---
title: "Asphalt Institute design: gravel road upgrade, 0.8×10³ ESA"
subject: "Transportation Engineering II"
chapter: 2
key: "semester_7/transportation_engineering_ii:asphalt-institute-design-0-8e3-esa"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-0-8e3-esa
site: BCT Engineering
---

# Asphalt Institute design: gravel road upgrade, 0.8×10³ ESA

Transportation Engineering II · Chapter 2: Highway Pavement

- Asked in: 2068 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-0-8e3-esa
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2

## Question

An existing two lane single carriageway gravel road has to be upgraded by bituminous pavement, to cater the growing traffic demand. Present traffic in terms of ESA is 0.8 × 10³ per day. The regional traffic growth rate is taken as 6.5% per annum. Data required for pavement design are as given below.
i) Design period = 10 years
ii) Construction period = 1 year
iii) 87.5 percentile CBR value of sub grade soil from 7 sample locations = 5%
iv) Elastic modulus of asphalt concrete for surface course Eac = 2000 MPa
v) Elastic modulus of crushed stone base Ebase = 350 MPa
vi) Elastic modulus of granular sub-base Esub-base = 250 MPa
You are required to design the pavement by Asphalt Institute Method. Draw the cross section of final pavement layers considering the minimum thickness of asphalt concrete on surface course equal to 50 mm.

## Short answer

Subgrade Mr = 50 MPa. Design lane ESAL = 800×0.5×1.065×365×[(1.065^10-1)/0.065] = 2.098×10^6. Asphalt concrete 140 mm, crushed stone base 100 mm, granular sub-base 170 mm (total 410 mm).

## Answer

#### Given data and assumptions
- Two-lane single carriageway gravel road upgraded to bituminous; present traffic 0.8 x 10^3 (800) ESA/day (both directions), growth 6.5 %, design period 10 years, construction period 1 year.
- 87.5th percentile subgrade CBR = 5 %; moduli: asphalt concrete 2000 MPa, crushed stone base 350 MPa, granular sub-base 250 MPa.
- Two lanes: design lane carries 50 percent of the total (Asphalt Institute). Minimum asphalt concrete 50 mm.
- 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
The design (87.5th percentile) CBR of the subgrade is given as 5 %.

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

#### Step 2 - Design traffic (ESAL)
Design-lane ESA/day at opening $=800\times0.5\times(1.065)^1=426.0$.

$$
ESAL=365\times426.0\times\frac{(1.065)^{10}-1}{0.065}=365\times426.0\times13.494=2.098\times10^6
$$

**Design ESAL = 2.098 $\times 10^6$**

#### 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).

| Protected layer | Modulus (MPa) | Required SN | Full-depth AC equivalent at $a_1=0.44$ (mm) |
|---|---|---|---|
| crushed stone base | 350 | 1.87 | 108 |
| granular sub-base | 250 | 2.14 | 124 |
| Subgrade | 50 | 3.96 | 229 |

#### 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 (2000 MPa): $a_1 = 0.344$
- crushed stone base (350 MPa): $a_2 = 0.195$
- granular sub-base (250 MPa): $a_3 = 0.196$

#### Step 5 - Layer thicknesses
- Asphalt concrete: $D_1 \ge SN_{base}/a_1 = 1.87/0.344=138$ mm, so provide **140 mm** (minimum 50 mm). $SN_1 = 1.90$.
- Crushed stone base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (2.14-1.90)/0.195 = 32$ mm, so provide **100 mm**. $SN_2 = 0.77$.
- Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (3.96-1.90-0.77)/0.196 = 168$ mm, so provide **170 mm**. $SN_3 = 1.31$.

Check: provided $SN = 1.90+0.77+1.31 = 3.98 \ge 3.96$ (required) - safe.

**Answer:** asphalt concrete 140 mm, crushed stone base 100 mm, granular sub-base 170 mm; total pavement thickness = 410 mm.

```text
   +-------------------------------------+
   | Asphalt concrete             140 mm |
   +-------------------------------------+
   | Crushed stone base           100 mm |
   +-------------------------------------+
   | Granular sub-base            170 mm |
   +-------------------------------------+
   | Subgrade, CBR 5.0 %                 |
   +-------------------------------------+
```

## Check yourself

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

1. How is the design-lane ESA/day at opening found?
   A. 800 x 1.065 = 852
   B. 800 x 0.5 x (1.065)^1 = 426
   C. 800 x 0.5 = 400
   D. 800 x (1.065)^10 = 1501
   Answer: B. Two-lane road gives a 0.5 lane factor, and the traffic is grown one year of construction at 6.5 %. (Key point: Design lane ESA at opening)

2. What is the subgrade resilient modulus for 87.5th percentile CBR of 5 %?
   A. 5 MPa
   B. 500 MPa
   C. 50 MPa
   D. 15 MPa
   Answer: C. M_R = 10 x CBR = 50 MPa. (Key point: Subgrade modulus)

3. The crushed stone base needs only 32 mm by calculation, yet 100 mm is provided. Why?
   A. The base must carry the full SN
   B. A very thin base cannot be built, so a practical minimum thickness is adopted
   C. The sub-base is weaker
   D. Design lane factor was 0.5
   Answer: B. A 32 mm layer is not constructible, so a practical thickness (100 mm) is provided, which makes the section safe. (Key point: Minimum practical thickness)

4. How is the sub-base thickness D3 calculated?
   A. D3 = SN_sg / a3
   B. D3 = SN_sub / a3
   C. D3 = (SN_sg - SN_1 - SN_2) / a3 using provided SN_1 and SN_2
   D. D3 = (SN_sg + SN_1 + SN_2) / a3
   Answer: C. The sub-base supplies what is left of the subgrade-required SN after the provided surface and base contributions. (Key point: Layer sequence formula)

## Related in Transportation Engineering II

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