GeotechnologyFundamental Theories of Rock and Soil Mechanics
GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = ???
GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = Force Area
GeotechnologyI. 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’)
GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
3. Secondary Forces (Human Induced)
Geotechnology3. Secondary Forces (Human Induced)
a. Excavation and Mining
Geotechnology3. Secondary Forces (Human Induced)
b. Loading
Geotechnology3. Secondary Forces (Human Induced)
c. Other
* Blasting
* Tunneling
* Pumping of Fluids
4. Stress (σn ) on a plane normal to Force
σn = Force / AreaWhere n = ‘normal’, or stress perpendicular To the cross sectional area
5. Stress on an inclined plane to Force
σ = Force / Area
Θ = angle to normal
Where inclined area = A = An/cos Θ
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 = ??
Fn = F cos ΘFs = F sin Θ
cos Θ = a = Fn h = F
sin Θ = o = Fs h = F
5. Stress on an inclined plane to Force
A reminder…
Fn = F cos ΘFs = F sin Θ A = An/cos Θ
5. Stress on an inclined plane to Force
Stress Normal = Force Normal / Areaσn = {F cos Θ} / {An/cos Θ}
Stress Shear = Force Shear / Areaτ = {F sin Θ} / {An/cos Θ}
A reminder…
Fn = F cos ΘFs = F sin Θ A = 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} = (10 lbs * cos 30)/(5 in2/cos 30) = lbs/in2
τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 30)/(5 in2/cos 30) = 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} = (10 lbs * cos 30)/(5 in2/cos 30) = 1.50 lbs/in2
τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 30)/(5 in2/cos 30) = 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} = (10 lbs * cos 45)/(5 in2/cos 45) =
τ = {F sin Θ} / {An/cos Θ} = (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} = (10 lbs * cos 45)/(5 in2/cos 45) = 1.00 lbs/in2
τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 45)/(5 in2/cos 45) = 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} = (10 lbs * cos 60)/(5 in2/cos 60) = 0.5 lbs/in2
τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 60)/(5 in2/cos 60) = 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’ via3 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
ε“which material is stronger?”
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 stressOR“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
Strainhardening
“under repeated loads”
II. Elastic Response
E. Hysteresis
Soft Rock
Hard Rock
‘delayed feedback’
II. Elastic Response
F. Stress-Strain in Soils
Limits of Proportionality (how much of the strain is Elastic?)
AssumesOM, MD
“under repeated loads”
II. Elastic Response
G. Repeated Loading of Soils (when rolled)
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
compression
tension
Steel is strong in tension;Transfer Load to more confined(stronger) rocks.
III. Time-Dependent Behavior – Strain
D. Mines
compression
tension
Steel is strong in tension;Transfer Load to more confined(stronger) rocks.
“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 StrengthA. 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 StrengthA. 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….
Φ
Φ
Φtan Φ = τ / σnormal
Φ
ϴ
Our governing equations…..
Φ
Φ
ϴ
Φ
Φ
ϴ
(cosΘ)*(cosΘ)
IV. Shearing Resistance and StrengthB. Triaxial Test for Soils & Mohrs Circles
“Strength of material ~ cohesion and angle of internal friction”
τ = c + σnormal * tanΦ
τ = shear stress on failure planec = cohesionσnormal = stress normal on failure planeΦ = angle of internal friction
σ3
σ1
0 10 20 30 40
10
0
stress(lbs/in2)
stress(lbs/in2)
Mohrs Circles
8 lbs/in2
33 lbs/in2
ϕ ϴ
Φ = angle of internal frictionϴ = angle between σ3 and
horizontal plane
OB = σ1
OA = σ3
OE = σnormal DE = τ (shear stress)
Φ
τ (s
hear
str
ess)
σ1σ3
An example……
An Example Problem:
The following Triaxial tests were performedon multiple samples of the same soil:
Test σ3 (psi) σ1 (psi)A 7 32B 17 61C 23 76D 31 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…..