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Civil Engineering 2022 Paper II 50 marks Solve

Paper II — Q2

(a) A construction company is planning to purchase an excavation equipment for operating a borrow soil pit that will last 5…

(a)

A construction company is planning to purchase an excavation equipment for operating a borrow soil pit that will last 5 years. The soil can be excavated by equipment 'A' or equipment 'B'. Equipment 'A' has an initial cost of ₹ 1·32 lakh and will have no salvage value at the end of the project. Equipment 'B' has an initial cost of ₹ 0·21 lakh. However, to provide the same capacity, 2 numbers of equipment 'B' are required and their operating cost per year together (for 2 numbers of equipment 'B') will be ₹ 0·18 lakh more than the equipment 'A'. Normal service life for equipment 'B' is 3 years with zero salvage value but a 2-year-old equipment 'B' can likely be sold for ₹ 0·05 lakh. If the interest rate is 15%, which equipment (equipment 'A' or 2 numbers of equipment 'B') should be preferred to purchase from the present worth point of view? The estimated salvage value for equipment 'B' must be included as a cash inflow at the end of year 5. 15 marks

(b)

Design a flexible pavement for a two-lane undivided carriageway using the following data: Subgrade CBR value = 8%, Lane distribution factor = 0·5, Design Life = 15 years, Planning and Construction period = 1·5 years. Present commercial traffic is as under: Bus (Gross Wt. 15T, 200 vehicles/day, Front Axle – Single, Rear Axle – Dual, Growth Rate 4%), Truck (Gross Wt. 20T, 1512 vehicles/day, Front Axle – Single, Rear Axle – Tandem, Growth Rate 8%). As per IRC 37, 2018 the following pavement composition is desired for CBR of 8% subgrade corresponding to different Design traffic: Design traffic (msa): BC / DBM / WMM / GSB (mm) — 5: 30/50/250/150, 10: 30/60/250/200, 20: 30/90/250/200, 30: 40/95/250/200, 40: 40/105/250/200, 50: 40/115/250/200. 20 marks

(c)

The following data were obtained from the spot speed study carried out at a city road. Suggest (i) Speed limit for regulation, (ii) Speed to check geometric design elements, (iii) Lower speed group causing congestion. Speed range (km/h): number of vehicles — 0–10: 0, 10–20: 11, 20–30: 30, 30–40: 105, 40–50: 233, 50–60: 216, 60–65: 68, 65–70: 24, 70–75: 10, >75: 6. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

एक निर्माण कम्पनी एक मिट्टी खदान के प्रचालन के लिए एक खुदाई उपकरण के क्रय की योजना बना रही है जो 5 वर्ष तक रहेगा। मिट्टी उपकरण 'A' या उपकरण 'B' के द्वारा खोदी जा सकती है। उपकरण 'A' की प्रारंभिक लागत ₹ 1·32 लाख है तथा परियोजना के अंत में निस्तारण मूल्य कुछ भी नहीं होगा। उपकरण 'B' की प्रारंभिक लागत ₹ 0·21 लाख है। यद्यपि समान क्षमता के लिए 'B' के दो उपकरणों की आवश्यकता होगी तथा उन दोनों की प्रचालन लागत प्रति वर्ष ('B' के दो उपकरण) उपकरण 'A' से ₹ 0·18 लाख अधिक होगी। उपकरण 'B' की सामान्य सेवा आयु 3 वर्ष है और निस्तारण मान शून्य है लेकिन एक 2 वर्ष पुराना उपकरण 'B' ₹ 0·05 लाख में विक्रय किया जा सकता है। यदि ब्याज दर 15% है, तो कौन-से उपकरण को (उपकरण 'A' या 2 संख्या में उपकरण 'B') वर्तमान मूल्य के दृष्टिकोण से क्रय करने हेतु वरीयता देनी चाहिए? उपकरण 'B' का अनुमानित निस्तारण मूल्य 5 वर्ष के अंत में नकदी अंतवाह के रूप में शामिल किया जाना चाहिए। (15 अंक)

(b)

निम्नलिखित आँकड़ों का उपयोग करते हुए एक अविभाजित दो-लेन मार्ग के लिए एक नम्य कुशिंग का अभिकल्पन कीजिए: सबग्रेड का सी.बी.आर. मान = 8%, लेन वितरण गुणक = 0·5, अभिकल्पन आयु = 15 वर्ष, परिकल्पन तथा निर्माण अवधि = 1·5 वर्ष। वर्तमान व्यावसायिक ट्रैफिक निम्न प्रकार है: बस (सकल भार 15T, 200 वाहन प्रतिदिन, अग्र धुरी – एकल, पश्च धुरी – दोहरी, वृद्धि दर 4%), ट्रक (सकल भार 20T, 1512 वाहन प्रतिदिन, अग्र धुरी – एकल, पश्च धुरी – अग्रपश्च (टैन्डम), वृद्धि दर 8%)। IRC 37, 2018 के अनुसार, सबग्रेड 8% के लिए विभिन्न अभिकल्पित ट्रैफिक के अनुसार निम्नलिखित कुंडिम संयोजन अपेक्षित है: [सारणी दिखाते हुए अभिकल्पित ट्रैफिक (एम.एस.ए.) बनाम परतें]। (15 अंक)

(c)

एक शहर की सड़क पर किए गए स्थानिक गति अध्ययन से निम्नलिखित आँकड़े प्राप्त किए गए। सुझाइए (i) नियमन के लिए गति सीमा, (ii) ज्यामितीय अभिकल्प घटकों की जाँच के लिए गति, (iii) संकुलन करने वाले निम्न गति समूह। [सारणी दिखाते हुए गति परास (km/hr) बनाम वाहनों की संख्या]। (15 अंक)

Q2 of the 2022 UPSC Mains Civil Engineering Paper II, as printed
The question as printed in the 2022 Civil Engineering paper
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The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(a) Table with two columns: 'Time (h)' and 'Ordinate of 4 hr UH (m3/s)'. Rows: 0, 0; 4, 30; 8, 120; 12, 200; 16, 250; 20, 210; 24, 130; 28, 75; 32, 50; 36, 20; 40, 10; 44, 0.

(b) Table: Pavement composition for CBR of 8% subgrade corresponding to different Design traffic. Columns: Design Traffic (msa), BC Wearing Course (mm), DBM Binder Course (mm), WMM Base Course (mm), GSB Sub-base Course (mm). Rows: 5, 30, 50, 250, 150; 10, 30, 60, 250, 200; 20, 30, 90, 250, 200; 30, 40, 95, 250, 200; 40, 40, 105, 250, 200; 50, 40, 115, 250, 200.

(c) Table with 3 columns and 5 rows (including header). Column 1 header: 'Activity' (Hindi: क्रिया). Column 2 header: 'Normal Time (Weeks)' (Hindi: सामान्य समय (सप्ताह)). Column 3 header: 'Normal Cost (₹ in lakh)' (Hindi: सामान्य लागत (₹ लाख में)). Row 1: '1 - 3', '9', '90'. Row 2: '1 - 2', '12', '180'. Row 3: '2 - 3', '18', '210'. Row 4: '3 - 4', '20', '360'.

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) Equipment A versus two units of equipment B: present-worth comparison

The option with the lower present worth of cost is preferred. Both options must deliver the same capacity over the 5-year project, so the comparison covers exactly 5 years. The part of A's operating cost that B also incurs is common to both and cancels out. Only the extra ₹ 0.18 lakh per year that B costs to operate is charged to B.

Interest factors at i = 15 %:

  • (P/F, 15 %, 3) = 1/1.15³ = 0.6575
  • (P/F, 15 %, 5) = 1/1.15⁵ = 0.4972
  • (P/A, 15 %, 5) = (1.15⁵ − 1)/(0.15 × 1.15⁵) = 3.3522

Equipment A

  • First cost ₹ 1.32 lakh at year 0; life covers the whole project; no salvage.
  • PW_A = ₹ 1.320 lakh

Two units of equipment B (cash flows in ₹ lakh)

  • Purchase of 2 units, year 0: 0.42 × 1.0000 = 0.4200
  • Replacement at the end of their 3-year life (salvage then = 0), year 3: 0.42 × 0.6575 = 0.2762
  • Extra operating cost, years 1–5: 0.18 × 3.3522 = 0.6034
  • Resale of the 2-year-old replacement units, 2 × 0.05 (inflow), year 5: −0.10 × 0.4972 = −0.0497

PW_B = 0.42 + 0.2762 + 0.6034 − 0.0497 = ₹ 1.250 lakh

Decision. PW_B (₹ 1.250 lakh) is less than PW_A (₹ 1.320 lakh), so two units of equipment B should be purchased. This saves about ₹ 0.07 lakh in present-worth terms.

The result does not depend on how the resale value is read. Even if the ₹ 0.05 lakh resale value applied to the pair rather than to each unit, PW_B would be ₹ 1.275 lakh, and B would still be preferred.

(b) Flexible pavement design (IRC:37-2018)

1. Design traffic formula

N = 365 × A × [(1 + r)ⁿ − 1]/r × D × F

where:

  • A = commercial vehicles per day at the end of construction = P(1 + r)ˣ, with x = 1.5 years
  • n = 15 years (design life)
  • D = 0.5 (lane distribution factor)
  • F = vehicle damage factor (VDF)

2. Vehicle damage factors

Assumptions:

  • Gross weight is shared between the axles as follows:
  • bus (15 t): front 5 t, rear 10 t
  • truck (20 t): front 5 t, rear tandem 15 t
  • Axle loads are converted by the fourth-power law using the IRC:37-2018 standard axle loads:
  • single axle, single wheel: 65 kN
  • single axle, dual wheel: 80 kN
  • tandem axle: 148 kN

Bus:

  • Front: 49.05 kN → (49.05/65)⁴ = 0.324
  • Rear: 98.1 kN → (98.1/80)⁴ = 2.261
  • VDF = 2.585

Truck:

  • Front: 0.324
  • Rear tandem: 147.15 kN → (147.15/148)⁴ = 0.977
  • VDF = 1.301

3. Cumulative standard axles

  • Bus: P = 200 CVPD, r = 4 %, A = 200 × 1.04^1.5 = 212.1, growth factor (1.04¹⁵ − 1)/0.04 = 20.02, N = 365 × 212.1 × 20.02 × 0.5 × 2.585 = 2.00 msa
  • Truck: P = 1512 CVPD, r = 8 %, A = 1512 × 1.08^1.5 = 1697.0, growth factor (1.08¹⁵ − 1)/0.08 = 27.15, N = 365 × 1697.0 × 27.15 × 0.5 × 1.301 = 10.94 msa
  • Total design traffic = 12.95 ≈ 13 msa

4. Pavement composition (CBR 8 %)

13 msa lies between the 10 msa and 20 msa rows of the given catalogue. Only the DBM thickness differs between these two rows, and interpolating it gives 60 + (90 − 60)(2.95/10) = 68.8 mm, rounded up to 70 mm.

  • BC (wearing course): 30 mm
  • DBM (binder course): 70 mm
  • WMM (base): 250 mm
  • GSB (sub-base): 200 mm
  • Total: 550 mm

Adopting the 20 msa row (DBM 90 mm) would be the conservative choice, particularly because heavily loaded trucks could make the actual VDF higher than assumed. Axle-load surveys should be used to confirm the VDF before construction.

(c) Spot speed study

Frequency table. The total sample is 703 vehicles. The class "> 75" is taken as 75–80 km/h.

  • 0–10 km/h: 0 vehicles (cumulative 0, 0 %)
  • 10–20: 11 (11, 1.56 %)
  • 20–30: 30 (41, 5.83 %)
  • 30–40: 105 (146, 20.77 %)
  • 40–50: 233 (379, 53.91 %)
  • 50–60: 216 (595, 84.64 %)
  • 60–65: 68 (663, 94.31 %)
  • 65–70: 24 (687, 97.72 %)
  • 70–75: 10 (697, 99.15 %)
  • > 75: 6 (703, 100 %)

Percentile speeds are read from the cumulative-percentage curve (the ogive, plotted against the upper limit of each class) by linear interpolation.

(i) Speed limit for regulation: 85th percentile speed

  • 85 % of 703 = 597.6 vehicles, which falls in the 60–65 class.
  • V₈₅ = 60 + (597.6 − 595)/68 × 5 = 60.2 km/h
  • Recommended speed limit: 60 km/h. Only 15 % of drivers exceed this speed, so it is both reasonable and enforceable.

(ii) Speed for checking geometric design elements: 98th percentile speed

  • 98 % of 703 = 689.0 vehicles, which falls in the 70–75 class.
  • V₉₈ = 70 + (689.0 − 687)/10 × 5 = 71.0 km/h
  • Sight distances, superelevation and curve radii should be checked for about 70 km/h.

(iii) Lower speed group causing congestion: 15th percentile speed

  • 15 % of 703 = 105.45 vehicles, which falls in the 30–40 class.
  • V₁₅ = 30 + (105.45 − 41)/105 × 10 = 36.1 km/h
  • Vehicles moving slower than about 36 km/h hold up the stream, causing congestion and unsafe overtaking. This value can serve as a minimum speed limit on the road.
Diagram: Cumulative speed distribution with 15th, 85th and 98th percentile speeds

How this answer will be evaluated

Framework: Present Worth Analysis (Engineering Economics)

What earns full marks: Precise calculations with correct factors, clear logic for equipment selection, accurate MSA derivation, and correct percentile identification.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Determine preferred equipment (A or 2x B) using Present Worth method at 15% interest. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Calculate PW of Equipment A (Cost 1.32L, 5 yrs, 0 salvage)
    • Calculate PW of 2x Equipment B (Cost 0.42L, 3-yr life, 0.05L salvage)
    • Account for B's replacement cycle and operating cost differential (0.18L/yr)
    • Compare final PW values to select the lower cost option

    Loses marks

    • Ignoring the 0.18L/yr operating cost difference
    • Treating B's life as 5 years without replacement logic
    • Omitting the year-5 resale value of the replacement B units

    Earns more

    • Correct application of Present Worth factor (P/A, 15%, 5)
    • Explicit handling of the year-5 resale of the 2-year-old replacement B units
    • Clear tabulation of cash flows for both options

    Extra mark

    • Sensitivity analysis on interest rate
  2. (b) Design flexible pavement thickness for 15-year life using IRC 37, 2018 data. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Calculate Design Traffic (MSA) using growth rates (4% bus, 8% truck)
    • Apply Lane Distribution Factor (0.5) and Planning Period (1.5 yrs)
    • Determine Equivalent Axle Loads (EAL) for Bus and Truck
    • Select pavement layer thicknesses from the provided IRC 37 table

    Loses marks

    • Ignoring growth during the 1.5-year construction period (A = P(1 + r)^x) or adding it to the 15-year design life
    • Ignoring the different growth rates for Bus and Truck
    • Selecting thickness without matching the calculated MSA

    Earns more

    • Correct calculation of cumulative standard axle loads
    • Explicit statement of assumptions for axle load conversion
    • Neat presentation of the final pavement composition

    Extra mark

    • Reference to specific IRC 37, 2018 clause numbers
  3. (c) Determine speed limit, design speed, and congestion group from spot speed data. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • 85th percentile speed (≈ 60 km/h) as the regulatory speed limit
    • 98th percentile speed (≈ 71 km/h) for checking geometric design elements
    • 15th percentile speed (≈ 36 km/h) as the lower speed group causing congestion
    • Show cumulative frequency calculation for each percentile

    Loses marks

    • Confusing 15th and 85th percentile definitions
    • Not using the correct total of 703 vehicles
    • Using the 15th percentile as the regulatory (upper) speed limit

    Earns more

    • Correct plotting or tabulation of cumulative percentage of vehicles
    • Clear distinction between regulatory speed and design speed
    • Logical justification for the chosen speed limit

    Extra mark

    • Graphical representation of the speed distribution curve
What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

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