Design and Analysis of Asphalt Pavemen

ABSTRACT

Chapter 13 highlights the design and analysis of asphalt pavements which is an important part of pavement engineering that deals with structural-empirical as well as empirical-mechanistic designs of asphalt pavements and the analysis of these pavements for stresses, strains, and deflections. The main inputs for empirical design of asphalt pavements include traffic loading, material properties, reliability, performance indicators, and pavement drainage quality. The mechanistic-empirical (M-E) design, on the other hand, takes into consideration the mechanistic analysis of the pavement structure and its performance over time regarding fatigue, rutting, and thermal cracking based on the available traffic, material, and environmental inputs. Consequently, the availability of this data to input in the M-E design defines three levels of M-E design. Level 3 provides the least level of accuracy in M-E design. This level may be used for less important roads or highways such as low-volume roads where the consequences of early failure are minimum. In this level, normally typical default values (such as average values) for the design inputs in the area are used and selected by the user. Level 2 offers an intermediate level of accuracy and uses values for the design inputs from available data sources of the agency, from limited testing, or based on correlations. The user selects these values to be used in the design. On the other hand, Level 1 provides the highest level of accuracy, and, therefore, has the least uncertainty (or error) level. This level, due to its high reliability and accuracy, is used for major highways and important roads with heavy traffic where the safety and economic consequences of early failure are high. In this level, the values of the design inputs are based on laboratory or field testing that requires more resources and time than other levels. The mechanistic analysis of asphalt pavements focuses on the determination of stresses, strains, and deflection on critical locations in the pavement structure under traffic loading. The analysis of asphalt pavements is composed of three different types: (1) the first one deals with one-layered pavement systems where the pavement structure is considered as one layer with infinite thickness, (2) the second type deals with two-layered pavement systems where the asphalt pavement structure is considered as two layers; one with finite thickness over another layer with infinite thickness, and (3) the third type considers asphalt pavements as three-layered systems; the first two layers with finite thickness over the last underneath layer with an infinite thickness. The second and third types are very close to reality and the real-life situation. The two-layered pavement systems represent full-depth asphalt pavements where there is an asphalt pavement layer over a subgrade layer. The three-layered pavement systems represent the conventional (traditional) asphalt pavements where there is an asphalt layer over base layers over a subgrade layer. The analysis of pavements goes in parallel with design due to its high correlation with design, particularly M-E design. The mechanistic analysis of an asphalt pavement provides an idea of how the different pavement layers would dissipate the high stresses and strains on the surface of the pavement as it gets deeper and deeper in the pavement structure. Pavement structures with high-quality materials or high-thickness layers would dissipate more stresses and strains than layers with low-quality materials or low-thickness layers. Therefore, reliability and cost play a big role in the entire design process. This section will provide questions and practical problems that deal with design as well as analysis of asphalt pavements. The analysis part will focus on multilayered pavement systems because they represent the real-world situation. 13.1

Select a typical or reasonable value from the set in the second column for each item (or property) of the group in the first column in Table 13.1.376

Solution:

See Table 13.2.377

13.2

Indicate the effect of the increase in the value of each of the following design inputs on the design thickness of the hot-mix asphalt (HMA) layer of a flexible pavement (see Table 13.3).

Solution:

See Table 13.4.

13.3

Order or rank the asphalt pavement structures subjected to different traffic loadings as shown in Figure 13.1 according to fatigue performance from best to worst assuming that the environmental conditions are the same.378 379

Solution:

All the pavement structures have the same subgrade layer with the same modulus (E₃). Therefore, the ranking will be based on the thickness of the HMA layer, the traffic loading, and the thickness of the base layer taking into consideration the following points:

Higher traffic loading results in lower pavement performance.

Higher number of load repetitions leads to lower performance.

Dual loads lead to higher pavement performance.

Higher HMA layer thickness leads to higher pavement performance.

Higher base-layer thickness also leads to higher pavement performance.

Higher modulus results in higher pavement performance.

The replacement of a base layer with an asphalt layer with a higher modulus leads to higher pavement performance.

Based on the above points, the following ranking is obtained (see Table 13.5).

Pavement Structure #6: load (13 ton), load repetitions (1000000), modulus of HMA layer (2E₁), HMA layer thickness (D₁+D₂), dual wheels.

Pavement Structure #5: same as #6 except for modulus: load (13 ton), load repetitions (1000000), modulus of HMA layer (E₁), HMA layer thickness (D₁+D₂), dual wheels.

Pavement Structure #2: load (13 ton), load repetitions (1000000), modulus of HMA and base layers (E₁ and E₂, respectively), two layers (HMA and base) with total thickness (D₁+D₂), dual wheels.

Pavement Structure #1: same as pavement structure #2 except for dual wheels: load (13 ton), load repetitions (1000000), modulus of HMA and base layers (E₁ and E₂, respectively), two layers (HMA and base) with total thickness (D₁+D₂), single wheels.

Pavement Structure #3: same as pavement structure #1 except for the load: load (20 ton), load repetitions (1000000), modulus of HMA and base layers (E₁ and E₂, respectively), two layers (HMA and base) with total thickness (D₁+D₂), single wheels.

Pavement Structure #4: same as pavement structure #3 except for the load repetitions: load (20 ton), load repetitions (2000000), modulus of HMA and base layers (E₁ and E₂, respectively), two layers (HMA and base) with total thickness (D₁+D₂), single wheels.380

13.4

State the mistake in each of the following flexible pavement structure designs and correct it (see Figure 13.2).

Solution:

Pavement Structure Design #1: the modulus of the HMA layer is smaller than the modulus of the base layer. A correction can be: E₁ = 300 ksi and E₂ = 100 ksi.

Pavement Structure Design #2: the thickness of the HMA layer is higher than the thickness of the base layer. A correction can be: D₁ = 4 in and D₂ = 5 in.

Pavement Structure Design #3: the modulus of the subgrade layer is very high (not typical). A correction can be: E₃ = 10 ksi (see Table 13.6).381

13.5

Compute the predicted total number of equivalent single-axle loads (ESALs) for a highway flexible pavement for a new six-lane rural highway with a first-year annual average daily traffic (AADT) of 3000 veh/day for both directions. The design lane factor (f_d) is 0.4, the traffic growth rate is 4%, the design period is 10 years, and the traffic mix is as shown in Figure 13.3. Ignore the ESALs of passenger cars. What is the truck factor of the truck in the third category? Which load is considered the critical load?382

Solution:

The total ESALs is calculated as shown in the following procedure:

13.1 ESAL i = ( AADT i ) ( 365 ) ( f d ) ( G ) ( N i ) ( f LE − i ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0001.tif"/>

Where:

ESAL_i = equivalent single-axle loads for traffic/axle category i

AADT_i= annual average daily traffic for category i

365 = number of days per year

f_d = lane distribution factor for the highway

G = growth factor

N_i = number of axles with the same type and load

f_LE-i = load equivalency factor for axle category i

The growth factor is calculated using the formula below:

13.2 G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0002.tif"/>

Where:

G = growth factor for the design period, n

r = traffic growth rate

n = design period (years)

Sample Calculation:

For the 12-kip single axle in Truck 1:

AADT=0 .20 (3000)=600 truck/day

f_d = 0.4 (given in the problem)

N = 2 (since there are two axles with the same type and load; two 12-kip single axles)

f_lE = 0.197 (determined below)

r = 4%

n = 10 years

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0003.tif"/>

⇒

G = ( 1 + 0.04 ) 10 − 1 0.04 = 12.006 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0004.tif"/>

ESAL i = ( AADT i ) ( 365 ) ( f d ) ( G ) ( N i ) ( f LE − i ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0005.tif"/>

⇒

ESAL 12 -kip Single = ( 600 ) ( 365 ) ( 0.4 ) ( 12.006 ) ( 2 ) ( 0.197 ) = 397556 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0006.tif"/>

The load equivalency factor is determined using the power model developed in this book for single-axle loads based on numbers in the table of the load equivalency factors from the Asphalt Institute’s Thickness Design Manual (MS-1), 9th Edition, 1982 as shown in the following relationship (see Figure 13.4).383

The best-fit model that describes the relationship between the load and the load equivalency factor for single-axle loads is a power model with high coefficient of determination (r²) as shown below:

13.3 f LE = 2.1159 × 10 − 17 ( Load ) 3.9120 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0007.tif"/>

Where:

f_LE = load equivalency factor for single-axle loads

Load = single-axle load in (lb)

⇒

f LE = 2.1159 × 10 − 17 ( 12000 ) 3.9120 = 0.192 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0008.tif"/>

For tandem-axle loads and tridem-axle loads, the following relationships are used, respectively.

The load equivalency factor is determined using the power model developed in this book for tandem-axle loads based on numbers in the table of the load equivalency factors from the Asphalt Institute’s Thickness Design Manual (MS-1), 9th Edition, 1982 as shown in the following relationship (see Figure 13.5).384

The best-fit model that describes the relationship between the load and the load equivalency factor for tandem-axle loads is a power model with high coefficient of determination (r²) as shown below:

13.4 f LE = 8.0863 × 10 − 19 ( Load ) 3.9937 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0009.tif"/>

Where:

f_LE = load equivalency factor for tandem-axle loads

Load = tandem-axle load in (lb)

For tridem-axle loads, the load equivalency factor is determined using the power model developed in this book based on numbers in the table of the load equivalency factors from the Asphalt Institute’s Thickness Design Manual (MS-1), 9th Edition, 1982 as shown in the following relationship (see Figure 13.6).385

The best-fit model that describes the relationship between the load and the load equivalency factor for tridem-axle loads is a power model with high coefficient of determination (r²) as shown below:

13.5 f LE = 6.7846 × 10 − 20 ( Load ) 4.0939 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0010.tif"/>

Where:

f_LE = load equivalency factor for tridem-axle loads

Load = tridem-axle load in (lb)

The ESAL results of the other axle types are summarized in Table 13.7 based on the same formula.

The truck factor is calculated using the following formula:

f T = ( N i ) ( f LE − i ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0011.tif"/>

Where:

Load_i = number of axles for axle category i

f_LE-i = the load equivalency factor for axle category i

⇒

f T = 1 × 0.094 + 1 × 0.192 + 1 × 0.694 = 0.980 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0012.tif"/>

The critical load is the load with the highest load equivalency factor. Therefore, in this case, the critical load is the 44-kip tridem axle.

The MS Excel worksheet used to compute the ESALs in this problem is shown in Figure 13.7.386

13.6

A highway flexible pavement is designed for a design period of 20 years. The traffic that is expected to use the highway pavement is of the type shown in Figure 13.8. If the first-year annual average daily traffic (AADT) of this type of trucks is 200 truck/day in both directions, the annual traffic growth rate is 2%, and the design lane factor is 0.45, determine the total equivalent single-axle loads (ESALs).

Solution:

The total ESALs is calculated as shown in the following procedure:

The growth factor is calculated using the formula below:

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0014.tif"/> 387

Sample Calculation:

For the 24-kip tandem axle:

AADT = 200 truck/day

f_d = 0.45 (given in the problem)

N = 4 (since there are two axles with the same type and load; two 24-kip tandem axles)

f_lE = 0.252 (determined below)

r = 2%

n = 20 years

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0015.tif"/>

⇒

G = ( 1 + 0.02 ) 20 − 1 0.02 = 24.30 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0016.tif"/>

⇒

ESAL 24 -kip Tandem = ( 200 ) ( 365 ) ( 0.45 ) ( 24.30 ) ( 4 ) ( 0.252 ) = 803811 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0018.tif"/>

The power model developed earlier for the load equivalency factor of tandem-axle loads will be used to compute f_LE for the 24-kip tandem-axle load as shown below:

f LE = 8.0863 × 10 − 19 ( Load ) 3.9937 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0019.tif"/>

⇒

f LE = 8.0863 × 10 − 19 ( 24000 ) 3.9937 = 0.252 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0020.tif"/>

The ESAL results of the other axle types are summarized in Table 13.8 based on the same formula.

The MS Excel worksheet used to compute the ESALs in this problem is shown in Figure 13.9.388

13.7

Design a flexible pavement for a new six-lane rural highway that connects two cities using the AASHTO design guide procedure to carry the traffic mix shown in Table 13.9. The base and subbase layers will be constructed using the same crushed stone material and therefore one combined thickness could be used for both layers. The material properties are also provided in Table 13.10.389

The annual average daily traffic (AADT) that is expected to use the highway in both directions is 5000 veh/day in the first year, the traffic growth rate is 5%, and the design life of the pavement is 10 years. Assume that the traffic growth rate will remain the same during the design life of the pavement.

Given Data:

S_o = 0.5

p _i = 4.5

p _t = 2.5

Drainage data: it is expected that it will take one week for water to be removed from the pavement, and the pavement structure will be exposed to moisture levels approaching saturation 20% of the time (see Figure 13.10).

Solution:

The following formula is used for the design of asphalt pavements following the AASHTO design method:

13.7 log 10 W 18 = Z R S 0 + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ Δ PSI ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 M R − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0021.tif"/> 390

Where:

W₁₈ = predicted number of 18,000 lb (80 kN) single-axle load applications

Z_R = standard normal deviation for a given reliability

S₀ = overall standard deviation

SN = structural number of the pavement

ΔPSI = p_i – p_t

p_i = initial serviceability index

p_t = terminal serviceability index

M_R = resilient modulus (psi)

For this pavement structure, there are three structural numbers (SN₁, SN₂, and SN₃) defined as shown below:

13.8 SN 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0022.tif"/>

Where:

SN₁ = Structural Number 1 above the base layer

a₁ = coefficient of Layer 1 (hot-mix asphalt (HMA) layer)

D₁ = thickness of Layer 1 (HMA layer)

13.9 SN 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0023.tif"/>

Where:

SN₂ = Structural Number 2 above the subbase layer

a₁ = coefficient of Layer 1 (hot-mix asphalt (HMA) layer)

a₂ = coefficient of Layer 2 (base layer)

D₁ = thickness of Layer 1 (HMA layer)

D₂ = thickness of Layer 2 (base layer)

SN 3 = a 1 D 1 + a 2 m 2 D 1 + a 3 m 3 D 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0024.tif"/>

Where:

SN₃ = Structural Number 3 above the subgrade layer

a₁ = coefficient of Layer 1 (hot-mix asphalt (HMA) layer)

a₂ = coefficient of Layer 2 (base layer)

a₃ = coefficient of Layer 3 (subbase layer)

D₁ = thickness of Layer 1 (HMA layer)

D₂ = thickness of Layer 2 (base layer)

D₃ = thickness of Layer 3 (subbase layer)

A polynomial of the third degree is used to fit the data of the elastic modulus of the HMA layer versus the layer coefficient (a₁). The data used to develop this relationship is extracted from the “Chart for Estimating Structural Layer Coefficient of Dense-Graded Asphalt Concrete Based on the Elastic (Resilient) Modulus” in the AASHTO Guide for Design of Pavement Structures (1993). This model can be used to determine the layer coefficient (a₁) at any modulus value (see Figure 13.11).391

Therefore,

13.11 a 1 = 0.0026 M R 3 − 0.0317 M R 2 + 0.1798 M R + 0.0407 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0025.tif"/>

Where:

a₁ = coefficient of Layer 1 (HMA layer)

M_R = resilient or elastic modulus of Layer 1 or HMA layer (10⁵ psi)

⇒

a 1 = 0.0026 ( 4.0 ) 3 − 0.0317 ( 4.0 ) 2 + 0.1798 ( 4.0 ) + 0.0407 = 0.42 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0026.tif"/>

The coefficient of Layer 2 (the base layer) is given by the following logarithmic formula (Source: AASHTO Guide for Design of Pavement Structures, 1993):

a 2 = 0.249 log ( E 2 ) − 0.977 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0027.tif"/>

Where:

a₂ = coefficient of Layer 2 (base layer)

E₂ = elastic modulus of Layer 2 or base layer (psi)

⇒

a 2 = 0.249 log ( 20000 ) − 0.977 = 0.094 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0028.tif"/>

The coefficient of Layer 3 (the subbase layer) is also given by the following logarithmic formula (Source: AASHTO Guide for Design of Pavement Structures, 1993):

a 3 = 0.227 log ( E 3 ) − 0.839 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0029.tif"/>

Where:

a₃ = coefficient of Layer 3 (subbase layer)

E₃ = elastic modulus of Layer 3 or subbase layer (psi)

⇒

a 3 = 0.227 log ( 20000 ) − 0.839 = 0.137 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0030.tif"/> 392

Based on the highway information given in the problem, the highway is a major principal rural highway that serves two main cities. Therefore, the recommended value for the reliability level would be 75% to 95% according to Table 13.11. The highest value is selected in this case (95%) to be in the safe side.

For a reliability (R) level of 95%, the Z-value is equal to −1.645 according to Table 13.12.

S₀ = 0.50

Δ PSI = p i − p t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0031.tif"/>

Where:

p_i = initial serviceability index

p_t = terminal serviceability index393

⇒

Δ PSI = 4.5 − 2.5 = 2.0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0032.tif"/>

Now the total ESALs should be determined as shown in the following procedure:

The growth factor is calculated using the formula below:

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0034.tif"/>

Sample Calculation for the ESALs:

For the 36-kip tandem axle in Truck 1:

AADT = 0.15 (5000) = 750 truck/day

Since the highway is a six-lane highway, the lane distribution factor is 40% (0.40) according to the Asphalt Institute (AI) values in Table 13.13. And based on the AASHTO recommended values in Table 13.14, for three lanes in each direction, f_d = 60-80% in one direction. The traffic in this problem is given for both directions; therefore, a value of 0.40 will be used for f_d.

Or:394

f_d = 0.40

N = 2 (since there are two axles with the same type and load; two 36-kip tandem axles)

f_lE = 1.271 (determined below)

The power model for the load equivalency factor of tandem-axle loads developed above in Problem 13.5 will be used:

⇒

f LE = 8.0863 × 10 − 19 ( 36000 ) 3.9937 = 1.271 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0036.tif"/>

r = 5 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0037.tif"/>

n = 10 years https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0038.tif"/>

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0039.tif"/>

⇒

G = ( 1 + 0.05 ) 10 − 1 0.05 = 12.58 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0040.tif"/>

⇒

ESAL 36 -kip Tandem = ( 750 ) ( 365 ) ( 0.4 ) ( 12.58 ) ( 2 ) ( 1.271 ) = 3801291 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0042.tif"/>

The results of the other axle types are summarized in Table 13.15 based on the same formula.395

Now using the resilient (elastic) modulus of the subgrade layer (10000 psi) and based on the design ESALs that the pavement structure should be capable of carrying, the structural number (SN₃) is determined using the formula below:

log 10 W 18 = Z R S 0 + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ Δ PSI ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 M R − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0043.tif"/>

⇒

log 10 ( 4981305 ) = − 1.282 ( 0.5 ) + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 ( 10000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0044.tif"/>

⇒

6.6973 = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.32 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.32 + 1 ) 5.19 ] + 2.32 log 10 ( 10000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0045.tif"/>

The MS Excel Solver tool is used as shown in the procedure below:

The left side of the equation is basically set to:

Left Side = Design ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0046.tif"/>

In this case,

Left Side = Design ESALs = 4981305 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0047.tif"/>

And the right side of the equation is set to:

Right Side = 10 [ − 1.282 ( 0.5 ) + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 ( 10000 ) − 8.07 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0048.tif"/>

An error is defined as the difference between the left side and the right side; and in this case the error is defined as follows:

13.17 Error = [ log ( Left Side ) − log ( right side ) ] 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0049.tif"/> 396

When the error is zero or minimum, that is when SN is equal to the correct value. Therefore, the Excel Solver will perform iterative approach by changing SN value until the error converges to a value of zero or minimum value possible. That will be the solution of SN. The MS Excel Solver is used three times; each time, only the resilient (elastic) modulus is changed to determine the SN. For SN₃, the resilient modulus of the subgrade is used. For SN₂, the resilient modulus of the subbase is used. And for SN₁, the resilient modulus of the base is used. The total ESALs is the same for the three cases (see Figure 13.12).

Based on that, the following results are obtained (see Table 13.16).

Now the thickness of each layer is determined as explained in the following procedure:

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0050.tif"/> 397

⇒

3.32 = 0.42 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0051.tif"/>

⇒

D 1 = 7.92 in ≅ 8 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0052.tif"/>

The drainage coefficients for the base and subbase layers (m₂ and m₃) are determined from Table 13.17 based on the given drainage data (for the condition “water removed within one week”, the quality of drainage is “Fair”, and for the condition “20% of the time the pavement structure is exposed to moisture levels approaching saturation”, the drainage coefficients value is 1.00−0.80. Therefore,

m₂ = m₃ = 0.90 (an average value in the range has been selected) (see Table 13.17).

S N 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0053.tif"/>

⇒

3.32 = 0.42 ( 8 ) + 0.094 ( 0.9 ) D 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0054.tif"/>

⇒

D 2 = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0055.tif"/>

S N 3 = a 1 D 1 + a 2 m 2 D 1 + a 3 m 3 D 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0056.tif"/>

⇒

4.23 = 0.42 ( 8 ) + 0.094 ( 0.9 ) ( 0 ) + 0.137 ( 0.9 ) D 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0057.tif"/>

⇒

D 3 = 7.1 in ≅ 8 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0058.tif"/> 398

Because the same crushed stone material (with the same modulus) is used for the base and the subbase layers, the thickness of the subbase layer comes out to be zero. In other words, only one layer (base) will be used for the pavement structure having a total thickness of 8.0 inches. The final design will be as follows (see Table 13.18 and Figure 13.13).

13.8

In the problem above, the pavement design engineer would like to reduce the thickness of the HMA layer and minimize the cost of the highway pavement. Therefore, another source for a base material has been searched. A new base material with an elastic modulus of 80000 psi will be used. The crushed stone material will only be used for the subbase layer. Use the same design inputs to determine the new HMA layer thickness and the thickness of the other pavement layers (see Table 13.19 and Figure 13.14).399

Given Data:

S_o = 0.5

R (%) = 95 %

p _i = 4.5

p _t = 2.5

Drainage coefficients: m₂ = m₃ = 0.9

f_d = 0.40

r = 5%

n = 10 years

Total predicted ESALs = 4981305

Solution:

Following the same procedure, and using the MS Excel Solver tool to determine the three structural numbers of the pavement (SN₁, SN₂, and SN₃), the following results are obtained (see Table 13.20).

The MS Excel worksheet along with the Excel Solver tool used to solve for the results of this problem are shown in Figure 13.15.400

a₁ is the same since the HMA material is the same; therefore,

a 1 = 0.42 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0059.tif"/>

a 2 = 0.249 log ( E 2 ) − 0.977 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0060.tif"/>

⇒

a 2 = 0.249 log ( 80000 ) − 0.977 = 0.244 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0061.tif"/>

a₃ is the same since the subbase material is the same; therefore,

a 3 = 0.137 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0062.tif"/>

The thickness of each layer is now determined as explained in the following procedure:

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0063.tif"/>

⇒

1.99 = 0.42 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0064.tif"/>

⇒401

D 1 = 4.7 in ≅ 5 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0065.tif"/>

S N 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0066.tif"/>

⇒

3.32 = 0.42 ( 5 ) + 0.244 ( 0.9 ) D 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0067.tif"/>

⇒

D 2 = 5.6 in ≅ 6 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0068.tif"/>

S N 3 = a 1 D 1 + a 2 m 2 D 1 + a 3 m 3 D 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0069.tif"/>

⇒

4.23 = 0.42 ( 5 ) + 0.244 ( 0.9 ) ( 6 ) + 0.137 ( 0.9 ) D 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0070.tif"/>

⇒

D 3 = 6.6 in ≅ 7 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0071.tif"/>

The final design will be as shown in Table 13.21.

13.9

Determine the number of equivalent single-axle loads (ESALs) that the asphalt pavement structure shown in Figure 13.16 is capable of carrying at a reliability level (R) of 90%, overall standard deviation (S₀) of 0.5, ΔPSI of 2.0, and m₂ = 1.0.402

Solution:

The following formula is used for the design of asphalt pavements:

For this pavement structure, there are two structural numbers (SN₁ and SN₂) as shown in Figure 13.17.

a 1 = 0.0026 M R 3 − 0.0317 M R 2 + 0.1798 M R + 0.0407 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0073.tif"/>

⇒

a 1 = 0.0026 ( 3.65 ) 3 − 0.0317 ( 3.65 ) 2 + 0.1798 ( 3.65 ) + 0.0407 = 0.401 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0074.tif"/>

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0075.tif"/>

⇒

S N 1 = 0.401 ( 5.3 ) = 2.12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0076.tif"/>

The coefficient of Layer 2 (the base layer) is given by the following logarithmic formula:

a 2 = 0.249 log ( E 2 ) − 0.977 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0077.tif"/>

⇒

a 2 = 0.249 log ( 30000 ) − 0.977 = 0.138 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0078.tif"/> 403

S N 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0079.tif"/>

⇒

S N 2 = 0.40 ( 5.3 ) + 0.138 ( 1.0 ) ( 8.6 ) = 3.31 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0080.tif"/>

For a reliability (R) level of 90%, the Z-value is equal to −1.282.

Now based on the pavement structural number (SN₂), the pavement structure will be able to carry ESALs calculated based on the design formula below:

⇒

log 10 W 18 = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.31 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.31 + 1 ) 5.19 ] + 2.32 log 10 ( 9000 ) − 8.07 = 6.074 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0082.tif"/>

⇒

W 18 = 1164377 ≅ 1.2 million ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0083.tif"/>

The MS Excel worksheet used to determine the ESALs in this problem is shown in Figure 13.18.404

13.10

A highway designer would like to set a limitation on the number of trucks that use a rural highway pavement that is designed to carry an equivalent single-axle loads (ESALs) of 0.5 million for a design period of 10 years. The truck traffic that is expected to use the highway pavement is of the type shown in Figure 13.19. Determine the maximum allowable first-year annual average daily traffic (AADT) of this type of trucks given that the traffic growth factor is 12.0 and the design lane factor is 0.45.

Solution:

The total number of ESALs is calculated using the formula below and based on the AADT data:

For the 12-kip single axle:

f_d = 0.45 (given in the problem)

N = 1 (since there is only one axle with the same type and load; one 12-kip single axle)405

f_lE = 0.192 (determined below)

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 12-kip single-axle load as shown below:

f LE = 2.1159 × 10 − 17 ( Load ) 3.9120 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0085.tif"/>

⇒

n = 10 years https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0087.tif"/>

G = 12 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0088.tif"/>

⇒

ESAL 12 -kip Single = ( AADT ) ( 365 ) ( 0.45 ) ( 12 ) ( 1 ) ( 0.192 ) = 378.4 AADT https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0090.tif"/>

For the 18-kip single axle:

f_d = 0.45 (given in the problem)

N = 1 (since there is only one axle with the same type and load; one 18-kip single axle)

f_lE = 1

n = 10 years

G = 12

⇒

ESAL 18 -kip Single = ( AADT ) ( 365 ) ( 0.45 ) ( 12 ) ( 1 ) ( 1 ) = 1971 AADT https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0092.tif"/>

Since the highway pavement is designed to carry design ESALs of 0.5 million, therefore,

378.4 AADT + 1971 AADT = 500000 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0093.tif"/>

⇒

AADT = 213 truck/day https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0094.tif"/>

The MS Excel worksheet used to perform the computations of this problem is shown in Figure 13.20.406

13.11

A highway flexible pavement is designed to carry an equivalent single-axle loads (ESALs) of 795000 for a design period (n). The traffic that is expected to use the highway pavement is of the type shown in Figure 13.21. If the first-year annual average daily traffic (AADT) of this type of trucks is 200 truck/day, the annual traffic growth rate is 2%, and the design lane factor is 0.45, determine the design life for this pavement in years.

Solution:

The total number of ESALs is calculated using the formula below and based on the AADT data:

For the 14-kip single axle:

f_d = 0.45 (given in the problem)

N = 1 (since there is only one axle with the same type and load; one 14-kip single axle)

f_lE = 0.351 (calculated below)407

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 14-kip single-axle load as follows:

⇒

f LE = 2.1159 × 10 − 17 ( 14000 ) 3.9120 = 0.351 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0097.tif"/>

⇒

ESAL 14 -kip Single = ( 200 ) ( 365 ) ( 0.45 ) ( G ) ( 1 ) ( 0.351 ) = 11526 G https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0099.tif"/>

For the 24-kip tandem axle:

f_d = 0.45 (given in the problem)

N = 4 (since there are four axles with the same type and load; four 24-kip tandem axles)

f_lE = 0.252 (computed below)

The power model developed earlier for the load equivalency factor of tandem-axle loads will be used to compute f_LE for the 24-kip tandem-axle load as follows:

⇒

ESAL 24 − kip Tandem = ( 200 ) ( 365 ) ( 0.45 ) ( G ) ( 4 ) ( 0.252 ) = 33082 G https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0103.tif"/>

Therefore, the estimated number of ESALs based on the AADT data is equal to:

Estimated ESALs = 11526 G + 33082 G https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0104.tif"/>

Since the highway pavement is designed to carry design ESALs of 795000, the estimated ESALs should equal to the design ESALs. Consequently,

11526 G + 33082 G = 795000 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0105.tif"/>

⇒

G = 17.82 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0106.tif"/>

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0107.tif"/> 408

⇒

G = ( 1 + 0.02 ) n − 1 0.02 = 17.82 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0108.tif"/>

⇒

n = 15.4 ≅ 16 years https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0109.tif"/>

The MS Excel worksheet used to perform the computations of this problem is shown in Figure 13.22.

13.12

The full-depth asphalt pavement shown in Figure 13.23 is designed to carry an equivalent single-axle loads (ESALs) of 6 million for a design period of 20 years at a reliability level of 90%. Determine the resilient modulus of the modified subgrade that should be used under the HMA layer if the overall standard deviation (S₀) is 0.5 and ΔPSI is 2.0 (see Figure 13.23).409

Solution:

The layer coefficient is determined for layer 1 (HMA layer) as below:

⇒

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0112.tif"/>

⇒

S N 1 = 0.401 ( 8.0 ) = 3.208 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0113.tif"/>

For a reliability (R) level of 90%, the Z-value is equal to −1.282.

⇒

log 10 ( 6 × 10 6 ) = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.208 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.208 + 1 ) 5.19 ] + 2.32 log 10 ( M R 2 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0115.tif"/>

⇒

6.778 = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.208 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.208 + 1 ) 5.19 ] + 2.32 log 10 ( M R 2 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0116.tif"/> 410

Solving the above equation for M_R2 provides the following solution:

M R 2 = 19907 psi https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0117.tif"/>

Therefore, a modified subgrade with a resilient (elastic) modulus of 19907 psi (≌ 137 MPa) should be used under the full-depth asphalt layer so that the pavement will be capable of carrying a 20-year expected traffic level of 6 million ESALs.

The MS Excel worksheet used to determine the ESALs in this problem is shown in Figure 13.24.

13.13

The flexible pavement structure shown in Figure 13.25 is designed for a four-lane highway based on the following inputs: reliability level (R) = 90%, overall standard deviation (S₀) = 0.5, ΔPSI = 2.0, m₂ = 1.0, traffic growth rate (r) = 4%, design period, n = 20 years, and design lane factor, f_d = 0.40. The traffic mix that will use the highway consists of 10% trucks of the type shown in Figure 13.26.411

If the annual average daily traffic (AADT) for the first year is 3000 veh/day, determine the following:

The ESALs after 20 years that this pavement structure can carry.

The load equivalency factor of the single axle of the design truck.

The maximum load that this single axle can carry.

Note: ignore the effect of the passenger cars.

Solution:

The layer coefficients are determined:

⇒

a 1 = 0.0026 ( 4.0 ) 3 − 0.0317 ( 4.0 ) 2 + 0.1798 ( 4.0 ) + 0.0407 = 0.419 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0119.tif"/>

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0120.tif"/>

⇒

S N 1 = 0.419 ( 4.8 ) = 2.01 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0121.tif"/>

The coefficient of Layer 2 (the base layer) is given by the following logarithmic formula:

a 2 = 0.249 log ( E 2 ) − 0.977 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0122.tif"/>

⇒

a 2 = 0.249 log ( 50000 ) − 0.977 = 0.193 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0123.tif"/>

S N 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0124.tif"/>

⇒

S N 2 = 0.419 ( 4.8 ) + 0.193 ( 1.0 ) ( 7.7 ) = 3.50 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0125.tif"/> 412

For a reliability (R) level of 90%, the Z-value is equal to −1.282.

Now based on the pavement structural number (SN₂), the pavement structure will be able to carry ESALs calculated based on the design formula below:

⇒

log 10 W 18 = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.50 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.50 + 1 ) 5.19 ] + 2.32 log 10 ( 9000 ) − 8.07 = 6.327 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0127.tif"/>

⇒

W 18 = 2125163 ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0128.tif"/>

But the estimated number of ESALs based on the AADT is calculated as below:

For the 12-kip single axle:

f_d = 0.40 (given in the problem)

N = 1 (since there is only one axle with the same type and load; one 12-kip single axle)

f_lE = 0.192 (computed below)

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 12-kip single-axle load as follows:

⇒

ESAL 12 -kip Single = ( 3000 × 0.10 ) ( 365 ) ( 0.40 ) ( 29.78 ) ( 1 ) ( 0.192 ) = 250393 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0133.tif"/> 413

For the other single axle:

f_d = 0.40 (given in the problem)

N = 1

f_lE = ?

⇒

ESAL x -Single = ( 3000 × 0.10 ) ( 365 ) ( 0.40 ) ( 29.78 ) ( 1 ) ( f LE − x ) = 1304280 f LE − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0135.tif"/>

The predicted number of ESALs based on AADT should be equal to the total number of design ESALs the pavement structure can carry in the design life; therefore,

W 18 = 2125163 = 250393 + 1304280 f LE − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0136.tif"/>

⇒

f LE − x = 1.437 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0137.tif"/>

Using the same power model developed earlier for the load equivalency factor of single-axle loads, the single-axle load that corresponds to a f_LE of 1.433 will be determined as shown below:

⇒

1.437 = 2.1159 × 10 − 17 ( Load ) 3.9120 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0139.tif"/>

Load = 20076 lb https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0140.tif"/>

The MS Excel worksheet used to determine the ESALs in this problem is shown in Figure 13.27.414

13.14

The full-depth asphalt pavement shown in Figure 13.28 is designed to carry an equivalent single-axle loads (ESALs) of 4.5 million for a design period of 20 years at a reliability level of 90%. Determine the terminal serviceability index (p_t) that this asphalt pavement is designed for so that it will carry the design ESALs knowing that the overall standard deviation (S₀) is 0.5 and the initial serviceability index (p_i) is 4.2.

Solution:

The layer coefficient is determined for layer 1 (HMA layer) as below:

⇒

a 1 = 0.0026 ( 3.8 ) 3 − 0.0317 ( 3.8 ) 2 + 0.1798 ( 3.8 ) + 0.0407 = 0.409 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0142.tif"/> 415

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0143.tif"/>

⇒

S N 1 = 0.409 ( 8.0 ) = 3.271 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0144.tif"/>

For a reliability (R) level of 90%, the Z-value is equal to −1.282.

⇒

log 10 ( 4.5 × 10 6 ) = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.271 + 1 ) − 0.20 + log 10 [ Δ PSI ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.271 + 1 ) 5.19 ] + 2.32 log 10 ( 16000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0146.tif"/>

⇒

MS Excel worksheet is used to solve the above equation for ΔPSI. The following solution is obtained:

Δ PSI = 2.2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0147.tif"/>

Δ PSI = p t − p i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0148.tif"/>

⇒

p i = p t − Δ PSI = 4.2 − 2.2 = 2.0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0149.tif"/>

Therefore, a terminal serviceability index (p_t) of 2.0 is used so that the given full-depth asphalt pavement structure with the given design inputs will be able to carry the design ESALs of 4.5 million.

The MS Excel worksheet used to determine ΔPSI and the terminal serviceability index (p_t) is shown in Figure 13.29.416

13.15

A rural highway asphalt pavement with a total structural number (SN) of 3.50 is designed for a design period of 20 years. The truck traffic that is expected to use the highway pavement is of the type shown in Figure 13.30. If the first-year annual average daily traffic (AADT) of this type of trucks is 350 truck/day, determine the load equivalency factor and the maximum allowable load for the tridem axle shown below given the following design inputs (note: ignore the effect of passenger cars):

Given Data:

R (%) = 95%

G = 33.0

f_d = 0.45

S₀ = 0.5

ΔPSI = 2.0

M_R-subgrade = 12000 psi (see Figure 13.30).417

Solution:

S N = 3.50 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0150.tif"/>

For a reliability (R) level of 95%, the Z-value is equal to −1.645.

Now based on the pavement structural number (SN), the pavement structure will be able to carry ESALs calculated based on the design formula below:

⇒

log 10 W 18 = − 1.645 ( 0.5 ) + 9.36 log 10 ( 3.50 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.50 + 1 ) 5.19 ] + 2.32 log 10 ( 12000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0152.tif"/>

⇒

W 18 = 4141294 ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0153.tif"/>

This number represents the design ESALs that the pavement structure can carry through the design life based on the empirical design formula shown above.

But the estimated number of ESALs based on the AADT is calculated as below:

Sample Calculation for the ESALs:

For the 10-kip single axle:

f_d = 0.45 (given in the problem)

N = 1

f_lE = 0.094 (calculated below)

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 10-kip single-axle load as follows:

⇒

f LE = 2.1159 × 10 − 17 ( 10000 ) 3.9120 = 0.094 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0156.tif"/> 418

⇒

ESAL 10 -kip Single = ( 350 ) ( 365 ) ( 0.45 ) ( 33.0 ) ( 1 ) ( 0.094 ) = 178477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0158.tif"/>

The results of the other axle loads are summarized in Table 13.22.

The predicted number of ESALs should be equal to the design ESALs the pavement structure can carry in the design life; therefore,

W 18 = 4141294 = 178477 + 2686655 + 1897088 f LE − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0159.tif"/>

⇒

f LE − x = 0.673 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0160.tif"/>

Using the same power model developed earlier for the load equivalency factor of tridem-axle loads, the tridem-axle load that corresponds to a f_LE of 0.673 will be determined as shown below:

f LE = 6.7846 × 10 − 20 ( Load ) 4.0939 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0161.tif"/>

⇒

0.673 = 6.7846 × 10 − 20 ( Load ) 4.0939 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0162.tif"/>

Load = 43666 lb https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0163.tif"/>

Therefore, the maximum allowable load for the tridem axle of the truck that will use this highway pavement with the given design data is about 44000 lb.

The MS Excel worksheet, used to calculate the estimated ESALs based on the AADT data, the design ESALs based on the empirical design formula using the given design inputs, and the maximum allowable load for the tridem axle of the truck that is expected to use the highway pavement, is shown in Figure 13.31.419

13.16

A rural highway flexible pavement is designed to carry an equivalent single-axle loads (ESALs) of 2 × 10⁶ for a design period of 20 years. The traffic that is expected to use the highway pavement is of the type shown in Figure 13.32. If the first-year annual average daily traffic (AADT) of this type of trucks is 275 trucks/day (the transportation agency in the area is planning to put a limitation on the maximum allowable load for the tridem axle in the trucks using the highway), determine the load equivalency factor and the maximum allowable load of the tridem axle given that the traffic growth factor is 29.8 and the design lane factor is 0.45.

Solution:

The asphalt pavement structure given in this problem is able to carry design ESALs of 2 million for the design period of 20 years.

The estimated number of ESALs based on the AADT is calculated as below:420

Sample Calculation for the ESALs:

For the 12-kip single axle:

f_d = 0.45 (given in the problem)

N = 1

f_lE = 0.192 (calculated below)

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 12-kip single-axle load as follows:

⇒

ESAL 12 -kip Single = ( 275 ) ( 365 ) ( 0.45 ) ( 29.8 ) ( 1 ) ( 0.192 ) = 258408 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0168.tif"/>

The results of the other axle loads are summarized in Table 13.23.

The predicted number of ESALs should be equal to the design ESALs the pavement structure can carry in the design life; therefore,

W 18 = 2000000 = 258408 + 1346029 + 1346029 f LE − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0169.tif"/>

⇒

f LE − x = 0.294 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0170.tif"/>

Using the same power model developed earlier for the load equivalency factor of tridem-axle loads, the tridem-axle load that corresponds to a f_LE of 0.294 will be determined as shown below:

⇒

0.294 = 6.7846 × 10 − 20 ( Load ) 4.0939 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0172.tif"/>

Load = 35669 lb https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0173.tif"/>

Therefore, the maximum allowable load for the tridem axle of the truck that will use this highway pavement with the given design data is about 36 kips.

The MS Excel worksheet used to perform all the computations to determine the maximum allowable load for the tridem axle of the truck that is expected to use the highway pavement is shown in Figure 13.33.

13.17

A highway designer would like to set a limitation on the first-year average annual daily traffic (AADT) for the asphalt pavement design shown in Figure 13.34 if the design is to be used for a multi-lane highway with a traffic mix consisting of 60% passenger cars and 40% trucks of the type shown in Figure 13.35. Determine the AADT given the following design data (note: ignore the effect of passenger cars):

Given Design Data:

R = 90%

Overall standard deviation (S₀) = 0.5

ΔPSI = 2.0

m₂ = 1.0

f_d = 0.40

r = 4%

n = 10 years422

Solution:

The layer coefficient of the HMA layer is determined first:

⇒

a 1 = 0.0026 ( 3.6 ) 3 − 0.0317 ( 3.6 ) 2 + 0.1798 ( 3.6 ) + 0.0407 = 0.398 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0175.tif"/>

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0176.tif"/>

⇒

S N 1 = 0.398 ( 4.0 ) = 1.59 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0177.tif"/>

The coefficient of Layer 2 (the base layer) is given by the following logarithmic formula:

a 2 = 0.249 log ( E 2 ) − 0.977 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0178.tif"/>

⇒

S N 2 = a 1 D 1 + a 2 m 2 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0180.tif"/> 423

⇒

S N 2 = 0.419 ( 4.0 ) + 0.244 ( 1.0 ) ( 6.0 ) = 3.06 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0181.tif"/>

For a reliability (R) level of 90%, the Z-value is equal to −1.282.

⇒

log 10 W 18 = − 1.282 ( 0.5 ) + 9.36 log 10 ( 3.06 + 1 ) − 0.20 + log 10 [ Δ PSI ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.06 + 1 ) 5.19 ] + 2.32 log 10 ( 16000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0183.tif"/>

S N = 3.50 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0184.tif"/>

For a reliability (R) level of 95%, the Z-value is equal to −1.645.

Now based on the pavement structural number (SN), the pavement structure will be able to carry ESALs calculated based on the design formula below:

⇒

log 10 W 18 = − 1.645 ( 0.5 ) + 9.36 log 10 ( 3.50 + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( 3.50 + 1 ) 5.19 ] + 2.32 log 10 ( 15000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0186.tif"/>

⇒

W 18 = 2281793 ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0187.tif"/> 424

This number represents the design ESALs that the pavement structure can carry through the design life based on the empirical design formula shown above.

But the estimated number of ESALs based on the AADT is calculated as below:

Sample Calculation for the ESALs:

For the 12-kip single axle:

f_d = 0.40

G = ( 1 + r ) n − 1 r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0189.tif"/>

⇒

G = ( 1 + 0.04 ) 10 − 1 0.04 = 12.01 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0190.tif"/>

N = 1

f_lE = 0.192 (calculated below)

The power model developed earlier for the load equivalency factor of single-axle loads will be used to compute f_LE for the 12-kip single-axle load as follows:

⇒

ESAL 10 -kip Single = ( AADT ) ( 365 ) ( 0.40 ) ( 12.01 ) ( 1 ) ( 0.192 ) = 134.6 AADT https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0194.tif"/>

The results of the other axle loads are summarized in Table 13.24.425

The predicted number of ESALs should be equal to the design ESALs the pavement structure can carry in the design life; therefore,

W 18 = 2281793 = 835.8 AADT https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0195.tif"/>

⇒

AADT ≅ 2730 veh/day https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0196.tif"/>

Therefore, the annual average daily traffic of the first year for this pavement structure is expected to be at maximum 2730 veh/day so that the pavement structure will be able to carry the given traffic mix under the existing conditions.

The MS Excel worksheet used to calculate the computations of this problem is shown in Figure 13.36.

13.18

A full-depth asphalt pavement is subjected to a circular single-wheel load of 5 kips with a contact pressure of 88 psi that corresponds to a single-axle with dual wheels as shown in Figure 13.37. If the maximum allowable horizontal tensile strain at the bottom of the hot-mix asphalt (HMA) layer is located at the mid-point between the two wheels as shown in the figure and its value is 220 micro-strains, then based on a mechanistic-empirical pavement design method and given the following design inputs, determine the minimum thickness of the HMA layer. Use the Asphalt Institute (AI) distress model (transfer function) for fatigue cracking shown below to perform the mechanistic analysis of the pavement.

N f = 0.0796 ( ε t ) − 3.291 ( E 1 ) − 0.854 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0197.tif"/> 426

Where:

N_f = number of load repetitions to fatigue

ε_t = critical tensile strain at the bottom of the HMA layer

E₁ = elastic modulus of the HMA layer

Given Design Data:

R = 90%

Overall standard deviation (S₀) = 0.5

ΔPSI = 2.0

Solution:

Based on the AI distress model given in the problem:

⇒

N f = 0.0796 ( 220 × 10 − 6 ) − 3.291 ( 420000 ) − 0.854 ≅ 1366927 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0199.tif"/>

This is the number of load repetitions that the pavement structure will be able to carry until fatigue failure based on the AI model.

The load of a single wheel is 5 kips, therefore, for a single-axle with dual wheels, the load would be 20 kips. The load equivalency factor of the 20-kip single-axle load is determined using the power model developed earlier for the load equivalency factor of single-axle loads:

⇒

f LE = 2.1159 × 10 − 17 ( 20000 ) 3.9120 = 1.416 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0201.tif"/>

N f = 1366927 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0202.tif"/>

This is the number of 20-kip load repetitions the pavement structure can carry until fatigue failure. Therefore,427

ESALs = ( N f ) ( f LE ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0203.tif"/>

⇒

ESALs = ( 1366926 ) ( 1.416 ) = 1935841 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0204.tif"/>

Using the empirical design formula for asphalt pavements shown below, the structural number (SN) of the pavement is determined:

log 10 W 18 = Z R S 0 + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ 2.0 ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 ( 12000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0205.tif"/>

Z_R that corresponds to a reliability level (R) of 90% is equal to −1.282.

W₁₈ is the ESALs that corresponds to the load repetitions determined using the mechanistic analysis above. In other words, W₁₈ is equal to 138743. Therefore,

⇒

log 10 ( 1935841 ) = ( − 1.282 ) ( 0.5 ) + 9.36 log 10 ( S N + 1 ) − 0.20 + log 10 [ Δ PSI ( 4.2 − 1.5 ) ] 0.40 + [ 1094 ( S N + 1 ) 5.19 ] + 2.32 log 10 ( 15000 ) − 8.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0207.tif"/>

Using the MS Excel Solver tool to solve for the SN in the above equation, the solution is obtained as follows:

S N = 2.98 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0208.tif"/>

The layer coefficient of the HMA layer is determined using the formula below:

⇒428

a 1 = 0.0026 ( 4.2 ) 3 − 0.0317 ( 4.2 ) 2 + 0.1798 ( 4.2 ) + 0.0407 = 0.429 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0210.tif"/>

S N 1 = a 1 D 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0211.tif"/>

⇒

D 1 = S N 1 a 1 = 2.98 0.429 = 6.9 in ≅ 7.0 in ( ≅ 18 cm ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0212.tif"/>

The MS Excel worksheet used to perform the mechanistic analysis of the asphalt pavement and the empirical design based on the mechanistic analysis is shown in Figure 13.38.

13.19

A circular loading of 4500 lb with a tire pressure of 60 psi is applied on a flexible pavement as shown in Figure 13.39.429

Determine the following:

The vertical compressive stress at point A

The tensile strain at point A

Solution:

The WinJULEA (WinLEA) program is used to perform the analysis of stresses and strains for this pavement.

The following inputs and assumptions are used in the program:

A slip value of 0 is used for full-friction layers or rough surfaces. A slip value of 100000 is used for full slip when there is no friction between layers.

The thickness of the last layer is entered as 0.

The Poisson ratio (PR) is assumed for each layer. A value of 0.35 is typically used for the HMA layer. A value of 0.40 is normally used for base and subbase layers. And a value of 0.45 is used for the subgrade layer.

The values of the applied loads are entered with the X-coordinate and Y-coordinate.

The contact area for each load is calculated using the formula below and entered in the program.

The X-coordinate and Y-coordinate along with the depth for the evaluation points are entered.

The values of the Poisson ratio for the asphalt layer and the subgrade layer are assumed to be 0.35 and 0.45, respectively. The contact area is calculated as follows:

A c = P q https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0213.tif"/>

Whe re:

A_c = contact area

P = applied loading

q = applied pressure

⇒

A c = 4500 60 = 75 in 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0214.tif"/>

The inputs needed for the program are shown in Figure 13.40.430

The evaluation point is located at x = 0, y = 0, and z = 5 in. The results of the analysis after conducting the calculations by hitting “Calculate” are summarized below (see Table 13.25).431

From the above results, the vertical stress at point A (at a depth, z = 5 in) is stress-z = 5.08 psi, and the tensile strain at point A is the horizontal strain-x = −0.238 × 10⁻³ in/in = 238 με. +ve is used for compression and –ve is used for tension.

The vertical compressive stress at point A is equal to 5.08 psi (≌ 35 kPa)

The tensile strain at point A is equal to 238 με.

13.20

A full-depth asphalt pavement is subjected to a circular single-wheel load of 10.2 kips with a contact pressure of 90.0 psi as shown in Figure 13.41. The elastic modulus of the hot-mix asphalt (HMA) layer is 150000 psi and the elastic modulus of the subgrade is 7500 psi. If the maximum allowable horizontal tensile strain at the bottom of the asphalt layer under the centerline of the wheel load is 360 microstrains, determine the following:

The radius of the contact area

The thickness of the asphalt layer (h₁) based on a mechanistic-empirical pavement design method

The number of load repetitions (applications) until fatigue failure using the Asphalt Institute (AI) distress model for fatigue cracking N f = 0.0796 ( ε t ) − 3.291 ( E 1 ) − 0.854 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0215.tif"/>

Solution:

The radius of the contact area is determined as follows:

A c = π a 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0216.tif"/>

⇒

a = A c π https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0217.tif"/>

Where:

A_c = contact area for applied loading

a = radius of contact area

A c = P q https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0218.tif"/>

⇒432

A c = 10200 90 = 113.3 in 2 ( ≅ 731 cm 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0219.tif"/>

⇒

a = 113.3 π = 6.0 in ( 15.2 cm ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0220.tif"/>

The thickness of the asphalt layer (h₁) is determined as shown in the procedure below:

The ratio E₁/E₂ is determined:

E 1 E 2 = 150000 7500 = 20 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0221.tif"/>

Using the critical tensile strain formula shown below:

ε = q E 1 F ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0222.tif"/>

Where:

ε = critical tensile strain at the bottom of the asphalt layer

q = applied wheel pressure

E₁ = elastic modulus of asphalt layer

Fε = tensile strain factor (determined from Figure 13.42)433

⇒

360 × 10 − 6 = 90 150000 F ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0223.tif"/>

⇒

F ε = 0.60 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0224.tif"/>

Using this value (0.60) for the strain factor and the value of 20 for the ratio E₁/E₂, the chart in Figure 13.42 provides the following value for h₁/a:

h 1 a = 1.75 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0225.tif"/>

⇒

h 1 = 1.75 a https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0226.tif"/>

⇒

h 1 = 1.75 ( 6.0 ) = 10.5 in ( 26.7 cm ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0227.tif"/>

The number of load repetitions (applications) until fatigue failure is determined using the AI distress model:

⇒

N f = 0.0796 ( 360 × 10 − 6 ) − 3.291 ( 150000 ) − 0.854 = 651238 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0229.tif"/>

The MS Excel worksheet used to solve this problem is shown in Figure 13.43.

434

13.21

A full-depth asphalt pavement is subjected to a circular single-wheel load of 10 kips with a contact pressure of 88 psi as shown in Pavement System #1 in Figure 13.44. The elastic modulus of the hot-mix asphalt (HMA) layer is 200000 psi and the elastic modulus of the subgrade is 4000 psi. Determine the critical tensile strain at the bottom of the asphalt layer. Based on a mechanistic-empirical pavement design method, what is the number of load repetitions (applications) the pavement structure can carry until fatigue failure using the Asphalt Institute (AI) distress model for fatigue cracking shown in the expression below.

Solution:

The radius of the contact area is determined as follows:

A c = π a 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0231.tif"/> 435

⇒

a = A c π https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0232.tif"/>

But:

A c = P p https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0233.tif"/>

⇒

A c = 10000 88 = 113.6 in 2 ( ≅ 733 cm 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0234.tif"/>

⇒

a = 113.6 π = 6.0 in ( 15.2 cm ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0235.tif"/>

The ratio E₁/E₂ is determined:

E 1 E 2 = 200000 4000 = 50 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0236.tif"/>

The ratio h₁/a is determined:

4.5 6.0 = 0.75 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0237.tif"/>

Using E₁/E₂ = 50 and h₁/a = 0.75, the chart in Figure 13.42 provides the following value for the strain factor (F_ε):

F ε = 2.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0238.tif"/>

Using the critical tensile strain formula shown below:

ε = q E 1 F ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0239.tif"/>

⇒

ε = 88 200000 ( 2.25 ) = 990 × 10 − 6 in/in = 990 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0240.tif"/>

The number of load repetitions (applications) until fatigue failure is determined using the AI distress model:

⇒

N f = 0.0796 ( 990 × 10 − 6 ) − 3.291 ( 200000 ) − 0.854 ≅ 18248 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0242.tif"/>

The MS Excel worksheet used to solve this problem is shown in Figure 13.45.436

13.22

If the load in the pavement system of Problem 13.21 above is to be applied over a set of dual wheels with a center-to-center spacing between the dual wheels of 11.5 inches as shown in Pavement System #2 in Figure 13.46, determine the critical tensile strain at the bottom of the asphalt layer. In this case, what is the number of load repetitions (applications) the pavement structure can carry until fatigue failure using the Asphalt Institute (AI) distress model for fatigue cracking N f = 0.0796 ( ε t ) − 3.291 ( E 1 ) − 0.854 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0243.tif"/> . What is the increase in the design life of the pavement based on the mechanistic analysis?

Solution:

The WinJULEA (WinLEA) program is used to perform the analysis of stresses and strains for this pavement.

The inputs used in the program are shown in Figure 13.47.437

The results of the WinLEA program run are summarized in the report below. Based on these results, the tensile strain at the bottom of the asphalt layer is calculated at three different locations with the coordinates (0, 0, 6), (5.75, 0, 6), and (11.5, 0, 6), where the first number is the X-coordinate, the second number corresponds to the Y-coordinate, and the third number is the Z-coordinate (depth) as shown in Table 13.26. In this case, the first point (0, 0, 6) is simply under the centerline of the first wheel load at a depth of 6 in (at the bottom of the asphalt layer), the second point (5.75, 0, 6) is at the mid-point between the two wheel loads at a depth of 6 in (at the bottom of the asphalt layer), and the third point (11.5, 0, 6) is under the centerline of the second wheel load at a depth of 6 in (at the bottom of the asphalt layer). The tensile strain results for these three points are summarized in Tables 13.27 and 13.28.438

As shown in Table 13.28, the critical tensile strain at the bottom of the asphalt layer is at the middle distance (midpoint) between the two wheels. The value is equal to 970 με. Comparing the critical tensile strain obtained in Pavement System #2 when the load is applied over dual wheels (970 με) with that obtained in Pavement System #1 when the load is applied on a single wheel (990 με). The value is decreased. If the same WinLEA program is used to perform the analysis for Pavement System #1, the critical tensile strain is obtained as 1040 με. In other words, the reduction in the tensile strain between the two cases is equal to:

Reduction in ε t = 1040 − 970 1040 × 100 = 6.7 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0244.tif"/> 439

For dual wheels:

⇒

N f = 0.0796 ( 970 × 10 − 6 ) − 3.291 ( 200000 ) − 0.854 ≅ 19515 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0246.tif"/>

For single wheel:

⇒

N f = 0.0796 ( 1040 × 10 − 6 ) − 3.291 ( 200000 ) − 0.854 ≅ 15516 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0248.tif"/>

⇒

Increase in N f = 19515 − 15516 15516 × 100 = 25.8 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0249.tif"/>

In conclusion, using an axle with dual wheels instead of single wheel will increase the fatigue design life of the asphalt pavement by about 26%.

The MS Excel worksheet used to perform the computations of this problem is shown in Figure 13.48.440

13.23

A circular loading with a radius of 4.0 inches is applied to the three-layer asphalt pavement shown in Figure 13.49. Calculate the critical pavement responses for fatigue cracking and for rutting in this pavement. Using the Asphalt Institute (AI) distress models for fatigue cracking and rutting as transfer functions in a mechanistic-empirical pavement design method, determine the number of load applications the pavement can sustain until fatigue failure and the number of load applications the pavement can carry until rutting failure.

N f = 0.0796 ( ε t ) − 3.291 ( E 1 ) − 0.854 for fatigue cracking https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0250.tif"/>

N f = 1.365 × 10 − 9 ( ε c ) − 4.477 for rutting https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0251.tif"/>

Solution:

The WinJULEA (WinLEA) program is used to perform the analysis of stresses and strains for this pavement.

The inputs used in the program are shown in Figure 13.50.441

The results of the WinLEA program run are summarized in the report below. Based on these results, the critical pavement responses are calculated at two different locations with the coordinates (0, 0, 5) and (0, 0, 15), where the first number is the X-coordinate, the second number corresponds to the Y-coordinate, and the third number is the Z-coordinate (depth). In this case, the first point (0, 0, 5) is simply under the centerline of the wheel load at a depth of 5 in (at the bottom of the asphalt layer), and the second point (0, 0, 15) is at the centerline of the wheel load at a depth of 15 in (at the top of the subgrade layer). The critical pavement response at the bottom of the asphalt layer is the horizontal tensile strain that is responsible for fatigue cracking. On the other hand, the critical pavement response at the top of the subgrade layer is the vertical compressive strain that is responsible for rutting in the asphalt pavement. The results are summarized in Tables 13.29 and 13.30.442

As shown in Table 13.30, the critical tensile strain at the bottom of the asphalt layer is equal to 170 με, and the vertical co pressive strain at the top of the subgrade layer is equal to 246 με.

Using the AI distress model for fatigue cracking shown below:

⇒

N f = 0.0796 ( 170 × 10 − 6 ) − 3.291 ( 600000 ) − 0.854 = 2354857 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0253.tif"/> 443

In other words, the asphalt pavement can carry up to approximately 2.4 million load applications before fatigue failure based on the AI fatigue distress model.

And using the AI distress model for rutting shown below:

N f = 1.365 × 10 − 9 ( ε c ) − 4.477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0254.tif"/>

⇒

N f = 1.365 × 10 − 9 ( 246 × 10 − 6 ) − 4.477 = 19629814 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0255.tif"/>

This means that the asphalt pavement can sustain about 19.6 million load applications before rutting failure based on the AI rutting distress model.

The computations in this problem are performed using the MS Excel worksheet shown in Figure 13.51.

13.24

For the three-layer asphalt pavement system shown in Figure 13.52, determine all the stresses and strains at the axis of symmetry at the interfaces of the pavement layers using the multi-layer elastic theory and the WinLEA program.444

Solution:

Prior to solving the problem, important theory and derivations for formulas that will be used in the solution of the problem will have to be presented below:

Two procedures will be used to solve for the stresses and strains in this problem: (1) the multi-layer elastic theory, and (2) the WinLEA program.

The multi-layer elastic theory and Jones’ tables (from Pavement Analysis and Design by Yang H. Huang, 2004) are used to determine the stresses and strain for three-layer pavement systems as described below:

The locations of all the interface stresses and strains are shown in the three-layered pavement system as illustrated in the Figures 13.53 and 13.54, respectively:445

The following ratios are determined:

k 1 = E 1 E 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0256.tif"/>

13.26 k 2 = E 2 E 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0257.tif"/>

13.27 A = a h 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0258.tif"/>

13.28 H = h 1 h 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0259.tif"/>

Where:

E₁ = elastic modulus of Layer #1 (asphalt layer)

E₂ = elastic modulus of Layer #2 (base layer)

E₃ = elastic modulus of Layer #3 (subgrade layer)

a = radius of contact area of applied load

h₁ = thickness of Layer #1

h₂ = thickness of Layer #2

Using the above ratios and Jones’ tables, the following are determined:

ZZ1, ZZ2, ZZ1−RR1, and ZZ2−RR2

These are stress factors used to calculate the following pavement stresses:

σ z 1 = q ( Z Z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0260.tif"/>

σ z 2 = q ( Z Z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0261.tif"/>

σ z 1 − σ r 1 = q ( Z Z 1 − R R 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0262.tif"/>

σ z 2 − σ r 2 = q ( Z Z 2 − R R 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0263.tif"/> 446

Where:

σ_z1 = vertical stress at Layer Interface 1 (bottom of Layer 1)

σ_z2 = vertical stress at Layer Interface 2 (bottom of Layer 2)

σ_r1= radial stress at Layer Interface 1 (bottom of Layer 1)

σ_r2 = radial stress at Layer Interface 2 (bottom of Layer 2)

q = applied wheel pressure

Since the Poisson ratio for all layers is assumed to be 0.5, the radial and tangential stresses are equal on the axis of symmetry, and using the generalized Hook’s law shown below, the following formulas are obtained for the strains:

ε x , y , or z = 1 E [ σ x , y , or z − v ( σ y , x , or x + σ z , z , or y ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0264.tif"/>

Where:

ε_x = strain in x-direction

σ_x = stress in x-direction

σ_y = stress in y-direction

σ_z = stress in z-direction

E = elastic modulus

Since the Poisson ratio (ν) is equal to 0.5 and σ_r and σ_t are equal on the axis of symmetry, therefore:

13.34 ε z = 1 E ( σ z − σ r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0265.tif"/>

ε r = 1 2 E ( σ r − σ z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0266.tif"/>

ε z = − 2 ε r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0267.tif"/>

Where:

ε_z = vertical strain

ε_r = radial strain

σ_z = vertical stress

σ_r = radial stress

E = elastic modulus

Using these formulas, the tangential strains and radial strains at Interface 1 and Interface 2 are determined as follows:

At Interface 1 (bottom of Layer #1), and since the point is located in Layer #1:

ε z = 1 E ( σ z − σ r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0268.tif"/>

⇒

ε z 1 = 1 E 1 ( σ z 1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0269.tif"/> 447

ε r = 1 2 E ( σ r − σ z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0270.tif"/>

⇒

ε r 1 = 1 2 E 1 ( σ r 1 − σ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0271.tif"/>

Where:

ε_z1 = vertical strain at Interface 1 and bottom of Layer #1

ε_r1 = radial strain at Interface 1 and bottom of Layer #1

σ_z1 = vertical stress at Interface 1 and bottom of Layer #1

σ_r1 = radial stress at Interface 1 and bottom of Layer #1

E₁ = elastic modulus of Layer #1

At Interface 2 (bottom of Layer #2), and since the point is located in Layer #2:

ε z = 1 E ( σ z − σ r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0272.tif"/>

⇒

ε z 2 = 1 E 2 ( σ z 2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0273.tif"/>

ε r = 1 2 E ( σ r − σ z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0274.tif"/>

⇒

13.40 ε r 2 = 1 2 E 2 ( σ r 2 − σ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0275.tif"/>

Where:

ε_z2 = vertical strain at Interface 2 and bottom of Layer #2

ε_r2 = radial strain at Interface 2 and bottom of Layer #2

σ_z2 = vertical stress at Interface 2 and bottom of Layer #2

σ_r2 = radial stress at Interface 2 and bottom of Layer #2

E₂ = elastic modulus of Layer #2

At Interface 1 (top of Layer #2), and since the point is located in Layer #2:

ε z = 1 E ( σ z − σ r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0276.tif"/>

⇒

13.41 ε ´ z 1 = 1 E 2 ( σ ´ z 1 − σ ´ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0277.tif"/> 448

ε r = 1 2 E ( σ r − σ z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0278.tif"/>

⇒

13.42 ε ´ r 1 = 1 2 E 2 ( σ ´ r 1 − σ ´ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0279.tif"/>

Where:

ε ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0280.tif"/> = vertical strain at Interface 1 and top of Layer #2

ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0281.tif"/> = radial strain at Interface 1 and top of Layer #2

σ ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0282.tif"/> = vertical stress at Interface 1 and top of Layer #2

σ ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0283.tif"/> = radial stress at Interface 1 and top of Layer #2

E₂ = elastic modulus of Layer #2

At Interface 2 (top of Layer #3), and since the point is located in Layer #3:

ε z = 1 E ( σ z − σ r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0284.tif"/>

⇒

13.43 ε ´ z 2 = 1 E 3 ( σ ´ z 2 − σ ´ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0285.tif"/>

ε r = 1 2 E ( σ r − σ z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0286.tif"/>

⇒

ε ´ r 2 = 1 2 E 3 ( σ ´ r 2 − σ ´ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0287.tif"/>

Where:

ε ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0288.tif"/> = vertical strain at Interface 2 and top of Layer #3

ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0289.tif"/> = radial strain at Interface 2 and top of Layer #3

σ ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0290.tif"/> = vertical stress at Interface 2 and top of Layer #3

σ ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0291.tif"/> = radial stress at Interface 2 and top of Layer #3

E₃ = elastic modulus of Layer #3

Using the continuity conditions at the interfaces, it implies that the radial strain at the bottom of one layer is equal to the radial strain at the top of the next layer. Therefore, the following formulas are obtained:

ε r 1 = ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0292.tif"/>

ε r 2 = ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0293.tif"/> 449

Where:

ε_r1 = radial strain at Interface 1 and bottom of Layer #1

ε_r2 = radial strain at Interface 2 and bottom of Layer #2

ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0294.tif"/> = radial strain at Interface 1 and top of Layer #2

ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0295.tif"/> = radial strain at Interface 2 and top of Layer #3

ε r 1 = ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0296.tif"/>

⇒

1 2 E 1 ( σ r 1 − σ z 1 ) = 1 2 E 2 ( σ ´ r 1 − σ ´ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0297.tif"/>

⇒

( σ ´ r 1 − σ ´ z 1 ) = ( σ r 1 − σ z 1 ) ( E 1 E 2 ) = ( σ r 1 − σ z 1 ) k 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0298.tif"/>

Or:

13.47 ( σ ´ r 1 − σ ´ z 1 ) = ( σ r 1 − σ z 1 ) k 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0299.tif"/>

ε r 2 = ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0300.tif"/>

⇒

1 2 E 2 ( σ r 2 − σ z 2 ) = 1 2 E 3 ( σ ´ r 2 − σ ´ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0301.tif"/>

⇒

( σ ´ r 2 − σ ´ z 2 ) = ( σ r 2 − σ z 2 ) ( E 2 E 3 ) = ( σ r 2 − σ z 2 ) k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0302.tif"/>

Or:

( σ ´ r 2 − σ ´ z 2 ) = ( σ r 2 − σ z 2 ) k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0303.tif"/>

Based on the continuity conditions, also the following equations are valid:

σ ´ z 1 = σ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0304.tif"/>

σ ´ z 2 = σ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0305.tif"/> 450

ε ´ z 1 = ε z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0306.tif"/>

13.52 ε ´ z 2 = ε z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0307.tif"/>

The sign convention used in the solution for the three-layer pavement systems is as follows: –ve for tension and +ve for compression.

k 1 = E 1 E 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0308.tif"/>

⇒

k 1 = 200000 10000 = 20 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0309.tif"/>

k 2 = E 2 E 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0310.tif"/>

⇒

k 2 = 10000 5000 = 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0311.tif"/>

A = a h 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0312.tif"/>

⇒

A = 4 10 = 0.4 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0313.tif"/>

H = h 1 h 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0314.tif"/>

⇒

H = 5 10 = 0.5 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0315.tif"/>

Using these values and Jones’ tables, the following stress factors are obtained:

Z Z 1 = 0.13480 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0316.tif"/>

Z Z 2 = 0.03998 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0317.tif"/>

Z Z 1 − R R 1 = 1.89817 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0318.tif"/>

Z Z 2 − R R 2 = 0.06722 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0319.tif"/>

σ z 1 = q ( Z Z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0320.tif"/>

σ z 2 = q ( Z Z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0321.tif"/> 451

σ z 1 − σ r 1 = q ( Z Z 1 − R R 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0322.tif"/>

σ z 2 − σ r 2 = q ( Z Z 2 − R R 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0323.tif"/>

⇒

σ z 1 = 100 ( 0.13480 ) = 13.48 psi ( 92.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0324.tif"/>

σ z 2 = 100 ( 0.03998 ) = 4.00 psi ( 27.6 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0325.tif"/>

σ z 1 − σ r 1 = 100 ( 1.89817 ) = 189.8 psi https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0326.tif"/>

⇒

σ r 1 = − 176.3 psi ( 1215.8 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0327.tif"/>

σ z 2 − σ r 2 = 100 ( 0.06722 ) = 6.72 psi https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0328.tif"/>

⇒

σ r 2 = − 2.72 psi ( − 18.8 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0329.tif"/>

ε z 1 = 1 E 1 ( σ z 1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0330.tif"/>

⇒

ε z 1 = 1 200000 ( 189.8 ) = 949 × 10 − 6 = 949 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0331.tif"/>

ε z 2 = 1 E 2 ( σ z 2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0332.tif"/>

⇒

ε z 2 = 1 10000 ( 6.72 ) = 672 × 10 − 6 = 672 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0333.tif"/>

ε r 1 = 1 2 E 1 ( σ r 1 − σ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0334.tif"/>

⇒

ε r 1 = 1 2 ( 200000 ) ( − 189.8 ) = 474.5 × 10 − 6 = 474.5 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0335.tif"/>

ε r 2 = 1 2 E 2 ( σ r 2 − σ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0336.tif"/>

⇒

ε r 2 = 1 2 ( 10000 ) ( − 6.72 ) = 336 × 10 − 6 = 336 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0337.tif"/>

( σ ´ r 1 − σ ´ z 1 ) = ( σ r 1 − σ z 1 ) k 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0338.tif"/> 452

⇒

( σ ´ r 1 − σ ´ z 1 ) = − 189.8 20 = − 9.49 psi https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0339.tif"/>

But due to continuity conditions at layer interfaces:

σ ´ z 1 = σ z 1 = 13.48 psi ( 92.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0340.tif"/>

⇒

σ ´ r 1 = 3.99 psi ( 27.5 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0341.tif"/>

⇒

( σ ´ r 2 − σ ´ z 2 ) = − 6.72 2 = − 3.36 psi https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0343.tif"/>

But due to continuity conditions at layer interfaces:

σ ´ z 2 = σ z 2 = 4.00 psi ( 27.6 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0344.tif"/>

⇒

σ ´ r 2 = 0.64 psi ( 4.4 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0345.tif"/>

ε ´ z 1 = 1 E 2 ( σ ´ z 1 − σ ´ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0346.tif"/>

⇒

ε ´ z 1 = 1 10000 ( 9.49 ) = 949 × 10 − 6 = 949 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0347.tif"/>

ε ´ z 2 = 1 E 3 ( σ ´ z 2 − σ ´ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0348.tif"/>

⇒

ε ´ z 2 = 1 5000 ( 3.36 ) = 672 × 10 − 6 = 672 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0349.tif"/>

ε ´ r 1 = 1 2 E 2 ( σ ´ r 1 − σ ´ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0350.tif"/>

⇒

ε ´ r 1 = 1 2 ( 10000 ) ( − 9.49 ) = − 474.5 × 10 − 6 = − 474.5 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0351.tif"/> 453

⇒

ε ´ r 2 = 1 2 ( 5000 ) ( − 3.36 ) = − 336 × 10 − 6 = − 336 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0353.tif"/>

The pavement response (stress and strain) results at the two-layer interfaces are summarized in Table 13.31.

The MS Excel worksheet shown in Figure 13.55 is used to perform all computations related to the pavement response analysis to determine the stresses and strains at the two-layer interfaces under the centerline of the wheel load.454

The WinLEA program is also used to perform the analysis of the asphalt pavement structure in this problem to compare the results. The same Poisson ratio is used (ν = 0.5) so that the comparison is one-to-one. The inputs used in the WinLEA program are shown in Figure 13.56.

The results obtained from the WinLEA program are shown in Table 13.32.455 456

The pavement response (stress and strain) results at the two-layer interfaces as obtained from the WinLEA program are summarized in Table 13.33.

The results obtained from the WinLEA program are compared with the results obtained from the multilayer elastic theory analysis and Jones’ tables; the results look similar and very comparable.

13.25

For the three-layer asphalt pavement system shown in Figure 13.57, at the axis of symmetry, the radial tensile strain at the bottom of the HMA layer is 285 με, the vertical compressive stress on the top of the base layer is 8.1 psi, the vertical compressive stress at the bottom of the base layer is 2.4 psi, and the radial tensile strain on the top of the subgrade layer is 202 με. Determine the following pavement responses at the axis of symmetry (under the centerline of the wheel load) using the continuity conditions at the layer interfaces and the multi-layer elastic theory formulas:

The vertical compressive strain at the bottom of the asphalt layer

The vertical compressive strain at the top of the base layer

The vertical compressive strain at the bottom of the base layer

The vertical compressive strain at the top of the subgrade layer

The radial tensile strain at the top of the base layer

The radial tensile strain at the bottom of the base layer457

The vertical compressive stress at the bottom of the asphalt layer

The vertical compressive stress at the top of the subgrade layer

The radial tensile stress at the bottom of the asphalt layer

The radial compressive stress at the top of the base layer

The radial tensile stress at the bottom of the base layer

The radial tensile stress at the top of the subgrade layer

Solution:

First, the given pavement responses are summarized based on the description and the sign convention:

ε_r1 = −285 με

σ ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0354.tif"/> = 8.1 psi

σ_z2 = 2.4 psi

ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0355.tif"/> = −202 με

Second, the requirements are determined based on the continuity conditions and the formulas derived from the generalized Hook’s law for strain:

The vertical compressive strain at the bottom of the asphalt layer is calculated as shown below:

This is ε_z1 and is calculated using the formula below:

ε z 1 = − 2 ε r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0356.tif"/>

⇒

ε z 1 = − 2 ( − 285 ) = 570 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0357.tif"/>

The vertical compressive strain at the top of the base layer is determined as below:

This is ε ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0358.tif"/> and is calculated using the formula below:

ε ´ z 1 = ε z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0359.tif"/>

⇒

ε ´ z 1 = 570 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0360.tif"/>

The vertical compressive strain at the bottom of the base layer is determined as follows:

This is ε_z2 and is calculated using the formula below:

ε z 2 = − 2 ε r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0361.tif"/>

And:

ε r 2 = ε ´ r 2 = − 202 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0362.tif"/>

⇒

ε z 2 = − 2 ( − 202 ) = 404 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0363.tif"/> 458

The vertical compressive strain at the top of the subgrade layer is computed as follows:

This is ε ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0364.tif"/> and is calculated using the formula below:

ε ´ z 2 = ε z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0365.tif"/>

⇒

ε ´ z 2 = 404 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0366.tif"/>

The radial tensile strain at the top of the base layer is determined as below:

This is ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0367.tif"/> and is calculated using the formula below:

ε ´ r 1 = ε r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0368.tif"/>

⇒

ε ´ r 1 = − 285 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0369.tif"/>

The radial tensile strain at the bottom of the base layer is calculated as follows:

This is ε_r2 and is calculated using the formula below:

ε r 2 = ε ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0370.tif"/>

⇒

ε r 2 = − 202 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0371.tif"/>

The vertical compressive stress at the bottom of the asphalt layer is computed as shown below:

This is σ_z1 and is calculated using the formula below:

σ z 1 = σ ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0372.tif"/>

⇒

σ z 1 = 8.1 psi ( 55.8 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0373.tif"/>

The vertical compressive stress at the top of the subgrade layer is determined as below:

This is σ ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0374.tif"/> and is calculated using the formula below:

σ ´ z 2 = σ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0375.tif"/>

⇒

σ ´ z 2 = 2.4 psi ( 16.5 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0376.tif"/> 459

The radial tensile stress at the bottom of the asphalt layer is calculated as below:

This is σ_r1 and is calculated using the formula below:

ε r 1 = 1 2 E 1 ( σ r 1 − σ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0377.tif"/>

⇒

− 285 × 10 − 6 = 1 2 ( 250000 ) ( σ r 1 − 8.1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0378.tif"/>

⇒

σ r 1 = − 134.4 psi ( − 926.7 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0379.tif"/>

Or:

ε z 1 = 1 E 1 ( σ z 1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0380.tif"/>

⇒

570 × 10 − 6 = 1 250000 ( 8.1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0381.tif"/>

⇒

σ r 1 = − 134.4 psi ( − 926.7 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0382.tif"/>

The radial compressive stress at the top of the base layer is calculated as below:

This is σ ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0383.tif"/> and is calculated using the formula below:

⇒

− 285 × 10 − 6 = 1 2 ( 12500 ) ( σ ´ r 1 − 8.1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0385.tif"/>

⇒

σ ´ r 1 = 0.975 psi ( 6.7 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0386.tif"/>

Or:

⇒

570 × 10 − 6 = 1 12500 ( 8.1 − σ ´ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0388.tif"/> 460

⇒

σ ´ r 1 = 0.975 psi ( 6.7 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0389.tif"/>

The radial tensile stress at the bottom of the base layer is computed as follows:

This is σ_r2 and is calculated using the formula below:

ε r 2 = 1 2 E 2 ( σ r 2 − σ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0390.tif"/>

⇒

− 202 × 10 − 6 = 1 2 ( 12500 ) ( σ r 2 − 2.4 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0391.tif"/>

⇒

σ r 2 = − 2.7 psi ( − 18.3 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0392.tif"/>

Or:

ε z 2 = 1 E 2 ( σ z 2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0393.tif"/>

⇒

404 × 10 − 6 = 1 12500 ( 2.4 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0394.tif"/>

⇒

σ r 2 = − 2.7 psi ( − 18.3 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0395.tif"/>

The radial tensile stress at the top of the subgrade layer is determined as shown below:

This is σ ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0396.tif"/> and is calculated using the formula below:

⇒

− 202 × 10 − 6 = 1 2 ( 6250 ) ( σ ´ r 2 − 2.4 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0398.tif"/>

⇒

σ ´ r 2 = − 0.125 psi ( − 0.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0399.tif"/> 461

Or:

⇒

404 × 10 − 6 = 1 6250 ( 2.4 − σ ´ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0401.tif"/>

⇒

σ ´ r 2 = − 0.125 psi ( − 0.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0402.tif"/>

The MS Excel worksheet shown in Figure 13.58 is used to perform all computations related to the pavement response analysis to determine the stresses and strains at the two-layer interfaces under the centerline of the wheel load.

13.26

For the three-layer asphalt pavement system shown in Figure 13.59, at the axis of symmetry, the radial tensile strain on the top of the base layer is 230 με, the vertical compressive stress at the bottom of the HMA layer is 8.5 psi, the vertical compressive stress on the top of the subgrade layer is 4.2 psi, and the radial tensile strain at the bottom of the base layer is 220 με. Determine the following pavement responses at the axis of symmetry (under the 462centerline of the wheel load) using the continuity conditions at the layer interfaces and the multi-layer elastic theory formulas:

The vertical compressive strain at the bottom of the asphalt layer

The vertical compressive strain at the top of the base layer

The vertical compressive strain at the bottom of the base layer

The vertical compressive strain at the top of the subgrade layer

The radial tensile strain at the bottom of the asphalt layer

The radial tensile strain at the top of the subgrade layer

The vertical compressive stress at the top of the base layer

The vertical compressive stress at the bottom of the base layer

The radial tensile stress at the bottom of the asphalt layer

The radial compressive stress at the top of the base layer

The radial tensile stress at the bottom of the base layer

The radial compressive stress at the top of the subgrade layer

Solution:

First, the given pavement responses are summarized based on the description and the sign convention:

ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0403.tif"/> = −230 με

σ_z1 = 8.5 psi

σ ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0404.tif"/> = 4.2 psi

ε_r2 = −220 με

Second, the requirements are determined based on the continuity conditions and the formulas derived from the generalized Hook’s law for strain:

The vertical compressive strain at the bottom of the asphalt layer is calculated as shown below:

This is ε_z1 and is calculated using the formula below:

ε z 1 = − 2 ε r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0405.tif"/> 463

But:

ε r 1 = ε ´ r 1 = − 230 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0406.tif"/>

⇒

ε z 1 = − 2 ( − 230 ) = 460 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0407.tif"/>

The vertical compressive strain at the top of the base layer is determined as below:

This is ε ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0408.tif"/> and is calculated using the formula below:

ε ´ z 1 = ε z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0409.tif"/>

⇒

ε ´ z 1 = 460 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0410.tif"/>

The vertical compressive strain at the bottom of the base layer is determined as follows:

This is ε_z2 and is calculated using the formula below:

ε z 2 = − 2 ε r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0411.tif"/>

⇒

ε z 2 = − 2 ( − 220 ) = 440 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0412.tif"/>

The vertical compressive strain at the top of the subgrade layer is computed as follows:

This is ε ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0413.tif"/> and is calculated using the formula below:

ε ´ z 2 = ε z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0414.tif"/>

⇒

ε ´ z 2 = 440 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0415.tif"/>

The radial tensile strain at the bottom of the asphalt layer is determined as below:

This is ε_r1 and is calculated using the formula below:

ε r 1 = ε ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0416.tif"/>

⇒

ε r 1 = − 230 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0417.tif"/> 464

The radial tensile strain at the top of the subgrade layer is calculated as follows:

This is ε_r2 and is calculated using the formula below:

ε ´ r 2 = ε r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0418.tif"/>

⇒

ε ´ r 2 = − 220 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0419.tif"/>

The vertical compressive stress at the top of the base layer is computed as shown below:

This is σ ´ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0420.tif"/> and is calculated using the formula below:

σ ´ z 1 = σ z 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0421.tif"/>

⇒

σ ´ z 1 = 8.5 psi ( 58.6 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0422.tif"/>

The vertical compressive stress at the bottom of the base layer is determined as below:

This is σ_z2 and is calculated using the formula below:

σ z 2 = σ ´ z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0423.tif"/>

⇒

σ z 2 = 4.2 psi ( 29.0 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0424.tif"/>

The radial tensile stress at the bottom of the asphalt layer is calculated as below:

This is σ_r1 and is calculated using the formula below:

ε r 1 = 1 2 E 1 ( σ r 1 − σ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0425.tif"/>

⇒

− 230 × 10 − 6 = 1 2 ( 300000 ) ( σ r 1 − 8.5 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0426.tif"/>

⇒

σ r 1 = − 129.5 psi ( 892.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0427.tif"/>

Or:

ε z 1 = 1 E 1 ( σ z 1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0428.tif"/> 465

⇒

460 × 10 − 6 = 1 300000 ( 8.5 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0429.tif"/>

⇒

σ r 1 = − 129.5 psi ( − 892.9 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0430.tif"/>

The radial compressive stress at the top of the base layer is calculated as below:

This is σ ´ r 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0431.tif"/> and is calculated using the formula below:

⇒

− 230 × 10 − 6 = 1 2 ( 15000 ) ( σ ´ r 1 − 8.5 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0433.tif"/>

⇒

σ ´ r 1 = 1.6 psi ( 11.0 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0434.tif"/>

Or:

⇒

460 × 10 − 6 = 1 15000 ( 8.5 − σ ´ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0436.tif"/>

⇒

σ ´ r 1 = 1.6 psi ( 11.0 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0437.tif"/>

The radial tensile stress at the bottom of the base layer is computed as follows:

This is σ_r2 and is calculated using the formula below:

ε r 2 = 1 2 E 2 ( σ r 2 − σ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0438.tif"/>

⇒

− 220 × 10 − 6 = 1 2 ( 15000 ) ( σ r 2 − 4.2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0439.tif"/> 466

⇒

σ r 2 = − 2.4 psi ( − 16.5 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0440.tif"/>

Or:

ε z 2 = 1 E 2 ( σ z 2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0441.tif"/>

⇒

440 × 10 − 6 = 1 15000 ( 4.2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0442.tif"/>

⇒

σ r 2 = − 2.4 psi ( − 16.5 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0443.tif"/>

The radial compressive stress at the top of the subgrade layer is determined as shown below:

This is σ ´ r 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0444.tif"/> and is calculated using the formula below:

⇒

− 220 × 10 − 6 = 1 2 ( 7500 ) ( σ ´ r 2 − 4.2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0446.tif"/>

⇒

σ ´ r 2 = 0.9 psi ( 6.2 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0447.tif"/>

Or:

⇒

440 × 10 − 6 = 1 7500 ( 4.2 − σ ´ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0449.tif"/>

⇒

σ ´ r 2 = 0.9 psi ( 6.2 kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0450.tif"/> 467

The MS Excel worksheet shown in Figure 13.60 is used to perform all computations related to the pavement response analysis to determine the stresses and strains at the two-layer interfaces under the centerline of the wheel load.

13.27

For the pavement structure shown in Figure 13.61, the vertical stress (kPa) and radial strain (με) under the centerline of the wheel load (at the axis of symmetry) are plotted versus depth within the pavement structure as shown in Figure 13.62.468

By using +ve for compression and −ve for tension as sign convention and assuming the Poisson ratio equal to 0.5 for all layers, determine the following:

The radial stress at the bottom of the asphalt layer

The radial stress at the top of the base layer

The radial stress at the bottom of the base layer

The radial stress at the top of the subgrade layer

The vertical strain at the bottom of the asphalt layer

The vertical strain at the bottom of the base layer

The applied wheel pressure

The radius of contact area if the applied wheel loading is 40 kN

Solution:

The radial stress at the bottom of the asphalt layer is determined as below:

At the bottom of the asphalt layer, the depth is equal to 15 cm, the following pavement responses are obtained from the figure:

σ z 1 = 410 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0451.tif"/>

ε r 1 = − 550 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0452.tif"/>

ε r 1 = 1 2 E 1 ( σ r 1 − σ z 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0453.tif"/>

⇒

− 550 × 10 − 6 = 1 2 ( 1500 × 10 3 ) ( σ r 1 − 410 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0454.tif"/> 469

⇒

σ r 1 = − 1240 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0455.tif"/>

The radial stress at the top of the base layer is determined as follows:

At the top of the base layer, the depth is equal to 15 cm, the following pavement responses are obtained from the figure:

σ ´ z 1 = σ z 1 = 410 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0456.tif"/>

ε ´ r 1 = ε r 1 = − 550 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0457.tif"/>

⇒

− 550 × 10 − 6 = 1 2 ( 350 × 10 3 ) ( σ ´ r 1 − 410 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0459.tif"/>

⇒

σ ´ r 1 = 25 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0460.tif"/>

The radial stress at the bottom of the base layer is calculated as shown below:

At the bottom of the base layer, the depth is equal to 65 cm, the following pavement responses are obtained from the figure:

σ z 2 = 25 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0461.tif"/>

ε r 2 = − 180 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0462.tif"/>

ε r 2 = 1 2 E 2 ( σ r 2 − σ z 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0463.tif"/>

⇒

− 180 × 10 − 6 = 1 2 ( 350 × 10 3 ) ( σ r 2 − 25 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0464.tif"/>

⇒

σ r 2 = − 101 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0465.tif"/> 470

The radial stress at the top of the subgrade layer is determined as below:

At the top of the subgrade layer, the depth is equal to 65 cm, the following pavement responses are obtained from the figure:

σ ´ z 2 = σ z 2 = 25 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0466.tif"/>

ε ´ r 2 = ε r 2 = − 180 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0467.tif"/>

⇒

− 180 × 10 − 6 = 1 2 ( 65 × 10 3 ) ( σ ´ r 2 − 25 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0469.tif"/>

⇒

σ ´ r 2 = 1.6 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0470.tif"/>

The vertical strain at the bottom of the asphalt layer is calculated as follows:

ε z 1 = 1 E 1 ( σ z 1 − σ r 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0471.tif"/>

⇒

ε z 1 = 1 1500 × 10 3 ( 410 − ( − 1240 ) ) = 1100 × 10 − 6 = 1100 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0472.tif"/>

The vertical strain at the bottom of the base layer is determined as below:

ε z 2 = 1 E 2 ( σ z 2 − σ r 2 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0473.tif"/>

⇒

ε z 2 = 1 350 × 10 3 ( 25 − ( − 101 ) ) = 504 × 10 − 6 = 504 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0474.tif"/>

The applied wheel pressure is determined as follows:

It is equal to the vertical stress at the surface of the pavement (at a depth of 0 cm). Therefore, from the figure at a depth of 0, the vertical stress is equal to 690 kPa.

q = 690 kPa https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0475.tif"/>

The radius of contact area if the applied wheel loading is 40 kN is determined using the formula below:471

A c = P q https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0476.tif"/>

⇒

A c = 40 690 = 0.058 m 2 = 580 cm 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0477.tif"/>

A c = π a 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0478.tif"/>

⇒

a = A c π = 580 π = 13.6 cm https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0479.tif"/>

The MS Excel worksheet shown in Figure 13.63 is used to perform the computations related to the results in this problem.

13.28

For the full-depth asphalt pavement structure shown in Figure 13.64, the relationship between the critical tensile strain at the bottom of the asphalt layer (ε_t) in micro-strains and the applied truck wheel pressure or load intensity (q) in psi is given by a linear model that has the following expression:472

ε t = 1.9 q + 178 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0480.tif"/>

and the relationship between the tensile strain at the bottom of the asphalt layer (ε_t) in micro-strains and the thickness of the asphalt layer (h₁) in inches is given by a logarithmic model as shown in the expression below:

ε t = 760 − 240 ln ( h 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0481.tif"/>

If the allowable truck loading is changed such that the applied pressure (q) is increased by 1.5 times from 80 psi to 120 psi; and based on that, the design thickness of the asphalt layer (h₁) is also increased by 1.5 times, (a) What is the current value of h₁? (b) What is the new value of h₁ after the increase? (c) Using a mechanistic-empirical pavement design procedure, is the new thickness of the asphalt layer enough for the new loading? And why? (Assume the relationships are still valid after the change in q and h₁.)

Solution:

The current value of h₁ is determined as follows:

At q = 80 psi, the tensile strain at the bottom of the asphalt layer is determined using the given relationship:

ε t = 1.9 q + 178 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0482.tif"/>

⇒

ε t = 1.9 ( 80 ) + 178 = 330 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0483.tif"/>

The thickness of the asphalt layer (h₁) is determined using the given relationship between ε_t and h₁:

ε t = 760 − 240 ln ( h 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0484.tif"/>

⇒

330 = 760 − 240 ln ( h 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0485.tif"/>

⇒

h 1 = 6.0 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0486.tif"/> 473

If the thickness of the asphalt layer (h₁) is increased by 1.5 times; therefore, the new value of the thickness is equal to:

h 1 = 1.5 ( 6.0 ) = 9.0 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0487.tif"/>

The new thickness (h₁ = 9.0 in) will imply that the tensile strain at the bottom of the asphalt layer is:

ε t = 760 − 240 ln ( h 1 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0488.tif"/>

⇒

ε t = 760 − 240 ln ( 9.0 ) = 232.7 μ ε ≅ 233 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0489.tif"/>

Therefore, the reduction value in the tensile strain is equal to:

Δ ε t = 330 − 233 = 97 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0490.tif"/>

After the increase in loading (q) from 80 psi to 120 psi, the new tensile strain is calculated as:

⇒

ε t = 1.9 ( 120 ) + 178 = 406 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0491.tif"/>

However, the increase in loading (from 80 to 120 psi) results in an increase in the tensile strain from 330 με to 406 με. In other words, the change in the tensile strain is

Δ ε t = 406 − 330 = 76 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0492.tif"/>

Since the reduction in the tensile strain due to the increase in the thickness of the asphalt layer is higher than the increase in the tensile strain as a result of the increase in loading, it implies that the new design thickness of the asphalt layer (9.0 inches) would be adequate for the new loading.

The MS Excel worksheet used to conduct the computations of this problem is shown Figure 13.65.474

13.29

A full-depth asphalt pavement is expected to carry a design traffic loading of 4 million ESALs for a design life of 10 years. The pavement will be subjected to a critical circular single-wheel load of 4.5 kips with a contact pressure of 100 psi as shown in Figure 13.66. Based on a mechanistic-empirical pavement design method, determine the maximum allowable initial tensile strain at the bottom of the asphalt layer (ε_t) and the required design thickness of the full-depth asphalt layer. Use the Asphalt Institute (AI) distress model for fatigue performance shown below:

Where:

N_f = number load applications to fatigue failure

ε_t = tensile strain at the bottom of asphalt layer

E₁ = elastic modulus of asphalt layer (psi)

(Assume that N_f is in ESALs.)475

Solution:

Using the AI distress model for fatigue performance, the tensile strain at the bottom of the asphalt layer is calculated based on the design ESALs the pavement structure is designed for:

⇒

4 × 10 6 = 0.0796 ( ε t ) − 3.291 ( 400000 ) − 0.854 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0495.tif"/>

Solving for ε_t provides the following solution:

ε t = 160.8 × 10 − 6 ≅ 161 μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0496.tif"/>

The tensile strain formula is used to determine the strain factor:

ε = q E 1 F ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0497.tif"/>

⇒

161 × 10 − 6 = 100 400000 F ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0498.tif"/>

⇒

F ε = 0.64 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0499.tif"/>

Using Figure 13.42 and the ratio E₁/E₂ = 400000/20000 = 20, the ratio h₁/a is determined as:

h 1 a = 1.6 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0500.tif"/>

But:

a = ( P q ) π https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0501.tif"/>

⇒

a = ( 4500 100 ) π = 3.78 in https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0502.tif"/>

⇒

h 1 = 1.6 ( 3.78 ) ≅ 6.0 in ( 15 cm ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0503.tif"/>

The MS Excel worksheet used to perform the computations in this problem is shown in Figure 13.67.476

13.30

Develop equivalent axle load factors (EALFs) for the pavement structure shown in Figure 13.68 using the following formula:

EALF = N f − 18 N f − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0504.tif"/>

Where:

EALF = equivalent axle load factor for single axles

N_f-18 = number of 18-kip single-axle load applications (repetitions) to failure (fatigue or rutting failure)

N_f-x = number of x-kip single-axle load applications (repetitions) to failure (fatigue or rutting failure)

The axle is a single axle with dual wheels.

Use the following Asphalt Institute (AI) distress models for fatigue and rutting performances:

N f = 0.0796 ( ε t ) − 3.291 ( E 1 ) − 0.854 For fatigue performance https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0505.tif"/>

N f = 1.365 × 10 − 9 ( ε c ) − 4.477 For rutting performance https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0506.tif"/> 477

The following single-axle load values are used in the analysis (see Table 13.34).

Determine the EALFs once using the AI fatigue distress model and another time using the AI rutting distress model. Compare the results obtained using the two models. Also compare the results with the AI’s EALFs. The WinLEA program will be run 13 times to perform the pavement analysis for the 13 different loadings.

Solution:

The WinLEA program is used to perform the pavement analysis for critical tensile strain at the bottom of the asphalt layer and critical compressive strain on the top of the subgrade layer.

The single-axle load is applied on a set of dual wheels as shown in Table 13.35. Therefore, the wheel load is equal to the given load divided by four since the axle has four wheels. The contact area is determined by simply dividing the single wheel load by the 478applied pressure (100 psi). The results are shown in Table 13.35. These results are used as inputs for the WinLEA program to run the analysis.

The WinLEA pavement response results are shown in Table 13.36.479

Based on the WinLEA pavement response results, the fatigue and rutting performances are determined using the AI distress models as shown below:

Sample Calculation:

For load = 5000 lb,

Based on the AI fatigue distress model, the EALF is calculated as follows:

EALF = N f − 18 N f − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0507.tif"/>

⇒

EALF = 0.0796 ( ε t − 18 ) − 3.291 ( E 1 ) − 0.854 0.0796 ( ε t − x ) − 3.291 ( E 1 ) − 0.854 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0508.tif"/>

⇒

EALF = ( ε t − 18 ) − 3.291 ( ε t − x ) − 3.291 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0509.tif"/>

Or:

EALF = ( ε t − 18 ε t − x ) − 3.291 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0510.tif"/>

Where:

EALF = equivalent axle load factor for single axles

ε_t-18 = maximum (critical) tensile strain at the bottom of the asphalt layer for the 18-kip single-axle load

ε_t-x = maximum (critical) tensile strain at the bottom of the asphalt layer for the x-kip single-axle load

Sulubstituting the tensile strains for x = 5000 lb and for the standard 18000 lb-load provides the following:

EALF = ( ε t − 18 ε t − 5 ) − 3.291 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0511.tif"/>

⇒

EALF = ( 234 98 ) − 3.291 = 0.05702 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0512.tif"/>

And based on the AI rutting distress model, the EALF is calculated as follows:

EALF = N f − 18 N f − x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0513.tif"/>

⇒

EALF = 1.365 × 10 − 9 ( ε c − 18 ) − 4.477 1.365 × 10 − 9 ( ε c − x ) − 4.477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0514.tif"/> 480

⇒

EALF = ( ε c − 18 ) − 4.477 ( ε c − x ) − 4.477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0515.tif"/>

Or:

EALF = ( ε c − 18 ε c − x ) − 4.477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0516.tif"/>

Where:

EALF = equivalent axle load factor for single axles

ε_c-18 = maximum (critical) vertical compressive strain on the top of the subgrade layer for the 18-kip single-axle load

ε_c-x = maximum (critical) vertical compressive strain on the top of the subgrade layer for the x-kip single-axle load

Substituting the vertical compressive strains for x = 5000 lb and for the standard 18000 lb-load provides the following:

EALF = ( ε c − 18 ε c − 5 ) − 4.477 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0517.tif"/>

⇒

EALF = ( 583 170 ) − 4.477 = 0.00402 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0518.tif"/>

The EALF results for all other loads based on fatigue performance and rutting performance are summarized in Tables 13.37 and 13.38, respectively.481

By comparing the results based on the analysis in this problem with the EALF values of the Asphalt Institute (AI), it is revealed that the results based on rutting performance provide more reasonable values compared to those based on fatigue performance.

The MS Excel worksheet that is used to perform the computations of this problem in an easy and efficient way is shown in Figure 13.69.482

13.31

Repeat Problem 13.30 above and perform a similar analysis but for the pavement structure shown in Figure 13.70 where a full-depth asphalt pavement has been used instead of a conventional asphalt pavement.

Solution:

The WinLEA program is used to perform the pavement analysis for critical tensile strain at the bottom of the asphalt layer and critical compressive strain on the top of the subgrade layer.

The single-axle load is applied on a set of dual wheels as shown in the diagram. Therefore, the wheel load is equal to the given load divided by four since the axle has four wheels. The contact area is determined by simply dividing the single wheel load over the applied pressure (100 psi). The results are shown in Table 13.39. These results are used as inputs for the WinLEA program to run the analysis.483

The WinLEA pavement response results are shown in Table 13.40.

Based on the WinLEA pavement response results, the fatigue and rutting performances are determined using the AI distress models as shown below:

Sample Calculation:

For load = 5000 lb,

Based on the AI fatigue distress model, the EALF is calculated as follows:

Substituting the tensile strains for x = 5000 lb and for the standard 18000 lb-load provides the following:

⇒

EALF = ( 264 86 ) − 3.291 = 0.02494 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0521.tif"/>

And based on the AI rutting distress model, the EALF is calculated as follows:

Substituting the vertical compressive strains for x = 5000 lb and for the standard 18000 lb-load provides the following:

⇒

EALF = ( 587 189 ) − 4.477 = 0.00626 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0524.tif"/>

The EALF results for all other loads based on fatigue performance and rutting performance are summarized in Tables 13.41 and 13.42, respectively.485 486

By comparing the results based on the analysis in this problem with the EALF values of the Asphalt Institute (AI), it can be seen that the results based on rutting performance provide again more reasonable values compared to those based on fatigue performance. However, the EALF values show some improvement when compared with the results for conventional asphalt pavement in the previous problem.

The MS Excel worksheet that is used to perform all the computations in this problem efficiently is shown in Figure 13.71.

Table 13.1 Typical Values and Flexible Pavement Properties for Problem 13.1 Property Typical Value Modulus of Elasticity for HMA 80 psi Resilient Modulus for Base Materials 12 inches Vertical Stress on top of subgrade 8,000 psi Dual Wheel Spacing 40 ksi Resilient Modulus for Subgrade 250 ksi Applied Single Wheel Loading 9,000 lb Applied Wheel Pressure for Trucks 10 psi Table 13.2 Matching of Typical Values for Flexible Pavement Properties for Problem 13.1 Property Answer Modulus of Elasticity for HMA 250 ksi Resilient Modulus for Base Materials 40 ksi Vertical Stress on top of subgrade 10 psi Dual Wheel Spacing 12 inches Resilient Modulus for Subgrade 8,000 psi Applied Single Wheel Loading 9,000 lb Applied Wheel Pressure for Trucks 80 psi Table 13.3 Effect of Design Inputs on HMA Design Thickness for Problem 13.2 Item Increase Decrease No Effect Base Resilient Modulus (M_R) Δpsi Reliability Level, R (%) S₀ HMA Modulus Subgrade CBR Terminal Serviceability Index, p_t Traffic ESALs Table 13.4 Identifying the Effect of Design Inputs on HMA Design Thickness for Problem 13.2 Item Increase Decrease No Effect Base Resilient Modulus (M_R) X Δpsi X Reliability Level, R (%) X S₀ X HMA Modulus X Subgrade CBR X Terminal Serviceability Index, p_t X Traffic ESALs X Figure 13.1 Flexible pavement structures under traffic loading to be ranked for Problem 13.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_1_B.tif"/> Table 13.5 Ranking of the Given Flexible Pavement Structures in Problem 13.3 Order or Ranking of Pavement Structures from Best to Worst 6 5 2 1 3 4 Figure 13.2 Mistakes in the design of flexible pavement structures for Problem 13.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_2_B.tif"/> Table 13.6 Identifying the Mistakes in the Design of Flexible Pavement Structures for Problem 13.4 Pavement Design Mistake Correction 1 E₁ = 100 ksi < E₂ = 300 ksi E₁= 300 ksi > E₂ = 100 ksi 2 D₁ = 5 in > D₂ = 4 in D₁ = 4 in < D₂ = 5 in 3 E₂ = 100 ksi (too high) E₂ = 10 ksi (typical) Figure 13.3 Truck traffic mix and loadings for the six-lane rural highway pavement in Problem 13.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_3_B.tif"/> Figure 13.4 Load versus load equivalency factors for flexible pavements (single-axle). Based on numbers in the table of the load equivalency factors from the Asphalt Institute’s <italic>Thickness Design Manual (MS-1)</italic>, 9th edition, 1982. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_4_B.tif"/> Figure 13.5 Load versus load equivalency factors for flexible pavements (tandem-axle). Based on numbers in the table of the load equivalency factors from the Asphalt Institute’s <italic>Thickness Design Manual (MS-1)</italic>, 9th edition, 1982. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_5_B.tif"/> Figure 13.6 Load versus load equivalency factors for flexible pavements (tridem-axle). Based on numbers in the table of the load equivalency factors from the Asphalt Institute’s <italic>Thickness Design Manual (MS-1)</italic>, 9th edition, 1982. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_6_B.tif"/> Table 13.7 ESALs Results for Problem 13.5 Axle Type AADT 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_1_B.tif"/> Single Axle 600 365 0.4 12.01 1 0.094 98,947 Single Axle 2 0.192 403,821 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_2_B.tif"/> Single Axle 450 1 0.094 74,210 Single Axle 1 0.192 151,433 Tandem Axle 1 1.271 1,002,819 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_3_B.tif"/> Single Axle 300 1 0.094 49,473 Single Axle 1 0.192 100,955 Tridem Axle 1 0.694 364,948 Total ESALs 2,246,606 Figure 13.7 Image of the MS Excel worksheet used for the computations of ESALs for Problem 13.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_7_B.tif"/> Figure 13.8 Type of truck traffic for the highway pavement of Problem 13.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_8_B.tif"/> Table 13.8 ESALs Computations and Results for Problem 13.6 Axle Type AADT 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_4_B.tif"/> Single Axle 200 365 0.45 24.30 1 0.351 280,055 Tandem Axle 4 0.252 803,811 Total ESALs 1,083,866 Figure 13.9 Image of the MS Excel worksheet used for the computations of ESALs for Problem 13.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_9_B.tif"/> Table 13.10 Material Properties for Designing the Highway Flexible Pavement in Problem 13.7 Material E (ksi) HMA 400 Base Course 20 Subbase Course 20 Subgrade 10 Figure 13.10 Sketch of the flexible pavement structure for Problem 13.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_10_B.tif"/> Table 13.9 Truck Traffic Mix and Loadings for Problem 13.7 Vehicle Type Percentage, % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_5_B.tif"/> 15 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_6_B.tif"/> 15 Figure 13.11 Resilient modulus versus layer coefficient of asphalt layer. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_11_B.tif"/> Table 13.11 Reliability Recommended Levels Based on the Functional Classification of the Highway Functional Classification Recommended Level of Reliability (%) Urban Rural Interstate and Other Freeways 85–99.9 80–99.9 Principal Arterials 80–99 75–95 Collectors 80–95 75–95 Local 50–80 50–80 Reproduced with Permission from AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, D.C, USA. Table 13.12 Standard Normal Deviates (Z-Values) at Various Levels of Reliability, R Reliability, R (%) Z_R 50 0.000 60 −0.253 70 −0.524 75 −0.674 80 −0.841 85 −1.037 90 −1.282 95 −1.645 97 −1.881 99 −2.327 99.9 −3.090 Reproduced with Permission from AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA. Table 13.13 Percentage of Total Truck Traffic in Design Lane (fd) Number of Traffic Lanes in Two Directions % of Trucks in Design Lane 2 50 4 45 (35–48) ≥ 6 40 (25–48) Reproduced with Permission from “Thickness Design-Asphalt Pavements for Highways and Streets MS-1”, 1999, by Asphalt Institute, Lexington, KY, USA. Table 13.14 Lane Distribution Factor (fd) Number of Lanes in Each Direction % of 18-kip ESAL in Design Lane 1 100 2 80–100 3 60–80 4 50–75 Reproduced with Permission from AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA. Table 13.15 ESALs Computations and Results for Problem 13.7 Truck Type Axle Type AADT 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_7_B.tif"/> Single 750 365 0.40 12.58 1 0.094 129,574 Tandem 2 1.271 3,501,927 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_8_B.tif"/> Single 750 1 0.094 129,574 Single 1 0.192 264,408 Tridem 1 0.694 955,822 Total ESALs 4,981,305 Figure 13.12 Image of the MS Excel worksheet used for the determination of SN values and design thicknesses for Problem 13.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_12_B.tif"/> Table 13.16 Values of SN1, SN2, and SN3 for the Flexible Pavement of Problem 13.7 Structural Number Value SN₁ 3.32 SN₂ 3.32 SN3 4.23 Table 13.17 Recommended Values of Drainage Coefficients (m) for Untreated Bases and Subbases in Flexible Pavements Quality of Drainage Percentage of Time Pavement Structure Is Exposed to Moisture Levels Approaching Saturation Rating Water removed within < 1% 1–5% 5–25% > 25% Excellent 2 hours 1.40−1.35 1.35−1.30 1.30−1.20 1.20 Good 1 day 1.35−1.25 1.25−1.15 1.15−1.00 1.00 Fair 1 week 1.25−1.15 1.15−1.05 1.00−0.80 0.80 Poor 1 month 1.15−1.05 1.05−0.80 0.80−0.60 0.60 Very Poor Never drain 1.05−0.95 0.95−0.75 0.75−0.40 0.40 Reproduced with Permission from AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA. Figure 13.13 Flexible pavement structure and design thicknesses for Problem 13.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_13_B.tif"/> Table 13.18 Design Thicknesses of Asphalt and Base Layers for Problem 13.7 Thickness in (cm) D₁ 8 (≌ 20) D₂ 8 (≌ 20) Table 13.19 Material Properties for Designing the Highway Flexible Pavement in Problem 13.8 Material E (ksi) HMA 400 Base Course 80 Subbase Course 20 Subgrade 10 Figure 13.14 Sketch of the flexible pavement structure for Problem 13.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_14_B.tif"/> Table 13.20 Values of SN1, SN2, and SN3 for the Flexible Pavement of Problem 13.8 Structural Number Value SN₁ 1.99 SN₂ 3.32 SN3 4.23 Figure 13.15 Image of the MS Excel worksheet used for the determination of SN values and design thicknesses for Problem 13.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_15_B.tif"/> Table 13.21 Final Design Thicknesses for the Flexible Pavement of Problem 13.8 Thickness in (cm) D₁ 5 (≌ 12.5 cm) D₂ 6 (≌ 15 cm)

₃

7 (≌ 18 cm) Figure 13.16 Flexible pavement structure for Problem 13.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_16_B.tif"/> Figure 13.17 Sketch for SN1 and SN2 on the flexible pavement structure for Problem 13.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_17_B.tif"/> Figure 13.18 Image of the MS Excel worksheet used for the determination of ESALs for Problem 13.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_18_B.tif"/> Figure 13.19 Type of truck traffic for Problem 13.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_19_B.tif"/> Figure 13.20 Image of the MS Excel worksheet used for determining the AADT of Problem 13.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_20_B.tif"/> Figure 13.21 Type of truck traffic for Problem 13.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_21_B.tif"/> Figure 13.22 Image of the MS Excel worksheet used for estimating G and N of Problem 13.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_22_B.tif"/> Figure 13.23 Full-depth asphalt pavement structure for Problem 13.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_23_B.tif"/> Figure 13.24 Image of the MS Excel worksheet used for determining MR2 of Problem 13.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_24_B.tif"/> Figure 13.26 Type of truck traffic for Problem 13.13. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_26_B.tif"/> Figure 13.25 Flexible pavement structure for Problem 13.13. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_25_B.tif"/> Figure 13.27 Image of the MS Excel worksheet used for determining the load of the rear single axle of the truck for Problem 13.13. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_27_B.tif"/> Figure 13.28 Full-depth asphalt pavement structure for Problem 13.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_28_B.tif"/> Figure 13.29 Image of the MS Excel worksheet used for determining ΔPSI and pt for Problem 13.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_29_B.tif"/> Figure 13.30 Type of truck traffic for Problem 13.15. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_30_B.tif"/> Table 13.22 Computations of the ESALs for Problem 13.15 Truck Type Axle Type AADT 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_9_B.tif"/> Single 350 365 0.45 33.0 1 0.094 178,477 Single 1 1.416 2,686,655 Tridem 1 ?? 1,897,088 f_LE-x Total ESALs Figure 13.31 Image of the MS Excel worksheet used for determining the load of the tridem axle of the truck for Problem 13.15. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_31_B.tif"/> Figure 13.32 Type of truck traffic for Problem 13.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_32_B.tif"/> Table 13.23 Computations of the ESALs for Problem 13.16 Truck Type Axle Type AADT 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_10_B.tif"/> Single 275 365 0.45 33.0 1 0.192 258,408 Single 1 1.000 1,346,029 Tridem 1 ?? 1,346,029 f_LE-x Total ESALs Figure 13.33 Image of the MS Excel worksheet used for determining the load of the tridem axle of the truck for Problem 13.16. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_33_B.tif"/> Figure 13.35 Type of truck traffic for Problem 13.17. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_35_B.tif"/> Figure 13.34 Flexible pavement structure for Problem 13.17. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_34_B.tif"/> Table 13.24 Computations of the ESALs for Problem 13.17 Truck Type Axle Type AADT %Trucks 365 f_d G N f_LE ESALs https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline13_11_B.tif"/> Single ?? 40 365 0.40 12.01 1 0.192 134.6 Single 1 1.000 701.2 Total ESALs (in terms of AADT) 835.8 AADT Figure 13.36 Image of the MS Excel worksheet used for determining the AADT for Problem 13.17. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_36_B.tif"/> Figure 13.37 Dual-wheel loading on a full-depth asphalt pavement structure for Problem 13.18. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_37_B.tif"/> Figure 13.38 Image of the MS Excel worksheet used for the mechanistic-empirical pavement design for Problem 13.18. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_38_B.tif"/> Figure 13.39 Single wheel circular loading on a two-layered flexible pavement for Problem 13.19. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_39_B.tif"/> Figure 13.40 Inputs for the WinJULEA program for Problem 13.19. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_40_B.tif"/> Table 13.25 WinJULEA Flexible Pavement Analysis Results for Problem 13.19 *** RESULTS

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DEPTH X-COORD Y-COORD STRESS-X STRESS-Y STRESS-Z SHEAR-XZ SHEAR-YZ SHEAR-XY STRAIN-X STRAIN-Y STRAIN-Z SHEAR-XZ SHEAR-YZ SHEAR-XY DISPLT-X DISPLT-Y DISPLT-Z P.STRESS1 P.STRESS2 P.STRESS3 P.STRAIN1 P.STRAIN2 P.STRAIN3 MAX.SHEAR OCT.STRESS OCT.SHEAR

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0.500E+01 0.000E+00 0.000E+00 0.200E+01 0.200E+01 0.508E+01 0.000E+00 0.000E+00 0.000E+00 −0.238E-03 −0.238E-03 0.657E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.216E-01 0.508E+01 0.200E+01 0.200E+01 0.657E-03 −0.238E-03 −0.238E-03 0.154E+01 0.302E+01 0.146E+01 Figure 13.41 Single wheel circular loading on a full-depth asphalt pavement (two-layered system) for Problem 13.20. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_41_B.tif"/> Figure 13.42 Strain factor for single wheel for flexible pavements. Huang, Yang H., <italic>Pavement Analysis and Design</italic>, 2nd edition ©2004. Reprinted by permission of Pearson Education, Inc., New York, New York, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_42_B.tif"/> Figure 13.43 Image of the MS Excel worksheet used for the computations of Problem 13.20. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_43_B.tif"/> Figure 13.44 Single wheel circular loading on a full-depth asphalt pavement for Problem 13.21 (Pavement System #1). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_44_B.tif"/> Figure 13.45 Image of the MS Excel worksheet used for the computations of Problem 13.21. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_45_B.tif"/> Figure 13.46 Dual wheel loading on a full-depth asphalt pavement for Problem 13.22 (Pavement System #2). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_46_B.tif"/> Figure 13.47 Inputs for the WinJULEA program for Problem 13.22. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_47_B.tif"/> Table 13.26 The Coordinates of the Three Points in Problem 13.22 Point Location (X, Y, Z) 1 (0, 0, 6) 2 (5.75, 0, 6) 3 (11.5, 0, 6) Table 13.27 WinJULEA Flexible Pavement Analysis Results for Problem 13.22 *** RESULTS

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0.450E+01 0.000E+00 0.000E+00 0.632E+01 0.542E+01 0.142E+02 −0.119E+01 −0.119E+01 0.000E+00 −0.625E-03 −0.949E-03 0.222E-02 −0.864E-03 0.000E+00 0.000E+00 −0.281E-02 0.000E+00 0.693E-01 0.143E+02 0.614E+01 0.542E+01 0.229E-02 −0.689E-03 −0.949E-03 0.446E+01 0.864E+01 0.405E+01 0.450E+01 0.575E+01 0.000E+00 0.751E+01 0.577E+01 0.139E+02 0.000E+00 0.000E+00 0.000E+00 −0.340E-03 −0.970E-03 0.199E-02 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.723E-01 0.139E+02 0.751E+01 0.577E+01 0.199E-02 −0.340E-03 −0.970E-03 0.408E+01 0.907E+01 0.351E+01 0.450E+01 0.115E+02 0.000E+00 0.632E+01 0.542E+01 0.142E+02 0.119E+01 0.119E+01 0.000E+00 −0.625E-03 −0.949E-03 0.222E-02 0.864E-03 0.000E+00 0.000E+00 0.281E-02 0.000E+00 0.693E-01 0.143E+02 0.614E+01 0.542E+01 0.229E-02 −0.689E-03 −0.949E-03 0.446E+01 0.864E+01 0.405E+01 Table 13.28 Tensile Strain Results for the Three Locations in Problem 13.22 using WinLEA Program Point Location Tensile Strain-X (με) Tensile Strain-Y (με) (0, 0, 6) Bottom of asphalt layer, centerline of 1st wheel load 625 949 (5.75, 0, 6) Bottom of asphalt layer, mid-point between the two wheel loads 340 970 (11.5, 0, 6) Bottom of asphalt layer, centerline of 2nd wheel load 625 949 Figure 13.48 Image of the MS Excel worksheet used for the computations of Problem 13.22. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_48_B.tif"/> Figure 13.49 Single wheel circular loading applied on a three-layered pavement system for Problem 13.23. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_49_B.tif"/> Figure 13.50 Inputs for the WinJULEA program for Problem 13.23. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_50_B.tif"/> Table 13.29 WinJULEA Flexible Pavement Analysis Results for Problem 13.23 *** RESULTS

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0.500E+01 0.000E+00 0.000E+00 0.139E+01 0.139E+01 0.148E+02 0.000E+00 0.000E+00 0.000E+00 −0.170E-03 −0.170E-03 0.457E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.948E-02 0.148E+02 0.139E+01 0.139E+01 0.457E-03 −0.170E-03 −0.170E-03 0.672E+01 0.587E+01 0.634E+01 0.150E+02 0.000E+00 0.000E+00 0.312E+00 0.312E+00 0.397E+01 0.000E+00 0.000E+00 0.000E+00 −0.108E-03 −0.108E-03 0.246E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.670E-02 0.397E+01 0.312E+00 0.312E+00 0.246E-03 −0.108E-03 −0.108E-03 0.183E+01 0.153E+01 0.172E+01 Table 13.30 Critical Pavement Response Results for the Two Locations using WinJULEA Program for Problem 13.23 Point Location Tensile Strain-X = Tensile Strain-Y (με) Compressive Strain-Z (με) (0, 0, 5) Bottom of asphalt layer, centerline of wheel load 170 (0, 0, 15) Top of subgrade layer, centerline of wheel load 246 Figure 13.51 Image of the MS Excel worksheet used of the computations for Problem 13.23. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_51_B.tif"/> Figure 13.52 Single wheel circular loading on a three-layered pavement system for Problem 13.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_52_B.tif"/> Figure 13.53 Location of the interface stresses in the three-layered pavement system for Problem 13.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_53_B.tif"/> Figure 13.54 Location of the interface strains in the three-layered pavement system for Problem 13.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_54_B.tif"/> Table 13.31 Pavement Response (Stress and Strain) Results at the Two-Layer Interfaces Using Multi-Layer Elastic Theory for Problem 13.24 Interface Pavement Layer Vertical Stress (σ_z) Radial Stress (σ_r) Vertical Strain (ε_z) Radial Strain (ε_r) 1 Bottom of layer #1 σ_z1 = 13.48 σ_r1 = −176.3 ε_z1 = 949 ε_r1 = −474.5 1 Top of layer #2 σ ´ z 1 = 13.48 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0525.tif"/> σ ´ r 1 = 3.99 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0526.tif"/> ε ´ z 1 = 949 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0527.tif"/> ε ´ r 1 = − 474.5 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0528.tif"/> 2 Bottom of layer #2 σ_z2 = 4.00 σ_r2 = −2.72 ε_z2 = 672 ε_r2 = −336 2 Top of layer #3 σ ´ z 2 = 4.00 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0529.tif"/> σ ´ r 2 = 0.64 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0530.tif"/> ε ´ z 2 = 672 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0531.tif"/> ε ´ r 2 = − 336 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0532.tif"/> Figure 13.55 Image of the MS Excel worksheet used for the flexible pavement analysis of Problem 13.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_55_B.tif"/> Figure 13.56 Inputs for the WinJULEA program for Problem 13.24. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_56_B.tif"/> Table 13.32 WinJULEA Flexible Pavement Analysis Results for Problem 13.24 *** RESULTS

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0.000E+00 0.000E+00 0.000E+00 0.236E+03 0.236E+03 0.999E+02 0.000E+00 0.000E+00 0.000E+00 0.340E-03 0.340E-03 −0.681E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.281E-01 0.236E+03 0.236E+03 0.999E+02 0.340E-03 0.340E-03 −0.681E-03 0.681E+02 0.191E+03 0.642E+02 0.499E+01 0.000E+00 0.000E+00 −0.175E+03 −0.175E+03 0.135E+02 0.000E+00 0.000E+00 0.000E+00 −0.472E-03 −0.472E-03 0.944E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.272E-01 0.135E+02 −0.175E+03 −0.175E+03 0.944E-03 −0.472E-03 −0.472E-03 0.944E+02 −0.112E+03 0.890E+02 0.501E+01 0.000E+00 0.000E+00 0.395E+01 0.395E+01 0.135E+02 0.000E+00 0.000E+00 0.000E+00 −0.479E-03 −0.479E-03 0.958E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.272E-01 0.135E+02 0.395E+01 0.395E+01 0.958E-03 −0.479E-03 −0.479E-03 0.479E+01 0.715E+01 0.452E+01 0.150E+02 0.000E+00 0.000E+00 −0.271E+01 −0.271E+01 0.400E+01 0.000E+00 0.000E+00 0.000E+00 −0.335E-03 −0.335E-03 0.670E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.200E-01 0.400E+01 0.271E+01 0.271E+01 0.670E-03 −0.335E-03 −0.335E-03 0.335E+01 −0.473E+00 0.316E+01 0.150E+02 0.000E+00 0.000E+00 0.640E+00 0.640E+00 0.399E+01 0.000E+00 0.000E+00 0.000E+00 −0.335E-03 −0.335E-03 0.669E-03 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.200E-01 0.399E+01 0.640E+00 0.640E+00 0.669E-03 −0.335E-03 −0.335E-03 0.167E+01 0.176E+01 0.158E+01 Table 13.33 Pavement Response (Stress and Strain) Results at the Two-Layer Interfaces Using WinJULEA Program for Problem 13.24 Interface Pavement Layer Vertical Stress (σ_z) Radial Stress (σ_r) Vertical Strain (ε_z) Radial Strain (ε_r) 1 Bottom of layer #1 σ_z1 = 13.5 σ_r1 = −175 ε_z1 = 944 ε_r1 = −472 1 Top of layer #2 σ ´ z 1 = 13.5 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0533.tif"/> σ ´ r 1 = 3.95 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0534.tif"/> ε ´ z 1 = 958 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0535.tif"/> ε ´ r 1 = − 479 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0536.tif"/> 2 Bottom of layer #2 σ_z2 = 4.00 σ_r2 = −2.71 ε_z2 = 670 ε_r2 = −335 2 Top of layer #3 σ ´ z 2 = 3.99 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0537.tif"/> σ ´ r 1 = 0.64 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0538.tif"/> ε ´ z 2 = 669 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0539.tif"/> ε ´ r 2 = − 335 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH013_eqn_0540.tif"/> Figure 13.57 Three-layered flexible pavement system for Problem 13.25. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_57_B.tif"/> Figure 13.58 Image of the MS Excel worksheet used for the three-layered flexible pavement analysis for Problem 13.25. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_58_B.tif"/> Figure 13.59 Three-layered asphalt pavement system for Problem 13.26. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_59_B.tif"/> Figure 13.60 Image of the MS Excel worksheet used for the three-layered flexible pavement analysis for Problem 13.26. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_60_B.tif"/> Figure 13.61 Single-wheel loading on an asphalt pavement structure for Problem 13.27. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_61_B.tif"/> Figure 13.62 Radial strain and vertical stress versus depth in the pavement structure for Problem 13.27. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_62_B.tif"/> Figure 13.63 Image of the MS Excel worksheet used for the analysis of Problem 13.27. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_63_B.tif"/> Figure 13.64 A single-wheel loading on a full-depth asphalt pavement structure for Problem 13.28. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_64_B.tif"/> Figure 13.65 Image of the MS Excel worksheet used for the computations of Problem 13.28. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_65_B.tif"/> Figure 13.66 A full-depth asphalt pavement for Problem 13.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_66_B.tif"/> Figure 13.67 Image of the MS Excel worksheet used for estimating the tensile strain and the asphalt layer thickness for Problem 13.29. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_67_B.tif"/> Figure 13.68 Dual wheel loading on a layered flexible pavement system for development of EALFs for Problem 13.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_68_B.tif"/> Table 13.34 Single-Axle Load Values used to Develop EALFs for Problem 13.30 Load # Load (lb) 1 5000 2 10000 3 12000 4 14000 5 16000 6 18000 7 20000 8 25000 9 30000 10 35000 11 40000 12 45000 13 50000 Table 13.35 Inputs for the WinJULEA Program to Perform the Analysis for Problem 13.30 Load # Load (lb) Single Wheel Load (lb) Contact Area (in²) 1 5000 1250 12.5 2 10000 2500 25.0 3 12000 3000 30.0 4 14000 3500 35.0 5 16000 4000 40.0 6 18000 4500 45.0 7 20000 5000 50.0 8 25000 6250 62.5 9 30000 7500 75.0 10 35000 8750 87.5 11 40000 10000 100.0 12 45000 11250 112.5 13 50000 12500 125.0 Table 13.36 Pavement Critical Response (εt and εc) using WinJULEA for Problem 13.30 Load (lb) ε_t (με) ε_c (με) 5000 98 170 10000 163 335 12000 183 399 14000 202 461 16000 219 523 18000 234 583 20000 248 645 25000 279 802 30000 307 956 35000 335 1110 40000 373 1260 45000 408 1400 50000 439 1550 Table 13.37 EALFs based on Fatigue Performance for Problem 13.30 Load (lb) ε_t (με) EALF AI’s EALF % Difference 5000 98 0.05702 0.00500 91 10000 163 0.3042 0.0877 71 12000 183 0.445 0.189 58 14000 202 0.616 0.360 42 16000 219 0.804 0.623 23 18000 234 1.000 1.000 0 20000 248 1.21 1.51 25 25000 279 1.78 3.53 98 30000 307 2.44 6.97 186 35000 335 3.26 12.50 283 40000 373 4.64 21.08 354 45000 408 6.23 34.00 446 50000 439 7.93 52.88 567 Table 13.38 EALFs based on Rutting Performance for Problem 13.30 Load (lb) ε_c (με) EALF AI’s EALF % Difference 5000 170 0.00402 0.00500 24 10000 335 0.0837 0.0877 5 12000 399 0.183 0.189 3 14000 461 0.350 0.360 3 16000 523 0.615 0.623 1 18000 583 1.000 1.000 0 20000 645 1.57 1.51 4 25000 802 4.17 3.53 15 30000 956 9.15 6.97 24 35000 1110 17.87 12.50 30 40000 1260 31.51 21.08 33 45000 1400 50.50 34.00 33 50000 1550 79.66 52.88 34 Figure 13.69 Image of the MS Excel worksheet used for the computations of Problem 13.30. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_69_B.tif"/> Figure 13.70 Dual wheel loading on a full-depth asphalt pavement system for development of EALFs for Problem 13.31. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_70_B.tif"/> Table 13.39 Inputs for the WinJULEA Program to Perform the Analysis for Problem 13.31 Load # Load (lb) Single Wheel Load (lb) Contact Area (in²) 1 5000 1250 12.5 2 10000 2500 25.0 3 12000 3000 30.0 4 14000 3500 35.0 5 16000 4000 40.0 6 18000 4500 45.0 7 20000 5000 50.0 8 25000 6250 62.5 9 30000 7500 75.0 10 35000 8750 87.5 11 40000 10000 100.0 12 45000 11250 112.5 13 50000 12500 125.0 Table 13.40 Pavement Critical Response (εt and εc) Using WinJULEA for Problem 3.31 Load (lb) ε_t (με) ε_c (με) 5000 86 189 10000 161 355 12000 189 417 14000 215 475 16000 240 532 18000 264 587 20000 288 640 25000 343 765 30000 403 883 35000 460 994 40000 514 1090 45000 566 1180 50000 615 1300 Table 13.41 EALFs Based on Fatigue Performance for Problem 13.31 Load (lb) ε_t (με) EALF AI’s EALF % Difference 5000 98 0.02494 0.00500 80 10000 163 0.1964 0.0877 55 12000 183 0.333 0.189 43 14000 202 0.509 0.36 29 16000 219 0.731 0.623 15 18000 234 1.000 1.000 0 20000 248 1.33 1.51 13 25000 279 2.37 3.53 49 30000 307 4.02 6.97 73 35000 335 6.22 12.50 101 40000 373 8.96 21.08 135 45000 408 12.30 34.00 176 50000 439 16.17 52.88 227 Table 13.42 EALFs based on Rutting Performance for Problem 13.31 Load (lb) ε_c (με) EALF AI’s EALF % Difference 5000 170 0.00626 0.00500 20 10000 335 0.1052 0.0877 17 12000 399 0.216 0.189 13 14000 461 0.388 0.360 7 16000 523 0.644 0.623 3 18000 583 1.000 1.000 0 20000 645 1.47 1.51 3 25000 802 3.27 3.53 8 30000 956 6.22 6.97 12 35000 1110 10.57 12.50 18 40000 1260 15.97 21.08 32 45000 1400 22.78 34.00 49 50000 1550 35.15 52.88 50 Figure 13.71 Image of the MS Excel worksheet used for the computations of Problem 13.31. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig13_71_B.tif"/>

Design and Analysis of Asphalt Pavements

ABSTRACT