# Talk:Contact mechanics

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## Too technical

In my opinion the page is too technical, I added the technical template to the top of the page.

• The introduction is quite long, and already contains a lot of details. It might try to focus more on the essential ideas.
• The distinction between non-adhesive and adhesive contact might be introduced separately.
• Classical solutions could be an entire top-level section by itself.
• Analytical and numerical solution techniques could also be discussed separately.
• The purposes, strengths and weaknesses of the various adhesive contact theories could be introduced in more general terms, before the theories are discussed in detail.

Edwinv1970 (talk) 09:20, 22 March 2011 (UTC)

## Line contact on a plane section

I think the integral formulas given in line contact on a plane section are incorrect. The dimensions don't match. Can someone confirm? I was reading contact mechanics by johnson and the formulas look a little different there. User:Blooneel 24 June, 2010

Johnson's book assumes a left-handed coordinate system with the $z$-axis pointing down. The results given in this article assume that the $z$-axis points up. That leads to the different relations. See Barber's book on elasticity for the form given in this article. Bbanerje (talk) 03:45, 25 June 2010 (UTC)
There seems to be an inconsitency between the (x,y) directions shown on the diagram and the use of z in the formulas. It needs to be clear what the directions are.Eregli bob (talk) 04:37, 30 August 2010 (UTC)

## Coordinate system

I am wondering about the coordinate system in the Chapter "Loading on a Half-Plane". The coordinate z seems to be the direction normal to the surface (as also in the chapter before). Does this chapter present a 3D solution for a point load given in the plane y=0? Than the term "Loading on a Half space" would be better. Or is a plane strain (plane stress) solution presented?

In any case: the appearance of the y coordinate in the figure ( (x,y) and σy ) is misleading. For the same reason y should also be replaced by z in the sentence following the formulae  : "for some point, (x,y), in the half-plane. " B Sadden (talk) 14:57, 30 May 2009 (UTC)

## Error in sphere on half-space?

I may be wrong, but I believe that there is a mistake here; the radius of the contact area is quoted as being sqrt (R * d), I think (from a bit of cursory mathematics) that is should actually be sqrt (2 * R * d), can anyone confirm this, I may be mistaken so I won't change this unless someone else confirms...

thanks,

Mike Strickland —Preceding unsigned comment added by 152.78.178.59 (talk) 16:59, 27 July 2010 (UTC)

The Hertz solution for the elastic displacements in the region of contact is
$u_1 + u_2 = \delta - A x^2 - B y^2$
where $x,y$ are coordinates of the contact surfaces projected on to the $x-y$-plane. For a circular contact area with radius $a$,
$A = B = \tfrac{1}{2}\left(\tfrac{1}{R_1} + \tfrac{1}{R_2}\right)$
If the second surface is a half-plane, $R_2 \rightarrow \infty$ and we have
$A = B = \tfrac{1}{2 R_1} = \tfrac{1}{2R}$
Therefore,
$u_1 + u_2 = \delta - \tfrac{1}{2 R} r^2$
where $r$ is the radial distance to a point in the contact region from the center of contact. The Hertzian pressure distribution
$p = p_0 \left[ 1 - (\tfrac{r}{a})^2\right]^{1/2}$
$u_1 = \left(\tfrac{1-\nu_1^2}{E_1}\right)\left(\tfrac{\pi p_0}{4a}\right)\left(2 a^2 - r^2\right) ~;~~ u_2 = \left(\tfrac{1-\nu_2^2}{E_2}\right)\left(\tfrac{\pi p_0}{4a}\right)\left(2 a^2 - r^2\right)$
Plugging these into the relation for $u_1+u_2$ gives
$\left(\tfrac{1}{E^*}\right)\left(\tfrac{\pi p_0}{4a}\right)\left(2 a^2 - r^2\right) = \delta - \tfrac{1}{2 R} r^2$
At $r = 0$
$\delta = \tfrac{\pi p_0 a}{2 E^*}$
For $r = a$ plugging in the expression for $\delta$ gives
$a = \tfrac{\pi p_0 R}{2 E^*}$
Therefore
$\tfrac{a}{\delta} = \tfrac{R}{a} \Leftrightarrow a^2 = R\delta \implies a = \sqrt{R\delta} \quad \square$
Bbanerje (talk) 00:00, 28 July 2010 (UTC)

## Error in rigid conical indenter and an elastic half-space?

The German Wikipedia has a and d switched in this formula: $a=\frac{2}{\pi}d\tan\theta$. And indeed, if one lets theta get towards 90° then only the switched version makes sense (radius gets towards 0). Peterthewall (talk) 17:55, 28 February 2013 (UTC)