---
title: "Asphalt Institute design: 1850 ESA/day four lane"
subject: "Transportation Engineering II"
chapter: 2
key: "semester_7/transportation_engineering_ii:asphalt-institute-design-1850-esa"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-1850-esa
site: BCT Engineering
---

# Asphalt Institute design: 1850 ESA/day four lane

Transportation Engineering II · Chapter 2: Highway Pavement

- Asked in: 2082 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-1850-esa
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2

## Question

Design a flexible pavement by using asphalt institute method from the following data of a stretch of existing four lane divided carriageway,
- Current traffic (sum of both direction) = 1850 esa per day
- Traffic growth rate = 6%
- Design period = 15 years
- Construction period = 16 months
- The load at 2.5 mm and 5 mm penetration during CBR test on subgrade soil is 68.5 kg and 80 kg respectively
- Elastic modulus of asphalt concrete surface course, bituminous treated base and granular sub-base course are 2,500 MPa, 1,200 MPa and 125 MPa respectively.

## Short answer

Using Asphalt Institute method with CBR 5% subgrade (Mr=50 MPa), design ESAL 7.64E+6. Layer thicknesses: 100 mm asphalt concrete (a1=0.376), 160 mm bituminous base (a2=0.328), 260 mm granular sub-base (a3=0.128). Total 520 mm.

## Answer

#### 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.

| Protected layer | Modulus (MPa) | Required SN | Full-depth AC equivalent, SN/$a_1$ (mm) |
|---|---|---|---|
| Bituminous treated base | 1200 | 1.42 | 96 |
| Granular sub-base | 125 | 3.47 | 234 |
| Subgrade | 50 | 4.82 | 326 |

#### 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.

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is CBR taken as 5.0 % and not 3.89 %?
   A. Lower value is always used
   B. The 2.5 mm value is higher than the 5.0 mm value, so it is adopted
   C. The average of both is used
   D. 5 mm is always ignored
   Answer: B. CBR at 2.5 mm = 68.5/1370 = 5.00 % is higher than at 5 mm (3.89 %), so 5.0 % is adopted. (Key point: CBR from penetration)

2. Resilient modulus of subgrade used is:
   A. M_R = 10 x CBR = 50 MPa
   B. M_R = 100 x CBR
   C. M_R = CBR/10
   D. M_R = 5 MPa
   Answer: A. M_R = 10 x 5.0 = 50 MPa. (Key point: Subgrade modulus)

3. How is traffic at opening in the design lane found?
   A. 1850 x (1.06)^15
   B. 1850 x 0.45 x (1.06)^1.33 = about 900 ESA/day
   C. 1850 x 0.5
   D. 1850 unchanged
   Answer: B. The design lane takes 45 percent of two-way traffic, grown for the 16-month construction period. (Key point: Design lane traffic)

4. How is the asphalt concrete thickness found?
   A. D1 >= SN required over base / a1 (1.42/0.376 = 96 mm, provide 100 mm)
   B. D1 = total SN x a1
   C. D1 = subgrade CBR x 10
   D. D1 equals base thickness
   Answer: A. Thickness = SN to be provided over the layer divided by its coefficient, rounded up. (Key point: Layer thickness by SN)

5. The total pavement thickness arrived at is:
   A. 260 mm
   B. 520 mm (100 + 160 + 260)
   C. 100 mm
   D. 800 mm
   Answer: B. AC 100 mm + treated base 160 mm + sub-base 260 mm = 520 mm. (Key point: Final design)

## Related in Transportation Engineering II

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- [Asphalt Institute design: 1.888×10³ ESAL/day, 10 years](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aasphalt-institute-design-1-888e3-esal) — Transportation Engineering II, chapter 2

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