Paper I — Q1
(a) A gripper is shown in the figure. A horizontal force F = 50 N is applied to the handle of the lever at E. The mean diameter…
A gripper is shown in the figure. A horizontal force F = 50 N is applied to the handle of the lever at E. The mean diameter of the single square threaded screw at C and E is 25 mm and lead is 5 mm. Determine the clamping force developed at G. The coefficient of static friction is μs = 0·3. 10 marks
The aluminium rod ABC having Young's modulus 70 GPa consisting of two cylindrical portions AB and BC, is to be replaced with a cylindrical steel rod DE with Young's modulus 200 GPa of same overall length. Determine minimum required diameter, d, of the steel rod if its vertical deformation is not to exceed the deformation of aluminium rod under the same load and if the allowable stress in the steel rod is not to exceed 165 MPa. 10 marks
A tank shown in the figure is filled with compressed air under pressure of 8 MPa. A torque of magnitude T = 12 kNm is applied at the end. The tank has an inner diameter of 180 mm and thickness of wall 12 mm. Determine the maximum normal stress and maximum shearing stress in the tank considering the cylinder is thin. 10 marks
With the help of a neat diagram, illustrate the microstructures of various phases of steel and exhibit the presence of the following: (10 marks) Ferrite
Austenite
Cementite
Pearlite
A pair of involute profile spur gears in mesh have to give a speed ratio of 2. The pressure angle is 20° and the module is 10 mm. The pinion has 24 teeth and drives the larger gear. If the addenda on pinion and gear wheels are equal to one module, determine: (10 marks) length of path of contact,
contact ratio, and
angle of action of the pinion.
हिंदी में प्रश्न पढ़ें
चित्र में एक पकड़ (ग्रिपर) दर्शाया गया है। लीवर के हस्ते (हैंडल) पर E पर एक क्षैतिज बल F = 50 N लगाया गया है। C तथा E पर एकल वर्ग चुड़ी वाले पेंच का औसत व्यास 25 mm तथा अग्रण (लीड) 5 mm है। G पर विकसित होने वाला बंधन बल (क्लैंपिंग फोर्स) निर्धारित कीजिए। स्थैतिक घर्षण गुणांक μs = 0·3 है। (10 अंक)
दो बेलनाकार भागों AB तथा BC से बनी 70 GPa यंग मापांक वाली एक ऐलुमिनियम छड़ ABC को एक 200 GPa यंग मापांक वाली समान समग्र लंबाई की बेलनाकार इस्पात छड़ DE के द्वारा प्रतिस्थापित किया जाना है। यदि इसका उद्वाधर विरूपण समान भार के अंतर्गत ऐलुमिनियम छड़ के विरूपण से अधिक न हो, एवं यदि इस्पात छड़ में स्वीकार्य प्रतिबल 165 MPa से अधिक न हो, तो इस्पात छड़ का न्यूनतम आवश्यक व्यास d निर्धारित कीजिए। (10 अंक)
चित्र में दर्शाई गई एक टंकी को 8 MPa के दबाव के तहत संपीडित वायु द्वारा भरा गया है। सिरे पर T = 12 kNm परिमाण का बल-आघूर्ण लगाया गया है। टंकी का आंतरिक व्यास 180 mm तथा दीवार की मोटाई 12 mm है। बेलन को पतला मानते हुए टंकी में अधिकतम सामान्य प्रतिबल तथा अधिकतम अपरूपण प्रतिबल निर्धारित कीजिए। (10 अंक)
एक साफ-सुथरे आरेख की सहायता से इस्पात के विभिन्न चरणों की सूक्ष्म संरचनाओं को चित्रित कीजिए तथा निम्नलिखित की उपस्थिति प्रदर्शित कीजिए: (10 अंक) फेराइट
ऑस्टेनाइट
सीमेंटाइट
पर्लाइट
एक जोड़ी प्रतिकेंद्रज प्रोफाइल में अंतर्नोजित स्पर गियर को 2 का गति अनुपात प्रदान करना है। दबाव कोण 20° है तथा प्रमात्रक (मॉड्यूल) 10 mm है। पिनियन में 24 दांते हैं तथा यह बड़े गियर को चलाती है। यदि पिनियन एवं गियर चक्रों पर युक्तक (ऐडेंडा) एक प्रमात्रक के बराबर है, तो निर्धारित कीजिए: (10 अंक) संपर्क मार्ग की लंबाई,
संपर्क अनुपात, और
पिनियन का क्रिया कोण।
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.
(a) A schematic diagram of a mechanical gripper mechanism. The assembly consists of a C-shaped frame with a horizontal top arm and a horizontal bottom arm. A vertical square-threaded screw passes through the top arm at point C and the bottom arm at point D. The screw extends downwards to a horizontal handle lever at point E. A horizontal force F is applied to the end of the handle at E. The handle length from the screw axis to point E is 125 mm. The top arm has a pivot point A and a clamping point C. The bottom arm has a pivot point B and a connection point D. The distance from the clamping point G (located at the left end of the top arm) to the pivot A is 200 mm. The distance from the pivot A to the screw axis at C is 200 mm. The screw is single square threaded with a mean diameter of 25 mm and a lead of 5 mm. The coefficient of static friction is 0.3. The applied force F is 50 N.
(b) Two vertical rod configurations are shown side-by-side. On the left is a stepped aluminium rod ABC fixed at the bottom end C. It consists of two cylindrical sections: the upper section AB has a length of 300 mm and a diameter of 38 mm; the lower section BC has a length of 450 mm and a diameter of 56 mm. A vertical downward force of 125 kN is applied at the top end A. On the right is a uniform cylindrical steel rod DE fixed at the bottom end E. It has a uniform diameter labeled 'd' and a total length equal to the sum of the lengths of sections AB and BC (750 mm). A vertical downward force of 125 kN is applied at the top end D.
(c) A vertical cylindrical tank with a flat top and bottom, mounted on a square base plate. A curved arrow labeled 'T' is drawn around the top rim of the cylinder, indicating a torque applied about the vertical axis. The cylinder is depicted with a shaded surface to suggest a 3D perspective.
Model answer
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(a)
Interpretation. Jaws GAC and BD are pinned to link AB at A and B. The screw is threaded (opposite-hand) into both jaws at C and D (the printed "C and E" is read as C and D, since E is the handle), so turning it drives C and D apart and closes the jaws at G. F acts at right angles to the 125 mm handle.
Torque applied to the screw: M = F × 0.125 = 50 × 0.125 = 6.25 N·m
Screw geometry: Lead angle: tan α = L/(π dₘ) = 5/(π × 25) = 0.06366, so α = 3.643° Friction angle: φs = tan⁻¹ 0.3 = 16.699° Tightening is a "raising-load" case, with tan(φs + α) = tan 20.342° = 0.37074.
Torque equilibrium of the screw. The same axial force W acts at C and at D, and both threads resist rotation: M = 2W(dₘ/2)tan(φs + α) 6.25 = 2 × W × 0.0125 × 0.37074, so W = 674.3 N
Since φs exceeds α, the screw is self-locking, and the clamp holds after the handle is released.
Moment equilibrium of jaw GAC about A. The screw pushes C up with W, the work reacts at G with F_G, and the two-force link AB acts vertically at A.
ΣM_A = 0: F_G × 200 = W × 200, so F_G = W = 674 N
The link AB then carries F_G + W = 1349 N in tension.
The clamping force at G is about 674 N. The equal 200 mm arms pass the screw force to G unchanged. If only one thread resisted the torque, the result would be 1.35 kN.
(b)
Given: P = 125 kN. For aluminium, E = 70 GPa; AB is 300 mm long and 38 mm in diameter, and BC is 450 mm long and 56 mm in diameter. For steel, E = 200 GPa, L = 750 mm and σ_allow = 165 MPa.
Deformation of the aluminium rod: A_AB = π(38)²/4 = 1134.1 mm² and A_BC = π(56)²/4 = 2463.0 mm² δ_Al = (P/E)(L_AB/A_AB + L_BC/A_BC) = (125 000/70 000)(300/1134.1 + 450/2463.0) = 1.7857 × (0.26452 + 0.18270) = 0.799 mm
Criterion 1 (stiffness): δ_st = PL/(A E_st) ≤ 0.799 mm A ≥ 125 000 × 750/(200 000 × 0.7986) = 586.9 mm², so d ≥ √(4 × 586.9/π) = 27.34 mm
Criterion 2 (strength): σ = P/A ≤ 165 MPa A ≥ 125 000/165 = 757.6 mm², so d ≥ √(4 × 757.6/π) = 31.06 mm
Selection. Both conditions must hold, so the larger value governs:
d_min ≈ 31.1 mm, set by the allowable stress.
At this diameter the steel shortens by only 0.62 mm, so strength, not stiffness, controls.
(c)
Given: p = 8 MPa, T = 12 kN·m, dᵢ = 180 mm, t = 12 mm. Since t/dᵢ = 1/15, thin-wall theory applies. The mean radius r = 90 + 6 = 96 mm is used, and the radial stress (zero at the outer surface) is neglected.
Membrane stresses: Hoop: σ_h = pr/t = 8 × 96/12 = 64 MPa Longitudinal (closed ends): σ_l = pr/2t = 32 MPa
Torsional shear (thin tube): τ = T/(2πr²t) = 12 × 10⁶/(2π × 96² × 12) = 17.27 MPa
Principal stresses (plane stress in the wall): σ₁,₂ = (σ_h + σ_l)/2 ± √[((σ_h – σ_l)/2)² + τ²] = 48 ± √(16² + 17.27²) = 48 ± 23.54
σ₁ = 71.5 MPa and σ₂ = 24.5 MPa. The principal planes lie at ½tan⁻¹(2 × 17.27/32) = 23.6° to the hoop direction.
Maximum shear stress: In-plane: τ = (σ₁ – σ₂)/2 = 23.5 MPa With the third principal stress σ₃ ≈ 0 at the outer surface, the absolute maximum is τ_max = σ₁/2 = 35.8 MPa. It acts on planes at 45° through the wall thickness.
Result: maximum normal stress ≈ 71.5 MPa; maximum shearing stress ≈ 35.8 MPa (in-plane 23.5 MPa). Using the inner radius gives 69.7 MPa and 34.9 MPa instead. The torque mainly rotates the principal axes away from the hoop and axial directions.
(d)
The phases of plain-carbon steel, as seen under an optical microscope after nital etching, are sketched below.
(i) Ferrite (α-iron) is an interstitial solid solution of carbon in BCC iron. It dissolves at most about 0.022% C, at 727 °C. In the microscope it shows as light, equiaxed polygonal grains with thin boundaries. It is soft, ductile and ferromagnetic, and it is the main constituent of low-carbon steel.
(ii) Austenite (γ-iron) is an interstitial solid solution of carbon in FCC iron. It dissolves up to about 2.1% C at 1147 °C. In plain-carbon steel it is stable only above 727 °C; at room temperature it survives only in austenitic alloy steels such as 18-8 stainless. It appears as polygonal grains with straight, parallel-sided annealing twins. It is non-magnetic, tough and ductile.
(iii) Cementite (Fe₃C) is an orthorhombic intermetallic compound containing 6.67% C. In hypereutectoid steels (above about 0.8% C) it separates as a white network along the prior austenite grain boundaries, and it also forms the plates within pearlite. It is very hard and brittle.
(iv) Pearlite is the eutectoid mixture formed when austenite of about 0.8% C cools through 727 °C. It consists of alternate lamellae of ferrite (about 88%) and cementite (about 12%) and has a characteristic "mother-of-pearl" look. Its strength and hardness lie between those of ferrite and cementite.
Significance. A hypoeutectoid steel shows ferrite plus pearlite, a eutectoid steel is fully pearlitic, and a hypereutectoid steel shows pearlite with a cementite network. This progression explains why hardness rises and ductility falls as carbon content increases.
(e)
Given: m = 10 mm, φ = 20°, pinion t = 24, speed ratio 2, so gear T = 48. Addendum = 1 module = 10 mm.
Radii: Pinion pitch radius r = mt/2 = 120 mm; gear pitch radius R = mT/2 = 240 mm Addendum radii: r_a = 130 mm; R_a = 250 mm Base radii: r_b = 120 cos 20° = 112.76 mm; R_b = 240 cos 20° = 225.53 mm
(i) Length of path of contact Path of approach (gear addendum) = √(R_a² – R_b²) – R sin φ = 107.879 – 82.085 = 25.794 mm Path of recess (pinion addendum) = √(r_a² – r_b²) – r sin φ = 64.688 – 41.042 = 23.645 mm Path of contact = 25.794 + 23.645 = 49.44 mm
(ii) Contact ratio Arc of contact = path/cos φ = 49.44/0.9397 = 52.61 mm Circular pitch p = πm = 31.416 mm Contact ratio = 52.61/31.416 = 1.67. Equivalently, it is the path divided by the base pitch: 49.44/29.52.
So two pairs of teeth are in contact for about 67% of the time.
(iii) Angle of action of the pinion θ = arc of contact/r = 52.61/120 = 0.4384 rad = 25.12°. Of this, 13.11° is approach and 12.01° is recess.
The contact ratio is comfortably above the usual minimum of about 1.2, so the drive transmits motion smoothly and continuously.
How this answer will be evaluated
What earns full marks: Rigorous application of governing equations with clear diagrams and physical interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine the clamping force developed at G. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Mechanism interpretation: opposite-hand threads at C and D (print says C and E); handle torque M = 50×0.125 = 6.25 N·m
- tanα = 5/(π×25) = 0.0637 (α = 3.64°); φ = tan⁻¹0.3 = 16.70°; tan(α+φ) = 0.3707
- Screw equilibrium: M = 2W r_m tan(α+φ) ⇒ W = 674.3 N; self-locking (φ > α)
- Moment about A of jaw GAC: F_G = W ≈ 674 N (link AB carries 1349 N in tension); labelled FBD
Loses marks
- Ignoring friction in screw calculation
- Incorrect moment arm identification
Earns more
- Correct calculation of friction angle
- Clear identification of reaction forces
Extra mark
- Check for self-locking condition
- (b) Determine minimum required diameter d of the steel rod. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Al deformation: δ_Al = (P/E)(L₁/A₁ + L₂/A₂) = 0.799 mm (A₁ = 1134.1, A₂ = 2463.0 mm²)
- Stiffness criterion: A ≥ PL/(E δ_Al) = 586.9 mm² ⇒ d ≥ 27.3 mm
- Strength criterion: A ≥ P/σ_allow = 757.6 mm² ⇒ d ≥ 31.1 mm
- Select d_min ≈ 31.1 mm (strength governs; steel δ = 0.62 mm < 0.799)
Loses marks
- Ignoring the stress limit
- Using wrong Young's modulus values
Earns more
- Correct area calculation for each segment
- Clear comparison of deformation vs stress limits
Extra mark
- Explicit statement of which criterion governs
- (c) Determine maximum normal and shearing stress in the tank. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Thin-wall: σ_h = pr/t = 64 MPa; σ_l = pr/2t = 32 MPa (mean r = 96 mm)
- τ = T/(2πr²t) = 17.27 MPa
- Principal: σ₁,₂ = 48 ± 23.5 ⇒ 71.5 and 24.5 MPa; planes at 23.6°
- τ_max = σ₁/2 = 35.8 MPa (absolute); in-plane 23.5 MPa
Loses marks
- Omitting the closed-end longitudinal stress or combining stresses without Mohr/transformation
- Ignoring torsional shear stress
Earns more
- Correct use of mean radius for thin wall
- Clear stress element diagram
Extra mark
- Mohr's circle construction
- (d) Illustrate microstructures of steel phases. 10 marks
describe— define → structure or process in order → labelled diagram → significance
Must cover
- Ferrite (BCC): light equiaxed grains; max ~0.022% C at 727 °C; soft, ductile, ferromagnetic
- Austenite (FCC): polygonal grains with annealing twins; up to 2.1% C at 1147 °C
- Cementite Fe₃C: orthorhombic, 6.67% C; hard, brittle; white network in hypereutectoid steels
- Pearlite: eutectoid lamellar mixture (≈88% ferrite/12% cementite) at 0.8% C/727 °C; four labelled diagrams provided
Loses marks
- Confusing Ferrite and Austenite structures
- Missing any of the four phases
Earns more
- Correct crystal structure labels
- Clear distinction between phases
Extra mark
- Mention of carbon content for each
- (e) Determine path of contact, contact ratio, and angle of action. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Pitch radii r = 120, R = 240 mm; base radii 112.76, 225.53 mm
- Path of approach = 25.79 mm; recess = 23.65 mm; path of contact = 49.44 mm
- Arc of contact = 52.61 mm; contact ratio = 1.67 (> 1.2 OK)
- Angle of action = 0.4384 rad = 25.12° (13.11° approach + 12.01° recess)
Loses marks
- Incorrect module application
- Confusing pressure angle with helix angle
Earns more
- Correct calculation of base circle radii
- Clear definition of contact ratio
Extra mark
- Diagram of gear meshing
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Where marks are lost
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