The objective of Eq. 4.1 and the nomograph in Fig. 4.5 is to determine a required structural number for given axle loadings, reliability, overall standard deviation, change in PSI, and soil resilient modulus. As previously mentioned, there are many pavement material combinations and thicknesses that will provide satisfactory pavement service life. The following equation can be used to relate individual material types and thicknesses to the structural number:
3 3 3 2 2 2 1
SN a1D a D M a D M (4.3)
where
a1, a2, a3 = structural-layer coefficients of the wearing surface, base, and subbase layers, respectively,
D1, D2, D3 = thickness of the wearing surface, base, and subbase layers in inches, respectively, and
M2, M3 = drainage coefficients for the base and subbase, respectively.
Values for the structural-layer coefficients for various types of material are presented in Table 4.5. Drainage coefficients are used to modify the thickness of the lower pavement layers (base and subbase) to take into account a material’s drainage characteristics. A value of 1.0 for a drainage coefficient represents a material with good drainage characteristics (a sandy material). A soil such as clay does not drain very well and, consequently, will have a lower drainage coefficient (less than 1.0) than a sandy material. The reader is referred to [AASHTO 1993] for further information on drainage coefficients.
Because there are many combinations of structural-layer coefficients and thicknesses that solve Eq. 4.3, some guidelines are used to narrow the number of solutions. Experience has shown that wearing layers are typically 2 to 4 inches thick, whereas subbases and bases range from 4 to 10 inches thick. Knowing which of the materials is the most costly per inch of depth will assist in the determination of an initial layer thickness.
4.4 Traditional AASHTO Flexible-Pavement Design Procedure
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Table 4.5 Structural-Layer Coefficients
Pavement component Coefficient Wearing surface
Sand-mix asphaltic concrete 0.35 Hot-mix asphaltic (HMA) concrete 0.44 Base
Crushed stone 0.14
Dense-graded crushed stone 0.18
Soil cement 0.20
Emulsion/aggregate-bituminous 0.30 Portland cement/aggregate 0.40
Lime-pozzolan/aggregate 0.40
Hot-mix asphaltic (HMA) concrete 0.40 Subbase
Crushed stone 0.11
EXAMPLE 4.1 FLEXIBLE PAVEMENT DESIGNʊSTRUCTURAL NUMBER DETERMINATION
A pavement is to be designed to last 10 years. The initial PSI is 4.2 and the TSI (the final PSI) is determined to be 2.5. The subgrade has a soil resilient modulus of 15,000 lb/in2. Reliability is 95% with an overall standard deviation of 0.4. For design, the daily car, pickup truck, and light van traffic is 30,000, and the daily truck traffic consists of 1000 passes of single-unit trucks with two single axles and 350 passes of tractor semi-trailer trucks with single, tandem, and triple axles. The axle weights are
cars, pickups, light vans = two 2000-lb single axles single-unit truck = 8000-lb steering, single axle
= 22,000-lb drive, single axle tractor semi-trailer truck = 10,000-lb steering, single axle
= 16,000-lb drive, tandem axle
= 44,000-lb trailer, triple axle
M2 and M3 are equal to 1.0 for the materials in the pavement structure. Four inches of hot-mix asphalt (HMA) is to be used as the wearing surface and 10 inches of crushed stone as the subbase. Determine the thickness required for the base if soil cement is the material to be used.
SOLUTION
Because the axle-load equivalency factors presented in Tables 4.1, 4.2, and 4.3 are a function of the structural number (SN), we have to assume an SN to start the problem (later we will arrive at a structural number and check to make sure that it is consistent with our assumed value). A typical assumption is to let SN = 4. Given this, the 18-kip–equivalent single-axle load for cars, pickups, and light vans is
2-kip single-axle equivalent = 0.0002 (Table 4.1)
This gives an 18-kip ESAL total of 0.0004 for each vehicle. For single-unit trucks, 8-kip single-axle equivalent = 0.041 (Table 4.1)
22-kip single-axle equivalent = 2.090 (Table 4.1)
This gives an 18-kip ESAL total of 2.131 for single-unit trucks. For tractor semi-trailer trucks,
10-kip single-axle equivalent = 0.102 (Table 4.1) 16-kip tandem-axle equivalent = 0.057 (Table 4.2) 44-kip triple-axle equivalent = 0.769 (Table 4.3)
This gives an 18-kip ESAL total of 0.928 for tractor semi-trailer trucks. Note the comparatively small effect of cars and other light vehicles in terms of the 18-kip ESAL.
This small effect underscores the nonlinear relationship between axle loads and pavement damage. For example, from Table 4.2 with SN = 4, a 36-kip single-axle load has 14.4 times the impact on pavement as an 18-kip single-axle load (twice the weight has 14.4 times the impact).
Given the computed 18-kip ESAL, the daily traffic on this highway produces an 18-kip ESAL total of 2467.8 (0.0004 u30,000 + 2.131 u 1000 + 0.928 u 350). Traffic (total axle accumulations) over the 10-year design period will be
2467.8 u 365 u 10 = 9,007,470 18-kip ESAL
With an initial PSI of 4.2 and a TSI of 2.5, ¨PSI = 1.7. Solving Eq. 4.3 for SN (using an equation solver on a calculator or computer) with ZR = 1.645 (which corresponds to R = 95%, as shown in Table 4.4) gives SN = 3.94 (Fig. 4.5 can also be used to arrive at an approximate solution for SN). Note that this is very close to the value that was assumed (SN = 4.0) to get the load equivalency factors from Tables 4.1, 4.2, and 4.3. If Eq. 4.1 gave SN = 5, we would go back and recompute total axle accumulations using the SN of 5 to read the axle-load equivalency factors in Tables 4.1, 4.2, and 4.3. Usually one iteration of this type is all that is needed. Later, Examples 4.3 and 4.5 will demonstrate this type of iteration.
Given that SN = 3.94, Eq. 4.3 can be applied with a1 = 0.44 (surface course, hot-mix asphalt, Table 4.5), a2 = 0.20 (base course, soil cement, Table 4.5), and a3 = 0.11 (subbase, crushed stone, Table 4.5), M2 = 1.0 (given), M3 = 1.0 (given), D1 = 4.0 inches (given), and D3 = 10.0 inches (given). We have
1 1 2 2 2 3 3 3
SN a D a D M a D M
3.94 = 0.44(4) + 0.20D2(1.0) + 0.11(10.0)(1.0)
Solving for D2 gives D2 = 5.4 inches. Using D2 = 5.5 inches would be a conservative estimate and allow for variations in construction. Rounding up to the nearest 0.5 inch is a safe practice.
4.4 Traditional AASHTO Flexible-Pavement Design Procedure
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EXAMPLE 4.2 FLEXIBLE PAVEMENT DESIGNʊRELIABILITY ASSESSMENTA flexible pavement is constructed with 4 inches of hot-mix asphalt (HMA) wearing surface, 8 inches of emulsion/aggregate-bituminous base, and 8 inches of crushed stone subbase. The subgrade has a soil resilient modulus of 10,000 lb/in2, and M2 and M3 are equal to 1.0 for the materials in the pavement structure. The overall standard deviation is 0.5, the initial PSI is 4.5, and the TSI is 2.5. The daily traffic has 1080 20-kip single axles, 400 24-kip single axles, and 680 40-kip tandem axles. How many years would you estimate this pavement would last (how long before its PSI drops below a TSI of 2.5) if you wanted to be 90% confident that your estimate was not too high, and if you wanted to be 99%
confident that your estimate was not too high?
SOLUTION
The pavement’s structural number is determined from Eq. 4.3 , using Table 4.5 to find a1 = 0.44, a2 = 0.30 and a3 = 0.11, and with D1 = 4, D2 = 8, D3 = 8, M2 = M3 = 1.0 (all given) as
1 1 2 2 2 3 3 3
SN a D a D M a D M
SN = 0.44(4) + 0.30(8)(1.0) + 0.11(8.0)(1.0) = 5.04
For the daily axle loads, the equivalency factors (reading axle equivalents from Tables 4.1 and 4.2 while using SN = 5, which is very close to the 5.04 computed above) are
20-kip single-axle equivalent = 1.51 (Table 4.1) 24-kip single-axle equivalent = 3.03 (Table 4.1) 40-kip tandem-axle equivalent = 2.08 (Table 4.2) Thus the total daily 18-kip ESAL is
Daily W18 = 1.51(1080) + 3.03(400) + 2.08(680) = 4257.2 18-kip ESAL
Applying Eq. 4.1, with So = 0.5, SN = 5.04, ¨PSI = 2.0 (4.5 2.5), and MR = 10,000 lb/in2, we find that at R = 90% (ZR = 1.282 for purposes of Eq. 4.1, as shown in Table 4.4), W18 is 26,128,077. Therefore, the number of years is
26,128,077 years
365 4257.2 16.82 yearsu
Similarly, with R = 99% (ZR = 2.326 for purposes of Eq. 4.1, as shown in Table 4.4), W18
is 7,854,299, so the number of years is
7,854,299 years
365 4257.2 5.05 yearsu
These results show that one can be 99% confident that the pavement will last (have a PSI above 2.5) at least 5.05 years, and one can be 90% confident that it will have a PSI above 2.5 for 16.82 years. This example demonstrates the large impact that the chosen reliability value can have on pavement design.