Confined nanorods: jamming due to helical buckling
aa r X i v : . [ c ond - m a t . s o f t ] D ec Confined nanorods: jamming due to helical buckling
Daniel Svenˇsek and Rudolf Podgornik
1, 2 Department of Physics, Faculty of Mathematics and Physics,University of Ljubljana, Jadranska 19, SI-1111 Ljubljana, Slovenia Department of Theoretical Physics, J. Stefan Institute, Jamova 39, SI-1000 Ljubljana, Slovenia (Dated: November 2, 2018)We investigate a longitudinally loaded elastic nanorod inside a cylindrical channel and showwithin the context of classical elasticity theory that the Euler buckling instability leads to a helicalpostbuckling form of the rod within the channel. The local pitch of the confined helix changes alongthe channel and so does the longitudinal force transmitted along the rod, diminishing away fromthe loaded end. This creates a possibility of jamming of the nanorod within the channel.
PACS numbers: 61.46.Fg,62.20.mq,81.07.De,46.32.+x
I. INTRODUCTION
Elastic instabilities of nanoscale rods and in particu-lar the Euler (buckling) instability have long been rec-ognized as being essential for structural as well as func-tional aspects of nano- and bio-systems. Some time ago[1] Manning proposed that DNA collapse in polyvalentsalt solutions proceeds through a type of Euler bucklingdriven by diminished electrostatic repulsive interactionsbetween charged phosphates along the DNA backbone.Though the details of his picture appear to be specula-tive, the general framework has been completely vindi-cated by later work [2]. Buckling instability appears toplay also an important part in the conformation of linkerDNA within the chromatin fiber [4], in the case of a grow-ing microtubule pushing against a microfabricated rigidbarrier [3] and in the stability of AFM-tips, nanotubes[5] and nanorods [6].The confinement of nanoscale rods to micro- andnano-channels presents yet another set of experimen-tal and theoretical problems [7] centered around thetechnological challenges of manufacturing nanopores andnanochannels for investigating and manipulating DNA.Here the main theoretical thrust is in the direction ofunderstanding the various contributions to the confine-ment free energy that depend on the size of the con-fining space and the intrinsic properties of the confinednanorods. The behavior of confined semiflexible poly-mers is particularly important in this context and hasbeen recently analyzed in detail [8, 9]. In many respectsthis type of problems are mainly centered upon trans-verse confinement, whereas elastic instabilities describedbefore are due to longitudinal confinement of elastic rods.Motivated by these phenomena we will try in what fol-lows to combine the two aspects of polymer confinementdescribed above, considering an elastic buckling insta-bility within a confining cylindrical channel. The con-finement of the elastic rod is thus twofold: longitudinal,leading to buckling, and transverse, leading straightfor-wardly to a helical postbuckling form. We will show thattransverse confinement with frictional walls furthermoreleads to a decay of the longitudinal force within the rod along the long axis of the confining channel. This will notbe too difficult to rationalize since we know from othersystems that longitudinal stresses can be taken up effec-tively by friction at confining walls. For a silo filled withgranular matter and compressed on one side by a force F , Janssen’s equation [10] gives an exponential decreaseof the compression force with the distance z from thecompressed side, F ( z ) = F exp( − λk fr z P / S ) , where λ = σ rr /σ zz is the ratio of horizontal and verticalstresses, P / S is the perimeter to cross-section area ratio, k fr is the static friction coefficient and we have neglectedgravity for the purpose. The same holds also for a regularelastic solid, i.e., a rigidly confined rod, if one neglectslateral deformation gradients in the rod [11]. There wehave F ( z ) = F exp[ − σk fr z P / S (1 − σ )] , where σ is Poisson’s ratio. On this simple basis one ex-pects also a conceptually similar effect, i.e., a decreaseof compression force with the distance along an elasticrod confined to a cylindrical channel in the post-bucklingregime. This will allow us to hypothesize on the existenceof a jamming regime for sufficiently long confined elasticrods.The plan of this paper is as follows: we shall firstpresent the classical elastic model for an Eulerian rodand derive the scaling relations valid for the onset of thebuckling instability. We shall solve an approximate formof the elastic equations within a cylindrically confinedchannel and show that the longitudinal force, at leastwithin the considered approximations, decays inverselyproportional to the length of the confining channel whenfriction at the cylindrical wall is taken into account. Toobtain exact solutions of the elastic equations we willproceed numerically and derive the minimal shape of theconfined rod and the magnitude of the transmitted lon-gitudinal force along the axis of the confining channel.At the end we will discuss what could be the conditionsthat would lead to a jamming of the elastic rod withinthe cylindrical confining channel. II. ELASTIC MODEL
We model the rod as a thin Eulerian elastic filament towhich the standard continuum theory of elasticity can beapplied. The confining channel is modeled as a straightcylindrical tube with rigid walls. The ratio
R/L of tuberadius R and undeformed filament length L is the rele-vant geometric parameter of the system.Let us briefly review the equilibrium equations for thinelastic rods [12]. The force balance readsd F d l + K = 0 , (1)where d l is the length element on the rod ( l is the nat-ural parameter—the length along the filament), F is theelastic force exerted on the leading surface of the rodelement (in the sense of increasing l ; the force on theopposite (trailing) side is − F , of course), and K is theexternal force per unit length. The torque balance readsd M d l + t × F = 0 , (2)where M is the elastic torque exerted about the center ofthe leading surface and t is the unit tangent of the rod.The elastic torque is related to the deformation of thefilament via M = Gτ t + EI t × ˙ t , (3)where E is the Young modulus and we assumed that therod is circular, i.e., both geometric moments of inertiaof the cross section are equal, I = R d S x = R d S y .The first term on the right describes torsion which weassume to be absent. One can show that in case of theisotropic moment of inertia, torsion is absent everywherein the rod when there is no twisting torque applied [12].In what follows, the twisting (torsional) torque will beabsent, i.e., the torque in the rod will have a vanishingtangential component, M = EI t × ˙ t . (4)From Eqs. (2) and (4) one then gets EI t × ¨ t + t × F = 0 (5)as the fundamental equation describing the shape of thebent filament. The solutions of this equation with variousboundary conditions are discussed in standard referenceson the theory of elasticity [12]. A. Buckling in confined geometry
It is well known that when a straight rod is compressed byan axial force, above threshold it undergoes a bucklinginstability—the so-called Euler buckling—and becomesbent. For a rod with hinged ends or ends that are lat-erally free (both situations are identical in this case), forexample, the threshold force is F crit = EIπ /L [12]. What happens when the (thin) rod is confined within arigid cylinder? Once the compressing force is sufficientlylarge so that the bent rod touches the wall of the cylinder(let it be bent in the diameter plane) and is further in-creased, there are two scenarios one can think of: the rodcan bend back-and-forth staying in the diameter plane, orit can become a helix touching the cylinder wall (spring-like configuration). In oil-drilling community it has beenrecognized for a long time that it is the latter case thateventually happens [13, 14, 15, 16] (In fact, this is theonly area of research to have studied helical buckling.)It turns out, as we will show in Section V, that there in-deed exists a secondary threshold above which the planarconfiguration is unstable with respect to the helix. B. Scaling
By introducing dimensionless quantites denoted withtilde, l = R ˜ l, F = EIR ˜F , K = EIR ˜K , M = EIR ˜M , (6)Eqs. (1)-(5) attain a universal form. Thus, the solutionsdepend only on the aspect ratio L/R . Upon rescaling thesystem size and preserving the aspect ratio, the solutionis unchanged if also the forces and torques of the bound-ary condition are rescaled correspondingly, i.e., accordingto Eq. (6). Similarly, taking a rod with a different EI and rescaling the forces and torques does not affect thesolution. III. ANALYTIC DESCRIPTION OF THE HELIXA. Simple model
Let us write down equilibrium equations for a helicallydeformed filament constrained to and lying on a cylindri-cal surface of radius R . The deformation is sustained byan external compressing force F z < r, φ, z ) the helix is given by r = R and the linear function φ ( l ). The tangent is thusgiven by t = d r d l = R ˙ φ ˆ e φ + q − ( R ˙ φ ) ˆ e z . (7)Further we put F = F r ˆ e r + F φ ˆ e φ + F z ˆ e z , where the firsttwo components are unknown and must be determined.The force of the cylinder per unit length is K = − K ˆ e r .Eq. (5) then reads (cid:20)q − ( R ˙ φ ) ( EI R ˙ φ − F φ ) + R ˙ φF z (cid:21) ˆ e r + q − ( R ˙ φ ) F r ˆ e φ − R ˙ φF r ˆ e z = 0 . (8)Eq. (8) requires F r = 0 , (9)while from Eq. (1) one gets F φ = − K ˙ φ (10)and hence for the radial component of Eq. (8) q − ( R ˙ φ ) (cid:16) EI R ˙ φ + K (cid:17) + R ˙ φ F z = 0 . (11)Now comes a crucial point. The equilibrium condition(11) involves both the helical deformation ˙ φ and the ex-ternal force K , so additional input is needed to determineone or the other. This comes from the boundary condi-tion, which is however inaccessible under the assumptionof fixed pitch.Let us digress a little and illustrate in physical termswhy the force of the cylinder, K , cannot be specified untilthe boundary condition is known. Besides the force F ,the boundary condition involves also the torque exertedon free ends (which must be normal to the tangent ofthe rod as we are not considering torsion). By usinga pair of wrenches and applying torque to the ends, onechanges the force exerted on the cylinder wall. Using justthe right torque, for example, one can make K vanish.Increasing the torque further, the rod detaches from thecylinder and forms a free helix with the radius smallerthan R [17]. According to Eqs. (4) and (7), the torquein the filament is M = EI R ˙ φ (cid:20) − q − ( R ˙ φ ) ˆ e φ + R ˙ φ ˆ e z (cid:21) . (12)In fact, with this torque applied to the ends (togetherwith F φ and F z ), the helical pitch is constant everywhere.By changing the torque, one changes the pitch, Eq. (12),and therewith the cylinder force K , Eq. (11). Note thatthere is no tangential force in the free (detached from thecylinder) helix, Eq. (10).When there is no torque applied to the ends, i.e., whencompressing the rod inside the cylinder by a pair of pis-tons, the part of the rod close to the end detaches fromthe wall while the tip (which is straight due to zerotorque) pushes against the wall with a discrete force (seenumeric solutions, Fig. 1). The torque due to this forceincreases as we move away from the tip and eventuallybecomes sufficiently large for the helical deformation.Let us return to the case of a constant helical pitch.One can invoke an energy argument to independently es-timate the dependence of ˙ φ on the compressing force F z . (a) (b) FIG. 1: (color online) A long-axis view of helical filamentswith lengths (a)
L/R = 20 and (b)
L/R = 640, buckled bya force | F z (0) | = 0 . EI/R , which is approximately (a) 10-times and (b) 10,000-times the critical force for Euler buckling F crit = EIπ /L . There is no torque applied to the ends. The elastic free energy of the compressed helix withoutthe contribution of the torsion is F = 12 EI Z L d l (cid:12)(cid:12) ˙ t (cid:12)(cid:12) − L z F z , (13)where L is the length of the filament and L z is thelength of its projection to the z axis, i.e., the lengthof the coil. Recall that F z <
0. With Eq. (7) andd z/ d l = q − ( R ˙ φ ) = L z /L , we have F = 12 EIL ( R ˙ φ ) + L q − ( R ˙ φ ) | F z | (14)and after minimizing, in the lowest order of R ˙ φ ,˙ φ = | F z | EI . (15)Hence, from Eq. (11) one gets a direct relation betweenthe load and the wall forces. To the lowest order the forceof the cylinder is K = R EI F z . (16)So far the friction has been absent. Let us now con-sider it as a perturbation and introduce a small frictioncoefficient k fr . One is aware, of course, that for finitefriction the constant pitch is not a solution, neither canone use the energy argument. We are interested in howthe compressing force | F z | is reduced by the friction go-ing from one end of the helix to the other. Let us assumethat the helix is compressed by the external force F z at l = 0 and that the friction is parallel to z , opposing theexternal force. Assuming the static friction is maximumeverywhere, it follows from the force balance (1) thatd | F z | d l = − k fr K. (17)Combining Eqs. (16) and (18) we finally get | F z ( l ) | = (cid:18) | F z | + k fr R EI l (cid:19) − . (18)The virtual point where F z would diverge lies at l c = − EI/k fr R | F z | , (19)which thus dictates the extent of the reduction of thecompression force. The smaller | l c | , the strongest is thedecay of the force. Increasing k fr , | F z | , or R reduces | l c | and moves the singular point closer to the beginning ofthe helix. Eq. (18) can be put into a universal form byintroducing the force unit F = 2 EI/k fr RL , | F z ( l ) | F = (cid:18) F | F z | + lL (cid:19) − . (20) Note in Eq. (6) that if the aspect ratio is constant, theforce scales as 1 /R (or equivalently, 1 /L ). This, ofcourse, always holds and is approximation-independent.In addition, the result (20) of our simple model givesa 1 /RL scaling of the force when the aspect ratio ischanged, which, however, is only an approximation. B. Exact analysis
For completness, let us derive exact equilibrium equa-tions for the helically buckled filament in the presenceof friction. We do not aim to solve them, yet they willbe helpful giving us insight into the solution. As before,the filament is assumed to lie on the cylinder everywhere, r = R , but now all variables are l -dependent, includingthe compressing force F z .Eq. (5) involves only two components as it is orthogo-nal to t . Therefore we project it to ˆ e r and the direction t × ˆe r . Writing ˙ φ ( l ) = ω ( l ), the former is EIR (1 − R ω ) / h ω (cid:0) − R ω (cid:1) − ¨ ω (cid:0) − R ω (cid:1) − R ω ˙ ω i − F φ p − R ω + F z Rω = 0 (21)and the latter − EIR dd l ( ω ) + F r = 0 . (22)Again we choose from an infinite number of solutions,which exist when the static friction is involved, the sim-plest and most symmetric one, in which the friction iseverywhere at its maximum and parallel to z . With K = K ( l )( − ˆ e r + k fr ˆ e z ) the three components of theforce balance (1) are (cid:16) ˙ F r − ωF φ − K (cid:17) ˆ e r = 0 , (23) (cid:16) ωF r + ˙ F φ (cid:17) ˆ e φ = 0 , (24) (cid:16) ˙ F z + k fr K (cid:17) ˆ e z = 0 . (25)Eqs. (21)-(25) represent a closed set of five ordinary dif-ferential equations for the five variables ω ( l ), K ( l ), and F ( l ). One can verify that in case ˙ ω = 0, Eqs. (9)-(11) arerecovered.One can expand the system (21)-(25) for small Rω .For k fr = 0, terms of third order must be included toget a solution (one can check that the second order gives only the second order part of the stationary solution (9)-(11) for k fr = 0, and no solution for k fr = 0). InsertingEq. (22) into Eq. (24) and integrating, one gets a partialresult F φ = A − EIω , where A is a constant. IV. NUMERICAL APPROACH
We will solve the fundamental equations for the actualdeformation of the compressed and confined filament nu-merically for it is only in this way that one can obtain acomplete solution of the problem. The filament is natu-rally allowed to detach from the wall, which essentiallytakes place near the two ends of the filament, and thehelix within the cylindrical pore can form spontaneouslywithout having been put in ”by hand”. Let us stressagain that the solution, including in particular the pitchof the bulk helix and the force exerted on the cylinderwall, is only determined when the ends of the filamentare taken into account, which furthermore inevitably in-volve detachment of the filament from the wall unlessthe boundary condition is very special, as explained inSec. III A, Eq. (12).For the purpose of numeric modelling that involves dis-cretization, it is generally better, if only possible, to startwith a discrete analogue of the continuum system andwrite down algebraic equations, rather than discretizingthe differential equations themselves, derived for the con-tinuum limit. In this spirit, the elastic filament will berepresented as a set of straight and stiff elements (links)of fixed length l with forces and torques acting betweenthem. The element i is described by its center of mass r i and a unit (tangent) vector t i giving its orientation.Dynamical evolution of the filament shape is obtainedby Newton’s laws for translation and rotation of eachelement. The joints between the links are rather cum- helical(cid:13)bent(cid:13) helical buckling(cid:13)Euler buckling(cid:13)straight(cid:13) (cid:13) (cid:13) L(cid:13) z(cid:13) |(cid:13)
F(cid:13) z(cid:13) |(cid:13)
FIG. 2: The external load dependence of the projection of fil-ament length onto the axis of the confining cylindrical pore.The weak force dependence of length in the straight configu-ration is due to the small artificial compressibility introducedby the penalty potential (26) (i.e., finite value of the springconstant k ). In the limit of zero compressibility this part ofthe functional dependence would be a straight horizontal line. bersome to model: they act as constraints for the endsof the two elements which must meet in a single point,and thus exert constraint forces that can be determinedonly implicitly, such that the elements satisfy the con-straints. To avoid this, we relax (soften) the constraintsand introduce a quadratic (bond) penalty potential V ( r + i , r − i +1 ) = k (cid:12)(cid:12) r − i +1 − r + i (cid:12)(cid:12) , (26)where r + i = r i + ( l / t i and r − i +1 = r i +1 − ( l / t i +1 are the two ends meeting at the joint. The coeffi-cient k should be sufficiently large so that the gap be-tween the ends is small compared to the element length, (cid:12)(cid:12) r − i +1 − r + i (cid:12)(cid:12) ≪ l . The side wall of the cylinder is mod-elled in a similar manner, introducing a wall penalty po- tential V w ( r ± i ) = (cid:26) k w (cid:0) | r ± i | − R (cid:1) ; | r ± i | > R | r ± i | ≤ R . (27)Again the coefficient k w should be sufficiently large sothat | r ± i |− R ≪ l . The forces on the ends of the elementsare F ± i = − ∂V∂ r ± i − ∂V w ∂ r ± i . (28)It makes little difference (i.e., no difference in the limit l →
0) whether the force of the wall actually acts onboth ends or just at the center of the element.Torques about the center of element i are of two kinds.One comes from elastic couples exerted by the two neigh-bouring elements and the other from the forces on bothends of element i : M i = C t i × ( t i − + t i +1 ) + l (cid:0) t i × F + i − t i × F − i (cid:1) , (29)where C is the bending stiffness which we shall connectto continuum parameters. In the continuum picture, thefirst part of the torque translates to the first term ofEq. (5), and the second part to the second term of Eq. (5).To get the evolution of the shape of the filament, wecoveniently assume overdamped dynamics: β d r i d t = F + i + F − i , (30) β d t i d t = β d ϕ i d t × t i = M i × t i , (31)where β is an arbitrary damping coefficient defining thetime scale.One can verify that the set of discrete equations (29)-(31) agrees with properly discretized continuum equa-tions with the following connection between the param-eters: C = EIl , (32)while β = l β ph , where β ph is the physical (continuum)damping coefficient. The latter is only important forproperly scaling the time when changing the elementlength l in the numerical model. Similarly, k = k ph /l to ensure that the spring constant k ph of the rod (whichis infinite in the continuum description (1)-(4) of a thinrod) is unaffected by changes of l .To avoid giant friction forces in certain discrete points(e.g., a few points arround the detachment region whichexert a large force on the confining wall), the frictionforce is properly capped, so that the Coulomb’s frictionlaw is violated in these few points but met everywhereelse. V. RESULTS
First let us take a look at numeric solutions of heli-cally buckled filaments. Fig. 1 shows two examples offilaments of different lengths buckled by the same axialforce. We use the most natural boundary condition: theupper (compressed) end is only pushed along z (the cylin-der axis), while the lower end is not allowed to move in z direction. Both ends are free to move laterally untilthey hit the confining wall. There is no torque appliedto them. To build up the torque exerted on the filamentcross section in the helical state, however, the ends mustleave the helical state as anticipated in Sec. III A. (a) (b)(c) FIG. 3: (color online) Planar configuration of the confinedfilament (a) is unstable (b) and is transformed into the heli-cal (c) upon perturbation. The helix can contain metastableirregularities as seen in the equilibrium state (c).
To numerically study the solution path from thestraight to the helical configuration, we introduce a tinyrandom perturbation to the straight compressed fila-ment. A numerically obtained state diagram is depictedin Fig. 2, where a suitable parameter distinguishing be-tween the configurations is simply the projection of thefilament length onto the z axis. There exist two thresholdforces beyond which the straight and the Euler-buckledstates are unstable, respectively. The helical state is aconsequence of the confinement and sets in after the fil-ament has touched the wall. Above the threshold, the planar state remains a solution but is unstable with re-spect to the helix, which is demonstrated in Fig. 3. Thereone also sees a defect in the helix. Such defects are ageneral feature subject to the initial condition (pertur-bation) and are metastable, i.e., they are stable whilethe compressing force is not reduced. (c)(cid:13) (b)(cid:13)(a)(cid:13) (a) (cid:13) k(cid:13) fr(cid:13) = 0.2(cid:13)(b) (cid:13) k(cid:13) fr(cid:13) = 0.4(cid:13)(c) (cid:13) k(cid:13) fr(cid:13) = 0.8(cid:13) |F(cid:13) z(cid:13) |(cid:13) l(cid:13) (cid:13) (cid:13) (cid:13) FIG. 4: (color online) Compression force profile | F z ( l ) | in afilament with length to radius ratio 320. The loading force is | F z (0) | = 0 . EI/R , which is approximately 2,600-times thecritical force for Euler buckling F crit = EIπ /L . Numericdata (black) is excellently fit (red) with the function f ( l ) = A ( l + l ) c (the curves are perfectly overlapping on the scaleof the figure): (a) A = 36 . l = 285 . c = − . A = 21 . l = 131 . c = − . A = 11 . l = 44 . c = − . We are furthermore interested in the force that has tobe applied to the lower end to keep it fixed, i.e., to sustainthe confined helix. As we are to show, this force can bemuch less than the load applied to the upper end due tothe action of friction and in fact approaches zero for verylong confining channels. This opens up the possibilityof jamming, i.e. of a stable static helical configurationof the rod that has been jammed against the confiningwalls via the surface friction and is sustained with onlya tiny (zero in the limit of very long channel) opposingforce at the other end of the channel.Fig. 4 shows the profiles of the longitudinal force | F z | as we move along the filament, and Fig. 5 shows the pro-file of the helical deformation ˙ φ . The force profiles areperfectly fit by a power law, yet with an exponent closeto -0.9 instead of -1 as suggested by Eq. (18). The va-lidity of the power law is remarkable though it deviatesfrom the approximate universal scaling of Eq. (20). Inother words, the solution is practically indistinguishablefrom a power-law, in spite of the evidence that it con-siderably departs from our simple model. The reason forthis remarkable validity of the power-law, yet with anexponent different than -1, is not clear. Fig. 6 shows theforce transmitted to the lower end as the function of the d f /d l l FIG. 5: Profile of d φ/ d l , the inverse of the helical pitch q =2 π/ (d φ/ d l ), for the filament of Fig. 4 (c). The boundarieswhere the pitch is not defined are not displayed. As expected,˙ φ gets smaller (the pitch gets larger) as the longitudinal forceis reduced. Its relative decrease is however smaller than thatof the force, as hinted by the quadratic dependence F z ∝ ˙ φ inEq. (15) of the simple model (which assumes constant pitch). length L of the filament, at a fixed load on the upperend. We see that for long filaments the transmitted forceis strongly reduced and asymptotically approaches zero. (cid:13) (cid:13) |(cid:13) F(cid:13) z(cid:13) (L)(cid:13) |(cid:13)
L/R(cid:13)
FIG. 6: Compression force, | F z ( L ) | , transmitted throughfilaments of lengths L ; the external compressing force is | F z (0) | = 0 . EI/R ; k fr = 0 .
8. The dotted line serves as aneye guide.
The physically interesting interval of loading forcestrengths depends on the aspect ratio of the filament. InFig. 4, for example, the loading force was selected suchthat the transmitted force was significantly reduced. Thereduction of the transmitted force is weaker if the load-ing force is smaller, Fig. 7, in accord with Eq. (19) of the simple model. On the other hand, increasing the loadingforce beyond a threshold (that scales as 1 /R ) results in acatastrophic event—a U bending of the filament, leadingto the escape out of the confining pore, Fig. 8. Hence, itis only for long and thin rods that the transmitted forcecan be strongly reduced. (cid:13) (cid:13) (cid:13) l(cid:13) |(cid:13) F(cid:13) z(cid:13) |(cid:13)
FIG. 7: (color online) Compression force profile | F z ( l ) | (black)in the filament with length to radius ratio 320 for a smallerloading force | F z (0) | = 0 . EI/R and k fr = 0 .
8, fit (red)with the function f ( l ) = A ( l + l ) c ; A = 15 . l = 2010, c = − . l c inEq. (19) moving further away. The exponent of the power law,however, remains nearly unchanged. The boundary regionsget wider as the load is reduced. VI. DISCUSSION
In this work we have analysed the behavior of an elas-tic filament that is confined in the longitudinal as wellas transverse direction. The longitudinal confinementleads to the well known Euler buckling instability thatdepends on the external loading and the length of the fil-ament. We have shown that the buckled configuration ontouching the confining walls of a cylindrical channel thenevolves through a planar deformed configuration towardsa helical state via an additional instability. Consideringthe effects of the friction on the walls of the cylindricalenclosure we have been able to demonstrate, that thelongitudinal force transmitted through the filament de-cays along its length. This phenomenon is not unrelatedto the decaying longitudinal stresses within a cylindricalgranular column anchored by the wall friction, or even aregular elastic solid with a finite Poisson ratio enclosedwithin a rigid hollow cylinder and again anchored by wallfriction.The important difference between the examples ofgranular and regular solids confined within cylindrical
FIG. 8: (color online) If the load is too large, the filamentmakes a U-bend and escapes out of the cylinder as shown onthe subsequent figures instead of forming the helix.
L/R =320, | F z (0) | = 0 . EI/R , only the upper half of the cylinderis shown. walls and the present case of thin elastic filament is thenature of the decay of the longitudinal force along thecylindrical enclosure. In the first two cases the decay isexponential and leads to a natural length scale for theproblem. This means that the jamming of the granularor classical elastic bodies depends only on intrinsic pa-rameters, describing the stress distributions within thebody and the magnitude of the friction forces betweenthe body and the cylindrical enclosure. In the case ofthe confined filament the longitudinal force transmittedalong the cylindrical enclosure decays algebraically. Inthis case the onset of jamming is determined by the in-trinsic elastic parameters of the filament as well as thelongitudinal force counteracting the loading from the op- posite side of the enclosure. This counter-force could bein principle very small and could even result from ther-modynamic fluctuations on the other end of the enclosureif the cylindrical pore is small enough. Nevertheless thefact that the onset of jamming in the case of a confinedelastic filament is scale free and thus depends on externalconstraints separates it fundamentally from the standardjamming in granular materials [18].What would be the systems that could exhibit thistype of jamming scenario? We have found out that thecriterion for the onset of helical postinstability shape isgeometrical: according to the curve in Fig. 6, the as-pect ratio of polymer length (or persistence length, if it issmaller) and the radius of the pore should be of the orderof 100 or more for a significant reduction of the compres-sion force that could in principle lead to jamming. Forcarbon nanotubes with a typical length of 100 µ m andYoung’s modulus of 10 Pa, the relevant pore radius is1 µ m or less, and the loading force is in the nN range. Allreasonable values. On the other hand, DNA appears tobe too flexible, i.e., its persistence length of 50 nm is tooshort as it should be confined to a sub nanometer pore,violating the structural integrity if DNA (the diameterof DNA is approximately 2 nm). It thus appears that atleast in principle one could observe the helical instabil-ity coupled to a jamming transition within the contextof confined microtubules. We hope that our theoreticalwork will provide enough motivation for experimentaliststo search for this interesting phenomenon. Acknowledgments
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