# Analytical Study of a Circular Thin Plate Contacting with an Elastic Sphere
## 2 Formulation and Governing Equations
### Contact Deformation of the Sphere
\begin{array}{l}u_{\rho}=\frac{A_{1}}{2G_{b}}\frac{\rho z}{R^{3}}+ \frac{A_{2}\rho}{2G_{b}R(R+z)}u_{z}=\frac{A_{1}}{2G_{b}}\Big{(}\frac{3-4\mu_{b }}{R}+\frac{z^{2}}{R^{3}}\Big{)}+\frac{A_{2}}{2G_{b}R}\\ \sigma_{\rho}=A_{1}\Big{[}\frac{(1-2\mu_{b})z}{R^{3}}-\frac{3\rho^{2}z}{R^{3}} \Big{]}+A_{2}\Big{[}\frac{z}{R^{3}}-\frac{1}{R(R+z)}\Big{]}\quad\sigma_{\varphi }=\frac{A_{1}(1-2\mu_{b})z}{R^{3}}+\frac{A_{2}}{R(R+z)}\\ \sigma_{z}=-A_{1}\Big{[}\frac{(1-2\mu_{b})z}{R^{3}}+\frac{3z^{3}}{R^{3}}\Big{]} -\frac{A_{2}z}{R^{3}}\qquad\qquad\qquad\qquad\tau_{z\rho}=\tau_{\rho z}=-A_{1 }\Big{[}\frac{(1-2\mu_{b})\rho}{R^{3}}+\frac{3\rho z^{2}}{R^{3}}\Big{]}-\frac{ A_{2}\rho}{R^{3}}\end{array} \right.. \tag{18}\]  
The boundary conditions on the free surface require that \(\sigma_{z}=\tau_{\rho z}=0\) everywhere except at the origin, and the summation of the total vertical force is equal to \(\eta\mathrm{d}s\). In Equation (18), note that the expression for \(\sigma_{z}\) vanishes on \(z=0\) but is indeterminate at the origin; thus, this relation will not directly provide a means to determine \(A_{1}\) and \(A_{2}\). Rather than trying to evaluate this singularity at the origin, we pursue the integrated condition on any typical plane \(z=\mathrm{constant}\)  
\[\int_{0}^{\infty}\sigma_{z}(2\pi\rho\mathrm{d}\rho)+q\mathrm{d}s=0. \tag{19}\]  
Invoking the boundary conditions determines the two constants  
\[A_{1}=\frac{q\mathrm{d}s}{2\pi}\qquad A_{2}=-\frac{(1-2\mu_{b})q\mathrm{d}s}{2 \pi}. \tag{20}\]Then, the results for the displacements and stresses are given by  
\[ \begin{cases}u_{\rho}=\frac{(1+\mu_{b})\text{d}\text{s}}{2\pi E_{b}R}\Big{[}\frac{ \rho z}{R^{2}}-\frac{(1-2\mu_{b})\rho}{R+z}\Big{]}&u_{z}=\frac{(1+\mu_{b})q \text{d}\text{s}}{2\pi E_{b}R}\Big{[}2(1-\mu_{b})+\frac{z^{2}}{R^{2}}\Big{]}\\ \sigma_{\rho}=\frac{q\text{d}\text{s}}{2\pi R^{2}}\Big{[}\frac{(1-2\mu_{b})R}{ R+z}-\frac{3\rho^{2}z}{R^{3}}\Big{]}&\sigma_{\rho}=\frac{(1-2\mu_{b})q\text{s}}{2 \pi R^{2}}\Big{(}\frac{z}{R}-\frac{R}{R+z}\Big{)}\\ \sigma_{z}=-\frac{3z^{2}q\text{d}\text{s}}{2\pi R^{3}}&\tau_{z\rho}=\tau_{\rho z }=-\frac{3\rho z^{2}q\text{d}\text{s}}{2\pi R^{3}}\end{cases} . \tag{21}\]  
Therefore, we can obtain the vertical displacement of a random point on the circular contact area under force element \(q\text{d}\text{s}\)  
\[(u_{z})_{z=0}=\frac{\big{(}1-\mu_{b}^{2}\big{)}q\text{d}\text{s}}{\pi E_{b} \rho}. \tag{22}\]  
Next, we consider the distributed load \(q(r)\) acting normal to the circular contact area. The vertical displacement of point \(M_{2}\) with the distance of \(r\) to the center of the circular contact area \(O_{p}\) can be obtained by integrating over the loading region. As shown in Figure 5, the area of the shadow region represented by ds is \(\rho\text{d}\text{y}\text{d}\rho\); then, we can obtain  
\[w_{2}=\big{(}u_{z}\big{)}_{M_{2}}=\frac{\big{(}1-\mu_{b}^{2}\big{)}}{\pi E_{b} }\iint_{S}q(r)\text{d}\text{y}\text{d}\rho. \tag{23}\]  
Note that the distance \(M_{1}M_{2}\) is equal to \(2\sqrt{b^{2}-r^{2}\sin^{2}\psi}\); thus, by substituting Equation (9) into Equation (23), the vertical displacement of point \(M_{2}\) can be expressed as  
\[w_{2}=\big{(}u_{z}\big{)}_{M_{2}}=\frac{2q_{0}\big{(}1-\mu_{b}^{2}\big{)}}{ \pi E_{b}b}\int_{0}^{\frac{\pi}{2}}\frac{\pi}{2}\bigg{(}\sqrt{b^{2}-r^{2}\sin ^{2}\psi}\bigg{)}^{2}d\psi=\frac{3P\big{(}1-\mu_{b}^{2}\big{)}\big{(}2b^{2}-r ^{2}\big{)}}{8E_{b}b^{3}}. \tag{24}\]