Page 1 CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Chapter 27 Flexible pavement design 27.1 Overview Flexible pavements are so named because the total pavement structure de ects, or exes, under loading. A exible pavement structure is typically composed of several layers of materials. Each layer receives loads from the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity material (and least expensive) on the bottom. 27.2 Design procedures For exible pavements, structural design is mainly concerned with determining appropriate layer thickness and composition. The main design factors are stresses due to trac load and temperature variations. Two methods of exible pavement structural design are common today: Empirical design and mechanistic empirical design. 27.2.1 Empirical design An empirical approach is one which is based on the results of experimentation or experience. Some of them are either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which is based on the results of experimentation or experience. An empirical analysis of exible pavement design can be done with or with out a soil strength test. An example of design without soil strength test is by using HRB soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio (CBR) test. CBR test is widely known and will be discussed. 27.2.2 Mechanistic-Empirical Design Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains, and de ections within a pavement structure and the physical causes are the loads and material properties of the pavement structure. The relationship between these phenomena and their physical causes are typically described using some mathematical models. Along with this mechanistic approach, empirical elements are used Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao Page 2 CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Chapter 27 Flexible pavement design 27.1 Overview Flexible pavements are so named because the total pavement structure de ects, or exes, under loading. A exible pavement structure is typically composed of several layers of materials. Each layer receives loads from the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity material (and least expensive) on the bottom. 27.2 Design procedures For exible pavements, structural design is mainly concerned with determining appropriate layer thickness and composition. The main design factors are stresses due to trac load and temperature variations. Two methods of exible pavement structural design are common today: Empirical design and mechanistic empirical design. 27.2.1 Empirical design An empirical approach is one which is based on the results of experimentation or experience. Some of them are either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which is based on the results of experimentation or experience. An empirical analysis of exible pavement design can be done with or with out a soil strength test. An example of design without soil strength test is by using HRB soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio (CBR) test. CBR test is widely known and will be discussed. 27.2.2 Mechanistic-Empirical Design Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains, and de ections within a pavement structure and the physical causes are the loads and material properties of the pavement structure. The relationship between these phenomena and their physical causes are typically described using some mathematical models. Along with this mechanistic approach, empirical elements are used Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 when dening what value of the calculated stresses, strains, and de ections result in pavement failure. The relationship between physical phenomena and pavement failure is described by empirically derived equations that compute the number of loading cycles to failure. 27.3 Trac and Loading There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement design. Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely used today for pavement design. Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load by number of repetitions of given axle load and its equivalent axle load factor. Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of groups and the stresses, strains, and de ections under each load group can be determined separately; and used for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and tyre contact area. 27.3.1 Equivalent single wheel load To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure, which produces same value of maximum stress, de ection, tensile stress or contact pressure at the desired depth. The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method, known as Boyd and Foster method, based on the following assumptions: equalancy concept is based on equal stress; contact area is circular; in uence angle is 45 o ; and soil medium is elastic, homogeneous, and isotropic half space. The ESWL is given by: log 10 ESWL = log 10 P + 0:301 log 10 ( z d=2 ) log 10 ( 2S d=2 ) (27.1) where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance between two wheels, and z is the desired depth. Example 1 Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center tyre spacing is 20cm and distance between the walls of the two tyres is 10cm. Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao Page 3 CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Chapter 27 Flexible pavement design 27.1 Overview Flexible pavements are so named because the total pavement structure de ects, or exes, under loading. A exible pavement structure is typically composed of several layers of materials. Each layer receives loads from the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity material (and least expensive) on the bottom. 27.2 Design procedures For exible pavements, structural design is mainly concerned with determining appropriate layer thickness and composition. The main design factors are stresses due to trac load and temperature variations. Two methods of exible pavement structural design are common today: Empirical design and mechanistic empirical design. 27.2.1 Empirical design An empirical approach is one which is based on the results of experimentation or experience. Some of them are either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which is based on the results of experimentation or experience. An empirical analysis of exible pavement design can be done with or with out a soil strength test. An example of design without soil strength test is by using HRB soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio (CBR) test. CBR test is widely known and will be discussed. 27.2.2 Mechanistic-Empirical Design Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains, and de ections within a pavement structure and the physical causes are the loads and material properties of the pavement structure. The relationship between these phenomena and their physical causes are typically described using some mathematical models. Along with this mechanistic approach, empirical elements are used Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 when dening what value of the calculated stresses, strains, and de ections result in pavement failure. The relationship between physical phenomena and pavement failure is described by empirically derived equations that compute the number of loading cycles to failure. 27.3 Trac and Loading There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement design. Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely used today for pavement design. Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load by number of repetitions of given axle load and its equivalent axle load factor. Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of groups and the stresses, strains, and de ections under each load group can be determined separately; and used for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and tyre contact area. 27.3.1 Equivalent single wheel load To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure, which produces same value of maximum stress, de ection, tensile stress or contact pressure at the desired depth. The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method, known as Boyd and Foster method, based on the following assumptions: equalancy concept is based on equal stress; contact area is circular; in uence angle is 45 o ; and soil medium is elastic, homogeneous, and isotropic half space. The ESWL is given by: log 10 ESWL = log 10 P + 0:301 log 10 ( z d=2 ) log 10 ( 2S d=2 ) (27.1) where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance between two wheels, and z is the desired depth. Example 1 Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center tyre spacing is 20cm and distance between the walls of the two tyres is 10cm. Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 No stress overlap (ESWL=P) No stress overlap Partial stress overlap Complete stress overlap (ESWL=2P) z 2s d/2 P P s d 2a 2a (log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) ) Figure 27:1: ESWL-Equal stress concept Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN. For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm, log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) =log 10 ESWL = log 10 2044+ 0:301log 10 ( 20 10=2 ) log 10 ( 220 10=2 ) =3.511. Therefore, ESWL = antilog(3.511)= 3244.49 kN 27.3.2 Equivalent single axle load Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels. But to carry large loads multiple axles are provided. Since the design of exible pavements is by layered theory, only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India, specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load and the design of pavement is based on the standard axle load. Repetition of axle loads: The deformation of pavement due to a single application of axle load may be small but due to repeated application of load there would be accumulation of unrecovered or permanent deformation which results in failure of pavement. If the pavement structure fails with N 1 number of repetition of load W 1 and for the same failure criteria if it requires N 2 number of repetition of load W 2 , then W 1 N 1 and W 2 N 2 are considered equivalent. Note that, W 1 N 1 and W 2 N 2 equivalency depends on the failure criterion employed. Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement by the i th type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings. The fatigue cracking model has the following form: N f = f 1 ( t ) f2 (E) f3 orN f / t f2 (27.2) where, N f is the number of load repetition for a certain percentage of cracking, t is the tensile strain at the bottom of the binder course, E is the modulus of elasticity, and f 1 ; f 2 ; f 3 are constants. If we consider fatigue Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao Page 4 CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Chapter 27 Flexible pavement design 27.1 Overview Flexible pavements are so named because the total pavement structure de ects, or exes, under loading. A exible pavement structure is typically composed of several layers of materials. Each layer receives loads from the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity material (and least expensive) on the bottom. 27.2 Design procedures For exible pavements, structural design is mainly concerned with determining appropriate layer thickness and composition. The main design factors are stresses due to trac load and temperature variations. Two methods of exible pavement structural design are common today: Empirical design and mechanistic empirical design. 27.2.1 Empirical design An empirical approach is one which is based on the results of experimentation or experience. Some of them are either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which is based on the results of experimentation or experience. An empirical analysis of exible pavement design can be done with or with out a soil strength test. An example of design without soil strength test is by using HRB soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio (CBR) test. CBR test is widely known and will be discussed. 27.2.2 Mechanistic-Empirical Design Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains, and de ections within a pavement structure and the physical causes are the loads and material properties of the pavement structure. The relationship between these phenomena and their physical causes are typically described using some mathematical models. Along with this mechanistic approach, empirical elements are used Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 when dening what value of the calculated stresses, strains, and de ections result in pavement failure. The relationship between physical phenomena and pavement failure is described by empirically derived equations that compute the number of loading cycles to failure. 27.3 Trac and Loading There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement design. Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely used today for pavement design. Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load by number of repetitions of given axle load and its equivalent axle load factor. Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of groups and the stresses, strains, and de ections under each load group can be determined separately; and used for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and tyre contact area. 27.3.1 Equivalent single wheel load To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure, which produces same value of maximum stress, de ection, tensile stress or contact pressure at the desired depth. The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method, known as Boyd and Foster method, based on the following assumptions: equalancy concept is based on equal stress; contact area is circular; in uence angle is 45 o ; and soil medium is elastic, homogeneous, and isotropic half space. The ESWL is given by: log 10 ESWL = log 10 P + 0:301 log 10 ( z d=2 ) log 10 ( 2S d=2 ) (27.1) where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance between two wheels, and z is the desired depth. Example 1 Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center tyre spacing is 20cm and distance between the walls of the two tyres is 10cm. Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 No stress overlap (ESWL=P) No stress overlap Partial stress overlap Complete stress overlap (ESWL=2P) z 2s d/2 P P s d 2a 2a (log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) ) Figure 27:1: ESWL-Equal stress concept Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN. For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm, log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) =log 10 ESWL = log 10 2044+ 0:301log 10 ( 20 10=2 ) log 10 ( 220 10=2 ) =3.511. Therefore, ESWL = antilog(3.511)= 3244.49 kN 27.3.2 Equivalent single axle load Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels. But to carry large loads multiple axles are provided. Since the design of exible pavements is by layered theory, only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India, specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load and the design of pavement is based on the standard axle load. Repetition of axle loads: The deformation of pavement due to a single application of axle load may be small but due to repeated application of load there would be accumulation of unrecovered or permanent deformation which results in failure of pavement. If the pavement structure fails with N 1 number of repetition of load W 1 and for the same failure criteria if it requires N 2 number of repetition of load W 2 , then W 1 N 1 and W 2 N 2 are considered equivalent. Note that, W 1 N 1 and W 2 N 2 equivalency depends on the failure criterion employed. Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement by the i th type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings. The fatigue cracking model has the following form: N f = f 1 ( t ) f2 (E) f3 orN f / t f2 (27.2) where, N f is the number of load repetition for a certain percentage of cracking, t is the tensile strain at the bottom of the binder course, E is the modulus of elasticity, and f 1 ; f 2 ; f 3 are constants. If we consider fatigue Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 cracking as failure criteria, and a typical value of 4 for f 2 , then: EALF = i std 4 (27.3) where, i indicate i th vehicle, and std indicate the standard axle. Now if we assume that the strain is proportional to the wheel load, EALF = W i W std 4 (27.4) Similar results can be obtained if rutting model is used, which is: N d = f 4 ( c ) f5 (27.5) where N d is the permissible design rut depth (say 20mm), c is the compressive strain at the top of the subgrade, and f 4 ; f 5 are constants. Once we have the EALF, then we can get the ESAL as given below. Equivalent single axle load, ESAL = m X i=1 F i n i (27.6) where,m is the number of axle load groups, F i is the EALF for i th axle load group, and n i is the number of passes of i th axle load group during the design period. Example 1 Let number of load repetition expected by 80 KN standard axle is 1000, 160 KN is 100 and 40 KN is 10000. Find the equivalent axle load. Solution: Refer the Table 27:1. The ESAL is given as P F i n i = 3225 kN Table 27:1: Example 1 Solution Axle No.of Load EALF Load Repetition i (KN) (n i ) (F i ) F i n i 1 40 10000 (40=80) 4 = 0.0625 625 2 80 1000 (80=80) 4 = 1 1000 3 160 100 (160=80) 4 = 16 1600 Example 2 Let the number of load repetition expected by 120 kN axle is 1000, 160 kN is 100, and 40 kN is 10,000. Find the equivalent standard axle load if the equivalence criteria is rutting. Assume 80 kN as standard axle load and the rutting model is N r = f 4 f5 c where f 4 =4.2 and f 5 =4.5. Solution Refer the Table 27:2. The ESAL is given as P F i n i = 8904:94 kN Introduction to Transportation Engineering 27.4 Tom V. Mathew and K V Krishna Rao Page 5 CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Chapter 27 Flexible pavement design 27.1 Overview Flexible pavements are so named because the total pavement structure de ects, or exes, under loading. A exible pavement structure is typically composed of several layers of materials. Each layer receives loads from the above layer, spreads them out, and passes on these loads to the next layer below. Thus the stresses will be reduced, which are maximum at the top layer and minimum on the top of subgrade. In order to take maximum advantage of this property, layers are usually arranged in the order of descending load bearing capacity with the highest load bearing capacity material (and most expensive) on the top and the lowest load bearing capacity material (and least expensive) on the bottom. 27.2 Design procedures For exible pavements, structural design is mainly concerned with determining appropriate layer thickness and composition. The main design factors are stresses due to trac load and temperature variations. Two methods of exible pavement structural design are common today: Empirical design and mechanistic empirical design. 27.2.1 Empirical design An empirical approach is one which is based on the results of experimentation or experience. Some of them are either based on physical properties or strength parameters of soil subgrade. An empirical approach is one which is based on the results of experimentation or experience. An empirical analysis of exible pavement design can be done with or with out a soil strength test. An example of design without soil strength test is by using HRB soil classication system, in which soils are grouped from A-1 to A-7 and a group index is added to dierentiate soils within each group. Example with soil strength test uses McLeod, Stabilometer, California Bearing Ratio (CBR) test. CBR test is widely known and will be discussed. 27.2.2 Mechanistic-Empirical Design Empirical-Mechanistic method of design is based on the mechanics of materials that relates input, such as wheel load, to an output or pavement response. In pavement design, the responses are the stresses, strains, and de ections within a pavement structure and the physical causes are the loads and material properties of the pavement structure. The relationship between these phenomena and their physical causes are typically described using some mathematical models. Along with this mechanistic approach, empirical elements are used Introduction to Transportation Engineering 27.1 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 when dening what value of the calculated stresses, strains, and de ections result in pavement failure. The relationship between physical phenomena and pavement failure is described by empirically derived equations that compute the number of loading cycles to failure. 27.3 Trac and Loading There are three dierent approaches for considering vehicular and trac characteristics, which aects pavement design. Fixed trac: Thickness of pavement is governed by single load and number of load repetitions is not considered. The heaviest wheel load anticipated is used for design purpose. This is an old method and is rarely used today for pavement design. Fixed vehicle: In the xed vehicle procedure, the thickness is governed by the number of repetitions of a standard axle load. If the axle load is not a standard one, then it must be converted to an equivalent axle load by number of repetitions of given axle load and its equivalent axle load factor. Variable trac and vehicle: In this approach, both trac and vehicle are considered individually, so there is no need to assign an equivalent factor for each axle load. The loads can be divided into a number of groups and the stresses, strains, and de ections under each load group can be determined separately; and used for design purposes. The trac and loading factors to be considered include axle loads, load repetitions, and tyre contact area. 27.3.1 Equivalent single wheel load To carry maximum load with in the specied limit and to carry greater load, dual wheel, or dual tandem assembly is often used. Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure, which produces same value of maximum stress, de ection, tensile stress or contact pressure at the desired depth. The procedure of nding the ESWL for equal stress criteria is provided below. This is a semi-rational method, known as Boyd and Foster method, based on the following assumptions: equalancy concept is based on equal stress; contact area is circular; in uence angle is 45 o ; and soil medium is elastic, homogeneous, and isotropic half space. The ESWL is given by: log 10 ESWL = log 10 P + 0:301 log 10 ( z d=2 ) log 10 ( 2S d=2 ) (27.1) where P is the wheel load, S is the center to center distance between the two wheels, d is the clear distance between two wheels, and z is the desired depth. Example 1 Find ESWL at depths of 5cm, 20cm and 40cm for a dual wheel carrying 2044 kg each. The center to center tyre spacing is 20cm and distance between the walls of the two tyres is 10cm. Introduction to Transportation Engineering 27.2 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 No stress overlap (ESWL=P) No stress overlap Partial stress overlap Complete stress overlap (ESWL=2P) z 2s d/2 P P s d 2a 2a (log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) ) Figure 27:1: ESWL-Equal stress concept Solution For desired depth z=40cm, which is twice the tyre spacing, ESWL = 2P=22044 = 4088 kN. For z=5cm, which is half the distance between the walls of the tyre, ESWL = P = 2044kN. For z=20cm, log 10 ESWL = log 10 P + 0:301log 10 ( z d=2 ) log 10 ( 2S d=2 ) =log 10 ESWL = log 10 2044+ 0:301log 10 ( 20 10=2 ) log 10 ( 220 10=2 ) =3.511. Therefore, ESWL = antilog(3.511)= 3244.49 kN 27.3.2 Equivalent single axle load Vehicles can have many axles which will distribute the load into dierent axles, and in turn to the pavement through the wheels. A standard truck has two axles, front axle with two wheels and rear axle with four wheels. But to carry large loads multiple axles are provided. Since the design of exible pavements is by layered theory, only the wheels on one side needed to be considered. On the other hand, the design of rigid pavement is by plate theory and hence the wheel load on both sides of axle need to be considered. Legal axle load: The maximum allowed axle load on the roads is called legal axle load. For highways the maximum legal axle load in India, specied by IRC, is 10 tonnes. Standard axle load: It is a single axle load with dual wheel carrying 80 KN load and the design of pavement is based on the standard axle load. Repetition of axle loads: The deformation of pavement due to a single application of axle load may be small but due to repeated application of load there would be accumulation of unrecovered or permanent deformation which results in failure of pavement. If the pavement structure fails with N 1 number of repetition of load W 1 and for the same failure criteria if it requires N 2 number of repetition of load W 2 , then W 1 N 1 and W 2 N 2 are considered equivalent. Note that, W 1 N 1 and W 2 N 2 equivalency depends on the failure criterion employed. Equivalent axle load factor: An equivalent axle load factor (EALF) denes the damage per pass to a pavement by the i th type of axle relative to the damage per pass of a standard axle load. While nding the EALF, the failure criterion is important. Two types of failure criterias are commonly adopted: fatigue cracking and ruttings. The fatigue cracking model has the following form: N f = f 1 ( t ) f2 (E) f3 orN f / t f2 (27.2) where, N f is the number of load repetition for a certain percentage of cracking, t is the tensile strain at the bottom of the binder course, E is the modulus of elasticity, and f 1 ; f 2 ; f 3 are constants. If we consider fatigue Introduction to Transportation Engineering 27.3 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 cracking as failure criteria, and a typical value of 4 for f 2 , then: EALF = i std 4 (27.3) where, i indicate i th vehicle, and std indicate the standard axle. Now if we assume that the strain is proportional to the wheel load, EALF = W i W std 4 (27.4) Similar results can be obtained if rutting model is used, which is: N d = f 4 ( c ) f5 (27.5) where N d is the permissible design rut depth (say 20mm), c is the compressive strain at the top of the subgrade, and f 4 ; f 5 are constants. Once we have the EALF, then we can get the ESAL as given below. Equivalent single axle load, ESAL = m X i=1 F i n i (27.6) where,m is the number of axle load groups, F i is the EALF for i th axle load group, and n i is the number of passes of i th axle load group during the design period. Example 1 Let number of load repetition expected by 80 KN standard axle is 1000, 160 KN is 100 and 40 KN is 10000. Find the equivalent axle load. Solution: Refer the Table 27:1. The ESAL is given as P F i n i = 3225 kN Table 27:1: Example 1 Solution Axle No.of Load EALF Load Repetition i (KN) (n i ) (F i ) F i n i 1 40 10000 (40=80) 4 = 0.0625 625 2 80 1000 (80=80) 4 = 1 1000 3 160 100 (160=80) 4 = 16 1600 Example 2 Let the number of load repetition expected by 120 kN axle is 1000, 160 kN is 100, and 40 kN is 10,000. Find the equivalent standard axle load if the equivalence criteria is rutting. Assume 80 kN as standard axle load and the rutting model is N r = f 4 f5 c where f 4 =4.2 and f 5 =4.5. Solution Refer the Table 27:2. The ESAL is given as P F i n i = 8904:94 kN Introduction to Transportation Engineering 27.4 Tom V. Mathew and K V Krishna Rao CHAPTER 27. FLEXIBLE PAVEMENT DESIGN NPTEL May 8, 2007 Table 27:2: Example 2 Solution Axle No.of Load EALF Load Repetition i (KN) (n i ) (F i ) F i n i 1 120 1000 (120=80) 4:5 = 6.200 6200 2 160 100 (160=80) 4:5 = 22.63 2263 3 40 10000 (40=80) 4:5 = 0.04419 441.9 Example 3 Let number of load repetition expected by 60kN standard axle is 1000, 120kN is 200 and 40 kN is 10000. Find the equivalent axle load using fatigue cracking as failure criteria according to IRC. Hint:N f = 2:21 10 4 ( t ) 3:89 (E) 0 :854 Solution Refer the Table 27:3. The ESAL is given as P F i n i = 6030:81 kN Table 27:3: Example 3 Solution Axle No.of Load EALF Load Repetition i (KN) (n i ) (F i ) F i n i 1 40 10000 (40=60) 3:89 = 0.2065 2065 2 60 1000 (60=60) 3:89 = 1 1000 3 120 200 (120=60) 3:89 = 14.825 2965.081 27.4 Material characterization It is well known that the pavement materials are not perfectly elastic but experiences some permanent deforma- tion after each load repetitions. It is well known that most paving materials are not elastic but experience some permanent deformation after each load application. However, if the load is small compared to the strength of the material and the deformation under each load repetition is almost completely recoverable then the material can be considered as elastic. The Figure 27:2 shows straining of a specimen under a repeated load test. At the initial stage of load applications, there is considerable permanent deformation as indicated by the plastic strain in the Figure 27:2. As the number of repetition increases, the plastic strain due to each load repetition decreases. After 100 to 200 repetitions, the strain is practically all-recoverable, as indicated by r in the gure. 27.4.1 Resilient modulus of soil The elastic modulus based on the recoverable strain under repeated loads is called the resilient modulus M R , dened as M R = d r : In which d is the deviator stress, which is the axial stress in an unconned compression Introduction to Transportation Engineering 27.5 Tom V. Mathew and K V Krishna RaoRead More

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