--- title: "AASHTO design: find initial commercial vehicles" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:aashto-flexible-design-initial-cvpd" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/aashto-flexible-design-initial-cvpd site: BCT Engineering --- # AASHTO design: find initial commercial vehicles Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2080 Bhadra, 8 marks - Video: [Civil PE Exam – Transportation – Find the Accumulated Equivalent Single-Axle Load (ESAL)](https://www.youtube.com/watch?v=eRc_YSnhoZM) — Civil Engineering Academy - 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-flexible-design-initial-cvpd - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question The flexible pavement was designed for two lane two way road as shown in figure. The drainage coefficient for base layer is 1, layer coefficient for surface and base layer are α1 = 0.323 log10 E1 − 1.42, and α2 = 0.249 log10 E2 − 0.977, where E1 and E2 are in PSi. Determine the initial commercial vehicle per day, the road was designed for. Traffic growth rate = 6%, construction period = 2 yrs, design life = 10 yrs, VDF = 2.5. [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 SN=3.18, solving AASHTO equation gives W18=900,251 ESALs. With growth factor 13.181 (10 yr, 6%), VDF=2.5, lane factor 0.5, opening traffic is 150 CVPD; back-casting 2 yr construction at 6% yields initial 133 CVPD at survey. ## Answer #### Given and assumptions Surface 5.3 in ($E_1=365$ ksi), base 8.6 in ($E_2=30$ ksi, $m_2=1$), subgrade $E_3=9$ ksi $=M_R=9000$ psi. Reliability, standard deviation and $\Delta PSI$ are not given for this part; the values of the companion problem are used: $R=90\%$ ($Z_R=-1.282$), $S_0=0.5$, $\Delta PSI=2.0$. Two-lane two-way road: design lane carries 50 percent of the total. #### Step 1 - Layer coefficients and SN $$ \begin{aligned} a_1&=0.323\log_{10}(365000)-1.42=0.377\\ a_2&=0.249\log_{10}(30000)-0.977=0.138\\ SN&=a_1D_1+a_2D_2m_2=0.377\times5.3+0.138\times8.6\times1=3.18 \end{aligned} $$ #### Step 2 - ESALs the section can carry (AASHTO-93) $$ \log_{10}W_{18}=Z_RS_0+9.36\log_{10}(SN+1)-0.20+\frac{\log_{10}[\Delta PSI/2.7]}{0.4+1094/(SN+1)^{5.19}}+2.32\log_{10}M_R-8.07 $$ Substituting $SN=3.18$, $M_R=9000$: $\log_{10}W_{18}=5.954$, so $W_{18}=900251$ ESALs. #### Step 3 - Initial commercial vehicles per day Growth factor for 10 years at 6 percent: $\dfrac{(1.06)^{10}-1}{0.06}=13.181$. $$ W_{18}=365\times A\times0.5\times VDF\times GF \ \Rightarrow\ A=\frac{900251}{365\times0.5\times2.5\times13.181}=149.7\ \text{CVPD} $$ This is the traffic in the year the road opens. The construction period is 2 years, so the traffic counted at the start is $$ A_0=\frac{149.7}{(1.06)^2}=133.2\ \text{CVPD} $$ **Answer:** the pavement was designed for an initial traffic of about **133 commercial vehicles per day** (at the time of the survey), which grows to about 150 CVPD at opening. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is SN computed for the section? A. SN = a1 D1 + a2 D2 m2 (= 3.18) B. SN = D1 + D2 C. SN = E1 + E2 D. SN = M_R/1500 Answer: A. 0.377 x 5.3 + 0.138 x 8.6 x 1 = 3.18. (Key point: Structural number) 2. Layer coefficient a1 uses E in: A. psi (365 ksi = 365000 psi) B. MPa C. kg/cm2 D. ksf Answer: A. The given formulas take E in psi, so 365 ksi becomes 365000 psi. (Key point: Layer coefficient) 3. What does the AASHTO equation give from SN and M_R? A. The ESALs W18 the section can carry (about 900,000) B. The thickness of base C. The CBR D. The VDF Answer: A. Substituting SN = 3.18 and M_R = 9000 gives W18 of about 900,251. (Key point: Capacity to ESAL) 4. How is the initial CVPD at opening obtained? A. A = W18 / (365 x 0.5 x VDF x growth factor) = 149.7 B. A = W18/365 C. A = W18 x 2.5 D. A = W18 x 13.181 Answer: A. The design lane factor is 0.5 for a two-lane road and the growth factor is 13.181 for 10 years at 6 percent. (Key point: Initial traffic) 5. Why is 149.7 divided by (1.06)^2 to give 133.2? A. To get traffic at the time of survey, 2 years before opening B. To account for lane factor C. To reduce the VDF D. To convert to msa Answer: A. The 149.7 is traffic at opening; discounting back over the 2-year construction gives the initial survey traffic. 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