Chapter 62: Highway and Airport Pavement Designfreeit.free.fr/The Civil Engineering...

62Highway and Airport

Pavement Design

62.1 Introduction62.2 Pavement Types and Materials

Flexible versus Rigid Pavement • Layered Structure of Flexible Pavement • Rigid Pavement • Considerations for Highway and Airport Pavements

62.3 Traffic-Loading Analysis for Highway Pavements Traffic Stream Composition • Traffic-Loading Computation • Directional Split • Design Lane Traffic Loading • Formula for Computing Total Design Loading

62.4 Traffic Loading Analysis for Airport PavementsTraffic Stream Composition • Computation of Traffic Loading • Equal Stress ESWL • Equal Deflection ESWL • Critical Areas for Pavement Design

62.5 Thickness Design of Flexible PavementsAASHTO Design Procedure for Flexible Highway Pavements • AI Design Procedure for Flexible Highway Pavements • FAA Design Procedure for Flexible Airport Pavements • Mechanistic Approach for Flexible Pavement Design

62.6 Structural Design of Rigid PavementsAASHTO Thickness Design for Rigid Highway Pavements • AASHTO Reinforcement Design for Rigid Highway Pavements • PCA Thickness Design Procedure for Rigid Highway Pavements • FAA Method for Rigid Airport Pavement Design

62.7 Pavement Overlay DesignAI Design Procedure for Flexible Overlay on Flexible Highway Pavement • AI Design Procedure for Flexible Overlay on Rigid Highway Pavement • PCA Design Procedure for Concrete Overlay on Concrete Highway Pavement • FAA Design Procedure for Flexible Overlay on Flexible Airport Pavement • FAA Design Procedure for Flexible Overlay on Concrete Airport Pavement • FAA Design Procedure for Concrete Overlay on Concrete Airport Pavement

62.1 Introduction

Pavements are designed and constructed to provide durable all-weather traveling surfaces for safe andspeedy movement of people and goods with an acceptable level of comfort to users. These functionalrequirements of pavements are achieved through careful considerations in the following aspects duringthe design and construction phases: (a) selection of pavement type, (b) selection of materials to be usedfor various pavement layers and treatment of subgrade soils, (c) structural thickness design for pavement

T. F. FwaNational University of Singapore

© 2003 by CRC Press LLC

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The Civil Engineering Handbook, Second Edition

layers, (d) subsurface drainage design for the pavement system, (e) surface drainage and geometric design,and (f ) ridability of pavement surface.

The two major considerations in the structural design of highway and airport pavements are materialdesign and thickness design. Material design deals with the selection of suitable materials for variouspavement layers and mix design of bituminous materials (for flexible pavement) or portland cementconcrete (for rigid and interlocking block pavements). These topics are discussed in other chapters ofthis handbook. This chapter presents the concepts and methods of pavement thickness design. As thename implies, thickness design refers to the procedure of determining the required thickness for eachpavement layer to provide a structurally sound pavement structure with satisfactory performance for thedesign traffic over the selected design life. Drainage design examines the entire pavement structure withrespect to its drainage requirements and incorporates facilities to satisfy those requirements.

62.2 Pavement Types and Materials

Flexible versus Rigid Pavement

Traditionally, pavements are classified into two categories, namely flexible and rigid pavements. The basisfor classification is the way by which traffic loads are transmitted to the subgrade soil through thepavement structure. As shown in Fig. 62.1, a flexible pavement provides sufficient thickness for loaddistribution through a multilayer structure so that the stresses and strains in the subgrade soil layers arewithin the required limits. It is expected that the strength of subgrade soil would have a direct bearingon the total thickness of the flexible pavement. The layered pavement structure is designed to takeadvantage of the decreasing magnitude of stresses with depth.

A rigid pavement, by virtue of its rigidity, is able to effect a slab action to spread the wheel load overthe entire slab area, as illustrated in Fig. 62.1. The structural capacity of the rigid pavement is largelyprovided by the slab itself. For the common range of subgrade soil strength, the required rigidity for aportland cement concrete slab (the most common form of rigid pavement construction) can be achieved

FIGURE 62.1 Flexible and rigid pavements.

(a) Typical Cross Section of Flexible Pavement

Wearing Course

Binder Course

Base Course

Subbase Course

Prepared SubgradeNatural Subgrade

Tack Coat

Prime Coat

(b) Load Transmission in Flexible Pavement

WheelLoad

(d) Load Transmission in Rigid Pavement

WheelLoad

(c) Typical Cross Section of Rigid Pavement

HighwayPavement

AirportPavement

Concrete Slab

Base or Subbase

Prepared SubgradeNatural Subgrade

6-12 in

4-6 in

6-12 in

10-24 in

4-12 in

9-18 in

(1 in = 25.4 mm)

HighwayPavement

AirportPavement

1-2 in

2-4 in

4-12 in

12-18 in

6-24 in

3-6 in

6-12 in

12-36 in

12-60 in

(1 in = 25.4 mm)

Highway and Airport Pavement Design

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without much variation in slab thickness. The effect of subgrade soil properties on the thickness of rigidpavement is therefore much less important than in the case of flexible pavement.

Layered Structure of Flexible Pavement

Surface Course

In a typical conventional flexible pavement, known as asphalt pavement, the surface course usuallyconsists of two bituminous layers — a wearing course and a binder course. To provide a durable,watertight, smooth-riding, and skid-resistant traveled surface, the wearing course is often constructed ofdense-graded hot mix asphalt with polish-resistant aggregate. The binder course generally has largeraggregates and less asphalt. The composition of the bituminous mixtures and the nominal top sizeaggregates for the two courses are determined by the intended use, desired surface texture (for the caseof wearing course), and layer thickness. A light application of tack coat of water-diluted asphalt emulsionmay be used to enhance bonding between the two courses. Table 62.1 shows selected mix compositionslisted in ASTM Standard Specification D3515 [1992]. Open-graded wearing courses, some with air voidexceeding 20%, have also been used to improve skid resistance and reduce splash during heavy rainfallby acting as a surface drainage layer.

Base Course

Base and subbase layers of the flexible pavement make up a large proportion of the total pavementthickness needed to distribute the stresses imposed by traffic loading. Usually base course also serves asa drainage layer and provides protection against frost action. Crushed stone is the traditional materialused for base construction to form what is commonly known as the macadam base course. In thisconstruction, choking materials consisting of natural sand or the fine product resulting from crushingcoarse aggregates are added to produce a denser structure with higher shearing resistance. Such basecourses are called by different names, depending on the construction method adopted.

Dry-bound macadam is compacted by means of rolling and vibration that work the choking materialsinto the voids of larger stones. For water-bound macadam, after spreading of the choking materials, wateris applied before the entire mass is rolled. Alternatively, a wet-mix macadam may be used by premixingcrushed stone or slag with a controlled amount of water. The material is spread by a paving machine

TABLE 62.1 Example Composition of Dense Bituminous Paving Mixtures

Mix Designation and Nominal Maximum Size of Aggregate

Sieve Size2 in.

(50 mm)1½ in.

(37.5 mm)1 in.

(25.0 mm)3/4 in.

(19.0 mm)1/2 in.

(12.5 mm)3/8 in.

(9.5 mm)

2½ in. 100 — — — — — 2 in. 90–100 90–100 100 — — — 1½ in. — 90–100 100 — — — 1 in. 60–80 — 90–100 100 — — 3/4 in. — 56–80 — 90–100 100 — 1/2 in. 35–65 — 56–80 — 90–100 1003/8 in. — — — 56–80 — 90–100No. 4 17–47 23–53 29–59 35–65 44–74 55–85No. 8 10–36 15–41 19–45 23–49 28–58 32–67No. 16 — — — — — — No. 30 — — — — — — No. 50 3–15 4–16 5–17 5–19 5–21 7–23No. 100 — — — — — — No. 200 0–5 0–6 1–7 2–8 2–10 2–10

Note: Numbers in table refer to percent passing by weight.Source: ASTM, Standard Specification D3515-84, Annual Book of ASTM Standards, Vol. 04.03 —

Road and Paving Materials; Travelled Surface Characteristics, 1992. With permission.

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and compacted by a vibrating roller. Table 62.2 shows specifications for unbound base and subbasematerials specified by AASHTO and ASTM.

Granular base materials may be treated with either asphalt or cement to enhance load distributioncapability. Bituminous binder can be introduced by spraying heated asphalt cement on consolidated androlled crushed stone layer to form a penetration macadam road base. Alternatively, bituminous roadbases can be designed and laid as in the case for bituminous surface courses. Cement-bound granularbase material is plant mixed with an optimal moisture content for compaction. It is laid by paver andrequires time for curing. Lean concrete base has also been used successfully under flexible pavements.Table 62.3 shows examples of grading requirements for these materials.

TABLE 62.2(a) Grading Requirements for Unbound Subbase and Base Materials — ASSHTO Designation M147-65 (1989)

Grading: Percentage Passing

Sieve Size A B C D E F

50 mm 100 10025 mm — 75–95 100 100 100 100

9.5 mm 30–60 40–75 50–85 60–100 — — 4.75 mm 25–55 30–60 35–65 50–85 55–100 70–100

2 mm 15–40 20–45 25–50 40–70 40–100 55–100425 mm 8–20 15–30 15–30 25–45 20–50 30–70

75 mm 2–8 5–20 5–15 5–20 6–20 8–25

Other requirements:1. Coarse aggregate (>2 mm) to have a percentage wear by Los Angeles

test not more than 50.2. Fraction passing 425-mm sieve to have a liquid limit not greater than

25% and a plasticity index not greater than 6%.Source: AASHTO Designation M147-65, AASHTO Standard Specifications

for Transportation Materials and Methods of Sampling and Testing, AmericanAssociation of State Highway and Transportation Officials, Washington, D.C.,1989. With permission.

TABLE 62.2(b) ASTM DesignationD2940-74 (Reapproved 1985)

Grading: Percentage Passing

Sieve Size Bases Subbases

50 mm 100 10037.5 mm 95–100 90–100

19 mm 70–92 — 9.5 mm 50–70 —

4.75 mm 35–55 30–60600 mm 12–25 —

75 mm 0–8 0–12

Other requirements:1. Fraction passing the 75-mm sieve not to

exceed 60% of the fraction passing the 600-mm sieve.

2. Fraction passing the 425-mm sieve shall have a liquid limit not greater than 25% and a plasticity index not greater than 4%.

Source: ASTM. 1992. ASTM Standard Specifi-cation D2940-74 (reapproved 1980), Annual Bookof ASTM Standards. Vol. 04.03 — Road and Pav-ing Materials; Travelled Surface Characteristics,1992. With permission.


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Subbase Course

The subbase material is of lower quality than the base material in terms of strength, plasticity, andgradation, but it is superior to the subgrade material in these properties. It may be compacted granularmaterial or stabilized soil, thus allowing building up of sufficient thickness for the pavement structureat relatively low cost. On a weak subgrade, it also serves as a useful working platform for constructingthe base course. Examples of grading requirements for subbase materials are given in Table 62.2. The

TABLE 62.3 Requirements for Stabilized Base Courses

Specification

Cement Treated Bituminous Treated

Lime TreatedClass A Class B Class C Class 1 Class 2

(a) Stabilized Base Courses for Flexible Pavements

Percent passing2½ in. 100 100 1003/4 in. — — 75–95No. 4 65–100 55–100 25–60No. 10 20–45 — 15–45No. 40 15–30 25–50 8–30No. 200 5–12 5–20 2–15

7-day fc (psi) 650–1000 300–650S (lb) — — — 750 min 500 min — F (0.01 in.) — — — 16 max 20 max — PI 12 max — — 6 max 6 max 6 max

Specification

Type A(Open

Graded)

Type B(Dense

Graded)

Type C(Cement Graded)

Type D(Lime

Treated)

Type E(Bituminous

Treated)Type F

(Granular)

(b) Base Materials for Concrete Pavement

Percent passing1½ in. 100 100 100 — — 1003/4 in. 60–90 85–100 — * * — No. 4 35–60 50–80 65–100 — — 65–100No. 40 10–25 20–35 25–50 — — 25–50No. 200 0–7 5–12 5–20 — — 0–15

(The minus No. 200 material should be held to a practical minimum)28-day fc (psi) — — 400–750 100 — — S (lb) — — — — 500 min — F (0.01 in.) — — — — 20 max — Soil constants:

LL 25 max 25 max — — — 25 maxPI* N.P. 6 max 10 max — 6 max 6 max

Notes:* To be determined by complete laboratory analysis, taking into consideration the ability of the

stabilized mixture to resist underslab erosion.fc = compressive strength as determined in unconfined compression tests on cylinders 4 inches in

diameter and 4 inches high. Test specimens should contain the same percentage of portlandcement and be compacted to the same density as achieved in construction.

S = Marshall stability.F = Marshall flow.

PI = plasticity index performed on samples prepared in accordance with AASHTO DesignationT-87 and applied to aggregate prior to mixing with the stabilizing admixture, except that, inthe case of lime-treated base, the value is applied after mixing.

LL= liquid limit.Source: AASHTO Interim Guide for Design of Pavement Structures, American Association of State

Highway and Transportation Officials, Washington, D.C., 1972. With permission.

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subbase course may be omitted if the subgrade soil satisfies the requirements specified for subbasematerial.

Prepared Subgrade

Most natural soils forming the roadbed for pavement construction require some form of preparation ortreatment. The top layer of a specified depth is usually compacted to achieve a desired density. The depthof compaction and the compacted density required depend on the type of soil and magnitudes of wheelloads and tire pressures. For highway construction, compaction to 100% modified AASHTO densitycovering a thickness of 12 in. (300 mm) below the formation level is commonly done. Compaction depthof up to 24 in. (600 mm) may be required for heavily trafficked pavements. For example, in the case ofcohesive subgrade, the Asphalt Institute [1991] requires a minimum of 95% of AASHTO T180(Method D) density for the top 12 in. (300 mm) and a minimum of 90% for all fill areas below the top12 in. (300 mm). For cohesionless subgrade, the corresponding compaction requirements are 100 and95%, respectively.

Due to the higher wheel loads and tire pressures of aircraft, many stringent compaction requirementsare found in airport pavement construction. Figure 62.2 shows an example of the compaction require-ments recommended by the FAA [1978].

In some instances it may be economical to treat or stabilize poor subgrade materials and reduce thetotal required pavement thickness. Portland cement, lime, and bitumen have all been used successfullyfor this purpose. The choice of the method of stabilization depends on the soil properties, improvementexpected, and cost of construction.

Rigid Pavement

Rigid pavements constructed of portland cement concrete are mostly found in heavy-traffic highwaysand airport pavements. To allow for expansion, contraction, warping, or breaks in construction of the

FIGURE 62.2 Subgrade compaction requirements for flexible airport pavements. (Source: Federal Aviation Admin-istration, Airport Pavement Design and Evaluation, Advisory Circular AC 150/5320-6C, 1978, p. 41. With permission.)

80 110 140 170 200 230

5

10

15

20

25

30

35

15

20

75

70

65

60

55

50

45

40

35

30

25

150 200 250 300 350 400GROSS AIRCRAFT WEIGHT-DUAL TANDEM GEAR-1000 POUNDS

GROSS AIRCRAFT WEIGHT-DUAL GEAR-1000 POUNDS

CO

MP

AC

TE

D S

UB

GR

AD

E D

EP

TH

-NO

NC

OH

ES

IVE

SO

ILS

-IN

CH

ES

CO

MP

AC

TE

D S

UB

GR

AD

E D

EP

TH

-CO

HE

SIV

E S

OIL

S-I

NC

HE

S

95% COHESIVE

90% COHESIVE

85% COHESIVE

80% COHESIVE

100% NONCOHESIVE

95% NONCOHESIVE

90% NONCOHESIVE

85% NONCOHESIVE


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concrete slabs, joints are provided in concrete pavements. The joint spacing, which determines the lengthof individual slab panels, depends on the use of steel reinforcements in the slab. The jointed plain concretepavement (JPCP), requiring no steel reinforcements and thus the least expensive to construct, is a popularform of construction. Depending on the thickness of the slab, typical joint spacings for plain concretepavements are between 10 and 20 ft (3 and 6 m). For slabs with joint spacing greater than 6 m, steelreinforcements have to be provided for crack control, giving rise to the use of jointed reinforced concretepavements (JRCP) and continuously reinforced concrete pavements (CRCP). Continuously reinforced con-crete pavements usually contain higher than 0.6% steel reinforcement to eliminate the need to providejoints other than construction and expansion joints.

The base course for rigid pavement, sometimes called subbase, is often provided to prevent pumping(ejection of foundation material through cracks or joints resulting from vertical movement of slabs undertraffic). The base course material must provide good drainage and be resistant to the erosive action ofwater. When dowel bars are not provided in short jointed pavements, it is common practice to constructcement-treated base to assist in load transfer across the joints.

Considerations for Highway and Airport Pavements

The two pavement types, flexible and rigid pavement, have been used for road and airport pavementconstruction. The choice of pavement type depends on the intended functional use of the pavement(such as operating speed and safety requirements), types of traffic loading, cost of construction, andmaintenance consideration.

The main differences in design considerations for highway and airport pavements arise from thecharacteristics of traffic using them. Over the typical design life span of 10 to 20 years for flexiblepavements, or 20 to 40 years for rigid pavements, a highway pavement will be receiving highly channelizedwheel load applications in the millions. Consideration of the effects of load repetitions — such ascumulative permanent deformation, crack propagation, and fatigue failure — becomes important. Thetotal number of load applications in the entire design life of a highway pavement must therefore beknown for pavement structural design. In contrast, the frequency of aircraft loading on airport pavementis much less. There are also the so-called wander effect of aircraft landing and taking off and the largevariation in the wheel assembly configurations and layout of different aircraft. These make wheel loadingon airport pavements less channelized than on highway pavements. Identification of the most criticalaircraft is therefore necessary for structural design of airport pavements.

Another important difference is in the magnitude of wheel loads. Airport pavements receive loads farexceeding those applied on the highway. An airport pavement may have to be designed to withstandequivalent single wheel loads of the order of 50 t (approximately 50 tons), whereas the maximum singlewheel load allowed on the road pavement by most highway authorities is about 10 t (approximately10 tons). Furthermore, the wheel tire pressure of an aircraft of about 1200 kPa (175 psi) is nearly twicethe value of a normal truck tire. These differences greatly influence the material requirements for thepavements.

62.3 Traffic Loading Analysis for Highway Pavements

Although it is convenient to describe the design life of a pavement in years, it is the total traffic loadingduring service that determines the actual design life of the pavement. It is thus more appropriate toassociate the design life of a pavement with the total design traffic loading. For example, a pavementdesigned for 20 years with an assumed traffic growth of 4% will reach the end of its design life soonerthan 20 years if the actual traffic growth is higher than 4%.

The ultimate aim of traffic analysis for pavement design is to determine the magnitudes of wheel loadsand the number of times each of these loads will be applied on the pavement during its design life. Forhighway pavements the computation of design traffic loading involves the following steps:

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1. Estimation of expected initial year traffic volume2. Estimation of expected annual traffic growth rate3. Estimation of traffic stream composition4. Computation of traffic loads5. Estimation of directional split of design traffic loads6. Estimation of design lane traffic loads

Information concerning the first two steps can be obtained from traffic surveys and forecasts basedon historical trends or prediction using transportation models. The analyses required for the remainingsteps are explained in the discussions that follow.

Traffic Stream Composition

The number of different types of vehicles — such as cars, buses, single-unit trucks, and multiple-unittrucks — expected to use the highways must be estimated. One may derive the vehicle-type distributionfrom results of classification counts made on similar highway type within the same region or from generaldata compiled by highway authorities, as illustrated in Table 62.4. However, as noted in the footnote ofthe table, individual situations may differ from the average values by 50% or more.

TABLE 62.4 Asphalt Institute Data for Truck Loading Computation

Average Trucks

Truck ClassInterstate

RuralOther Rural

All Rural

AllUrban

AllSystem

(a) Average Distribution on Different Classes of Highways (U.S.)

Single-unit trucks

2 axle, 4 tire 39 58 47 61 492 axle, 6 tire 10 11 10 13 113 axle or more 2 4 2 3 3All single-unit 51 73 59 77 63

Multiple-unit trucks3 axle 1 1 1 1 14 axle 5 3 4 4 45 axle or more 43 23 36 18 32All multiple-unit 49 27 41 23 37

All trucks 100 100 100 100 100

(b) Average Truck Factors (TF) for Different Classes of Highways and Vehicles (U.S.)

Single-unit trucks

2 axle, 4 tire 0.02 0.02 0.03 0.03 0.022 axle, 6 tire 0.19 0.21 0.20 0.26 0.213 axle or more 0.56 0.73 0.67 1.03 0.73All single-unit 0.07 0.07 0.07 0.09 0.07

Multiple-unit trucks3 axle 0.51 0.47 0.48 0.47 0.484 axle 0.62 0.83 0.70 0.89 0.735 axle or more 0.94 0.98 0.95 1.02 0.95All multiple-unit 0.93 0.97 0.94 1.00 0.95

All trucks 0.49 0.31 0.42 0.30 0.40

Note: Individual situations may differ from these average values by 50% ormore.

Source: Asphalt Institute, Asphalt Technology and Construction Practices. Edu-cational Series ES-1, 1983b. pp. J5–J7. With permission.


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Traffic-Loading Computation

Two aspects of traffic loading are of concern in the structural design of highway pavements, namely, thenumber of applications and the magnitude of each load type. A traffic count survey that classifies vehiclesby axle configuration, as shown in Table 62.5, enables one to compute the number of repetitions by axletype (i.e., by single axle, tandem axle, and tridem axle). With this information, one must further subdivideeach axle type by load magnitude to arrive at a traffic-loading table such as that illustrated in Table 62.6.

The combined loading effects of different axle types on pavements cannot be easily analyzed. In thelate 1950s, AASHO [Highway Research Board 1962] conducted the now well-known AASHO road testto provide, among other information, equivalency factors to convert one pass of any given single- ortandem-axle load to equivalent passes of an 18-kip (80 kN) single-axle load. The single-axle load of18 kip (80 kN) was arbitrarily chosen in the AASHO road test as the standard axle with a damaging effectof unity. The equivalency factor, known as the equivalent single-axle load (ESAL) factor, was derivedbased on the relative damaging effects of various axle loads. Table 62.7 presents the ESAL factors of axleloads for different thicknesses of flexible pavements with a terminal serviceability index of 2.5. Table 62.8presents the corresponding ESAL factors for rigid pavements.

Another approach to computing the combined effect of mixed traffic is to adopt the hypothesis ofcumulative damage. For a given form of pavement damage, the allowable number of repetitions by eachvehicle type or load group is established separately. A damage ratio for vehicle type or load group i isdefined as

(62.1)

TABLE 62.5 Vehicle Classification by Axle Configuration

VehicleClass

AxleConfiguration

Total No.of Axles

Number ofSingle Axles

Number of Tandem Axles

1 2 2

2 2 2

3 2 2

4 2 2

5 3 3

6 3 1 1

7 3 3

8 4 4

9 4 2 1

10 4 2 1

11 4 2 1

12 5 1 2

13 5 5

14 6 4 1

Source: Fwa, T.F. and Sinha, K.C., 1985. A Routine Maintenance and Pavement Perfor-mance Relationship Model for Highways, Report JHRP-85-11, Purdue University, WestLafayette, IN, 1985. With permission.

Di ni Ni§( )=

62-10 The Civil Engineering Handbook, Second Edition

where ni is the design repetitions and Ni the allowable repetitions. The total level of damage caused bythe mixed traffic is computed as the sum of damage ratios of all vehicle types or load groups.

Example 62.1

This example involves computation of the ESAL contribution of a passenger car, a bus, and a combinationtruck. The axle loads of the three fully laden vehicles are given as follows:

Car. Front single axle = 2 kips; rear single axle = 2 kips.Bus. Front single axle = 10 kips; rear single axle = 8 kips.Truck. Front single axle = 12 kips; middle single axle = 18 kips; rear tandem axle = 32 kips.

Assuming a terminal serviceability index of 2.5, the ESAL contributions of the three vehicles can becomputed for a flexible pavement with structural number SN = 5.0 [see Eq. (62.17) for definition of SN]and a rigid pavement of slab thickness equal to 10 in.

For the ESAL on flexible pavement, Table 62.7 is used to obtain the ESAL factor for each axle. TheESAL contribution of the passenger car is (0.0002 + 0.0002) = 0.0004. The ESAL contribution of the busis (0.088 + 0.034) = 0.122. The contribution of the truck is (0.189 + 1.00 + 0.857) = 2.046. Table 62.8is used for the ESAL computation in the case of rigid pavement. The ESAL contributions are (0.0002 +0.0002) = 0.0004 for the car, (0.081 + 0.032) = 0.113 for the bus, and (0.175 + 1.00 + 1.50) = 2.675 forthe truck.

The ratios of ESAL contributions are (car):(bus):(truck) = 5012:305:1 for flexible pavement and6688:283:1 for rigid pavement. It can be seen from this example that the damaging effects of a truck anda bus are, respectively, more than 5000 and 280 times that of a passenger car. This explains why passengercar volumes are often ignored in traffic-loading computation for pavement design.

Example 62.2

This example involves ESAL computation based on axle load data. Calculate the total daily ESAL of thetraffic-loading data of Table 62.6 for (a) a flexible pavement with structural number SN = 5.0 [seeEq. (62.17) for definition of SN] and (b) a rigid pavement with slab thickness of 10 in. The design terminalserviceability index for both pavements is 2.5.

The data in Table 62.6 are repeated in columns (1) and (2) of the following table. The ESAL factorsin column (3) are obtained from Table 62.7 (second part) for SN = 5.0, and those in column (5) areobtained from Table 62.8 (second part) for slab thickness of 10 in. The ESAL contribution by each axle

TABLE 62.6 Examples of Axle-Load Data Presentation

Single Axle Tandem Axle

Axle Load (kips) No. Axles/Day Axle Load (kips) No. Axles/Day

Less than 3 1438 9–11 20933–5 3391 11–13 18675–7 3432 13–15 12987–9 6649 15–17 14659–11 9821 17–19 1734

11–13 2083 19–21 187013–15 946 21–23 267415–17 886 23–25 287917–19 472 25–27 235919–21 299 27–29 210421–23 98 29–31 1994

31–33 177933–35 86235–37 65937–39 39539–41 46

Highway and Airport Pavement Design 62-11

group is computed by multiplying its ESAL factor by the number of axles per day. The total ESAL of thetraffic loading is 12,642 for the flexible pavement and 19,309 for the rigid pavement.

Axle Load No. Axles Flexible Pavement Rigid Pavement

(kips) per Day ESAL Factor ESAL ESAL Factor ESAL(1)* (2) (3) (2) ¥ (3) (5) (2) ¥ (5)

S2 1438 0.0002 0.2876 0.0002 0.2876S4 3391 0.002 6.782 0.002 6.782S6 3432 0.01 34.32 0.01 34.32S8 6649 0.034 226.066 0.032 212.768S10 9821 0.088 864.248 0.081 795.501S12 2083 0.189 393.687 0.175 364.525S14 946 0.36 340.56 0.338 319.748S16 886 0.623 551.978 0.601 532.486S18 472 1.00 472.00 1.00 472.00S20 299 1.51 451.49 1.58 472.42S22 98 2.18 213.64 2.38 233.24T10 2093 0.007 14.651 0.012 25.116T12 1867 0.014 26.138 0.025 46.675T14 1298 0.027 35.046 0.047 61.006T16 1465 0.047 68.855 0.081 118.665T18 1734 0.077 133.518 0.132 228.888T20 1870 0.121 226.27 0.204 381.48T22 2674 0.18 481.32 0.305 815.57T24 2879 0.26 748.54 0.441 1269.639T26 2359 0.364 858.676 0.62 1462.58T28 2104 0.495 1041.48 0.85 1788.4T30 1994 0.658 1312.052 1.14 2273.16T32 1779 0.857 1524.603 1.5 2668.5T34 862 1.09 939.58 1.95 1680.9T36 659 1.38 909.42 2.48 1634.32T38 395 1.7 671.5 3.12 1232.4T40 46 2.08 95.68 3.87 178.02

Total = 12,642.38 Total = 19,309.39

* In column (1), the prefix S stands for single axle and T stands for tandem axle.

TABLE 62.7 AASHTO Load Equivalency Factors for Flexible Pavements

Axle Load (kips)

Pavement Structural Number (SN)

1 2 3 4 5 6

(a) Single Axles and pt of 2.5

2 .0004 .0004 .0003 .0002 .0002 .00024 .003 .004 .004 .003 .002 .0026 .011 .017 .017 .013 .010 .0098 .032 .047 .051 .041 .034 .031

10 .078 .102 .118 .102 .088 .08012 .168 .198 .229 .213 .189 .17614 .328 .358 .399 .388 .360 .34216 .591 .613 .646 .645 .623 .60618 1.00 1.00 1.00 1.00 1.00 1.0020 1.61 1.57 1.49 1.47 1.51 1.5522 2.48 2.38 2.17 2.09 2.18 2.3024 3.69 3.49 3.09 2.89 3.03 3.2726 5.33 4.99 4.31 3.91 4.09 4.4828 7.49 6.98 5.90 5.21 5.39 5.9830 10.3 9.5 7.9 6.8 7.0 7.8


TABLE 62.7 (continued) AASHTO Load Equivalency Factors for Flexible Pavements

Axle Load (kips)


1 2 3 4 5 6

32 13.9 12.8 10.5 8.8 8.9 10.034 18.4 16.9 13.7 11.3 11.2 12.536 24.0 22.0 17.7 14.4 13.9 15.538 30.9 28.3 22.6 18.1 17.2 19.040 39.3 35.9 28.5 22.5 21.1 23.042 49.3 45.0 35.6 27.8 25.6 27.744 61.3 55.9 44.0 34.0 31.0 33.146 75.5 68.8 54.0 41.4 37.2 39.348 92.2 83.9 65.7 50.1 44.5 46.550 112. 102. 79. 60. 53. 55.

(b) Tandem Axles and pt of 2.5

2 .0001 .0001 .0001 .0000 .0000 .00004 .0005 .0005 .0004 .0003 .0003 .00026 .002 .002 .002 .001 .001 .0018 .004 .006 .005 .004 .003 .003

10 .008 .013 .011 .009 .007 .00612 .015 .024 .023 .018 .014 .01314 .026 .041 .042 .033 .027 .02416 .044 .065 .070 .057 .047 .04318 .070 .097 .109 .092 .077 .07020 .107 .141 .162 .141 .121 .11022 .160 .198 .229 .207 .180 .16624 .231 .273 .315 .292 .260 .24226 .327 .370 .420 .401 .364 .34228 .451 .493 .548 .534 .495 .47030 .611 .648 .703 .695 .658 .63332 .813 .843 .889 .887 .857 .83434 1.06 1.08 1.11 1.11 1.09 1.0836 1.38 1.38 1.38 1.38 1.38 1.3838 1.75 1.73 1.69 1.68 1.70 1.7340 2.21 2.16 2.06 2.03 2.08 2.1442 2.76 2.67 2.49 2.43 2.51 2.6144 3.41 3.27 2.99 2.88 3.00 3.1646 4.18 3.98 3.58 3.40 3.55 3.7948 5.08 4.80 4.25 3.98 4.17 4.4950 6.12 5.76 5.03 4.64 4.86 5.2852 7.33 6.87 5.93 5.38 5.63 6.1754 8.72 8.14 6.95 6.22 6.47 7.1556 10.3 9.6 8.1 7.2 7.4 8.258 12.1 11.3 9.4 8.2 8.4 9.460 14.2 13.1 10.9 9.4 9.6 10.762 16.5 15.3 12.6 10.7 10.8 12.164 19.1 17.6 14.5 12.2 12.2 13.766 22.1 20.3 16.6 13.8 13.7 15.468 25.3 23.3 18.9 15.6 15.4 17.270 29.0 26.6 21.5 17.6 17.2 19.272 33.0 30.3 24.4 19.8 19.2 21.374 37.5 34.4 27.6 22.2 21.3 23.676 42.5 38.9 31.1 24.8 23.7 26.178 48.0 43.9 35.0 27.8 26.2 28.880 54.0 49.4 39.2 30.9 29.0 31.782 60.6 55.4 43.9 34.4 32.0 34.884 67.8 61.9 49.0 38.2 35.3 38.1


TABLE 62.7 (continued) AASHTO Load Equivalency Factors for Flexible Pavements

Axle Load (kips)


1 2 3 4 5 6

86 75.7 69.1 54.5 42.3 38.8 41.788 84.3 76.9 60.6 46.8 42.6 45.690 93.7 85.4 67.1 51.7 46.8 49.7

(c) Triple Axles and pt of 2.5

2 .0000 .0000 .0000 .0000 .0000 .00004 .0002 .0002 .0002 .0001 .0001 .00016 .0006 .0007 .0005 .0004 .0003 .00038 .001 .002 .001 .001 .001 .001

10 .003 .004 .003 .002 .002 .00212 .005 .007 .006 .004 .003 .00314 .008 .012 .010 .008 .006 .00616 .012 .019 .018 .013 .011 .01018 .018 .029 .028 .021 .017 .01620 .027 .042 .042 .032 .027 .02422 .038 .058 .060 .048 .040 .03624 .053 .078 .084 .068 .057 .05126 .072 .103 .114 .095 .080 .07228 .098 .133 .151 .128 .109 .09930 .129 .169 .195 .170 .145 .13332 .169 .213 .247 .220 .191 .17534 .219 .266 .308 .281 .246 .22836 .279 .329 .379 .352 .313 .29238 .352 .403 .461 .436 .393 .36840 .439 .491 .554 .533 .487 .45942 .543 .594 .661 .644 .597 .56744 .666 .714 .781 .769 .723 .69246 .811 .854 .918 .911 .868 .83848 .979 1.015 1.072 1.069 1.033 1.00550 1.17 1.20 1.24 1.25 1.22 1.2052 1.40 1.41 1.44 1.44 1.43 1.4154 1.66 1.66 1.66 1.66 1.66 1.6656 1.95 1.93 1.90 1.90 1.91 1.9358 2.29 2.25 2.17 2.16 2.20 2.2460 2.67 2.60 2.48 2.44 2.51 2.5862 3.09 3.00 2.82 2.76 2.85 2.9564 3.57 3.44 3.19 3.10 3.22 3.3666 4.11 3.94 3.61 3.47 3.62 3.8168 4.71 4.49 4.06 3.88 4.05 4.3070 5.38 5.11 4.57 4.32 4.52 4.8472 6.12 5.79 5.13 4.80 5.03 5.4174 6.93 6.54 5.74 5.32 5.57 6.0476 7.84 7.37 6.41 5.88 6.15 6.7178 8.83 8.28 7.14 6.49 6.78 7.4380 9.92 9.28 7.95 7.15 7.45 8.2182 11.1 10.4 8.8 7.9 8.2 9.084 12.4 11.6 9.8 8.6 8.9 9.986 13.8 12.9 10.8 9.5 9.8 10.988 15.4 14.3 11.9 10.4 10.6 11.990 17.1 15.8 13.2 11.3 11.6 12.9

Source: AASHTO Guides for Design of Pavement Structures, American Associationof State Highway and Transportation Officials, Washington, D.C., 1993. Withpermission.


TABLE 62.8 AASHTO Load Equivalency Factors for Rigid Pavements

Axle Load (kips)

Slab Thickness, D (inches)

6 7 8 9 10 11 12 13 14

(a) Single Axles and pt of 2.5

2 .0002 .0002 .0002 .0002 .0002 .0002 .0002 .0002 .00024 .003 .002 .002 .002 .002 .002 .002 .002 .0026 .012 .011 .010 .010 .010 .010 .010 .010 .0108 .039 .035 .033 .032 .032 .032 .032 .032 .032

10 .097 .089 .084 .082 .081 .080 .080 .080 .08012 .203 .189 .181 .176 .175 .174 .174 .173 .17314 .376 .360 .347 .341 .338 .337 .336 .336 .33616 .634 .623 .610 .604 .601 .599 .599 .599 .59818 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.0020 1.51 1.52 1.55 1.57 1.58 1.58 1.59 1.59 1.5922 2.21 2.20 2.28 2.34 2.38 2.40 2.41 2.41 2.4124 3.16 3.10 3.22 3.36 3.45 3.50 3.53 3.54 3.5526 4.41 4.26 4.42 4.67 4.85 4.95 5.01 5.04 5.0528 6.05 5.76 5.92 6.29 6.61 6.81 6.92 6.98 7.0130 8.16 7.67 7.79 8.28 8.79 9.14 9.35 9.46 9.5232 10.8 10.1 10.1 10.7 11.4 12.0 12.3 12.6 12.734 14.1 13.0 12.9 13.6 14.6 15.4 16.0 16.4 16.536 18.2 16.7 16.4 17.1 18.3 19.5 20.4 21.0 21.338 23.1 21.1 20.6 21.3 22.7 24.3 25.6 26.4 27.040 29.1 26.5 25.7 26.3 27.9 29.9 31.6 32.9 33.742 36.2 32.9 31.7 32.2 34.0 36.3 38.7 40.4 41.644 44.6 40.4 38.8 39.2 41.0 43.8 46.7 49.1 50.846 54.5 49.3 47.1 47.3 49.2 52.3 55.9 59.0 61.448 66.1 59.7 56.9 56.8 58.7 62.1 66.3 70.3 73.450 79.4 71.7 68.2 67.8 69.6 73.3 78.1 83.0 87.1

(b) Tandem Axles and pt of 2.5

2 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .00014 .0006 .0006 .0005 .0005 .0005 .0005 .0005 .0005 .00056 .002 .002 .002 .002 .002 .002 .002 .002 .0028 .007 .006 .006 .005 .005 .005 .005 .005 .005

10 .015 .014 .013 .013 .012 .012 .012 .012 .01212 .031 .028 .026 .026 .025 .025 .025 .025 .02514 .057 .052 .049 .048 .047 .047 .047 .047 .04716 .097 .089 .084 .082 .081 .081 .080 .080 .08018 .155 .143 .136 .133 .132 .131 .131 .131 .13120 .234 .220 .211 .206 .204 .203 .203 .203 .20322 .340 .325 .313 .308 .305 .304 .303 .303 .30324 .475 .462 .450 .444 .441 .440 .439 .439 .43926 .644 .637 .627 .622 .620 .619 .618 .618 .61828 .855 .854 .852 .850 .850 .850 .849 .849 .84930 1.11 1.12 1.13 1.14 1.14 1.14 1.14 1.14 1.1432 1.43 1.44 1.47 1.49 1.50 1.51 1.51 1.51 1.5134 1.82 1.82 1.87 1.92 1.95 1.96 1.97 1.97 1.9736 2.29 2.27 2.35 2.43 2.48 2.51 2.52 2.52 2.5338 2.85 2.80 2.91 3.03 3.12 3.16 3.18 3.20 3.2040 3.52 3.42 3.55 3.74 3.87 3.94 3.98 4.00 4.0142 4.32 4.16 4.30 4.55 4.74 4.86 4.91 4.95 4.9644 5.26 5.01 5.16 5.48 5.75 5.92 6.01 6.06 6.0946 6.36 6.01 6.14 6.53 6.90 7.14 7.28 7.36 7.4048 7.64 7.16 7.27 7.73 8.21 8.55 8.75 8.86 8.9250 9.11 8.50 8.55 9.07 9.68 10.14 10.42 10.58 10.6652 10.8 10.0 10.0 10.6 11.3 11.9 12.3 12.5 12.754 12.8 11.8 11.7 12.3 13.2 13.9 14.5 14.8 14.9


TABLE 62.8 (continued) AASHTO Load Equivalency Factors for Rigid Pavements

Axle Load (kips)


6 7 8 9 10 11 12 13 14

56 15.0 13.8 13.6 14.2 15.2 16.2 16.8 17.3 17.558 17.5 16.0 15.7 16.3 17.5 18.6 19.5 20.1 20.460 20.3 18.5 18.1 18.7 20.0 21.4 22.5 23.2 23.662 23.5 21.4 20.8 21.4 22.8 24.4 25.7 26.7 27.364 27.0 24.6 23.8 24.4 25.8 27.7 29.3 30.5 31.366 31.0 28.1 27.1 27.6 29.2 31.3 33.2 34.7 35.768 35.4 32.1 30.9 31.3 32.9 35.2 37.5 39.3 40.570 40.3 36.5 35.0 35.3 37.0 39.5 42.1 44.3 45.972 45.7 41.4 39.6 39.8 41.5 44.2 47.2 49.8 51.774 51.7 46.7 44.6 44.7 46.4 49.3 52.7 55.7 58.076 58.3 52.6 50.2 50.1 51.8 54.9 58.6 62.1 64.878 65.5 59.1 56.3 56.1 57.7 60.9 65.0 69.0 72.380 73.4 66.2 62.9 62.5 64.2 67.5 71.9 76.4 80.282 82.0 73.9 70.2 69.6 71.2 74.7 79.4 84.4 88.884 91.4 82.4 78.1 77.3 78.9 82.4 87.4 93.0 98.186 102. 92. 87. 86. 87. 91. 96. 102. 108.88 113. 102. 96. 95. 96. 100. 105. 112. 119.90 125. 112. 106. 105. 106. 110. 115. 123. 130.

(c) Triple Axles and pt of 2.5

2 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .00014 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .00036 .001 .001 .001 .001 .001 .001 .001 .001 .0018 .003 .002 .002 .002 .002 .002 .002 .002 .002

10 .006 .005 .005 .005 .005 .005 .005 .005 .00512 .011 .010 .010 .009 .009 .009 .009 .009 .00914 .020 .018 .017 .017 .016 .016 .016 .016 .01616 .033 .030 .029 .028 .027 .027 .027 .027 .02718 .053 .048 .045 .044 .044 .043 .043 .043 .04320 .080 .073 .069 .067 .066 .066 .066 .066 .06622 .116 .107 .101 .099 .098 .097 .097 .097 .09724 .163 .151 .144 .141 .139 .139 .138 .138 .13826 .222 .209 .200 .195 .194 .193 .192 .192 .19228 .295 .281 .271 .265 .263 .262 .262 .262 .26230 .384 .371 .359 .354 .351 .350 .349 .349 .34932 .490 .480 .468 .463 .460 .459 .458 .458 .45834 .616 .609 .601 .596 .594 .593 .592 .592 .59236 .765 .762 .759 .757 .756 .755 .755 .755 .75538 .939 .941 .946 .948 .950 .951 .951 .951 .95140 1.14 1.15 1.16 1.17 1.18 1.18 1.18 1.18 1.1842 1.38 1.38 1.41 1.44 1.45 1.46 1.46 1.46 1.4644 1.65 1.65 1.70 1.74 1.77 1.78 1.78 1.78 1.7946 1.97 1.96 2.03 2.09 2.13 2.15 2.16 2.16 2.1648 2.34 2.31 2.40 2.49 2.55 2.58 2.59 2.60 2.6050 2.76 2.71 2.81 2.94 3.02 3.07 3.09 3.10 3.1152 3.24 3.15 3.27 3.44 3.56 3.62 3.66 3.68 6.6854 3.79 3.66 3.79 4.00 4.16 4.26 4.30 4.33 4.3456 4.41 4.23 4.37 4.63 4.84 4.97 5.03 5.07 5.0958 5.12 4.87 5.00 5.32 5.59 5.76 5.85 5.90 5.9360 5.91 5.59 5.71 6.08 6.42 6.64 6.77 6.84 6.8762 6.80 6.39 6.50 6.91 7.33 7.62 7.79 7.88 7.9364 7.79 7.29 7.37 7.82 8.33 8.70 8.92 9.04 9.1166 8.90 8.28 8.33 8.83 9.42 9.88 10.17 10.33 10.4268 10.1 9.4 9.4 9.9 10.6 11.2 11.5 11.7 11.970 11.5 10.6 10.6 11.1 11.9 12.6 13.0 13.3 13.572 13.0 12.0 11.8 12.4 13.3 14.1 14.7 15.0 15.2


Example 62.3

This example entails ESAL computation based on Asphalt Institute truck distribution and truck factors.Consider a two-lane rural highway with a design lane daily directional traffic of 2000 vehicles per dayduring the first year. The traffic growth factor is 4.5% and there are 16% trucks in the traffic. The totalnumber of trucks in the 20-year design traffic is computed by the following geometric sum equation:

The following table summarizes the procedure for ESAL computation. The numbers of vehicles incolumn (2) are calculated based on the distribution of “other rural” in Table 62.4(a) and the truck factorsin column (3) are for “other rural” given in Table 62.4(b).

Directional Split

It is common practice to report traffic volume of a highway to include flows for all lanes in both directions.To determine the design traffic loading on the design lane, one must split the traffic by direction anddistribute the directional traffic by lanes. An even split assigning 50% of the traffic to each directionappears to be the norm. In circumstances where an uneven split occurs, pavements are designed basedon the heavier directional traffic loading.

TABLE 62.8 (continued) AASHTO Load Equivalency Factors for Rigid Pavements

Axle Load (kips)


6 7 8 9 10 11 12 13 14

74 14.6 13.5 13.2 13.8 14.8 15.8 16.5 16.9 17.176 16.5 15.1 14.8 15.4 16.5 17.6 18.4 18.9 19.278 18.5 16.9 16.5 17.1 18.2 19.5 20.5 21.1 21.580 20.6 18.8 18.3 18.9 20.2 21.6 22.7 23.5 24.082 23.0 21.0 20.3 20.9 22.2 23.8 25.2 26.1 26.784 25.6 23.3 22.5 23.1 24.5 26.2 27.8 28.9 29.686 28.4 25.8 24.9 25.4 26.9 28.8 30.5 31.9 32.888 31.5 28.6 27.5 27.9 29.4 31.5 33.5 35.1 36.190 34.8 31.5 30.3 30.7 32.5 34.4 36.7 38.5 39.8

Source: AASHTO Guides for Design of Pavement Structures, American Association of State Highway and TransportationOfficials, Washington, D.C., 1993. With permission.

Vehicle Number of Truck ESALType Vehicles Factor Contribution(1) (2) (3) (4)

Single-Unit Trucks

2-axle, 4-tire 2,125,220 0.02 45,2002-axle, 6-tire 403,060 0.21 84,6403-axle or more 146,570 0.73 107,000All singles 2,674,850 236,840

Tractor Semitrailers and Combinations

3-axle 36,640 0.48 17,5904-axle 109,930 0.70 76,9505-axle or more 842,760 0.95 800,620All multiple units 989,330 895,160

Total design ESAL = 1,132,000

2000 365 16%¥¥( ) 1 0.01 4.5¥+( )20 1–0.01 4.5¥

--------------------------------------------------◊ 3 664 180, ,=


Design Lane Traffic Loading

The design lane for pavement structural design is usually the slow lane (lane next to the shoulder in mostcases), in which a large proportion of the directional heavy-vehicle traffic is expected to travel. Somehighway agencies assign 100% of the estimated directional heavy-vehicle traffic to the design lane for thepurpose of structural design. This leads to overestimation of traffic loading on roads with more thanone lane in each direction. Studies have shown that — depending upon road geometry, traffic volume,and composition — as much as 50% of the directional heavy vehicles may not travel on the design lane[Fwa and Li, 1994]. Figure 62.3 shows the lane-use distributions recommended by a number of organi-zations. It is noted that while most agencies apply lane-use factors to traffic volume, the factors inAASHTO recommendations are for lane distributions of ESAL. The latter tends to provide a betterestimate of traffic loading in cases involving a higher concentration of heavily loaded vehicles in the slowlane.

Formula for Computing Total Design Loading

Depending on the information available, the computations of the design load for pavement structuraldesign may differ slightly. Assuming that the initial-year total ESAL is known and a constant growth ofESAL at a rate of r% per annum is predicted, the design lane loading for an analysis period of n yearscan be computed by the following equation:

(62.2)

where (ESAL)T = total design lane ESAL for n years(ESAL)0 = initial-year design lane ESAL

r = annual growth rate of ESAL in percentfD = directional split factorfL = lane-use distribution factor

FIGURE 62.3 Percentage of truck traffic in design lane.

(b) Recommendation by AsphaltInstitute [1991]

Number of Lanes % Trucks inper Direction Design Lane

1 100%2 90% (70-96%)

3 or more 80% (50-96%)

(c) Recommendation by AASHTO [1993]

Number of Lanes % ESAL inper Direction Design Lane

1 100%2 80-100%3 60-80%4 50-75%

Note: ESAL stands for equivalent 80 kNsingle axle load

0.5 0.6 0.7 0.8 0.9 1.0

Proportion of Trucks in Right Lane

100

60

20

10

6

4

2

0

AD

T (

One

Dire

ctio

n), T

hous

ands

2 Lanes inOne Direction

3 Lanes inOne Direction

ESAL( )T ESAL( )01 0.01r+( )n 1–

0.01r------------------------------------- fD fL◊ ◊ ◊=


For design methods that rely on damage ratio computation [see Eq. (62.1)], instead of computingcumulative ESAL, the total number of repetitions of each vehicle type or axle type is calculated.

Example 62.4

The daily ESAL computed in Example 62.2 is based on the initial traffic estimate for both directions oftravel in an expressway with three lanes in each direction. The annual growth of traffic is estimated at3%. The directional split is assumed to be 45 to 55%. The expressway is to be constructed of flexiblepavement. Calculate the design lane ESAL for a service life of 15 years.

Adopting the AASHTO lane distribution factor shown in Fig. 62.3(b), we have fL = 0.7. Given fD =0.55, n = 15, and i = 3, and from Example 62.2 (ESAL)0 = (12,642 ¥ 365), the total design lane ESAL is

62.4 Traffic Loading Analysis for Airport Pavements

The procedure of traffic loading analysis for airport pavements differs slightly from that for highwaypavements due to differences in traffic operations and functional uses of the pavements. The basic stepsare:

1. Estimation of expected initial year traffic volume2. Estimation of expected annual traffic growth rate3. Estimation of traffic stream composition4. Computation of traffic loading5. Estimation of design traffic loading for different functional areas

Information concerning the first two steps is usually obtained from the planning forecast of the airportauthority concerned.

Traffic Stream Composition

The weight of an aircraft is transmitted to the pavement through its nose gear and main landing gears.Figure 62.4 shows the wheel configurations commonly found on the main legs of landing gear of civilaircraft. Since the gross weight and exact arrangement of wheels differ among different aircraft, there isa need to identify the types of aircraft, landing gear details, and their respective frequencies of arrival forthe purpose of pavement design.

Computation of Traffic Loading

For pavement design purposes, the maximum takeoff weights of the aircraft are usually considered. It isalso common to assume that 95% of the gross weight is carried by the main landing gears and 5% bythe nose gear. In the consideration of mixed traffic loading, both the equivalent load concept and Miner’shypothesis have been used. For example, the FAA method [Federal Aviation Administration, 1978]converts the annual departure of all aircraft into the equivalent departures of a selected design aircraftusing the factors in Table 62.9. In establishing the thickness design curves for flexible airport pavements,the concept of equivalent single-wheel load (ESWL) is adopted by the FAA. The concept of ESWL iswidely used in airport pavement design to assess the effect of multiple-wheel landing gears. The valueof ESWL of a given landing gear varies with the control response selected for ESWL computation,

ESAL( )T 12 642 365¥,( ) 1 0.01 3¥+( )15 1–0.01 3¥

---------------------------------------------- 0.7 0.55◊ ◊ ◊=

33.04 106¥=


thickness of pavement, and the relative stiffness of pavement layers. For airport pavement design, ESWLcomputations based on equal deflection (at surface or at pavement–subgrade interface) or equal stress(at bottom face of bound layer) are commonly used.

FIGURE 62.4 Typical wheel configurations of a main leg of aircraft landing gear.

TABLE 62.9 Conversion Factors for Computing Annual Departures

Aircraft Type Design Aircraft Conversion Factor F

Single-wheel Dual-wheel 0.8Single-wheel Dual-tandem 0.5Dual-wheel Dual-tandem 0.6Double dual-tandem Dual-tandem 1.0Dual-tandem Single-wheel 2.0Dual-tandem Dual-wheel 1.7Dual-wheel Single-wheel 1.3Double dual-tandem Dual-wheel 1.7

Note: Multiply the annual departures of given aircraft type bythe conversion factor to obtain annual departures in design air-craft landing gear.

Source: Federal Aviation Administration. 1978. Airport Pave-ment Design and Evaluation. Reprinted from FAA Advisory Cir-cular. Report FAA/AC-150/5320-6C. 7 December 1978; NTISAccession No. AD-A075 537/1.

Landing GearLanding Gear

NoseGear

NoseGear

Dual TandemLanding Gear

(Example : B767-200)

Dual WheelLanding Gear

(Example : DC-6)

Landing Gear LandingGear

NoseWheel

NoseWheel

Single WheelLanding Gear

(Example : DC-3)

Double Dual TandemLanding Gear

(Example : B747-100)


Example 62.5

This example concerns the representation of annual departures of designed aircraft. An airport pavementis to be designed for the following estimated traffic:

In this example, the 727-200 requires the greatest pavement thickness and is therefore the designaircraft. The conversion factors are obtained from Table 62.9. The entries in the last column are theproducts of the conversion factors and the estimated annual departures.

Equal Stress ESWL

An elaborate procedure for computing the ESWL would call for both a proper analytical solution for therequired stress produced by the wheel assembly of interest and a trial-and-error process to identify themagnitude of the single wheel that will produce identical stress. As this procedure is time consuming,simplified methods have been employed in practice.

Figure 62.5(a) presents a simplified procedure for estimating the equal subgrade stress ESWL of a setof dual wheels for flexible pavement design. With the assumed 45˚ spread of applied pressure, the ESWLis equal to one wheel load P if the pavement thickness is less than or equal to d/2, where d is the smallestedge-to-edge distance between the tire imprints of the dual wheels. The method further assumes thatESWL = 2P for any pavement equal to or thicker than 2S, where S is the center-to-center spacing of thedual wheels. For pavement thicknesses between d/2 and 2S, ESWL is determined, as shown in Fig. 62.5(a),by assuming a linear log–log relationship between ESWL and pavement thickness. Note that d is equalto (S – 2a), where a is the radius of tire imprint given by

(62.3)

where p is the tire pressure.This simplified procedure provides an approximate ESWL estimation for flexible pavements. In the

case of rigid pavements, computation of stresses for equal stress ESWL should be based on rigid slabanalysis such as the well-known Westergaard formulas, which give the maximum bending stress smax andthe maximum deflection dmax as shown below. smax and dmax under interior loading [Westergaard, 1926]are given as

(62.4)

(62.5)

where b = (1.6a2 + h2)0.5 – 0.675h, a < 1.724h = a, a > 1.724h

P = total applied loadm = slab Poisson’s ratio

AircraftLanding

Gear TypeEst. AnnualDepartures

Max. Wt.(kips)

ConversionFactor

Converted Annual Departures

727-100 Dual 4500 160 1.0 4500727-200 Dual 9900 190.5 1.0 9900707-320B Dual tandem 3200 327 1.7 5440DC-9-30 Dual 5500 108 1.0 5500747-100 Dual DT 60 700 1.7 102

a Pp p------=

smax3P 1 m+( )

2ph2------------------------ 2L

b------Ë ¯

Ê ˆln 0.5 g–+3P 1 m+( )

64h2------------------------ b

L---Ë ¯

Ê ˆ 2

+=

dmaxP

8kL2----------- 1 a2

2pL2------------ a

2L------Ë ¯

Ê ˆln g 1.25–++Ó ˛Ì ˝Ï ¸

=


(b)

FIGURE 62.5 Computation of ESWL. (a) Equal subgrade stress ESWL of dual wheels for flexible pavement design.(b) One-layer deflection factor for equal-deflection ESWL computation. (Source: Yoder, E.J. and Witczak, M.W.,Principles of Pavement Design, 2nd ed., John Wiley & Sons, New York, 1975, p. 138. With permission.)

d/2 z 2S

P

2PP P

ds

z = depth of pavementstructure

Log (Depth below pavement surface)

(ESWL)

Log

(Loa

d)

(a)

0 1 2 3 4 5 6 7 8 9 10 11 12

1.5

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.18

0.16

0.14

0.12

0.10

0.09

0.08

0.07

0.06

0.05

0.04

0.03

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

5.0

6.0

7.0

8.0

9.010.0

12.0

14.0

16.0

17.5

0.0

Offset radii

Radii =Single-tire contact area

p

Depth (radii)

Def

lect

ion

fact

or


h = slab thicknessL = radius of relative stiffnessa = radius of loaded areak = modulus of subgrade reactiong = Euler’s constant = 0.577216

smax and dmax under edge loading [Westergaard, 1926, 1933, 1948] are given as

(62.6)

(62.7)

where E = slab elastic modulus and all other variables are as defined in Eqs. (62.1) and (62.2). smax andd max under corner loading [Westergaard 1926] are given as

(62.8)

(62.9)

Example 62.6

This example addresses equal subgrade stress ESWL. The total load on a set of dual wheels is 45,000 lb.The tire pressure of the wheels is 185 psi. The center-to-center spacing of the wheels is 34 in. Calculatethe equal stress ESWL if the thickness of pavement structure is (a) h = 30 in. and (b) h = 70 in.

Load per wheel = 22,500 lb., radius of tire imprint a = = 6.22 in., S = 34 in., and d =(S – 2a) = 21.56. By means of a log–log plot as shown in Fig. 62.5(a), ESWL is determined to be 33,070lb. for h = 30 in. For h = 70 in., since h > 2S, ESWL = 2(22,500) = 45,000 in.

Equal Deflection ESWL

Equal deflection ESWL can be derived by assuming either constant tire pressure or constant area of tireimprint. A simplified method for computing ESWL on flexible pavement, based on the Boussinesq one-layer theory [Boussinesq, 1885], is presented in this section. It computes the ESWL of an assembly of nwheels by equating the surface deflection under the ESWL to the maximum surface deflection caused bythe wheel assembly, that is,

(62.10)

where P = gross load on each tire of the wheel assemblyE = stiffness modulus of the soil

K, K1, K2, Kn = Boussinesq deflection factor given by Fig. 62.5(b)

With the assumption of constant tire pressure, Eq. (62.2) can be solved for ESWL by the followingiterative procedure: (1) compute (K1 + K2 + L + Kn)max at the point of maximum surface deflection;(2) assume a, the radius of tire imprint for ESWL; (3) determine K from Fig. 62.5(b) with zero horizontal

smax3 1 m+( )P

p 3 m+( )h2-------------------------- Eh3

100k a( )4---------------------ln 1.18 a

L---Ë ¯

Ê ˆ 1 2m+( ) 2.34 116-----m–+ +

Ó ˛Ì ˝Ï ¸

=

dmax P 2 1.2m+( )Eh3k

-------------------------0.5

1 0.76 0.4m+( ) aL---Ë ¯

Ê ˆ–Ó ˛Ì ˝Ï ¸

=

smax3P

h2------ 1 1.4142a

L-------------------Ë ¯

Ê ˆ0.6

–Ë ¯Ê ˆ=

dmaxP

kL2-------- 1.1 0.88 1.4142a

L-------------------Ë ¯

Ê ˆ–Ë ¯Ê ˆ=

22 500 /185p,( )

ESWL( )0.5

pE--------------------------K P0.5

pE-------- K1 K2 L Kn+ + +( )max=


offset; (4) compute ESWL from Eq. (62.2); (5) calculate new a = and (6) if new a does notmatch the assumed a, return to step (3) and repeat the procedure with the new a until convergence.

Deflections of rigid pavements under loads are computed by means of Westergaard’s theory [seeEqs. (62.4–62.9)] or more elaborate analysis using the finite-element method. Improved deflection com-putations using thick-plate theory [Shi et al., 1994; Fwa et al., 1993] could also be used for the purposeof ESWL evaluation.

Example 62.7

Calculate the equal subgrade-deflection ESWL for the dual wheels in Example 62.6 for h = 30 in. (h/a) =4.82. For a point directly below one of the wheels, (r/a)1 = (34/6.22) = 5.47, K1 = 0.15 [from Fig. 62.5(b)];and (r/a)2 = 0, K2 = 0.31. For the point on the vertical line midway between the two wheels, (r/a)1 =(r/a)2 = 2.73, and K1 = K2 = 0.24 [from Fig. 62.5(b)]. The critical (K1 + K2) = 0.48. The ESWL is obtainedby trial and error as follows. The ESWL equals 35,950 lb.

Critical Areas for Pavement Design

Runway ends, taxiways, aprons, and turnoff ramp areas receive a concentration of aircraft movementswith maximum loads. They are designated as the critical areas for pavement design purposes. Reducedthickness may be used for other areas.

62.5 Thickness Design of Flexible Pavements

The thickness design of flexible pavements is a complex engineering problem involving a large numberof variables. Most of the design methods in use today are largely empirical or semiempirical proceduresderived from either full-scale pavement tests or performance monitoring of in-service pavements. Thissection presents the methods of the Asphalt Institute and AASHTO for flexible highway pavements andthe FAA method for flexible airport pavements. A brief description of the development of the mechanisticapproach to flexible pavement design is also presented.

AASHTO Design Procedure for Flexible Highway Pavements

The AASHTO design procedure [AASHTO, 1993] was developed based on the findings of the AASHOroad test [Highway Research Board, 1962]. It defines pavement performance in terms of the presentserviceability index (PSI), which varies from 0 to 5. The PSIs of newly constructed flexible pavementsand rigid pavements were found to be about 4.2 and 4.5, respectively. For pavements of major highways,the end of service life is considered to be reached when PSI = 2.5. A terminal value of PSI = 2.0 may beused for secondary roads. Serviceability loss, given by the difference of the initial and terminal service-ability, is required as an input parameter. Pavement layer thicknesses are designed using the nomographin Fig. 62.6. The design traffic loading in ESAL is computed by Eq. (62.2). Other input parameters arediscussed in this section.

Reliability

The AASHTO guide incorporates in the design a reliability factor R% to account for uncertainties intraffic prediction and pavement performance. R% indicates the probability that the pavement designedwill not reach the terminal serviceability level before the end of the design period. The AASHTO suggested

Trial a (h/a) K ESWL by Eq. (62.10) New a by Eq. (62.3)

6.5 in. 4.615 0.3177 51,361 lb. 9.408.0 in. 3.75 0.3865 34,704 lb. 7.737.85 in. 3.8217 0.3797 35,954 lb. 7.86

P/pp( ) ;


ranges of R% are 85 to 99.9%, 80 to 99%, 80 to 95%, and 50 to 80% for urban interstates, principalarterials, collectors, and local roads, respectively. The corresponding ranges for rural roads are 80 to99.9%, 75 to 95%, 75 to 95%, and 50 to 80%. The overall standard deviation, so, for flexible and rigidpavements developed at the AASHO road test is 0.45 and 0.35, respectively.

Effective Roadbed Soil Resilient Modulus

Determination of Subgrade Resilient Modulus. The total pavement thickness requirement is a functionof the resilient modulus, Mr, of subgrade soil. Methods for the determination of Mr for granular materialsand fine-grained soils are described in AASHTO Test Method T274 [AASHTO, 1989].

Since many laboratories are not equipped to perform the resilient modulus test for soils, it is commonpractice to estimate Mr through empirical correlation with other soil properties. Equation (62.11) is onesuch correlation suggested by AASHTO for fine-grained soils with soaked CBR of 10 or less.

(62.11)

Other correlations are also found in the literature, such as in the work by Van Til et al. [1972].

Determination of Effective Mr . To account for seasonal variations of subgrade soil resilient modulus,AASHTO defines an effective roadbed soil Mr to represent the combined effect of all the seasonal modulusvalues. This effective Mr is a weighted value that would give the correct equivalent annual pavementdamage for design purpose. The steps in computing the effective Mr are as follows:

1. Divide the year into equal-length time intervals, each equal to the smallest season. AASHTOsuggests that the smallest season should not be less than one-half month.

2. Estimate the relative damage uf corresponding to each seasonal modulus by the following equation:

(62.12)

where Mr is expressed in 103 psi.3. Sum the uf of all seasons and divide by the number of seasons to give the average seasonal damage.4. Substitute the average seasonal damage into Eq. (62.12) and calculate Mr to arrive at the effective

roadbed soil Mr .

FIGURE 62.6 AASHTO design chart for flexible highway pavements. (Source: AASHTO Guides for Design of Pave-ment Structures, American Association of State Highway and Transportation Officials, Washington, D.C., 1993. Withpermission.)

99.9

99

90

80

706050

50

105.0

1.0

.5

.1

.05

40

20

10

5

1

Rel

iabi

lity,

R(%

)

Ove

rall

Sta

ndar

d D

evia

tion,

S0

.6

.4

.2

Est

imat

ed T

otal

18-

kip

Equ

ival

ent

Sin

gle

Axl

e Lo

ad A

pplic

atio

ns, W

18 (m

illio

ns)

Effe

ctiv

e R

oadb

ed S

oil

Res

ilien

t Mod

ulus

, MR

(ks

i)

Design Structural Number, SN

9 8 7 6 5 4 3 2 1

Design Serviceability Loss, DPSI

.05

1.01.52.0

TL TL

3.0

Mr psi( ) 1500 CBR¥=

uf 1.18 108¥ Mr2.32–¥=


Example 62.8

This example examines effective roadbed soil resilient modulus. The resilient moduli of a roadbed soildetermined at 24 half-month intervals are 6000, 20,000, 20,000, 4000, 4500, 5000, 6000, 6000, 5000, 5000,5000, 6000, 6000, 6500, 6500, 6500, 6500, 6500, 6000, 6000, 5500, 5500, 5500, and 6000. The total relativedamage u computed by Eq. (62.12) is

The mean u = 0.232. Applying Eq. (62.12) again, the effective Mr is 5655 psi.

Pavement Layer Modulus

Structural thicknesses required above other pavement layers are also determined based on their respec-tive Mr values. For bituminous pavement layers, Mr may be tested by the repeated load indirect tensiletest described in ASTM Test D-4123 [ASTM, 1992]. Figure 62.7 shows a chart developed by Van Til et al.[1972] relating Mr of hot-mix asphalt mixtures to other properties.

For unbound base and subbase materials, Mr may be estimated from the following correlations:

(62.13)

FIGURE 62.7 Correlation charts for estimating resilient modulus of asphalt concrete. (Source: Van Til, C.J. et al.,Evaluation of AASHO Interim Guides for Design of Pavement Structures, NCHRP Report 128, Highway Research Board,Washington, D.C., 1972.)

0.6

0.5

0.4

0.3

0.2

18002000

1600

1400

1200

1000

800

600

400

400

300

200

175

150

125

100

1.0

1.5

2.0

2.53.0

4.0

4.55.0

6.0

7.08.09.0

10.0

Str

uctu

ral l

ayer

coe

ffici

ent

Mar

shal

l Sta

bilit

y (1

Ib)

Coh

esio

met

er a

t 140

∞

Mod

ulus

(3) 1

05 -psi

Str

uctu

ral c

oeffi

cien

t (a 2

)

Mar

shal

l sta

bilit

y (1

Ib)

Mod

ulus

-105

psi

4.0

3.02.5

2.0

1.5

1.0

1900

0.40

0.30

0.20

0.10

1700

1500

1300

1100

900

700

500300

200

100

(a) Surface Course (b) Base Course

u 0.2026 0.0124 0.0124 0.5189 0.3948 0.3092 0.2026+ + + + + +=

0.2026 0.3092 0.3092 0.3092 0.2026 0.2026 0.1682+ + + + + + +

0.1682 0.1682 0.1682 0.1682 0.2026 0.2026 0.2479+ + + + + + +

0.2479 0.2479 0.2026+ + +

5.568=

Mr psi( ) 740 CBR for q¥ 100 psi= =


(62.14)

(62.15)

(62.16)

where q is the sum of principal stresses, (s1 + s2 + s3).

Thickness Requirements

Using the input parameters described in the preceding sections, the total pavement thickness requirementis obtained from the nomograph in Fig. 62.6 in terms of structural number SN. SN is an index numberequal to the weighted sum of pavement layer thicknesses, as follows:

(62.17)

where a1, a2, and a3 are numbers known as layer coefficients; D1, D2, and D3 are layer thicknesses; andm2 and m3 are layer drainage coefficients. SN can be considered a form of equivalent thickness, and layercoefficients and drainage coefficients are applied to actual pavement thicknesses to account for theirstructural and drainage properties, respectively.

Drainage coefficients are determined from Table 62.10. Coefficient a1 can be estimated from Fig. 62.8.Coefficients a2 and a3 of granular base and subbase layers can be obtained from the following correlations:

(62.18)

(62.19)

where q is the stress state in psi, and k1 and k2 are regression constants. Recommended valuesare given in Table 62.11.

The thicknesses of individual pavement layers are determined by means of the layer analysis conceptdepicted in Fig. 62.9. The total structural number required above the subgrade soil, denoted SN1, is

TABLE 62.10 Base and Subbase Stress States

Asphalt Concrete Thickness (inches)

Roadbed Soil Resilient Modulus (psi)

3000 7500 15,000

(a) Stress State for Base Course

Less than 2 201 25 302–4 10 15 204–6 5 10 15

Greater than 6 5 5 5

Asphalt Concrete Thickness (inches) Stress State (psi)

(b) Stress State for Subbase (6–12 in.)

Less than 2 10.02–4 7.5

Greater than 4 5.0

Source: AASHTO Guide for Design of PavementStructures, American Association of State Highway andTransportation Officials, Washington, D.C., 1993.With permission.




SN a1D1 a2D2m2 a3D3m3+ +=

a2 0.249 log10 Mr( ) 0.977–=

a3 0.227 log10 Mr( ) 0.839–=

Mr k1 q( )k2 ,=


FIGURE 62.8 Chart for estimating structural layer coefficient of dense-graded asphalt concrete. (Source: AASHTOGuides for Design of Pavement Structures, American Association of State Highway and Transportation Officials,Washington, D.C., 1993. With permission.)

TABLE 62.11 Recommended mi Value for Modifying Structural Layer Coefficient of Untreated Base and Subbase Materials in Flexible Pavements

Percent of Time Pavement Structure Is Exposed to Moisture Levels Approaching Saturation

Quality of Drainage

Less than 1% 1–5% 5–25%

Greater than 25%

Excellent 1.40–1.35 1.35–1.30 1.30–1.20 1.20Good 1.35–1.25 1.25–1.15 1.15–1.00 1.00Fair 1.25–1.15 1.15–1.05 1.00–0.80 0.80Poor 1.15–1.05 1.05–0.80 0.80–0.60 0.60Very Poor 1.05–0.95 0.95–0.75 0.75–0.40 0.40

Source: AASHTO. 1993. AASHTO Guides for Design of PavementStructures, American Association of State Highway and Transporta-tion Officials, Washington, D.C., 1993. With permission.

FIGURE 62.9 The concept of layer analysis.

0 100,000 200,000 300,000 400,000 500,000

0.5

0.4

0.3

0.2

0.1

0.0

Elastic Modulus, EAC (psi), of

Asphalt Concrete (at 68∞F)

Str

uctu

ral L

ayer

Coe

ffici

ent a

1 fo

r

Asp

halt

Con

cret

e S

urfa

ce C

ours

e

Design Requirements inStructural Number

Requirements inLayer Thickness

SN3

SN2

SN1 SurfaceCourse

BaseCourse

SubbaseCourse

Subgrade

D1

D2

D3

a1D1 ‡ SN1

a1D1 + a2m2D2 ‡ SN2

a1D1 + a2m2D2 + a3m3D3 ‡ SN3


determined from the nomograph in Fig. 62.6, with the effective roadbed soil Mr as input. SN2 and SN3

are determined likewise by replacing Mr with E3 (stiffness modulus of subbase material) and E2 (stiffnessmodulus of base course material), respectively. All pavement layer thicknesses are then derived by solvingthe following inequalities:

(62.20)

(62.21)

(62.22)

Environmental Effects

The moisture effect on subgrade strength has been considered in the computation of effective roadbedsoil Mr. Other environmental impacts such as roadbed swelling, frost heave, aging of asphalt mixtures,and deterioration due to weathering could result in considerable serviceability loss. This loss in service-ability can be added to that caused by traffic loading for design purposes.

Minimum Thickness Requirements

It is impractical to construct pavement layers less than a certain minimum thickness. AASHTO [1993]recommends minimum thicknesses for different layers as a function of design traffic, which are given inTable 62.12(a).

Example 62.9

On the subgrade examined in Example 62.8 is to be constructed a pavement to carry a design lane ESALof 5 ¥ 106. The elastic moduli of the surface, base, and subbase courses are, respectively, E1 = 360,000,E2 = 30,000, and E3 = 13,000 psi. The drainage coefficients of the base and subbase courses are m2 = 1.20and m3 = 1.0, respectively. The design reliability is 95% and the standard deviation so is 0.35. Provide athickness design for the pavement if the initial serviceability level is 4.2 and the terminal serviceabilitylevel is 2.5.

For R = 95%, so = 0.35, ESAL = 5 ¥ 106, Mr = 5655 psi, and DPSI = 4.2 – 2.5 = 1.7, to obtain SN3 =5.0 from Fig. 62.6. Repeat the procedure with E3 = 13,000 to obtain SN2 = 3.8, and with E2 = 30,000to obtain SN1 = 2.7. From Fig. 62.8, a1 = 0.40. By Eq. (62.18), a2 = 0.249(log 30,000) – 0.977 = 0.138,and by Eq. (62.19), a3 = 0.227(log 13,000) – 0.839 = 0.095. The layer thicknesses are D1 = (2.7/0.4) =6.75 in.; D2 = {3.8 – (0.4 ¥ 6.75)}/(0.138 ¥ 1.20) = 6.64 or 6.75 in.; and D3 = {5.0 – (0.4 ¥ 6.75) – (0.138 ¥1.20 ¥ 6.75)}/(0.095 ¥ 1.0) = 12.4 or 12.5 in.

TABLE 62.12(a) Minimum Thickness of Pavement Layers — AASHTO Thickness Requirements in Inches

Traffic, ESAL Asphalt Concrete Aggregate Base

Less than 50,000 1.0 (or surface treatment) 450,001–150,000 2.0 4150,001–500,000 2.5 4500,001–2,000,000 3.0 62,000,001–7,000,000 3.5 6Greater than 7,000,000 4.0 6

Source: AASHTO Guides for Design of Pavement Structures, AmericanAssociation of State Highway and Transportation Officials, Washington,D.C., 1993. With permission.

D1SN1

a1

-----------≥

D2SN2 a1D1–

a2m2

----------------------------≥

D3SN3 a1D1– a2D2m2–

a3m3

-----------------------------------------------------≥


AI Design Procedure for Flexible Highway Pavements

The Asphalt Institute [1991] promotes the use of full-depth pavements in which asphalt mixtures areemployed for all courses above the subgrade. Potential benefits of full-depth pavements derive from thehigher load bearing and spreading capability and moisture resistance of asphalt mixtures as compared tounbound aggregates. Thickness design charts are provided for full-depth pavements, pavements withemulsified asphalt base, and untreated aggregate base. These charts are developed based on two designcriteria: (1) maximum tensile strains induced at the underside of the lowest asphalt-bound layer and(2) maximum vertical strains induced at the top of the subgrade layer. The design curves have incorporatedthe effects of seasonal variations of temperature and moisture on the subgrade and granular base materials.

Computation of Design ESAL

When detailed vehicle classification and weight data are available, the design lane ESAL is computedaccording to Eq. (62.2). The AASHTO ESAL factors for SN = 5 and terminal serviceability index = 2.5are used. When such data are not available, estimates can be made based on the information in Table 62.4.The truck factor in the table refers to the total ESAL contributed by one pass of the truck in question.

Example 62.10

Calculate the design lane 20-year ESAL by the AI procedure for a 3-lane rural interstate with an initialdirectional AADT of 600,000. The predicted traffic growth is 3% per annum and the percent truck trafficis 16%.

From Fig. 62.3(b), the design lane is to be designed to carry 80% of the directional truck traffic. Totaldesign lane truck volume = 600,000 ¥ 16% ¥ 80% ¥ {(1 + 0.03)20 – 1}/0.03 = 2,063,645. The total ESALis computed as follows.

Subgrade Resilient Modulus

The Asphalt Institute design charts require subgrade resilient modulus Mr as input. However, Mr can beestimated by performing the CBR test [ASTM Method D1883, 1992] or the R-value test [ASTM MethodD2844, 1992] and applying the following relationships:

TABLE 62.12(b) Asphalt Institute Requirements

Traffic, ESAL Asphalt Concrete Thickness

(a) Minimum Thickness of Asphalt Concrete on Aggregate Base

Less than 10,000 1 in. (25 mm)Less than 100,000 1.5 in. (40 mm)Greater than 100,000 2 in. (50 mm)

Source: Asphalt Institute, Asphalt Technology and Con-struction Practices, Educational Series ES-1, 1983b, p. J25.With permission.

Traffic, ESAL

Asphalt Concrete Thickness

Type I Base Type II and III Base

(b) Minimum Thickness of Asphalt Concrete over Emulsified Asphalt Bases

£104 1 in. (25 mm) 2 in. (50 mm)£105 1.5 in. (40 mm) 2 in. (50 mm)£106 2 in. (50 mm) 3 in. (75 mm)£107 2 in. (50 mm) 4 in. (100 mm)£107 2 in. (50 mm) 5 in. (130 mm)


(62.23)

(62.24)

For each soil type, six to eight tests are recommended for the purpose of selecting the design subgraderesilient modulus by the following procedure: (1) arrange all Mr values in ascending order; (2) for eachtest value, compute y = percent of test values equal to or greater than it; (3) plot y against Mr; and (4) readfrom the plot the design subgrade strength at an appropriate percentile value. The design subgradepercentile value is selected according to the magnitude of design ESAL as follows: 60th percentile forESAL £ 104, 75th percentile if 104 < ESAL < 106, and 87.5th percentile if ESAL ≥ 106.

Example 62.11

Eleven CBR tests on the subgrade for the pavement in Example 62.10 yield the following results: 7, 5, 7,2, 8, 6, 5, 3, 4, 3, and 6. Determine the design subgrade resilient modulus by the AI method.

From Example 62.10, ESAL = 987,041; hence, use the 75th percentile according to the Asphalt Instituterecommendation. Next, arrange the test values in ascending order.

The design CBR (75th-percentile value) is 4%. By Eq. (62.23), the corresponding design Mr = 1500 ¥4 = 6000 psi.

Pavement Thickness Requirements

Thickness requirement charts are developed for three different designs of pavement structure: (1) Fig. 62.10for full-depth pavements, (2) Figs. 62.11–62.13 for pavements with emulsified asphalt base, and (3) Figs. 62.14and 62.15 for pavements with untreated aggregate base. These charts are valid for mean annual airtemperature (MAAT) of 60∞F (15.5∞C). Corresponding charts are also prepared by the Asphalt Institutefor MAATs of 45∞F (7∞C) and 75∞F (24∞C). Type I, II, and III emulsified asphalt mixes differ in theaggregates used. Type I mixes are made with processed dense-graded aggregates; type II are made withsemiprocessed, crusher-run, pit-run, or bank-run aggregates; and type III are made with sands or siltysands. The minimum thicknesses of full-depth pavements at different traffic levels are indicated inFig. 62.10. The minimum thicknesses of asphalt concrete surface course for pavements with other basecourses are given in Table 62.12(b).

Example 62.12

With the data in Examples 62.10 and 62.11, design the required thickness for (1) a full-depth pavement,(2) a pavement with type II emulsified base, and (3) a pavement with 12-in. aggregate base.

(1) With ESAL = 987,041 and Mr = 6000 psi, the required full-depth pavement thickness is 9.5 in.,according to Fig. 62.10. (2) The total required thickness is 11.5 in., from Fig. 62.12. The minimum asphaltconcrete surface course is 3 in., according to Table 62.12(b). Hence, the thickness of the emulsified base =11.5 – 3 = 8.5 in. (3) The required thickness of asphalt concrete is 7.5 in. Use 2 in. of asphalt concretesurface, according to Table 62.12(b), and (7.5 – 2) = 5.5 in. of asphalt base layer.

Truck Type% Share from Table 62.4(a)

Truck Factor from Table 62.4(b)

ESAL Contribution (Col. 2 ¥ Col. 3)

SU 2-axle, 4-tire 39 0.02 16,096SU 2-axle, 6-tire 10 0.19 29,209SU 3-axle or more 2 0.56 23,113MU 3-axle 1 0.51 10,525MU 4-axle 5 0.62 63,973MU 5-axle or more 43 0.94 834,125

Total ESAL = 987,041

CBR 2 3 4 5 6 7 8% ≥ 100 91 73 64 45 27 9

Mr MPa( ) 10.3 CBR or Mr psi( ) 1500 CBR= =

Mr MPa( ) 8.0 3.8 R or Mr+ 1155 555 R+= =


FIGURE 62.10 Thickness design curves for full-depth asphalt concrete. (Source: Asphalt Institute. 1991. ThicknessDesign — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

FIGURE 62.11 Thickness design curves for type I emulsified asphalt mix. (Source: Asphalt Institute. 1991. ThicknessDesign — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

FIGURE 62.12 Thickness design curves for type II emulsified asphalt mix. (Source: Asphalt Institute. 1991. ThicknessDesign — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

Sub

grad

e R

esili

ent M

odul

us,

MR (

psi)

Equivalent 18,000 lb Single Axle Load

Full-Depth Asphalt Concrete MAAT = 60°F

104

103

103 104

105

4

2

4

2

42 105 1064242 107 4242108

4 in, minimum

5in 6in 7in 8in 9in10

in

11in

12in

13in

14in

15in

16in

17in

18in

19in

20in

Sub

grad

e R

esili

ent M

odul

us,

MR (

psi)


Emulsified Asphalt Mix Type I MAAT = 60°F

104

103

103 104

105

4

2

4

2

42 105 1064242 107 4242 108

4 in, minimum

5in 6in 7in 8in 9in10

in

11in

12in

13in

14in

15in

16in

17in

18in

19in

20in

Emulsified Asphalt Type II MAAT = 60°F

Sub

grad

e R

esili

ent M

odul

us,

MR (

psi)

104

103

105

4

2

4

2


103 10442 105 1064242 107 4242 108

6 in5 in

23in

24in

21in

7in 8in 10in

11in

12in

13in

14in

15in

16in

17in

18in

19in

20in

22in

4 in, minimum

9in


FIGURE 62.13 Thickness design curves for type III emulsified asphalt mix. (Source: Asphalt Institute. 1991. ThicknessDesign — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

FIGURE 62.14 Thickness design curves for asphalt pavement with 6-in. untreated aggregate base. (Source: AsphaltInstitute. 1991. Thickness Design — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

FIGURE 62.15 Thickness design curves for asphalt pavement with 12-in. untreated aggregate base. (Source: AsphaltInstitute. 1991. Thickness Design — Asphalt Pavements for Highways and Streets. Manual Series MS-1. With permission.)

Emulsified Asphalt Type III MAAT = 60°FS

ubgr

ade

Res

ilien

t Mod

ulus

,M

R (

psi)

104

103

105

4

2

4

2


103 10442 105 1064242 107 4242 108

6 in5 in

7in

23in

25in

26in

22in

24in

8in 9in 10in11in

12in

13in

14in

15in

16in

17in

18in

19in

20in

21in

4 in, minimum

Untreated Aggregate Base 6.0 in. MAAT = 60°F

Sub

grad

e R

esili

ent M

odul

us,

MR (

psi)

104

103

105

4

2

4

2

Equivalent 18,000 lb Single Axle Load103 10442 105 1064242 107 4242 108

3 in, minimum 4 in, minimum

minimum

5 in 6 in 7in 8in 9in 10in 11in12in

13in14in

15in16in

Untreated Aggregate Base 12.0 in. MAAT = 60°F

Sub

grad

e R

esili

ent M

odul

us,

MR (

psi)

104

103

105

4

2

4

2

Equivalent 18,000 lb Single Axle Load103 10442 105 1064242 107 4242 108

3 in, minimum

4 in, minimum

minimum

5 in 6in 7in 8in 9in 10in11in

12in13

in

14in

15in


FAA Design Procedure for Flexible Airport Pavements

Based on the California bearing ratio (CBR) method of design, FAA [1978] developed — through testtrack studies and observations of in-service pavements — pavement thicknesses that are necessary toprotect pavement layers with various CBR values from shear failure. In establishing the thickness require-ments, the equivalent single-wheel loads of wheel assemblies were computed based on deflection con-sideration. The design assumes that 95% of the gross aircraft weight is carried on the main landing gearassembly and 5% on the nose gear assembly. Generalized design curves are available for single, dual, anddual-tandem main landing gear assemblies. Design curves for specific wide-body aircraft have also beendeveloped.

Computation of Design Loading

The FAA design charts are based on the equivalent annual departures of a selected design aircraft. Theannual departures are assumed to occur over a 20-year life. The following steps are involved in theselection of design aircraft and determination of equivalent annual departures:

1. Obtain forecasts of annual departures by aircraft type.2. Determine for each aircraft type the required pavement thickness using the appropriate design

curve with the forecast number of annual departures for that aircraft. The aircraft requiring thegreatest pavement thickness is selected as the design aircraft.

3. Convert the annual departures of all aircraft to equivalent annual departures of the design aircraftby the following formula:

(62.25)

where Req = equivalent annual departures by the design aircraftRi = annual departures of aircraft type iFi = conversion factor obtained from Table 62.9W = wheel load of the design aircraftWi = wheel load of aircraft i

In the computation of equivalent annual departures, each wide-body aircraft is treated as a 300,000-lb(136,100-kg) dual-tandem aircraft.

Example 62.13

This example entails computation of equivalent annual departures. The equivalent annual departures indesign aircraft for the design traffic of Example 62.5 are computed by means of Eq. (62.25), as follows.

Pavement Thickness Requirements

Figures 62.16–62.22 are the FAA design charts for different aircraft types. The charts have incorporatedthe effects of load repetitions, landing gear assembly configuration, and the “wandering” (lateral distri-bution) effect of aircraft movements. With subgrade CBR, gross weight, and total equivalent annualdepartures of design aircraft as input, the total pavement thickness required can be read from the

AircraftSingle-Wheel Load

Wi (lbs.)(R1 ¥ Fi)

from Example 62.5 Req by Eq. (62.25)

727-100 38,000 4500 2229727-200 45,240 9900 9900707-320B 38,830 5440 2890DC-9-30 25,650 5500 655747-100 35,625 102 61

Total equivalent design annual departures = 15,735

Reqlog Ri Fi¥( ) Wi

W------

Ó ˛Ì ˝Ï ¸

¥0.5

log=


appropriate chart. Each design chart also indicates the required thickness of bituminous surface course.The minimum base course thickness is obtained from Fig. 62.23.

The FAA requires stabilized base and subbase courses to be used to accommodate jet aircraft weighing100,000 lb or more. These stabilized courses may be substituted for granular courses using the equivalencyfactors in Table 62.13.

FIGURE 62.16 Critical area flexible pavement thickness for single-wheel gear. (Source: Federal Aviation Adminis-tration. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

THICKNESS - BITUMINOUSSURFACES

4-IN. CRITICAL AREAS3-IN NONCRITICAL AREAS

NOTE CURVES BASED ON 20-YEARPAVEMENT LIFE

GROSS AIRCRAFT

WEIGHT, LB

30,000

45,000

60,000

75,000

1 in. = 2.54 cm1 1b. = 0.454 kg

ANNUAL DEPARTURES

25,000

15,000

6000

3000

1200

3 4 5 6 7 8 9 10

THICKNESS, IN.

15 20 30 40 50

3 4 5 6 7 8 9 10

CBR

15 20 25 30 40 50


The FAA [1978] suggests that the full design thickness T be used at critical areas where departingtraffic will be using the pavement, 0.9T be used at areas receiving arriving traffic such as high-speedturnoffs, and 0.7T be used where traffic is unlikely. These reductions in thickness are applied to base andsubbase courses. Figure 62.24 shows a typical cross section for runway pavements.

For pavements receiving high traffic volumes and exceeding 25,000 departures per annum, the FAArequires that the bituminous surfacing be increased by 1 in. (3 cm) and the total pavement thickness be

FIGURE 62.17 Critical area flexible pavement thickness for dual-wheel gear. (Source: Federal Aviation Administra-tion. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

NOTE: CURVES BASED ON 20-YEARPAVEMENT LIFE


4-IN. CRITICAL AREAS3-IN. NON CRITICAL AREAS

CBR3 4 5 6 7 8 9 10 15 20 30 40 50

3 4 5 6 7 8 9 10

THICKNESS. IN.

15 20 30 40 50

GROSS AIRCRAFT

WEIGHT, LB200,000150,000100,00075,000

50,000

1 in. = 2.54 cm1 1b. = 0.454 kg

ANNUAL DEPARTURES

1200

3000

6000

15,000

25,000


increased as follows: 104, 108, 110, and 112% of design thickness (based on 25,000 annual departures)for annual departures of 50,000, 100,000, 150,000, and 200,000, respectively.

Example 62.14

For the design traffic in Example 62.13, determine the thickness requirements for a pavement withsubgrade CBR = 5 and subbase CBR = 20.

FIGURE 62.18 Critical area flexible pavement thickness for dual-tandem gear. (Source: Federal Aviation Adminis-tration. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

3 4 5 6 7 8 9 10

CBR

THICKNESS, IN.

15 20 30 40 50

3 4 5 6 7 8 9 10 15 20 30 40 50


4-IN. CRITICAL AREAS3-IN NONCRITICAL AREAS

GROSS AIRCRAFT

WEIGHT, LB

400,000300,000200,000150,000100,000

1 in. = 2.54 cm1 1b. = 0.454 kg

ANNUAL DEPARTURES

25,000

15,000

6000

3000

1200

NOTE: CURVES BASED ON 20-YEARPAVEMENT LIFE


The design aircraft has dual-wheel landing gear and a maximum weight of 190,500 lb. Figure 62.17gives the total thickness requirement as 45 in. above subgrade and 18 in. above subbase. Minimum asphaltconcrete surface for the critical area is 4 in. Thickness of base = 18 – 4 = 14 in. Thickness of subbase =45 – 4 – 14 = 27 in. Since the design aircraft weighs more than 100,000 lb, stabilized base and subbaseare needed. Use bituminous base course with equivalency factor of 1.5 [see Table 62.13(a)] and cold-laid

FIGURE 62.19 Critical area flexible pavement thickness for B-747-100, SR, 200B, 200C, and 200F. (Source: FederalAviation Administration. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C.With permission.)

3 4 5 6 7 8 9 10

THICKNESS, IN.

15 20 30 40 50

3 4 5 6 7 8 9 10CBR

15 20 25 30 40 50

NOTE: CURVES BASED ON 20-YEARPAVEMENT LIFECONTACT AREA = 245 SQ IN.DUAL SPACING = 44 IN.TANDEM SPACING = 58 IN.


5-IN. CRITICAL AREAS4-IN NONCRITICAL AREASGROSS AIRCRAFT

WEIGHT, LB

850,000800,000700,000600,000500,000400,000

300,000

1 in. = 2.54 cm1 1b. = 0.454 kg

ANNUAL DEPARTURES

25,000

15,000

6000

3000

1200


bituminous base course with equivalency factor of 1.5 [see Table 62.13(b)]. The required stabilized basethickness = (14/1.5) = 9 in., and the required subbase thickness = (27/1.5) = 18 in., both for criticalareas. In the case of noncritical areas, the asphalt concrete surface thickness is 3 in., and the correspondingbase and subbase thicknesses are (9 ¥ 0.9) = 8 in. and (18 ¥ 0.9) = 16 in.

FIGURE 62.20 Critical area flexible pavement thickness for B-747-SP. (Source: Federal Aviation Administration.1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

NOTE: CURVES ARE BASED ON 20-YEARPAVEMENT LIFE.CONTACT AREA = 210 SQ IN.DUAL SPACING = 43.25 IN.TANDEM SPACING = 54 IN.


5-IN. CRITICAL AREAS4-IN. NONCRITICAL AREAS

GROSS AIRCRAFT

WEIGHT, LB700,000600,000500,000400,000300,000

1 in. = 2.54 cm1 Ib. = 0.454 kg

ANNUAL DEPARTURES

1200

3000

6000

15,000

25,000

3 4 5 6 7 8 9 10 15 20 30 40 50

3 4 5 6 7 8 9 10 15 20 25 30 40 50

THICKNESS, IN.

CBR


Mechanistic Approach for Flexible Pavement Design

The methods described in the preceding sections provide evidence of the continued effort and progressmade by engineers toward adopting theoretically sound approaches with fundamental material propertiesin pavement design. For example, the Asphalt Institute method described is a complete revision that uses

FIGURE 62.21 Critical area flexible pavement thickness for DC 10-10, 10CF. (Source: Federal Aviation Administra-tion. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

NOTE: CURVES BASED ON 20-YEARPAVEMENT LIFECONTACT AREA = 294 SQ IN.DUAL SPACING = 54 IN.TANDEM SPACING = 64 IN.


5-IN. CRITICAL AREAS4-IN. NONCRITICAL AREAS

GROSS AIRCRAFT

WEIGHT, LB450,000

1 in. = 2.54 cm1 Ib. = 0.454 kg

ANNUAL DEPARTURES

1200

3000

6000

15,000

25,000

3 4 5 6 7 8 9 10 15 20 30 40 50

THICKNESS, IN.

3 4 5 6 7 8 9 10 15 20 3025 40 50

CBR

400,000300,000

200,000


analyses based on elastic theory to generate pavement thickness requirements against two failure criteria:a fatigue-cracking criterion for the asphalt layer and a rutting criterion for the subgrade.

More comprehensive mechanistic procedures, capable of handling the following aspects in pavementdesign, are available in the literature: (a) viscoelastic behavior of bituminous materials, (b) nonlinearresponse of untreated granular and cohesive materials, (c) aging of bituminous materials, (d) materialvariabilities, (e) dynamic effect of traffic loading, (f ) effect of mixed traffic loading, and (g) interdependency

FIGURE 62.22 Critical area flexible pavement thickness for DC 10-30, 30CF, 40, and 40CF. (Source: Federal AviationAdministration. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

NOTE: CURVES BASED ON 20-YEARPAVEMENT LIFECONTACT AREA = 331 SQ IN.DUAL SPACING = 54 IN.TANDEM SPACING = 64 IN.CENTER-LINE GEAR SPACING = 37.5 IN.


5-IN. CRITICAL AREAS4-IN. NONCRITICAL AREASGROSS AIRCRAFT

WEIGHT, LB600,000500,000400,000

300,000

200,000

1 in. = 2.54 cm1 ib. = 0.454 kg

ANNUAL DEPARTURES

1200

3000

6000

15,000

25,000

3 4 5 6 7 8 9 10 15 20 30 40 50

THICKNESS, IN.

3 4 5 6 7 8 9 10 15 20 25 30 40 50

CBR


of the development of different distresses, including pavement roughness. Unfortunately, quantificationof necessary material properties and analysis of pavement response by these procedures are often com-plicated, time consuming, skill demanding, and costly. Although there are great potentials for theseprocedures when fully implemented, simplifications such as that adopted by the Asphalt Institute methodwill have to be applied for pavement design in practice.

62.6 Structural Design of Rigid Pavements

Structural design of rigid pavements includes thickness and reinforcement designs. Two major forms ofthickness design methods are being used today for concrete pavements. The first form is an approach

FIGURE 62.23 Minimum base course thickness requirements. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C, p. 53. With permission.)

TO

TAL

PA

VE

ME

NT

TH

ICK

NE

SS

, IN

.

(cm

)

100

90

80

70

60

50

45

40

35

30

25

20

15

106 7 8 9 10 15

MINIMUM BASE COURSE THICKNESS, IN.

20 25

70

60

50

40

30

(cm)55

250

200

175

150140

130

120

110

100

90

80

5045403530252017

SUBGRADE CBR

4

5

8

7

6

9

20

15

12

10


that relies on empirical relationships derived from performance of full-scale test pavements and in-servicepavements. The design procedure of AASHTO [1993] is an example. The second form develops relation-ships in terms of the properties of pavement materials as well as load-induced and thermal stresses andcalibrates these relationships with pavement performance data. The PCA [1984] and the FAA [1978]methods of design adopt this approach. Thickness design procedures by AASHTO, PCA, and FAA arediscussed in this section. Reinforcement designs by the AASHTO and FAA procedures will be presented.

AASHTO Thickness Design for Rigid Highway Pavements

The serviceability-based concept of the AASHTO design procedure for rigid pavements [AASHTO, 1993]is similar to its design procedure for flexible pavements. Pavement thickness requirements are establishedfrom data of the AASHO road test [Highway Research Board, 1962]. Input requirements such as reliabilityinformation and serviceability loss for design have been described in the section on AASHTO flexiblepavement design. Details for other input requirements are described in this section.

Pavement Material Properties

The elastic modulus Ec and modulus of rupture Sc of concrete are required input parameters. Ec isdetermined by the procedure specified in ASTM C469. It could also be estimated using the followingcorrelation recommended by ACI [1977]:

(62.26)

where fc = the concrete compressive strength in psi as determined by AASHTO T22, T140 [AASHTO,1989] or ASTM C39 [ASTM, 1992].

TABLE 62.13 FAA-Recommended Equivalency Factors for Stabilized Base and Subbase

MaterialEquivalency

Factor

(a) Equivalency Factors for Stabilized Base Course

Bituminous surface course 1.2–1.6Bituminous base course 1.2–1.6Cold-laid bituminous base course 1.0–1.2Mixed-in-place base course 1.0–1.2Cement-treated base course 1.2–1.6Soil cement base course N/ACrushed aggregate base course 1.0Subbase course N/A

(b) Equivalency Factors for Stabilized Subbase Course

Bituminous surface course 1.7–2.3Bituminous base course 1.7–2.3Cold-laid bituminous base course 1.5–1.7Mixed-in-place base course 1.5–1.7Cement-treated base course 1.6–2.3Soil cement base course 1.5–2.0Crushed aggregate base course 1.4–2.0Subbase course 1.0

Source: Federal Aviation Administration. 1978. AirportPavement Design and Evaluation. Reprinted from FAAAdvisory Circular. Report FAA/AC-150/5320-6C.7 December 1978; NTIS Accession No. AD-A075 537/1.

Ec psi( ) 57 000 fc( )0.5,=


Sc is the mean 28-day modulus of rupture determined using third-point loading as specified byAASHTO T97 [AASHTO, 1989] or ASTM C39 [ASTM, 1992].

Modulus of Subgrade Reaction

The value of modulus of subgrade reaction k to be used in the design is affected by the depth of bedrockand the characteristics of the subbase layer, if used. Figure 62.25 is first applied to account for the presenceof subbase course and obtain the composite modulus of subgrade reaction. Figure 62.26 is next used toinclude adjustment for the depth of rigid foundation. It is noted from Fig. 62.25 that the subgrade soilproperty required for input is the resilient modulus Mr.

Example 62.15

This example entails computation of composite subgrade reaction. A concrete pavement is constructedon a 6-in.-thick subbase with elastic modulus of 20,000 psi. The resilient modulus of the subgrade soilis 7000 psi. The depth of subgrade to bedrock is 5 ft.

Entering Fig. 62.25, with DSB = 6 in., ESB = 20,000 psi, and Mr = 7000 psi, obtain k• = 400 pci. Withbedrock depth of 5 ft., composite k = 500 pci, from Fig. 62.26.

Effective Modulus of Subgrade Reaction

Like the effective roadbed soil resilient modulus Mr for flexible pavement design, an effective k is computedto represent the combined effect of seasonal variations of k. The procedure is identical to the computationof effective Mr , except that the relative damage u is now computed as

(62.27)

Instead of solving the above equation, u can be obtained from Fig. 62.27.

FIGURE 62.24 Typical plan and cross section for runway pavements. (Source: Federal Aviation Administration.1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C, p. 35. With permission.)

200’(61m)

200’(61m)

200’(61m)

200’(61m)

200’(61m)

RUNWAY WIDTHS IN ACCORDANCE WITHAPPLICABLE ADVISORY CIRCULAR.

TRANSVERSE SLOPES IN ACCORDANCEWITH APPLICABLE ADVISORY CIRCULAR.

SURFACE, BASE, PCC, ETC., THICKNESSAS INDICATED ON DESIGN CHART.

MINIMUM 12" (30 cm) UP TO 30" (90cm)ALLOWABLE.

FOR RUNWAYS WIDER THAN 150’ (45.7 m)THIS DIMENSION WILL INCREASE.

NOTES

A

ATRANSITIONS

RUNWAY WIDTH

6@ 25’ (7.6 m)

BASE PCC

SUBBASE

THICKNESS = T

THICKNESS TAPERS = T O.7 T

THICKNESS = 0.9 T

THICKNESS = 0.7 T

LEGEND

SUBBASE

SURFACE2"(1 cm) MINIMUMSURFACE THICKNESS

1

1

2

3

44

555

23

4

TRANSITIONS

P1 P1

u D0.75 0.39k0.25–( )3.42=


Depending on the type of subbase and subgrade materials, the effective k must be reduced accordingto Fig. 62.28 to account for likely loss of support by foundation erosion and/or differential soil move-ments. Suggested values of LS in Fig. 62.28 are given in Table 62.14.

Example 62.16

The value of composite k values determined at 1-month intervals are 400, 400, 450, 450, 500, 500, 450,450, 450, 450, 450, and 450. Projected slab thickness is 10 in. and LS = 1.0. Determine effective k.

By means of Eq. (62.27) or Fig. 62.27, the relative damage for each k can be determined. Hence, totalu = 100 + 100 + 97 + 97 + 93 + 93 + 94 + 94 + 94 + 94 + 94 + 94 = 1144. Average u = 95.3 and averagek = 470 pci. Entering Fig. 62.28 with LS = 1.0 and D = 10 in., read effective k = 150 pci.

Load Transfer Coefficient

Load transfer coefficient J is a numerical index developed from experience and stress analysis. Table 62.15presents the J values for the AASHO road test conditions. Lower J values are associated with pavementswith load transfer devices (such as dowel bars) and those with tied shoulders. For cases where a rangeof J values applies, higher values should be used with low k values, high thermal coefficients, and largevariations of temperature. When dowel bars are used, the AASHTO guide recommends that the dowel

FIGURE 62.25 Chart for estimating composite k •. (Source: AASHTO. 1993. AASHTO Guides for Design of PavementStructures. Copyright 1993 by the American Association of State Highway and Transportation Officials, Washington,D.C. Used by permission.)

Subbase ElasticModulus, ESB (psi) k•(pci)

(Assumes Semi— infinite Subgrade

Depth)200015001000800600500400300200100

50

Subbase Thickness, DSB (inches)

RoadbedSoil Resilient

Modulus, MR (psi)

18 16 14 12 10 8 6

15,00030,00050,00075,000

100,000200,000400,000600,000

1,000,000

1000

10,000

20,000

12,00016,000

2000

3000

50007000

(Turning Line)

Composite Modulus ofSubgrade Reaction,


diameter should be equal to the slab thickness multiplied by 1/8, with normal dowel spacing and lengthof 12 in. and 18 in., respectively.

Drainage Coefficient

To allow for changes in thickness requirement due to differences in drainage properties, pavement layers,and subgrade, a drainage coefficient Cd was included in the design. Setting Cd = 1 for conditions at theAASHO road test, Table 62.16 shows the Cd values for other conditions. The percentage of time duringthe year that the pavement structure would be exposed to moisture levels approaching saturation can beestimated from the annual rainfall and the prevailing drainage condition.

Thickness Requirement

The required slab thickness is obtained using the nomograph in Fig. 62.29. If the environmental effectsof roadbed swelling and frost heave are important, they are considered in the same way as for flexiblepavements.

Example 62.17

Apply the AASHTO procedure to design a concrete pavement slab thickness for ESAL = 11 ¥ 106. Thedesign reliability is 95%, with a standard deviation of 0.3. The initial and terminal serviceability levelsare 4.5 and 2.5, respectively. Other design parameters are Ec = 5 ¥ 106, S ¢c = 650 psi, J = 3.2, and Cd = 1.0.

Design PSI loss = 4.5 – 2.5 = 2.0. From Fig. 62.29, D = 10 in.

FIGURE 62.26 Chart for k as a function of bedrock depth. (Source: AASHTO. 1993. AASHTO Guides for Design ofPavement Structures. Copyright 1993 by the American Association of State Highway and Transportation Officials,Washington, D.C. Used by permission.)

Subgrade Depth to RigidFoundation, DSG (ft.)

Roadbed Soil Resilient Modulus, MR (psi)

2

510

50 100 200 300 400

500

600

700

800

1000

1200

1400

Modulus of Subgrade Reaction, k•(pci)Assuming Semi-infinite Subgrade Depth

Modulus of Subgrade Reaction, k (pci)(Modified to account for presence of

rigid foundation near surface)

20,000 15,000 10,000 5,000 500 1000 1500 20000


AASHTO Reinforcement Design for Rigid Highway Pavements

Reinforcements are introduced into concrete pavements for the purpose of crack-width control. Theyare designed to hold cracks tightly closed so that the pavement remains an integral structural unit. Theamount of reinforcement required is a function of slab length (or joint spacing) and thermal propertiesof the pavement material.

Reinforcements are not required in jointed plain concrete pavements (JPCP) whose lengths are rela-tively short. As a rough guide proposed by AASHTO, the joint spacing (in feet) for JPCP should notgreatly exceed twice the slab thickness (in inches), and the ratio of slab width to length should not exceed1.25.

Reinforcement Design for JRCP

The percentage of steel reinforcement (either longitudinal or transverse reinforcement) required forjointed reinforced concrete pavement (JRCP) is given by

(62.28)

FIGURE 62.27 Chart for estimating relative damage to rigid pavements. (Source: AASHTO. 1993. AASHTO Guidesfor Design of Pavement Structures. Copyright 1993 by the American Association of State Highway and TransportationOfficials, Washington, D.C. Used by permission.)

Projected SlabThickness(inches)

Rel

ativ

e D

amag

e, u

1000

500

100

50

10

5

110 50 100 500

Composite k-value (pci)

1000 2000

14

12

10

9

8

7

6

Projected SlabThickness(inches)

PsL F◊2fs

---------- 100%¥=


where L = slab length in feet, F = friction factor between the bottom of the slab and the top of theunderlying subbase or subgrade, and fs = allowable working stress of steel reinforcement in psi. AASHTO’srecommended values for F are given in Table 62.17. The allowable steel working stress is equal to 75%of the steel yield strength. For grade 40 and grade 60 steel, fs is equal to 30,000 and 45,000 psi, respectively.For welded wire fabric, fs is 48,750 psi. Equation (62.28) is also applicable to the design of transversesteel reinforcement for continuously reinforced concrete pavement (CRCP).

FIGURE 62.28 Correction of effective modulus of subgrade reaction for potential loss of subbase support. (Source:AASHTO. 1993. AASHTO Guides for Design of Pavement Structures. Copyright 1993 by the American Association ofState Highway and Transportation Officials, Washington, D.C. Used by permission.)

TABLE 62.14 Typical Ranges of Loss of Support (LS) Factors for Various Types of Materials

Type of MaterialLoss of Support

(LS)

Cement treatment granular base (E = 1,000,000 to 2,000,000 psi)

0.0–1.0

Cement aggregate mixtures (E = 500,000 to 1,000,000 psi)

0.0–1.0

Asphalt treated base(E = 350,000 to 1,000,000 psi)

0.0–1.0

Bituminous stabilized mixtures(E = 40,000 to 300,000 psi)

0.0–1.0

Lime stabilized(E = 20,000 to 70,000 psi)

1.0–3.0

Unbound granular materials(E = 15,000 to 45,000 psi)

1.0–3.0

Fine-grained or natural subgrade materials (E = 3000 to 40,000 psi)

2.0–3.0

Source: AASHTO. 1993. AASHTO Guides for Design of PavementStructures. Copyright 1993 by the American Association of State High-way and Transportation Officials, Washington, D.C. Used by permission.

Effe

ctiv

e M

odul

us o

f Sub

grad

e R

eact

ion,

K (

pci)

(Cor

rect

ed fo

r P

oten

tial L

oss

of S

uppo

rt)

1000

500

100

50

10

5

15 10 50 100

Effective Modulus of Subgrade Reaction, K (pci)

500 1000 2000

LS=3.0LS=2.0LS=1.0LS=0

(170)

(540)


Example 62.18

Determine the longitudinal steel reinforcement requirement for a 30-ft-long JRCP constructed on crushedstone subbase.

From Table 62.17, F = 1.5. Percentage of steel reinforcement Ps = (30 ¥ 1.5)/(2 ¥ 30,000) = 0.075%.

Longitudinal Reinforcement Design for CRCP

The design of longitudinal reinforcement for CRCP is an elaborate process. The amount of reinforcementselected must satisfy limiting criteria in the following three aspects: (a) crack spacing, (b) crack width,and (c) steel stress.

CRCP Reinforcements Based on Crack Spacing: (Pmin )1 and (Pmax )1. The amount of steel reinforce-ment provided should be such that the crack spacing is between 3.5 ft (1.1 m) and 8 ft (2.4 m). Thelower limit is to minimize punchout and the upper limit to minimize spalling. For each of these twocrack spacings, Fig. 62.30 is used to determine the percent reinforcement P required, resulting in twovalues of P that define the range of acceptable percent reinforcement: (Pmax)1 and (Pmin)1.

The input variables for determining P are the thermal coefficient of portland cement concrete ac, thethermal coefficient of steel as, diameter of reinforcing bar, concrete shrinkage Z at 28 days, tensile stresssw due to wheel load, and concrete tensile strength ft at 28 days. Values of a c and Z are given in Table 62.18.A value of as = 5.0 ¥ 10–6 in./in./˚F may be used. Steel bars of 5/8- and 3/4-in. diameter are typicallyused, and the 3/4-in. bar is the largest practical size for crack-width control and bond requirements. Thenominal diameter of a reinforcing bar, in inches, is simply the bar number divided by 8. Meanwhile, sw

is the tensile stress developed during initial loading of the constructed pavement by either construction

TABLE 62.15 Recommended Load Transfer Coefficient for Various Pavement Types and Design Conditions

Shoulder Asphalt Tied P.C.C.

Load Transfer Device Yes No Yes No

Pavement Type

Plain Jointed and Jointed reinforced

3.2 3.8–4.4 2.5–3.1 3.6–4.2

CRCP 2.9–3.2 N/A 2.3–2.9 N/A

Source: AASHTO. 1993. AASHTO Guides for Design of Pave-ment Structures. Copyright 1993 by the American Association ofState Highway and Transportation Officials, Washington, D.C.Used by permission.

TABLE 62.16 Recommended Value of Drainage Coefficient, Cd, for Rigid Pavement Design

Quality of Drainage

Percent of Time Pavement Structure Is Exposedto Moisture Levels Approaching Saturation

Less than 1% 1–5% 5–25%

Greater than 25%

Excellent 1.25–1.20 1.20–1.15 1.15–1.10 1.10Good 1.20–1.15 1.15–1.10 1.10–1.00 1.00Fair 1.15–1.10 1.10–1.00 1.00–0.90 0.90Poor 1.10–1.00 1.00–0.90 0.90–0.80 0.80Very Poor 1.00–0.90 0.90–0.80 0.80–0.70 0.70

Source: AASHTO. 1993. AASHTO Guides for Design of PavementStructures. Copyright 1993 by the American Association of State High-way and Transportation Officials, Washington, D.C. Used by permission.


equipment or truck traffic. It is determined using Fig. 62.31 based on the design slab thickness, themagnitude of the wheel load, and the effective modulus of subgrade reaction. Likewise, ft is the concreteindirect tensile strength determined by AASHTO T198 or ASTM C496. It can be assumed as 86% of themodulus of rupture Sc used for thickness design.

CRCP Reinforcements Based on Crack Width: (Pmin)2. Crack width in CRCP is controlled to within0.04 in. (1.0 mm) to prevent spalling and water infiltration. The minimum percent steel (Pmin)2 thatwould produce crack widths of 0.04 in. can be determined from Fig. 62.32 with a selected bar size andinput variables sw and ft .

FIGURE 62.29 Rigid pavement thickness design chart. (Source: AASHTO. 1993. AASHTO Guides for Design ofPavement Structures. Copyright 1993 by the American Association of State Highway and Transportation Officials,Washington, D.C. Used by permission.)

TABLE 62.17 Recommended Friction Factors

Type of Material beneath Slab

Friction Factor

Surface treatment 2.2Lime stabilization 1.8Asphalt stabilization 1.8Cement stabilization 1.8River gravel 1.5Crushed stone 1.5Sandstone 1.2Natural subgrade 0.9

Source: AASHTO. 1993. AASHTOGuides for Design of Pavement Structures.Copyright 1993 by the American Asso-ciation of State Highway and Transpor-tation Officials, Washington, D.C. Usedby permission.

99.9 99 95 90 80 70 60 50

Concrete Elastic Modulus, Ec(106pci)

1200

1100

1000

900

800

700

600

500

800 500 100 50 10

Effective Modulus of SubgradeReaction, k (pci)

4.04.5

1.31.10.9

0.70.6

3.53.0

2.52.2

Load

Tra

nsfe

rC

oeffi

cien

t, J

Dra

inag

e C

oeffi

cien

t, C

d

0

10

20

30

40

50

60

70

80

90

100

Mat

ch L

ine

Des

ign

Ser

vice

abili

ty L

oss,

∆P

SI

1000 500 100 50 10 5 10 5 1 05

Estimated Total 18−kip Equivalent Single AxleLoad (ESAL) Applications, W8 (millions)

Mea

n C

oncr

ete

Mod

ulus

of R

uptu

re, S

′ c (p

si)

NOTE : Application of reliablilityin this chart requiresthe use of mean valuesfor all the input variables.

TL TL

.51.02.03.0

Design slab thickness, D (inches)

Reliability, R (%)

23

4 5 6

Overall Standard Deviation, S0

76543

14

TL

13 12 11 10 9 8 7 6 55


FIGURE 62.30 Minimum percent reinforcement to satisfy crack-spacing criteria. (Source: AASHTO. 1993. AASHTOGuides for Design of Pavement Structures. Copyright 1993 by the American Association of State Highway and Trans-portation Officials, Washington, D.C. Used by permission.)

TABLE 62.18 Shrinkage and Thermal Coefficient of Portland Cement Concrete

Indirect Tensile Strength (psi) Shrinkage (in./in.)

(a) Approximate Relations between Shrinkage and Indirect Tensile Strength of Portland Cement Concrete

300 (or less) 0.0008400 0.0006500 0.00045600 0.0003700 (or greater) 0.0002

Type of Coarse AggregateConcrete Thermal

Coefficient (10–6/˚F)

(b) Recommended Value of the Thermal Coefficient of Concrete as a Function of Aggregate Types

Quartz 6.6Sandstone 6.5Gravel 6.0Granite 5.3Basalt 4.8Limestone 3.8

Source: AASHTO. 1993. AASHTO Guides for Design ofPavement Structures. Copyright 1993 by the American Asso-ciation of State Highway and Transportation Officials, Wash-ington, D.C. Used by permission.

12.0

11.0

10.0

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1/2

5/8

1.50

1.00.75.50

.9

.8

.7

.6

.5

.4

.0004

.0006

.0008

280800

700

500

400

240

160

120

80

TL TL TL TL

Und

esira

ble

Und

esira

ble

Per

cent

Ste

el, P

Cra

ck S

paci

ng,

X (

ft.)

2.0

a s/a

c R

atio

Bar

Dia

met

er, f

(in.

)

3/4

Con

cret

e S

hrin

kage

at 2

8 D

ays,

Z (

in./i

n.)

.0002

Tens

ile S

tres

s D

ue to

Whe

el L

oad,

sw(p

si)

200

600

Con

cret

e Te

nsile

Str

engt

h at

28

Day

s, f t

(ps

i)


CRCP Reinforcements Based on Steel Stress: (Pmin)3. To guard against steel fracture and excessive per-manent deformation, a minimum amount of steel (Pmin)3 is determined according to Fig. 62.33. Inputvariables Z, sw, and ft have been determined earlier. For the steel stress ss, a limiting value equal to 75%of the ultimate tensile strength is recommended. Table 62.19 gives the allowable steel working stress forgrade 60 steel meeting ASTM A615 specifications. The determination of (Pmin)3 also requires the com-putation of a design temperature drop given by

(62.29)

where TH = the average daily high temperature during the month the pavement is constructedTL = the average daily low temperature during the coldest month of the year

FIGURE 62.31 Chart for estimating wheel load tensile stress s w . (Source: AASHTO. 1993. AASHTO Guides forDesign of Pavement Structures. Copyright 1993 by the American Association of State Highway and TransportationOfficials, Washington, D.C. Used by permission.)

13

12

11

10

9

8

7

6

5

4

600

550

500

450

400

350

300

250

200

150

EXAMPLE:

D = 9.5 in. WHEEL LOAD = 20,000 lb. k = 170 pci

SOLUTION:σw = 230 psi

6,000

8,000

10,000

12,000

14,000

16,000

18,000

20,00022,000

24,000

WHEEL LOADMAGNITUDE(POUNDS)

Des

ign

Sla

b T

hick

ness

, D (

inch

es)

Whe

el L

oad

Tens

ile S

tres

s σ w

(ps

i)

EFFECTIVEMODULUS OFSUBGRADEREACTION k (pci)

500

400

300

200

100

50

DTD TH TL–=


FIGURE 62.32 Minimum percent steel reinforcement to satisfy crack-width criteria. (Source: AASHTO. 1993.AASHTO Guides for Design of Pavement Structures. Copyright 1993 by the American Association of State Highwayand Transportation Officials, Washington, D.C. Used by permission.)

FIGURE 62.33 Minimum percent reinforcement to satisfy steel–stress criteria. (Source: AASHTO 1993. AASHTOGuides for Design of Pavement Structures. Copyright 1993 by the American Association of State Highway and Trans-portation Officials, Washington, D.C. Used by permission.)

.10

.09

.08

.07

.06

.05

.04

.03

.02

.01

3/45/81/2

800

700

600

500

400

.9

.8

.7

.6

.5

.4

TLTL

Cra

ck W

idth

, CW

(in

.)

Bar

Dia

met

er, f

(in.

)

Tens

ile S

tres

s D

ue to

Whe

el L

oad,

sw (

psi)

Con

cret

e Te

nsile

Str

engt

h, f t

(ps

i)

Per

cent

Ste

el, P

280

240

200

160

120

80

30

20

355575

4565100

300 800

700

600

500

400

220

140

60

.0002

.9

.8

.7

.6

.5

.4

.0005

.0008

40

50

60

70

80

90

100

110

120

Und

esira

ble

Ste

el S

tres

s, σ

s (k

si)

Des

ign

Tem

pera

ture

Dro

p, D

TD (

°F)

Con

cret

e S

hrin

kage

at 2

8 D

ays,

Z- (

in/in

)

Ten

sile

Str

ess

Due

to W

heel

Loa

d, σ

W (

psi)

Con

cret

e T

ensi

le S

tren

gth,

f t (

psi)

Per

cent

Ste

el, P

TLTL TL


Reinforcement Design. Based on the three criteria discussed above, the design percent steel should fallwithin Pmax and Pmin given by

(62.30)

(62.31)

If Pmax is less than Pmin, a design revised by changing some of the input parameters is required. WithPmax greater than Pmin, the number of reinforcing bars or wires required, N, is given by Nmin £ N £ Nmax,where Nmin and Nmax are computed by

(62.32)

(62.33)

where Ws = the width of pavement in inchesD = the slab thickness in inchesf = the reinforcing bar or wire diameter in inches

Example 62.19

This example concerns longitudinal steel reinforcement design for CRCP. Design data are D = 9 in., as =5 ¥ 10–6 in./in./˚F, ac = 5.3 ¥ 10–6 for concrete with granite coarse aggregate (see Table 62.18), ft = 600 psiat 28 days, Z = 0.0003 (see Table 62.18), maximum construction wheel load = 18,000 lb, effective k =100 pci, and design temperature drop = 55˚F.

Try steel bar diameter f = 0.5 in. (a) Crack spacing control. sw = 240 psi, from Fig. 62.31. With (as /ac) =0.94, obtain (Pmin)1 < 0.4% for X = 8 ft, and (Pmax)2 = 0.49% for X = 3.5 ft from Fig. 62.30. (b) Crackwidth control. Obtain (Pmin)2 = 0.40% from Fig. 62.32. (c) Steel stress control. ss = 67 psi, fromTable 62.19. Obtain (Pmin)3 = 0.45%. Hence, overall (Pmin) = 0.4% and (Pmax) = 0.49%. For a pavementwidth of 12 ft, apply Eqs. (62.32) and (62.33), Nmin = 29.7, Nmax = 32.3. Use 30 numbers of 0.5-in. bars.

PCA Thickness Design Procedure for Rigid Highway Pavements

The thickness design procedure published by PCA [1984] was developed by relating theoretically com-puted values of stress, deflection, and pressure to pavement performance criteria derived from data of(1) major road test programs, (2) model and full-scale tests, and (3) performance of normally constructedpavements subject to normal mixed traffic.

Traffic-loading data in terms of axle load distribution are obtained in the usual way as described earlierin this chapter. Each axle load is further multiplied by a load safety factor (LSF) according to the followingrecommendations: (1) LSF = 1.2 for interstate highways and other multilane projects with uninterrupted

TABLE 62.19 Allowable Steel Working Stress, ksi

Indirect Tensile Strength of Concrete at 28 days, psi

Reinforcing Bar Size

No. 4 No. 5 No. 6

300 (or less) 65 57 54 400 67 60 55 500 67 61 56 600 67 63 58 700 67 65 59 800 (or greater) 67 67 60

Source: AASHTO. 1993. AASHTO Guides for Design ofPavement Structures. Copyright 1993 by the American Asso-ciation of State Highway and Transportation Officials, Wash-ington, D.C. Used by permission.

Pmax Pmax( )1=

Pmin max Pmin( )1 Pmin( )2 Pmin( )3, ,{ }=

Nmin 0.01273Pmin WsD f§ 2=

Nmax 0.01273PmaxWs D f§ 2=


traffic flow and high volumes of truck traffic; (2) LSF = 1.1 for highways and arterial streets with moderatevolumes of truck traffic; and (3) LSF = 1.0 for roads, residential streets, and other streets with smallvolumes of truck traffic. The flexural strength of concrete is determined by the 28-day modulus of rupturefrom third-point loading according to ASTM Test Method C78. Subgrade and subbase support is definedin terms of the modulus of subgrade reaction k.

The design procedure consists of a fatigue analysis and an erosion analysis, which are considered separatelyusing different sets of tables and design charts. The final thickness selected must satisfy both analyses.

Fatigue Design

Fatigue design is performed with the aim to control fatigue cracking. The slab thickness based on fatiguedesign is the same for JRCP, for JPCP with doweled and undoweled joints, and for CRCP. This is becausethe most critical loading position is near midslab and the effect of joints is negligible. The presence of atied concrete shoulder, however, must be considered since it significantly reduces the critical edge stress.The analysis is based on the concept of cumulative damage given by

(62.34)

where m = the total number of axle load groupsni = the predicted number of repetitions for the ith load groupNi = the allowable number of repetitions for the ith load group

The steps in the design procedure are:

1. Multiply the load of each design axle load group by the appropriate LSF.2. Assume a trial slab thickness.3. Obtain from Table 62.20(a) or (b) the equivalent stress for the input slab thickness and k, and

calculate the stress ratio factor as

(62.35)

4. For each axle load i, obtain from Fig. 62.34 the allowable load repetitions Ni.5. Compute D from Eq. (62.34). If D exceeds 1, select a greater trial thickness and repeat steps 3

through 5. The trial thickness is adequate if D is less than or equal to 1.

Erosion Design

PCA requires erosion analysis in pavement thickness design to control foundation and shoulder erosion,pumping, and faulting. Since the most critical deflection occurs at the corner, the presence of shoulderand the type of joint construction will both affect the analysis. The concept of cumulative damage asdefined by Eq. (62.34) is again applied. The steps are:

1. Multiply the load of each design axle load group by the LSF.2. Assume a trial slab thickness.3. Obtain from Table 62.21(a), (b), (c), or (d) the erosion factor for the input slab thickness and k.4. For each axle load i, obtain from Fig. 62.35(a) or (b) the allowable load repetitions Ni.5. Compute D from Eq. (62.34). If D exceeds 1, select a greater trial thickness and repeat steps 3

through 5. The trial thickness is adequate if D is less than or equal to 1.

Example 62.20

Determine the required slab thickness for an expressway with the design traffic shown in the table below.The pavement is to be constructed with doweled joint but without concrete shoulder. Concrete modulusof rupture is 650 psi. The subgrade k is 130 pci.

Dni

Ni

-----i 1=

m

Â=

Stress ratio factor Equivalent stressConcrete flexural strength--------------------------------------------------------------=


A trial-and-error approach is needed by assuming slab thickness. The solution is shown only for slabthickness h = 9.5 in.

For an expressway, the LSF = 1.2. The design load is equal to (1.2 ¥ axle load). From Table 62.20,equivalent stress for single axle is 206 and for tandem axle is 192. The corresponding stress ratios are

TABLE 62.20 Equivalent Stress for Fatigue Analysis

Slab Thickness

(in.)

k of Subgrade–Subbase (pci)

50 100 150 200 300 500 700

(a) Equivalent Stress — No Concrete Shoulder (Single Axle/Tandem Axle)

4 in. 825/679 726/585 671/542 634/516 584/486 523/457 484/4434.5 in. 699/586 616/500 571/460 540/435 498/406 448/378 417/3635 in. 602/516 531/436 493/399 467/374 432/349 390/321 363/3075.5 in. 526/461 464/387 431/353 409/331 379/305 343/278 320/2646 in. 465/416 411/348 382/316 362/296 336/271 304/246 285/2326.5 in. 417/380 367/317 341/286 324/267 300/244 273/220 256/2077 in. 375/349 331/290 307/262 292/244 271/222 246/199 231/1867.5 in. 340/323 300/268 279/241 265/224 246/203 224/181 210/1698 in. 311/300 274/249 255/223 242/208 225/188 205/167 192/1558.5 in. 285/281 252/232 234/208 222/193 206/174 188/154 177/1439 in. 264/264 232/218 216/195 205/181 190/163 174/144 163/1339.5 in. 245/248 215/205 200/183 190/170 176/153 161/134 151/124

10 in. 228/235 200/193 186/173 177/160 164/144 150/126 141/11710.5 in. 213/222 187/183 174/164 165/151 153/136 140/119 132/11011 in. 200/211 175/174 163/155 154/143 144/129 131/113 123/10411.5 in. 188/201 165/165 153/148 145/136 135/122 123/107 116/9812 in. 177/192 155/158 144/141 137/130 127/116 116/102 109/9312.5 in. 168/183 147/151 136/135 129/124 120/111 109/97 103/8913 in. 159/176 139/144 129/129 122/119 113/106 103/93 97/8513.5 in. 152/168 132/138 122/123 116/114 107/102 98/89 92/8114 in. 144/162 125/133 116/118 110/109 102/98 93/85 88/78

(b) Equivalent Stress — Concrete Shoulder (Single Axle/Tandem Axle)

4 in. 640/534 559/468 517/439 489/422 452/403 409/388 383/3844.5 in. 547/461 479/400 444/372 421/356 390/338 355/322 333/3165 in. 475/404 417/349 387/323 367/308 341/290 311/274 294/2675.5 in. 418/360 368/309 342/285 324/271 302/254 276/238 261/2316 in. 372/325 327/277 304/255 289/241 270/225 247/210 234/2036.5 in. 334/295 294/251 274/230 260/218 243/203 223/188 212/1807 in. 302/270 266/230 248/210 236/198 220/184 203/170 192/1627.5 in. 275/250 243/211 226/193 215/182 201/168 185/155 176/1488 in. 252/232 222/196 207/179 197/168 185/155 170/142 162/1358.5 in. 232/216 205/182 191/166 182/156 170/144 157/131 150/1259 in. 215/202 190/171 177/155 169/146 158/134 146/122 139/1169.5 in. 200/190 176/160 164/146 157/137 147/126 136/114 129/108

10 in. 186/179 164/151 153/137 146/129 137/118 127/107 121/10110.5 in. 174/170 154/143 144/130 137/121 128/111 119/101 113/9511 in. 164/161 144/135 135/123 129/115 120/105 112/95 106/9011.5 in. 154/153 136/128 127/117 121/109 113/100 105/90 100/8512 in. 145/146 128/122 120/111 114/104 107/95 99/86 95/8112.5 in. 137/139 121/117 113/106 108/99 101/91 94/82 90/7713 in. 130/133 115/112 107/101 102/95 96/86 89/78 85/7313.5 in. 124/127 109/107 102/97 97/91 91/83 85/74 81/7014 in. 118/122 104/103 97/93 93/87 87/79 81/71 77/67

Source: Portland Cement Association. 1984. Thickness Design for Concrete Highway andStreet Pavements. With permission.


0.317 and 0.295. N1 for fatigue analysis is obtained from Fig. 62.34. From Table 62.21, the erosion factoris 2.6 for single axle and is 2.8 for tandem axle. N2 for erosion analysis is obtained from Fig. 62.35(a).The results show that the design is satisfactory.

FAA Method for Rigid Airport Pavement Design

Both the thickness design and reinforcement design procedures by the FAA [1978] are presented in thissection.

FIGURE 62.34 Allowable repetitions for fatigue analysis. (Source: Portland Cement Association. 1984. ThicknessDesign for Concrete Highway and Street Pavements, p. 15. With permission.)

Axle Load(kips)

Design Load(kips)

Design Fatigue Erosion

n N1 (n/N1) N2 (n/N2)

52T 62.4T 3,100 800,000 0.004 800,000 0.00450T 60.0T 32,000 2,000,000 0.016 1,000,000 0.03048T 57.6T 32,000 10,000,000 0.0032 1,200,000 0.02746T 55.2T 48,000 Unlimited 0 1,700,000 0.02844T 52.8T 158,000 Unlimited 0 2,000,000 0.07942T 50.4T 172,000 Unlimited 0 2,800,000 0.06140T 48.0T 250,000 Unlimited 0 3,500,000 0.07130S 36.0S 3,100 25,000 0.124 1,700,000 0.00228S 33.6S 3,100 70,000 0.044 2,200,000 0.00126S 31.2S 9,300 200,000 0.045 3,000,000 0.00224S 28.8S 545,000 800,000 0.682 5,000,000 0.03322S 26.4S 640,000 10,000,000 0.064 9,000,000 0.071

Total 0.982 0.41

SINGLE AXLE LOAD, KIPS

8 12 14 16 18 22 24 26 28 32 34 36 38 42 44 46 48 52 54 56 58

20 30 40 50 60120

110

100

9080706050403020

16

10

2 4 6 8 2 4 6 8 2 4 6 8 2 4 6 2 4 68

100

1000

10,000

100,000

1,000,000

10,000,000

ALLOWABLE LOAD REPETITIONS

1.50 1.00

0.90

0.80

0.70

0.60 0.50 0.40 0.30 0.25 0.20

0.15

TANDEM AXLE LOAD, KIPS

STRESS RATIO FACTOR


TABLE 62.21 Erosion Factors for Erosion Analysis

SlabThickness

(in.)


50 100 200 300 500 700

(a) Erosion Factors — Doweled Joints, No Concrete Shoulder (Single Axle/Tandem Axle)

4 3.74/3.83 3.73/3.79 3.72/3.75 3.71/3.73 3.70/3.70 3.68/3.674.5 3.59/3.70 3.57/3.65 3.56/3.61 3.55/3.58 3.54/3.55 3.52/3.535 3.45/3.58 3.43/3.52 3.42/3.48 3.41/3.45 3.40/3.42 3.38/3.405.5 3.33/3.47 3.31/3.41 3.29/3.36 3.28/3.33 3.27/3.30 3.26/3.286 3.22/3.38 3.19/3.31 3.18/3.26 3.17/3.23 3.15/3.20 3.14/3.176.5 3.11/3.29 3.09/3.22 3.07/3.16 3.06/3.13 3.05/3.10 3.03/3.077 3.02/3.21 2.99/3.14 2.97/3.08 2.96/3.05 2.95/3.01 2.94/2.987.5 2.93/3.14 2.91/3.06 2.88/3.00 2.87/2.97 2.86/2.93 2.84/2.908 2.85/3.07 2.82/2.99 2.80/2.93 2.79/2.89 2.77/2.85 2.76/2.828.5 2.77/3.01 2.74/2.93 2.72/2.86 2.71/2.82 2.69/2.78 2.68/2.759 2.70/2.96 2.67/2.87 2.65/2.80 2.63/2.76 2.62/2.71 2.61/2.689.5 2.63/2.90 2.60/2.81 2.58/2.74 2.56/2.70 2.55/2.65 2.54/2.62

10 2.56/2.85 2.54/2.76 2.51/2.68 2.50/2.64 2.48/2.59 2.47/2.5610.5 2.50/2.81 2.47/2.71 2.45/2.63 2.44/2.59 2.42/2.54 2.41/2.5111 2.44/2.76 2.42/2.67 2.39/2.58 2.38/2.54 2.36/2.49 2.35/2.4511.5 2.38/2.72 2.36/2.62 2.33/2.54 2.32/2.49 2.30/2.44 2.29/2.4012 2.33/2.68 2.30/2.58 2.28/2.49 2.26/2.44 2.25/2.39 2.23/2.3612.5 2.28/2.64 2.25/2.54 2.23/2.45 2.21/2.40 2.19/2.35 2.18/2.3113 2.23/2.61 2.20/2.50 2.18/2.41 2.16/2.36 2.14/2.30 2.13/2.2713.5 2.18/2.57 2.15/2.47 2.13/2.37 2.11/2.32 2.09/2.26 2.08/2.2314 2.13/2.54 2.11/2.43 2.08/2.34 2.07/2.29 2.05/2.23 2.03/2.19

(b) Erosion Factors — Aggregate–Interlock Joints, No Concrete Shoulder (Single Axle/Tandem Axle)

4 3.94/4.03 3.91/3.95 3.88/3.89 3.86/3.86 3.82/3.83 3.77/3.804.5 3.79/3.91 3.76/3.82 3.73/3.75 3.71/3.72 3.68/3.68 3.64/3.655 3.66/3.81 3.63/3.72 3.60/3.64 3.58/3.60 3.55/3.55 3.52/3.525.5 3.54/3.72 3.51/3.62 3.48/3.53 3.46/3.49 3.43/3.44 3.41/3.406 3.44/3.64 3.40/3.53 3.37/3.44 3.35/3.40 3.32/3.34 3.30/3.306.5 3.34/3.56 3.30/3.46 3.26/3.36 3.25/3.31 3.22/3.25 3.20/3.217 3.26/3.49 3.21/3.39 3.17/3.29 3.15/3.24 3.13/3.17 3.11/3.137.5 3.18/3.43 3.13/3.32 3.09/3.22 3.07/3.17 3.04/3.10 3.02/3.068 3.11/3.37 3.05/3.26 3.01/3.16 2.99/3.10 2.96/3.03 2.94/2.998.5 3.04/3.32 2.98/3.21 2.93/3.10 2.91/3.04 2.88/2.97 2.87/2.939 2.98/3.27 2.91/3.16 2.86/3.05 2.84/2.99 2.81/2.92 2.79/2.879.5 2.92/3.22 2.85/3.11 2.80/3.00 2.77/2.94 2.75/2.86 2.73/2.81

10 2.86/3.18 2.79/3.06 2.74/2.95 2.71/2.89 2.68/2.81 2.66/2.7610.5 2.81/3.14 2.74/3.02 2.68/2.91 2.65/2.84 2.62/2.76 2.60/2.7211 2.77/3.10 2.69/2.98 2.63/2.86 2.60/2.80 2.57/2.72 2.54/2.6711.5 2.72/3.06 2.64/2.94 2.58/2.82 2.55/2.76 2.51/2.68 2.49/2.6312 2.68/3.03 2.60/2.90 2.53/2.78 2.50/2.72 2.46/2.64 2.44/2.5912.5 2.64/2.99 2.55/2.87 2.48/2.75 2.45/2.68 2.41/2.60 2.39/2.5513 2.60/2.96 2.51/2.83 2.44/2.71 2.40/2.65 2.36/2.56 2.34/2.5113.5 2.56/2.93 2.47/2.80 2.40/2.68 2.36/2.61 2.32/2.53 2.30/2.4814 2.53/2.90 2.44/2.77 2.36/2.65 2.32/2.58 2.28/2.50 2.25/2.44

(c) Erosion Factors — Doweled Joints, Concrete Shoulder (Single Axle/Tandem Axle)

4 3.28/3.30 3.24/3.20 3.21/3.13 3.19/3.10 3.15/3.09 3.12/3.084.5 3.13/3.19 3.09/3.08 3.06/3.00 3.04/2.96 3.01/2.93 2.98/2.915 3.01/3.09 2.97/2.98 2.93/2.89 2.90/2.84 2.87/2.79 2.85/2.775.5 2.90/3.01 2.85/2.89 2.81/2.79 2.79/2.74 2.76/2.68 2.73/2.65


FAA Thickness Design Procedure for Rigid Airport Pavements

The FAA thickness design method is based on the Westergaard analysis of an edge-loaded slab on a denseliquid foundation. The design curves in Figs. 62.36–62.42 have been developed with the assumption thatthe landing gear assembly is either tangent to a longitudinal joint or perpendicular to a transverse joint —whichever produces the largest stress. It is also assumed that 95% of the gross aircraft weight is carriedon the main landing gear assembly. The design curves provide slab thickness T for the critical areasdefined earlier in this chapter. The thickness of 0.9T for noncritical areas applies to the concrete slab

TABLE 62.21 (continued) Erosion Factors for Erosion Analysis

SlabThickness

(in.)


50 100 200 300 500 700

6 2.79/2.93 2.75/2.82 2.70/2.71 2.68/2.65 2.65/2.58 2.62/2.546.5 2.70/2.86 2.65/2.75 2.61/2.63 2.58/2.57 2.55/2.50 2.52/2.457 2.61/2.79 2.56/2.68 2.52/2.56 2.49/2.50 2.46/2.42 2.43/2.387.5 2.53/2.73 2.48/2.62 2.44/2.50 2.41/2.44 2.38/2.36 2.35/2.318 2.46/2.68 2.41/2.56 2.36/2.44 2.33/2.38 2.30/2.30 2.27/2.248.5 2.39/2.62 2.34/2.51 2.29/2.39 2.26/2.32 2.22/2.24 2.20/2.189 2.32/2.57 2.27/2.46 2.22/2.34 2.19/2.27 2.16/2.19 2.13/2.139.5 2.26/2.52 2.21/2.41 2.16/2.29 2.13/2.22 2.09/2.14 2.07/2.08

10 2.20/2.47 2.15/2.36 2.10/2.25 2.07/2.18 2.03/2.09 2.01/2.0310.5 2.15/2.43 2.09/2.32 2.04/2.20 2.01/2.14 1.97/2.05 1.95/1.9911 2.10/2.39 2.04/2.28 1.99/2.16 1.95/2.09 1.92/2.01 1.89/1.9511.5 2.05/2.35 1.99/2.24 1.93/2.12 1.90/2.05 1.87/1.97 1.84/1.9112 2.00/2.31 1.94/2.20 1.88/2.09 1.85/2.02 1.82/1.93 1.79/1.87

12.5 1.95/2.27 1.89/2.16 1.84/2.05 1.81/1.98 1.77/1.89 1.74/1.8413 1.91/2.23 1.85/2.13 1.79/2.01 1.76/1.95 1.72/1.86 1.70/1.80

13.5 1.86/2.20 1.81/2.09 1.75/1.98 1.72/1.91 1.68/1.83 1.65/1.7714 1.82/2.17 1.76/2.06 1.71/1.95 1.67/1.88 1.64/1.80 1.61/1.74

(d) Erosion Factors — Aggregate–Interlock Joints, Concrete Shoulder(Single Axle/Tandem Axle)

4 3.46/3.49 3.42/3.39 3.38/3.32 3.36/3.29 3.32/3.26 3.28/3.244.5 3.32/3.39 3.28/3.28 3.24/3.19 3.22/3.16 3.19/3.12 3.15/3.095 3.20/3.30 3.16/3.18 3.12/3.09 3.10/3.05 3.07/3.00 3.04/2.975.5 3.10/3.22 3.05/3.10 3.01/3.00 2.99/2.95 2.96/2.90 2.93/2.866 3.00/3.15 2.95/3.02 2.90/2.92 2.88/2.87 2.86/2.81 2.83/2.776.5 2.91/3.08 2.86/2.96 2.81/2.85 2.79/2.79 2.76/2.73 2.74/2.687 2.83/3.02 2.77/2.90 2.73/2.78 2.70/2.72 2.68/2.66 2.65/2.617.5 2.76/2.97 2.70/2.84 2.65/2.72 2.62/2.66 2.60/2.59 2.57/2.548 2.69/2.92 2.63/2.79 2.57/2.67 2.55/2.61 2.52/2.53 2.50/2.488.5 2.63/2.88 2.56/2.74 2.51/2.62 2.48/2.55 2.45/2.48 2.43/2.439 2.57/2.83 2.50/2.70 2.44/2.57 2.42/2.51 2.39/2.43 2.36/2.389.5 2.51/2.79 2.44/2.65 2.38/2.53 2.36/2.46 2.33/2.38 2.30/2.33

10 2.46/2.75 2.39/2.61 2.33/2.49 2.30/2.42 2.27/2.34 2.24/2.2810.5 2.41/2.72 2.33/2.58 2.27/2.45 2.24/2.38 2.21/2.30 2.19/2.2411 2.36/2.68 2.28/2.54 2.22/2.41 2.19/2.34 2.16/2.26 2.14/2.2011.5 2.32/2.65 2.24/2.51 2.17/2.38 2.14/2.31 2.11/2.22 2.09/2.1612 2.28/2.62 2.19/2.48 2.13/2.34 2.10/2.27 2.06/2.19 2.04/2.1312.5 2.24/2.59 2.15/2.45 2.09/2.31 2.05/2.24 2.02/2.15 1.99/2.1013 2.20/2.56 2.11/2.42 2.04/2.28 2.01/2.21 1.98/2.12 1.95/2.0613.5 2.16/2.53 2.08/2.39 2.00/2.25 1.97/2.18 1.93/2.09 1.91/2.0314 2.13/2.51 2.04/2.36 1.97/2.23 1.93/2.15 1.89/2.06 1.87/2.00

Source: Portland Cement Association. 1984. Thickness Design for Concrete Highway and StreetPavements. With permission.


thickness. As in the case for flexible pavement design, stabilized subbase is required to accommodateaircraft weighing 100,000 lb or more.

Design Loading. The same method of selecting a design aircraft and computing design annual departuresis followed as for the FAA flexible airport pavement design.

FIGURE 62.35 Allowable repetitions for erosion analysis. (Source: Portland Cement Association. 1984. ThicknessDesign for Concrete Highway and Street Pavements. With permission.)

FIGURE 62.36 Rigid pavement thickness for single-wheel gear. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

60

50

40

30

20

18

16

14

12

9

8 16

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18

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25

2.0

2.2

2.4

2.6

2.8

3.0

3.2

3.4

3.6

3.8

4.0

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GLE

AX

LE L

OA

D, K

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TAN

DE

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XLE

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AD

, KIP

S

ER

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ION

FA

CT

OR

ALL

OW

AB

LE L

OA

D R

EP

ET

ITIO

NS

WITHOUT CONCRETE SHOULDER

100,000,000 864

2

10,000,00086

4

2

10,000,00086

4

2

100,00086

4

2

10,000

1000

86

4

2

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AD

, KIP

S

ER

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OW

AB

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OA

D R

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ET

ITIO

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WITH CONCRETE SHOULDER

100,000,00042

6

6

4

2

4

10,000,000

1,000,000

2

8

100,00086

4

2

8

6

4

10,000

1000

2

30,00

0

45,00

0

60,0

00

500300

200

100k=50 pci

75,0

00 lb

s

0178 E

14

1415

13

12

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10

9

13

12

11

10

9

8

7

6

NOTE:1 inch = 2.54 cm 1 psi = 0.0069 MN/m2

1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

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SLA

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ES

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CO

NC

RE

TE

FLE

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RA

L S

TR

EN

GT

H, p

si

1200900

800

700

600

500

3000 6000 15,000 25,000

ANNUAL DEPARTURES

8


Concrete Flexural Strength. The 28-day flexural strength of concrete is determined by ASTM TestMethod C78. A 90-day flexural strength may be used. It can be taken to be 10% higher than the 28-daystrength, except when high early strength cement or pozzolanic admixtures are used.

Foundation Modulus. The subgrade modulus k is determined by the test method specified in AASHTOT222. When a layer of subbase is used, the design k is obtained from Fig. 62.43 for unstabilized subbaseand Fig. 62.44 for stabilized subbase.

High Traffic Volumes. For airports with design traffic exceeding 25,000 annual departures, the FAAsuggests using thicker pavements as follows: 104, 108, 110, and 112% of design thickness for 25,000 annualdepartures for annual departure levels of 50,000, 100,000, 150,000, and 200,000, respectively. This sugges-tion is based on a logarithmic relationship between percent thickness and departures.

Example 62.21

Determine the thickness of concrete pavement required for the design traffic of Example 62.13. Thesubgrade k = 100 pci.

Since the design aircraft exceeds 100,000 lb in gross weight, use 6 in. stabilized subbase. From Fig. 62.44,effective k = 210 pci. Using concrete with flexural strength of 650 psi, the slab thickness required is 18in., from Fig. 62.37.

FAA Joint Spacing and Reinforcement Design for Rigid Airport Pavements

The recommended maximum joint spacings are shown in Table 62.22. Tie bars are used across longitudinaljoints. They are deformed bars 5/8 in. (16 mm) in diameter, 30 in. (76 cm) long, and spaced 30 in. (76 cm)on centers. Dowel bars are used at transverse joints to prevent relative vertical displacement of adjacentslab ends. Table 62.23 indicates the dowel dimensions and spacings for various slab thicknesses.

The area of steel required for a reinforced concrete pavement is determined by

FIGURE 62.37 Rigid pavement thickness for dual-wheel gear. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

300

DUAL WHEEL GEAR

50,000

0579 E

75,000

100,0

00

150,

000

200,

000

Ibs500

200

100k = 50 pci

NOTE:1 inch = 2.54 cm 1 psi = 0.0069 MN/m2

1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

2223

22

21

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7 7

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2426 27

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1200 3000 500015,00025,000ANNUAL DEPARTURES

SLA

B T

HIC

KN

ES

S ,i

n

FLE

XU

RA

L S

TR

EN

GT

H, p

si

900

850

800

750

700

650

600

550

500

1312

1112


(62.36)

where As = area of steel per foot width or length, in square inchesL = length or width of slab, in feett = thickness of slab, in inchesfs = allowable tensile stress in steel, in psi, taken as two-thirds of the yield strength of the steel

The minimum percentage of steel reinforcement is 0.05%. The maximum allowable slab length regard-less of steel percentage is 75 ft (23 m).

Example 62.22

An 18-in.-thick concrete airport pavement has a slab length of 50 ft. Determine the longitudinal steelrequirement.

Using grade 60 steel, fs = (60,000) = 40,000 psi. By Eq. (62.36), As = 3.7(50) /40,000 =0.14 in.2 per ft. This is equal to 0.016% steel, satisfying the minimum requirement of 0.05%.

62.7 Pavement Overlay Design

As a pavement reaches the end of its service life, a new span of service life can be provided by either areconstruction or an application of overlay over the existing pavement. There are three common formsof overlay construction — bituminous overlay on flexible pavement, bituminous overlay on concretepavement, and concrete overlay on concrete pavement. The Asphalt Institute method of flexible overlaydesign for highway pavement, the Portland Cement Association method of concrete overlay design forhighway pavement, and the Federal Aviation Administration method of overlay design for airport pave-ment are described in this section.

FIGURE 62.38 Rigid pavement thickness for dual-tandem gear. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

900

850

800

750

700

650

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550

500

500

300

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100

k=50 pci

100,000

150,

000

200,

000

300,

00040

0,00

0 Ib

s

1200 3000 5000 15,000 25,000

ANNUAL DEPARTURES

DUAL TANDEM GEAR

0478 E

NOTE:

1 inch = 2.54 cm 1 psi = 0.0069 MN/m2

1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

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FLE

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EN

GT

H ,

psi

As3.7 L( ) Lt( )0.5

fs

------------------------------=

23-- 50( ) 18( )


AI Design Procedure for Flexible Overlay on Flexible Highway Pavement

The Asphalt Institute [1983] presents two different approaches to flexible overlay design — one basedon the concept of effective thickness and the other based on deflection analysis.

AI Effective Thickness Approach

This approach evaluates the so-called effective thickness Te of the existing pavement and determines therequired overlay thickness TOL as

(62.37)

where T is the required thickness of a new full-depth pavement if constructed on the existing subgrade,to be determined from Fig. 62.10.

The Asphalt Institute recommends two methods for evaluating effective pavement thickness. The firstmethod involves the use of a conversion factor C based on the PSI (present serviceability index) of theexisting pavement plus the use of conversion factors E for converting various pavement layers intoequivalent thickness of asphalt concrete. That is,

(62.38)

where n is the total number of pavement layers. C is obtained from either line A or line B in Fig. 62.45.Line A assumes that the overlaid pavement would exhibit a reduced rate of change in PSI compared tobefore overlay. Line B represents a more conservative design, assuming that the rate of change in PSI

FIGURE 62.39 Rigid pavement thickness for B-747-100, SR, 200B, 200C, and 200F. (Source: Federal AviationAdministration. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With per-mission.)

900

850

800

750

700

650

600

550

500

1200 3000 6000 15,000 25,000ANNUAL DEPARTURES

0378 E

NOTE:

1 inch = 2.54 cm 1 psi = 0.0069 MN/m2

1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

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psi

B747-100, SR,200 B,C,FCONTACT AREA = 245 sq. in.DUAL SPACING = 44 in.TANDEM SPACING = 53 in.

300,000

400,

000

500,

00060

0,00

0700,

000

800,

000

850,

000

Ibsk=50 pci

100200

300

500

TOL T Te–=

Te C h i E i{ }i 1=

n

Â=


would remain unchanged after overlay. PSI is usually estimated from correlation with pavement roughnessmeasurements. Equivalency factors Ei are obtained from Table 62.24.

Example 62.23

An old pavement has 3-in. asphalt surface course and 8.5-in. type II emulsified asphalt base (seeExample 62.12). Its current PSI is 2.8. Provide an overlay to the pavement to carry the design traffic ofExample 62.10.

With PSI = 2.8, C = 0.75 by line A of Fig. 62.45. The thickness of new full-depth asphalt pavementrequired is 9.5 in. (see Example 62.12). The equivalency factor of type II emulsified base is 0.83, fromTable 62.24. Overlay thickness Te = 9.5 – 0.75{(3 ¥ 1.0) + (8.5 ¥ 0.83)} = 2 in.

The second recommended method relies on component analysis that assigns conversion factors Ci

from Table 62.25 to individual pavement layers based on their respective physical conditions. The effectivethickness for the existing pavement structure is given by

(62.39)

where hi = the layer thickness of layer in = the total number of layers in the existing pavement

Example 62.24

For the old pavement in Example 62.23, it is observed that the asphalt concrete surface exhibits appreciablecracking and the emulsified asphalt base has some fine cracking and slight deformation in the wheelpaths. Design an overlay for the same traffic as in Example 62.23.

FIGURE 62.40 Rigid pavement thickness for B-747-SP. (Source: Federal Aviation Administration. 1978. AirportPavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

ANNUAL DEPARTURES

SLA

B T

HIC

KN

ES

S, i

n.

FLE

XU

RA

L S

TR

EN

GT

H ,

psi

900

850

800

750

700

650

600

550

500

500

300

200

100

k=50 pci 700,

000

lbs

600,

000

500,

000

400,

000

300,000

1200 3000 6000 15,000 25,000

B 747 SP

CONTACT AREA = 210 sq. in.

DUAL SPACING = 43.25 in.

TANDEM SPACING = 54 in.

0278 E

NOTE: 1 inch = 2.54 cm 1 psi = 0.0069 MN/m2

1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

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7

22

21

20

19

18

17

16

15

14

12

11

10

9

8

7

21

20

19

18

17

16

15

14

13

12

11

10

9

8

7

13

Te h i C i

i 1=

n

Â=


From Table 62.25, the conversion factors for the surface and base courses are both 0.6. TOL = 9.5 –{(0.6 ¥ 3) + (0.6 ¥ 8.5)} = 2.6 in. Use 3 in.

AI Deflection-Based Approach

This approach is based on the correlation between wheel load, repetitions of wheel loads, and themagnitude of pavement rebound deflection. Rebound deflections are measured using the Benkelmanbeam on the outer wheel path at a minimum of 10 locations within the test section or a minimum of20 measurements per mile (12 per km). The Benkelman beam is a 12-ft (3.66-m) beam pivoted at apoint 8 ft (2.44 m) from the probe end. The probe is positioned at the test point between the dual tiresof a rear wheel of a loaded truck that has an 18-kip (80-kN) load equally distributed on its two dualwheels of the rear axle. The amount of vertical rebound at the test point after the truck moves away isrecorded as the rebound deflection.

The deflection measurements are used to determine a representative rebound deflection d r:

(62.40)

where d m = the mean of rebound deflection measurementss = the standard deviationF = the temperature adjustment factorc = the critical period adjustment factor

The factor c converts the measured deflection to the maximum deflection that would have occurredif the test were performed at the most critical time of the year. Numerically, it is equal to the ratio ofmeasured deflection to the corresponding deflection measurement if it were to be made during the critical

FIGURE 62.41 Rigid pavement thickness for DC 10-10, 10CF. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

ANNUAL DEPARTURES

SLA

B T

HIC

KN

ES

S ,i

n

FLE

XU

RA

L S

TR

EN

GT

H ,

psi

900

850

800

750

700

650

600

550

500

500

200

300

k=50 pci

200,000

300,

00

400,

000

450,

000

Ibs.

1200 3000 6000 15,000 25,000

DC 10-10, 10 CFCONTACT AREA = 294 sq. in.DUAL SPACING = 54 in.TANDEM SPACING = 64 in.

0278 E


1 Ib = 0.454 kg 1 pci = 0.272 MN/m3

22 23 2426

26

27

25

24

23

22

21

20

19

18

17

16

15

14

13

12

11

10

9

8

25

24

23

22

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12

11

10

9

8

23

22

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13

12

11

10

9

8

7

22

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11

10

9

8

7

21

20

19

18

17

16

15

14

13

12

11

10

9

8

7

100

dr dm 2s–( ) Fc=


period. It can be established from historical records or derived from engineering judgment when norecord is available.

F is determined from Fig. 62.46 with two inputs: thickness of untreated granular base and mean pavementtemperature. The estimation of mean pavement temperature requires information of the pavement surfacetemperature at the time of test and the 5-day mean air temperature computed from the maximum andminimum air temperature for each of the 5 days prior to the date of deflection testing. Fig. 62.47 is usedto obtain temperature at the middepth and bottom of the pavement. Next, the surface temperature, mid-depth temperature, and bottom temperature are averaged to provide the mean pavement temperature.

Having computed the representative rebound deflection, Fig. 62.48 is used to determine the requiredoverlay thickness. The design ESAL is estimated by means of the procedure described under the heading oftraffic-loading computation.

Example 62.25

Rebound deflection measurements made at 12 randomly selected locations on an old asphalt pavementusing Benkelman beam produced the following net rebound deflections in in.: 0.038, 0.035, 0.039, 0.039,0.039, 0.039, 0.044, 0.044, 0.037, and 0.036. The temperature of pavement surface was found to be 131˚F.The extreme air temperatures in the previous 5 days were (88˚F, 75˚F), (86˚F, 75˚F), (90˚F, 77˚F), (88˚F,77˚F), and (88˚F, 75˚F). The thickness of the asphalt layer was 6 in. The thickness of untreated granularbase was 12 in. Determine the overlay thickness required to carry additional ESAL of 5 ¥ 106.

Mean deflection d m = 0.0391 in. and standard deviation s = 0.0029. Five-day mean air temperature =81.9˚F. From Fig. 62.47, pavement layer middepth temperature T1 = 105˚F and bottom temperature T2 =100˚F. Mean pavement temperature = 112˚F. From Fig. 62.46, F = 0.82.

Assuming a critical period factor of c = 0.9, dr = {0.0391 + 2(0.0029)} ¥ (0.82)(0.9) = 0.0331 in. Fordesign ESAL of 5 ¥ 106, read from Fig. 62.48, the overlay thickness is 3 in.

FIGURE 62.42 Rigid pavement thickness for DC 10-30, 30CF, 40, and 40CF. (Source: Federal Aviation Administra-tion. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. With permission.)

FLE

XU

RA

L S

TR

EN

GT

H ,

psi

900

850

800

750

700

650

600

550

500k=50 pci

200,000

300,

000

400,

000

500,

000

600,

000

Ibs

DC-10-30, 30 CF, -40, 40 CFCONTACT AREA = 331 sq. in.DUAL SPACING = 54 in.TANDEM SPACING = 64 in.CENTER GEAR SPACING = 37.5 in.

0478 E

100

200

300

500

ANNUAL DEPARTURES1200 3000 600015,000 25,000

SLA

B T

HIC

KN

ES

S, i

n.

26

27

25

24

23

22

21

20

19

18

17

16

15

14

13

12

11

10

9

8

26

25

24

23

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21

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19

18

17

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14

13

12

11

10

9

8

24

23

22

21

20

19

18

17

16

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14

13

12

11

10

9

8

7

23

22

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20

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12

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10

9

8

7

22

21

20

19

18

17

16

15

14

13

12

11

10

9

8

7


1 Ib =0.454 kg 1 pci = 0.272 MN/m3


AI Design Procedure for Flexible Overlay on Rigid Highway Pavement

Two design procedures are presented by the Asphalt Institute [1983], namely, the effective thicknessprocedure and the deflection procedure.

AI Effective Thickness Procedure

The component analysis procedure described earlier for asphalt overlay on flexible pavement also appliesfor the design of asphalt overlay on concrete pavement. The same table (Table 62.25) is used for both.

FIGURE 62.43 Effect of subbase on subgrade modulus. (Source: Federal Aviation Administration. 1978. AirportPavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C, p. 25. With permission.)

EF

FE

CT

IVE

K O

N T

OP

OF

SU

BB

AS

E

( M

N/m

3 )

450

100

200

300

400

500

5 6 7 8 9 10THICKNESS OF SUBBASE, INCHES

BANK - RUN SAND & GRAVEL (P1 < 6)

11 12 13

34323028262422

( cm )

2018161412

14

125

100

( M

N /

m3

)

LB /

IN3

75

50

40

30

20

15

450

100

200

300

400

500

5 6 7 8 9 10THICKNESS OF SUBBASE, INCHES

WELL - GRADED CRUSHED AGGREGATE

11 12 13

343230282624

K = 300 (81)

K = 200 (54)

K = 100 (27)

SUBGRADE K = 50 (14)

K = 300 (81)

K = 200 (54)

K = 100 (27)

SUBGRADE K = 50 (14)

22

( cm )

2018161412

14

125

100

LB /

IN3

75

50

40

30

20

15


AI Deflection-Based Procedure

Deflection measurements are made using Benkelman beam or other devices at the following locations:(a) the outside edge on both sides of two-lane highways; (b) the outermost edge of divided highways;and (c) corners, joints, cracks, and deteriorated pavement areas.

FIGURE 62.44 Effect of stabilized subbase on subgrade modulus. (Source: Federal Aviation Administration. 1978.Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C, p. 64. With permission.)

TABLE 62.22 FAA Recommended Maximum Joint Spacings

Slab Thickness Transverse Spacing Longitudinal Spacing

<9 in. (23 cm) 15 ft (4.6 m) 12.5 ft (3.8 m)9–12 in. (23–31 cm) 20 ft (6.1 m) 20 ft (6.1 m)

>12 in. (31 cm) 25 ft (7.6 m) 25 ft (7.6 m)

Source: Federal Aviation Administration. 1978. Airport PavementDesign and Evaluation. Reprinted from FAA Advisory Circular. ReportFAA/AC-150/5320-6C. 7 December 1978; NTIS Accession No. AD-A075537/1.

TABLE 62.23 Dowel Bar Dimensions and Spacings

Slab Thickness Diameter Length Spacing

6–7 in. (15–18 cm) 0.75 in. (20 mm) 18 in. (46 cm) 12 in. (31 cm)8–12 in. (21–31 cm) 1 in. (25 mm) 19 in. (48 cm) 12 in. (31 cm)

13–16 in. (33–41 cm) 1.25 in. (30 mm)* 20 in. (51 cm) 15 in. (38 cm)17–20 in. (43–51 cm) 1.50 in. (40 mm)* 20 in. (51 cm) 18 in. (46 cm)21–24 in. (54–61 cm) 2 in. (50 mm)* 24 in. (61 cm) 18 in. (46 cm)

* Dowels may be a solid bar or high-strength pipe. High-strength pipe dowelsmust be plugged on each end with a tight-fitting plastic cap or with bituminousor mortar mix.

Source: Federal Aviation Administration. 1978. Airport Pavement Design andEvaluation. Reprinted from FAA Advisory Circular. Report FAA/AC-150/5320-6C.7 December 1978; NTIS Accession No. AD-A075 537/1.

K O

N T

OP

OF

SU

BB

AS

E L

B/I

N3

500

400

300

200

10090

80

70

60

50

( M

N /

m3

)

50

40

35

30

25

20

4 5 6 7 8THICKNESS OF SUBBASE, INCHES

9 10 11 12

15

12 14 16 18 20

( CM )

22 24 26 28 30

120

1009080

70

60

K = 300 (81)

K = 200 (54)

K = 100 (27)

SUBGRADE K = 50 (1

4)


For JPCP and JRCP the differential vertical deflection at joints should be less than 0.05 mm (0.002 in.)and the mean deflection should be less than 0.36 mm (0.014 in.). For CRCP, Dynaflect deflections of15 to 23 mm (0.0006 to 0.0009 in.) or greater lead to excessive cracking and deterioration. Undersealingor stabilization is required when the deflection exceeds 15 mm (0.0006 in.).

Dense-graded asphalt concrete overlay can reduce deflections by 0.2% per mm (5% per in.) ofthickness. However, depending on the mix type and environmental conditions, deflection may be as highas 0.4 to 0.5% per mm (10 to 12% per in.). If a reduction of 50% or more of deflection reduction isrequired, it is more economical to apply undersealing before overlay is considered. For a given slab lengthand mean annual temperature differential, the required overlay thickness is selected from Fig. 62.49. Thethicknesses are provided to minimize reflective cracking by taking into account the effects of horizontaltensile strains and vertical shear stresses.

The design chart has three sections — A, B, and C. In section A, a minimum thickness of 100 mm(4 in.) is recommended. This thickness should reduce the deflection by an estimated 20%. In sections Band C, the thicknesses may be reduced if the pavement slabs are shortened by breaking and seating(denoted as alternative 2 in Fig. 62.49) to reduce temperature effects. This is recommended as an overlaythickness approaches the 200- to 225-mm (8- to 9-in.) range. Another alternative is the use of a crackrelief layer (denoted as alternative 3 in Fig. 62.49). A recommended crack relief structure is a 3.5-in.-thick layer of coarse, open-graded hot mix containing 25 to 35% interconnecting voids and made up of100% crushed material. It is overlain by a dense-graded asphalt concrete surface course (at least 1.5 in.thick) and a dense-graded asphalt concrete leveling course (at least 2 in. thick).

Example 62.26

The vertical deflections measured by a Benkelman beam test at a joint of a portland cement concretepavement are 0.042 and 0.031 in. The pavement has a slab length of 40 ft. Design an asphalt concreteoverlay on the concrete pavement. The design temperature differential is 80˚F.

FIGURE 62.45 PSI-based conversion factors for determining effective thickness. (Source: Asphalt Institute. 1983a.Asphalt Overlays for Highway and Street Rehabilitation. Manual Series MS-17, p. 51. With permission.)

TABLE 62.24 Asphalt Institute Equivalency Factors for Converting Layers of Other Material Types to Equivalent Thickness of Asphalt Concrete

Material Type Equivalency Factor Ei

Asphalt concrete 1.00Type I emulsified asphalt base 0.95Type II emulsified asphalt base 0.83Type III emulsified asphalt base 0.57

Source: Asphalt Institute. 1983. Asphalt Overlays for High-way and Street Rehabilitation. MS-17. p. 52. With permission.

CO

NV

ER

SIO

N F

AC

TO

R0.9

0.8

0.7

0.6

0.5

1.52.0

LINE ALINE B

2.5PRESENT SERVICEABILITY INDEX (PSI)

3.03.5


Mean vertical deflection is 0.0365 in., and the differential deflection is 0.009 in. Alternative 1. Thickoverlay: From Fig. 62.49, more than 9 in. of overlay is required. Use either alternative 2 or 3. Alternative2. Break and seat to reduce slab length: Break slab into 20-ft sections. From Fig. 62.49, 5.5 in. of overlayis required. For the overlaid pavement, mean vertical deflection = 0.0365 – {(5.5 ¥ 5%) ¥ 0.0365} =0.0265 > 0.014 in., and vertical differential deflection = 0.009 – {(5.5 ¥ 5%) ¥ 0.009} = 0.0025 > 0.002 in.Undersealing is needed. Alternative 3. Crack relief layer: Use 3.5-in. crack relief course with 1.5-in. surfacecourse and 2-in. leveling course, giving a total of 7-in. asphalt concrete courses. Similar procedure ofdeflection checks to those for alternative 2 indicates that undersealing is required.

PCA Design Procedure for Concrete Overlay on Concrete Highway Pavement

Depending on the bonding between the overlay and the existing pavement slab, concrete overlays can beclassified into three types: bonded, unbonded, and partially bonded. Bonded overlay is achieved byapplying a thin coating of cement grout before overlay placement. The construction of unbonded overlayinvolves the use of an unbonding medium at the surface of the existing pavement. Asphaltic concreteand sand asphalt are common unbonding media. Partially bonded overlay refers to a construction in

TABLE 62.25 Conversion Factors C for Determining Effective Thickness

Case Description Factor C

I (a) Native subgrade in all cases. 0.0(b) Improved subgrade, predominantly granular material, may contain some silt and clay but have

P.I. of 10 or less.(c) Lime-modified subgrade constructed from high-plasticity soils, P.I. greater than 10.

II Granular subbase or base, reasonably well-graded, hard aggregates with some plastic fines and CBR not less than 20. Use upper part of range if P.I. is 6 or less, lower part of range if P.I. is more than 6.

0.1–0.2

III Cement or lime–fly ash stabilized subbases and bases constructed from low-plasticity soils, P.I. of 10 or less.

0.2–0.3

IV (a) Emulsified or cutback asphalt surfaces and bases that show extensive cracking, considerable raveling or aggregate degradation, appreciable deformation in the wheel paths, and lack of stability.

0.3–0.5

(b) Portland cement concrete pavements (including those under asphalt surfaces) that have been broken into small pieces 2 ft (0.6 m) or less in maximum dimension prior to overlay construction. Use upper part of range when subbase is present, lower part of range when slab is on subgrade.

(c) Cement or lime–fly ash stabilized bases that have developed pattern cracking, as shown by reflected surface cracks. Use upper part of range when cracks are narrow and tight, lower part of range with wide cracks, pumping, or evidence of instability.

V (a) Asphalt concrete surface and base that exhibit appreciable cracking and crack patterns. 0.5–0.7(b) Emulsified or cutback asphalt surface and bases that exhibit some fine cracking, some raveling or

aggregate degradation, and slight deformation in the wheel paths but remain stable.(c) Appreciably cracked and faulted portland cement concrete pavement (including such under asphalt

surfaces) that cannot be effectively undersealed. Slab fragments, ranging in size from approximately 10 to 160 ft2 (1 to 4 m2), have been well seated on the subgrade by heavy pneumatic-tired rolling.

VI (a) Asphalt concrete surfaces and bases that exhibit some fine cracking, have small intermittent cracking patterns and slight deformation in the wheel paths but remain stable.

0.7–0.9

(b) Emulsified or cutback asphalt surface and bases that are stable, generally uncracked, show no bleeding, and exhibit little deformation in the wheel paths.

(c) Portland cement concrete pavements (including such under asphalt surfaces) that are stable and undersealed, have some cracking but contain no pieces smaller than about 10 ft2 (1 m2).

VII (a) Asphalt concrete, including asphalt concrete base, generally uncracked, and with little deformation in the wheel paths.

0.9–1.0

(b) Portland cement concrete pavement that is stable, undersealed, and generally uncracked.(c) Portland cement concrete base, under asphalt surface, that is stable, nonpumping, and exhibits

little reflected surface cracking.

Source: Asphalt Institute. 1983a. Asphalt Overlays for Highway and Street Rehabilitation. Manual Series MS-17, pp. 54–55.With permission.


which the overlay is placed directly on the existing pavement without the application of a bonding orunbonding medium.

Design of Unbonded Overlay

The procedure selects an overlay thickness that, under the action of an 18-kip (80-kN) single-axle load,would have an edge stress in the overlay equal to or less than the corresponding edge stress in an adequately

FIGURE 62.46 Chart for determining temperature correction factor F. (Source: Asphalt Institute. 1983a. AsphaltOverlays for Highway and Street Rehabilitation. Manual Series MS-17. With permission.)

FIGURE 62.47 Estimation of pavement temperature. (Source: Asphalt Institute. 1983a. Asphalt Overlays for Highwayand Street Rehabilitation. Manual Series MS-17. With permission.)

UNTREATED GRANULARBASE THICKNESS

DEFLECTION ADJUSTMENT FACTORS FORBENKELMAN BEAM TESTING

TEMPERATURE ADJUSTMENT FACTOR (F)

ME

AN

PA

VE

ME

NT

TE

MP

ER

AT

UR

E

300.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4

0¢¢ 3¢¢ 25¢¢

25¢¢

40

50

60

70

80

90

100

110

120

0¢¢3¢¢

6¢¢12¢¢19¢¢

TE

MP

ER

AT

UR

E A

T D

EP

TH

, ∞F

TE

MP

ER

AT

UR

E A

T D

EP

TH

, ∞C

-10

-10 0 10 20 30 40 50 60 70 80 90 100 110 120

PAVEMENT SURFACE TEMPERATURE PLUS 5-DAY MEAN AIR TEMPERATURE, ∞C

PAVEMENT SURFACE TEMPERATURE PLUS 5-DAY MEAN AIR TEMPERATURE, ∞F

0

10

20

30

40

50

60

700 20 40 60 80 100 120 140 160 180 200

DEPTH IN PAVEMENT 25mm (1 in.)

50mm (2 in.)

150mm(6 in.)

200mm(8 in.)

300mm(12 in.)

100mm (4 in.)

220 240 260160

140

120

100

80

60

40

20


designed new pavement under the same load. Design charts in Fig. 62.50 are provided for the followingthree cases:

Case 1. Existing pavement exhibiting a large amount of midslab and corner cracking; poor load transferat cracks and joints.

Case 2. Existing pavement exhibiting a small amount of midslab and corner cracking; reasonably goodload transfer across cracks and joints; localized repair performed to correct distressed slabs.

Case 3. Existing pavement exhibiting a small amount of midslab cracking; good load transfer acrosscracks and joints; loss of support corrected by undersealing.

The design charts were obtained from computer analysis of pavements assuming modulus of elasticityof 5 ¥ 106 psi (35 GPa) for overlays and 3 ¥ 106 and 4 ¥ 106 psi (21 and 28 GPa) for existing pavements.If a tied shoulder is provided, the thickness of the overlay may be reduced by 1 in. (25 mm) subject tothe minimum thickness requirement of 6 in. (150 mm).

Example 62.27

Design a concrete overlay for an existing 10-in.-thick concrete pavement if the required new single-slabthickness is 10 in.

For case 1, TOL = 9 in., from Fig. 62.50(a). For case 2, TOL = 6 in., from Fig. 62.50(b). For case 3, whereTOL < 6 in., from Fig. 62.50(c), use minimum 6 in.

Design of Bonded Overlay

The same structural equivalency concept as for unbonded overlay is adopted in the design for bondedoverlay, except that the comparison is now made between the stress-to-strength ratios of the new andthe overlaid pavements. The design chart (Fig. 62.51) has three curves for three different ranges of moduliof rupture Sc of the existing concrete. Sc may be estimated from the effective splitting tensile strength fte

as follows:

FIGURE 62.48 Design chart for overlay thickness. (Source: Asphalt Institute. 1983a. Asphalt Overlays for Highwayand Street Rehabilitation. Manual Series MS-17. With permission.)

10,000,000ESAL

5,000,000

2,000,000

1,000,000

500,000

200,000

100,000

50,00020,00010,0005,000

16

14

12

10

8

6

4

2

0

0 2 4 6 8 10 12 14 16 18

Representative Rebound Deflection (0.01 in.)

Ove

rlay

Thi

ckne

ss (

in.)

20,000,00050,000,000


(62.41)

(62.42)

where ft is the average value of splitting tensile strength of cored specimens determined according toASTM Test Method C496, s is the standard deviation of splitting tensile strength, and A is a regressionconstant ranging from 1.35 to 1.55. An A value of 1.45 is suggested in the absence of local information.One core should be taken every 300 to 500 ft (91 to 152 m) at midslab and about 2 ft (0.6 m) from theedge of the outside lane. The 0.9 factor in Eq. (62.42) relates the strength of the concrete specimens tothat near or at the edge.

Example 62.28

A 9-in.-thick concrete pavement is to be strengthened to match the capacity of a new 10-in.-thick concretepavement. What is the required thickness of bonded overlay if the existing concrete has a flexural strengthof 450 psi?

Using curve 3 of Fig. 62.51, the thickness of existing pavement plus overlay = 11.5 in. Overlay thickness= 11.5 – 9 = 2.5 in.

FAA Design Procedure for Flexible Overlay on Flexible Airport Pavement

The design method of FAA [1978] is similar in the equivalent thickness concept to the PCA method thatadopts the component analysis. The FAA equivalency factors are shown in Table 62.13. A high-qualitymaterial may be converted to a lower-quality material — for example, surfacing to base and base to

FIGURE 62.49 Thickness of asphalt overlay on concrete pavement. (Source: Asphalt Institute. 1983a. Asphalt Over-lays for Highway and Street Rehabilitation. Manual Series MS-17, p. 79. With permission.)

TEMPERATURE DIFFERENTIAL ∑ (∞F)

TEMPERATURE DIFFERENTIAL (∞C)

SlabLength

(Ft)

10or Less

15

30

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

115mm(4.5 in.)

125mm(5 in.)

150mm(6 in.)

20

25

30

35

40

45

50

60

SlabLength

(m)

3

4.5

6

7.5

9

10.5

12

13.5

15

18

17

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

200mm(8 in.)

175mm(7 in.)

150mm(6 in.)

140mm(5.5 in.)

115mm(4.5 in.)

100mm(4 in.)

40

22

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

125mm(5 in.)

150mm(6 in.)

175mm(7 in.)

190mm(7.5 in.)

215mm(8.5 in.)

UseAlternative

2 or 3

50

28

100mm(4 in.)

100mm(4 in.)

100mm(4 in.)

125mm(5 in.)

150mm(6 in.)

225mm(9 in.)

200mm(8 in.)

175mm(7 in.)

UseAlternative

2 or 3

UseAlternative

2 or 3

60

33

100mm(4 in.)

100mm(4 in.)

125mm(5 in.)

150mm(6 in.)

175mm(7 in.)

215mm(8.5 in.)

UseAlternative

2 or 3

UseAlternative

2 or 3

UseAlternative

2 or 3

UseAlternative

2 or 3

70

39

100mm(4 in.)

100mm(4 in.)

140mm(5.5 in.)

175mm(7 in.)

200mm(8 in.)

UseAlternative

2 or 3

UseAlternative

2 or 3

UseAlternative

2 or 3

UseAlternative

2 or 3

UseAlternative

2 or 3

80

44

fte f t 1.65s–=

Sc 0.9A fte=


FIGURE 62.50 PCA design charts for unbonded overlays. (Source: Tayabji, S.D. and Okamoto, P.A., Proceedings 3rdInt. Conf. on Concrete Pavement Design and Rehabilitation, Purdue University, April 23–25, 1985, pp. 367–379. Withpermission.)

8 9 10 11 12 13 14

k = 100—300 pciOverlay

Thickness (in.)

Case 2

BaseLine

4 5 6

Existing Pavement Thickness (in.)

New Full-Depth Slab Thickness (in.)

( b ) Case 2

7 8 9 10

6

78

9101112

OverlayExisting

8 9 10 11 12 13 14

k = 100—300 pciOverlayThickness (in.)

Case 1

BaseLine

4 5 6


New Full-Depth Slab Thickness (in.)

( a ) Case 1

7 8 9 10

6

7

89

101112

OverlayExisting


New Full-Depth Slab Thickness (in.)8 9 10 11 12 13 14

4 5 6 7 8 9 10

(c) Case 3

Case 3

OverlayThickness (in.)

76

8

91011

BaseLine

k = 100—300 pci

OverlayExisting


subbase. A material may not be converted to a higher-quality material. The overlay thickness is equal tothe difference between the total equivalent layer thicknesses of the existing pavement and the correspond-ing required layer thickness of a new pavement. The minimum overlay thickness allowed is 3 in. (75 mm).

Example 62.29

An existing asphalt concrete airport pavement has 4-in. bituminous surface course, 7-in. base course,and 14-in. subbase. The CBR of the subgrade is 8 and that of the subbase is 12. Provide an overlay tostrengthen the pavement for 6000 annual departures of a design aircraft (dual-wheel landing gear) withmaximum weight of 100,000 lb.

From Fig. 62.17, a new pavement requires 30 in. total thickness based on subgrade CBR of 8 and athickness of 17 in. above subbase, based on subbase CBR of 12. Using 4 in. of asphalt concrete surfacecourse, the base layer is (17 – 4) = 13 in. The deficiency of thickness of the existing pavement is all inthe base layer. Assuming that the existing asphalt concrete surface course can be converted to base at anequivalency ratio of 1.4 to 1 (see Table 62.13), the thickness of asphalt base required = (13 – 7)/1.4 =4.3 in. An additional 0.3 in. of asphalt concrete base is needed. The total thickness of overlay = (4 in. ofnew surface course) + 0.3 = 4.3 in. Use a 4.5-in. overlay.

FAA Design Procedure for Flexible Overlay on Concrete Airport PavementThe equation for computing bituminous overlay thickness T is

(62.43)

where F = factor to be obtained from Fig. 62.52

h = single thickness of rigid pavement required for design condition, in inches (use the exactvalue without rounding off )

FIGURE 62.51 PCA design charts for bonded overlays. (Source: Tayabji, S.D. and Okamoto, P.A., Proceedings 3rdInt. Conf. on Concrete Pavement Design and Rehabilitation, Purdue University, April 23–25, 1985, pp. 367–379. Withpermission.)

8 9 10 11 12 13 14

8 9 10 11 12 13 14

1 2 3

Full Depth Stab Thickness, In

Total Thickness of Existing Pavement and Resurfacing. in.

Base Line

Curve No.

123

Existing PavementFlexural Strength, psi

526 to 575476 to 525426 to 475

ResurfacingExisting Pavement

T inches( ) 2.5 F( h Cb he )–=


Cb = condition factor for base pavement, 0.75 £ Cb £ 1.0he = thickness of existing rigid pavement, in inches

The F factor is related to the degree of cracking that will occur in the base pavement. It has a valueless than one, indicating that the entire single concrete slab thickness is not needed because a bituminousoverlay pavement is allowed to crack and deflect more than a conventional rigid pavement. Cb is anassessment of the structural integrity of the existing pavement. Cb is 1.0 when the existing slabs containnominal initial cracking and 0.75 when the slabs contain multiple cracking.

Example 62.30

An existing 10-in. concrete pavement has a condition factor Cb of 0.8. The subgrade k is 200 pci. Providea bituminous overlay to strengthen the pavement to be equivalent to a single rigid pavement thicknessof 12 in. for a design traffic of 3000 annual departures.

From Fig. 62.52, F = 0.92. By Eq. (62.43), overlay thickness t = 2.5{(0.92 ¥ 12) – (0.8 ¥ 10)} = 7.6 in.Use an 8-in.-thick overlay.

FAA Design Procedure for Concrete Overlay on Concrete Airport Pavement

The design of concrete overlay requires an assessment of the structural integrity of the existing pavementand the thickness of a new concrete pavement on the existing subgrade. The design equations are

(62.44)

(62.45)

(62.46)

where Cr = 1.0 for existing pavement in good condition — some minor cracking evident but nostructural defects

Cr = 0.75 for existing pavement containing initial corner cracks due to loading but no progres-sive cracking or joint faulting

Cr = 0.35 for existing pavement in poor structural condition — badly cracked or crushed andfaulted joints.

FIGURE 62.52 Graph for determination of F factor. (Source: Federal Aviation Administration. 1978. Airport Pave-ment Design and Evaluation. Advisory Circular AC No. 150/5320-6C, p. 105. With permission.)

F -

FA

CT

OR

MODULUS OF SUBGRADE REACTION

0.6

0.7

0.8

0.9

1.00 100 200

pci

300 400

25000

1500060003000

[100][75][50]

MN/m3

[25][0]

1200 ANNUALDEPARTURES

Unbonded overlay T h 2 Cr h e2–( )1 2§=

Partially bonded overlay T h1.4 C r he1.4–( )1 1.4§=

Bonded overlay T h he–=


The variables h and he are the thicknesses of new and existing pavements, respectively. The use ofpartially bonded overlay — which is constructed directly on an existing pavement without debondingmedium (such as a bituminous leveling course) — is not recommended for an existing pavement withCr less than 0.75. Bonded overlays should be used only when the existing rigid pavement is in goodcondition. The minimum bonded overlay thickness is 3 in. For partially and unbonded overlays, theminimum thickness is 5 in.

Example 62.31

An existing 10-in.-thick concrete airport pavement with Cr = 1.0 is to be strengthened to match thecapacity of a new 12-in. rigid pavement. Determine the required thickness of bonded, partially bonded,and unbonded overlays.

Unbonded overlay. T = = 6.6 in. Use 7 in.

Partially bonded overlay. T = = 4.1 in. Use 5 in. (min).

Bonded overlay. T = 12 – 10 = 2 in. Use 3 in. (min).

Defining Terms

Asphalt pavement (asphalt concrete pavement, bituminous pavement) — The most commonform of flexible pavement in which the surface course is constructed of asphaltic (or bituminous)mixtures.

Base course — The layer of selected material in a pavement structure placed between a subbase and asurface course.

Concrete pavement — The most common form of rigid pavement, in which the top slab is constructedof portland cement concrete.

Flexible pavement — A pavement structure that does not distribute traffic load to the subgrade bymeans of slab action but mainly through spreading of the load by providing sufficient thicknessof the pavement structure.

Overlay — A new surface layer laid on an existing pavement to improve the latter’s load-carryingcapacity.

Pavement structure — A structure consisting of one or more layers of selected materials constructedon prepared subgrade to designed strength and thickness(es) for the purpose of supportingtraffic.

Rigid pavement — A pavement structure that distributes traffic loads to the subgrade by means of slabaction through its top layer of high-bending resistance.

Subbase — The layer of selected material in a pavement structure placed between the subgrade and thebase or surface course.

Subgrade — The top surface of graded foundation soil, on which the pavement structure is constructed.Surface course — The top layer of a pavement structure placed on the base course, the top surface of

which is in direct contact with traffic loads.

References

AASHTO. 1972. AASHTO Interim Guide for Design of Pavement Structures. American Association of StateHighway and Transportation Officials, Washington, D.C.

AASHTO. 1989. Standard Specifications for Transportation Materials and Methods of Sampling and Testing.Part I and II. American Association of State Highway and Transportation Officials, Washington,D.C.

AASHTO. 1993. AASHTO Guides for Design of Pavement Structures. American Association of StateHighway and Transportation Officials, Washington, D.C.

ACI. 1977. Building Code Requirements for Reinforced Concrete. American Concrete Institute, Detroit, MI.

122 102–

121.4 101.4–1.4


Asphalt Institute. 1983a. Asphalt Overlays for Highway and Street Rehabilitation. Manual Series No. 17.Lexington, KY.

Asphalt Institute. 1983b. Asphalt Technology and Construction Practices. Educational Series ES-I, 2nd ed.Lexington, KY.

Asphalt Institute. 1991. Thickness Design — Asphalt Pavements for Highways & Streets. Manual Series No. 1.Lexington, KY.

ASTM. 1992. Annual Books of ASTM Standards. American Society for Testing and Materials, Philadelphia,PA.

Boussinesq, J. 1885. Application des Potentiels a l’etude de l’equilibre et du Mouvement des Solids Elastiques.Gauthier-Villars, Paris.

FAA. 1978. Airport Pavement Design and Evaluation. Advisory Circular AC No. 150/5320-6C. FederalAviation Administration.

Fwa, T.F., and Li, S. 1994. Estimation of lane distribution of truck traffic for pavement design. Paperaccepted for publication. Journal of Transportation Engineering.

Fwa, T.F., Shi, X.P., and Tan, S.A. 1993. Load-Induced Stresses and Deflections in Concrete Pavement —Analysis by Rectangular Thick-Plate Model. CTR Technical Report CTR-93-5. Centre for Transpor-tation Research, Faculty of Engineering, National University of Singapore.

Fwa, T.F., and Sinha, K.C. 1985. A Routine Maintenance and Pavement Performance Relationship Modelfor Highways. Joint Highway Research Project Report JHRP-85-11. Purdue University, West Lafay-ette, IN.

Highway Research Board. 1962. The AASHO Road Test, Report 5 — Pavement Research. HRB SpecialReport 61E. Washington, D.C.

PCA. 1984. Thickness Design for Concrete Highway and Street Pavements. Portland Cement Association,Skokie, IL.

Shi, S.P., Tan, S.A., and Fwa, T.F. 1994. Rectangular plate with free edges on a Pasternak foundation.Journal of Engineering Mechanics. 120(5):971–988.

Tayabji, S.D. and Okamoto, P.A. 1985. Thickness design of concrete resurfacing. Proc. 3rd Int. Conf. onConcrete Pavement Design and Rehabilitation, April 23–25, Purdue University, West Lafayette, IN,pp. 367–379.

Van Til, C.J., McCullough, B.F., Vallerga, B.A., and Hicks, R.G. 1972. Evaluation of AASHO Interim Guidesfor Design of Pavement Structures. NCHRP Report 128. Highway Research Board, Washington, D.C.

Westergaard, H.M. 1926. Stresses in concrete pavements computed by theoretical analysis. Public Roads.7(2):25–35.

Westergaard, H.M. 1933. Analytical tools for judging results of structural tests of concrete pavements.Public Roads. 14(10).

Westergaard, H.M. 1948. New formulas for stresses in concrete pavements of airfield. ASCE Transactions.Vol. 113.

Yoder, E.J., and Witczak, M.W. 1975. Principles of Pavement Design, 2nd ed. John Wiley & Sons, New York.

Further Information

A widely quoted reference to the basics of practical design of highway and airport pavements is Principlesof Pavement Design, by E.J. Yoder and M.W. Witczak. Although the described design methods by variousagencies are outdated, the book is still a valuable reference on the requirements of pavement constructionand design.

Detailed descriptions of pavement design methods, pavement material, and construction requirementsby various organizations are available in their respective publications. The Asphalt Institute publishes amanual series addressing bituminous pavement-related topics — including thickness design, pavementrehabilitation and maintenance, pavement drainage, hot-mix design, and paving technology. Additionalinformation concerning topics related to portland cement concrete pavement is found in publicationsby the Portland Cement Association and American Concrete Institute.


The latest developments in various aspects of pavement design are reported in a number of technicaljournals in the field. The most important are the Journal of Transportation Engineering, publishedbimonthly by the American Society of Civil Engineers, and Transportation Research Records, publishedby the Transportation Research Board. There are about 40 issues of Transportation Research Recordspublished each year, each collecting a group of technical papers addressing a specialized area of trans-portation engineering.

There are several major conferences that focus on highway and airport pavements. The InternationalConference on Structural Design of Asphalt Pavements has been held once every five years since 1962.The seventh conference, in 1992, was named International Conference on Asphalt Pavements: Design,Construction and Performance to reflect the added scope of the conference. The proceedings of theconferences document advances in areas of asphalt pavement technology. Another conference, the Inter-national Conference on Concrete Pavement Design and Rehabilitation, focuses on the development ofconcrete pavement technology. It has been organized once every four years since 1977 by Purdue Uni-versity. There is also the International Conference on the Bearing Capacity of Roads and Airfields, heldat intervals of four years since 1982. Other related publications are the Proceedings of the World RoadCongress, published by the Permanent International Association of Road Congress, and the Proceedingsof the Road Congress of the International Road Federation.

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