Question 3 of 7: Flexible pavement design by the 1993 AASHTO method

Question 3: Flexible pavement design by the 1993 AASHTO method

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The traffic, serviceability, reliability and subgrade inputs of the 1993 AASHTO flexible design procedure, for a four-lane divided highway (two lanes in each direction).

Find. The structural number required over the design period and a layer thickness combination that provides it, for the design (outside) lane, with the inside lane checked separately.

Approach. Accumulate the daily ESALs over the design period with the compound growth factor, apportion them to the design lane, solve the 1993 AASHTO flexible equation for the structural number, then distribute that structural number across asphalt, base and subbase using standard layer coefficients.

1. Read the growth input correctly. The paper prints a traffic growth of 3.5 percent, which is an annual rate, not the cumulative AASHTO growth factor. It cannot be the cumulative factor: that quantity is $\mathrm{GF} = [(1+g)^n - 1]/g$, whose smallest possible value over a 20-year period is $n = 20$ itself, at zero growth. Accumulating 20 years of compound growth at 3.5 percent gives $$\mathrm{GF} = \frac{(1 + g)^n - 1}{g} = \frac{(1.035)^{20} - 1}{0.035} = 28.280$$
2. Accumulate the directional traffic. Nine hundred ESALs per day in one direction is $$W_{18,\text{dir}} = 900 \times 365 \times \mathrm{GF} = 328\,500 \times 28.280 = 9.289 \times 10^{6}\ \text{ESALs}$$ over the twenty years. Because the paper already gives the traffic per direction, no further directional distribution factor is applied — applying one as well would halve the design traffic and is the commonest error in this question.
3. Apportion to the design lane. The outside lane carries 80 percent of the directional traffic and governs the design: $$W_{18} = D_L\,W_{18,\text{dir}} = 0.80 \times 9.289 \times 10^{6} = \boxed{7.43 \times 10^{6}\ \text{ESALs}}$$ The inside lane sees $0.20 \times 9.289 \times 10^{6} = 1.86 \times 10^{6}$ ESALs and is checked at the end.
4. Assemble the remaining terms. At 90 percent reliability the standard normal deviate is $Z_R = -1.282$; the serviceability loss is $\Delta \mathrm{PSI} = p_i - p_t = 4.3 - 2.5 = 1.8$; and the roadbed modulus must be converted to the units in which the equation was calibrated, $$M_R = 30\ \text{MPa} \times 145.04 = 4\,351\ \text{psi}$$ Substituting 30 directly returns nonsense, because the constant 2.32 multiplies $\log_{10} M_R$ in psi.
5. Solve the 1993 AASHTO flexible equation for the structural number. $$\log_{10} W_{18} = Z_R S_o + 9.36 \log_{10}(SN + 1) - 0.20 + \frac{\log_{10}\!\left[\dfrac{\Delta \mathrm{PSI}}{4.2 - 1.5}\right]}{0.40 + \dfrac{1094}{(SN+1)^{5.19}}} + 2.32 \log_{10} M_R - 8.07$$ The equation is implicit in $SN$ and is solved by bisection. With $\log_{10}(7.43\times10^{6}) = 6.871$ the root is $$\boxed{SN_{\text{required}} = 5.47}$$ Substituting 5.47 back reproduces $7.43 \times 10^{6}$ ESALs, which is the check that the root is the right one.
6. Choose layer coefficients and drainage coefficients. The paper supplies none, so under its Note 2 the standard AASHTO values for good-quality Canadian materials are adopted: $a_1 = 0.44$ for dense-graded asphalt concrete, $a_2 = 0.14$ for a crushed granular base, $a_3 = 0.11$ for a granular subbase, with drainage coefficients $m_2 = m_3 = 1.00$ for a structure that drains within a day and is saturated a small fraction of the year.
7. Distribute the structural number. The structural number is assembled as $$SN = a_1 D_1 + a_2 m_2 D_2 + a_3 m_3 D_3$$ with the thicknesses in inches. Trying 150 mm of asphalt concrete over 300 mm of base and 300 mm of subbase, $$\begin{aligned}a_1 D_1 &= 0.44 \times (150/25.4) = 2.598 \\a_2 m_2 D_2 &= 0.14 \times 1.00 \times (300/25.4) = 1.654 \\a_3 m_3 D_3 &= 0.11 \times 1.00 \times (300/25.4) = 1.299\end{aligned}$$ so that $$SN_{\text{provided}} = 2.598 + 1.654 + 1.299 = \boxed{5.55 \ge 5.47}$$ a margin of about 1.5 percent, over a total pavement thickness of 750 mm.
8. Check the inside lane. Repeating the solution at $W_{18} = 1.86 \times 10^{6}$ gives $SN_{\text{required}} = 4.50$, about 82 percent of the outside-lane requirement even though it carries a quarter of the traffic — the equation is strongly logarithmic in load. Holding the asphalt and base constant, that structural number would need only 56 mm of subbase, which is below any practical placing and compaction thickness. The inside lane is therefore built to the same section as the outside lane, which is normal practice: the saving from thinning one lane of a two-lane carriageway is not worth the longitudinal construction joint and the difference in future performance.

It is worth naming what the structural number is and is not. It is a dimensionless index of the total load-spreading capacity of the bound and unbound layers above the subgrade, not a thickness: the section that delivers $SN = 5.55$ here is 750 mm deep. Expressed in the millimetre convention used by several prairie agencies it would be quoted as $5.47 \times 25.4 = 139$ mm, which is the same number in different units and still not a depth. Any combination of layers reaching the required index is acceptable to the equation; it is minimum practical thicknesses, the requirement that each layer protect the one beneath it, and cost that decide which combination is built.