Mechanical Engineering 2023 Paper I 50 marks Solve

Q2

(a) In Figure 2(a), AB is a rigid bar, CE is a steel rod of 20 mm diameter and DF is a copper rod of 20 mm diameter. If the end A is hinged and a load of 30 kN acts at the free end B, find the stresses in the steel and copper rods and the reaction at A. Given : E_S = 200 GN/m² and E_C = 100 GN/m² (20 marks) (b) The length of each arm of a Porter governor is 350 mm. The lower arms are attached to a sleeve at a distance of 25 mm from the axis and the upper arms are pivoted on the axis of rotation. The load on the sleeve is 250 N and the weight of each ball is 25 N. If the radius of rotation of the balls is 125 mm at a speed of 225 rpm, then calculate the following : (20 marks) (i) Speed of governor after the sleeve has lifted 55 mm (ii) Effort (iii) Power of the governor (c) Discuss three reasons why ferrous alloys are used so extensively and also discuss three characteristics of ferrous alloys that limit their utilization. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) चित्र 2(a) में, AB एक दृढ़ छड़ है, CE, 20 mm व्यास की इस्पात की छड़ तथा DF, 20 mm व्यास की ताँबे की छड़ है । यदि छोर A हिंजड़ है तथा 30 kN का एक भार स्वतंत्र छोर B पर लग रहा है, तो इस्पात और ताँबे की छड़ों का प्रतिबल तथा A पर प्रतिक्रिया ज्ञात कीजिए । दिया गया है : Eₛ = 200 GN/m² तथा E꜀ = 100 GN/m² (20 अंक) (b) एक पोर्टर गवर्नर की प्रत्येक भुजा की लंबाई 350 mm है । निचली भुजाएं घूर्णन अक्ष से 25 mm की दूरी पर स्लीव से संलग्न हैं तथा ऊपरी भुजाएं घूर्णन अक्ष पर घूर्णायस्थ हैं । स्लीव पर 250 N का भार है तथा प्रत्येक गेंद का वजन 25 N है । यदि गेंदों की घूर्णन त्रिज्या 225 rpm पर 125 mm है, तो निम्नलिखित को परिकलित कीजिए : (20 अंक) (i) गवर्नर की गति जब स्लीव 55 mm उठ गई हो (ii) प्रयास (iii) गवर्नर की शक्ति (c) लोहे मिश्रधातु के बहुतायत उपयोग होने के तीन कारणों की विवेचना कीजिए तथा लोहे मिश्रधातु के तीन लक्षणों की भी विवेचना कीजिए जो इनकी उपयोगिता को सीमित करते हों । (10 अंक)

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Approach

Solve part (a) by applying compatibility of deformations and moment equilibrium for the statically indeterminate rigid bar system, allocating ~40% time. For part (b), apply Porter governor geometry and equilibrium equations to find speed, effort and power, allocating ~40% time. Conclude with part (c) discussing ferrous alloys' advantages (cost, strength, availability) and limitations (corrosion, density, brittleness at low temperatures) in ~20% time. Present all derivations systematically with clear free-body diagrams.

Key points expected

  • Part (a): Compatibility equation δ_C/a = δ_D/b relating steel and copper rod deformations; moment equilibrium about hinge A
  • Part (a): Stress in steel rod σ_S = 80 MPa (tensile) and copper rod σ_C = 40 MPa (tensile); reaction at A = 10 kN upward
  • Part (b)(i): New equilibrium geometry after 55 mm sleeve lift; speed N₂ = 240.5 rpm using tan(α₂) = (125+55)/√(350²-180²)
  • Part (b)(ii): Effort = 357.5 N calculated from (F₁-F₂)tan(α) with proper angle determination
  • Part (b)(iii): Power = Effort × sleeve lift = 357.5 × 0.055 = 19.66 N.m or 19.66 W
  • Part (c): Three reasons—low cost due to abundant iron ore (India: Odisha, Jharkhand), excellent mechanical properties (strength, hardness, wear resistance), mature processing technology
  • Part (c): Three limitations—susceptibility to corrosion (requires protective coatings), high density (heavier than Al alloys, affects automotive fuel efficiency), poor corrosion resistance in coastal/humid Indian climates

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies statically indeterminate system in (a) with compatibility of deformations; applies Porter governor equilibrium with proper force polygon and instantaneous centre method in (b); distinguishes ferritic, austenitic and martensitic structures in (c) with specific Indian applications (railway tracks, bridges).Uses correct basic principles but confuses rigid bar rotation with simple elongation in (a); applies governor formula directly without geometric analysis in (b); lists generic ferrous properties without alloy-specific examples in (c).Treats (a) as statically determinate or ignores compatibility; uses Watt governor formula for Porter governor in (b); confuses ferrous with non-ferrous alloys or gives incorrect chemical compositions in (c).
Numerical accuracy20%10All calculations precise: (a) σ_S = 80 MPa, σ_C = 40 MPa, R_A = 10 kN; (b)(i) N₂ = 240-241 rpm, (ii) Effort ≈ 357 N, (iii) Power ≈ 19.7 W; unit conversions correct (GN/m² to N/mm², rpm to rad/s where needed).Correct final answers in (a) and (b)(i) but minor arithmetic errors in effort/power calculation; consistent unit handling but occasional slip (e.g., mm vs m in area calculation).Major errors: wrong area (uses diameter instead of radius), incorrect moment arm geometry, or order-of-magnitude errors (MPa vs GPa); no unit consistency.
Diagram quality20%10Clear FBD of rigid bar AB in (a) showing hinge at A, load at B, rod forces at C and D with correct geometry; Porter governor force polygon in (b) with angles α, β, weight vectors and centrifugal forces labelled; free-hand sketches acceptable but proportionate and labelled.Diagrams present but poorly proportioned; missing some labels (e.g., angles not marked in governor); FBD in (a) shows forces but not moment arms.No diagrams or incomprehensible sketches; missing critical elements like hinge symbol, load direction arrows, or governor sleeve position.
Step-by-step derivation20%10Systematic derivation: (a) shows ΣM_A = 0, compatibility δ_S/a = δ_C/b, stress-strain relations, solves simultaneous equations; (b) derives tan(α) and tan(β) from geometry, establishes F₁ = mg/2 + Mg/2 × (1+tan(β)/tan(α)), then finds speed relation; all algebraic steps visible.Key equations stated but skips intermediate algebra (e.g., jumps from compatibility to final stresses); governor speed formula quoted without geometric derivation of angle changes.Final answers only with no derivation; or incorrect starting equations leading to wrong answers despite some working.
Practical interpretation20%10Interprets (a) results: steel carries higher stress due to higher E, design implication for composite structures; (b) comments on governor sensitivity and stability, industrial applications (steam turbines, hydroelectric plants in India); (c) cites specific examples: TMT bars in construction (corrosion issue in coastal Mumbai, Chennai), high-speed rail bogies requiring fatigue-resistant steels.Brief mention of design implications without specific examples; notes governor controls speed but no discussion of sensitivity coefficient.No interpretation; treats all parts as pure calculation exercises; no connection to real engineering applications or Indian industrial context.

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