--- title: "AASHTO: ESALs a pavement can carry" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:aashto-esal-capacity-reliability-90" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/aashto-esal-capacity-reliability-90 site: BCT Engineering --- # AASHTO: ESALs a pavement can carry Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2080 Baisakh, 8 marks - Video: [ESAL Example](https://www.youtube.com/watch?v=UWb60ekR_Fc) — wu2ube - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/aashto-esal-capacity-reliability-90 - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Figure below shows the section of a flexible asphalt pavement designed using AASHTO method. Determine the number of equivalent single-axle loads (ESALs) that the asphalt pavement structure is capable of carrying at a reliability level (R) of 90% [Z-value is equal to -1.282] overall standard deviation (S0) of 0.5, ΔPSI of 2.0, and m2 = 1.0. Assume material coefficient a1 = 0.401 and coefficient of Layer 2 (the base layer) given by the following formula a2 = 0.249 log(E2) − 0.977 $$ \log_{10} W_{18} = Z_R S_0 + 9.36\log_{10}(SN+1) - 0.20 + \frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.4+\frac{1094}{(SN+1)^{5.19}}} + 2.32\log_{10} M_R - 8.07 $$ [Figure: HMA surface of thickness 5.3 in with E1 = 365 ksi on a base of thickness 8.6 in with E2 = 30 ksi, over subgrade with E3 = 9 ksi] ## Short answer The base layer coefficient a2 = 0.138 from a2=0.249log(E2)-0.977. Structural number SN = a1D1 + a2D2 = 3.310. Substituting into the AASHTO design equation with MR=9000 psi yields logW18 = 6.066, so the pavement capacity is approximately 1.16 million ESALs. ## Answer #### Given data Surface $D_1=5.3$ in, $a_1=0.401$; base $D_2=8.6$ in, $E_2=30$ ksi, $m_2=1$; subgrade $M_R=E_3=9$ ksi $=9000$ psi; $R=90\%$, $Z_R=-1.282$, $S_0=0.5$, $\Delta PSI=2.0$. #### Step 1 - Layer coefficient of base $$ a_2=0.249\log_{10}(E_2)-0.977=0.249\log_{10}(30000)-0.977=0.138 $$ #### Step 2 - Structural number $$ SN=a_1D_1+a_2D_2m_2=0.401\times5.3+0.138\times8.6\times1.0=2.125+1.185=3.310 $$ #### Step 3 - ESALs from the AASHTO equation $$ \begin{aligned} \log_{10}W_{18}&=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}\left[\frac{\Delta PSI}{4.2-1.5}\right]}{0.4+\frac{1094}{(SN+1)^{5.19}}}+2.32\log_{10}M_R-8.07\\ &=(-1.282)(0.5)+9.36\log_{10}(4.310)-0.20+\frac{\log_{10}(2.0/2.7)}{0.4+1094/(4.310)^{5.19}}+2.32\log_{10}(9000)-8.07 \end{aligned} $$ Terms: $Z_RS_0=-0.641$; $9.36\log_{10}(SN+1)=5.939$; fatigue term $=-0.136$; $2.32\log_{10}M_R=9.174$. $$ \log_{10}W_{18}=6.066\ \Rightarrow\ W_{18}=10^{6.066}=1163446 $$ **Answer:** the pavement can carry about **1.16 $\times10^6$ ESALs** (1.16 million) at 90 percent reliability. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the structural number of this section obtained? A. SN = a1 D1 + a2 D2 m2 using the layer coefficients and thicknesses in inches B. SN = E1 + E2 + E3 C. SN = a1 + a2 + M_R D. SN = D1 + D2 without any coefficients Answer: A. SN is the sum of each layer's coefficient times thickness (and drainage coefficient), here 0.401x5.3 + 0.138x8.6x1.0 = 3.31. (Key point: Structural number formula) 2. To find a2, which value of E2 is substituted in a2 = 0.249 log(E2) - 0.977? A. 30 psi B. 30 000 psi (30 ksi converted to psi) C. 9000 psi D. 365 000 psi Answer: B. The base modulus is 30 ksi = 30 000 psi, giving a2 = 0.249 log(30000) - 0.977 = 0.138. (Key point: Layer coefficient of base) 3. Which value is used as M_R in the AASHTO equation? A. The base modulus 30 000 psi B. The surface modulus 365 000 psi C. The subgrade modulus E3 = 9 ksi = 9000 psi D. The SN value 3.31 Answer: C. M_R is the roadbed (subgrade) resilient modulus, taken as E3 = 9000 psi. (Key point: Subgrade modulus in equation) 4. With Z_R = -1.282 and S0 = 0.5, what does the reliability term do, and what is the result? A. Adds +0.641 to log W18, giving about 10^7 ESAL B. Subtracts 0.641 from log W18; final capacity about 1.16 x 10^6 ESAL C. Multiplies SN by 0.5; capacity about 3.3 ESAL D. Has no effect; capacity is 10^9 ESAL Answer: B. Z_R x S0 = -0.641 reduces log W18 for safety, and log W18 = 6.066 gives W18 of about 1.16 million ESAL. (Key point: Reliability term and final result) ## Related in Transportation Engineering II - [Flexible design: 4-lane single carriageway, 2000 CVPD](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aflexible-design-four-lane-single-2000) — Transportation Engineering II, chapter 2 - [Flexible pavement life with 140 cvpd traffic](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aflexible-design-life-years-140-cvpd) — Transportation Engineering II, chapter 2 - [Types of rigid pavement; standard axle load](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afour-rigid-pavement-types-standard-axle) — Transportation Engineering II, chapter 2 - [Cumulative standard axle load; rigid pavement advantages](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Acumulative-standard-axle-rigid-advantages) — Transportation Engineering II, chapter 2 - [Asphalt Institute design: 1.888×10³ ESAL/day, 10 years](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aasphalt-institute-design-1-888e3-esal) — Transportation Engineering II, chapter 2 - [Flexible pavement design with 8 CBR values](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Aflexible-design-eight-cbr-340-cvd) — Transportation Engineering II, chapter 2 ## Related in other subjects - [Single equal angle in compression, 50 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Acompression-single-equal-angle-50kn) — Design of Steel and Timber Structure, chapter 6 - [Single-phase Cu cable for three loads](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asingle-phase-cable-three-loads) — Industrial Power Distribution and Illumination, chapter 5 - [Why sheet pile is a flexible structure](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Asheet-pile-flexible-why) — Foundation Engineering, chapter 5 - [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2