Flexible pavement design Notes | EduRev

: Flexible pavement design Notes | EduRev

 Page 1


CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Chapter 27
Flexible pavement design
27.1 Overview
Flexible pavements are so named because the total pavement structure de
ects, or 
exes, under loading. A

exible pavement structure is typically composed of several layers of materials. Each layer receives loads from
the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be
reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum
advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the
highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity
material (and least expensive) on the bottom.
27.2 Design procedures
For 
exible pavements, structural design is mainly concerned with determining appropriate layer thickness and
composition. The main design factors are stresses due to trac load and temperature variations. Two methods
of 
exible pavement structural design are common today: Empirical design and mechanistic empirical design.
27.2.1 Empirical design
An empirical approach is one which is based on the results of experimentation or experience. Some of them are
either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which
is based on the results of experimentation or experience. An empirical analysis of 
exible pavement design can
be done with or with out a soil strength test. An example of design without soil strength test is by using HRB
soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate
soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio
(CBR) test. CBR test is widely known and will be discussed.
27.2.2 Mechanistic-Empirical Design
Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as
wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains,
and de
ections within a pavement structure and the physical causes are the loads and material properties of
the pavement structure. The relationship between these phenomena and their physical causes are typically
described using some mathematical models. Along with this mechanistic approach, empirical elements are used
Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao
Page 2


CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Chapter 27
Flexible pavement design
27.1 Overview
Flexible pavements are so named because the total pavement structure de
ects, or 
exes, under loading. A

exible pavement structure is typically composed of several layers of materials. Each layer receives loads from
the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be
reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum
advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the
highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity
material (and least expensive) on the bottom.
27.2 Design procedures
For 
exible pavements, structural design is mainly concerned with determining appropriate layer thickness and
composition. The main design factors are stresses due to trac load and temperature variations. Two methods
of 
exible pavement structural design are common today: Empirical design and mechanistic empirical design.
27.2.1 Empirical design
An empirical approach is one which is based on the results of experimentation or experience. Some of them are
either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which
is based on the results of experimentation or experience. An empirical analysis of 
exible pavement design can
be done with or with out a soil strength test. An example of design without soil strength test is by using HRB
soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate
soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio
(CBR) test. CBR test is widely known and will be discussed.
27.2.2 Mechanistic-Empirical Design
Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as
wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains,
and de
ections within a pavement structure and the physical causes are the loads and material properties of
the pavement structure. The relationship between these phenomena and their physical causes are typically
described using some mathematical models. Along with this mechanistic approach, empirical elements are used
Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
when dening what value of the calculated stresses, strains, and de
ections result in pavement failure. The
relationship between physical phenomena and pavement failure is described by empirically derived equations
that compute the number of loading cycles to failure.
27.3 Trac and Loading
There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement
design.
Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not
considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely
used today for pavement design.
Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a
standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load
by number of repetitions of given axle load and its equivalent axle load factor.
Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so
there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of
groups and the stresses, strains, and de
ections under each load group can be determined separately; and used
for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and
tyre contact area.
27.3.1 Equivalent single wheel load
To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly
is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure,
which produces same value of maximum stress, de
ection, tensile stress or contact pressure at the desired depth.
The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method,
known as Boyd and Foster method, based on the following assumptions:
 equalancy concept is based on equal stress;
 contact area is circular;
 in
uence angle is 45
o
; and
 soil medium is elastic, homogeneous, and isotropic half space.
The ESWL is given by:
log
10
ESWL = log
10
P +
0:301 log
10
(
z
d=2
)
log
10
(
2S
d=2
)
(27.1)
where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance
between two wheels, and z is the desired depth.
Example 1
Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center
tyre spacing is 20cm and distance between the walls of the two tyres is 10cm.
Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao
Page 3


CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Chapter 27
Flexible pavement design
27.1 Overview
Flexible pavements are so named because the total pavement structure de
ects, or 
exes, under loading. A

exible pavement structure is typically composed of several layers of materials. Each layer receives loads from
the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be
reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum
advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the
highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity
material (and least expensive) on the bottom.
27.2 Design procedures
For 
exible pavements, structural design is mainly concerned with determining appropriate layer thickness and
composition. The main design factors are stresses due to trac load and temperature variations. Two methods
of 
exible pavement structural design are common today: Empirical design and mechanistic empirical design.
27.2.1 Empirical design
An empirical approach is one which is based on the results of experimentation or experience. Some of them are
either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which
is based on the results of experimentation or experience. An empirical analysis of 
exible pavement design can
be done with or with out a soil strength test. An example of design without soil strength test is by using HRB
soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate
soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio
(CBR) test. CBR test is widely known and will be discussed.
27.2.2 Mechanistic-Empirical Design
Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as
wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains,
and de
ections within a pavement structure and the physical causes are the loads and material properties of
the pavement structure. The relationship between these phenomena and their physical causes are typically
described using some mathematical models. Along with this mechanistic approach, empirical elements are used
Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
when dening what value of the calculated stresses, strains, and de
ections result in pavement failure. The
relationship between physical phenomena and pavement failure is described by empirically derived equations
that compute the number of loading cycles to failure.
27.3 Trac and Loading
There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement
design.
Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not
considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely
used today for pavement design.
Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a
standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load
by number of repetitions of given axle load and its equivalent axle load factor.
Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so
there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of
groups and the stresses, strains, and de
ections under each load group can be determined separately; and used
for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and
tyre contact area.
27.3.1 Equivalent single wheel load
To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly
is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure,
which produces same value of maximum stress, de
ection, tensile stress or contact pressure at the desired depth.
The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method,
known as Boyd and Foster method, based on the following assumptions:
 equalancy concept is based on equal stress;
 contact area is circular;
 in
uence angle is 45
o
; and
 soil medium is elastic, homogeneous, and isotropic half space.
The ESWL is given by:
log
10
ESWL = log
10
P +
0:301 log
10
(
z
d=2
)
log
10
(
2S
d=2
)
(27.1)
where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance
between two wheels, and z is the desired depth.
Example 1
Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center
tyre spacing is 20cm and distance between the walls of the two tyres is 10cm.
Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
No stress overlap
(ESWL=P)
No stress overlap
Partial stress overlap
Complete stress overlap (ESWL=2P)
z 2s
d/2
P P
s
d 2a 2a
(log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
)
Figure 27:1: ESWL-Equal stress concept
Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN.
For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm,
log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
=log
10
ESWL = log
10
2044+
0:301log
10
(
20
10=2
)
log
10
(
220
10=2
)
=3.511. Therefore, ESWL
= antilog(3.511)= 3244.49 kN
27.3.2 Equivalent single axle load
Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement
through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels.
But to carry large loads multiple axles are provided. Since the design of 
exible pavements is by layered theory,
only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate
theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum
allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India,
specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load
and the design of pavement is based on the standard axle load.
Repetition of axle loads: The deformation of pavement due to a single application of axle load may
be small but due to repeated application of load there would be accumulation of unrecovered or permanent
deformation which results in failure of pavement. If the pavement structure fails with N
1
number of repetition
of load W
1
and for the same failure criteria if it requires N
2
number of repetition of load W
2
, then W
1
N
1
and W
2
N
2
are considered equivalent. Note that, W
1
N
1
and W
2
N
2
equivalency depends on the failure criterion
employed.
Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement
by the i
th
type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the
failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings.
The fatigue cracking model has the following form:
N
f
= f
1
(
t
)
f2
 (E)
f3
orN
f
/ 
t
f2
(27.2)
where, N
f
is the number of load repetition for a certain percentage of cracking, 
t
is the tensile strain at the
bottom of the binder course, E is the modulus of elasticity, and f
1
; f
2
; f
3
are constants. If we consider fatigue
Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao
Page 4


CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Chapter 27
Flexible pavement design
27.1 Overview
Flexible pavements are so named because the total pavement structure de
ects, or 
exes, under loading. A

exible pavement structure is typically composed of several layers of materials. Each layer receives loads from
the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be
reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum
advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the
highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity
material (and least expensive) on the bottom.
27.2 Design procedures
For 
exible pavements, structural design is mainly concerned with determining appropriate layer thickness and
composition. The main design factors are stresses due to trac load and temperature variations. Two methods
of 
exible pavement structural design are common today: Empirical design and mechanistic empirical design.
27.2.1 Empirical design
An empirical approach is one which is based on the results of experimentation or experience. Some of them are
either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which
is based on the results of experimentation or experience. An empirical analysis of 
exible pavement design can
be done with or with out a soil strength test. An example of design without soil strength test is by using HRB
soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate
soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio
(CBR) test. CBR test is widely known and will be discussed.
27.2.2 Mechanistic-Empirical Design
Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as
wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains,
and de
ections within a pavement structure and the physical causes are the loads and material properties of
the pavement structure. The relationship between these phenomena and their physical causes are typically
described using some mathematical models. Along with this mechanistic approach, empirical elements are used
Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
when dening what value of the calculated stresses, strains, and de
ections result in pavement failure. The
relationship between physical phenomena and pavement failure is described by empirically derived equations
that compute the number of loading cycles to failure.
27.3 Trac and Loading
There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement
design.
Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not
considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely
used today for pavement design.
Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a
standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load
by number of repetitions of given axle load and its equivalent axle load factor.
Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so
there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of
groups and the stresses, strains, and de
ections under each load group can be determined separately; and used
for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and
tyre contact area.
27.3.1 Equivalent single wheel load
To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly
is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure,
which produces same value of maximum stress, de
ection, tensile stress or contact pressure at the desired depth.
The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method,
known as Boyd and Foster method, based on the following assumptions:
 equalancy concept is based on equal stress;
 contact area is circular;
 in
uence angle is 45
o
; and
 soil medium is elastic, homogeneous, and isotropic half space.
The ESWL is given by:
log
10
ESWL = log
10
P +
0:301 log
10
(
z
d=2
)
log
10
(
2S
d=2
)
(27.1)
where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance
between two wheels, and z is the desired depth.
Example 1
Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center
tyre spacing is 20cm and distance between the walls of the two tyres is 10cm.
Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
No stress overlap
(ESWL=P)
No stress overlap
Partial stress overlap
Complete stress overlap (ESWL=2P)
z 2s
d/2
P P
s
d 2a 2a
(log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
)
Figure 27:1: ESWL-Equal stress concept
Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN.
For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm,
log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
=log
10
ESWL = log
10
2044+
0:301log
10
(
20
10=2
)
log
10
(
220
10=2
)
=3.511. Therefore, ESWL
= antilog(3.511)= 3244.49 kN
27.3.2 Equivalent single axle load
Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement
through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels.
But to carry large loads multiple axles are provided. Since the design of 
exible pavements is by layered theory,
only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate
theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum
allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India,
specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load
and the design of pavement is based on the standard axle load.
Repetition of axle loads: The deformation of pavement due to a single application of axle load may
be small but due to repeated application of load there would be accumulation of unrecovered or permanent
deformation which results in failure of pavement. If the pavement structure fails with N
1
number of repetition
of load W
1
and for the same failure criteria if it requires N
2
number of repetition of load W
2
, then W
1
N
1
and W
2
N
2
are considered equivalent. Note that, W
1
N
1
and W
2
N
2
equivalency depends on the failure criterion
employed.
Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement
by the i
th
type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the
failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings.
The fatigue cracking model has the following form:
N
f
= f
1
(
t
)
f2
 (E)
f3
orN
f
/ 
t
f2
(27.2)
where, N
f
is the number of load repetition for a certain percentage of cracking, 
t
is the tensile strain at the
bottom of the binder course, E is the modulus of elasticity, and f
1
; f
2
; f
3
are constants. If we consider fatigue
Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
cracking as failure criteria, and a typical value of 4 for f
2
, then:
EALF =


i

std

4
(27.3)
where, i indicate i
th
vehicle, and std indicate the standard axle. Now if we assume that the strain is proportional
to the wheel load,
EALF =

W
i
W
std

4
(27.4)
Similar results can be obtained if rutting model is used, which is:
N
d
= f
4
(
c
)
f5
(27.5)
where N
d
is the permissible design rut depth (say 20mm), 
c
is the compressive strain at the top of the subgrade,
and f
4
; f
5
are constants. Once we have the EALF, then we can get the ESAL as given below.
Equivalent single axle load, ESAL =
m
X
i=1
F
i
n
i
(27.6)
where,m is the number of axle load groups, F
i
is the EALF for i
th
axle load group, and n
i
is the number of
passes of i
th
axle load group during the design period.
Example 1
Let number of load repetition expected by 80 KN standard axle is 1000, 160 KN is 100 and 40 KN is 10000.
Find the equivalent axle load.
Solution: Refer the Table 27:1. The ESAL is given as
P
F
i
n
i
= 3225 kN
Table 27:1: Example 1 Solution
Axle No.of Load EALF
Load Repetition
i (KN) (n
i
) (F
i
) F
i
n
i
1 40 10000 (40=80)
4
= 0.0625 625
2 80 1000 (80=80)
4
= 1 1000
3 160 100 (160=80)
4
= 16 1600
Example 2
Let the number of load repetition expected by 120 kN axle is 1000, 160 kN is 100, and 40 kN is 10,000. Find
the equivalent standard axle load if the equivalence criteria is rutting. Assume 80 kN as standard axle load and
the rutting model is N
r
= f
4

f5
c
where f
4
=4.2 and f
5
=4.5.
Solution Refer the Table 27:2. The ESAL is given as
P
F
i
n
i
= 8904:94 kN
Introduction to Transportation Engineering 27.4 Tom V. Mathew and K V Krishna Rao
Page 5


CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Chapter 27
Flexible pavement design
27.1 Overview
Flexible pavements are so named because the total pavement structure de
ects, or 
exes, under loading. A

exible pavement structure is typically composed of several layers of materials. Each layer receives loads from
the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be
reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum
advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the
highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity
material (and least expensive) on the bottom.
27.2 Design procedures
For 
exible pavements, structural design is mainly concerned with determining appropriate layer thickness and
composition. The main design factors are stresses due to trac load and temperature variations. Two methods
of 
exible pavement structural design are common today: Empirical design and mechanistic empirical design.
27.2.1 Empirical design
An empirical approach is one which is based on the results of experimentation or experience. Some of them are
either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which
is based on the results of experimentation or experience. An empirical analysis of 
exible pavement design can
be done with or with out a soil strength test. An example of design without soil strength test is by using HRB
soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate
soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio
(CBR) test. CBR test is widely known and will be discussed.
27.2.2 Mechanistic-Empirical Design
Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as
wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains,
and de
ections within a pavement structure and the physical causes are the loads and material properties of
the pavement structure. The relationship between these phenomena and their physical causes are typically
described using some mathematical models. Along with this mechanistic approach, empirical elements are used
Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
when dening what value of the calculated stresses, strains, and de
ections result in pavement failure. The
relationship between physical phenomena and pavement failure is described by empirically derived equations
that compute the number of loading cycles to failure.
27.3 Trac and Loading
There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement
design.
Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not
considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely
used today for pavement design.
Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a
standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load
by number of repetitions of given axle load and its equivalent axle load factor.
Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so
there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of
groups and the stresses, strains, and de
ections under each load group can be determined separately; and used
for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and
tyre contact area.
27.3.1 Equivalent single wheel load
To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly
is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure,
which produces same value of maximum stress, de
ection, tensile stress or contact pressure at the desired depth.
The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method,
known as Boyd and Foster method, based on the following assumptions:
 equalancy concept is based on equal stress;
 contact area is circular;
 in
uence angle is 45
o
; and
 soil medium is elastic, homogeneous, and isotropic half space.
The ESWL is given by:
log
10
ESWL = log
10
P +
0:301 log
10
(
z
d=2
)
log
10
(
2S
d=2
)
(27.1)
where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance
between two wheels, and z is the desired depth.
Example 1
Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center
tyre spacing is 20cm and distance between the walls of the two tyres is 10cm.
Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
No stress overlap
(ESWL=P)
No stress overlap
Partial stress overlap
Complete stress overlap (ESWL=2P)
z 2s
d/2
P P
s
d 2a 2a
(log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
)
Figure 27:1: ESWL-Equal stress concept
Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN.
For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm,
log
10
ESWL = log
10
P +
0:301log
10
(
z
d=2
)
log
10
(
2S
d=2
)
=log
10
ESWL = log
10
2044+
0:301log
10
(
20
10=2
)
log
10
(
220
10=2
)
=3.511. Therefore, ESWL
= antilog(3.511)= 3244.49 kN
27.3.2 Equivalent single axle load
Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement
through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels.
But to carry large loads multiple axles are provided. Since the design of 
exible pavements is by layered theory,
only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate
theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum
allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India,
specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load
and the design of pavement is based on the standard axle load.
Repetition of axle loads: The deformation of pavement due to a single application of axle load may
be small but due to repeated application of load there would be accumulation of unrecovered or permanent
deformation which results in failure of pavement. If the pavement structure fails with N
1
number of repetition
of load W
1
and for the same failure criteria if it requires N
2
number of repetition of load W
2
, then W
1
N
1
and W
2
N
2
are considered equivalent. Note that, W
1
N
1
and W
2
N
2
equivalency depends on the failure criterion
employed.
Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement
by the i
th
type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the
failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings.
The fatigue cracking model has the following form:
N
f
= f
1
(
t
)
f2
 (E)
f3
orN
f
/ 
t
f2
(27.2)
where, N
f
is the number of load repetition for a certain percentage of cracking, 
t
is the tensile strain at the
bottom of the binder course, E is the modulus of elasticity, and f
1
; f
2
; f
3
are constants. If we consider fatigue
Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
cracking as failure criteria, and a typical value of 4 for f
2
, then:
EALF =


i

std

4
(27.3)
where, i indicate i
th
vehicle, and std indicate the standard axle. Now if we assume that the strain is proportional
to the wheel load,
EALF =

W
i
W
std

4
(27.4)
Similar results can be obtained if rutting model is used, which is:
N
d
= f
4
(
c
)
f5
(27.5)
where N
d
is the permissible design rut depth (say 20mm), 
c
is the compressive strain at the top of the subgrade,
and f
4
; f
5
are constants. Once we have the EALF, then we can get the ESAL as given below.
Equivalent single axle load, ESAL =
m
X
i=1
F
i
n
i
(27.6)
where,m is the number of axle load groups, F
i
is the EALF for i
th
axle load group, and n
i
is the number of
passes of i
th
axle load group during the design period.
Example 1
Let number of load repetition expected by 80 KN standard axle is 1000, 160 KN is 100 and 40 KN is 10000.
Find the equivalent axle load.
Solution: Refer the Table 27:1. The ESAL is given as
P
F
i
n
i
= 3225 kN
Table 27:1: Example 1 Solution
Axle No.of Load EALF
Load Repetition
i (KN) (n
i
) (F
i
) F
i
n
i
1 40 10000 (40=80)
4
= 0.0625 625
2 80 1000 (80=80)
4
= 1 1000
3 160 100 (160=80)
4
= 16 1600
Example 2
Let the number of load repetition expected by 120 kN axle is 1000, 160 kN is 100, and 40 kN is 10,000. Find
the equivalent standard axle load if the equivalence criteria is rutting. Assume 80 kN as standard axle load and
the rutting model is N
r
= f
4

f5
c
where f
4
=4.2 and f
5
=4.5.
Solution Refer the Table 27:2. The ESAL is given as
P
F
i
n
i
= 8904:94 kN
Introduction to Transportation Engineering 27.4 Tom V. Mathew and K V Krishna Rao
CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007
Table 27:2: Example 2 Solution
Axle No.of Load EALF
Load Repetition
i (KN) (n
i
) (F
i
) F
i
n
i
1 120 1000 (120=80)
4:5
= 6.200 6200
2 160 100 (160=80)
4:5
= 22.63 2263
3 40 10000 (40=80)
4:5
= 0.04419 441.9
Example 3
Let number of load repetition expected by 60kN standard axle is 1000, 120kN is 200 and 40 kN is 10000.
Find the equivalent axle load using fatigue cracking as failure criteria according to IRC. Hint:N
f
= 2:21
10
4
(
t
)
3:89
(E)
0
:854
Solution Refer the Table 27:3. The ESAL is given as
P
F
i
n
i
= 6030:81 kN
Table 27:3: Example 3 Solution
Axle No.of Load EALF
Load Repetition
i (KN) (n
i
) (F
i
) F
i
n
i
1 40 10000 (40=60)
3:89
= 0.2065 2065
2 60 1000 (60=60)
3:89
= 1 1000
3 120 200 (120=60)
3:89
= 14.825 2965.081
27.4 Material characterization
It is well known that the pavement materials are not perfectly elastic but experiences some permanent deforma-
tion after each load repetitions. It is well known that most paving materials are not elastic but experience some
permanent deformation after each load application. However, if the load is small compared to the strength of
the material and the deformation under each load repetition is almost completely recoverable then the material
can be considered as elastic.
The Figure 27:2 shows straining of a specimen under a repeated load test. At the initial stage of load
applications, there is considerable permanent deformation as indicated by the plastic strain in the Figure 27:2.
As the number of repetition increases, the plastic strain due to each load repetition decreases. After 100 to 200
repetitions, the strain is practically all-recoverable, as indicated by 
r
in the gure.
27.4.1 Resilient modulus of soil
The elastic modulus based on the recoverable strain under repeated loads is called the resilient modulus M
R
,
dened as M
R
=

d
 r
: In which 
d
is the deviator stress, which is the axial stress in an unconned compression
Introduction to Transportation Engineering 27.5 Tom V. Mathew and K V Krishna Rao
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