--- title: "Flexible design: 4-lane single carriageway, 2000 CVPD" subject: "Transportation Engineering II" chapter: 2 key: "semester_7/transportation_engineering_ii:flexible-design-four-lane-single-2000" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/flexible-design-four-lane-single-2000 site: BCT Engineering --- # Flexible design: 4-lane single carriageway, 2000 CVPD Transportation Engineering II · Chapter 2: Highway Pavement - Asked in: 2075 Asoj, 8 marks - Video: [Flexible Pavement Design Solved Problem using IRC 37:2018| Pavement Thickness and Layer Composition](https://www.youtube.com/watch?v=tVWccYaLNbk) — Master Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/flexible-design-four-lane-single-2000 - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-2 ## Question Design the flexible pavement for 4-lane single carriage way road with the following parameters: i) Initial traffic in each direction = 2000 CVPD ii) Design life = 15 years iii) Construction period = 3 years iv) Traffic growth rate = 8% v) Design CBR value = 6% vi) Modulus of elasticity of asphalt concrete surface course = 2500 MPa vii) Modulus of elasticity of bituminous treated base = 1200 MPa viii) Modulus of elasticity of granular subbase course = 125 MPa ix) Axle load distribution of commercial vehicles on the road is as follows: | Axle Load (kN) | No. of Axles (%) | |---|---| | 10 | 15 | | 30 | 15 | | 50 | 20 | | 70 | 30 | | 90 | 10 | | 110 | 10 | ## Short answer Design ESAL = 32.7 × 10⁶. Provide asphalt concrete 125 mm, bituminous treated base 200 mm, granular sub-base 230 mm (total 555 mm) over subgrade CBR 6%. ## Answer #### Given data and assumptions - Four-lane single carriageway, initial traffic 2000 CVPD in each direction (4000 CVPD both ways), growth 8 %, design life 15 years, construction period 3 years, design CBR 6 %. - Moduli: asphalt concrete 2500 MPa, bituminous treated base 1200 MPa, granular sub-base 125 MPa. - Assumptions: each commercial vehicle has 2 axles on average; design lane carries 45 percent of the total two-way traffic (Asphalt Institute, four lanes); standard axle 80 kN and equivalency by the fourth power law. - 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 Design CBR is given as 6 %. Resilient modulus of subgrade: $M_R = 10\times CBR = 10\times 6.0 = 60$ MPa. #### Step 2 - Design traffic (ESAL) Equivalent axle load factors from the axle distribution (average EALF per axle): | Axle load (kN) | % axles | EALF $(L/80)^4$ | % x EALF | |---|---|---|---| | 10 | 15 | 0.0002 | 0.0000 | | 30 | 15 | 0.0198 | 0.0030 | | 50 | 20 | 0.1526 | 0.0305 | | 70 | 30 | 0.5862 | 0.1759 | | 90 | 10 | 1.6018 | 0.1602 | | 110 | 10 | 3.5745 | 0.3574 | | **Total** | 100 | | **0.7270** | ESAL per day at present (design lane) $=4000\times0.45\times2\times0.7270=2617.2$. At opening (3 years): $2617.2\times(1.08)^3=3296.9$ per day. $$ ESAL=365\times3296.9\times\frac{(1.08)^{15}-1}{0.08}=365\times3296.9\times27.152=32.674\times10^6 $$ **Design ESAL = 32.674 $\times 10^6$** #### 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 | 1.84 | 106 | | granular sub-base | 125 | 4.36 | 252 | | Subgrade | 60 | 5.58 | 322 | #### 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: $D_1 \ge SN_{base}/a_1 = 1.84/0.376=124$ mm, so provide **125 mm** (minimum 50 mm). $SN_1 = 1.85$. - Bituminous treated base: $D_2 \ge (SN_{sub}-SN_1)/a_2 = (4.36-1.85)/0.328 = 195$ mm, so provide **200 mm**. $SN_2 = 2.58$. - Granular sub-base: $D_3 \ge (SN_{sg}-SN_1-SN_2)/a_3 = (5.58-1.85-2.58)/0.128 = 228$ mm, so provide **230 mm**. $SN_3 = 1.16$. Check: provided $SN = 1.85+2.58+1.16 = 5.59 \ge 5.58$ (required) - safe. **Answer:** asphalt concrete 125 mm, bituminous treated base 200 mm, granular sub-base 230 mm; total pavement thickness = 555 mm. ```text +-------------------------------------+ | Asphalt concrete 125 mm | +-------------------------------------+ | Bituminous treated base 200 mm | +-------------------------------------+ | Granular sub-base 230 mm | +-------------------------------------+ | Subgrade, CBR 6.0 % | +-------------------------------------+ ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the subgrade resilient modulus obtained from CBR 6 %? A. M_R = 10 x CBR = 60 MPa B. M_R = 6 MPa C. M_R = 600 MPa D. M_R = CBR/10 Answer: A. M_R = 10 CBR (MPa). (Key point: Subgrade modulus) 2. How is the average EALF per axle computed from the axle distribution? A. Sum of (percent of axles x (L/80)^4), giving 0.727 B. Average of axle loads C. Highest EALF D. Sum of percentages Answer: A. The weighted sum of fourth-power factors is 0.727. (Key point: Average EALF) 3. What lane share is assumed for a four-lane single carriageway? A. 45 percent of total two-way traffic B. 75 percent C. 100 percent D. 20 percent Answer: A. The solution assumes 45 percent of the 4000 CVPD two-way traffic. (Key point: Design lane traffic) 4. Why is the present traffic multiplied by (1.08)^3? A. To grow traffic over the 3-year construction period to the year of opening B. To account for design life C. To convert to msa D. To apply a lane factor Answer: A. Traffic is counted now and the road opens after 3 years. (Key point: Construction period growth) 5. How is the thickness of the AC layer determined? A. D1 >= SN required over the base / a1, 1.84/0.376, about 124 mm so 125 mm B. D1 = SN subgrade C. D1 = 50 mm always D. D1 = a1 x 1.84 Answer: A. The surface must supply the SN required to protect the base, divided by its coefficient. 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