Question 3 of 5: Design ESAL and AASHTO-93 flexible pavement for a six-lane urban freeway

Question 3: Design ESAL and AASHTO-93 flexible pavement for a six-lane urban freeway (20 marks: a 10, b 10)

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. A six-lane urban freeway carrying a first-year AADT of 25,000 vehicles per day, growing at 2 % per annum over a 20-year design life, to be surfaced as a conventional three-layer flexible pavement.

Find. (a) the cumulative design-lane ESAL over 20 years; (b) the thicknesses of the asphalt, base and subbase layers.

Approach. Build the first-year ESAL class by class from the truck-factor table, escalate it with the growth factor, then run the AASHTO-93 flexible equation three times — once for each layer modulus — and convert the resulting structural numbers to thicknesses from the top down.

1. (a) Fix the design-lane distribution factor. The AADT quoted is two-way, so half of it belongs to the design direction; on a facility with three lanes in each direction AASHTO assigns 70 % of the directional truck traffic to the outer, design lane. Hence $$DF = 0.50 \times 0.70 = \boxed{0.35}$$ Reading the 25,000 as a one-direction volume, or forgetting the lane factor, are the two commonest ways to be out by a factor of two here.
2. Take the truck factor for each class from Table IV-5. The facility is an urban freeway but is not called an Interstate, so the Urban — Other Freeways column applies: 0.015 for two-axle four-tire single units, 0.13 for two-axle six-tire single units, and 0.74 for single units of three axles or more — the row into which the “six-axle or more single unit” class falls, since the table does not subdivide single units beyond three axles. Passenger cars carry no entry in the table and their damage is taken as negligible, which is the standard assumption (a car ESAL is of order 0.0003 and contributes less than 0.5 % here).
3. Assemble the first-year ESAL in the design lane. The appendix gives $ESAL = AADT \times HVP \times DF \times TF \times TDY$ with $TDY = 365$. Class by class:

   First-year design-lane ESAL by vehicle class

   Class	Share	TF	ESAL / year
   Passenger cars	50 %	≈ 0	—
   Single unit, 2-axle, 4-tire	28 %	0.015	13,414
   Single unit, 2-axle, 6-tire	18 %	0.130	74,734
   Single unit, 6-axle or more	4 %	0.740	94,535
   Total, first year			182,683

   Note how the traffic mix inverts the vehicle counts: the 4 % of multi-axle single units does more damage than the 46 % of lighter trucks put together, because the fourth-power law weights axle load so heavily.
4. Escalate over the 20-year life. With a uniform 2 % growth rate the appendix growth factor is $$GF = \frac{(1+g)^{t}-1}{g} = \frac{(1.02)^{20}-1}{0.02} = \frac{0.4859}{0.02} = 24.30$$ so the cumulative design ESAL is $$W_{18} = 182\,683 \times 24.30 = \boxed{4.44 \times 10^{6}\ \text{ESAL}}$$ Because a single growth rate applies to every class over the whole life, no continuity factor between growth stages is needed here.
5. (b) Solve the flexible performance equation for the structural number at each interface. The equation printed on appendix pages 1 and 4 is $$\log_{10}W_{18} = Z_R S_0 + 9.36\log_{10}(SN+1) - 0.20 + \frac{\log_{10}\!\left[\dfrac{\Delta PSI}{4.2-1.5}\right]}{0.40 + \dfrac{1094}{(SN+1)^{5.19}}} + 2.32\log_{10}M_R - 8.07$$ with $Z_R = -1.282$, $S_0 = 0.50$ and $\Delta PSI = 4.5 - 2.5 = 2.0$. Solving it three times, once for the modulus of each supporting material: $$SN_1 = 2.63\ (M_R = 30\,000\ \text{psi}), \quad SN_2 = 3.18\ (18\,000), \quad SN_3 = \boxed{3.91}\ (10\,000)$$ $SN_3$ is the total structure the subgrade requires; $SN_1$ and $SN_2$ are what must sit above the base and the subbase respectively so that neither of those layers is overstressed.
6. Read the layer coefficients and the drainage modifier off the charts. Figure 2.5 at $E_{AC} = 450{,}000\ \text{psi}$ gives $a_1 = 0.44$; Figure 2.6 at a base modulus of 30,000 psi gives $a_2 = 0.14$; Figure 2.7 at a subbase modulus of 18,000 psi gives $a_3 = 0.13$. For the drainage modifier, water that clears in about one day is good drainage on the AASHTO scale, and the structure is saturated 15 % of the time, which is the 5–25 % column of Table 2.4; that cell reads 1.15 down to 1.00, and the conservative lower bound is adopted: $$m_2 = m_3 = 1.00$$
7. Convert the structural numbers into thicknesses from the top down. The asphalt must by itself deliver $SN_1$: $$D_1 \ge \frac{SN_1}{a_1} = \frac{2.63}{0.44} = 5.99\ \text{in} \rightarrow \boxed{D_1 = 6.0\ \text{in}}, \quad SN_1^{*} = 0.44 \times 6.0 = 2.64$$ The base then makes up the difference to $SN_2$: $$D_2 \ge \frac{SN_2 - SN_1^{*}}{a_2 m_2} = \frac{3.18 - 2.64}{0.14} = 3.82\ \text{in}$$ but the appendix minimum-thickness table requires 6 in of aggregate base for traffic between 2 and 7 million ESAL, so $$\boxed{D_2 = 6\ \text{in}}, \quad SN_2^{*} = 0.14 \times 6 = 0.84$$ Finally the subbase closes the gap to $SN_3$: $$D_3 \ge \frac{SN_3 - SN_1^{*} - SN_2^{*}}{a_3 m_3} = \frac{3.91 - 2.64 - 0.84}{0.13} = 3.31\ \text{in} \rightarrow \boxed{D_3 = 6\ \text{in}}$$ where the 6 in is the practical minimum lift for a placed and compacted granular subbase, not a structural requirement.
8. Check the assembled structure. $$SN_{\text{provided}} = a_1D_1 + a_2m_2D_2 + a_3m_3D_3 = 2.64 + 0.84 + 0.78 = 4.26 \ \ge \ 3.91 \ \checkmark$$ Two layers out of three are on their minimum thicknesses rather than on a structural requirement. That is not an error to be corrected by thinning them — it is what happens when a stiff 450,000 psi asphalt with $a_1 = 0.44$ carries 62 % of the structural number in its first six inches. Say so explicitly rather than inventing thinner granular layers that would violate the construction minima.