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Civil Engineering 2025 Paper I 50 marks Calculate

Paper I — Q7

(a) A braced cut 7·0 m deep and 3·0 m wide is proposed in a cohesionless sand deposit. Assume that the first row of struts to be…

(a)

A braced cut 7·0 m deep and 3·0 m wide is proposed in a cohesionless sand deposit. Assume that the first row of struts to be located at 0·5 m below ground level and the spacing of strut as 2·5 m as shown in the diagram. In the plan, the struts are placed at spacing of 2 m centre to centre. Using Peck's empirical relation for pressure diagram, determine the design loads in the struts. The properties of sand are as follows: Angle of shearing resistance = 30° Bulk density = 16·5 kN/m³

(b)

A 5 m thick clay layer is subjected to drained condition both at top and bottom. It has few sand drains in square pattern. The spacing of sand drains are 3 m centre to centre. The coefficient of consolidation in vertical and radial directions are same and equal to 5 × 10⁻³ m²/day. The radius of the sand drains is 0·25 m. Assuming that there is no smear at the periphery of drain wells, it has been estimated that a given uniform surcharge would cause a total consolidation settlement of 200 mm without sand drains. Find the consolidation settlement of clay layer with same surcharge and sand drains, at times of 6 months, 9 months and one year. Draw the variation of settlement with time.

(c)
(i)

A liquid whose specific gravity is 0·8 and dynamic viscosity is 1·8 poise, flows in a vertical pipe of 8 cm diameter. Pressure gauges in the pipe located 20 m apart indicate a pressure of 180 kPa at the upper end and a pressure of 360 kPa at the lower end. Calculate the flow rate and find the direction of the flow in the pipe. (Use Hagen-Poiseuille equation.)

(ii)

Water flows from A to B through a tapering pipe. The following data is given at section A and B:

SectionAB
Diameter of pipe12 cm10 cm
Elevation100.000 m101.000 m
Gauge pressure30 kPa20 kPa

Estimate the discharge in the pipe line. (Assume zero loss of energy between two sections.)

हिंदी में प्रश्न पढ़ें
(a)

7·0 m गहरी और 3·0 m चौड़ी एक बंधनयुक्त काट एक संसजन रहित बालू निक्षेप में प्रस्तावित है। मान लीजिए कि टेकों की प्रथम पंक्ति धरातल से 0·5 m नीचे लगानी है और आरेख में दर्शाए अनुसार टेकों का अंतराल 2·5 m है। अनुविन्यास में, टेकों को केंद्र से केंद्र के 2 m के अंतराल पर रखा गया है। दाब आरेख के लिए पैक के अनुभाविक संबंध का उपयोग करते हुए, टेकों में अभिकल्प भारों को निर्धारित कीजिए। बालू के गुणधर्म इस प्रकार हैं : अपरूपण प्रतिरोध का कोण = 30° स्थूल घनत्व = 16·5 kN/m³

(b)

एक 5 m मोटी मृत्तिका परत शीर्ष और तली पर अपवाहित अवस्था में है। इसमें कुछ बालू नालियाँ वर्गाकार प्रारूप में हैं। बालू नालियों की केंद्र से केंद्र की दूरी 3 m है। संघनन गुणांक ऊर्ध्वाधर और त्रिजीय दिशाओं में समान है और 5 × 10⁻³ m²/दिन के बराबर है। बालू नालियों की त्रिज्या 0·25 m है। यह मानते हुए कि नाली कुओं की परिधि पर कोई स्मीयर नहीं है, यह आकलन किया गया कि एक प्रदत एकसमान अधिभार बालू नालियों के बिना 200 mm का सकल संघनन निष्पदन करेगा। इसी अधिभार और बालू नालियों के साथ मृत्तिका परत का संघनन निष्पदन 6 माह, 9 माह और 1 वर्ष के समय पर ज्ञात कीजिए। निष्पदन का समय के साथ विचरण दर्शाइए।

(c)
(i)

एक द्रव, जिसका विशिष्ट घनत्व 0·8 और गतिक श्यानता 1·8 पॉइज़ है, 8 cm व्यास के एक ऊर्ध्वाधर पाइप में प्रवाहित होता है। पाइप में एक दूसरे से 20 m की दूरी पर लगे दाबमापी ऊपरी सिरे पर दाब 180 kPa और निचले सिरे पर दाब 360 kPa दर्शाते हैं। पाइप में प्रवाह दर की गणना कीजिए और प्रवाह की दिशा ज्ञात कीजिए। (हैगन-पॉइज़ुइल समीकरण का उपयोग कीजिए।)

(ii)

एक शुंडाकार पाइप में जल A से B तक प्रवाहित होता है। परिच्छेद A और B पर निम्नलिखित आँकड़े प्रदत हैं :

परिच्छेदAB
पाइप का व्यास12 cm10 cm
ऊँचाई100.000 m101.000 m
गेज दाब30 kPa20 kPa

पाइप लाइन में निस्सरण का आकलन कीजिए। (दो परिच्छेदों के बीच ऊर्जा ह्रास शून्य मान लीजिए।)

Q7 of the 2025 UPSC Mains Civil Engineering Paper I, as printed
The question as printed in the 2025 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.

(a) Continuous beam with three supports: fixed support at left end A, roller support at B, and fixed support at right end C. Span AB is 4 m total, divided into two 2 m segments by a point load of 4 kN applied downward at the midpoint between A and B. Span BC is 6 m. EI is constant throughout the beam. The beam is labeled A, B, C at the supports.

Rectangular column cross-section, width 500 mm (horizontal dimension) and depth 300 mm (vertical dimension). Longitudinal reinforcement: 6 bars of 25 mm diameter (6-25 phi), arranged as 3 bars on the top face and 3 bars on the bottom face. Transverse reinforcement: 8 mm diameter ties (8 phi ties). Clear cover 40 mm. The section is subjected to loading eccentricity with respect to the major axis alone.

Model answer

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(a)

Peck’s empirical pressure diagram for cohesionless sand is a uniform apparent pressure: K_a = tan²(45° − φ/2) = tan²(45° − 30°/2) = tan²30° = 1/3. p = 0.65 γ H K_a = 0.65 × 16.5 × 7 × 1/3 = 25.025 kN/m². Strut rows at depths 0.5 m, 3.0 m and 5.5 m. Horizontal spacing S = 2.0 m. Tributary heights:

  • Top strut: h₁ = (0.5 + 3.0)/2 − 0 = 1.75 m.
  • Middle strut: h₂ = (3.0 + 5.5)/2 − (0.5 + 3.0)/2 = 4.25 − 1.75 = 2.50 m.
  • Bottom strut: h₃ = 7.0 − 4.25 = 2.75 m. Design loads:
  • Top: P₁ = p S h₁ = 25.025 × 2 × 1.75 = 87.59 kN
  • Middle: P₂ = 25.025 × 2 × 2.50 = 125.13 kN
  • Bottom: P₃ = 25.025 × 2 × 2.75 = 137.64 kN

Check: total = 87.59 + 125.13 + 137.64 = 350.36 kN per 2 m bay = p S H. Valid for dry cohesionless sand with Peck’s uniform diagram. Alternative (hinge method, sheeting assumed hinged at strut B and at the base): A = 1.8p × 2 = 90.1 kN, B = (1.2p + 1.25p) × 2 = 122.6 kN, C = (1.25p + 0.75p) × 2 = 100.1 kN, with 37.5 kN taken by the soil below the base. The tributary-area values above are more conservative for strut C and are adopted for design.

(b)

H = 5 m, drained at top and bottom, so H_d = 5/2 = 2.5 m. For square pattern, d_e = 1.13 s = 1.13 × 3 = 3.39 m. R_e = d_e/2 = 1.695 m, r_w = 0.25 m, n = R_e/r_w = 6.78. F(n) = [n²/(n²−1)] ln n − (3n²−1)/(4n²) = 1.212. T_v = C_v t / H_d² = 0.005 t / 6.25 = 0.0008 t. T_r = C_r t / d_e² = 0.005 t / 11.4921. U_v = 1 − (8/π²)[exp(−π²T_v/4) + (1/9)exp(−9π²T_v/4) + …] U_r = 1 − exp(−8T_r/F(n)) Combined: U = 1 − (1 − U_v)(1 − U_r) Settlement S_t = 200 U mm.

Using 1 month = 30 days for 6 and 9 months, and 1 year = 365 days:

  • 6 months, t = 180 d: T_v = 0.144, U_v = 0.428; T_r = 0.0783, U_r = 0.404; U = 0.659; S = 131.8 mm
  • 9 months, t = 270 d: T_v = 0.216, U_v = 0.524; T_r = 0.1175, U_r = 0.540; U = 0.781; S = 156.1 mm
  • 1 year, t = 365 d: T_v = 0.292, U_v = 0.606; T_r = 0.1588, U_r = 0.650; U = 0.862; S = 172.3 mm

If 1 year is taken as 360 days, S ≈ 171.7 mm. Plot S against t: curve rises steeply initially, then flattens and approaches the ultimate settlement of 200 mm.

Diagram: Consolidation settlement against time for a 5 m clay layer with sand drains at 3 m square spacing

(c)(i)

μ = 1.8 poise = 0.18 Pa·s, ρ = 0.8 × 1000 = 800 kg/m³, r = 0.04 m, L = 20 m. Piezometric head h = p/(ρg) + z. At lower end: h_lower = 360000/(800 × 9.81) + 0 = 45.87 m. At upper end: h_upper = 180000/(800 × 9.81) + 20 = 42.94 m. Since h_lower > h_upper, flow is from lower end to upper end, i.e. upward.

Effective driving pressure: Δp = p_lower − p_upper − ρgL = 360000 − 180000 − 800 × 9.81 × 20 = 23040 Pa. Hagen-Poiseuille: Q = π r⁴ Δp / (8 μ L) = π(0.04)⁴(23040)/(8 × 0.18 × 20) = 6.434 × 10⁻³ m³/s = 6.434 L/s. Reynolds number Re = ρVD/μ ≈ 455 < 2000, so laminar flow is valid.

(c)(ii)

A_A = π(0.12)²/4 = 0.01131 m² A_B = π(0.10)²/4 = 0.007854 m² Continuity: V_B = V_A(A_A/A_B) = V_A(0.12/0.10)² = 1.44 V_A. Bernoulli from A to B: (30/9.81 + 100) + V_A²/(2g) = (20/9.81 + 101) + (1.44V_A)²/(2g). Thus (1.44² − 1)V_A²/(2g) = 10/9.81 − 1 = 0.01937 m. V_A² = 2 × 9.81 × 0.01937 / 1.0736 = 0.35394 V_A = 0.595 m/s. Q = A_A V_A = 0.01131 × 0.595 = 6.73 × 10⁻³ m³/s = 6.73 L/s.

How this answer will be evaluated

Framework: Peck's empirical relation for braced cuts; Terzaghi's consolidation theory with sand drains; Hagen-Poiseuille equation; Bernoulli's principle

What earns full marks: All parts show complete method with correct formulas, unit consistency, and final values with checks

Evaluation rubric

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

  1. (a) Determine design loads in struts A, B, and C using Peck's empirical pressure diagram. 15 marks

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

    Must cover

    • Ka = tan²(45 − φ/2) = 1/3; Peck apparent pressure p = 0.65 γ H Ka = 25.03 kN/m² (uniform, sand)
    • Strut rows at 0.5, 3.0, 5.5 m; tributary heights 1.75, 2.50, 2.75 m
    • Design loads (2 m plan spacing): top 87.6 kN, middle 125.1 kN, bottom 137.6 kN by tributary area (hinge method 90.1 / 122.6 / 100.1 kN also accepted)
    • Check: total 350.4 kN = p S H per bay

    Loses marks

    • Uses active earth pressure instead of Peck's diagram
    • Ignores strut spacing in load calculation
    • Final values without showing pressure distribution

    Earns more

    • Draws pressure diagram with correct shape
    • Shows tributary area calculation for each strut
    • States assumptions clearly
    • Carries units through all steps

    Extra mark

    • References IS code for braced cut design
    • Provides neat labelled sketch of pressure distribution
  2. (b) Find consolidation settlement at 6 months, 9 months, and 1 year with sand drains. 15 marks

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

    Must cover

    • Hd = 2.5 m; de = 1.13 × 3 = 3.39 m; n = 6.78; F(n) = 1.212
    • Tv = 0.0008t, Tr = 0.005t/11.4921; Uv (Terzaghi series), Ur = 1 − exp(−8Tr/F(n)); U = 1 − (1−Uv)(1−Ur)
    • Settlements: 131.8 mm (6 months), 156.1 mm (9 months), 172.3 mm (1 year)
    • Settlement–time curve drawn, approaching 200 mm asymptotically

    Loses marks

    • Ignores radial drainage from sand drains
    • Uses H=5m instead of H=2.5m for double drainage
    • Final settlement values without showing U calculation

    Earns more

    • Shows effective radius calculation for square pattern
    • Draws settlement-time variation curve
    • States no smear assumption clearly
    • Carries units in time and length calculations

    Extra mark

    • References Terzaghi's consolidation theory
    • Provides neat labelled settlement-time graph
  3. (c)(i) Calculate flow rate and direction using Hagen-Poiseuille equation. 10 marks

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

    Must cover

    • μ = 0.18 Pa·s; ρ = 800 kg/m³; piezometric heads: lower 45.87 m > upper 42.94 m → flow upward
    • Driving Δp = (360 − 180) − ρgL = 23.04 kPa (static column head subtracted)
    • Hagen–Poiseuille: Q = πr⁴Δp/(8μL) = 6.43 × 10⁻³ m³/s (6.43 L/s)
    • Re = ρVD/μ ≈ 455 < 2000 — laminar flow valid

    Loses marks

    • Ignores elevation difference in pressure gradient
    • Uses wrong viscosity units in calculation
    • Final flow rate without showing equation application

    Earns more

    • Shows unit conversions clearly
    • States flow direction explicitly
    • Carries units through all steps
    • Verifies laminar flow assumption

    Extra mark

    • Calculates Reynolds number to verify laminar flow
    • References Hagen-Poiseuille equation explicitly
  4. (c)(ii) Estimate discharge in tapering pipe using Bernoulli's equation. 10 marks

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

    Must cover

    • Continuity: VB = VA(Aa/Ab) = 1.44 VA
    • Bernoulli: (30/9.81 + 100) + VA²/2g = (20/9.81 + 101) + VB²/2g
    • Solve: VA = 0.595 m/s
    • Q = AA VA = 6.73 × 10⁻³ m³/s (6.73 L/s)

    Loses marks

    • Ignores elevation difference in energy equation
    • Does not use continuity equation to relate velocities
    • Final discharge without showing Bernoulli application

    Earns more

    • Shows area calculations for both sections
    • States zero energy loss assumption
    • Carries units through all steps
    • Verifies flow direction A to B

    Extra mark

    • References Bernoulli's principle explicitly
    • Provides neat labelled pipe section diagram
What "Calculate" is asking you to do

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Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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