topic 6 - ai flexible pavement design

8/7/2019 Topic 6 - AI Flexible Pavement Design

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Topic 6 Asphalt Institute DesignProcedure

Dr. Christos Drakos

University of Florida

Introduction to Pavement Design

1. Introduction

Establish Layer Thicknesses:

To limit distress (acceptable levels)

For anticipated loading & environmental conditions

Using available/selected materials

1.1 Elements to be Defined/Identified for Design

Conditions:

Traffic loading (volume, frequency,magnitude ESALs)

Environment (temperature, moisture)

Material Properties: Subgrade varies w/ season (existing material)

Pavement Structure (engineered materials)


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1.1 Elements to be Defined/Identified for Design (cont.)

Performance Criteria:

Conditions that define failure

Performance Relationship

TRAFFIC ENVIRONMENT SUBGRADE MATL PROPERTIES

LAYER THICKNESSES

PAVEMENTPERFORMANCE

PERFORMANCERELATION


2. Design Approach

TRIAL

MATERIALS

TRIAL

THICKNESSES

PERFORMANCE

RELATION

NO

YESLIFE-COST

CYCLE

PERFORMANCE

PERFORMANCE

CRITERIA

TRAFFIC

ENVIRONMENT

SUBGRADE

MATERIAL

PROPERTIES

A Pavement Performance Model is an equation that relatessome extrinsic time factor (age, or number of loadapplications) to a combination of intrinsic factors (structuralresponses, drainage, etc) and performance indicators


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3. Empirical Vs Mechanistic-Empirical

Difference is in the nature of Performance Relation

3.1 Empirical

Statistical/Experimental (based on road tests)

Limited conditions/environment

3.2 Mechanistic-Empirical

Relate analytical response to performance:

More reliable/robust than empirical

Integrates the structural aspects of a pavement to thematerial/mix design properties of the pavement layers!!!

Improve the relation by understanding the mechanics


4. Response and Performance

4.1 Response = Reaction to an action

Response = Pavement & Material response to applied loads

(traffic & environment)

AC

BASE

1 &2Pavement Responses

1

2

3

&Material Responses

1

2

What are Pavement & Material Responses?

element


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4.1 Response = Reaction to an action

Predict load responses with structural response models:

Vary in sophistication:

Linear Elastic

Non-linear Elastic

Viscoelastic

etc

Predict temperature responses with thermal response models:

th = fnc (material, temperature, cooling rate, dimensions)


4.2 Performance

Performance is the measurable adequacy ofSTRUCTURAL&FUNCTIONALservice over a specified design period

Structural Functional (user defined)

Number of loads the pavementcan support before it reachesunacceptable level ofstructural/functional distress

Roughness Ride quality

Friction Geometry Appearance

Surface cracking

Loss of color


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Topic 6 Asphalt Institute Design Procedure

ASPHALT INSTITUTE (AI)

US based association of international asphalt producers thatpromotes the use of petroleum asphalt products

http://www.asphaltinstitute.org/

Design method based on computer model DAMA Computes amount of damage (cracking & rutting) based

on traffic in a specific environment Multilayer elastic theory; used correction factors to

account for base non-linearity Used three temperature regimes; representing three

climatic regions in the US NY(45), SC(60) & AZ(75) Developed design charts from the results

1. Development


Two types of strains are considered critical in design of asphaltpavements:

Horizontal tensile strain, t @ the bottom of AC layer Vertical compressive stain, c @ the top of the subgrade

2. Design Criteria

2.1 Fatigue Cracking

AC t

Basic equation:

32

1

ff

tfEfN

=

Where: Nf= Number of cycles to failure t = Tensile strain @ bottom of AC layer f1 = Field correlation shift factor f2 & f3 = Laboratory determined values


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2.1 Fatigue Cracking (cont)

32

1ff

tf EfN =

Asphalt Institute calibrated the field shift factor using data fromthe AASHO road test

f1 = 0.0796

2.1.1 Fatigue tests

t

Why 3rd-point loading?

To have an even distributionof M; we know the value of

M, no matter where thespecimen failsV

M


2.1.1 Fatigue tests (cont)


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2.1.2 Constant Stress Fatigue Test

Apply constant stress Failure occurs when the material fractures

0

Stress,

Number of Cycles, N

Strain,

Number of Cycles, N

2.1.3 Constant Strain Fatigue Test

Apply constant strain (rate of deformation) Failure occurs when E=E0

Stress,

Number of Cycles, N

Strain,

Number of Cycles, N

0

0

0

0

0

02

1

= ; 021

=


2.1.4 Fatigue Test Analysis

Plot the strain Vs number of repetitions to failure on log scales C1 & C2 curves for the same material @ different temperature

Strain,Logt

Number of Cycles, Log Nf

Which curve has the highest stiffness?

Check: Select a strain level Find the corresponding Nf Higher stiffness will have less

number of cycles to failure

C1

C2

Nf1Nf2

Low

High

From the graph: Stiffness of the material will depend on time of the year (temperature) t depends on the material properties (E) So, the cycles to failure Nfwill also depend on the temperature

Must use cumulative damage approach to evaluate failure


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2.2 Damage Ratio

Dr=Actual # of Load Repetitions

Allowable # of Load RepetitionsPavement has failed if Dr=1

= =

=p

i

m

j ji

ji

N

nDr

1 1 ,

, Where:m = no. of load types = 1 for AIp = no. of periods in analysis = 12 for a year

2.2.1 Damage ratio example

Damage Ratio

Actual Traffic

Allowable Traffic

E4, t4E3, t3E2, t2E1, t1Material properties

4321Periods (Seasons)

Dr=Dri i.e. Dr=0.1; Design Life = 1/Dr = 10 years

Nf4Nf3Nf2Nf1

n1 n2 n3 n4

Dr1= n1/Nf1 Dr2= n2/Nf2 Dr3= n3/Nf3 Dr4= n4/Nf4


2.3 Permanent Deformation

Only SUBGRADE rutting considered, as governed bycompressive strain

5

4

f

cd fN

=

477.4910365.1

= cdN

AI calibrated the equationusing AASHO road test data

Consider the following two pavements

E1

E2

E1

E2

cA cBE3A E3B

Similar structure E3A>> E3B Assume cA= cB

Assume cA= cB

BUT:

c @ P =c

E3cA> NdB

So,


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3. Environment

Nf& Nd vary with time of the year because of change inmaterial properties with the weather

64

34

4

11 +

+

++=

zzMM

AP

Where: MP = Mean pavement temperature

MA= Mean monthly air temperature z = Depth below the surface (1/3 of AC layer depth)

AI procedure considers the environment based on: Mean monthly temperature Monthly variable material modulus

3.1 Asphalt Concrete

Then we can use: ( ) ( )PMELog 01.048.61 =


Four distinct periods: Freeze Thaw Recovery Normal

Table 11.9 shows the suggested conditions to represent frosteffects on the subgrade

3.2 Subgrade

Normal MR

Frozen MR

Thaw MR


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4. Traffic

Calculate design ESALs (Topic 4)

5. Design Procedure

5.1 Objective

DETERMINE THE REQUIRED STRUCTURAL THICKNESS FOR

EXPECTED TRAFFIC, SUBGRADE CONDITIONS, AND

ENVIRONMENT SUCH THAT:

Rutting < in

Fatigue Cracking < 20% of Area

OVER THE DESIGN LIFE (as defined by traffic)


5.2 Pavement Types

5.2.1 Full-Depth HMA

Pavement constructed completely from HMA Figure 11.11; includes both surface and base course thickness

HMA BASE

HMA SURFACE

Thickness Use Subgrade MR& ESALs Read thickness off the chart

Example: Subgrade MR= 11,000 psi Traffic = 1.1x106 ESAL Thickness = ?

For multiple HMA within a layer usecomposite modulus

( ) ( )

+

+=

BA

BBAA

hh

EhEhE

11

3/1

11

3/1

11

h1A

h1B


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5.2.1 Full-Depth HMA (cont)

Thickness = 8in


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) 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)


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5.2.2 HMA over Emulsified Asphalt Base (cont)

Minimum HMA thickness required (ESAL & Base Type) Table 11.12

HMA SURFACE hHMA

EMULSIFIED BASEhEMUL

TYPE I Fig 11.12 TYPE II Fig 11.13 TYPE III Fig 11.14

hEMUL from the graph Determine hHMA


5.2.2 HMA over Emulsified Asphalt Base (cont)


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5.2.3 HMA over Untreated Aggregate Base

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

AGGREGATE BASE (known)

Determine the required HMA thicknessfor the specific base thickness

Fig 11.15-11.20


5.2.3 HMA over Untreated Aggregate Base (cont)


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5.2.4 HMA on Asphalt Emulsion over Untreated Aggregate Base

Design charts do not exist Have to determine substitution ratio between HMA &

emulsified asphalt base

Substitution Ratio (SR) Thickness of emulsified asphalt base required to substitute a unit

thickness of HMA

HMA Surface 2

2HMA Surface

Full Depth HMA

hHMA

Figure 11.11

Emulsified Base

hEMUL

Figure 11.12-11.14

THMA=hHMA-2TEMUL=hEMUL-2

HMA

EMUL

T

TSR =


5.2.4 HMA on Asphalt Emulsion over Untreated Aggregate Base

1. Design pvt using full-depth HMA Fig 11.11 Assume 2 HMA surface Determine THMA

2. Design pvt using Emulsified Asphalt Mix Fig 11.12-11.14 Assume 2 HMA surface Determine TEMUL

3. Calculate SR=TEMUL/THMA4. Design pvt using HMA on Aggregate Base

Select aggregate base thickness Fig 11.15-11.20

5. Determine minimum HMA thickness Table 11.12

6. Determine HMA thickness to be replaced by Emulsified Mix Design thickness (step 4) Min HMA (step 5)

7. Determine thickness of Emulsified Mix Thickness (step 6) * SR (step 3)

First threesteps todetermineSR

Last threesteps toperform thesubstitution

ActualDesign


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5.2.5 Combined Design Example

Given: ESUB = 10,000 psi Design ESAL = 1,000,000

Need to design a pavement with HMA surface, emulsified mixType I base, and 8 aggregate subbase

WORK EXAMPLE ON THE BOARD


6. Planned Stage Construction

Based on the concept of remaining life Second stage constructed before first shows significant

distress

Why? What are the advantages/reasons for planned stage?

1. When funds are insufficient2. When traffic is unpredictable (Utah Olympics example)

3. May detect weak spots during 1st stage (organic peat insubgrade)

Apply successive HMA layers according to predeterminedschedule:


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6.1 Relative Damage

1

1

1N

nDr =

Where: n1 = Actual (predicted) ESALs for Stage 1 N1 = Allowable ESALs for initial thickness h1

Dr = 1 so our pavement will fail at the end of stage 1!BUT, we want to construct the second stage before the first onestarts showing signs of distress

What happens if n1=N1?

Stage 1 = X-amount of years. So, n1 is the predicted traffic forthe specific location for X-amount of years

N1 is the DESIGN life ESALs for h1. Meaning that the pavementwill fail (20% cracking / rutting) after N1 applications of loads


Stage 1X-years & n1 (actual) loads

Remaininglife

Dr0 0.6 1

Design Life for N1 loads

1. Define a relative damage for the end of the first stage (AIsuggests 0.6)

6.2 Planned Stage Procedure

2. Assume Dr1 = 0.6 at the end of the 1st stage (after X-

amount of years & n1 loads) the pavement reached 60% of itslife span

3. By dividing n1 with 0.6 we get a design N1 that allows somuch traffic, that by the end of Stage 1 we reach a damageratio of 0.6


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6.2 Planned Stage Procedure (cont.)

Stage 1:Purpose is to select an initial thickness that will have someremaining life after the initial applied (n1) ESALs

Specify Dr1 after Stage 1(AI suggests Dr1 = 0.6)

1

1

1Dr

nN =

Use N1 to obtain thickness h1 that will provide sufficientprotection, so that after n1 loads the relative damage will beequal to 0.6

N1 = allowable ESALs for Stage 1


6.2 Planned Stage Procedure (cont.)

Stage 2:For the 2nd stage design we need to consider the existingstructure from Stage 1; the remaining life that carries over tothe 2nd stage is Dr2

Use N2 to obtain thickness h2 that will provide sufficientprotection for the expected traffic, n2

hoverlay = h2 h1

For the 2nd stage we expect to have n2 ESALs over Y-amountof years

)1( 1

2

2Dr

nN

=

Dr2 = 1-Dr1


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6.2.1 Planned Stage Construction Example

Given: Full-depth HMA pavement to undergo two-stage construction ESUB=10,000 psi & Dr1=0.6 First Stage: 5 years, n1=150,000 ESAL Second Stage: 15 years, n2=850,000 ESAL

WORK EXAMPLE ON THE BOARD

Determine h1 & h2 (hoverlay)


7. Material Characterization

Calculate subgrade MR(Topic 5): Confining stress: 1=2=2 psi Deviator stress: d=6 psi

8. Variability/Reliability

Subgrade MRvalues 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)

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


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1. 104 ESAL or less, design using MR60 60% probability that MR>MR60

2. 104-106, design using MR75 75% probability that MR>MR75

3. 106 or more, design using MR87 87.5% probability that MR>MR87

8.1 Three Levels of Reliability

MR(avg)M

R

(max)MR

(min)MR60MR87 MR75

50% of values greater than MRAVG50% of values less than MRAVG


8.2 Variability/Reliability Method

1. Need to get at least eight subgrade samples

x2x

xx

xxxx

2. Evaluate the samples and rank in descending MRorder3. Calculate percent equal or greater than

C2C1 C3

2C3C =valuesof#

100%

C1= MRvalues indescending order

C2= # of values equalto or greater than


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8.2 Variability/Reliability Method (cont)

4. Plot Percent Greater/Equal Than Vs Resilient Modulus

Which value is the most conservative estimate?

topic 6 - ai flexible pavement design

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