Construcción del chasis y procedimientos de acoplamiento y operación

1637.01 trucks / day0.75

- W

W PDL

EXAMPLE 4.5 FLEXIBLE AND RIGID PAVEMENT DESIGN-LIFE EQUIVALENCE A new flexible pavement was designed for four lanes of traffic (conservatively designed for load distribution among the lanes). The design is for a total directional daily traffic of 967 10-kip single axles and 1935 30-kip tandem axles. The pavement has an 8-in hot-mix asphaltic (HMA) surface, 10-in dense-graded crushed stone base and a 10-in crushed-stone subbase (the drainage coefficients are 0.9 for the base and 0.78 for the subbase). The soil CBR is 2, the reliability used was 95%, the overall standard deviation was 0.4, initial PSI was 4.5, and the TSI was 2.5. Determine the required slab thickness for a rigid pavement designed to last the same number of years as the flexible pavement, but with only three lanes (instead of four) in the design direction (again, conservatively designing for the load distribution among the lanes). The design is to use the same truck traffic, reliability, soil, initial PSI, TSI, and overall standard deviations as the flexible pavement. In addition, the rigid pavement is to have a modulus of rupture of 800 lb/in², a concrete modulus of elasticity of 5.5 u 10⁶ lb/in², a load transfer coefficient of 3.0, and a drainage coefficient of 0.9.

SOLUTION

We will first need to determine the design life of the existing flexible pavement and then use this value to determine the required slab thickness of the rigid pavement.

The flexible pavement’s structural number is determined from Eq. 4.3, using Table 4.5 to find a1 = 0.44, a2 = 0.18 and a3 = 0.11, and with D1 = 8, D2 = 10, D3 = 10, M2 = 0.9 and M3 = 0.78 (all given) as

1 1 2 2 2 3 3 3

SN a D a D M a D M

SN = 0.44(8) + 0.18(10)(0.9) + 0.11(10)(0.9) = 5.998 § 6

With a CBR = 2 (given), Eq. 4.2 is applied to give MR = 3000 (1500 u 2). For other elements required to solve Eq. 4.1 (or alternatively Fig. 4.5), ¨PSI = 2 (initial PSI of 4.5

minus the TSI of 2.5), So = 0.4 (given), and ZR = 1.645 (which corresponds to R = 95%, as shown in Table 4.4). With these values, applying Eq. 4.1 (or using Fig. 4.5) gives W18 = 5,703,439.

For the daily axle loads, the equivalency factors (reading axle equivalents from Tables 4.1 and 4.2 while using SN = 6) are

10-kip single-axle equivalent = 0.08 (Table 4.1) 30-kip tandem-axle equivalent = 0.633 (Table 4.2) Thus the total daily 18-kip ESAL is

Daily W18 = 0.08(967) + 0.633(1935) = 1302.215 18-kip ESAL/day

From Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives

18 18 18

design lane - W PDLudirectional W = 976.66 W / day or 356,481 W18/yr (976.66 u 365). So the design life of the flexible pavement is:

18 18

design design life (in years) =

daily 5,703,439

356,481 16 years

W W

For the rigid pavement, we have that all of the design parameters are the same and are given Scc = 800 lb/in², Ec = 5.5 u 10⁶ lb/in², J = 3.0, Cd = 0.9. In addition, the modulus of subgrade reaction, k, is 100 lb/in³ from Table 4.9 (with CBR = 2). Note that the accumulated W18 for the rigid pavement will not be the same as the daily axle loads for the flexible pavement because the load equivalency factors will be different. To determine the accumulated W18 for the rigid pavement, we assume a slab thickness of 10 inches (D = 10 inches) and read axle equivalents from Tables 4.6 and 4.7 as

10-kip single-axle equivalent = 0.08 (Table 4.6) 30-kip tandem-axle equivalent = 1.14 (Table 4.7) Thus the total daily 18-kip ESAL is

Daily W18 = 0.08(967) + 1.14(1935) = 2283.26 18-kip ESAL/day

From Table 4.10, the PDL for a conservative design on a three-lane highway is 0.8, so applying Eq. 4.4 gives a required slab thickness of 10.28 inches (Figs. 4.7 and 4.8 can also be used to arrive at an approximate solution for slab thickness). For design, this can be rounded up to 10.5 inches. It can be shown by the reader that going back to the load equivalency factors and assuming 10.5 inches instead of the previously assumed 10 inches will result in the same, correct slab-thickness solution of 10.5 inches.

4.6 Traditional AASHTO Rigid-Pavement Design Procedure

125

A roadway is determined to have 400 18-kip single axles, 200 24-kip tandem axles and 100 40-kip triple axles per day. The subgrade CBR is 2 and the roadway pavement is designed for an overall standard deviation of 0.4, a reliability of 99 percent and the initial PSI is 4.5 and the TSI is 2.5. One newly constructed section of this roadway is a rigid pavement designed with a 9-inch slab with a modulus of rupture of 700 lb/in², a modulus of elasticity of 4.0 u 10⁶ lb/in², and a joint transfer coefficient of 3.0. Another newly constructed section of the same roadway is a flexible pavement with a 5-in hot-mix asphalt (HMA) surface, 10-in dense-graded crushed stone base and a 9-10-in crushed-stone subbase. If the roadway has four lanes in each direction and is conservatively designed, which of the pavement sections will last longer and by how many years (all drainage coefficients are 1.0)?

SOLUTION

We first determine the total design W18 for the flexible pavement. To do this, the flexible pavement’s structural number is computed from Eq. 4.3, using Table 4.5 to find a1 = 0.44, a2 = 0.18 and a3 = 0.11, and with D1 = 5, D2 = 10, D3 = 9, M2 = 1.0 and 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(5) + 0.18(10)(1.0) + 0.11(9)(1.0) = 4.99 § 5

With a CBR = 2 (given), Eq. 4.2 is applied to give MR = 3000 (1500 u 2). For other elements required to solve Eq. 4.1 (or alternatively Fig. 4.5), So = 0.4 (given), ZR = 2.326 (which corresponds to R = 99%, as shown in Table 4.4), and ¨PSI = 2 (initial PSI of 4.5 minus the TSI of 2.5). With these values, applying Eq. 4.1 (or using Fig. 4.5) gives W18 = 775,133. To determine the total truck traffic, the equivalency factors (reading axle equivalents from Tables 4.1, 4.2 and 4.3 while using SN = 5) are

18-kip single-axle equivalent = 1.0 (Table 4.1) 24-kip tandem-axle equivalent = 0.260 (Table 4.1)

40-kip triple-axle equivalent = 0.487 (Table 4.2)

which gives a total W18 of 500.7 18-kip ESAL/day (1.0 u 400 + 0.260 u 200 + 0.487 u 100).

From Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives the design-lane W18 = 375.53 18-kip ESAL/day (0.75 u 500.7). So the design life for the flexible pavement is

18 18

total design - lane design life (in years) =

daily design - lane 365 days/yr 775,133

375.53 365 5.655 years

W W u

u

For the rigid pavement, we are given a slab thickness of 9 inches (D = 9 inches) Scc = 900 lb/in², Ec = 4.0 u 10⁶ lb/in², J = 3.0, Cd = 1.0 and the modulus of subgrade reaction, k, is 100 lb/in³ from Table 4.9 (with CBR = 2), and all other design parameters are the same as those for the flexible pavement. With these values, applying Eq. 4.4 (or using Figs. 4.7 and

4.8) gives W18 = 2,364,522. To determine the total truck traffic, the equivalency factors (reading axle equivalents from Tables 4.7, 4.8, and 4.9 while using D = 9) are

18-kip single-axle equivalent = 1.0 (Table 4.6) 24-kip tandem-axle equivalent = 0.444 (Table 4.7)

40-kip triple-axle equivalent = 1.17 (Table 4.8)

which gives a total W18 of 605.8 18-kip ESAL/day (1.0 u 400 + 0.444 u 200 + 1.17 u 100).

As in the flexible-pavement case, from Table 4.10, the PDL for a conservative design on a four-lane highway is 0.75, so applying Eq. 4.4 gives the design-lane W18 = 454.35 18-kip ESAL/day (0.75 u 605.8). So the design life for the rigid pavement is

18 18

total design - lane design life (in years) =

daily design - lane 365 days/yr 2,364,522

454.35 365 14.258 years

W W u

u

So the rigid pavement will last 8.603 years longer (14.258 – 5.655).

4.7

MEASURING PAVEMENT QUALITY AND PERFORMANCE

The design procedure for pavements originally focused on the pavement serviceability index (PSI) as a measure of pavement quality. However, the pavement serviceability index is based on the opinions of a panel of experts (as discussed in Section 4.4.1), which can introduce some variability into their determination. As a result, efforts have been undertaken to develop quantitative measures of pavement condition that provide additional insights into pavement quality and performance and that correlate with the traditional pavement serviceability index. Some factors that are regularly measured by highway pavement agencies now include the International Roughness Index, friction measurements, and rut depth.