Question 2 of 7: Design ESAL for a 25-year urban freeway pavement

Question 2: Design ESAL for a 25-year urban freeway pavement (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. A four-lane urban freeway carrying 65,000 vehicles per day in its opening year, with the vehicle mix, growth rates and design life set out below and truck factors read from the appendix page 5 table, urban systems, Other Freeways column.

Find. The meaning of the equivalent single axle load, the cumulative 18-kip ESAL applications in the design lane over 25 years, and the way the mechanistic-empirical method replaces that single number.

Approach. Convert the first-year mix into design-lane 18-kip equivalent applications class by class, build a growth factor for each class (a two-stage factor for the class whose rate changes), and sum.

1. Explain what an equivalent single axle load is. Pavements are not damaged in proportion to load; damage rises roughly as the fourth power of axle load, so one heavy axle can do the work of thousands of car axles. The AASHO Road Test measured, for each axle load and configuration, how many repetitions were needed to drive a test section from its initial serviceability to its terminal serviceability, and expressed that as a ratio against the number of repetitions of a standard 80 kN (18,000 lb) single axle on dual tyres producing the same loss. That ratio is the load equivalency factor. Multiplying every axle in the traffic stream by its factor and summing converts a mixed stream of cars, buses and multi-axle trucks into a single number, the cumulative equivalent single axle loads, which is the only traffic input the 1993 design equation accepts. The equivalency factor depends weakly on the structural number (or slab thickness) and the terminal serviceability, so the design is formally iterative, and a truck factor is simply the sum of the equivalency factors of all axles on one vehicle of a given class.
2. Fix the design-lane distribution factor. AADT is a two-way volume, so half of it runs in each direction. A four-lane freeway has two lanes per direction, for which AASHTO recommends 80 to 100 % of the directional truck traffic in the design lane; taking 90 %, $$DF = 0.50 \times 0.90 = \boxed{0.45}$$
3. Convert the first year of each class to design-lane ESALs. Using the appendix relation with 365 days per year, $$ESAL_{1} = AADT \times HVP \times DF \times TF \times 365$$ For the two-axle four-tire single units, $$65\,000 \times 0.20 \times 0.45 \times 0.015 \times 365 = 32\,029$$ For the two-axle six-tire single units, $$65\,000 \times 0.15 \times 0.45 \times 0.13 \times 365 = 208\,187$$ For the tractor semi-trailers and other multiple units, $$65\,000 \times 0.15 \times 0.45 \times 0.98 \times 365 = 1\,569\,409$$ Passenger cars carry a truck factor of order 0.0002 and are not tabulated; they are neglected, which is the standard convention and changes the total by well under one per cent. The first-year design-lane total is $$\boxed{ESAL_{1} = 1\,809\,624\ \text{applications}}$$ Note how the traffic stream is dominated by the multiple units: they are 15 % of the vehicles and 87 % of the damage.
4. Build the growth factor for the class whose rate changes. A single-rate growth factor is the sum of a geometric series, $GF = [(1+g)^{n} - 1]/g$. When the rate steps from g1 to g2 after t1 years, the second stage must start from the volume the first stage has already reached, so a continuity factor is needed: $$GF = \frac{(1+g_1)^{t_1} - 1}{g_1} + (1+g_1)^{t_1 - 1}(1+g_2)\,\frac{(1+g_2)^{t_2} - 1}{g_2}$$ With g1 = 0.015 over 5 years and g2 = 0.03 over the remaining 20 years, $$GF_{4\text{-tire}} = 5.152 + (1.015)^{4}(1.03)\,(26.870) = 5.152 + 29.375 = \boxed{34.53}$$ Dropping the continuity factor would give 32.02, understating this class by 8 %.
5. Build the growth factor for every other class. These grow at a single rate of 2 % for the full 25 years: $$GF_{other} = \frac{(1.02)^{25} - 1}{0.02} = \boxed{32.03}$$
6. Accumulate the design ESAL. Applying each factor to its own class, $$W_{18} = 32\,029(34.53) + \bigl(208\,187 + 1\,569\,409\bigr)(32.03)$$ $$W_{18} = 1.106\times10^{6} + 5.694\times10^{7} = \boxed{5.80 \times 10^{7}\ \text{ESALs}}$$ For design purposes this is quoted as 58 million 18-kip equivalent single axle load applications in the design lane over 25 years — a heavy urban freeway loading that will drive the structural number well above 5.
7. Describe the mechanistic-empirical replacement for ESAL. The AASHTO Mechanistic-Empirical Pavement Design Guide (now AASHTOWare Pavement ME) abandons the single damage number entirely and works from the axle load spectra: for each vehicle class and each axle configuration (single, tandem, tridem, quad) it carries the full distribution of measured axle weights, together with monthly and hourly volume adjustment factors, class distribution, axle-per-truck ratios and lateral wander. The design then runs a layered-elastic or finite-element analysis of the trial structure for every load level, in every month, at the temperature and moisture state predicted by an integrated climatic model, computes the incremental damage from each pass, and accumulates distress (rutting, bottom-up and top-down fatigue cracking, thermal cracking, faulting, roughness) directly through calibrated transfer functions. The differences that matter are these: ESAL compresses everything into one number using a fourth-power rule calibrated on one soil, one climate and one set of 1950s materials, whereas the spectra approach keeps the loads separate and lets each one damage the actual trial structure; ESAL cannot distinguish a flexible from a rigid response to the same axle, whereas the mechanistic method computes the response of the structure it is designing; and the ME output is a predicted distress-versus-time curve at a chosen reliability, not a pass or fail thickness.