CH-19-21 Design of Flexible Pavements – Traffic and Highway Engineering : Book Guide

This Document Contains Chapters 19 to 21 Chapter 19 Design of Flexible Pavements 19-1 Discuss the structural components of a flexible pavement. A flexible pavement consists of the subgrade, the subbase, the base and the wearing surface. The subgrade is usually the natural material located along the horizontal alignment of the pavement and is prepared as a roadbed to serve as the foundation of the pavement structure. The subbase is located immediately above the subgrade and consists of material of a superior quality to that which generally is used for subgrade construction. When the quality of the subgrade material meets the requirements of the subbase material, the subbase component may be omitted. The base course lies immediately above the subbase (or subgrade) and usually consists of granular materials such as crushed stone, crushed or uncrushed slag, crushed or uncrushed gravel, and sand. The requirements for plasticity, gradation, and strength increase through the subgrade, subbase, and base courses. The surface course is the upper course of the road pavement and is constructed immediately above the base course. The surface course in flexible pavements usually consists of a mixture of mineral aggregates and asphaltic materials. 19-2 Describe the purpose of soil stabilization and discuss at least three methods of achieving it. Soil stabilization is performed to improve the engineering properties of natural soils. There are two categories of soil stabilization, mechanical and chemical. Mechanical stabilization is the blending of different grades of soils to obtain the required grade. Chemical stabilization is the blending of the natural soil with chemical agents. Cement stabilization usually involves the addition of 5% to 14% Portland cement by volume of the compacted mixture to the soil being stabilized. This type of stabilization is usually used to obtain the required engineering properties for base course materials. The process of stabilizing soils with cement involves pulverizing the soil, mixing the required quantity of cement with the pulverized soil, compacting the soil-cement mixture, and curing the compacted layer. Bituminous stabilization is carried out to achieve waterproofing of natural materials and/or binding of natural materials. Waterproofing aids in maintaining the correct moisture content and helps prevent water from seeping into the subgrade, protecting the subgrade from failure due to an increase in moisture content. Binding improves the durability characteristics of the natural soil by providing an adhesive characteristic, whereby the soil particles adhere to each other, increasing cohesion. Several types of soils can be stabilized with bituminous materials, although it is generally required that less than 25% of the material passes the No. 200 sieve. Lime stabilization can be used for stabilizing both base and subbase materials. The materials most commonly used for lime stabilization are calcium hydroxide and dolomite. Clayey materials are most suitable for lime stabilization, but the materials should have plasticity index values less than 10 for the stabilization to be most effective. 19-3 Discuss the process of lime stabilization, and indicate the soil characteristics for which it is most effective. Lime stabilization can be used for stabilizing both base and subbase materials. The materials most commonly used for lime stabilization are calcium hydroxide and dolomite. Clayey materials are most suitable for lime stabilization, but the materials should have plasticity index values less than 10 for the stabilization to be most effective. The reaction of the material with the lime reduces the tendency to swell as a result of an increase in moisture content; the PI value of the soil is also reduced. Desirable changes also include a pozzolanic reaction may also occur resulting in some cementation of the soil and in an increase in strength; soils with high silica or alumina content may exhibit significant increases in strength over long periods of time; and, finally, flocculation occurs, increasing the effective grain size of the soil. 19-4 An axle weight study on a section of highway gave the following data on axle load distribution:
Axle Load Group: Single (1000 lb) No. of axles per 1000 vehicles Axle Load Group: Tandem (1000 lb) No. of axles per 1000 vehicles
<4 678 <6 18
4-8 775 6-12 236
8-12 500 12-18 170
12-16 150 18-24 120
16-18 60 24-30 152
18-20 40 30-32 66
20-22 7 32-34 30
22-24 4 34-36 12
24-26 3 36-38 4
38-40 1
Determine the truck factor for this section of highway. Assume SN = 4 and pt = 2.5. The following table shows the given data and the load equivalency factors taken, by interpolation if necessary, from Table 19.3.
Axle Load Group: Single # of axles/ 1000 vehicles Factor Col. (2) × Col. (3)
<4 678 0.0016 1.0848
4-8 775 0.013 10.075
8-12 500 0.102 51
12-16 150 0.388 58.2
16-18 60 0.8225 49.35
18-20 40 1.235 49.4
20-22 7 1.78 12.46
22-24 4 2.49 9.96
24-26 3 3.4 10.2
Total for single axle load group = 251.730
Axle Load Group: Tandem # of axles/ 1000 vehicles Factor Col. (2) × Col. (3)
<6 18 0.0003 0.0054
6-12 236 0.065 15.34
12-18 170 0.045 7.65
18-24 120 0.174 20.88
24-30 152 0.4675 71.06
30-32 66 0.791 52.206
32-34 30 0.9985 29.955
34-36 12 1.245 14.94
36-38 4 1.53 6.12
38-40 1 1.855 1.855
Total for tandem axle load group = 220.011 Total = 251.730 + 220.011 = 471.741 Truck factor = 471.741/1000 Truck factor = 0.4717 19-5 How does an ESAL differ from a truck factor? A truck factor is defined as the number of 18,000 lb single load applications caused by a single passage of a vehicle. The ESAL, equivalent single axle load, is the number of repetitions of an 18,000 lb single-axle load applied to the pavement on two sets of dual tires. 19-6 A six-lane divided highway is to be designed to replace an existing highway. The present AADT (both directions) of 6000 vehicles is expected to grow at 5% per annum. Assume SN = 4 and pt = 2.5. The percent of traffic on the design lane is 45%. Determine the design ESAL if the design life is 20 years and the vehicle mix is:
Passenger cars (1000 lb/axle) = 60%
2-axle single-unit trucks (5000 lb/axle) = 30%
3-axle single-unit trucks (7000 lb/axle) = 10%
Use Equation 19.2 to determine ESALs: ESALi = (fd)(Gjt)(AADTi)(365)(Ni)(FEi) The design lane use factor, fd, is given as 0.45 From Table 19.4, the growth factor, Gjt is 33.06 for a growth rate of 5% and design life of 20 yr From Table 19.3, by interpolation, the load equivalency factors are: axle SU truck = 0.008 axle SU truck = 0.025 ESAL2-axle SU truck= (0.45)(33.06)(6000)(0.3)(365)(2)(0.008) ESAL2-axle SU truck = 1.56 × 105 ESAL3-axle SU truck = (0.45)(33.06)(6000)(0.1)(365)(3)(0.025) ESAL3-axle SU truck = 2.44 × 105 Note that the contribution of passenger car traffic to ESALs is negligible and therefore not considered in the calculations. Total ESAL = 4.00 × 105 19-7 A section of a two-lane rural highway is to be realigned and replaced by a four-lane highway with a full-depth asphalt pavement. The AADT (both ways) on the existing section can be represented by 500 ESAL. It is expected that construction will be completed 5 years from now. If the traffic growth rate is 5% and the effective CBR of the subgrade on the new alignment is 85, determine a suitable depth of the asphalt pavement using the AASHTO method. Take the design life of the pavement as 20 years. The resilient modulus of the asphalt (EAC) is 400,000 lb/in2. Assume mi for the subgrade is 1 and the percent of traffic on the design lane is 45%. Use a reliability level of 90%, a standard deviation of 0.45, and a design serviceability loss of 2.0. First, determine design ESAL From Table 19.4, the growth factor, Gjt, for 25 years (5 years of construction plus 20 year design life) = 47.73 and the growth factor, Gjt, for 5 years (to subtract the ESAL for the construction period) = 5.53 Design ESAL = (500)(47.73 – 5.53)(365)(0.45) Design ESAL = 3.47 × 106 Determine Mr from Figure 19.3, Mr = 19,000 psi Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 90% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 3.47 × 106 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 19 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 3.8 Determine structural coefficients for each layer From Figure 19.5, with EAC = 400,000 lb/in2 a1= 0.40 Since this pavement is being designed with full-depth asphalt concrete, the asphalt concrete must provide all of the structural support. Therefore, equation 19.6 can be modified for use in this problem as: SN = a1D1 = 3.8 = (0.40)D1 D1 = 9.5 inches. 19-8 A section of a two-lane rural highway is to be realigned and replaced by a four-lane highway with a full-depth asphalt pavement. The AADT (both ways) on the existing section can be represented by 750 ESAL. It is expected that construction will be completed 5 years from now. If the traffic growth rate is 2% and the effective CBR of the subgrade on the new alignment is 90, determine a suitable depth of the asphalt pavement using the AASHTO method. Take the design life of the pavement as 20 years. The resilient modulus of the asphalt (EAC) is 450,000 lb/in2. Assume mi for the subgrade is 1 and the percent of traffic on the design lane is 45%. Use a reliability level of 90%, a standard deviation of 0.45, and a design serviceability loss of 2.0. First, determine design ESAL From Table 19.4, the growth factor, Gjt, for 25 years (5 years of construction plus 20 year design life) = 32.03 and the growth factor, Gjt, for 5 years (to subtract the ESAL for the construction period) = 5.20 Design ESAL = (750)(32.03 – 5.20)(365)(0.45) Design ESAL = 3.31 × 106 Determine Mr from Figure 19.6, Mr = 19,000 psi Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 90% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 3.31 × 106 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 19 x 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 2.9 Determine structural coefficients for each layer From Figure 19.5, with EAC = 450,000 lb/in2 a1= 0.43 Since this pavement is being designed with full-depth asphalt concrete, the asphalt concrete must provide all of the structural support. Therefore, equation 19.6 can be modified for use in this problem as: SN = a1D1 = 2.9 = (0.43)D1 D1 = 6.74inches. Therefore, the depth of the asphalt concrete should be 7.0 inches. 19-9 The predicted traffic mix of a proposed four-lane urban non-interstate freeway: Passenger cars = 78% Single-unit trucks 2-axle, 5,000 lb/axle = 12% 2-axle, 9,000 lb/axle = 4% 3-axle or more, 23,000 lb/axle = 3% Tractor semitrailers and combinations 3-axle, 20,000 lb/axle = 3% The projected AADT during the first year of operation is 3900 (both directions). If the traffic growth rate is estimated at 3% and the CBR of the subgrade is 75, determine the depth of a full-asphalt pavement using the AASHTO method and n = 20 years. The resilient modulus of the asphalt (EAC) is 320,000 lb/in Assume mi for the subgrade is 1 and the percent of traffic on the design lane is 42%, pt = 2.5 and SN = 4. Use a reliability level of 90%, a standard deviation of 0.45, and a design serviceability loss of 2.0. Assume the design lane factor fd is 0.42. Use Equation 19.2 to determine design ESALs. ESALi = (fd)(Gjt)(AADTi)(365)(Ni)(fEi) From Table 19.4, Gjt = 27.04 From Table 19.3, the following truck factors are obtained: f1 = 0.008 (2-axle, 5,000 lb/axle) f2 = 0.0715 (2-axle, 9,000 lb/axle) f3 = 0.2495 (3-axle or more, 23,000 lb/axle) f4 = 0.141 (3-axle, 20,000 lb/axle) ESAL1 = (0.42)(27.04)(3900)(0.12)(365)(2)(0.008) = 3.10 × 104 ESAL2 = (0.42)(27.04)(3900)(0.04)(365)(2)(0.0715) = 9.25 × 104 ESAL3 = (0.42)(27.04)(3900)(0.03)(365)(3)(0.2495) = 3.63 × 105 ESAL4 = (0.42)(27.04)(3900)(0.03)(365)(3)(0.141) = 2.05 × 105 Total ESAL = 6.92 × 105 Determine Mr from Figure 19.6, for a CBR of 75, Mr = 18,500 psi Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 90% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 6.92 × 105 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 18.5 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 2.4 Determine structural coefficients for each layer From Figure 19.5, with EAC = 320,000 lb/in2 a1= 0.36 Since this pavement is being designed with full-depth asphalt concrete, the asphalt concrete must provide all of the structural support. Therefore, equation 19.6 can be modified for use in this problem as: SN = a1D1 = 2.4 = (0.36)D1 D1 = 6.67 inches  round up to 7.0 inches. 19-10 A rural principal arterial is expected to carry an ESAL of 0.188x106 during the first year of operation with an expected annual growth of 6% over the 20-year design life. If the subgrade has a resilient modulus of 15,000 lb/in2, design a suitable pavement consisting of a granular subbase with a layer coefficient of 0.13, a granular base layer with a layer coefficient of 0.14, and an asphalt concrete surface with an elastic modulus of 400,000 lb/in2. Assume all mi values =1, the percent of traffic on the design lane is 47%, and SN = 4. Use a reliability level of 85%, a standard deviation of 0.45, and a design serviceability loss of 2.0. First, determine design ESAL The design lane use factor, fd, is 0.47 From Table 19.4, Gjt = 36.79 Total ESALi = (0.47)(36.79)(0.188 × 106) ESAL = 3.25 × 106 Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 85% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 3.25 × 106 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 15 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 3.9 Determine structural coefficients for each layer From Figure 19.5, with EAC = 400,000 lb/in2 a1= 0.40 Since this pavement is being designed with three layers, equation 19.6 can be modified for use in this problem as: SN = a1D1 + a2D2m2 + a3D3m3 3.9 = (0.40)D1 + (0.14)D2(1) + (0.13)D3(1) From Table 19.9, for 3.25 × 106 ESALs, a recommended minimum thickness of asphalt concrete is 3.5 inches and 6 inches for aggregate base. The depth of subbase can then be found: 3.9 = (0.40)(3.5) + (0.14)(6)(1) + (0.13)D3(1) D3 = 12.77 inches  round up to 13 inches. 19-11 Using the information given in Problem 19-6, design a suitable pavement consisting of an asphalt mixture surface with an elastic modulus of 250,000 lb/in2, a granular base layer with a structural coefficient of 0.14 on a subgrade having a CBR of 10. Assume all mi values = 1, and the percent of traffic on the design lane is 45%. Use a reliability level of 85%, a standard deviation of 0.45, and a design serviceability loss of 2.0. From Problem 19-6, ESAL = 4.00 × 105 Using Equation 19.3, Mr = 1,500(CBR) = 15,000 lb/in2 Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 85% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 4.00 × 105 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 15 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 3.2 Determine structural coefficients for each layer From Figure 19.5, with EAC = 250,000 lb/in2 a1= 0.33 Since this pavement is being designed with two layers, equation 19.6 can be modified for use in this problem as: SN = a1D1 + a2D2m2 3.2 = (0.33)D1 + (0.14)D2(1) + (0.13)D3(1) From Table 19.9, for 4.00 × 105 ESALs, a recommended minimum thickness of asphalt concrete is 2.5 inches and 4 inches for aggregate base. The structural number provided using these minimums can then be found: SN = (0.33)(2.5) + (0.14)(4)(1) = 1.385 Since this SN is below that required, layer thicknesses will need to be increased in order to provide sufficient support. If the asphalt concrete thickness is held at 2.5 inches, the required thickness of the base course can then be found: 3.2 = (0.33)(2.5) + (0.14)D2(1) D2 = 16.9 inches  round up to 17 inches. 19-12 Repeat Problem 19-11 for a pavement consisting of an asphalt mixture surface with an elastic modulus of 280,000 lb/in2, and 6” of granular subbase with a Resilient Modulus of 18 × 103 lb/in2. From Problem 19-6, ESAL = 4.00 × 105 Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 85% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 4.00 × 105 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 18 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 2.0 Step 5: Draw vertical line to read SN of 2.1 Determine structural coefficients for each layer From Figure 19.5, with EAC = 280,000 lb/in2 a1= 0.35 Since this pavement is being designed with two layers, equation 19.6 can be modified for use in this problem as: SN = a1D1 + a2D2m2 2.1 = (0.35)D1 + (0.14)D2(1) + (0.13)D3(1) From Table 19.9, for 4.00 × 105 ESALs, a recommended minimum thickness of asphalt concrete is 2.5 inches and 4 inches for aggregate base. The structural number provided using these minimums can then be found: SN = (0.35)(2.5) + (0.14)(4)(1) = 1.435 Since this SN is below that required, layer thicknesses will need to be increased in order to provide sufficient support. If the asphalt concrete thickness is held at 2.5 inches, the required thickness of the base course can then be found: 2.1 = (0.35)(2.5) + (0.14)D2(1) D2 = 8.75 inches  round up to 9 inches. 19-13 Repeat Problem 19-7 using two different depths of untreated aggregate bases of 6” and 12”. Highway contractors in your area can furnish rates for providing and properly laying an asphalt concrete surface and untreated granular base. Assume a structural coefficient of 0.12 for the base course. If these rates are available, determine the cost for constructing the different pavement designs if the highway section is 5 miles long and the lane width is 12 ft. Which design will you select for construction? From Problem 19-7, SN = 3.8. Since the only change from Problem 19-7 is that a base course of 6 or 12 inches is added, the same SN value can be used in this problem. Since this pavement is being designed with two layers, equation 19.6 can be modified for use in this problem as: SN = a1D1 + a2D2m2 3.8 = (0.40)D1 + (0.12)D2(1) For the 6 inch base course: 3.8 = (0.40)D1 + (0.12)(6)(1) D1 = 7.7 inches. For the 12 inch base course: 3.8 = (0.40)D1 + (0.12)(12)(1) D1 = 5.9 inches. 19-14 The traffic on the design lane of a proposed four-lane rural interstate highway consists of 40% trucks. If classification studies have shown that the truck factor can be taken as 0.45, design a suitable flexible pavement using the 1993 AASHTO procedure if the AADT on the design lane during the first year of operation is 1000, pi = 4.2, and pt = 2.5 . Growth rate = 4% Design life = 20 years Reliability level = 95% Standard deviation = 0.45 The pavement structure will be exposed to moisture levels approaching saturation 20% of the time, and it will take about one week for drainage of water. Effective CBR of the subgrade material is 7. CBR of the base and subbase are 70 and 22, respectively, and Mr for the asphalt mixture, 450,000 lb/in2. Calculate ESALs using Equation 19.2, ESALi = (fd)(Gjt)(AADTi)(365)(fi) ESAL = (0.40)(29.78)(1000)(0.45)(365) ESAL = 1.957 × 106 Calculate Mr of subgrade using Equation 19.4, Mr = 1500 CBR Mr = (1500)(7) = 1.05 × 104 Calculate serviceability loss (ΔPSI), ΔPSI = pi – pt ΔPSI = 4.2 – 2.5 = 1.7 Using the nomograph in Figure 19.8, Step 1: Connect Reliability of 95% to standard deviation of 0.45 and extend to first turning line to create “A” Step 2: Connect “A” to ESAL of 2.0 × 106 and extend to second turning line to create “B” Step 3: Connect “B” to Mr of 10 × 103 and extend to design serviceability loss chart to create “C” Step 4: Connect “C” to ΔPSI of 1.7 Step 5: Draw vertical line to read SN of 3.7 Determine structural coefficients for each layer From Figure 19.5, with Mr = 4.5 × 105 (asphalt surface course) a1= 0.44 From Figure 19.4, with CBR = 70 (base course) a2= 0.13 From Figure 19.3, with CBR = 22 (subbase) a3= 0.10 Determine drainage index, mi, using Tables 19.5 and 19.6 For fair drainage and 20% saturation, mi = 0.9 Knowing that each course must meet the requirements of the course above it, we can determine the required depth of each course, using various forms of Equation 19.6. Beginning with the surface course, SN1 can be found using the Mr for the base course. From Figure 19.8, SN = 2.7 D1*= SN1/a1 = 2.7/0.44 = 6.14 in. Use 6.5 in. SN1*= a1D1* = 0.44(6.5) = 2.86 Repeating this procedure for the base course: Mr = 1.35 × 104 (Subbase) From Figure 19.8, SN = 3.3 D2* = (3.3 – 2.86) / [(0.13)(0.9)] = 3.76 in. From Table 19.9, use recommended minimum of 6 in. SN2* = (a2)(m2)(D2*) + SN1* = (0.13)(0.9)(6) + 2.86 = 3.56 D3* = (3.7 – 3.56) / [(0.1)(0.9)] = 1.56 in. From Table 19.9, use recommended minimum of 4 in. Check design: SN = 0.44(6.5) + (0.13)(0.9)(6) + (0.1)(0.9)(4) = 3.92 (which is greater than SN of 3.7), so the design is satisfactory. The pavement will consist of 6.5 in. of asphalt concrete surface, 6 in. of granular base, and 4 in. of subbase. 19-15 Repeat Problem 19-14 with the subgrade Mr values for each month from January through December being 18,000, 18,000, 12,000, 8000, 8000, 9000, 9000, 9000, 9500, 9500, 12,000, and 20,000 lb/in2, respectively. The pavement structure will be exposed to moisture levels approaching saturation for 20% of the time, and it will take about 4 weeks for drainage of water from the pavement. Use untreated sandy gravel with Mr of 15 × 103 lbs/in2 for subbase and untreated granular material with Mr of 28 × 103 lbs/in2 for the base course. First determine the relative damage each month. Using Figure 19.6 or the following equation and table:
Month Mr uf
Jan. 18000 0.016
Feb. 18000 0.016
March 12000 0.04
April 8000 0.10
May 8000 0.10
June 9000 0.08
July 9000 0.08
Aug. 9000 0.08
Sept. 9500 0.07
Oct. 9500 0.07
Nov. 12000 0.04
Dec. 20000 0.012
Σuf = 0.704
Average uf = Σuf / n = 0.704/12 = 0.0587 Roadbed soil, Mr = 10800 lb/in2 From Problem 19-13, ΔPSI = 1.7 Using Figure 19.8, SN = 3.8 Determine the structural layer coefficient for each layer: From Figure 19.5, EAC = 4.5 × 105 (asphalt concrete, from Problem 20-14) a1 = 0.44 From Figure 19.4, Mr = 28 × 103 (untreated gravel base) a2 = 0.13 From Figure 19.3, Mr = 15 × 103 (sandy gravel subbase) a3 = 0.11 Find layer thicknesses, starting with surface course: SN of surface course is dependent on base course Mr, therefore enter Figure 19.8 with Mr = 28 × 103 SN = 2.5 D1*= 2.5/0.44 = 5.68 in. Use 6 in. SN1*= a1D1* = 0.44(6) = 2.64 Repeating this procedure for the base course: Mr= 1.5 × 104 (Subbase) From Figure 19.8, SN = 3.2 D2* = (3.2 – 2.64) / [(0.13)(0.7)] = 6.15 in. Use 6.5 in. SN2* = (a2)(m2)(D2*) + SN1* = (0.13)(0.7)(6.5) + 2.64 = 3.23 D3* = (3.8 – 3.23) / [(0.11)(0.7)] = 7.4 in. Use 7.5 in. Check design: SN = 0.44(6) + (0.13)(0.7)(6.5) + (0.11)(0.7)(7.5) = 3.81 (which is greater than SN of 3.8), so the design is satisfactory The pavement will consist of 6 in. of asphalt concrete surface, 6.5 in. of untreated gravel base, and 7.5 in. of sandy gravel subbase. 19-16 An existing two-lane rural highway is to be replaced by a four-lane divided highway on a new alignment. Construction of the new highway will commence two years from now and is expected to take three years to complete. The design life of the pavement is 20 years. The present ESAL is 150000. Design a flexible pavement consisting of an asphalt concrete surface and lime-treated base using the California (Hveem) method. The results of a stabilometer test on the subgrade soil are as follows. Moisture Content R Value Exudation Expansion Pressure (%) Pressure (lb/in2) Thickness (ft) 19.8 55 575 1.00 22.1 45 435 0.15 24.9 16 165 0.10 Total ESAL = 150000 × (47.73 – 5.53) = 6.33 × 106 Calculate traffic index using Equation 19.8, TI = 9.0(6.33)0.119 = 11.21 Next, calculate thickness above base using Equation 19.9: GE = 0.0032(TI)(100 – R) GE = 0.0032(11.21)(100 – 80) GE = 0.717 ft. From Table 19.10, Gf for asphalt concrete with TI of 11.21 = 1.71 Actual depth of AC = 0.717/1.71 = 0.419 ft = 5 in. Next, calculate base thickness To determine GE at 300 lb/in2, Plot GE vs. Exudation pressure At 300 lb/in2, GE = 2.5 From Table 19.10, for lime-treated base, Gf = 1.2 GE of base layer = 2.5 – (5/12)(1.717) = 1.785 Base thickness = 1.785 / 1.2 = 1.4896 ft. = 17.85 in. Use 18 in. Check design: Plot GE vs. Moisture content to determine moisture content at 300 lb/in2 From plot above, right, Moisture content = 23.5% 19-17 Briefly describe the main steps in the Mechanistic-Empirical Pavement Design method as given in the MEPDG. The method of pavement design applied in the MEPDG consists of five basic steps. The first step is to select a trial design based on another method (such as the 1993 AASHTO Pavement Design Guide, agency policies or standards, or a design based on engineering judgment and experience). The second step is selection of acceptable thresholds for performance criteria related to pavement distresses of rutting, transverse cracking, longitudinal cracking, alligator cracking, and smoothness. The third step involves the collection of all input data pertaining to project information and to traffic pavement structure, material properties, and climate. Fourth, the trial design is evaluated, typically through the MEPDG software. Finally, the trial design is revised and then evaluated until an acceptable design is achieved. 19-18 Determine the expected rut depth of the fourth layer of a 4” thick HMA flexible pavement surface divided into four equal sub-layers for the following conditions: Resilient or elastic strain calculated by the structural response model at the mid-depth of each HMA sub-layer, in/in = 45 × 10-6 Number of axle load repetitions = 1.5 × 104 Mixture field calibration constants: β1r, = 0.99 β2r, = 0.98 β3 = 0.95 Use Equation 19.10: Δ =ε h =β εk 10k1r nk2 2rβrTk3 3rβr p HMA( ) p HMA( ) HMA 1r z r HMA( ) Compute kz = (C1 + C2 D)0.0328196D C1 = –0.1039(HHMA)2 + 2.4868HHMA – 17.342 = –0.1039(4)2 + 2.4868(4) – 17.342 = –9.0572 C2 = 0.01712(HHMA)2 – 1.7331HHMA + 27.428 = 0.01712(4)2 – 1.7331(4) + 27.428 = 20.7695 kz = (C1 + C2 D)0.0328196D kz = (-9.0572 + 20.7695 x 3)0.03281963 = 0.00199 Δ =β εk 10k1r nk2 2rβrTk3 3rβr p HMA( ) 1r z r HMA( ) = 0.99 × 0.00199 × 45 × 10–6 × 10–3.35412 × (1.5)(104) × 0.4791 × 0.98 × 851.5606 × 0.95 = 2.59 × 10–6 19-19 Repeat problem 19-18 for axle load repetitions of 2 × 104. Based on your answers for this problem and problem 19-18, discuss the effect of axle load repetitions on expected rut of HMA flexible pavements. From Problem 19-18, kz = 0.00199 Δp HMA( ) =β ε1r z r HMAk ( )10k1r nk2 2rβrTk3 3rβr = 0.99 × 0.00199 × 45 × 10–6 × 10–3.35412 × (2)(104) × 0.4791 × 0.98 × 851.5606 × 0.95 = 2.96 × 10–6 Compared with Problem 19-18, an increase of 33% in axle load repetitions (from 15,000 to 20,000) resulted in an increase of 14% in rut depth.
Chapter 20 Design of Rigid Pavements 20-1 Portland cement concrete consists of what four primary elements? The four primary elements Portland cement concrete consists of are: Portland cement, coarse aggregate, fine aggregate, and water. 20-2 List and briefly describe the five main types of Portland cement as specified by AASHTO. The AASHTO specifications list five main types of Portland Cement: Type I is suitable for general concrete construction, where no special properties are required. Type II is suitable for use in general concrete construction, where the concrete will be exposed to moderate action of sulphate or where moderate heat of hydration is required. Type III is suitable for concrete construction that requires a high concrete strength in a relatively short time and is sometimes referred to as high early strength cement. Type IV is suitable for projects where low heat of hydration is necessary. Type V is used in concrete construction projects where the concrete will be exposed to high sulphate action. 20-3 What is the main requirement for the water used in Portland cement? Water used in Portland cement concrete should also be suitable for drinking. This requires that the quantity of organic matter, oil, acids, and alkalis should not be greater than the allowable amount in drinking water. 20-4 Describe the basic types of highway concrete pavements and give the conditions under which each type will be constructed. Rigid highway pavements can be divided into three general types: plain concrete pavements, simply reinforced concrete pavements, and continuously reinforced concrete pavements. Plain concrete pavements have no temperature steel or dowels for load transfer. This type of pavement is used mainly on low-volume highways or when cement stabilized soils are used as subbase material. Joints are typically spaced at intervals of 10 ft to 20 ft to control cracking. This type is sometimes referred to as jointed plain concrete pavement (JPCP). Simply reinforced concrete pavements have dowels for the transfer of traffic loads across joints, with these joints spaced at larger distances than with plain pavements, ranging from 30 ft to 100 ft. Temperature steel is used throughout the slab. Continuously reinforced concrete pavements have no transverse joints, except construction joints or expansion joints when they are necessary at specific positions, such as at bridges. This type of pavement is typically used on highvolume, high-speed roadways. In design this type is often simply referred to as CRCP. 20-5 List and briefly describe the four type of joints used in concrete pavements. There are four different types of joints placed in concrete pavements: expansion joints, contraction joints, hinge joints, and construction joints. Expansion joints are usually placed transversely, at regular intervals, to provide adequate expansion space for the slab to expand when the pavement is subjected to an increase in temperature. Contraction joints are placed transversely at regular intervals across the width of the pavement to release some of the tensile stresses that are induced due to a decrease in temperature. Hinge joints are placed longitudinally mainly to reduce cracking along the centerline of highway pavements. Construction joints are placed transversely across the pavement width to provide suitable transition between concrete placed at different times or on different days. 20-6 Define the phenomenon of pumping and its effects on rigid pavements. Pumping is the discharge of water and subgrade (or subbase) material through joints, cracks, and along the pavement edge. It is primarily caused by the repeated deflection of the pavement slab in the presence of accumulated water beneath it. Pumping action results in a relatively large void space formed underneath the concrete slab. This results in faulting of the joints and eventually the formation of transverse cracks or the breaking of the corner of the slab. Joint faulting and cracking is therefore progressive, since formation of cracks facilitates the pumping action. 20-7 List and describe the types of stresses that are developed in rigid pavements. There are typically three categories of stresses developed in rigid pavements. The first category of stress is induced by traffic wheel loads. There are three critical locations of the wheel load on the concrete pavements that induce stress; at the corner, at the interior, and at the edge of the slab. The second category of stress is due to temperature changes. One type of this stress is caused by the tendency of the slab edges to curl downward during the day and upward during the night as a result of temperature gradients (between the base material and the air) and is resisted by the weight of the slab itself. Another type of this stress is caused by expansion and contraction of the slab due to daily temperature gradients. The third category of stress is induced by bending. These stresses are caused by the yielding of the sub-base or sub-grade supporting the concrete pavement. 20-8 Determine the tensile stress imposed at night by a wheel load of 750 lb located at the edge of a concrete pavement with the following dimensions and properties. Pavement thickness = 8.5 inches μ = 0.15 Ec = 4.2 × 106 lb/in2 k = 200 lb/in3 Radius of loaded area = 3 inches Determine the stress on the edge of the slab using Equation 20.8. σe = 0.5722 P[4log10( )l + log10 b] h b First, solve for the radius of equivalent distribution of pressure, b. Since the radius of contact area, a = 3 in. < 1.724h = 1.724(8.5) = 14.654, b = (1.6a2 + h2)1/2 – 0.675h b = (1.6(3)2 + (8.5)2)1/2 – (0.675)(8.5) b = 3.57 inches Next, solve for the radius of relative stiffness, l. l = (Ech3 / (12(1 – μ2)k))1/4 l = (((4.2 × 106)(8.5)3 ) / (12(1 – (0.15)2)200))1/4 l = 32.38 Now use Equation 20.8 to determine the tensile stress imposed. σe = 0.5722 P[4log10( )l + log10 b] h b σe = ((0.572)(750) / (8.5)2) [4log10(32.38 / 3.57) + log10(3.57)] σe = 26.0 lb/in2 Therefore, the tensile stress imposed on this slab will be 26.0 lb/in2. 20-9 Repeat Problem 20-8 for the load located at the interior of the slab. Determine the stress at the interior of the slab using Equation 20.10. σi = 0.3162 P[4log10( )l +1.069] h b From Problem 20-8, b = 3.57 inches l = 32.38 Use Equation 20.10 to determine the tensile stress imposed. σi = 0.3162 P[4log10( )l +1.069] h b σi = ((0.316)(750) / (8.5)2) [4log10(32.38 / 3.57) + 1.069)] σi = 3.280 [3.830 + 1.069] σi = 16.1 lb / in2 Therefore, the tensile stress imposed on this slab will be 16.1 lb/in2. 20-10 Determine the tensile stress imposed at day by a wheel load of 800 lb located at the edge of a concrete pavement with the following dimensions and properties. Pavement thickness = 10 inches μ = 0.15 Ec = 4.6 × 106 lb/in2 k = 160 lb/in3 Radius of loaded area = 3 inches Determine the stress on the edge of the slab using Equation 20.8. σe = 0.5722 P[4log10( )l + log10 b] h b First, solve for the radius of equivalent distribution of pressure, b. Since the radius of contact area, a = 3 in. < 1.724h = 1.724(10) = 17.24, b = (1.6a2 + h2)1/2 - 0.675h b = (1.6(3)2 + (10)2)1/2 – (0.675)(10) b = 3.95 inches Next, solve for the radius of relative stiffness, l. l = (Ech3 / (12(1 – μ2)k))1/4 l = (((4.6 × 106)(10)3 ) / (12(1 – (0.15)2)160))1/4 l = 39.57 Now use Equation 20.8 to determine the tensile stress imposed. σe = 0.5722 P[4log10( )l + log10 b] h b σe = ((0.572)(800) / (10)2) [4log10(39.57 / 3.95) + log10(3.95)] σe = 21.0 lb/in2 Therefore, the tensile stress imposed on this slab will be 21.0 lb/in2. 20-11 Repeat Problem 20-10 for the load located at the interior of the slab. Determine the stress at the interior of the slab using Equation 20.10. σi = 0.3162 P[4log10( )l +1.069] h b From Problem 20-10, b = 3.95 inches l = 39.57 Use Equation 20.10 to determine the tensile stress imposed. σi = 0.3162 P[4log10( )l +1.069] h b σi = ((0.316)(800) / (10)2) [4log10(39.57 / 3.95) + 1.069)] σi = 2.528 [4.003 + 1.069] σi = 12.8 lb / in2 Therefore, the tensile stress imposed on this slab will be 12.8 lb/in2. 20-12 Determine the maximum distance of contraction joints for a plain concrete pavement if the maximum allowable tensile stress in the concrete is 50 lb/in2 and the coefficient of friction between the slab and the subgrade is 1.7. Assume uniform drop in temperature. Weight of concrete is 144 lb/ft3. Use Equation 20.18 to determine length of pavement between contraction joints. L = 288pc / fγc L = (288)(50) / (1.7)(144) L = 58.82 ft Therefore, the maximum distance between contraction joints should be 58 ft. 20-13 Repeat Problem 20-12, with the slab containing temperature steel in the form of welded wire mesh consisting of 0.125 in2 steel/ft width. The modulus of steel, Es, is 30 × 106 lb/in2, Ec = 5 × 106 lb/in2, and h = 6 inches. Use Equation 20.20 to determine length of pavement between contraction joints. L = 24pc (12h + nAs ) hγc f First, determine the modular ratio, n. n = Es / Ec n = (30 × 106) / (5 × 106) = 6 Use Equation 20.20 to determine length of pavement between contraction joints. L = 24pc (12h + nAs ) hγc f L = [(24)(50) / (6)(1.7)(144)] [(12)(6) + (6)(0.125)] L = (1200 / 1468.8) (72.75) L = 59.4 ft Therefore, the maximum distance between contraction joints should be 59 ft. 20-14 A concrete pavement is to be constructed for a 4-lane urban expressway on a subgrade with an effective modulus of subgrade reaction k of 100 lb/in3. The accumulated equivalent axle load for the design period is 3.25 × 106. The initial and terminal serviceability indices are 4.5 and 2.5, respectively. Using the AASHTO design method determine a suitable thickness of the concrete pavement, if the working stress of the concrete is 600 lb/in2 and the modulus of elasticity is 5 × 106 lb/in2. Take the overall standard deviation, So, as 0.30, the load transfer coefficient, J, as 3.2, the drainage coefficient as 0.9, and R=95%. First, determine the design serviceability loss design serviceability loss = 4.5 – 2.5 = 2.0 From the nomographs provided in Figures 20.13 and 20.14 and using the factors above, the required thickness of the concrete slab is 9.4 inches, which for design may be rounded up to 9.5 inches, the nearest half-inch for design. 20-15 A six-lane concrete roadway is being designed for a metropolitan area. This roadway will be constructed on a subgrade with an effective modulus of subgrade reaction k of 170 lb/in3. The ESALs used for the design period is 2.5 × 106. Using the AASHTO design method, determine a suitable thickness of the concrete pavement (to the nearest 1/2 inch), provided that the working stress of the concrete to be used is 650 lb/in2 and the modulus of elasticity is 5 × 106 lb/in2. Assume the initial serviceability is 4.75 and the terminal serviceability is 2.5. Assume the overall standard deviation, So, is 0.35, the load transfer coefficient J as 3.2, the drainage coefficient, Cd, is 1.15, and R = 95%. First, determine the design serviceability loss design serviceability loss = 4.75 – 2.5 = 2.25 From the nomographs provided in Figures 20.13 and 20.14 and the factors above, the required thickness of the concrete slab is 7.8 inches, which for design may be rounded up to 8 inches. 20-16 Revisit Problem 20-15 under the conditions in which a revised traffic analysis indicates that 3.5 × 106 ESALs are expected during the design life and the desired reliability has been revised to 99%. Design serviceability loss = 4.75 – 2.5 = 2.25 ESALs = 3.5 × 106 R = 99% From the nomographs provided in Figures 20.13 and 20.14 and the factors above, the required thickness of the concrete slab is 9.0 inches. 20-17 Determine the required slab thickness, according to the AASHTO Alternate Design Method, for a rigid pavement being designed near Jacksonville, Florida, for the following input data: Estimated future traffic load = 10,000,000 ESALs Design reliability = 95% Overall standard deviation = 0.39 Initial serviceability index = 4.5 Terminal serviceability index = 2.5 Effective subgrade modulus = 250 lb/in2 Mean concrete modulus of rupture = 700 lb/in2 Mean concrete elastic modulus = 4,200,000 lb/in2 Joint spacing = 20 ft Base modulus = 25,000 lb/in2 Base thickness = 6 inches Slab width = 12 ft supported by asphalt concrete shoulders Solution: Rigid pavement type: jointed plain concrete pavement (JPCP) according to Table 15 (Page 26), Jacksonville, FL has: mean annual wind speed = WIND = 8.1 mph, mean annual temperature = TEMP = 68°F, mean annual precipitation = PRECIP = 52.8 in assume slab thickness D = 11.5 in effective positive TD = 0.962 – 52.181/D + 0.341 WIND +0.184 TEMP – 0.00836 PRECIP = 11.25 According to Table 16 (Page 28), the required thickness of the concrete slab is 11.9 inches. 20-18 Determine whether the slab thickness obtained in Problem 20-17 is adequate for the tensile stress that will occur for joint loading. First, solve for the radius of relative stiffness, l. l = (EcD3 / (12(1 – μ2)k))1/4, assuming μ=0.20 = 39.59 Assume the JPCP is doweled, and thus, the midslab is the critical fatigue damage location. Midslab tensile stress due to loading (Equation 44 from Page 20): σl = 18000/D2 {4.227 – 2.381 (180/l )2 – 0.0015 (Eb Hb /1.4k)0.5 – 0.155[Hb(Eb/Ec)0.75]0.5 = 116.6 lb/in2 Since σl = 116.6 lb/in2 < flexural strength = 700 lb/in2, the slab thickness D = 11.9 in obtained from Problem 20-17 is adequate. 20-19 An existing rural 4-lane highway is to be replaced by a 6-lane divided expressway (3 lanes in each direction). Traffic volume data on the highway indicate that the AADT (both directions) during the first year of operation is 24,000 with the following vehicle mix and axle loads. Passenger cars = 50 percent 2-axle single-unit trucks (12,000 lb/axle) = 40 percent 3-axle single-unit trucks (16,000 lb/axle) = 10 percent The vehicle mix is expected to remain the same throughout the design life of 20 years, although traffic is expected to grow at a rate of 3.5 percent annually. Using the AASHTO design procedure, determine the minimum depth of concrete pavement required for the design period of 20 years. Pi = 4.5 J = 3.2 Pt = 2.5 Cd = 1.0 S'c = 650 lb/in2 So = 0.3 Ec = 5 × 106 lb/in2 R = 95% k = 130 lb/in3 First, determine the growth factor for 20 years at 3.5%. Growth factor = [(1 + 0.035)20 – 1] / 0.035 Growth factor = 28.28 Next, assume a pavement thickness and estimate the ESALs for the 20 year period. Assume depth of 9 inches From Table 20.7: ESAL of 12 kips = 0.176 ESAL of 16 kips = 0.604 ESAL for 2-axle single unit trucks: ESAL = 0.176(2)(0.4)(28.28)(24,000)(365)(0.40) ESAL = 13.95 × 106 ESAL for 3-axle single unit trucks: ESAL = 0.604(3)(0.1)(28.28)(24,000)(365)(0.40) ESAL = 17.96 × 106 Total ESAL = 31.91 × 106 Next, determine the design serviceability loss design serviceability loss = 4.5 – 2.5 design serviceability loss = 2.0 From the nomographs provided in Figures 20.13 and 20.14 and the factors above, the required thickness of the concrete slab is 12.3 inches, which is far from the assumed value of 9 inches. Hence, 9 inches cannot be used as the design depth. Assume depth of 12 inches From Table 20.7: ESAL of 12 kips = 0.174 ESAL of 16 kips = 0.599 ESAL for 2-axle single unit trucks: ESAL = 0.174(2)(0.4)(28.28)(24,000)(365)(0.40) ESAL = 13.79 × 106 ESAL for 3-axle single unit trucks: ESAL = 0.599(3)(0.1)(28.28)(24,000)(365)(0.40) ESAL = 17.81 × 106 Total ESAL = 31.60 × 106 From the nomographs provided in Figures 20.13 and 20.14 and the factors above, the required thickness of the concrete slab is 12.2 inches, which is close to the assumed value of 12 inches. Hence, the rounded value of 12 inches will be used as the design depth. 20-20 Repeat Problem 20-19 for a pavement containing doweled joints, 6 inch untreated subbase, and concrete shoulders, using the PCA design method. First, determine the estimated ESALs: 2-axle (12,000 lb) single unit trucks: ESAL = 2(0.4)(28.28)(24,000)(365)(0.40) ESAL = 79.27 × 106 3-axle (16,000 lb) single unit trucks ESAL = 3(0.1)(28.28)(24,000)(365)(0.40) ESAL = 29.73 × 106 Subgrade k = 130 lb/in3 For 6 inch untreated subbase, determine effective k from Table 20.10: k = 140 + [(230 – 140) / 100] × 30 = 167 lb/in3 Assume a slab 6 inches thick with doweled joints and concrete shoulders. From Table 20.12, obtain equivalent stress values and then interpolate for effective k, Equivalent stress = 304 – ((304 – 289) × (33 / 50)) = 298.9 Stress Ratio = 298.9 / 650 = 0.460 From Table 20.15, determine the erosion factor of 2.72 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions (ESALs): Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 7 × 105 2.6 × 106 The 6 inch slab is inadequate since the allowable ESALs are less than the expected ESALs. Assume a slab 7 inches thick with doweled joints and concrete shoulders. From Table 20.12, Equivalent stress = 248 – ((248 – 236) × (33 / 50)) = 243.9 Stress Ratio = 243.92 / 650 = 0.38 From Table 20.15, determine the erosion factor of 2.54 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions: Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 unlimited 100 × 106 Damage percent = 29.73 × 106 / 100 × 106 = 29.73% The 7 inch slab is adequate, check a 6.5 inch slab. Assume a slab 6.5 inches thick with doweled joints and concrete shoulders. From Table 20.12, equivalent stress = 274 – ((274 – 260) × (33 / 50)) = 269.24 Stress Ratio = 269.24 / 650 = 0.41 From Table 20.15, determine the erosion factor of 2.62 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions: Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 unlimited 10 × 106 The 6.5 inch slab is inadequate, use a 7 inch slab depth. 20-21 Repeat Problem 20-19 using the PCA design method, for a pavement containing doweled joints and concrete shoulders. The modulus of rupture of the concrete used is 600 lb/in2. First, determine the estimated ESALs: 2-axle (12,000 lb) single unit trucks: ESAL = 2(0.4)(28.28)(24,000)(365)(0.40) ESAL = 79.27 × 106 3-axle (16,000 lb) single unit trucks ESAL = 3(0.1)(28.28)(24,000)(365)(0.40) ESAL = 29.73 × 106 Subgrade k = 130 lb/in3 For 6 inch untreated subbase, determine effective k from Table 20.10: k = 140 + [(230 – 140) / 100] × 30 = 167 lb/in3 Assume a slab 7 inches thick with doweled joints and concrete shoulders. From Table 20.12, obtain equivalent stress values and then interpolate for effective k, Equivalent stress = 248 – ((248 – 236) × (33 / 50)) = 243.9 Stress Ratio = 243.9 / 600 = 0.41 From Table 20.15, determine the erosion factor of 2.54 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Now, calculate the allowable repetitions (ESALs): Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 7 × 105 100 × 106 Damage percent = 29.73 × 106 / 100 × 106 = 29.73% The 7 inch slab is adequate, check a 6.5 inch slab. Assume a slab 6.5 inches thick with doweled joints and concrete shoulders.From Table 20.12, Equivalent stress = 274 – ((274 – 260) × (33 / 50)) = 269.24 Stress Ratio = 269.24 / 600 = 0.45 From Table 20.15, determine the erosion factor of 2.62 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Now, calculate the allowable repetitions: Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 unlimited 8 × 106 The 6.5 inch slab is inadequate since the allowable ESALs are less than the expected ESALs. The 6.5 inch slab is inadequate, use a 7 inch slab depth. 20-22 Repeat Problem 20-19, using the PCA design method, if the subgrade k value is 50 and a 6-inch stabilized subbase is used. The modulus of rupture of the concrete is 600 lb/in2 and the pavement has aggregate interlock joints (no dowels) and a concrete shoulder. First, determine the estimated ESALs: 2-axle (12,000 lb) single unit trucks: ESAL = 2(0.4)(28.28)(24,000)(365)(0.40) ESAL = 79.27 × 106 3-axle (16,000 lb) single unit trucks ESAL = 3(0.1)(28.28)(24,000)(365)(0.40) ESAL = 29.73 × 106 Subgrade k = 50 lb/in3 For 6 inch stabilized subbase, determine effective k from Table 21.10: k = 230 lb/in3 Assume a slab 6.5 inches thick with aggregate interlock joints and concrete shoulders. From Table 20.12, obtain equivalent stress values and then interpolate for effective k, Equivalent stress = 260 – ((260 – 243) × (30 / 100)) = 254.9 Stress Ratio = 254.9 / 600 = 0.42 From Table 20.16, determine the erosion factor of 2.80 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions (ESALs): Load Fatigue analysis Erosion analysis 14,400 unlimited 60 × 106 19,200 unlimited 1.2 × 106 The 6.5 inch slab is adequate under fatigue analysis but not for the erosion criterion. Assume a slab 8 inches thick with aggregate interlock joints and concrete shoulders. From Table 20.16, determine the erosion factor of 2.56 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions: Load Erosion analysis 14,400 unlimited 19,200 4 × 106 The 8 inch slab is adequate, check a 7.5 inch slab. Assume a slab 7.5 inches thick with aggregate interlock joints and concrete shoulders. From Table 20.16, determine the erosion factor of 2.56 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions: Load Erosion analysis 14,400 unlimited 19,200 6 × 106 The 7.5 inch slab is inadequate, use a 8 inch slab depth. 20-23 Repeat Problem 20-22, using the PCA design method, assuming the pavement has doweled joints and no concrete shoulders. First, determine the estimated ESALs: 2-axle (12,000 lb) single unit trucks: ESAL = 2(0.4)(28.28)(24,000)(365)(0.40) ESAL = 79.27 × 106 3-axle (16,000 lb) single unit trucks ESAL = 3(0.1)(28.28)(24,000)(365)(0.40) ESAL = 29.73 × 106 Subgrade k = 50 lb/in3 For 6 inch stabilized subbase, determine effective k from Table 20.10: k = 230 lb/in3 Assume a slab 8 inches thick with doweled joints and no concrete shoulders. From Table 20.11, obtain equivalent stress values and interpolate for k, Equivalent stress = 242 – ((242 – 225) × (30 / 100)) = 236.9 Stress Ratio = 236.9 / 600 = 0.39 From Table 20.13, determine the erosion factor of 2.80 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions (ESALs): Load Fatigue analysis Erosion analysis 14,400 unlimited unlimited 19,200 unlimited 20 × 106 The 8 inch slab is adequate under fatigue analysis but not for the erosion criterion. Assume a slab 8.5 inches thick with aggregate interlock joints and concrete shoulders. From Table 20.13, determine the erosion factor of 2.72 12,000(1.2) = 14,400 16,000(1.2) = 19,200 Calculate the allowable repetitions (ESALs): Load Erosion analysis 14,400 unlimited 19,200 40 × 106 Damage percent = 29.73 × 106 / 40 × 106 = 74.3% Use a 8.5 inch slab depth. 20-24 Briefly describe the steps involved in the MEPDG method for JPCP. The method of pavement design applied in the MEPDG consists of five basic steps. The first step is to select a trial design based on another method (such as the 1993 AASHTO Pavement Design Guide, agency policies or standards, or a design based on engineering judgment and experience. The second step is selection of acceptable thresholds for performance criteria related to pavement distresses of mean transverse joint faulting, transverse slab cracking (bottom-up and top-down), and smoothness. The third step involves the collection of all input data pertaining to project information and to traffic, pavement structure, material properties, and climate. Fourth, the trial design is evaluated, typically through the MEPDG software, by examining the performance indicators. Finally, the structural viability of the trial design is evaluated and the design revised and evaluated until an acceptable design is achieved. 20-25 What are the main input parameters used in the MEPDG for JPCP? The input parameters for design of jointed plain concrete pavement are similar to those used in flexible pavement design. These include traffic, foundation and subgrade soils and material characteristics. Traffic is characterized by the distribution of axle weights, or loads, also known as load spectra. This differs from traditional empirical approaches that are based on converting traffic loads to equivalent single axle loads (ESALs). 20-26 List and define each of the main criteria that are used for evaluating a trial JPCP structure in the MEPDG method. The performance criteria used to evaluate a JPCP design are mean transverse joint faulting, transverse slab cracking (bottom-up and top-down), and smoothness. Threshold values deemed acceptable for each criterion are selected initially. The values that result from analysis of a trial design using the MEPDG are then compared to the acceptable thresholds and the design evaluated accordingly. Transverse joint faulting refers to the extent to which transverse joint have faulted such that continuity between slabs is lost. This is evaluated on a monthly, incremental basis for the pavement design life. Transverse slab cracking is evaluated as the percentage of slab expected to have transverse cracks at the end of the design life of the proposed pavement structure. Smoothness pertains to the profile of the pavement; the greater the deviation in the profile from its design, the less smooth the pavement is said to be. 20-27 Determine the expected predicted transverse cracking for the five traffic levels and associated applied stresses given on Table 20.19 for a given condition of age, season, and temperature difference if the Modulus of Rupture of the PCC is 650 lb/in2. Table 20.19 Data for Problem 20-27
Load Level Number of Applications (n) Applied Stress (σi,j,k,l,m,n) (lb/in2) Axle
(1) (2) (3)
1 0.02 × 106 300
2 0.05 × 106 280
3 0.08 × 106 275
4 0.10 × 106 270
5 0.12 × 106 260
Solution:
Load Level (1) Number of Applications (n) (2) Applied Stress (σi,j,k,l,m,n) (lb/in2) Axle (3) Allowable Number of Load Applications (N) n/N
1 0.02 × 106 300 0.137 × 106 0.15
2 0.05 × 106 280 0.387 × 106 0.13
3 0.08 × 106 275 0.515 × 106 0.16
4 0.10 × 106 270 0.694 × 106 0.14
5 0.12 × 106 260 1.308 × 106 0.09
Sum 0.67
Determine the allowable number of load applications for each axle level load using Eq.20.24. See column 4 in the Table. Determine the ratio of number of load applications to allowable number of load applications for each axle load level. See column 5 in the Table. Determine total fatigue damage from Equation 20.23 DIF = 0.13 + 0.16 + 0.14 + 0.09 = 0.67 Determine predicted amount of transverse cracking using Equation 20.22 CRK = = = 0.312
Chapter 21 Pavement Management 21-1 What is meant by the term pavement management? Describe three strategies used by public agencies to develop restoration and rehabilitation programs. Pavement management refers to the various strategies that can be used to select a program for pavement restoration. Three strategies used by public agencies to develop programs include: a “squeaky wheel” approach in which projects attracting the greatest attention are selected; a program in which all roads are repaired on a regular schedule regardless of relative project costs; and a program in which minimum standards are set and a program of priorities established within budget limitations. 21-2 What are the three principal uses for pavement condition data? The three principal uses for pavement condition data are: to establish project priorities; establish rehabilitation options; and forecast performance. 21-3 What is the difference between PSI and PSR? PSI (present serviceability index) is a value for pavement condition determined as a surrogate for PSR (present serviceability rating) and is based on physical measurements. PSR is a number grade given to a pavement section based on its ability to serve traffic. It is established by observation and requires judgment by trained personnel. Therefore, PSR is subjective, while PSI is objective. 21-4 Draw a sketch showing the relationship between pavement condition (expressed as PSI) and time for a service life of 20 years, if the PSI values range from 4.5 to 2.5 in a six year period, and then the pavement is resurfaced such that the PSI is increased to 4.0. After another 6 years, the PSI has reached 2.0. With rehabilitation, PSI is increased to 4.5. At the end of its service life the PSI value is 3.4. . 21-5 Describe the four characteristics of pavement condition used to evaluate whether a pavement should be rehabilitated, and if so, determine the appropriate treatment. The four characteristics of pavement condition used to evaluate whether a pavement should be rehabilitated are: Pavement Roughness is a measurement of the extent to which a road surface deviates from a plane. Pavement Distress refers to the condition of a pavement in terms of its general appearance. Distress is a combination of fractured, distorted, or disintegrated pavement. Pavement Deflection refers to the structural adequacy of a pavement section. Tests may be dynamic, static, or destructive. Skid Resistance describes the effectiveness of a pavement to prevent or reduce skidrelated crashes. 21-6 A given pavement rating method uses six distress types to establish the DR. These are: corrugation, alligator cracking, raveling, longitudinal cracking, rutting, and patching. For a section of highway, the number of points assigned to each category were 5, 3, 3, 3, 4, and 3. If the weighting factors are 1, 1, 1, 1, 2, and 1.5, respectively, determine the DR for the section. Distress rating (DR) can be calculated using Equation 21.1, DR = 100 -  (di × wi) DR = 100 – [(5)(1) + (3)(1) + (3)(1) + (3)(1) + (4)(2) + (3)(1.5)] DR = 100 – 26.5 DR = 73.5 Therefore, the DR for this pavement section is 73.5. 21-7 What are the problems with distress surveys, and how can such problems be solved? One of the major problems with distress surveys is the variability in results due to the subjective procedures used, and the concerns about the safety of pavement surveyors and the hazards associated with their being on the road. The best solution lies in the use of automated techniques for evaluating distress. This would remove the survey crew from the hazardous road conditions, and allow for more consistent results. Other sources of error are variations in the condition of the highway segment observed, changes in evaluation procedure, and changes in observed location from year to year. Some of these variations can be minimized by using the same pavement location each year by observing pavement condition at regular intervals of about one mile. 21-8 How are computers used in pavement management? Computer vision is an area of artificial intelligence (AI) research that involves the use of sensors and computers to emulate human vision. As applied to pavement management, the objective is to develop an automated approach capable of identifying the type, severity, and extent of pavement distresses. The process generally involves the following four basic steps: image acquisition, image digitization, image processing, and pattern recognition. For pavement management, the image is usually captured using a video camera. Digitization then creates a numerical representation of the image (digitized image) suitable for computer processing in the form of an array of numbers where the integer value of each element in the array represents the color or the gray tone of the corresponding area in the original image. Values range from 0, representing black, to 255, representing white. The cracked region will have low values due to crack shadows being much darker than the surrounding background. The image processing step seeks to extract the different distresses from the background. The image is first transformed into what is known as a “binary image”, where the distresses are indicated by black pixels and the background by white. Algorithms are also applied to clean up the image and remove the “noise.” Counting the number of black pixels will thus yield an idea of the distressed area. Finally, the pattern recognition step classifies the distress in a pavement surface image as a particular type of distress. 21-9 Describe the methods used to determine static and dynamic deflection of pavements. To what extent are these tests used in pavement rehabilitation management? The methods used to determine the static and dynamic deflection of pavements are measurement of static deflections (Benkelman Beam), and measurement of dynamic or repeated loads (Dynaflect). These methods are rarely used to evaluate pavement condition, but deflection data are used for design purposes and to develop strategies. 21-10 A 4000-lb load is placed on two tires, which are then locked in place. A force of 2100 lb is necessary to cause the trailer to move at a speed of 20 mi/h. Determine the value of the skid number. If treaded tires were used, characterize the pavement type. First, calculate the skid number using Equations 21.2 and 21.3, SK = 100 × (L / N) SK = 100 × (2100 / 4000) = 52.5 Therefore, the skid number would be 52.5. Based on the skid number determined above and Figure 21.13, the pavement can be characterized as having a surface type (5), coarse-textured and gritty. 21-11 The PMS database has accumulated information regarding the performance of a pavement section before being overlaid as well as the performance of the overlay. The following data were recorded. Develop a linear prediction model (PCR=a + b × Age) for the two cases. If a criterion was established that maintenance or rehabilitation should occur when the PCR reaches a value of 81, when should such actions take place in the two cases? Before Overlay Overlay Performance PCR 100 0.5 0.5 98 1.5 96 1.5 94 3.5 92 3.5 91 4.5 86 5.5 89 5.5 83 6.5 86 6.5 80 7.5 83 8.0 Using regression analysis, the prediction model for the pavement section before being overlaid is: PCR = 101.47 – 2.31(AGE) (with an R2 = 0.997) Using regression analysis, the prediction model for the overlay is: PCR = 100.96 – 2.72(AGE) (with an R2 = 0.975) Therefore, the original pavement section will reach a PCR value of 81 at age: 81 = 101.47 – 2.31(AGE) AGE = (101.47 – 81) / 2.31 AGE = 8.9 years The overlay will reach a PCR value of 81 after: 81 = 100.96 – 2.72(AGE) AGE = (100.96 – 81) / 2.72 AGE = 7.3 years The first action should take place after 8.9 years for the original section, and after 7.3 years for the overlay. 21-12 An agency uses the following formula to determine pavement condition rating: PCR = 95.6 – 5.51 (5.0-ROUGH) – 1.59 LNALL – 0.221 AVGOUT – 0.0306 LONG – 0.531 TRAN where: ROUGH = roughness measured by present serviceability rating (PSR), scale of 0 to 5 LNALL = natural logarithm of alligator cracking, ft per mile AVGOUT = outer wheel path rutting (all locations averaged), 0.01 inch LONG = longitudinal cracking, ft per mile TRAN = transverse cracking, number of cracks per mile And, present serviceability rating is determined as: PSR5e.0051118IRI 0.0016027 A pavement section has the following distress data: IRI = 105 in/mi Alligator cracking = 110 ft/mi Average rutting = 0.20 inch Longitudinal cracks = 130 ft/mi Transverse cracks = 12/mi Determine the current PCR for this section. Then, assume a new roughness measurement is expected soon. If the PCR falls below 60, the section will be scheduled for rehabilitation. Determine the new IRI value that will trigger the rehabilitation. First, determine PSR, IRI = 105 in/mi PSR = 2.92 Using the PCR equation as shown above, PCR = 95.6 – 5.51(5.0 – 2.92) – 1.59Ln(110) – 0.221(20) – 0.0306(130) – 0.531(12) = 95.6 – 11.46 – 7.47 – 4.42 – 3.98 – 6.37 = 61.9 Therefore, the current PCR for this section is 61.9. Assume other distress values besides IRI remain unchanged, PCR = 95.6 – 5.51 (5.0 – PSR) – 7.47 – 4.42 – 3.98 – 6.37 60.0 = 95.6 – 5.51 (5.0 – PSR) – 7.47 – 4.42 – 3.98 – 6.37 PSR = 2.575 PSR5e.0051118IRI 0.0016027 IRI = 129.5 Therefore, the IRI of 129.5 in/mi will trigger the rehabilitation. 21-13 Referring to the Markovian transition matrix in Table 21.6, what is the probability that a section with a PCR value of 77 in the current year will have a PCR value between 60 and 69 (a) one year later, and (b) two years later. According to the transition matrix, the section is currently in state 7. Therefore the probability of dropping to state 6 (PCR between 60 and 69) in the next year is 0.30. To predict the condition two years ahead, one will have to apply the matrix twice, and consider all possibilities as follows: A section currently in state 7 can in one year: remain in state 7 with a probability of 0.60 drop to state 6 with the probability of 0.30 drop to state 5 with a probability of 0.10 For the second year: A section in state 7 can drop to state 6 with a probability of 0.30 A section in state 6 can remain in state 6 with a probability of 0.50 A section in state 5 can move to state 6 with a probability of 0.00 Therefore, the probability of that section currently in state 7 of being in state 6 two years later can be calculated as: Probability = (0.60)(0.30) + (0.30)(0.50) Probability = 0.33 The probability of the given section being in state 6 in two years is 0.33. 21-14 Differentiate between corrective and preventive rehabilitation techniques. Cite three examples of surface treatments in each category. What is the best preventive maintenance technique for subsurface maintenance? Corrective rehabilitation involves the permanent or temporary repair of deficiencies on an as-needed basis. Techniques include patching, crack filling, joint sealing, and seal coat (with aggregate). Preventive rehabilitation involves surface applications of either structural or non-structural improvements intended to keep the quality of the pavement above a predetermined level. Techniques include fog seal asphalt rejuvenators, joint sealing, and seal coat (with aggregate). The best preventative technique for subsurface maintenance is proper and functional drainage. 21-15 Describe the techniques used to repair flexible and rigid pavements and their effectiveness for the following treatment types: (a) patching, (b) crack maintenance, and (c) overlays. Patching Techniques: Temporary, permanent, spot seal (spray), cold mix, hot mix, and level. Effectiveness: When properly done, patching is considered to be effective. Temporary patching is considered moderately effective to serve for a short period until permanent repairs can be made. Crack Maintenance Techniques: Crack cleaning, sealing and filling Effectiveness: Relatively effective. Fairly short life. Must be repeated often (1- 2 years) (c) Overlays Techniques: Thick or thin. Pavement reinforcing, fabric overlays, inverted overlays Effectiveness: Considered to be an effective technique. 21-16 What is the basic difference between an expert system and a conventional computer program? The basic difference lies in the fact that an expert system utilizes knowledge or experience, and therefore they can address problems that can only be solved through heuristics or subjective knowledge. Conventional computer programs, on the other hand, are basically quantitative and computational in nature, and address problems where a defined answer exists. 21-17 Discuss the differences between condition and priority assessment models used in developing pavement improvement programs. Condition assessment is a method for developing single-year programs. The pavement condition is compared against an established criterion or “trigger point” to decide whether a deficiency or need exists. Consequently, all sections are separated into two groups: “now needs” and “later needs”. The condition assessment model is only concerned with the “now needs”. Having decided upon which sections need an action, the next step is to select appropriate treatments. This could involve the use of simple economic analysis methods. If the needs exceed the budget, a ranking of the potential projects is performed to determine which of the needs could be deferred to the next year. Priority assessment, on the other hand, is a method for developing multiyear programs, and therefore, may be considered an extension of the condition assessment method. In this method, performance prediction models are used to predict when each road section will reach its trigger point. That is, instead of separating the network into two groups as with the condition assessment method, this method separates the pavement sections into a number of groups according to the year where action will be needed. The process from this point is essentially the same as the condition assessment method. 21-18 Describe six methods that can be used to select a program of pavement rehabilitation. The six methods that can be used to select a program of pavement rehabilitation are: 1) Cost effectiveness, 2) Economic cost analysis (Present Worth), 3) Allocation process, 4) Sufficiency ratings, 5) Visual inspections, and 6) Management plans. The method selected will depend on data availability and viewpoints of transportation agencies concerning the relevant factors for selecting projects. Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157