Subgrade Characterization
For Concrete
Pavement Design
Fig. 2. Approximate interrelationships of soil classification and bearing values
Concrete Pavements: Slab-on-gradetype Structures
Slab is treated as an elastic plate where the response of the
supporting soil medium can have considerable effect on the
analysis.
The inherent complexity of real soil characterizations has led to
idealized models which provide certain aspects of pavement
response under specific loading and boundary conditions.
Most Common Foundation Models
Winkler Foundation
P
P(x,y)=k w(x,y)
w
Figure 1
Elastic Foundation
P
(a)
Figure 2
Winkler
Plate
k
Elastic Solid
Plate
Eo
These theories:
1. Provide significantly different responses of the pavement
system.
2. Represent different assumptions to characterize soil support.
3. Use parameters (E and k) that have different units.
E: F/L2
k: F/L2/L
Considerable effort has been expended in attempting to establish
correlation between k and Eo of the soil.
Winkler
Elastic Solid
Es
ko 
2 1   s2 
D
k 4
k
Deflection (Concentrated load; infinitely large plate)
P 2k
w
8D
P 2e
w
3 3D
Subgrade Stress
σz 
P
8 2k
σz 
2D
e
C
Eo
C
1  o2
;

3
P 3
P

9 2e 3 3 2e
Bending Stress
σ r  6m r /h
mr 
2


 1
2 k
P 
1


Log

γ

1





 2
4π 
γ



P
σ r  0.366 1    2
h


 Eh 3 
Log

0.266



3 
k
b


 o 
Design Tip
Part II - The modulus of Subgrade Reaction, k
In our previous discussion of the AASHTO pavement design method we discussed
the influence of reliability as defined by the 1986 AASHTO Guide for Design of Pavement
Structures on pavement design. The variable used in the Rigid pavement thickness
determination that accounts for the amount of support supplied by the roadbed and subbase is
called the Modulus of Subgrade Reaction or “k”. For this discussion the variables names and
definitions we will be using are as follows:
The Roadbed Resilient Modulus,
MR (psi)
The Subbase Elastic Modulus,
Esb (psi)
The Modulus of Subgrade Reaction, k (pci)
The Composite k value,
kc (pci)
k value on Rigid Foundations
k’ (pci)
Loss of Support
LS (no units)
Each variable is used to describe the quality of the foundation or a factor
contributing to the quality of the foundation beneath the pavement slab.
What exactly is the purpose of a foundation system? According to the AASHTO
guide, a foundation system should provide a uniform, stable, and permanent support system
for the pavement. The support system minimizes the damaging effects of frost action,
prevents pumping of fine-grained soils at joints, cracks, and edges of the pavement slab,
and provides a working platform for construction equipment. Subbase layers are frequently
used to increase the total strength of the foundation system.
The determination of k of a soil according to ASTM requires the placement of a
30 inch diameter rigid steel plate on the soil and the application of a static load. Knowing
the amount of weight with which the plate is loaded and the measured deflection of the
plate into the soil, k can be calculated as follows:
k=
Pressure on the soil
Deflection of the soil
Where:
The Pressure on the soil is equal to the loading of the rigid steel plate divided by
its area.
From this calculation you can see why k is expressed in units of pounds per cubic
inch, (pci). It would probably make more sense if k was expressed as pounds per square
inch deflection, (psi/in2/in).
The history associated with the letter notation of k goes back to basic physics.
You may remember, k is the variable name of the spring constant; and when Dr.
Westergaard modeled the interaction of a rigid slab resting on a soil, he treated the soil as a
bed of springs with a stiffness of “k.”
With that brief history, you are probably wondering why the roadbed resilient
modulus is listed? After all, it is also a property indicating the soil strength. The reason
for this is that the new AASHTO Guide depends heavily on MR for the design of
bitummous pavements. Many of the new mechanistic design methods use MR to
characterize the strength of the soils rather than k.
Since the resilient modulus can be calculated from a laboratory test and
correlated to k, the new AASHTO Design Guide uses MR. This value replaces the former
Soil Support Value used to describe soil strength in the previous editions of the guide. The
analytical relationship between a soils resilient modulus and it Modulus, of Subgrade
Reaction is a follows:
k=MR/19.4
The actual k used in the design of concrete pavements is modified to reflect
increased support resulting from frozen subgrade, high-quality subbases and loss of
support to develop a “composite k” (kc).
For equal deflections:
P 2k
w
8D
P 2e
w
3 3D
k
h
D
D1/3
C
E
100
200
400
8
8
8
174,595,055
555.4
19,829
23,258
39,116
10,397
17,490
29,400
100
200
400
10
341,005,967
654.3
15,638
26,300
44,232
11,760
19,775
33,257
100
200
400
12
12
12
589,258,311
832.7
18,738
31,514
53,000
14,090
23,690
39,850
100
200
400
14
14
14
935,720,375
971.4
21,033
35,375
59,491
15,815
26,600
44,730
8 2e  3 3 2k
k
3 3D
4/3
C
8D4/3 2 2/3
C 4/3
k  0.1673 1/3
D

C  5.977D1/3 k
3/4
From the AASHTO Design Guide
Eo
k
19.4
Generally, Eo =1500 CBR
=1500 (5)=7500 psi
k = 385 psi/in
from PCA (OJ Porter chart), CBR=5 yields k=140 psi
In these designs, NO correlation between Eo and k is apparent
Es vs k
Solid lines are for Equal Slab deflection used on slab Theory 42
50
1. Westergaard for Dense Liquid
2. Hogg for Elastic Solid
40
h=14”
h=12”
h=10”
h=8”
Slab
Thickness
30
20
10
0
0
100
200
300
k (psi/in)
400
From Boussinesq for a circular loaded area (on subgrade)
P
a
Eo

ho=
r
z
E o  1.5a
P
 1.5ak
w
(no plate stiffness involved)
Eo constant
E,k
1.5pa
w
for   0.5
Eo
k constant
For a=30”/2 & k=100
Eo=2250= psi
a
Therefore Eo and k can be correlated only if the size of the loaded area is
taken into account.
E o  C 1  o2 
C
For
2D
3
e
Ec=4,000,000 psi
c=0.15 & o=.5
C=13,896
Eo=0.75(13,896)=10,422 psi
Difference is in the slab stiffness and the effect on the loaded area.
For a concrete slab on subgrade, the size of the loaded area is a
function of slab stiffness
Equating max Deflection:
P 2e
P 2k

8D 3 3D
Equal Deflections
0
0
1
2
1
3
4
2
3
0.05
w
w
0.10
0.15
0.192
5
D
(Hertz)
Pl 2
l  1.241lo
0.125 (Hertz)
6
r/lo  x o
r/l  x
D
(Elast.Solid)
2
Plo
Equal Subgrade Stress
0
0.05
0
1
1
2
2
3
3
4
lo2
p (Elast.solid)
P
l2
p (Hertz)
P
0.10
0.15
0.192
l  0.806lo
0.125 (Hertz)
5
6
r/lo  x o
r/l  x
Note: the rel.
between l & lo is
diff depending on
the respondnce !
Equal Bending Stress
Mr
0
0
0.05
M
P
1
2
3
4
Mt
0.10
=0.3
0.15
5
6
x=xo
Wester
RLE W, T, or Q: Elastic

IF we=wk : w, same max Defl
IF r= b : T, same max bending
IF z= sg : Q, Same max subgrade stress\
ESW______ : Elastic Eo
DeFES_____ : Elastic Surf. Defl
SSES______ : Elastic Subg Stress
BSES______ : Elastic layer radial E
EQRE______ : Equivalent ‘k’ (P/w)
EQRL______ : Equivalent ‘kl’
Different load area/slab stiffness relationships result from
equating
• max deflection
• max subgrade stress
•max Bending stress
Soil Modulus (ksi)
18
16
14
12
Conditions
Same Max Defl
10
8
Same Max B S
* Same Max S S
6
4
k=100
2
0
6
7
8
9
10
11
12
Pavement Thickness (in.)
Stiffness Effect on Es vs. K
13
14
Maximum Subgrade Stress
ELASTIC MODULUS OF SOIL (psi)
(Thousands)
E(S) vs. h
14
13
12
11
10
9
8
7
6
5
4
3
2
1
6
k=50
8
Thickness (in)
k=100
10
k=200
12
k=400
Maximum Bending Stress
ELASTIC MODULUS OF SOIL (psi)
(Thousands)
E(S) vs. h
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
6
k=50
8
Thickness (in)
k=100
10
k=200
12
k=400
Maximum Deflection
E(S) vs. h
ELASTIC MODULUS OF SOIL (psi)
(Thousands)
45
40
35
30
25
20
15
10
5
6
k=50
8
Thickness (in)
k=100
k=200
10
12
k=400
Same Maximum Deflection
250
Pavement Stress (psi)
200
150
Westergaard
Elastic Solid
Elastic Layer
100
50
0
k= 100, 8"
K=400, 8"
k= 100, 12"
Foundation Modulus (pci)
K=400, 12"
Same Maximum Deflection
5
4.5
4
Subgrade Stress (psi)
3.5
3
Westergaard
Elastic Solid
2.5
Elastic Layer
2
1.5
1
0.5
0
k= 100, 8"
K=400, 8"
k= 100, 12"
Foundation Modulus (pci)
K=400, 12"
0.10
W=0.001 inches in one min
K=
P
10
=
=133 psi/in
W 0.075
Deflection, inches
0.075
Alternate Time Deflection
Curves for Plate
0.050
KD =
P
10
=
=250 psi/in
W 0.040
0.025
10 Sec 1
2
3
4
Time, Min
Deflection-Time plots for Typical Plate
Load Test on Cohesive Soils
5