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 $\u2212d<x<\u221e$, and the surface displacement gradient is defined byDisplay Formula

$dvdx=\u22121Rx,\u2212d\u2264x\u22640$

(24)

Display Formulai.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 of

Display Formula$p(x)=dsin2\lambda \pi AR\pi (s\u2215d)\lambda [11\u2212\lambda +1\u2212(s\u2215d)\lambda F12(1,\lambda ;1+\lambda ;(s\u2215d))](s\u2215d)<1$

(26)

Display Formula$=dsin2\lambda \pi AR\pi (1\u2212\lambda )(2\u2212\lambda )(s\u2215d)\lambda \u22121F12(1,1\u2212\lambda ;3\u2212\lambda ;1(s\u2215d))(s\u2215d)>1$

(27)

where

$s=x+d$ and

$F12(.)$ is a standard hypergeometric function. We note that, when

$(s\u2215d)\u226b1$,

$p(x)\u223cs\lambda \u22121$, so that the asymptotic form given by Eq.

27 applies, and calibration shows that the generalized stress intensity factor is given by

Display Formula$KN=sin2\lambda \pi d2\u2212\lambda AR\pi (1\u2212\lambda )(2\u2212\lambda )$

(28)

This 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 as

Display Formula$d1\u2212\lambda p(s)KN=(2\u2212\lambda )(s\u2215d)\lambda [1+(1\u2212\lambda )\lambda (1\u2212(s\u2215d))F12(1,\lambda ;1+\lambda ;(s\u2215d))](s\u2215d)<1$

(29)

Display Formula$=(s\u2215d)\lambda \u22121F12(1,1\u2212\lambda ;3\u2212\lambda ;1(s\u2215d))(s\u2215d)>1$

(30)

whereas the relative slip displacement is given by

Display Formula$UU0=dR\Psi (sd,f,\beta )$

(31)

Display Formula$\Psi (sd,f,\beta )=f[1\u2212(sd)]+\beta (1+f2)\pi (1+f2\beta 2)(sd)\lambda [1(1\u2212\lambda )+1\lambda [1\u2212(sd)]F12(1,\lambda ;1+\lambda ;(sd))]$

(32)

and, employing the contact law for this problem,

Display Formula$d=(KNAR\pi (1\u2212\lambda )(2\u2212\lambda )sin2\lambda \pi )1\u2215(2\u2212\lambda )$

(33)

gives

Display Formula$UU0=(KNAR\lambda \u22121)1\u22152\u2212\lambda (\pi (1\u2212\lambda )(2\u2212\lambda )sin2\lambda \pi )1\u2215(2\u2212\lambda )\Psi (sd,f,\beta )$

(34)

Display Formula$=(2aR)(1\u2212\lambda )\u2215(2\u2212\lambda )(po\mu (1\u2212\nu ))1\u2215(2\u2212\lambda )((1\u2212\lambda )(2\u2212\lambda )sin\lambda \pi )1\u2215(2\u2212\lambda )\Psi (sd,f,\beta )$

(35)

Therefore there is a violation in slip direction when

$U\u2215U0>1$, i.e., when

Display Formula$(R2a)\lambda \u22121(po\mu (1\u2212\nu ))>(sin\lambda \pi (1\u2212\lambda )(2\u2212\lambda ))[\Psi (sd,f,\beta )]\lambda \u22122$

(36)

The violation is more likely to occur for cases when

$R\u22152a$ is very small and the load

$po\u2215\mu $ is relatively high. However, unlike the perfectly sharp solution, the function

$\Psi (s\u2215d,f,\beta )$ does not exhibit a monotonically increasing value as

$s\u21920$ but shows a maximum. This cannot be found analytically, but has been found numerically for the extreme example case used earlier (

$f=0.8$,

$\nu =0.2$) so that the maximum of

$\Psi (s\u2215d,f,\beta )$ occurs at

$s\u2215d=0.133$ and is of magnitude

$1.002.$ Therefore, using the same load as before,

$po\u2215\mu =0.001$, we find that the paradox will occur if

Display Formula$R2a<5.9\xd710\u22126$

(37)

If 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.