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Soil Compaction and Pavement Design

Soil Compaction and Pavement Design. pneumatic rubber-tired roller. vibratory steel-wheeled roller. Sheeps foot roller. Overview of Soil Compaction Compaction (concept): the densification of soil by removal of air. Requires mechanical energy Densification increases with help of water.

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Soil Compaction and Pavement Design

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  1. Soil Compaction and Pavement Design pneumatic rubber-tired roller vibratory steel-wheeled roller Sheeps foot roller

  2. Overview of Soil Compaction • Compaction (concept): the densification of soil by removal of air. • Requires mechanical energy • Densification increases with help of water “water acts as softening agent and allows soil particles to slip over one another, thereby increasing the packing factor”

  3. There is an optimal moisture content that maximizes densification, or maximum density • “optimum moisture, maximum density” Unit weight has units of density * gravity

  4. Overview B. Factors Affecting Compaction • Grain size • Grain shape • Sorting

  5. II. Laboratory Methods for Determining OM and MD The Proctor Test (after Ralph R. Proctor, 1933)

  6. II. Laboratory Methods for Determining OM and MD The Proctor Test (after Ralph R. Proctor, 1933)

  7. II. The Method The Proctor Test (after Ralph R. Proctor, 1933)

  8. II. The Method The Proctor Test (after Ralph R. Proctor, 1933)

  9. Unit weight has units of density * gravity

  10. Porosity: = volume of voids n = Vv volume total of material Vt III. Terms to “Reckon with” Moisture Content: = weight of waterw = Ww weight of dry soil Ws Unit Weight: (φw) = weight of soil and waterφw = Ws+Ww = Ww (moist) volume total of soil Vt Vt (lbs/ft3) Unit Weight (φd) : = unit weight (wet) (φd)= φw (dry) 1 + (moisture content /100) 1+(w/100) (lbs/ft3)

  11. IV. Field Methods of Determining if OM & MD are achieved A. Sand Cone Method

  12. IV. Field Methods of Determining if OM & MD have been achieved A. Sand Cone Method Unit Weight: = weight of soil and waterγw = Ws+Ww = Ww (moist) volume total of soil Vt Vt Moisture Content: = weight of waterw = Ww weight of soil Ws

  13. IV. Field Methods of Determining if OM & MD have been achieved B. Nuclear Density Meter

  14. V. Pavement Design A. Overview Degree of curvature “Principal cause of pavement failure shown above—not the blacktop”

  15. V. Pavement Design B. California Bearing Ratio (CBR) “How to build a road!”

  16. V. Pavement Design B. California Bearing Ratio (CBR) 1. The California bearing ratio (CBR) is a penetration test for evaluation of the mechanical strength of road subgrades and basecourses. It was developed by the California Department of Transportation.

  17. V. Pavement Design B. California Bearing Ratio (CBR) • The California bearing ratio (CBR) is a penetration test for evaluation of the mechanical strength of road subgrades and basecourses. It was developed by the California Department of Transportation. 2. The test is performed by measuring the pressure required to penetrate a soil sample with a plunger of standard area. The measured pressure is then divided by the pressure required to achieve an equal penetration on a standard crushed rock material.

  18. V. Pavement Design B. California Bearing Ratio (CBR) 3. “The Test” Take load readings at penetrations of: “the result” 0.025” ……………70 psi 0.05”……………...115 psi 0.1”……………….220 psi 0.2”……………….300 psi 0.4”……………….320 psi 6” mold “Achieve OM &MD” Penetrations of 0.05” per minute

  19. 4. Plot the Data

  20. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. “The Gold Standard” for CBR for 0.1” of penetration, 1000 psi for 0.2” of penetration, 1500 psi Example above: for 0.1” of penetration, 220 psi for 0.2” of penetration, 300 psi The standard material for this test is crushed California limestone

  21. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. Example psi = CBR Standard psi 220 psi = .22, or 22% 1000 psi 300 psi = .20, or 20% 1500 psi CBR of material = 22% “The Gold Standard” for CBR for 0.1” of penetration, 1000 psi for 0.2’ of penetration, 1500 psi Example above: for 0.1” of penetration, 220 psi for 0.2” of penetration, 300 psi

  22. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. • In General: • The harder the surface, the higher the CBR rating. • A CBR of 3 equates to tilled farmland, • A CBR of 4.75 equates to turf or moist clay, • Moist sand may have a CBR of 10. • High quality crushed rock has a CBR over 80. • The standard material for this test is crushed California • limestone which has a value of 100. Example psi = CBR Standard psi 220 psi = .22, or 22% 1000 psi 300 psi = .20, or 20% 1500 psi CBR of material = 22%, or “22” “The Gold Standard” for CBR for 0.1” of penetration, 1000 psi for 0.2’ of penetration, 1500 psi Example above: for 0.1” of penetration, 220 psi for 0.2” of penetration, 300 psi

  23. Potential Corrections to the Stress-Penetration Curves

  24. V. Pavement Design C. The Mechanics of the Design

  25. V. Pavement Design C. The Mechanics of the Design Determine The CBR values of the subgrade The type of use expected (runways vs. taxiways)

  26. V. Pavement Design C. The Mechanics of the Design Determine The CBR values of the subgrade The type of use expected (runways vs. taxiways) The expected wheel load during service Types of CBR materials available for the construction

  27. V. Pavement Design C. The Mechanics of the Design 2. Primary Goals • Total strength of each layer only as good as what is beneath it • Therefore, must meet minimum thickness requirements • “Don’t break the bank” • Use less inexpensive CBR materials when allowed while not shortchanging the project’s integrity

  28. V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load)?

  29. “ a point on the curve for a given CBR material represents the minimum thickness of pavement courses that will reside above it, in order to maintain stability

  30. CBR subbase of 8, Taxiway, and wheel load of 40,000 lb 23 inches

  31. V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load) What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? 23 inches

  32. V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load) What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? 23 inches Wearing Surface 0-15k…….....2” >15k-40k…..3” >40k-55k…..4” >55k-70k…..5” >70k……..…6” Wheel Pound LoadsCBR Value 15,000 or less 50 15k-40k 65 40k-70k 80 70k-150k 80+

  33. V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load) What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? 23 inches 3 inches 6 inches of CBR 65/80 Wearing Surface 0-15k…….....2” >15k-40k…..3” >40k-55k…..4” >55k-70k…..5” >70k……..…6” Wheel Pound LoadsCBR Value 15,000 or less 50 >15k-40k 65 >40k-70k 80 >70k-150k 80+

  34. 3” 6” CBR = 80 V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load) What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? 23 inches 3 inches 6 inches of CBR 65/80 Wearing Surface 0-15k…….....2” >15k-40k…..3” >40k-55k…..4” >55k-70k…..5” >70k……..…6” Wheel Pound LoadsCBR Value 15,000 or less 50 >15k-40k 65 >40k-70k 80 >70k-150k 80+

  35. 3” 6” CBR = 80 V. Pavement Design C. The Mechanics of the Design 3. An example A compacted subgrade has a CBR value of 8. What is the minimum pavement thickness if it is to support a taxiway pavement designed to support a 80,000 lb airplane (40,000 wheel load) What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? What can we use for the remainder of thickness? 23 inches 3 inches 6 inches of CBR 65/80 Wearing Surface 0-15k…….....2” >15k-40k…..3” >40k-55k…..4” >55k-70k…..5” >70k……..…6” Wheel Pound LoadsCBR Value 15,000 or less 50 >15k-40k 65 >40k-70k 80 >70k-150k 80+

  36. Need = 9” minimum thickness

  37. CBR = 27 for remainder of base (14”)

  38. Given: Same CBR subgrade as before Materials available of: CBR=30, 80 Determine: Optimal thickness of each layer while minimizing costs

  39. Given: Same CBR subgrade as before Materials available of: CBR=30, 80 Determine: Optimal thickness of each layer while minimizing costs CBR of 30 needs minimum of 9” of pavement courses above it.

  40. Given: Same CBR subgrade as before Materials available of: CBR=30, 80 Determine: Optimal thickness of each layer while minimizing costs CBR of 30 needs minimum of 9” of pavement courses above it. 3” of wearing surface 6” of CBR 80 in upper 6”

  41. Given: Same CBR subgrade as before Materials available of: CBR=30, 80 Determine: Optimal thickness of each layer while minimizing costs CBR of 30 needs minimum of 9” of pavement courses above it. 3” of wearing surface 6” of CBR 80 in upper 6” 14” of CBR 30

  42. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs

  43. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 3” of wearing surface 6” of CBR 80 in upper 6”

  44. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 3” of wearing surface 6” of CBR 80 in upper 6” A CBR of 15 requires X” above it

  45. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 3” of wearing surface 6” of CBR 80 in upper 6” A CBR of 15 requires 15” above it

  46. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 3” of wearing surface 6” of CBR 80 in upper 6” A CBR of 15 requires 15” above it A CBR of 30 requires X” above it

  47. Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 3” of wearing surface 6” of CBR 80 in upper 6” A CBR of 15 requires 15” above it A CBR of 30 requires 9” above it

  48. Your turn…. Subbase of CBR=7, 50,000 lb loads for a taxiway CBR materials available: 80, 30, 15 Design the pavement with attention paid to optimizing costs and stability

  49. Your turn…. Sub base of CBR=7, 50,000 lb loads for a taxiway CBR materials available: 80, 30, 15 Design the pavement with attention paid to optimizing costs and stability Solution: Total Thickness: 28” Wearing Surface Thickness: 4” Upper 6” of CBR=80 CBR 30 of 7” CBR 15 of 11”

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