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JNTU B.Tech II THEORY OF MACHINES Previous Question Paper (March - 2022 - R13)

B.Tech | II | II-II | R13 | March - 2022

Answer PaperPaper Code: 114CJMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Describe the working of ‘Ellipse Trammels’ mechanism with a neat sketch.

7 marks

Answer

An Ellipse Trammel is a kinematic mechanism consisting of a slotted plate (fixed link), two sliders, and a connecting bar. The mechanism is designed to trace an elliptical path. The fixed link contains two perpendicular slots. Two sliders move within these slots, and they are connected by a rigid bar (the trammel). As the sliders move along the perpendicular axes, any point on the connecting bar (or its extension) traces an ellipse.

If the distance of a point P on the bar from the slider in the vertical slot is 'a' and from the slider in the horizontal slot is 'b', the coordinates of point P are x = b cos θ and y = a sin θ. This satisfies the equation of an ellipse: (x²/b²) + (y²/a²) = 1. The sketch consists of a frame with two perpendicular grooves, a rod with two pins sliding in these grooves, and a point P on the rod tracing the ellipse.

b

In a slider crank mechanism, the crank is 480 mm long and rotates at 20 rad/s in the counter clockwise direction. Length of the connecting rod is 1.6 m. When the crank turns 60° from the inner-dead center, determine the velocity and acceleration of the slider.

8 marks

Answer

Given: Crank length (r) = 0.48 m, Connecting rod length (l) = 1.6 m, angular velocity (ω) = 20 rad/s, crank angle (θ) = 60°. Ratio (n) = l/r = 1.6 / 0.48 = 3.333.

Velocity of the slider (v) is given by: v = ωr * (sin θ + (sin 2θ / 2n)). Substituting values: v = 20 * 0.48 * (sin 60° + (sin 120° / (2 * 3.333))) = 9.6 * (0.866 + (0.866 / 6.666)) = 9.6 * (0.866 + 0.1299) = 9.56 m/s.

Acceleration of the slider (a) is given by: a = ω²r * (cos θ + (cos 2θ / n)). Substituting values: a = (20)² * 0.48 * (cos 60° + (cos 120° / 3.333)) = 400 * 0.48 * (0.5 + (-0.5 / 3.333)) = 192 * (0.5 - 0.15) = 192 * 0.35 = 67.2 m/s².

215 marks
a

Define the ‘Velocity of Sliding’ and derive an expression for the same.

7 marks

Answer

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b

In an epicyclic gear train shown in Figure, an annular wheel ‘A’ having 54 teeth meshes with a planet wheel ‘B’ which gears with a sun wheel ‘C’. The wheels ‘A’ and ‘C’ are co–axial. The wheel ‘B’ is carried on a pin fixed on one end of the arm ‘P’ which rotates about the axes of the wheels ‘A’ and ‘C’. If ‘A’ makes 20 rpm clockwise, the arm rotates at 100 rpm ccw, and the wheel ‘C’ has 24 teeth, determine the speed and sense of rotation of the wheel ‘C’.

8 marks

Answer

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315 marks
a

Explain the Tabular method to find the Train Value or Velocity Ratio of epicyclic gear Train.

7 marks

Answer

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b

A standard 20° pressure angle involute teeth gear with 32 teeth and module of 6 mm meshes with a rack without interference. Find (i) the addendum of the rack, and (ii) the minimum number of teeth on the gear wheel, if the rack has a standard addendum.

8 marks

Answer

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415 marks
a

A single plate clutch has friction surfaces of 300 mm external diameter and 240 mm internal diameter. The intensity of normal pressure is limited to 80 kPa of the contact surface area, and the coefficient of friction is 0.28. Find the maximum HP that can be transmitted by the clutch at 2200 rpm, if both sides of the plate are effective.

8 marks

Answer

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b

What is uniform pressure theory? Deduce an expression for the friction torque considering this theory for a flat collar.

7 marks

Answer

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515 marks
a

What are centrifugal governors? How do they differ from inertia governors? Explain.

8 marks

Answer

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b

Each arm of a porter governor is 400 mm long. The upper arms are pivoted on the axis of the sleeve and the lower arms are attached to the sleeve at a distance of 40 mm from the axis. Each ball has a mass of 6 kg and the weight on the sleeve is 50 kg. Find the range of speed of the governor if the extreme radii of rotation of the balls are 260 mm and 300 mm.

7 marks

Answer

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615 marks
a

Derive an expression for the maximum fluctuation of energy in a flywheel and discuss the salient points.

7 marks

Answer

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b

In a turning moment diagram, the areas above and below the mean torque line taken in order are 4400, 1150, 1300, and 4550 mm2 respectively. The scales of the turning moment diagram are: Turning moment: 1 mm = 100 N-m; Crank angle: 1 mm = 1°. Find the mass of the flywheel required to keep the speed between 297 and 303 rpm, if the radius of the gyration is 0.525m.

8 marks

Answer

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715 marks
a

Explain the different types of follower motion for cams along with their applications.

7 marks

Answer

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b

Four masses A, B, C and D are attached to a shaft and revolve in the same plane. The masses are 12 kg, 10 kg, 18 kg and 15 kg respectively and their radii of rotations are 40 mm, 50 mm, 60 mm and 30 mm. The angular position of the masses B, C and D are 60°, 135° and 270° from mass A. Find the magnitude and position of the balancing mass at a radius of 100 mm.

8 marks

Answer

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8

Write short notes on:

15 marks
a

Instantaneous center of motion

5 marks

Answer

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b

Coefficient of fluctuation of speed

5 marks

Answer

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c

Partial primary balancing of reciprocating masses.

5 marks

Answer

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