# Machine Dynamics Questions and Answers – Flywheel in a Punching Press

This set of Machine Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Flywheel in a Punching Press”.

1. In a punching press, which of the following quantity is constant?
b) Torque
c) Angular velocity
d) Angle of rotation

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. In a punching press, load is 0 at the time of punching.
a) True
b) False

Explanation: In a punching press, the load acts only during the time of punching and remains 0 for the rest of the time.

3. The maximum shear force required for punching depends on ________
a) Sheared area
b) Length of the plate
c) Speed of the flywheel

Explanation: Maximum shear force is given by the equation
Fs = (Sheared area) x ultimate shear stress.

4. 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

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. 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

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.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. If the stroke punch is 100mm, find the radius of the crank in mm.
a) 200
b) 100
c) 50
d) 400

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. 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

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. Energy during actual punching operation is same as the energy supplied by the motor.
a) True
b) False

Explanation: Energy during actual punching (E2) operation is different as the energy supplied by the motor, E2 = E1x(t/4r).

9. 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

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. 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

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.

Sanfoundry Global Education & Learning Series – Machine Dynamics.

To practice all areas of Machine Dynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]