Mechanical Engineering 2022 Paper I 50 marks Calculate

Q3

(a) A thin cylinder of diameter 200 mm and length 1000 mm is subjected to an internal pressure of 10 MPa. The allowable stress of the material is 200 MPa and Young's modulus is 200 GPa. Determine the thickness, hoop and longitudinal strains under the given pressure. (10 marks) (b) A beam carries a uniformly distributed load of 360 N/m over the entire span together with a concentrated load of 400 N at the extreme left. The beam is having a span of 10 m and is supported at two points 7 m apart. The supports are so chosen, that each support carries half the total load. Draw the shear force and bending moment diagrams after obtaining the maximum bending moment and points of contraflexure. (20 marks) (c) In a Hartnell governor, the lengths of ball and roller arms of the bell crank lever are 100 mm and 80 mm respectively. Each ball has a mass of 1·5 kg. The extreme radii of rotation of the balls are 90 mm and 140 mm. The minimum equilibrium speed is 840 rpm and the maximum equilibrium speed is 5% greater than this. Assuming the sleeve to be of negligible mass and neglecting friction and obliquity of arms, determine: (i) Spring stiffness (ii) Initial compression of the central spring (iii) Equilibrium speed corresponding to radius of rotation of 130 mm (20 marks)

हिंदी में प्रश्न पढ़ें

(a) एक 200 mm व्यास व 1000 mm लंबाई के पतले बेलन पर 10 MPa का आंतरिक दाब लगा हुआ है । पदार्थ का अनुज्ञेय प्रतिबल 200 MPa व यंग मापांक 200 GPa है । दिए गए दाब पर मोटाई, परिधीय व अनुदैर्ध्य विकृतियाँ ज्ञात कीजिए । (10 अंक) (b) एक धरन संपूर्ण विस्तृति पर 360 N/m के एकसमान वितरित भार के साथ चरम बायीं और एक 400 N का संकेन्द्रित भार वहन करती है । धरन की विस्तृति की लम्बाई 10 m है तथा दो बिन्दुओं, जो परस्पर 7 m की दूरी पर हैं, पर वो आलम्बित है । आलम्ब इस तरह से चुने गए हैं कि प्रत्येक आलम्ब कुल भार का आधा वहन करता है । अधिकतम बंकन आघूर्ण व प्रतिनमन बिन्दु प्राप्त करने के पश्चात् अपरूपण बल आरेख व बंकन आघूर्ण आरेख खींचिए । (20 अंक) (c) एक हार्टनेल गवर्नर में बेल क्रैंक लीवर की बॉल व रोलर भुजाएँ क्रमशः 100 mm व 80 mm हैं । प्रत्येक बॉल का द्रव्यमान 1·5 kg है । बॉलों की घूर्णन की चरम त्रिज्याएँ 90 mm व 140 mm हैं । न्यूनतम साम्यावस्था चाल 840 rpm व अधिकतम साम्यावस्था चाल इससे 5% अधिक है । स्लीव को नगण्य द्रव्यमान का मानते हुए तथा घर्षण व भुजाओं की तिर्यकता की उपेक्षा करते हुए निम्न को ज्ञात कीजिए : (i) स्प्रिंग दृढ़ता (ii) केन्द्रीय स्प्रिंग का प्रारम्भिक संपीडन (iii) 130 mm घूर्णन अर्धव्यास की संगता की साम्यावस्था चाल (20 अंक)

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How this answer will be evaluated

Approach

Calculate the required quantities for each sub-part systematically: for (a) apply thin cylinder stress formulas and strain relations; for (b) first determine support positions using moment equilibrium about supports condition, then draw SFD/BMD with contraflexure points; for (c) apply Hartnell governor equilibrium equations at minimum and maximum speeds. Allocate approximately 20% time to part (a), 40% to part (b) due to complex loading and diagram construction, and 40% to part (c) with its three sub-requirements. Present derivations stepwise with clear free-body diagrams for the beam and governor lever systems.

Key points expected

  • Part (a): Thickness t = pD/(2σ_allow) = 10×200/(2×200) = 5 mm; hoop strain ε_h = (pD/4tE)(2-ν) with ν=0.3 assumed or stated; longitudinal strain ε_l = (pD/4tE)(1-2ν)
  • Part (b): Support positions determined from ΣM=0 condition that each carries half total load (2300 N each); reactions found, SFD drawn with linear and constant segments, BMD parabolic with maximum value and location identified
  • Part (b): Points of contraflexure found where BM=0, typically two points for this overhang configuration with interior supports
  • Part (c): Spring stiffness k determined from equilibrium at r₁=90mm (N₁=840 rpm) and r₂=140mm (N₂=882 rpm) using moment balance about pivot
  • Part (c): Initial compression x₀ from minimum speed equilibrium: mg + (k x₀)(b/a) = mω₁²r₁
  • Part (c): Equilibrium speed at r=130mm found by interpolating or solving governor equation with known k and x₀
  • All parts: Proper units (mm, MPa, GPa, N, m, rad/s, N/m) and significant figures maintained

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies thin cylinder theory with σ_h = pD/2t and σ_l = pD/4t for (a); recognizes overhanging beam with unequal overhangs for (b) and uses ΣM=0 about support to find positions; applies Hartnell governor moment equilibrium mω²r = mg + F_s(b/a) with correct arm ratio for (c)Uses correct formulas but with minor errors like treating cylinder as thick or assuming symmetric supports for (b); governor moment equation correct but arm ratio confusedUses wrong stress formulas (thick cylinder Lame's equations) for (a); treats beam as simply supported at ends for (b); confuses Porter/Hartnell governor principles for (c)
Numerical accuracy20%10All calculations precise: (a) t=5mm, strains ~0.000425 and ~0.000125; (b) support positions at 2m and 9m from left, reactions correct, M_max≈-800 N.m; (c) k≈12.5 N/mm, x₀≈45mm, N at 130mm≈865 rpmCorrect final answers for major quantities but arithmetic slips in intermediate steps (e.g., ω conversion from rpm, or moment arm calculations)Order-of-magnitude errors (e.g., GPa vs MPa confusion, rpm not converted to rad/s in governor), or missing critical calculations entirely
Diagram quality20%10Clear SFD with sign convention, values at key points (supports, ends, load changes); BMD with parabolic curves, maximum value labelled, contraflexure points marked; Hartnell governor schematic with force components shownDiagrams present but missing key labels (values, dimensions) or rough freehand sketches without scale indicationNo diagrams drawn for (b), or SFD/BMD shapes incorrect (e.g., constant BM under UDL); no governor sketch for (c)
Step-by-step derivation20%10Complete derivations: (a) stress from equilibrium, strain from Hooke's law; (b) FBD of entire beam, moment equations to find support positions, section-by-section SF and BM expressions; (c) two equilibrium equations at extreme radii solved simultaneously for k and x₀Key steps shown but some shortcuts taken (e.g., direct formula for beam reactions without derivation, governor equations stated without free-body diagram)Final answers only with no working; or working present but logically disconnected steps with no clear progression
Practical interpretation20%10Interprets (a) cylinder design for pressure vessels (IS 2825 relevance); (b) discusses why contraflexure matters for reinforcement placement in RC beams, overhang design implications; (c) explains governor sensitivity, isochronism, and speed regulation in IC engines/steam turbinesBrief mention of practical relevance for one part only, or generic statements without specific applicationNo interpretation; treats all parts as pure mathematical exercises with no engineering context

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