The laws of rolling Текст научной статьи по специальности «Физика»

УДК 531.45 + 531.13 + 531.8 + 514.85

The laws of rolling

G.P. Cherepanov

The New York Academy of Sciences, New York, USA

The man-made transport was born when the wheel was invented. Since then the study of rolling has started. In 1781 the problem of rolling was mathematically formulated by Ch. Coulomb who offered Coulomb's law of rolling, but the science of rolling has been purely empirical still until recently. In this paper the exact laws of rolling are analytically derived in terms of elastic and geometric properties of rolling bodies and foundations. Using the mathematical theory of elasticity and the CH-rule, the rolling resistance coefficient is calculated in the cases of: (i) an elastic cylinder rolling over another elastic cylinder of another material, in particular, over an elastic half-space, and an elastic wheel rolling over the rail of another elastic material; (ii) an elastic ball rolling over another elastic ball of another material, particularly over an elastic half-space; (iii) an elastic torus rolling over an elastic half-space of another material, and (iv) a cylinder, or a ball, or a torus rolling over a tightly stretched membrane or over a thin elastic plate. Empirical results of the measurement of the rolling resistance coefficient gained earlier by the railroad and automobile engineers appeared to be in excellent agreement with the results of this analytical calculation based on the suggested rule of rolling. The effect of adhesion was also studied using the exemplary case of an elastic cylinder rolling over an elastic half-space.

Keywords: Coulomb's law of rolling, Hertz problem, Riemann problem, rolling resistance coefficient, tribology, rolling cylinders, balls, wheels and tori, rolling resistance coefficient of tires and railway cars, rolling friction, rolling drag, adhesion effect DOI 10.24411/1683-805X-2018-15004

Законы качения

Г.П. Черепанов

Нью-Йоркская Академия наук, Нью-Йорк, США

Изобретение колеса ознаменовало собой рождение транспорта и начало изучения явления качения. В 1781 г. Ш. Кулон сформулировал проблему качения и предложил свой эмпирический закон качения, и до недавнего времени наука о качении колеса была чисто эмпирической. В предлагаемой статье точные законы качения выведены аналитически в зависимости от геметрических и физических параметров тел качения и основания. При помощи СН-правила и теории упругости коэффициент сопротивления качению вычислен в случаях: а) качения упругого цилиндра по другому упругому цилиндру из другого материала, в частности, по упругому полупространству, а также качения упругого колеса по упругому рельсу из другого материала; б) качения упругого шара по другому упругому шару из другого материала, в частности, по упругому полупространству; в) качения упругого тора по упругому полупространству из другого материала; г) качения цилиндра, шара или тора по туго натянутой мембране, а также по тонкой упругой пластине. Полученные результаты практически совпадают с эмпирическими результатами измерения коэффициента сопротивления качению, полученными ранее для железнодорожного и автомобильного транспорта. Эффект адгезионного прилипания был также изучен на примере качения упругого цилиндра по упругому полупространству.

Ключевые слова: закон качения Кулона, задача Герца, проблема Римана, коэффициент сопротивления качению, трибология, качение цилиндров, шаров, колес и торов, коэффициент сопротивления качению для автомобильного и железнодорожного транспорта, трение качения, эффект сцепления

1. Introduction

Wheel was one of the earliest and most important inventions of the man because, except for sleds on ice and snow, it was much easier to wheel a weight than to drag it. That is why the man-made transport was born when the wheel was invented. This happened long before the horse domestication. In [1] A. Erman told us that wheeled carts carried by oxen were common in Ancient Egypt about two

thousand years before the conquest of Egypt by proto-Indo-European Hyksos who were the first to invent chariots and breed horses for war.

Since the dawn of human history the problem of rolling resistance has existed. Rolling resistance sometimes called also rolling friction or rolling drag is the force resisting the motion when a body rolls on a surface. In mathematical terms, this problem sounds as follows: What is the value of

© Cherepanov G.P., 2018

the rolling resistance coefficient Crr in the following equation

F = CrrN ? (1)

Here N is the load on the wheel, e.g. the weight of the vehicle, and F is the rolling resistance force.

Very many scientists tried to solve this problem but failed so that test seemed to be the only way to find out the rolling resistance coefficient—see, e.g., [2-9], among many thousands of other papers.

For example, in the case of cast iron mine car wheels on steel rails, the following empirical equation is used [2, 8] Crr = 0.0048(9/ R)^2 (100/ N )1/4. (2)

Here R is the wheel radius in inches, and N is the load on the wheel in lbs.

Coulomb's law and Dupuit's equation are, probably, the most known theories of rolling that say:

Crr = KC/R (Coulomb), (3)

Crr = *d/R2 (Dupuit). (4)

Here KC and KD are some empirical constants. These equations work in some cases and do not work in others as it is common for empirical theories.

To get in with what follows, consider at first the elementary problem of equilibrium of an absolutely rigid heavy cylinder of radius R lying in a very small and shallow cylindrical socket of the same radius in a half-space y < 0 of an absolutely rigid material. Let us assume there is no adhesion and apply the thrust force T to the cylinder center directed along the half-space surface y = 0, with the gravity force N of the cylinder being directed perpendicularly. Evidently, without a socket, a however small thrust force makes the cylinder roll. It is the socket that provides resistance to rolling so that, if thrust T is less than aN/R, no rolling occurs (2a is the width of the cylindrical socket, a << R). However, the rolling starts on when TR > aN so that TR = aN is the critical state.

In reality, a socket is formed by an elastic deformation of the foundation and cylinder caused by the weight and other forces acting upon the cylinder. The socket moves along the surface together with the rolling cylinder. The magnitude of sockets made by rolling bodies should be found from the solution of the corresponding mixed boundary value problems of the elasticity theory. To solve some of these problems the Kolosov-Muskhelishvili representations [10] and the Riemann works are used in this paper: namely, the Riemann boundary value problem solved by F.D. Gakhov in 1940 [11].

2. Rolling of an elastic cylinder over another elastic cylinder, particularly, over an elastic half-space

We study, at first, the mathematical problem of the elasticity theory, and then, based on its solution, we take account of the rolling.

2.1. The contact problem of the theory of elasticity

Let two elastic cylinders of radii R1 and R2 generally made of different materials contact each other so that the stresses and strains in both are under plane strain conditions. We confine ourselves by small deformations, which is the most important practical case, so that the width of the contact area 2a is always much smaller than R1 and R2, that is R1 >> a. We assume also that | R21 > R1 > 0.

When R2 ^ ro, the second cylinder turns into a halfspace which is the most important particular case. When R2 < 0, it is the infinite elastic space with a cylindrical hole of radius | R21, inside which an elastic cylinder of radius R1 is in contact.

Let us use the Cartesian rectangular coordinate system Oxy in a plane normal to the parallel axes of the cylinders so that its origin O is chosen to be at the center of the contact area, and y be the axis of symmetry of the problem directed to the center of the cylinder of radius R1. This cylinder is pressed to the other cylinder by force N applied to its center.

The boundary value conditions of this plane strain contact problem of the elasticity theory can be formulated as follows:

|x| > a: (ay - * )± = 0, i = 7-1, (5)

y = 0, y = 0,

|x| < a:

[ay - iTxy ] = 0

du .dv ( 1 1 1

— +1 — = -ix --1--

dx dx t r1 r2 j

+ C.

(6)

Here (u, v) is the displacement vector, ax, ay and Txy are the stress tensor components, constant C characterizes the difference in displacement u of opposite surfaces of the cylinders which forms in the process of compression before the adhesion on contact, and z = x + iy is the complex variable.

Besides, the following designation is used everywhere in this paper:

A± = lim A, when z ^ x ± i0, [A] = A+- A- (7) The second equation in Eqs. (6) means that neither rupture nor sliding occur on the contact area beyond of some previous local distortion.

Let us use the following Kolosov-Muskhelishvili presentations for elastic stresses and strains in the plane elasticity theory [10] :

a x + a y = 4ReOj (z), j = 1, 2, (8)

= Oj (z) + Oj (z) + zO' (z) + Wj (z), (9)

ay - iTxy

i du .dv 2^ J — +1 —

' dx dx

= Kj Oj (z) - Oj (z) - zOj (z) - Wj (z). (10)

Here j = 1 and j = 2 are subscripts of the upper and lower half-planes, correspondingly, O(z) and W(z) with subscripts 1 or 2 are some functions of the complex variable z

o± = -of - xo;+ -Yf, o± = -o2 - x02m -? 2.

which are analytical in the corresponding half-planes, |J.j and vj are the shear modulus and Poisson's ratio of the corresponding material, and Kj = 3 - 4Vj for plane strain.

The boundary value problem of Eqs. (5)-(10) unsolved by Muskhelishvili [10] or elsewhere is being solved here by the approach of papers [12, 13].

Let us use the following analytical continuation: o (z) = -o (z) - zo;(z) - ? (z), J = Imz < 0, (11)

02(z) = -o2(z) - zo2(z) -?2(z), y = Imz > 0. (12) When y = 0, we get from here:

' 2 2 2 (13)

(14)

Here either the upper or the lower signs hold.

In view of [ay - /t ] = 0 everywhere at y = 0, we get from Eq. (9) that

o+ + o+ + xoi+ + ?+ = o- + o- + xo2- + . (15) Using Eqs. (13) and (14) we can rewrite Eq. (15) as follows

o+ - o- = o- - o+, (16)

or (o +o2) + = (o +o2)-. (17)

Based on the principle of analytical continuation we conclude that

o^ z) = -o2( z). (18)

Because of Eqs. (5), (15) and (16) we have

oj+ = oj- as |x| > a, y = 0. (19)

Hence functions o^z) and o2(z) are analytical in the whole plane z cut along y = 0, -a < x < a.

Using Eqs. (13) and (14) the second boundary condition in Eqs. (6) is written at y = 0, -a < x < a as follows

^2 (Kxo+ + o-) = ^ (^o+ + o2) -

(

2ix^2

1 1 —+—

R R

\

+ .

(20)

Using Eq. (18) we come to the following Riemann problem for one pair of functions [11-13]:

o+ + mo- = 2isx - 2^2C (^ + ^2K1)-1, (21)

y = 0, |x| < a. Here

( 1

m = s = -

^2

1 1 -+-

R R

(22)

+ (X2 K + (X2 Kj

The general solution to this Riemann problem in the class of functions limited at z ^ ±a and z ^ ro can be written as follows

o2( z) =

2is

1 + m 2^2

2is 1 + m C

F (z) -

(1 + m)( Kj)

where

F(z) = (z - a)1/2-'8 (z + a)1/2+iS, 1

(25)

S =— ln m. 2n

Here F(z) is the single-valued analytical branch of the function in the z plane cut along y = 0, -a < x < a which is equal to

F (z) = z + 2iaS - — (1 + 4S2) + O( z "2) 2z

(26)

as z ^ ro.

Constants a and C are determined by the following behavior of function o2(z) at infinity [12, 14-16]:

iN

o2 (z) =- as z ^ ro. (27)

2nz

Here N is the value of the equivalent force of the pressure on the contact area per unit of the cylinder length acting upon the lower half-plane, which is equal to the corresponding weight of the upper cylinder of radius R1.

According to Eqs. (23) and (26), as z ^ ro we have

f

z + iaS---

2z

^ . . 2is 2is

o2(z) = 7~z

1 + m 1 + m

2^2 C (1 + m)( 2Kj) By comparing Eqs. (27) and (28), we get

2a2S2^

2sa2 + 8sa2S2 = N (1 + m), C = aS n

(

1 1 — + —

R R

(28)

(29)

Using Eqs. (21) and (29) we find the width of the contact area

2 = N ^(1+ K2) + ^2(1 + K1)( 11

2n MiM^G + 4 S2)

+

R R

(30)

It is more convenient to use this equation in terms of common technical elastic constants as follows

a2 =

4 N

(

n(1 + 4 S2)

1 -v2 1 -v

2 v

1

R+R

-1

(31)

Here E and v are Young's modulus and Poisson's ratio, with subscripts corresponding to the relative cylinder.

Using Eqs. (23)-(25) and (29) we come to the final solution:

o2(z) = 1 Ai\z-(z2 -a2)^2[ ^

Here

(

A =

1

R+R

V

1 -v2 1 -v

2 y

+ iaS [. (32)

(33)

E e2

Using Eqs. (9) and (32) the stresses on the contact area y = 0, -a < x < a can be written as follows

= -1 Ach (tcS)Va2 - x2 x

y 2

xcos(Sln |(a + x)/(a-x)|),

(34)

= -^ Ash (tcS)a/a2 -

x2 x

Xsin(8ln |(a + x)/(a- x)|). (35)

Equations (31)-(35) provide the solution of the stated problem in most convenient form. For the particular case 8 = 0, it was given in [17, 18].

2.2. The problem of rolling

Now, let us try to roll the upper cylinder of radius R1 over the horizontal foundation, i.e. over the lower cylinder of radius R2 = ro. Suppose T is the thrust or driving force per unit of the cylinder length applied to the center of the upper cylinder in the horizontal direction perpendicular to the vertical direction of force N, particularly, the weight of the upper cylinder. If both the cylinder and foundation are rigid, neither socket nor dent can form, and no contact area can be made so that the cylinder starts on rolling at a however small thrust because the rolling resistance force F is equal to zero.

But, if either the cylinder or foundation is elastic, a dent forms so that a nonzero contact area provides some resistance to the rolling. If thrust T is small enough so that vector (T, N) of the resultant force points to somewhere inside the contact area, the equilibrium holds and no rolling occurs. This is evident because some change of the contact area size Aa due to elasticity has the order of a (T/N) for small values of T/N and, hence, can be ignored.

As soon as vector (T, N) of the resultant force points to somewhere outside the contact area, the equilibrium gets broken and an accelerated rolling begins. This follows from the equilibrium equation of the moments of forces with respect to pointy = 0, x = a even for the imaginable situation when both elastic bodies become rigid for small values of T. For real elastic bodies, the loss of balance is even more significant because of an additional elastic reaction of the lower foundation.

Thus, we come to the following rule: Rolling can start only when the resultant force vector (T, N) points to the front of the contact area so that we have

T/N = a/Ry. (36)

This is the necessary rule of rolling called the "chief head" rule, or the CH-rule, for short. It is central for the understanding of the rolling processes and rolling resistance [17, 18].

The state characterized by the CH-rule of Eq. (36) corresponds to the rolling at constant speed when thrust T is equal to the rolling resistance force F. If T/N < a/R1, no rolling occurs, but if T/N > a/R1, the upper cylinder of mass M rolls and moves along the x axis at acceleration dV/dt so that

MdV/dt = T - aN/Rl. (37)

Based on the CH-rule the rolling resistance coefficient is equal to

Crr = a/R1. (38)

Here the value of a is completely determined by Eq. (31) written in terms ofN,E1, E2, R1, R2, v1, v2 and parameter 8.

Therefore, Coulomb's problem is solved, and the correct value of Crr proves to be very different from both Coulomb's law and Dupuit's equation as well as from the empirical law of Eq. (2).

In the important particular case when 8 = 0, E1 = E2 = = E, v1 =v2 =v, R2 =ro and R1 = R, the expressions in Eqs. (31)-(36) and Eq. (38) reduce to the following equations:

O( z) = z -Vz^O7, (39)

2ER

1 -v2

a y

y = 0, |x| < a.

■■4a2-

Txy = 0,

Crr =—=a=2J2N(1 -v).

N R V nER

(40)

(41)

To demonstrate the accuracy of the present analysis based on the CH-rule let us calculate, as an example, the rolling resistance coefficient for railroad steel wheels on steel rails for passenger railcars with eight 36-inch diameter wheels on the 63 kg/m rails used in New York Central Railroad System.

In this case, the rail head width being contacted with the railcar wheel is equal to 3 inch = 7.62 cm so that using Eqs. (41) we get the following rolling resistance coefficient: for 10-ton railcar Crr = 0.00113 and for 40-ton rail-car Crr = 0.00226 (v1 = v2 = 0.29, E1 = E2 = 200 GPa, 8 = = 0, R2 = ro, and R1 = 45.72 cm).

The elastic constants of both the rail and wheel are very close. The plane strain condition holds well in the process zone of the contact area because its width is much smaller than the 3 -inch rail head or any other dimension of the structure. Indeed, according to Eq. (41) we have a = 0.5 mm for 10-ton railcar and a = 1.0 mm for 40-ton railcar.

This result of calculation of the rolling resistance coefficient coincides with the corresponding standard data, see, e.g. [2], according to which Crr = 0.0010 to 0.0024 for railroad steel wheel on steel rail and Crr = 0.002 for passenger railcars.

Similarly, we can calculate that for ordinary car tires on concrete or asphalt the rolling resistance coefficient is equal to Crr = 0.010 to 0.015 in accordance with available data from [2].

Let us estimate the practical significance of real values of parameter 8 responsible for some anomalous distortion. Since according to Eq. (22) we have 1/3 < m < 3, we can derive from Eq. (6) that

-0.175 <8 = —lnm <0.175.

2n

For common values of Poisson's ratio in the range (1/4,1/3) the value of S varies in even more narrow interval (-0.11, +0.08). Small parameter S causes some specific variations significant only at the very ends of the contact area. It means that when m ^ 1 the violation of the condition of nonpenetration occurs only in the small neighborhood of the ends so that it can be ignored despite this makes the problem ill-posed [17-20]. In fact, these variations indicate that when S ^ 0 some local sliding zones much smaller than a form near the ends of the contact area.

All mathematical calculations of this section are also valid for plane stress by means of the well-known substitution of constants Kj in Eqs. (10). This is the case of a very thin wheel rolling over a very thin rail. Unfortunately, this case is of almost no practical value for the rolling because the contact area width a is already very small while the thickness of the plane stress wheel should be much less than even this width [16, 21].

3. The effect of adhesion upon the rolling

Because of atomic forces of attraction all solid surfaces experience the effect of adhesion while getting in contact. This effect can prevail over usual mechanical forces for objects of sufficiently small dimensions. Besides, it can be used in practice for some control of rolling processes by means of special adhesives and lubricants.

Let us study this effect using the problem of the previous section, when S = 0, i.e. m = 1 so that 1^(1 -v 2) = = ^2(1 -v1), or in most common case when ^ = ^ 2 = ^ and v1 = v2 = v. By applying the procedure of Sect. 2 we come to the following boundary value problem similar to Eq. (21)

o+ + o- = 2isx as y = 0, -b < x < a. (42)

Here S is given by Eq. (22).

As a reminder, the vector of the equivalent force of normal tractions on the contact area, which magnitude is equal, for example, to the weight of the upper cylinder per unit length, is directed along the y axis of this Oxy system. As a result of adhesion, the positive thrust force T applied to the center of this cylinder can create some extension and concentration of tensile stresses near the back point x = -b of the contact area.

The rolling starts when the adhesive bond at point x = -b is broken, that is when the stress intensity factor at this point achieves a certain limiting value kIC characterizing this bond so that [12, 14-16]

a = kic

y V2ne

(43)

as e ^ 0, where y = 0, x = -b + e.

This constant called the adhesion toughness or bond toughness is expressed in terms of the specific adhesion bond energy y as follows

167 = ^^4. (44)

Ma( K2 -1)

The solution of the Riemann problem (42), which is limited at point x = a and at infinity but singular at point x = -b, has the following shape [11, 12, 14-16]

o2( z) = si( z -V (z - a)( z + b)) - bM . (45)

V z + b

Here a, b, and B are some real constants to be found. As z —^ ro the chosen branches of the root functions behave as follows:

V( z - a)( z + b) = = z +1/2 (b - a) -1/(8z) (a + b)2 + O(z"2), (46)

r- a . a + b -2. .„_.

— = 1 + O(z2). (47)

1 z+b 2z

The tensile bond of adhesion at z = -b is broken, if the

critical state described by Eq. (43) is achieved due to the

rolling. In terms of function o 2 (z) it says:

o2(z) =1 „ as z — -b. (48)

2>/2tc( z + b)

Based on Eqs. (45) and (48) we can conclude that

2B^J 2n(a + b) = kIC. (49)

From Eqs. (27), and (45)-(47) we can derive that

2B = s(a - b), ns(a + b)(3b - a) = 4N. (50)

The equation system (49) and (50) serves to determine constants a, b, and B characterizing the effect of adhesion upon the process of rolling.

Based on Eqs. (6) and (45) the stresses on the contact area y = 0, -b < x < a are equal to

= -2iJ(a - x)(b + x) + 25 fc—x, Vb + x

, Txy 0. (51)

lb + x

The solution of the equation system (49) and (50) is reduced to the following equation

T2 -PT/2 -1 + 0, (52)

T = L + L. L = 2^, r = 1 kIC (-f N"* (53)

L L Vns 2 ^s J

The dimensionless number r determines the effect of adhesion like the Reynolds number does the effect of viscosity in hydrodynamics.

Particularly, for a comparatively small effect of adhesion upon the rolling we get

N , N

a = (1 + F>, — b = — as r<< 1. (54)

V ns V ns

Here are some values of the monotonously growing function t = T(r) (see table).

In the most important scenario, the process proceeds in three stages. Using the CH-rule, let us consider it in the case when the second cylinder is a lower half-space.

On the first stage, the elastic cylinder of radius R presses into the half-space of another material by normal

Table

r 0 1.02 2.12 4.62 7.5

T 1 1.5 2 3 4

force N so that a symmetrical dent forms along the contact area (-b, +b) with its width being determined by Eq. (54) at r = 0. On the next stage, thrust T is being applied to the center of the cylinder and increasing but having no effect until its value becomes equal to Nb/R1 so that vector (T,N) points at the front of the contact area x = +b.

On the third stage, the adhesion bond at x = —b is stretching on and the tensile stress singularity is growing at this point while thrust T increases, and the front of the contact area x = +a moves in order to balance the normal force N. On this stage, the process is stable so that value of a slowly increases while T grows, with vector (T, N) always pointing to the front x = +a according to the CH-rule.

When the growing stress singularity at x = — b achieves the critical value defined by Eq. (49), the rolling begins. At this critical state, the equation system (49) and (50) becomes valid, with the value of the critical thrust being determined by the same CH-rule T = a N/R1. Based on Eq. (54), for small values of r the CH-rule provides the following law of rolling

T = (1 + r)

N

3/2

RjVn"

c„ = (1+r)

1 N

R1 \ ns

(55)

This scenario plays only when, at first, load N increases while T = 0 and, then, thrust T grows, with the normal load being constant. For other loading paths, the results will be different because this process is, evidently, path-dependent.

4. Rolling of an elastic ball over an elastic half-space

Let us consider the problem of an elastic ball of radius Rj rolling over another elastic ball of another elastic material of greater radius R2. When R2 the latter is an elastic half-space, and when R2 < 0 the latter is an infinite elastic space with a spherical hole of radius | R21 > Rj.

For this purpose, we can use Hertz's solution for the static problem of an axially symmetric contact of two elastic balls of different materials [6]:

( R1+R2 ^V3 r1r2

r1 + R2 r 2

3 N

4

1-V2 1-V

2 y

2/3

2R1R2

3N r

az = -——.1 -—, R >> a, |R2I > R > 0, 2na V a

a3 = 3 N

1 -v2 1 -v

2 y

2R1R2

R1 + R2

(56)

(57)

(58)

Here r is the distance from the z axis which is the axis of symmetry in this problem, a is the radius of the contact area, N is the resultant force of pressure | az | on the contact area z = 0, r < a (here shear stress is equal to zero), and w is the sum of the z components of the elastic displacements at opposite points of the ball surfaces.

Hertz's solution is commonly used to study the impact of two elastic balls. We will use it to study the equilibrium of two heavy elastic balls resting one on the other in the vertical field of gravitation before rolling. The cases when R2 ^ro or R2 < 0 are of most interest. They relate to the ball rolling on the surface of a half-space or to the ball rolling on the surface of a spherical cavity in an elastic space.

Let us slowly increase the tangential thrust force T applied to the center of the smaller ball of weight N. This force is perpendicular to the z axis. If vector (T, N) points inside the contact area, no motion occurs. In this case, there are no substantial changes in the size of the contact area because T/N is very small.

However, based on the CH-rule this ball starts on rolling when vector (T, N) points to the circular front of the contact area so that Eq. (36) becomes valid. From here, using Eqs. (1) and (58) we find the rolling resistance coefficient and the law of rolling T

N R

R

N

1

( 1 -v2

1-v

2 y

r1r2

R1 + R2

1/3

(59)

In the case of a ball rolling over a half-space we get from here

C = — = — =

N R

(

4R,2

-N

1 -vi 1 -v

2 Y

13

(60)

Using Eq. (60) we find, e.g., that the rolling resistance coefficient of hardened steel ball bearings on steel is equal to about 0.001 to 0.0015 within the practical range of parameters, which is in excellent agreement with known data of tests [2, 7].

5. Rolling of an elastic torus over an elastic half-space

Let us consider, first, the static contact problem of the elasticity theory of the pressure of an elastic solid torus to an elastic half-space of another material. The torus surface is formed by rotation of a circumference of radius r around an axis lying in the plane of this circumference, but not intersecting the latter, so that the circumference center forms another circumference of greater radius R > r. The center of the latter is the torus center. Let this solid torus press on the boundary of an elastic half-space z < 0 so that the plane of the major circle of the torus is perpendicular to the halfspace boundary z = 0. The pressing force N is applied to

the torus center and directed normal to the half-space boundary (this "load on the wheel" can be transferred to the torus via spokes or a connecting disk).

In the vicinity of a small contact area, the torus surface coincides with the surface of the following elliptic paraboloid

2 2

z = x + (61)

2(r + R) 2R

Here the directions x and y and the corresponding principal curvature radii r + R and r of this paraboloid coincide with the directions and principal curvature radii of the torus at the initial contact point x = y = z = 0.

Accordingly, our problem is reduced to Hertz's problem of the contact of two different elastic paraboloids [6], one of which is the lower half-space z < 0. According to Hertz's solution, the small contact area in this problem is the interior of the following ellipse on the plane z = 0:

y

= 1, L > b.

(62)

L b

Here 2L and 2b are the major and minor axes of the ellipse. The stresses on the contact area are [6]

3N

a, =--

2nbL

2 2 1 - L. - L.

L2 b2'

T x, =T y, =

(63)

Here the values of b and L are determined by the following equations

(1 - «2) *(e)

B(e) R + r

L3 = 1

1 -v2 1 -v

1 - « 2 = L

L2

2 ^

ND(e)( R + r ).

(64)

(65)

The subscripts 1 and 2 refer to the torus and base materials, correspondingly.

As a reminder, B(e) and D(e) are the following elliptic integrals:

, f cos2 9 ,

B(e) = J

0 J1 - e2sin2 m ¿2 2 (66)

n.. f sin2 m ,

D(e) = J i 2 2 dm.

0 \J1 - e sin m

Let us apply the thrust force T to the center of torus in the direction of the major axis x and consider the problem of its rolling over the surface z = 0 of the elastic half-space of another material.

As long as force T is sufficiently small so that vector (T, N) points to somewhere inside the contact area, no rolling occurs. However, based on the CH-rule, when vector (T, N) points to the boundary of the contact area at x = L, y = 0, the torus starts on rolling.

Therefore, the law of rolling of an elastic torus over an elastic half-space of another material says that the rolling resistance coefficient in this case is equal to

c =T

rr N R + r

1

R + r

,f

n

L V

1 -v2 1 -v

2 ^

1/3

ND(e)( R + r)

(67)

e2

Here function e = e(r/R) is determined by Eq. (64).

This law of rolling is valid as far as Hertz's solution (63)-(65) is valid. Its violation can be due to the local stick and slip zones on the contact area studied in [17, 22].

6. Rolling of a rigid cylinder over a sticky membrane

Let us study the problem of a rigid cylinder of radius R rolling over the horizontal surface of a flat membrane shell, or film, tightly stretched in all directions by the tension force y which is equal to the product of the tensile stress in the shell and the shell thickness. At first, we consider the static problem, with the force N per unit length of the cylinder, e.g. its weight, being applied to its center in the vertical direction. Let 2a be the width of a small contact area so that a << R.

The force N is balanced by the film pressure on the contact area and by the tension forces of adhesion applied to the edge of this area. From equilibrium equation of the film Yd w /dx2 = p, it follows that y = pR because w = x21(2R) on the contact area -a < x < a where the shape of the film w = w(x) coincides with the shape of the cylinder (p is the film pressure on the cylinder).

From the energy conservation law, it follows [22] that

Y(1 - cosp) =Tc. (68)

Here rc is the specific energy of adhesion of the film and cylinder materials, and P is the angle between the film and the plane tangential to the cylinder at the edge of the contact area. The value of Tc can be controlled by special glues or lubricants.

For small P, we get from Eq. (68)

YP2 = 2rc. (69)

The balance of forces acting on the cylinder is written as follows

N = 2Y(a - P). (70)

a Q /2rc

R S Y From Eqs. (70) and (71), it follows that R

a = — ( N + 2V2YT7.

(71)

(72)

Now, let us apply the small thrust force T (per unit length of the cylinder) to the cylinder center in the horizontal direction. It does not cause a rolling, if vector (T, N) points to somewhere inside the contact area.

By the CH-rule, the cylinder starts on rolling, when this vector points to the front of the contact area so that based on Eq. (72) the law of rolling and the rolling resistance

coefficient are as follow

crr = T=a=-NY+

N R 2y

Y

(73)

In the case if the adhesion can be ignored so that N2 >> 8Yrc, we can get Crr = N/2y , where y >> N. In the opposite case, when the weight is supported by the adhesion only so that N2 << 8yFc, we have Crr = ^/2Tjy .

7. Rolling of a rigid ball over a sticky membrane

Let us study the problem of a rigid ball of radius R rolling over a flat, sticky membrane, or film, tightly stretched by tension y in all directions. Again, we need, at first, to solve the corresponding static problem, with the force N, being applied to the ball center (e.g. its weight). Let Orz be the cylindrical coordinate system, the z axis of which is the vertical coinciding with the axis of symmetry of the problem. The contact area forms a circle of small radius a << R.

The displacement w(r) of the film is described by the following equation:

Y

( d2w

dr

1 dw

2+r dT

= ^0(r).

(74)

/

Here 6(r) = 0 when r > a and 6(r) = 1 when r < a, andp is the pressure in the contact area to be found. From Eq. (74) we can find

r

w = C ln—, when r > a, a

w = — (r 2 - a2), when r < a.

4y

(75)

Here C is a constant to be also found.

For r < a function w(r) coincides with the surface of the ball so that using Eq. (75) we get

pR = 2y. (76)

Similarly to the problem of Sect. 6, because of the adhesion with the ball surface the film forms the angle P with the plane tangential to the ball at any point of the front of the contact area at z = 0, r = a. This angle is determined by Eq. (68) or by Eq. (69) for small P.

The force N, e.g. the weight of the ball, is balanced by the film tension so that

(

N = 2naY(a - P) = 2tc—Y

R

2T

Y

(77)

C = a(a — P). (78)

Using Eq. (77) and the CH-rule, we find the radius of the contact area, the law of rolling and the rolling resistance coefficient

C = — = — =

Crr N R

N rc -+ —^ +

2rcyR 2y

(79)

In the case of a negligibly small adhesion when nRrc << N, we get a simpler relation Crr N/(2rcyR) from Eq. (79). In the opposite case of a very strong adhesion when nRrc >> N, the formula Crr = yj2rc/Y is valid.

It should be noticed that the notion of membrane shell, or film, is rather about the stress state in the shell than about its mechanical property. If tensile stresses in a shell are much greater than bending stresses, the shell can be treated as a membrane, even if it is made, for example, of the hardest steel.

8. Rolling of a rigid torus over a membrane

Let us study the problem of a rigid torus rolling over a flat membrane shell, tightly stretched by tension y in all directions so that the size of the contact area is very small as compared to the main radii r and R of the torus. In this case, the contact problem for the torus is reduced to the contact problem for the elliptic paraboloid given by Eq. (61), the main radii of curvature at the apex of this paraboloid being equal to r and r + R.

Let us solve the contact problem for this rigid paraboloid pressing by its apex onto the flat membrane shell tightly stretched by a very strong tension so that the contact area is small compared to the minor radius of curvature at the apex [19]. We use the Oxyz coordinate system, with its origin being at the apex and the z axis being the symmetry line inside this paraboloid, see Eq. (61). Each cross section z = = const of the paraboloid is an ellipse, with its major axis lying on the x axis. The resistance force N is directed along the z axis because the paraboloid moves in the opposite direction (no rolling so far!).

The purpose of our next calculation is to find the shape and size of the unknown contact area D on the xy plane, which is evidently symmetrical with respect to the x and y axes. The displacement w of the membrane shell satisfies the following equation

-P a.

3/ Y

Here p is the pressure on the paraboloid inside domain D where a = 1, while a = 0 outside domain D.

Inside the contact area D, displacement w coincides with z in Eq. (61). From here and Eq. (80), we find

p=y ~2~~+~Rt <8»

r (r + R)

Let us introduce the complex variable £ = x + iy. The general solution of Eq. (80) outside domain D is

w = Re f (£). (82)

Here f (£) is an analytical function to be found.

The derivative of this function is

f = ^ " i ^ • (83)

ox oy

The functions dw/dx and dw/dy are, evidently, continuous on the sought contour CD of the contact area because the normal and tangential derivatives of the membrane displacement on this contour should coincide with the corresponding values of the paraboloid. Using Eqs. (61) and (83) it is easy to derive the equation of the boundary

d 2 w d 2 w

Hx2:

(80)

value problem on the unknown contour CD

f =

-+R - ¿Z, ^ Cd.

r + R r

(84)

Let us apply the conformal mapping £ = ro(Z) which converts the unknown domain D into the exterior of the unit circle | Z| > 1 on the parametric plane of the complex variable Z, with the x axis being converted to the real axis on the Z plane. According to the Riemann theorem, the analytical function ro(Z) is uniquely defined by this way.

Since ro(Z) = x + iy, we can transform Eq. (84) into the following boundary value problem for the exterior of the unit circle |Z| > 1 on the parametric Z plane:

2r(r + R)F(Z) = (R + 2r)ro(Z), |Z|= 1, (85)

F(Z) = f'(ro(Z)) + R p.m(Z). (86)

2r (r + R)

We arrived at the boundary value problem (85) for two unknown analytical functions ro(Z) and F(Z). Both have a first order pole at infinity.

Let us apply the method of functional equations [12, 23-25] and continue Eq. (85) analytically from the unit circle onto the whole Z plane. As a result, we get the following functional equation in the Z plane

(87)

2r (r + R) F (Z) = (R + 2r )ro(Z-1). Let us show that this functional equation has the following exact solution

®(Z) = oZ + C1Z-1, (88)

2r (r + R)F (Z) = (R + 2r )(c1Z + CoZ-1 ).

(89)

It satisfies Eq. (87) for any values of coefficients C0 and C1.

First, study the behavior of function at infinity where function f (£) — 0. Hence, according to Eqs. (86), (88) and (89) we get

Rc0 = (R + 2r )c1. (90)

From the equilibrium equation of the shell at infinity, it follows that

f U) =

N 2ny£

as £ ^ ■

(91)

Now, study the pole at £ — 0 and eliminate singularity in Eq. (86) using Eqs. (88), (89) and (91). As a result, we get one more relation

( R + 2r )c0 = CjR + r (r + R) N.

kyc0

From Eqs. (90) and (92), we can find that

2 R + 2r

c0 =-

4ny

N, c =

R

R + 2r0

(92)

(93)

0

The conformal mapping is done by the following func-

tion

®(Z) =

R + 2r 4nY

N

Z+

R 1

R + 2r Z

(94)

The displacement field of the membrane shell has the following shape

dw . dw = 1 I N dx dy Z \TCY(R + 2r).

(95)

Formulae of Eqs. (94) and (95) provide the solution to the problem of a heavy body lying on the membrane shell. It can be a torus, a paraboloid or any smooth symmetrical body with a small elliptical contact area.

As a matter of fact, this solution fits also for the hydro-dynamic problem of a heavy body lying on the surface of a liquid in the case of no wetting for small values of the di-mensionless number SR2/y where R is a specific dimension of the body, S is the specific weight of the liquid, and Y is the surface tension of the liquid.

For Z = exp (iv) which corresponds to contour CD of the contact area, it is easy to derive the following equation of this contour from Eq. (94)

N (96)

2 2 x + y

(R + r )2 r2 tcy (R + 2r)

The contact area represents the ellipse which major radius is equal to

a = ( R + r )

N

, (97)

[tcy(R + 2r)

Now, let us consider the original problem of the torus rolling over the flat, tightly stretched membrane shell under the action of the thrust force applied to the torus center along the major axis of the ellipse on the contact area. Based on the CH-rule, the steady-state rolling takes place when the resultant force vector (T, N) points at the frontal edge x = a, y = z = 0 of the contact area. Using Eq. (97), from here we get the law of rolling of the torus over membrane and the rolling resistance coefficient

C = t =

a

N R + r

N

tcy( R + 2r )

(98)

9. Rolling of a cylinder over an elastic plate

Plates considered in this and next sections are assumed to be isotropic, elastic, thin, of constant thickness, and to have small lateral deflection w which must be less than a few tenths of thickness t. As distinct from membranes, plates carry loads essentially by bending stresses.

Contact problems for plates have some peculiarities that do not allow to solve or use them. For example, in the case of the problem of flat stamp, plates can carry a heavy stamp only by concentrated forces applied at the edges of the stamp, with no load being carried on flat sides.

Even worse, it can be proven that the contact problem of indentation of any heavy parabolic cylinder onto an elastic plate has no solutions with a contact area, except for two trivial ones. In the first solution, all weight is carried by concentrated forces at the ends of the contact area, with

the distance between the ends being undetermined [19]. In the second solution, all weight is carried by one contact point at the cylinder vertex so that the zero thrust force can make the parabolic cylinder of any weight roll, which is absurd.

Therefore, the simple solution of the contact problem given below is of principal interest. The point is that the cross-sectional shape of the rolling cylinder on the contact area should be replaced not by a parabola but by a more complicated figure described by the following equation

=*L+XL

y 2R 8R3!

x <<

VR.

(99)

This follows from the exact equation of the cylinder surface

(

y = R

1 1 -

r2

(100)

Equation (99) represents two first terms of the expansion of functiony = y(x) in Eq. (100) for small x/R so that we get a more precise, second-order approximation of the cylinder of radius R in the contact area. In the contact problems of cylinders on plates, it is insufficient to use only the first term of this equation, like it was done for solids and membranes.

Let us write down the equations of flat plates (very wide beams):

nd4 w nd2w

^TT = ^ M = -dx4 dx2

«=- - &

(101)

Here q, M and Q are the load, bending moment and transverse shear force, and D = EI (where I is the moment of inertia of the cross section)

1

D = — EtJ(1 -v2) 12

2\-1

(102)

Let us press the rigid cylinder of Eq. (99) onto the plate by force N per unit length of the cylinder so that a contact area of width 2a forms. We assume that the y axis is the symmetry line of the problem, and that R >> a >> t. The plate deflection w = y(x) in the contact area —a < x < a is determined by Eq. (99) so that using the first equation in Eqs. (101) we can get the load on the plate

q = 3D/R3, - a < x < a. (103)

From the equilibrium equation and Eq. (103), it follows that the width of the contact area is equal to

NR_ 6D

(104)

Then, using the CH-rule and Eq. (104) we derive the law of rolling and the rolling resistance coefficient for a rigid cylinder of radius R rolling over an elastic plate

C = L=a=NRL

rr = N = R = 6D

2NR2 (1 -v2) Et2 .

(105)

The elastic line of the plate outside the contact area can be found from the first equation in Eqs. (101)

N 3 x2( NaR

w =--x +—I 1 +-

12D 2RI 2D

Na 3D

(106)

Here wm is defined by the boundary condition at the end support. At x = ±a deflection w(x) and its first two derivatives coincide with function y(x) and its corresponding derivatives, see Eq. (99).

10. Rolling of a ball over an elastic plate

To solve the problem of a rigid ball of radius R rolling over an elastic plate we apply the approach of Sect. 9. To this purpose, we replace the ball by the following axially symmetric body which for small r/R approximates the shape of the ball more accurately than any paraboloid r2 r4

z = — + —t-. (107)

2R 8R

The axially symmetric problems of plate bending are described by the following equation for their lateral deflection w(r)

1 d

—<r

r dr I dr

1A ( r—

r dr I dr

D.

(108)

Since w(r) = z(r) in the contact area, using Eqs. (107) and (108) we can find

q = (109)

In this case, the contact area is circular. Using Eq. (109) and the equilibrium equation we get its radius

a2 = NR? = 3(1 -v2) N(R "3

, , (110) 8nD 2n E | t 1

By means of Eq. (110) and the CH-rule, we obtain the law of rolling and the rolling resistance coefficient in this

case

C = l=a = /3(1 - v )nr

N R

2nEt3

(111)

The lateral deflection of the plate outside the contact area can be found by the help of Eq. (108) similarly to Sect. 9. The homogeneous equation (108) has four evident particular solutions, namely, lnr, r2, r2 ln r and a constant. Hence, the general solution (108) can be written as their superposition

w = C1r2lnr + C2r2 + C3 ln r + C4. (112)

Here Cj, C2, C3 and C4 are some constants. The first three of them are, evidently, defined by two boundary conditions at the edge of the contact area and by the equilibrium equation. Constant C4 depends only on the support outside the contact area.

If the support is very far from the contact area, then based on Saint-Venant's principle, at the distance much greater than a, there exists the following intermediate field

corresponding to the concentrated force N at the center, [14-16]:

N

Mr = D

16nD

2

d w

2r2ln r + c2 -r2

dr2

+ v dw r dr

N_ 4n

1 + (1 + v) ln—

c

Qr = a

dw

N 2nr

1 -v

Mr - Mfi =-N.

(113)

(114)

(115)

dr 2nr 4n

Here Mr, M0, and Qr are the corresponding bending moments and transverse shear force (A is Laplace's operator).The lateral deflection in Eq. (113) counts from the support at r = c where it is assumed that dw/dr = 0.

11. Rolling of a rigid torus over an elastic plate

In the case of a torus rolling over an elastic plate, it can be shown that, similarly to more simple problems of Sects. 9 and 10, it is insufficient to approximate the surface of the torus by an elliptic paraboloid of Eq. (61) which represents only the first order expansion of the torus surface equation at and near the contact area. Although we can construct two mathematical solutions for the first order approximation, they do not satisfy some evident physical requirements.

In the first solution, with the one point contact at the base of the rolling torus, the torus of any weight can be rolled over any elastic plate by zero thrust force, which is absurd. In the second solution, the load-free contact area represents an ellipse of some constant eccentricity but of any undetermined dimension, with the weight being carried by concentrated transverse shear forces distributed along the contour of the ellipse [19, 20]. The latter can be proven by a simple dimensional consideration.

To solve the problem of rolling we need to use the following second-order approximation of the torus surface at and near the contact area

y

y

- 3 3 (116) 2(R + r) 2r 8( R + r )3 8r3

Here the fourth-order terms are much less than the second-order terms so that this change of the surface of the ellipsoidal paraboloid is very small.

The lateral deflection of the plate should satisfy the bi-harmonic equation

A2 w = q/D. (117)

Therefore, inside the contact area of this body where z(x) = : w(x), the load is equal to 1

q = 3D

(L + 1 ^

r3 ( R + r)3

v v ' J

(118)

A very small change in the shape of the pressing body caused a very significant effect, namely, a redistribution of the bearing load from the concentrated transverse shear force at the edge of the contact area to the constant load throughout all contact area. This is a bad signal for the solution with concentrated transverse shear forces—it suf-

fers a big change by a small variation of boundary conditions and is easily "washed away" by the continuous solution [14-16, 19, 20]. Such unstable solutions are called incorrect and usually ignored. It is the way we follow as well.

Unfortunately, the exact solution of the contact problem for a body described by Eq. (116) is very complicated and cumbersome for practice. Instead, we apply an approximate solution which is much simpler and easier to use.

Indeed, we approximate the sought contour of the contact area by an ellipse which is coaxial and similar to the basic ellipse defined by the first two terms of Eq. (116), and which supports weight N of the torus by the distributed load of Eq. (117). As a result, we get the following two equations:

nabq = N, (19)

R + r

(120)

Here a and b are, respectively, the major and minor radii of the sought ellipse (nab is its area).

Using Eqs. (119) and (120) we find the major radius of the contact area that determines the rolling law of the torus

( N V/2

nq

v 1 J

1+R

1/4

(121)

Substituting q by Eq. (117), we find the major radius of the contact area in terms of original data

N

3nD

yv +R

1

-r+

1

(r + R)3

-12

(122)

From here, it follows that the law of rolling of a torus and the rolling resistance coefficient can be written as follows

C = — = a

r N R + r

N

3nD

12 r 5/4

(r + R)

3/4

r3 + (r + R)3

(123)

Here D is given by Eq. (102) in terms of Young's modulus, Poisson's ratio and the thickness of the plate. As a reminder, r and R are the minor and major radii of the torus.

12. Conclusion

This paper provides the exact laws of rolling and the accurate rolling resistance coefficients, so much needed in industries and for so long gained only in very expensive tests, have gotten now known for all basic round bodies. This knowledge should become common to all professionals and graduated students of, at least, mechanical and automotive engineering.

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Genady P. Cherepanov, Prof., Honorary Life Member, The New York Academy of Sciences, USA, [email protected]