Mechanical Engineering 2022 Paper I 50 marks Compulsory Solve

Q1

(a) A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in Figure 1(a). The crank is rotating with a uniform angular speed of 10 rad/s, counterclockwise. For the given configuration, determine the speed of the slider. A and B in Figure 1(a) are at the same horizontal level. (10 marks) (b) In a complex two-dimensional stress system, the maximum and minimum principal stresses are found to be 160 MPa tensile and 80 MPa compressive. The material elastic limit is 300 MPa in a simple tension test. Find factor of safety using the following theories: (i) Maximum principal stress theory (ii) Maximum shear stress theory (iii) Maximum distortion energy theory (10 marks) (c) A pipe with external forces is shown in Figure 1(c). The loads 300 N, 200 N, and 900 N are acting at the centres of pipe sections as shown in the figure. Find the resultant of force system at point A shown in the figure. (10 marks) (d) Compare thermosetting and thermoplastic types of plastics in terms of properties, response to heating and applications. (10 marks) (e) Two meshing spur gears with pressure angle of the involute teeth being 20° have addendum equal to one module. The pinion has 14 teeth and the larger gear has 54 teeth. Does the interference occur? If it occurs, what should be the change in the pressure angle in order to eliminate interference? Take standard module = 10 mm. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) एक स्लाइडर-क्रैंक यंत्रावली जिसमें क्रैंक की त्रिज्या 60 mm तथा संयोजी दंड की लम्बाई 240 mm है, को चित्र 1(a) में दर्शाया गया है। क्रैंक एकसमान कोणीय चाल 10 rad/s से वामावर्त दिशा में घूर्णन कर रहा है। दिए गए विन्यास के लिए, स्लाइडर की चाल ज्ञात कीजिए। चित्र 1(a) में A तथा B एक ही क्षैतिज तल पर हैं। (10 अंक) (b) एक जटिल द्वि-विमीय प्रतिबल प्रणाली में, अधिकतम व न्यूनतम मुख्य प्रतिबल क्रमशः 160 MPa तनन व 80 MPa संपीडन पाए जाते हैं। एक सामान्य तनन परीक्षण में पदार्थ की प्रत्यास्थ सीमा 300 MPa है। निम्नलिखित सिद्धांतों का प्रयोग करके सुरक्षा कारक ज्ञात कीजिए: (i) अधिकतम मुख्य प्रतिबल सिद्धांत (ii) अधिकतम अपरूपण प्रतिबल सिद्धांत (iii) अधिकतम विकृति ऊर्जा सिद्धांत (10 अंक) (c) एक पाइप जिसमें बाह्य बल लगे हुए हैं उसे चित्र 1(c) में दर्शाया गया है। चित्रानुसार भार 300 N, 200 N एवं 900 N पाइप खंडों के मध्य लगे हुए हैं। बिन्दु A पर चित्र में दर्शाए अनुसार बल निकाय का परिणामी ज्ञात कीजिए। (10 अंक) (d) ताप-दृढ़ प्लास्टिक व ताप-सुग्राह्य प्लास्टिक प्रकारों की तुलना गुणों, तापन-अनुक्रिया व अनुप्रयोगों के पदों में कीजिए। (10 अंक) (e) दो मिले हुए स्पर गियर जिनके प्रतिकेन्द्रज दाँतों का दाब कोण 20° है, उनका ऐडेंडम एक मॉड्यूल के बराबर है। पिनियन में 14 दाँते हैं व बड़े गियर में 54 दाँते हैं। क्या व्यतिकरण होता है? यदि यह होता है, तो व्यतिकरण को दूर करने के लिए दाब कोण में कितना बदलाव करना चाहिए? मानक मॉड्यूल = 10 mm लीजिए। (10 अंक)

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Approach

Solve each sub-part systematically: for (a) apply velocity polygon or analytical method for slider-crank; for (b) compute factor of safety using all three failure theories with proper stress transformations; for (c) resolve forces and moments about point A; for (d) construct a comparative table of thermosetting vs thermoplastic polymers; for (e) check interference using minimum teeth and contact ratio criteria, then recalculate pressure angle if needed. Allocate approximately 15% time to (a), 20% to (b), 15% to (c), 20% to (d), and 30% to (e) due to its analytical complexity.

Key points expected

  • (a) Slider speed using velocity analysis: v_B = ω × r = 10 × 0.06 = 0.6 m/s tangential; velocity polygon or analytical solution gives slider velocity ≈ 0.5196 m/s or 0.52 m/s (using n = l/r = 4, θ = 90°)
  • (b)(i) Maximum principal stress theory: σ1 = 160 MPa < 300 MPa, FOS = 300/160 = 1.875
  • (b)(ii) Maximum shear stress theory: τ_max = (160-(-80))/2 = 120 MPa, allowable τ = 300/2 = 150 MPa, FOS = 150/120 = 1.25
  • (b)(iii) Maximum distortion energy theory: σ_eq = √(160² + (-80)² - 160×(-80)) = √44800 ≈ 211.66 MPa, FOS = 300/211.66 ≈ 1.417
  • (c) Resultant force and moment at A: resolve 300N, 200N, 900N into components; calculate resultant force vector and resultant moment about A
  • (d) Comparison table: thermoplastics (PE, PP, PVC) - linear chains, reversible softening, recyclable, lower strength; thermosets (Bakelite, epoxy) - cross-linked, irreversible curing, higher heat resistance, used in composites and electrical fittings
  • (e) Interference check: minimum teeth for 20° full depth = 18 (pinion has 14 < 18), interference occurs; new pressure angle φ' where cos φ' = (t_p/2)sin²φ/(1+2/t_g) or using limiting condition, φ' ≈ 23.5° to 25° range

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies velocity analysis for slider-crank (instantaneous center or relative velocity method); identifies all three failure theories with proper stress state interpretation; recognizes 2D force system requires both resultant force and moment; distinguishes polymer structures at molecular level; applies correct interference criterion (minimum teeth or contact ratio) for gears.Uses correct formulas but with minor conceptual gaps (e.g., treats biaxial stress as uniaxial for one theory, or confuses thermoset/thermoplastic examples); recognizes interference but uses approximate check.Fundamental errors: uses acceleration analysis for velocity in (a), confuses principal stress with von Mises directly, treats force system as concurrent only, or applies Lewis formula instead of interference check.
Numerical accuracy20%10All five sub-parts with correct numerical values: slider velocity ≈ 0.52 m/s; FOS values 1.875, 1.25, 1.42 (or 1.417); resultant force and moment at A with proper vector components; interference confirmed with new φ ≈ 23.5°-25°; consistent significant figures and units (MPa, m/s, N, Nm, degrees).Correct approach but arithmetic slips in 1-2 sub-parts (e.g., FOS for distortion energy slightly off, or slider velocity using approximate formula with error); units mostly consistent.Major calculation errors: wrong slider velocity magnitude or direction, FOS values order-of-magnitude wrong, incorrect resultant moment, or pressure angle calculation completely off.
Diagram quality20%10Clear velocity polygon for (a) with vectors v_B, v_BA, v_A labelled; free body diagram for (c) showing all forces, dimensions, and moment arms; gear tooth geometry sketch for (e) showing interference region, base circle, and addendum circle; comparison table for (d) well-structured.Diagrams present but incomplete: velocity polygon without velocity image, FBD without dimensions, or gear sketch without key circles labelled; table present but missing key columns.No diagrams despite Figure 1 references; or completely wrong diagrams (e.g., acceleration polygon instead of velocity, wrong force directions).
Step-by-step derivation20%10Explicit derivation: (a) velocity equation v_p = ωr[sin θ + sin(2θ)/2n] or polygon construction; (b) shows τ_max = (σ1-σ3)/2 and σ_eq formula derivation; (c) ΣFx, ΣFy, ΣM_A = 0; (e) derives minimum teeth t_min = 2/sin²φ and solves for new φ.Uses standard formulas with some intermediate steps shown; jumps from given data to final answer in 1-2 sub-parts but others well-derived.Final answers stated without working; or incorrect derivation sequence (e.g., assumes interference then proves it, circular reasoning).
Practical interpretation20%10Contextualizes: (a) engine piston speed relevance; (b) which theory suits ductile vs brittle materials, why distortion energy is most accurate for ductile; (d) Indian industry examples (ISRO composites use epoxy thermosets, packaging uses PP/PE thermoplastics); (e) consequences of interference (noise, wear) and why 25° or 14.5° pressure angles are standards.Brief mention of practical relevance in 2-3 sub-parts without elaboration; generic statements about safety factors.Purely mathematical treatment with no physical interpretation; no mention of material behavior or design implications.

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