Mechanical Engineering 2025 Paper I 50 marks Solve

Q4

(a) A car is moving in a straight line with a velocity of v = (0·6t² + 2t) m/s for a short duration, where t is in seconds. Take initial time t = 0, s = 0. Find : (i) distance travelled in 4 s, and (ii) acceleration at 4 s. (10 marks) (b) A solid shaft AB rotates at 450 rpm and transmits 20 kW from the motor M to machine tools connected to gears F and G. A power of 8 kW is taken off at gear F and 12 kW is taken off at gear G. The allowable shear stress is 55 MPa. Determine the smallest permissible diameter of the shaft AB. (20 marks) (c) (i) In an epicyclic gear train of the sun and planet type shown in the figure below, the annular gear 'A' meshes internally. The three identical planet wheels 'P' of equal size, mesh with annular gear 'A' and the sun wheel 'S'. The planet wheels are carried by a star shaped spider 'C'. The size of the different toothed wheels are such that the spider 'C' which carries the planet wheels is to make one revolution for every 5 rotations of the spindle carrying the sun wheel 'S', when the gear 'A' is stationary. If the minimum number of teeth on any wheel is 14, determine the number of teeth for all the wheels. Further, if the driving torque on the sun wheel is 200 N-m, determine the fixing torque required to keep the annular gear 'A' stationary. (10 marks) (ii) In a four cylinder symmetrical engine, the intermediate cranks are at 90° and each has a reciprocating mass of 500 kg. The engine is in complete primary balance. The centre distance between intermediate cranks is 600 mm and between extreme cranks is 1800 mm. The lengths of the connecting rods and cranks are 800 mm and 200 mm, respectively. Determine the masses fixed to the extreme cranks along with their relative angular positions. If the engine speed is 150 rpm, find the magnitude of secondary unbalanced forces. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) अल्पकाल के लिए एक कार सीधी रेखा में v = (0·6t² + 2t) m/s के वेग से चल रही है, जहाँ t सेकेंड में है। प्रारंभिक समय t = 0, s = 0 लीजिए। ज्ञात कीजिए : (i) 4 s में तय की गई दूरी, और (ii) 4 s पर त्वरण। (10 अंक) (b) एक ठोस शाफ्ट AB 450 rpm पर घूमता है तथा मोटर M से गियर F एवं G द्वारा जुड़े मशीन टूल्स को 20 kW संचारित करता है। गियर F पर 8 kW की शक्ति हटाई गई है तथा गियर G पर 12 kW की शक्ति हटाई गई है। अनुमेय (स्वीकार्य) अपरूपण प्रतिबल 55 MPa है। शाफ्ट AB का सबसे छोटा अनुमेय व्यास निर्धारित कीजिए। (20 अंक) (c) (i) सूर्य और ग्रह प्रकार की एक अधिचक्रिक गियर माला नीचे चित्र में दर्शाई गई है। वलयाकार गियर 'A' आंतरिक रूप से अन्तयोजित है। तीन समान आकार के एकसमान ग्रह चक्र 'P', वलयाकार गियर 'A' तथा सूर्य (सन) चक्र 'S' के साथ अन्तयोजित है। ग्रह चक्रों को एक तारे (स्टार) के आकार के लूता (स्पाइडर) 'C' द्वारा ले जाया जाता है। विभिन्न दांतदार चक्रों के आकार इस प्रकार हैं कि सूर्य चक्र 'S' को ले जाने वाले तर्कु (स्पिंडल) के प्रत्येक 5 घूर्णन पर, ग्रह चक्रों को ले जाने वाला लूता 'C' एक परिक्रमण करता है, जबकि गियर 'A' स्थिर होता है। यदि किसी भी चक्र पर दांतों की न्यूनतम संख्या 14 हो, तो सभी चक्रों पर दांतों की संख्या निर्धारित कीजिए। यदि सूर्य चक्र पर चालन बल-आघूर्ण 200 N-m हो, तो वलयाकार गियर 'A' को स्थिर बनाए रखने के लिए आवश्यक स्थिरीकरण बल-आघूर्ण (फिक्सिंग टार्क) निर्धारित कीजिए। (10 अंक) (ii) एक चार सिलिंडर सममित इंजन में, मध्यवर्ती क्रैंक 90° पर है तथा प्रत्येक का प्रत्यागामी द्रव्यमान 500 kg है। इंजन पूर्ण प्राथमिक संतुलन में है। मध्यवर्ती क्रैंकों के बीच केंद्र दूरी 600 mm तथा चरम क्रैंकों के बीच 1800 mm है। कनेक्टिंग रॉडों तथा क्रैंकों की लंबाई क्रमशः: 800 mm तथा 200 mm है। चरम क्रैंकों पर निर्धारित द्रव्यमानों को उनकी सापेक्ष कोणीय स्थितियों के साथ निर्धारित कीजिए। यदि इंजन की गति 150 rpm है, तो द्वितीयक असंतुलित बलों का परिमाण ज्ञात कीजिए। (10 अंक)

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Approach

Solve all five sub-parts systematically, allocating time proportionally to marks: ~12 minutes for (a), ~24 minutes for (b), ~12 minutes for (c)(i), and ~12 minutes for (c)(ii). Begin each part with the governing equation, show complete derivation with proper units, and conclude with numerical answers. For (c)(i), draw the epicyclic gear train diagram; for (c)(ii), sketch the crank arrangement and secondary force polygon.

Key points expected

  • Part (a): Integrate v = 0.6t² + 2t to get s = 0.2t³ + t²; evaluate s(4) = 28.8 m; differentiate to get a = 1.2t + 2; evaluate a(4) = 6.8 m/s²
  • Part (b): Calculate torque T = P/(2πN/60) = 424.4 N·m; apply τ_max = 16T/(πd³) with τ_allow = 55 MPa; solve for d = 34.4 mm; round up to 35 mm or next standard size
  • Part (c)(i): Apply epicyclic train equation (N_S - N_C)/(N_A - N_C) = -T_A/T_S with N_A = 0, N_C/N_S = 1/5; derive T_A = T_S + 2T_P; with T_P = 14 minimum, solve T_S = 21, T_A = 49; fixing torque = 200 × (5-1) = 800 N·m
  • Part (c)(ii): Apply primary balance conditions Σmr cosθ = 0 and Σmr sinθ = 0; for symmetrical engine, m₁ = m₄, m₂ = m₃ = 500 kg; solve m₁ = 250 kg at 180° to intermediate cranks; secondary unbalanced force = 2mrω²(λ/2)cos2θ with λ = r/l = 0.25; compute F_sec = 1542 N
  • Units and significant figures: All answers carry proper SI units (m, m/s², mm, N·m, N, kg); intermediate steps retain 3-4 significant figures

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies kinematic relations (a = dv/dt, s = ∫v dt) in (a); uses torsion formula τ = Tr/J with proper power-torque conversion in (b); applies tabular method or relative velocity method for epicyclic train in (c)(i); recognizes primary and secondary balancing conditions with correct λ = r/l ratio in (c)(ii)Correct final formulas but minor conceptual slips (e.g., confuses angular velocity units in power-torque relation, or uses wrong sign convention in epicyclic equation)Fundamental errors: treats v = ds/dt as algebraic rather than calculus relation; uses bending stress formula instead of torsion for shaft; applies simple gear train formula to epicyclic train; confuses primary and secondary forces
Numerical accuracy20%10All five numerical answers correct: s(4) = 28.8 m, a(4) = 6.8 m/s²; d = 34-35 mm; T_S = 21, T_P = 14, T_A = 49, fixing torque = 800 N·m; m_extreme = 250 kg at 180°, F_sec ≈ 1540-1550 N; arithmetic shown with proper significant figuresCorrect method but one calculation error (e.g., integration constant error, power unit conversion 20 kW vs 20000 W, or tooth count rounding); final answers within 5-10% of correct valueMultiple calculation errors or order-of-magnitude mistakes (e.g., shaft diameter in meters, torque in kN·m without conversion, secondary force missed by factor of 2 or λ)
Diagram quality20%10Clear epicyclic gear train diagram in (c)(i) showing sun, planet, annular gears and spider with rotation arrows; crank arrangement diagram in (c)(ii) showing 4-cylinder inline configuration with angles 0°, 90°, 90°, 180° and mass vectors; free-body diagram for shaft in (b) with torque directionsDiagrams present but incomplete (e.g., epicyclic train missing spider label, crank diagram without angle markings, or shaft without torque arrows)No diagrams despite (c) explicitly requiring figure interpretation; or irrelevant sketches (e.g., velocity-time graph for (a) when not asked)
Step-by-step derivation20%10Complete working: integration with limits 0 to 4 shown for (a); torque calculation from P = 2πNT/60, then d³ = 16T/(πτ) for (b); tabular method or speed ratio equation with algebraic solution for tooth counts in (c)(i); primary balance equations written and solved for m₁, θ₁ in (c)(ii), then secondary force formula derivedKey steps shown but some shortcuts (e.g., jumps from ∫v dt to final answer without intermediate s(t), or states epicyclic formula without derivation)Final answers only with no derivation; or incorrect formulas applied blindly without explanation
Practical interpretation20%10Comments on physical meaning: for (a), notes acceleration increases with time (non-uniform motion); for (b), selects next standard shaft size (35 mm or 36 mm) with justification; for (c)(i), explains torque multiplication in epicyclic trains (used in automatic transmissions); for (c)(ii), discusses secondary shaking force at 2× engine speed and its effect on engine mountsBrief mention of practical relevance (e.g., 'shaft must be safe', 'balancing reduces vibration') without specific connection to calculated valuesNo interpretation; treats all parts as pure mathematics without engineering context

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