Elastocapillary Coiling of an Elastic Rod Inside a Drop

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Elastocapillary Coiling of an Elastic Rod Inside a Drop

Hervé Elettro, Paul Grandgeorge, Sébastien Neukirch

To cite this version:

Hervé Elettro, Paul Grandgeorge, Sébastien Neukirch. Elastocapillary Coiling of an Elastic Rod Inside a Drop. Journal of Elasticity, Springer Verlag, 2016, 127 (2), pp.235-247. �10.1007/s10659-016-9611-4�.

�hal-01445679�

(will be inserted by the editor)

Elastocapillary coiling of an elastic rod inside a drop

Herv´e Elettro · Paul Grandgeorge · S´ebastien Neukirch

Received: date / Accepted: date

Abstract Capillary forces acting at the surface of a liquid drop can be strong enough to deform small objects and recent studies have provided several examples of elas- tic instabilities induced by surface tension. We present such an example where a liquid drop sits on a straight fiber, and we show that the liquid attracts the fiber which thereby coils inside the drop. We derive the equilibrium equations for the system, compute bifurcation curves, and show the packed fiber may adopt several possible configurations inside the drop. We use the energy of the system to discrim- inate between the different configurations and find a intermittent regime between two-dimensional and three-dimensional solutions as more and more fiber is driven inside the drop.

Keywords capillarity·bifurcation·packing

Mathematics Subject Classification (2000) 74K10·74F10·74G65

1 Introduction

The packaging of elastic filaments in cavities [28] is a model system for a large variety of physical phenomena, for example the ejection of DNA from viral capsids [20, 19, 17], carbone nanotubes compaction [6], or the windlass mechanism in spider capture threads [30]. In the case of a mechanical wire spooled in a sphere, or DNA in a capsid, the presence of a motor is necessary for the packing process, the energy to bend the filament being provided by this external actuator. In the case the cavity is a liquid drop (or a bubble in a liquid medium) surface tension may provide the actuation energy:

if the affinity of the filament for the liquid is stronger than that of the filament for the surrounding gas, then the bending energy required for packing could be provided by the difference of surface energies. As always, surface energy prevails at small

Sorbonne Universit´es, UPMC Univ Paris 06, CNRS, UMR 7190, Institut Jean Le Rond d’Alembert, F- 75005 Paris, France

E-mail: sebastien.neukirch@upmc.fr

ez

T

A B

O

`_in

`_out

Fig. 1 A cylindrical rigid rod partially immersed in a liquid drop. As affinity of the rod for the liquid is stronger than that with air, a tensionTis applied to prevent complete immersion. A capillary force, applied by the liquid at pointA, is balancing this tensionT.

scale and carbon nanotubes adopting ring shapes in cavitation bubbles have been experimentally observed [22] and theoretically analyzed [7]. For larger drop-on-fiber systems [21, 10], typically of millimeter or centimeter sizes, the competition between capillary and elastic forces is not automatically won by the former: a threshold length emerges [7] and separates systems in which packaging is possible from those in which the fiber remains straight. This threshold length, called elastocapillary length [26], plays a central role in problems in which surface tension bends or buckles slender rods [14] or thin elastic sheets [25].

Computations of configurations of a filament packaged in a spherical cavity have been performed using Finite Elements [29], molecular mechanics [3], or statisti- cal physics [1] approaches. Here we present a model for the buckling and coiling of an elastic rod in a rigid spherical cavity. In section 3 we derive the equilibrium equations for the elastic rod using an energy approach [27, 5]. In Section 4 we non- dimensionalize the equations and set the boundary-value problem which we numer- ically solve. In Section 5 we plot bifurcation curves and compare the energy of the different coiling solutions. We finally present experimental configurations of coiled systems involving elastomeric beams and oil droplets.

2 Surface tension and interface energy

In this section we recall the link between surface tension and surface energy. Any in- terface between two mediums costs energy as molecules lying at the interface are in a less favorable state than molecules deeply buried in the medium. Theses molecules at the interface are then slightly scattered and therefore in a state of tension: surface energy yields surface tension [15]. We illustrate this link in Figure 1 which shows a drop sitting on a wall in the absence of gravity. A cylindrical rigid rod is then par-

tially immersed in the liquid drop. The rod’s material has a stronger affinity with the liquid than with air and consequently the rod is attracted toward the liquid. One has then to pull the rod at its right end to equilibrate the system. We compute this pulling tensionT by writing the potential energy of the system. The solid-liquid in- terface has energyγ_SLper unit area, yielding a total surface energy`inPγ<sub>SL</sub>+Σ γ<sub>SL</sub> whereP is the perimeter and Σ the area of the cylinder cross-section, and`in the immersed length. Similarly the total surface energy of the emerged part (of length

`out) is`outPγSV+Σ γ_SV. The liquid-vapor surface energy is 2πR²γLVwhereRis the (constant) radius of the liquid drop. Summing up all these energies, adding the work

−T z(B)of the external tensionT, and dropping out constant terms, we write the total potential energy of the system as

E=`inPγSL+`outPγSV−T z(B). (1) Replacing`in=L−`outandvz(B) =R+`outwe finally arrive at

E(`out) =`out(P[γ_SV−γ_SL]−T) +const. (2) Equilibrium is then achieved for dE/d`out=0 that is

T =P(γ_SV−γSL)>0. (3) In the language of forces we say that the external tensionTis balanced by a capillary forceF_γ=P(γ_SV−γSL), applied by the liquid on the rod at pointA and oriented toward the center of the drop.

3 Variational approach

We consider a rod that is naturally straight and has a linear elastic response to bending and twisting. In addition, we work under the assumption that the rod is inextensible and unshearable. The position of the center line of the rod isR(S)whereSis the arc length along the rod. The arc length runs from 0 toL,Lbeing the total contour length of the rod. To follow the deformation of the rod material around the center line, we use a set of Cosserat orthonormal directors{d₁(S),d₂(S),d₃(S)}whered₃(S)is the tangent to the rod center line and{d₁(S),d₂(S)}register the rotation of the rod cross section about the tangent. Orthonormality of the Cosserat frame implies the existence of a Darboux vectorU(S)such that

d_i⁰(S) =U(S)×d_i(S), i=1,2,3. (4) The componentsUi=U·d_iare used to write the strain energy of the rod

E_strain= Z L

0 (1/2)

K₁U₁²(S) +K₂U₂²(S) +K₃U₃²(S)

dS, (5)

whereK₁andK₂are the bending rigidities,K₃is the twist rigidity,U₁andU₂are the curvature strains, andU₃is the twist strain[2, 4].

A liquid drop of massM and radiusRis attached to the rod. The rod enters the drop at meniscus pointAand exits the drop at meniscus pointB. WritingS_Athe arc

e

O e

e

e

d1(L)

C

S = 0 S = L

z =L

z =0

∆

B A

αL

flip

d1(0)

-α0

Fig. 2 Equilibrium configuration of an elastic rod buckled by a liquid spherical drop. The rod is clamped at both extremities, and an end-shortening∆is imposed. The configuration is symmetric about aπ-rotation about the axis joining the centerCof the sphere and the rod middle pointS=L/2. Meniscus forces, applied on the rod at meniscus pointsAandB, together with soft-wall barrier forces are represented. Their sum is equal to the weight of the drop, see Equation (22).

length of pointAandS_Bthe arc length of pointB, we divide the rod in three regions:

(I) whereS∈[0;S_A), (II) whereS∈(S_A;S_B), and (III) whereS∈(S_B;L], with region (II) lying inside the liquid drop. In regions (I) and (III) the solid-vapor interface has an energyγ_SV per unit area. In region (II) the solid-liquid interface has energyγ_SL per unit area. The total surface energy of the rod then scales linearly with the contour length:

E_surface=P[γ_SVS_A+γSL(S_B−S_A) +γSV(L−S_B)], (6) wherePis the perimeter of the cross-section of the rod. If the drop was free-standing it would adopt a spherical shape in order to minimize its surface energy. Neverthe- less due to(i)its own weight,(ii)contact pressure coming from the coiled rod, and (iii)meniscus contact angles, the drop is non-spherical. But as we deal with(i)drops which are smaller than the capillary length,(ii)rods which are flexible compared to surface tension, and(iii)drops with radii much larger than the diameter of the cross- section of the rod, to a first approximation we consider the drop to be spherical. The surface energy 4πR²γLVcorresponding to the liquid-vapor interface (i.e.the drop sur- face) is then constant and therefore discarded. As in [13] we use a soft-wall potential to prevent the rod from exiting the sphere elsewhere than at the meniscus pointsA andB:

V_wall(R(S),R_C) = V0

1+ρ−(1/R)p

(R(S)−R_C)², (7) whereR_C= (X_C,YC,Z_C)^T is the position of the center of the spherical drop, andV₀ is an energy per unit length withV₀→0 corresponding to the hard-wall limit. The small dimensionless parameter ρ is introduced to avoid the potential to diverge at

the meniscus pointsAandB, where the rod enters and exits the sphere. As we are working with sub-millimetric systems we neglect the weight or the rod, but take into account the weight of the drop by adding the termMgY_C into the energy. The total potential energy of the system is then, up to constant terms,

Etot=Estrain+ Z _SB

SA

VwalldS−F_γ(S_B−SA) +MgYC, (8) where we have introduced the surface tension forceF_γ=P(γ_SV−γSL)>0. We con- sider boundary conditions of clamping, where the rod positionR(S)and rotation are restrained at both sides:

R(0) =0, R(L) = (0,0,L−∆)^T, (9a) d₁(0) =cosα0e_x+sinα0e_y, d₂(0) =−sinα0e_x+cosα₀e_y, d₃(0) =e_z, (9b) d₁(L) =cosα_Lex+sinα_Ley, d₂(L) =−sinα_Lex+cosα_Ley, d₃(L) =ez, (9c) where∆is the prescribed end-shortening,α₀andα_Lare fixed angles, and{ex,e_y,e_z} is our reference frame. We minimizeEtotunder the following constraints

R⁰(S) =d₃(S) (inextensibility), (10a) d_i(S)·d_j(S) =δi j (orthonormality), (10b) U1=d₂⁰·d₃, U2=d₃⁰·d₁, U3=d₁⁰·d₂ (Darboux relations), (10c) (R(S_A)−R_C)²=R²= (R(S_B)−R_C)² (position of meniscus), (10d) where⁰≡d/ds. We cope with constraints by introducing Lagrange multipliers and considering the Lagrangian

L(U₁,U2,U3,R,d₁,d₂,d₃,S_A,S_B,RC) = Z L

0

h(1/2) K₁U₁²+K₂U₂²+K₃U₃²

−M₁ U₁−d₂⁰·d₃

−M₂ U₂−d₃⁰·d₁

−M3 U3−d₁⁰·d₂

−ν₁₂(d₁·d₂)−ν₁₃(d₁·d₃)−ν₂₃(d₂·d₃)

−(1/2)ε1(d₁·d₁−1)−(1/2)ε2(d₂·d₂−1)−(1/2)ε3(d₃·d₃−1)i dS + (F_A/2R)h

(R(S_A)−R_C)²−R²i

+ (F_B/2R)h

(R(S_B)−R_C)²−R²i

−Fγ(S_B−SA)−W·R_C+ Z _SB

S_A

VwalldS +

Z _SA 0

N^I· R⁰−d₃ dS+

Z _SB SA

N^II· R⁰−d₃ dS+

Z _L SB

N^III· R⁰−d₃

dS, (11) whereW= (0,−Mg,0)^T is the weight of the drop, and whereM_i(S),νi j(S),εi(S), N^I,II,III(S), and F_A,B are Lagrange multipliers. As we will see in the following, the multipliers M_i(S)correspond to the components of the internal moment M, M= M₁d₁+M₂d₂+M₃d₃. It can be shown that, as in the planar case [13],M(S)does not experience any discontinuity at the meniscus pointsS=S_AandS=S_B. In the same manner, the multipliers N^I,II,III(S)will be interpreted as the internal force in each

region of the system. Since the internal force in the rod does experience discontinu- ities at the meniscus points, we have therefore introduced three different multipliers N^I,II,III(S).

First variation

We notew= (U₁,U2,U3,R,d₁,d₂,d₃,S_A,SB,R_C)and we consider the conditions for a statew_eto minimize the energyE_tot. Calculus of variations shows that a necessary condition is

L⁰(w_e)·w= d

dεL(w_e+εw) _ε=0

=0 ∀w, (12)

wherewis a variation of the variablew. Boundary conditions (9) imply that

R(0) =0, R(L) =0, (13) d_i(0,L) =0, d_i(L) =0 ∀i, (14) Noting that _dε^{d R}₀^a+εaf(x)dx

ε=0=a f(a)we calculate the first variation (12) L⁰(w_e)·w=

Z _L 0

h

K₁U₁U₁+K₂U₂U₂+K₃U₃U₃

−M1

U1−d₂⁰·d₃−d₂⁰·d₃

−M2

U2−d₃⁰·d₁−d₃⁰·d₁

−M₃

U₃−d₁⁰·d₂−d₁⁰·d₂

−ν12 d₁·d₂+d₁·d₂

−ν13 d₁·d₃+d₁·d₃

−ν₂₃ d₂·d₃+d₂·d₃

−ε1d₁·d₁−ε2d₂·d₂−ε3d₃·d₃i dS + (F_A/R)

(R(S_A)−R_C)· R(S_A) +S_AR⁰(S_A)−R_C + (F_B/R)

(R(S_B)−R_C)· R(S_B) +S_BR⁰(S_B)−R_C

−F_γ(S_B−S_A)−W·R_C +

Z _SB SA

∂V_wall

∂R ·R+∂V_wall

∂R_C ·R_C

dS+S_BV_wall(S_B)−S_AV_wall(S_A) +

Z _SA 0

N^I·

R⁰−d₃ dS+

Z _SB SA

N^II·

R⁰−d₃ dS+

Z _L SB

N^III·

R⁰−d₃ dS. (15) Note that we have used (10a) to eliminate several terms. We first require (15) to vanish for allU_iand obtain

M_i=K_iU_i. (16)

Equation (16) appears as the bending/twisting constitutive relation of the elastic rod provided the Lagrange multipliersMiare seen as the components of the internal mo- ment. We next perform integrations by part for the terms involvingd_i⁰. Boundary

conditions (14) implies that the boundary terms vanish. Requiring the result to vanish for alld_iyields:

M₂d₃⁰−M₃⁰d₂−M₃d₂⁰−ν12d₂−ν₁₃d₃−ε1d₁=0, (17a) M3d₁⁰−M₁⁰d₃−M₁d₃⁰−ν23d₃−ν₁₂d₁−ε2d₂=0, (17b)

−N^I,II,III+M1d₂⁰−M₂⁰d₁−M₂d₁⁰−ν13d₁−ν₂₃d₂−ε3d₃=0. (17c) We take the scalar product of each of these three equations withd₁,d₂, andd₃and combine them to eliminate theν_{i j}and theε_ito finally obtain

M₁⁰ =M₂U3−M3U2+N₂^I,II,III, (18a) M₂⁰ =M₃U₁−M₁U₃−N₁^I,II,III, (18b)

M₃⁰ =M₁U₂−M₂U₁, (18c)

which is the component version of the moment balance equation for the elastic rod, M⁰=N×R⁰. The Lagrange multiplierNis then seen as the internal force in the rod.

We now perform integrations by parts for the terms involvingR⁰and find that (F_A/R) (R(SA)−R_C)·R(S_A) + (F_B/R) (R(SB)−R_C)·R(S_B)

+ Z S_B

SA

∂V_wall

∂R ·R+

N^I·RSA

0 +

N^II·RSB

SA+

N^III·RL SB

− Z S_A

0

N^I(S)⁰·RdS− Z S_B

S_A

N^II(S)⁰·RdS− Z L

S_B

N^III(S)⁰·RdS =0 ∀R. (19) This condition implies the balance equations for the rod:

N^I(S)⁰=0 ∀S∈[0;S_A), (20a) N^II(S)⁰−∂V_wall

∂R =0 ∀S∈(S_A;S_B), (20b) N^III(S)⁰=0 ∀S∈(S_B;L], (20c) whereN^I,II,IIIare the internal forces in each region of the system. The force per unit length−∂V_wall/∂Rcorresponds to the soft-wall repulsion, from the drop interface, on the rod. Boundary conditions (13) make the boundary terms atS=0 andS=L vanish, but arbitrariness of the variationsR(S_A)andR(S_B)require that

N^II(S_A)−N^I(S_A) +F_A[R_C−R(S_A)]/R=0, (21a) N^III(S_B)−N^II(S_B) +F_B[R_C−R(S_B)]/R=0. (21b) These equations correspond to forces balance at the meniscus points. We see that at meniscus pointsAandB, the total force from the drop on the rod is oriented toward the center of the spherical dropand has intensityF_AandF_Brespectively. Requiring (15) to vanish for allR_Cyields

FA[R_C−R(S_A)]/R+F_B[R_C−R(S_B)]/R=W+ Z _SB

S_A

∂Vwall/∂RdS, (22)

where we have used the identity∂V_wall/∂R_C=−∂V_wall/∂R. Equation (22) tells us that the total meniscus force from the drop on the rod (the right hand side of (22)) is equal to the weightW of the drop plus the opposite of the integrated soft-wall repulsion. We therefore see that the soft-wall repulsion applied inside the drop is balanced by the meniscus force. Accordingly, using (20b), (21), and (22) we find

N^III(S_B)−N^I(S_A) +W=0. (23) That is, the total force from the liquid on regions I and III of the rod is simply the weight of the drop. Finally the conditions for (15) to vanish for allS_AandS_Bare

(F_A/R) [RC−R(S_A)]·R⁰(S_A) =F_γ−V_wall(S_A), (24a) (F_B/R) [RC−R(S_B)]·R⁰(S_B) =−F_γ+V_wall(S_B). (24b) These conditions can be seen as a way to compute the intensityF_AandF_B of the meniscus force. In particular we see thatF_A(F_B) depends on the relative orientation of the rod’s tangent and the radial vector at the meniscus pointsA(B), as also shown in [24]. Introducing the tension in rodT(S):=N·d₃and using (21), we find that

T^I(S_A)−T^II(S_A) =F_γ−V_wall(S_A), (25a) T^III(S_B)−T^II(S_B) =F_γ−V_wall(S_B), (25b) where we see that, in the limitV₀→0, the jump in the rod’s tension is equal to the surface tension forceF_γ.

4 Boundary value problem

We restrict our attention to rods with isotropic bending behavior,K₁=K₂=K₀, where U=M/K0+ (1−K₃/K0)U₃d₃withU₃= (M_xd_3x+M_yd_3y+M_zd_3z)/K3. In this case (18c) shows that dU3/dS≡0∀S, and consequentlyUvanishes for the equations [18, 23]. We additionally focus here on the weightless,W=0, and symmetricalS_A= L/2−Σ, SB=L/2+Σ case. We use the diameter D=2R of the spherical drop as unit length, and the buckling loadK0/D²as unit force, that is we introduce the following dimensionless quantities

s=S−L/2

D ; `= L

D; (x,y,z) =(X,Y,Z−(L−∆)/2)

D ; (26a)

σ=Σ

D; k₃=K₃ K0

; δ=∆

D; u=UD; f_γ=FγD² K0

; (26b)

n=ND²

K₀ ; m=MD

K₀ ; (v,v₀) =(V_wall,V0)D²

K₀ . (26c)

We further restrict the study to equilibrium shapes being invariant by a rotation of an- gleπabout the line passing through the middle point of the rod,r(s=0), and directed along the axise_y. Such configurations are sometimes referred to as flip-symmetric [9].

The quantitiesx(s),z(s),d_1y(s),d_2x(s),d_2z(s),d_3y(s),m_y(s), andn_y(s)are then odd functions ofs, whiley(s),d_1x(s),d_1z(s),d_2y(s),d_3x(s),d_3z(s),m_x(s),m_z(s),n_x(s),

andn_z(s)are even functions ofs. The center of the spherical drop consequently lies on the flip-symmetry axis, that isx_C=0 andz_C=0, and the end-rotation angles are such thatα₀=−α_L. Making use of this symmetry, we only integrate the equilibrium equations fors∈[0, `/2], which read

x⁰(s) =d_3x,y⁰(s) =d_3y,z⁰(s) =d_3z, (27a) d_1x⁰ (s) =d1zm_y−d1ym_z+ (1−k3)u3(d_1zd3y−d_1yd3z), (27b) d_1y⁰ (s) =d_1xm_z−d_1zm_x+ (1−k₃)u₃(d_1xd_3z−d_1zd_3x), (27c) d_1z⁰ (s) =d1ymx−d_1xmy+ (1−k3)u₃(d_1yd3x−d1xd3y), (27d) d_3x⁰ (s) =d_3zm_y−d_3ym_z, d_3y⁰ (s) =d_3xm_z−d_3zm_x,d_3z⁰ (s) =d_3ym_x−d_3xm_y, (27e) m⁰_x(s) =d_3zn_y−d_3yn_z,m⁰_y(s) =d_3xn_z−d_3zn_x,m⁰_z(s) =d_3yn_x−d_3xn_y, (27f) n⁰_x(s) =χ ∂v/∂x+2f_Bx(σ)ds−σ, (27g) n⁰_y(s) =χ ∂v/∂y+2f_B[y(σ)−y_C]ds−σ, (27h) n⁰_z(s) =χ ∂v/∂z+2f_Bz(σ)ds−σ, (27i) where ds−s₀is the Dirac distribution centered ons₀, andv=v₀

1+ρ−2p

x²+ (y−y_C)²+z²−1

. Fors∈[0;σ)the rod lies inside the spherical drop and we haveχ =1, otherwise

χ=0. We setα=0 in (9) and considerv0,ρ, fγ,`, andk3as fixed parameters. We look for equilibrium solutions by integrating (27) with initial conditions

x(0) =0,y(0) =y₀,z(0) =0, (28a) d_3x(0) =sinθ0,d_3y(0) =0, d_3z(0) =cosθ0, (28b) d_1x(0) =cosθ₀,d_1y(0) =0,d_1z(0) =−sinθ₀, (28c) n_x(0) =n_x0,n_y(0) =0,n_z(0) =n_z0, (28d) mx(0) =mx0, my(0) =0, mz(0) =mz0 (28e) wherey₀,θ₀,n_x0,n_z0,m_x0,m_z0along withy_C, f_B, andσ are 9 unknowns which are balanced by the following 8 conditions. We restrict to cases where the directord₁is aligned with thee_xaxis at both extremities of the rod, that is we setα₀=α_L=0.

Then, using (9), we write 5 boundary conditions for the right extremity of the rod, s=`/2

x(`/2) =0 ;y(`/2) =0 ;d_3x(`/2) =0 ;d<sub>3y</sub>(`/2) =0 ;d_1y(`/2) =0 ; (29) At the meniscus pointB, we have 3 conditions, adapted from (10d), (23), and (24b)

q

x²(σ) + (y(σ)−y_C)²+z²(σ) =1/2, (30a)

n_y(σ) =0, (30b)

2f_B[−x(σ)d_3x(σ) + (y_C−y(σ))d_3y(σ)−z(σ)d_3z(σ)] =−fγ+v(σ). (30c) The solution set is thus a 9−8=1 dimensional manifold and we plot in Section 5 different solution paths.

õ õ ô ô

à à

á á æ

æ ç ç

ì ì

í í

0.0 0.5 1.0 1.5 2.0 2.5

∆ Π

0 5 10 15 20 25

t

A

A

P

P

L

L

3D 3D

Fig. 3 Force-displacement post-buckling curves for f_γ=20 andR=0, wheret is the applied tension andδ the end-shortening. Paths of 3D and 2D configurations are plotted with solid and dashed lines respectively. Points marked with empty and filled circles, squares, diamonds, and triangles correspond to configurations shown in Figures 5, 6, 7, and 8 respectively. Points marked triangles belong to paths not shown here.

5 Bifurcation diagram

We numerically solve the boundary-value problem defined in Section 4, using ei- ther a shooting method or the AUTO collocation method [8]. Pseudo-arc-length con- tinuation then enables us to follow the solutions as parameters are varied. In order to compare present three-dimensional (3D) results with two dimensional (2D) solu- tions studied in [13], we use the same parameters k₃=0.9,`=10, f<sub>γ</sub>=20, and v<sub>0</sub>=0.02/`²=2×10⁻⁴, but reduce ρ toρ =0.1 thereby bringing the meniscus points closer to the bounding sphere.

We start with a straight configuration subject to a large applied tensiont=n(`/2)·

d₃(`/2). We define the end rotationRas the angle betweend<sub>1</sub>(`/2)andd₁(−`/2), R=α_L−α0=2α_Land we first restrict to configurations withR=0. The straight configuration is consequently twistless. (For the present case of clamped-clamped configurations this end-rotationRis closely related to the topological linkLk, seee.g.

[16].) As the tensiontis decreased under a threshold valuet'3.7, the rod buckles and the post-buckling regime first involves 2D configurations, this is the path Ax introduced in [13]. The dimensionless end-shorteningδ is then gradually increased, and atδ '0.24πlies a pitchfork bifurcation and a secondary path, consisting of 3D configurations, emerges and progresses toward the plateau valuet_P=f_γ−2 [12], see Figure 3. Later along the path, forδ'1.9π, another pitchfork bifurcation occurs and, for a limitedδ interval, the equilibrium configurations become planar again: The 3D path merges with the pathL₂, which consists in configurations which are planar and

õ ô õ ô

à à

á á

æ æ

ç ì ç

ì í í

0.0 0.5 1.0 1.5 2.0 2.5

∆ Π

0 10 20 30 40 50

E `

D  K

A

A

P

3D

3D L

Fig. 4 Energy as function of the end-shorteningδfor the post-buckling curves of Figure 3.

Fig. 5 3D configurations corresponding to the filled and empty circles on the post-buckling diagram of Figures 3 and 4.

looping twice inside the sphere. After yet another pitchfork bifurcation atδ'2.15π,

Fig. 6 2D configurations corresponding to the filled and empty squares on the post-buckling diagram of Figures 3 and 4.

theL₂and 3D paths split again.