Asphalt Institute design: 1500 pcu/day, 5% growth

Given data and assumptions

• Two-lane two-way road; traffic 1500 pcu/day (all taken as commercial vehicles of damage factor 2.5), growth 5 %, design life 15 years, construction period 18 months (1.5 years), subgrade CBR 5 %.
• Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa.
• Design lane carries 50 percent of the two-way traffic (Asphalt Institute, two lanes).
• 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)

ESAL/day now (design lane) $=1500\times2.5\times0.5=1875$. At opening: $1875\times(1.05)^{1.5}=2017.4$ per day.

$$ESAL=365\times2017.4\times\frac{(1.05)^{15}-1}{0.05}=365\times2017.4\times21.579=15.889\times10^6$$

Design ESAL = 15.889 $\times 10^6$

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: $D_1 \ge SN_{base}/a_1 = 1.62/0.376=110$ mm, so provide 110 mm (minimum 50 mm). $SN_1 = 1.63$.
• Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (3.90-1.63)/0.328 = 176$ mm, so provide 180 mm. $SN_2 = 2.32$.
• Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (5.35-1.63-2.32)/0.128 = 279$ mm, so provide 280 mm. $SN_3 = 1.41$.

Check: provided $SN = 1.63+2.32+1.41 = 5.36 \ge 5.35$ (required) - safe.

Answer: asphalt concrete 110 mm, bituminous treated base 180 mm, granular sub-base 280 mm; total pavement thickness = 570 mm.

   +-------------------------------------+
   | Asphalt concrete             110 mm |
   +-------------------------------------+
   | Bituminous treated base      180 mm |
   +-------------------------------------+
   | Granular sub-base            280 mm |
   +-------------------------------------+
   | Subgrade, CBR 5.0 %                 |
   +-------------------------------------+