−

1 1−

(I.16)

⎛⎝⎜

⎞⎠⎟

ga

α g α θ θ ga

α aA′ − ( − cos )A − +′

2A + N + (L ) = 0,A+

−+ −

1−

−−

1 +(I.17)

and

ga

α g α ga

α g aB′ + ( B − B ) − N + L = 0,1

−+

+−

−

−+ −

1 (I.18)

⎛⎝⎜

⎞⎠⎟

ga

α g θ θ a θ(B )′ + (sin B − cos N ) − −′

2B + (L ) = 0,B

+−

+1 1 −

+

(I.19)

⎛⎝⎜

⎞⎠⎟

ga

α α g θ θ ga

α g aB′ + ( − cos )B + −′

2B − N + (L ) = 0.B−

−− +

1+

−+

1 −(I.20)

Next, projecting (I.11) on c±, it is straightforward to show that aA − N = B−+

− and aB − N = A+−

+ the substitution of whichtogether with (I.12) in (I.15) and (I.18) show that both equations are equivalent and both read

ga

α g α(P )′ + ( A − B ) + L = 0.1−

++

−− 1 (I.21)

Finally, using the definitions (3.26) and (6.30), the identities (6.18) and the relations (6.12)–(6.15), it is a matter ofstraightforward algebra to show that

θ aθ

θ aθ

P + P cos − Nsin

= − A andP + P cos − N

sin= − B .

+ −−

+

− ++

− (I.22)

Substituting (I.22) in (I.21) allows to recover Eqs. (6.36). Similarly, the substitution of (I.22) in (I.19) and (I.16) leads to Eqs. (6.38)and (6.37). In conclusion, Eqs. (6.36)–(6.38) can be viewed as a sub-set of the projection of (I.13) and (I.14) on c c c{ , − , }1

−+ and

c c c{ , , }1+

− respectively.Gathering Eqs. (I.4)–(I.6), (I.16)–(I.17), (I.19)–(I.20), (I.21), (6.39) in which we substitute (I.22) and (I.12) as necessary, the full

projection of N F 0′ + = and (I.13) and (I.14) together with (6.39) reads

ga

α g αN′ + ( N − N ) + F = 0,1

−+

+−

− 1 (I.23)

⎛⎝⎜

⎞⎠⎟

ga

α g θ θ(N )′ + sin N − −′

2N + F = 0,+

−+

1 −+

(I.24)

⎛⎝⎜

⎞⎠⎟

ga

α α g θ θN′ + ( − cos )N + −′

2N + F = 0,−

−− +

1+

−(I.25)

⎛⎝⎜

⎞⎠⎟

θ ga

α θ(N )′ + +′

2N − sin N + F = 0,−

+

−−

1−

(I.26)

⎛⎝⎜

⎞⎠⎟

ga

α g α θ θN′ − ( − cos )N − +′

2N + F = 0,+

−+ −

1−

+(I.27)

ga

α g α(P )′ + ( A − B ) + L = 0,1−

++

−− 1 (I.28)

⎛⎝⎜

⎞⎠⎟

θ ga

α θ a θ(A )′ + +′

2A + (cos N − sin P) + (L ) = 0,A

−+

−−

1 1−

(I.29)

⎛⎝⎜

⎞⎠⎟

ga

α g α θ θ ga

α aA′ − ( − cos )P − +′

2A + N + (L ) = 0,A+

−+ −

1−

−−

1 +(I.30)


194

⎛⎝⎜

⎞⎠⎟

ga

α g θ θ a θ(B )′ + (sin P − cos N ) − −′

2B + (L ) = 0,B

+−

+1 1 −

+

(I.31)

⎛⎝⎜

⎞⎠⎟

ga

α α g θ θ ga

α g aB′ + ( − cos )P + −′

2B − N + (L ) = 0,B−

−− +

1+

−+

1 −(I.32)

ga

α g α ga

α g α(Q )′ + ( A + B ) − 2 (Q + Q ) + 2l + h = 0,1

−+

+−

−

−

++

−−

1 1 (I.33)

where F is the line density of external forces applied on the birod, LA and LB are the line density of the total moment of externalforces respectively about r− and r+, and h1 is the difference of line density of moment of external forces along c1. The set of 7 equations(6.33)–(6.39) expressing the equilibrium of forces and momenta of the birod can be replaced by the set of 11 equations (I.23)–(I.33). According to (I.12), the constitutive laws that were found for the components P1, P− and P+ – see discussion in Section 6.3, orexplicitly (6.49)–(6.51) – translate to constitutive laws respectively for P1, A− and B+.

Regarding the variable count, the 4 extra equations in (I.23)–(I.33) compared to (6.33)–(6.39) are balanced by the additional 4mechanical variables N+, N−, A+ and B−.

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