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 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
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
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 :
the power developed in kw
diameter and width of runner at inlet
guide blade angle at inlet
inlet angle at runner vane
Assume that the discharge at outlet is radial. 20 marks
हिंदी में प्रश्न पढ़ें
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
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
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 :
उत्पन्न शक्ति, kw में
अंतर्गम पर चक्राल का व्यास और चौड़ाई
अंतर्गम पर निर्देशक ब्लेड कोण
चक्राल वेन पर अंतर्गम कोण
निर्गम पर विसर्जन त्रिज्यीय मान लीजिए। 20 marks
Model answer
Written 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.
(a) For a smooth vertical wall, passive case, Kₚ = tan²(45° + φ/2) = tan²(52.5°) = 1.6984.
At depth z: pₚ = Kₚ(q + γz) + 2c√Kₚ, with q = 40 kN/m², c = 50 kN/m², γ = 18 kN/m³.
At top, z = 0: pₚ₀ = 1.6984×40 + 2×50×√1.6984 = 67.94 + 130.32 = 198.26 kN/m².
At base, z = 8 m: pₚ₈ = 1.6984×(40 + 18×8) + 130.32 = 1.6984×184 + 130.32 = 442.83 kN/m².
Diagram: the passive pressure distribution is a linear trapezoid, increasing from 198.26 kN/m² at top to 442.83 kN/m² at base. It is equivalent to a uniform pressure of 198.26 kN/m² plus a triangular pressure of 244.57 kN/m², zero at top and maximum at base.
Resultant thrust per metre length: P = 198.26×8 + 1/2×1.6984×18×8² = 1586.07 + 978.28 = 2564.34 kN/m.
Moment about base: M = 1586.07×4 + 978.28×(8/3) = 6344.27 + 2608.74 = 8953.01 kN·m/m.
Point of application above base: z̄ = M/P = 8953.01/2564.34 = 3.491 m.
Final (a): The resultant thrust is 2564.34 kN/m, acting 3.491 m above the base, i.e. 4.509 m below the top. Valid for drained c–φ soil, horizontal backfill, and no wall friction.
(b) Mohr–Coulomb criterion: σ₁ = σ₃Nφ + 2c√Nφ, where Nφ = tan²(45° + φ/2).
Using the initial failure σ₁ = 600 kN/m², σ₃ = 200 kN/m², and new σ₃′ = 300 kN/m²: σ₁′ = σ₃′Nφ + (600 − 200Nφ) = 600 + (300 − 200)Nφ = 600 + 100Nφ.
(b)(i) φ = 35°: Nφ = tan²(45° + 17.5°) = tan²(62.5°) = 3.69017. σ₁′ = 600 + 100×3.69017 = 969.02 kN/m².
(b)(ii) φ = 0°: Nφ = tan²45° = 1. σ₁′ = 600 + 100×1 = 700 kN/m².
Final (b): (i) 969.02 kN/m²; (ii) 700 kN/m².
(c) Take g = 9.81 m/s², ρ = 1000 kg/m³. Let S = √(2gH) = √(2×9.81×30) = 24.261 m/s.
u₁ = 0.9S = 21.835 m/s, Vf₁ = 0.3S = 7.278 m/s.
(c)(i) Water power = ρgQH = 1000×9.81×10×30 = 2943 kW. Using overall efficiency: Power developed = 0.80×2943 = 2354.4 kW. (Hydraulic power at runner = 0.90×2943 = 2648.7 kW.)
Hydraulic efficiency: ηₕ = (u₁Vw₁ − u₂Vw₂)/(gH). Outlet is radial, so Vw₂ = 0. Thus Vw₁ = ηₕgH/u₁ = 0.90×9.81×30/21.835 = 12.131 m/s.
(c)(iii) Guide blade angle at inlet α: tan α = Vf₁/Vw₁ = 7.278/12.131 = 0.600. α = tan⁻¹0.600 = 30.96°.
(c)(iv) Inlet angle at runner vane β₁: tan β₁ = Vf₁/(u₁ − Vw₁) = 7.278/(21.835 − 12.131) = 0.750. β₁ = tan⁻¹0.750 = 36.87°.
(c)(ii) Diameter at inlet: u₁ = πD₁N/60 D₁ = 60u₁/(πN) = 60×21.835/(π×300) = 1.390 m.
Width at inlet: Q = πD₁b₁Vf₁ b₁ = Q/(πD₁Vf₁) = 10/(π×1.390×7.278) = 0.3146 m.
Final (c): (i) 2354.4 kW; (ii) D₁ = 1.390 m, b₁ = 0.3146 m; (iii) α = 30.96°; (iv) β₁ = 36.87°.
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.
How this answer will be evaluated
Approach
Framework: Geotechnical Engineering (Soil Mechanics) and Fluid Mechanics (Hydraulic Turbines). (a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: All calculations correct with proper diagrams and units
Key points expected
- Calculate passive earth pressure coefficient Kp
- Determine depth of tensile crack z0
- Calculate total passive thrust Pp
- Locate point of application from base
- Apply Mohr-Coulomb failure criterion
- Calculate major stress for phi = 35 deg
- Calculate major stress for phi = 0 deg
- Use correct relationship between principal stresses
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Passive pressure distribution diagram and point of application of resultant thrust. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate passive earth pressure coefficient Kp
- Determine depth of tensile crack z0
- Calculate total passive thrust Pp
- Locate point of application from base
Loses marks
- Using active pressure coefficient Ka
- Ignoring tensile crack depth z0
- Omitting surcharge contribution
Earns more
- Correctly identifies passive state (wall pushed)
- Includes surcharge effect in pressure diagram
- Shows calculation of resultant force components
Extra mark
- Neatly labelled pressure distribution diagram
- (b) Major principal stress for two different friction angle scenarios. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Apply Mohr-Coulomb failure criterion
- Calculate major stress for phi = 35 deg
- Calculate major stress for phi = 0 deg
- Use correct relationship between principal stresses
Loses marks
- Confusing major and minor principal stresses
- Incorrect application of friction angle
- Arithmetic errors in stress calculation
Earns more
- Correctly identifies failure envelope parameters
- Shows step-by-step substitution in formula
Extra mark
- Sketch of Mohr's circle showing failure state
- (c) Power, dimensions, and angles for inward flow reaction turbine. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate power developed using efficiency
- Determine runner diameter and width
- Calculate guide blade angle alpha1
- Calculate inlet vane angle beta1
Loses marks
- Incorrect efficiency application
- Confusing guide blade and vane angles
- Missing velocity triangle relationships
Earns more
- Correct velocity triangle construction
- Uses given velocity coefficients properly
- Consistent units throughout calculation
Extra mark
- Neat velocity triangle diagram at inlet
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