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Civil Engineering 2023 Paper I 50 marks Solve

Paper I — Q6

(a) A flow of 9.0 m³/s occurs in a long rectangular channel of 3.0 m width with 1.5 m depth of water flow. There is a smooth…

(a)
(i)

A flow of 9.0 m³/s occurs in a long rectangular channel of 3.0 m width with 1.5 m depth of water flow. There is a smooth constriction in the channel to 2.0 m width in the downstream direction. Answer the following : What depths are to be expected in and just upstream of the constriction, if losses are neglected ?

(ii)

Classify the gradually varied flow profile upstream of the constriction, with proper justification. 15 marks

(b)
(i)

A two-dimensional incompressible flow field is given by V = 2xy î + (x² − y²) ĵ, where î and ĵ are the unit vectors along x and y axes, respectively. Answer the following : Determine the magnitude and the angle the velocity vector makes with x-axis at x = 3 m and y = 1 m.

(ii)

Is the flow physically possible ? If so, determine an expression for stream function.

(iii)

What is the discharge between the streamlines passing through (1, 0) and (0, 1) ?

(iv)

Is the flow irrotational ? Justify your answer with appropriate reasons. 15 marks

(c)

A retaining wall is shown in the figure below : [Figure: smooth vertical wall retaining Layer 1 above the ground water table, 3 m thick, γ = 17 kN/m³, φ′ = 28°, C = 0; Layer 2 below the ground water table, 4 m thick, γsat = 20 kN/m³, φ′ = 35°, C = 0; z measured downward from the top] Assuming that the wall can yield sufficiently, determine the Rankine active force per unit length of the wall and also determine the location of the resultant line of action. 20 marks

हिंदी में प्रश्न पढ़ें
(i)

(क) 3.0 m चौड़ी एक लम्बी आयताकार वाहिका में 1.5 m की जल प्रवाह की गहराई पर 9.0 m³/s का एक प्रवाह होता है । वाहिका में, अनुप्रवाह की दिशा में 2.0 m की चौड़ाई तक का एक मसृण संकुचन है । निम्नलिखित के उत्तर दीजिए : यदि हानियाँ नगण्य हैं, तो संकुचन में और संकुचन के ठीक प्रतिप्रवाह पर प्रत्याशित गहराइयाँ क्या हैं ?

(ii)

संकुचन के प्रतिप्रवाह पर क्रमशः-परिवर्ती प्रवाह प्रोफाइल का वर्गीकरण उचित औचित्य देते हुए कीजिए । (15 अंक)

(ख) एक द्वि-विमीय असंपीड्य प्रवाह क्षेत्र V = 2xy î + (x² − y²) ĵ द्वारा दिया गया है, जहाँ î और ĵ क्रमशः x और y अक्षों के साथ एकक सदिश हैं । निम्नलिखित के उत्तर दीजिए :

(i)

x = 3 m और y = 1 m पर वेग सदिश का परिमाण और इसके द्वारा x-अक्ष के साथ बनाए जाने वाले कोण का निर्धारण कीजिए ।

(ii)

क्या प्रवाह भौतिक रूप में सम्भव है ? यदि हाँ, तो धारा फलन का व्यंजक निर्धारित कीजिए ।

(iii)

(1, 0) और (0, 1) से गुज़रने वाली धारा रेखाओं के बीच निस्सरण क्या है ?

(iv)

क्या प्रवाह अघूर्णी है ? उचित कारणों के साथ अपने उत्तर का औचित्य सिद्ध कीजिए । (15 अंक)

(ग) एक प्रतिधारक भित्ति नीचे चित्र में दर्शाई गई है : [चित्र: चिकनी ऊर्ध्वाधर भित्ति; परत 1, भौम जल स्तर के ऊपर, 3 m मोटी, γ = 17 kN/m³, φ′ = 28°, C = 0; परत 2, भौम जल स्तर के नीचे, 4 m मोटी, γsat = 20 kN/m³, φ′ = 35°, C = 0; z ऊपर से नीचे की ओर] यह मानते हुए कि भित्ति का पराभव पर्याप्त रूप से हो सकता है, रेन्किन का सक्रिय बल भित्ति की प्रति एकक लम्बाई पर ज्ञात कीजिए और क्रिया की परिणामी रेखा की स्थिति भी ज्ञात कीजिए । (20 अंक)

Q6 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.

(c) A cross-section of a retaining wall with a vertical face. The wall retains two soil layers. The total height of the wall is 7 m, divided into two sections. The top section is 3 m high, and the bottom section is 4 m high. The ground water table is located at the interface between the two layers, 3 m below the top of the wall. Layer 1 (top layer, 0 to 3 m depth): Unit weight gamma = 17 kN/m3, friction angle phi' = 28 degrees, cohesion C = 0. Layer 2 (bottom layer, 3 m to 7 m depth): Saturated unit weight gamma_sat = 20 kN/m3, friction angle phi' = 35 degrees, cohesion C = 0. A vertical axis labeled 'z' is shown on the left side, pointing downwards.

A vertical cross-section of a retaining wall. The wall is a vertical line on the left. The backfill soil is divided into two horizontal layers. A vertical dimension line on the left indicates the depth 'z' from the top surface. The top layer, labeled 'Layer 1', has a thickness of 3 m. Its properties are listed as: gamma = 17 kN/m^3, phi' = 28 degrees, C = 0. A horizontal dashed line at the bottom of Layer 1 is labeled 'Ground Water Table' with a standard water level symbol. The bottom layer, labeled 'Layer 2', has a thickness of 4 m. Its properties are listed as: gamma_sat = 20 kN/m^3, phi' = 35 degrees, C = 0. The total height of the wall is 7 m.

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) Channel constriction

Take g = 9.81 m/s², a horizontal bed and no loss. In the 3 m channel q₁ = 9/3 = 3 m²/s, V = 2 m/s, so E₁ = 1.5 + 2²/(2 × 9.81) = 1.704 m. Fr = 2/√(9.81 × 1.5) = 0.52, so the approach flow is subcritical.

(i) In the 2 m throat q₂ = 4.5 m²/s. Critical depth y_c = (q₂²/g)^(1/3) = (20.25/9.81)^(1/3) = 1.273 m and E_min = 1.5 y_c = 1.910 m. Since E₁ = 1.704 m is less than 1.910 m, the throat cannot pass the flow at the existing energy; it chokes, and critical flow occurs in it. Depth in the constriction = 1.27 m.

The upstream energy rises to E_min: y + 3²/(2 × 9.81 y²) = 1.910, i.e. y + 0.4587/y² = 1.910. The subcritical root is y = 1.762 m. Depth just upstream = 1.76 m (a rise of 0.26 m over the original 1.5 m).

(ii) Taking the long-channel depth of 1.5 m as normal depth y_n, and y_c = (3²/9.81)^(1/3) = 0.972 m in the 3 m reach, y_n > y_c, so the slope is mild. Just upstream of the constriction y = 1.762 m is above y_n. For y > y_n, S_f < S₀, and for y > y_c, Fr < 1, so dy/dx = (S₀ − S_f)/(1 − Fr²) is positive: depth increases downstream. This is an M1 (mild-slope backwater) profile.

Diagram: Specific energy diagram showing choking at the channel constriction

(b) Flow field V = 2xy î + (x² − y²) ĵ

(i) At (3, 1): u = 6 m/s, v = 9 − 1 = 8 m/s. Magnitude = √(36 + 64) = 10 m/s, angle with the x-axis = tan⁻¹(8/6) = 53.13°.

(ii) Continuity for 2-D incompressible flow: ∂u/∂x + ∂v/∂y = 2y − 2y = 0. The flow is physically possible. With u = ∂ψ/∂y: ψ = xy² + f(x). With v = −∂ψ/∂x: −y² − f′(x) = x² − y², so f′(x) = −x² and ψ = xy² − x³/3 (+ constant), in m²/s.

(iii) Discharge per unit depth = |ψ₂ − ψ₁|. ψ(1, 0) = −1/3 and ψ(0, 1) = 0, so q = 1/3 m²/s per metre depth.

(iv) Vorticity ω_z = ∂v/∂x − ∂u/∂y = 2x − 2x = 0 everywhere, so the flow is irrotational (a velocity potential φ = x²y − y³/3 exists).

(c) Rankine active force on the retaining wall

Take a smooth vertical wall, horizontal backfill, c = 0 and γ_w = 9.81 kN/m³. K_a1 = tan²(45° − 14°) = 0.361 and K_a2 = tan²(45° − 17.5°) = 0.271. Depth z is measured from the top.

Layer 1: at z = 3 m, σ′_v = 17 × 3 = 51 kN/m², p_a = 0.361 × 51 = 18.41 kN/m². F₁ = ½ × 3 × 18.41 = 27.62 kN/m, at 2.00 m below the top.

Layer 2: just below z = 3 m, p_a = 0.271 × 51 = 13.82 kN/m². With γ′ = 20 − 9.81 = 10.19 kN/m³, σ′_v at 7 m = 51 + 40.76 = 91.76 kN/m², p_a = 24.87 kN/m². Rectangle = 13.82 × 4 = 55.28 kN/m at 5.00 m; triangle = ½ × 4 × 11.05 = 22.09 kN/m at 5.67 m.

Water pressure: u at 7 m = 9.81 × 4 = 39.24 kN/m², F_w = ½ × 4 × 39.24 = 78.48 kN/m at 5.67 m.

Soil active force = 27.62 + 55.28 + 22.09 = 105.0 kN/m. Total with water = 183.5 kN/m. Moment about the top = 27.62 × 2.00 + 55.28 × 5.00 + 22.09 × 5.67 + 78.48 × 5.67 = 901.6 kN·m/m.

Line of action = 901.6/183.5 = 4.91 m below the top, 2.09 m above the base. The Rankine soil force alone is 105.0 kN/m, acting 4.35 m below the top.

Answer: total thrust 183.5 kN/m (105.0 kN/m soil + 78.5 kN/m water) at 2.09 m above the base.

Diagram: Rankine active earth pressure and water pressure on the retaining wall with two soil layers

How this answer will be evaluated

What earns full marks: Complete working with all steps shown, correct units, and proper justification for all parts.

Evaluation rubric

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

  1. (a) Find the depths in and just upstream of the constriction and classify the gradually varied profile upstream. 15 marks

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

    Must cover

    • E₁ = 1.704 m, less than E_min = 1.910 m, so the throat chokes
    • Critical depth in throat y_c = 1.27 m (q = 4.5 m²/s)
    • Upstream depth from E = E_min: y = 1.76 m
    • y > y_n > y_c on a mild slope gives an M1 backwater profile

    Loses marks

    • Giving the throat depth from the original energy without checking choking.

    Earns more

    • Marks for the energy check, both depths and the profile with reasoning.

    Extra mark

    • Stating the assumption that 1.5 m is the normal depth.
  2. (b) Answer the four parts on the flow field V = 2xy î + (x² − y²) ĵ. 15 marks

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

    Must cover

    • At (3, 1): V = 10 m/s at 53.13° to x-axis
    • Continuity satisfied, so the flow is possible; ψ = xy² − x³/3
    • Discharge between the streamlines = 1/3 m²/s per metre
    • Vorticity is zero, so the flow is irrotational

    Loses marks

    • Sign errors in ψ or stating discharge without units.

    Earns more

    • Marks for each of (i) to (iv) with working.

    Extra mark

    • Giving the velocity potential φ = x²y − y³/3.
  3. (c) Find the Rankine active force on the wall and the location of its resultant. 20 marks

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

    Must cover

    • K_a1 = 0.361 and K_a2 = 0.271
    • Soil pressure diagrams: 27.62 + 55.28 + 22.09 = 105.0 kN/m
    • Water thrust 78.48 kN/m; total 183.5 kN/m
    • Resultant at 4.91 m below top, 2.09 m above base

    Loses marks

    • Omitting water pressure or using γ_sat for effective stress below the water table.

    Earns more

    • Marks for Ka values, pressure diagrams, total force and its location.

    Extra mark

    • Stating the soil-only force and its arm separately.
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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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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