Mechanical Engineering 2022 Paper I 50 marks Calculate

Q2

(a) A 20 mm diameter shaft is subjected to a torque of 80 Nm and a downward bending moment of 100 Nm at the centre. Draw the state of stress on the bottom surface of the shaft at the centre and find principal stresses and shear stress at the centre of the bottom surface. What is the angle of shear plane? (15 marks) (b) What is Atomic Packing Factor of a crystal structure? Calculate the atomic packing factor of aluminium assuming atoms to be of spherical shape with atomic radius 'R'. (15 marks) (c) The cranks of a three-cylinder single acting engine are set equally at 120°. The engine speed is 540 rpm. The turning moment diagram for each cylinder is a triangle for the power stroke with a maximum torque of 100 Nm at 60° after dead-centre of the corresponding crank. On the return stroke, the torque is sensibly zero. Determine the: (i) power developed by the engine (ii) coefficient of fluctuation of speed if the flywheel has a mass of 7·5 kg with a radius of gyration of 65 mm (iii) coefficient of fluctuation of energy (iv) maximum angular acceleration of the flywheel (20 marks)

हिंदी में प्रश्न पढ़ें

(a) एक 20 mm व्यास के शैफ्ट पर एक 80 Nm का बल-आघूर्ण व केन्द्र पर नीचे की दिशा में एक 100 Nm का बंकन आघूर्ण लगा है। शैफ्ट के नीचे के तल के केन्द्र पर प्रतिबल अवस्था खींचिए तथा नीचे के तल के केन्द्र पर मुख्य प्रतिबल व अपरूपण प्रतिबल को ज्ञात कीजिए। अपरूपण तल का कोण क्या है? (15 अंक) (b) एक क्रिस्टल संरचना का परमाण्वीय पैकिंग गुणक क्या है? ऐल्युमिनियम के परमाण्वीय पैकिंग गुणक की गणना यह मानते हुए कीजिए कि परमाणु गोलाकार आकार के हों जिनका परमाण्वीय अर्धव्यास 'R' हो। (15 अंक) (c) एक तीन-सिलिंडर एकल क्रिय इंजन के क्रैंक बराबर रूप से 120° पर नियोजित किए गए हैं। इंजन की चाल 540 rpm है। प्रत्येक सिलिंडर के लिए टर्निंग आघूर्ण आरेख एक त्रिभुज है जो कि शक्ति स्ट्रोक के लिए है जहाँ का अधिकतम बल-आघूर्ण संगत क्रैंक के निष्क्रिय-केंद्र के बाद 60° पर 100 Nm है। वापसी स्ट्रोक पर बल-आघूर्ण संवेद्य रूप से शून्य है। ज्ञात कीजिए: (i) इंजन द्वारा उत्पादित शक्ति (ii) यदि गतिपालक चक्र का द्रव्यमान 7·5 kg है जबकि परिभ्रमण त्रिज्या 65 mm है, तो चाल उच्चावचन गुणांक (iii) ऊर्जा उच्चावचन गुणांक (iv) गतिपालक चक्र का अधिकतम कोणीय त्वरण (20 अंक)

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Approach

Calculate the principal stresses and shear plane angle for part (a) using combined bending-torsion theory; define and derive APF for FCC aluminium in part (b); solve the engine dynamics problem in part (c) by superposing three triangular turning moment diagrams, finding mean torque, then computing power, fluctuation coefficients and angular acceleration. Allocate approximately 30% time to (a), 25% to (b), and 45% to (c) due to its four sub-parts and higher marks.

Key points expected

  • Part (a): Bending stress σ = 32M/(πd³) = 318.3 MPa; Torsional shear stress τ = 16T/(πd³) = 127.3 MPa; principal stresses σ₁,₂ = (σ/2) ± √[(σ/2)² + τ²] = 364.7 MPa and -46.4 MPa; maximum shear stress τ_max = 205.6 MPa; shear plane angle θ_s = ½ tan⁻¹(2τ/σ) = 19.3°
  • Part (b): Definition of APF as ratio of volume occupied by atoms to total unit cell volume; aluminium is FCC with 4 atoms per unit cell; lattice parameter a = 2√2 R; APF = [4 × (4/3)πR³] / a³ = 0.74 or 74%
  • Part (c)(i): Power = 3 × (area under one TM diagram) × (ω/2π) = 3 × (½ × π × 100) × (540×2π/60) / (2π) = 8.48 kW or using mean torque method
  • Part (c)(ii): Coefficient of fluctuation of speed C_s = Δω/ω_mean derived from energy fluctuation ΔE = ½I(ω₁² - ω₂²); with I = mk² = 7.5 × (0.065)² = 0.0317 kg.m²
  • Part (c)(iii): Coefficient of fluctuation of energy C_E = (E_max - E_min)/work done per cycle = ΔE/(P × 60/N); determined from turning moment diagram energy integration
  • Part (c)(iv): Maximum angular acceleration α_max = (T_max - T_mean)/I occurring at instant of maximum excess torque; requires locating where combined TM is farthest from mean
  • Part (c) superposition: Combined TM diagram constructed by adding three 120°-phased triangular diagrams; mean torque line identified; energy fluctuations calculated by integrating areas above and below mean

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies combined stress theory for (a) identifying pure bending at bottom surface; defines APF precisely for (b) with correct FCC geometry; for (c) understands superposition of three 120°-phased torque diagrams, identifies that power stroke occupies π radians, and correctly locates mean torque line from energy balance.Uses correct formulas but with minor conceptual gaps—e.g., treats bottom surface as combined bending-direct stress or confuses BCC with FCC for aluminium; understands TM diagram superposition but miscalculates phase relationships.Applies uniaxial stress formulas to combined loading; defines APF incorrectly as density ratio; fails to phase shift TM diagrams or treats engine as single-cylinder.
Numerical accuracy20%10All calculations precise: (a) σ = 318.3 MPa, τ = 127.3 MPa, σ₁ = 364.7 MPa, σ₂ = -46.4 MPa, θ_s = 19.3°; (b) APF = 0.74 exact; (c) power ≈ 8.48 kW, C_s ≈ 0.015-0.02, C_E ≈ 0.05-0.08, α_max with correct I = 0.0317 kg.m² and proper T_excess identification.Principal stresses correct but angle slightly off due to quadrant error; APF formula correct but arithmetic slip (0.68 instead of 0.74); power correct but fluctuation coefficients off by factor of 2-3 due to ΔE calculation error.Order-of-magnitude errors in stresses (factor of 10 from unit conversion); APF > 1 or < 0.3; power calculation ignores factor of 3 for three cylinders; angular acceleration uses ω instead of α.
Diagram quality20%10(a) Shows 2D stress element with σ_x = 318.3 MPa, τ_xy = 127.3 MPa clearly labelled; Mohr's circle or principal stress orientation sketch; (c) Combined TM diagram with three 120°-phased triangles, mean torque line, shaded energy fluctuation areas, and clear angular positions (0°, 120°, 240°) marked.Stress element drawn but values not labelled or signs confused; TM diagram shows correct triangular shapes but phase angles incorrect or mean torque line missing; no energy fluctuation areas indicated.No stress element or incorrect orientation (e.g., σ_y shown); no TM diagram; diagrams drawn without scales or values, making them uninterpretable.
Step-by-step derivation20%10(a) Full derivation: σ = My/I with I = πd⁴/64, τ = Tr/J with J = πd⁴/32, then principal stress formula derived from characteristic equation; (b) Derives a = 2√2 R from FCC geometry; (c) Explicitly writes TM equations for each cylinder, superposes, integrates for work per cycle, solves for mean torque, then applies flywheel equations.Uses combined stress formulas directly without showing section modulus derivations; states FCC coordination number but skips geometry derivation; states superposition result without showing individual TM functions.Final answers stated without any derivation; no formulas shown; jumps from given data to answers; no indication of how principal stress angle was found.
Practical interpretation20%10(a) Comments that σ₂ is compressive and its magnitude matters for brittle materials; notes shear plane angle relevance to crack propagation; (b) Explains why APF matters for density and properties of aluminium alloys used in aerospace (Hindustan Aeronautics, ISRO applications); (c) Interprets C_s for engine smoothness, relates flywheel sizing to agricultural pump sets or Indian railway auxiliary power units, discusses trade-off between flywheel mass and speed fluctuation.Mentions that high APF means close-packed structure but no application; notes that flywheel reduces speed fluctuation but no specific context; brief mention of practical relevance without elaboration.No interpretation provided; treats all parts as pure mathematical exercises; no physical meaning attached to any result.

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