Geology 229Engineering Geology

Lecture 4

Engineering Properties of Rocks(West, Ch. 6)

Outline of this Lecture

1. Rock properties in general2. Density, specific gravity, porosity, and void ratio3. Rock strength

• Definition of stress, normal and shear stress• stress units, conversion between difference unit

system• Rock strength: tensile, shear, and compressive• Failure criteria: tensile, compressive• Failure criterion: shear failure

Rock Properties for Engineering

Rock are significant for two major reasons in engineering:

(1) As building materials for constructions;

(2) As foundations on which the constructions are setting;

For the consideration of rocks as construction material the engineers concern about:

(a) Density to some extent (for calculating the weight, load to the foundation, etc.);

(b) Strength;

(c) Durability;

For the consideration of rocks as the construction foundation the engineers concern about:

(a)Density;

(b) Strength;

(c) Compressibility;

So the major difference is for material we want the durable (not have to be hard) rock, and for foundation we want hard rock (hard ones usually durable especially in a stable subsurface condition).

The same rock properties measured in lab and in field may have different values. This is because that in lab the rock properties are measured on small sized samples, but rocks in situ usually contains weakness planes (foliations, joints, cracks and fractures, etc.) In general the measured values on density, strength, seismic velocity, etc usually have a smaller value in field than in lab.

Rock mass (in situ) properties are fundamentally controlled by weakness planes.

So it is a common practice to have a room for the variability when use lab measured value to field engineering projects.

Density, specific gravity, porosity, and void ratio

Specific gravity:

vm

=ρ

Density:

w

ssG

ρρ

=

Porosity: Void ratio:

total

pore

vv

n =skeleton

pore

vv

e =

Rock with higher density usually corresponds to higher strength and better material for engineering construction.

Schematic representation of porous medium indicating relationship between air (A), solid skeleton (B) and water (C).

porosity, and void ratio

The limits of the value of porosity:

0)min(,1)max( ====total

pore

total

pore

vv

nvv

n

The limits of the value of Void ratio:

0)min(,)max( ==∞→=skeleton

pore

skeleton

pore

vv

evv

e

So

nne

een

−=

+=

1,

1

Rock Strength

The strength is the stress required to break down the rock sample. So we first need define the concept of stress.What is stress? Stress is the force per area applied on the object, it is in the same unit as pressure. In SI nit system the unit of stress is Pa for Pascal. The pressure is a concept in fluid: it corresponds to the normal stress when we expand the concept of pressure into the case for solid. Like the fluid, solid can sustain a force normal to its surface.Nevertheless, unlike the fluid, solid can also sustain the forces parallel to its surface (see the figure below for illustration). Thus for solid, we can further classify the stress into normal stress and shear stress.

x

r

y

z

Fv

Fh

Normal Stress and Shear Stress

We can use the Greek letter σn to denote the normal stress, and σs for the shear stress:

A A

Fh

Fv

v hn s

F FA A

σ σ= =

The normal stress can be either a positive or a negative value, means either a tensile normal stress, or a compressive normal stress. The left sketch in the above figure shows a case of compressive normal stress. In earth science we define the compressive normal stress is the positive one since the rocks in depth are constantly experiencing compression.

After the stress is acted on a piece of material, the material will deform. The microscopic description of the deformation of the material is called strain.

Tensional strainCompressive strain

Shear strain

Actually, since the force is a vector, and can act on any direction, it can be dissolved into 3 orthogonal directions (e.g., x, y, z, in Cartesian coordinate); a surface in or on the solid can face any direction, too. The normal of that surface is also a function of x, y, z. Consequently, a complete stress should be a tensor with 9 elements.

xx xy xz

yx yy yz

zx zy zz

σ σ σσ σ σσ σ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

σ

y

z

x

x

z

σ xz

σ yz

σ zz

yyσ

σ xy

σ zy

σ xx

σ zx

σyx

Stress Units

In traditional geology, the unit of pressure is in bars,

1 bar = 1 standard atmospheric pressure at sea level

It is also about equal to the pressure under 10 meters of water.For pressures deep in the earth we use the kilo-bar, equal to 1000 bars. The pressure beneath 10 km of water, or at the bottomof the deepest oceanic trenches, is about 1 kilo-bar. Beneath the Antarctic ice cap (maximum thickness about 5 km) the pressure is about half a kilobar at greatest.

Normal stress caused by self-gravity (overburden)

Pressure (normal stress) = density x gravity x depth= ρgz

Stress Units (cont.)

In the SI System, the fundamental unit of length is in meters, time in seconds, and mass in kilograms (basic units) . The unit in SI system for stress/Pressure is in Pascals

1 Pascal = 1 Newton/m2 or 1 kg/(m-sec2).

A force of one Newton spread out over a square meter is a pretty feeble force. Atmospheric pressure (1 bar) is about 100,000 Pascals. A manila file folder (35 g, 700 cm2 area) exerts a pressure of about 5 Pascals.

1 g = 0.001 kg, 1 cm = 0.01 m, 1 cm2 = 0.0001 m2

So

Pa905.481.95.081.90001.0001.0

70035

=×=××

Exchange between SI system and other units

By comparison the SI with traditional pressure units, we have

1 bar = 100,000 pascals = 0.1 MPa1 MPa = 1 megapascal = 10 bars 1 GPa = 1 gigapascal = 10 kilobars

The normal stress (pressure) at 1 km depth is about 25 MPa. By comparing the SI unit with the more familiar British psi (pound per square inch) units, we have

1 psi = 6895 Pa = 0.006895 MPa1 Pa = 0.000145 psi

So 1 ATM=1 bar = 100,000 Pa = 14.5 psiYou can find many more useful constants and conversion factors in the first and last pages inside the cover of the textbook.

Rock Strength

Again, the rock strength is the value of the stress required to break down the rock sample. So the strength has the same unit with stress, i.e., in Pascals (Pa). We have learned that stress can be divided into normal stress and shear stress, and normal stress can be further divided into compressive and tensile stress. Similarly, the rock strength should also have 3 types of strength:

1) compressive strength C;2) shear strength S0;3) tensile strength T.

Rock Strength (cont.)

For the same kind of rock usually we have

T < S0 < C

The tensile strength is the least, the shear strength the intermediate, and the compressive strength the maximum, for the same kind of rock. Usually, the tensile strength T is only about 10% of compressive strength C. Tensile strength governs rock behavior when a rock is under bending stresses.

F

Rock Strength and Failure Criteria

Quantitatively we can express the failure criteria by using the different rock strength.1) tensile failure will occur if we have

|σ| > T

2) compressive failure will occur if we have

|σ| > C

Rock Strength and Failure Criteria (cont.)

However, the shear failure is not as simple as the tensile and compressive failure criteria. (Because it involves both normal and shear stresses). Laboratory experiments in rock mechanics shows that the shear failure is more complicated and follows the Coulomb Criterion stated as

τ > S0 + µσ

The Coulomb Criterion shows that shear stress tends to cause failure across a plane, it is resisted by the sum of the shear strength (cohesion) of the material and a constant times the normal stress across the plane. This constant is the coefficient of friction. When the shear stress is equal to or greater than that sum, shear failure occurs.

Let’s illustrate the Coulomb criterion for shear failure by a very common test (see the sketch below). First, assume that there is no inherent shear strength (cohesion) between the block (with a mass of m) and the underlying slope. The contact area is A, and the slope on which the block is possible sliding has a dip angle of θ. We have the following relations:

cohesion: S0 = 0,normal stress: σ = mgcosθ/A,shear stress: τ = mgsinθ/A,

Put into the Coulomb criterion we get

mgsinθ/A = S0 + µ σ = µmgcosθ/A,

And we finally get

µ= tan θ = tan φ

φ is called the angle of shear resistance or frictional angleQuestion is: θ > φ? or θ < φ? Problem: of eraser on ruler.

m

Now we can generalize the shear and normal stress to include any causes, not only the gravitational force and make the inherent shear strength (cohesion) not have to be zero, we can apply this criterion to any shear deformation to check if a rock has reached the point to have a shear failure. The critical parameter in the Coulomb criterion is the coefficient of friction µ. For the rocks in the crust, the value of µ has a range of 0.6-1.0.

If the cohesion is not equal to zero, a greater shear stress is needed to make shear failure occur when the coefficient of friction keeps the same.

Readings:

Ch. 6

Homework:

Chapter 6: Problems 2,3,4