Geology 229Engineering Geology

Lecture 5

Engineering Properties of Rocks(West, Ch. 6)

Common mechanic properties:

Density;

Elastic properties:

- elastic modulii

1. Uniaxial rock mechanics test1. Deformation and strain2. Elastic modulus, Poisson’s ratio3. Hooke’s law – the linear elastic constitutive relation

2. Triaxial rock mechanics test• Mohr circle• Combination of Coulomb shear failure criterion and

Mohr circle• Expressing Coulomb criterion by principal stresses

Outline of this Lecture

Review of Rock Strength and Failure Criteria

1) tensile failure will occur if we have

|σ| > T

2) compressive failure will occur if we have

|σ| > C

3) shear failure will occur if we have (Coulomb Criterion)

τ > S0 + µσ

Typical rock strength values

1 psi = 6.895 kPa

Uni-axial Compression Test

In a uni-axial compression test, the direction of the load is called the maximum principal direction and there are no other loads (forces) working on other direction. Attention should be exercised to the fact that the convention for defining the principal direction and principal stress may be different from earth science and physics. In physics, it is usually define the tensile stress, the extensional deformation as positive, whereas in earth science it is the opposite. We define compressive stress, and compressional deformation as positive,simply because the nominal status in the crust is compressive and compressional (think about a diver at the depth of 100 m, but the material is not water but rock now, and the normal compressive stress is 2.5 MPa at depth of 100 m of rock).

Rock DeformationFundamental Definitions

First we need define deformation. The deformation is the change of shape and size of a material under loading. The elastic deformation is the part or the kind of deformation that can be recoverable, i.e., after the load is removed, the material changes back to its original shape and size, The part or kind of deformation that cannot be recovered is the plastic or ductile deformation.

Correspondingly, the property of the material of elastic deformation is called elasticity; the property of plastic deformation is the plasticity.

In a relatively loose definition, load is the external force acting of the material to cause deformation, so load and deformation is a pair of terms, one is the reason, the other is the result. In a more specific or more quantitative way there is another pair of terms to describe it. This is the stress and strain.

stress:force per unit area;

Special case: stress in fluid is called pressure, the stress at a given point in all directions are the same – isotropic stress.

Strain: deformation in a unit length, area, or volume.

The stress-strain relation of rock deformationFor a uni-axial loading test

STRAIN

Change in shape or size of an object in response to an applied stress.

= DeformationThree Types of Strain

•Elastic

•Ductile (Plastic)

•Brittle (Rupture)

Elastic DeformationA temporary change in shape or size that is recovered when the applied stress is removed.

If the response of the material to the load/unload is instantaneous, it is a pure elastic material;

If the response of the material to the load/unload needs finite time, it is a visco-elastic material;

Ductile (Plastic) Deformation

• A permanent change in shape or size that is not recovered when the stress is removed.

• i.e. it flows or bends

Rupture is a kind of Brittle Deformation

• the loss of cohesion of a body under the influence of deforming stress.

• i.e. “it breaks”

Let’s concentrate only on the elastic properties of the earth materials

These material properties are described by elastic modulii.

Young’s modulus E

Young’s modulus is the stress needed to compress the solid to shorten in a unit strain.

Poisson’s ratio ν

Poisson’s measures the relativity of the expansion in the lateral directions and compression in the direction in which the uni-axial compression applies.

zzE

/1

∆=

σ

zzrr

//

∆∆

−=ν

Shear Modulus µ (some books use the letter G)

The shear modulus describes how difficult it is to deform a cube of the material under an applied shearing force. For example, imagine you have a cube of material firmly cemented to a table top. Now, push on one of the top edges of the material parallel to the table top. If the material has asmall shear modulus, you will be able to deform the cube in the direction you are pushing it so that the cube will take on the shape of a parallelogram. If the material has a large shear modulus, it will take a large force applied in this direction to deform the cube. Gases and fluids can not support shear forces. That is, they have shear modulus of zero. From the relation given above, notice that this implies that fluids and gases do not allow the propagation of the shear motion carried by the seismic S-waves.

Shear Modulus µ

The shear modulus describes how difficult it is to deform a cube of the material under an applied shearing force. For example, imagine you have a cube of material firmly cemented to a table top. Now, push on one of the top edges of the material parallel to the table top. If the material has asmall shear modulus, you will be able to deform the cube in the direction you are pushing it so that the cube will take on the shape of a parallelogram. If the material has a large shear modulus, it will take a large force applied in this direction to deform the cube. Gases and fluids can not support shear forces. That is, they have shear modulii of zero. From the equations given above, notice that this implies that fluids and gases do not allow the propagation of S waves.

yxAF

//

∆=µ

Shear Modulus µ (cont.)

yxAF

//

∆=µ

Bulk Modulus K

Imagine you have a small cube of the material making up the medium and that you subject this cube to pressure by squeezing it on all sides. If the material is not very stiff, you can image that it would be possible to squeeze the material in this cube into a smaller cube. The bulk modulus describes the ratio of the pressure applied to the cube to the amount of volume change that the cube undergoes. If k is very large, then the material is very stiff, meaning that it doesn't compress very much even under large pressures. If K is small, then a small pressure can compress the material by large amounts. For example, gases have very small Bulk Modulus . Solids and liquids have large Bulk Modulus.

vvAFK//

∆= ∆v

σ

Hooke’s law in macroscopic form

kuf −=

Hooke’s law in microscopic form

εσ M=

The linear relationship between the stress and strain in the elastic part of the deformation process can be well described by the Hooke’s law in a simple form of σ = Mε, and a complete form of:

λ is called the Lame’s constant and it related to Yong’s modulus E and Poisson’s ratio ν.

Any change in rock or soil property that causes ρ, µ, or K to change will cause seismic wave speed to change. For example, going froman unsaturated soil to a saturated soil will cause both the density and the bulk modulus to change. The bulk modulus changes because air-filled pores become filled with water. Water is much more difficult to compress than air. In fact, bulk modulus changes dominate this example. Thus, the P wave velocity changes a lot across water table while S wave velocities change very little.

Although this is a single example of how seismic velocities can change in the subsurface, you can imagine many other factors causing changes in velocity (such as changes in lithology, changes in cementation, changes in fluid content, changes in compaction, etc.). Thus, variations in seismic velocities offer the potential of being able to map many different subsurface features.

Seismic velocity vs material’s mechanic properties

Seismic Velocities related to material propertiesVp- P-wave (compressive wave) velocityVs- S-wave (shear wave) velocity

So, seismic velocities are determined by the mechanic properties of the materials in which the seismic waves propagate through.

From: Sheriff and Geldart, Exploration Seismology, p69.

Rock C (psi) S (psi) T (psi) Ex106 (psi) φ (deg) ν Gs n(%)

Readings:

Ch. 6

Homework:

1, what is the seismic S-wave velocity in the near surface earth given:Density = 2500 kg/(m3), the shear modulus = 1010 Pa.2, if the Poisson’s ratio is 0.25 (this is known as the Poisson condition which can be a nominal value for the Poisson’s ratio of earth materials), what is the P-wave velocity in the same material as in Question 1 (check the relations of elastic parameters in the table).