Engineering Questions with Answers - Multiple Choice Questions
Machine Dynamics MCQ – Critical or Whirling Speed of a Shaft
1 - Question
Assuming the shaft to be freely supported. Calculate the whirling speed of the shaft : 2 cm diameter and 60 cm long carrying a mass of 1 kg at its mid-point. The density of the shaft material is 40 Mg/m3, and Young’s modulus is 200 GN/m2.
a) 2598
b) 2434
c) 2756
d) 2634
View Answer
Answer: a
Explanation: I = πd4/64 = 7.855×10-9 m4
d due to 1 kg mass = 28×10-6 m
d due to mass of the shaft = 0.133×10-3m
substituting these values into the frequency relation of transverse vibrations
we get
f = 43.3 Hz
Therefore Nc = 43.3×60 = 2598 rpm.
2 - Question
When the centre of gravity of the rotor lies between the centre line of the shaft and the centre line of the bearing, e is taken positive.
a) True
b) False
View Answer
Answer: b
Explanation: If the centre of gravity(G) of the rotor does not lie between the centre line of the shaft and the centre line of the bearing, then the value of e is taken positive.
3 - Question
The speed at which the shaft runs so that the additional deflection from the axis of rotation of the shaft becomes infinite, is known as _________
a) Whirling speed
b) Rotational speed
c) Stabilizing speed
d) Reciprocating speed
View Answer
Answer: a
Explanation: The rotational speed at which the shaft runs so that the additional deflection of the shaft from the axis of rotation becomes infinite, is known as critical or whirling speed.
4 - Question
From the following data, calculate the critical speed of the shaft in rps.
Shaft diameter = 5mm
length = 200mm
Mass of disc = 50Kg at centre of shaft
E = 200GN/m2
Centre of disc at 0.25m away from centre of axis of shaft.
a) 8.64
b) 9.64
c) 10.64
d) 11.64
View Answer
Answer: a
Explanation: I = πd4/64 = 30.7×10-12 m4
d = 3.33×10-3m
substituting these values into the frequency relation of transverse vibrations
we get
Nc = 8.64 rps.
5 - Question
From the following data, calculate the static deflection in mm.
Critical speed = 8.64 rps
Mass of disc = 100Kg at centre of shaft
E = 100GN/m2
Centre of disc at 0.25m away from centre of axis of shaft.
a) 3.32
b) 9.64
c) 10.64
d) 11.64
View Answer
Answer: a
Explanation: I = πd4/64 = 30.7×10-12 m4
Nc = 8.64 rps
substituting these values into the frequency relation of transverse vibrations
we get
d = 3.32mm.
6 - Question
If the static deflection is 1.665×10-3m, calculate the critical speed of the shaft in rps.
Centre of disc at 0.25m away from centre of axis of shaft.
a) 8.64
b) 9.64
c) 10.64
d) 12.2
View Answer
Answer: d
Explanation: d = 1.665×10-3m
substituting these values into the frequency relation of transverse vibrations
we get
Nc = 12.2 rps.
7 - Question
A shaft supported in ball bearings is assumed to be a simply supported.
a) True
b) False
View Answer
Answer: a
Explanation: A shaft supported in short bearings is assumed to be a simply supported shaft while the shaft supported in journal bearings is assumed to have both ends fixed.
8 - Question
From the following data, calculate the static deflection in mm.
Shaft diameter = 5mm
length = 200mm
Mass of disc = 100Kg at centre of shaft
E = 100GN/m2
Centre of disc at 0.25m away from centre of axis of shaft.
a) 4.32
b) 9.64
c) 10.64
d) 11.64
View Answer
Answer: a
Explanation: I = πd4/64 = 30.7×10-12 m4
d = 13.32×10-3m.