Contact

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Concentrated Force on a Half-Plane[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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}

The displacement gradient is


-\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[edit]

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[edit]

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[edit]

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).

Related Content[edit]

Introduction to Elasticity