# Contact

## Concentrated Force on a Half-Plane

 Concentrated force on a half plane

From the Flamant Solution

\begin{align} F_1 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\cos\theta d\theta & = 0 \\ F_2 + 2\int_{\alpha}^{\beta} \left(\frac{C_1\cos\theta - C_3\sin\theta}{a}\right)a\sin\theta d\theta & = 0 \end{align}

and

$\sigma_{rr} = \frac{2C_1\cos\theta}{r} + \frac{2C_3\sin\theta}{r} ~;~~ \sigma_{r\theta} = \sigma_{\theta\theta} = 0$

If $\alpha = -\pi$ and$\beta = 0$, we obtain the special case of a concentrated force acting on a half-plane. Then,

\begin{align} F_1 + 2\int_{-\pi}^{0} \left(C_1\cos^2\theta - \frac{C_3}{2}\sin(2\theta)\right) d\theta & = 0 \\ F_2 + 2\int_{-\pi}^{0} \left(\frac{C_1}{2}\sin(2\theta) - C_3\sin^2\theta\right) d\theta & = 0 \end{align}

or,

\begin{align} F_1 + \pi C_1 & = 0 \\ F_2 - \pi C_3 & = 0 \end{align}

Therefore,

$C_1 = - \frac{F_1}{\pi} ~;~~ C_3 = \frac{F_2}{\pi}$

The stresses are

$\sigma_{rr} = -\frac{2F_1\cos\theta}{\pi r} - \frac{2F_2\sin\theta}{\pi r} ~;~~ \sigma_{r\theta} = \sigma_{\theta\theta} = 0$

The stress $\sigma_{rr}$ is obviously the superposition of the stresses due to $F_1$ and $F_2$, applied separately to the half-plane.

### Problem 1: Stresses and displacements due to $F_2$

The tensile force $F_2$ produces the stress field

$\sigma_{rr} =- \frac{2F_2\sin\theta}{\pi r} ~;~~ \sigma_{r\theta} = \sigma_{\theta\theta} = 0$
 Stress due to concentrated force $F_2$ on a half plane

The stress function is

$\varphi = \frac{F_2}{\pi} r\theta\cos\theta$

Hence, the displacements from Michell's solution are

\begin{align} 2\mu u_r & = \frac{F_2}{2\pi}\left[(\kappa-1)\theta\cos\theta + \sin\theta - (\kappa+1)\ln(r)\sin\theta\right] \\ 2\mu u_{\theta} & = \frac{F_2}{2\pi}\left[-(\kappa-1)\theta\sin\theta - \cos\theta - (\kappa+1)\ln(r)\cos\theta\right] \end{align}

At $\theta = 0$, ($x_1 > 0$, $x_2 = 0$),

\begin{align} 2\mu u_r = 2\mu u_1 & = 0 \\ 2\mu u_{\theta} = 2\mu u_2 & = \frac{F_2}{2\pi}\left[-1 - (\kappa+1)\ln(r)\right] \end{align}

At $\theta = -\pi$, ($x_1 < 0$, $x_2 = 0$),

\begin{align} 2\mu u_r = -2\mu u_1 & =\frac{F_2}{2\pi}(\kappa-1)\\ 2\mu u_{\theta} = -2\mu u_2 & = \frac{F_2}{2\pi}\left[1 + (\kappa+1)\ln(r)\right] \end{align}

where

\begin{align} \kappa = 3 - 4\nu & & \text{plane strain} \\ \kappa = \frac{3 - \nu}{1+\nu} & & \text{plane stress} \end{align}

Since we expect the solution to be symmetric about $x = 0$, we superpose a rigid body displacement

\begin{align} 2\mu u_1 & = \frac{F_2}{4\pi}(\kappa-1)\\ 2\mu u_2 & = \frac{F_2}{2\pi} \end{align}

The displacements are

\begin{align} u_1 & = \frac{F_2(\kappa-1)\text{sign}(x_1)}{8\mu} \\ u_2 & = - \frac{F_2(\kappa+1)\ln|x_1|}{4\pi\mu} \end{align}

where

$\text{sign}(x) = \begin{cases} +1 & x > 0 \\ -1 & x < 0 \end{cases}$

and $r = |x|$ on $y = 0$.

### Problem 2: Stresses and displacements due to $F_1$

The tensile force $F_1$ produces the stress field

$\sigma_{rr} =- \frac{2F_2\cos\theta}{\pi r} ~;~~ \sigma_{r\theta} = \sigma_{\theta\theta} = 0$
 Stress due to concentrated force $F_1$ on a half plane

The displacements are

\begin{align} u_1 & = - \frac{F_1(\kappa+1)\ln|x_1|}{4\pi\mu} \\ u_2 & = - \frac{F_1(\kappa-1)\text{sign}(x_1)}{8\mu} \end{align}

### Stresses and displacements due to $F_1 + F_2$

Superpose the two solutions. The stresses are

$\sigma_{rr} = -\frac{2F_1\cos\theta}{\pi r} - \frac{2F_2\sin\theta}{\pi r} ~;~~ \sigma_{r\theta} = \sigma_{\theta\theta} = 0$

The displacements are

\begin{align} u_1 & = - \frac{F_1(\kappa+1)\ln|x_1|}{4\pi\mu} + \frac{F_2(\kappa-1)\text{sign}(x_1)}{8\mu} \\ u_2 & = - \frac{F_2(\kappa+1)\ln|x_1|}{4\pi\mu} - \frac{F_1(\kappa-1)\text{sign}(x_1)}{8\mu} \end{align}

## Distributed Force on a Half-Plane

 Distributed force on a half plane
• Applied load is $p(\xi)$ per unit length in the $x_2$ direction.
• We already know the stresses and displacements due to a concentrated force. The stresses and displacements due to the distributed load can be found by { superposition}.
• The Flamant solution is used as a Green's function, i.e., the distributed load is taken as the limit of a set of point loads of magnitude $p(\xi)\delta\xi$.

At the point $P$

$u_2 = - \frac{(\kappa+1)}{4\pi\mu} \int_A p(\xi)\ln|x - \xi|~d\xi$

As $x \rightarrow \infty$, $u_2$ is unbounded. However, if we are interested in regions far from $A$, we can apply the distributed force as a statically equivalent concentrated force and get displacements using the concentrated force solution.

The avoid the above issue, contact problems are often formulated in terms of the { displacement gradient}

$\frac{du_2}{dx_1} = - \frac{(\kappa+1)}{4\pi\mu} \int_A \frac{p(\xi)}{x - \xi}~d\xi$

If the point $P$ is inside $A$, then the integral is taken to be the sum of the integrals to the left and right of $P$.

## Indentation due to a Frictionless Rigid Flat Punch

 Indentation by a plat rigid punch
• Start with uneven surface profile $u_0(x_1)$.
• Unsymmetric load $F$, but sufficient for complete contact over the area $A$.

Displacement in $x_2$ direction is

$u_2 = - u_0(x_1) + C_1 x_1 + C_0$

where $C_0$ is a rigid body translation and $C_1 x_1$ is a rigid body rotation.

Rigid body motions can be determined using a statically equivalent set of forces and moments

\begin{align} \int_A p(\xi)~d\xi & = -F \\ \int_A p(\xi)\xi~d\xi & = -Fd \end{align}

$-\frac{du_0}{dx_1} +C_1 = - \frac{(\kappa+1)}{4\pi\mu} \int_{-a}^a \frac{p(\xi)}{x - \xi}~d\xi ~;~~ -a < x < a$

Integral is a { Cauchy Singular Integral} that appears often and very naturally when the problem is solved using complex variable methods.

Note that the only thing we are interested in is the distribution of contact forces $p(\xi)$.If we change the variables so that $x = a \cos\phi$ and $\xi = a \cos\theta$, then

$\frac{1}{a\sin\phi}\frac{du_0}{d\phi}+C_1 = - \frac{(\kappa+1)}{4\pi\mu} \int_0^{\pi} \frac{p(\theta)\sin\theta}{\cos\phi - \cos\theta}~d\theta ~;~~ 0 < \phi < \pi$

If we write $p(\theta)$ and $du_0/d\phi$ as

\begin{align} p(\theta) & = \sum_0^{\infty} \frac{p_n \cos(n\theta)}{\sin\theta} \\ \frac{du_0}{d\phi} & = \sum_1^{\infty} u_n \sin(n\phi) \end{align}

and do some algebra, we get

\begin{align} p_0 & = -\frac{F}{\pi a} \\ p_1 & = -\frac{F d}{\pi a^2} \\ p_n & = -\frac{4\mu u_n}{(\kappa+1)a} ~;~~ n > 1 \end{align}

### Flat Punch with Symmetric Load: $u_0 = C$

In this case,

$\frac{du_0}{d\phi} = 0 \Rightarrow u_n = 0 ~;~~ n = 1 {\infty}$

Also, $d = 0$ (origin at the center of $A$), hence $p_1 = 0$. Therefore,

$p(x) = \frac{p_0}{\sin\phi} = -\frac{F}{\pi \sqrt{a^2 - x^2}}$

At $x = \pm a$, the load is infinite, i.e. there is a singularity.

## The Hertz Problem: Rigid Cylindrical Punch

 Hertz indentation
• The contact length $a$ depends on the load $F$.
• There is no singularity at $x = \pm a$.
• The radius of the cylinder ($R$) is large.

We have,

$\frac{d^2 u_0}{dx^2} = -\frac{1}{R}$

Hence,

$u_0 = C_0 - \frac{x^2}{2R} = C_0 - \frac{a^2\cos(2\phi)}{4R}-\frac{a^2}{4R}$

and

$\frac{d u_0}{d\phi} = -\frac{a^2\sin(2\phi)}{2R}$

Therefore,

$u_1 = 0 ~;~~ u_2 = \frac{a^2}{2R} ~;~~ u_n = 0 ~(n > 2)$

and

$p_0 = -\frac{F}{\pi a} ~;~~ p_1 = 0 ~;~~ p_2 = \frac{2\mu a}{R(\kappa+1)} ~;~~ p_n = 0 ~(n > 2)$

Plug back into the expression for $p(\theta)$ to get

$p(\theta) = \left(-\frac{F}{\pi a} + \frac{2\mu a}{R(\kappa+1)}\cos(2\theta) \right)/\sin\theta$

This expression is singular at $\theta=0$ and $\theta=\pi$, unless we choose

$\frac{F}{\pi a} = \frac{2\mu a}{R(\kappa+1)} \Rightarrow a = \sqrt{\frac{F(\kappa+1)R}{2\pi\mu}}$

Plugging $a$ into the equation for $p(\theta)$,

$p(\theta) = -\frac{2F\sin\theta}{\pi a} \Rightarrow p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2}$

### Two deformable cylinders

If instead of the half-plane we have an cylinder; and instead of the rigid cylinder we have a deformable cylinder, then a similar approach can be used to obtain the contact length $a$

$a = \sqrt{\frac{FR_1R_2}{2\pi(R_1+R_2)}\left( \frac{\kappa_1+1}{\mu_1} + \frac{\kappa_2+1}{\mu_2}\right)}$

and the force distribution $p$

$p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2}$

Many other problems are discussed in the texts by Timoshenko and Goodier (Elasticity) and K.L. Johnson (Contact Mechanics, 1985).

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