Design and Analysis of Rigid Pavements

ABSTRACT

Chapter 14 discusses an important subject in pavement engineering which focuses on the design and analysis of rigid pavements. The behavior of concrete material is different than the behavior of asphalt material in the pavement structure. The main inputs for empirical design of rigid pavements include traffic loading, material properties, reliability, performance indicators, pavement drainage quality, and load transfer coefficient. The properties of the concrete slab considered in the design include the modulus of elasticity and the modulus of rupture; whereas, the property of the subgrade layer under the rigid slab which is considered in the design is the effective modulus of subgrade reaction. The estimation of the effective modulus of subgrade reaction takes into account the monthly values of the soil resilient modulus, the subbase elastic (resilient) modulus and thickness, the depth of the rock foundation beneath the subgrade surface, the subbase loss of support due to erosion, and the projected rigid slab thickness. All these factors are actually taken into consideration in the design procedure of rigid pavements directly or indirectly. Reliability and therefore cost are a crucial element in the design process. The load transfer coefficient is one of the inputs in the design of rigid pavements that accounts for the ability of the concrete slab to transfer the load across joints and cracks. Therefore, the quality of the joint between rigid pavement slabs and the use of load transfer devices and tied pavement shoulders improve the load transfer across the joints for rigid pavements. On the other hand, the analysis of rigid pavements emphasizes the response and the behavior of rigid slabs under traffic loading and temperature changes. Consequently, curling stresses due to temperature and moisture changes will be highlighted. In addition, total bending (curling) stresses in rigid slabs are introduced on the basis of infinite plate assumption. The analysis of stresses and deflections due to traffic loading at critical locations on the concrete slab (interior, corner, and edge) are presented in this section. Single-wheel versus dual-wheel loading is considered as well in the analysis of stresses and deflections due to traffic loading. 14.1

The 20-year design slab thickness for a rigid pavement of a rural highway is 10.0 inches. Determine the ESALs that this rigid pavement can carry over the design life given the following design input data:

The effective modulus of subgrade reaction = 100 pci

The mean concrete modulus of rupture = 600 psi

The concrete modulus of elasticity = 5000000 psi

The load transfer coefficient = 3.2

The drainage coefficient = 1.0

The initial serviceability index = 4.5

The terminal serviceability index = 2.5

The reliability value = 95%

The overall standard deviation = 0.29

Solution:

The following formula is used for the design of rigid pavements following the AASHTO design method:488

log 10 W 18 = Z R S 0 + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 Δ PSI 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 p t log 10 S ′ c C D 215.63 J D 0.75 − 1.132 D 0.75 − 18.42 E c k 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0001.tif"/>

Where:

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

Z_R = standard normal variant for a given reliability

S₀ = overall standard deviation

D = thickness of rigid pavement (concrete slab) to the nearest half inch

ΔPSI = p_i – p_t

p_i = initial serviceability index

p_t = terminal serviceability index

E_c = elastic modulus of the concrete to be used in construction

k = effective modulus of subgrade reaction

S ′ c https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0002.tif"/> = modulus of rupture of concrete to be used in construction

J = load transfer coefficient (typical value = 3.2)

C_d = drainage coefficient

For a reliability level (R) of 95%, the Z_R value is −1.645. The S₀ value for rigid pavements is 0.29.

p_i = 4.5 (AASHTO recommended value for rigid pavements)

p_t = 2.5 (AASHTO recommended value for major highways)

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-CH014_eqn_0003.tif"/>

⇒489

log 10 W 18 = − 1.645 ( 0.29 ) + 7.35 log 10 ( 10 + 1 ) − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 10 + 1 8.46 + 4.22 − 0.32 ( 2.5 ) log 10 600 ( 1 ) 215.63 ( 3.2 ) 10 0.75 − 1.132 10 0.75 − 18.42 5 × 10 6 100 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0004.tif"/>

⇒

W 18 = 5904387 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0005.tif"/>

In other words, the 10-in rigid pavement with the design data given in this problem can carry a traffic level of approximately 5.9 million ESALs over the design life of the pavement.

The MS Excel worksheet as shown in Figure 14.1 is used to solve for D in the above equation.

14.2

Design a rigid pavement for an urban highway to carry a total number of equivalent single-axle loads (ESALs) of 3.6 million over a design life of 20 years given the following input data:490

The effective modulus of subgrade reaction = 75 pci

The mean concrete modulus of rupture = 650 psi

The concrete modulus of elasticity = 5000000 psi

The load transfer coefficient = 3.2

The drainage coefficient = 1.0

The initial serviceability index = 4.5

The terminal serviceability index = 2.5

The reliability value = 95%

The overall standard deviation = 0.29

Solution:

The same formula is used for the design of rigid pavements:

⇒

log 10 3.6 × 10 6 = − 1.645 ( 0.29 ) + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 ( 2.5 ) log 10 650 ( 1 ) 215.63 ( 3.2 ) D 0.75 − 1.132 D 0.75 − 18.42 5 × 10 6 75 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0007.tif"/> 491

The MS Excel Solver is used to solve for the thickness of the concrete slab (D) in the above equation as shown in the procedure below:

The left side of the equation is 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-CH014_eqn_0008.tif"/>

In this case,

Left Side = Design ESALs = 3 .5 × 10 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-CH014_eqn_0009.tif"/>

And the right side of the equation is set to:

Right side = 10 − 1.645 0.29 + 7.35 log 10 D + 1 − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 ( 2.5 ) log 10 650 ( 1 ) 215.63 ( 3.2 ) D 0.75 − 1.132 D 0.75 − 18.42 5 × 10 6 75 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0010.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:

14.3 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-CH014_eqn_0011.tif"/>

The objective in Excel Solver is set to the “error” cell. The cell to be changed is the “D” cell. The Excel Solver keeps changing the value of D through an iterative approach until the error value is zero or minimum. That is the numerical solution for the thickness of the pavement (D). Solving the above equation provides the following solution:

D = 9.0 in ( ≅ 23 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-CH014_eqn_0012.tif"/>

The MS Excel worksheet used to solve for D is shown in Figure 14.2.492

14.3

Design a rigid pavement (see Figure 14.3) for a new four-lane rural highway that connects two major cities using the AASHTO design procedure to carry the traffic mix shown in Figure 14.4. The elastic modulus (E_c) of the Portland cement concrete (PCC) used for the rigid slab is 5000000 psi and the effective modulus of subgrade reaction (k) of the slab over the subgrade layer and a granular base layer with 6-in thickness is 150 pci.

The annual average daily traffic (AADT) that is expected to use the highway in both directions is 2500 veh/day in the first year, the traffic growth rate is 2%, and the design life of the pavement is 20 years. Assume that the traffic growth rate will remain the same during the design life of the pavement.

Given Data:

The mean concrete modulus of rupture = 650 psi

The load transfer coefficient = 3.2

The initial serviceability index = 4.5

The terminal serviceability index = 2.5

The overall standard deviation = 0.29

Drainage data: it is expected that it will take one day for water to be removed from the pavement, and the pavement structure will be exposed to moisture levels approaching saturation 30% of the time.493

Solution:

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

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 75–95% according to Table 14.1. The highest value is selected in this case (95%) to be on the safe side (see Table 14.1).

For a reliability (R) level of 95%, the Z-value is equal to −1.645 according to Table 14.2.

S₀ = 0.29 (for rigid pavements)

14.4 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-CH014_eqn_0014.tif"/>

⇒

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-CH014_eqn_0015.tif"/> 495

The drainage coefficient is determined from Table 14.3 based on the given drainage data. For the condition “water removed within one day”, the quality of drainage is “Good”, and for the condition “30% of the time the pavement structure is exposed to moisture levels approaching saturation”, the drainage coefficient (C_d) value is 1.00 (Table 14.3). Therefore,

C_d = 1.00.

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

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-CH014_eqn_0016.tif"/>

The growth factor is calculated using the formula below:

r = 2%

n = 20 years

14.6 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-CH014_eqn_0017.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-CH014_eqn_0018.tif"/>

The load equivalency factor is determined using the power model developed in this book for single-axle loads for rigid pavements based on numbers in the table of equivalent axle load factors from the AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA, as shown in the following relationship (see Figure 14.5).496

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:

14.7 f LE = 5.2199 × 10 − 18 ( P ) 4.0627 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0019.tif"/>

Where:

f_LE = load equivalency factor for single-axle loads

P = single-axle load in (lb)

For tandem-axle loads, 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 equivalent axle load factors from the AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA, as shown in the following relationships (see Figures 14.6 and 14.7).497

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:

For load ≤ 50000 lb:

f LE = 1.2385 × 10 − 17 ( P ) 3.7859 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0020.tif"/>

For load > 50000 lb:

14.9 f LE = 1.8382 × 10 − 19 ( P ) 4.1877 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0021.tif"/>

Where:

f_LE = load equivalency factor for single-axle loads

P = tandem-axle load in (lb)

For tridem-axle loads, the load equivalency factor is determined using the power model developed in this booluk based on numbers in the table of equivalent axle load factors from the AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA, as shown in the following relationship (see Figures 14.8 and 14.9).498

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:

For load ≤ 50000 lb:

14.10 f LE = 7.4004 × 10 − 17 ( P ) 3.5018 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0022.tif"/>

For load > 50000 lb:

14.11 f LE = 6.5203 × 10 − 19 ( P ) 3.9698 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0023.tif"/>

Where:

f_LE = load equivalency factor for single-axle loads

P = tridem-axle load in (lb)499

Sample Calculation for the ESALs:

For the 36-kip tandem axle in truck 1:

AADT = 0.10 ( 2500 ) = 250 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-CH014_eqn_0024.tif"/>

Since the highway is a four-lane highway, the lane distribution factor is 45% (0.45) according to the Asphalt Institute (AI) values in Table 14.4. And based on the AASHTO recommended values in Table 14.5, for two lanes in each direction, f_d = 80–100% in one direction. The traffic in this problem is given for two directions; therefore, a value of 0.45 is used for f_d.

Or:

f_d = 0.45

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

f_lE = 2.212 (calculated below)

The power model (for load ≤ 50000 lb) for the load equivalency factor of tandem-axle loads developed above will be used:

f LE = 1.2385 × 10 − 17 ( P ) 3.7859 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0025.tif"/>

⇒

f LE = 1.2385 × 10 − 17 ( 36000 ) 3.7859 = 2.212 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0026.tif"/>

⇒

ESAL 36-kip Tandem = ( 250 ) ( 365 ) ( 0.45 ) ( 24.3 ) ( 2 ) ( 2.212 ) = 4391737 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0028.tif"/>

The results of the other axle types are summarized in Table 14.6 based on the same formula.

The modulus of subgrade reaction is determined using the following relationship:

14.12 k = q w 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-CH014_eqn_0029.tif"/>

This relationship is based on an analysis of a plate-bearing test. The ratio of the applied pressure (q) in the plate test and the resulting deflection (w₀) of the plate on a solid foundation is equal to the modulus of subgrade reaction. Therefore:

14.13 k = 2 M R π ( 1 − ν 2 ) 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-CH014_eqn_0030.tif"/>

Where:

k = modulus of subgrade reaction

q = applied pressure

w₀ = plate deflection

M_R = resilient modulus of subgrade

ν = Poisson ratio of foundation (assumed as 0.45)

a = radius of plate (15 in)

Therefore, by substituting the above values, the formula becomes:

14.14 k = M R 18.8 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0031.tif"/>

In this problem, the effective modulus of subgrade reaction that is used in the design formula is provided as 150 pci.501

Now using the design formula for rigid pavements shown below:

⇒

log 10 ( 6116847 ) = ( − 1.645 ) ( 0.29 ) + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 × 2.5 log 10 650 ( 1.00 ) 215.63 ( 3.2 ) D 0.75 − 1.132 D 0.75 − 18.42 5 × 10 6 150 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0033.tif"/>

The MS Excel Solver is used to solve for D in the above equation as shown in the procedure below:

The left side of the equation is 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-CH014_eqn_0034.tif"/> 502

In this case,

Left Side = Design ESALs = 6116847 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0035.tif"/>

And the right side of the equation is set to:

Right side = 10 = − 1.645 0.29 + 7.35 log 10 D + 1 − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 × 2.5 log 10 650 1.00 215.63 3.2 D 0.75 − 1.132 D 0.75 − 18.42 5 × 10 6 150 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0036.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:

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-CH014_eqn_0037.tif"/>

The MS Excel Solver performs an iterative approach to find the numerical solution of the rigid pavement thickness (D). The Solver keeps doing iteration by changing the value of the cell that corresponds to D and targeting the cell that corresponds to the error defined above. When the error is close to zero or reaches a minimum value, the solution converges to the objective D value.

D = 9.5 in ( ≅ 24 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-CH014_eqn_0038.tif"/>

The MS Excel worksheet used to perform the computations and the solver solution is shown in Figure 14.10.503

14.4

The 20-year design slab thickness for a rural highway rigid pavement is 8.0 inches. The truck traffic that is expected to use the highway pavement is of the type shown in Figure 14.11. Determine the ESALs that this rigid pavement can carry and the maximum allowable load that can be carried by the tandem axle shown below given the following design input data (note: ignore the effect of passenger cars):

The traffic growth factor = 29.8

The design lane factor = 0.45

The 1st year AADT = 200 truck/day

The effective modulus of subgrade reaction = 180 pci

The mean concrete modulus of rupture = 620 psi

The concrete modulus of elasticity = 5,000,000 psi

The load transfer coefficient = 3.2

The drainage coefficient = 1.0

The initial serviceability index = 4.5

The terminal serviceability index = 2.5

The reliability value = 95%

The overall standard deviation = 0.29

Solution:

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

Using the design formula shown below, the ESALs that the rigid pavement can carry can be determined:

Substituting all design inputs into the above equation

⇒

log 10 W 18 = ( − 1.645 ) ( 0.29 ) + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 4.5 − 2.5 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 × 2.5 log 10 620 ( 1.0 ) 215.63 ( 3.2 ) D 0.75 − 1.132 D 0.75 − 18.42 5 × 10 6 180 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0040.tif"/>

Solving for W₁₈ provides the following solution:

W 18 = 5494255 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-CH014_eqn_0041.tif"/>

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

For the 12-kip single axle:

f_d = 0.45 (given)

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

f_LE = 0.195 (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:

f LE = 5.2199 × 10 − 18 ( P ) 4.0627 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0043.tif"/>

⇒

f LE = 2.1159 × 10 − 17 ( 12000 ) 3.9120 = 0.195 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0044.tif"/>

⇒

ESAL 12-kip Single = ( 200 ) ( 365 ) ( 0.45 ) ( 29.8 ) ( 1 ) ( 0.195 ) = 190943 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0046.tif"/> 505

For the other single axle:

f_d = 0.45 (given)

N = 1

f_LE is unknown.

⇒

ESAL x-Single = ( 200 ) ( 365 ) ( 0.45 ) ( 29.8 ) ( 1 ) ( f LE − x ) = 978930 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-CH014_eqn_0048.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 = 5494255 = 190943 + 978930 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-CH014_eqn_0049.tif"/>

⇒

f LE − x = 5.417 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0050.tif"/>

Using the power model developed earlier for the load equivalency factor of tandem-axle loads for rigid pavements, the tandem-axle load that corresponds to a f_LE of 5.417 is determined as shown below:

For load ≤ 50000 lb:

f LE = 1.2385 × 10 − 17 ( P ) 3.7859 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0051.tif"/>

⇒

5.417 = 1.2385 × 10 − 17 ( P ) 3.7859 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0052.tif"/>

⇒

P = 45670 lb ( ≅ 203 kN ) ≤ 50000 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-CH014_eqn_0053.tif"/> , which means the power model used above is the correct one.

Therefore, the maximum allowable load that can be carried on the tandem-axle of the truck using the 8-in rigid pavement under the existing conditions is about 46 kips.

The MS Excel worksheet used to determine the ESALs that the rigid pavement can carry and the maximum allowable tandem-axle load of the truck given in this problem is shown in Figure 14.12.506

14.5

A rigid concrete pavement is designed using the AASHTO design procedure to carry a predicted number of 18-kip ESALs equal to 2.6 million for 20 years. The following input values are given:

The effective modulus of subgrade reaction (k) = 120 pci

The mean concrete modulus of rupture ( S ′ c https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0054.tif"/> ) = 700 psi

The concrete modulus of elasticity (E_c) = 4000000 psi

The load transfer coefficient (J) = 3.0

The drainage coefficient (C_d) = 0.7

The initial serviceability index (p_i) = 4.5

The terminal serviceability index (p_t) = 2.5

Reliability (R) = 90 percent

Overall standard deviation (S₀) = 0.30

Determine the design thickness (D). If the reliability level is 60% instead of 90%, what would be the design thickness of the rigid pavement? What is the reduction in thickness in this case?507

Solution:

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

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

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-CH014_eqn_0056.tif"/>

⇒

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-CH014_eqn_0057.tif"/>

The following design inputs are given:

S₀ = 0.30

C_d = 0.7

J = 3.0

E_c = 4000000 psi

S ′ c https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0058.tif"/> = 700 psi

k = 120 pci

Design ESALs = 2.6 × 10⁶

⇒508

log 10 2.6 × 10 6 = ( − 1.282 ) ( 0.30 ) + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 × 2.5 log 10 700 ( 0.7 ) 215.63 ( 3.0 ) D 0.75 − 1.132 D 0.75 − 18.42 4 × 10 6 120 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0059.tif"/>

The MS Excel Solver is used to solve for D in the above equation; the solution obtained using the Excel Solver tool is:

D = 9.0 in ( ≅ 23 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-CH014_eqn_0060.tif"/>

If the reliability level is changed from 90% to 60%, only the Z_R used in the design formula will be different (in this case, Z_R = -0.253). Substituting the design inputs into the formula provides:

log 10 2.6 × 10 6 = ( − 0.253 ) ( 0.30 ) + 7.35 log 10 ( D + 1 ) − 0.06 + log 10 2.0 4.5 − 1.5 1 + 1.624 × 10 7 D + 1 8.46 + 4.22 − 0.32 × 2.5 log 10 700 ( 0.7 ) 215.63 ( 3.0 ) D 0.75 − 1.132 D 0.75 − 18.42 4 × 10 6 120 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0061.tif"/> 509

Using the MS Excel Solver tool, this provides a solution for D:

D = 8.0 in ( ≅ 20.5 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-CH014_eqn_0062.tif"/>

The reduction in the design thickness is 1.0 inch and the percentage of the reduction is estimated as:

Reduction in D = 100 × ( 9 − 8 ) 9 ≅ 11 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0063.tif"/>

The MS Excel worksheet used to perform the computations and the solver solution is shown in Figure 14.13.

14.6

An interior circular loading with a radius of 6 inches is applied on a concrete slab with a radius of relative stiffness of 40 inches. The modulus of subgrade reaction (k) of the slab is 100 pci, the modulus of concrete is 4000000 psi, and the Poisson ratio is 0.15. If the maximum deflection due to this loading is 0.012 inch, determine the interior loading and the thickness of the slab.

Solution:

Westergaard formula for the deflection due to interior loading is used to solve for the interior loading in this problem as shown below:

Δ i = P 8 k l 2 { 1 + 1 2 π [ ln ( a 2 l ) − 0.673 ] [ a l ] 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-CH014_eqn_0064.tif"/> 510

Where:

Δ_i = deflection in the interior of the slab

P = applied load

k = modulus of subgrade reaction

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

0.012 = P 8 ( 100 ) ( 40 ) 2 { 1 + 1 2 π [ ln ( 6 2 ( 40 ) ) − 0.673 ] [ 6 40 ] 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-CH014_eqn_0065.tif"/>

Solving the above equation for P provides:

P = 15541.6 lb ≅ 15.5 kips ( ≅ 69 kN ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0066.tif"/>

The thickness of the slab is determined from the formula of the radius of relative stiffness of the slab as follows:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0067.tif"/>

Where:

l = radius of relative stiffness of concrete slab

E = modulus of elasticity of concrete slab

h = thickness of concrete slab

ν = Poisson ratio (typical value = 0.15)

k = modulus of subgrade reaction

⇒

40 = [ ( 4 × 10 6 ) h 3 12 ( 1 − ( 0.15 ) 2 ) 100 ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0068.tif"/>

Solving the above equation for h ⇒

h = 9.1 in ( ≅ 23 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-CH014_eqn_0069.tif"/>

The MS Excel worksheet is used to perform the computations of this problem in a rapid and efficient way as shown in Figure 14.14.511

14.7

An interior circular loading of 9 kips with a contact radius of 6 inches is applied on a 10-inch concrete slab with a modulus of subgrade reaction (k) of 100 pci. The modulus of concrete (E) is 4000000 psi, and the Poisson ratio is 0.15. Determine the maximum stress and deflection due to this loading.

Solution:

Westergaard formula for the stress due to interior loading is used as shown below:

σ i = 0.316 P h 2 [ 4 log ( l b ) + 1.069 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0070.tif"/>

Where:

σ_i = tensile stress in the interior of the slab

P = applied load

h = thickness of the slab

l = radius of relative stiffness of concrete slab

b = a if a ≥ 1.724 h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0071.tif"/>

14.19 b = [ 1.6 a 2 + h 2 ] − 0.675 h if a < 1.724 h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0072.tif"/>

Where:

a = radius of contact area

Since a = 6 in < 1.724 h = 17.24 in, therefore, the value of b is determined as below:

b = [ 1.6 a 2 + h 2 ] − 0.675 h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0073.tif"/> 512

⇒

b = [ 1.6 ( 6 ) 2 + ( 10 ) 2 ] − 0.675 ( 10 ) = 5.804 ( 14.7 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-CH014_eqn_0074.tif"/>

The radius of relative stiffness is also determined:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0075.tif"/>

⇒

l = [ ( 4 × 10 6 ) ( 10 ) 3 12 ( 1 − ( 0.15 ) 2 ) 100 ] 0.25 = 43.0 in ( 109.2 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-CH014_eqn_0076.tif"/>

Now the maximum stress due to the interior loading on the concrete slab is calculated using Westergaard formula:

⇒

σ i = 0.316 ( 9000 ) ( 10 ) 2 [ 4 log ( 43.0 5.804 ) + 1.069 ] = 129.3 psi ( 892 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-CH014_eqn_0078.tif"/>

The deflection due to the interior loading on the concrete slab is computed using Westergaard formula for deflection as follows:

⇒

Δ i = 9000 8 ( 100 ) ( 43.0 ) 2 { 1 + 1 2 π [ ln ( 6 2 ( 43.0 ) ) − 0.673 ] [ 6 43.0 ] 2 } = 0.006 in ( 0.015 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-CH014_eqn_0080.tif"/>

The MS Excel worksheet used to perform the computations in this problem to determine the maximum stress and deflection due to the interior loading on the concrete slab is shown in Figure 14.15.513

14.8

A concrete slab is subjected to a corner circular loading of 12 kips. The thickness of the slab is 8 inches and the modulus of subgrade reaction (k) = 100 pci. If the maximum stress due to this loading is 220 psi, the radius of the circular area is 6 inches, determine the maximum stress and deflection due to the corner loading (assume the elastic modulus of concrete is 4000000 psi and the Poisson ratio is 0.15).

Solution:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0081.tif"/>

⇒

l = [ ( 4 × 10 6 ) ( 8 ) 3 12 ( 1 − ( 0.15 ) 2 ) 100 ] 0.25 = 36.35 in ( 92.3 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-CH014_eqn_0082.tif"/>

Westergaard formula for the maximum stress due to corner loading is used as shown below:

σ c = 3 P h 2 [ 1 − ( a 2 l ) 0.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0083.tif"/>

Where:

σ_c = stress at the corner of the slab

P = applied load

h = thickness of concrete slab

a = radius of contact area

l = radius of relative stiffness of concrete slab514

⇒

σ c = 3 ( 12000 ) ( 8 ) 2 [ 1 − ( 6 2 36.35 ) 0.6 ] = 327.5 psi ( ≅ 2258 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-CH014_eqn_0084.tif"/>

14.21 Δ c = P k l 2 [ 1.1 − 0.88 ( a 2 l ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0085.tif"/>

Where:

Δ_c = deflection at the corner of the slab

P = applied load

k = modulus of subgrade reaction

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

Δ c = 12000 100 ( 36.35 ) 2 [ 1.1 − 0.88 ( 6 2 36.35 ) ] = 0.0812 in ( 0.206 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-CH014_eqn_0086.tif"/>

Using MS Excel worksheets makes the solution always easy and fast. The MS Excel worksheet used to perform the computations in this problem to determine the maximum stress and deflection due to the corner loading on the concrete slab is shown in Figure 14.16.515

14.9

A corner circular loading with a contact radius of 4 inches is applied on a concrete slab with a modulus of subgrade reaction (k) of 150 pci creating a corner deflection of 0.0155 inch. The modulus of concrete is 4000000 psi, and the Poisson ratio is 0.15. If the radius of relative stiffness (l) for the concrete slab is 44.5 inches, determine the thickness of the slab, the corner loading, and the corner stress due to this loading.

Solution:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0087.tif"/>

⇒

44.5 = [ ( 4 × 10 6 ) h 3 12 ( 1 − ( 0.15 ) 2 ) 150 ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0088.tif"/>

Solving the above equation for h provides the following solution:

h = 12.0 in ( ≅ 30 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-CH014_eqn_0089.tif"/>

Westergaard formula for corner deflection is used to determine the applied load:

Δ c = P k l 2 [ 1.1 − 0.88 ( a 2 l ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0090.tif"/>

⇒

0.0155 = P 150 ( 44.5 ) 2 [ 1.1 − 0.88 ( 4 2 44.5 ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0091.tif"/>

⇒

P = 4659 lb ( ≅ 20.7 kN) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0092.tif"/>

The corner stress is calculated using Westergaard formula for corner stress:

σ c = 3 P h 2 [ 1 − ( a 2 l ) 0.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0093.tif"/>

⇒

σ c = 3 ( 4659 ) ( 12 ) 2 [ 1 − ( 4 2 44.5 ) 0.6 ] = 69.0 psi ( ≅ 476 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-CH014_eqn_0094.tif"/>

The MS Excel worksheet used to perform the computations in this problem makes the solution rapid and efficient as shown in Figure 14.17.516

14.10

A concrete slab with a modulus of subgrade reaction (k) of 120 pci is subjected to an edge circular loading with a radius of 6 inches resulting in a deflection of 0.012 inch. If the modulus of elasticity of the concrete is 4000000 psi, the Poisson ratio is 0.15, and the radius of relative stiffness of the concrete slab is 40 inches, determine the applied edge loading, the thickness of the slab, and the stress due to the edge loading.

Solution:

Westergaard formula for deflection due to edge circular loading is used:

Δ e-circle = 0.431 P k l 2 [ 1 − 0.82 ( a l ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0095.tif"/>

Where:

Δ_e-circle = deflection due to edge circular loading

P = applied load

k = modulus of subgrade reaction

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

0.012 = 0.431 P 120 ( 40 ) 2 [ 1 − 0.82 ( 6 40 ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0096.tif"/>

⇒517

P = 6095 lb ( ≅ 27 kN) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0097.tif"/>

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0098.tif"/>

⇒

40 = [ ( 4 × 10 6 ) h 3 12 ( 1 − ( 0.15 ) 2 ) 120 ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0099.tif"/>

Solving the above equation for h provides the following solution:

h = 9.7 in ( ≅ 25 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-CH014_eqn_0100.tif"/>

Westergaard formula for stress due to edge circular loading is used to calculate the edge stress on the concrete slab:

σ e-circle = 0.803 P h 2 [ 4 log ( l a ) + 0.666 ( a l ) − 0.034 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0101.tif"/>

Where:

σ_e-circle = deflection due to edge circular loading

P = applied load

h = thickness of concrete slab

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

σ e-circle = 0.803 ( 6095 ) ( 9.7 ) 2 [ 4 log ( 40 6 ) + 0.666 ( 6 40 ) − 0.034 ] = 176.4 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-CH014_eqn_0102.tif"/>

( ≅ 1216 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-CH014_eqn_0103.tif"/>

The MS Excel worksheet used to perform the computations related to this problem to provide a fast and efficient solution is shown in Figure 14.18.518

14.11

If the same loading in Problem 14.10 is applied at the corner of the concrete slab, determine the resulting stress in this case. What is the reduction in the stress between edge and corner?

Solution:

σ c = 3 P h 2 [ 1 − ( a 2 l ) 0.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0104.tif"/>

⇒

σ c = 3 ( 6095 ) ( 9.7 ) 2 [ 1 − ( 6 2 40 ) 0.6 ] = 117.7 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-CH014_eqn_0105.tif"/>

The reduction in the stress between the edge and the corner is equal to:

Reduction in Stress = 176.4 − 117.7 = 58.7 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-CH014_eqn_0106.tif"/>

The percentage of reduction in the stress between the edge and the corner is:

% Reduction in Stress = 100 × 58.7 176.4 ≅ 33 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0107.tif"/>

The MS Excel worksheet shown in Figure 14.19 is used to perform the computations in this problem.519

14.12

In Problem 14.10, if the edge loading is a semi-circular loading, determine the applied edge loading, the thickness of the slab, and the stress due to the edge loading.

Solution:

Westergaard formula for the deflection due to edge semi-circular loading is used:

Δ e-semicircle = 0.431 P k l 2 [ 1 − 0.349 ( a l ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0108.tif"/>

Where:

Δ_e-semicircle = deflection due to edge semi-circular loading

P = applied load

k = modulus of subgrade reaction

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

0.012 = 0.431 P 120 ( 40 ) 2 [ 1 − 0.349 ( 6 40 ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0109.tif"/>

⇒

P = 5641 lb ( ≅ 25 kN) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0110.tif"/> 520

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0111.tif"/>

⇒

Solving the above equation for h provides the following solution:

h = 9.7 in ( ≅ 25 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-CH014_eqn_0113.tif"/>

Westergaard formula for stress due to edge semi-circular loading is used to calculate the edge stress on the concrete slab:

σ e-semicircle = 0.803 P h 2 [ 4 log ( l a ) + 0.282 ( a l ) + 0.650 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0114.tif"/>

Where:

σ_e-semicircle = deflection due to edge semi-circular loading

P = applied load

h = thickness of concrete slab

l = radius of relative stiffness of concrete slab

a = radius of contact area

⇒

σ e-circle = 0.803 ( 6095 ) ( 9.7 ) 2 [ 4 log ( 40 6 ) + 0.282 ( 6 40 ) + 0.650 ] = 193.7 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-CH014_eqn_0115.tif"/>

( ≅ 1335 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-CH014_eqn_0116.tif"/>

The MS Excel worksheet shown in Figure 14.20 is used to perform the computations related to this problem to provide a fast and efficient solution.521

14.13

A concrete slab is subjected to a corner loading of 14 kips on a set of dual wheels as shown in Figure 14.21. The thickness of the slab is 8 inches. If the stress due to this loading is 338.4 psi, the equivalent radius of the contact area for the dual wheels is 8 inches, and the radius of a single circular contact area is 6 inches, determine the spacing between the dual wheels from center to center, the radius of relative stiffness of the concrete slab, and the modulus of subgrade reaction.

Solution:

When the load is applied on a set of dual wheels, an equivalent radius of contact area is determined using the formula below:522

14.26 a = 0.8521 P d q π + S d π ( P d 0.5227 q ) 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-CH014_eqn_0117.tif"/>

Where:

P_d = the applied load on one wheel (tire)

q = applied contact pressure

S_d = spacing between the two wheels

The applied pressure is calculated as below:

q = P A c https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0118.tif"/>

⇒

q = 14000 π ( 6 ) 2 = 123.8 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-CH014_eqn_0119.tif"/>

To determine the spacing between the dual wheels, the formula for the equivalent radius of contact area is applied:

⇒

8 = 0.8521 ( 7000 ) 123.8 π + S d π ( 7000 0.5227 × 123.8 ) 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-CH014_eqn_0120.tif"/>

Solving for S_d provides the following solution:

⇒

S d = 14.7 in ( ≅ 37 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-CH014_eqn_0121.tif"/>

Since the stress at the corner of the slab is given, the radius of relative stiffness is determined using the stress formula as shown below:

σ c = 3 P h 2 [ 1 − ( a 2 l ) 0.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0122.tif"/>

⇒

338.4 = 3 ( 14000 ) ( 8 ) 2 [ 1 − ( 8 2 l ) 0.6 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0123.tif"/>

Solving this equation for l provides the following solution:

l = 37.9 in ( ≅ 96 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-CH014_eqn_0124.tif"/>

To determine the modulus of subgrade reaction, the formula for the radius of relative stiffness is applied:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0125.tif"/> 523

⇒

37.9 = [ 4 × 10 6 ( 8 ) 3 12 ( 1 − ( 0.15 ) 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0126.tif"/>

Solving the above equation for k provides the solution below:

k ≅ 85 pci ( ≅ 23 MPa/m) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0127.tif"/>

The MS Excel worksheet (see Figure 14.22) is used to perform the computations related to this problem to provide a fast and efficient solution particularly back-calculations are needed to determine the required parameters in this problem.

14.14

An interior circular loading with a contact radius of 6 inches is applied on a concrete slab with a modulus of subgrade reaction (k) of 100 pci. The modulus of concrete is 4000000 psi, and the Poisson ratio is 0.15. If the maximum stress and deflection due to this loading are 259.0 psi and 0.0120 inch, respectively, determine the thickness and the radius of relative stiffness of the concrete slab, and the interior loading.

Solution:

Westergaard formulas for stress and deflection due to interior loading are applied:

a = 6 in, assume that a = 6 in < 1.724 h, therefore, the value of b is determined as below:

b = [ 1.6 a 2 + h 2 ] − 0.675 h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0129.tif"/>

⇒

b = [ 1.6 ( 6 ) 2 + h 2 ] − 0.675 h https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0130.tif"/>

⇒

A 259.0 = 0.316 P h 2 [ 4 log ( l [ 1.6 ( 6 ) 2 + h 2 ] − 0.675 h ) + 1.069 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0131.tif"/>

⇒

0.0120 = P 8 ( 100 ) l 2 { 1 + 1 2 π [ ln ( 6 2 l ) − 0.673 ] [ 6 l ] 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-CH014_eqn_0133.tif"/>

And:

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0134.tif"/>

⇒

l = [ ( 4 × 10 6 ) h 3 12 ( 1 − ( 0.15 ) 2 ) 100 ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0135.tif"/>

The above three equations (A), (B), and (C) are solved simultaneously to determine the values of h, l, and P. This task is very difficult (or impossible) to do manually. Therefore, the MS Excel Solver tool is used to obtain the solution using iterative approach as shown in Figure 14.23.

h = 9.6 in ( ≅ 24 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-CH014_eqn_0136.tif"/>

l = 41.7 in ( ≅ 106 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-CH014_eqn_0137.tif"/>

P = 16838 lb ( ≅ 75 kN ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0138.tif"/>

The MS Excel worksheet with the Solver tool set up are shown in Figure 14.23.525

14.15

A 20 ft × 10 ft × 8 in-concrete slab with a modulus of subgrade reaction (k) of 100 pci is subjected to a temperature differential of 25°F. Determine the maximum curling stress in the interior and at the edge of the slab if the coefficient of thermal expansion of concrete (α_t) is 5 × 10⁻⁶ in/in/°F, the elastic modulus of concrete is 4 × 10⁶ psi, and the Poisson ratio is 0.15 (see Figure 14.24).

Solution:

The curling stresses are explained by theory of plate on a Winkler or liquid foundation expressed by a series of springs under the rigid slab.

The curling stresses in the interior of a finite slab are determined using the following formulas:526

σ x = E α t Δ t 2 ( 1 − ν 2 ) ( C x + ν C 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-CH014_eqn_0139.tif"/>

σ y = E α t Δ t 2 ( 1 − ν 2 ) ( C y + ν C 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-CH014_eqn_0140.tif"/>

Where:

σ_x = total stress in the x-direction

σ_y = total stress in the y-direction

E = modulus of concrete

α_t = coefficient of thermal expansion of concrete

Δt = temperature differential between the top and the bottom of the slab

ν = Poisson ratio of concrete

C_x = stress correction factor in the x-direction

C_y = stress correction factor in the y-direction

To determine C_x and C_y, L_x/l and L_y/_l should be determined first. L_x is the length of the slab and L_y is the width of the slab, and l is the radius of relative stiffness.

l = [ E h 3 12 ( 1 − ν 2 ) k ] 0.25 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0141.tif"/>

⇒

l = [ 4 × 10 6 ( 8 ) 3 12 ( 1 − ( 0.15 ) 2 ) 100 ] 0.25 = 36.7 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-CH014_eqn_0142.tif"/>

⇒

L x l = 20 × 12 36.7 = 6.5 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0143.tif"/>

L y l = 10 × 12 36.7 = 3.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0144.tif"/>

The following figure is reproduced based on numbers from the chart of the stress correction factors in Pavement Analysis and Design by Yang H. Huang, 2004. It provides the stress corrections factors for finite slabs. The fifth-order polynomial is developed in this book based on numbers from the stress correction factors chart shown below with a high coefficient of determination (see Figure 14.25).527

From the figure or from the developed polynomial model, C_x and C_y are determined using the values L_x/l and L_y/l, respectively:

C x = 1.07 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0145.tif"/>

C y = 0.36 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0146.tif"/>

⇒

σ x = ( 4 × 10 6 ) ( 5 × 10 − 6 ) ( 25 ) 2 ( 1 − ( 0.15 ) 2 ) ( 1.07 + 0.15 ( 0.36 ) ) = 287.5 psi ( ≅ 1982 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-CH014_eqn_0148.tif"/>

⇒

σ y = ( 4 × 10 6 ) ( 5 × 10 − 6 ) ( 25 ) 2 ( 1 − ( 0.15 ) 2 ) ( 0.36 + 0.15 ( 1.07 ) ) = 133.1 psi ( ≅ 918 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-CH014_eqn_0150.tif"/>

Therefore, the maximum curling stress in the interior of the slab is equal to 287.5 psi (≌ 1982 kPa).528

The curling stress at the edge of the slab is given by the following formula:

σ x or y = C x or y E α t Δ t 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-CH014_eqn_0151.tif"/>

Where:

σ_{x or y} = stress at the edge of the slab in the x- or y-direction

E = modulus of concrete

α_t = coefficient of thermal expansion of concrete

Δt = temperature differential between the top and the bottom of the slab

C_{x or y} = stress correction factor in the x- or y-direction

In this case, the maximum stress at the edge is σ_x since C_x is higher than C_y. Therefore, the edge stress is equal to:

σ x = ( 1.07 ) ( 4 × 10 6 ) ( 5 × 10 − 6 ) ( 25 ) 2 = 267.5 psi ( ≅ 1844 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-CH014_eqn_0152.tif"/>

14.16

Determine the curling (bending) stresses in the interior of the slab in Problem 14.15 if the concrete slab is assumed to be an infinite plate.

Solution:

In case of an infinite plate, the total stress in the x-direction or y-direction in the interior of the slab is given by the following formula:

σ x or y = E α t Δ t 2 ( 1 − ν ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH014_eqn_0153.tif"/>

Where:

σ_{x or y} = stress in the interior of infinite slab in the x- or y-direction

E = modulus of concrete

α_t = coefficient of thermal expansion of concrete

Δt = temperature differential between the top and the bottom of the slab

ν = Poisson ratio of concrete

⇒

σ x or y = ( 4 × 10 6 ) ( 5 × 10 − 6 ) ( 25 ) 2 ( 1 − 0.15 ) = 294.1 psi ( ≅ 2028 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-CH014_eqn_0154.tif"/>

Figure 14.1 MS Excel worksheet image for the computations of Problem 14.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_1_B.tif"/> Figure 14.2 MS Excel worksheet image for the computations of Problem 14.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_2_B.tif"/> Figure 14.3 Designing a rigid pavement over a 6-in granular subbase layer. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_3_B.tif"/> Figure 14.4 Traffic mix expected to use the given 4-lane rural highway. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_4_B.tif"/> Table 14.1 Reliability Level for Roadways Based on Functional Classification 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, DC, USA. Table 14.2 Standard Normal Deviates (Z- Value) 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 14.3 Recommended Values of Drainage Coefficients (Cd) for Rigid 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 1.25–1.20 1.20–1.15 1.15–1.10 1.10 Good 1 day 1.20–1.15 1.15–1.10 1.10–1.00 1.00 Fair 1 week 1.15–1.10 1.10–1.00 1.00–0.90 0.90 Poor 1 month 1.10–1.00 1.00–0.90 0.90–0.80 0.80 Very Poor Never drain 1.00–0.90 0.90–0.80 0.80–0.70 0.70 Reproduced with permission from AASHTO Guide for Design of Pavement Structures, 1993, by AASHTO, Washington, DC, USA. Figure 14.5 Load versus load equivalency factor for rigid pavements (Single-axle load, D = 9 in, pt = 2.5). Based on numbers in the table of equivalent axle load factors from <italic>AASHTO Guide for Design of Pavement Structures,</italic> 1993, by AASHTO, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_5_B.tif"/> Figure 14.6 Load versus load equivalency factor for rigid pavements (Tandem-axle load ≤ 50000 lb, D = 9 in, pt = 2.5). Based on numbers in the table of equivalent axle load factors from <italic>AASHTO Guide for Design of Pavement Structures</italic>, 1993, by AASHTO, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_6_B.tif"/> Figure 14.7 Load versus load equivalency factor for rigid pavements (Tandem-axle load > 50000 lb, D = 9 in, pt = 2.5). Based on numbers in the table of equivalent axle load factors from <italic>AASHTO Guide for Design of Pavement Structures</italic>, 1993, by AASHTO, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_7_B.tif"/> Figure 14.8 Load versus load equivalency factor for rigid pavements (Tridem-axle load ≤ 50000 lb, D = 9 in, pt = 2.5). Based on numbers in the table of equivalent axle load factors from <italic>AASHTO Guide for Design of Pavement Structures</italic>, 1993, by AASHTO, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_8_B.tif"/> Figure 14.9 Load versus load equivalency factor for rigid pavements (Tridem-axle load > 50000, D = 9 in, pt = 2.5). Based on numbers in the table of equivalent axle load factors from <italic>AASHTO Guide for Design of Pavement Structures</italic>, 1993, by AASHTO, Washington, DC, USA. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_9_B.tif"/> Table 14.4 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, Kentucky, USA. Table 14.5 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 14.6 ESALs Computations for Problem 14.3 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/inline14_1_B.tif"/> Single 250 365 0.45 24.30 1 0.093 92,783 Tandem 2 2.201 4,391,737 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/inline14_2_B.tif"/> Single 250 1 0.093 92,783 Single 1 0.195 194,606 Tridem 1 1.348 1,344,939 Total ESALs 6,116,847 Figure 14.10 MS Excel worksheet image for the computations of Problem 14.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_10_B.tif"/> Figure 14.11 Type of truck traffic expected to use the highway for Problem 14.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_11_B.tif"/> Figure 14.12 MS Excel worksheet image for the computations of Problem 14.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_12_B.tif"/> Figure 14.13 MS Excel worksheet image for the computations of Problem 14.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_13_B.tif"/> Figure 14.14 MS Excel worksheet image for the computations of Problem 14.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_14_B.tif"/> Figure 14.15 MS Excel worksheet image for the computations of Problem 14.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_15_B.tif"/> Figure 14.16 MS Excel worksheet image for the computations of Problem 14.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_16_B.tif"/> Figure 14.17 MS Excel worksheet image for the computations of Problem 14.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_17_B.tif"/> Figure 14.18 MS Excel worksheet image for the computations of Problem 14.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_18_B.tif"/> Figure 14.19 MS Excel worksheet image for the computations of Problem 14.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_19_B.tif"/> Figure 14.20 MS Excel worksheet image for the computations of Problem 14.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_20_B.tif"/> Figure 14.21 A concrete slab subjected to 14-kip corner loading on dual wheels. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_21_B.tif"/> Figure 14.22 MS Excel worksheet image for the computations of Problem 14.13. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_22_B.tif"/> Figure 14.23 MS Excel worksheet image for the computations of Problem 14.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_23_B.tif"/> Figure 14.24 A concrete slab subjected to a temperature differential of 25°F. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig14_24_B.tif"/> Figure 14.25 Stress correction factors for concrete slab. Huang, Yang H., <italic>Pavement Analysis and Design</italic>, 2nd Ed., ©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/fig14_25_B.tif"/>