Design of Rigid Pavements 2.pdf

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Design ProcedureThe objective of the design is to determine the thickness of the concrete pavementthat is adequate to carry the projected design ESAL. The basic equation developed inthe 1986 AASHTO design guide for the pavement thickness is given as

(20.21)

where

ZR standard normal variant corresponding to the selected level of relia-bility

So overall standard deviation (see Chapter 19)

W18 predicted number of 18 kip ESAL applications that can be carried bythe pavement structure after construction

D thickness of concrete pavement to the nearest half-inchPSI design serviceability loss pi pt

pi initial serviceability indexpt terminal serviceability indexEc elastic modulus of the concrete to be used in construction (lb/in

2)Sc modulus of rupture of the concrete to be used in construction (lb/in

2)J load transfer coefficient 3.2 (assumed)

Cd drainage coefficient

14.22 0.32Pt 2log10e ScCd215.63J

a D.75 1.132D

.75318.42> 1Ec>k 2.25 4b f

log10W18 ZRSo 7.35 log101D 1 2 0.06 log10 3PSI> 14.5 1.52 41 3 11.624 107 2> 1D 1 28.46 4

1104 Part 5 Materials and Pavements

Table 20.8 ESAL Factors for Rigid Pavements, Tandem Axles, and ptof 2.5

Slab Thickness, D (in.)

Axle

Load (kip) 6 7 8 9 10 11 12 13 14

2 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .0001 .00014 .0006 .0006 .0005 .0005 .0005 .0005 .0005 .0005 .0005

6 .002 .002 .002 .002 .002 .002 .002 .002 .0028 .007 .006 .006 .005 .005 .005 .005 .005 .005

10 .015 .014 .013 .013 .012 .012 .012 .012 .01212 .031 .028 .026 .026 .025 .025 .025 .025 .02514 .057 .052 .049 .048 .047 .047 .047 .047 .04716 .097 .089 .084 .082 .081 .081 .080 .080 .08018 .155 .143 .136 .133 .132 .131 .131 .131 .13120 .234 .220 .211 .206 .204 .203 .203 .203 .20322 .340 .325 .313 .308 .305 .304 .303 .303 .30324 .475 .462 .450 .444 .441 .440 .439 .439 .439

(Continued)


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Table 20.8 ESAL Factors for Rigid Pavements, Tandem Axles, and ptof 2.5 (continued)

Slab Thickness, D (in.)

Axle

Load (kip) 6 7 8 9 10 11 12 13 14

26 .644 .637 .627 .622 .620 .619 .618 .618 .61828 .855 .854 .852 .850 .850 .850 .849 .849 .84930 1.11 1.12 1.13 1.14 1.14 1.14 1.14 1.14 1.1432 1.43 1.44 1.47 1.49 1.50 1.51 1.51 1.51 1.5134 1.82 1.82 1.87 1.92 1.95 1.96 1.97 1.97 1.9736 2.29 2.27 2.35 2.43 2.48 2.51 2.52 2.52 2.5338 2.85 2.80 2.91 3.03 3.12 3.16 3.18 3.20 3.2040 3.52 3.42 3.55 3.74 3.87 3.94 3.98 4.00 4.0142 4.32 4.16 4.30 4.55 4.74 4.86 4.91 4.95 4.9644 5.26 5.01 5.16 5.48 5.75 5.92 6.01 6.06 6.09

46 6.36 6.01 6.14 6.53 6.90 7.14 7.28 7.36 7.4048 7.64 7.16 7.27 7.73 8.21 8.55 8.75 8.86 8.9250 9.11 8.50 8.55 9.07 9.68 10.14 10.42 10.58 10.6652 10.8 10.0 10.0 10.6 11.3 11.9 12.3 12.5 12.754 12.8 11.8 11.7 12.3 13.2 13.9 14.5 14.8 14.956 15.0 13.8 13.6 14.2 15.2 16.2 16.8 17.3 17.558 17.5 16.0 15.7 16.3 17.5 18.6 19.5 20.1 20.460 20.3 18.5 18.1 18.7 20.0 21.4 22.5 23.2 23.662 23.5 21.4 20.8 21.4 22.8 24.4 25.7 26.7 27.364 27.0 24.6 23.8 24.4 25.8 27.7 29.3 30.5 31.366 31.0 28.1 27.1 27.6 29.2 31.3 33.2 34.7 35.768 35.4 32.1 30.9 31.3 32.9 35.2 37.5 39.3 40.570 40.3 36.5 35.0 35.3 37.0 39.5 42.1 44.3 45.972 45.7 41.4 39.6 39.8 41.5 44.2 47.2 49.8 51.774 51.7 46.7 44.6 44.7 46.4 49.3 52.7 55.7 58.076 58.3 52.6 50.2 50.1 51.8 54.9 58.6 62.1 64.878 65.5 59.1 56.3 56.1 57.7 60.9 65.0 69.0 72.380 73.4 66.2 62.9 62.5 64.2 67.5 71.9 76.4 80.282 82.0 73.9 70.2 69.6 71.2 74.7 79.4 84.4 88.884 91.4 82.4 78.1 77.3 78.9 82.4 87.4 93.0 98.186 102.0 92.0 87.0 86.0 87.0 91.0 96.0 102.0 108.088 113.0 102.0 96.0 95.0 96.0 100.0 105.0 112.0 119.090 125.0 112.0 106.0 105.0 106.0 110.0 115.0 123.0 130.0

SOURCE: Adapted from AASHTO Guide for Design of Pavement Structures, American Association of State Highway andTransportation Officials, Washington, D.C., 1993. Used with permission.

Chapter 20 Design of Rigid Pavements 1105

Equation 20.21 can be solved for the thickness of the pavement (D) in inchesby using either a computer program or the two charts in Figures 20.13 and 20.14(page 1108). The use of a computer program facilitates the iteration necessary, sinceD has to be assumed to determine the effective modulus of subgrade reaction and theESAL factors used in the design.


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Example 20.3 Designing a Rigid Pavement Using the AASHTO Method

The use of the charts is demonstrated with the example given in Figure 20.13. In thiscase, input values for Segment 1 of the chart (Figure 20.13) are

Effective modulus of subgrade reaction, k 72 lb/in3

Mean concrete modulus of rupture, Sc 650 lb/in2

Load transfer coefficient,J 3.2

Drainage coefficient, Cd 1.0


Table 20.9 Recommended Values for Drainage Coefficient, Cd, for Rigid Pavements

Percent of Time Pavement Structure is Exposed

to Moisture Levels Approaching Saturation

Quality of Greater Than

Drainage Less Than 1% 15% 525% 25%

Excellent 1.21.20 1.20 1.15 1.151.10 1.10Good 1.20 1.15 1.151.10 1.10 1.00 1.00Fair 1.151.10 1.10 1.00 1.00 0.90 0.90Poor 1.10 1.00 1.00 0.90 0.90 0.80 0.80Very poor 1.00 0.90 0.90 0.80 0.80 0.70 0.70

SOURCE: Adapted from AASHTO Guide for Design of Pavement Structures, American Association of State Highway andTransportation Officials, Washington, D.C., 1993. Used with permission.

Figure 20.13 Design Chart for Rigid Pavements Based on Using Values for Each InputVariable (Segment 1)

SOURCE: Redrawn fromAASHTO Guide for Design of Pavement Structures, American Association of StateHighway and Transportation Officials, Washington, D.C., 1993. Used with permission.


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Chapter 20 Design of Rigid Pavements 1107

These values are used to determine a value on the match line as shown inFigure 20.13 (solid line ABCDEF). Input parameters for Segment 2 (Figure 20.14)of the chart are

Match line value determined in segment 1 (74)

Design serviceability loss, PSI 4.5 2.5 2.0

Reliability, R% 95% (ZR 1.645)

Overall standard deviation, So 0.29

Cumulative 18 kip ESAL (5 106)

Solution: The required thickness of the concrete slab is then obtained, as shown inFigure 20.14, as 10 in. (nearest half-inch).

Note that when the thickness obtained from solving Eq. 20.21 analytically or byuse of Figures 20.13 and 20.14 is significantly different from that originally assumedto determine the effective subgrade modulus and to select the ESAL factors, the

whole procedure has to be repeated until the assumed and designed values areapproximately the same, emphasizing the importance of using a computer programto facilitate the necessary iteration.

Example 20.4 Evaluating the Adequacy of a Rigid Pavement Using the AASHTO Method

Using the data and effective subgrade modulus obtained in Example 20.2, determine

whether the 9 in. pavement design of Example 20.2 will be adequate on a ruralexpressway for a 20-year analysis period and the following design criteria

Pi 4.5

Pt 2.5

ESAL on design lane during first year of operation 0.2 106

Traffic growth rate 4%

Concrete elastic modulus, Ec 5 106 lb/in2

Mean concrete modulus of rupture 700 lb/in2

Drainage conditions are such that Cd 1.0

R 0.95 (ZR 1.645)

So 0.30 (for rigid pavements So 0.3 0.4)

Growth factor 29.78 (from Table 19.6)

k 170 (from Example 20.2)

Assume D 9 in. (from Example 20.2)

ESAL over design period 0.2 106 29.78 6 106

Solution: The depth of concrete required is obtained from Figures 20.13 and20.14. The dashed lines represent the solution, and a depth of 9 in. is obtained. Thepavement is therefore adequate.


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Figure 20.14 Design Chart for Rigid Pavements Based on Using Mean Values for Each Input Variable(Segment 2)

SOURCE: Redrawn fromAASHTO Guide for Design of Pavement Structures, American Association of State Highway andTransportation Officials, Washington, D.C., 1993. Used with permission.

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