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, 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.

Question

pi = 4.2, and pt = 2.5.
Growth rate = 4%
Design life = 20 years
Reliability level = 95%
Standard deviation = 0.45

What will be an ideal response?

Answer 1

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 19.3,
Mr = 1500 CBR
Mr = (1500)(7) = 1.05 × 104
Calculate serviceability loss (?PSI),
?PSI = pi – pt
?PSI = 4.2 – 2.5 = 1.7
Using the nomograph in Figure 19.10,
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 10.5 × 103 and extend to design
serviceability loss chart to create “C”
Step 4: Connect “C” to ?PSI of 1.7
Step 5: Draw vertical line to read SN of 3.7
Determine structural coefficients for each layer
From Figure 19.7, with Mr = 4.5 × 105 (asphalt surface course)
a1= 0.44
From Figure 19.6, with CBR = 70 (base course)
a2= 0.13
From Figure 19.5, with CBR = 22 (subbase)
a3= 0.10
Determine drainage index, mi, using Table 19.8
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 19.10, SN = 2.5
D1*= SN1/a1 = 2.5/0.44 = 5.68 in. Use 6.0 in.
SN1*= a1D1* = 0.44(6.0) = 2.64
Repeating this procedure for the base course:
Mr = 1.35 × 104 (Subbase)
From Figure 19.10, SN = 3.3
D2* = (3.3 – 2.64) / [(0.13)(0.9)] = 5.64 in.
From Table 19.11, use recommended minimum of 6 in.
SN2* = (a2)(m2)(D2*) + SN1* = (0.13)(0.9)(6) + 2.64 = 3.34
D3* = (3.5 – 3.34) / [(0.1)(0.9)] = 1.78 in.
From Table 19.11, use recommended minimum of 4 in.
Check design:
SN = 0.44(6.0) + (0.13)(0.9)(6) + (0.1)(0.9)(4) = 3.58 (which is greater
than SN of 3.5), so the design is satisfactory.
The pavement will consist of 6.0 in. of asphalt concrete surface, 6 in. of granular
base, and 4 in. of subbase.