In this article, the problem of axisymmetric contact between dissimilar elastic bodies is formulated mainly to treat non-slipping cases. When two elastically dissimilar bodies are incrementally pressed into contact, a relative radial displacement along the contact interface emerges because of the material property mismatch. To find the contact relative radial displacement separately for non-adhesive and adhesive conditions, a direct and generic approach is developed. The explicit solutions of the interface relative radial displacement for non-slipping contacts are derived in a rigorous manner, enabling a thorough analysis. In addition, based on the present formulation and approach, as an example, a typical adhesive contact problem with an applied radial mismatch strain is re-examined. This shows that radial mismatch strain may significantly influence the pull-off force. These results may be used to model more accurately contacts at macro, micro and nanoscales.

1. Introduction

Non-slipping contact can be simply described as follows. When two dissimilar elastic bodies are compressed towards each other under the influence of a gradually increasing external force, and if there is no reciprocal slippage along the interface in this process, a relative radial displacement appears in the contact zone, owing to the material property mismatch of the bodies and their surfaces. This contact zone displacement will not vary with progressive loading. The path of a surface particle subject to non-slipping contact is illustrated in figure 1, where the indenter is supposed rigid. This relative contact radial displacement is believed to reflect intrinsically the characteristic of non-slipping contacts.

Path of surface particle of non-slipping contact during incremental normal load. — Figure 1. Path of surface particle of non-slipping contact during incremental normal load (Adopted from the reference paper by Spence [1]).

To model more accurately practical contact problems, such as fretting fatigue [2], nano-indentation [3–5] and bio-adhesion [6–9], non-slipping contact models were adopted, in which the relative contact radial displacement is taken into account. Finding this displacement plays a vital role in the process of solving contact problems. Moreover, apart from these practical demands, this research has an evident theoretical significance for contact mechanics [7,10,11], and is proving an intriguing challenge, attracting a lot of research (see, e.g. [5,12–17]).

Non-slipping contact study can be traced back to Mossakovskii [18,19] and Goodman [20]. In Mossakovskii’s model, a rigid-to-elastic axisymmetric contact with no slippage was initially studied. A step-by-step incremental approach was used, and the contact stresses were evaluated simultaneously with the evolution of the contact radius. Mossakovskii presented solutions for a flat-ended cylinder and a parabolic punch. Goodman [20] proposed an approximate solution to the non-slipping Hertzian contact problem, in which the contact pressure was assumed to be the same as the elastically similar case. Afterwards, based on an inspectional analysis, Spence [1] proposed a self-similarity approach and made a significant contribution to this problem. He extended the range of rigid indenter profiles considered to include polynomials. Non-slipping contact problems were formulated with an integral equation which was solved using the Wiener–Hopf technique [21]. Later, the self-similarity approach was used to explore symmetric partial-slipping contact with finite friction for the cases of a two-dimensional problem [22,23]. This approach is not limited to these relative simple contact problems, which has been extended to other more complex frictional contact problems [13,24–26]. Mossakovskii’s incremental approach and Spence’s self-similarity approach are the key tools in treating non-slipping contact problems, but they are rather complicated in application. From a general point of view, it should be kept in mind that such non-slipping contact problems belong to the classical elastic mixed boundary-value problems, and that contact relative radial displacement is entirely dependent on the geometric profiles of both contact surfaces; independent of loading history. Therefore, a variety of suitable techniques developed in elastic mechanics should be readily employed to tackle them in a rigorous manner. Spence’s self-similarity approach, based on an inspectional analysis, is expected to be alternatively addressed. Clearly, a direct and elaborate analysis on these problems is required.

In addition, at the micro or nanoscale, the significant surface adhesion effect, i.e. the Van der Waals force, should be taken in to account when studying contacts [27]. These small-scale surface adhesion problems have become a popular research topic, and include work on non-indentation [25,28–30] and bio-adhesion [31–33]. The relative radial displacement in adhesive non-slipping contact is very different from the common non-slipping contacts at macroscale, which highlights the need for more careful consideration of the relative contact radial displacement associated with them.

It should be clarified that contacts with an ‘adhesive boundary condition’ in early literatures such as [1,18,19,21,34] are actually referred to as non-slipping, with no surface molecular adhesion. By contrast, as mentioned above, surface molecular adhesion should be considered in many contacts at microscale [26,35]. The molecular adhesion condition is referred to as ‘the surface adhesion condition’ in this study, and will be treated using the assumption of JKR theory [27]. Thus, according to the surface molecular adhesion effect, non-slipping contacts are classified into two groups: with adhesion and with non-adhesion. We further clarify that the non-slipping contact assumption used here means that the contacts are pure elastic processes, which are unrelated to the friction coefficient regardless of whether the contact surfaces have molecular adhesion or not.

The aim of the present study is to: (i) formulate the axisymmetric contact problems and then to establish a rigorous and generic approach to treating non-slipping contact problems; (ii) seek the explicit solutions for the relative contact radial displacement of axisymmetric non-slipping contacts with an arbitrary profiled indenter (especially those with surface adhesion effect); (iii) directly extend the conventional rigid-to-elastic contact models cited in the literature to general elastic-to-elastic non-slipping contacts (see figure 2), striving for a broad application; (iv) re-explore the influence of applied radial mismatch strain on the pull-off force. It is expected that this study will provide useful formulation for axisymmetric contact problems and shed some light on the subject of elastic contacts where non-slipping conditions are observed.

Non-slipping contact between two axisymmetric solids. — Figure 2. Non-slipping contact between two axisymmetric solids, in which #1 stands for the upper solid and #2 stands for the lower solid material.

The remainder of this article is laid out as follows. In §2, a basic formulation for axisymmetric contact between dissimilar solids is derived, and boundary conditions and external load conditions for non-slipping contact are presented. In §3, to solve the governing dual equations in §2, they are transformed by using an Abel transform, or Fourier sine and cosine transforms. Subsequently, contacts between similar materials and dissimilar materials are respectively treated in §§4 and 5. To demonstrate this proposed approach in solving the relative radial displacement, a non-slipping conical contact is analysed separately for the non-adhesion condition and adhesion condition in §6. Then, to demonstrate the application of the proposed formulation and the approach, a typical adhesion contact problem, with an applied radial mismatch strain, is studied in §7. Finally, a concise conclusion is given in §8.

2. General formulation for axisymmetric contacts

In this section, the governing equations for axisymmetric contacts are derived and presented, and the contact boundary conditions are discussed.

(a) The Papkovich–Neuber function potential formulae

The harmonic potential function method has significant advantages in solving axisymmetric contact problems [34,36,37]. The Papkovich–Neuber functions are more convenient to be used to formulate these mixed boundary-value problems. The elastic fields due to contact may be expressed in terms of harmonic Papkovich–Neuber potential functions $φ (r, z)$ (the scalar potential) and $χ (r, z)$ (the vector potential with three components), $\nabla^{2} φ (r, z) = 0$ , $\nabla^{2} χ (r, z) = 0$ . Owing to the problem’s symmetry, it is possible to reduce the number of functions to only two, so that the vector potential possesses only one component, $χ (r, z) = χ (r, z) e_{z}$ , where $e_{z}$ is the unit vector along the z-axis. In terms of a pair of harmonic functions $φ_{j} (r, z)$ and $χ_{j} (r, z),$ in which $j = 1$ represents the upper solid and $j = 2$ stands for the lower solid in figure 2, the displacement and stress components of axisymmetric problems can be written as [38,39]

{\begin{cases} 2 μ_{j} u_{r} = - φ_{j, r} - z χ_{j, r} \\ 2 μ_{j} u_{z} = κ_{j} χ_{j} - φ_{j, z} - z χ_{j, z} \\ σ_{z z} = \frac{1 + κ_{j}}{2} χ_{j, z} - φ_{j, z z} - z χ_{j, z z} \\ σ_{r z} = \frac{κ_{j} - 1}{2} χ_{j, r} - φ_{j, r z} - z χ_{j, r z} \end{cases}, (j = 1, 2),

(2.1)

where $μ_{j}$ is the shear modulus, $κ_{j} = (3 - 4 ν_{j})$ is the Kolosov constant and $ν_{j}$ is Poisson’s ratio. The subscript letters following the comma indicate differentiation with respect to the indicated coordinates, e.g. $χ_{j, r z} = \partial^{2} χ_{j} / \partial r \partial z$ , and

\nabla^{2} φ_{j} (r, z) = 0, \nabla^{2} χ_{j} (r, z) = 0,

(2.2)

where $\nabla^{2} = \frac{\partial^{2}}{\partial^{2} r} + \frac{1}{r} \frac{\partial}{\partial r} + \frac{\partial^{2}}{\partial^{2} z}$ is the Laplacian operator.

(b) The dual integral equations for axisymmetric contacts

To solve the Papkovich–Neuber potentials in equation (2.2), the Hankel transform and its inverse transform will be employed. The Hankel transform of order n of a function $f (r)$ is defined as [40]

F_{n} (λ) = H_{n} [f (r); r \to λ] = \int_{0}^{\infty} r f (r) J_{n} (λ r) d r,

(2.3)

and its inverse as

f (r) = H_{n} [F_{n} (λ); λ \to r] = \int_{0}^{\infty} λ F_{n} (λ) J_{n} (λ r) d λ,

(2.4)

where $J_{n} (*)$ denotes the Bessel function of the first kind, of order n. Thus, the harmonic equation (2.2) of the two independent variables r and z can be reduced to two ordinary differential equations of the variable z. Here the Hankel transforms of order zero of the Papkovich–Neuber potentials are the solutions of the ordinary differential equations:

{\begin{cases} H_{0} [\nabla^{2} φ_{j} (r, z); r \to λ] = \frac{\partial^{2}}{\partial z^{2}} Φ_{j} (λ, z) - λ^{2} Φ_{j} (λ, z) = 0 \\ H_{0} [\nabla^{2} χ_{j} (r, z); r \to λ] = \frac{\partial^{2}}{\partial z^{2}} X_{j} (λ, z) - λ^{2} X_{j} (λ, z) = 0 \end{cases}, (j = 1, 2),

(2.5)

with

{\begin{cases} Φ_{j} (λ, z) = H_{0} [φ_{j} (r, z); r \to λ] = \int_{0}^{\infty} r φ_{j} (r, z) J_{0} (λ r) d r \\ X_{j} (λ, z) = H_{0} [χ_{j} (r, z); r \to λ] = \int_{0}^{\infty} r χ_{j} (r, z) J_{0} (λ r) d r \end{cases}, (j = 1, 2) .

(2.6)

Solutions to equation (2.5) are

{\begin{cases} Φ_{1} (λ, z) = A_{1} e^{- λ z} \\ X_{1} (λ, z) = B_{1} e^{- λ z} \end{cases}, (z > 0), {\begin{cases} Φ_{2} (λ, z) = A_{2} e^{λ z} \\ X_{2} (λ, z) = B_{2} e^{λ z} \end{cases}, (z < 0),

(2.7)

where $A_{i} = A_{i} (λ), B_{i} = B_{i} (λ), (i = 1, 2)$ . With equation (2.7), $φ_{j} (r, z), χ_{j} (r, z)$ become

\begin{array}{ll} {\begin{cases} φ_{1} (r, z) = H_{0} [Φ_{1} (λ, z); λ \to r] = \int_{0}^{\infty} A_{1} e^{- λ z} λ J_{0} (λ r) d λ \\ χ_{1} (r, z) = H_{0} [X_{1} (λ, z); λ \to r] = \int_{0}^{\infty} B_{1} e^{- λ z} λ J_{0} (λ r) d λ \end{cases}, (z > 0), \\ {\begin{cases} φ_{2} (r, z) = H_{0} [Φ_{2} (λ, z); λ \to r] = \int_{0}^{\infty} A_{2} e^{λ z} λ J_{0} (λ r) d λ \\ χ_{2} (r, z) = H_{0} [X_{2} (λ, z); λ \to r] = \int_{0}^{\infty} B_{2} e^{λ z} λ J_{0} (λ r) d λ \end{cases}, (z < 0) . \end{array}

(2.8)

By substituting equation (2.8) into equation (2.1), one deduces

\begin{array}{l} (z > 0) \end{array} {\begin{array}{ll} 2 μ_{1} u_{r} = \int_{0}^{\infty} A_{1} e^{- λ z} λ^{2} J_{1} (λ r) d λ + z \int_{0}^{\infty} B_{1} e^{- λ z} λ^{2} J_{1} (λ r) d λ \\ 2 μ_{1} u_{z} = k_{1} \int_{0}^{\infty} B_{1} e^{- λ z} λ J_{0} (λ r) d λ + \int_{0}^{\infty} A_{1} e^{- λ z} λ^{2} J_{0} (λ r) d λ + z \int_{0}^{\infty} B_{1} e^{- λ z} λ^{2} J_{0} (λ r) d λ \\ σ_{z z} = - \frac{1 + k_{1}}{2} \int_{0}^{\infty} B_{1} e^{- λ z} λ^{2} J_{0} (λ r) d λ - \int_{0}^{\infty} A_{1} e^{- λ z} λ^{3} J_{0} (λ r) d λ - z \int_{0}^{\infty} B_{1} e^{- λ z} λ^{3} J_{0} (λ r) d λ \\ σ_{r z} = - \frac{k_{1} - 1}{2} \int_{0}^{\infty} B_{1} e^{- λ z} λ^{2} J_{1} (λ r) d λ - \int_{0}^{\infty} A_{1} e^{- λ z} λ^{3} J_{1} (λ r) d λ - z \int_{0}^{\infty} B_{1} e^{- λ z} λ^{3} J_{1} (λ r) d λ \end{array},

(2.9a)

\begin{array}{l} (z < 0) \end{array} {\begin{cases} 2 μ_{2} u_{r} = \int_{0}^{\infty} A_{2} e^{λ z} λ^{2} J_{1} (λ r) d λ + z \int_{0}^{\infty} B_{2} e^{λ z} λ^{2} J_{1} (λ r) d λ \\ 2 μ_{2} u_{z} = k_{2} \int_{0}^{\infty} B_{2} e^{λ z} λ J_{0} (λ r) d λ - \int_{0}^{\infty} A_{2} e^{λ z} λ^{2} J_{0} (λ r) d λ - z \int_{0}^{\infty} B_{2} e^{λ z} λ^{2} J_{0} (λ r) d λ \\ σ_{z z} = \frac{1 + k_{2}}{2} \int_{0}^{\infty} B_{2} e^{λ z} λ^{2} J_{0} (λ r) d λ - \int_{0}^{\infty} A_{2} e^{λ z} λ^{3} J_{0} (λ r) d λ - z \int_{0}^{\infty} B_{2} e^{λ z} λ^{3} J_{0} (λ r) d λ . \\ σ_{r z} = - \frac{k_{2} - 1}{2} \int_{0}^{\infty} B_{2} e^{λ z} λ^{2} J_{1} (λ r) d λ + \int_{0}^{\infty} A_{2} e^{λ z} λ^{3} J_{1} (λ r) d λ + z \int_{0}^{\infty} B_{2} e^{λ z} λ^{3} J_{1} (λ r) d λ \end{cases}

(2.9b)

The continuity of tractions $σ_{z z}^{+} = σ_{z z}^{-} = σ_{z z}, σ_{r z}^{+} = σ_{r z}^{-} = σ_{r z}, (z = 0)$ along the contacted interface requires

{\begin{cases} A_{1} λ = - \frac{1}{2} (B_{2} + κ_{1} B_{1}) \\ A_{2} λ = \frac{1}{2} (κ_{2} B_{2} + B_{1}) \end{cases} .

(2.10)

Thus, the displacement and traction (equation (2.9)) of either solid surface can be written as

\begin{array}{l} T h e u p p e r s u r f a c e : \\ {\begin{cases} u_{r} = - \int_{0}^{\infty} \frac{(B_{2} + κ_{1} B_{1})}{4 μ_{1}} λ J_{1} (λ r) d λ, \\ u_{z} = \int_{0}^{\infty} \frac{(κ_{1} B_{1} - B_{2})}{4 μ_{1}} λ J_{0} (λ r) d λ, \\ σ_{z z} = \int_{0}^{\infty} \frac{(B_{2} - B_{1})}{2} λ^{2} J_{0} (λ r) d λ, \\ σ_{r z} = \int_{0}^{\infty} \frac{(B_{2} + B_{1})}{2} λ^{2} J_{1} (λ r) d λ . \end{cases} \end{array} \begin{array}{l} T h e l o w e r s u r f a c e : \\ {\begin{cases} u_{r} = \int_{0}^{\infty} \frac{(κ_{2} B_{2} + B_{1})}{4 μ_{2}} λ J_{1} (λ r) d λ, \\ u_{z} = \int_{0}^{\infty} \frac{(κ_{2} B_{2} - B_{1})}{4 μ_{2}} λ J_{0} (λ r) d λ, \\ σ_{z z} = \int_{0}^{\infty} \frac{(B_{2} - B_{1})}{2} λ^{2} J_{0} (λ r) d λ, \\ σ_{r z} = \int_{0}^{\infty} \frac{(B_{1} + B_{2})}{2} λ^{2} J_{1} (λ r) d λ . \end{cases} \end{array}

(2.11)

When two bodies are pressed together, deformation must occur so that the deformed bodies conform within the contact. The derivative of the overlapping displacements ( $Δ u_{r}$ and $Δ u_{z}$ ) with respect to the radial coordinate (r-axis) within the contact can be evaluated as

(Δ u_{r} + i Δ u_{z}) = (u_{r} + i u_{z})_{u p p e r s u r f a c e} - (u_{r} + i u_{z})_{l o w e r s u r f a c e}, (r < a) .

(2.12)

In terms of equation (2.11), condition equation (2.12) can be expressed as

{\begin{cases} Δ u_{r} = - \frac{1}{4} \int_{0}^{\infty} [(\frac{1}{μ_{1}} + \frac{κ_{2}}{μ_{2}}) B_{2} + (\frac{κ_{1}}{μ_{1}} + \frac{1}{μ_{2}}) B_{1}] λ J_{1} (λ r) d λ \\ Δ u_{z} = \frac{1}{4} \int_{0}^{\infty} [(\frac{κ_{1}}{μ_{1}} + \frac{1}{μ_{2}}) B_{1} - (\frac{1}{μ_{1}} + \frac{κ_{2}}{μ_{2}}) B_{2}] λ J_{0} (λ r) d λ \end{cases} .

(2.13)

If one denotes $ϕ = ϕ (λ) = (B_{2} - B_{1}) λ / 2, ψ = ψ (λ) = (B_{2} + B_{1}) λ / 2$ , the traction on the contact surface and equation (2.13) can be used as input in a pair of dual integral equations as

{\begin{cases} \int_{0}^{\infty} (β ϕ + ψ) J_{1} (λ r) d λ = - \frac{1}{A} Δ u_{r} = g (r), (0 \leq r < a) \\ \int_{0}^{\infty} (β ψ + ϕ) J_{0} (λ r) d λ = - \frac{1}{A} Δ u_{z} = f (r), (0 \leq r < a) \\ \int_{0}^{\infty} ϕ λ J_{0} (λ r) d λ = σ_{z z} = 0, (a < r < \infty) \\ \int_{0}^{\infty} ψ λ J_{1} (λ r) d λ = σ_{r z} = 0, (a < r < \infty) \end{cases},

(2.14)

where $Δ u_{r}$ and $Δ u_{z}$ are sometimes called gap functions, and

A = \frac{(k_{1} + 1)}{4 μ_{1}} + \frac{（ k_{2} + 1 ）}{4 μ_{2}}, β = \frac{μ_{1} (κ_{2} - 1) - μ_{2} (κ_{1} - 1)}{μ_{1} (κ_{2} + 1) + μ_{2} (κ_{1} + 1)} .

(2.15)

The inverse Hankel transform of equation (2.14) is

{\begin{cases} ϕ (λ) = \int_{0}^{\infty} σ_{z z} (r) r J_{0} (λ r) d r \\ ψ (λ) = \int_{0}^{\infty} σ_{r z} (r) r J_{1} (λ r) d r \end{cases}, {\begin{cases} ϕ (λ) = \frac{1}{(1 - β^{2})} λ \int_{0}^{\infty} f (r) r J_{0} (λ r) d r - \frac{β}{(1 - β^{2})} λ \int_{0}^{\infty} g (r) r J_{1} (λ r) d r \\ ψ (λ) = \frac{1}{(1 - β^{2})} λ \int_{0}^{\infty} g (r) r J_{1} (λ r) d r - \frac{β}{(1 - β^{2})} λ \int_{0}^{\infty} f (r) r J_{0} (λ r) d r \end{cases} .

(2.16)

The first two equations suggest that $ϕ (λ)$ is an even function, and $ψ (λ)$ is an odd function. Then equation (2.8) can be written as

{\begin{array}{ll} φ_{1} (r, z) & = - \frac{1}{2} \int_{0}^{\infty} [(1 - k_{1}) ϕ + (1 + k_{1}) ψ] λ^{- 1} e^{- λ z} J_{0} (λ r) d λ \\ χ_{1} (r, z) & = \int_{0}^{\infty} (ψ - ϕ) e^{- λ z} J_{0} (λ r) d λ \end{array}, (z > 0), {\begin{array}{ll} φ_{2} (r, z) & = \frac{1}{2} \int_{0}^{\infty} [(κ_{2} - 1) ϕ + (κ_{2} + 1) ψ] λ^{- 1} e^{λ z} J_{0} (λ r) d λ \\ χ_{2} (r, z) & = \int_{0}^{\infty} (ϕ + ψ) e^{λ z} J_{0} (λ r) d λ \end{array}, (z < 0) .

(2.17)

In addition, inserting the first two expressions of equation (2.16) into the first two of equation (2.14) gives

{\begin{cases} β \int_{0}^{a} σ_{z z} t [\int_{0}^{\infty} J_{0} (λ t) J_{1} (r) d λ] d t + \int_{0}^{a} σ_{r z} t [\int_{0}^{\infty} J_{1} (λ t) J_{1} (λ r) d λ] d t = - \frac{Δ u_{r}}{A} = g (r) \\ β \int_{0}^{a} σ_{r z} t [\int_{0}^{\infty} J_{1} (λ t) J_{0} (λ r) d λ] d t + \int_{0}^{a} σ_{z z} t [\int_{0}^{\infty} J_{0} (λ t) J_{0} (λ r) d λ] d t = - \frac{Δ u_{z}}{A} = f (r) \end{cases} (0 \leq r < a),

(2.18)

which can be alternatively written as

\begin{matrix} {\begin{cases} β \int_{0}^{a} σ_{z z} t H_{01} (t, r) d t + \int_{0}^{a} σ_{r z} t H_{11} (t, r) d t = - \frac{Δ u_{r}}{A} = g (r) \\ β \int_{0}^{a} σ_{r z} t H_{10} (t, r) d t + \int_{0}^{a} σ_{z z} t H_{00} (t, r) d t = - \frac{Δ u_{z}}{A} = f (r) \end{cases}, (0 \leq r < a), \\ H_{i j} (t, r) = \int_{0}^{\infty} J_{i} (λ t) J_{j} (λ r) d λ, \end{matrix}

(2.19)

with [41]

\begin{array}{ll} H_{00} (t, r) = {\begin{array}{ll} \frac{2}{π t} K (\frac{r}{t}), r < t \\ \frac{2}{π r} K (\frac{t}{r}), r > t \end{array}, H_{01} (t, r) = {\begin{cases} 0, r < t \\ \frac{1}{r}, r > t \end{cases}, \\ H_{10} (t, r) = {\begin{cases} \frac{1}{t}, r < t \\ 0, r > t \end{cases}, H_{11} (t, r) = {\begin{cases} \frac{2}{π r} [K (\frac{r}{t}) - E (\frac{r}{t})], r < t \\ \frac{2}{π t} [K (\frac{t}{r}) - E (\frac{t}{r})], r > t \end{cases}, \end{array}

(2.20)

where K and E are the complete elliptic integrals of the first and second kinds.

The governing equation for axisymmetric contact problems, equation (2.19), however, is not easy to solve directly [42]. For completeness, an alternative governing equation in an integral formulation is derived and presented in appendix D; this may be used to model other axisymmetric contact problems such as frictionless ones. Here, we stick on the original dual equation (2.14) and try to solve through systematic transform manipulations [40].

(c) Non-slipping contact conditions

Loading history-independent condition for non-slipping contact
In general, the function $f (r)$ in equation (2.14) is directly determined by the profiles of the two contact bodies, but the function $g (r)$ is difficult to specify, and is to be determined. Importantly, their derivatives with respect to the radial coordinate r will not change once the correspondent point pair has entered the contact as the contact radius a successively increases or decreases:
$\frac{\partial}{\partial a} f^{'} (r) = 0, \frac{\partial}{\partial a} g^{'} (r) = 0, (r < a) .$ (2.21)
It has been shown in the literature that finding the solution of $g (r)$ is the most challenging issue in the scope of non-slipping contacts. Only after obtaining it can the non-slipping contact problems be explicitly posed and solved.
Remote loading condition. To solve the boundary value problem, the remote loading condition must be known, so that the contact radius a may be specified. The external load is equal to the resultant force of interfacial traction, given by
$P = 2 π \int_{0}^{a} σ_{z z} r d r .$ (2.22)
Traction behaviour at contact interface rim. For non-slipping contact with no adhesion, we suppose that the traction at the rim will vanish, while for non-slipping contact with adhesion, we suppose that the traction at the contact rim will be singular. These conditions will be used to determine the contact traction behaviour in the following sections.

3. Transform approaches to treating the dual equation problems

In this section, we transform the dual integral equation (2.14) into solvable forms by using Abel and Fourier sine and cosine transforms, which provide a basis for further treatment of equation (2.13) in §§4 and §5.

First, some useful relationships and identities are introduced. To treat the dual integral equation (2.14), the following relationships,

H_{0} [ϕ (λ); λ \to r] = \frac{1}{r} \frac{d}{d r} r H_{1} [λ^{- 1} ϕ (λ); λ \to r],

(3.1a)

H_{1} [ϕ (λ); λ \to r] = - \frac{d}{d r} H_{0} [λ^{- 1} ϕ (λ); λ \to r],

(3.1b)

where $ϕ (λ)$ is an arbitrary function, and the identities

{\begin{cases} J_{0} (λ x) = \frac{2}{π} \int_{0}^{x} \frac{c o s (λ t) d t}{\sqrt{x^{2} - t^{2}}} = \frac{2}{π} \int_{x}^{\infty} \frac{s i n (λ t) d t}{\sqrt{t^{2} - x^{2}}} \\ \frac{d}{d x} \int_{0}^{x} \frac{J_{0} (λ r) - 1}{\sqrt{x^{2} - r^{2}}} r d r = c o s λ x - 1 \\ r J_{1} (r x) = \frac{2}{π} \int_{0}^{r} \frac{s s i n (x s) d s}{\sqrt{r^{2} - s^{2}}} \\ \frac{s i n λ x}{λ} = \int_{0}^{x} \frac{r J_{0} (λ r) d r}{\sqrt{x^{2} - r^{2}}} = \int_{x}^{\infty} \frac{x J_{1} (λ r) d r}{\sqrt{r^{2} - x^{2}}} \end{cases}, (x > 0, λ > 0),

(3.2)

play an important role in the following manipulations. In addition, the Abel operators $A_{1}$ and $A_{2}$ , as well as Fourier sine and cosine transforms operators $F_{s}$ and $F_{c}$ , are given in appendix A. They will be repetitively used here. By virtue of equations (3.1) and (3.2) and the Abel operators $A_{1}$ and $A_{2}$ , we can prove that [43]

A_{1} [r H_{0} [ϕ (λ); λ \to r]; r \to x] = F_{s} [ϕ (λ); λ \to x],

(3.3a)

A_{1} [H_{1} [ϕ (λ); λ \to r]; r \to x] = x^{- 1} {F_{c} [ϕ (λ); λ \to 0] - F_{c} [ϕ (λ); λ \to x]},

(3.3b)

A_{2} [\frac{d}{d r} r H_{1} [λ^{- 1} ϕ (λ); λ \to r]; r \to x] = F_{c} [ϕ (λ), λ \to x],

(3.3c)

A_{2} [\frac{d}{d r} H_{0} [λ^{- 1} ϕ (λ); λ \to r]; r \to x] = - x^{- 1} F_{s} [ϕ (λ), λ \to x] .

(3.3d)

Now, let us handle the last two equations in equation (2.14). Using equation (3.1), they can be written as

{\begin{cases} σ_{z z} (r) = 0 = H_{0} [ϕ; λ \to r] = \frac{1}{r} \frac{d}{d r} r H_{1} [λ^{- 1} ϕ (λ); λ \to r], (a < r < \infty) \\ σ_{r z} (r) = 0 = H_{1} [ψ; λ \to r] = - \frac{d}{d r} H_{0} [λ^{- 1} ψ (λ); λ \to r], (a < r < \infty) \end{cases}

(3.4)

{\begin{cases} r σ_{z z} (r) = 0 = \frac{d}{d r} r H_{1} [λ^{- 1} ϕ (λ); λ \to r], (a < r < \infty) \\ - σ_{r z} (r) = 0 = \frac{d}{d r} H_{0} [λ^{- 1} ψ (λ); λ \to r], (a < r < \infty) \end{cases} .

(3.5)

By applying equation (3.3c) to the first relation, and equation (3.3d) to the second relation in equation (3.5) , it follows that

{\begin{cases} A_{2} [r σ_{z z} (r); r \to x] = F_{c} [ϕ (λ), λ \to x] = 0, (a < x < \infty) \\ A_{2} [- σ_{r z} (r); r \to x] = - x^{- 1} F_{s} [ψ (λ), λ \to x] = 0, (a < x < \infty) \end{cases} .

(3.6)

Next, we operate on the first relation in equation (2.14) by taking the operator $(\frac{1}{r} \frac{d}{d r} r)$ to both sides as

(\frac{1}{r} \frac{d}{d r} r) g (r) = (\frac{1}{r} \frac{d}{d r} r) H_{1} [λ^{- 1} (β ϕ + ψ); λ \to r] .

(3.7)

Considering equation (3.1a), one has

(\frac{1}{r} \frac{d}{d r} r) g (r) = (\frac{1}{r} \frac{d}{d r} r) H_{1} [λ^{- 1} (β ϕ + ψ); λ \to r] = H_{0} [(β ϕ + ψ); λ \to r] .

(3.8)

By multiplying by r and applying the Abel operator $A_{1}$ to both sides of equation (3.8), and then considering equation (3.3a), one can show that

\begin{matrix} A_{1} [(\frac{d}{d r} r) g (r); r \to x] = A_{1} [r H_{0} [(β ϕ + ψ); λ \to r]; r \to x] \\ = F_{s} [(β ϕ + ψ); λ \to x] . \end{matrix}

(3.9)

Finally, let us deal with the second relation in equation (2.14). By taking the derivative of both sides with respect to r, multiplying both sides by −1 and then using equation (3.1b), it can be reformulated as

- f^{'} (r) = - \frac{d}{d r} H_{0} [λ^{- 1} (β ψ + ϕ); λ \to r] = H_{1} [(β ψ + ϕ); λ \to r], (0 \leq r < a) .

(3.10)

Making use of the operator $A_{1}$ to both sides in equation (3.10), and (equation (3.3b)), one can obtain

\begin{array}{ll} A_{1} [- f^{'} (r); r \to x] = A_{1} [H_{1} [(β ψ + ϕ); λ \to r]; r \to x] \\ = x^{- 1} {F_{c} [(β ψ + ϕ); λ \to 0] - F_{c} [(β ψ + ϕ); λ \to x]}, \\ (0 \leq x < a) . \end{array}

(3.11)

Now, we collect equations (3.6), (3.9) and (3.11), together as

{\begin{cases} Φ_{c} (x) = 0, (a < x < \infty) \\ Ψ_{s} (x) = 0, (a < x < \infty) \\ β Φ_{s} (x) + Ψ_{s} (x) = G (x), (0 \leq x < a) \\ β Ψ_{c} (x) + Φ_{c} (x) = β Ψ_{c} (0) + Φ_{c} (0) + x F (x), (0 \leq x < a), \end{cases}

(3.12)

where

{\begin{cases} Φ_{c} (x) = F_{c} [ϕ (λ), λ \to x] \\ Φ_{s} (x) = F_{s} [ϕ (λ), λ \to x] \end{cases}, {\begin{cases} Ψ_{c} (x) = F_{c} [ψ (λ), λ \to x] \\ Ψ_{s} (x) = F_{s} [ψ (λ), λ \to x] \end{cases}, {\begin{cases} G (x) = A_{1} [\frac{d}{d r} [r g (r)]; r \to x] = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{1}{\sqrt{x^{2} - r^{2}}} \frac{d}{d r} [r g (r)] d r \\ F (x) = A_{1} [f^{'} (r); r \to x] = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{f^{'} (r)}{\sqrt{x^{2} - r^{2}}} d r \end{cases} .

(3.13)

After a series of transformations, equation (2.14) is now transformed into equation (3.12). In the following two sections, the transformed dual equation (3.12) will be separately treated according to the value of the Dundurs parameter $β$ .

4. The contact problems with $β = 0$

In this section, a degenerate case (uncoupled contact) will be treated.

As $β = 0$ , equation (3.12) degenerates to

{\begin{cases} Ψ_{s} (x) = 0, (a < x < \infty) \\ Ψ_{s} (x) = G (x) = 0, (0 \leq x < a) \\ Φ_{c} (x) = 0, (a < x < \infty) \\ Φ_{c} (x) = Φ_{c} (0) + x F (x), (0 \leq x < a) \end{cases},

(4.1)

because

ϕ (λ) = \sqrt{\frac{2}{π}} \int_{0}^{\infty} Φ_{c} (x) c o s (x λ) d x = \sqrt{\frac{2}{π}} \int_{0}^{a} Φ_{c} (x) c o s (x λ) d x,

(4.2)

and using equations (4.1), (4.2), (B 3) and (B 4) the third relation from equation (2.14) can be reversed as

\begin{matrix} σ_{z z} (r) = \int_{0}^{\infty} ϕ λ J_{0} (λ r) d λ = \sqrt{\frac{2}{π}} \int_{0}^{a} Φ_{c} (x) [\int_{0}^{\infty} c o s (x λ) λ J_{0} (λ r) d λ] d x \\ = \frac{1}{r} \frac{d}{d r} r {\sqrt{\frac{2}{π}} \int_{0}^{a} Φ_{c} (x) d x [\int_{0}^{\infty} c o s (x λ) J_{1} (λ r) d λ]} \\ = - \frac{1}{r} \frac{d}{d r} {\sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x Φ_{c} (x)}{\sqrt{x^{2} - r^{2}}} d x} \\ = - \frac{1}{r} \frac{d}{d r} {\sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x [Φ_{c} (0) + x F (x)]}{\sqrt{x^{2} - r^{2}}} d x} \\ = \sqrt{\frac{2}{π}} \frac{Φ_{c} (0)}{\sqrt{a^{2} - r^{2}}} - \frac{1}{r} \frac{d}{d r} {\sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x^{2} F (x)}{\sqrt{x^{2} - r^{2}}} d x} . \end{matrix}

(4.3)

Generally, we suppose that the gap function $f (r)$ can be described in a general form as

f (r) = - \frac{1}{A} \sum_{n = 0}^{N} f_{n} r^{n}

(4.4)

and then

\begin{matrix} F (x) = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{f^{'} (r)}{\sqrt{x^{2} - r^{2}}} d r = - \frac{1}{A} \sqrt{\frac{2}{π}} \sum_{n = 0}^{N} f_{n} n \int_{0}^{x} \frac{r^{n - 1}}{\sqrt{x^{2} - r^{2}}} d r \\ = - \frac{1}{A} \sqrt{\frac{2}{π}} \sum_{n = 0}^{N} f_{n} \frac{n \sqrt{π} Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} x^{n - 1} . \end{matrix}

(4.5)

The external load force can be calculated as

\begin{array}{ll} P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r = - 2 π \int_{0}^{a} \frac{d}{d r} {\sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x [Φ_{c} (0) + x F (x)]}{\sqrt{x^{2} - r^{2}}} d x} \\ = \sqrt{8 π} \int_{0}^{a} [Φ_{c} (0) + x F (x)] d x . \end{array}

(4.6)

Inserting equation (4.5) into equations (4.3) and (4.6), and considering the adhesion or non-adhesion condition, the problem can be completely solved.

In the following, let us take the Hertzian contact as an example to illustrate the contact traction analysis. For the Hertzian contact profile, the gap function becomes

f (r) = - \frac{1}{A} (Δ - \frac{k}{2} r^{2})

(4.7)

and

F (x) = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{f^{'} (r)}{\sqrt{x^{2} - r^{2}}} d r = \frac{k}{A} \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{r}{\sqrt{x^{2} - r^{2}}} d r = \sqrt{\frac{2}{π}} \frac{k}{A} x .

(4.8)

Substituting equation (4.8) into (4.3) and (4.6), respectively, yields

σ_{z z} (r) = \sqrt{\frac{2}{π}} \frac{1}{\sqrt{a^{2} - r^{2}}} [Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} r^{2}] - \frac{2 k}{π A} \sqrt{a^{2} - r^{2}}

(4.9)

and

\begin{array}{ll} P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r = \sqrt{8 π} \int_{0}^{a} [Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} x^{2}] d x \\ = \sqrt{8 π} {Φ_{c} (0) a + \sqrt{\frac{2}{π}} \frac{k}{A} \frac{a^{3}}{3}} . \end{array}

(4.10)

The parameter $Φ_{c} (0)$ will be specified according to the adhesion or non-adhesion condition as below.

(a) Case 1: regular contact without surface adhesion effect

For this boundary condition, the traction distribution along the contact rim should vanish as

\underset{r \to a}{l i m} σ_{z z} (r) = 0,

(4.11)

from equation (4.9), which requires

Φ_{c} (0) = - \sqrt{\frac{2}{π}} \frac{k}{A} a^{2} .

(4.12)

Thus

σ_{z z} (r) = - \frac{4}{π} \frac{k}{A} \sqrt{a^{2} - r^{2}},

(4.13)

P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r = \sqrt{8 π} \int_{0}^{a} [Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} x^{2}] d x = - \frac{8}{3} \frac{k}{A} a^{3} .

(4.14)

This result is completely consistent with the one solved by other approaches (see e.g. [10,42]).

(b) Case 2: contact with surface adhesion effect

For this boundary condition, it is always supposed that the traction distribution is singular along the contact rim; as that of a crack-tip. Being consistent with the linear fracture mechanics concept, the stress intensity factor at the rim of the contact can be analogically defined as

\begin{array}{ll} K_{I} & = \underset{r \to a}{l i m} \sqrt{2 π (a - r)} σ_{z z} (r) \\ = \underset{r \to a}{l i m} \sqrt{2 π (a - r)} {\sqrt{\frac{2}{π}} \frac{1}{\sqrt{a^{2} - r^{2}}} [Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} r^{2}] - \frac{2 k}{π A} \sqrt{a^{2} - r^{2}}} \\ = \frac{\sqrt{2}}{\sqrt{a}} [Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} a^{2}] . \end{array}

(4.15)

We note that the interface contact rim is always at a critical energy balance state when in non-slipping adhesive contact, as in the JKR model, regardless of whether the interface is closing or peeling. By employing the Griffith energy balance at the interfacial crack tip, the critical interface energy release rate at the adhesive contact rim can be written in terms of the stress intensity factor in equation (4.15) as [41,44–46]

γ = \frac{A}{4 c o s h^{2} π ε} {| K |}^{2} = \frac{A (1 - β^{2})}{4} {| K |}^{2} = \frac{A (1 - β^{2})}{4} K_{I}^{2},

(4.16)

where $γ$ is the critical interface energy release rate, which is a material parameter when the pair of contact materials is given; parameters $A$ and $β$ are given in equation (2.15) and $K$ is the complex stress intensity factor at an interfacial crack-tip. Thus one can obtain

γ = \frac{A}{2 a} {[Φ_{c} (0) + \sqrt{\frac{2}{π}} \frac{k}{A} a^{2}]}^{2} .

(4.17)

Combining equations (4.10), (4.17) yields

\frac{2 γ}{A} a = {(\frac{P_{0}}{a \sqrt{8 π}} + \frac{2}{3} \sqrt{\frac{2}{π}} \frac{k}{A} a^{2})}^{2} .

(4.18)

Thereby, the relation between the external load $P_{0}$ and the contact radius a can be expressed as

P_{0} = 4 \sqrt{\frac{γ π}{A}} a^{\frac{3}{2}} - \frac{8}{3} \frac{k}{A} a^{3} .

(4.19)

The pull-off can be determined by applying the pull-off condition $\partial P_{0} / \partial a = 0$ as

\frac{\partial P_{0}}{\partial a} = \frac{\partial}{\partial a} (4 \sqrt{\frac{γ π}{A}} a^{\frac{3}{2}} - \frac{8}{3} \frac{k}{A} a^{3}) = 6 \sqrt{\frac{γ π}{A}} a^{\frac{1}{2}} - 8 \frac{k}{A} a^{2} = 0,

(4.20)

which gives

a_{c} = {(\frac{9}{16} \frac{γ π A}{k^{2}})}^{\frac{1}{3}} .

(4.21)

Inserting it into equation (4.19), one has

P_{c} = \frac{3}{2} \frac{γ π}{k} .

(4.22)

This solution is completely consistent with the result of the JKR model [27].

It can be proved that the solutions developed for this decoupled case $(β = 0)$ are consistently applicable to the axisymmetric frictionless contact problems. The contact deformation can be easily derived for arbitrary polynomial indenter profiles. The preceding analysis procedure for non-adhesion contact and adhesion contact also guides us for treating the coupled case ( $β \neq 0$ ) in the next section.

5. The contact problems with $β \neq 0$

In this section, we develop an approach to solve dual integral equation (3.12). This case is the general one and also the main challenge in this study. Fortunately, it has been demonstrated that the dual integral equations can be transformed into a singular integral equation by virtue of the properties of the Hilbert transform [43] and, there exist a couple of techniques in the literature used to solve the singular integral equation. These provide a good basis for us to tackle equations equations (3.12).

(a) Reduction to a singular integral equation and its formal solution

The solution to the dual integral equation (3.12) may be expressed in the form

Φ_{c} (x) = r (x) H (a - x), Ψ_{s} (x) = s (x) H (a - x),

(5.1)

where $H (*)$ is the Heaviside unit function, $r (x)$ is an even extended function to $[- a, a]$ and $s (x)$ is an odd extended function to $[- a, a]$ . Equation (5.1) automatically satisfies the first two relations in equations (3.12). By using the Hilbert transform results (see appendix A), one finds

Φ_{s} (x) = - \frac{1}{π} \int_{- a}^{a} \frac{r (t) d t}{t - x}, Ψ_{c} (x) = \frac{1}{π} \int_{- a}^{a} \frac{s (t) d t}{t - x} .

(5.2)

The last two relations in equations (3.12) can be cast as

{\begin{cases} s (x) - \frac{β}{π} \int_{- a}^{a} \frac{r (t) d t}{t - x} = G (x) \\ β \frac{1}{π} \int_{- a}^{a} \frac{s (t) d t}{t - x} + r (x) = β Ψ_{c} (0) + Φ_{c} (0) + x F (x) = C_{r} + x F (x) \end{cases}, (0 \leq x < a),

(5.3)

in which $C_{r} = β Ψ_{c} (0) + Φ_{c} (0)$ is a real number.

Now, by letting

ω (x) = s (x) + i r (x),

(5.4)

equation (5.3) can be assembled into a Fredholm integral of the second kind as

ω (x) - \frac{β}{i π} \int_{- a}^{a} \frac{ω (t) d t}{t - x} = G (x) + i [C_{r} + x F (x)] = V (x), (- a \leq x < a),

(5.5)

where $F (x)$ and $G (x)$ are also extended to as odd functions to $(- a \leq x < a)$ , because of equation (5.3). If the function $ω (x) = s (x) + i r (x)$ is inverted using equation (5.5), we see that the coupled contact problem is solved.

Now, introducing the complex variable $ζ = x + z i$ and

M (ζ) = \frac{1}{2 π i} \int_{- a}^{a} \frac{ω (t) d t}{t - ζ}

(5.6)

In terms of the Plemelj formulae [47], one can derive

{\begin{cases} M (x^{+}) = \frac{1}{2 π i} \int_{- a}^{a} \frac{ω (t) d t}{t - x} + \frac{ω (x)}{2} \\ M (x^{-}) = \frac{1}{2 π i} \int_{- a}^{a} \frac{ω (t) d t}{t - x} - \frac{ω (x)}{2} \end{cases} .

(5.7)

Equation (5.5) can be cast as

M (x^{+}) - \frac{(1 + β)}{(1 - β)} M (x^{-}) = \frac{V (x)}{(1 - β)}, (- a \leq x < a) .

(5.8)

To solve equation (5.8), a multivalued function is introduced as

X (ζ) = {(ζ - a)}^{- δ} {(ζ + a)}^{δ} .

(5.9)

Thus

\frac{(1 + β)}{(1 - β)} = \frac{X (x^{+})}{X (x^{-})} = e^{- i 2 δ π}, δ = i \frac{1}{2 π} l n \frac{(1 + β)}{(1 - β)} = i ε,

(5.10)

where

\begin{array}{l} X (x^{+}) = e^{ε π} {(a - x)}^{- i ε} {(x + a)}^{i ε} = \sqrt{\frac{1 + β}{1 - β}} {(\frac{x + a}{a - x})}^{i ε}, \\ X (x^{-}) = e^{- ε π} {(a - x)}^{- i ε} {(x + a)}^{i ε} = \sqrt{\frac{1 - β}{1 + β}} {(\frac{x + a}{a - x})}^{i ε}, \end{array}

(5.11)

and ‘+’ represents $ζ \to x + i 0^{+} = x^{+}$ from the upper side, while ‘−’ represents $ζ \to x + i 0^{-} = x^{-}$ from the lower side of $ζ \in (- a, a)$ .

Hence equation (5.8) can be recast as

\frac{M (x^{+})}{X (x^{+})} - \frac{M (x^{-})}{X (x^{-})} = \frac{1}{(1 - β)} \frac{V (x)}{X (x^{+})}, (- a < x < a) .

(5.12)

Its solution is

M (ζ) = \frac{X (ζ)}{(1 - β)} \frac{1}{2 π i} \int_{- a}^{a} \frac{V (t)}{X (t^{+})} \frac{1}{t - ζ} d t + P (ζ) X (ζ),

(5.13a)

where $P (ζ)$ is a polynomial function. It is noted that the function $X (ζ)$ has a pure imaginary exponent, and the function $M (ζ)$ is bounded at the end-points and vanishes as $ζ \to \infty$ , so that the polynomial function $P (ζ)$ is taken to be zero. It follows that

M (ζ) = \frac{X (ζ)}{(1 - β)} \frac{1}{2 π i} \int_{- a}^{a} \frac{V (t)}{X (t^{+})} \frac{1}{t - ζ} d t .

(5.13b)

As noted above, the arbitrary gap functions can be put in polynomial form

f (r) = - \frac{1}{A} \sum_{n = 0}^{N} f_{n} r^{n}, g (r) = - \frac{1}{A} \sum_{n = 1}^{N} g_{n} r^{n},

(5.14)

where the function $f (r)$ is pre-given, and the relative radial displacement function $g (r)$ is to be specified, in which the coefficient $g_{n}$ must be independent of the contact radius a to satisfy the boundary condition of equation (2.21).

Inserting equation (5.14) into equation (3.13) leads to

\begin{array}{ll} F (x) = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{f^{'} (r)}{\sqrt{x^{2} - r^{2}}} d r = - \frac{1}{A} \sqrt{\frac{2}{π}} \sum_{n = 1}^{N} f_{n} n \int_{0}^{x} \frac{r^{n - 1}}{\sqrt{x^{2} - r^{2}}} d r \\ = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} f_{n} \frac{n Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} x^{n - 1}, \\ G (x) = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{1}{\sqrt{x^{2} - r^{2}}} \frac{d}{d r} [r g (r)] d r = - \frac{1}{A} \sqrt{\frac{2}{π}} \sum_{n = 1}^{N} g_{n} (n + 1) \int_{0}^{x} \frac{r^{n}}{\sqrt{x^{2} - r^{2}}} d r \\ = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} x^{n} . \end{array}

(5.15)

Because $F (x)$ and $G (x)$ are extended as odd functions to $(- a \leq x < a)$ in equation (5.5), so equation (5.15) should be modified to become

{\begin{cases} F (x) = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} f_{n} \frac{n Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} s g n (x^{n}) x^{n - 1} \\ G (x) = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} s g n (x^{n + 1}) x^{n} \end{cases},

(5.16)

and thus

\begin{array}{ll} V (x) = G (x) + i [C_{r} + x F (x)] \\ = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} [g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} s g n (x^{n + 1}) + i f_{n} \frac{n Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} s g n (x^{n})] x^{n} + i C_{r} . \end{array}

(5.17)

Substituting equation (5.17) into equation (5.13b) gives us the formal solution of equation (5.6).

(b) Traction along the contact surface

From equation (5.13) one may deduce

\begin{array}{l} M (x^{+}) = \frac{X (x^{+})}{(1 - β)} {\frac{1}{2 π i} \int_{- a}^{a} \frac{V (t)}{X (t^{+})} \frac{1}{t - x} d t + \frac{1}{2} \frac{V (t)}{X (t^{+})}}, \\ M (x^{-}) = \frac{X (x^{-})}{(1 - β)} {\frac{1}{2 π i} \int_{- a}^{a} \frac{V (t)}{X (t^{+})} \frac{1}{t - x} d t - \frac{1}{2} \frac{V (t)}{X (t^{+})}} . \end{array}

(5.18)

Then equation (5.4) can be written as

\begin{array}{ll} ω (x) = s (x) + i r (x) = M (x^{+}) - M (x^{-}) \\ = \frac{2 β}{(1 - β^{2})} \frac{X (x^{+})}{2 π i} \int_{- a}^{a} \frac{V (t)}{X (t^{+})} \frac{1}{t - x} d t + \frac{V (x)}{(1 - β^{2})} \\ = \frac{β {(a - x)}^{- i ε} {(x + a)}^{i ε}}{i (1 - β^{2})} \frac{1}{π} \int_{- a}^{a} \frac{{(a - t)}^{i ε} {(t + a)}^{- i ε} V (t)}{t - x} d t + \frac{V (x)}{(1 - β^{2})} . \end{array}

(5.19)

Now, by expanding the function $V (t)$ in a Jacobi polynomial series, as

V (a T) = \sum_{n = 0}^{\infty} d_{n} P_{n}^{(i ε, - i ε)} (T), (- 1 \leq T \leq 1),

(5.20)

equation (5.19) can be further simplified in terms of equation (C 11), as

\begin{array}{ll} ω (x) = s (x) + i r (x) \\ = \frac{β {(a - x)}^{- i ε} {(x + a)}^{i ε}}{i (1 - β^{2})} \frac{1}{π} \int_{- 1}^{1} \frac{{(1 - T)}^{i ε} {(1 + T)}^{- i ε} V (a T)}{T - \frac{x}{a}} d T + \frac{V (x)}{(1 - β^{2})} \\ = \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{1 - β^{2}}} \sum_{n = 0}^{\infty} d_{n} P_{n}^{(- i ε, i ε)} (\frac{x}{a}) \\ = \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{1 - β^{2}}} V_{2} (x), \end{array}

(5.21)

where

{\begin{cases} V_{2} (x) = \sum_{n = 0}^{\infty} d_{n} P_{n}^{(- i ε, i ε)} (\frac{x}{a}) \\ d_{n} = \frac{(2 n + 1) Γ (n + 1) Γ (n + 1)}{2 Γ (n + i ε + 1) Γ (n - i ε + 1)} \int_{- 1}^{1} V (a T) {(1 - T)}^{i ε} {(1 + T)}^{- i ε} P_{n}^{(i ε, - i ε)} (T) d T \end{cases} .

(5.22)

From equation (5.1) and the inverse of equation (3.13), one finds

{\begin{cases} ϕ (λ) = \sqrt{\frac{2}{π}} \int_{0}^{\infty} r (x) H (a - x) c o s (x λ) d x = \sqrt{\frac{2}{π}} \int_{0}^{a} r (x) c o s (x λ) d x \\ ψ (λ) = \sqrt{\frac{2}{π}} \int_{0}^{\infty} s (x) H (a - x) s i n (x λ) d x = \sqrt{\frac{2}{π}} \int_{0}^{a} s (x) s i n (x λ) d x \end{cases} .

(5.23)

and finally from equation (3.4), taking $(0 \leq r < a)$ , and considering equation (5.23) and equations (B 1), (B 2) the contact tractions are found to be

\begin{array}{ll} σ_{z z} (r) = \int_{0}^{\infty} ϕ λ J_{0} (λ r) d λ = \sqrt{\frac{2}{π}} \int_{0}^{a} r (x) d x \int_{0}^{\infty} c o s (x λ) λ J_{0} (λ r) d λ \\ = \frac{1}{r} \frac{d}{d r} r {\sqrt{\frac{2}{π}} \int_{0}^{a} r (x) d x [\int_{0}^{\infty} c o s (x λ) J_{1} (λ r) d λ]} \\ = - \frac{1}{r} \frac{d}{d r} {\sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x r (x)}{\sqrt{x^{2} - r^{2}}} d x}, \end{array}

(5.24)

\begin{array}{ll} σ_{r z} (r) = \int_{0}^{\infty} ψ λ J_{1} (λ r) d λ = \sqrt{\frac{2}{π}} \int_{0}^{a} s (x) d x \int_{0}^{\infty} s i n (x λ) λ J_{1} (λ r) d λ \\ = - \frac{d}{d r} \sqrt{\frac{2}{π}} \int_{0}^{a} s (x) d x \int_{0}^{\infty} s i n (x λ) J_{0} (λ r) d λ \\ = - \frac{d}{d r} \sqrt{\frac{2}{π}} \int_{r}^{a} \frac{s (x)}{\sqrt{x^{2} - r^{2}}} d x . \end{array}

(5.25)

Now, by assembling equations (5.24), (5.25) in the form

\begin{matrix} σ_{z z} (r) - i σ_{r z} (r) = - \frac{1}{r} \frac{d}{d r} \sqrt{\frac{2}{π}} \int_{r}^{a} \frac{x r (x)}{\sqrt{x^{2} - r^{2}}} d x + \frac{d}{d r} \sqrt{\frac{2}{π}} \int_{r}^{a} \frac{i s (x)}{\sqrt{x^{2} - r^{2}}} d x \\ = - \sqrt{\frac{2}{π}} I m [\int_{r}^{a} \frac{ω (x)}{(x + r) \sqrt{x^{2} - r^{2}}} d x] + i \sqrt{\frac{2}{π}} \frac{d}{d r} \int_{r}^{a} \frac{ω (x)}{\sqrt{x^{2} - r^{2}}} d x \\ = - \sqrt{\frac{2}{π}} I m [\int_{r}^{a} \frac{ω (x)}{\sqrt{x - r} {(x + r)}^{\frac{3}{2}}} d x] + i \sqrt{\frac{2}{π}} \frac{d}{d r} \int_{r}^{a} \frac{ω (x)}{\sqrt{x - r} \sqrt{x + r}} d x \end{matrix}

(5.26)

and replacing $ω (x)$ with equation (5.21), one finds

\begin{array}{ll} σ_{z z} (r) - i σ_{r z} (r) = - \frac{1}{\sqrt{1 - β^{2}}} \sqrt{\frac{2}{π}} I m [\int_{r}^{a} \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{x - r} {(x + r)}^{\frac{3}{2}}} V_{2} (x) d x] \\ + i \frac{1}{\sqrt{1 - β^{2}}} \sqrt{\frac{2}{π}} \frac{d}{d r} \int_{r}^{a} \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{x - r} \sqrt{x + r}} V_{2} (x) d x . \end{array}

(5.27)

Similarly, the external load force is found to be

\begin{matrix} P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r = \sqrt{8 π} \int_{0}^{a} r (x) d x = \sqrt{8 π} I m [\int_{0}^{a} ω (x) d x] \\ = \frac{\sqrt{8 π}}{\sqrt{1 - β^{2}}} I m [\int_{0}^{a} {(a - x)}^{- i ε} {(x + a)}^{i ε} V_{2} (x) d x] . \end{matrix}

(5.28)

Next, by considering the boundary traction behaviour around the contact rim involved in equation (5.27), the coefficients $g_{n}$ in equation (5.14) can be specified, as we did in §4. Similarly in accordance with either the non-adhesion or adhesion conditions, this coupled problem can be also classified into two cases as follows.

(c) Case 1: regular contact without surface adhesion effect

For this boundary condition the traction distribution vanishes around the contact rim as

\underset{r \to a}{l i m} [σ_{z z} (r) - i σ_{r z} (r)] = 0.

(5.29)

Inserting equation (5.27) into equation (5.29), and after a lengthy but straightforward manipulation, we obtain the corresponding constraint condition as

\sum_{n = 0}^{\infty} d_{n} P_{n}^{(- i ε, i ε)} (1) = V_{2} (a) = 0.

(5.30)

In consideration of equation (5.22) and an identity

P_{n}^{(- i ε, i ε)} (1) = (\begin{matrix} n - i ε \\ n \end{matrix}) = \frac{Γ (n - i ε + 1)}{Γ (n + 1) Γ (1 - i ε)},

(5.31)

equation (5.30) can be rewritten as

\int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T = 0,

(5.32)

with

h (T) = \sum_{n = 0}^{\infty} \frac{(2 n + 1) Γ (n + 1)}{Γ (n + i ε + 1)} P_{n}^{(i ε, - i ε)} (T) .

(5.33)

Subsequently, in view of equation (5.17), equation (5.32) can be expressed as

\int_{- 1}^{1} [G (a T) + i F (a T) a T] {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T + i C_{r} 2 Γ (1 - i ε) = 0

(5.34)

i C_{r} - \frac{\sqrt{2}}{2 A Γ (1 - i ε)} \sum_{n = 1}^{N} [\begin{matrix} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} \int_{- 1}^{1} s g n (T^{n + 1}) T^{n} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \\ + i f_{n} \frac{n Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} \int_{- 1}^{1} s g n (T^{n}) T^{n} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \end{matrix}] a^{n} = 0.

(5.35)

Because the coefficients $g_{n}$ are independent of the value of the contact radius a, $g_{n}$ and $C_{r}$ are real numbers which can be determined by constructing equations for the real and imaginary parts of equation (5.35) and comparing the order of $a^{n}$ . In addition, using the external load condition (equation (5.28)), the contact radius a can be specified. This point will be demonstrated further with an example in §6.

(d) Case 2: contact with surface adhesion effect

For this boundary condition, the traction distribution at the contact rim is similar to the one at an interfacial-crack-tip. To characterize the traction intensity along the contact rim, according to the conventional definition, the stress intensity factor is defined as

K = K_{I} - i K_{I I} = \underset{r \to a}{l i m} \sqrt{2 π} {(a - r)}^{\frac{1}{2} + i ε} [σ_{z z} (r) - i σ_{r z} (r)] .

(5.36)

By inserting equation (5.27) into equation (5.35), again, after a lengthy manipulation, one obtains

\begin{array}{ll} K = K_{I} - i K_{I I} = \underset{r \to a}{l i m} \sqrt{2 π} {(a - r)}^{\frac{1}{2} + i ε} [σ_{z z} (r) - i σ_{r z} (r)] \\ = - i \frac{2 \sqrt{π}}{\sqrt{1 - β^{2}}} \frac{Γ (1 - i ε)}{Γ (\frac{1}{2} - i ε)} {(2 a)}^{- \frac{1}{2} + i ε} V_{2} (a) \\ = - i \frac{2 \sqrt{π}}{\sqrt{1 - β^{2}}} \frac{Γ (1 - i ε)}{Γ (\frac{1}{2} - i ε)} {(2 a)}^{- \frac{1}{2} + i ε} \sum_{n = 0}^{\infty} d_{n} P_{n}^{(- i ε, i ε)} (1) \\ = - i \frac{\sqrt{π}}{\sqrt{1 - β^{2}}} \frac{{(2 a)}^{- \frac{1}{2} + i ε}}{Γ (\frac{1}{2} - i ε)} \int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \end{array} .

(5.37)

In this manipulation process, the following identity is used

\int_{0}^{1} \frac{{(1 - t)}^{- i ε}}{\sqrt{t}} d t = \frac{\sqrt{π} Γ (1 - i ε)}{Γ (\frac{3}{2} - i ε)} .

(5.38)

Inserting equation (5.37) into equation (4.16) leads to

\begin{array}{ll} γ = \frac{A}{4 c o s h^{2} π ε} | K |^{2} = \frac{A (1 - β^{2})}{4} | K_{c} |^{2} \\ = \frac{A c o s h π ε}{8 a} [\int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T] [\int_{- 1}^{1} \bar{V} (a T) {(\frac{1 - T}{1 + T})}^{- i ε} \bar{h} (T) d T], \end{array}

(5.39)

where

V (a T) = i C_{r} - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} [g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} s g n (T^{n + 1}) + i f_{n} \frac{n Γ (\frac{n}{2})}{2 Γ (\frac{1 + n}{2})} s g n (T^{n})] T^{n} a^{n} .

(5.40)

Furthermore, by forming equations from the real and imaginary parts of equation (5.37) and considering equation (5.39), then comparing the order of arbitrary value of $a^{n}$ , the real coefficients $g_{n}$ can be finally determined. The value of contact radius a can be determined from equation (5.28), which now can be expressed as

\begin{array}{ll} P_{0} = \frac{\sqrt{8 π}}{\sqrt{1 - β^{2}}} I m [\int_{0}^{a} {(a - x)}^{- i ε} {(x + a)}^{i ε} V_{2} (x) d x] \\ = \frac{a \sqrt{8 π}}{\sqrt{1 - β^{2}}} I m [\sum_{n = 0}^{\infty} d_{n} \int_{0}^{1} {(1 - X)}^{- i ε} {(X + 1)}^{i ε} P_{n}^{(- i ε, i ε)} (X) d X] \\ = \sqrt{\frac{π}{2}} \frac{a}{\sqrt{1 - β^{2}}} I m [\int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} \sum_{n = 1}^{\infty} \frac{(2 n + 1) Γ (n + 1) Γ (n + 1)}{Γ (n + 1 + i ε) Γ (n + 1 - i ε)} \frac{1}{n} P_{n - 1}^{(1 - i ε, 1 + i ε)} (0) P_{n}^{(i ε, - i ε)} (T) d T] . \end{array}

(5.41)

Note that, in this analysis, as we mentioned above, one of the main aims is to identify the contact relative radial displacement. Once it has been obtained, the other quantities can be derived or evaluated by following a previous procedure. In the next section, a typical contact will be studied to elucidate how to evaluate $g_{n}$ as well as the parameter $C_{r}$ in detail.

6. Relative contact radial displacement in a non-slipping conical contact

In this section, to demonstrate how to identify the relative contact radial displacement, without loss of generality, a non-slipping conical contact is studied in accordance with the non-adhesion and adhesion conditions, respectively. Note that the non-slipping conical contact between a rigid indenter and an elastic space was studied by Spence [1] and Borodich et al. [48]. Its corresponding frictionless case was studied by Shtaerman [49] and Chaudhri [50].

(a) Non-adhesion

If there is no adhesion within the non-slipping conical contact, it conforms with equation (5.29), and equation (5.16) can be adapted to give

F (x) = - \frac{1}{A} \sqrt{\frac{π}{2}} f_{1} s g n (x), G (x) = - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} s g n (x^{n + 1}) x^{n},

(6.1)

where $f_{1}$ is known, and $g_{n}$ is to be identified. For this case, $g_{1}$ can be expressed in terms of equation (5.35) as

\begin{matrix} g_{1} = \frac{C_{r}}{a} \frac{i \sqrt{π} A Γ (1 - i ε)}{\sqrt{2} \int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T} - f_{1} \frac{π}{4} \frac{i \int_{- 1}^{1} s g n (T) T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T}{\int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T} \\ - \frac{\sqrt{π}}{2} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} \frac{\int_{- 1}^{1} s g n (T^{n + 1}) T^{n} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T}{\int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T} a^{n - 1} . \end{matrix}

(6.2)

For convenience, we denote

{\begin{cases} c_{1} + i c_{2} = \int_{- 1}^{1} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T, c_{3} + i c_{4} = \int_{- 1}^{1} s g n (T) T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \\ η_{1} + i υ_{1} = \int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T, η_{n} + i υ_{n} = \int_{- 1}^{1} s g n (T^{n + 1}) T^{n} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \\ Γ (1 - i ε) = b_{1} + i b_{2}, h (T) = \sum_{n = 0}^{\infty} \frac{(2 n + 1) Γ (n + 1)}{Γ (n + i ε + 1)} P_{n}^{(i ε, - i ε)} (T) \end{cases},

(6.3)

where $b_{1}, b_{2}$ , $c_{i} (i = 1, 2, 3, 4)$ and $η_{n}, υ_{n} (n = 1... N)$ are all known real numbers, for example, some values can be approximated as follows:

{\begin{cases} c_{1} + i c_{2} = \int_{- 1}^{1} {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \approx 2 + 1.15443 i ε + O (ε^{2}) \\ η_{1} + i υ_{1} = \int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \approx 2 - 2.84556 i ε + O (ε^{2}) \\ c_{3} + i c_{4} = \int_{- 1}^{1} s g n (T) T {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \approx 2 - 1.54359 i ε + O (ε^{2}) \end{cases} .

(6.4)

These substitutions allow equation (6.2) to be put in simple form as

g_{1} = \frac{i \sqrt{π} A C_{r}}{\sqrt{2} a} \frac{(b_{1} + i b_{2})}{η_{1} + i υ_{1}} - f_{1} \frac{π}{4} \frac{i (c_{3} + i c_{4})}{η_{1} + i υ_{1}} - \frac{\sqrt{π}}{2} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} \frac{(η_{n} + i υ_{n})}{η_{1} + i υ_{1}} a^{n - 1} .

(6.5)

Splitting equation (6.5) into the real and imaginary parts gives rise to

{\begin{cases} η_{1} g_{1} = - \frac{\sqrt{π} A C_{r}}{\sqrt{2} a} b_{2} + f_{1} \frac{π}{4} c_{4} - \frac{\sqrt{π}}{2} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} η_{n} a^{n - 1} \\ υ_{1} g_{1} = \frac{\sqrt{π} A C_{r}}{\sqrt{2} a} b_{1} - f_{1} \frac{π}{4} c_{3} - \frac{\sqrt{π}}{2} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} υ_{n} a^{n - 1} \end{cases} .

(6.6)

By solving the parameter $C_{r}$ from the second relation in equation (6.6) and then substituting it into the first one, the coefficient $g_{1}$ can be solved as

g_{1} = \frac{π}{4} \frac{(b_{1} c_{4} - b_{2} c_{3})}{(b_{2} υ_{1} + b_{1} η_{1})} f_{1} - \frac{\sqrt{π}}{2} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} \frac{(b_{1} η_{n} + b_{2} υ_{n})}{(b_{2} υ_{1} + b_{1} η_{1})} a^{n - 1} .

(6.7)

The quantities $f_{1}$ and $g_{n}$ are independent of the contact radius $a$ but equation (6.7) must be true for all values of a, and so

g_{n} = 0, (n \geq 2) .

(6.8)

Thereafter, from equation (6.6), one can obtain

g_{1} = \frac{π}{4} \frac{(b_{1} c_{4} - b_{2} c_{3})}{(b_{2} υ_{1} + b_{1} η_{1})} f_{1}, C_{r} = \sqrt{\frac{π}{8}} \frac{a}{A} \frac{(c_{4} υ_{1} + c_{3} η_{1})}{(b_{2} υ_{1} + b_{1} η_{1})} f_{1} .

(6.9)

We emphasize, here, that our approach is able to provide an explicit solution of the relative radial displacement for conical contact in contrast with the Spence’s approach [1]. Now, the basic parameters for the non-slipping conical contact without adhesion are at hand, and thus the contact problem can be solved accordingly, by following the procedure listed in §5.

We further add, that for other non-slipping contacts with a polynomial profile, for example, $f (r) = - f_{n} r^{n} / A$ , the corresponding relative radial displacement $g (r) = - g_{n} r^{n} / A$ can be fixed in a similar way. More importantly, the above manipulation actually provides a rigorous proof that the contact relative radial displacement $g (r)$ must be a function of the radius having the same order as the gap function $f (r)$ . This was taken with an inspectional analysis in the Spence’s self-similarity approach [1].

(b) With adhesion

If the surface adhesion effect along the interface of the non-slipping conical contact is considered, the traction around the contact rim would be singular, as described previously in §5. The relative radial displacement may be determined by combining the stress intensity factor (equation (5.36)) and the energy release rate (equation (5.38)).

Similarly, by adopting equation (6.1), equation (5.16) becomes

V (a T) = i C_{r} - i \frac{\sqrt{2}}{A} f_{1} \frac{\sqrt{π}}{2} s g n (T) T a - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} s g n (T^{n + 1}) T^{n} a^{n} .

(6.10)

Thus

\int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T = [\begin{matrix} i C_{r} (c_{1} + i c_{2}) - i \frac{f_{1}}{A} \sqrt{\frac{π}{2}} (c_{3} + i c_{4}) a \\ - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} a^{n} (η_{n} + i υ_{n}) \end{matrix}],

(6.11)

where $c_{i} (i = 1,2, 3,4)$ and $η_{n}, υ_{n} (n = 1 . . . N)$ are given in equation (6.3).

Inserting equation (6.11) into equation (5.36) gives

\begin{matrix} K_{c} = K_{I} - i K_{I I} = - i \frac{\sqrt{π}}{\sqrt{1 - β^{2}}} \frac{{(2 a)}^{- \frac{1}{2} + i ε}}{Γ (\frac{1}{2} - i ε)} \int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T \\ = - i \frac{\sqrt{π}}{\sqrt{1 - β^{2}}} \frac{{(2 a)}^{- \frac{1}{2} + i ε}}{Γ (\frac{1}{2} - i ε)} [\begin{matrix} i C_{r} (c_{1} + i c_{2}) - i \frac{f_{1}}{A} \sqrt{\frac{π}{2}} (c_{3} + i c_{4}) a \\ - \frac{\sqrt{2}}{A} \sum_{n = 1}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} a^{n} (η_{n} + i υ_{n}) \end{matrix}], \end{matrix}

(6.12)

which can be rearranged as

\begin{array}{ll} \frac{C_{r}}{a} (c_{1} + i c_{2}) - \frac{f_{1}}{A} \sqrt{\frac{π}{2}} (c_{3} + i c_{4}) + \frac{\sqrt{2}}{A} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} a^{n - 1} (i η_{n} - υ_{n}) \\ + \frac{1}{A} \sqrt{\frac{8}{π}} (i η_{1} - υ_{1}) g_{1} - \frac{\sqrt{2} \sqrt{1 - β^{2}}}{\sqrt{a π}} Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{I} - i K_{I I}) = 0. \end{array}

(6.13)

Note that, according to the principles of interfacial fracture mechanics, $K_{I}$ , $K_{I I}$ may be functions of a but they should satisfy equation (5.38). Separating the real and imaginary parts of equation (6.13) gives

\begin{array}{ll} \frac{C_{r}}{a} c_{1} - \frac{f_{1}}{A} \sqrt{\frac{π}{2}} c_{3} - \frac{\sqrt{2}}{A} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} a^{n - 1} υ_{n} \\ - \frac{1}{A} \sqrt{\frac{8}{π}} υ_{1} g_{1} - \frac{\sqrt{2} \sqrt{1 - β^{2}}}{\sqrt{a π}} R e [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{I} - i K_{I I})] = 0, \end{array}

(6.14a)

\begin{array}{ll} \frac{C_{r}}{a} c_{2} - \frac{f_{1}}{A} \sqrt{\frac{π}{2}} c_{4} + \frac{\sqrt{2}}{A} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} a^{n - 1} η_{n} \\ + \frac{1}{A} \sqrt{\frac{8}{π}} η_{1} g_{1} - \frac{\sqrt{2} \sqrt{1 - β^{2}}}{\sqrt{a π}} I m [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{I} - i K_{I I})] = 0. \end{array}

(6.14b)

Combining equations (6.14a) and (6.14b) we have

\begin{array}{ll} \frac{1}{A} \sqrt{\frac{π}{2}} {f_{1} (\frac{c_{2} c_{3}}{c_{1}} - c_{4}) + \frac{4}{π} (\frac{c_{2} υ_{1}}{c_{1}} + η_{1}) g_{1}} + \frac{\sqrt{2}}{A} \sum_{n = 2}^{N} g_{n} \frac{(n + 1) Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} (\frac{c_{2} υ_{n}}{c_{1}} + η_{n}) a^{n - 1} \\ + \frac{\sqrt{2} \sqrt{1 - β^{2}}}{\sqrt{a π}} {\frac{c_{2}}{c_{1}} R e [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{1} + i K_{2})] - I m [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{I} - i K_{I I})]} = 0. \end{array}

(6.15)

Because $|K_{I} - i K_{I I}| = |K_{c}| = c o n s t$ via equation (5.38), and a can be an arbitrary value in equation (6.15), each group is independent of the others, which therefore requires

{\begin{cases} f_{1} (\frac{c_{2} c_{3}}{c_{1}} - c_{4}) + \frac{4}{π} (\frac{c_{2} υ_{1}}{c_{1}} + η_{1}) g_{1} = 0 \\ g_{n} = 0, (n \geq 2) \\ \frac{c_{2}}{c_{1}} R e [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{1} + i K_{2})] - I m [Γ (\frac{1}{2} - i ε) {(2 a)}^{- i ε} (K_{I} - i K_{I I})] = 0 \end{cases} .

(6.16)

Thus, as the surface adhesion effect is taken into account, the contact relative radial displacement of the conical contact is

{\begin{cases} g_{1} = \frac{π}{4} (\frac{c_{4} c_{1} - c_{2} c_{3}}{c_{2} υ_{1} + η_{1} c_{1}}) f_{1} \\ g_{n} = 0, (n \geq 2) \\ K_{I I} = - K_{I} \frac{{c_{2} Γ_{1} - c_{1} Γ_{2} + (Γ_{2} c_{2} + Γ_{1} c_{1}) t a n [ε l n (2 a)]}}{{c_{2} Γ_{2} + c_{1} Γ_{1} - (Γ_{1} c_{2} - Γ_{2} c_{1}) t a n [ε l n (2 a)]}} \end{cases},

(6.17)

with $Γ (1 / 2 - i ε) = (Γ_{1} + i Γ_{2}) \approx \sqrt{π} + 3.48032 i ε .$

Also, equation (6.10) is updated as

V (a T) = (i C_{r} - i \frac{1}{A} f_{1} \sqrt{\frac{π}{2}} s g n (T) T a - \frac{1}{A} g_{1} \sqrt{\frac{8}{π}} T a)

(6.18)

and equation (6.11) becomes

\int_{- 1}^{1} V (a T) {(\frac{1 - T}{1 + T})}^{i ε} h (T) d T = [i C_{r} (c_{1} + i c_{2}) - i \frac{f_{1}}{A} \sqrt{\frac{π}{2}} (c_{3} + i c_{4}) a - \frac{g_{1}}{A} \sqrt{\frac{8}{π}} (η_{1} + i υ_{1}) a] .

(6.19)

Thus equation (5.39) becomes

\frac{8 a γ}{A c o s h π ε} = [\begin{matrix} C_{r}^{2} (c_{1}^{2} + c_{2}^{2}) - \frac{f_{1}}{A} C_{r} \sqrt{2 π} (c_{3} c_{1} + c_{4} c_{2}) a + \frac{g_{1}}{A} \sqrt{\frac{32}{π}} C_{r} (η_{1} c_{2} - υ_{1} c_{1}) a \\ + 4 \frac{f_{1} g_{1}}{A^{2}} (υ_{1} c_{3} - η_{1} c_{4}) a^{2} + \frac{8}{π} \frac{g_{1}^{2}}{A^{2}} (η_{1}^{2} + υ_{1}^{2}) a^{2} + \frac{π}{2} \frac{f_{1}^{2}}{A^{2}} (c_{3}^{2} + c_{4}^{2}) a^{2} \end{matrix}]

(6.20)

and equation (5.41) can be written as

\begin{array}{ll} P_{0} = \sqrt{\frac{π}{2}} \frac{a}{\sqrt{1 - β^{2}}} I m [\int_{- 1}^{1} (i C_{r} - i \frac{1}{A} f_{1} \sqrt{\frac{π}{2}} s g n (T) T a - \frac{1}{A} g_{1} \sqrt{\frac{8}{π}} T a) {(\frac{1 - T}{1 + T})}^{i ε} q (T) d T] \\ = \sqrt{\frac{π}{2}} \frac{a}{\sqrt{1 - β^{2}}} [ϑ_{1} C_{r} - \frac{1}{A} f_{1} a \sqrt{\frac{π}{2}} ς_{1} - \frac{1}{A} g_{1} a \sqrt{\frac{8}{π}} ω_{2}], \end{array}

(6.21)

with

{\begin{cases} ϑ_{1} + i ϑ_{2} = \int_{- 1}^{1} {(\frac{1 - T}{1 + T})}^{i ε} q (T) d T, ς_{1} + i ς_{2} = \int_{- 1}^{1} s g n (T) T {(\frac{1 - T}{1 + T})}^{i ε} q (T) d T \\ ω_{1} + i ω_{2} = \int_{- 1}^{1} T {(\frac{1 - T}{1 + T})}^{i ε} q (T) d T \\ q (T) = \sum_{n = 1}^{\infty} \frac{(2 n + 1) Γ (n + 1) Γ (n + 1)}{Γ (n + 1 + i ε) Γ (n + 1 - i ε)} \frac{1}{n} P_{n - 1}^{(1 - i ε, 1 + i ε)} (0) P_{n}^{(i ε, - i ε)} (T) \end{cases} .

(6.22)

By solving for $C_{r}$ from equation (6.21), substituting it into equation (6.20), then differentiating equation (6.20) with respect to the radius a, and considering the pull-off condition $\frac{\partial P_{0}}{\partial a} = 0$ , we can get a new equation for the critical pull-off contact radius $a_{c}$ . In sequence, the pull-off force can be obtained by substituting $a_{c}$ into equation (6.21). Because this is a rather lengthy and cumbersome process, its details are not presented here.

For other polynomial profile non-slipping adhesive contact problems, the corresponding contact relative radial displacement can be derived in a similar way. In addition, it can be verified that all the preceding formulae will degenerate to the ones of corresponding to similar material contacts as $ε \to 0$ , and automatically $g (r) = 0$ .

7. A non-slipping Hertzian contact subjected to a constant radial mismatch strain load

In this section, to illustrate the contact traction analysis with the formulation and approach developed in §5, the Hertzian contact, subjected to a constant applied radial mismatch strain, as shown in figure 3, is re-examined. Because the applied radial mismatch strain is often much larger than that resulting from a successive normal loading, for simplicity, the latter is ignored [32]. This model is often used to understand bio-adhesion deformation behaviour [9,32,51,52]. The constant applied radial mismatch strain may be imposed because of a uniform thermal load, pre-strain or prestressing. We intend to develop a more rigorous and accurate analysis.

Figure 3. Non-slipping Hertzian adhesive contact between two axisymmetric solids, in which #1 stands for a Hertzian indenter, #2 stands for the lower solid material subjected to a radial constant mismatch strain $ε_{0}$ and P is the external normal loading.

For this Hertzian contact, the gap function can be expressed as

f (r) = - \frac{1}{A} (Δ - \frac{k}{2} r^{2}), g (r) = - \frac{1}{A} ε_{0} r

(7.1)

and

F (x) = \frac{k}{A} \sqrt{\frac{2}{π}} x, G (x) = - \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} x,

(7.2)

where $ε_{0}$ stands for the constant radial mismatch strain.

In consideration of equation (7.2), equation (5.17) can be written as

V (x) = G (x) + i [C_{r} + x F (x)] = i C_{r} - \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} x + i \frac{k}{A} \sqrt{\frac{2}{π}} x^{2} .

(7.3)

Inserting equation (7.3) into equation (5.18), and using the identities in appendix C, we have

\begin{array}{ll} ω (x) = s (x) + i r (x) = M (x^{+}) - M (x^{-}) \\ = \frac{β {(a - x)}^{- i ε} {(x + a)}^{i ε}}{i (1 - β^{2})} \frac{1}{π} \int_{- a}^{a} \frac{{(a - t)}^{i ε} {(t + a)}^{- i ε} V (t)}{t - x} d t + \frac{V (x)}{(1 - β^{2})} \\ = i \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{1 - β^{2}}} [C_{r} + i \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} (x - 2 i a ε) + \frac{k}{A} \sqrt{\frac{2}{π}} (x^{2} - 2 a i ε x - 2 a^{2} ε^{2})] \\ = i \frac{{(a - x)}^{- i ε} {(x + a)}^{i ε}}{\sqrt{1 - β^{2}}} [(C_{r} + 2 a ε \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} - \frac{k}{A} \sqrt{\frac{8}{π}} a^{2} ε^{2}) + \frac{i}{A} \sqrt{\frac{8}{π}} (ε_{0} - k a ε) x + \frac{k}{A} \sqrt{\frac{2}{π}} x^{2}] . \end{array}

(7.4)

And, by inserting the above into equation (5.27), we obtain

\begin{array}{ll} P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r = \sqrt{8 π} \int_{0}^{a} r (x) d x = \sqrt{8 π} I m [\int_{0}^{a} ω (x) d x] \\ = \frac{\sqrt{8 π}}{\sqrt{1 - β^{2}}} (C_{r} a + \frac{ε}{A} \sqrt{\frac{8}{π}} ε_{0} a^{2} + \sqrt{\frac{2}{π}} \frac{k}{3 A} a^{3}) + O (ε^{2}) . \end{array}

(7.5)

As $ε \to 0$ , this solution automatically degenerates to equation (4.10). Ignoring the higher order of $ε$ in equation (7.5), $C_{r}$ can be isolated as

C_{r} = \frac{\sqrt{1 - β^{2}}}{a \sqrt{8 π}} P_{0} - \frac{ε}{A} \sqrt{\frac{8}{π}} ε_{0} a - \sqrt{\frac{2}{π}} \frac{k}{3 A} a^{2} .

(7.6)

Inserting equation (7.4) into equation (5.25) and then from equation (5.35), with a lengthy but straightforward manipulation, we obtain

\begin{array}{ll} K = \underset{r \to a}{l i m} \sqrt{2 π} {(a - r)}^{\frac{1}{2} + i ε} [σ_{z z} (r) - i σ_{r z} (r)] \\ = \frac{\sqrt{π} 2^{\frac{1}{2} + i ε} a^{- \frac{1}{2} + i ε} [C_{0} + i C_{1} a + C_{2} a^{2}]}{\sqrt{1 - β^{2}}} \frac{Γ (1 - i ε)}{Γ (\frac{1}{2} - i ε)}, \end{array}

(7.7)

with

C_{0} = (C_{r} + 2 a ε \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} - \frac{k}{A} \sqrt{\frac{8}{π}} a^{2} ε^{2}), C_{1} = \frac{1}{A} \sqrt{\frac{8}{π}} (ε_{0} - k a ε), C_{2} = \frac{k}{A} \sqrt{\frac{2}{π}} .

(7.8)

Now, substituting equation (7.7) into equation (4.16), we get

γ = \frac{A π}{2 a} [{(C_{0} + a^{2} C_{2})}^{2} + {(a C_{1})}^{2}] {\frac{Γ (1 - i ε)}{Γ (\frac{1}{2} - i ε)} \frac{Γ (1 + i ε)}{Γ (\frac{1}{2} + i ε)}} .

(7.9)

From equations (7.6), (7.8) and (7.9) by discarding the terms with higher order of $ε$ , we obtain

\begin{array}{ll} γ = \frac{A π}{2 a} [{(C_{0} + a^{2} C_{2})}^{2} + {(a C_{1})}^{2}] {\frac{Γ (1 - i ε)}{Γ (\frac{1}{2} - i ε)} \frac{Γ (1 + i ε)}{Γ (\frac{1}{2} + i ε)}} \\ = \frac{A}{2 a} [{(C_{r} + 2 a ε \frac{ε_{0}}{A} \sqrt{\frac{8}{π}} + a^{2} \frac{k}{A} \sqrt{\frac{2}{π}})}^{2} + {(\frac{a}{A} \sqrt{\frac{8}{π}} (ε_{0} - k a ε))}^{2}] \\ = \frac{A}{2 a} [{(\frac{\sqrt{1 - β^{2}}}{a \sqrt{8 π}} P_{0} + ε_{0} \sqrt{\frac{8}{π}} \frac{a ε}{A} + \frac{2}{3} \sqrt{\frac{2}{π}} \frac{k}{A} a^{2})}^{2} + {(\frac{a}{A} \sqrt{\frac{8}{π}} (ε_{0} - k a ε))}^{2}] . \end{array}

(7.10)

Equation (7.10) can be rearranged as

\frac{\sqrt{1 - β^{2}}}{8} P_{0} = a {[\frac{π}{4} \frac{a}{A} γ - \frac{1}{A^{2}} {(a ε_{0} - k a^{2} ε)}^{2}]}^{\frac{1}{2}} - ε_{0} \frac{ε}{A} a^{2} - \frac{1}{3} \frac{k}{A} a^{3} .

(7.11)

By introducing the non-dimensional and normalizing parameters

a = {(A γ R^{2})}^{\frac{1}{3}} \bar{a}, ε_{0} = {(\frac{γ A}{R})}^{\frac{1}{3}} {\bar{ε}}_{0}, P_{0} = γ R {\bar{P}}_{0},

(7.12)

equation (7.11) can be expressed as

{\bar{P}}_{0} = 8 c o s h (π ε) [{\bar{a}}^{\frac{3}{2}} {(\frac{π}{4} - \bar{a} {\bar{ε}}_{0}^{2} + 2 {\bar{a}}^{2} {\bar{ε}}_{0} ε)}^{\frac{1}{2}} - {\bar{ε}}_{0} ε {\bar{a}}^{2} - \frac{1}{3} {\bar{a}}^{3}] .

(7.13)

Before proceeding, let us consider a special scenario for which analytical solution exists. If material mismatch in the contact is ignored (i.e. $ε = 0$ ) but the applied radial mismatch strain ${\bar{ε}}_{0}$ is retained, equation (7.13) degenerates to

{\bar{P}}_{0} = 8 [{\bar{a}}^{\frac{3}{2}} {(\frac{π}{4} - \bar{a} {\bar{ε}}_{0}^{2})}^{\frac{1}{2}} - \frac{1}{3} {\bar{a}}^{3}] .

(7.14)

Applying the critical pull-off criterion

\frac{\partial}{\partial \bar{a}} {\bar{P}}_{0} = 0

(7.15)

to (equation (7.14)), we get

{\bar{ε}}_{0}^{2} {\bar{a}}^{4} - \frac{π}{4} {\bar{a}}^{3} + 4 {\bar{ε}}_{0}^{4} {\bar{a}}^{2} - \frac{3 π}{2} {\bar{ε}}_{0}^{2} \bar{a} + \frac{9 π^{2}}{64} = 0.

(7.16)

This is a quartic equation and the important, relevant root is

{\bar{a}}_{c 0} = \frac{h}{2} + \frac{π}{16 {\bar{ε}}_{0}^{2}} - \frac{1}{32} \sqrt{\frac{π}{h} (512 + \frac{π^{2}}{{\bar{ε}}_{0}^{6}}) - 256 w - \frac{8 (2 \times 2^{\frac{1}{3}} m {\bar{ε}}_{0}^{2} + 512 {\bar{ε}}_{0}^{6} - 3 π^{2})}{3 {\bar{ε}}_{0}^{4}}},

(7.17)

where

\begin{array}{ll} m = (243 π^{4} + 6912 π^{2} {\bar{ε}}_{0}^{6} + 131072 {\bar{ε}}_{0}^{12} + 27 π^{2} \sqrt{81 π^{4} + 3584 π^{2} {\bar{ε}}_{0}^{6} + 65536 {\bar{ε}}_{0}^{12}})^{\frac{1}{3}}, \\ w = \frac{9 π^{2} {\bar{ε}}_{0}^{2} + 256 {\bar{ε}}_{0}^{8}}{3 \times 2^{\frac{1}{3}} m {\bar{ε}}_{0}^{2}}, h = \sqrt{w + \frac{3 π^{2} + 4 \times 2^{\frac{1}{3}} m {\bar{ε}}_{0}^{2} - 512 {\bar{ε}}_{0}^{6}}{192 {\bar{ε}}_{0}^{4}}} . \end{array}

By inserting equation (7.17) into equation (7.14), the pull-off force can be obtained. The pull-off force versus ${\bar{ε}}_{0}$ is plotted as the solid curve shown in figure 4.

As ${\bar{ε}}_{0} \to 0$ in equation (7.17), one finds

{\bar{a}}_{c 0} = {(\frac{9 π}{16})}^{\frac{1}{3}},

(7.18)

which is consistent with the solution of equation (4.21).

Now we consider the general case of equation (7.13). Again, applying the pull-off condition to equation (7.13), one has

\begin{array}{ll} 8 ε \bar{ε_{0}} {\bar{a}}^{5} - (4 {\bar{ε_{0}}}^{2} + 68 ε^{2} {\bar{ε}}_{0}^{2}) {\bar{a}}^{4} + (π + 64 ε {\bar{ε}}_{0}^{3} + 32 ε^{3} {\bar{ε}}_{0}^{3}) {\bar{a}}^{3} \\ - (11 π ε {\bar{ε}}_{0} + 16 {\bar{ε}}_{0}^{4} + 16 ε^{2} {\bar{ε}}_{0}^{4}) {\bar{a}}^{2} + (6 π {\bar{ε}}_{0}^{2} + 4 π ε^{2} {\bar{ε}}_{0}^{2}) \bar{a} - \frac{9 π^{2}}{16} = 0. \end{array}

(7.19)

As ${\bar{ε}}_{0} \to 0$ , equation (7.16) is reproduced. Because the solution to equation (7.19) cannot be solved in an explicit form, by applying a perturbation method, one may write

{\bar{a}}_{c} = {\bar{a}}_{c 0} (1 + \sum_{n = 1}^{\infty} C_{n} ε^{n}),

(7.20)

where $C_{n}$ are to be determined via substitution of equations (7.20) into equation (7.19) and comparison of the power exponent of $ε$ as ${\bar{ε}}_{0}$ is treated as a parameter, ( $ε \leq \frac{1}{2 π} \ln (3) \approx 0.1784$ ). This is a straightforward manipulation, and ignoring higher orders of $ε$ , one may find the pull-off radius:

{\bar{a}}_{c} = {\bar{a}}_{c 0} (1 + C_{1} ε) = {\bar{a}}_{c 0} (1 + \frac{8 {\bar{a}}_{c 0}^{3} {\bar{ε}}_{0} (8 {\bar{a}}_{c 0}^{3} - 11 π + 64 {\bar{a}}_{c 0} {\bar{ε}}_{0}^{2}) ε}{9 π^{2} - 40 {\bar{a}}_{c 0}^{3} π + 192 {\bar{a}}_{c 0}^{4} {\bar{ε}}_{0}^{2} - 144 {\bar{a}}_{c 0} π {\bar{ε}}_{0}^{2} + 512 {\bar{a}}_{c 0}^{2} {\bar{ε}}_{0}^{4}}) .

(7.21)

By inserting this relation into equation (7.13), the pull-off force ${\bar{P}}_{p u l l o f f}$ can be obtained. The pull-off force ${\bar{P}}_{p u l l o f f}$ against the applied radial mismatch strain ${\bar{ε}}_{0}$ is plotted in figure 4, with two extremum values $ε = 0$ , $ε = 0.1784$ and a mid-value $ε = 0.0892$ . It may be observed from figure 4 that the pull-off force ${\bar{a}}_{c}$ drastically decreases with the increase of the applied radial mismatch strain ${\bar{ε}}_{0}$ . Generally, if the stiffness ratio between two solids is less than 10, provided that they have an identical Poisson ratio ( $0.3$ , for example), the index $ε$ is less than $0.0660$ . Thus, it can be inferred from figure 4 that the influence of material mismatch on the pull-off force is quite small. In other words, the material mismatch can be ignored when modelling the most adhesive contacts, where the stiffness ratio of two contacted components is not extremely large.

8. Concluding remarks

The problem of axisymmetric non-slipping contact between dissimilar elastic bodies has been formulated. The intriguing relative radial interface displacement entrained within non-slipping contacts has been studied with a direct and generic approach. For the cases of both non-adhesive and adhesive surface conditions, and for different contact profiles, explicit solutions can be obtained, which enable a thorough study to be made of a general elastic-to-elastic non-slipping contact model. Finally, based on the formulation and approach, a typical adhesion contact problem with an applied radial mismatch strain load has been studied. It has been demonstrated that the applied radial mismatch strain may significantly influence the pull-off force.

These results provide useful formulation for axisymmetric contact problems where non-slipping conditions are observed, and may be used to model the axisymmetric contacts at macro, micro and nanoscales.

Data accessibility

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Declaration of AI use

We have not used AI-assisted technologies in creating this article.

Authors’ contributions

L.M.: conceptualization, formal analysis, investigation, writing—original draft; A.M.K.: supervision, writing—review and editing; D.A.H.: supervision, writing—review and editing.

All authors gave final approval for publication and agreed to be held accountable for the work performed therein.

Conflict of interest declaration

We declare we have no competing interests.

Funding

This work is also partially supported by National Natural Science Foundation of China (grant no. 12072254).

Acknowledgements

The support of the Sir Joseph Pope Fellowship from Nottingham University is much appreciated. The authors thank the EPSRC for the support under the Prosperity Partnership Grant/Cornerstone: Mechanical Engineering Science to Enable Aero Propulsion Futures, Grant ref: EP/R004951/1.

Appendix

Appendix A. Abel transforms, Fourier sine and cosine transform, and Hilbert transform

The Abel transforms $A_{1}$ $A_{2}$ , and their inverse transforms are defined as

{\begin{cases} {\hat{f}}_{1} (x) \equiv A_{1} [f_{1} (r); r \to x] = \sqrt{\frac{2}{π}} \int_{0}^{x} \frac{f_{1} (r) d r}{\sqrt{x^{2} - r^{2}}} \\ f_{1} (r) = A_{1}^{- 1} [{\hat{f}}_{1} (x); x \to r] = \sqrt{\frac{2}{π}} \frac{d}{d r} \int_{0}^{r} \frac{x {\hat{f}}_{1} (x) d x}{\sqrt{r^{2} - x^{2}}}, \end{cases}

(A 1)

{\begin{cases} {\hat{f}}_{2} (x) \equiv A_{2} [f_{2} (r); r \to x] = \sqrt{\frac{2}{π}} \int_{x}^{\infty} \frac{f_{2} (r) d r}{\sqrt{r^{2} - x^{2}}} \\ f_{2} (r) = A_{2}^{- 1} [{\hat{f}}_{2} (x); x \to r] = - \sqrt{\frac{2}{π}} \frac{d}{d r} \int_{r}^{\infty} \frac{x {\hat{f}}_{2} (x) d x}{\sqrt{x^{2} - r^{2}}}, \end{cases}

(A 2)

and the Fourier sine and cosine transform are defined as

{\begin{cases} F_{c} (ω) = F_{c} [f (t); t \to ω] = \sqrt{\frac{2}{π}} \int_{0}^{\infty} f (t) c o s (ω t) d t \\ a s f (- t) = f (t), \\ f (t) = F_{c}^{- 1} [F_{c} (ω); ω \to t] = \sqrt{\frac{2}{π}} \int_{0}^{\infty} F_{c} (ω) c o s (ω t) d ω \end{cases}

(A 3)

{\begin{array}{l} F_{s} (ω) = F_{s} [f (t); t \to ω] = \sqrt{\frac{2}{π}} \int_{0}^{\infty} f (t) \sin ω t d t \\ f (t) = F_{s}^{- 1} [F_{s} (ω); ω \to t] = \sqrt{\frac{2}{π}} \int_{0}^{\infty} F_{s} (ω) \sin ω t d ω \end{array} as f (- t) = - f (t) .

(A 4)

The Hilbert transform and its inverse transform are defined as

{\begin{array}{l} g (y) = H [f (x); x \to y] = \frac{1}{π} P V \int_{- \infty}^{\infty} \frac{f (x)}{x - y} d x \\ f (x) = H^{- 1} [f (x); x \to y] = - \frac{1}{π} P V \int_{- \infty}^{\infty} \frac{g (y)}{y - x} d y \end{array},

(A 5)

where the Cauchy principal value is taken in each of the integrals, denoted by ‘PV’. Two typical results are listed below:

{\begin{cases} g (y) = H [\sin (x); x \to y] = \frac{1}{π} P V \int_{- \infty}^{\infty} \frac{\sin (x)}{x - y} d x = \cos y \\ g (y) = H [\cos (x); x \to y] = \frac{1}{π} P V \int_{- \infty}^{\infty} \frac{\cos (x)}{x - y} d x = - \sin y \end{cases} .

(A 6)

In terms of equation (A 6), one may prove

F_{s} (x) = - \frac{1}{π} \int_{- \infty}^{\infty} \frac{F_{c} (ω) d ω}{ω - x},

(A 7)

if a primitive function $f (x), x \in (0, \infty)$ is extended as an odd function to the domain $x \in (- \infty, \infty)$ ; and

F_{c} (x) = \frac{1}{π} \int_{- \infty}^{\infty} \frac{F_{s} (ω) d ω}{ω - x},

(A 8)

if a primitive function $f (x), x \in (0, \infty)$ is extended as an even function to the domain $x \in (- \infty, \infty)$ , where $F_{s} (x)$ and $F_{c} (x)$ are given in equations (A 3) and (A 4).

Appendix B. Some integrals involving Bessel functions, sine and cosine functions

Some useful integral identities related to the Jacobi polynomial and sine (or cosine) function are useful in the formulation [53,54]

\begin{aligned} \int_{0}^{\infty} J_{0} (r λ) \sin (ω λ) d λ = {\begin{array}{l} 0, (ω < r) \\ \frac{1}{\sqrt{ω^{2} - r^{2}}}, (ω > r) \end{array}, \\ \int_{0}^{\infty} J_{0} (r λ) \cos (ω λ) d λ = {\begin{array}{l} \frac{1}{\sqrt{r^{2} - ω^{2}}}, (ω < r) \\ 0, (ω > r) \end{array}, \end{aligned}

(B 1)

\begin{aligned} \int_{0}^{\infty} J_{1} (r λ) \sin (ω λ) d λ = {\begin{array}{l} \frac{ω}{r \sqrt{r^{2} - ω^{2}}}, (ω < r) \\ 0, (ω > r) \end{array}, \\ \int_{0}^{\infty} J_{1} (r λ) \cos (ω λ) d λ = {\begin{cases} \frac{1}{r}, (ω < r) \\ \frac{1}{r} [1 - ω {(ω^{2} - r^{2})}^{- \frac{1}{2}}], (ω > r) . \end{cases} \end{aligned}

(B 2)

Appendix C. Singular integrals

The following singular integral can be frequently encountered in contact problems:

I_{1} (x) = \frac{1}{π} \int_{- 1}^{1} \frac{(1 - t)^{α} (1 + t)^{β} Q_{1} (t)}{t - x} d t, (- 1 < x < 1),

(C 1)

where $- 1 < R e (α, β) < 1, α + β = \pm 1, 0$ , $Q_{1} (t)$ can be any polynomial expressions. equation (C 1) should be interpreted in the sense of the Cauchy principal value integral.

The following formula for Jacobi polynomials is useful in solving the above integral [55–57]:

\frac{1}{π} \int_{- 1}^{1} \frac{{(1 - t)}^{α} {(1 + t)}^{β} P_{n}^{(α, β)} (t)}{t - x} d t = c o t (π α) {(1 - x)}^{α} {(1 + x)}^{β} P_{n}^{(α, β)} (x) - \frac{2^{(α + β)}}{s i n α π} P_{n + (α + β)}^{(- α, - β)} (x),

(C 2)

where $R e (α, β) > - 1, α + β = \pm 1, 0; α \neq 0, 1$ , and the Jacobi polynomials can be expressed with the hypergeometric function as [58]

P_{n}^{(α, β)} (z) = \frac{Γ (α + 1 + n)}{Γ (α + 1) Γ (n + 1)} F ⟮ - n, 1 + α + β + n; α + 1; \frac{1 - z}{2} ⟯, (n \geq 0) .

(C 3)

It follows that

P_{n}^{(α, β)} (t) = 0, (n < 0),

(C 4)

and

\begin{aligned} P_{0}^{(α, β)} (t) = 1, P_{1}^{(α, β)} (x) = (α + 1) + \frac{(α + β + 2)}{2} (x - 1), \\ P_{2}^{(α, β)} (x) = \frac{(α + 1) (α + 2)}{2} + \frac{(α + 2) (α + β + 3)}{2} (x - 1) + \frac{(α + β + 3) (α + β + 4)}{2} {(\frac{x - 1}{2})}^{2} . \end{aligned}

(C 5)

The recurrence relation for the Jacobi polynomials is

\begin{array}{l} P_{n}^{(α, β)} (x) = \frac{(2 n + α + β - 1) {(2 n + α + β) (2 n + α + β - 2) x + α^{2} - β^{2}}}{2 n (n + α + β) (2 n + α + β - 2)} P_{n - 1}^{(α, β)} (x) \\ - \frac{2 (n + α - 1) (n + β - 1) (2 n + α + β)}{2 n (n + α + β) (2 n + α + β - 2)} P_{n - 2}^{(α, β)} (x), \end{array}

(C 6)

with $(n = 2, 3, . . .),$

and

P_{n}^{(α, β)} (- x) = {(- 1)}^{n} P_{n}^{(β, α)} (x) .

(C 7)

The polynomial $Q_{1} (t)$ in equation (C 1) can be put into a series of Jacobi polynomials in the form of

Q_{1} (t) = \sum_{n = 0}^{M} d_{n} P_{n}^{(α, β)} (t) .

(C 8)

The coefficients $d_{n}$ can be found by the Jacobi polynomial’s orthogonal relationship:

\int_{- 1}^{1} {(1 - t)}^{α} {(1 + t)}^{β} P_{m}^{(α, β)} (t) P_{n}^{(α, β)} (t) d t = \frac{2^{α + β + 1}}{(2 n + α + β + 1)} \frac{Γ (n + α + 1) Γ (n + β + 1)}{Γ (n + α + β + 1) Γ (n + 1)} δ_{m n},

(C 9)

d_{n} = \frac{(2 n + α + β + 1)}{2^{α + β + 1}} \frac{Γ (n + α + β + 1) Γ (n + 1)}{Γ (n + α + 1) Γ (n + β + 1)} \int_{- 1}^{1} Q_{1} (t) {(1 - t)}^{α} {(1 + t)}^{β} P_{n}^{(α, β)} (t) d t .

(C 10)

Inserting equation (C 8) into equation (C 1) leads to

\begin{array}{ll} I_{1} (x) & = \frac{1}{π} \int_{- 1}^{1} \frac{(1 - t)^{α} (1 + t)^{β} Q_{1} (t)}{t - x} d t \\ = \cot (π α) (1 - x)^{α} (1 + x)^{β} Q_{1} (x) - \frac{2^{(α + β)}}{\sin (α π)} Q_{2} (x), (- 1 < x < 1) \end{array}

(C 11)

where

Q_{2} (x) = \sum_{n = 0}^{M} d_{n} P_{n + (α + β)}^{(- α, - β)} (x) .

(C 12)

By use of equation (C 11), the following identities can be verified:

\frac{1}{π} \int_{- 1}^{1} \frac{(1 - t)^{α} (1 + t)^{β}}{t - x} d t = \cot (π α) (1 - x)^{α} (1 + x)^{β} - \frac{2^{(α + β)}}{\sin α π} P_{(α + β)}^{(- α - β)} (x), (- 1 < x < 1),

(C 13)

\begin{array}{l} \frac{1}{π} \int_{- 1}^{1} \frac{t (1 - t)^{α} (1 + t)^{β}}{t - x} d t & = \cot (π α) (1 - x)^{α} (1 + x)^{β} x \\ - \frac{2^{α + β}}{\sin α π} [\frac{2}{α + β + 2} P_{1 + α + β}^{(- α, - β)} (x) + \frac{β - α}{α + β + 2} P_{α + β}^{(- α, - β)} (x)], (- 1 < x < 1) \end{array}

(C 14)

\begin{array}{ll} \frac{1}{π} \int_{- 1}^{1} \frac{t^{2} (1 - t)^{α} (1 + t)^{β}}{t - x} d t & = \cot (π α) (1 - x)^{α} (1 + x)^{β} x^{2} \\ - \frac{2^{α + β}}{\sin α π} [\frac{2 + α + β + (α - β)^{2}}{(2 + α + β) (3 + α + β)} P_{(α + β)}^{(- α, - β)} (x) - \frac{4 (α - β) P_{1 + (α + β)}^{(- α, - β)} (x)}{(2 + α + β) (4 + α + β)} + \frac{8 P_{2 + (α + β)}^{(- α, - β)} (x)}{(3 + α + β) (4 + α + β)}], (- 1 < x < 1), \end{array}

(C 15)

Appendix D. An integral formulation for axisymmetric contacts

Here, we provide another general integral formulation for axisymmetric contact problems. Equation (2.17) can be written as

{\begin{matrix} \int_{0}^{a} σ_{z z} t H_{0} [λ^{- 1} J_{0} (λ t), λ \to r] d t + β \int_{0}^{a} σ_{r z} t H_{0} [λ^{- 1} J_{1} (λ t), λ \to r] d t = f (r) \\ β \int_{0}^{a} σ_{z z} t H_{1} [λ^{- 1} J_{0} (λ t), λ \to r] d t + \int_{0}^{a} σ_{r z} t H_{1} [λ^{- 1} J_{1} (λ t), λ \to r] d t = g (r) \\ (0 \leq r < a) \end{matrix} .

(D 1)

Operating $\frac{d}{d r}$ to the first equation and $(\frac{1}{r} \frac{d}{d r} r)$ to the second relation in equation (D 1), respectively, yields

{\begin{cases} \int_{0}^{a} σ_{z z} t \frac{d}{d r} H_{0} [λ^{- 1} J_{0} (λ t), λ \to r] d t + β \int_{0}^{a} σ_{r z} t \frac{d}{d r} H_{0} [λ^{- 1} J_{1} (λ t), λ \to r] d t = \frac{d}{d r} f (r) \\ β \int_{0}^{a} σ_{z z} t (\frac{1}{r} \frac{d}{d r} r) H_{1} [λ^{- 1} J_{0} (λ t), λ \to r] d t \\ + \int_{0}^{a} σ_{r z} t (\frac{1}{r} \frac{d}{d r} r) H_{1} [λ^{- 1} J_{1} (λ t), λ \to r] d t = (\frac{1}{r} \frac{d}{d r} r) g (r) \\ (0 \leq r < a) \end{cases} .

(D 2)

In view of equation (3.1) this can be written as

{\begin{cases} \int_{0}^{a} σ_{z z} t H_{1} [J_{0} (λ t), λ \to r] d t + β \int_{0}^{a} σ_{r z} t H_{1} [J_{1} (λ t), λ \to r] d t = - \frac{d}{d r} f (r) \\ β \int_{0}^{a} σ_{z z} t {r H_{0} [J_{0} (λ t), λ \to r]} d t + \int_{0}^{a} σ_{r z} t {r H_{0} [J_{1} (λ t), λ \to r]} d t = \frac{d}{d r} (r g (r)) \end{cases}, (0 \leq r < a) .

(D 3)

By applying equation (3.3b) to the first relation of equation (D 3), and applying equation (3.3a) to the second relation of equation (D 3), one finds

{\begin{cases} \int_{0}^{a} σ_{z z} t {F_{c} [J_{0} (λ t); λ \to 0] - F_{c} [J_{0} (λ t); λ \to x]} d t \\ + β \int_{0}^{a} σ_{r z} t {F_{c} [J_{1} (λ t); λ \to 0] - F_{c} [J_{1} (λ t); λ \to x]} d t = - x A_{1} [\frac{d}{d r} f (r); r \to x] \\ β \int_{0}^{a} σ_{z z} t F_{s} [J_{0} (λ t); λ \to x] d t \\ + \int_{0}^{a} σ_{r z} t F_{s} [J_{1} (λ t); λ \to x] d t = A_{1} [\frac{d}{d r} (r g (r)); r \to x] \end{cases} .

(D 4)

In view of identities (B 1) and (B 2), and equations (A 1), equation (D 4) can be finally written as

{\begin{cases} β \int_{0}^{x} \frac{σ_{z z} t}{\sqrt{x^{2} - t^{2}}} d t + \int_{x}^{a} \frac{x σ_{r z}}{\sqrt{t^{2} - x^{2}}} d t = \int_{0}^{x} \frac{1}{\sqrt{x^{2} - r^{2}}} \frac{d}{d r} [r g (r)] d r \\ \int_{x}^{a} \frac{σ_{z z} t}{\sqrt{t^{2} - x^{2}}} d t - β \int_{0}^{x} \frac{σ_{r z} x}{\sqrt{x^{2} - t^{2}}} d t = x \int_{0}^{x} \frac{1}{\sqrt{x^{2} - r^{2}}} \frac{d}{d r} f (r) d r + \int_{0}^{a} σ_{z z} d t \end{cases},

(D 5)

with an external condition

P_{0} = 2 π \int_{0}^{a} σ_{z z} (r) r d r .

(D 6)

Equations (D 6) and (D 7) comprise an alternative integral formulation for axisymmetric contacts. Here, we note that a more general and systematic formulation for the axisymmetric mixed boundary-value contact problems, in terms of the singular integral equation approach, can be found in recent work [59], and the related key relations regarding the Abel operator and the singular operator generated by the Cauchy kernel can be found in the works [60–62].

Footnotes

Published by the Royal Society under the terms of the Creative Commons Attribution License http://creativecommons.org/licenses/by/4.0/, which permits unrestricted use, provided the original author and source are credited.

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Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences cover image

January 2025

Volume 481Issue 2305

Article Information

DOI:https://doi.org/10.1098/rspa.2024.0459
Published by:Royal Society
Online ISSN:1471-2946

History:

Manuscript received18/06/2024
Manuscript accepted18/09/2024
Published online08/01/2025

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mechanical engineering

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Formulation for axisymmetric non-slipping contacts between dissimilar elastic solids

Abstract

1. Introduction

2. General formulation for axisymmetric contacts

(a) The Papkovich–Neuber function potential formulae

(b) The dual integral equations for axisymmetric contacts

(c) Non-slipping contact conditions

3. Transform approaches to treating the dual equation problems

4. The contact problems with β=0

(a) Case 1: regular contact without surface adhesion effect

(b) Case 2: contact with surface adhesion effect

5. The contact problems with β≠0

(a) Reduction to a singular integral equation and its formal solution

(b) Traction along the contact surface

(c) Case 1: regular contact without surface adhesion effect

(d) Case 2: contact with surface adhesion effect

6. Relative contact radial displacement in a non-slipping conical contact

(a) Non-adhesion

(b) With adhesion

7. A non-slipping Hertzian contact subjected to a constant radial mismatch strain load

8. Concluding remarks

Data accessibility

Declaration of AI use

Authors’ contributions

Conflict of interest declaration

Funding

Acknowledgements

Appendix

Appendix A. Abel transforms, Fourier sine and cosine transform, and Hilbert transform

Appendix B. Some integrals involving Bessel functions, sine and cosine functions

Appendix C. Singular integrals

Appendix D. An integral formulation for axisymmetric contacts

Footnotes

This Issue

4. The contact problems with $β = 0$

5. The contact problems with $β \neq 0$