**4. Examples**

*4.1. Contact with Parabolic Profiles with Arbitrary Cross-Section* Considerprofile

> *z* = *f*(*<sup>r</sup>*, *ϕ*) =

 a

Thus

$$f\_0(r) = r^2 \overline{\psi} \text{ and } \delta f(r, \varphi) = r^2 \left(\psi(\varphi) - \overline{\psi}\right) \tag{33}$$

*r*<sup>2</sup>*ψ*(*ϕ*)

where *ψ* is the averaged value of *ψ*(*ϕ*) over all angles:

$$\overline{\Psi} = \frac{1}{2\pi} \int\_0^{2\pi} \psi(\varphi) \mathrm{d}\varphi \tag{34}$$

(32)

The corresponding angle-dependent "MDR profile" is

$$\mathcal{g}\_{\mathcal{P}}(a) = 2a^2 \psi(\mathcal{q}) \tag{35}$$

For all necessary auxiliary functions, we ge<sup>t</sup>

$$\log\_0(a) = 2\overline{\Psi}a^2,\ \delta\mathfrak{g}\_\varphi(a) = 2a^2(\psi(\varphi) - \overline{\Psi}),\ \mathfrak{g}\_0'(a) = 4\overline{\Psi}a,\tag{36}$$

$$\mathcal{G}\_0(a) = \frac{2\overline{\psi}a^3}{3}, \; \delta \mathcal{G}\_\varphi(a) = \frac{2a^3}{3} (\psi(\varphi) - \overline{\psi}).\tag{37}$$

The effective contact radius is given by Equation (27):

$$a\_0 = \sqrt{\frac{d}{2\overline{\psi}}}\tag{38}$$

and the contact area is given by the relation (18)

$$a(\wp) = a\_0 \left(\frac{5}{3} - \frac{2}{3} \frac{\psi(\wp)}{\overline{\wp}}\right) \tag{39}$$

To find the pressure distribution, we use Equation (31)

$$p(r,\boldsymbol{\varphi}) = \frac{E^\*}{\pi} \int\_r^{a(\boldsymbol{\varphi})} \frac{\widetilde{a}(\boldsymbol{\varphi})}{\sqrt{\widetilde{a}(\boldsymbol{\varphi})^2 - r^2}} \frac{1}{\widetilde{a}\_0} \frac{\mathrm{d}\mathfrak{g}\_0(\widetilde{a}\_0)}{\mathrm{d}\widetilde{a}(\boldsymbol{\varphi})} \mathrm{d}\widetilde{a}(\boldsymbol{\varphi}) = \frac{2}{\pi} \mathrm{E}^\* \left(2d \cdot \overline{\boldsymbol{\varphi}}\right)^{1/2} \sqrt{1 - \left(\frac{r}{a(\boldsymbol{\varphi})}\right)^2} \tag{40}$$

#### *4.2. Contact with a Paraboloid*

As a special case of a general parabolic profile, let us consider a paraboloid

$$z = f(\mathbf{x}, y) = \frac{\mathbf{x}^2}{2R\_1} + \frac{y^2}{2R\_2} = \frac{r^2}{2} \left( \frac{\cos^2 \varrho}{R\_1} + \frac{\sin^2 \varrho}{R\_2} \right) \tag{41}$$

In this case,

$$\psi(\varphi) = \frac{1}{2} \left( \frac{\cos^2 \varphi}{R\_1} + \frac{\sin^2 \varphi}{R\_2} \right) = \frac{1}{4} \left( \frac{1}{R\_1} + \frac{1}{R\_2} \right) + \frac{1}{4} \left( \frac{1}{R\_1} - \frac{1}{R\_2} \right) \cos 2\varphi \tag{42}$$

and

$$\overline{\psi} = \frac{1}{4} \left( \frac{1}{R\_1} + \frac{1}{R\_2} \right) \tag{43}$$

Equations (38)–(40) take the form

$$a\_0 = \sqrt{\frac{2d \cdot R\_1 R\_2}{R\_1 + R\_2}}\tag{44}$$

$$a(r, \varphi) = a\_0 \left[ 1 - \frac{2}{3} \frac{R\_2 - R\_1}{R\_1 + R\_2} \cos 2\varphi \right] \tag{45}$$

and

$$p(r,\varphi) = \frac{2}{\pi} E^\* \left( \frac{d \cdot (R\_1 + R\_2)}{2R\_1 R\_2} \right)^{1/2} \sqrt{1 - \left(\frac{r}{a(\varphi)}\right)^2} \tag{46}$$

The total normal force is equal to

$$F\_N = \frac{4}{3} E^\* \left(\frac{2R\_1R\_2}{R\_1 + R\_2}\right)^{1/2} d^{3/2} \left(1 - \left(\frac{2}{3} \frac{R\_2 - R\_1}{R\_1 + R\_2}\right)^2\right) \tag{47}$$

Equation (45) describes in the approximation of small eccentricities an ellipse with half-axes

$$a = a\_0 \left[ 1 + \frac{2}{3} \frac{R\_2 - R\_1}{R\_1 + R\_2} \right], \text{ and } b = a\_0 \left[ 1 - \frac{2}{3} \frac{R\_2 - R\_1}{R\_1 + R\_2} \right] \tag{48}$$

For the eccentricity of the contact area, we ge<sup>t</sup>

$$\varepsilon = \sqrt{1 - \frac{b^2}{a^2}} = \frac{2}{\sqrt{3}} \varepsilon\_{\%} \frac{\sqrt{1 - \varepsilon\_{\%}^2 / 2}}{1 - \varepsilon\_{\%}^2 / 6} \tag{49}$$

where

$$
\varepsilon\_{\mathcal{S}} = \sqrt{1 - \frac{R\_1}{R\_2}}\tag{50}
$$

is the eccentricity of cross-sections of the indenter. For small eccentricities, the ratio of eccentricities of contact area and cross-sections of the ellipsoid is constant and equal to 2/√3, which is the exact asymptotic value according to the Hertzian solution [9] (p. 35). Equation (49) provides a very good approximation for the exact solution for eccentricities up to 0.7 (see Figure 1).

**Figure 1.** Eccentricity *e* of the contact area as a function of *e*g according to Equation (49) (dashed line) and comparison with the value according to the Hertzian solution [9] (solid line).

*4.3. Contact of a Conical Indenter with an Arbitrary Cross-Section*

Consider a general conical profile with an "almost axisymmetric" shape

$$f(r, \varphi) = r \cdot \psi(\varphi) = f\_0(r) + \delta f(r, \varphi) \tag{51}$$

Thus

$$f\_0(r) = r\overline{\psi} \text{ and } \delta f(r, \varphi) = r(\psi(\varphi) - \overline{\psi}) \tag{52}$$

where *ψ* is the averaged value of *ψ*(*ϕ*) over all angles according to (34). For all necessary auxiliary functions, we ge<sup>t</sup>

$$\log(a) = \frac{\pi}{2}a\overline{\psi}, \; \delta g\_{\overline{\varphi}}(a) = \frac{\pi}{2}a(\psi(\varphi) - \overline{\psi}), \; g\_{0}{}^{\prime}(a) = \frac{\pi}{2}\overline{\psi} \tag{53}$$

$$\mathcal{G}\_0(a) = \frac{\pi}{4} a^2 \overline{\Psi}, \; \delta G\_\varphi(a) = \frac{\pi}{4} a^2 (\psi(\varphi) - \varepsilon) \tag{54}$$

The contact area is given by the relation (18)

$$a(\varphi) = \frac{2}{\pi} \frac{d}{\overline{\psi}} \left( \frac{5}{2} - \frac{3}{2} \frac{\psi(\varphi)}{\overline{\psi}} \right) \tag{55}$$

To find the pressure distribution, we use Equation (31)

$$p(r,\boldsymbol{\varphi}) = \frac{E^\*}{\pi} \int\_r^{a(\rho)} \frac{1}{\sqrt{\tilde{a}(\rho)^2 - r^2}} \frac{\tilde{a}(\rho)}{\tilde{a}\_0} \frac{\mathrm{d}\mathfrak{g}\_0(\check{a}\_0)}{\mathrm{d}\check{a}(\rho)} \mathrm{d}\check{a}(\rho) = \frac{E^\*}{2} \overline{\boldsymbol{\psi}} \cdot \ln\left(\frac{a(\rho)}{r} + \sqrt{\left(\frac{a(\rho)}{r}\right)^2 - 1}\right) \tag{56}$$

*4.4. Contact of a Pyramid with Square Cross-Section*

> Let's consider a linear profile

$$f(r, \varphi) = r \tan \alpha \cos \varphi , -\frac{\pi}{4} < \varphi < \frac{\pi}{4} \tag{57}$$

and similar equations for the angles *π*/4 < *ϕ* < 3*π*/4, 3*π*/4 < *ϕ* < 5*π*/4 and −3*π*/4 < *ϕ* < −*<sup>π</sup>*/4, which describes a pyramid with square cross-sections and the angle *α* between the inclined planes of the pyramid and the horizon. In this case,

$$\psi(\varphi) = \tan a \cos \varphi,\ \overline{\psi} = \frac{2\sqrt{2}}{\pi} \tan a \tag{58}$$

The boundary of the contact area is given by the relation (55) (see Figure 2)

$$a(\varphi) = \frac{d}{\tan \alpha} \left( \frac{5}{2\sqrt{2}} - \frac{3\pi}{8} \cos \varphi \right) \text{ for } -\frac{\pi}{4} < \varphi < \frac{\pi}{4} \text{ usw.} \tag{59}$$

**Figure 2.** The normalized contact boundary line (*a*(*ϕ*)tan *α*/*d*) according to Equation (59) (blue line) and the result of numerical simulation with Boundary Element Method (gray figure). Numerical data have been obtained by Qiang Li [12] with the method described in [13].

For the normal force, we ge<sup>t</sup>

$$F\_N = \frac{2}{\pi} E^\* \frac{d^2}{\overline{\psi}} = \frac{E^\* d^2}{\sqrt{2} \tan a} \tag{60}$$

Numerical simulation gives the solution *FN* = 0.7395*E*∗*d*2/ tan *α* which is 4.6% larger than the above analytical result.

For the pressure distribution in the contact area, we obtain, according to (56),

$$p(r,\varphi) = \frac{E^\*}{2}\overline{\psi} \cdot \ln\left(\frac{a(\varphi)}{r} + \sqrt{\left(\frac{a(\varphi)}{r}\right)^2 - 1}\right) \tag{61}$$

*4.5. Contact of Power-Law Shapes*

> Consider a profile having the form

$$f(r, \varphi) = r^n \cdot \psi(\varphi) \tag{62}$$

which means that all vertical cross-sections are self similar differing only by the scaling factor *ψ*(*ϕ*). The decomposition into a rotationally symmetric part and deviation is as follows

$$f\_0(r) = r^n \cdot \overline{\Psi}, \; \delta f(r, \varphi) = r^n \cdot \left(\psi(\varphi) - \overline{\Psi}\right) \tag{63}$$

where *ψ* is the average value of *ψ*(*ϕ*), Equation (34).

For all necessary auxiliary functions, we ge<sup>t</sup>

$$\text{G.g.}(a) = \psi(\varphi) \cdot \gamma(a), \text{ G.g.}(a) = \psi(\varphi) \cdot (a), \, ^{\prime}(a) = \gamma(a), \, \text{g.}(a) = \overline{\psi} \cdot \gamma(a) \tag{64}$$

with

$$\gamma(a) = a \int\_0^a \frac{nr^{n-1}}{\sqrt{a^2 - r^2}} \mathrm{d}r = \kappa\_n a^n, \; \kappa\_n = \int\_0^1 \frac{\xi^{n-1} \mathrm{d}\xi}{\sqrt{1 - \xi^2}} = \frac{\sqrt{\pi}}{2} \frac{n \Gamma\left(\frac{\mu}{2}\right)}{\Gamma\left(\frac{\mu}{2} + \frac{1}{2}\right)}\tag{65}$$

while <sup>Γ</sup>(...) is the gamma function.

> For the deviations, we have

$$\delta g\_{\boldsymbol{\uprho}}(\boldsymbol{a}) = \kappa\_{\boldsymbol{n}} a^{\boldsymbol{n}} (\psi(\boldsymbol{\uprho}) - \overline{\boldsymbol{\uprho}}), \\ \delta \mathcal{G}\_{\boldsymbol{\uprho}}(\boldsymbol{a}) = \kappa\_{\boldsymbol{n}} \frac{a^{\boldsymbol{n}+1}}{n+1} (\psi(\boldsymbol{\uprho}) - \overline{\boldsymbol{\uprho}}) \tag{66}$$

The boundary of the contact area is given by the relation (28)

$$a(\varphi) = a\_0 \left( 1 + \frac{n+2}{n(n+1)} \left( 1 - \frac{\psi(\varphi)}{\overline{\psi}} \right) \right) \tag{67}$$

where *a*0 is determined by the condition *ψκna*0*<sup>n</sup>* = *d*:

$$a\_0 = \left(\frac{d}{\kappa\_n \overline{\psi}}\right)^{1/n} \tag{68}$$

The normal force is equal to

$$F\_N(a\_0) = 2E^\* \int\_0^{a\_0} (d - \overline{\Psi} \kappa\_n a^n) \mathbf{d}a = 2E^\* da\_0 \frac{n}{n+1} = \frac{2n}{n+1} E^\* d^{\frac{n+1}{n}} \left(\kappa\_n \overline{\Psi}\right)^{-1/n} \tag{69}$$

The average pressure is equal to

$$\overline{p} = \frac{F\_N(a\_0)}{\pi a\_0^2} = \frac{2n}{n+1} \frac{E^\* d^{\frac{n-1}{n}} \left(\kappa\_n \overline{\Psi}\right)^{1/n}}{\pi} \tag{70}$$

For the pressure distribution in the contact area, normalized by the average pressure, we obtain, according to (31),

$$p(r,\boldsymbol{\varrho}) = \frac{(n+1)}{2} \int\_{\mathbb{T}}^{1} \frac{\mathbb{S}^{n-1}}{\sqrt{\mathbb{S}^{2} - \mathbb{T}^{2}}} \mathrm{d}\boldsymbol{\xi} \tag{71}$$

with *r* = *<sup>r</sup>*/*a*(*ϕ*). It is again the same result as that for an axisymmetric contact ([10], p. 78) with the substitution *<sup>r</sup>*/*<sup>a</sup>*0 → *<sup>r</sup>*/*a*(*ϕ*).

#### **5. Extended Method of Dimensionality Reduction (MDR) for Slightly Non-Axisymmetric Contacts**

The above-sketched solution procedure can be considered as an extension of the Method of Dimensionality Reduction [11] to non-axisymmetric profiles. Let us summarize the above findings in the form of procedure that has to be applied for the solution of such problems.

We consider an "arbitrary" profile *z* = *f*(*<sup>r</sup>*, *ϕ*) underlying the following restrictions:


Under these conditions, the following procedure can be applied.

I. In the first step, an "equivalent axisymmetric profile" is determined. In the first-order approximation, it was shown to be just the profile averaged over the angles:

$$f\_0(r) = \frac{1}{2\pi} \int\_0^{2\pi} f(r, \varphi) \mathrm{d}\varphi \tag{72}$$

II. The profile can now be decomposed into an axisymmetric part and the small deviation

$$f(r, \boldsymbol{\varphi}) = f\_0(r) + \delta f(r, \boldsymbol{\varphi}) \tag{73}$$

with

$$
\delta f(r, \varphi) = f(r, \varphi) - f\_0(r) \tag{74}
$$

by definition.

III. With the equivalent axisymmetric profile (72), the usual MDR solution procedure is applied. In particular, the MDR-transformed profile is determined as

$$\mathbf{g}\_0(a) = a \int\_0^a \frac{f \mathbf{o}'(r)}{\sqrt{a^2 - r^2}} \mathbf{d}r \tag{75}$$

IV. The effective radius *a*0 is determined by the condition

$$d = \mathcal{g}\_0(a\_0) \tag{76}$$

V. The normal force is given by the equation

$$F\_N = 2E^\* \int\_0^{a\_0} (d - g\_0(\mathbf{x})) \mathbf{dx} \tag{77}$$

VI. The effective pressure distribution in the contact area is given by the usual MDR equation [14] (p. 9)

$$p\_0(r/a\_0) = -\frac{1}{\pi} \int\_r^\infty \frac{q'(x)}{\sqrt{x^2 - r^2}} dx \tag{78}$$

with

$$q(\mathbf{x}) = E^\* \begin{cases} |d - \mathbf{g}(\mathbf{x})\_\prime|\mathbf{x}| < a\_0 \\ 0, |\mathbf{x}| > a\_0 \end{cases} \tag{79}$$

#### VII. The true non-axisymmetric contact area is given by the equation

$$a(\varphi) = a\_0 + \delta a(\varphi) \tag{80}$$

where *<sup>δ</sup>a*(*ϕ*) is determined by (28). VIII. Finally,thetruepressuredistributionisgivenby

$$p(r,\varphi) = p\_0(r/a(\varphi))\tag{81}$$

The last equation can be considered as a generalization of Fabrikant's ansatz for flatended punches. Fabrikant states that the pressure distribution under a non-axisymmetric punch is equal to that under an axisymmetric punch but "rescaled" to the true shape of the contact area. Similarly, Equation (81) states that the contact pressure is equal to that under an "equivalent axisymmetric indenter" but rescaled to the true contact area. In the present paper, we have shown that Equation (81) is the exact first-order approximation for the power-law profiles (with arbitrary cross-section)—independently of the exponent of the power law. Even while it was not proven in a general case, the independence of the exponent gives hope that it could be a good approximation for arbitrarily shaped profiles. Testing of this hypothesis (e.g., through comparison with a numerical solution with BEM) is an important task of further studies.
