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Civil Engineering 2024 Paper I 50 marks Solve

Paper I — Q8

(a) A retaining wall 8 m high, with a smooth vertical back is pushed against a soil mass having c = 50 kN/m², φ = 15° and unit…

(a)

A retaining wall 8 m high, with a smooth vertical back is pushed against a soil mass having c = 50 kN/m², φ = 15° and unit weight 18 kN/m³. It carries a surcharge of 40 kN/m² uniformly on its top surface. Draw the passive pressure distribution diagram and find the point of application of the resultant thrust. 15 marks

(b)

A particular soil failed under a major principal stress of 600 kN/m² with a corresponding minor principal stress of 200 kN/m². If for the same soil, the minor principal stress had been 300 kN/m², determine what the major principal stress would have been if (i) φ = 35° and (ii) φ = 0°. 15 marks

(c)

An inward flow reaction turbine works under a head of 30 m and discharge of 10 m³/s. The speed of runner is 300 r.p.m. At the inlet tip of runner vane, the peripheral velocity of wheel is 0.9√2gH and the radial velocity of flow is 0.3√2gH, where H is the head on the turbine. If the overall efficiency and the hydraulic efficiency of the turbine are 80% and 90% respectively, determine :

(i)

the power developed in kw

(ii)

diameter and width of runner at inlet

(iii)

guide blade angle at inlet

(iv)

inlet angle at runner vane

Assume that the discharge at outlet is radial. 20 marks

हिंदी में प्रश्न पढ़ें

(क) एक मसृण ऊर्ध्वाधर पृष्ठ वाली 8 m ऊँची एक प्रतिधारक भित्ति को एक मृदा संहति जिसका c = 50 kN/m², φ = 15° एवं एकक भार 18 kN/m³ है, के विरुद्ध धक्का दिया जाता है। यह अपने शीर्ष सतह पर 40 kN/m² का सम अधिभार वहन करती है। निष्क्रिय-दाब-वितरण-आरेख बनाइए और परिणामी प्रणोद के लिए प्रयोग बिंदु भी ज्ञात कीजिए। (15 अंक)

(ख) एक विशिष्ट मृदा, 600 kN/m² के एक उच्च मुख्य प्रतिबल और 200 kN/m² के संगत निम्न मुख्य प्रतिबल पर विफल हो जाती है। यदि इसी मृदा नमूने के लिए निम्न मुख्य प्रतिबल 300 kN/m² होता तो निर्धारित कीजिए कि उच्च मुख्य प्रतिबल कितना होता, यदि (i) φ = 35° एवं (ii) φ = 0° है। (15 अंक)

(ग) एक अंतर्मुख प्रवाही प्रतिक्रिया टरबाइन 30 m की दाबोच्चता और 10 m³/s के निस्सरण पर कार्यरत है। चक्राल (रनर) की गति 300 r.p.m है। चक्राल वेन के अंतर्गम अग्र पर चक्र का परिधीय वेग 0.9√2gH एवं प्रवाह का त्रिज्य वेग 0.3√2gH है, जहाँ H टरबाइन पर दाबोच्चता है। यदि टरबाइन की कुल दक्षता एवं द्रवीय (हाइड्रॉलिक) दक्षता क्रमशः: 80% एवं 90% है तो, ज्ञात कीजिए :

(i)

उत्पन्न शक्ति, kw में

(ii)

अंतर्गम पर चक्राल का व्यास और चौड़ाई

(iii)

अंतर्गम पर निर्देशक ब्लेड कोण

(iv)

चक्राल वेन पर अंतर्गम कोण

निर्गम पर विसर्जन त्रिज्यीय मान लीजिए। (20 अंक)

Q8 of the 2024 UPSC Mains Civil Engineering Paper I, as printed
The question as printed in the 2024 Civil Engineering paper
Evaluate your answer to this question → See all 2024 Civil Engineering questions

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) Rankine passive pressure on a smooth vertical wall in c–φ soil with surcharge q:

p_p = K_p(q + γz) + 2c√K_p, where K_p = tan²(45° + φ/2) = tan²52.5° = 1.698 and √K_p = 1.303.

  • At the top (z = 0): p_p = 1.698 × 40 + 2 × 50 × 1.303 = 67.9 + 130.3 = 198.3 kN/m².
  • At the base (z = 8 m): p_p = 1.698 × (40 + 18 × 8) + 130.3 = 312.5 + 130.3 = 442.8 kN/m².

The diagram is a trapezium: a uniform part K_p q + 2c√K_p = 198.3 kN/m² plus a triangle rising to K_p γH = 244.6 kN/m² at the base. Cohesion adds to the passive resistance, so the pressure is compressive over the full height and, unlike the active case, there is no tension zone or crack.

Resultant per metre run:

  • Rectangle: 198.3 × 8 = 1586 kN/m, acting at 4.0 m above the base.
  • Triangle: ½ × 244.6 × 8 = 978 kN/m, acting at 8/3 = 2.67 m above the base.
  • P_p = 2564 kN/m.
  • Moments about the base: ȳ = (1586 × 4 + 978 × 2.67)/2564 = 8953/2564 = 3.49 m above the base (4.51 m below the top).

(b) At failure (Mohr–Coulomb): σ₁ = σ₃N_φ + 2c√N_φ, with N_φ = tan²(45° + φ/2). For the same soil the envelope is fixed, so changing σ₃ by Δσ₃ changes σ₁ by N_φ Δσ₃:

σ₁′ = σ₁ + N_φ(σ₃′ − σ₃) = 600 + 100N_φ.

  • (i) φ = 35°: N_φ = tan²62.5° = 3.69, so σ₁′ = 600 + 369 = 969 kN/m².
  • (ii) φ = 0°: N_φ = 1 and c = (600 − 200)/2 = 200 kN/m², so σ₁′ = 300 + 2 × 200 = 700 kN/m² (the deviator stress stays at 400 kN/m², as expected for a purely cohesive soil).

Note: with φ = 35° the first failure state implies 2c√N_φ = 600 − 200 × 3.69 < 0, i.e. a negative intercept, which is not physically possible; the given stresses actually fit φ ≈ 30° with c = 0. The answer 969 kN/m² follows the question's data as set.

(c) Data: H = 30 m, Q = 10 m³/s, N = 300 rpm, η_o = 0.80, η_h = 0.90; radial discharge at outlet (V_w2 = 0).

  • √(2gH) = √(2 × 9.81 × 30) = 24.26 m/s; u₁ = 0.9 × 24.26 = 21.84 m/s; V_f1 = 0.3 × 24.26 = 7.28 m/s.

(i) Power developed: P = η_o ρgQH = 0.80 × 1000 × 9.81 × 10 × 30 / 1000 = 2354 kW (water power 2943 kW).

Whirl velocity from the hydraulic efficiency: η_h = V_w1u₁/(gH), so V_w1 = 0.90 × 9.81 × 30/21.84 = 12.13 m/s.

(ii) Runner at inlet:

  • D₁ = 60u₁/(πN) = 60 × 21.84/(π × 300) = 1.39 m.
  • b₁ = Q/(πD₁V_f1) = 10/(π × 1.39 × 7.28) = 0.315 m (vane thickness neglected).

(iii) Guide blade angle: tan α = V_f1/V_w1 = 7.28/12.13 = 0.600, so α = 30.96°.

(iv) Runner vane angle at inlet: tan θ = V_f1/(u₁ − V_w1) = 7.28/9.70 = 0.750, so θ = 36.87°. Since V_w1 < u₁, the vane is inclined backwards (θ < 90°).

Diagram: Passive pressure diagram with surcharge and cohesion, Mohr circles at failure, and inlet velocity triangle of an inward-flow reaction turbin

How this answer will be evaluated

Framework: Rational Design Method

What earns full marks: All calculations correct with proper units, complete diagrams, and clear assumptions stated.

Evaluation rubric

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

  1. (a) Passive pressure distribution diagram and point of application of resultant thrust. 15 marks

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

    Must cover

    • K_p = tan²(45° + φ/2) = 1.698 (√K_p = 1.303)
    • p_p = K_p(q + γz) + 2c√K_p: 198.3 kN/m² at the top and 442.8 kN/m² at the base; compressive throughout (no tension zone in the passive case)
    • P_p = 198.3 × 8 + ½ × 244.6 × 8 = 1586 + 978 = 2564 kN/m
    • Point of application ȳ = (1586 × 4 + 978 × 8/3)/2564 = 3.49 m above the base

    Loses marks

    • Introducing a tension crack or subtracting 2c√K_p (both belong to the active case)
    • Incorrect Kp formula for cohesive soil

    Earns more

    • Correctly identifies surcharge contribution
    • Splits the diagram into a rectangle (K_p q + 2c√K_p) and a triangle (K_p γH) to locate the resultant
    • Uses Rankine's theory explicitly

    Extra mark

    • Neatly labelled pressure distribution diagram
  2. (b) Major principal stress for two different friction angle scenarios. 15 marks

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

    Must cover

    • σ₁ = σ₃N_φ + 2c√N_φ with N_φ = tan²(45° + φ/2)
    • Same soil and envelope: σ₁′ = σ₁ + N_φ(σ₃′ − σ₃) = 600 + 100N_φ
    • φ = 35°: N_φ = 3.69, σ₁′ ≈ 969 kN/m²
    • φ = 0: N_φ = 1 (c = 200 kN/m²), σ₁′ = 700 kN/m²

    Loses marks

    • Confuses major and minor principal stress
    • Incorrect application of failure criterion

    Earns more

    • Uses correct relationship between principal stresses
    • Shows intermediate calculation steps

    Extra mark

    • Mentions Mohr circle construction
  3. (c) Turbine performance parameters: power, dimensions, and blade angles. 20 marks

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

    Must cover

    • √(2gH) = 24.26 m/s; u₁ = 21.84 and V_f1 = 7.28 m/s; P = η_o ρgQH = 0.8 × 2943 = 2354 kW
    • Radial discharge: η_h = V_w1u₁/(gH) gives V_w1 = 12.13 m/s
    • D₁ = 60u₁/(πN) = 1.39 m; b₁ = Q/(πD₁V_f1) = 0.315 m
    • Guide blade angle α = tan⁻¹(V_f1/V_w1) = 30.96°; vane angle θ = tan⁻¹[V_f1/(u₁ − V_w1)] = 36.87°

    Loses marks

    • Incorrect efficiency application
    • Ignores radial discharge assumption

    Earns more

    • Uses velocity triangle correctly
    • Applies Euler's turbine equation
    • Considers radial discharge condition

    Extra mark

    • Draws velocity triangle at inlet
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.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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