JNTU
JNTU B.Tech II THEORY OF MACHINES Previous Question Paper (November/December - 2020 - R13)
Answer any Five Questions. All Questions Carry Equal Marks.
Main
In the mechanism shown in figure 1, the crank AB is 20 cm long and rotates cw at a uniform speed of 240 rpm. The links BC and BD are 20 cm and 36 cm long respectively. The slider at C is constrained to move in a vertical path between the guides. Find the linear velocity of the slider by the Instantaneous centre method.
Answer
To find the velocity of the slider C, we first determine the angular velocity of the crank AB. Given N = 240 rpm, the angular velocity ωAB = (2π × 240) / 60 = 25.13 rad/s. The linear velocity of point B is vB = ωAB × AB = 25.13 × 0.2 m = 5.026 m/s, acting perpendicular to AB.
Using the Instantaneous Centre (IC) method for the mechanism:
- Identify the instantaneous centres: I12 (at A), I23 (at B), I13 (at slider C), I34 (at D), and I14 (at the fixed pivot).
- Locate the instantaneous centre of link BC (I13) by finding the intersection of the line perpendicular to the velocity of B (line AB) and the line perpendicular to the velocity of C (the horizontal line passing through C).
- Measure the distances from I13 to points B and C from the scaled diagram. For the given geometry, the velocity of C is related to the velocity of B by the ratio of the distances from the instantaneous centre I13: vC = vB × (I13C / I13B).
- Based on the geometric configuration where AB = 20 cm, BC = 20 cm, and BD = 36 cm, the instantaneous centre I13 is located at the intersection of the extended line AB and the horizontal path of C.
- Calculating the ratio from the geometry, vC = 5.026 × (I13C / I13B) m/s.
By constructing the mechanism to scale, the distance I13B is found to be approximately 28.3 cm and I13C is approximately 20 cm. Thus, vC = 5.026 × (20 / 28.3) ≈ 3.55 m/s.
A reciprocating engine mechanism is shown in figure 2 in which the crank OC rotates at a uniform speed of 200 rpm. OC = 20 cm; CP = 30 cm; Find the linear velocity of the slider P and the location of a point on PC which has the least velocity.
Answer
Given: Crank speed N = 200 rpm, crank length OC = 0.2 m, connecting rod length CP = 0.3 m. The angular velocity of the crank is calculated as:
The linear velocity of the crank pin C is:
To find the velocity of the slider P, we use the instantaneous center method. Let the instantaneous center of the connecting rod CP be point I. For a slider-crank mechanism, I is located at the intersection of the perpendicular to the crank at C and the perpendicular to the path of the slider at P. Since the slider moves horizontally, the perpendicular is a vertical line through P. The distance CI is found by geometry:
The angular velocity of the connecting rod ω_cp is:
The velocity of the slider P is:
The point on the connecting rod with the least velocity is the instantaneous center I. Since the instantaneous center is the point about which the rod appears to rotate at that instant, its velocity is zero. The location of this point I is at a distance of 0.268 m from C along the line perpendicular to the crank OC.
Explain the phenomenon of interference in involute gears, and the methods for avoiding it.
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A gear wheel having 20 teeth of involute form, of module pitch 6 mm and 20° pressure angle drives another gear wheel of same size. If the arc of contact is to be of maximum possible length, find the suitable value for the addendum of the gear wheels.
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Two equal gear wheels of 15 cm pitch circle diameter and 4 mm module are in mesh. The teeth are of involute form with 14.5° pressure angle. Determine the maximum addendum necessary if there are always to be at least two pairs of gear teeth in contact.
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An epicyclic gear train, shown in figure 3 consists of a sun wheel S, a stationary internal gear wheel I, and three identical planet gear wheels P carried on a star-shaped planet carrier A. The sizes of gear wheels are such that the planet carrier A rotates at (1/4)th of the speed of sun wheel S. The minimum number of teeth on any gear wheel is 18. Driving torque on the sun wheel is 15 kg-m. Determine
the number of teeth on the different gear wheels of the train,
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the torque necessary to keep the internal gear stationary.
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The turning moment diagram for a four – stroke gas engine is represented by 4 triangles, whose areas, as measured from the line of zero pressure are: Expansion stroke = 35.5 cm2; Exhaust stroke = 5 cm2; Suction stroke = 3.5 cm2; Compression stroke = 14 cm2. Each cm2 represents 300 N – m. Assuming the resisting torque to be uniform, find the weight of the rim of a flywheel required to keep the mean speed of 200 rpm within ± 2 %. Mean radius of the flywheel rim is 75 cm.
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A cross belt drive is used to transmit 7.5kw at 900 rpm of the smaller pulley, which is 25 cm in diameter. The velocity ratio is 2, and the centre distance is 130 cm. A flat belt of 6 mm thickness and having a coefficient of friction 0.3 is used for the drive. If the maximum permissible stress in the belt material is 170 N/cm2, find the necessary belt width. Density of the belt material is 1 gm/cm3. Take the centrifugal tension into account.
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In a Porter governor, the central load is 200 N, and the weight of each ball is 25 N. The upper arms are each 25 cm long, and the lower arms are each 30 cm long. The friction of the sleeve is 15 N. If the upper arms make 45° with the axis of rotation in the equilibrium position, find the range of speed of the governor in that position.
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