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Soil Compaction and Pavement Design pneumatic rubber-tired roller vibratory steel-wheeled roller Sheeps foot roller I. Overview of Soil Compaction A. 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” • There is an optimal moisture content that maximizes densification, or maximum density – “optimum moisture, maximum density” Unit weight has units of density * gravity I. Overview B. Factors Affecting Compaction 1. Grain size 2. Grain shape 3. Sorting II. Laboratory Methods for Determining OM and MD The Proctor Test (after Ralph R. Proctor, 1933) II. Laboratory Methods for Determining OM and MD The Proctor Test (after Ralph R. Proctor, 1933) II. The Method The Proctor Test (after Ralph R. Proctor, 1933) II. The Method The Proctor Test (after Ralph R. Proctor, 1933) Unit weight has units of density * gravity III. Terms to “Reckon with” Porosity: = volume of voids n = Vv volume total of material Vt Moisture Content: = weight of water w = Ww weight of dry soil Ws Unit Weight: (φw) = weight of soil and water φw = 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) IV. Field Methods of Determining if OM & MD are achieved A. Sand Cone Method IV. Field Methods of Determining if OM & MD have been achieved A. Sand Cone Method Unit Weight: = weight of soil and water γw = Ws+Ww = Ww (moist) volume total of soil Vt Vt Moisture Content: = weight of water w = Ww weight of soil Ws IV. Field Methods of Determining if OM & MD have been achieved B. Nuclear Density Meter V. Pavement Design A. Overview Degree of curvature “Principal cause of pavement failure shown above—not the blacktop” V. Pavement Design B. California Bearing Ratio (CBR) “How to build a road!” 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. 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. 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. 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 4. Plot the Data 350 300 250 Load on Piston (psi) 200 150 100 50 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Penetration (inches) 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. 350 300 250 Load on Piston (psi) 200 150 100 50 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Penetration (inches) “The Gold Standard” for CBR Example above: for 0.1” of penetration, 1000 psi for 0.1” of penetration, 220 psi for 0.2” of penetration, 1500 psi for 0.2” of penetration, 300 psi The standard material for this test is crushed California limestone 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. 350 300 Example psi = CBR Standard psi 250 Load on Piston (psi) 200 220 psi = .22, or 22% 150 1000 psi 300 psi = .20, or 20% 100 1500 psi 50 CBR of material = 22% 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Penetration (inches) “The Gold Standard” for CBR Example above: for 0.1” of penetration, 1000 psi for 0.1” of penetration, 220 psi for 0.2’ of penetration, 1500 psi for 0.2” of penetration, 300 psi 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2 penetration. Example psi = CBR In General: Standard psi •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, 220 psi = .22, or 22% •Moist sand may have a CBR of 10. 1000 psi •High quality crushed rock has a CBR over 80. 300 psi = .20, or 20% •The standard material for this test is crushed California 1500 psi limestone which has a value of 100. CBR of material = 22%, or “22” “The Gold Standard” for CBR Example above: for 0.1” of penetration, 1000 psi for 0.1” of penetration, 220 psi for 0.2’ of penetration, 1500 psi for 0.2” of penetration, 300 psi Potential Corrections to the Stress-Penetration Curves 350 300 250 Load on Piston (psi) 200 150 100 50 0 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 Penetration (inches) V. Pavement Design C. The Mechanics of the Design V. Pavement Design C. The Mechanics of the Design 1. Determine • The CBR values of the subgrade • The type of use expected (runways vs. taxiways) V. Pavement Design C. The Mechanics of the Design 1. 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 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 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)? “ 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 CBR subbase of 8, Taxiway, and wheel load of 40,000 lb 23 inches 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) 23 inches What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? 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) 23 inches What is the optimal pavement thickness (wearing surface)? What is the optimal CBR value of upper 6 inches? Wheel Pound Loads CBR Value Wearing Surface 15,000 or less 50 0-15k…….....2” 15k-40k 65 >15k-40k…..3” 40k-70k 80 >40k-55k…..4” 70k-150k 80+ >55k-70k…..5” >70k……..…6” 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) 23 inches What is the optimal pavement thickness (wearing surface)? 3 inches What is the optimal CBR value of upper 6 inches? 6 inches of CBR 65/80 Wheel Pound Loads CBR Value Wearing Surface 15,000 or less 50 0-15k…….....2” >15k-40k 65 >15k-40k…..3” >40k-70k 80 >40k-55k…..4” >70k-150k 80+ >55k-70k…..5” >70k……..…6” 3” 6” V. Pavement Design CBR = 80 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) 23 inches What is the optimal pavement thickness (wearing surface)? 3 inches What is the optimal CBR value of upper 6 inches? 6 inches of CBR 65/80 Wheel Pound Loads CBR Value Wearing Surface 15,000 or less 50 0-15k…….....2” >15k-40k 65 >15k-40k…..3” >40k-70k 80 >40k-55k…..4” >70k-150k 80+ >55k-70k…..5” >70k……..…6” 3” 6” V. Pavement Design CBR = 80 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) 23 inches What is the optimal pavement thickness (wearing surface)? 3 inches What is the optimal CBR value of upper 6 inches? 6 inches of CBR 65/80 What can we use for the remainder of thickness? Wheel Pound Loads CBR Value Wearing Surface 15,000 or less 50 0-15k…….....2” >15k-40k 65 >15k-40k…..3” >40k-70k 80 >40k-55k…..4” >70k-150k 80+ >55k-70k…..5” >70k……..…6” Need = 9” minimum thickness CBR = 27 for remainder of base (14”) Given: Same CBR subgrade as before Materials available of: CBR=30, 80 Determine: Optimal thickness of each layer while minimizing costs 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. 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” 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 Another Example: Given: Same CBR subgrade as before Materials available of: CBR=15, 30, 80 Determine: Optimal thickness of each layer while minimizing costs 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” 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 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 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 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 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 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” Homework: Subbase of CBR=15, 70,000 lb loads for a runway CBR materials available: 80, 40, 20 Design the pavement with attention paid to optimizing costs and stability