---
title: "Flexible design: four-lane, AADT 12,000"
subject: "Transportation Engineering II"
chapter: 2
key: "semester_7/transportation_engineering_ii:flexible-design-aadt-12000-four-lane"
source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/flexible-design-aadt-12000-four-lane
site: BCT Engineering
---

# Flexible design: four-lane, AADT 12,000

Transportation Engineering II · Chapter 2: Highway Pavement

- Asked in: 2076 Asoj, 8 marks
- Video: [ESAL Example](https://www.youtube.com/watch?v=UWb60ekR_Fc) — wu2ube
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/flexible-design-aadt-12000-four-lane
- Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2

## Question

A 4-lane divided highway is to be constructed on a new alignment. Traffic volume forecasts indicate that the AADT in both directions during the first year of operation will be 12,000 with 50% Passenger cars (5 kN/axle), 33% of two axle single unit trucks (25 kN/axle) and 17% of three axle single unit trucks (30 kN/axle) = 17%. The vehicle mix is expected to remain the same throughout the design life of the pavement. If the expected annual traffic growth rate is 4% for all vehicles, design pavement with AC of thickness 7.5 cm, base and sub base. CBR of subgrade = 6.5%, E_sub base = 275 MPa, E_base = 500 MPa and E_surface = 2400 MPa.

## Short answer

Design ESAL is 0.86 million; required structural number is 3.10. Asphalt concrete 7.5 cm (SN 1.09), base 15 cm (500 MPa, SN 1.38) and sub-base 15 cm (275 MPa, SN 1.21) provide total SN 3.68, so the pavement consists of AC 7.5 cm over base and sub-base each 15 cm.

## Answer

#### Assumptions
Design method: AASHTO-93 structural number with layer coefficients from the given moduli. Not stated in the problem, so assumed: design life 20 years (new alignment, no construction period), 50:50 directional split, design-lane factor 0.8 for two lanes in one direction, reliability 90 % ($Z_R=-1.282$), $S_0=0.45$, $\Delta PSI=4.2-2.5=1.7$, drainage coefficients 1.0. Standard axle 80 kN, fourth power law for ESAL.

#### Step 1 - ESAL per day (first year, both directions)
| Vehicle | Number/day | Axles | Axle load | Per-axle factor $(P/80)^4$ | ESAL/day |
|---|---|---|---|---|---|
| Passenger car | 6000 | 2 | 5 kN | 0.000015 | 0.18 |
| 2-axle truck | 3960 | 2 | 25 kN | 0.0095 | 75.5 |
| 3-axle truck | 2040 | 3 | 30 kN | 0.0198 | 121.0 |
| **Total** | | | | | **196.7** |

#### Step 2 - Design ESAL
$$
W_{18}=196.7\times0.5\times0.8\times365\times\frac{(1.04)^{20}-1}{0.04}=196.7\times0.4\times365\times29.778=855345
$$

#### Step 3 - Layer coefficients (E in psi) and required SN
- Subgrade $M_R=10\times6.5=65$ MPa $=9427$ psi.
- $a_1=0.323\log_{10}(2400\times145)-1.42=0.370$, $a_2=0.249\log_{10}(500\times145)-0.977=0.233$, $a_3=0.227\log_{10}(275\times145)-0.839=0.205$.
- Required $SN$ over base (protecting $M_R$ = 500 MPa): $SN_1=1.38$; over sub-base (275 MPa): $SN_2=1.77$; over subgrade: $SN_3=3.10$.

#### Step 4 - Thicknesses
- AC is specified as 7.5 cm (2.95 in): $SN_{AC}=0.370\times2.953=1.09$. A layer-wise check over the base would ask for $SN_1=1.38$ (about 95 mm of AC). However, the base modulus is 500 MPa = 72 500 psi, and AASHTO-93 states that this layered check should not be used above a base or sub-base with a modulus above 40 000 psi; the AC over such a stiff base is fixed by cost and minimum-thickness considerations instead. For 0.5-2 million ESAL the AASHTO minimum AC thickness is 3.0 in (about 75 mm), so the specified 7.5 cm is acceptable. (The sub-base, 275 MPa = 39 900 psi, is below the limit, so the check above the sub-base is valid.)
- Base: $D_2\ge(SN_2-SN_{AC})/a_2=(1.77-1.09)/0.233=2.92$ in $=74$ mm; this is below the AASHTO minimum base thickness (6 in for 0.5-2 million ESAL), so provide **15 cm** (5.91 in): $a_2D_2=1.38$.
- Sub-base: $D_3\ge(SN_3-SN_{AC}-a_2D_2)/a_3=(3.10-1.09-1.38)/0.205=3.08$ in $=78$ mm; provide the minimum practical **15 cm**: $a_3D_3=1.21$.
- Check: SN above the sub-base $=1.09+1.38=2.47\ge1.77$, and total $SN=1.09+1.38+1.21=3.68\ge3.10$ - safe. The base and sub-base are governed by minimum thicknesses because the design traffic is light.

```text
  +--------------------------+
  | Asphalt concrete  7.5 cm |
  | Base (500 MPa)   15.0 cm |
  | Sub-base (275)   15.0 cm |
  | Subgrade CBR 6.5 %       |
  +--------------------------+
```

**Answer:** AC 7.5 cm, base 15 cm, sub-base 15 cm (total 37.5 cm), for a design traffic of about 0.86 million ESAL. The passenger cars and light trucks cause very little damage; the 3-axle trucks produce most of the ESAL.

## Check yourself

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

1. Why do passenger cars contribute almost nothing to ESAL here?
   A. Damage follows the fourth power of axle load, and 5 kN is tiny compared with 80 kN
   B. Cars are excluded by law
   C. They have more axles
   D. The CBR is high
   Answer: A. (5/80)^4 = 0.000015, so car damage is negligible. (Key point: ESAL per vehicle)

2. Which vehicles produce most of the ESAL?
   A. Passenger cars
   B. Three-axle single unit trucks
   C. Two-axle trucks only
   D. All equally
   Answer: B. 3-axle trucks give 121 ESAL/day of the total 196.7. (Key point: Which vehicles dominate)

3. Which factors convert first-year daily ESAL to design ESAL?
   A. Directional split 0.5, lane factor 0.8, 365 days and the growth sum factor over design life
   B. Only multiply by 365
   C. Divide by growth rate
   D. Multiply by CBR
   Answer: A. W18 = ESAL/day x 0.5 x 0.8 x 365 x [(1.04)^20 - 1]/0.04. (Key point: Design ESAL factors)

4. How is the required base thickness found?
   A. D2 >= (SN2 - SN_AC)/a2, with SN2 the number needed to protect the sub-base
   B. D2 = SN3/a3
   C. D2 = CBR x 2.5
   D. D2 = SN1 x a1
   Answer: A. The layer must supply the SN not provided by the layer above, divided by its coefficient. (Key point: Thickness by layers)

5. Why is the final design considered safe?
   A. Provided SN = 3.68 is at least the required SN over subgrade 3.10
   B. Because the AC is 7.5 cm
   C. Because traffic is zero
   D. Because the SN is less than required
   Answer: A. The sum 1.09 + 1.38 + 1.21 = 3.68 exceeds 3.10. (Key point: Final check)

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