Geostatic Stess & Strain: Basic Principals

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Geostatic Stress/strain revolves completely around forces on a particular point in the ground. Figure 1 will help illustrate the properties involved. For example, take point A below the ground.

  

Figure 1.

From point A there is a vertical distance, h, to the ground surface. Assuming the ground (i.e. rock or soil) is the same and consistent in the diagram, we can say that at any point the rock will have a particular unit weight. The unit weight, γ, of rock is force per cubic area. Now let’s try adding in that vertical height. By multiplying the unit weight of the rock by the vertical distance to the particular point, A, below the ground we get a force per square area. The resultant is a pressure force or stress.

There are also many more additional properties that are involved in geostatic stresses and strains. Plate tectonics, or the slabs of solid ground that move around on the plastic mantle part of the earth cause many different scenarios in geostatics. The forces involved include vertical pressures and lateral compressive or tensional forces. These forces relate back to plate tectoics where the crust of the earth moves on top of a plastic mantle. The movement of the these plates causes compressive and tensional forces on rocks and soils.

Kehew Stress Cube

Figure 2.

Figure 2 shows a block diagram describing stresses and a resultant plane of shear. This is a simplified version of a cube of rock that taken from below the ground. The three stresses,
σ1, σ2, σ3, are primary stresses in the vertical and horizontal directions. These stresses will produce a resultant shear, τ, and normal force on a plane within the cube.

Calculations:

I. Stress at a point below the ground surface:    σ1 = γh
This equation is used when the subsurface material is homogeneous from the ground surface to the point in question. In the event that there are several layers of different material (sandstone >> limestone >> conglomerate) the equation changes to sum the vertical distances multiplied by the unit weights for each of the materials.

II. Shear Plane and Normal Stress from lab testing:
Shear on an inclined plane with known σ1, σ3, and an angle β:    τ = ½(σ1 - σ3) sin(22 β)

Normal Stress on the inclined normal plane with known σ1, σ3, and an angle β:    σn = ½[(σ1 + σ3) + (σ1 - σ3) cos(2 β)]

Let's do an example to demonstrate the calculations for I and II.