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…..