Pressure on walls of granite shaft

[SOLVED] Pressure on walls of granite shaft

Homework Statement

A friend said "In the deeper mines rock can literally explode if struck or impacted b/c of the pressure of the rock above it." This brought to mind the formula for hydrostatic pressure and nothing else. The density of granite is given as 2.75 g/cm^3. I want to find the pressure the rock is under at the bottom of a 1,000 foot mine shaft.

Homework Equations

P = rho * g * h + Pa

The Attempt at a Solution

(2750 kg/m^3)*(9.8 m/s^2)*(1000 feet) + (101,325 Pa) is 1,206 psi.

I know bed rock has sizable voids because there are aquifers, caves and caverns; further assuming there are voids on smaller scales the density of granite could be seen as an upper bound -- I would expect granite chunks to just float in a liquid with a density of 2.75 g/cm^3. Given this insight I decided that hydrostatic pressure probably sanely gives an upper bound. Was I right?

Last edited:

Related Introductory Physics Homework Help News on Phys.org
Check your units. The hydrostatic formula is in metric (i.e. meters), not feet.

Check your units. The hydrostatic formula is in metric (i.e. meters), not feet.
It also yields pascals and not psi. Was it wrong to assume converting feet to meters and pascals to psi goes without saying? If so here is a version you should like better:

2.75 g/cm^3 is 2750 kg/m^3
1,000 feet is 304.8 m
(2750 kg/m^3)*(9.8 m/s^2)*(304.8 m) + (101,325 Pa) is 8,315,685 pascals (1,206 psi)
8,315,685 pascals is 1,206 psi

mgb_phys
Homework Helper
Hydrostatic pressure in a mine is very real. You normally use screening (strong wire mesh) held in place with long steel bolts drilled into the rock together with sprayed on concrete, it's called 'ground control'

You also have to be careful to back-fill with rock and concrete any mined out spaces before mining new regions near them.

It also yields pascals and not psi. Was it wrong to assume converting feet to meters and pascals to psi goes without saying? If so here is a version you should like better:

2.75 g/cm^3 is 2750 kg/m^3
1,000 feet is 304.8 m
(2750 kg/m^3)*(9.8 m/s^2)*(304.8 m) + (101,325 Pa) is 8,315,685 pascals (1,206 psi)
8,315,685 pascals is 1,206 psi
Oh sorry, you're correct. I didn't know google automatically converts units.

Hydrostatic pressure in a mine is very real.
Thanks for the reply mgb_phys! I don't doubt the pressure is very real. I am just worried about whether I made a serious blunder in assuming the formula for hydrostatic pressure would be very useful for predicting the pressures involved in solids; so I thought I would post the question here. It sounds like you agree that treating solid rock as if it were a fluid is useful in this case?

You normally use screening (strong wire mesh) held in place with long steel bolts drilled into the rock together with sprayed on concrete, it's called 'ground control'
Hehe 'ground control.' Is that from personal experience in a mine? If so how deep was it and what sort of bed rock was the mine situated in?

Oh sorry, you're correct. I didn't know google automatically converts units.
I thought you made a valid point. I was intending to include a Google link but I cannot link to anything until I have made 15 posts. In lieu of the link I should probably have shown the conversions explicitly to avoid confusion.

mgb_phys
Homework Helper
Yes I think treating it as a fluid is probably reasonable. Basically whenever the applied pressure is much greater than the maximum tensile stress of the material it is a fluid.
It's the same modeling shell impacts or implosions, at those pressures steel behaves like a liquid and so the subject is called hydrodynamics.

It's a bit nervous when small bits of rock ping off a face and you have to be especially carefull under any deep shafts because even a small rock fragment falling a few 100m can be dangerous. Most of the ground control is to prevent larger lumps falling off the ceiling. A few square meters of granite falling off the roof of a tunnel could really ruin your day.

Yes I think treating it as a fluid is probably reasonable. Basically whenever the applied pressure is much greater than the maximum tensile stress of the material it is a fluid.