--- title: "Asphalt Institute design: 1500 pcu/day, 5% growth" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:asphalt-institute-design-1500-pcu" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-1500-pcu site: BCT Engineering --- # Asphalt Institute design: 1500 pcu/day, 5% growth Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2073 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/asphalt-institute-design-1500-pcu - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Design a flexible pavement by using Asphalt Institute Method for a two lane two way pavement carrying traffic of 1500pcu/day with growth rate of traffic 5% per annum. The design life is 15 years. The vehicle damage factor is 2.5 and CBR value of sub grade soil is 5%. The modulus of asphalt concrete surface course, bituminous treated base course and granular sub-base course are 2500MPa, 1200MPa and 125 MPa respectively. Assume construction period of 18 months. Draw a neat sketch of pavement layers. ## Short answer Design ESAL = 15.889 × 10⁶ using Asphalt Institute method with subgrade MR = 50 MPa. Required layer thicknesses: asphalt concrete 110 mm, bituminous treated base 180 mm, granular sub-base 280 mm; total pavement thickness 570 mm. ## Answer #### Given data and assumptions - Two-lane two-way road; traffic 1500 pcu/day (all taken as commercial vehicles of damage factor 2.5), growth 5 %, design life 15 years, construction period 18 months (1.5 years), subgrade CBR 5 %. - Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa. - Design lane carries 50 percent of the two-way traffic (Asphalt Institute, two lanes). - Method: Asphalt Institute (AI) thickness design as taught for IOE (and used in the Nepal DoR Pavement Design Guidelines); design reliability 90 percent, terminal serviceability 2.5. #### Step 1 - Subgrade strength Subgrade CBR is given as 5 %. Resilient modulus of subgrade: $M_R = 10\times CBR = 10\times 5.0 = 50$ MPa. #### Step 2 - Design traffic (ESAL) ESAL/day now (design lane) $=1500\times2.5\times0.5=1875$. At opening: $1875\times(1.05)^{1.5}=2017.4$ per day. $$ ESAL=365\times2017.4\times\frac{(1.05)^{15}-1}{0.05}=365\times2017.4\times21.579=15.889\times10^6 $$ **Design ESAL = 15.889 $\times 10^6$** #### Step 3 - Thickness required over each layer The AI charts give the thickness of asphalt concrete (full depth) needed to protect a layer of a given modulus for the design ESAL. The same requirement is expressed here as a structural number $SN$ (AASHTO-93 relation with R = 90 percent, $S_0$ = 0.45, $\Delta PSI$ = 1.7), which closely approximates the AI chart readings (if the chart is supplied in the exam, read the thickness from it and use the same layer-coefficient steps below). | Protected layer | Modulus (MPa) | Required SN | Full-depth AC equivalent at $a_1=0.44$ (mm) | |---|---|---|---| | bituminous treated base | 1200 | 1.62 | 94 | | granular sub-base | 125 | 3.90 | 225 | | Subgrade | 50 | 5.35 | 309 | #### Step 4 - Layer coefficients Using $a_1=0.323\log_{10}E-1.42$, $a_2=0.249\log_{10}E-0.977$, $a_3=0.227\log_{10}E-0.839$ with $E$ in psi: - Asphalt concrete (2500 MPa): $a_1 = 0.376$ - bituminous treated base (1200 MPa): $a_2 = 0.328$ - granular sub-base (125 MPa): $a_3 = 0.128$ #### Step 5 - Layer thicknesses - Asphalt concrete: $D_1 \ge SN_{base}/a_1 = 1.62/0.376=110$ mm, so provide **110 mm** (minimum 50 mm). $SN_1 = 1.63$. - Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (3.90-1.63)/0.328 = 176$ mm, so provide **180 mm**. $SN_2 = 2.32$. - Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (5.35-1.63-2.32)/0.128 = 279$ mm, so provide **280 mm**. $SN_3 = 1.41$. Check: provided $SN = 1.63+2.32+1.41 = 5.36 \ge 5.35$ (required) - safe. **Answer:** asphalt concrete 110 mm, bituminous treated base 180 mm, granular sub-base 280 mm; total pavement thickness = 570 mm. ```text +-------------------------------------+ | Asphalt concrete 110 mm | +-------------------------------------+ | Bituminous treated base 180 mm | +-------------------------------------+ | Granular sub-base 280 mm | +-------------------------------------+ | Subgrade, CBR 5.0 % | +-------------------------------------+ ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the present design-lane ESAL per day found? A. 1500 x 0.5 = 750 B. 1500 x 2.5 x 0.5 = 1875 C. 1500 x 2.5 = 3750 D. 1500 x 2.5 x 1.05 = 3937 Answer: B. PCU is multiplied by the vehicle damage factor 2.5 to get ESAL, then by 0.5 for the design lane. (Key point: Design lane ESAL per day) 2. In ESAL = 365 x 2017.4 x [(1.05)^15 - 1]/0.05, what does the term [(1.05)^15 - 1]/0.05 = 21.579 represent? A. Growth during the 18-month construction B. The vehicle damage factor C. The sum of annual growth over the 15-year design life D. The share of traffic in the design lane Answer: C. It is the cumulative growth factor that adds up the yearly traffic, growing at 5 %, over 15 years. (Key point: Cumulative growth factor) 3. Why is 1875 multiplied by (1.05)^1.5? A. To bring traffic to the level at opening after 18 months of construction B. To bring traffic to the end of design life C. To convert 1.5 lanes to one lane D. To apply the damage factor Answer: A. The construction period is 1.5 years, so the traffic at opening is 1875 x 1.05^1.5 = 2017.4 per day. (Key point: Construction period) 4. Why is the design considered safe after choosing 110, 180 and 280 mm layers? A. Because the total thickness exceeds 500 mm B. Because the asphalt layer exceeds 50 mm C. Because each layer thickness is a multiple of 10 mm D. Because the provided SN (5.36) is at least the required subgrade SN (5.35) Answer: D. The sum of layer SNs (1.63 + 2.32 + 1.41 = 5.36) meets or exceeds the SN required to protect the subgrade. 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