Soil Compaction and
Pavement Design
pneumatic rubber-tired roller
Sheeps foot roller
vibratory steel-wheeled roller
I.
A.
•
•
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”
•
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
volume total of material
n = Vv
Vt
Moisture Content: = weight of water
weight of dry soil
w = Ww
Ws
Unit Weight: (φw) = weight of soil and water
(moist)
volume total of soil
(lbs/ft3)
φw = Ws+Ww = Ww
Vt
Vt
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
(moist)
volume total of soil
Moisture Content: = weight of water
weight of soil
γw = Ws+Ww = Ww
Vt
Vt
w = Ww
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
Penetrations of 0.05” per minute
“Achieve OM &MD”
4. Plot the Data
350
300
Load on Piston (psi)
250
200
150
100
50
0
0
0.05
0.1
0.15
0.2
0.25
Penetration (inches)
0.3
0.35
0.4
0.45
5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2
penetration.
350
300
Load on Piston (psi)
250
200
150
100
50
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Penetration (inches)
“The Gold Standard” for CBR
for 0.1” of penetration, 1000 psi
for 0.2” of penetration, 1500 psi
The standard material for this test is
crushed California limestone
Example above:
for 0.1” of penetration, 220 psi
for 0.2” of penetration, 300 psi
0.4
0.45
5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2
penetration.
350
300
220 psi = .22, or 22%
1000 psi
300 psi = .20, or 20%
1500 psi
CBR of material = 22%
250
Load on Piston (psi)
Example psi = CBR
Standard psi
200
150
100
50
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Penetration (inches)
“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
0.4
0.45
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
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.
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
Potential Corrections to the
Stress-Penetration Curves
350
300
Load on Piston (psi)
250
200
150
100
50
0
0
0.05
0.1
0.15
0.2
0.25
Penetration (inches)
0.3
0.35
0.4
0.45
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
15,000 or less
15k-40k
40k-70k
70k-150k
CBR Value
50
65
80
80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>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
15,000 or less
>15k-40k
>40k-70k
>70k-150k
CBR Value
50
65
80
80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
3”
6”
V. Pavement Design
C. The Mechanics of the Design
3. An example
CBR = 80
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
15,000 or less
>15k-40k
>40k-70k
>70k-150k
CBR Value
50
65
80
80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
3”
6”
V. Pavement Design
C. The Mechanics of the Design
3. An example
CBR = 80
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
15,000 or less
>15k-40k
>40k-70k
>70k-150k
CBR Value
50
65
80
80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>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