Engineering Questions with Answers - Multiple Choice Questions
Machine Dynamics MCQ – Flywheel in a Punching Press
1 - Question
In a punching press, which of the following quantity is constant?
a) Load
b) Torque
c) Angular velocity
d) Angle of rotation
View Answer
Answer: b
Explanation: The punching press works on the principle that the input torque acting remains constant when the action is performed whereas the load is varied.
2 - Question
In a punching press, load is 0 at the time of punching.
a) True
b) False
View Answer
Answer: b
Explanation: In a punching press, the load acts only during the time of punching and remains 0 for the rest of the time.
3 - Question
The maximum shear force required for punching depends on ________
a) Sheared area
b) Length of the plate
c) Speed of the flywheel
d) Total load
View Answer
Answer: a
Explanation: Maximum shear force is given by the equation
Fs = (Sheared area) x ultimate shear stress.
4 - Question
A machine punching 38 mm holes in 32 mm thick plate requires 7 N-m of energy per sq. mm of sheared area, find the maximum shear force required.
a) 26.7 kN
b) 53.4 kN
c) 13.35 kN
d) 106.8 kN
View Answer
Answer: a
Explanation: Given: d = 38 mm; t = 32 mm; 2 E1 = 7 N-m/mm of sheared area
A = π.38.32 mm2
F = 7.A N = 26.7 kN.
5 - Question
The relation between stroke punch s and radius of crank r is ______
a) s=r
b) s=2r
c) s=4r
d) s=r/2
View Answer
Answer: b
Explanation: Stroke punch is the distance travelled from top to bottom at the time of punching. Hence is equal to the diameter of the crank.
6 - Question
If the stroke punch is 100mm, find the radius of the crank in mm.
a) 200
b) 100
c) 50
d) 400
View Answer
Answer: c
Explanation: Stroke punch is the distance travelled from top to bottom at the time of punching. Hence is equal to the diameter of the crank. Now s=100mm therefore r=s/2mm.
7 - Question
In a punching press, the requirement is to punch 40 mm diameter holes in a plate having a thickness of 15 mm. The rate at which the holes should be punched is 30 holes/min. Energy requirement is of 6 N-m per mm2 of sheared area. If the punching takes time 1/10 of a second and the speed of the flywheel varies from 160 to 140 rpm, determine the mass of the flywheel having radius of gyration of 1 m.
a) 327 Kg
b) 654 Kg
c) 163.5 Kg
d) 200 Kg
View Answer
Answer: a
Explanation: N = (160+140)/2 = 150 rpm
E1 = 11 310 N-m (A x Shear force)
E2 = 565.5 N-m (energy supplied by motor in 1/10 of a second)
ΔE = 10744.5 N-m
therefore m = 327 Kg.
8 - Question
Energy during actual punching operation is same as the energy supplied by the motor.
a) True
b) False
View Answer
Answer: b
Explanation: Energy during actual punching (E2) operation is different as the energy supplied by the motor, E2 = E1x(t/4r).
9 - Question
The balance energy required for punching is supplied by the flywheel by ________
a) Increase in its kinetic energy
b) Decrease in its kinetic energy
c) Decrease in its potential energy
d) By variation of mass
View Answer
Answer: b
Explanation: The balance energy is to be supplied by the flywheel by the decrease in its kinetic energy when its speed falls from maximum to minimum.
10 - Question
When the length of the connecting rod is unknown then the value (θ2 –θ1)/2π is equal to ________
a) t/s
b) t/2s
c) t/2r
d) t/r
View Answer
Answer: b
Explanation: The values of θ1 and θ2 may be determined only if the crank radius (r), length of connecting rod (l) and the relative position of the job with respect to the crankshaft axis are known, in their absence we use the relation (θ2 –θ1)/2π = t/2s.