Investigating and Analyzing the Pavement
Design (30%)
Roadway and Pavement Design
CVL 461
Semester: FALL 2024
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TABLE OF CONTENT
1 INTRODUCTION ____________________________________________ 3
1.1 FLEXIBLE PAVEMENT _________________________________________________________________________________ 3
1.2 RIGID PAVEMENTS ___________________________________________________________________________________ 5
2 FLEXIBLE PAVEMENT DESIGN _______________________________ 6
2.1 ESAL CALCULATION__________________________________________________________________________________ 6
2.2 TIME CONSTRAINS ASSUMPTION_________________________________________________________________________ 6
2.3 RELIABILITY _______________________________________________________________________________________ 7
2.4 SERVICEABILITY _____________________________________________________________________________________ 8
2.5 PAVEMENT LAYER MATERIALS CHARACTERIZATION _____________________________________________________________ 9
2.6 DRAINAGE COEFFICIENTS _____________________________________________________________________________ 10
2.7 DESIGN PART OF THE PAVEMENT ________________________________________________________________________ 11
2.8 DETERMINATION OF LAYER THICKNESS FOR THE STRUCTURE NUMBER _______________________________________________ 12
2.8.1 Layer Thickness calculation ______________________________________________________________________ 13
2.9 DIAGRAM _______________________________________________________________________________________ 14
3 RIGID PAVEMENT DESIGN __________________________________ 15
3.1 TIME CONSTRAINTS _________________________________________________________________________________ 15
3.2 RELIABILITY AND STANDARD DEVIATION ____________________________________________________________________ 15
3.3 SERVICEABILITY ____________________________________________________________________________________ 15
3.4 MATERIAL PROPERTIES _______________________________________________________________________________ 16
3.4.1 PCC Elastic Modulus (Ec) ________________________________________________________________________ 16
3.4.2 PCC Modulus of Rupture (S’c) ____________________________________________________________________ 16
3.5 PAVEMENT STRUCTURAL CHARACTERISTIC __________________________________________________________________ 16
3.5.1 Drainage ____________________________________________________________________________________ 16
3.5.2 Load Transfer _________________________________________________________________________________ 17
3.5.3 Loss of Support _______________________________________________________________________________ 17
3.5.4 Reinforcement Variables ________________________________________________________________________ 17
3.5.5 Friction Factor ________________________________________________________________________________ 18
3.6 DESIGN _________________________________________________________________________________________ 18
3.7 DETERMINE REQUIRED SLAB THICKNESS ___________________________________________________________________ 19
3.8 DIAGRAM _______________________________________________________________________________________ 21
4 DRAINAGE SYSTEM DESIGN ________________________________ 22
4.1 DESIGN _________________________________________________________________________________________ 22
4.2 __________________________________________________________________________________________________ 23
4.3 CALCULATE THE NUMBER OF CATCH BASINS ________________________________________________________________ 23
4.4 DESIGN OF TRANSVERSE SLOPE _________________________________________________________________________ 23
4.5 DESIGN OF LONGITUDINAL SLOPE _______________________________________________________________________ 23
4.6 DIAGRAM _______________________________________________________________________________________ 24
5 REFERENCE _______________________________________________ 25
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1 Introduction
1.1 Flexible Pavement
Flexible Pavements are constructed from bituminous or unbound material and the stress is transmitted to the
sub-grade through the lateral distribution of the applied load with depth.
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Flexible pavement is composed of a bituminous material surface course and underlying base and subbase
courses. The bituminous material is more often asphalt whose viscous nature allows significant plastic
deformation. Most asphalt surfaces are built on a gravel base, although some 'full depth' asphalt surfaces are
built directly on the subgrade. Depending on the temperature at which it is applied, asphalt is categorized as
hot mix asphalt (HMA), warm mix asphalt, or cold mix asphalt. Flexible Pavement is so named as the
pavement surface reflects the total deflection of all subsequent layers due to the traffic load acting upon it.
The flexible pavement design is based on the load distributing characteristics of a layered system.
1.1.1 Figure
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1.2 Rigid pavements
Rigid pavements have sufficient flexural strength to transmit the wheel load stresses to a wider area below. A
typical cross section of the rigid pavement is shown in Figure 3. Compared to flexible pavement, rigid
pavements are placed either directly on the prepared sub-grade or on a single layer of granular or stabilized
material. Since there is only one layer of material between the concrete and the sub-grade, this layer can be
called as base or sub-base course.
Figure : Typical Cross section of Rigid pavement
In rigid pavement, load is distributed by the slab action, and the pavement behaves like an elastic plate
resting on a viscous medium. Rigid pavements are constructed by Portland cement concrete (PCC) and
should be analyse by plate theory instead of layer theory, assuming an elastic plate resting on viscous
foundation. Plate theory is a simplified version of layer theory that assumes the concrete slab as a medium
thick plate which is plane before loading and to remain plane after loading. Bending of the slab due to wheel
load and temperature variation and the resulting tensile and flexural stress.
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2 FLEXIBLE PAVEMENT DESIGN
2.1 ESAL Calculation
Number of truck
vehicles per year
Growth Directional Lane
Vehicl ((ADT)0 * T Truck
factor distribution distribution ESAL
e * 365) factor
(G)(Y) factor (D) factor (L)
Type (Tf)
Single Unit trucks
2 axles,4 tires 50000 0.003 20.02 0.5 0.6 328828
2 axles,6 tires 20000 0.25 20.02 0.5 0.6 547500
3 axles or more 1000 0.86 20.02 0.5 0.6 1885283
Total 2761611
2.2 Time Constrains Assumption
Analysis Period - 20 years
Performance Period - 20 years
Annual growth rate - 4 %
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ESAL20 = 21864528
2.3 Reliability
Number Stage including initial pavement structure = 2
Overall reliability level = 0.9 %
R stage = 0.90
Overall standard deviation =0.35
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2.4 Serviceability
Initial service ability index : Po = 4.2
Terminal service ability index : Pt = 2.2
The design loss of serviceability : ∆PSI = Po -Pt = 2
Effective Roadbed soil resilient modulus
Soil Resilient Modulus CBR = 12 %
Modulus of subgrade = 9600 psi
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2.5 Pavement Layer Materials Characterization
Asphalt Concrete : EAC = 450000 psi
Granular Base : EBS = 30000 psi
Soil Aggregate Subbase : ESB = 14000 psi
Layer Coefficients
Asphalt Concrete(a1)= 0.44 psi
AASHTO II-22
Granular Base layer coefficient (a2)=0.138 psi
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AASHTO II-22
Soil Aggregate Subbase(a3) = 0.10 psi
2.6 Drainage Coefficients
The drainage coefficients for “good” drainage and the percent of time pavement structure layer coefficients of
untreated base and subbase for 5-25% is chosen
average coefficient of drainage m = 1 for Good (5-25)%
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2.7 Design part of the Pavement
From AASHTO-1993 page I-5,I-6
1st Stage
Analysis Period 20 years
Modulus of Subgrade = 9600 psi
R stage = 90 %
Overall standard deviation S0 = 0.35
Two way ESAL applications for 15 years = 21864528veh.
Yields an initial structural number : SNinitial = 5.6
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2.8 Determination of Layer Thickness for the Structure Number
AASHTO II-35
AASHTO II-36
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2.8.1 Layer Thickness calculation
Determination SN1
Two way ESAL applications for 20 years = 21864528 veh.
The design loss of serviceability PSI =2
Granular Base : EBS = 30000 psi
R stage : = 90 %
Structural Number 1 : SN1 = 3.6
Determine Asphalt Thickness (D1*)
D1*=SN1/a1=3.6/0.44=8.18 in
use D1* =9 in
Determine Actual SN1(SN1*)
D1*x a1 = 3.96 in use 4 in
Determination SN2
Two way ESAL applications for 20 years = 21864528 veh.
The design loss of serviceability PSI = 2
Soil Aggregate Subbase EBS = 14000 psi
R stage : = 90%
Structural Number 2 :
SN2 = 5.2
Determine Base Thickness (D2*)
Base Thickness : D2* = (SN2 - SN1*) / (a2 x m2) = 8.69 in.
use D2*= 11 in.
Determine Actual SN2 (SN2*)
Actual SN2 = SN2* = D2* x a2 x m2 = 1.242 in.
use 1.3 in.
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Yields an initial structural number =SN initial =SN3= 5.6 in
Determine Subbase Thickness (D3*)
D3* = (SN3 - (SN1* + SN2*)) / (a3 x m3 ) = 3 in
use D3 *= 5 in
2.9 Diagram
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3 RIGID PAVEMENT DESIGN
3.1 Time Constraints
Analysis Period - 20 years
Performance Period - 35 years
Traffic
Two way ESAL applications during the first year (W'18) - 71000 vehicle
Annual growth rate - 4 %
ESAL20 = 21864528
3.2 Reliability And standard deviation
For the arterial highway in urban area, level of reliability is 80-99 %. For design, the overall
reliability (R) is 90 percent from Table 2.2, AASHTO Guide for Design of Pavement Structures
1993, page II-9
R overall = 90%
AASHTO Guide for Design of Pavement Structures 1993, page II-10, Overall Standard Deviation
(S0) value is equal 0.35 for rigid pavement.
due to lack of available soil data neglect environment impact
3.3 Serviceability
Initial service ability index : Po = 4.3
Terminal service ability index : Pt = 2.5
The design loss of serviceability : ∆PSI = Po -Pt = 2.2
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3.4 Material Properties
3.4.1 PCC Elastic Modulus (Ec)
Based on AASHTO Guide for Design of Pavement Structures 1993, page II-16
EC=57000(fc)0.5
EC = PCC elastic modulus (psi)
f’C= PCC compressive strength (psi
compressive strength of concrete 5,000 psi is assumed for design.
E c=57000(5000)0.5=4030509 psi
Elastic Modulus of Granular Sub base Layer (ESB) = 14,000 psi
3.4.2 PCC Modulus of Rupture (S’c)
3.5 Pavement Structural Characteristic
3.5.1 Drainage
considered as good and approaches 15 % saturation of the time, take Cd=1.1 - 1
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3.5.2 Load Transfer
determine by using AASHTO Guide for Design of Pavement Structure. Thus, load transfer
coefficient (J) = 3.1
3.5.3 Loss of Support
unbound granular materials which has LS value 1.0 to 3.0. Thus, LS value 1.0 was selected for
design.
3.5.4 Reinforcement Variables
AASTHO guide manual for grade 40 steel and 60, the allowable working stress are 30,000 and
45,000 psi, respectively (AASTHO 1993, II-28) and 45,000 psi was choose for the design.
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3.5.5 Friction Factor
frictional factor 1.5 choose for design
3.6 Design
Develop Effective Modulus of Subgrade Reaction
From II-44
K=MR/19.4=495 psi
page II-42 is use for correct value of effective modulus of subgrade reaction, k. Hence, the k value becomes
170 psi and 150 psi in order without and with subbase layer.
From II-39
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From II-42
3.7 Determine Required Slab Thickness
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The slab thickness is 12 in (without subbase) and 14 in (using subbase 7 in)
(without subbase) 1
K-value =170 psi
Concrete modulus rupture =566 psi
Concrete elastic modulus =4,030,509 psi
Load transfer coefficient =3
Drainage coefficient =1
Loss of serviceability =2.2
Reliability =95 %
Overall standard deviation =0.35
ESAL for design period 20 years = ESAL20 = 21864528
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(without subbase) 2
K-value =170 psi
Concrete modulus rupture = 566 psi
Concrete elastic modulus =4,030,509 psi
Load transfer coefficient =3.1
Drainage coefficient =1.1
Loss of serviceability =2.2
Reliability =95 %
Overall standard deviation =0.35
ESAL for design period 20 years = ESAL20 = 21864528
3.8 Diagram
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4 Drainage System Design
Estimate Runoff Using the Rational Method
Q=C.I.A
Q: flow rate
C Runoff coefficient =0.9
I: Rainfall intensity (mm/hr) for the design storm =50 mm/hr=0.05 mm/hr
A: Drainage area=4000x60 + 15000x23= 585000 m2
Q= 26325m3/hr=7.3125 m3/s
4.1 Design
Using manning’s formula
Assume a trapezoidal channel with base width b=1 m, side slope s=1:1 and depth d=0.5 m
Q channel = 2.83 m3/s
Trapezoidal section with b=0.5 m, d=0.3 m, provide of 0.36 m³/s per channel is sufficient to manage surface
water runoff.
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4.2
4.3 Calculate the Number of Catch Basins
Abasin=100m×30m=3,000m2
Q basin=C ×I × Abasin
Substitute the values
=0.9×0.05m/hr×3,000m2 = 0.0375 m3/s
Number of Catch Basins=Q/Q basin =7.3125 = 195 take 250
Need 250 catch basins in total, install 125 catch basins per side of the runway.
4.4 Design of Transverse Slope
runoff from the centre of the runway to the edges apply 1.5% transverse slope.
Water Flow Rate Due to Transverse Slope
Drop = Width x Slope =30m x 0.015 = 0.45 cm for Runway
1.5% slope 45 cm for Runway from the centre to edge.
Drop = Width x Slope = 11.5 m x 0.005 = 17 cm for taxiway
1.5% slope 17 cm for taxiway from the centre to edge
4.5 Design of Longitudinal Slope
A minimum longitudinal slope of 0.5% is necessary to prevent water from ponding along the runway.
Slope Calculation
Total length = 4,000 meters
Longitudinal Slope = 0.5%
Calculate the elevation difference (Δh) required the runway length 4000m
Δh= Length × Slope=4,000 m×0.005 =20m
20-meter elevation difference from one end of the runway to the other is need 0.5% longitudinal slope.
For taxiway length 15000 m
15000 x 0.005 = 75m
75-meter elevation difference from one end of the taxiway to the other is need 0.5% longitudinal slope.
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4.6 Diagram
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5 Reference
(I)AASHTO GUIDE For Design of Pavement Structures
https://www.civil.iitb.ac.in/tvm/1100_LnTse/401_lnTse/plain/plain.html
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