flexiblepavementpresentation-161120101101 (1).pptx

FLEXIBLE PAVEMENT
THEORY AND DESIGN
Guide :
Assistant prof. Civil engineering department.
UNIVERSAL COLLEGE OF ENGINEERING

THEORY OF
FLEXIBLE PAVEMENT :
2
Presented by:
⦁ SAGAR J LUTYA

What is pavement ?
3
⦁ A structure consisting of superimposed layers of
processed materials above the natural soil sub-
grade, whose primary function is to distribute the
applied vehicle loads to the sub-grade.

Types of Pavement
PAVEMENT
FLEXIBLE PAVEMENT RIGID PAVEMENT
4

Flexible pavement:
5
⦁ Flexible pavements are those which on a whole have
low or negligible flexural strength and rather flexible
in their structural action under load.

Load transfer:
⦁ Load is transferred to the lower layer by grain to
grain distribution as shown in the figure given below;
6

Load Transfer (continue …)
⦁ The wheel load acting on the pavement will be
distributed to a wider area, and the stress decreases
with the depth. Flexible pavement layers reflect the
deformation of the lower layers on to the surface
layer
7

TYPICAL LAYERS OF A FLEXIBLE
PAVEMENT :
⦁ Typical layers of a conventional flexible pavement
includes seal coat, surface course, tack coat, binder
course, prime coat, base course, sub-base course,
compacted sub-grade, and natural sub-grade.
8

PAVEMENT
9
⦁ Seal coat is a thin surface treatment used to water-
proof the surface and to provide skid resistance.
⦁ Tack coat is a very light application of asphalt
emulsion diluted with water. And It provides bonding
between two layers of binder course.
⦁ Prime coat is an application of low viscous cutback
bitumen to an absorbent surface like granular bases
on which binder layer is placed and provides
bonding between two layers.

PAVEMENT (Continue ….)
10
⦁ Surface course is the layer directly in contact with
traffic loads and are constructed with dense graded
asphalt concrete.
⦁ Binder course purpose is to distribute load to the
base course. Binder course requires lesser quality of
mix as compared to course above it.
⦁ Base course provides additional load distribution
and contributes to the sub-surface drainage

PAVEMENT (Continue ….)
11
⦁ Sub-base course the primary functions are to
provide structural support, improve drainage, and
reduce the intrusion of fines from the sub-grade in
the pavement structure
⦁ Sub-grade The top soil or sub-grade is a layer of
natural soil prepared to receive the stresses from the
layers above

FACTORS AFFECTING PAVEMENT DESIGN
12
⦁ 1. Design Wheel Load
⦁ Max. Wheel load
⦁ Axle configuration
⦁ Contact pressure
⦁ ESWL.
⦁ Repetition of loads
⦁ 2. Climatic Factor
⦁ 3. Pavement component material

Design Wheel Load.
13
⦁ Max. Wheel load - It is used to determine the depth
of the pavement required to ensure that the
subgrade soil does not fail.
⦁ Contact pressure - It determines the contact area and the
contact pressure between the wheel and the pavement
surface. For simplicity elliptical contact area is consider to
be circular.

Design Wheel Load (Continue)
⦁ Axle configuration - the axle configuration is important to
know the way in which the load is applied on the pavement
surface.
14

⦁ Equivalent single wheel load (ESWL)
15

16
⦁ Repetition of loads :
⦁ Each load application causes some deformation and the
total deformation is the summation of all these.
⦁ Although the pavement deformation due to single axle
load is very small, the cumulative effect of number of load
repetition is significant.
⦁ Therefore, modern design is based on total number of
standard axle load (usually 80 KN single axle)

Climatic Factor
17
⦁ 1. Temperature -
⦁ Wide temperature variations may cause damaging
effects.
⦁ Pavement becomes soft in hot weather and brittle in very
cold weather.
⦁ 2. Variation in moisture condition –
⦁ It depends on type of the pavement, type of soil type,
ground water variation etc.
⦁ It can be controlled by providing suitable surface and sub-
surface drainage.

Characteristic of Pavement material
18
⦁ 1. California bearing ratio- It determines the strength
of soil sub-grade, sub-base or base and it is used for
the design of pavement.
⦁ 2. Elastic modulus -It measures the materials
resistance to being deformed elastically upon
application of the wheel load.
⦁ 3. Poisson Ratio – It is the ratio of lateral strain to the
axial strain caused by a load parallel axis along axial
strain.
⦁ 4. Resilient modulus- The elastic modulus based on
the recoverable strain under repeated loads is called
the resilient modulus MR =σd/σr .

Characteristic of Pavement material
(Continue ….)
19
⦁ The following material properties are consider for
both flexible and rigid pavements.
⦁ When pavements are considered as linear elastic, the
elastic moduli and poisson ratio are specified.
⦁ If the elastic modulus of a material varies with the time of
loading, then the resilient modulus is selected.

Design procedures for flexible pavements:
Design Procedures
Empirical Design Mechanistic-
Empirical Design
Mechanistic
Design
20
IRC:37-2012 is based on Mechanistic-Empirical
Design

Mechanistic-empirical design
21
⦁ 1. It can be used for both existing pavement
rehabilitation and new pavement construction
⦁ 2. It can accommodate changing load types
⦁ 3. It uses material proportion that relates
better with actual pavement performance
⦁ 4. It provides more reliable performance
predictions

Failures of flexible pavements:
22
⦁ Different types of failure encountered in flexible
pavements are as follow.
⦁ 1. Alligator cracking or Map cracking (Fatigue)
⦁ 2. Consolidation of pavement layers (Rutting)
⦁ 3. Shear failure cracking
⦁ 4. Longitudinal cracking
⦁ 5. Frost heaving
⦁ 6. Lack of binding to the lower course
⦁ 7. Reflection cracking
⦁ 8. Formation of waves and corrugation
⦁ 9. Bleeding
⦁ 10. Pumping

1. ALLIGATOR OR MAP CRACKING
(FATIGUE CRACKING)
⦁ Followings are the primary causes of
this type of failure.
⦁ Relative movement of pavement
layer material
⦁ Repeated application of heavy
wheel loads
⦁ Swelling or shrinkage of subgrade
or other layers due to moisture
variation
23

2. CONSOLIDATION OF PAVEMENT
LAYERS (RUTTING)
⦁ Formation of ruts falls in this
type of failure.
⦁ A rut is a depression or
groove worn into a road by
the travel of wheels.
⦁ This type of failure is caused
due to following reasons.
⦁ •Repeated application of
load along the same
wheel path resulting
longitudinal ruts.
⦁ •Wearing of the surface
course along the wheel
path resulting shallow
ruts.
24

3. SHEAR FAILURE CRACKING:
⦁ Shear failure causes
upheaval of pavement
material by forming a
fracture or cracking.
⦁ Followings are the primary
causes of shear failure
cracking.
⦁ •Excessive wheel loading
⦁ •Low shearing resistance of
pavement mixture
25

4. LONGITUDINAL CRACKING:
⦁ This types of cracks extents to the
full thickness of pavement.
⦁ The following are the primary
causes of longitudinal cracking.
⦁ Differential volume changes in
subgrade soil
⦁ Settlement of fill materials
⦁ Sliding of side slopes
26

5. FROST HEAVING:
⦁ F
r
o
s
t
heaving
upheaval
causes
of localized
a pavement.
portion of
The extent
heaving depends
the ground water
of frost
upon
table
and climatic condition.
27

6. LACK OF BINDING WITH LOWER LAYER
(POTHOLES & SLIPPAGE)
⦁ When there is lack of
binding between surface
course and underlying
layer, some portion of
surface course looses up
materials creating patches
and potholes.
⦁ Slippage cracking is one
form of this type of failure.
⦁ Lack of prime coat or tack
coat in between two layers
is the primary reason
behind this type of failure.
28

7. REFLECTION CRACKING:
⦁ This type of
occurs,
bituminous
failure
when
surface
course is laid over the
existing
concrete
cement
pavement
with some cracks. This
the same
crack is reflected in
pattern on
bituminous surface.
29

8. FORMATION OF WAVES &
CORRUGATION :
⦁ Transverse
undulations appear
at regular intervals
due to the unstable
surface course
caused by stop-and-
go traffic.
30

9. BLEEDING:
⦁ Excess bituminous
binder occurring on the
pavement surface
causes bleeding.
Bleeding causes a shiny,
glass-like, reflective
surface that may be
tacky to the touch.
Usually found in the
wheel paths.
31

10. PUMPING:
⦁ Seeping or ejection
of water and fines
from beneath the
pavement through
cracks is called
pumping
32

FAILURES OF FLEXIBLE PAVEMENTS
DESIGN CONSIDERATION:
33
⦁ The design of flexible pavement as per IRC is
based on two major failure that are, fatigue
cracking and rutting failure.

IRC METHOD OF DESIGN OF FLEXIBLE
PAVEMENTS (IRC: 37-2012)
34
⦁ 1. IRC:37-1970
⦁ based on California Bearing Ratio (CBR) of subgrade
⦁ Traffic in terms of commercial vehicles (more than 3
tonnes laden weight)
⦁ 2. IRC:37-1984
⦁ based on California Bearing Ratio (CBR) of subgrade
⦁ Design traffic was considered in terms of cumulative
number of equivalent standard axle load of 80 kN in
millions of standard axles (msa)
⦁ Design charts were provided for traffic up to 30 msa using
an empirical approach.
⦁ .

Continue … .
35
⦁ 3. IRC:37-2001
⦁ based on Mechanistic-Empirical method
⦁ Pavements were required to be designed for traffic as
high as 150 msa.
⦁ The limiting rutting is recommended as 20 mm in 20 per
cent of the length for design traffic
⦁ 4. IRC:37-2012
⦁ based on Mechanistic-Empirical method
⦁ The limiting rutting is recommended as 20 mm in 20 per
cent of the length for design traffic up to 30 msa and 10
per cent of the length for the design traffic beyond

Guidelines for Design by IRC: 37: 2012
⦁ Design Traffic:
⦁ The recommended method considers design traffic
in terms of the cumulative number of standard
axles (80 kN) to be carried by the pavement during
the design life.
⦁ Only the number of commercial vehicles having
gross vehicle weight of 30 kN or more and their axle-
loading is considered for the purpose of design of
pavement.
⦁ Assessment of the present day average traffic
should be based on seven-day-24-hour count made
in accordance with IRC: 9-1972 "Traffic Census on
36
Non-Urban Roads".

Traffic growth rate (r):
37
⦁ Estimated by Analyzing:
⦁ The past trends of traffic growth,
⦁ Change in demand of Traffic by factors like specific
development, Land use changes etc.
⦁ If the data for the annual growth rate of commercial
vehicles is not available or if it is less than 5 per
cent, a growth rate of 5 per cent should be used
(IRC:SP:84-2009).

Design life (n)
38
⦁ The design life is defined in terms of the cumulative
number of standard axles in msa that can be carried
before a major strengthening, rehabilitation or
capacity augmentation of the pavement is
necessary.
⦁ Depending upon road type, Design traffic is ranges
from 10 to 15 years.

Vehicle damage factor (VDF)
39
⦁ It is defined as equivalent number of standard axles
per commercial vehicle.
⦁ The Vehicle Damage Factor (VDF) is a multiplier to
convert the number of commercial vehicles of
different axle loads and axle configuration into the
number of repetitions of standard axle load of
magnitude 80 kN.
🞂

Sample Size for Axle Load Survey:
42

Lane distribution factor
43
⦁ Distribution of commercial traffic in each direction
and in each lane is required for determining the total
equivalent standard axle load applications to be
considered in the design.
⦁ In the absence of adequate and conclusive data, the
following distribution may be assumed until more
reliable data on placement of commercial vehicles
on the carriageway lanes are available:

Lane distribution calculation:
44
⦁ 1) Single-lane roads:
⦁ 2) Two-lane single carriageway roads:
⦁ 3) Four-lane single carriageway roads:
⦁ 4) Dual carriageway roads:

Computation of Design traffic:
⦁ The design traffic in terms of the cumulative number
of standard axles to be carried during the design life
of the road should be computed using the following
equation:
45

Sub-grade
46
⦁ Requirements of CBR: Sub grade is made up of in-
situ material, select soil or stabilized soil.
⦁ Compacted to a minimum of 97% of laboratory dry
density achieved with heavy compaction.
⦁ Minimum CBR of 8% for traffic > 450 CVPD
⦁ CBR can also be determined from Dynamic Cone
Penetrometer (60º cone) by ..
⦁ Log10 CBR = 2.465-1.12log10 N
⦁ Where, N = mm/blow

Sub-grade (Continue…)
⦁ Where different types of soils are used in sub grade
minimum 6 to 8 average value for each type is required.
⦁ 90th percentile for high volume and 80th percentile for
other category of road is adopted as design CBR .
⦁ Maximum permissible variation
⦁ Where variation is more average CBR should be average
of 6 samples and not three.
47

Effective CBR
⦁ Where there is significant difference between the
CBRs of the select sub grade and embankment
soils, the design should be based on effective CBR.
The effective CBR of the subgrade can be
determined from Fig.
48

Lab procedure for CBR calculation:
49
⦁ The test must always be performed on remoulded
samples of soils in the laboratory.
⦁ The pavement thickness should be based on 4-day
soaked CBR value of the soil, remoulded
at placement density and moisture content
ascertained from the compaction curve.
⦁ In areas with rainfall less than 1000 mm, four day
soaking is too severe a condition for well protected
sub-grade with thick bituminous layer and the
strength of the sub-grade soil may be
underestimated.

Continue … .
50
⦁ If data is available for moisture variation in the
existing in-service pavements of a region in different
seasons, molding moisture content for the CBR test
can be based on field data.
⦁ Wherever possible the test specimens should be
prepared by static compaction. Alternatively dynamic
compaction may also be used.

Resilient Modulus:
⦁ Resilient modulus is the measure of its elastic
behavior determined from recoverable deformation
in the laboratory tests.
⦁ The modulus is an important parameter for design
and the performance of a pavement.
⦁ The relation between resilient modulus and the
effective CBR is given as:
51

Continue … .
52
⦁ The CBR of the sub-grade should be determined as
per IS: 2720 (Part 16) (36) at the most critical
moisture conditions likely to occur at site.

Principle of pavement design:
⦁ Pavement Model:
⦁ Modeled as linear elastic
multilayer structure.
⦁ Stress Analysis is based on
IITPave software
⦁ Critical parameters for
analysis are
⦁ 1. Tensile strain at the bottom
of bituminous layer
⦁ 2. Vertical sub-grade strain at
the top of sub-grade.
⦁ Failure of pavement is
considered due to cracking
and rutting
53

Check for Fatigue:
54
⦁ Micro cracks at the bottom of bituminous layer are
developed with every load repetition
⦁ These cracks goes on expending till they propagate
to the surface due to the large load repetition
⦁ In these guidelines, cracking in 20 per cent area has
been considered for traffic up to 30 msa and 10 per
cent for traffic beyond that.

Check for Fatigue (Continue….)
⦁ Two fatigue equations developed based on
performance data collected during various study are
(80 %
⦁ Nf= 2.21 * 10-04x [1/εt]3.89* [1/MR]0.854
reliability)…(a)
⦁ Nf= 0.711 * 10-04x [1/εt]3.89* [1/MR]0.854
reliability)...(b)
⦁ Where,
⦁ Nf= fatigue life in number of standard axles,
⦁ εt= Maximum Tensile strain at the bottom of the
bituminous layer, and
⦁ MR= resilient modulus of the bituminous layer.
(90 %

Check for Fatigue (Continue….)
⦁ To consider the effect of volume of the bitumen and air
voids equation (b) is modified as follows
⦁ Nf =0.5161 * C * 10-04 x [1/ εt]3.89 * [1/MR]0.854………(c)
⦁ Va= per cent volume of air void and Vb= per cent volume
of bitumen in a given volume of bituminous mix.
⦁ Nf= fatigue life, єt= maximum tensile strain at the bottom
of DBM.
⦁ MR= Resilient modulus of bituminous mix.
⦁ For traffic < 30 msa consider equation (a); For traffic >
30msa equation (c) is recommened.
56

Check for Rutting:
57
⦁ Rutting is the permanent deformation in pavement
usually occurring longitudinally along the wheel path.
⦁ Causes –
⦁ 1. Deformation in sub grade /non-bituminous layer
⦁ 2. Secondary compaction and shear deformation of
bituminous layer
⦁ Limiting value
⦁ 20 mm in 20% length for upto 30 msa
⦁ 20 mm in 10% length for > 30 msa
⦁ Rutting affects the serviceability of pavement.

Rutting (Continue …)
58
⦁ Based on various studies the two equation develops
are;
⦁ N = 4.1656 x 10-08[1/εv]4.5337 (80 per cent reliability)
⦁ N = 1.41x 10-8x [1/εv]4.5337 (90 per cent reliability)
⦁ Where,
⦁ N = Number of cumulative standard axles, and
⦁ εv= Vertical strain in the sub-grade

Pavement composition as per IRC:
⦁ A flexible pavement covered in these guidelines
consists of different layers as shown in figure;
59

SUB-BASE LAYER
60
⦁ UNBOUND SUB-BASE LAYER
⦁ Sub-base materials may consist of natural sand,
moorum, gravel, laterite, kankar, brick metal,
crushed stone, crushed slag
⦁ Sub-base materials passing 425 micron sieve when
tested in accordance with IS:2720 (Part 5) should
have liquid limit and plasticity index of not more than
25 and 6 respectively.

SUB-BASE LAYER(Unbound SB Continue…)
61
⦁ When coarse graded sub-base is used as a drainage
layer, Los Angeles abrasion < 40
⦁ Required permeability; fines passing 0.075 mm
should be less than 2 per cent.
⦁ Sub-base is constructed in two layers, the lower
layer forms the separation/filter layer to prevent
intrusion of subgrade soil into the pavement and the
upper GSB forms the drainage layer to drain away
any water
⦁ Resilient modulus (MR) for granular sub-base
🞂 MRgsb = 0.2 h0.45 * MR subgrade
⦁ Where, h = thickness of sub-base layer in mm

SUB-BASE LAYER
62
⦁ Bound Sub base
⦁ Material for bound sub-base may consist of soil,
aggregate or soil aggregate mixture modified with
chemical stabilizers such as cement, lime-flyash.
⦁ The drainage layer of the sub-base may consist of
coarse graded aggregates bound with about 2 per
cent cement while retaining the permeability.
⦁ Drainage and separation layers are essential when
water is likely to enter into pavements from the
shoulder, median or through the cracks in surface
layer.

SUB-BASE LAYER(Unbound SB Continue…)
63
⦁ Strength Parameter:
⦁ Elastic Modulus E of bound sub-bases is
⦁ Ecgsb = 1000 * UCS
⦁ Where UCS = 28 day strength of the
cementitious granular material

BASE LAYER
64
⦁ UNBOUND BASE LAYER
⦁ Base layer may consist of wet mix macadam, water
bound macadam, crusher run macadam, reclaimed
concrete etc.
⦁ Resilient modulus of the granular base is given as..
🞂 MR granular = 0.2 * h0.45 MR subgrade
⦁ Where h = thickness of granular sub-base and base,
mm
⦁ Poisson's ratio of granular bases and sub-bases
is recommended as 0.35.

BASE LAYER(Continue..)
65
⦁ CEMENTITIOUS BASES :
⦁ Cemented base layers may consist of aggregates or
soils or both stabilized with chemical stabilizers, to
give a minimum strength of 4.5 to 7 MPa in 7/28
days.
⦁ Default values of modulus of rupture are
recommended for cementitious bases (MEPDG).
⦁ Cementitious stabilized aggregates -
⦁ Lime—flyash-soil -
⦁ Soil cement -
1.40 MPa
1.05 MPa
0.70 MPa
⦁ Poisson's ration of the cemented layers may be
taken as 0.25.

Criteria for selecting Bitumen grade.
⦁ The recommended resilient modulus values of the
bituminous materials with different binders are:
66

Continue … . .
67
⦁ The Poisson’s ratio of bituminous layer depends upon the
pavement temperature and a value of 0.35 is
recommended for temperature up to 35°C and value of
0.50 for higher temperatures.
⦁ Higher viscosity of bituminous binders, which can be
achieved either by using higher viscosity grade bitumen
or modified bitumen will improve both fatigue and rutting
behavior of mixes as compared to mixes with normal
bitumen.
⦁ Fatigue equation at any pavement temperature from
20°C to 40°C can be evaluated by substituting the
appropriate value of the resilient modulus of the
bituminous mix, air void and volume of bitumen.
Catalogue of designs has been worked out for a
temperature of 35°C.

Drainage Layer
68
⦁ Improvement of drainage can significantly reduce the
magnitude of seasonal heave. The desirable
requirements are:
⦁ (a). Provision must be made for the lateral drainage of the
pavement structural section. The granular sub-base/base
should accordingly be extended across the shoulders
⦁ (b). No standing water should be allowed on either side of
the road embankment.
⦁ (c). A minimum height of1 m between the subgrade level
and the highest water level

Drainage Layer(Continue…)
⦁ Some typical drainage system is illustrated in
following Figs….
⦁ Fig.1 Pavement along a Slope
69

⦁ Fig. 2 Pavement with Filter and Drainage Layers
70

⦁ Criteria to be satisfied:
⦁ The filter/separation layer should satisfy the following
criteria:
⦁ To prevent entry of soil particles into the drainage layer:
⦁ D85 means the size of sieve that allows 85 per cent by
weight of the material to pass through it.
⦁ Similar is the meaning of D50 and D15.
71

DESIGN OF
FLEXIBLE PAVEMENT :
72
Presented by:
⦁ SAGAR J LUTYA

What is design ?
73
⦁ Design of pavement includes deciding
the number of layers, its composition and
thickness for selected material, to
support traffic load safely without failure.

Various cases in design.
74
⦁ The flexible pavement with different combinations of
traffic loads and material properties.
⦁ 1) Granular base and Granular sub-base.
⦁ 2) Cementitious base and sub-base with agg.
Interlayer.
⦁ 3) Cementitious base and sub-base with SAMI.
⦁ 4) RAP agg. Over cemented sub-base
⦁ 5) Cemented base and Granular sub-base

Problem statement.
75
⦁ Design the pavement for construction of a
new flexible pavement with the following data:
⦁ Four lanes divided National Highway.
⦁ Design life is 15 years.

Data collection
76
⦁ Material properties :
⦁ California Bearing Ratio (CBR)
⦁ Resilient Modulus (MR)
⦁ Modulus of Elasticity (E)
⦁ Poisson’s ratio (µ)

Material properties
77
⦁ CBR : The CBR values are calculated after
every kilometre on selected stretch of 10 km
having the same type of soil. Suppose the
values obtained are: 3.8, 2.8, 4.5, 3.9, 4.2, 2.9,
4.7, 4.3, 4.0 and 4.6%. Based on the
collected data the design CBR (90th percentile
CBR) is calculated as below:

Solution :
78
⦁ Arrange in ascending order : 2.8, 2.9, 3.8, 3.9, 4.0,
4.2, 4.3, 4.5, 4.6 and 4.7.
⦁ Calculate the percentage greater than equal of the
value as follows:
⦁ For CBR of 3.8, percentage of values greater than
equal to 3.8 = (8/10) x100 = 80%
⦁ Similarly for 2.8 % is 100%, 4.5% CBR is 80% and
so on.
⦁ Now a plot is made between Percentages of values
greater than equal to the CBR values versus the
CBR as follows.

Continue …
RESULT : The 90th Percentile CBR value is 2.90%
79

Effective CBR:
⦁ (Figure 5.1, Page 11, IRC: 37: 2012)
80

Poisson’s ratio
81
⦁ Poisson’s ratio µ is define as the ratio of lateral strain
(ɛl) to the axial strain (ɛa), caused by load parallel to
the axis along which ɛa is measured.
⦁ It is found that for most of the pavement structures,
the influence of µ value is normally small.
⦁ For most of cement treated materials (soil cement,
cement treated base, lean concrete and PCC), the
value of µ normally lies between 0.10 and 0.25.
⦁ Unbound granular material lie between 0.2 and 0.5
and those for bituminous mixes range from 0.35 to
0.50

Elastic modulus
82
⦁ Elastic moduli of various pavement materials
are obtained either through tests or through
the recommendations available in
the guidelines.
⦁ Repeated flexure or indirect tensile tests are
carried out to determine the dynamic modulus
Ed of bituminous mixes.

Resilient modulus
83
⦁ Resilient modulus is the measure of its elastic
behaviour determined from recoverable deformation
in the laboratory tests.
⦁ The behaviour of the subgrade is essentially elastic
under the transient traffic loading with negligible
permanent deformation in a single pass.
⦁ This can be determined in the laboratory by
conducting tests.

Calculation of MR for Sub-grade.
84
⦁ The resilient modulus is calculated as follow;
⦁ MR (Mpa) = 10 x CBR …………. For CBR 5
= 17.6 x CBR0.64 ………For CBR > 5
⦁ (From equation 5.2, Page no. 12, IRC: 37: 2012)

Calculation of MR for Granular base and
85
sub-base.
⦁ The resilient modulus is calculated as follow;
⦁ MRgsb = 0.20 x h0.45 x MR subgrade
⦁ h = Thickness of sub-base layer in mm, …… sub-
base,
🞂 = Cumulative thickness of Base layer and Sub-
base layer in mm ... for base

Traffic count
86
⦁ Assessment of average daily traffic should be normally
based on 7 day-24hr count made in accordance with
IRC: 9 “Traffic census on non-urban roads”.
⦁ Classify traffic into different categories such as two
wheelers, three wheelers, passenger cars, trucks etc.
⦁ But only commercial vehicle with laden weight > 3 tonne
is taken into consideration of design.
⦁ Commercial vehicles are further categorised as single
axle single wheel, single axel dual wheel, Tandem axle
dual wheel and Tridem axle dual wheel.
⦁ Where no traffic count data is available, data from roads
of similar classification and importance may be used to
predict the design traffic

Calculation of Design factor
87
⦁ 1) Design Traffic,
⦁ 2) Axle load survey,
⦁ 3) Vehicle Damage Factor
⦁ 4) Lane Distribution Factor

Design Traffic:
88
⦁ Initial traffic after construction in terms of number of
Commercial Vehicles per day (CVPD).
⦁ Traffic growth rate during the design life in
percentage.
⦁ Design life in number of years.
⦁ Spectrum of axle loads.
⦁ Vehicle Damage Factor (VDF).
⦁ Distribution of commercial traffic over the
carriageway.

Calculation of Design traffic:
⦁ For our case the number of heavy commercial vehicle
per day is taken as 7 day average for 24 hour count
comes to be 2792 vehicle per day as per the last count.
⦁ i. e. P = 2792 cvpd, r = 7 %, and x = 10 years
⦁ A = 2792 (1+0.07)10 = 5000 cvpd.
⦁ RESULT: Traffic in the year of completion of construction
is 5000 cvpd in both the directions.
89

Axle load survey :
90
⦁ Required for VDF calculation and Fatigue damage
analysis of cementitious base.
⦁ The axle load spectrum is formulated by considering
10 kN, 20 kN and 30 kN intervals for single, tandem
and tridem axle respectively.
⦁ RESULT: As per study the percentage of Single,
Tandom and Tridom axle are 45%, 45% and 10%
respectively

Axle load spectrum
Axle load Class Percentage of Axle load Class Percentage of Axle load Class Percentage of
(KN) Axles (KN) Axles (KN) Axles
185-195 0.64 390-410 1.85 585-615 1.40
175-185 0.80 370-390 2.03 555-585 1.60
165-175 0.80 350-370 2.03 525-555 1.60
155-165 2.58 330-350 2.08 495-525 1.80
145-155 2.58 310-330 2.08 465-495 1.80
135-145 5.80 290-310 4.17 435-465 4.40
125-135 5.80 270-290 4.17 405-435 4.40
115-125 11.82 250-270 12.67 375-405 13.10
105-115 11.82 230-250 12.67 345-375 13.10
95-105 12.90 210-230 10.45 315-345 10.90
85-95 12.16 190-210 10.45 285-315 10.40
< 85 32.30 170-190 7.05 255-285 7.15
<170 28.28 <255 28.33
Total 100 100 100
Single Axle Load Tandem Axle Load
91
Tridem Axle Load

Vehicle damage factor
⦁ The formula to calculate VDF is given as follows:
⦁ W1, W2, ….. are the mean values of the various axle load
groups.
⦁ V1, V2, …. are the respective traffic volumes.
⦁ Ws is the standard axle load.
⦁ Standard axle load for Single axle, Tandem axle and
Tridem axle is 80 KN, 148 KN and 224 KN as per
IRC: 37:2012 (Page 7)
⦁ RESULT: The VDF for Single axle load, Tandem axle
load and Tridem axle load is 4.11, 8.37 and 7.51.
92

Vehicle Damage factor (Continue.)
⦁ Were sufficient information on axle loads are not
available or the small size of project does not
warrant an axle load survey the default values of
VDF may be adopted as given in the table given
below.
93

Lane distribution factor.
94
⦁ Distribution of commercial traffic in each direction
and in each lane is required for determining the total
equivalent standard axle load applications to be
considered in the design.
⦁ Single lane road : Total vehicle in both direction.
⦁ Two lane single carriage way : 50% of total vehicle in
both direction.
⦁ Four lane single carriage way : 40% of total vehicle
in both direction.
⦁ Dual carriage way: Two lane 75%, Three lane 60%,
Four lane 45% of number of CV in each direction.

Lane distribution factor (Continue….)
95
⦁ RESULT: In the present design problem we are
given to design a four lane divided highway,
therefore the Lane distribution factor is 75 percent of
number of commercial vehicle in each direction.

Million standard axle
⦁ The design traffic is calculated in terms of cumulative
number of standard axle of 80 kN carried during the
design life of the road.
⦁ r = 7.5 %,
⦁ n = 20 yr. ( Expressway and Urban roads), 15 yr (NH
and SH), In this problem we have to design National
highway take n as 15 years,
⦁ A is 5000cvpd in both direction and 2500 in one
direction
96

Calculation of Million std. axle.
97
A : 0.10 x 2500 = 250, F : 7.51
⦁ Single axle load (N1): 45 percent vehicles are of single
axle.
A : 0.45 x 2500 = 1125, F : 4.11
N1 = 33.06 x 106 = 33.06 msa
⦁ Tandem axle load (N2): 45 percent vehicles are of
tandem axle.
A : 0.45 x 2500 = 1125, F : 8. 37
N2 = 67.33 x 106 = 67.33 msa
⦁ Tridem axle load (N3): 10 percent vehicles are of tridem
axle.

Calculation of Million std. axle. (Continue…)
98
⦁ Total msa (N1+N2+N3)
🞂 = 33.06 + 67.33 + 13.42
= 113.81 ̴ 150 msa (Aprox.)
⦁ RESULT: The cumulative million standard axles to
be consider for design is 150 msa.

Determination pavement thickness
⦁ Case 1 : Bituminous pavement with untreated
granular layer
99

Determination of thickness for Case 1
⦁ The thickness of various layers is determined with
the help pavement design catalogue given in IRC:
37: 2012 from page 26 to 28, for various values of
effective CBR.
100

(Continue ….)
101
⦁ RESULT:
⦁ For design traffic of 150msa and CBR of 7%
⦁ Thickness of subbase (GSB) is 230 mm,
⦁ Thickness of base (G. Base) is 250 mm,
⦁ Thickness of Dense Bitumen macadam (DBM) is 140
mm,
⦁ Thickness of Bituminous concrete (BC) is 50 mm

Case 2 : Bituminous pavement with
cemented base and cemented sub-base
with aggregate inter layer of 100mm
102

Determination of thickness for case 2.
104
⦁ RESULT:
⦁ For design traffic of 150msa and CBR of 7%
⦁ Thickness of Cementitious sub-base (CT Subbase)
is 250 mm,
⦁ Thickness of Cementitious base (CT Base) is 120
mm, Aggregate interlayer is 100mm
mm
⦁ Thickness of Bituminous concrete (BC) is 50 mm are
⦁ Obtained by interpolating the thickness of CBR 5%
and 10%.

Calculation of Resilient Modulus (MR) for
case 2
105
⦁ MR subgrade = 17.6 x CBR0.64 = 17.6 x 70.64 = 61.15
Mpa.
⦁ MR Bituminous layer = 3000 Mpa (From table 7.1
Resilienent Modulus of Bituminous Mixes, page 23,
IRC: 37: 2012)
⦁ Pavement composition for 90 per cent Reliability is
BC + DBM = 100 mm,
⦁ Aggregate interlayer = 100 mm (MR = 450 MPa),
⦁ Cemented base = 120 mm (E = 5000 MPa),
⦁ Cemented subbase = 250 mm (E = 600 Mpa)

cemented base and cemented sub-base with
SAMI layer over cemented base.
106

Continue … .
PAGE 33 AND 34 OF IRC: 37: 2012
107

108
RESULT:
⦁ Design traffic of 150 msa and CBR of 7%
⦁ thickness of Cementitious sub-base (CT Subbase) is
250 mm,
⦁ Thickness of Cementitious base (CT Base) is 165
mm,
mm
are
⦁ obtained by interpolating the thickness of CBR 5%
and 10%.

Case 4 Bituminous pavement with base of
fresh aggregate or RAP treated with foamed
bitumen/ Bitumen emulsion and cemented
sub-base
109

Continue …
PAGE 36 AND 37 OF IRC: 37: 2012
110

Determination of thickness for case 4
⦁ RESULT:
⦁ Thickness of Cementitious sub-base (CT Subbase) is
250 mm,
⦁ Thickness of Treater reclaimed aspalt pavement (Treated
RAP) is 180 mm,
⦁ Thickness of Dense Bitumen macadam (DBM) is 50 mm
⦁ Thickness of Bituminous concrete (BC) is 50 mm are
⦁ Obtained by interpolating the thickness of CBR 5% and
10%.
⦁ Instead of RAP base of fresh aggregates treated with
bitumen emulsion/ foamed bitumen can be used to obtain
11
s
1tronger base.

cemented base and granular sub-base with
100mm WMM layer over cemented base:
112

Determination of thickness for case 5
114
⦁ RESULT:
⦁ Thickness of Granulated Subbase (GSB) is 250 mm
⦁ Cementitious sub-base (CT Subbase) is 195 mm,
⦁ Thickness of aggregate layer is 100 mm, Thickness
of Dense Bitumen macadam (DBM) is 50 mm
⦁ Obtain by interpolating the thickness of CBR 5% and
10%.
⦁ The upper 100 mm of granular sub-base should be
open graded so that its permeability is about 300
mm/day or higher for quick removal of water entering
from surface.

Calculation of Resilient Modulus (MR) and
Modulus of Elasticity (E):
115
⦁ For traffic of 150 msa, Subgrade CBR 7%,
⦁ MR subgrade = 17.6 x CBR0.64 = 17.6 x 70.64 = 61.15
Mpa.
⦁ MR Bituminous layer = 3000 Mpa (From table 7.1
Resilienent Modulus of Bituminous Mixes, page 23,
IRC: 37: 2012)
⦁ MR Aggregate = 450 Mpa and
⦁ E of cemented base is 5000 MPa,
⦁ E Granular subbase = MR subgrade x 0.20 x h0.45
⦁ Where, h = Thickness of GSB = 250 mm
🞂 = 61.15 x 0.20 x 2500.45 = 146.72 Mpa.

Design check
116
⦁ To check the suitability of pavement design
discussed above we carry out checks, which ensure
safety against the failure of designed pavement.
⦁ The flexible pavement is checked for two types of
failures i.e. Rutting in pavement and Fatigue in
bottom layer of bituminous surfacing.
⦁ The following condition should be satisfied for the
design to be satisfactory
⦁ Design strain < Allowable strain
⦁ Allowable strain = Obtained by fatigue model and
rutting model
🞂 Design strain = IITpave software

Design of Drainage layer
⦁ Design a granular drainage layer for a four lane
heavy duty divided highway for an annual
precipitation of 1200 mm. Longitudinal slope = 3 per
cent, Camber = 2.5 percent.
⦁ Crack Infiltration Method
117

Continue ...
⦁ Depth of drainage layer = 450 mm (WMM 250mm
and Sub-base 200mm) By design.
⦁ Width of drainage layer : Calculate
⦁ AB = 8.5+1+2x0.45 = 10.4 m (1m unpave shoulder)
⦁ AC = 10.4 x(3/2) = 12.48 m.
⦁ AD = 16.24 m
⦁ (hypotenious of AB and AC)
⦁ Elevation drop :
⦁ Along AC: 12.48x3% = 0.374m
⦁ Along CD: 10.40x2.5% = 0.26m
⦁ Total drop = 0.634
118

Continue … .
119
⦁ Hydraulic gradient = [Elevation drop/ length AD]
🞂 = [0.634/16.24] =0.039
⦁ Infiltration rate calculation:
⦁ qi = Ic [Nc/Wp + Wc / (Wp.Cs)]
⦁ Ic = 0.223 cub. m/day/meter
⦁ Nc = 3
⦁ Wp = 10.4 m
⦁ Wc = Wp,
⦁ Cs = 12 m
🞂 q = 0.083 Cub.meter/day/meter

Continue.
120
⦁ Amount of water infiltrated (Q);
⦁ Q = 0.083 x 1 x 16.24 = 1.35 Cub.meter/ day.
⦁ Compare with
⦁ Q = KIA
⦁ A = Area of cress section = 1 x 0.1 = 0.1 sq.m
⦁ K = Coeff of permeability (Unknown)
⦁ I = Hydraulic gradient (0.039)
⦁ 1.35 = K x 0.039 x 0.1
⦁ K = 346.62 m/day
⦁ This value of K is useful for deciding gradation.

(Decide grade by using table)
% Passing
121
Sieve
Opening,
Mm
Grading 1 Grading 2 Grading 3 Grading 4 Grading 5 Grading 6
20 100 100 100 100 100 100
12.5 85 84 83 81.5 79.5 75
9.5 77.5 76 74 72.5 69.5 63
4.76 58.3 56 52.5 49 43.5 32
2.36 42.5 39 34 29.5 22 5.8
2.00 39 35 30 25 17 0
0.84 26.5 22 15.5 9.8 0 0
0.42 18.2 13.3 6.3 0 0 0
0.25 13.0 7.5 0 0 0 0
0.10 6 0 0 0 0 0
0.075 0 0 0 0 0 0
Coeff. Of
permeability
m/day
3 35 100 350 850 950
Provide Grading 4 for K 346.62 m/day = 350m/day

Recommendation
122
⦁ Specifications should be modified according to local
condition. In wet climate wearing course should be
impermeable.
⦁ long duration and low intensity rainfall causes more
damage as compare with rainfall of small duration
and more density.
⦁ If DBM and SDBC/BC are designed properly (4% air
voids and protected shoulder) impermeably can be
ensure.
⦁ Adequate provision for sub-surface drainage prevent
pavement damage.

Recommendations.
123
⦁ Thickness charts with BC/ SDBC are valid for all
rainfall area.
⦁ For pavement carrying heavy traffic wearing course
laid over WBM shows better performance.
⦁ For low traffic (upto 5 msa) bitumen surfacing with
two coats is found to be suitable.

Conclusion
124
⦁ Time to time revisions of code provision are needed
keeping in view changes in traffic pattern and
development of new technologies. Further with the
gain of experience in the design as well as
construction procedure of flexible pavement have
demanded certain changes.
⦁ Hence by considering the above factors IRC: 37:
2012 includes some conceptual changes in the
design of flexible pavement such as inclusion of
Resilience moduli and consideration of strain in
design.

Conclusion .
125
⦁ This code also encourages the use IIT pave software
which is newly recommended.
⦁ Since the use of semi-mechanistic approach, the
design is not only based on the experience but it
also gives parameters (strain parameter) to check
the obtained design.
⦁ Solution to the above pavement design problem
shows that the thickness design varies with the
variation in various factors.

References
126
⦁ [1] IRC: 37: 2012, “Guidelines for Design of Flexible
pavement”, second revision.
⦁ [2] IRC: 37: 2001, “Tentative guidelines for Design of
Flexible pavement”

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