Stress on Rock

[vtmike]

Hi,

I have a configuration in which a peice of rock is under hydrostatic pressure (see attached figure) and i'm trying to figure out what the stress on this rock will be.
The boundary conditions on the Steel end plates are such that the right and left plates are restricted to move in the -ve X and +ve X directions respectively.

According to my calculations:
From FBD, -(6000 X 41.3) + (6500 X 31.6) = -42,400 lbf
Stress = 42400/41.3 = 1027 psi Compressive in axial direction

But I am not sure about my approach here. I did not account for the force due to 5,850 psi in the rock because it cancels out but I am not sure if it should cancel out or taken into account as an additional compressive load on the rock?


Thanks,
Mike

 

[JedClampett]

First of all, you need a geologist. Us structural types worry more about the stresses in the stuff we build, not the rocks underneath. But in looking at the stresses, they seem very high. If you did a Mohr's Circle analysis on this, there would be very high shears. Unless the rock is more stout the concrete, you're going to have a problem. But you do have a confined section, so that helps.

 

[vtmike]

Actually I wanted to know if my approach to find the load acting on the rock is correct. I know it's a simple mechanics problem, but I feel like i'm missing something here.

 

[dhengr]

What is ‘ve X’? What does the ‘ve’ mean? I do understand X & Y coordinates, what about Z? While this may be a simple mechanics problem, there is something wrong with the sum of the horiz. forces, they aren’t equal. What are the dimensions on this FBD, what does it do, how are the forces applied and why, is the rock fully confined, what are all the mechanical properties of the rock? A problem well defined, with sufficient basic information, is usually a problem half solved. You can achieve these kinds of pressures with hydraulic fluid, and you’ll never get there with a weak rock material, not well confined. Look up Rock Mechanics.

 

[IRstuff]

positive X --> +ve X
negative X --> -ve X

TTFN

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[vtmike]

IRstuff is right about +ve X & -ve X. Sorry for using abbreviated terms.

This has been tested in this exact same configuration with the rock being intact. So assume the rock is strong enough. The two Steel end plates are circular with cross sectional areas as shown in figure. The cross setional area of the rock is 41.3 in^2 and it is 30" long.

The rock is fully confined and the pressure is applied using a pressure vessel.

 

[dhengr]

Instead of +ve X and -ve X, why not use +bighoumk X and -bighoumk X? Does the ‘ve’ actually stand for something, or could it be ignored since we are talking about the coordinate direction X, and he kindly shows his coordinate system on his FBD? Does ‘ve’ stand for vexing eccentricity, vibrantly exquisite, or something else? And, now we’ve added a new exquisite vexation (a ‘ev’?), wherein –> relates to +ve X and the very same –> relates to -ve X also. Never say what you mean, when an abbreviation or acronym will perfectly confuse the communication, or cause the reader to spend his time trying to decipher its meaning.

 

[Teguci]

if you are applying 6 ksi to the rock and the rock is reacting with 5.85 ksi back you do not have equilibrium.

As I understand it, you have an open 3 1/2" diameter hole to rock under significant pressure and you want to know what local stresses in the rock are near the hole?

If you had a 3 1/2" diameter sample of rock, the rock with expand through the hole by delta L which could be calculated using poissons ratio. However, in this example, the rock restrains itself. at the hole there will be shears and arching action. Very difficult problem.

 

[JStephen]

On the diagram, it looks like the steel areas are cross-sectional areas, but you're multiplying them by stress applied in the X direction, which doesn't make sense.

Normally, you don't know what all the boundary stresses are, that's what you're trying to find.

 

[vtmike]

Teguci,
There is no 3-1/5" dia hole. It is one solid Steel component. The blue components on both ends are solid Steel parts.

JStephen,
I'm multiplying the pressure applied on those cross-sectional areas to get forces on the rock.

 

[Teguci]

n max = 1/2 (n1+n2) + tau max
tau max = (1/2(n1 - n2)^2 + tau12^2)^.5
ny = nz = 6500 psi
nx = 6000 psi
tau12 = 0
n max = 6,600 psi
tau max = 350 psi

 

[ishvaaag]

What we lack here (I think) is a complete statement of the problem. Are the extreme faces fixed? Are both steel parts as well submerged and subject to the hydrostatic pressure? Inspection shows that if on the hydrostatic pressure applied annular surface to the left, but not to the right, you don't have horizontal equilibrium, as someone has noted. Equilibrium will be established on the boundary conditions, that stay somewhat undefined. So we have a number of cases

1. You have the whole outfit at the hydrostatic pressure, left side fixed, and then you apply mechanically 6000 additional pressure on the right surface.

The only differential pressure is that mechanically applied. Then the overall hydrostatic compression will be compounded with the additional axial compression from the right.

2. You have the right side excluded from hydrostatic pressure and apply there 6000 psi, whereas everything else is subject to hydrostatic force and the left side free to move axially but tight to rigid bearing on the left.

Then since the force from the right (247.8 kips) exceeds the hydrostatic force to the right (205.4 kips), the outfit will get compressed axially for a stess condition very similar to number 1 case, since the hydrostatic pressure applies to the whole rock test item and still has to support the axial compression.

3. As 2, but left surface welded to the rigid bearing. A very similar condition, again: hydrostatic loading plus the effect of axial loading.

If we dismiss the effects of volume change, the three previous cases amount to sigmax=sigmay<sigmaz, all sigmas compression and principal stresses.

As long you consider different boundary conditions and interfaces, plus you allow for more detail, different answers will follow.

 

[ishvaaag]

Of course my x is not the x of the drawing. sigmaz is the one aligned with the x of the drawing.