Chat with us
Civil Engineering 2025 Paper I 50 marks Solve

Paper I — Q4

(a) Analyse the frame shown in the figure by the moment distribution method. Draw the Bending Moment Diagram (BMD). Joints 'B'…

(a)

Analyse the frame shown in the figure by the moment distribution method. Draw the Bending Moment Diagram (BMD). Joints 'B' and 'C' are rigid; 'A' and 'D' are fixed. 20 marks

(b)

A horizontal steel bar of 60 mm diameter is rigidly tied at each end, the ties being 1·25 m apart. A rigid bracket is fixed to the middle of the bar. Determine the maximum radial length of the bracket at which a vertical load of 1400 N can be suspended, if the deflection of the load is not to exceed 0·5 mm. Take E = 2 × 10⁵ N/mm²; G = 7·6 × 10⁴ N/mm². 10 marks

(c)
(i)

For a continuous beam ABCDEFG, show the pattern of loading for the live load for the following : (A) max –ve bending moment at support C (B) max +ve bending moment in span CD (C) max –ve bending moment in span CD (D) max +ve bending moment at support C

(ii)

Design a (tread and riser) dog-legged staircase of an office building, given the following data : Height between floors = 3·2 m Riser = 160 mm Tread = 270 mm Width of flight = Landing width = 1·25 m Live load = 5 kN/m² Finishes load = 0·6 kN/m² Consider the landing to be supported only on two edges perpendicular to the risers. Take density of concrete = 25 kN/m³.

Pt/100 = Ast/bd = fck/2fy [1 - √(1 - 4.598 R/fck)] where R = Mu/bd². 14 marks

हिंदी में प्रश्न पढ़ें
(a)

चित्र में दर्शाए गए फ्रेम का विश्लेषण आघूर्ण वितरण विधि द्वारा कीजिए । बंकन आघूर्ण आरेख (बी.एम.डी.) बनाइए । जोड़ 'B' और 'C' दृढ़ हैं; 'A' और 'D' आबद्ध हैं । 20 अंक

(b)

60 mm व्यास की एक क्षैतिज इस्पात छड़ प्रत्येक सिरे पर दृढ़ता से आबद्ध है, आबद्धक एक दूसरे से 1·25 m की दूरी पर हैं । एक दृढ़ ब्रैकेट छड़ के मध्य पर आबद्ध है । ब्रैकेट की अधिकतम त्रिज्यीय लंबाई निर्धारित कीजिए जिस पर 1400 N का एक उल्लंबधर भार लटकाया जा सके, यदि भार का विस्थापन 0·5 mm से अधिक नहीं हो पाए । E = 2 × 10⁵ N/mm²; G = 7·6 × 10⁴ N/mm² लीजिए । 10 अंक

(c)
(i)

निम्नलिखित के लिए, एक सतत धरन ABCDEFG के लिए चल भार के भारण के स्वरूप को दर्शाइए : (A) आलम्ब C पर अधिकतम ऋणात्मक बंकन आघूर्ण (B) विस्तृति CD में अधिकतम धनात्मक बंकन आघूर्ण (C) विस्तृति CD में अधिकतम ऋणात्मक बंकन आघूर्ण (D) आलम्ब C पर अधिकतम धनात्मक बंकन आघूर्ण

(ii)

एक कार्यालय भवन की प्रतिवर्ती सीढ़ी (ट्रेड और राइजर) की अभिकल्पना कीजिए, निम्नलिखित आँकड़े प्रदत हैं : मंजिलों के बीच ऊँचाई = 3·2 m राइजर = 160 mm ट्रेड = 270 mm फ्लाइट की चौड़ाई = लैंडिंग चौड़ाई = 1·25 m चल भार = 5 kN/m² परिष्कृतियों का भार = 0·6 kN/m²

लैंडिंग को राइजर के लम्बवत केवल दो सिरों पर आलम्बित मान लीजिए । कंक्रीट का घनत्व = 25 kN/m³ लीजिए ।

Pt/100 = Ast/bd = fck/2fy [1 - √(1 - 4.598 R/fck)] जहाँ R = Mu/bd² है । 14 अंक

Q4 of the 2025 UPSC Mains Civil Engineering Paper I, as printed
The question as printed in the 2025 Civil Engineering paper
Evaluate your answer to this question → See all 2025 Civil Engineering questions

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) A rectangular frame with four joints labeled A, B, C, and D. Joint A is at the bottom-left, B at the top-left, C at the top-right, and D at the bottom-right. Supports at A and D are fixed supports (indicated by hatching). The vertical member AB has a length of 3 m and a moment of inertia labeled 'I' inside a circle. The vertical member CD has a length of 3 m and a moment of inertia labeled 'I' inside a circle. The horizontal member BC connects the top of the vertical members. A vertical downward point load of 40 kN is applied to member BC. The load is positioned 2 m from joint B and 3 m from joint C. The moment of inertia for the entire member BC is labeled '2I' inside a circle. Joints B and C are rigid connections.

(c) A 2D plan view of a dog-legged staircase landing. The landing is a rectangular slab with a total width of 2600 mm (vertical dimension in the drawing) and a total length of 3680 mm (horizontal dimension in the drawing). The slab is supported on the top and bottom edges (indicated by hatching), which are perpendicular to the direction of the risers. The slab is divided into three sections along its length: a left section of width 1250 mm (composed of two 625 mm segments), a central section of width 2430 mm, and a right section of width 1250 mm (composed of two 625 mm segments). The central section contains the stair treads, indicated by vertical lines. A horizontal line divides the central section into two flights of stairs, each with a vertical dimension of 1250 mm. The thickness of the slab is indicated as 100 mm. The supports are shown as fixed or continuous along the top and bottom edges.

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) Portal frame ABCD by moment distribution

Data: columns AB = CD = 3 m (moment of inertia I); beam BC = 5 m (2I); a 40 kN load on BC at 2 m from B (a = 2 m, b = 3 m). A and D are fixed; B and C are rigid joints.

Fixed-end moments (clockwise positive):

  • M_FBC = −Pab²/L² = −40 × 2 × 3²/5² = −28.8 kN·m
  • M_FCB = +Pa²b/L² = +40 × 2² × 3/5² = +19.2 kN·m
  • Columns carry no load: FEMs are zero.

Stiffness and distribution factors: k_AB = k_CD = I/3 = 0.333I; k_BC = 2I/5 = 0.4I. At B (and at C): DF_BA = 0.333/0.733 = 0.455 and DF_BC = 0.4/0.733 = 0.545. Carry-over factor = ½; the fixed ends A and D receive carry-overs only.

Step 1: non-sway analysis (joints balanced with sway prevented).

  • First cycle at B: unbalanced +28.8 → BA +13.09, BC +15.71; carry-over AB +6.55, CB +7.85.
  • At C: unbalanced 19.2 + 7.85 = 27.05 → CB −14.76, CD −12.29; carry-over BC −7.38, DC −6.15.
  • Second cycle: B → BA +3.36, BC +4.02 (carry-over AB +1.68, CB +2.01); C → CB −1.10, CD −0.91 (carry-over BC −0.55, DC −0.46).
  • Third cycle: B → BA +0.25, BC +0.30 (AB +0.13, CB +0.15); C → CB −0.08, CD −0.07 (DC −0.04).
  • Non-sway end moments: M_AB = +8.36, M_BA = +16.70, M_BC = −16.70, M_CB = +13.27, M_CD = −13.27, M_DC = −6.65 kN·m.

Step 2: sway check. The load is not at mid-span, so the frame sways. Horizontal shear in the columns: H_A = (M_AB + M_BA)/3 = 8.35 kN and H_D = (M_CD + M_DC)/3 = −6.64 kN. Their sum, 1.71 kN, is not zero, so a sway force S = 1.71 kN is needed to hold the frame.

Step 3: sway correction. Apply an arbitrary sway giving equal column FEMs of −10 kN·m at A, B, C and D (the columns are identical) and distribute as above. This gives M_AB = M_DC ≈ −8.23 and M_BA = M_CD ≈ −6.44 kN·m, with M_BC = +6.44 and M_CB = +6.44 kN·m; the corresponding column shears total −9.78 kN. The correction factor is k = 1.71/9.78 = 0.175.

Final end moments (non-sway + 0.175 × sway):

  • M_AB = +6.9, M_BA = +15.6, M_BC = −15.6, M_CB = +14.4, M_CD = −14.4, M_DC = −8.1 kN·m.
  • Check: H_A + H_D = (6.9 + 15.6)/3 + (−14.4 − 8.1)/3 = 7.5 − 7.5 = 0 ✓; joints B and C are in balance ✓.

BMD: the beam hogs 15.6 kN·m at B and 14.4 kN·m at C. The free (simply supported) moment under the load is Pab/L = 48 kN·m; the hogging line under the load is 15.6 − (15.6 − 14.4) × 2/5 = 15.1 kN·m, so the net sagging moment under the load ≈ 32.9 kN·m. Each column varies linearly between its end moments, changing sign along its height (6.9 at A to 15.6 at B; 8.1 at D to 14.4 at C). If the frame is braced against sway, the non-sway values from Step 1 apply.

Diagram: Portal frame ABCD: sway-corrected bending moment diagram by moment distribution

(b) Radial length of the bracket on a bar fixed at both ends

Data: d = 60 mm, L = 1250 mm, P = 1400 N, allowable deflection of the load δ = 0.5 mm, E = 2 × 10⁵ N/mm², G = 7.6 × 10⁴ N/mm².

  • I = πd⁴/64 = 6.362 × 10⁵ mm⁴; J = πd⁴/32 = 1.272 × 10⁶ mm⁴.

The load P at radius r produces (i) a central point load P on a fixed–fixed bar and (ii) a torque T = Pr at mid-span, shared equally by the two halves.

  • Bending: δ_b = PL³/(192EI) = 1400 × 1250³/(192 × 2 × 10⁵ × 6.362 × 10⁵) = 0.112 mm.
  • Torsion: each half (length L/2) carries T/2 = Pr/2, so the twist at mid-span is θ = (Pr/2)(L/2)/(GJ) = PrL/(4GJ). The load moves down by rθ = Pr²L/(4GJ).
  • Condition: 0.112 + Pr²L/(4GJ) = 0.5, so Pr²L/(4GJ) = 0.388 mm.
  • r² = 0.388 × 4 × 7.6 × 10⁴ × 1.272 × 10⁶/(1400 × 1250) = 85 700 mm², so r ≈ 293 mm.

(c)(i) Live-load patterns for the continuous beam ABCDEFG

(spans AB, BC, CD, DE, EF, FG)

  • (A) Maximum negative moment at support C: load the two spans adjacent to C (BC and CD) and then alternate spans beyond them — EF loaded; AB, DE and FG unloaded.
  • (B) Maximum positive moment in span CD: load CD and alternate spans — AB and EF loaded; BC, DE and FG unloaded.
  • (C) Maximum negative moment in span CD: leave CD unloaded and load the adjacent spans and alternate spans beyond — BC, DE and FG loaded; AB, CD and EF unloaded.
  • (D) Maximum positive moment at support C: leave both spans adjacent to C unloaded and load alternate spans beyond — AB, DE and FG loaded; BC, CD and EF unloaded. With dead load acting on all spans, this pattern gives the least hogging moment at C (a net sagging moment at C is uncommon). These follow from the influence lines (Müller-Breslau): for a support moment the influence line is negative over the adjacent spans and alternates in sign beyond; for a span moment it is positive over that span and alternates in sign.

(c)(ii) Dog-legged staircase (M25, Fe 500 assumed)

Geometry. Height 3.2 m ÷ 0.16 m = 20 risers, i.e. two flights of 10 risers; each flight has 9 treads, so going = 9 × 270 = 2430 mm. Width of flight = landing width = 1.25 m.

Flight.

  • Effective span (IS 456 cl. 33.1(b), landings spanning parallel to the risers): going + half the landing width at each end = 2430 + 625 + 625 = 3680 mm.
  • Waist thickness: span/25 ≈ 150 mm; d = 150 − 20 − 5 = 125 mm.
  • Loads per m² of plan: waist 0.15 × 25 × √(160² + 270²)/270 = 4.36 kN; steps 0.5 × 0.16 × 25 = 2.0 kN; finishes 0.6 kN; live load 5.0 kN. Total 11.96 kN/m²; factored w_u = 1.5 × 11.96 = 17.9 kN/m per metre width.
  • M_u = 17.9 × 3.68²/8 = 30.4 kN·m/m; R = M_u/(bd²) = 30.4 × 10⁶/(1000 × 125²) = 1.95 N/mm² (below R_lim = 3.34 for Fe 500).
  • p_t = (25/1000)[1 − √(1 − 4.598 × 1.95/25)] = 0.50% → A_st = 0.005 × 1000 × 125 = 620 mm²/m. 10 mm bars @ 125 c/c (628 mm²) satisfy strength; the deflection check below governs the final choice.
  • Distribution steel: 0.12% × 1000 × 150 = 180 mm²/m → 8 mm @ 250 c/c.
  • Shear: V_u = 17.9 × 3.68/2 = 33 kN; τ_v = 33 000/(1000 × 125) = 0.26 N/mm² < τ_c (≈0.49 for p_t = 0.5%) ✓.
  • Deflection: span/d = 3680/125 = 29.4. With 10 mm @ 125 c/c, f_s = 0.58 × 500 × 620/628 ≈ 286 N/mm² and the modification factor is only about 1.25 (allowable 20 × 1.25 = 25), which is not enough. Provide 10 mm @ 100 c/c (785 mm²): f_s ≈ 229 N/mm², p_t = 0.63%, modification factor ≈ 1.5, allowable ≈ 30 ≥ 29.4 ✓.

Landing (supported on the two edges perpendicular to the risers, so it spans across the stair width, 2 × 1.25 + 0.1 gap ≈ 2.6 m).

  • Own load: 0.15 × 25 + 0.6 + 5.0 = 9.35 kN/m² → factored 14.0 kN/m².
  • Load from the flights: each flight delivers half its going load, 17.9 × 2.43/2 = 21.8 kN per metre width, spread over the 1.25 m landing width = 17.4 kN/m².
  • Total ≈ 31.4 kN/m²; M_u = 31.4 × 2.6²/8 = 26.5 kN·m/m; R = 1.70; p_t ≈ 0.43% → A_st ≈ 540 mm²/m. Provide 10 mm @ 140 c/c across the landing, with 8 mm @ 250 c/c distribution steel. Main bars of the flight continue into the landings and are lapped with the landing steel at the junctions.

How this answer will be evaluated

Framework: null

What earns full marks: Complete, accurate calculations with clear diagrams and correct application of all formulas.

Evaluation rubric

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

  1. (a) Determine final end moments and draw the Bending Moment Diagram (BMD) for the frame. 20 marks

    analyse— intro → causes → effects → stakeholders/linkages → way forward

    Must cover

    • FEM: M_BC = −Pab²/L² = −28.8, M_CB = +Pa²b/L² = +19.2 kN·m; K_AB = I/3, K_BC = 2I/5 → DFs 0.455/0.545
    • Non-sway distribution: +8.36, +16.70, −16.70, +13.27, −13.27, −6.65 kN·m; sway check H_A + H_D = 1.71 kN ≠ 0
    • BMD: beam hogging 15.6 (B) and 14.4 (C), net sagging ≈ 32.9 kN·m under the load; columns linear, changing sign
    • Sway correction (k ≈ 0.175) → final +6.9, +15.6, −15.6, +14.4, −14.4, −8.1 kN·m; ΣH = 0 check

    Loses marks

    • Incorrect stiffness calculation for the 2I member
    • Failure to balance moments at rigid joints B and C
    • BMD drawn without numerical values at key points

    Earns more

    • Correctly identifies stiffness of member BC as 2I
    • Shows carry-over factors (COF) in the distribution table
    • Checks equilibrium at joints B and C

    Extra mark

    • Calculates horizontal reactions at supports A and D
  2. (b) Determine the maximum radial length of the bracket for the given load and deflection limit. 10 marks

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

    Must cover

    • I = 6.36 × 10⁵ mm⁴; J = 1.27 × 10⁶ mm⁴; bending deflection δb = PL³/192EI = 0.112 mm
    • Torsion: θ = T(L/2)/(GJ) with T = Pr/2, so the rotation contributes rθ = Pr²L/(4GJ) = 0.388 mm
    • Solve r² = 0.388 × 4GJ/(PL) → r ≈ 293 mm
    • Check: 0.112 + 0.388 = 0.5 mm = allowable

    Loses marks

    • Omitting the torsional component of deflection
    • Unit mismatch between N, mm, and m
    • Incorrect formula for deflection of a fixed-fixed beam

    Earns more

    • Correctly identifies the 1.25 m span as the length L
    • Uses consistent units (N, mm) throughout the calculation

    Extra mark

    • Verifies the final deflection is exactly 0.5 mm
  3. (c)(i) Show the live load patterns for maximum positive and negative bending moments. 6 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Max −ve at C: load BC + CD and alternate spans (DE off, EF on, FG off; AB off)
    • Max +ve in CD: load CD and alternate spans (AB on, BC off, DE off, EF on)
    • Max −ve in CD: CD unloaded; BC, DE and FG loaded (AB, EF unloaded)
    • Max +ve at C: load alternate far spans (AB + DE + FG) with BC and CD unloaded (rare case)

    Loses marks

    • Loading adjacent spans for maximum positive moment in a span
    • Confusing the loading pattern for support moment vs span moment

    Earns more

    • Clear diagrams showing loaded and unloaded spans
    • Correct application of influence line sign conventions

    Extra mark

    • Brief explanation of the influence line logic for each case
  4. (c)(ii) Design the dog-legged staircase, determining the required reinforcement for the flight and landing. 14 marks

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

    Must cover

    • 20 risers (2 flights × 10); going = 9 × 270 = 2430 mm; effective flight span = 2430 + 625 + 625 = 3680 mm
    • Loads: waist 4.36 + steps 2.0 + finishes 0.6 + live 5.0 = 11.96 kN/m²; wu = 1.5 × 11.96 = 17.9 kN/m
    • Mu = 30.4 kN·m/m; R = 1.95; pt ≈ 0.50% → Ast = 620 mm²/m; deflection check governs → 10 mm @ 100 c/c
    • Landing (span 2.6 m) carries own load + flight reaction ≈ 31.4 kN/m²: Mu = 26.5 kN·m/m → 10 mm @ 140 c/c; distribution 8 mm @ 250

    Loses marks

    • Ignoring the self-weight of the concrete in the load calculation
    • Incorrect effective span calculation for the inclined flight
    • Failure to use the provided formula for Ast calculation

    Earns more

    • Correctly calculates the effective span for the flight
    • Considers the landing as a one-way slab supported on two edges
    • Checks for minimum reinforcement requirements

    Extra mark

    • Provides a neat sketch of the reinforcement layout
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.

All UPSC directive words, compared →

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Civil Engineering 2025 Paper I

Write this answer and get it marked in about 20 seconds

Free account: 5 AI evaluations a month, marked against the rubric above.

Start free

Already have an account? Log in