Geotechnology
Fundamental Theories of Rock and Soil Mechanics
Geotechnology
I. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = ???
Geotechnology
I. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = Force
Area
Geotechnology
I. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = Force
Area
versus
A. Stress
2. Primary Forces (natural)
A. Stress
2. Primary Forces (natural)
a. Gravitational Forces (overlying
materials and upslope activity)
A. Stress
2. Primary Forces (natural)
b. Tectonic Forces
“Important for Virginia and the Eastern Seaboard?”
A. Stress
2. Primary Forces (natural)
c. Fluid Pressures (‘quick conditions’)
Geotechnology
I. Theory of Rock and Soil Mechanics
A. Stress
3. Secondary Forces (Human Induced)
Geotechnology
3. Secondary Forces (Human Induced)
a. Excavation and Mining
Geotechnology
3. Secondary Forces (Human Induced)
b. Loading
Geotechnology
3. Secondary Forces (Human Induced)
c. Other
* Blasting
* Tunneling
* Pumping of Fluids
4. Stress (σn ) on a plane normal to Force
σn = Force / Area
Where n = ‘normal’, or stress perpendicular
To the cross sectional area
5. Stress on an inclined plane to Force
σ = Force / Area
Where inclined area = A = An/cos Θ
Θ = angle to normal
5. Stress on an inclined plane to Force
σ = Force / Area
Where is 1)Normal Force and 2)Shear Force = ??
5. Stress on an inclined plane to Force
σ = Force / Area
Where Normal Force and Shear Force = ??
cos Θ = a
h
sin Θ = o
h
5. Stress on an inclined plane to Force
σ = Force / Area
Where Normal Force and Shear Force = ??
cos Θ = a = Fn
h=F
sin Θ = o = Fs
h=F
5. Stress on an inclined plane to Force
σ = Force / Area
Where Normal Force and Shear Force = ??
cos Θ = a = Fn
h=F
sin Θ = o = Fs
h=F
Fn = F cos Θ
Fs = F sin Θ
5. Stress on an inclined plane to Force
A reminder…
Fn = F cos Θ
Fs = F sin Θ
A = An/cos Θ
5. Stress on an inclined plane to Force
A reminder…
Fn = F cos Θ
Fs = F sin Θ
A = An/cos Θ
Stress Normal = Force Normal / Area
σn = {F cos Θ} / {An/cos Θ}
Stress Shear = Force Shear / Area
τ = {F sin Θ} / {An/cos Θ}
5. Limits:
(max) σn when Θ = 0
(min) σn when Θ = 90
(max) τ when Θ = 45
(min) τ when Θ = 0 or 90
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
Θ = 30°, Θ = 45°, and Θ = 60°
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 30°
σn = {F cos Θ} / {An/cos Θ}
= (10 lbs * cos 0)/(5 in2/cos 0)
=
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * sin 0)/(5 in2/cos 0)
=
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 30°
σn = {F cos Θ} / {An/cos Θ}
= (10 lbs * cos 0)/(5 in2/cos 0)
= 2 lbs/in2
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * sin 0)/(5 in2/cos 0)
= 0 lbs/in2
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 30°
σn = {F cos Θ} / {An/cos Θ}
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * cos 30)/(5 in2/cos 30) = (10 lbs * sin 30)/(5 in2/cos 30)
= lbs/in2
= lbs/in2
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 30°
σn = {F cos Θ} / {An/cos Θ}
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * cos 30)/(5 in2/cos 30) = (10 lbs * sin 30)/(5 in2/cos 30)
= 1.50 lbs/in2
= 0.87 lbs/in2
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 45°
σn = {F cos Θ} / {An/cos Θ}
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * cos 45)/(5 in2/cos 45) = (10 lbs * sin 45)/(5 in2/cos 45)
=
=
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 45°
σn = {F cos Θ} / {An/cos Θ}
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * cos 45)/(5 in2/cos 45) = (10 lbs * sin 45)/(5 in2/cos 45)
= 1.00 lbs/in2
= 1.00 lbs/in2
Example Problem
Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °,
and Θ = 60°
σn = {F cos Θ} / {An/cos Θ}
τ = {F sin Θ} / {An/cos Θ}
= (10 lbs * cos 60)/(5 in2/cos 60) = (10 lbs * sin 60)/(5 in2/cos 60)
= 0.5 lbs/in2
= 0.87 lbs/in2
5. Limits:
(max) σn when Θ = 0
(min) σn when Θ = 90
(max) τ when Θ = 45
(min) τ when Θ = 0 or 90
Do your answers conform to the
trends shown here?
6. Stress (σ) in 3 dimensions
Stress at any point can be ‘resolved’ via
3 mutually perpendicular stresses:
σ1 , σ2 , σ3
Where σ1 > σ2 > σ3
B. Strain
“your ideas??”
B. Strain
1. Strain Effects
B. Strain
1. Strain Effects
a. Stress produces deformation
Strain = dL
L
B. Strain
1. Strain Effects
a. Stress produces deformation
“phi”
B. Strain
1. Strain Effects
a. Strain Ellipse
Maximum Shear Stress:
Where σ1 - σ3
2
2. Stress – Strain Diagrams
σ
“which material is stronger?”
ε
II. Elastic Response
A. Young’s Modulus (E)
“best shown in rocks”
E = stress σ
strain ε
“The greater E is, ……?
“elastic limit”
II. Elastic Response
A. Young’s Modulus (E)
“best shown in rocks”
E = stress σ
strain ε
“The greater E is, the less deformation per unit stress
OR
“the stronger the material”
An Example:
II. Elastic Response
B. Poisson’s Ratio (ν)
ν = lateral strain
length strain
In compression
In tension
II. Elastic Response
C. Ideal Elastic Behavior
II. Elastic Response
D. Non-Ideal
Elastic Behavior
Strain
hardening
II. Elastic Response
E. Hysteresis
‘delayed feedback’
“under repeated loads”
Hard Rock
Soft Rock
II. Elastic Response
F. Stress-Strain in Soils
Limits of Proportionality (how much of the strain is Elastic?)
Assumes
OM, MD
II. Elastic Response
G. Repeated Loading of Soils (when rolled)
“under repeated loads”
Increment of permanent strain decreases (densification)
III. Time-Dependent Behavior – Strain
A. Creep – under static loads
Elastic response occurs instantaneously
Collapsed Culvert,
Cincinnati, OH
III. Time-Dependent Behavior – Strain
A. Creep – under static loads
III. Time-Dependent Behavior – Strain
B. Specific Rocks
III. Time-Dependent Behavior – Strain
C. Griggs Relationship
III. Time-Dependent Behavior – Strain
D1. Pavements
“The Benkelman Beam measures the
deflection of a flexible pavement under
moving wheel loads.”
III. Time-Dependent Behavior – Strain
D2. Mines
Steel is strong in tension;
Transfer Load to more confined
(stronger) rocks.
compression
tension
III. Time-Dependent Behavior – Strain
D. Mines
Steel is strong in tension;
Transfer Load to more confined
(stronger) rocks.
compression
tension
IV. Shearing Resistance and Strength
A. Introduction
•Internal Friction
•Cohesion
“One of the most important engineering properties of soil is their shearing strength,
or its ability to resist sliding along internal surfaces within a mass.”
IV. Shearing Resistance and Strength
A. Introduction
•Internal Friction
•Cohesion
“One of the most important engineering properties of soil is their shearing strength,
or its ability to resist sliding along internal surfaces within a mass.”
An example of basic principles of friction between two bodies….
Φ
Φ
An example of basic principles of friction between two bodies….
Φ
Φ
Φ
An example of basic principles of friction between two bodies….
tan Φ = τ / σnormal
Φ
Φ
Φ
ϴ
Φ
Our governing equations…..
Φ
ϴ
Φ
(cosΘ)*(cosΘ)
Φ
ϴ
Φ
IV. Shearing Resistance and Strength
B. Triaxial Test for Soils & Mohrs Circles
“Strength of material ~ cohesion and angle of
internal friction”
τ = c + σnormal * tanΦ
τ = shear stress on failure plane
c = cohesion
σnormal = stress normal on failure plane
Φ = angle of internal friction
Mohrs Circles
σ3
σ1
8 lbs/in2
33 lbs/in2
τ (shear stress)
Φ = angle of internal friction
ϴ = angle between σ3 and
horizontal plane
OB = σ1
OA = σ3
OE = σnormal
DE = τ (shear stress)
ϕ
ϴ
σ3
σ1
Φ
An example……
An Example Problem:
The following Triaxial tests were performed
on multiple samples of the same soil:
Test
A
B
C
D
σ3 (psi)
7
17
23
31
σ1 (psi)
32
61
76
92
IF:
A minimum confining load (σ3) is required
to stabilize a vertical load of 70 psi
DETERMINE:
• σn
•τ
• angle of internal friction
• cohesion
An example to get you started…
The slides that follow are extra material for
your review as needed…..