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 (σ n ) on a plane normal to Force σ n = Force / Area Where 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 = A n /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 Θ Fs = 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 Θ Fs = F sin Θ A = An/cos Θ

5. Stress on an inclined plane to Force Stress Normal = Force Normal / Area σ n = {F cos Θ} / {A n /cos Θ} Stress Shear = Force Shear / Area τ = {F sin Θ} / {A n /cos Θ} A reminder… Fn = F cos Θ Fs = 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 in 2 Determine σn and τ when Θ = 0 °, Θ = 30°, Θ = 45°, and Θ = 60°

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 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 in 2 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/in 2 τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 0)/(5 in2/cos 0) = 0 lbs/in 2

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 Determine σn and τ when Θ = 0 °, and Θ = 30° σn = {F cos Θ} / {An/cos Θ} = (10 lbs * cos 30)/(5 in 2 /cos 30) = lbs/in 2 τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 30)/(5 in 2 /cos 30) = lbs/in 2

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 Determine σn and τ when Θ = 0 °, and Θ = 30° σn = {F cos Θ} / {An/cos Θ} = (10 lbs * cos 30)/(5 in 2 /cos 30) = 1.50 lbs/in 2 τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 30)/(5 in 2 /cos 30) = 0.87 lbs/in 2

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 Determine σn and τ when Θ = 0 °, and Θ = 45° σn = {F cos Θ} / {An/cos Θ} = (10 lbs * cos 45)/(5 in 2 /cos 45) = τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 45)/(5 in 2 /cos 45) =

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 Determine σn and τ when Θ = 0 °, and Θ = 45° σn = {F cos Θ} / {An/cos Θ} = (10 lbs * cos 45)/(5 in 2 /cos 45) = 1.00 lbs/in 2 τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 45)/(5 in 2 /cos 45) = 1.00 lbs/in 2

Example Problem Given: Force = 10 lbs, Area (normal) = 5 in 2 Determine σn and τ when Θ = 0 °, and Θ = 60° σn = {F cos Θ} / {An/cos Θ} = (10 lbs * cos 60)/(5 in 2 /cos 60) = 0.5 lbs/in 2 τ = {F sin Θ} / {An/cos Θ} = (10 lbs * sin 60)/(5 in 2 /cos 60) = 0.87 lbs/in 2

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: σ 1, σ 2, σ 3 Where σ 1 > σ 2 > σ 3

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 σ 1 - σ 3 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 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

“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?) Assumes OM, 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 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

Φ Φ Φ

Φ Φ Φ tan Φ = τ / σ normal

Φ

Our governing equations….. Φ Φ

Φ Φ (cosΘ)*(cosΘ)

IV. Shearing Resistance and Strength B. Triaxial Test for Soils & Mohrs Circles “Strength of material ~ cohesion and angle of internal friction” τ = c + σ normal * tanΦ τ = shear stress on failure plane c = cohesion σ normal = stress normal on failure plane Φ = angle of internal friction

σ3σ3 σ1σ1 Mohrs Circles 8 lbs/in2 33 lbs/in2

ϕ Φ = angle of internal friction = angle between σ 3 and horizontal plane OB = σ 1 OA = σ 3 OE = σ normal DE = τ (shear stress) Φ τ (shear stress) σ1σ1 σ3σ3

An example……

An Example Problem: The following Triaxial tests were performed on multiple samples of the same soil: Testσ3 (psi) σ1 (psi) A732 B1761 C2376 D3192 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…..