Question

A flexible pavement is designed with 4 inches of sand-mix asphalt wearing surface, 6 inches

of dense-graded crushed stone base, and 8 inches of crushed stone subbase. All drainage

coefficients are 1.0. The initial PSI is 4.5, the TSI is 2.5, the soil resilient modulus is 12,000

lb/in2, reliability is 90%, the overall standard deviation is 0.4, and the design life is 15 years.

A design mistake was made that ignored 100 daily passes of trucks with 22-kip single and

30-kip tandem axles. What crushed stone subbase thickness should have been used? [Do

not interpolate the axle equivalency factors, use the closest values]

Answer 1

Given,

Thickness of Sand-mix asphaltic surface, D1 = 4 in

Layer coefficient for Wearing Surface, a1 = 0.35

Thickness of dense-graded crushed stone base, D2 = 6 in

Layer coefficient for base, a2 = 0.18

Thickness of crushed stone sub-base, D3 = 8 in

Layer coefficient for sub-base, a3 = 0.11

Drainage coefficients, m2, m3 = 1.0

Structural Number, SN = a1D1 + a2D2m2 + a3D3m3 = 0.35 * 4 + 0.18 * 6 * 1.0 + 0.11 * 8 * 1.0 = 3.36

Initial PSI = 4.5

TSI = 2.5

Soil Resilient Modulus = 12000 psi

Overall Standard Deviation = 0.4

Reliability = 90 %

ZR = -1.282

log10(W18) = ZR * So + 9.36 * log10(SN + 1) - 0.20 + ((log10(ΔPSI/4.2 – 1.5))/(0.40 + (1094/(SN+1)5.19))) + 2.32 * log10(MR) – 8.07

log10(W18) = -1.282 * 0.4 + 9.36 * log10(3.36 + 1) - 0.20 + ((log10(4.5 – 2.5/4.2 – 1.5))/(0.40 + (1094/(3.36+1)5.19))) + 2.32 * log10(12000) – 8.07

log10(W18) = -0.5128 + 5.9856 – 0.20 + (-0.1303/0.9249) + 9.4637 – 8.07

log10(W18) = 6.5256

W18 = 106.5256 = 3354285.31 ESAL

Total no. of ESAL, this pavement lasts = 3354285.31 ~ 3354286 ESAL

Design life = 15 years

Design lane Annual 18-kip ESAL, W18 = 3354286/(15 * 365) = 612.655 ~ 613 ESAL

Ignored Daily Traffic:

100 trucks (each with one 22-kip single axle and one 30-kip tandem axles) = 100 * (1 * 2.17 + 1 * 0.703) = 287.3 ~ 288 ESAL

Total Traffic = 613 + 288 = 901 ESAL

Total no. of ESAL = 901 * 365 * 15 = 4,932,975 ESAL

log10(4932975) = -1.282 * 0.4 + 9.36 * log10(SN + 1) - 0.20 + ((log10(4.5 – 2.5/4.2 – 1.5))/(0.40 + (1094/(SN+1)5.19))) + 2.32 * log10(12000) – 8.07

6.6931 = -0.5128 + 9.36 * log10(SN + 1) – 0.20 + (-0.1303/0.40 + (1094/(SN+1)5.19))) + 9.4637 – 8.07

6.0122 = 9.36 * log10(SN + 1) + (-0.1303/0.40 + (1094/(SN+1)5.19)))

By trial and error,

SN = 3.57

Structural Number, SN = a1D1 + a2D2m2 + a3D3m3 = 0.35 * 4 + 0.18 * 6 * 1.0 + 0.11 * x * 1.0 = 3.57

X = 9.9 in ~ 10 in

Crushed stone sub-base thickness = 10 in