Lecture 3

Geometrical properties of rock and their effects

on rock mechanical behaviour
Triaxial Setting, 1 Million Pound Machine and Data
Center
Large Lab Triaxial and Portable Field Triaxial Testing
System
Large Scale Drilling Structure
CT Scan, High Pressure Pump, Reservoir Simulators
60000 psi and 12” Sample System and Mud Pump
System
Core-Log Integration
Core Analysis and Petrographical Image Analysis
Rock mechanics
Rock mechanics is part of the broader scientific field of
geomechanics which refers to the science that studies the mechanical
behaviour of all earth materials, including soils. In the other hand,
rock mechanics is the theoretical and applied science of the
mechanical behaviour of rock. It is that branch of mechanics
concerned with the response of rock to the force fields of its physical
environment.
The rock elastic properties
Determining the Rock elastic properties primarily include Poisson’s
ratio, Young’s modulus and rock properties analysis. Well log
sonic/seismic velocity and density data can be used to calculate the
dynamic properties of either the Poisson’s ratio or Young’s modulus.
The results were then being converted into static Young’s modulus
for engineering applications. Rock properties can also be calculated
using DT/velocity and/or porosity data.
Mechanical Properties of Rocks

➢Uniaxial Compressive Strength/ Unconfined Compression Strength

➢ Triaxial Compressive Strength

➢Tensile Strength

➢ Shear Strength

➢Poisson’s Ratio

➢ Young’s Modulus
Rock Strength
Rock can be subjected to three primary types of stress
1. Compressive stress decreases the volume of the rock
by forces acting inward and directly opposite to each
other

2. Shear stress is caused by two equal forces acting on

opposite directions as a couple

3. Tensile forces tend to pull a substance apart by

outward acting, equally opposing forces
• Strength of intact rock depends on component mineral strengths
and the way they are bound together – by interlocking or
cementation.

• Rock failure is generally in shear.

• For determining the above strength values the tests are conducted
either on intact rock specimens in the laboratory tests or on rock
mass in the field, i.e., in-situ strength tests.
 Uniaxial Compressive Strength/Unconfined
Compression Strength
Unconfined Compression Strength (qu, UCS) is a strength under
uniaxial load in unconfined state. UCS of dry rock is standard for
defining rock strength. UCS of saturated rock is less than of dry
rock.

The strength in general

P – the failure load in newtons

A – the cross-sectional area of the sample in square meters
Engineering Classification of Intact Rock

Class Level of Strength UCS, MPa

A Very High 220
B High 110 – 222
C Medium 55 – 110
D Low 27.5 – 55
E Very Low 27.5
 Triaxial Compressive Strength
Triaxial Compressive Strength is measured when the rock sample is
confined. Generally, the strength increases with an increase in
confining stress. This strength is rarely measured in rocks, more
important for soil.
Cylinder of rock loaded axially
(σ1) with equal confining
stresses on radial axes due to
fluid bath pressure (σ3). The
data plotted on Mohr diagram
to determine angle of internal
friction (φ) and cohesion (c).
 Tensile Strength of Rocks

• Tensile strength (T0, St) is rarely measured or

applied directly. Tensile strength of rock is generally
about UCS/20 to UCS/8.

• For rock stability, the tensile strength is not as

significant parameter as the UCS.
 Shear Strength and Peak Strength of Rock
• Shear Strength of Rock (Ss) is resistance to direct shear when
unconfined.
• General relationship
• Shear strength typically ranges from UCS/6 in strong rock to
UCS/2 in soft clay.

• Shear Strength is particularly important for layered rock

Peak Strength on initial shearing declines to residual strength
along the sheared surface; there is no accepted measure of rock
brittleness (post peak decline of strength). Shear strength depends on
the cohesion due to cementation and the internal friction angle. The
friction component increases with the normal stress.
Stress Strain Relationship for Some Rock
Poisson’s ratio
Poisson’s ratio is “the ratio of
transverse contraction strain to
longitudinal extension strain in the
direction of the stretching force.”

Poisson Effect
When a material is stretched in one direction, it tends to compress in
the direction perpendicular to that of force application and vice
versa. The measure of this phenomenon is given in terms of Poisson’s
ratio. For example, a rubber band tends to become thinner when
stretched.
• For tensile deformation, Poisson’s ratio is positive.
• For compressive deformation, it is negative.
• Here, the negative Poisson ratio suggests that the material will
exhibit a positive strain in the transverse direction, even though the
longitudinal strain is positive as well.
• For most materials, the value of Poisson’s ratio lies in the range, 0
to 0.5.
• The Dynamic Poisson’s ratio can be calculated based on the elastic
wave theory. In the calculation the elastic compressional and shear
velocities obtained from sonic logs (Real time LWD Sonic data).
 Deformability Properties of Rocks
Modulus of elasticity or Young’s modulus is a slope of stress-strain
diagram and refers to the recoverable (elastic) deformation. The non-
recoverable part is termed plastic deformation.
▪ Commonly, the elastic deformation is directly proportional to the stress state
▪ For unconfined compression: E=σ/ε
▪ The proper value of the modulus is the one which best relates to the strain
level and confinement level of the rock under load
▪ The accepted practice in rock engineering is to use the tangent modules at
50% maximum load
▪ Rock should be loaded in the same direction as will occur during loading of
the engineering structure
Ei – initial tangent modulus
Es – secant modulus (at a particular point)
Et – tangent modulus (at a particular point)
• The Dynamic Young’s modulus can be calculated based on the
elastic wave theory. In the calculation the elastic shear velocity
and/or density obtained from sonic logs, seismic surveys and/or
Poisson’s ratio are needed.

• The Static Young Modulus was calculated by converting the

Dynamic Young’s modulus dataset into a Static Young’s modulus
dataset.
 Mohr’s Circle
Mohr's circle is a two-dimensional graphical representation of the
transformation law for the Cauchy stress tensor.
Mohr's circle is often used in calculations relating to mechanical
engineering for materials' strength, geotechnical engineering for
strength of soils, and structural engineering for strength of built
structures. It is also used for calculating stresses in many planes by
reducing them to vertical and horizontal components. These are
called principal planes in which principal stresses are calculated;
Mohr's circle can also be used to find the principal planes and the
principal stresses in a graphical representation, and is one of the
easiest ways to do so.
• The transformation equations for plane stress can be represented
in graphical form by a plot known as Mohr’s Circle.

• This graphical representation is extremely useful because it enables

you to visualize the relationships between the normal and shear
stresses acting on various inclined planes at a point in a stressed
body.
The abscissa (X axis) and ordinate (Y axis) ( σn,τn) of each point on
the circle are the magnitudes of the normal stress and shear stress
components, respectively, acting on the rotated coordinate system. In
other words, the circle is the locus of points that represent the state of
stress on individual planes at all their orientations, where the axes
represent the principal axes of the stress element.
 Stress Transformation Equations
(Mohr circle Equation)

From equilibrium of forces on the infinitesimal element, the

magnitudes of the normal Ϭn and the shear stress τn are given by

 бy y1
Y
τxy
x1
τxy бy1
Y
τx1y1 бx1
бx τx1y1
бx X
τx1y1 ϴ
X
τxy τx1y1

τxy бx1 бy1

бy
 Derivation of Mohr’s Circle
• If we vary θ from 0° to 360°, we will get all possible values of σx1
and τx1y1 for a given stress state.

• Eliminate θ by squaring both sides of 1 and 2 equation and adding

the two equations together.
 Mohr’s Circle Equation
The circle with that equation is called a Mohr’s Circle, named after
the German Civil Engineer Otto Mohr. He also developed the
graphical technique for drawing the circle in 1882.

Where the equation with center (σavg,0) and radius R.

 Constructing Mohr’s Circle
1. Draw a set of coordinate axes with σx1 as positive to the right and τx1y1 as
positive downward.
2. Locate point A, representing the stress conditions on the x face of the
element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note that
point A on the circle corresponds to θ = 0°.
3. Locate point B, representing the stress conditions on the y face of the
element by plotting its coordinates σx1 = σy and τx1y1 = -τxy. Note that
point B on the circle corresponds to θ = 90°.
4. Draw a line from point A to point B, a diameter of the circle passing
through point c (center of circle). Points A and B are at opposite ends of the
diameter (and therefore 180° apart on the circle).
5. Using point c as the center, draw Mohr’s circle through points A and B. This
circle has radius R. The center of the circle c at the point having coordinates
σx1 = σavg and τx1y1 = 0.
 Mohr’s Circle
бy B бy
τxy
B(ϴ=90) τxy
-τxy

Y
бx1 бx
C R бx
τxy

X
A(ϴ=0) τxy
A
бavg τxy
бx
τx1y1 бy
 τxy B
бy
Principle stresses Y τxy
бx
B(ϴ=90) бx X
τxy
A

2ϴp2 τxy

бy
бx1 Y
C б2

R 2ϴp1 P2 б1
ϴp2
P1
A(ϴ=0) X
2ϴp1

б1 б2

Principle stresses are stresses that act on a principle

τx1y1 surface. This surface has no shear force components
(that means τx1y1=0)
 Maximum shear stress бy
τxy B
B(ϴ=90) Y τxy
бx
бx X
τmin τxy
2ϴs A

бx1 τxy

C бy
τmax R Y
бs
A(ϴ=0)
τmax бs
The point of maximum τmax
Ţx1y1value is the maximum τmax
ϴs
X

бs shear stress value Ţmax. τmax

Note carefully the direction бs бs

τx1y1
of the shear forces
 Example
1. Draw the Mohr’s Circle of the stress element shown below.
2. Determine the principle stresses and the maximum shear stresses.
3. Find the stress components when it is inclined at 30° clockwise.
4. Draw the corresponding stress elements.
• What we know:
σx = -80 Mpa σy = +50 Mpa τxy = 25 Mpa
• Coordinates of Points
A: (-80,25) B: (50,-25)
1. Mohr’ cycle

B 50 MPa

Y A(ϴ=0)

80 MPa 80 MPa C
X б
A R
B(ϴ=90)

50 MPa
τ
2. The principle stresses

A(ϴ=0)
2ϴ2
б2 C б1
б
2ϴ1 R

B(ϴ=90)
2ϴ
τ
 Define бavg and R
бx + бy бx + бy
бavg = R= + τxy 2
2 2

бx + бy -80 + 50
C = бavg = = = - 15
2 2

2
R= ( -80 + 50
2
) +(25)
2
= -65 2 + 25 2 = 69.6
 Б1,2 = c ± R Y
Б1,2 = -15 ± 69.6 54.6 MPa
Б1 = -15 + 69.6 = 54.6 MPa
Б2 = -15 - 69.6 = -84.6 MPa 100.5° 84.6 MPa

X
84.6 MPa 10.5°

25 54.6 MPa
tan2ϴ 2 = = 0.3846
80-15
2ϴ2 = 21.0°
2ϴ1 = 21.0° + 180° = 201°
ϴ1 = 100.5° ϴ2 = 10.5°
The maximum shear stresses
τsmin

A(ϴ=0)
2ϴsmin
2ϴ
C
2ϴ2 б

2ϴ2 = 21.0° 2ϴsmax R

2ϴsmin = -(90-21.0) = 69.0° B(ϴ=90)
ϴsmin = -34.5° 2ϴ2 = 21.0°
Taking sign convention 2ϴsmax = 90-21.0 = 111.0°
into account. τsmax τ ϴsmax = 55.5°
 Y
15 MPa
15 MPa
100.5°

55.5°
X
-34.5°

15 MPa 15 MPa

69.6 MPa
The stress components when it is inclined at 30° clockwise.
C (ϴ=-30)

A(ϴ=0) -60

-60 + 180 B(ϴ=90)

ϴ = -30°
D (ϴ=-30 + 90) 2ϴ
τ 2ϴ = -60°
 бx + бy б x - бy
бx1 - = Cos2ϴ + τxy sin2ϴ
2 2
τx1y1 = бx + бy
sin2ϴ + τxy cos2ϴ
2 Y
25.8 MPa 4.15 MPa
σx = -25.8 Mpa
X
σy = -4.15 Mpa -30°

Txy = 68.8 MP 4.15 MPa 25.8 MPa

69.8 MPa
Any Questions

Thank you