AASHTO-93 flexible pavement design for a four-lane divided highway: 16-Civ-B7 Transportation Planning and Engineering, December 2014, Q3

Question 3: AASHTO-93 flexible pavement design for a four-lane divided highway (20 marks)

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 design inputs tabulated above. The chart supplied on page 3 is drawn for exactly the serviceability pair and standard deviation this question specifies, and for the 30 MPa roadbed modulus, so it is the intended design tool; only the design-lane ESAL total and the reliability level have to be brought to it.

Find. The required structural number for the design lane, and a layer combination of asphalt concrete, granular base and granular subbase that provides it.

Approach. Accumulate the daily directional ESALs over the design period with the compound growth factor, apportion them to the outside lane, enter the supplied chart on the 90 percent reliability curve to read the required structural number in millimetres, confirm that reading against the 1993 AASHTO flexible design equation, and then distribute the structural number over a practical layer combination.

Compute the traffic growth factor over the design period. With traffic growing at a compound rate $g$ over $n$ years, the sum of the annual traffic expressed as a multiple of the first year is the uniform-series compound-amount factor $$G = \frac{(1+g)^n - 1}{g} = \frac{(1.03)^{20} - 1}{0.03} = \frac{1.806111 - 1}{0.03} = 26.870$$
Accumulate the design-lane ESALs. The 888 ESALs per day are already directional, so only the annualisation, the growth factor and the lane distribution remain. The outside lane governs, since it takes 85 percent of the directional traffic: $$W_{18} = (\text{ESAL/day})\times 365 \times D_L \times G = 888 \times 365 \times 0.85 \times 26.870$$ $$W_{18} = \boxed{7.40 \times 10^{6}\ \text{ESALs}}$$ The inside lane, by the same arithmetic with $D_L = 0.15$, accumulates only $1.31\times10^{6}$ ESALs.
Read the structural number from the supplied chart. Entering the page-3 chart at a design ESAL value of $7.40\times10^{6}$ and rising to the 90 percent reliability curve gives a structural number of $$SN \approx \boxed{144\ \text{mm}}$$ Note that this chart expresses $SN$ in millimetres, which is Alberta and general Canadian practice: the customary AASHTO structural number in inches is simply multiplied by 25.4, so 144 mm corresponds to $SN = 5.67$ in the units used in the American literature.
Confirm the chart reading against the AASHTO-93 design equation. A chart read by eye should always be checked, and the governing equation is available in the open-book examination: $$\log_{10}W_{18} = Z_R S_o + 9.36\log_{10}(SN+1) - 0.20 + \frac{\log_{10}\!\left[\dfrac{\Delta PSI}{4.2-1.5}\right]}{0.40 + \dfrac{1094}{(SN+1)^{5.19}}} + 2.32\log_{10}M_R - 8.07$$ with $SN$ in inches and $M_R$ in psi. Here $Z_R = -1.282$ for 90 percent reliability, $S_o = 0.45$, $\Delta PSI = 4.2 - 2.5 = 1.7$, and $M_R = 30\ \text{MPa} \times 145.04 = 4351\ \text{psi}$. Solving for $SN$ at $W_{18} = 7.40\times10^{6}$ returns $SN = 5.670$ in, that is $$SN = 5.670 \times 25.4 = \boxed{144.0\ \text{mm}}$$ which reproduces the chart reading to better than a millimetre. As an independent check on the chart itself, the same equation with $Z_R = 0$ (50 percent reliability) returns 119.4 mm, and 119 mm is exactly where the 50 percent curve sits at this ESAL level. The chart and the equation are consistent, so the design value is secure.
Distribute the structural number over the layers. With the structural number expressed in millimetres the layer equation keeps the familiar AASHTO coefficients and takes the thicknesses directly in millimetres: $$SN = a_1D_1 + a_2m_2D_2 + a_3m_3D_3$$ Taking Alberta’s customary values $a_1 = 0.44$ for asphalt concrete, $a_2 = 0.14$ for a 20 mm crushed granular base course and $a_3 = 0.11$ for granular subbase, with drainage coefficients $m_2 = m_3 = 1.0$, and choosing thicknesses that respect normal minimums and lift practice: $$SN = 0.44(150) + 0.14(250) + 0.11(400) = 66.0 + 35.0 + 44.0 = \boxed{145.0\ \text{mm}}$$ which exceeds the required 144.0 mm.
Check the layers individually and the total depth. The 150 mm of asphalt concrete would be placed in two or three lifts and is at the usual minimum for a highway carrying more than about $5\times10^{6}$ ESALs. The base and subbase satisfy the AASHTO layered-analysis principle that each successive layer must itself be thick enough to protect the one beneath it. The total granular thickness of 650 mm below the asphalt gives a structure 800 mm deep, which for Edmonton is also close to what frost protection demands: with a design frost penetration of roughly 1.8 to 2.4 m, full frost protection is uneconomic, and Alberta practice is instead to provide a substantial non-frost-susceptible granular depth and to accept partial protection.
Report the inside lane. Repeating steps 2 to 5 with $D_L = 0.15$ gives $W_{18} = 1.31\times10^{6}$ and a required $SN$ of 112 mm, about 22 percent less than the outside lane. It is nonetheless normal to build both lanes of a carriageway to the same section: the saving is small, the paving operation is far simpler, and lane closures for future rehabilitation are unpredictable. The outside-lane design therefore governs the whole carriageway.

Check — assumptions declared under Note 2 of the paper. The question supplies no layer coefficients, no drainage coefficients and no minimum thicknesses, so the values above are assumed and stated: $a_1 = 0.44$, $a_2 = 0.14$, $a_3 = 0.11$, $m_2 = m_3 = 1.0$. They are the standard Alberta Transportation values and are the ones consistent with the supplied chart. If the drainage coefficients were reduced to $m = 0.90$ to reflect a poorly drained granular layer, the same three thicknesses would deliver $0.44(150)+0.9[0.14(250)+0.11(400)] = 137.1$ mm and would no longer be adequate; the base would then be increased to 300 mm, giving 143.4 mm, or the asphalt to 165 mm. The design is therefore sensitive to drainage but not to the reliability level in any dramatic way — raising reliability from 90 to 95 percent moves the requirement only from 144 to 151 mm.

Figure 3.1 — The adopted flexible pavement section for the design (outside) lane. The structural-number contribution of each layer is shown on the right; the three layers together deliver 145.0 mm against the 144.0 mm required.

Quantity	Value
Traffic growth factor, 20 years at 3 percent	26.870
Design ESALs, outside (design) lane	7.40 × 10⁶
Design ESALs, inside lane	1.31 × 10⁶
Required structural number, R = 90 percent	144.0 mm (SN = 5.67 in)
Asphalt concrete pavement	150 mm
Granular base course	250 mm
Granular subbase	400 mm
Structural number provided	145.0 mm
Total pavement structure	800 mm

Topic. Empirical flexible pavement design to the 1993 AASHTO Guide, with the structural number expressed in metric units as used in the Canadian prairie provinces.

Key relations. Growth factor $G = [(1+g)^n-1]/g$; design-lane traffic $W_{18} = \text{ESAL/day}\times 365\times D_L\times G$; the AASHTO-93 flexible design equation quoted in Step 4; and the layer equation $SN = \sum a_i m_i D_i$.

Why this works. The AASHTO-93 method is an empirical regression on the AASHO Road Test of 1958 to 1960, in which the loss of serviceability of real pavement sections was correlated with accumulated equivalent single axle loads. The structural number is not a physical quantity but an index that collapses the whole layered structure into one number weighted by how effectively each material resists load spreading, and the design equation simply states the traffic that a given index will carry before the serviceability falls from $p_i$ to $p_t$. The reliability term $Z_RS_o$ shifts the design to the conservative side of that regression: $Z_R$ is the standard normal deviate for the chosen probability of survival and $S_o$ is the standard deviation of the whole prediction, so at 90 percent reliability the design carries the traffic that would be predicted for a structure roughly $1.282\times0.45 = 0.58$ log cycles — a factor of about 3.8 in traffic — heavier than the mean prediction. That is why the chart on page 3 is a family of parallel curves rather than a single line, and it is why the check in Step 4 reproducing both the 90 percent and the 50 percent readings is such a strong validation: matching two curves fixes both the position and the reliability shift of the chart.

Common pitfall. The most frequent error is forgetting the lane distribution factor and designing the outside lane for the full directional traffic, which here would inflate the ESALs to $8.71\times10^{6}$ and the requirement to 147 mm. The second is mixing unit systems: the AASHTO equation demands $M_R$ in psi, and substituting 30 directly for the modulus in MPa returns a nonsensical structural number. The third is treating the millimetre structural number as though it were a thickness — 144 mm is an index, not a depth, and the pavement it implies here is 800 mm deep.

Source. AASHTO, Guide for Design of Pavement Structures (1993), Part II, Ch. 2 (Design Requirements) and Ch. 3 (Flexible Pavement Design); Alberta Transportation and Utilities, Pavement Design Manual, for the metric structural-number charts and the provincial layer coefficients; Garber and Hoel, 5th ed., Ch. 20.

Question 3 of 7: AASHTO-93 flexible pavement design for a four-lane divided highway

Question 3: AASHTO-93 flexible pavement design for a four-lane divided highway (20 marks)