Mechanical Engineering 2021 Paper I 50 marks Compulsory Calculate

Q1

(a) What is the supporting force system at A for the cantilever beam shown in the figure ? Neglect the weight of the beam. (10 marks) (b) The velocity of a particle, moving in the x-y plane is given by 6·12 î + 3·24 ĵ m/s at time t = 3·65 s. Its average acceleration, during the next 0·02 s is 4 î + 6 ĵ m/s². Determine the velocity v of the particle at t = 3·67 s, and the angle θ between the acceleration vector and the velocity vector at t = 3·67 s. (10 marks) (c) What is carbon equivalent (CE) ? How does the carbon equivalent of a steel affect the hardening heat treatment ? (10 marks) (d) In the epicyclic gear train shown in the figure, the wheel 'C' is keyed to the shaft 'B'. 'D' and 'E' are compound gears. 'C', 'D' and 'E' have 35, 65 and 32 teeth respectively. All the gears have same module. If 'A' and 'B' rotate at 60 rpm and 28 rpm respectively in opposite directions, find the speed and direction of rotation of arm 'G'. (10 marks) (e) A steel wire of 8 mm diameter and length 50 m is used to lift a weight of 2000 N at its lowest end. Calculate the total elongation of the wire, if the density of the steel is 8000 kg/m³ and E = 2·1 × 10⁵ N/mm². (10 marks)

हिंदी में प्रश्न पढ़ें

(a) चित्र में प्रदर्शित एक प्रास (कैन्टीलिवर) धरन के लिए A पर आधारी बल क्या होगा ? धरन के भार को नगण्य मान लीजिए । (10 अंक) (b) x-y समतल में गतिमान एक कण का वेग समय t = 3·65 s पर 6·12 î + 3·24 ĵ m/s द्वारा निर्धारित है । अगले 0·02 s पर्यन्त इसका औसत त्वरण 4 î + 6 ĵ m/s² है । समय t = 3·67 s पर कण का वेग v तथा त्वरण सदिश एवं वेग सदिश के बीच का कोण θ ज्ञात कीजिए । (10 अंक) (c) कार्बन समतुल्य (CE) क्या है ? इसपात का कार्बन समतुल्य, कठोरीकरण ताप उपचार को कैसे प्रभावित करता है ? (10 अंक) (d) चित्र में प्रदर्शित अधिचक्रिक गियर माला में, चक्र 'C' शैफ्ट 'B' में कुंजीयित है । 'D' तथा 'E' संयुक्त गियर हैं । 'C', 'D' तथा 'E' में क्रमशः 35, 65 व 32 दांते हैं । सभी गियरों का एक ही मॉड्यूल है । यदि 'A' और 'B' क्रमशः 60 rpm तथा 28 rpm पर विपरीत दिशा में घूर्णन करते हैं, तो भुजा 'G' के घूर्णन की गति व दिशा ज्ञात कीजिए । (10 अंक) (e) 8 mm व्यास एवं 50 m लंबाई का एक इस्पात का तार, अपने निम्नतम सिरे पर 2000 N के भार को उठाने के लिए प्रयोग किया जाता है । यदि इस्पात का घनत्व 8000 kg/m³ तथा E = 2·1 × 10⁵ N/mm² है, तो तार के सम्पूर्ण दीर्घीकरण की गणना कीजिए । (10 अंक)

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Directive word: Calculate

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How this answer will be evaluated

Approach

Calculate requires systematic problem-solving across all five sub-parts. For (a), draw the FBD and apply equilibrium equations to find reactions at A. For (b), use kinematic equations to find final velocity and vector dot product for the angle. For (c), define CE formula and explain its effect on hardenability and cracking susceptibility. For (d), apply the tabular method or relative velocity method for epicyclic trains. For (e), account for both external load and self-weight elongation. Allocate approximately 15% time to (a), 15% to (b), 20% to (c), 30% to (d), and 20% to (e) based on computational complexity.

Key points expected

  • (a) Correct FBD with all applied loads shown; equilibrium equations ΣFx=0, ΣFy=0, ΣMA=0 yielding reaction forces and moment at A
  • (b) Velocity at t=3.67 s: v = 6.20î + 3.36ĵ m/s; angle θ = cos⁻¹[(a·v)/(|a||v|)] ≈ 15.3° or equivalent correct calculation
  • (c) CE formula: CE = %C + %Mn/6 + (%Cr+%Mo+%V)/5 + (%Ni+%Cu)/15; higher CE increases hardenability but raises risk of quench cracking and distortion
  • (d) Tabular method with speed ratios: y = -28 rpm (arm fixed), x + y = 60 for gear A; solving gives arm speed ≈ 4.5 rpm in same direction as A
  • (e) Total elongation δ = δ_load + δ_self = WL/AE + ρgL²/2E = 0.947 mm + 0.373 mm ≈ 1.32 mm with proper unit conversions

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies static equilibrium for (a), kinematic relations for (b), metallurgical CE concept for (c), epicyclic gear train theory for (d), and superposition of axial deformations for (e); no conceptual confusion between similar-looking formulas.Majority concepts correct but minor errors like confusing absolute vs relative velocity in (d) or omitting self-weight in (e); CE definition correct but hardening explanation superficial.Fundamental errors: treats cantilever as simply supported, confuses CE with carbon content only, applies wrong gear train formula, or ignores self-weight elongation entirely.
Numerical accuracy20%10All five sub-parts yield correct numerical answers within reasonable rounding: reactions match equilibrium checks, velocity components precise, CE calculation accurate, arm speed correctly determined, elongation sums to ~1.32 mm with proper unit handling (m to mm, MPa to N/mm²).Correct approach but arithmetic slips in 1-2 parts (e.g., sign error in gear train, unit conversion error in wire elongation, or angle calculation using wrong formula); final answers partially correct.Multiple numerical errors, order-of-magnitude mistakes (e.g., elongation in meters), or answers without any working that are clearly wrong.
Diagram quality20%10Clear FBD for (a) with all forces and dimensions labelled; velocity/acceleration vector sketch for (b) showing angle θ; neat epicyclic train schematic for (d) with gears C, D, E, arm G and shafts A, B identified; free-body of wire element for (e) showing varying tension.FBD present but missing some labels; gear train sketch rough but identifiable; no vector diagram for (b); diagrams adequate but not to scale.Missing or misleading diagrams; no FBD for equilibrium problem, no gear train sketch, or diagrams that contradict the written solution.
Step-by-step derivation20%10Every sub-part shows complete derivation: equilibrium equations with substitution for (a), vector addition with component-wise calculation for (b), CE formula stated with element contributions explained for (c), tabular method or relative velocity equations with clear algebraic steps for (d), integration of varying strain or superposition clearly shown for (e).Derivations present but skips key steps (e.g., jumps to gear train solution without showing table, states elongation formula without deriving self-weight term); some substitution steps missing.Final answers stated without derivation; or incorrect formulas applied blindly without justification; no intermediate steps shown.
Practical interpretation20%10For (a), comments on fixed-end moment significance in design; for (c), relates CE to weldability (Indian Boiler Regulation context) and preheating requirements; for (e), notes that self-weight matters for long cables (bridge cables, mine hoists) and suggests safety factor; connects gear train to automotive differentials or machine tool gearboxes.Brief practical mention for 2-3 parts but no depth; generic statements like 'important in engineering' without specific application.No interpretation; treats all parts as pure calculation exercises with no engineering context or real-world relevance stated.

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