Flexible design: 4-lane single carriageway, 2000 CVPD

Given data and assumptions

• Four-lane single carriageway, initial traffic 2000 CVPD in each direction (4000 CVPD both ways), growth 8 %, design life 15 years, construction period 3 years, design CBR 6 %.
• Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa.
• Assumptions: each commercial vehicle has 2 axles on average; design lane carries 45 percent of the total two-way traffic (Asphalt Institute, four lanes); standard axle 80 kN and equivalency by the fourth power law.
• Method: Asphalt Institute (AI) thickness design as taught for IOE (and used in the Nepal DoR Pavement Design Guidelines); design reliability 90 percent, terminal serviceability 2.5.

Step 1 - Subgrade strength

Design CBR is given as 6 %.

Resilient modulus of subgrade: $M_R = 10\times CBR = 10\times 6.0 = 60$ MPa.

Step 2 - Design traffic (ESAL)

Equivalent axle load factors from the axle distribution (average EALF per axle):

ESAL per day at present (design lane) $=4000\times0.45\times2\times0.7270=2617.2$. At opening (3 years): $2617.2\times(1.08)^3=3296.9$ per day.

$$ESAL=365\times3296.9\times\frac{(1.08)^{15}-1}{0.08}=365\times3296.9\times27.152=32.674\times10^6$$

Design ESAL = 32.674 $\times 10^6$

Step 3 - Thickness required over each layer

The AI charts give the thickness of asphalt concrete (full depth) needed to protect a layer of a given modulus for the design ESAL. The same requirement is expressed here as a structural number $SN$ (AASHTO-93 relation with R = 90 percent, $S_0$ = 0.45, $\Delta PSI$ = 1.7), which closely approximates the AI chart readings (if the chart is supplied in the exam, read the thickness from it and use the same layer-coefficient steps below).

Step 4 - Layer coefficients

Using $a_1=0.323\log_{10}E-1.42$, $a_2=0.249\log_{10}E-0.977$, $a_3=0.227\log_{10}E-0.839$ with $E$ in psi:

• Asphalt concrete (2500 MPa): $a_1 = 0.376$
• bituminous treated base (1200 MPa): $a_2 = 0.328$
• granular sub-base (125 MPa): $a_3 = 0.128$

Step 5 - Layer thicknesses

• Asphalt concrete: $D_1 \ge SN_{base}/a_1 = 1.84/0.376=124$ mm, so provide 125 mm (minimum 50 mm). $SN_1 = 1.85$.
• Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (4.36-1.85)/0.328 = 195$ mm, so provide 200 mm. $SN_2 = 2.58$.
• Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (5.58-1.85-2.58)/0.128 = 228$ mm, so provide 230 mm. $SN_3 = 1.16$.

Check: provided $SN = 1.85+2.58+1.16 = 5.59 \ge 5.58$ (required) - safe.

Answer: asphalt concrete 125 mm, bituminous treated base 200 mm, granular sub-base 230 mm; total pavement thickness = 555 mm.

   +-------------------------------------+
   | Asphalt concrete             125 mm |
   +-------------------------------------+
   | Bituminous treated base      200 mm |
   +-------------------------------------+
   | Granular sub-base            230 mm |
   +-------------------------------------+
   | Subgrade, CBR 6.0 %                 |
   +-------------------------------------+