I had gone through too many papers and am unable to find an exact or rough value of pressure at depth 8-12 m generated by rocks like granite, sandstone, basalt and quarzite. Could you please suggest how much pressure is generated by rocks like granite, sandstone, basalt and quarzite in nature at depth of 8-12 m?
I had calculated from mathematical equation pgh, where p= density of rock, g= gravity (acceleration) and h= depth. For different rock, stress generated by rocks is different, can we get apprx. vertical stress generated by every rock at depth of 10-12 m?
Unit weight of most of rock varies between 2400 to 2800 kg per cubic meter, multiply this with depth and acceleration due to gravity you will get the pressure at desired depth.
Dear Mr. Singh,
your question is hard to be placed within one of my geoscientific disciplines. Could you give us full particulars so as to see whether there is still another possibility to solve the problem.
Best regards
Harald G.Dill
Hope you find at once handy approx stress below
For case 1.
1' foot of water column exerted 0.433psi/ft=0.03kg=30gm.
(Therefore 1 meter =3.3'x30gm=100gm aprrox=0.1kg)
General Handy Calculation of pressure of hydrostatic gradient with depth.
hydrostatic column meter kg psi bar
1 meter 0.1kg
10meter 1kg 14.2psi 1
100metrs 10kg 142psi 10
1000meters 100kg 1420psi 100
1500 meters 150kg 2130psi 150
2000 meters 200kg 2840psi 200
same way you can calculate for Litho static gradient & Litho pressure putting below value.
Litho static gradient = (rocks+ water both) =0.533psi/ft.
Litho pressure = (rocks only) =1.08psi/ft.
What I would normally ask the students to do is to work on the basis of knowing the density of the rocks, the depth of interests and the acceleration due to gravity. This will give a very good approximation. The other thing to be careful about are the units (SI) -that they are consistent.
For the entire 10 to 12 meter depth find out the individual lithological column including soil / debris and find out the density (p)of those lit ho units (both saturated and unsaturated and average it out). find out the depth to water column or Water Level. and then calculate vertical pressure (pgh)at each point of interest both above & below water table mathematically described else where.(All test under confined condition will give rough estimate of vertical stress) It is a good research item, carry it on, which may help us in estimating tunneling condition at depth. Best wishes.
To calculate the pressure at that depth you need to know also the proportions of all these rocks or how they are presented down. Without knowledge from the geophysics or from other sources it will be difficult to estimate the total lithostatic pressure.
At 8 to 12 m you should precise if you looking for rock's pressure or pore pressure if existent. But supposing that at this depth no water table is present the pressure should be relative to the density of there rocks only: lthostatic pressure.
Pressure (in bar) = (D x H) / 10.2
D: Density of your sample (is better to measure real rock's density from samples)
H: Vertical depth.
For Granite (D ~ 2.65) and Basalt (D ~ 2.9) you can use these average density from books ! but they could be slightly different in reality (they still averages!).
If the only force is gravity at shallow depth (12 m), vertical pressure is nominally just the rock density times height, a straightforward calculation for any rock if you know its density. Moisture content must be included in the density. Take care with the units. In some situations water pressure or even air pressure may have to be considered. For a large enough area, density times height will give a good average value for vertical pressure but the details may be more complicated.
In stiff rock such as granite, sandstone, basalt or quartzite, the actual pressure at a point depends on the fracture condition of the rock. If the rock is fractured, horizontal pressure or vertical pressure in the rock may be zero across open fractures or some large value (more than the nominal pressure) at contact points along the fractures. If water fills open fractures or pore spaces, then only water pressure exists in the open fractures or pore spaces.
If the rock is solid, homogeneous, and isotropic and if there is no tectonic force, horizontal pressure in the rock (Ph) may be related to vertical pressure (Pv) using Poisson's ratio (pr) by the formula:
Ph = Pv * pr / (1 - pr)
For most rock Poisson's ratio is around 0.2 or 0.3.
Maximum pr value is 0.5 for any fluid so Ph = Pv * 0.5 / (1 - 0.5) = Pv in fluid.
If there is some tectonic force or an artificial force such as the weight of a large building or dam, the vertical and horizontal stresses (pressure) can be almost anything.
Some rocks at depth or even near the surface contain a fossil or trapped stress from a previous tectonic condition. Pore fluid pressure adds another complication but usually not at shallow 12 m depth.
Dear Singh
Your answer is right if you mean the lithostatic pressure which equals to pgh. In this case the important variable factor is the difference in the density of the different rocks e.g. granite, sandstone....etc. If there is a pore fluid pressure it will be subtracted. For further details you can refer to structural geology texts....
I am thanking to everyone for valuable information and I got answer from above discussion.
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