geotechnology fundamental theories of rock and soil mechanics
TRANSCRIPT
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GeotechnologyFundamental Theories of Rock and Soil Mechanics
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GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = ???
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GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = Force Area
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GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
1. Concept
Stress = Pressure = Force Area
versus
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A. Stress
2. Primary Forces (natural)
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A. Stress
2. Primary Forces (natural)
a. Gravitational Forces (overlying
materials and upslope activity)
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A. Stress
2. Primary Forces (natural)
b. Tectonic Forces
“Important for Virginia and the Eastern Seaboard?”
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A. Stress
2. Primary Forces (natural)
c. Fluid Pressures (‘quick conditions’)
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GeotechnologyI. Theory of Rock and Soil Mechanics
A. Stress
3. Secondary Forces (Human Induced)
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Geotechnology3. Secondary Forces (Human Induced)
a. Excavation and Mining
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Geotechnology3. Secondary Forces (Human Induced)
b. Loading
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Geotechnology3. Secondary Forces (Human Induced)
c. Other
* Blasting
* Tunneling
* Pumping of Fluids
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4. Stress (σn ) on a plane normal to Force
σn = Force / AreaWhere n = ‘normal’, or stress perpendicular To the cross sectional area
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5. Stress on an inclined plane to Force
σ = Force / Area
Θ = angle to normal
Where inclined area = A = An/cos Θ
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5. Stress on an inclined plane to Force
σ = Force / Area
Where is 1)Normal Force and 2)Shear Force = ??
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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
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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
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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
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5. Stress on an inclined plane to Force
A reminder…
Fn = F cos ΘFs = F sin Θ A = An/cos Θ
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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
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5. Limits:(max) σn when Θ = 0(min) σn when Θ = 90
(max) τ when Θ = 45(min) τ when Θ = 0 or 90
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Example Problem Given: Force = 10 lbs, Area (normal) = 5 in2
Determine σn and τ when Θ = 0 °, Θ = 30°, Θ = 45°, and Θ = 60°
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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) =
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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
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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
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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
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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) =
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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
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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
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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?
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6. Stress (σ) in 3 dimensions
Stress at any point can be ‘resolved’ via3 mutually perpendicular stresses:
σ1 , σ2 , σ3
Where σ1 > σ2 > σ3
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B. Strain
“your ideas??”
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B. Strain
1. Strain Effects
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B. Strain 1. Strain Effects a. Stress produces deformation
Strain = dL L
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B. Strain 1. Strain Effects a. Stress produces deformation
“phi”
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B. Strain 1. Strain Effects a. Strain Ellipse
Maximum Shear Stress:Where σ1 - σ3
2
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2. Stress – Strain Diagrams
σ
ε“which material is stronger?”
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II. Elastic Response A. Young’s Modulus (E) “best shown in rocks”
E = stress σ strain ε
“The greater E is, ……?
“elastic limit”
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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”
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An Example:
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II. Elastic Response B. Poisson’s Ratio (ν)
ν = lateral strain length strain
In compression
In tension
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II. Elastic Response C. Ideal Elastic Behavior
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II. Elastic Response D. Non-Ideal Elastic Behavior
Strainhardening
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“under repeated loads”
II. Elastic Response
E. Hysteresis
Soft Rock
Hard Rock
‘delayed feedback’
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II. Elastic Response
F. Stress-Strain in Soils
Limits of Proportionality (how much of the strain is Elastic?)
AssumesOM, MD
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“under repeated loads”
II. Elastic Response
G. Repeated Loading of Soils (when rolled)
Increment of permanent strain decreases (densification)
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III. Time-Dependent Behavior – Strain
A. Creep – under static loads
Elastic response occurs instantaneously
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Collapsed Culvert, Cincinnati, OH
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III. Time-Dependent Behavior – Strain A. Creep – under static loads
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III. Time-Dependent Behavior – Strain
B. Specific Rocks
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III. Time-Dependent Behavior – Strain
C. Griggs Relationship
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III. Time-Dependent Behavior – Strain
D1. Pavements
“The Benkelman Beam measures the deflection of a flexible pavement under moving wheel loads.”
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III. Time-Dependent Behavior – Strain
D2. Mines
compression
tension
Steel is strong in tension;Transfer Load to more confined(stronger) rocks.
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III. Time-Dependent Behavior – Strain
D. Mines
compression
tension
Steel is strong in tension;Transfer Load to more confined(stronger) rocks.
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“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
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“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.”
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An example of basic principles of friction between two bodies….
Φ
Φ
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An example of basic principles of friction between two bodies….
Φ
Φ
Φ
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An example of basic principles of friction between two bodies….
Φ
Φ
Φtan Φ = τ / σnormal
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Φ
ϴ
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Our governing equations…..
Φ
Φ
ϴ
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Φ
Φ
ϴ
(cosΘ)*(cosΘ)
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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
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σ3
σ1
0 10 20 30 40
10
0
stress(lbs/in2)
stress(lbs/in2)
Mohrs Circles
8 lbs/in2
33 lbs/in2
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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
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An example……
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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…
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