Asphalt Institute design: 1850 ESA/day four lane

Given data and assumptions

• Current traffic (both directions) 1850 ESA/day, growth 6 %, design period 15 years, construction period 16 months (1.33 years).
• Four-lane divided carriageway: the design lane carries 45 percent of the total two-way traffic (AI lane factor for 4 lanes).
• Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa.
• Method: Asphalt Institute (AI) layered 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

CBR at 2.5 mm = $\frac{68.5}{1370}\times100=5.00\ \%$; CBR at 5.0 mm = $\frac{80}{2055}\times100=3.89\ \%$ (standard loads 1370 kg and 2055 kg). The 2.5 mm value is higher, so CBR = 5.0 % is adopted.

Resilient modulus of the subgrade: $M_R = 10\times CBR = 10\times 5.0 = 50$ MPa.

Step 2 - Design traffic (ESAL)

Traffic at opening: $1850\times0.45\times(1.06)^{1.33} = 899.8$ ESA/day (design lane).

$$\begin{aligned} ESAL &= 365\times 899.8 \times \frac{(1.06)^{15}-1}{0.06} \\ &= 365\times899.8\times23.276 = 7.644\times10^6 \end{aligned}$$

Design ESAL = 7.644 $\times 10^6$

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. Here the same requirement is expressed as a structural number $SN$ (AASHTO-93 relation with R = 90 percent, $S_0$ = 0.45, $\Delta PSI$ = 1.7), which approximates the chart readings. If the AI chart is supplied in the exam, read the thickness from it and use the same layer 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: $D_1 \ge SN_{base}/a_1 = 1.42/0.376 = 3.78$ in $= 96$ mm, so provide 100 mm (minimum 50 mm). $SN_1 = 1.48$.
• Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (3.47-1.48)/0.328 = 6.07$ in $= 154$ mm, so provide 160 mm. $SN_2 = 2.07$.
• Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (4.82-1.48-2.07)/0.128 = 9.92$ in $= 252$ mm, so provide 260 mm. $SN_3 = 1.31$.

Check: provided $SN = 1.48+2.07+1.31 = 4.86 \ge 4.82$ (required) - safe.

Answer: asphalt concrete 100 mm, bituminous treated base 160 mm, granular sub-base 260 mm; total pavement thickness = 520 mm.

   +-------------------------------------+
   | Asphalt concrete             100 mm |
   +-------------------------------------+
   | Bituminous treated base      160 mm |
   +-------------------------------------+
   | Granular sub-base            260 mm |
   +-------------------------------------+
   | Subgrade, CBR 5.0 %                 |
   +-------------------------------------+