It is shown by Popov in Section 5.1(a) that a rigid sphere (radius $R$) in contact with an elastic half-space (elastic modulus $E$, Poisson ratio $\nu$) with a circular contact patch of radius $a$ generates a pressure distribution

$p(r) = p_0 \sqrt{1 - (\frac{r}{a})^2}$

where $r$ is the distance from the center of the contact patch and $0 \le r \le a$. The total normal force $F_z$ is also computed as

$F_z = \frac{2}{3} p_0 \pi a^2$

In section 5.2, the parameters $p_0$ and $a$ are solved in terms of other parameters and the maximum penetration depth $d$

$a^2 = Rd$

$E^* = \frac{E}{1 - \nu^2}$

$p_0 = \frac{2}{\pi} E^* \sqrt{\frac{d}{R}}$

so the normal force can be expressed as

$F_z = \frac{4}{3} E^* R^{0.5} d^{1.5}$

In Section 8.2.4, the case of torsional tangential loading is presented for a rigid sphere (radius $R$) in contact with an elastic half-space (elastic modulus $E$, Poisson ratio $\nu$) with a circular contact patch of radius $a$. That example uses a shear stress distribution corresponding to "stiction", a uniform rotation of all components of the contact patch.

This analysis uses a different shear stress distribution that corresponds to the maximum torsional moment during sliding, with a direct relation to the normal pressure distribution.

The torsional loading at any point in the contact patch (specified in polar coordinates by radius $r$ and angle $\phi$) consists of tangential forces directed perpendicular to the polar radius $r$ with stresses

$\sigma_{zx} = -\tau(r) \sin \phi$

$\sigma_{zy} = \tau(r) \cos \phi$

The stresses at a given position are combined into the vector

$\boldsymbol{\sigma}_z = (\sigma_{zx},\sigma_{zy},0)^T$

and with the incremental area at a given point defined as

$dA = r dr d\phi$

the incremental force at a given point is

$d\textbf{F}_z = \boldsymbol{\sigma}_z dA$.

The position vector

$\textbf{r} = (r \cos \phi,r \sin \phi,0)^T$

is then used to compute the incremental torsional moment with respect to the center of the contact patch

$dM_z = \textbf{r} \times d\textbf{F}_z = r^2 \tau(r) dr d\phi$

The full torsional moment is computed by integrating over the entire circular area of the contact patch:

$M_z = \int dM_z = \int_0^{2\pi}\int_0^{a} r^2 \tau(r) dr d\phi$

with the following simplification due to angular symmetry:

$M_z = 2\pi \int_0^{a} r^2 \tau(r) dr$

For sliding torsion, $\tau(r)$ has a similar form to the normal pressure $p(r)$ generated by a rigid sphere contacting an elastic half-space (see the previous section):

$\tau(r) = \tau_0 \sqrt{1 - (\frac{r}{a})^2}$

The torsional moment integral then becomes:

$M_z = 2\pi \tau_0 \int_0^{a} r^2 \sqrt{1 - (\frac{r}{a})^2} dr$

and is simplified with a substitution $s = \frac{r}{a}$ and $dr = a ds$:

$M_z = 2\pi \tau_0 \int_0^{1} (as)^2 \sqrt{1 - s^2} a ds$

or

$M_z = 2 a^3 \pi \tau_0 \int_0^{1} s^2 \sqrt{1 - s^2} ds$

From integrals.wolfram.com:

$\int s^2 \sqrt{1 - s^2} ds = \frac{1}{8} (s \sqrt{1-s^2} (2s^2 -1) + \sin^{-1} s)$

The first term evaluates to $0$ at both $s=0$ and $s=1$, so only the second term needs to be considered. The torsional moment then evaluates to

$M_z = 2 a^3 \pi \tau_0 \frac{1}{8} (\sin^{-1} 1 - \sin^{-1} 0)$

or

$M_z = 2 a^3 \pi \tau_0 \frac{1}{8} (\pi / 2)$

or

$M_z = \frac{\pi}{8} a^3 \pi \tau_0$

The ratio between the torsional moment and normal force $M_z / F_z$ is then computed as

$M_z = \frac{\pi}{8} a^3 \pi \tau_0$

$F_z = \frac{2}{3} p_0 \pi a^2$

$\frac{M_z}{F_z} = \frac{3\pi}{16} a \frac{\tau_0}{p_0}$

and labelling the stress ratio $\frac{\tau_0}{p_0}$ as $\mu$,

$\frac{M_z}{F_z} = \frac{3\pi}{16} a \mu$

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