Chapter 4 Pavement Design 1.1 INTRODUCTION

• Over 3 million miles (4.8 million kilometers) of highways in US (55% are paved). • Several states (Pennsylvania, Texas, Illinois, and California) have pavement

construction and rehabilitation budgets that exceed a billion dollars per year 1.2 PAVEMENT TYPES Why pavements?

• Typical soil-bearing capacities can be less than 50 lb/in2 (345 kPa) and in some cases as low as 2 to 3 lb/in2 (14 to 21 kPa).

• Typical automobile weighs approximately 2700 lb (12 kN), with tire pressures of 35 lb/in2 (241 kPa).

• Typical tractor semi-trailer truck that can weigh up to 80,000 lb (355.8 kN), the legal limit in many states, on five axles with tire pressures of 100 lb/in2 (690 kPa) or higher.

1.2.1 Flexible Pavements Figure 4.1 Typical flexible-pavement cross section. Components:

• Top layer made of asphaltic concrete, which is a mixture of asphalt cement and aggregates.

• The purpose of the wearing layer: • Protect the base layer from wheel abrasion and to waterproof the

entire pavement structure. • Provides a skid-resistant surface for vehicle performance.

• Other layers (Fig. 4.1) – thicknesses vary with the type of axle loading, available materials, and expected pavement design life.

1.2.2 Rigid Pavements Rigid pavement is constructed with Portland cement concrete (PCC) and aggregates:

Figure 4.2 Typical rigid-pavement cross section. 1.3 PAVEMENT SYSTEM DESIGN:

PRINCIPLES FOR FLEXIBLE PAVEMENTS Assumed load distribution: Figure 4.3 Distribution of load on a flexible pavement.

1.3.1 Calculation of Flexible Pavement Stresses and Deflections

• Boussinesq theory started by assuming that the pavement is one layer thick and the material is elastic, homogeneous, and isotropic. With point load:

Figure 4.4 Point load on a one-layer pavement. The basic equation for the stress at a point in the system is

2z

PK =zσ (4.1)

Where: σz = stress at a point in lb/in2 (kPa), P = wheel load in lb (N), z = depth of the point in question in inches (mm), and K = variable defined as

( )[ ] 2/52/1

123

zr

π K

+= (4.2)

Where: r = radial distance in inches from the centerline of the point load to the point in question.

• More realistic approach to assume an elliptical area that represents a tire footprint instead of a point load.

• The tire footprint can be defined by an equivalent circular area with a radius calculated by:

πp

P = a (4.3)

Where: a = equivalent load radius of the tire footprint in inches, P = tire load in lb, and p = tire pressure in lb/in2. • Ahlvin and Ulery provided solutions for the evaluation of stresses, strains, and

deflections at any point. • The Ahlvin and Ulery equation for the calculation of vertical stress, is

( )B Ap =z +σ (4.4)

Where: σz = vertical stress in lb/in2, p = pressure due to the load in lb/in2,

A and B = function values, as presented in Table 4.1, that depend on z/a and r/a, the depth in radii and offset distance in radii, respectively, Where:

z = depth of the point in question in inches (mm), r = radial distance in inches (mm) from the centerline of the point load to the

point in question, and a = equivalent load radius of the tire footprint in inches (mm).

The equation for radial-horizontal stress: ( )[ ]FμμA + C + = pσ r 212 − (4.5)

Where: μ = Poisson ratio and, σr = radial-horizontal stress in lb/in2 (kPa).

And the equation for deflection, Δz, is ( ) ( ) ⎥⎦

⎤⎢⎣⎡ −Δ HμA + az

E a + μp 11 =z

(4.6)

Where: p = pressure due to the load in lb/in2, a = equivalent load radius of the tire footprint in inches, E = modulus of elasticity in lb/in2, and C, F, and H = function values, as presented in Table 4.1, that depend on z/a and r/a,

Where: z = depth of the point in question in inches, r = radial distance in inches from the centerline of the point load to the point in

question, and a = equivalent load radius of the tire footprint in inches.

Table 4.1 One-Layer Elastic Function Values

Problem 4.1 A tire carries a 5000-lb load and has a pressure of 100 lb/in2. The pavement that the tire is on is constructed with a modulus of elasticity of 43,500 lb/in2. A deflection of 0.016 inches is observed at a point at the pavement surface, 0.8 inches from the center of the tire load. Using Ahlvin and Ulery equations, what is the radial horizontal stress at this point?.

1.4 THE AASHTO FLEXIBLE-PAVEMENT DESIGN PROCEDURE

• AASHO Road Test in Illinois from 1958 to 1960 • Current standards are based on 1993 AASHTO guideline • Updates of these guidelines have been rumored for decades

1.4.1 Serviceability Concept

• Based on Pavement Serviceability Index (PSI) • Attempts to link physical pavement factors with public expectations. • A 1 to 5 scale is often used, and based on observed pavement deterioration, the

following curve can be used:

Figure 4.5 Pavement performance trends.

• The present serviceability index is usually measured by a panel of raters who drive over the pavement section and rate the pavement performance on a scale of one to five with five being the smoothest ride.

• Accumulated traffic loads cause the pavement to deteriorate and the serviceability rating drops.

• At some point, a terminal serviceability index (TSI) is reached. • New pavements usually have an initial PSI rating of approximately 4.2 to 4.5. • Interstate highways or principal arterials have TSIs of 2.5 or 3.0, while local roads

can have TSIs of 2.0. 1.4.2 Flexible-Pavement Design Equation

( )[ ] [ ]

( )[ ]07.8log 32.2

1 SN1094 40.07.2/PSIlog

0.20 - 1 SNlog9.36 =log

10

19.5 10

10o1810

−+

++

Δ+++

R

R

M

SZW (4.7)

Where:

W18 = 18-kip-equivalent single-axle load, ZR = reliability (z-statistic from the standard normal curve), So = overall standard deviation of traffic, SN = structural number, ΔPSI = loss in serviceability from when the pavement is new until it

reaches its TSI, and MR = soil resilient modulus of the subgrade in lb/in2.

W18 Handles all traffic with a standard 18-kip (80.1-kN)- equivalent single-axle load (ESAL). The idea is to determine the impact of any axle load on the pavement in terms of the

equivalent amount of pavement impact that an 18-kip single-axle load would have. The axle-load equivalency factors for flexible pavement design, with a TSI of 2.5, are

presented in Tables 4.2 (for single axles), 4.3 (for tandem axles), and 4.4 (for triple axles). Table 4.2 Axle-Load Equivalency Factors for Flexible Pavements, Single Axles, and TSI = 2.5 Table 4.3 Axle-Load Equivalency Factors for Flexible Pavements, Tandem Axles, and TSI =

2.5 Table 4.4 Axle-Load Equivalency Factors for Flexible Pavements, Triple Axles, and TSI = 2.5

ZR Estimates the probability that the pavement will perform at or above the TSI level during the

design period. It accounts for the inherent uncertainty in design. Equation 4.7 uses the z-statistic, which is obtained from the cumulative probabilities of the

standard normal distribution (a normal distribution with mean equal to zero and variance equal to one). See Table 4.5.

In the flexible-pavement-design nomograph (Fig. 4.7), the probabilities (in percent) are used directly (instead of the ZR as in the case of Eq. 4.7) and these percent probabilities are denoted R, the reliability (see Table 4.5).

Typical reliability values for interstate highways are 90% or higher, whereas local roads can have a reliability as low as of 50%.

Table 4.5 Cumulative Percent Probabilities of Reliability, R, of the Standard Normal

Distribution, and Corresponding ZR

So The overall standard deviation, So, takes into account the designers' inability to accurately estimate the variation in future 18-kip (80.1-kN)-equivalent axle loads, and the statistical error in the equations resulting from variability in materials and construction practices. Typical values of So are in the order of 0.30 to 0.50.

SN The structural number, SN, represents the overall structural requirement needed to sustain

the design’s traffic loadings. ΔPSI The amount of serviceability loss, over the life of the pavement, initial PSI minus TSI. MR The soil resilient modulus, MR, is used to reflect the engineering properties of the subgrade

(the soil). The resilient modulus can be determined by AASHTO test method T274.

It is related to the California bearing ratio, CBR, the ratio of the load-bearing capacity of the soil to the load-bearing capacity of a high-quality aggregate, multiplied by 100. The relationship, used to provide a very basic approximation of MR (in lb/in2) from a known CBR, is

MR = 1500 × CBR (4.8) The coefficient of 1500 in Eq. 4.8 is used for CBR values less than 10. Caution must be

exercised when applying this equation to higher CBRs because the coefficient (the 1500 shown in Eq. 4.8) has a range of 750 to 3000.

1.4.3 Structural Number There are many pavement material combinations and thicknesses that will provide satisfactory pavement service life. The following equation can be used to relate individual material types and thickness to the structural number: 33322211SN MDaMDaDa ++= (4.9)

Where: a1, a2, and a3 = structural-layer coefficients of the wearing surface, base, and subbase

layers, respectively, D1, D2, and D3 = thickness of the wearing surface, base, and subbase layers in inches,

respectively, and M2 and M3 = drainage coefficients for the base and subbase, respectively.

• Values for the structural-layer coefficients for various types of material are presented in

Table 4.6. Table 4.6 Structural-Layer Coefficients (a's)

• Drainage coefficients are used to modify the thickness of the lower pavement layers (base and subbase) to take into account a material’s drainage characteristics. A value of 1.0 for a drainage coefficient represents a material with good drainage characteristics (a sandy material). Thickness:

wearing layers are typically 2 to 4 inches (50.8 to 101.6 mm) thick, subbases and bases range from 4 to 10 inches (101.6 to 254.0 mm) thick.

Figure 4.7 Design chart for flexible pavements based on using mean values for each input.

Problem 4.15 A flexible pavement has a 4-inch sand-mix asphalt wearing surface, 10-inch soil cement base, and a 10 inch crushed stone subbase. It is designed to withstand 400 20-kip single axle loads and 900 35-kip tandem axle loads per day. The subgrade CBR is 8, overall standard deviation is 0.45, initial PSI is 4.2 and the final PSI is 2.5. What is the probability that this pavement will have a PSI above 2.5 after 25 years? (Drainage coefficients are 1.0). Table 4.6 Structural-Layer Coefficients (a's) 33322211SN MDaMDaDa ++= (4.9) So, with D1 = 4, D2 = 10, D3 = 10 and M2 = M3 = 1 SN = (0.35)(4) + (0.20)(10) + 0.11(10) = 4.5 Axle load equivalency Tables 4.2 and 4.3, with SN =4.5: Single 20kip (Table 4.2): interpolate between SN =4 and SN =5: Axle SN 3 4 5 20 1.49 1.47 1.51 So, single ESAL is 1.49. Axle load equivalency Tables 4.2 and 4.3, with SN =4.5: Tandem 35kip (Table 4.3): interpolate between 34kip and 36kip and SN =4 and SN =5: Axle SN 3 4 5 34 1.11 1.11 1.09 →1.10 36 1.38 1.38 1.38 →1.38 So, tandem ESAL is 1.24. Total ESAL for 20kip single axle is 400 × 1.49 = 596 Total ESAL for 35kip tandem axle is 900 × 1.24 = 1116

W18 = (1116 + 596) × 365 × 25 = 15, 622,000 With:

• 15,622,000 (calculated) • S0 = 0.45 (given) • CBR = 8 (given) so, Mr =1500 CBR = 12,000psi (Eq. 4.8) • ΔPSI = 4.2 -2.5 =1.7 (given) • SN =4.5 (calculated)

Gives R=90% (Figure 4.7) If use Eq. 4.7: Solving for ZR gives -1.265. Using Table 4.5: Gives about 89.7%. 1.5 PAVEMENT SYSTEM DESIGN: PRINCIPLES FOR RIGID PAVEMENTS

• Rigid pavements distribute wheel loads by the beam action of the Portland cement concrete (PCC) slab,

• Material has a high modulus of elasticity, on the order of 4 to 5 million lb/in2 (27.6 to 34.5 GPa).

Figure 4.8 Beam action of a rigid pavement. 1.5.1 Calculation of Rigid-Pavement Stresses and Deflections

• H. M. Westergaard [1926] presented an important theoretical analysis for rigid pavements. • New term: modulus of the subgrade reaction (which is related to the subgrade CBR and soil

resilient modulus),

pk

=Δ (4.10)

Where: k = modulus of subgrade reaction in pounds per cubic inch (lb/in3), p = reactive pressure in lb/in2, and Δ = slab deflection in inches.

• The modulus k is assumed to be constant at each point under the slab and independent of deflection.

• The Westergaard equations were developed for three loading cases: an interior load, an edge load, and a corner load, as shown in Fig. 4.9.

• Ioannides, Thompson, and Barenberg [1985] reconsidered the Westergaard solutions and compared them with finite element analysis. T

Figure 4.9 Westergaard loading cases. For the interior loading,

( ) ( ) 2

22 6413502ln

213

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛

la

h + μP + γ. +

al

πh + μP = σ i

(4.11)

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

2

2 45

2ln

211

8Δ

la + γ

la

π +

klP = i

(4.12)

Where: σi = bending stress in lb/in2, Δi = slab deflection in inches, P = total load in lb, μ = Poisson ratio, h = slab thickness in inches, k = modulus of subgrade reaction in lb/in3, a = radius of circular load in inches (the tire footprint radius), γ = Euler’s constant and is equal to 0.577215, and l = radius of relative stiffness (a measure of the slab thickness in inches) and is defined

as

( )

250

2

3

112

.

kμEhl = ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

(4.13)

Where: E = modulus of elasticity in lb/in2, and Other terms as defined previously.

Note that the radius of the tire footprint, in US Customary units is

pπPa = (4.14)

Where:

a = equivalent load radius of the tire footprint in inches, P = tire load in lb, and p = tire pressure in lb/in2.

For the edge loading, ( ) ⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛= 710log54015290 4

3

102 .kaEh

hPμ. + .σe

(4.15)

( ) ⎟⎠⎞

⎜⎝⎛Δ 24014080

klP. + . = e μ (4.16)

For the corner loading,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

720

2 13 .l

c la

hP = σ (4.17)

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−Δ

la

..klP = l

c 69020512 (4.18)

Where: al = the distance to the point of action of the resulting load on a common angle bisection

at the slab corner as shown in Fig. 4.9, and is equal to a2 and, Other terms as defined above.

1.6 THE AASHTO RIGID-PAVEMENT DESIGN PROCEDURE

• The design procedure for rigid pavements presented in the AASHTO design guide is also based on the field results of the AASHO Road Test.

• The design procedure for rigid pavements is based on a selected reduction in serviceability and is similar to the design procedure followed for flexible pavements.

• However, instead of measuring pavement strength by using a structural number, the thickness of the PCC slab is the measure of strength.

• Equation to determine the thickness of a rigid-pavement PCC slab is

( )[ ] [ ]

( )[ ]( ) [ ]

( )[ ]{ }⎟⎟⎠⎞

⎜⎜⎝

⎛

−

−−+

+×+

Δ+++=

25075.0

75.0

10

46.8 710

10oR1810

421863.215132.1

logTSI32.022.4

1 10624.1 10.3/PSIlog

0.06 - 1 log7.35 log

.c

d'c

kE.DJDCS

DDSZW

(4.19)

Where: W18 = 18-kip-equivalent single-axle load, ZR = reliability (z-statistic from the standard normal curve), So = overall standard deviation of traffic, D = PCC slab thickness in inches, TSI = pavement’s terminal serviceability index, ΔPSI = loss in serviceability from when the pavement is new until it reaches its TSI,

'CS = concrete modulus of rupture in lb/in2,

Cd = drainage coefficient, J = load transfer coefficient, Ec = concrete elastic modulus in lb/in2, and k = modulus of subgrade reaction.

Figure 4.10 Segment 1 of the design chart for rigid pavement based on using mean values for each input variable. Fig 4.11 below. Table 4.7 Axle-Load Equivalency Factors for Rigid Pavements, Single Axles, and TSI = 2.5 Table 4.8 Axle-Load Equivalency Factors for Rigid Pavements, Tandem Axles, and TSI = 2.5 Table 4.9 Axle-Load Equivalency Factors for Rigid Pavements, Triple Axles, and TSI = 2.5 New terms used in Eq. 4.19 and Figs. 4.10 and 4.11 are presented here.

'cS The concrete modulus of rupture, '

CS , is a measure of the tensile strength of the concrete and is determined by loading a beam specimen, at the third points, to failure. The test method is ASTM C 78, Flexural Strength of Concrete. Because concrete gains strength with age, the average 28-day strength is used for design purposes. Typical values are 500 to 1200 lb/in2 (3,450 to 8,270 kPa).

Cd The drainage coefficient, Cd, is slightly different from those used in flexible-pavement

design. In rigid-pavement design, it accounts for the drainage characteristics of the subgrade. A value of 1.0 for a drainage coefficient represents a material with good drainage characteristics (such as a sandy material). Other soils, with less than ideal drainage characteristics, will have drainage coefficients that are less than 1.0.

J The load transfer coefficient, J, is a factor that is used to account for the ability of pavement

to transfer a load from one PCC slab to another across the slab joints. Pavements with dowel bars at the joints are typically designed with a J value of 3.2.

Ec The concrete modulus of elasticity, Ec is derived from the stress-strain curve as taken in the

elastic region. Typical values of Ec for Portland cement concrete are between 3 and 7 million lb/in2 (20.7 and 48.3 GPa).

k The modulus of subgrade reaction, k, depends upon several different factors including the

moisture content and density of the soil. Typical values for k range from 100 to 800 lb/in3 (0.0271 to 0.2165 N/mm3).

Table 4.10 Relationship Between California Bearing Ratio (CBR) and Modulus of Subgrade Reaction, k

Problem 4.24 You have been asked to design the pavement for an access highway to a major truck terminal. The design daily truck traffic consists of the following: 80 single axles at 22,500 lb each, 570 tandem axles at 25,000 lb each, 50 tandem axles at 39,000 lb each, and 80 triple axles at 48,000 lb each. The highway is to be designed with rigid pavement having a modulus of rupture of 600 lb/in2 and a modulus of elasticity of 5 million lb/in2. The reliability is to be 95%, the overall standard deviation is 0.4, the drainage coefficient is 0.9, ΔPSI is 1.7 (with a TSI of 2.5), and the load transfer coefficient is 3.2. The modulus of subgrade reaction is 200 lb/in3. If a 20-year design life is to be used, determine the required slab thickness. Solution: Assume D=11"

Axle loads: Single 22,500 = 2.675 (interpolating Table 4.7) Tandem 25kip = 0.5295 (interpolating Table 4.8) Tandem 39kip = 3.55 (interpolating Table 4.8) Triple 48kip = 2.58 (Table 4.9) 2.675× 80 + 0.5295× 570 + 3.55× 50 + 2.58× 80 = 899.715/day W18 = 899.715 × 365 × 20 = 6,567,920 With: W18 = 899.715 × 365 × 20 = 6,567,920 Sc = 600 Ec = 5,000,00 So =0.4 Cd = 0.9 J = 3.2 k =200 ΔPSI = 1.7 R = 95% Figure 4.10 Segment 1 of the design chart for rigid pavement based on using mean values for each input variable. Fig 4.11 below. D=11.43 inches

Traffic distribution among lanes: • Traffic tends to concentrate in the right lane (particularly heavy vehicles), so the distribution of

axle loads is not as simple as dividing the W18 by the number of lanes. • In equation form,

design-lane W18 = PDL(directional W18) (4.20) Where: W18 = 18-kip equivalent single axle loads (ESALs) and, PDL = proportion of directional W18 assumed to be in the design lane. • AASHTO-recommended values for PDL are given in Table 4.11.

Table 4.11 Proportion of Directional W18 Assumed to Be in the Design Lane Problem 4.22 A rigid pavement is designed with an 11-inch slab thickness, 90% reliability, Ec = 4 million lb/in2, modulus of rupture of 600 lb/in2, 2.8 load transfer coefficient, initial PSI = 4.8, final PSI = 2.5, overall standard deviation of 0.35, and a drainage coefficient of 0.8. The pavement has a 20-year design life. The pavement has three lanes and is conservatively designed for trucks that have one 20,000 lb single axle, one 26,000 lb tandem axle, and one 34k triple axle. What is the daily estimated truck traffic on the three lanes? Solution: D = 11; Sc = 600; Ec = 4,000,00; So =0.35; Cd = 0.8; J = 2.8 k =150 ΔPSI = 2.3 (4.8 – 2.5) R = 90% Figure 4.10 Segment 1 of the design chart for rigid pavement based on using mean values for each input variable. Fig 4.11 below. Gives: W18 = 11,162,000 Axle loads: Single 20kip = 1.58 (Table 4.7) Tandem 26kip = 0.619 (Table 4.8) Triple 34kip = 0.593 (Table 4.9) or 2.792 per truck Design lane traffic = W18 / (365× 20× 2.792) = 570.2 trucks/day But with 3 lanes, it is assumed that 80% of trucks are in the design lane (Table 4.11), so 570.2 / 0.80 = 712.8 trucks per day over all 3 lanes.

4.7 MEASURING PAVEMENT QUALITY AND PERFORMANCE

• International Roughness Index (IRI) • Friction Measurements • Rut Depth

4.7.1 International Roughness Index (IRI) Widely accepted measure of pavement condition IRI procedures were developed by the World Bank in Brazil Measures suspension movement over some longitudinal distance (in/mi) IRI correlates with vertical passenger acceleration and tire load

Table 4.12 Relationship Between the International Roughness Index (IRI) and Perceptions of Pavement Quality for Interstate Highways

Table 4.13 Relationship Between the International Roughness Index (IRI) and Perceptions of Pavement Quality for Non - Interstate Highways

4.7.2 Friction Measurements • Standardized test is conducted under wet conditions using either a treaded or smooth tire • Generally conducted at 40mi/h (65km/h) using a friction-testing trailer in which the wheel is

locked on the wetted road surface, and the torque developed from this wheel locking is used to measure friction number

• Gives an approximation of the coefficient of road adhesion under wet conditions (as shown in Table 2.4) and is multiplied by 100 to produce a value between 0 and 100

• The friction number with a treaded tire (FNt) attempts to measure pavement microtexture which is a function of the aggregate quality and composition.

• The friction number with a smooth tire (FNs) provides a measure pavement macrotexture which is critical in providing a water drainage escape path between the pavement and tire.

• Guidelines used by some states suggest that values of FNt < 30 or FNs < 15 indicate that poor friction may be contributing to wet-weather accidents.

• In Indiana FNs < 20 indicates need to resurface

4.7.3 Rut Depth • Rut depth, which is a measure of pavement surface deformation in the wheel paths, can affect

roadway safety because the ruts will accumulate water and increasing the possibility of vehicle hydroplaning (which results in the tire skimming over a film of water, greatly reducing braking and steering effectiveness)

• Because of this, rut depths are considered to be unacceptably high when their values reach 0.5 inches (12.5 mm) to 1 in (25 mm), indicating that corrective action is warranted.