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Geotechnology Fundamental Theories of Rock and Soil Mechanics

Geotechnology Fundamental Theories of Rock and Soil Mechanics. Geotechnology. Theory of Rock and Soil Mechanics Stress 1. Concept. Stress = Pressure = ???. Geotechnology. Theory of Rock and Soil Mechanics Stress 1. Concept. Stress = Pressure = Force Area.

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Geotechnology Fundamental Theories of Rock and Soil Mechanics

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  1. GeotechnologyFundamental Theories of Rock and Soil Mechanics

  2. Geotechnology • Theory of Rock and Soil Mechanics • Stress 1. Concept Stress = Pressure = ???

  3. Geotechnology • Theory of Rock and Soil Mechanics • Stress 1. Concept Stress = Pressure = Force Area

  4. Geotechnology • Theory of Rock and Soil Mechanics • Stress 1. Concept Stress = Pressure = Force Area versus

  5. A. Stress 2. Primary Forces (natural)

  6. A. Stress 2. Primary Forces (natural) a. Gravitational Forces (overlying materials and upslope activity)

  7. A. Stress 2. Primary Forces (natural) b. Tectonic Forces “Important for Virginia and the Eastern Seaboard?”

  8. A. Stress 2. Primary Forces (natural) c. Fluid Pressures (‘quick conditions’)

  9. Geotechnology • Theory of Rock and Soil Mechanics • Stress 3. Secondary Forces (Human Induced)

  10. Geotechnology 3. Secondary Forces (Human Induced) a. Excavation and Mining

  11. Geotechnology 3. Secondary Forces (Human Induced) b. Loading

  12. Geotechnology 3. Secondary Forces (Human Induced) c. Other * Blasting * Tunneling * Pumping of Fluids

  13. 4. Stress (σn ) on a plane normal to Force σn = Force / Area Where n = ‘normal’, or stress perpendicular To the cross sectional area

  14. σ = Force / Area 5. Stress on an inclined plane to Force Where inclined area = A = An/cos Θ Θ = angle to normal

  15. σ = Force / Area 5. Stress on an inclined plane to Force Where is 1)Normal Force and 2)Shear Force = ??

  16. σ = Force / Area 5. Stress on an inclined plane to Force Where Normal Force and Shear Force = ?? cos Θ = a h sin Θ = o h

  17. σ = Force / Area 5. Stress on an inclined plane to Force Where Normal Force and Shear Force = ?? cos Θ= a =Fn h = F sin Θ = o = Fs h=F

  18. σ = Force / Area 5. Stress on an inclined plane to Force 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

  19. A reminder… Fn = F cos Θ Fs = F sin Θ A = An/cos Θ 5. Stress on an inclined plane to Force

  20. 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 σn = {F cos Θ} / {An/cos Θ} Stress Shear = Force Shear / Area τ = {F sin Θ} / {An/cos Θ}

  21. 5. Limits: (max) σn when Θ = 0 (min) σn when Θ = 90 (max) τ when Θ = 45 (min) τ when Θ = 0 or 90

  22. Example Problem Given: Force = 10 lbs, Area (normal) = 5 in2 Determine σn and τ when Θ = 0 °, Θ = 30°, Θ = 45°, and Θ = 60°

  23. 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) =

  24. 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

  25. 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

  26. 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

  27. 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) =

  28. 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

  29. 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

  30. Do your answers conform to the trends shown here? 5. Limits: (max) σn when Θ = 0 (min) σn when Θ = 90 (max) τ when Θ = 45 (min) τ when Θ = 0 or 90

  31. Stress at any point can be ‘resolved’ via 3 mutually perpendicular stresses: σ1 , σ2 , σ3 Where σ1 > σ2 > σ3 6. Stress (σ) in 3 dimensions

  32. B. Strain “your ideas??”

  33. B. Strain 1. Strain Effects

  34. B. Strain 1. Strain Effects a. Stress produces deformation Strain = dL L

  35. B. Strain 1. Strain Effects a. Stress produces deformation “phi”

  36. B. Strain 1. Strain Effects a. Strain Ellipse Maximum Shear Stress: Where σ1 - σ3 2

  37. 2. Stress – Strain Diagrams σ “which material is stronger?” ε

  38. II. Elastic Response A. Young’s Modulus (E) “best shown in rocks” E = stressσ strain ε “elastic limit” “The greater E is, ……?

  39. II. Elastic Response A. Young’s Modulus (E) “best shown in rocks” E = stressσ strain ε “The greater E is, the less deformation per unit stress OR “the stronger the material”

  40. An Example:

  41. II. Elastic Response B. Poisson’s Ratio (ν) ν = lateral strain length strain In compression In tension

  42. II. Elastic Response C. Ideal Elastic Behavior

  43. II. Elastic Response D. Non-Ideal Elastic Behavior Strain hardening

  44. ‘delayed feedback’ II. Elastic Response E. Hysteresis Hard Rock “under repeated loads” Soft Rock

  45. Assumes OM, MD II. Elastic Response F. Stress-Strain in Soils Limits of Proportionality (how much of the strain is Elastic?)

  46. “under repeated loads” II. Elastic Response G. Repeated Loading of Soils (when rolled) Increment of permanent strain decreases (densification)

  47. III. Time-Dependent Behavior – Strain A. Creep – under static loads Elastic response occurs instantaneously

  48. Collapsed Culvert, Cincinnati, OH

  49. III. Time-Dependent Behavior – Strain A. Creep – under static loads

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