Civil Engineering 2024 Paper I 50 marks Solve

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 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

(a) 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 बहन करती है। निष्क्रिय-दाब-वितरण-आरेख बनाइए और परिणामी प्रणोद के लिए प्रयोग बिंदु भी ज्ञात कीजिए। (15 marks) (b) 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 marks) (c) 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 marks)

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Approach

Solve this multi-part numerical problem by allocating approximately 30% time to part (a) passive earth pressure, 30% to part (b) Mohr-Coulomb failure analysis, and 40% to part (c) turbine calculations. Begin each part with the relevant governing equation, show complete derivations with proper units, draw required diagrams to scale with clear labeling, and conclude with physically verified numerical answers.

Key points expected

  • Part (a): Calculate passive earth pressure coefficient Kp using Rankine theory, determine pressure distribution with surcharge component, draw trapezoidal pressure diagram, and locate centroid for resultant thrust application point
  • Part (b): Establish Mohr-Coulomb failure envelope from given stress state, determine cohesion c and friction angle φ, then calculate major principal stress for modified minor principal stress conditions for both φ = 35° and φ = 0° cases
  • Part (c)(i): Calculate power developed using P = η₀ × ρgQH with proper unit conversion to kW
  • Part (c)(ii): Determine inlet diameter from peripheral velocity u₁ = πD₁N/60 and width from continuity equation Q = πD₁B₁Vf₁
  • Part (c)(iii): Find guide blade angle α from tanα = Vf₁/u₁ using velocity triangle geometry
  • Part (c)(iv): Calculate runner inlet angle β from tanβ = Vf₁/(u₁ - Vw₁) using hydraulic efficiency to find Vw₁
  • Verify all calculations: passive pressure increases with depth, failure envelope is consistent, turbine velocities satisfy √2gH relationships, and radial discharge implies Vw₂ = 0

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies Rankine passive earth pressure theory for part (a) with proper Kp formula; establishes Mohr-Coulomb failure criterion correctly in part (b) distinguishing between total and effective stress; applies Euler turbine equations and velocity triangle relationships accurately for part (c) with clear understanding of radial discharge conditionUses correct basic formulas but makes minor errors in passive pressure coefficient application or confuses active/passive conditions; establishes failure envelope with calculation errors; applies turbine formulas with some velocity triangle confusionApplies active earth pressure instead of passive; fails to establish Mohr-Coulomb envelope or uses wrong failure theory; fundamentally misunderstands reaction turbine velocity components or applies centrifugal turbine equations
Numerical accuracy20%10All calculations precise to 2-3 significant figures: Kp = 1.70, passive pressure at base = 617 kN/m², resultant at 3.08 m from base; part (b) c = 173.2 kN/m², σ₁ = 937 kN/m² (φ=35°), σ₁ = 600 kN/m² (φ=0°); part (c) power = 2354 kW, D₁ = 1.27 m, B₁ = 0.296 m, α = 18.4°, β = 24.8°Correct methodology with arithmetic errors in 1-2 parts; final answers within 10% of correct values; minor unit conversion errors (kN vs kPa)Major calculation errors exceeding 20% deviation; wrong formulas leading to meaningless results; consistent unit errors (g = 9.81 vs 10 m/s² not compensated)
Diagram quality20%10Part (a) shows trapezoidal passive pressure distribution with clear labels: σp at top = 114 kN/m², σp at bottom = 503 kN/m² (or combined with surcharge 114+40Kp and 503+40Kp if surcharge converted), resultant vector marked with h/3 from base for triangular component; part (c) velocity triangles drawn to scale showing u₁, V₁, Vr₁, Vf₁, Vw₁ with angles α and β clearly markedDiagrams present but poorly scaled or missing key labels; pressure diagram shows correct shape but wrong values; velocity triangles drawn but angles not markedMissing required passive pressure diagram; velocity triangles absent or completely wrong; diagrams without any labels or dimensions
Step-by-step derivation20%10Each part shows complete derivation: (a) Kp = tan²(45+φ/2), σp = Kpσv + 2c√Kp, integration for resultant; (b) Mohr circle construction or direct Mohr-Coulomb equations to find c, then σ₁ = σ₃tan²(45+φ/2) + 2ctan(45+φ/2); (c) sequential calculation of velocities, power, dimensions, angles with explicit formula substitutionShows key steps but skips some algebraic manipulation; jumps from formula to answer without intermediate values; partial derivations in some partsOnly final answers with no working; or incorrect derivations with formula misapplication; missing essential steps like velocity triangle construction
Practical interpretation20%10Interprets passive pressure application for basement walls and bridge abutments; explains why φ=0° represents undrained clay condition relevant to Indian soft soil deposits; discusses turbine design implications—specific speed calculation, cavitation checks, and comparison with standard Francis turbine ranges for Indian hydropower projects like Nathpa JhakriBrief mention of practical relevance without specific examples; generic statements about soil mechanics or turbine applications; no connection to Indian engineering contextNo practical interpretation; purely mathematical solution; or incorrect physical interpretation (e.g., suggesting passive pressure for retaining wall design without safety factor)

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