---
title: "Asphalt Institute design: 5.11×10⁴ ESAL/year, 12 years"
subject: "Transportation Engineering II"
chapter: 2
key: "semester_7/transportation_engineering_ii:asphalt-institute-design-5-11e4-esal"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-5-11e4-esal
site: BCT Engineering
---

# Asphalt Institute design: 5.11×10⁴ ESAL/year, 12 years

Transportation Engineering II · Chapter 2: Highway Pavement

- Asked in: 2070 Chaitra (old course), 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-5-11e4-esal
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2

## Question

An existing single lane road has to be upgraded by bituminous pavement for a certain length by the following considerations:
i) Base traffic of 80 kN equivalent single axle load (ESAL) = 5.11×10⁴ ESAL per year
ii) Design period = 12 years
iii) Construction period = 1 year
iv) Traffic growth rate = 6%
v) 87.5th percentile CBR value of sub-grade soil from 7 sample locations = 4%
vi) Elastic modulus of asphalt concrete for surface course, Eac = 2900 MPa
vii) Elastic modulus of emulsified stabilized base, Eb = 1600 MPa
viii) Elastic modulus of granular sub-base, Esb = 125 MPa
You are required to design the pavement from Asphalt Institute Method. Draw the cross section of final pavement layers considering the thickness of asphalt concrete on surface course is not less than 50 mm. (Full depth AC curve attached herewith)

## Short answer

Subgrade Mr = 10×CBR = 40 MPa. Design ESAL = 51100×1.06×[(1.06^12-1)/0.06] = 0.914×10^6. Asphalt concrete 55 mm, emulsified stabilised base 120 mm, granular sub-base 250 mm (total 425 mm).

## Answer

#### Given data and assumptions
- Single-lane road; base traffic 5.11 x 10^4 (51 100) ESAL (80 kN) per year, design period 12 years, construction period 1 year, growth 6 %.
- 87.5th percentile subgrade CBR = 4 %; moduli: asphalt concrete 2900 MPa, emulsified stabilised base 1600 MPa, granular sub-base 125 MPa.
- Single lane: all traffic is in the design lane (factor 1.0). 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 4 %.

Resilient modulus of subgrade: $M_R = 10\times CBR = 10\times 4.0 = 40$ MPa.

#### Step 2 - Design traffic (ESAL)
ESAL at opening $=51\,100\times(1.06)^1=54166$ per year.

$$
ESAL=54166\times\frac{(1.06)^{12}-1}{0.06}=54166\times16.870=0.914\times10^6
$$

**Design ESAL = 0.914 $\times 10^6$**

#### 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) |
|---|---|---|---|
| emulsified stabilised base | 1600 | 0.79 | 46 |
| granular sub-base | 125 | 2.44 | 141 |
| Subgrade | 40 | 3.77 | 218 |

#### 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 (2900 MPa): $a_1 = 0.397$
- emulsified stabilised base (1600 MPa): $a_2 = 0.359$
- granular sub-base (125 MPa): $a_3 = 0.128$

#### Step 5 - Layer thicknesses
- Asphalt concrete: $D_1 \ge SN_{base}/a_1 = 0.79/0.397=51$ mm, so provide **55 mm** (minimum 50 mm). $SN_1 = 0.86$.
- Emulsified stabilised base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (2.44-0.86)/0.359 = 112$ mm, so provide **120 mm**. $SN_2 = 1.70$.
- Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (3.77-0.86-1.70)/0.128 = 242$ mm, so provide **250 mm**. $SN_3 = 1.26$.

Check: provided $SN = 0.86+1.70+1.26 = 3.81 \ge 3.77$ (required) - safe.

**Answer:** asphalt concrete 55 mm, emulsified stabilised base 120 mm, granular sub-base 250 mm; total pavement thickness = 425 mm.

```text
   +-------------------------------------+
   | Asphalt concrete              55 mm |
   +-------------------------------------+
   | Emulsified stabilised base   120 mm |
   +-------------------------------------+
   | Granular sub-base            250 mm |
   +-------------------------------------+
   | Subgrade, CBR 4.0 %                 |
   +-------------------------------------+
```

## Check yourself

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

1. Why is no lane distribution factor (0.5) applied to the 5.11 x 10^4 ESAL per year?
   A. Because the traffic is already in the design lane
   B. Because it is a single-lane road, so all traffic uses the design lane
   C. Because ESAL is already halved for single lanes
   D. Because the construction period is one year
   Answer: B. On a single lane road all traffic travels in that lane, so the factor is 1.0. (Key point: Single lane factor)

2. Which calculation gives the design ESAL of 0.914 x 10^6?
   A. 51 100 x 12 = 0.61 x 10^6
   B. 51 100 x 1.06 x 12
   C. 54166 x [(1.06)^12 - 1]/0.06 with 54166 = 51100 x 1.06
   D. 51 100 x (1.06)^12
   Answer: C. Traffic is grown one year of construction to 54166 per year, then multiplied by the 12-year cumulative growth factor 16.87. (Key point: Design ESAL calculation)

3. What subgrade modulus is used for design CBR = 4 %?
   A. 4 MPa
   B. 400 MPa
   C. 14 MPa
   D. 40 MPa
   Answer: D. M_R = 10 x CBR = 40 MPa. (Key point: Subgrade modulus)

4. SN_base/a1 = 0.79/0.397 gives 51 mm. What does the answer provide and why?
   A. 55 mm, as the thickness must be at least the computed and at least 50 mm minimum, rounded to practical size
   B. 40 mm, since SN is small
   C. 50 mm exactly, as the minimum governs
   D. 100 mm, as a safety margin
   Answer: A. The computed 51 mm is rounded up to a practical 55 mm, which also respects the 50 mm minimum. (Key point: Surface layer rounding)

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