Keywords

1 Introduction

The rolling mechanics of the elastic wheel on the drum is the same as when the wheel is rolling on a flat rigid support surface [1,2,3,4,5,6,7,8,9].

In the papers [10,11,12,13,14,15,16,17],  the equations for calculation of the power loss due to friction in the contact and hysteresis in the material of the wheel relative loss (slipping) wheel speed value were obtained under considering the mechanics of rolling elastic wheels on rigid drum. Wheel speed value is occurring in the contact tangent forces due to rolling resistance coefficient of the driven wheels. However, the question of the points of application of tangent and normal forces in contact with the drum and, accordingly, the shoulders of these forces relative to the axis of the wheel (Fig. 1). These values are necessary, in particular, when considering the rolling of cars in tests on drum stands.

Fig. 1.
figure 1

Normal Fn and tangential Ft forces in the contact of the wheel with the drum

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This work is devoted to determining the point of application of the tangent force in contact and its shoulder relative to the axis of the wheel.

2 Main Part

The elementary tangent forces dFτ = qtdx2b distributed along the arc of the wheel-drum contact have a resultant, defined as the geometric sum of the elementary tangent forces (Fig. 2), and located outside the surface of the drum.

Fig. 2.
figure 2

Linear elementary tangential forces τ′ and τ″, implemented at sites remote from the center of contact at a distance ±x, and their resultant τ

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To find the magnitude and the line of action of this force, consider two elementary sites of length dx each, located symmetrically with respect to the centerline OδOК (Fig. 2) and distant from the center of contact at a distance ±x. Linear elementary tangential forces realized at these sites, denote τ′ and τ″. The resultant of these forces passes through the point of intersection of their lines of action, lying on the inter-axial line OδOК, and can be found by the “cosine theorem”:

$$\tau = \sqrt {(\tau^{\prime})^{2} + (\tau^{\prime\prime})^{2} + 2\tau^{\prime}\tau^{\prime\prime}\cos 2\alpha }$$
(1)

where α = x/rδ is the angle that determines the position of the symmetric sites under consideration.

When the selected elementary sites are located on the coupling site, the values τ′(″) are determined in accordance with the equation [1, 2, 10,11,12, 14, 15] τ′(″) = qtdx = λξ(a − 2x)dx. If the platforms are located on the slip area, then τ′(″) = q cк.t dx = μqndx = μ\(q_{{n_{0} }}\) (a2 − x2)dx.

Considering that x ≪ rδ and the angle α is small, the expression (1) after some transformations with power series expansion can be represented as:

$$\tau = \left( {\tau^{\prime} + \tau^{\prime\prime}} \right)\left( {1 - \frac{{\tau^{\prime}\tau^{\prime\prime}}}{{\left( {\tau^{\prime} + \tau^{\prime\prime}} \right)^{2} }} 2\alpha^{2} } \right)$$

Let us consider further two cases—when the maximum traction force is realized by coupling the wheel with the drum, and when the traction force realized in contact is small.

In the first calculated case, when the traction force is realized in the contact, the linear elementary tangential forces realized on two symmetrically arranged elementary platforms of length dx are the same τ′ = τ″ = μqndx. Therefore, their resultant is perpendicular to the center line OδOК and according to Eq. (2) it will be equal to:

$$\tau = 2\mu q_{n} \cos \alpha {\text{d}}x = 2\mu q_{n} \left( {1 - \frac{{\alpha^{2} }}{2}} \right){\text{d}}x = 2\mu q_{{n_{0} }} (a^{2} - x^{2} )\left( {1 - \frac{{x^{2} }}{{2r_{\delta }^{2} }}} \right){\text{d}}x$$
(2)

The picture of the distribution of the resultant linear elementary tangent forces is presented in Fig. 3.

Fig. 3.
figure 3

Distributions of the resultant linear elementary tangent forces τ and their resultant tangent forces Fτ

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The resulting tangent force is found as:

$$F_{\tau } = 2b\mathop \int \limits_{0}^{{y_{0} }} \tau {\text{d}}y = 4b\mu q_{{n_{0} }} \mathop \int \limits_{0}^{{y_{0} }} \left( {a^{2} - 2r_{\delta } y} \right)\left( {1 - \frac{y}{{r_{\delta } }}} \right)\frac{{\sqrt {yr_{\delta } } }}{2y}{\text{d}}y = \frac{8}{3}\mu ba^{3} q_{{n_{0} }} \left( {1 - \frac{{a^{2} }}{{10r_{\delta }^{2} }}} \right)$$
(3)

where \(y = \frac{{r_{\delta } }}{\cos \alpha } - r_{\delta } \approx \frac{{x^{2} }}{{2r_{\delta } }}\) is the coordinate of the point of application of the elementary tangent force τ (the origin in this case is located at point B); y0 = a2/2rδ corresponds to the position of point A.

According to [1, 2, 10,11,12, 14, 15], the circumferential thrust force equal to the algebraic sum of all specific tangent forces in contact is defined as:

$$F_{t} = 2b\left[ {\frac{\lambda \xi }{2}(a - x_{\text{S}} )^{2} \pm \frac{1}{3}\mu q_{{n_{0} }} \left( {2a^{3} + 3a^{2} x_{\text{S}} - x_{\text{S}}^{3} } \right)} \right]$$
(4)

where xS is the coordinate of the boundary of the clutch and slip areas in contact, b is the width of the contact spot.

In the implementation of the limit on the coupling of the circumferential traction force, the value of the latter is equal to \(F_{\text{t}} = \frac{8}{3}\mu ba^{3} q_{{n_{0} }}\). Comparing the value of this force with the value Fτ, obtained above, it can be concluded that the resulting tangent force Fτ is less than the thrust force Ft, but their difference is so small that it can be neglected.

The line of action of the tangent force Fτ is perpendicular to the center line OδOК and is located (Fig. 4a) above the circumference of the drum: \(\frac{AB}{2} = \frac{{a^{2} }}{{4r_{\delta } }} > y_{R} > \frac{AB}{3} = \frac{{a^{2} }}{{6r_{\delta } }}\).

Fig. 4.
figure 4

Plots of longitudinal tangential stresses qt and normal pressures qn in the longitudinal section of the contact spot; coordinates of the centers of gravity of the tangential stresses plot

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In the second case, at low thrust, the area of sliding contact is small and the coordinate boundaries of the grip and slide close to x = −a.  So, to simplify, it is assumed that the plot of the distribution of tangential stress along the entire length of the contact patch takes the triangle shape. As a result, considering in this case x as an arithmetic quantity, we get:

$$t^{\prime} = \lambda \xi \left( {a - x} \right)d\quad t^{\prime\prime} = \lambda \xi \left( {a + x} \right){\text{d}}x$$
$$\tau = 2\lambda \xi a\left( {1 - \frac{{a^{2} - x^{2} }}{{2a^{2} r_{\delta }^{2} }}x^{2} } \right){\text{d}}x$$

The picture of distribution τ, constructed by the last equation, is presented in Fig. 3b. The lengths of the vectors \(A\bar{C} = \bar{\tau }^{\prime\prime}\) and \(B\bar{D} = \bar{\tau^{\prime}}\) for x equal to x = a and x = 0 are the same. In view of the smallness of the second term in parentheses of the last equation with a sufficient degree of accuracy, it can be assumed that the vector of the resulting tangent force Fτ of the system of forces τ passes through the middle of the segments AB and CD, and the modulus of force is equal to:

$$F_{\tau } = 2b\mathop \smallint \limits_{0}^{{y_{0} }} \tau \sqrt {2\left( {1 + \cos \frac{{\alpha_{0} }}{2}} \right)} {\text{d}}y \approx 2b\mathop \smallint \limits_{0}^{{y_{0} }} 2\tau \left( {1 - \alpha_{0}^{2} /16} \right){\text{d}}y = 4ba^{2} \left( {1 - \frac{{a^{2} }}{{16r_{\delta }^{2} }}} \right)\lambda \xi .$$
(5)

where, as in the previous case, \(y = \frac{{r_{\delta } }}{\cos \alpha } - r_{\delta } \approx \frac{{x^{2} }}{{2r_{\delta } }}\). is the coordinate of the point of application of the elementary tangent force τ (the origin in this case is also located at point B).

Considering the assumed smallness of the thrust force and xS = −a from the formula (4), which determines the magnitude of the thrust force, we obtain Ft = 4ba2λξ. Comparing the last expression with the dependency for Fτ, it can be written as:

$$F_{\tau } = F_{\text{t}} \left( {1 - \frac{{a^{2} }}{{15r_{\delta }^{2} }}} \right)^{2} ,$$

 i.e. the tangential force, which acts from the side of the drum on the wheel, is the resultant of elementary tangential forces, which is less than the circumferential thrust force realized in contact; however, this difference is very small, and it can practically be neglected.

From the equality of the moments produced by forces Ft and Fτ, Mt = Ft rδ = Fδ rδ, get the shoulder action of the resultant of the tangential forces Fτ equal to rτ = rδ Ft/Fτ. Despite of the shoulder exceeds  the value of rτ and the value of rδ, and the difference between these values  is very small, and it can be considered that rτ ≈ rδ.

The angle of inclination Fτ at low thrust forces is approximately φ ≈ α0/2, which corresponds to xFτ ≈ −a/2.

It should be noted that if we consider the difference xS ≠ −a, the picture of the distribution of the resultant elementary tangent forces will change slightly in the area of point A (Fig. 3b) and the total resultant tangent force will be located slightly lower than previously accepted, and with a smaller angle of inclination.

With the growth, which is realized in contact with the thrust force Ft, tangential force Fτ falls, as expected, below the angle between the center line OδOк and the line of action of Fτ. This angle will be closer to 90°. The coordinate xFτ, determines the angle ατ = xFτ/rδ, the slope of the line of action of this force will change from xFτ = −a/2 to xFτ → 0.

In further calculations, as already shown, we can assume that Ft = Fτ, and the coordinate xFτ to take equal to xFτ ≈ −0.5a.

However, given that, xFτ still depends on the realized thrust force, it is desirable to obtain an analytical dependence xFτ = f(Ft).

At a ≪ rδ, the elementary tangential forces qt2bdx, as well as the tangential stresses qt, can be conditionally considered acting in one direction due to the smallness of the angles α of their inclination.

Consider the plot of tangential stresses (Fig. 4), the area of is numerically equal to the linear thrust force Ft/2b.

The coordinate of the center of gravity of this plot can be considered the point of application of the thrust force and, under accepted conventions, the point of application of the resulting tangent force.

The plot can be divided into two figures, one of which corresponds to the tangential stresses on the clutch section, and the other to the tangential stresses on the slip section.

It is known that the coordinate of the center of gravity of the figure is calculated as:

$$x_{\text{c}} = \frac{{\mathop \smallint \nolimits_{{x_{1} }}^{{x_{2} }} xy{\text{d}}x}}{A},$$
(6)

where A is the area of the figure; y is its current ordinate; x1 and x2 are the coordinates of the figure boundary.

For the first figure

$$y = q_{{t_{x} }} = \lambda \xi \left( {a - x} \right){;}\quad x_{1} = x_{\text{s}}{;}\quad x_{2} = + a;$$
$$A_{1} = \int\limits_{{x_{\text{S}} }}^{ + a} {q_{{t_{x} }} {\text{d}}x} = \lambda \xi (a - x_{\text{S}} )^{2} /2$$

For the second figure, y = μqn; x1 = −a; x2 = xS

$$A_{2} = \int\limits_{ - a}^{{x_{\text{S}} }} {\mu q_{n} {\text{d}}x}$$

Then,

$$x_{{{\text{c}}_{1} }} = \frac{{\int\nolimits_{{x_{\text{S}} }}^{ + a} {x\lambda \xi (a - x){\text{d}}x} }}{{\int\nolimits_{{x_{\text{S}} }}^{ + a} {\lambda \xi (a - x){\text{d}}x} }};\quad x_{{c_{2} }} = \frac{{\int\nolimits_{ - a}^{{x_{\text{S}} }} {x\mu q_{n} {\text{d}}x} }}{{\int\nolimits_{ - a}^{{x_{\text{S}} }} {\mu q_{n} {\text{d}}x} }}$$

The coordinate of the center of gravity of the entire figure can be found as:

$$x_{{F_{t} }} = \frac{{\mathop \sum \nolimits A_{i} x_{{c_{i} }} }}{{\sum A_{i} }}$$

or, substituting the previous expressions:

$$x_{{F_{\text{t}} }} = \frac{{A_{1} x_{{{\text{c}}_{1} }} + A_{2} x_{{{\text{c}}_{2} }} }}{{A_{1} + A_{2} }} = \frac{{\int\limits_{ - a}^{{x_{S} }} {x\mu q_{n} {\text{d}}x} + \int\limits_{{x_{\text{S}} }}^{ + a} {x\lambda \xi (a - x){\text{d}}x} }}{{\int\limits_{ - a}^{{x_{S} }} {\mu q_{n} {\text{d}}x} + \int\limits_{{x_{\text{S}} }}^{ + a} {\lambda \xi (a - x){\text{d}}x} }}$$

After the transformations, considering that the expression in the denominator is the linear thrust force Ft/2b [10,11,12, 14, 15], we obtain:

$$x_{{F_{\text{t}} }} = - \frac{{\mu q_{{n_{0} }} b}}{{6F_{t} }}\left( {a^{2} - x_{\text{s}}^{2} } \right),$$
(7)

If to substitute the expression received earlier [10,11,12, 14] to the last equation

$$x_{\text{s}} = a\left( {1 - 2\sqrt[3]{{1 - \frac{{F_{\text{t}} }}{{\mu F_{\text{z}} }}}}} \right)$$

we will finally get:

$$x_{{F_{\text{t}} }} = a\left( {1 - \frac{{\mu F_{\text{z}} }}{{F_{\text{t}} }}} \right)\left( {1 - \sqrt[3]{{1 - \frac{{F_{\text{t}} }}{{\mu F_{\text{z}} }}}}} \right)$$
(8)

For small tangent forces compared to the coupling limit, the dependence (8) can be represented in a simplified form if decomposed \(\sqrt[3]{{1 - \frac{{F_{\text{t}} }}{{\mu F_{\text{z}} }}}}\) into a power series:

$$x_{{F_{\text{t}} }} = a\left( {1 - \frac{{\mu F_{\text{z}} }}{{F_{\text{t}} }}} \right)\frac{{F_{\text{t}} }}{{3\mu F_{\text{z}} }} = \frac{a}{3}\left( {\frac{{F_{\text{t}} }}{{\mu F_{\text{z}} }} - 1} \right)$$
(9)

In the expressions (8) and (9), the value of the Ft force is taken modulo.

From the dependencies obtained for xFτ, it can be seen that for Ft → μFz \(x_{{F_{\text{t}} }} \to 0;\)

figure a

The shoulder of the action of the tangent force relative to the center of the wheel is for geometric reasons:

$$h_{\tau } = O_{\delta } O_{k} \cos \alpha_{\tau } - r_{\delta } = \left( {r_{\text{d}} + r_{\delta } } \right)\cos \alpha_{\tau } - r_{\delta }$$
(10)

where angle is

$$\alpha_{\tau } = x_{{F_{\tau } }} /r_{\delta }$$
(11)

after the transformations, we get

$$h_{\tau } = r_{\text{d}} \left( {1 - x_{{F_{\tau } }}^{2} /2r_{\text{d}} r_{\delta } } \right)$$
(12)

3 Conclusion

The equations for determining the point of application of the tangent force in contact and its shoulder relative to the wheel axis are derived from the consideration of the mechanics of interaction between an elastic wheel and a drum. These dependencies have a simple form and can be applied when considering the rolling of a car on a drum stand. The study of the roller wheel with a tilt of the wheel and the influence of the bulge of the treadmill of the wheel is given in the works [17,18,19,20,21,22,23,24,25,26,27,28].