Wavelength selection in a buckling garden hose
WWavelength selection in a buckling garden hose
Tianyi Guo ∗ Advanced Materials and Liquid Crystal Institute, Kent State University, Kent, OH, USA
Xiaoyu Zheng † Department of Mathematical Sciences, Kent State University, Kent, OH, USA
Peter Palffy-Muhoray ‡ Advanced Materials and Liquid Crystal Institute, Kent State University, Kent, OH, USA
September 16, 2020
Abstract
We consider sinusoidal undulations which appear on certain garden hoses under normal use. Wepropose an explanation and provide a model for this phenomenon, and make a connection with biolog-ical structures as well as self-buckling. We compare observations with model predictions, and suggestpotential applications in the area of shape-changing materials.
An intriguing summertime phenomenon is the sinusoidal shape assumed by certain garden hoses left onlawns. One example is shown in Fig. 1 The pattern, which can extend the length of the hose, is apparentlyspontaneously formed during normal use. We have studied this curious phenomenon and, in this brief note,we consider the underlying physics and propose a mechanism for the wavelength selection.Figure 1: Example of undulations in a garden hose. ∗ [email protected] † [email protected] ‡ mpalff[email protected] a r X i v : . [ c ond - m a t . s o f t ] S e p The appearance of the deformation
Garden hoses are typically connected to a pressurized domestic water system via a shutoff valve at the inletend and are terminated with a spray nozzle with a control valve at the outlet end. In the normal process ofwatering, it is first ensured that the control valve by the nozzle is closed. Next, the inlet valve is opened; atthis point, water flows into the hose, pressurizing it. The outlet valve is then opened; water flows throughthe hose and sprays out of the nozzle, which is directed at whatever is to be watered. When watering iscompleted, the outlet valve by nozzle is typically closed first, followed by closing the inlet valve. At thispoint, the hose, still pressurized, is lying on the grass often in a fairly straight line. If the outlet valve isnow opened to relieve the pressure in the hose, many hoses exhibit a remarkable phenomenon: they visiblyelongate as water is expelled, and form sinusoidal undulations along their length. The same phenomenonoccurs due to simple leakage of water from the hose. Two questions immediately arise: why does the hoseelongate, and what is the mechanism of wavelength selection? We attempt to answer both of these questionsbelow.
Garden hoses in the distant past were made of rubber, which is still a popular option today. Rubber gardenhoses, however, exhibit neither the elongation nor the sinusoidal undulations described above. Elementaryconsiderations show that the longitudinal stress σ l in a rubber tube of radius r and wall thickness w underexcess pressure P is, to a good approximation σ l = P r w , (1)while the azimuthal stress σ a is σ a = P r w . (2)Since strain is proportional to stress, the longitudinal strain ε l is, in the linear regime, ε l = P E r w , (3)while the azimuthal strain ε a is ε a = P E r w , (4)where E is the Young’s modulus. Remarkably, for a cylindrical tube, the azimuthal strain is twice as largeas the longitudinal one.A simple rubber hose thus expands both in diameter and length when pressurized, and contracts in bothwhen the pressure is reduced. It is interesting to ask then why certain garden hoses show a dramaticallydifferent behavior, and lengthen, rather then shorten, when their internal pressure is reduced.With advances in materials technology, a wide variety of materials are available for the construction oftubing in addition to rubber, such as polyurethane, polypropylene, polyvinyl chloride and others. More tothe point, a variety of architectures are also available, and composite structures with linings and reinforcingmeshes are frequently employed.A popular current design consists of an inner polymer lining inside an elastomer tube, containing helicallywound and nearly inextensible filaments with opposite helicity. A typical garden hose[1] with such structureis shown in Fig. 2.A key element in such a hose is the reinforcing helical mesh which plays a key role in the unusual behaviorconsidered here. In all the subsequent discussions, it will be assumed that the hose maintains its circularcross-section everywhere unless explicitly noted otherwise. Consider a helix, winding about the z -axis. The position r of a point on the helix, in cylindrical coordinates,is given by r = ( r cos φ, r sin φ, aφ ) . (5)2igure 2: Camco TastePURE 50ft. Premium drinking water garden hose for domestic use.The pitch of the helix is p = 2 πa (6)and the radius is r . The length of the helix s for one period can be obtained by integrating along itslength; since ds = (cid:113) r sin φdφ + r cos φdφ + a dφ , (7)and the length of the helix for one period is s = (cid:112) (2 πr ) + p . (8)We note that we can write ps = cos θ (9)and 2 πr s = sin θ, (10)which defines θ , the helix angle; the angle that the helix makes with its axis.We are interested in deformations of the hose allowed by the constraints due to the reinforcing helicalstructure. It consists of pairs of fibers forming helices of opposite handedness but equal pitch, as can be seenin Fig. 2. Typically, a number of such fiber pairs are present, forming a helical fiber array. This architectureresists twisting along the longitudinal center line, since such a twist would reduce the pitch for one of thehelices, but increase it for the other. If there is no such twist and the fibers are inextensible, then s is aconstant; it is the distance along the helix for one period (where φ has changed by 2 π ), regardless of thepitch p . Since the total number of turns N traced out by either helix is a constant, the length L of the ofhose is L = N p, (11)where N is the number of pitches present. Since the length of a fiber L f = N s is fixed, N is a constant ofthe hose. We then have L f = ( N πr ) + L ; (12)the constraint linking the radius and the length of the hose. This is the principle underlying the Chinesefinger traps[2]; elongation causes narrowing, and vice versa.The volume of the cavity enclosed by the hose is V = N pπr , (13)where we have assumed, for simplicity, that the wall thickness is negligible. Eliminating r and using Eq.(8), we have V = N π s (cid:18) ps − ( ps ) (cid:19) (14)3nd we see that changing the volume corresponds to changing the pitch and the length of the hose as wellas its radius.We note that dVdL = 14 π s (1 − θ ) (15)and if cos θ < √ θ > . o , then increasing the volume lengthens and narrows the hose, while ifcos θ > √ , (17)then increasing the volume shortens and widens the hose. In this way, the helix angle determines whetherthe tube lengthens or shortens when its volume is increased. At the critical helix angle θ c = 54 . o , andcos θ c = 1 √ , (18) dV /dL = 0, and the volume of the tube is the maximum.The experimentally determined helix angle of our garden hose is θ = 36 . o , placing it firmly into theregime where reducing the volume results in elongation. The reason for the particular choice of helix angleis not clear to us. (The manufacturer, Camco Manufacturing, did not respond to our queries.) The elastic modulus of the hose is of paramount importance in determining stability against buckling. Wetherefore consider the elastic energy cost of shape deformations asociated with changing the helix angle. Thestructure of the hose material with the reinforcing helical filament array is clearly anisotropic and hence afull description of the elastic properties would require at least 5 elastic constants [3]. Here we provide thesimplest description consistent with the salient aspects of the elastic response.We show that there are two distinct elastic constants in the system, and we derive expressions enablingtheir experimental determination.The elastic energy density of the hose associated with the deformation ∆ θ can be estimated in terms ofthe strains as [4] E = G ( ε l + ε a + ( ε l + ε a ) ) , (19)where G is the shear modulus of the material in which the helical array is embedded. The longitudinal strainis ε l = ∆ pp = − tan θ ∆ θ, (20)and the azimuthal strain along the circumference of the hose is ε a = ∆ r r = cot θ ∆ θ, (21)and we have assumed volume conservation of the hose material. The Poisson’s ratio η v = − ε a ε l = cot θ (22)is evidently determined by the helix angle.The elastic energy density associated with deformations correseponding to changing the helix angle is E = 2 G (cid:18) θ cos θ − (cid:19) (∆ θ ) . (23)4riting the energy density in terms of the strains ε l and ε a , we get E = 2 G cot θ (cid:18) θ cos θ − (cid:19) ε l (24)and E = 2 G tan θ (cid:18) θ cos θ − (cid:19) ε a . (25)We can then regard the effective elastic modulus associated with elongation of the hose as E l = 4 G cot θ ( 1sin θ cos θ −
3) (26)and the effective modulus associated with increasing the radius as E a = 4 G tan θ (cid:18) θ cos θ − (cid:19) . (27)We note that E l tan θ = E a . (28)It is interesting to note that the moduli are independent of the details of the geometry, and E l and E a areboth functions of the helix angle. Through governing the Poisson ratio, the helical fiber array acts as atransformer of the shear modulus G , giving rise to the two effective elastic moduli. At θ = θ c , E l = 3 G ,while E a = 12 G . In the limiting case, when θ = 0, E l = ∞ and E a = 4 G , while if θ = π/ E l = 4 G and E a = ∞ .The longitudinal elastic modulus may be determined by simply stretching the empty hose without regardto its volume. Then E tot = V mat E l (cid:18) ∆ LL (cid:19) , (29)where V mat is the volume of the hose material, and since the stretching force F is proportional to ∆ L , E l = F πr w L ∆ L . (30)Alternately, the longitudinal elastic modulus may be determined from the volume change of the hose dueto an applied pressure P . The hose volume V = πr L can be written as V = N π s cos θ sin θ, (31)and the relative change in volume is ∆ VV = 2 − θ cos θ sin θ ∆ θ. (32)The relation between the relative volume change and the relative length change is∆ VV = 1 − θ − cos θ ε l . (33)The total elastic energy of the hose is E T ot = V mat E = ∈ N s wG cos θ sin θ (cid:18) θ cos θ − (cid:19) (∆ θ ) . (34)The pressure P = 2 E T ot / ∆ V is then P = 8 G wr (1 − θ sin θ )(2 − θ ) ∆ VV , (35)5hich allows determination of the elastic constants.Explicitly, in terms of the pressure and the volume change, we have E l = 12 r w P ∆ V /V ( (cid:18) − θ (cid:19) (36)and E a = 12 r w P ∆ V /V (cid:18) − θ (cid:19) . (37) Helical fiber-array reinforced fluid filled tubes, so-called hydroskeletons, consisting of helical filaments ofopposite handedness and with a variety of helix angles are abundant in nature, ranging from the stems ofsunflowers to the tubefeet of starfish and skins of sharks [5], [6]. Their utility lies in their ability to becomemore or less rigid without rigid components.If a helical fiber-array reinforced tube, with helix angle different from critical value, is filled with waterat zero excess pressure, it will be in stress free equilibrium. If the ends are sealed, and the angle is less thancritical, the tube cannot be elongated, but a compressive force on the ends can shorten the tube; but sincethis would increase the volume, the tube cross-section cannot remain circular, and the tube will flatten. Ifthe angle is greater than critical, the tube cannot be shortened, but a tensile force at the ends can elongateit. Again, this increases the volume, and the tube must flatten. In either case, when the helix angle is notcritical, even though the tube with circular cross-section is completely filled with an incompressible fluid, itwill not be fully rigid, since it can become flattened and more easily bend. Flatworms, for eample, makeuse of this controllable rigidity in their locomotion [5]. However, when the angle is critical, the volume is amaximum, neither elongation nor shortening of the tube increases the volume, so the filled tube with closedends is nearly perfectly rigid.Until fairly recently, it was believed that all hydroskeletons in nature consisted of structures with helicalfiber arrays [5]. However, in 1997, D.A. Kelly demonstrated that this is not the case; she showed thatreinforcing muscle fiber arrays in hydroskeletons in the penises of armadillos were axial and orthogonal[7];that is, either parallel to the tube axis, or perpendicular to it. (Strictly speaking, these are two helical arraysin the two limiting cases when θ = 0 and θ = π/ Ignoring the effects of gravity, if the hose is full of water, but under no excess pressure, it is undeformed,with zero stress and strain. If the hose is then pressurized, the volume of the hose increases, in our case bygetting wider and shorter. The hose has to undergo an elastic deformation, subject to the inextensibilityconstraint of the helical fibers. Elastic energy is stored in the deformed strained hose wall. If the inlet valveis next closed and the spray nozzle is opened (or the hose simply leaks at a joint), water leaves the hose, thepressure drops, and the hose elongates. The observed bucking is a direct consequence of the elongation ofthe hose.To understand the proposed selection mechanism, we first imagine a straigth section of the hose of length2 l and consider its elongation assuming that the ends are free to move. Due to symmetry, the center of masswill not move during the elongation; instead the two ends will move away from the center. Since the hose fullof water, it has considerable weight, and there will be kinetic friction between the hose and the grass/earth6upporting its weight. Each element of length of the hose must have a net force acting on it to overcomethe kinetic friction. The longitudinal stress in the hose due to friction must therefore be a linear function ofposition; maximum in the middle, decreasing linearly and vanishing at the ends. If µ k is the kinetic frictioncoefficient, the friction force per length acting on the hose is f = µ k ρg, (38)where ρ is linear mass density and g is the acceleration of gravity. The longitudinal stress in the walls of thetube at the midpoint of must therefore be σ l = µ k ρgl/ (2 πr w ) . (39)Clearly, stress in the wall increases with distance from the end, and there must be a point where the tubebecomes unstable against buckling.If only one-half of the tube is considered, with stress linearly increasing with distance from the free end,its stress is equivalent to that in a column loaded by its own weight. The stability of such a clamped-freecolumn has been examined by Euler [9],[10],[11]. Remarkably, Euler erred in his first calculation ; it wasshown by Dinnik [12] that the column will buckle at the critical length l c l c = α EIF c , (40)where α = 7 . F c is the total load, E is Young’s modulus and I is the moment of area of the cross-section.In our example, F c = µ k ρgl c (41)and we obtain l c = α EIµ k ρg . (42)We therefore expect the wavelength of the deformation λ = 2 l c , or λ = 2( α EIµ k ρg ) / . (43)Since the friction force per length of the water filled hose is f = µ k ρg, (44)the expected wavelength of the deformation is λ = 2( α EIf ) / . (45)This is our key result. If the buckling mechanism is as proposed above, then the amplitude of the sinusoidal deformation is simplyrelated to the elongation of the hose; the length of the deformed hose in one period must equal the wavelengthtimes the longitudinal stretch 1 + ε l .If the shape of the curve traced out by the hose is sinusoidal, that is, of the form y = A cos(2 π xλ ) , (46)then λ (1 + ε l ) = (cid:90) p (cid:112) y (cid:48) + 1 dx. (47) for history see Wikipedia entry on self-buckling. y (cid:48) , we have λ (1 + ε l ) (cid:39) (cid:90) λ (1 + 12 A ((2 π λ ) sin (2 π xλ )) dx = λ (1 + 14 A (2 π λ ) ) (48)and the amplitude is given, approximately, by A (cid:39) π λ √ ε l . (49) The physical dimensions of the hose were determined by disecting the hose and measuring the diameter andwall thickness with a measuring tape. The outside diameter was found to be 20 mm and the inside diameter16 mm .The moment of area of a thin tube is I = πr w, (50)where w is the wall thickness. In our case, I = 3 . × − m . (51)Key quantities determining the wavelength of the instability are the elastic modulus and the kineticfriction.The friction force per length f was measured a number of times by dragging a length of hose and measuringthe required force with a spring balance. The required force can vary considerably due to different conditionsof humidity, temperature and grass height, wetness and stiffness. We found f = 27 ± N/m. (52)Another key measurement is the determination of the relevant elastic constant. In the original calculation[11],[12], an isotropic material was considered, with Young’s modulus E . Our hose is anisotropic, with twodifferent effective elastic constants. We assume that the relevant elastic constant in our case is E l whichrelates the total elastic energy to changes in length, since both bulk compression and bend are associatedprimarily with longitudinal strains.The first and most direct method to determine the longitudinal elastic constant E l is to stretch the emptyhose by applying a force manually, and to measure the displacement ∆ L of the end of the hose with a meterstick and the force with a spring balance. We found that the displacement ∆ L was proportional to the force F , F ∆ L = 73 . N/m (53)and for our L = 30 . m hose, Eq. (30) gives E l = 22 . M P a. (54)We remark here that the force to stretch the hose by fixed length is time dependent; there is a relaxationtime on the scale of seconds. We took force readings ∼ s after the displacement, then the force readingremained nearly constant.The second and less direct method of determining the longitudinal elastic constant consisted filling thehose with pressurized water, and measuring the total volume V of the water in the hose. Subtracting fromthis volume the volume of the empty hose gives the extra volume ∆ V of the hose due to pressure P . Wefound that from measurements on a 30 m hose that∆ VV = 0 . ± . . (55)8here is considerable uncertainty in these measurements due to errors associated in part with the dynamicresponse of the hose, and in part with accurately measuring the volume of the water in the hose. The waterpressure at supply was P = 426 kP a , and from Eq. (36) we obtain E l = 27 . M P a (56)in fair agreement with the result from the stretching method. We assume that a reasonable estimate of theeffectdive longitudinal elastic modulus is the average of the two measurements, with (cid:104) E l (cid:105) = 24 . M P a .We note that longitudinal strain under pressure can be obtained from Eq. (33) to give ε l = 0 . . (57)We can now calculate the wavelength of the modulation; from Eq. (45) we have λ = 2( α E l If ) / (58)giving the theoretical prediction λ th = 0 . m. (59)Our measured values have considerable scatter likely due again to humidity, temperature and grassconditions. The average experimental value is λ exp = 0 . ± . m. (60)A typical example is shown in Fig. 3 below.Figure 3: A typical example of an undulation with λ = 66 cm . The measuring tape shown was used tomeasure the wavelengths.The amplitudes of the sinusoidal undulations, which were clearly affected by surface topography, werenot studied carefully. The theoretical estimate from Eq. (49) is A (cid:39) . m (61)and from our observations (see Fig. 3), we estimate A exp (cid:39) . ± . m. (62)
10 Results and conclusions
We have studied the spontaneous buckling instability exhibited by certain garden hoses. We have foundthat the instability occurs as a result the hose elongation when water is released, and the internal pressureis reduced. The elongation is associated with the constraint on shape change associated with a helical arrayof fibers with helix angles less than θ c = 54 . o . During elongation, kinetic friction between the hose andthe supporting surface results in stress buildup in the hose, leading to buckling. The instability is similar to9he self-buckling of columns, enabling prediction of the selected wavelength. A simple model for the elasticresponse of the hose enabled determination of the relevant elastic modulus of the hose, and this, together withthe measured friction force allowed comparison of the predicted and observed wavelengths and undulationamplitudes. We have found reasonable agreement between of theory and experiment.In addition to predictingthe instability and its details, our simple model provided useful insights into the effects of the helical fiberarray on the material response. The constraint of the array couples longitudinal and azimuthal strains, andthe helix angle changes not only Poisson’s ratio, but also the effective longitudinal and azimuthal elasticmoduli of the otherwise isotropic tube. The ability to tune elastic moduli in this way may be useful in thedesign of active shape-changing materials such as photomechanical elastomers for a variety of applications.
11 Acknowledgments
This work was supported by the Office of Naval Research [ONRN00014-18-1-2624].