Wheel rolling on deformable ground with slippage

A characteristic feature of wheeled earth-moving machines and tractors is their operation on deformable soils. This makes it necessary to consider the features of the interaction of elastic wheels with the ground, including such issues as the depth of the track and power losses for rutting, wheel movement along the laid track, rolling with pull, forces and moments acting on the wheel, the effect of wheel parameters and rolling mode etc. When a pneumatic wheel with a high internal air pressure rolls on a deformable soil, the nature of the interaction of the wheels with the soil is such that the wheel can be considered rigid, since its normal deformation is small. This greatly simplifies the calculations associated with the analysis of the operation of wheeled machines taking into account the transforming properties of the wheeled propulsion unit, in particular, the determination of the kinematic and power parameters of the wheel. In this paper, using the motion reversal method, the picture of physical phenomena in the contact of the wheel with the ground has been considered, which made it possible to obtain relatively simple expressions for calculating the circumferential thrust force and the coefficient of tangential elasticity of the pair “wheel-deformable soil”.


Introduction
Wheeled tractors and wheeled earth-moving machines are widely used in road construction and municipal services. A large and varied group of wheeled vehicles in terms of purpose and design are general transport vehicles used for the transportation of various types of bulk cargo, and specialized vehicles that are designed to perform narrow target functions.
A characteristic feature of these types of wheeled vehicles is their ability to work not only on roads with asphalt-concrete pavement, but also on deformable soil surfaces.
When the wheels roll on the ground, the power consumption required to ensure the movement of the wheeled machine increases sharply, and questions arise related to the possibility of the machines, their controllability and loads in the transmission [1][2][3][4][5][6][7][8][9]. In this regard, it becomes necessary to consider various aspects of the movement of machines in these conditions, in particular, the determination of the forces realized in the contact of the wheels with the ground. In this work, to determine the circumferential force realized in contact, the authors used a proven methodology previously applied to the case of interaction of a wheel with a rigid support surface [10][11][12].

Materials and methods
Let us consider the rolling of a radially rigid wheel with circumferential compliance on deformable soil (which corresponds to the wheel model "elastic band on a rigid rim").
In this case, we will take the assumption about the elastic properties of the soil, i.e. we will assume that up to a certain limit zone there is a linear relationship between the tangential (with respect to the wheel surface) soil displacements and the limiting tangential forces in the contact.
For the convenience of solving the problem, we use an inverted mechanism (i.e., the wheel axis is stationary, and the base -the ground -moves at a speed -V in the direction opposite to the actual direction of the wheel movement, as shown in the Figure 1, similar to how it was used in the works [10][11][12]. We will consider the movement of point K, which belongs to both the wheel and the ground in the contact zone at the adhesion section. In the time corresponding to the rotation of the wheel from the moment the pointКenters contact with the ground to its position determined by the angle α, the tangential displacement of the ground point will be UGR, and the tangential displacement of the wheel point (relative to the beam on which this point was located before entering the contact) -Ux. The speed of the point K of the wheel surface in the reversed mechanism in the absence of tangential displacements is equal to The actual speed of the point under consideration in the presence of tangential displacement can be represented as follows: (1) where dUК -increment of the tangential displacement of the wheel point during the time dt=dx/V (for the driving wheel it is negative, for the driven and brake ones -positive), dx -elementary movement (increment) of the base (ground) in the horizontal direction in the reversed mechanism.
Similarly, we can represent the same speed of the same point, referring it to the base: Here dUGR is the increment in the tangential displacement of a point on the ground surface over the same period of time; . . . = − sin is the speed of a given base point in a reverse gear, i.e. in the absence of tangential displacements. Since expressions (1) and (2) represent the same speed of point K, then, equating their right-hand sides, after transformations, we get: Considering that , the expression 1/sinα, after expanding it in a power series and discarding quantities of the second and more orders of smallness, can be represented in the following form: 1/sinα≈1+x 2 /2r 2 Putting the last expression into (3), we obtain: As a result, for the position of point K, determined by theхcoordinate, in the adhesion area Taking into account the boundary conditions (atх=аU=0), after finding the constant c, we will finally have: Specific tangential forces (tangential stresses) acting at a given point on the wheel and on the ground are equal in magnitude, but opposite in direction: / = / * = −/ . Under the accepted assumption about the proportionality of the specific tangential forces to tangential displacements, the last equality can be represented as: where λk and λGR are coefficients of tangential stiffness of the wheel and soil.
Expressing the value UGR from equality (7) and taking into account that U= UGR-UW, after transformations we obtain an expression for the tangential displacements of points on the wheel surface, and then the soil: is the reduced coefficient of tangential stiffness of the wheel-soil pair. Expressions (5) and (9) determine the tangential displacements and specific tangential forces, respectively, due to both the implementation of the circumferential force in the contact and the geometry of the contact zone.
With the adopted wheel rolling model, even with the implementation of significant forces close to the limiting adhesion forces, the coordinate of the boundary of the adhesion and sliding sections is хB=0 ( Figure 2 shows the stress distribution along the sweep of the contact arc), which greatly simplifies the calculations, in particular, the determination of tangential strength.
We will use this expression to find the rolling radius r k 0 , at which Mt=0. Substituting equation (5) into (11) and equating the expression obtained after transformations to zero, we will have: r r a r r a r r H k 0 Taking this formula into account, from (5) we obtain a dependence that determines the law of change in tangential displacements in the contact zone, at which the torque on the wheel, due to the presence of tangential stresses, would be equal to zero: Subtracting the value U0, from the total tangential displacement, we obtain a component due only to the action of the moment Mt: where As a result, the tangential stresses in the adhesion section, due to the implementation of the thrust force in the contact, will be represented by the dependencies: The corresponding torque on the wheel is found as: After substitution of dependence (16) and subsequent integration, we obtain: will be called the circumferential traction.
Elementary tangents dFτ=qtdx2b, distributed along the arc of contact, have the resulting Fτ, located outside the arc of contact.
Let's find the magnitude and line of action of this force. Figure 3 shows that each pair of elementary tangential forces «τ′» and «τ″», acting at points whose coordinates (abscissas) are symmetric about the midpoint with the coordinate => =(π/2+α0)/2 has its resulting: passing through the intersection of the lines of action τ′and τ″, located at an angleϕrelative to each other.
We transform expression (20) by presenting it in the following form: For the extreme and close to them points of the considered contact area τ″>>τ′, therefore, even with a significant immersion depth of the wheel (for example, whenϕ ≈ 1) the value of the square root in the last expression differs little from unity. For points located in the middle part of the contact, as before,τ″>τ′ (except for the point with α=αmid, for which τ″=τ′ and, in addition (which is more important), ϕ → 0 (ϕ = 0 at α = αmid). Therefore, for the middle section of the contact, the value of the square root is close to unity. This allows us to take as a first approximation τ = τ′+τ″.
Considering that α π = ≈ − arccos x r The diagram of the resulting elementary tangential forces is shown in Figure 4. Summing up these resulting forces, we can find the magnitude and direction of the total tangential force.
Since the diagram is a system of forces converging at point А,as shown in the Figure 4, then the total tangential force also passes through this point, being perpendicular to the line ОкА, i.e. at an angleψt to the horizontal. From geometric considerations ψt=ψτ/2=(π/2-α0)/4 (27)  Figure 4 shows that the moment arm rτ=ОкА of force Fτ and the coordinate x F t , defining the point of application of this force will be found (), as: When determining the magnitude of the total tangential force, summing up the resulting elementary tangential forces, one should take into account the angles between these resulting forces. However, due to the small value of these angles (which are in the range from 0 0 to 2ψt ) their cosines are close to unity (for example, if we take the average value of the indicated angle, equal toψt, then at an angle α0=π/6, corresponding to immersion of the wheel into the ground to a depth of 0.5r, the cosine of the angle ψt will be≈0.97). This allows the shear force to be calculated using the following formula: Neglecting the indicated angles gives a slightly overestimated value of the tangential force, calculated by the last formula, but this overestimation does not exceed several percent.
Thus, the actual value of the total tangential force Fτ is slightly less than the circumferential traction force Ft, calculated by formula (19), and its arm of force (see formula (30)) is slightly larger than the radius r. However, these differences are small, and in the future, in our calculations, we will assume Ft=Fτ and rτ=r.
Having a dependence for calculating the tangential force, one can obtain a dependence for finding the rolling radius.
It follows from the equilibrium condition of the driving wheel that ? @ = ? cos C = ? ' + ?
-free rolling radius when there is no longitudinal force on the wheel axis. The presented formulas have the same notation with the formula obtained by E.A. Chudakov [13]. In this regard, the expression represents the coefficient of tangential elasticity of the pair "wheel -deformable soil". Formulas (32) and (33) are also valid for the cases of wheel rolling in the braking or driven mode, if the forces Fx and Ft are substituted into the indicated formulas with a minus sign.
Realization of a tangential force in the contact leads to a change in the normal pressures in the contact by an amount tqt cosα = qtx/r. With a driving wheel, for which ξ>0, the value of normal pressures increases, for a brake wheel -decreases (since ξ,<0): Here qn is the normal contact pressure, determined by the Berstein-Letoshnev formula: qn = С(H -h) m , С and m are soil parameters determined experimentally ( Сis the deformation modulus, m is a parameter characterizing the law of change in soil resistance to indentation), Htrack depth, as it sown in the Figure 1, h -depth of immersion of the wheel in the ground at a distanceхfrom the vertical axis.
In the slip section, taking into account the above, the dependence for the distribution of normal pressures will have the following form: The coordinate of the border of the adhesion and sliding sections can be found from the equality q q t n = µ α / sin .
An increase in normal pressures in the contact of the driving wheel and a decrease in them in the contact of the brake wheel leads to the fact that at the same value of the relative speed loss at . M > 0 a greater tangential force is realized in the contact of the driving wheel than in the rolling of the brake wheel. Accordingly, there will be more friction losses in the contact of the drive wheel, and, therefore, larger power costs required for rolling the wheel on the ground. This is clearly confirmed by the dependence obtained experimentally by Yu.V. Pirkovsky [13], it is shown in Figure 5. The dependences obtained above are valid for the case when the shear stress in the soil is less than the shear stress value, at which the upper soil surfaces are sheared.
If the soil shear occurs earlier than the tangential stresses reach the limiting adhesion, then the picture of the phenomena occurring in contact and their mathematical description changes. When cutting the soil, it must move into the zone of increasing normal pressures. In this case, with an increase in normal pressures, the permissible stress increases. Due to the movement of the soil layers, there will be friction between them, while there will be no friction losses between the wheels and the soil. The value of the realized tangential force, with the same value of the relative speed loss, will be even less than in the previous case.
Denoting by τ0=C+qntgρ (where ρ is the angle of friction between soil particles) the shear stress at which the shear of soil layers begins, we find the coordinate of the beginning of the shear from the equation: The realizable tangential force is found from the equation:

Conclusion
The studies of rolling on deformable ground of a rigid wheel in the radial direction performed in this work allow us to determine the circumferential traction force realized by the wheel using relatively simple expressions, taking into account the properties of the soil and wheel slippage.