Civil Engineering 2022 Paper I 50 marks Solve

Q2

(a) Determine the forces in all the members of a pin-jointed truss shown in the figure below, with a vertical force of 20 kN and a horizontal force of 10 kN acting at C : (15 marks) (b) A 3 m high square column is effectively held in position but not restrained against rotation at both ends. The size of the column is restricted to 400 mm. Design and detail the column to carry a factored axial load of 2000 kN. Use M25 grade of concrete and Fe500 grade of steel. Use limit state method. (20 marks) (c) A solid circular shaft is subjected to a bending moment of 10×10³ N-m and a twisting moment of 13 kN-m. In a simple uniaxial tensile test of the same material, it gave the following data: σᵧ = 300 N/mm², E = 200×10³ N/mm², Factor of safety (FOS) = 3, ν = 0·25. Determine the least diameter required using the following: (i) Maximum principal stress theory (ii) Maximum shear stress theory (15 marks)

हिंदी में प्रश्न पढ़ें

(a) नीचे चित्र में दर्शाई गई पिन जोड़ वाली एक कैंची, जिसमें C पर 20 kN का एक उद्वधर बल और 10 kN का एक क्षैतिज बल लगा है, के सभी अवयवों में बलों को निर्धारित कीजिए : (15 अंक) (b) एक 3 m ऊँचा वर्गाकार स्तम्भ दोनों सिरों पर स्थिति में प्रभावी रूप से आबद्ध परन्तु घूर्णन के प्रति आबद्ध नहीं है। स्तम्भ का आमाप 400 mm तक सीमित है। एक 2000 kN के गुणित अक्षीय भार को वहन करने के लिए स्तम्भ का अभिकल्पन कीजिए एवं विवरण दीजिए। M25 ग्रेड कंक्रीट और Fe500 ग्रेड इस्पात का उपयोग कीजिए। सीमित अवस्था विधि का उपयोग कीजिए। (20 अंक) (c) एक ठोस वृत्ताकार शाफ्ट पर 10×10³ N-m का बंकन आघूर्ण और 13 kN-m का ऐंठन आघूर्ण लगा है। इसी पदार्थ पर किए गए एक साधारण एक-अक्षीय तनन परीक्षण से निम्नलिखित आँकड़े प्राप्त हुए: σᵧ = 300 N/mm², E = 200×10³ N/mm², सुरक्षा गुणक (FOS) = 3, ν = 0·25। निम्नलिखित का उपयोग करते हुए आवश्यक न्यूनतम व्यास का निर्धारण कीजिए: (i) अधिकतम मुख्य प्रतिबल सिद्धान्त (ii) अधिकतम अपरूपण प्रतिबल सिद्धान्त (15 अंक)

Evaluate your answer to this question → See all 2022 Civil Engineering questions

Directive word: Solve

This question asks you to solve. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

Solve all three sub-parts systematically, allocating approximately 30% time to part (a) truss analysis, 40% to part (b) column design (highest marks), and 30% to part (c) shaft design. Begin with clear free body diagrams for each part, show complete calculations with IS code references where applicable, and conclude with practical design implications for the column and shaft.

Key points expected

  • Part (a): Correct identification of zero-force members, application of method of joints or method of sections to determine forces in all truss members with proper sign convention (tension positive, compression negative)
  • Part (b): Calculation of effective length (L_eff = 3.0 m for pinned-pinned condition), determination of minimum eccentricity as per IS 456, design of column section using interaction formula or SP-16 design charts, and detailing of longitudinal and transverse reinforcement
  • Part (c): Conversion of bending and twisting moments to equivalent stresses, calculation of principal stresses and maximum shear stress, application of both failure theories with proper FOS consideration
  • Correct application of limit state method principles for column design including material safety factors (γ_m = 1.5 for concrete, 1.15 for steel)
  • Proper unit conversions throughout (kN to N, m to mm) and final presentation of results with appropriate significant figures
  • IS 456:2000 compliance for column design including minimum and maximum reinforcement percentages, spacing requirements, and cover specifications

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies truss analysis methodology (method of joints/sections), applies limit state design philosophy for column with proper effective length factor (L_eff = L for pinned-pinned), and selects appropriate failure theories for combined loading; cites IS 456:2000 provisions accuratelyUses correct basic concepts but makes minor errors in effective length determination or failure theory selection; partial IS code awarenessConfuses method of joints with method of sections, uses working stress method instead of limit state, or applies wrong failure theory (e.g., maximum strain theory)
Numerical accuracy25%12.5All calculations accurate: truss member forces with correct magnitudes and signs, column reinforcement area within 2% of optimal, shaft diameter correctly determined for both theories with proper FOS application; no arithmetic errorsCorrect approach with 1-2 calculation errors or unit conversion mistakes; final answers within 10% of correct valueMajor calculation errors, incorrect formula substitution, or consistent unit confusion leading to unrealistic results
Diagram quality15%7.5Clear free body diagram for truss with all dimensions and loading shown, column cross-section detailing with reinforcement layout (longitudinal bars + ties), and properly labeled stress element diagram for shaft; all diagrams neat, dimensioned, and IS-compliantBasic diagrams present but lacking dimensions or incomplete reinforcement detailing; truss diagram without member labelsMissing diagrams, poorly sketched figures, or incorrect representation of support conditions and loading
Step-by-step derivation25%12.5Systematic stepwise solution: for (a) shows equilibrium equations at each joint; for (b) presents complete design steps from load calculation to final detailing with IS 456 checks; for (c) derives equivalent bending/twisting moments and shows theory application clearlySome steps shown but skips critical intermediate calculations or jumps to final answers without justificationNo derivation shown, only final answers stated, or completely disorganized presentation preventing verification
Practical interpretation15%7.5Interprets truss force results for member selection (tension vs compression members), justifies column size restriction with practical construction constraints, compares both failure theories for shaft and recommends conservative design; mentions Indian construction practicesBrief mention of practical implications without elaboration; generic statements about safetyPurely mathematical solution with no physical interpretation, or unrealistic design choices ignoring constructability

Practice this exact question

Write your answer, then get a detailed evaluation from our AI trained on UPSC's answer-writing standards. Free first evaluation — no signup needed to start.

Evaluate my answer →

More from Civil Engineering 2022 Paper I