Introduction to Pavement Design

1. Introduction
• Establish Layer Thicknesses:
– To limit distress (acceptable levels)
– For anticipated loading & environmental conditions
Asphalt Institute Design Procedure – Using available/selected materials
1.1 Elements to be Defined/Identified for Design
• Conditions:
Dr. Christos Drakos – Traffic loading (volume, frequency,magnitude … ESALs)
– Environment (temperature, moisture)
University of Florida • Material Properties:
– Subgrade – varies w/ season (existing material)
– Pavement Structure (engineered materials)

Introduction to Pavement Design Introduction to Pavement Design

1.1 Elements to be Defined/Identified for Design (cont.) 2. Design Approach
• Performance Criteria:
– Conditions that define failure
• Performance Relationship NO

PERFORMANCE
• TRAFFIC y
y
TRAFFIC
ENVIRONMENT
• ENVIRONMENT y SUBGRADE TRIAL TRIAL PERFORMANCE
y MATERIALS THICKNESSES RELATION
• SUBGRADE
MATERIAL
PROPERTIES
• MATL PROPERTIES PERFORMANCE
PERFORMANCE PAVEMENT CRITERIA

RELATION PERFORMANCE LIFE-COST YES

CYCLE
• LAYER THICKNESSES

A Pavement Performance Model is an equation that relates

some extrinsic ‘time factor’ (age, or number of load
applications) to a combination of intrinsic factors (structural
responses, drainage, etc) and performance indicators

Introduction to Pavement Design Introduction to Pavement Design

3. Empirical Vs Mechanistic-Empirical 4. Response and Performance
Difference is in the nature of Performance Relation 4.1 Response = “Reaction to an action”
3.1 Empirical Response = Pavement & Material response to applied loads
• Statistical/Experimental (based on road tests) What are Pavement & Material Responses?
(traffic & environment)
• Limited conditions/environment
3.2 Mechanistic-Empirical δ1 & δ2
Pavement Responses σ1
Improve the relation by understanding the mechanics σ&ε
δ2 Material Responses
• Relate analytical response to performance: σ3
δ1
– More reliable/robust than empirical AC
element σ2
– Integrates the structural aspects of a pavement to the
material/mix design properties of the pavement layers!!! BASE
Introduction to Pavement Design Introduction to Pavement Design
4.1 Response = “Reaction to an action” 4.2 Performance
Predict load responses with structural response models: Performance is the measurable adequacy of STRUCTURAL &
• Vary in sophistication: FUNCTIONAL service over a specified design period
– Linear Elastic
– Non-linear Elastic Structural Functional (user defined)
– Viscoelastic Number of loads the pavement • Roughness
can support before it reaches – Ride quality
– … etc • Friction
unacceptable level of
structural/functional distress • Geometry
Predict temperature responses with thermal response models: • Appearance
– Surface cracking
• σth = fnc (material, temperature, cooling rate, dimensions)
– Loss of color

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
ASPHALT INSTITUTE (AI) 2. Design Criteria
US based association of international asphalt producers that Two types of strains are considered critical in design of asphalt
promotes the use of petroleum asphalt products pavements:
• http://www.asphaltinstitute.org/ • Horizontal tensile strain, εt @ the bottom of AC layer
• Vertical compressive stain, εc @ the top of the subgrade
1. Development
Design method based on computer model DAMA 2.1 Fatigue Cracking
• Computes amount of damage (cracking & rutting) based
on traffic in a specific environment Basic equation:
• Multilayer elastic theory; used correction factors to N f = f1 ⋅ ε t
− f2
⋅ E − f3
account for base non-linearity AC εt
• Used three temperature regimes; representing three Where:
• Nf = Number of cycles to failure
climatic regions in the US – NY(45), SC(60) & AZ(75) • εt = Tensile strain @ bottom of AC layer
• Developed design charts from the results • f1 = Field correlation shift factor
• f2 & f3 = Laboratory determined values

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
2.1 Fatigue Cracking (cont) 2.1.1 Fatigue tests (cont)
− f2
N f = f1 ⋅ ε t ⋅E − f3

Asphalt Institute calibrated the field shift factor using data from
the AASHO road test
• f1 = 0.0796
2.1.1 Fatigue tests

Why 3rd-point loading?

εt
To have an even distribution
of M; we know the value of
V M, no matter where the
specimen fails

M
Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
2.1.2 Constant Stress Fatigue Test 2.1.4 Fatigue Test Analysis
• Apply constant stress • Plot the strain Vs number of repetitions to failure on log scales
• Failure occurs when the material fractures • C1 & C2 curves for the same material @ different temperature

σ0 C1 Which curve has the highest stiffness?

Stress, σ

Strain, ε
ε0
Low

Strain, Log εt
C2
Check:
• Select a strain level
Number of Cycles, N Number of Cycles, N High • Find the corresponding Nf
2.1.3 Constant Strain Fatigue Test • Higher stiffness will have less
Nf2 Nf1 number of cycles to failure
• Apply constant strain (rate of deformation)
Number of Cycles, Log Nf
• Failure occurs when E=½E0 εσ = 12 × σε ; σ = 2 × σ
1 0
0
0 0
From the graph:
ε0 • Stiffness of the material will depend on time of the year (temperature)
Stress, σ

Strain, ε

σ0 • εt depends on the material properties (E)

• So, the cycles to failure Nf will also depend on the temperature
Must use cumulative damage approach to evaluate failure
Number of Cycles, N Number of Cycles, N

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
2.2 Damage Ratio 2.3 Permanent Deformation

Dr=
Actual # of Load Repetitions
Pavement has ‘failed’ if Dr=1 Only SUBGRADE rutting considered, as governed by
Allowable # of Load Repetitions compressive strain
p m
n − f5
Dr = ∑∑
Where:
i, j
m = no. of load types = 1 for AI
Nd = f4 ⋅ε c AI calibrated the equation
i =1 j =1 Ni, j p = no. of periods in analysis = 12 for a year using AASHO road test data
−4.477
N d = 1.365 ×10 ⋅ ε c −9
2.2.1 Damage ratio example
Consider the following two pavements
Periods (Seasons) 1 2 3 4
• Similar structure
Material properties E1, εt1 E2, εt2 E3, εt3 E4, εt4 • E3A >> E3B
Allowable Traffic E1 E1 • Assume σcA = σcB
Nf1 Nf2 Nf3 Nf4
E2 E2
Actual Traffic n1 n2 n3 n4 BUT:
E3A εcA E3B εcB σc So, εcA << εcB
Damage Ratio
εc @ P =
Dr1= n1/Nf1 Dr2= n2/Nf2 Dr3= n3/Nf3 Dr4= n4/Nf4 E3
Assume σcA = σcB THUS:
Dr=ΣDri Æ i.e. Dr=0.1; Design Life = 1/Dr = 10 years
NdA >> NdB

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
3. Environment 3.2 Subgrade
• Nf & Nd vary with time of the year because of change in • Four distinct periods: Frozen MR

material properties with the weather – Freeze

– Thaw Normal MR

3.1 Asphalt Concrete – Recovery Thaw MR

• AI procedure considers the environment based on: – Normal

– Mean monthly temperature • Table 11.9 shows the suggested conditions to represent frost
– Monthly variable material modulus effects on the subgrade
⎛ 1 ⎞ ⎛ 34 ⎞
M P = M A ⋅ ⎜1 + ⎟−⎜ ⎟+6
⎝ z +4⎠ ⎝ z +4⎠
• Where:
– MP = Mean pavement temperature
– MA = Mean monthly air temperature
– z = Depth below the surface (1/3 of AC layer depth)

• Then we can use: Log (E1 ) = 6.48 − 0.01(M P )

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
4. Traffic 5.2 Pavement Types
Calculate design ESALs (Topic 4) 5.2.1 Full-Depth HMA
5. Design Procedure • Pavement constructed completely from HMA
• Figure 11.11; includes both surface and base course thickness
5.1 Objective
DETERMINE THE REQUIRED STRUCTURAL THICKNESS FOR
HMA SURFACE h1A
EXPECTED TRAFFIC, SUBGRADE CONDITIONS, AND • Use Subgrade MR & ESALs
Thickness • Read thickness off the chart
ENVIRONMENT SUCH THAT:
HMA BASE h1B
• Rutting < ½ in
For multiple HMA within a layer use
• Fatigue Cracking < 20% of Area composite modulus
OVER THE DESIGN LIFE (as defined by traffic) Example: ⎛ h ⋅ (E )1/ 3 + h1B ⋅ (E1B )1/ 3 ⎞
• Subgrade MR = 11,000 psi E = ⎜⎜ 1 A 1 A ⎟
⎟
⎝ h1 A + h1B ⎠
• Traffic = 1.1x106 ESAL
• Thickness = ?

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
5.2.1 Full-Depth HMA (cont) 5.2.2 HMA over Emulsified Asphalt Base
Emulsified Asphalt:
• Mixture of asphalt cement, water and emulsifying agent
• Run through a colloid mill that produces asphalt droplets (5-10 microns)
Thickness = 8in • Suspended in in the mixture by electrical charge
• Upon contact with aggregate it ‘sets’ or ‘breaks’; water is squeezed out or
evaporated
• Anionic emulsified asphalts – Negatively charged; compatible with
aggregate with positive charge (limestone)
• Cationic emulsified asphalts – Positively charged; compatible with
aggregate with negative charge (siliceous aggregates)
• Rapid, Medium and Slow setting

Emulsified Base:
• TYPE I – Dense Graded (Crushed Rock)
• TYPE II – Gap Graded (Rounded Gravel)
• TYPE III – Uniform Graded (Sand Asphalt)

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
5.2.2 HMA over Emulsified Asphalt Base (cont) 5.2.2 HMA over Emulsified Asphalt Base (cont)

HMA SURFACE hHMA • TYPE I – Fig 11.12

• hEMUL from the graph
hEMUL • TYPE II – Fig 11.13
EMULSIFIED BASE • Determine hHMA
• TYPE III – Fig 11.14

Minimum HMA thickness required

• ƒ(ESAL & Base Type) Table 11.12
Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
5.2.3 HMA over Untreated Aggregate Base 5.2.3 HMA over Untreated Aggregate Base (cont)
• Select the thickness of the aggregate base first
• Figures 11.15-11.20 – design charts for HMA surface courses
on aggregate base courses of 4,6,8,10,12 and 18 in

HMA SURFACE hHMA

• Determine the required HMA thickness
for the specific base thickness
• Fig 11.15-11.20
AGGREGATE BASE (known)

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
5.2.4 HMA on Asphalt Emulsion over Untreated Aggregate Base 5.2.4 HMA on Asphalt Emulsion over Untreated Aggregate Base
• Design charts do not exist 1. Design pvt using full-depth HMA
• Have to determine substitution ratio between HMA & • Fig 11.11
• Assume 2” HMA surface
emulsified asphalt base First three
• Determine THMA
steps to
Substitution Ratio (SR) determine 2. Design pvt using Emulsified Asphalt Mix
SR • Fig 11.12-11.14
• Thickness of emulsified asphalt base required to substitute a unit
• Assume 2” HMA surface
thickness of HMA
• Determine TEMUL
HMA Surface 2”
3. Calculate SR=TEMUL/THMA
Actual
4. Design pvt using HMA on Aggregate Base
HMA Surface 2” Design
hHMA hEMUL • Select aggregate base thickness
Full Depth HMA THMA=hHMA-2 Emulsified Base
TEMUL=hEMUL-2 • Fig 11.15-11.20
5. Determine minimum HMA thickness
Last three • Table 11.12
Figure 11.11 Figure 11.12-11.14 steps to 6. Determine HMA thickness to be replaced by Emulsified Mix
perform the • Design thickness (step 4) – Min HMA (step 5)
TEMUL substitution
SR = 7. Determine thickness of Emulsified Mix
THMA • Thickness (step 6) * SR (step 3)

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
5.2.5 Combined Design Example 6. Planned Stage Construction
Given: Apply successive HMA layers according to predetermined
• ESUB = 10,000 psi schedule:
• Design ESAL = 1,000,000
• Based on the concept of remaining life
Need to design a pavement with HMA surface, emulsified mix • Second stage constructed before first shows significant
Type I base, and 8” aggregate subbase distress
Why? What are the advantages/reasons for planned stage?
WORK EXAMPLE ON THE BOARD
1. When funds are insufficient
2. When traffic is unpredictable (Utah Olympics example)
3. May detect weak spots during 1st stage (organic peat in
subgrade)
Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
6.1 Relative Damage 6.2 Planned Stage Procedure
n1 Where: 1. Define a relative damage for the end of the first stage (AI
Dr1 = • n1 = Actual (predicted) ESALs for Stage 1 suggests 0.6)
N1 • N1 = Allowable ESALs for initial thickness h1
0 0.6 1
Dr
Stage 1 = X-amount of years. So, n1 is the predicted traffic for
the specific location for X-amount of years Stage 1 Remaining
X-years & n1 (actual) loads life
N1 is the DESIGN life ESALs for h1. Meaning that the pavement
will fail (20% cracking / ½” rutting) after N1 applications of loads
Design Life for N1 loads

What happens if n1=N1? 2. Assume Dr1 = 0.6 Æ at the end of the 1st stage (after X-
amount of years & n1 loads) the pavement reached 60% of its
Dr = 1 … so our pavement will fail at the end of stage 1! life span
BUT, we want to construct the second stage before the first one 3. By dividing n1 with 0.6 we get a design N1 that allows so
starts showing signs of distress much traffic, that by the end of Stage 1 we reach a damage
ratio of 0.6

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
6.2 Planned Stage Procedure (cont.) 6.2 Planned Stage Procedure (cont.)
Stage 1: Stage 2:
Purpose is to select an initial thickness that will have some For the 2nd stage design we need to consider the existing
remaining life after the initial applied (n1) ESALs structure from Stage 1; the remaining life that carries over to
the 2nd stage is Dr2
• Specify Dr1 after Stage 1(AI suggests Dr1 = 0.6)
• Dr2 = 1-Dr1
n1
N1 = • For the 2nd stage we expect to have n2 ESALs over Y-amount
Dr1
of years
• N1 = allowable ESALs for Stage 1 n2
N2 =
(1 − Dr1 )
• Use N1 to obtain thickness h1 that will provide sufficient
protection, so that after n1 loads the relative damage will be • Use N2 to obtain thickness h2 that will provide sufficient
equal to 0.6 protection for the expected traffic, n2
• hoverlay = h2 – h1

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
6.2.1 Planned Stage Construction Example 7. Material Characterization
Given: Calculate subgrade MR (Topic 5):
• Full-depth HMA pavement to undergo two-stage construction • Confining stress: σ1=σ2=2 psi
• ESUB=10,000 psi & Dr1=0.6 • Deviator stress: σd=6 psi
• First Stage: 5 years, n1=150,000 ESAL
• Second Stage: 15 years, n2=850,000 ESAL 8. Variability/Reliability
Determine h1 & h2 (hoverlay) • Subgrade MR values WILL vary within a design unit (segment)
• If material and test method remain the same, we may
assume that MR is normally distributed with mean MR(avg)
WORK EXAMPLE ON THE BOARD

MR(min) MR(avg) MR(max)

Topic 6 – Asphalt Institute Design Procedure Topic 6 – Asphalt Institute Design Procedure
8.1 Three Levels of Reliability 8.2 Variability/Reliability Method
1. 104 ESAL or less, design using MR60 1. Need to get at least eight subgrade samples
• 60% probability that MR>MR60
2. 104-106, design using MR75 2’ x x x x
x x x x
• 75% probability that MR>MR75
3. 106 or more, design using MR87 2. Evaluate the samples and rank in descending MR order
• 87.5% probability that MR>MR87
3. Calculate percent equal or greater than

• C1= MR values in
C1 C2 C3 descending order
• C2= # of values equal
MR(min) MR(avg) MR(max)
MR87 MR75 MR60 to or greater than
50% of values less than MRAVG 50% of values greater than MRAVG 100%
• C3 = × C2
# of values

Topic 6 – Asphalt Institute Design Procedure

8.2 Variability/Reliability Method (cont)
4. Plot Percent Greater/Equal Than Vs Resilient Modulus

Which value is the most conservative estimate?