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?
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 - Question

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 - Question

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 - 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

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

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

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

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

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

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