Analytical Study and Design of Flexible Pavement

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Bedground

1. Subgrade (Road Bed)

The subgrade is usually the natural material located along the horizontal alignment of the road and serves as the foundation of the pavement structure.

2. Subbase Course

Located immediately above the subgrade, the subbase component consists of material of a superior quality to that which is generally used for subgrade construction.

3. Base Course

Located immediately above the subbase, if a subbase course is not used. This course usually consists of granular materials such as crushed stone, crushed or uncrushed slag, crushed or uncrushed gravel, and sand.

4. Surface Course

Constructed immediately above the base course, usually consists a mixture of aggregates and asphalt. It should be capable of withstanding high tire pressures, resisting abrasive forces due to traffic, providing a skid-resistant driving surface, and preventing the penetration of surface water into the underlying layers.

I. GENERAL PRINCIPLES OF FLEXIBLE PAVEMENT DESIGN

Assumed initially that the subgrade layer is infinite in both the horizontal and vertical directions, whereas the other layers are finite in the vertical direction and infinite in the horizontal direction.

In the design of flexible pavements, the pavement structure usually is considered as a multilayered elastic system, with the material that may include the modulus of elasticity (E), the resilient modulus (R), and the Poisson ratio ( ν ) .

Applying of a wheel load causes a stress distribution; the maximum vertical stresses are compressive and occur directly under the wheel load. These decrease with an increase in depth from the surface.

(a) Typical Cross Section of Flexible Pavement (b) Load Transmission in Flexible Pavement

II. AASHTO DESIGN METHOD CONSIDERATIONS

a) Pavement Performance

  • Structural performance is related to the physical condition of the pavement with respect to factors that hurt the capability of the pavement to carry the traffic load.

  • These factors include cracking, faulting, unraveling, and so forth.

Functional performance is an indication of how effectively the pavement serves the user.

  • The main factor considered under functional performance is riding comfort.
  • To quantify pavement functional performance, a concept known as the serviceability performance was developed.

And Under this concept, a procedure was developed to determine the present serviceability index (PSI) of the pavement, based on its roughness and distress, which were measured in terms of the extent of cracking, patching, and rut depth for flexible pavements.

Two serviceability indices are used in the design procedure:

  1. The initial serviceability index (pi) is the serviceability index immediately after the Construction of the pavement.

  2. The terminal serviceability index (pt) is the minimum acceptable value before reconstruction is necessary.

b) Traffic Load

In the AASHTO design method, the traffic load is determined in Terms of the number of repetitions of an 18,000-lb (80 KN) the Single-axle load applied to the pavement.

This is usually referred to as: The Equivalent Single-Axle Load (ESAL).

A general equation for the accumulated ESAL for each category of axle load is

E S A L = A A D T × N × F E i × G r n × 3 6 5 × f d

Where:

ESAL = Equivalent Accumulated 18,000-lb (80 KN) single-axle load.

AADT = First year annual average daily traffic.

N = Number of axles on each vehicle.

F E i = load equivalency factor for Single and Tandem axle.

G r n = Growth factor for a given growth rate r and design period n.

365 = Convert from day to year.

f d = Design lane factor.

NameEquation
ESAL Design laneAADT × N × FEi × Grn × 365 × fd
Total ESALAADT × N × FEi × Grn × 365
First Year ESAL Design laneAADT × N × FEi × 365 × fd
Total ESAL First YearAADT × N × FEi × 365
Daily ESAL Design LaneAADT × N × FEi × fd
Total Daily ESALAADT × N × FEi

Table 19.4: Growth Factors

Design Period, Years (n)Annual Growth Rate, Percent (r)
No Growth24567810
11.01.01.01.01.01.01.01.0
22.02.022.042.052.062.072.082.10
33.03.063.123.153.183.213.253.31
44.04.124.254.314.374.444.514.64
55.05.205.425.535.645.755.876.11
66.06.316.636.806.987.157.347.72
77.07.437.908.148.398.658.929.49
88.08.589.219.559.9010.2610.6411.44
99.09.7510.5811.0311.4911.9812.4913.58
1010.010.9512.0112.5813.1813.8214.4915.94
1111.012.1713.4914.2114.9715.7816.6518.53
1212.013.4115.0315.9216.8717.8918.9821.38
1313.014.6816.6317.7118.8820.1421.5024.52
1414.015.9718.2919.1621.0122.5524.2127.97
1515.017.2920.0221.5823.2825.1327.1531.77
1616.018.6421.8223.6625.6727.8930.3235.95
1717.020.0123.7025.8428.2130.8433.7540.55
1818.021.4125.6528.1330.9134.0037.4545.60
1919.022.8427.6730.5433.7637.3841.4551.16
2020.024.3029.7833.0636.7941.0045.7657.28
2525.032.0341.6547.7354.8663.2573.1198.35
3030.040.5756.0866.4479.0694.46113.28164.49
3535.049.9973.6590.32111.43138.24172.32271.02

EX: If the value of design period years n = 9 , and annual growth rate r = 5 % , Find G r n ?

  • From above table G r n = 11.03

  • From Equation G r n = ( 1 + r ) n 1 r = ( 1 + 0.05 ) 9 1 0.05 = 11.03

  • F E i From table 19.3a & 19.3b

Table 19.3a: Axle Load Equivalency Factors for Flexible Pavements, Single Axles, and p t of 2.5

Axle Load (kips)Pavement Structural Number (SN)
123456
2.0004.0004.0003.0002.0002.0002
4.003.004.004.003.002.002
6.011.017.017.013.010.009
8.032.047.051.041.034.031
10.078.102.118.102.088.080
12.168.198.229.213.189.176
14.328.358.399.388.360.342
16.591.613.646.645.623.606
181.001.001.001.001.001.00
201.611.571.491.471.511.55
222.482.382.172.092.182.30
243.693.493.092.893.033.27
265.334.994.313.914.094.48
287.496.985.905.215.395.98
3010.39.57.96.87.07.8
3213.912.810.58.88.910.0
3418.416.913.711.311.212.5
3624.022.017.714.413.915.5
3830.928.322.618.117.219.0
4039.335.928.522.521.123.0
4249.345.035.627.825.627.7
4461.355.944.034.031.033.1
4675.568.854.041.437.239.3
4892.283.965.750.144.546.5
50112.0102.079.060.053.055.0
Ex 1:

Traffic (AADT) in both directions on the Highway during the first year of operation will be 12,000 with the following vehicle mix and axle loads.

Ex 2:

The presentAADT (in both directions) of 6000 vehicles is expected to grow at 5 % per annum. Assume SN = 4 and the percent of the traffic on the design lane is 55 % , the design life is 20 years.

If the vehicle mix is:

(3-axle tandem, 4-single axle) trucks (10000 lb/axle) = 60%

2-axle single-unit trucks (5000 lb/axle) = 30%

(3-axle single-unit, 2-axle tandem) trucks (7000 lb/axle) = 10%

Determine:

  • a-Daily Design ESAL
  • b- Total ESAL for the First year
  • c- Total ESAL
  • d- Design ESAL for (10000 lb/axle)
  • e- Daily Total ESAL for (7000 lb/axle)
Solution:

1- 10000 lb/axle = 10 kips/axle

F Ei 1 single = 0.102

F Ei 1 tandem = 0.009

[5000 , lb/axle = 5 , kips/axle \longrightarrow \quad F_{ Ei2 } , single = 0.003 + 0.013 2 = 0.008]

7000 lb/axle = 7 kips/axle F Ei2 single = 0.013 + 0.041 2 = 0.027

F Ei 2 tandem = 0.001 + 0.004 2 = 0.0025

2-AADT1 = 60 % × 6000 = 3600

AADT2 = 30% × 6000 = 1800

AADT3 = 10% × 6000 = 600

3 n = 20 y e a r s

r = 5 % G r n = 33.06

a- Daily Design ESAL

Daily Design ESAL = AADT × N × FEi × fd

DDESAL = [3600(3× 0.009 + 4 × 0.102) + 1800(2 × 0.008)

+ 6 0 0 ( 3 × 0. 0 2 7 + 2 × 0. 0 0 2 5 ) ] 0. 4 5 = 7 4 1
b- Total ESAL for First year

Total ESAL the First Year = AADT × N × F Ei × 365

= [ 3600 ( 3 × 0.009 + 4 × 0.102 ) + 1800 ( 2 × 0.008 ) + 600 ( 3 × 0.027 + 2 × 0.0025 ) ] × 36 = 600936
c- Total ESAL

Total ESAL = AADT × N × FEi × Grn × 365

= [ 3 6 0 0 ( 3 × 0. 0 0 9 + 4 × 0. 1 0 2 ) + 1 8 0 0 ( 2 × 0. 0 0 8 ) + 6 0 0 ( 3 × 0. 0 2 7 + 2 × 0. 0 0 2 5 ) ] 3 6 5 3 3. 0 6 = 1 9 8 6 6 9 4 4 = 1 9. 9 × 1 0 6

d- Design ESAL for (10000 lb/axle)

1 0 0 0 0 l b / a x l e = 1 0 k i p s / a x l e F E i 1 single = 0.1 0 2
D E S A L = A A D T 1 0 × N × F E i × G r n × 3 6 5 × f d = 3 6 0 0 ( 3 × 0. 0 0 9 + 4 × 0. 1 0 2 ) × 3 3. 0 6 × 3 6 5 × 0. 4 5 = 8 5 0 3 5 4 4 = 8. 5 × 1 0 6

e- Daily Total ESAL for (7000 lb/axle)

TotalESAL = A A D T 7 × N × F E i × G r n × 3 6 5 = 6 0 0 ( 3 × 0.0 2 7 + 2 × 0.0 0 2 5 ) × 3 3.0 6 × 3 6 5 = 6 2 2 6 5 2 = 0.6 2 × 1 0 6

c) Roadbed Soils (Subgrade Material)

AASHTO Code uses the Resilient Modulus (Mr) of the soil to define its property and convert the CBR value of the soil to an equivalent Mr value using the following conversion factor Mr ( lb / in 2 ) = 1500 CBR (for CBR of 10 or less).

d) Materials of Construction

The materials used for the construction of pavement can be classified under three general groups:

i. Subbase Construction Materials

The quality of the material used is determined in terms of the layer coefficient (a3), which is used to convert the actual thickness of the subbase to an equivalent Structure Number (SN).

ii. Base Course Construction Materials

Materials selected should satisfy the general requirements for base course materials, A structural layer coefficient, a2, for the material used also should be determined.

iii. Surface Course Construction Materials

The structural layer coefficient (a1) relates to a dense graded asphalt concrete surface with its resilient modulus at 68 F .

Figure 19.5 2 Structural Layer Coefficient, a 1 , for Asphalt Concrete Surface Course

EX: Elastic Modulus of Asphalt Concrete E AC = 250 , 000 lb / in 2 Find Asphalt layer Coefficient a 1 ?

a 1 = 0. 3 3

e) Environment

Temperature and rainfall are the two main environmental factors used in evaluating pavement performance in the AASHTO method.

The effects of temperature on asphalt pavements include stresses induced by thermal action and the impact of freezing and thawing water in the subgrade.

The effect of temperature, particularly about to the weakening of the underlying material during the thaw period, is considered a significant factor in determining the strength of the underlying materials used in the design. The effect of rainfall is due mainly to the penetration of the surface water into the underlying material, if penetration occurs, the properties of the underlying materials may be altered significantly.

The resilient modulus of materials susceptible to frost action can reduce by 50 percent to 80 percent during the thaw period, and it is likely that the strength of the material will be affected during the periods of heavy rains.

The AASHTO guide suggests a method for determining the effective, resilient modulus. In this method, a relationship is then used to determine the resilient modulus for each season based on the estimated in situ moisture content and Relative damage during the period of time.

The relative damage u f for each period is determined from the following chart, using the vertical scale or the equation given in the chart. The mean comparable damage u f then computed, and the effective subgrade resilient modulus is determined using the Chart and value of u f .

MonthRoadbed Soil Modulus M r (lb/in.2)Relative Damage u f
Jan.220000.01
Feb.220000.01
Mar.55000.25
Apr.50000.30
May50000.30
June80000.11
July80000.11
Aug.80000.11
Sept.85000.09
Oct.85000.09
Nov.60000.20
Dec.220000.01
Summation: u f = 1.59
Roadbed Soil Resilient Modulus, M1(103lb/in.2)
.005
.01
.05
.10
.50
1.0
5.0
10.0
13.0
Ex:
  • 1- Find Roadbed Resilient Modulus M r When Relative Damage u u f 7.23?

  • 2- Find Roadbed Resilient Modulus M r For mean Relative damage u u f ?

  • Using the Equation: u f = 1.18 × 10 8 × M r 2.32

  • 1- 0.723=1.18 × 10 8 × M r 2.32 Solving → M r = 1285 lb / in 2

  • 2-Mean u u f = 0.133 = 1.18 × 10 8 × M r 2.32 M r = 7193 lb / in 2

-Using vertical scale:

f) Drainage. ( m i )

The effect of drainage on the performance of flexible pavement is considered to the effect water has on the strength of the base material and roadbed soil, and The approach used is to provide for the rapid drainage of the water from the pavement structure by providing a suitable drainage layer and by modifying the structural layer coefficient.

The modification is carried out by adding a factor m i for the base and subbase layer coefficients ( a 2 and a 3 ) . The m i factors are based both on the percentage of time during which the pavement structure will be nearly saturated, and on the quality of drainage, which is dependent on the time it takes to drain the base layer to 50 percent of saturation.

Ex1:

A flexible pavement takes one day for water to be drained from within it and the pavement structure will be exposed to moisture levels approaching saturation for 7 % of the time. Find the pavement drainage coefficient?

Table 19.5: Definition of Drainage Quality

Quality of DrainageWater Removed Within*
Excellent 2 Good <2 hours 1 day <
3 Fair Poor Very poor1 week 1 month (water will not drain)
Time required to drain the base layer to 7% saturation.
Table 19.6: Recommended m_i Values
Percent of Time Pavement Structure Is Exposed to Moisture Levels Approaching Saturation4
Quality of DrainageLess Than 1%1 to 5%5 to 25% Greater Than 25%
Excellent1.40-1.351.35-1.301.30-1.20 1.20
Good1.35-1.251.25-1.151.15-1.00 1.00
Fair1.25-1.151.15-1.051.00-0.80 0.80
Poor1.15-1.051.05-0.800.80-0.60 0.60
Very poor1.05-0.950.95-0.750.75-0.40 0.40

Pavement drainage coefficient m = ( 1.15 1 )

g) Reliability. R%

The cumulative ESAL is an essential input to any pavement design method. However, the determination of this input is usually based on assumed growth rates which may not be accurate.

AASHTO guide proposes the use of a reliability factor that considers the possible uncertainties in traffic prediction and pavement performance prediction.

For example, a 50 % reliability design level implies 50 % chance for successful pavement performance.

Table 19.7 shows suggested reliability levels based on the AASHTO guide.

Table 19.7: Suggested Levels of Reliability for Various Functional Classifications

Recommended Level of Reliability
Functional ClassificationUrbanRural
Interstate and other freeways85–99.980–99.9
Other principal arterials80–9975–95
Collectors80–9575–95
Local50–8050–80

III. FLEXIBLE PAVEMENT STRUCTURAL DESIGN

SN = a 1 × D 1 + a 2 × D 2 × m 2 + a 3 × D 3 × m 3

m 2 , 3 = drainage coefficient for the base, subbase layer a 1 , 2 , 3 = layer coefficient of surface, base, and subbase Course

D 1 , 2 , 3 = actual thickness in inches of the surface, base, and subbase courses

1-ByEquation

log 10 ( W 18 ) = Z R S o + 9.36 log 10 ( SN + 1 ) 0.2 + log 10 ( Δ P S I 2.7 ) 0.4 + 1094 ( S N + 1 ) 5.19 + 2.32 log 10 ( M R ) 8.07

Where:

W 18 = predicted number of 18,000-lb (80 KN) single-axle load applications

Δ P S I = p i p t

SN = structural number indicative of the total pavement thickness

Z R = standard normal deviation for a given reliability

S o = overall standard deviation

Flexible pavements

Rigid pavements

Standard Deviation, S o

0. 4 0 0. 5 0
0. 3 0 0. 4 0

Table 19.8: Standard Normal Deviation ( Z R ) Values Corresponding to Selected Levels of Reliability

R% given → Reliability (R%) → Standard Normal Deviation, ZR
150-0.000
60-0.253
70-0.524
75-0.674
80-0.841
85-1.037
90-1.282
91-1.340
92-1.405
93-1.476
94-1.555
95-1.645
96-1.751
97-1.881
98-2.054
99-2.327
99.9-3.090
99.99-3.750
2-By Figure 19.8

Ex: Designing a Flexible Pavement Using the AASHTO Method

A flexible pavement for an urban interstate highway R = 95 % , and Standard deviation S 0 = 0.45 is to be designed to carry a design ESAL of 2 × 10 6 . It is estimated that it takes about a week for the water to be drained from within the pavement, and the pavement structure will be exposed to moisture levels approaching saturation for 30 % of the time. The following additional information is available:

Initial serviceability index p i = 4.5

Final serviceability index p t = 2.5

Resilient modulus of asphalt concrete at 68 F = 450 , 000 lb / in 2

CBR value of base course material = 100 , M r = 31 , 000   lb / in 2

CBR value of subbase course material = 22 , M r = 13 , 500 lb / in 2

CBR value of subgrade material = 5

M r = 1500 CBR = 1500 5 = 7500 lb / in 2

Solution:

1- Find Drainage Coefficient m i from table 19.5, 19.6

Table 19.5: Definition of Drainage Quality

Quality of DrainageWater Removed Within*
Excellent2 hours
Good1 day
Fair1 week
Poor1 month
Very poor(water will not drain)

Table 19.6: Recommended m i Values

Quality of DrainagePercent of Time Pavement Structure Is Exposed to Moisture Levels Approaching Saturation
Less Than 1%1 to 5%5 to 25%Greater Than 25%
Excellent1.40-1.351.35-1.301.30-1.201.20
Good1.35-1.251.25-1.151.15-1.001.00
Fair1.25-1.151.15-1.051.00-0.800.80
Poor1.15-1.051.05-0.800.80-0.600.60
Very poor1.05-0.950.95-0.750.75-0.400.40

Drainage Coefficient m i for base and subbase layer = 0.8

2- Find Layer Coefficient a 1 , a 2 , a 3 from figure 19.3, 19.4, 19.5

a 2 = 0.14

CBR = 100

E A C = 450 , 000 lb / in 2

3- Find the Structure Number for each layer from figure 19.8

SN1=2.45 SN2=3.4 SN3=4
S N 1 = a 1 × D 1
2. 4 5 = 0. 4 4 × D 1
D 1 = 5. 5 7 = 6 i n
S N 1 = 0. 4 4 × 6 = 2. 6 4
S N 2 = S N 1 + a 2 × D 2 × m 2
3.4 = 2.64 + 0.14 × D 2 × 0.8
D 2 = 6.79 = 8 in
S N 2 = 2. 6 4 + 0. 1 4 × 8 × 0. 8 = 3. 5 4
SN 3 = SN 2 + a 3 × D 3 × m 3
4 = 3. 5 4 + 0. 1 × D 2 × 0. 8
D 2 = 5. 7 5 = 6 i n
S N 3 = 3. 5 4 + 0. 1 × 6 × 0. 8 = 4. 0 2
  1. Principles of Pavement Design Yoder Ej (ch1,2,3).

  2. Highway and Traffic Engineering N. Garber (ch19,20).

Funding

No external funding was declared for this work.

Conflict of Interest

The authors declare no conflict of interest.

Ethical Approval

No ethics committee approval was required for this article type.

Data Availability

Not applicable for this article.

How to Cite This Article

Mousa Al-Asakereh. 2026. "Analytical Study and Design of Flexible Pavement". Global Journal of Research in Engineering - E: Civil & Structural GJRE-E Volume 22 (GJRE Volume 22 Issue E2).

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Flexible pavement study on resilient surface design for engineering journals.
Journal Specifications

Crossref Journal DOI 10.17406/gjre

Print ISSN 0975-5861

e-ISSN 2249-4596

Keywords
Classification
GJRE-E Classification DDC Code: 738.52095694 LCC Code: NA3760
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v1.2

Issue date
November 15, 2022

Language
English
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Analytical Study and Design of Flexible Pavement

Mousa Al-Asakereh
Mousa Al-Asakereh