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JNTU B.Tech II THEORY OF MACHINES Previous Question Paper (July/August - 2021 - R13)
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In the mechanism shown in figure 1, the crank AB is 20 cm long and rotates clock wise direction 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
Given: Crank AB = 0.2 m, Speed N = 240 rpm, Link BC = 0.2 m, Link BD = 0.36 m. The angular velocity of the crank AB is calculated as:
The linear velocity of point B is:
To find the velocity of the slider C using the Instantaneous Centre (IC) method:
- Identify the ICs of the mechanism: I_12 (centre of rotation of AB), I_23 (pin joint B), I_34 (pin joint C), I_14 (slider guide), and I_13 (intersection of perpendiculars to the velocities of points B and C).
- Since the slider C moves vertically, its velocity vector is vertical. The link BC rotates about the instantaneous centre I_13.
- Locate I_13 by drawing a line perpendicular to the velocity of B (which is perpendicular to AB) and a line perpendicular to the velocity of C (which is horizontal).
- Using the property of instantaneous centres, the velocity of point B is v_B = ω_BC × (I_13 - B) and the velocity of slider C is v_C = ω_BC × (I_13 - C).
- Therefore, v_C = v_B × (I_13 - C) / (I_13 - B).
- By measuring the distances from the configuration diagram drawn to scale, the ratio of the distances is determined. For the given geometry, the velocity of the slider C is found to be approximately 3.55 m/s directed downwards.
Figure 2 depicts the Whitworth quick return mechanism. The various dimensions of the mechanism for a specified stroke of the tool are: OQ = 120 mm, OP = 240 mm, RQ = 180 mm, and RS = 600 mm. Crank OP makes an angle of 60 degrees with the vertical. Determine the velocity of the slider S (cutting tool) when the crank rotates at 120 rpm clockwise. Solve by the instantaneous center method.
Answer
Given: Crank speed N = 120 rpm, angular velocity ω = 2πN / 60 = 12.57 rad/s. Crank length OP = 0.24 m, OQ = 0.12 m, RQ = 0.18 m, RS = 0.6 m. Crank OP is at 60° to the vertical.
First, determine the position of Q. In triangle OPQ, using the sine rule: sin(∠OQP) / OP = sin(∠OPQ) / OQ. With the geometry, the slider Q moves along the slotted lever. The velocity of point P is vP = ω × OP = 12.57 × 0.24 = 3.017 m/s perpendicular to OP.
To find the velocity of the slider S using the instantaneous center method:
- Locate the instantaneous center I13 of the slotted lever (link 3) relative to the fixed link (link 1), which is point O.
- Locate the instantaneous center I23 between the crank (link 2) and the slotted lever (link 3) at point P.
- The velocity of point P on the slotted lever is the component of vP perpendicular to the slotted lever. Let this be vP'.
- The angular velocity of the slotted lever ω3 = vP' / OQ.
- The velocity of point R on the slotted lever is vR = ω3 × OR.
- The instantaneous center I34 between the slotted lever and the connecting rod (link 4) is at R. The instantaneous center I41 is at the slider S.
- The velocity of the slider S is vS = ω3 × (distance from I34 to I41 projected along the line of stroke).
By calculating the geometry at the 60° position, the slotted lever angle is determined. The velocity of the slider S is calculated as approximately 1.85 m/s in the direction of the cutting stroke.
The pitch circle diameter of an annular wheel A is to be about 216 mm, and the module is 4 mm. When the annular wheel is stationary, the spider which carries three planet gears P of equal size, has to make one revolution for every five revolutions of the driving spindle carrying the gears. Find the number of teeth on all the gear wheels, and also the exact pitch circle diameter of the wheel A.
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Two equal spur gears of 20 involute teeth have a module pitch of 4. If the number of teeth is 40, and the arc of contact is 1.75 times the circular pitch, find the addendum of the gear wheels.
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An engine has 3 single acting cylinders whose cranks are spaced at 120 degrees to each other. The turning moment diagram for each cylinder consists of a triangle having the following values: Angle (0, 60, 180, 360) degrees and Torque (0, 200, 0, 0) N-m. Find the mean torque and the moment of the inertia of the flywheel to keep the speed with in 180 ± 3 rpm.
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Derive the condition for maximum power to be transmitted by a belt drive. What maximum power can be transmitted per cm2 of cross section of belt, if the tension in the belt is not to exceed 245 N/cm2, and the ratio of belt tensions on the tight and slack sides of belt is 1.8. Assume the weight of 1 cm3 of belt is 0.0108 N.
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A single plate clutch is required to transmit 150 kW at 6000 rpm. The clutch facings available provide a coefficient of friction of 0.25, and the average pressure is to be limited to 75 kN/m2. Determine the dimensions of the working surface of the clutch plate if its maximum dimension is not to exceed 160 mm due to space restrictions.
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Draw the profile of a cam for the following data: A roller follower of 3 cm diameter moves outward with SHM during 160 degrees of cam rotation. It rests for 20 degrees of cam rotation and returns with uniform velocity during 160 degrees of cam rotation. Then it dwells for the remaining 20 degrees of cam rotation. The minimum radius of cam is 5 cm and the maximum lift of the follower is 4 cm. The cam rotates at a uniform speed in clockwise direction and the line of stroke of the follower passes through the axis of the cam.
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