# Contact Mechanics: Deformation of a ball on a surface

• phantomvommand
phantomvommand
Homework Statement
See picture below
Relevant Equations
Y = Stress / Strain
A solid ball of radius R, density ρ, and Young’s modulus Y rests on a hard table. Because of its weight, it deforms slightly, so that the area in contact with the table is a circle of radius r. Estimate r, assuming that it is much smaller than R.

I have no issues understanding the estimation of stress, but the estimation of strain greatly confuses me.

1. How did the author know that "at heights greater than r, the pressure will be smaller"? Isn't r a horizontal quantity? What does it have to do with heights?

2. "Since stress is proportional to strain, that means that the part of the ball that is significantly strained has height r." Again, how does this have anything to do with r, given that r is a horizontal quantity and that the strain is vertical (ish).

3. Even if assuming that the previous 2 claims are true, how was the equation ##\frac {\delta} {r} \sim \frac {r} {R}## obtained?

phantomvommand said:
1. How did the author know that "at heights greater than r, the pressure will be smaller"? Isn't r a horizontal quantity? What does it have to do with heights?
At height r above the flat base, the radius is ##\sqrt{2r\sqrt{R^2-r^2}}## (check my algebra). If r<<R that's about ##\sqrt{2Rr}##. The author is taking that as constituting a value significantly greater than r.
phantomvommand said:
3. Even if assuming that the previous 2 claims are true, how was the equation ##\frac {\delta} {r} \sim \frac {r} {R}## obtained?
In the diagram, the total extent of compression is ##\delta=R-\sqrt{R^2-r^2}\approx \frac{r^2}{2R}##.

phantomvommand
haruspex said:
At height r above the flat base, the radius is ##\sqrt{2r\sqrt{R^2-r^2}}## (check my algebra). If r<<R that's about ##\sqrt{2Rr}##. The author is taking that as constituting a value significantly greater than r.

In the diagram, the total extent of compression is ##\delta=R-\sqrt{R^2-r^2}\approx \frac{r^2}{2R}##.
But how does ##\sqrt {2Rr} >> r## mean that pressures at heights above r will be smaller?

Sorry, I wasnt clear about my final question. It should be why ##strain \sim \frac {\delta} {r}##. But I suppose if the claim that pressures above r are substantially smaller, then r will be the effective length used in calculation of strain, instead of the more obvious R?

phantomvommand said:
But how does ##\sqrt {2Rr} >> r## mean that pressures at heights above r will be smaller?
It is not that there is anything magic about the height being less than or more than r.
At the base, the force is spread over area ##\pi r^2##. At height r it is spread over ##2\pi Rr##. If R>>r then ##Rr>>r^2##. Therefore the stress is significantly less by the time you get to that height.
phantomvommand said:
Sorry, I wasnt clear about my final question. It should be why ##strain \sim \frac {\delta} {r}##. But I suppose if the claim that pressures above r are substantially smaller, then r will be the effective length used in calculation of strain, instead of the more obvious R?
Yes, that is the reasoning.

phantomvommand

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