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Flexible pavement design: four lane, ADT 8000 trucks

Short answer

Use 90th percentile CBR, giving design CBR 5%. Traffic after 3 years: 9800 CVPD; design lane 5390 CVPD; cumulative standard axles about 150 msa with VDF 4.0, LDF 0.75, r=7%, n=15. Pavement: 50 mm BC + 170 mm DBM + 250 mm WMM + 300 mm GSB, total 770 mm.

A new four lane divided highway is to be for the ADT of 8000 truck traffic based on the last count. The directional split of traffic is 55:45. Vehicle damage factor based on axle load survey was 4.0. Design a suitable pavement section for a design life of 15 years. The last traffic count was taken 1 year back and the project would be completed in 2 years from now. Growth rate of traffic = 7%.

The results of subgrade soil CBR (%) at seven locations obtained in a certain stretch of road are given below.

S.No.ChainageTest CBR results in (%)
10+5012
20+3508
30+5007
40+6508
50+8006
60+9505
71+5005

Design method

Four-lane divided highway, so IRC:37-2012 (Guidelines for the Design of Flexible Pavements) is used with its CBR-based catalogue; Nepal DoR adopts the same approach (cumulative msa and subgrade CBR). The traffic of 8000 is taken as commercial vehicles per day (CVPD) in both directions.

Step 1 - Design subgrade CBR

For a major highway the 90th percentile CBR is used (the value that 90 percent of the results equal or exceed):

CBR (%)No. of values equal or greaterPercent equal or greater
57100.0
6571.4
7457.1
8342.9
12114.3

The 90th percentile lies between 5 percent (100 percent of values) and 6 percent (71.4 percent of values); interpolation gives about 5.3 percent, so the design CBR is conservatively taken as CBR = 5 %.

Step 2 - Design traffic

  • The traffic count was one year ago and construction takes 2 years, so x=1+2=3x = 1+2 = 3 years.
  • Traffic at completion (both ways): Atotal=8000(1.07)3=9800A_{total}=8000(1.07)^3 = 9800 CVPD.
  • Heavier direction 55 percent: A=0.55×9800=5390A = 0.55\times9800=5390 CVPD.
  • Lane distribution factor for a dual two-lane carriageway (IRC:37-2012): D=0.75D = 0.75 of the directional traffic.
  • VDF F=4.0F = 4.0, design life n=15n = 15 years, r=7%r=7\%.
Ns=365[(1+r)n−1]r A D F=365×(1.07)15−10.07×5390×0.75×4.0=365×25.129×5390×0.75×4.0=148.3×106 standard axles\begin{aligned} N_s &= \frac{365[(1+r)^n-1]}{r}\,A\,D\,F\\ &= 365\times\frac{(1.07)^{15}-1}{0.07}\times 5390\times0.75\times4.0 \\ &= 365\times25.129\times5390\times0.75\times4.0 = 148.3\times10^{6}\ \text{standard axles} \end{aligned}

Design traffic is about 150 msa (the upper limit of the IRC:37-2012 catalogue).

Step 3 - Pavement composition (IRC:37-2012 plate for CBR 5 %, 150 msa)

Read the layer thicknesses from the design plate for CBR 5 % at 150 msa (use the chart supplied in the exam if one is given):

LayerThickness (mm)
Bituminous concrete (BC)50
Dense bituminous macadam (DBM)170
Wet mix macadam (WMM) base250
Granular sub-base (GSB, drainage + filter)300
Total770
  +-----------------------------+
  | BC               50 mm      |
  | DBM             170 mm      |
  +-----------------------------+
  | WMM base        250 mm      |
  +-----------------------------+
  | GSB sub-base    300 mm      |
  +-----------------------------+
  | Subgrade (CBR 5 %)          |
  +-----------------------------+

Answer: design traffic ≈148\approx 148 msa (150 msa); pavement = 50 mm BC + 170 mm DBM + 250 mm WMM + 300 mm GSB = 770 mm total. The sub-base should be extended across the full formation width for drainage.

✓ Answer checked by Claude Opus 5.5 (high effort)

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