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Mechanical Engineering 2023 Paper I 50 marks Calculate

Paper I — Q4

(a) Two involute gears in mesh have pressure angle of 20° and module of 6 mm. The number of teeth on pinion is 28 and the larger…

(a)

Two involute gears in mesh have pressure angle of 20° and module of 6 mm. The number of teeth on pinion is 28 and the larger gear has 52 teeth.

Calculate the following : 20 marks

(i)

Contact ratio

(ii)

Angle of action of gear wheel and pinion

(iii)

Ratio of sliding velocity to rolling velocity at

(I) Starting of contact,

(II) Pitch point,

(III) End of contact.

Consider the addenda on pinion and gear wheel as equal to one module.

(b)

A solid circular shaft made of steel is subjected to a bending moment of 12 kNm and a twisting moment of 16 kNm. In a simple uniaxial tensile test of the same material it gave the stress at yield point = 300 N/mm². Assuming factor of safety = 2, estimate the minimum diameter required for the circular shaft using (i) Maximum Principal Stress theory, and (ii) Maximum Shear Stress theory. 20 marks

(c)

Write short notes on the following : 10 marks

(i)

Ceramics

(ii)

Nano-materials

हिंदी में प्रश्न पढ़ें
(a)

दो अंतर्वलित गियर (मेश में) का दाब कोण 20° तथा प्रमाणक (मॉड्यूल) 6 mm है । पिनियन पर दांतों की संख्या 28 तथा बड़े गियर पर 52 दांते हैं ।

निम्नलिखित की गणना कीजिए : (20 अंक)

(i)

संपर्क अनुपात

(ii)

गियर पहिये और पिनियन का क्रिया कोण

(iii)

सर्पण वेग और रोलिंग वेग का अनुपात

(I) संपर्क के प्रारंभ में,

(II) पिच बिंदु पर,

(III) संपर्क की समाप्ति पर ।

पिनियन और गियर व्हील पर ऐडेंडा एक प्रमाणक (मॉड्यूल) के बराबर मान लीजिए ।

(b)

एक इस्पात के ठोस वृत्तीय शाफ्ट पर 12 kNm का बंकन आघूर्ण और 16 kNm का ऐंठन आघूर्ण लगाया जाता है । एक सामान्य एक-अक्षीय तनन परीक्षण में इस पदार्थ का प्रारंभ बिंदु प्रतिबल 300 N/mm² पाया गया । सुरक्षा गुणक को 2 मानते हुए इस वृत्तीय शाफ्ट के न्यूनतम आवश्यक व्यास का आकलन (i) अधिकतम मुख्य प्रतिबल सिद्धांत, तथा (ii) अधिकतम अपरूपण प्रतिबल सिद्धांत का प्रयोग करते हुए कीजिए । (20 अंक)

(c)

निम्नलिखित पर संक्षिप्त टिप्पणियाँ लिखिए : (10 अंक)

(i)

मृत्तिका-शिल्प

(ii)

नैनो-पदार्थ

Q4 of the 2023 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2023 Mechanical Engineering paper
Evaluate your answer to this question → See all 2023 Mechanical Engineering questions

Model answer

Written 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.

(a)(i) Using involute gear contact-path geometry. For the pinion (1) and gear wheel (2): r1 = m z1/2 = 6×28/2 = 84 mm, r2 = 6×52/2 = 156 mm. ra1 = r1 + m = 90 mm, ra2 = r2 + m = 162 mm. rb1 = r1 cos20° = 84×0.9396926 = 78.934 mm, rb2 = 156×0.9396926 = 146.592 mm.

Path of approach = √(ra2² − rb2²) − r2 sin20° = √(162² − 146.592²) − 156×0.342020 = 68.955 − 53.355 = 15.600 mm.

Path of recess = √(ra1² − rb1²) − r1 sin20° = √(90² − 78.934²) − 84×0.342020 = 43.237 − 28.730 = 14.507 mm.

Total path of contact L = 15.600 + 14.507 = 30.107 mm. Base pitch = π m cos20° = π×6×0.9396926 = 17.713 mm.

Contact ratio = L / base pitch = 30.107 / 17.713 = 1.700.

(a)(ii) For involute gears, angle of action θ = L / rb.

Pinion: θp = 30.107 / 78.934 = 0.3814 rad = 21.85°. Gear wheel: θg = 30.107 / 146.592 = 0.2054 rad = 11.77°.

(a)(iii) For external gears, sliding velocity vs rolling velocity is given by vs/vr = x(1/r1 + 1/r2), where x is distance from pitch point along path of contact. Here 1/r1 + 1/r2 = 1/84 + 1/156 = 5/273 mm⁻¹.

  • (I) Starting of contact: x = 15.600 mm. vs/vr = 15.600×5/273 = 0.2857.
  • (II) Pitch point: x = 0. vs/vr = 0.
  • (III) End of contact: x = 14.507 mm. vs/vr = 14.507×5/273 = 0.2657.

If the gear wheel is taken as driver, the two extreme ratios interchange.

(b)(i) Using Maximum Principal Stress theory. For solid circular shaft: σb = 32M/(πd³), τ = 16T/(πd³). M = 12×10⁶ Nmm, T = 16×10⁶ Nmm. Allowable principal stress = σy/FOS = 300/2 = 150 N/mm².

σ1 = σb/2 + √((σb/2)² + τ²) = 16/(πd³) [M + √(M² + T²)]. √(M² + T²) = √(12² + 16²)×10⁶ = 20×10⁶ Nmm.

Thus σ1 = 16/(πd³) (12×10⁶ + 20×10⁶) = 512×10⁶/(πd³).

Set σ1 = 150: d³ = 512×10⁶/(π×150) = 1.0865×10⁶ mm³. d = 102.80 mm.

(b)(ii) Using Maximum Shear Stress theory. Maximum shear stress: τmax = √((σb/2)² + τ²) = 16/(πd³) √(M² + T²). Allowable shear stress = σy/(2FOS) = 300/(2×2) = 75 N/mm².

τmax = 16/(πd³) × 20×10⁶ = 320×10⁶/(πd³). Set τmax = 75: d³ = 320×10⁶/(π×75) = 1.3581×10⁶ mm³. d = 110.74 mm.

(c)(i) Ceramics Ceramics are inorganic, non-metallic solids, usually oxides, carbides, nitrides or silicates. They have strong ionic/covalent bonds, giving high hardness, high compressive strength, good heat and chemical resistance, and electrical insulation. However, they are brittle, weak in tension and poor in shock resistance. Examples: alumina, silicon carbide, zirconia, porcelain. Applications include refractory linings, cutting tools, spark plugs, insulators, turbine components and bio-implants. They are generally shaped by powder compaction, slip casting and sintering.

(c)(ii) Nano-materials Nano-materials have at least one dimension in the range 1–100 nm. Their very high surface-area-to-volume ratio, quantum confinement and large grain-boundary fraction give properties different from bulk materials, such as enhanced strength, catalytic activity, optical, electrical, magnetic and thermal behaviour. Examples are nanoparticles, nanotubes, graphene, quantum dots and nanocomposites. They are used in drug delivery, sensors, catalysts, composites, electronics, coatings and energy storage. Main concerns are toxicity, agglomeration and safe handling.

What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

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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How this answer will be evaluated

Approach

(a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) write short notes: define > 3-4 key features > one example > one-line significance Full marks: All calculations correct with clear steps, units, and assumptions; notes are precise with examples and significance.

Key points expected

  • Calculate pitch radii and addendum circles from module and teeth count
  • Determine length of arc of contact using pressure angle and geometry
  • Apply formula for contact ratio (arc of contact / circular pitch)
  • Calculate sliding velocity ratio at start, pitch, and end of contact
  • State given moments (M=12 kNm, T=16 kNm) and yield stress (300 N/mm²)
  • Apply Maximum Principal Stress theory formula for combined loading
  • Apply Maximum Shear Stress theory formula for combined loading
  • Incorporate factor of safety (FOS=2) in allowable stress calculation

Evaluation rubric

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

  1. (a) Determine contact ratio, angle of action, and sliding/rolling velocity ratios for the given gear pair. 20 marks

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

    Must cover

    • Calculate pitch radii and addendum circles from module and teeth count
    • Determine length of arc of contact using pressure angle and geometry
    • Apply formula for contact ratio (arc of contact / circular pitch)
    • Calculate sliding velocity ratio at start, pitch, and end of contact

    Loses marks

    • Omitting the pressure angle in contact length calculation
    • Confusing sliding velocity with relative velocity at pitch point
    • Incorrect identification of start/end of contact points

    Earns more

    • Correct calculation of base circle radii
    • Explicit statement of addendum as one module
    • Clear distinction between pinion and gear wheel values
    • Consistent use of units (mm, degrees)

    Extra mark

    • Schematic diagram of meshing gears with contact points marked
    • Verification of contact ratio > 1 for continuous motion
  2. (b) Estimate minimum shaft diameter using Maximum Principal Stress and Maximum Shear Stress theories. 20 marks

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

    Must cover

    • State given moments (M=12 kNm, T=16 kNm) and yield stress (300 N/mm²)
    • Apply Maximum Principal Stress theory formula for combined loading
    • Apply Maximum Shear Stress theory formula for combined loading
    • Incorporate factor of safety (FOS=2) in allowable stress calculation

    Loses marks

    • Using yield stress directly without applying factor of safety
    • Incorrect formula for equivalent moment in combined loading
    • Unit mismatch between kNm and N/mm²

    Earns more

    • Correct conversion of kNm to Nmm for unit consistency
    • Explicit statement of governing equations for each theory
    • Comparison of diameters obtained from both theories
    • Selection of larger diameter as final answer

    Extra mark

    • Schematic of shaft with bending and twisting moments indicated
    • Mention of von Mises criterion as alternative check
  3. (c) Provide concise notes on Ceramics and Nano-materials. 10 marks

    write short notes— define → 3-4 key features → one example → one-line significance

    Must cover

    • Define ceramics with key structural/property features
    • Define nano-materials with size range (1-100 nm) and unique properties
    • Give one application example for each material type
    • State one-line significance of each in engineering context

    Loses marks

    • Vague definitions without specific properties or applications
    • Confusing ceramics with glass or polymers
    • Omitting the size range for nano-materials

    Earns more

    • Mention of brittleness and high-temperature resistance for ceramics
    • Reference to quantum effects or high surface area for nano-materials
    • Clear distinction between bulk and nano-scale behavior
    • Use of technical terms like 'crystalline' or 'colloidal'

    Extra mark

    • Mention of specific ceramic types (e.g., alumina, zirconia)
    • Reference to specific nano-materials (e.g., carbon nanotubes, graphene)

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