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Civil Engineering 2023 Paper I 50 marks Solve

Paper I — Q8

(a) At a site, fine sand exists to a depth of 10 m and below this lies a soft clay layer 7.0 m thick. Water table is 4.0 m below…

(a)

At a site, fine sand exists to a depth of 10 m and below this lies a soft clay layer 7.0 m thick. Water table is 4.0 m below the ground surface. Saturated unit weight of sand is 20.0 kN/m³ and the wet unit weight above the water table is 18 kN/m³. The water content of the normally consolidated clay is 42%, liquid limit is 46% and the specific gravity of the solid particles is 2.75. The proposed construction will transmit a net stress of 130 kN/m² at the centre of the clay layer. Find the average settlement of the clay layer. 15 marks

(b)

A strip footing of width 2.8 m as shown in the figure is founded at a depth of 2.5 m below the ground surface in a C–φ soil (φ = 30°, γ = 18 kN/m³, C = 40 kN/m²). Water table is at a depth of 6 m below the ground surface. The average moist weight of soil above the water table is 18 kN/m³. Determine the ultimate bearing capacity, net ultimate bearing capacity, net allowable bearing pressure and the load/m for a factor of safety of 2.5. Use the general shear failure theory of Terzaghi. Given : For φ = 30°, Nc = 37.2, Nq = 22.5, Nγ = 19.7. What will be the percent decrease in ultimate bearing capacity if during the flooding, water level rises 2 m above ground surface ? 15 marks

(c)
(i)

Water at 20°C flows through a pipe of inlet diameter of 10 cm and passes further through a circular nozzle of diameter 2.5 cm, exits into the air as a jet, and strikes a vertical plate as shown in the figure. A force, F = 100 N is required to hold the plate stationary. Assuming steady, frictionless, one-dimensional flow and densities of water and mercury as 1000 kg/m³ and 13550 kg/m³ respectively, answer the following : Determine the velocities at sections ① and ②.

(ii)

Determine the mass flow rate of water.

(iii)

Determine the mercury manometer reading 'h'. [Figure: a mercury U-tube manometer connects the taps at sections ① and ②; h is the difference in mercury levels] 20 marks

हिंदी में प्रश्न पढ़ें

(क) एक स्थल पर, 10 m गहराई तक महीन बालू है और इसके नीचे 7.0 m मोटी मृदु मृत्तिका परत है । भौम जल स्तर भूमि तल से 4.0 m नीचे है । बालू का संतृप्त एकक भार 20.0 kN/m³ है और भौम जल स्तर से ऊपर आर्द्र एकक भार 18 kN/m³ है । सामान्य रूप से संघनित मृत्तिका का जलांश 42%, द्रव सीमा 46% है और ठोस कणों का विशिष्ट घनत्व 2.75 है । प्रस्तावित निर्माण, मृत्तिका परत के मध्य पर 130 kN/m² का निवल प्रतिबल प्रेषित करेगा । मृत्तिका परत का औसत निषदन ज्ञात कीजिए । (15 अंक)

(ख) 2.8 m चौड़ी एक पट्टी पाद (फुटिंग) को चित्र में दर्शाए अनुसार एक C–φ मृदा (φ = 30°, γ = 18 kN/m³, C = 40 kN/m²) में भूमि तल से 2.5 m नीचे आधारित किया गया है । भौम जल स्तर भूमि तल से नीचे 6 m गहराई पर है । भौम जल स्तर से ऊपर मृदा का औसत आर्द्र भार 18 kN/m³ है । 2.5 के एक सुरक्षा गुणक के लिए चरम धारण क्षमता, निवल चरम धारण क्षमता, निवल अनुज्ञेय धारण दाब और भार/मीटर निर्धारित कीजिए । टेरज़ाघी के सामान्य अपरूपण विफलन सिद्धांत का उपयोग कीजिए । प्रदत्त : φ = 30° के लिए, Nc = 37.2, Nq = 22.5, Nγ = 19.7 । यदि बाढ़ के दौरान, जल स्तर भूमि तल से 2 m ऊपर हो जाता है तो चरम धारण क्षमता में प्रतिशत कमी क्या होगी ? (15 अंक)

(i)

(ग) 10 cm के अन्तर्गम व्यास के एक पाइप से 20°C पर जल प्रवाहित है और आगे यह 2.5 cm व्यास के एक वृत्ताकार नोज़ल से गुजरते हुए वायु में एक जैट की तरह बाहर निकलता है और चित्र में दर्शाए अनुसार एक ऊर्ध्वाधर प्लेट से टकराता है । प्लेट को स्थिर रखने के लिए एक बल, F = 100 N आवश्यक है । प्रवाह को अपरिवर्ती, घर्षण रहित, एक-विमीय प्रवाह और जल तथा पारे के घनत्व क्रमशः 1000 kg/m³ और 13550 kg/m³ मानते हुए, निम्नलिखित के उत्तर दीजिए : परिच्छेद ① और ② पर वेग निर्धारित कीजिए ।

(ii)

जल की द्रव्यमान प्रवाह दर निर्धारित कीजिए ।

(iii)

पारा मैनोमीटर पाठ्यांक 'h' निर्धारित कीजिए । [चित्र: परिच्छेद ① और ② की टोंटियों को जोड़ने वाला पारे का U-नली मैनोमीटर; h पारे के तलों का अंतर है] (20 अंक)

Q8 of the 2023 UPSC Mains Civil Engineering Paper I, as printed
The question as printed in the 2023 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.

(b) A cross-sectional diagram showing a pile group embedded in a three-layer cohesive soil profile:

  • Ground level and water table: The ground level is at the top with a groundwater table symbol located at the ground surface.
  • Soil strata (depths shown on the vertical 'Depth (m)' axis on the left):
  1. Top layer: Firm Clay of thickness 3 m (from depth 0 to 3 m), with properties Cu = 50 kN/m^2, alpha = 0.9, and gamma = 18 kN/m^3.
  2. Middle layer: Soft Clay of thickness 9 m (from depth 3 m to 12 m), with properties Cu = 30 kN/m^2, alpha = 1.0, and gamma = 16 kN/m^3.
  3. Bottom layer: Stiff Clay of thickness 5 m (from depth 12 m to 17 m), with properties Cu = 90 kN/m^2, alpha = 0.5, and gamma = 20 kN/m^3.
  • Pile group:
  • A pile cap is embedded in the top layer, with its base (cutoff level) at a depth of 2 m below the ground level (indicated by a vertical dimension of 2 m).
  • A column extends upwards from the pile cap above the ground level.
  • Three piles of the group are shown in elevation, extending from the cutoff level (depth 2 m) to the base of the stiff clay layer (depth 17 m), giving a total embedded pile length of 15 m (1 m in Firm Clay, 9 m in Soft Clay, and 5 m in Stiff Clay).
  • Labelled on the right as 'Pile group'.

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) Settlement of the normally consolidated clay layer

Clay saturated: e₀ = wG_s = 0.42 × 2.75 = 1.155. γ_sat = (G_s + e₀)γ_w/(1 + e₀) = 3.905 × 9.81/2.155 = 17.78 kN/m³, so γ′ = 17.78 − 9.81 = 7.97 kN/m³.

Effective stress at the clay centre (13.5 m): 18 × 4 + (20 − 9.81) × 6 + 7.97 × 3.5 = 72 + 61.14 + 27.90 = 161.0 kN/m².

For a normally consolidated clay, C_c = 0.009(LL − 10) = 0.009 × 36 = 0.324 (Terzaghi–Peck).

S = [C_c H/(1 + e₀)] log₁₀[(σ′₀ + Δσ)/σ′₀] = (0.324 × 7.0/2.155) × log₁₀(291.0/161.0) = 1.0525 × 0.257 = 0.27 m (about 270 mm), taking the centre-of-layer stress as representative of the whole layer.

(b) Strip footing by Terzaghi's general shear theory

γ = 18 kN/m³, D_f = 2.5 m, B = 2.8 m. The water table is 3.5 m below the base, more than B, so it does not affect the failure zone. Surcharge q = 18 × 2.5 = 45 kN/m².

q_u = cN_c + qN_q + 0.5γBN_γ = 40 × 37.2 + 45 × 22.5 + 0.5 × 18 × 2.8 × 19.7 = 1488 + 1012.5 + 496.4 = 2997 kN/m².

Net ultimate q_nu = 2997 − 45 = 2952 kN/m². Net allowable pressure = 2952/2.5 = 1181 kN/m². Load per metre = 1181 × 2.8 = 3306 kN/m (net). If the gross allowable pressure 1181 + 45 = 1226 kN/m² is used, the load is 3432 kN/m.

Flooding (water 2 m above ground): the whole soil is submerged. Using γ′ = 18 − 9.81 = 8.19 kN/m³ (taking 18 kN/m³ as the saturated weight), q = 8.19 × 2.5 = 20.5 kN/m². q_u = 1488 + 20.5 × 22.5 + 0.5 × 8.19 × 2.8 × 19.7 = 1488 + 460.7 + 225.9 = 2175 kN/m².

Decrease = (2997 − 2175)/2997 × 100 = 27.4%.

Diagram: Strip footing in c-phi soil: normal water table and flooded condition

(c) Jet on a plate with a venturi-type nozzle

The jet strikes the plate normally and loses all its horizontal momentum, so F = ρA₂V₂². A₂ = π × 0.025²/4 = 4.909 × 10⁻⁴ m², so V₂² = 100/(1000 × 4.909 × 10⁻⁴) = 203.7 and V₂ = 14.27 m/s.

(i) Continuity: V₁ = V₂(D₂/D₁)² = 14.27 × (1/4)² = 0.892 m/s, and V₂ = 14.27 m/s.

(ii) ṁ = ρA₂V₂ = 1000 × 4.909 × 10⁻⁴ × 14.27 = 7.01 kg/s (also F/V₂ = 100/14.27).

(iii) Bernoulli from ① to ② at the same level, with p₂ = 0 gauge at the free jet: p₁ = ½ρ(V₂² − V₁²) = 500 × (203.7 − 0.80) = 101.5 kPa. The manometer gives p₁ − p₂ = (ρ_Hg − ρ_w)gh, so h = 101 460/(12 550 × 9.81) = 0.824 m of mercury.

Diagram: Pipe, nozzle and jet striking a plate, with a mercury manometer across sections 1 and 2

How this answer will be evaluated

Framework: Terzaghi's Bearing Capacity Theory, 1D Consolidation Theory, Bernoulli's Equation

What earns full marks: All steps shown with correct formulas, units, and final values; clear sketches.

Evaluation rubric

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

  1. (a) Find the average consolidation settlement of the 7 m normally consolidated clay layer. 15 marks

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

    Must cover

    • e₀ = wG_s = 1.155 and γ′_clay = 7.97 kN/m³
    • σ′₀ at clay centre = 161 kN/m²
    • C_c = 0.009(LL − 10) = 0.324
    • S = C_cH/(1 + e₀) log[(σ′₀ + Δσ)/σ′₀] ≈ 0.27 m

    Loses marks

    • Using total stress or γ_sat of the sand without subtracting γ_w below the water table.

    Earns more

    • Marks for e₀, effective stress at mid-depth, C_c and settlement.

    Extra mark

    • Stating that mid-layer stress is taken as representative.
  2. (b) Find the ultimate, net ultimate and net allowable pressure and the load per metre for the strip footing; then the percentage decrease on flooding. 15 marks

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

    Must cover

    • q_u = 1488 + 1012.5 + 496.4 = 2997 kN/m²
    • q_nu = 2952 kN/m²; net allowable 1181 kN/m²; load 3306 kN/m
    • Flooded case with γ′ = 8.19 kN/m³: q_u = 2175 kN/m²
    • Decrease = 27.4%

    Loses marks

    • Using γ′ in the dry case or leaving out the surcharge reduction on flooding.

    Earns more

    • Marks for each bearing-capacity term, the safe load and the flooded-case comparison.

    Extra mark

    • Noting that the water table at 6 m does not affect the failure zone.
  3. (c) Find the velocities, the mass flow rate and the mercury manometer reading. 20 marks

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

    Must cover

    • F = ρA₂V₂², so V₂ = 14.27 m/s
    • V₁ = 0.892 m/s by continuity
    • ṁ = 7.01 kg/s
    • p₁ = ½ρ(V₂² − V₁²) = 101.5 kPa; h = 0.824 m of mercury

    Loses marks

    • Using ρ_Hg alone or ignoring the water column in the limbs.

    Earns more

    • Marks for momentum, continuity, mass flow, Bernoulli and the manometer equation.

    Extra mark

    • Using (ρ_Hg − ρ_w) in the manometer equation.
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