4. I. 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.
5. 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.
6. 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
Penetrations of 0.05” per minute
“Achieve OM &MD”
6” mold
7. 4. Plot the Data
0
50
100
150
200
250
300
350
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45
Penetration (inches)
Load
on
Piston
(psi)
8. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2
penetration.
0
50
100
150
200
250
300
350
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45
Penetration (inches)
Load
on
Piston
(psi)
“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
9. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2
penetration.
0
50
100
150
200
250
300
350
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45
Penetration (inches)
Load
on
Piston
(psi)
“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
Example psi = CBR
Standard psi
220 psi = .22, or 22%
1000 psi
300 psi = .20, or 20%
1500 psi
CBR of material = 22%
10. 5. Determine the percent of compacted crushed stone values for the 0.1 and 0.2
penetration.
0
50
100
150
200
250
300
350
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45
Penetration (inches)
Load
on
Piston
(psi)
“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
Example psi = CBR
Standard psi
220 psi = .22, or 22%
1000 psi
300 psi = .20, or 20%
1500 psi
CBR of material = 22%
Use 0.1” of penetration, unless 0.2” is the greater value.
•If so, then rerun the test, taking the higher of the two values from this second trial
11. 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
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”
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.
13. I. Pavement Design
C. The Mechanics of the Design
1. Determine
• The CBR values of the subgrade
• The type of use expected (runways vs.
taxiways)
14. I. 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
15. I. 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
16. I. 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)?
17. “ 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
18. CBR subbase of 8,
Taxiway, and wheel load
of 40,000 lb
23 inches
“23 inches of
total earth material
and pavement”
19. I. 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
20. I. 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?
Wheel Pound Loads CBR Value
15,000 or less 50
15k-40k 65
40k-70k 80
70k-150k 80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
21. I. 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?
Wheel Pound Loads CBR Value
15,000 or less 50
>15k-40k 65
>40k-70k 80
>70k-150k 80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
23 inches
3 inches
6 inches of CBR 65/80
22. 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?
Wheel Pound Loads CBR Value
15,000 or less 50
>15k-40k 65
>40k-70k 80
>70k-150k 80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
23 inches
3 inches
6 inches of CBR 65/80
CBR = 80
3” 6”
23. 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?
Wheel Pound Loads CBR Value
15,000 or less 50
>15k-40k 65
>40k-70k 80
>70k-150k 80+
Wearing Surface
0-15k…….....2”
>15k-40k…..3”
>40k-55k…..4”
>55k-70k…..5”
>70k……..…6”
23 inches
3 inches
6 inches of CBR 65/80
CBR = 80
3” 6”
26. Given: Same CBR subgrade as
before
Materials available of:
CBR=30, 80
Determine:
Optimal thickness of each
layer while minimizing costs
27. 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.
28. Given: Same CBR subgrade as
before
Materials available of:
CBR=30, 80
Determine:
Optimal thickness of each
layer while minimizing costs
3” of wearing surface
6” of CBR 80 in upper 6”
CBR of 30 needs minimum of 9”
of pavement courses above it.
29. Given: Same CBR subgrade as
before
Materials available of:
CBR=30, 80
Determine:
Optimal thickness of each
layer while minimizing costs
3” of wearing surface
6” of CBR 80 in upper 6”
14” of CBR 30
CBR of 30 needs minimum of 9”
of pavement courses above it.
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
31. 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”
32. 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
33. 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
34. 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
35. 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
