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Mechanical Engineering 2021 Paper I 50 marks Calculate

Paper I — Q2

(a) In a slider-crank mechanism, the lengths of the crank and connecting rod are 150 mm and 600 mm respectively. Locate all the…

(a)

In a slider-crank mechanism, the lengths of the crank and connecting rod are 150 mm and 600 mm respectively. Locate all the I-centres of the mechanism for the position when the crank has turned 30° from IDC. Also, find the velocity of the slider and the angular velocity of the connecting rod, if the crank rotates at 30 rad/s. 15 marks

(b)

A mass weighing 100 N is suspended from a spring of constant k = 4000 N/m. At time t = 0, it has a downward velocity of 1 m/s as it passes through the position of static equilibrium. Determine the following : (i) The static spring deflection. (ii) The natural frequency of the system. (iii) The displacement (x) of the mass as a function of time, where x is measured from the position of static equilibrium. (iv) The maximum acceleration attained by the mass. 15 marks

(c)

Write the equations for shearing force and bending moment for various sections and draw SFD and BMD for the beam supported at A and B as shown in the figure. 20 marks

हिंदी में प्रश्न पढ़ें
(a)

एक सर्पी-क्रैंक शेप्ट यंत्र युक्ति में क्रैंक तथा संयोजी दंड की लंबाई क्रमशः 150 mm तथा 600 mm है। क्रैंक के IDC से 30° घूमने के बाद की स्थिति के लिए यंत्र युक्ति के सभी I-केंद्रों का स्थान निर्धारित कीजिए। यदि क्रैंक 30 rad/s पर घूर्णन कर रहा हो, तो स्पर्क (स्लाइडर) का वेग तथा संयोजी दंड का कोणीय वेग भी ज्ञात कीजिए। (15 अंक)

(b)

एक द्रव्यमान जिसका भार 100 N है, एक ऐसी स्प्रिंग से लटका है जिसका स्थिरांक k = 4000 N/m है। समय t = 0 पर स्थैतिक स्थिरता (संतुलन) के स्थान से गुजरते समय इसका नीचे की ओर वेग 1 m/s है। निम्न का मान ज्ञात कीजिए : (i) स्थैतिक स्प्रिंग विस्थापन। (ii) तंत्र की स्वाभाविक आवृत्ति। (iii) समय के फलन के रूप में द्रव्यमान का विस्थापन (x), जहाँ x का मापन स्थैतिक संतुलन के स्थान से किया जाता है। (iv) द्रव्यमान द्वारा लब्ध अधिकतम त्वरण। (15 अंक)

(c)

चित्र में दर्शाए अनुसार टेक A और B पर आधारित (स्थित) एक धरन के विभिन्न काटों (सेक्शन्स) के लिए अपरूपक बल और बंकन आघूर्ण हेतु समीकरण लिखिए तथा SFD एवं BMD आरेखित कीजिए। (20 अंक)

Q2 of the 2021 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2021 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 horizontal beam AE of total length 7 m supported at points A and B. A is at the left end with a support. Point C is 2 m to the right of A, where a clockwise concentrated moment of 2 kN-m is applied. Point D is 2 m to the right of C (4 m from A), where a downward vertical point load of 1 kN is applied. Point B is a support located 2 m to the right of D (6 m from A). The beam extends as an overhang from B to point E for a length of 1 m. At the right free end E (7 m from A), a downward vertical point load of 1 kN is applied. The span intervals between points A-C, C-D, D-B, and B-E are dimensioned as 2 m, 2 m, 2 m, and 1 m respectively.

A horizontal beam of total length 7 m with points labelled from left to right as A, C, D, B, and E. Supports are located at point A (left end) and point B, both shown with upward vertical support arrows. The span between A and C is 2 m, between C and D is 2 m, between D and B is 2 m, and between B and E (overhang) is 1 m. At point C, there is an applied clockwise concentrated moment of 2 kN-m. At point D, there is a downward vertical point load of 1 kN. At point E (the rightmost free end), there is a downward vertical point load of 1 kN.

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) I-centres, slider velocity and rod angular velocity

Take links 1 = frame, 2 = crank OA (r = 0.15 m), 3 = rod AB (l = 0.60 m), 4 = slider B. Place the crank at θ = 30° to the line of stroke, measured from the dead-centre position on the slider side (the usual crank-angle convention), with O = (0, 0). Then A = (0.1299, 0.0750) m and B = (r cos θ + √(l² − r² sin²θ), 0) = (0.7252, 0) m. Also sin φ = r sin θ/l = 0.125, so φ = 7.18°.

Six I-centres, by Kennedy's theorem: I₁₂ = O; I₂₃ = A; I₃₄ = B; I₁₄ is at infinity, perpendicular to the line of stroke; I₁₃ is the intersection of OA produced with the vertical through B, at (0.7252, 0.4187) m; I₂₄ is the intersection of AB produced with the vertical through O, at (0, 0.0914) m.

Diagram: Instantaneous centres of a slider-crank mechanism at 30 degrees crank angle

Velocities: v_A = ωr = 30 × 0.15 = 4.5 m/s. I₁₃A = √(0.5953² + 0.3437²) = 0.6874 m, so

ω₃ = v_A/I₁₃A = 4.5/0.6874 = 6.55 rad/s.

v_slider = ω₃ × I₁₃B = 6.55 × 0.4187 = 2.74 m/s.

Check: v = ωr(sin θ + sin 2θ/2n), n = l/r = 4, gives 4.5 × 0.6082 = 2.74 m/s.

Answer: slider velocity 2.74 m/s (towards the crank axis); connecting-rod angular velocity 6.55 rad/s, opposite in sense to the crank.

(b) Spring–mass system

m = W/g = 100/9.81 = 10.19 kg.

(i) Static deflection δ = W/k = 100/4000 = 0.025 m = 25 mm.

(ii) ωₙ = √(k/m) = √(4000/10.19) = 19.81 rad/s, so fₙ = ωₙ/2π = 3.15 Hz.

(iii) With x(0) = 0 and ẋ(0) = 1 m/s, x(t) = (ẋ₀/ωₙ) sin ωₙt = 0.0505 sin(19.81t) m, measured downward from the static equilibrium position.

(iv) a_max = ωₙ²X = ωₙẋ₀ = 19.81 m/s².

(c) Overhanging beam: reactions, SFD and BMD

Positions from A: C at 2 m (2 kN·m clockwise couple), D at 4 m (1 kN down), B at 6 m (support), E at 7 m (1 kN down).

ΣM_A = 0: 6R_B − 1(4) − 1(7) − 2 = 0, so R_B = 13/6 = 2.167 kN upward. ΣF_y = 0: R_A = 2 − 13/6 = −1/6 kN, i.e. 0.167 kN downward.

Shear force (x from A, kN): 0 < x < 4: V = −1/6; 4 < x < 6: V = −7/6; 6 < x < 7: V = +1. It drops 1 kN at D, rises 13/6 kN at B, and returns to zero at E after the 1 kN load.

Bending moment (kN·m, sagging positive): 0 ≤ x < 2: M = −x/6; at C the couple adds +2; 2 < x < 4: M = 2 − x/6; 4 < x < 6: M = 6 − 7x/6; 6 < x ≤ 7: M = x − 7.

Key values: M_A = 0, M_C⁻ = −1/3, M_C⁺ = 5/3, M_D = 4/3, M_B = −1, M_E = 0.

The SFD is a step diagram. The BMD is piecewise linear with a jump of 2 at C. Maximum sagging moment is 5/3 kN·m just right of C; maximum hogging moment is 1 kN·m at B.

Diagram: Loading, SFD and BMD of the overhanging beam

How this answer will be evaluated

What earns full marks: Complete working with all diagrams, correct equations, proper units, and physical interpretation throughout.

Evaluation rubric

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

  1. (a) Locate all I-centres of the slider-crank at 30° and find slider velocity and connecting-rod angular velocity 15 marks

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

    Must cover

    • All six I-centres located with I₁₃ and I₂₄ by Kennedy's theorem
    • Connecting-rod angular velocity about 6.55 rad/s
    • Slider velocity about 2.74 m/s

    Loses marks

    • Missing I₁₄ at infinity
    • Using the wrong crank angle convention without stating it

    Earns more

    • Geometry of A and B at 30° computed correctly
    • v_A = ωr and I₁₃A, I₁₃B distances used correctly

    Extra mark

    • Cross-checks with v = ωr(sin θ + sin 2θ/2n)
  2. (b) Find static deflection, natural frequency, displacement-time law and maximum acceleration of the spring-mass system 15 marks

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

    Must cover

    • δ = 25 mm and ωₙ = 19.81 rad/s (fₙ = 3.15 Hz)
    • x(t) = 0.0505 sin(19.81t) m
    • a_max = ωₙ²X = 19.8 m/s²

    Loses marks

    • Using the weight in newtons as the mass
    • Including a cosine term despite x(0) = 0

    Earns more

    • Mass obtained from weight with g = 9.81
    • Initial conditions applied to the free-vibration solution

    Extra mark

    • Notes that the amplitude is ẋ₀/ωₙ
  3. (c) Write shear force and bending moment equations and draw SFD and BMD for the overhanging beam 20 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • R_B = 13/6 kN up and R_A = 1/6 kN down
    • Section-wise SF and BM equations with the jump at C
    • SFD and BMD with maximum values 5/3 kN·m at C and −1 kN·m at B

    Loses marks

    • Ignoring the couple at C
    • Wrong sign convention for the bending moment

    Earns more

    • Moment equilibrium including the 2 kN·m couple with the correct sense
    • Diagram labelled with key values

    Extra mark

    • Identifies zero crossing of bending moment points
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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