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Mechanical Engineering 2022 Paper I 50 marks Calculate

Paper I — 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…

(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)
(i)

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: 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)
(i)

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 : स्प्रिंग दृढ़ता

(ii)

केन्द्रीय स्प्रिंग का प्रारम्भिक संपीडन

(iii)

130 mm घूर्णन अर्धव्यास की संगता की साम्यावस्था चाल (20 अंक)

Q3 of the 2022 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2022 Mechanical Engineering paper
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The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(c) A 3D diagram of a pipe assembly fixed to a vertical wall at point A. The coordinate system is defined with the origin at A, the y-axis pointing vertically upwards, the x-axis pointing horizontally to the right, and the z-axis pointing outwards from the wall to the left. The pipe consists of three segments: 1) A horizontal segment of length 20 m extending from A along the negative z-axis. 2) A vertical segment of length 8 m extending downwards from the end of the first segment. 3) A horizontal segment of length 10 m extending from the bottom of the vertical segment, parallel to the x-axis. Several forces are applied to the pipe: A 900 N force acts vertically downwards at the midpoint of the 20 m segment. A 200 N force acts vertically downwards at the midpoint of the 8 m segment. At the free end of the 10 m segment, four forces are applied: 600 N acting horizontally to the left (negative z-direction), 400 N acting horizontally to the right (positive x-direction), 300 N acting vertically downwards, and 500 N acting vertically downwards.

Table titled 'AREAS UNDER THE STANDARD NORMAL PROBABILITY DISTRIBUTION'. A small bell curve diagram is shown above the table with the area between 0 and z shaded. The table lists values representing the proportion of area under the normal curve between the mean (mu = 0) and a positive value of z. The rows are labeled by z values from 0.0 to 3.0 in increments of 0.1. The columns are labeled by the second decimal place of z: .00, .01, .02, .03, .04, .05, .06, .07, .08, .09. The table contains the following data: z=0.0: 0.0000, 0.0040, 0.0080, 0.0120, 0.0160, 0.0199, 0.0239, 0.0279, 0.0319, 0.0359. z=0.1: 0.0398, 0.0438, 0.0478, 0.0517, 0.0557, 0.0596, 0.0636, 0.0675, 0.0714, 0.0753. z=0.2: 0.0793, 0.0832, 0.0871, 0.0910, 0.0948, 0.0987, 0.1026, 0.1064, 0.1103, 0.1141. z=0.3: 0.1179, 0.1217, 0.1255, 0.1293, 0.1331, 0.1368, 0.1406, 0.1443, 0.1480, 0.1517. z=0.4: 0.1554, 0.1591, 0.1628, 0.1664, 0.1700, 0.1736, 0.1772, 0.1808, 0.1844, 0.1879. z=0.5: 0.1915, 0.1950, 0.1985, 0.2019, 0.2054, 0.2088, 0.2123, 0.2157, 0.2190, 0.2224. z=0.6: 0.2257, 0.2291, 0.2324, 0.2357, 0.2389, 0.2422, 0.2454, 0.2486, 0.2517, 0.2549. z=0.7: 0.2580, 0.2611, 0.2642, 0.2673, 0.2703, 0.2734, 0.2764, 0.2794, 0.2823, 0.2852. z=0.8: 0.2881, 0.2910, 0.2939, 0.2967, 0.2995, 0.3023, 0.3051, 0.3078, 0.3106, 0.3133. z=0.9: 0.3159, 0.3186, 0.3212, 0.3238, 0.3264, 0.3289, 0.3315, 0.3340, 0.3365, 0.3389. z=1.0: 0.3413, 0.3438, 0.3461, 0.3485, 0.3508, 0.3531, 0.3554, 0.3577, 0.3599, 0.3621. z=1.1: 0.3643, 0.3665, 0.3686, 0.3708, 0.3729, 0.3749, 0.3770, 0.3790, 0.3810, 0.3830. z=1.2: 0.3849, 0.3869, 0.3888, 0.3907, 0.3925, 0.3944, 0.3962, 0.3980, 0.3997, 0.4015. z=1.3: 0.4032, 0.4049, 0.4066, 0.4082, 0.4099, 0.4115, 0.4131, 0.4147, 0.4162, 0.4177. z=1.4: 0.4192, 0.4207, 0.4222, 0.4236, 0.4251, 0.4265, 0.4279, 0.4292, 0.4306, 0.4319. z=1.5: 0.4332, 0.4345, 0.4357, 0.4370, 0.4382, 0.4394, 0.4406, 0.4418, 0.4429, 0.4441. z=1.6: 0.4452, 0.4463, 0.4474, 0.4484, 0.4495, 0.4505, 0.4515, 0.4525, 0.4535, 0.4545. z=1.7: 0.4554, 0.4564, 0.4573, 0.4582, 0.4591, 0.4599, 0.4608, 0.4616, 0.4625, 0.4633. z=1.8: 0.4641, 0.4649, 0.4656, 0.4664, 0.4671, 0.4678, 0.4686, 0.4693, 0.4699, 0.4706. z=1.9: 0.4713, 0.4719, 0.4726, 0.4732, 0.4738, 0.4744, 0.4750, 0.4756, 0.4761, 0.4767. z=2.0: 0.4772, 0.4778, 0.4783, 0.4788, 0.4793, 0.4798, 0.4803, 0.4808, 0.4812, 0.4817. z=2.1: 0.4821, 0.4826, 0.4830, 0.4834, 0.4838, 0.4842, 0.4846, 0.4850, 0.4854, 0.4857. z=2.2: 0.4861, 0.4864, 0.4868, 0.4871, 0.4875, 0.4878, 0.4881, 0.4884, 0.4887, 0.4890. z=2.3: 0.4893, 0.4896, 0.4898, 0.4901, 0.4904, 0.4906, 0.4909, 0.4911, 0.4913, 0.4916. z=2.4: 0.4918, 0.4920, 0.4922, 0.4925, 0.4927, 0.4929, 0.4931, 0.4932, 0.4934, 0.4936. z=2.5: 0.4938, 0.4940, 0.4941, 0.4943, 0.4945, 0.4946, 0.4948, 0.4949, 0.4951, 0.4952. z=2.6: 0.4953, 0.4955, 0.4956, 0.4957, 0.4959, 0.4960, 0.4961, 0.4962, 0.4963, 0.4964. z=2.7: 0.4965, 0.4966, 0.4967, 0.4968, 0.4969, 0.4970, 0.4971, 0.4972, 0.4973, 0.4974. z=2.8: 0.4974, 0.4975, 0.4976, 0.4977, 0.4977, 0.4978, 0.4979, 0.4979, 0.4980, 0.4981. z=2.9: 0.4981, 0.4982, 0.4982, 0.4983, 0.4984, 0.4984, 0.4985, 0.4985, 0.4986, 0.4986. z=3.0: 0.4987, 0.4987, 0.4987, 0.4988, 0.4988, 0.4989, 0.4989, 0.4989, 0.4990, 0.4990.

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a) Thickness and strains of the thin cylinder

Diagram: Thin cylinder stresses

Given d = 200 mm, L = 1000 mm, p = 10 MPa, σ_allow = 200 MPa, E = 2 × 10⁵ MPa. Thin closed cylinder, linear elastic, Poisson ratio ν = 0.3 assumed (not given).

Thickness: the hoop stress σh = pd/(2t) is twice σl = pd/(4t), so it governs: t = pd/(2σ_allow) = (10 × 200)/(2 × 200) = 5 mm (d/t = 40, thin-shell theory valid).

Stresses: σh = 200 MPa, σl = 100 MPa. Hoop strain: εh = (σh − νσl)/E = (200 − 30)/(2 × 10⁵) = 8.5 × 10⁻⁴ Longitudinal strain: εl = (σl − νσh)/E = (100 − 60)/(2 × 10⁵) = 2.0 × 10⁻⁴

Diameter increases by 0.17 mm and length by 0.20 mm, volumetric strain 2εh + εl = 1.9 × 10⁻³.

(b) SFD and BMD of the overhanging beam

Diagram: Overhanging beam SFD BMD

Total load W = 360 × 10 + 400 = 4000 N, so R₁ = R₂ = 2000 N. Let the left support be at a from the left end, the right one at a + 7. Moments about the left end: 2000a + 2000(a + 7) = 3600 × 5 ⟹ a = 1 m. Supports at x = 1 m and 8 m (overhangs 1 m and 2 m).

Shear force (x from the left end, upward on the left positive): 0 to 1 m: V = −400 − 360x, −400 to −760 N 1 to 8 m: V = 1600 − 360x, +1240 to −1280 N, zero at x = 4.444 m 8 to 10 m: V = 3600 − 360x, +720 N to 0

Bending moment (sagging positive): 0 to 1 m: M = −400x − 180x², M(1) = −580 N·m 1 to 8 m: M = −180x² + 1600x − 2000 At the right support: M(8) = −360 × 2²/2 = −720 N·m

Maximum bending moment at x = 4.444 m: M_max = −180(4.444)² + 1600(4.444) − 2000 = +1555.6 N·m (sagging), larger than the 720 N·m hogging at the right support.

Points of contraflexure: 180x² − 1600x + 2000 = 0 ⟹ x = (1600 ± 1058.3)/360, x = 1.505 m and 7.384 m from the left end (0.505 m right of the left support and 0.616 m left of the right support).

The SFD is linear with jumps of 2000 N at each support, and the BMD is parabolic, hogging over the supports and sagging between the contraflexure points.

(c) Hartnell governor

Diagram: Hartnell governor

x = 100 mm, y = 80 mm, m = 1.5 kg, r₁ = 90 mm, r₂ = 140 mm, N₁ = 840 rpm, N₂ = 882 rpm. Moments about the fulcrum (sleeve mass, friction and obliquity neglected): Fc·x = (S/2)·y, so S = 2Fc(x/y), with Fc = mω²r.

ω₁ = 87.96 rad/s, Fc1 = 1.5 × 87.96² × 0.09 = 1044.6 N, S₁ = 2611.5 N ω₂ = 92.36 rad/s, Fc2 = 1.5 × 92.36² × 0.14 = 1791.5 N, S₂ = 4478.7 N Sleeve lift h = (r₂ − r₁)(y/x) = 50 × 0.8 = 40 mm

(c)(i) Spring stiffness k = (S₂ − S₁)/h = 1867.2/40 = 46.68 N/mm

(c)(ii) Initial compression δ₀ = S₁/k = 2611.5/46.68 = 55.9 mm

(c)(iii) Speed at r = 130 mm Lift = (130 − 90)(0.8) = 32 mm, S = 2611.5 + 46.68 × 32 = 4105.3 N, Fc = S(y/x)/2 = 1642.1 N, ω² = 1642.1/(1.5 × 0.13) = 8421, ω = 91.77 rad/s, N = 876.3 rpm, lying between 840 and 882 rpm and rising with radius, so the governor is stable.

How this answer will be evaluated

What earns full marks: Rigorous application of governing equations with clear diagrams and correct units.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Determine thickness, hoop strain, and longitudinal strain for a thin cylinder. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • State thin cylinder assumptions (t << d)
    • Apply hoop stress formula σh = pd/2t
    • Solve for thickness t using allowable stress
    • t = 5 mm, εh = 8.5 × 10⁻⁴, εl = 2.0 × 10⁻⁴ (ν = 0.3 assumed)

    Loses marks

    • Plugging numbers without governing equations
    • Confusing hoop and longitudinal stress formulas

    Earns more

    • Show dimensional consistency check
    • State Poisson's ratio assumption (e.g., 0.3)

    Extra mark

    • Provide a labelled schematic of the cylinder
  2. (b) Draw SF and BM diagrams, finding max BM and contraflexure points. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Locate supports (1 m and 8 m) so that R₁ = R₂ = 2000 N
    • Derive Shear Force equation for each segment
    • Derive Bending Moment equation for each segment
    • M_max = +1555.6 N·m at 4.44 m, contraflexure at 1.505 m and 7.384 m

    Loses marks

    • Incorrect sign convention for shear/moment
    • Missing points of contraflexure in the diagram

    Earns more

    • Label all key values on the diagrams
    • Calculate maximum bending moment explicitly

    Extra mark

    • Include a clear free-body diagram of the beam
  3. (c) Determine spring stiffness, initial compression, and equilibrium speed. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Apply force balance at min and max radii
    • Calculate centrifugal force at extreme speeds
    • Solve for spring stiffness k
    • k ≈ 46.7 N/mm, initial compression ≈ 55.9 mm, speed at 130 mm ≈ 876 rpm

    Loses marks

    • Ignoring the 5% speed increase for max speed
    • Incorrect lever arm ratio application

    Earns more

    • State assumption of negligible sleeve mass
    • Show calculation for speed at 130 mm radius

    Extra mark

    • Draw a schematic of the Hartnell governor
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