ESALs carried by an in-service pavement, and the implied LEF: 16-Civ-A6 Highway Design, Construction, and Maintenance, December 2017, Q6

Question 6: ESALs carried by an in-service pavement, and the implied LEF (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 data (Question 6)
Quantity	Symbol	Value
Wearing surface	$D_1$, $a_1$	4 in hot-mix asphalt, 0.44
Base	$D_2$, $a_2$	6 in dense-graded crushed stone, 0.18
Subbase	$D_3$, $a_3$	10 in crushed stone, 0.11
Effective roadbed resilient modulus	$M_R$	3800 psi
Present serviceability index	$p_t$	3.0
Assumed initial serviceability	$p_0$	4.2 (AASHTO default)
Age	$n$	10 years
Truck volume (two-way)	—	200 trucks/day

Given. An in-service flexible pavement of 4 in hot-mix asphalt, 6 in dense-graded crushed-stone base and 10 in crushed-stone subbase on a 3800 psi clay subgrade, ten years old and now at $PSI=3.0$, carrying 200 trucks per day each with two 20-kip single axles.

Find. (a) the ESALs the pavement has carried; (b) the implied load equivalency factor of the 20-kip single axle.

The as-built section. Note the base is dense-graded crushed stone, $a_2=0.18$, not the plain crushed stone $a_2=0.14$ used for a subbase.

Approach. Sum the layer contributions into $SN$, run the AASHTO design equation forwards at $R=50\ \%$ to obtain $W_{18}$, then divide by the number of 20-kip axle applications the traffic record implies.

Compute the structural number of the section as built. Reading the coefficients off the attached table and taking $m_2=m_3=1.0$ in the absence of drainage data, $$ SN=0.44(4)+0.18(6)+0.11(10)=1.76+1.08+1.10=\boxed{3.94} $$
Fix the serviceability loss actually observed. The pavement has fallen from the AASHTO default initial value to its present condition, so $$ \Delta PSI=p_0-p_t=4.2-3.0=1.2. $$
Choose the reliability appropriate to a back-calculation. The question asks what the pavement has carried, which is a mean prediction rather than a design with a safety margin. That is $R=50\ \%$, for which Table 4.5 gives $Z_R=0$, so the $Z_RS_0$ term drops out entirely.
Evaluate the design equation forwards. With $\log_{10}(1.2/2.7)=-0.35218$, $(SN+1)^{5.19}=4.94^{5.19}=3982.9$ and $2.32\log_{10}(3800)=8.30510$, $$ \log_{10}W_{18}=9.36\log_{10}(4.94)-0.20 +\frac{-0.35218}{0.40+\frac{1094}{3982.9}}+8.30510-8.07 $$ $$ =6.49329-0.20-0.52200+8.30510-8.07=6.00626 $$ $$ W_{18}=10^{6.00626}=\boxed{1.015\times10^{6}\ \text{ESALs}} $$
Count the 20-kip axle applications that produced it. On a two-lane two-way highway the design lane carries one direction, so the AASHTO lane-distribution factor is 0.5. Over ten years, $$ N=200\times0.5\times2\ \text{axles}\times365\times10=730{,}000 \ \text{applications of the 20-kip single axle}. $$
Back out the load equivalency factor. By definition $W_{18}=N\times LEF$, so $$ LEF=\frac{W_{18}}{N}=\frac{1.015\times10^{6}}{730{,}000} =\boxed{1.39} $$ Table 4.2 lists 1.47 for a 20-kip single axle at $SN=4$ and $p_t=2.5$, so the back-calculated value sits 5.5 % below the published one — excellent agreement given that the table is entered at $p_t=2.5$ while this pavement is still at 3.0, and that $p_0$ had to be assumed.

Check: two data items are not supplied and have been assumed under NOTE 2. First, the initial serviceability is taken as the AASHTO default $p_0=4.2$ for flexible pavements; had the pavement been built to $p_0=4.5$ as in Question 5, $\Delta PSI$ would be 1.5 and $W_{18}$ would rise to $1.41\times10^{6}$. Second, the 200 trucks are read as the two-way volume, so the design lane receives half of them; if instead the 200 trucks are already the design-lane volume, the axle count doubles to 1,460,000 and the implied $LEF$ halves to 0.695, which is far below the tabulated 1.47 and would suggest the traffic record, not the pavement, is at fault. The two-way reading is adopted because it reproduces the published factor to within 6 %.

Question 6 — final results
Quantity	Result
Structural number of the section	$SN=3.94$
Serviceability loss	$\Delta PSI=1.2$
(a) ESALs carried in 10 years	$W_{18}=1.015\times10^{6}$
20-kip axle applications (design lane)	730,000
(b) Back-calculated load equivalency factor	$LEF=1.39$
Published value, Table 4.2 at $SN=4$	1.47 (5.5 % higher)

Topic. Using the AASHTO design equation in reverse, as a pavement evaluation tool, and the definition of the load equivalency factor.

Key relations. $SN=a_1D_1+a_2D_2m_2+a_3D_3m_3$; $\Delta PSI=p_0-p_t$; the design equation with $Z_R=0$ at $R=50\ \%$; $W_{18}=N\times LEF$, i.e. $LEF=W_{18}/N$.

Why this works. The design equation is a performance model, not merely a sizing formula: given the section, the subgrade and the serviceability actually lost, it returns the number of standard 18-kip axle passes that would produce that loss. Dividing by the count of real axles then recovers how many standard axles each real axle is worth, which is exactly what an $LEF$ means.

Common pitfall. Leaving a design reliability such as 90 % in the equation when back-calculating — that deliberately conservative $Z_R$ understates the traffic the pavement has really taken. Also, forgetting that each truck contributes two axles, and omitting the lane-distribution factor on a two-way volume.

Source. AASHTO, Guide for Design of Pavement Structures, 1993, Part II §2.1 and Appendix D; Garber & Hoel, Traffic and Highway Engineering, 5th ed., §20.2 (equivalent single axle loads) and §20.3.

Question 6 of 7: ESALs carried by an in-service pavement, and the implied LEF

Question 6: ESALs carried by an in-service pavement, and the implied LEF (20 marks — (a) 10, (b) 10)