Discussion: “A Paradox in Sliding Contact Problems With Friction” (Adams, G. G., Barber, J. R., Ciavarella, M., and Rice, J. R., 2005, ASME J. Appl. Mech., 72, pp. 450–452)

The punch is moving at velocity $U0$ in the positive $x$ direction, relative to the half plane (with the coordinate system $(x,y)$ moving with it), and the surface normal displacement, $v(x)$⁠, and tangential displacement, $u(x)$⁠, are given bywhere$ν$ being the Poisson’s ratio and $μ$ the modulus of rigidity of the half plane. The slip velocity of particles on the punch surface relative to particles on the half-plane surface, $U(x)$⁠, is given in the original paper byAs the contact is sliding the tractions are related everywhere by $q(x)=fp(x)$⁠, where $f$ is the coefficient of friction, so that Eq. 1 becomesand from Eq. 2 we haveTherefore, the surface normal displacements and surface tangential displacements are given byand hence the relative slip velocity isThe slip direction is reversed when $U∕Uo>1$⁠.

Because the region of apparent reverse slip is so small the problem can conveniently be readdressed using an asymptote which gives the problem additional simplicity and applicability. Suppose that a rigid quarter plane is pressed onto the elastic half plane and slid in the positive $x$ direction, with friction. The local contact pressure, $p(s)$⁠, and shearing traction, $q(s)$⁠, may be written in the form (1)where $KN$ is a generalized stress intensity factor and s is a coordinate measured from the contact edge. The exponent, $λ$⁠, is given by the characteristic equationWe now turn our attention to the finite, flat-ended rigid punch, of half-width $a$⁠, to which a normal load, $P$⁠, is applied, together with a force sufficient to cause sliding in the positive $x$ direction. This develops a contact pressure, $p(x)$⁠, (2) given bysuch that $p(x)$ is positive in compression. If we apply the change of coordinate $s∕a=1+x∕a$ we getwhere we have introduced an average pressure $po=P∕2a$⁠, and henceWe now apply the relationship for the slip velocity $U(x)$ to this solution where, of course, $dv∕dx=0$⁠, and the region where the implied slip direction is reversed is wheni.e., over a region $s<e$ whereTherefore, for this specific geometry, when we employ the calibration for $KN$⁠, the reversal in slip direction occurs over a region $s<e$ whereThus, the phenomenon presented in the original paper occurs whenever the half plane has a finite compressiblity $(ν<1∕2)$⁠, even if no shearing tractions arise $(f=0)$⁠, but clearly the region of violation increases in size with (a) friction, (b) reduced Poisson effect, (c) dimensionless contact pressure $(po∕μ)$⁠. The most extreme values one might expect to encounter in practice might be a Poisson’s ratio of $0.2$⁠, a coefficient of friction of $0.8$⁠, and a mean contact pressure of $po∕μ=0.001$⁠, giving $λ=0.41$⁠, so thatwhich is itself tiny, and readily swamped by local plasticity or the effects of rounding.

The Muskhelishvili potential for a rigid punch on a half plane is given bywhere $z=s+iy$ is a complex coordinate in the half plane (and $i=−1$⁠). The second invariant of the stress tensor as described by von Mises’ equivalent stress is (along the interface, $y=0$⁠)and yield is expected to occur when $σe=σY=3τY.$ This condition is satisfied over a region $s<rp,$ where $rp$ isand again using the calibration for $KN$ from the finite geometrywhich for $ν=0.2,f=0.8,$ and $σY∕μ=3×10−3$ (hence $po∕σY=13$⁠) givesThus, for this contact pressure, there will be plasticity over a region roughly $105$ times larger than the zone over which the paradox occurs.

To probe the effect of edge rounding we consider sliding contact between another rigid semi-infinite punch pressed onto the half plane and sliding. This time, contact is assumed to extend from $−d<x<∞$⁠, and the surface displacement gradient is defined byi.e., the indenter has the form of a parabolic arc of equivalent radius $R$ to the left of the origin, and is flat to the right of the origin. This profile is substituted into integral Eqs. 1 and 2 and solved, giving a pressure along the interface ofwhere $s=x+d$ and $F12(.)$ is a standard hypergeometric function. We note that, when $(s∕d)≫1$⁠, $p(x)∼sλ−1$⁠, so that the asymptotic form given by Eq. 27 applies, and calibration shows that the generalized stress intensity factor is given byThis scaling factor serves both to provide a connection between the applied load $(KN)$ and the extent of contact in the radiused portion, $d$⁠, and it enables us to write the contact pressure aswhereas the relative slip displacement is given byand, employing the contact law for this problem,givesTherefore there is a violation in slip direction when $U∕U0>1$⁠, i.e., whenThe violation is more likely to occur for cases when $R∕2a$ is very small and the load $po∕μ$ is relatively high. However, unlike the perfectly sharp solution, the function $Ψ(s∕d,f,β)$ does not exhibit a monotonically increasing value as $s→0$ but shows a maximum. This cannot be found analytically, but has been found numerically for the extreme example case used earlier (⁠$f=0.8$⁠, $ν=0.2$⁠) so that the maximum of $Ψ(s∕d,f,β)$ occurs at $s∕d=0.133$ and is of magnitude $1.002.$ Therefore, using the same load as before, $po∕μ=0.001$⁠, we find that the paradox will occur ifIf we have a contact of total width $20mm$ then in order to avoid the paradox occurring we require $R>120nm$⁠. This is such a tiny radius that for all contacts of practical importance the paradox will not exist.