Flexible Pavement Design by the AASHTO Method: 16-Civ-B7 Transportation Planning and Engineering, May 2016, Q4

Question 4: Flexible Pavement Design by the AASHTO Method (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.

Quantity	Symbol	Value
One-way average daily traffic	ADT	8000 veh/day
Truck proportion	—	5 %
Annual traffic growth	g	2 %
Design period	n	20 years
Lanes in each direction	—	2
Subgrade resilient modulus	M_R	34.5 MPa = 5000 psi
Average load equivalency per truck	—	2 ESAL
Reliability / overall standard deviation	R / S_o	90 % (Z_R = −1.282) / 0.49
Initial and terminal serviceability	p₀, p_t	4.5, 2.6 (ΔPSI = 1.9)
Trial section	—	100 mm AC (a = 0.42), 300 mm base (a = 0.14), 400 mm subbase (a = 0.10)

Find. (a) the cumulative number of trucks in the design lane over 20 years, (b) the corresponding 80 kN equivalent single-axle load applications, and (c) whether the trial section has the structural number the AASHTO 1993 equation requires for that traffic.

Trial flexible pavement section and its structural number. The provided SN of 4.88 falls short of the 5.23 required for 6.39 million ESAL at 90 per cent reliability.

Approach. Grow the daily truck volume over the design period with the compound-growth factor, take the design-lane share, convert trucks to ESAL with the given equivalency, compute the structural number the trial section provides from its layer coefficients and thicknesses, then solve the AASHTO 1993 flexible-pavement equation for the structural number the traffic requires and compare.

Part (a) — daily truck volume in one direction. The stated ADT is already one-way, so $$\text{trucks/day} = 0.05(8000) = 400\ \text{trucks/day}$$
Growth factor over the design period. With traffic growing at a constant annual rate, the cumulative multiplier over n years is $$GF = \frac{(1+g)^n - 1}{g} = \frac{(1.02)^{20}-1}{0.02} = \frac{1.48595-1}{0.02} = 24.297$$ Note that this factor must exceed the design period itself — 24.297 against 20 years — which is the quick check that a growth rate has not been confused with a growth factor.
Design-lane distribution. With two lanes in each direction the trucks are not shared evenly; AASHTO gives a lane-distribution factor between 0.80 and 1.00, and 0.90 is adopted here for a four-lane highway. The cumulative truck volume in the design lane is then $$T = 400\,(365)\,(24.297)\,(0.90)$$ $$\boxed{T = 3.19\times 10^{6}\ \text{trucks}}$$ more precisely 3 192 674 trucks over the twenty years.
Part (b) — design ESAL. Multiplying by the average load equivalency of 2 ESAL per truck, $$W_{18} = 2.0\,(3\,192\,674)$$ $$\boxed{W_{18} = 6.39\times 10^{6}\ \text{ESAL}}$$ (6 385 349 applications of the 80 kN standard axle in the design lane.)
Part (c) — structural number provided by the trial section. The structural number is the sum of each layer's coefficient times its thickness in inches, with drainage coefficients taken as m = 1.0 for well-drained granular layers: $$SN = a_1D_1 + a_2D_2m_2 + a_3D_3m_3$$ Converting thicknesses, 100 mm = 3.937 in, 300 mm = 11.811 in and 400 mm = 15.748 in, so the contributions are $0.42(3.937)=1.654$, $0.14(11.811)=1.654$ and $0.10(15.748)=1.575$, giving $$SN_{\text{provided}} = 1.654+1.654+1.575 = 4.88$$
The AASHTO 1993 design equation. The required structural number satisfies $$\begin{aligned} \log_{10}W_{18} =\ & Z_R S_o + 9.36\log_{10}(SN+1) - 0.20 + 2.32\log_{10}M_R - 8.07 \\ & + \frac{\log_{10}\!\left[\Delta PSI/(4.2-1.5)\right]}{0.40 + \dfrac{1094}{(SN+1)^{5.19}}} \end{aligned}$$ with $Z_R = -1.282$ for 90 per cent reliability, $S_o = 0.49$, $\Delta PSI = 4.5-2.6 = 1.9$ and $M_R = 5000$ psi. The equation cannot be inverted in closed form, so it is solved for SN numerically.
Solve for the required structural number. Iterating on SN until the right-hand side reproduces $\log_{10}(6.385\times10^6) = 6.805$ gives $$\boxed{SN_{\text{required}} = 5.23}$$ Substituting the provided SN = 4.88 into the same equation returns a capacity of only $3.87\times10^6$ ESAL — about 61 per cent of the design traffic.
Verdict and the deficiency expressed as thickness. Since $SN_{\text{provided}} = 4.88 < SN_{\text{required}} = 5.23$, the trial section is not adequate for 6.39 million ESAL over twenty years. The shortfall of 0.35 in structural number corresponds to $$\Delta D = \frac{0.35}{0.42} = 0.84\ \text{in} = 21\ \text{mm of additional asphalt concrete}$$ so thickening the asphalt layer from 100 mm to 125 mm (a round 25 mm increase, giving SN = 5.30) makes the section adequate. Equivalent alternatives are a further 150 mm of granular base (0.83 in of SN per 150 mm) or, if the traffic estimate itself is conservative, a re-examination of the lane-distribution factor.
Sensitivity of the verdict to the assumed lane factor. With $D_L = 0.80$ the design traffic falls to $5.68\times10^6$ ESAL and the requirement to SN = 5.15; with $D_L = 1.00$ it rises to $7.10\times10^6$ and SN = 5.31. The provided 4.88 is short in every case, so the conclusion does not depend on the assumption.

Check: lane-distribution and drainage factors. Neither is given. $D_L = 0.90$ and $m = 1.0$ are adopted under Note 2. Poor drainage ($m = 0.90$ on both granular layers) would reduce the provided SN to 4.56 and worsen the deficiency, so the assumption made here is the favourable one and the "not adequate" verdict is conservative in the right direction.

Quantity	Result
Trucks per day, one direction	400
Growth factor over 20 years at 2 %	24.297
(a) Total design-lane trucks (D_L = 0.90)	3 192 674 ≈ 3.19 × 10⁶
(b) Design ESAL, W₁₈	6 385 349 ≈ 6.39 × 10⁶
SN provided by the trial section	4.88
SN required (R = 90 %, S_o = 0.49, M_R = 5000 psi)	5.23
Traffic the trial section can carry	3.87 × 10⁶ ESAL
(c) Verdict	inadequate — add 25 mm of asphalt concrete

Topic. Traffic loading in equivalent single-axle loads and the AASHTO 1993 flexible-pavement structural number.

Key relations. $GF = [(1+g)^n-1]/g$; design-lane traffic $=\text{ADT}\times\%\text{trucks}\times365\times GF\times D_L$; $W_{18} = \text{ESAL per truck}\times$ design-lane trucks; $SN=\sum a_iD_im_i$ with $D_i$ in inches; and the AASHTO 1993 equation above.

Why this works. Pavement damage is not proportional to vehicles but to axle load raised to roughly the fourth power, so all traffic is first converted to a common currency — the 80 kN single axle — before any thickness is chosen. The structural number is the matching abstraction on the supply side: it collapses a layered section into a single index in which a stiffer material earns a larger coefficient per unit thickness, which is why 100 mm of asphalt at 0.42 contributes as much as 300 mm of granular base at 0.14. The design equation then relates the loading the pavement will see to the index it must have, with the reliability term $Z_RS_o$ acting as a deliberate shift toward a heavier section to absorb the scatter in materials, construction and traffic forecasting. Because SN is an index rather than a depth, many different sections satisfy the same requirement, and the choice among them is economic.

Common pitfall. Forgetting the lane-distribution factor, or applying a directional split to an ADT that is already one-way and so halving the traffic twice. Inside the design equation, the two recurring errors are substituting the resilient modulus in MPa rather than psi — which changes the answer by more than a full structural number — and reading $Z_R$ with the wrong sign; the reliability term must always reduce the traffic the section is credited with.

Source. AASHTO, Guide for Design of Pavement Structures (1993), Part II Ch. 2 (design requirements) and Part III Ch. 1 (flexible pavement design); Garber & Hoel, 5th ed., §19.4 (AASHTO flexible pavement design).

Question 4 of 7: Flexible Pavement Design by the AASHTO Method

Question 4: Flexible Pavement Design by the AASHTO Method (20 marks)