Flexible Pavement Design-Asphalt Institute Method - Icivil

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Flexible Pavement Thickness
Design / Asphalt Institute Method
Source:
Chapter 20: Traffic & Highway Engineering by Nicholas Garber and Lester Hoel, Third
Edition, Brooks/Cole.
Chapter 16: Highway Engineering, by Paul Wright & Karen Dixon, 7th Edition, Wiley &
sons
Instructor:
Dr. TALEB M. AL-ROUSAN
Pavement Types
Flexible Pavement:

Pavement constructed of bituminous & granular
materials.

A structure that maintains intimate contact with
subgrade and distribute loads to it, and depends on
aggregate interlock, particle friction, and cohesion
for stability.
Rigid pavement:

Pavement constructed of Portland cement concrete.

It is assumed to posses considerable flexural
strength that will permit it to act as a beam and allow
it to bridge minor irregularities in base and subgrade.
Typical Cross Section for
Conventional Flexible Pavement
Principles of Flexible Pavement
Design





Pavement structure is considered as a
multilayered elastic system.
Materials in each layers is characterized by
certain physical properties (Mr, E,….).
It assumes that subgrade is infinite in both the
vertical and horizontal directions.
Other layers are finite the vertical direction and
infinite in the horizontal direction.
The application of the wheel load causes a
stress distribution (See Figure 20.2)..
Principles of Flexible Pavement
Design Cont.

The maximum vertical stresses are compressive and
occur directly under the wheel load.
 Stresses decrease with increase in depth from the
surface.
 The maximum horizontal stress also occurs directly
under the wheel load but can be either tensile or
compressive.
 When the load and pavement thickness are within
certain ranges, horizontal compressive stresses will
occur above the neutral axis, whereas horizontal
tensile stresses will occur below the neutral axis.
 The temperature distribution within the pavement
structure will also have an effect on the magnitude of
the stresses
Figure 20.2 Typical stresses and temperature
distribution in flexible pavements under wheel load
Principles of Flexible Pavement
Design

The design of the pavement is therefore
generally based on on strain criteria that limit
both horizontal and vertical strain below those
that will cause excessive cracking and
permanent deformation.
 These criteria are considered in terms of
repeated load applications.
 Most commonly used methods:



Asphalt Institute Method
AASHTO method
California method
Elements of Thickness Design
1.
2.
3.
4.
Traffic Loading
Climate or Environment
Material Characteristics
Others: Cost, Construction,
Maintenance, Design period.
Traffic Loading

Pavement must withstand the large umber of
repeated loads of variable magnitudes
Primary loading factors:

1.
2.
3.


Magnitude of axle loads (controlled by legal load limits).
Volume & composition of axle load (Traffic survey, load
meters, & growth rate).
Tire pressure & contact area.
Equivalent Standard Axle Load ESAL (80 kN (18,000
lb or 18 kips) single axle load.
The total no. of ESAL is used as a traffic loading
input in the design of pavement structure.
Climate or Environment

1.
2.
Climate or environment affect the behavior &
performance of materials used in pavements
Temperature: high temp. cause asphalt to
loose stability, low temp. cause asphalt to
become hard & stiff, and frost heave.
Moisture: Frost related damage, volume
changes due to saturation, chemical stability
problems with moisture existence (Stripping).
Material Characteristics

1.
2.
3.



Required materials characteristics:
Asphalt surface: Material should be strong & stable
to resist repeated loading (fatigue).
Granular base & subbase: gradation, stable & strong
to resist shears from repeated loading.
Subgrade: soil classification, strong & stable.
Various standard tests are available for
determination of desired properties.
CBR, Marshal stability, Resilient Modulus, Shear
strength.
Mr (psi) = 1500 CBR or Mr (Mpa) = 10.3 CBR
Asphalt Institute Method

Method is based on two assumed stress –strain
conditions:
1.
Wheel load (W) is transmitted to the pavement
surface through the tire at a uniform vertical
pressure (Po). The stresses are then spread through
the pavement structure to produce a reduced max.
vertical stress (P1) at the subgrade surface. (See
Figure 20.3 in text )
The wheel load (W) causes the pavement structure
to deflect creating both compressive & tensile
stresses in the pavement structure. (See Figure
20.4 in text ).
2.
Figure 20.3 Spread of wheel load pressure through
pavement structure
Figure 20.4 Tensile and compressive stresses in
pavement structure
Asphalt Institute Method Cont.

1.
2.

This method considers the following strains
as being responsible for the most common
traffic related distresses:
Max. Horizontal tensile strains (Et) on the
bottom of the asphalt layer (causes fatigue
cracking).
Max. Vertical compressive strains (Ec) on the
top of subgrade (causes permanent
deformation).
Et & Ec are used as failure criteria
Design Procedure for Asphalt
Institute Method




Asphalt Institute thickness design manual was
prepared using a computer program and suitable data.
The charts have been prepared for a range of
traffic load, which are usually adequate for normal
traffic volume encountered in practice, when this
range is exceeded the computer version should be
used.
The manual includes charts for six types of pavement
structures, and three sets of environmental conditions
based on the mean annual air temp. (45o, 60o, and 75o
F).
Example design chart is shown in the coming slide.
Pavement Types in The Asphalt
Institute Method
Full depth asphalt concrete. (see Fig. 20.5)
Asphalt concrete surface and emulsified asphalt
base.
1.
2.
1.
2.
3.
Type I: Emulsified asphalt mixes made with processed
dense-graded aggregates. (see Fig. 20.6)
Type II: Emulsified asphalt mixes made with semiprocessed, crusher-run, pit-run, or bank-run aggregates.
(see Fig. 20.7)
Type III: Emulsified asphalt mixes made with sands or silty
sands. (see Fig. 20.8)
Asphalt concrete and untreated aggregate base.
3.
1.
2.
Base thickness of 6”. (see Fig. 20.9)
Base thickness of 12”. (see Fig. 20.10)
Design Procedure for Asphalt
Institute Method
1.
2.
3.
4.
5.
Select or determine Input data.
Select surface and base material.
Determine minimum thickness required for
input data.
Evaluate feasibility of stage construction and
prepare stage construction plan.
Carry our economic analysis of alternative
design and select the best design.
Step 1: Determine Design Inputs

Design Inputs are:
Traffic characteristics.
 Subgrade engineering properties.
 Subbase and base engineering properties.

Traffic Characteristics
Determined in terms of number of
repetitions of an 18,000 lb (80)kN single
axle load applied to the pavement on two
sets of dual tires (Equivalent Single
Axle Load (ESAL)).
 ٍSee next slides for the determination of
the ESAL.

Traffic Analysis
Estimate the number of vehicles of
different types (Passenger cars, single
unit trucks, multi unit trucks of various
sizes) expected to use the pavement
over the design period.
In case data are not available, estimates
can be made from Table 20.4 in text
which gives representative values for
the united states.
1.
Traffic Analysis Cont.
2. Estimate the (%) of total truck traffic expected
to use the design lane.
Design lane: Lane expected to receive the
severe service.
 % of trucks is found by observation

In case data are not available, estimates can
be made from Table 20.4 which gives
representative values for truck distribution in
the united states.
 (see Table 16.1 also in the Reference text).
Table 20.4 Distribution of Trucks
Traffic Analysis Cont.
3. When the axle load of each vehicle type is known,
these can be converted to ESAL using the equivalency
factors given in Table 20.3 in text or Table 16.3 in Ref
If the axle load is unknown, the ESAL can also be
found from the vehicle types by using a truck factor for
that vehicle type.
Truck Factor (TF): The no. of ESALs contributed by
passage of a vehicle.
For each weight class, determine the truck factor.
Traffic Analysis Cont.
TF = [SUM (No. of axles in each wt. class X EALF)] / Total No. of
vehicles





Truck factor can be estimated Using Table 20.5 in text or
Table16.2 from ref.
Equivalent Axle Load factor or Load equivalency factor (EALF)
presented in Table 20.3 in text or Table 16.3 in Ref.
EALF: Defines the damage per pass to a pavement by the axle of
question relative to the damage per pass of a standard axle load
(80 kN or 18-kip)
EALF depends on type of pavement, thickness or structural
capacity, and failure conditions (based on experience).
See Truck Factor Example provided in Figure 16.8 Ref. and
example in Table 20.8 in text.
Axle & Wheel Configurations
Single Axle with
Single Tire
Tandem Axles with Dual
Tires
Single Axle with Dual Tires
Tridem Axles with Dual
Tires
Truck Factor Example
Traffic Analysis Cont.
4. Multiply (Tf) by the no. of vehicles in
each group and get the sum for all
groups.
ESAL = Sum (TF X No. of vehicles) all
groups.
See Example provided in next slides.
Example on Computation of ESAL
Total ESAL Calculation

The total ESAL applied on the highway during its
design period can be determined only if the following
are known:





Design period
Traffic growth factor
Traffic growth factor is estimated using historical
records or comparable facilities or obtained from
studies made by specialized agencies.
It is advisable to determine annual growth rates for
trucks and passenger cars separately.
Design period: Number of years the pavement will
effectively continue to carry the traffic load without
requiring an overlay. (usually 20 years).
Expected Traffic Volume During
Design Period
See Table 20.6 for growth factors, or
calculate it using:
Gjt = ( (1 + j)t -1)/ j )
j: Rate of growth.
t: Design period (yrs).
Table 20.6 for growth factors
Computing Design ESAL (Projected)
Total ESAL Calculation Cont.







the portion of the ESAL acting on the design lane is
used in the determination of pavement thickness.
Either lane of a two-lane highway is a design lane.
In multilane highways the outer lane is the design
lane.
See Table 20.7 for percentage of total truck traffic on
design lane.
The initial daily traffic is in two directions over all traffic
lanes.
Must be multiplied by direction distribution & Lane
distribution to obtain initial traffic on design lane.
Traffic to be used in design is the average traffic
during design period (i.e. multiply by growth factor).
Table 20.7 for percentage of total
truck traffic on design lane
Total ESAL Calculation Cont.
ESALi = (AADTi) (Fd) (Gjt) (Ni) (FEi) (365)
ESALi : ESAL for axle category i
AADTi: First year annual average daily traffic for axle
category i.
(Fd): Design lane factor
(Gjt): growth rate factor for a given growth rate j and
design period t.
(Ni): number of axles on each vehicle in category i.
(FEi): load equivalency factor for axle category i.
Total ESAL Calculation Cont.
When truck factors are used
ESALi = (AADTi) (Fd) (Gjt) (fi) (365)
ESALi : ESAL for axle category i
AADT: First year annual average daily traffic for axle
category i.
(Fd): Design lane factor
(Gjt): growth rate factor for a given growth rate j and
design period t.
(Ni): number of axles on each vehicle in category i.
(fi): Truck factor for vehicle in truck category i.
Total ESAL Calculation Cont.
When truck factors are used
ESAL = SUM [ESALi ]
from i =1 to n
n= number of truck categories
ESAL : ESAL for all vehicles during the design
period.
ESAL Example

An 8-lane divided highway is to be constructed
on a new alignment. Traffic volume forecast
indicates that AADT in both direction during
the first year of operation will be 12,000 with
the following vehicle mix:
Passenger cars (1000 lb/axle) = 50%
 2-axle single unit trucks (6000 lb/axle) = 33%
 3-axle single unit trucks (10,000 lb/axle) = 17%
If the expected annual traffic growth rate is 4% for all
vehicles,
Determine the design ESAL for a design period of 20
years.

ESAL Example
Solution
 Growth Factor = Gjt = [(1 + j)t - 1]/ j = [(1 + 0.04)20 - 1]/ 0.04
= 29.78 (or see Table 20.6)


% truck volume on design lane = 45 ( assumed,
Table 20.7)
Load equivalency Factors (Table 20.3)
 Passenger cars (1000 lb/axle) = 0.00002
(negligible)
 2-axle single unit trucks (6000 lb/axle) = 0.01043
 3-axle single unit trucks (10,000 lb/axle) = 0.0877
ESAL Example

Solution
ESALi = (AADTi) (Fd) (Gjt) (Ni) (FEi) (365)
For passenger cars…… ESAL = 0 or negligible
For 2-axle single unit trucks
ESAL = (12,000 X 0.33) X 0.45 X 29.78 X 2 X 0.01043 X 365
= 0.4041 X 106
For 3-axle single unit trucks
ESAL = (12,000 X 0.17) X 0.45 X 29.78 X 3 X 0.0877 X 365
= 0.2.6253 X 106
Total ESAL = 3.0294 X 106
Subgrade Engineering Properties
Materials Evaluation

The main engineering property required for the
subgrade is its Resilient Modulud (Mr).
 The design subgrade (Mr) should be based on
expected level of traffic expressed in ESALs.
 To ensure more conservative design, lower value of
(Mr) is used for higher volumes of traffic.
 It is recommended that (Mr) is found for (6 to 8)
samples of subgrade.
 Arrange Mr values in descending order.
 Plot as cumulative distribution.
 Chose design subgrade (Mr) from the curve as
follows:
Design Subgrade Mr
Mr test
Value
13500
% values
Value >=
>=
1
Subgrade Design Limits
12.5
11900
2
25
11300
3
37.5
10000
4
50
9500
5
62.5
8800
6
75
7800
7
87.5
6200
8
100
Traffic Level
Design Percentile
ESAL
Value
<= 10,000
60
10000 to 1000,000
75
> 1000,000
87.5
Design Mr
120
100
% >=
80
60
40
20
0
0
2000
4000
6000
8000
Mr (psi)
10000
12000
14000
16000
Resilient Modulus (Mr)





AASHTO T292 (Resilient Modulus of Subgrade soils).
(Mr) can be found using repeated loading procedure test such as
(unconfined compression test or triaxial compression test).
0.1 sec loading and 1 to 3 sec. unloading.
Linear Variable Displacement Transducers (LVDTs) are used to
measure strains.
Elastic modulus based on the recoverable strain under repeated
loading is called the resilient modulus







(Mr) = (Deviator stress/ Recoverable axial strain)
Deviator stress = Axial stress – confining pressure
Recoverable axial strain = Max strain – permanent strain
See Fig. 16.2 in ref.
Mr = 10.342 CBR (MPa) or
Mr = 1500 CBR (psi)
These formulas can be used when only when Mr < 30000 psi….. CBR < 20
Subgrade Mr Seasonal Variation
Freeze Time
Recovery
Time
Subbase & Base engineering
Properties

Certain requirements are needed, which
are given in terms of
PI
 % passing sieve # 200
 Min. sand equivalent.


See Table 20.9 in texts for quality
requirements of untreated base and
subbase.
Table 20.9 Quality Requirements for
Base and Subbase Materials
Step 2: Select Surface and Base
materials




Designer is free to select either an asphalt concrete
surface or an emulsified asphalt surface along with an
asphalt concrete base, an emulsified asphalt base, or
an untreated aggregate base for the underlying layer.
The choice will depend on the material that is
economically available.
Asphalt Institute recommend certain grades of asphalt
cement that should be used for different temperature
conditions (mean annual temp.)
See Table 20.10 in text.
Table 20.10 Recommended Asphalt Grades for Different
Temperature Conditions
Step 3: Determine Min. Thickness
Requirements

The minimum thickness required for the design ESAL
and the type of surface, base, and subbase selected is
obtained either by:





Computer program DAMA.
Charts: using design ESAL and subgrade Mr.
See Figure 20.5-10 in text
See Table 20.11 for min thicknesses of asphalt
concrete over types II & III emulsified asphalt base.
See Table 20.12 for min thicknesses of asphalt
concrete over untreated aggregate base.
Table 20.11 Min. thicknesses of asphalt concrete over
emulsified asphalt bases
Table 20.12 Min. thicknesses of asphalt concrete over
Untreated Aggregate Base
Step 4: Feasibility of Planning stage
construction

Planned Stage Construction involves
successive application of HMA layers
according to a predetermined time schedule.
 Beneficial when:


Funds are insufficient for constructing a pavement
with long design life.
Great amount of uncertainty in estimating traffic.
• Concept: Remaining life which implies that the
second stage will be constructed before the
first stage shows serious signs of distress.
Planned Stage Construction Cont.
Pavement is designed for initial traffic &
next stage can be designed using traffic
projections based on traffic in service.
 Stage construction allows weak spots
that develop in the first stage to be
detected and repaired in the second
stage.

Planned Stage Construction Cont.
n1: Actual ESAL for stage 1
N1: Allowable ESAL for initial thickness (h1) selected for stage 1.
Then The damage ratio (Dr) at the end of stage 1 is:
Dr = n1/ N1
Dr < 1.0 ………. When Dr =1.0 pavement fails.



(1-Dr) = Remaining life in the existing pavement at the end of
stage 1.
h1 is obtained based on Dr =1.0.
To keep some life, h1 should be determined based on adjusted
ESAL (N1) > ESAL (n1)
N1= n1/Dr
Planned Stage Construction Cont.
n2: Design ESAL for stage 2.
N2: Allowable or adjusted ESAL to permit selection of (h2) that will carry
traffic n2 and use the remaining life in stage 2.
Then The damage incurred in stage 2 should not exceed the remaining life.
n2/ N2 = (1-Dr)
N2 = n2/ (1-Dr)


hs = h2 –h1 = Additional thickness required in stage 2 (overlay).
MS-1 recommended (5 – 10 yrs) stage 1 with 60% Dr.
Planned Stage Construction
Example
Given:
 Full-depth asphalt pavement
 subgrade Mr = 10,000 psi
 Use two stage to construct this pavement
 Stage 1: 5 yrs, ESAL = 150,000 , Dr = 60% at
the end of stage 1.
 Stage 2: 15 yrs, ESAL = 850,000
 Required:


Determine thickness of HMA required for first 5yrs.
Thickness of overlay required to accommodate the
additional traffic expected during the next 15 yrs.
Planned Stage Construction
Example Cont.
Solution:
 n1=150,000 &
Dr = 0.6
Find N1 = n1/Dr = 150,000/ 0.6 = 250,000
From design Chart with N1 & Mr find
h1=7.3 in use 7.5 in.

n2 = 850,000
& 1-Dr = 0.40
Find N2 = n2/ (1-Dr) = 850,000/ 0.4 = 2,100,000
From design chart with N2 & Mr find
h2 =11.0


First stage thickness = h1 = 7.5 in
Overlay thickness hs = h2 –h1 = 11.0 – 7.5 = 3.5 in.
Planned Stage Construction
Example Cont.
Solution:
 If the design was not divided into 2 stages, the
thickness of the pavement using (Mr = 10,000
psi & ESAL = 1000,000) is :
9.8 in use 10 in.
 The use of stage construction decreased the
thickness in first stage by (10.0 -2.5 = 2.5 in),
but increased the total thickness by
(11.0 - 10.0 = 1.0 in).
Step 5: Economic Analysis & Design
Selection
Find several alternative designs for the
same design ESAL and subgrade Mr.
 Carry out an economic evaluation of
these alternatives.
 Determine best alternative.

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