**1. Introduction**

In 1882, Heinrich Hertz solved the problem of elastic contact of parabolic bodies [1]. A solution for the contact of an arbitrary body of revolution with a compact contact area was found much later, in 1941–1942 by Föppl and Schubert [2–4]. An attempt to overcome the restriction of axial symmetry was undertaken in 1990 by Barber and Billings [5]. Their approach is based on using Betti's reciprocity theorem as suggested by Shield in 1967 [6] and the extremal principle found by Barber [7]. However, in [5] Barber and Billings merely illustrated their method on examples of contacts with "linear profiles" (pyramids with polygonal cross-sections) since an analytical solution is possible only for this case. In the present paper, we apply the extremal principle of Barber to another case where a largely analytical treatment is possible: To contacts of profiles that are not axially symmetrical but deviate only slightly from the axial symmetry.

#### **2. Barber's Extremal Principle**

In [6], Shield used Betti's reciprocity theorem to show that the normal force *FN*(*A*) appearing due to indentation of the profile *f*(*<sup>x</sup>*, *y*) to a depth *d* (while *A* is the contact area in this state) is given by the equation

$$F\_N(A) = \frac{1}{d^\*} \iint\limits\_A p^\*(\mathbf{x}, y)(d - f(\mathbf{x}, y)) \mathbf{dx} dy \tag{1}$$

where in the pressure distribution, *p*<sup>∗</sup>(*<sup>x</sup>*, *y*) is that under a flat punch with the crosssectional shape *A* which is indented by *d*<sup>∗</sup>. Note that the indentation *d*∗ is an auxiliary

parameter used solely for determining the pressure *p*<sup>∗</sup>(*<sup>x</sup>*, *y*); it has nothing to do with the true indentation depth *d*. Of course, the integral (1) does not depend on *d*∗ as pressure *p*<sup>∗</sup>(*<sup>x</sup>*, *y*) under an arbitrary flat punch is proportional to *d*<sup>∗</sup>. In [7], Barber has shown that the area *A* fulfilling the usual contact conditions (the pressure inside the contact area is positive and there is no interpenetration, outside the contact, the pressure is zero and the distance between surfaces is positive), corresponds to the maximum force at a given indentation depth. (For axisymmetric profiles, this extremal principle has already been used by Shield [6]).

To be able to look constructively for a solution, Barber and Billings propose to use Fabrikant's approximation [8] for the pressure distribution under a flat-ended punch. Fabrikant's hypothesis is that the stress distribution is given in a good approximation by the equation

$$p = \frac{2E^\*d}{L} \frac{a(\varphi)}{\sqrt{a(\varphi)^2 - r^2}}\tag{2}$$

where *a*(*ϕ*) is the equation for the contact boundary in polar coordinates and

$$L = \int\_0^{2\pi} a(\varphi) \mathrm{d}\varphi \tag{3}$$

The motivation for the ansatz (2) is straightforward. It is known to be exact for elliptical punches with arbitrary eccentricity, and it provides the correct kind of asymptotic behavior in the vicinity of the boundary, which must be fulfilled for a flat-ended punch of arbitrary shape [9].

The origin for the polar coordinate system has to be taken as the centroid of the area *A* provided there are no tilting moments around the origin (which we will assume in this paper).

With Fabrikant's pressure (2), Equation (1) becomes

$$F\_N(A) = \frac{2E^\*}{L} \int\_0^{2\pi} \int\_0^{a(\varphi)} \frac{a(\varphi)(d - f(r, \varphi)) r \mathrm{d}r \mathrm{d}\varphi}{\sqrt{a(\varphi)^2 - r^2}} \tag{4}$$

Introducing definitions

$$\mathrm{g}\_{\varphi}(a) = a \int\_{0}^{a} \frac{f'(r, \varphi)}{\sqrt{a^2 - r^2}} \mathrm{d}r, \; \frac{\mathrm{d}G\_{\varphi}(a)}{\mathrm{d}a} = \mathrm{g}\_{\varphi}(a), \tag{5}$$

the force (4) can be rewritten as

$$F\_N(A) = \frac{2E^\*}{L} \int\_0^{2\pi} \left(d \cdot a(\varphi)^2 - a(\varphi)G\_\theta(a(\varphi))\right) \mathrm{d}\varphi\tag{6}$$

#### **3. Finding the Force Maximum**

*3.1. Axially Symmetric Profiles*

In the case of an axially symmetric profile, *a*(*ϕ*) = *a*0, *L* = 2*πa*0, the force equation simplifies to

$$F\_N(a\_0) = 2E^\*(d \cdot a\_0 - G(a\_0))\tag{7}$$

The condition for its maximum reads d*FN*(*<sup>a</sup>*0)/d*<sup>a</sup>*0 = 2*E*∗(*d* − *g*(*<sup>a</sup>*0)) = 0 or

> *d*

$$\mathbf{x} = \mathbf{g}(a\_0) \tag{8}$$

and the normal force is equal to

$$F\_N(a\_0) = E^\* \int\_{-a\_0}^{a\_0} (d\_0 - \mathfrak{g}(a)) \mathrm{d}a \tag{9}$$

The pressure distribution can be calculated by integrating contributions (2) during indentation from the first touch up to the given indentation depth, as described in detail in [10] (Appendix B),

$$p(r) = \frac{E^\*}{\pi} \int\_r^a \frac{1}{\sqrt{\tilde{a}^2 - r^2}} \frac{\mathrm{d}\mathcal{G}(\hat{\vec{a}})}{\mathrm{d}\vec{a}} \,\mathrm{d}\vec{a} \tag{10}$$

Equations (8)–(10) are the well-known equations of the Method of Dimensionality Reduction (MDR) [11].

## *3.2. General Profiles*

In the general case of non-axisymmetric profiles, let us first search for the maximum of functional (6) at a constant value of *L*. This can be done, according to Lagrange, by searching for an unconditional extremum of the functional

$$\frac{2E^\*}{L} \int\_0^{2\pi} \left( d \cdot a(\varphi)^2 - a(\varphi) G\_\varphi(a(\varphi)) \right) \mathrm{d}\varphi - D \left( \int\_0^{2\pi} a(\varphi) \mathrm{d}\varphi - L \right) \tag{11}$$

where *D* is a Lagrange multiplier. At the maximum of this functional, the first variation must vanish identically

$$\frac{2E^\*}{L}\left(2d \cdot a(\varphi) - G\_{\varphi}(a(\varphi)) - a(\varphi)g\_{\varphi}(a(\varphi))\right) = D \tag{12}$$

This equation determines implicitly the contact boundary *<sup>a</sup>*(*ϕ*), while *D* is connected with *L* through the condition (3). After substitution of *a*(*ϕ*) into (6), the force will remain a function of the still undefined *L*, which should finally be found as a value giving the maximum to the force. All stated operations, beginning with the solution of Equation (12) with respect to *<sup>a</sup>*(*ϕ*), can be carried out analytically without approximations only in two cases: Either the above case of axisymmetric profiles (leading to the MDR) or in the case of linear (conical) profiles. The latter has already been done in the paper [5] for polygonal cross-sections. To be able to constructively realize the above program for more general shapes, let us consider profiles that only slightly deviate from axisymmetric ones. We will show below that this is the third case where a largely analytical treatment is possible.

#### *3.3. Profiles Slightly Deviating from Axisymmetric Ones*

Let us consider a profile that is not axially symmetrical but deviates only slightly from axial symmetry:

$$f(r, \varphi) = f\_0(r) + \delta f(r, \varphi) \tag{13}$$

where *δ f*(*<sup>r</sup>*, *ϕ*) is a small deviation. We define the position of the origin of coordinates in such a way that

$$f(0, \varphi) = f\_0(0) = \delta f(0, \varphi) = 0 \tag{14}$$

The functions *<sup>g</sup>ϕ*(*a*) and *<sup>G</sup>ϕ*(*a*) will correspondingly consist of two parts

$$\mathcal{G}\_{\boldsymbol{\uprho}}(\boldsymbol{a}) = \mathcal{G}\_{0}(\boldsymbol{a}) + \delta \mathcal{G}\_{\boldsymbol{\uprho}}(\boldsymbol{a}),\\\mathcal{g}\_{\boldsymbol{\uprho}}(\boldsymbol{a}) = \mathcal{g}\_{0}(\boldsymbol{a}) + \delta \mathcal{g}\_{\boldsymbol{\uprho}}(\boldsymbol{a}),\\\mathcal{g}\_{\boldsymbol{\uprho}}'(\boldsymbol{a}) = \mathcal{g}\_{0}'(\boldsymbol{a}) + \delta \mathcal{g}\_{\boldsymbol{\uprho}}'(\boldsymbol{a})\tag{15}$$

with

$$\log(a) = a \int\_0^a \frac{f\_0'(r)}{\sqrt{a^2 - r^2}} \mathrm{d}r,\ \delta g\_{\varphi}(a) = a \int\_0^a \frac{\delta f\_0'(r, \varphi)}{\sqrt{a^2 - r^2}} \mathrm{d}r\tag{16}$$

and correspondingly

$$G\_0{}'(a) = \mathcal{g}\_0(a), \; \delta G\_{\; \!\!/}(a) = \delta \mathcal{g}\_{\!\!\!/}(a) \tag{17}$$

The contact boundary will also be almost a circle with a small perturbation

$$a(\varphi) = a\_0 + \delta a(\varphi) \tag{18}$$

Substituting (18) into (12), expanding all functions with respect to small deviation *<sup>δ</sup>a*(*ϕ*) and neglecting all terms of the second or higher orders, we ge<sup>t</sup> for the deviation of the contact contour from a circle in the first approximation

$$\delta a(\varphi) = \frac{\frac{D\pi a\_0}{E^\*} - 2da\_0 + G\_{\varphi}(a\_0) + a\_0 \mathfrak{g}\_{\varphi}(a\_0)}{2d - 2\mathfrak{g}\_{\varphi}(a\_0) - a\_0 \mathfrak{g}\_{\varphi}'(a\_0)}\tag{19}$$

With (15), we obtain

$$\begin{aligned} &\delta a(\boldsymbol{\varrho})[2d - 2\mathbf{g}\_0(\boldsymbol{a}\_0) - a\_0 \mathbf{g}\_0'(\boldsymbol{a}\_0)]^2 \\ &= \left[\frac{D\pi a\_0}{E^\*} - 2da\_0 + \mathbf{G}\_0(\boldsymbol{a}\_0) + a\_0 \mathbf{g}\_0(\boldsymbol{a}\_0)\right][2d - 2\mathbf{g}\_0(\boldsymbol{a}\_0) - a\_0 \mathbf{g}\_0'(\boldsymbol{a}\_0)] \\ &+ [2d - 2\mathbf{g}\_0(\boldsymbol{a}\_0) - a\_0 \mathbf{g}\_0'(\boldsymbol{a}\_0)]\left[\delta \mathbf{G}\_\boldsymbol{\varrho}(\boldsymbol{a}\_0) + a\_0 \delta \mathbf{g}\_\varrho(\boldsymbol{a}\_0)\right] \\ &+ \left[\frac{D\pi a\_0}{E^\*} - 2da\_0 + \mathbf{G}\_0(\boldsymbol{a}\_0) + a\_0 \mathbf{g}\_0(\boldsymbol{a}\_0)\right]\left[2\delta \mathbf{g}\_\varrho(\boldsymbol{a}\_0) - a\_0 \delta \mathbf{g}\_\varrho'(\boldsymbol{a}\_0)\right] \end{aligned} \tag{20}$$

To determine the Lagrange multiplier *D* we require

$$\int\_0^{2\pi} \delta a(\varphi) \mathrm{d}\varphi = 0 \tag{21}$$

Note that the separation (13) into an axisymmetric part and perturbation is not unique. We can use this freedom to essentially simplify the following relations. Let us define the deviation in such a way that

$$\int\_{0}^{2\pi} \delta \mathbf{g}\_{\varphi}(a\_0) \mathbf{d}\varphi = 0 \tag{22}$$

for all *a*0. This will automatically mean that

$$\int\_0^{2\pi} \delta \mathbf{G}\_{\varphi}(a\_0) \mathbf{d}\varphi = 0 \text{ and } \int\_0^{2\pi} \delta \mathbf{g}\_{\varphi}{}'(a\_0) \mathbf{d}\varphi = 0 \tag{23}$$

and also guarantee that

$$\int\_{0}^{2\pi} \left[ \delta G\_{\varphi}(a\_0) + a\_0 \delta g\_{\varphi}(a\_0) \right] \mathbf{d}\varphi = 0 \tag{24}$$

and thus, that integral over *ϕ* of the term in the third line in (20) vanishes. If we now chose the constant *D* such that *Dπa*0 *E*∗ − 2*da*0 + *<sup>G</sup>*0(*<sup>a</sup>*0) + *<sup>a</sup>*0*g*0(*<sup>a</sup>*0) = 0, then the integrals over *ϕ* of the terms in the second and fourth lines of (20) vanish identically and the condition (21) is fulfilled. We then obtain from (20)

$$\delta a(\varphi) = \frac{\left[\delta G\_{\varphi}(a\_0) + a\_0 \delta g\_{\varphi}(a\_0)\right]}{\left[2d - 2g\_0(a\_0) - a\_0 g\_0 \prime (a\_0)\right]} \tag{25}$$

The force (6) can now be written as

$$\begin{split} F\_{\rm N}(A) &= \frac{E^\*}{\pi a\_0} \int\_0^{2\pi} \left( d \cdot a(\varphi)^2 - a(\varphi) G\_{\varphi}(a(\varphi)) \right) \mathrm{d}\varphi \\ &= \frac{E^\*}{\pi a\_0} \int\_0^{2\pi} \left( d \cdot a\_0^2 - a\_0 G\_0(a\_0) \right) \mathrm{d}\varphi + \frac{E^\*}{\pi a\_0} \int\_0^{2\pi} [2d \cdot a\_0 - G\_0(a\_0) - a\_0 g\_0(a\_0)] \delta a(\varphi) \mathrm{d}\varphi - \frac{E^\*}{\pi} \int\_0^{\cdot} \delta G\_{\varphi}(a\_0) \mathrm{d}\varphi \end{split} \tag{26}$$

The second and the third term are equal to zero due to relations (22) and (23) so that only the first term remains. Its maximization leads to the usual MDR result [11]

$$d = \mathcal{g}\_0(a\_0) \tag{27}$$

This means that Equation (25) can finally be rewritten as follows:

$$\delta a(\varphi) = -\frac{\delta G\_{\varphi}(a\_0) + a\_0 \delta g\_{\varphi}(a\_0)}{a\_0 g\_0 \prime (a\_0)}\tag{28}$$

This equation provides the explicit solution for the deviation of the contact boundary from the circle with radius *a*0. The normal force is given by the Equation (9)

$$F\_N(a\_0) = 2E^\* \int\_0^{a\_0} (d - g\_0(a)) \mathrm{d}a \tag{29}$$

#### *3.4. Pressure Distribution for Profiles Slightly Deviating from Axisymmetric Ones*

Let us take a close look at the process of indentation from the first touch to the final indentation depth *d* and denote the current values of the force, the indentation depth and the effective contact radius respectively by *F N*, *d* and *a*0. The entire process consists of changing the indentation depth from *d* = 0 to *d* = *d*, whereby the contact radius changes from *a* = 0 to *a*0 = *a*0 and the contact force from *F N* = 0 to *F N* = *FN*. An infinitesimal indentation by d*d* of the area, which is given by the contour equation *r* = *<sup>a</sup>*(*ϕ*), produces the following contribution to the pressure distribution

$$\mathrm{d}p = \frac{E^\*}{\pi a\_0} \frac{a(\,\,\rho)}{\sqrt{a(\,\,\rho)^2 - r^2}} \mathrm{d}\tilde{d} \text{ for } r < a(\,\,\rho) \tag{30}$$

The pressure distribution at the end of the indentation process is equal to the sum of the incremental pressure distributions:

$$p(r,\boldsymbol{\varrho}) = \frac{E^\*}{\pi} \int\_{\tilde{d}(r)}^d \frac{1}{\tilde{a}\_0} \frac{\tilde{a}(\boldsymbol{\varrho})}{\sqrt{\tilde{a}(\boldsymbol{\varrho})^2 - r^2}} \mathrm{d}\tilde{d} = \frac{E^\*}{\pi} \int\_r^{\tilde{a}(\boldsymbol{\varrho})} \frac{1}{\sqrt{\tilde{a}(\boldsymbol{\varrho})^2 - r^2}} \frac{\tilde{a}(\boldsymbol{\varrho})}{\tilde{a}\_0} \frac{\mathrm{d}\boldsymbol{\varrho}\_0(\tilde{a}\_0)}{\mathrm{d}\tilde{a}(\boldsymbol{\varrho})} \mathrm{d}\tilde{a}(\boldsymbol{\varrho}) \tag{31}$$

where *a*0 must be considered here as a function of *<sup>a</sup>*(*ϕ*).
