Asphalt Institute design: 0.95×10³ EAL/day

Given data and assumptions

• Existing two-lane road; current traffic 0.95 x 10^3 (950) EAL of 80 kN per day in both directions, growth 7.5 %, design period 15 years, construction period 16 months (1.33 years), subgrade CBR 5 %.
• Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa.
• Two-lane road: design lane carries 50 percent of the total traffic (Asphalt Institute).
• 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

Subgrade CBR is given as 5 %.

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

Step 2 - Design traffic (ESAL)

Traffic in design lane at opening: $950\times0.5\times(1.075)^{1.33}=523.1$ EAL/day.

$$ESAL=365\times523.1\times\frac{(1.075)^{15}-1}{0.075}=365\times523.1\times26.118=4.987\times10^6$$

Design ESAL = 4.987 $\times 10^6$

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: $D_1 \ge SN_{base}/a_1 = 1.31/0.376=88$ mm, so provide 90 mm (minimum 50 mm). $SN_1 = 1.33$.
• Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (3.23-1.33)/0.328 = 147$ mm, so provide 150 mm. $SN_2 = 1.94$.
• Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (4.53-1.33-1.94)/0.128 = 251$ mm, so provide 260 mm. $SN_3 = 1.31$.

Check: provided $SN = 1.33+1.94+1.31 = 4.57 \ge 4.53$ (required) - safe.

Answer: asphalt concrete 90 mm, bituminous treated base 150 mm, granular sub-base 260 mm; total pavement thickness = 500 mm.

   +-------------------------------------+
   | Asphalt concrete              90 mm |
   +-------------------------------------+
   | Bituminous treated base      150 mm |
   +-------------------------------------+
   | Granular sub-base            260 mm |
   +-------------------------------------+
   | Subgrade, CBR 5.0 %                 |
   +-------------------------------------+