Engineering Questions with Answers - Multiple Choice Questions

# Machine Dynamics MCQ – Torsionally Equivalent Shaft

1 - Question

From the following data, calculate the equivalent length of shaft in m.
l1=0.6m, l2=0.5m, l3=0.4m
d1=0.095m, d2=0.06m, d3=0.05m
a) 8.95
b) 7.95
c) 6.95
d) 5.95

Explanation: We know that torsional equivalent shaft length is given by
l1/(d1)4 + l2/(d2)4 + l3/(d3)4
substituting the values we get
L = 8.95 m.

2 - Question

In a system with different shaft parameters, the longest shaft is taken for calculations.
a) True
b) False

Explanation: In a system with different shaft parameters the equivalent shaft length is taken which depends on the length and diameters of each shaft.

3 - Question

From the following data, calculate natural frequency of free torsional vibrations in Hz.
l1=0.6m, l2=0.5m, l3=0.4m
d1=0.095m, d2=0.06m, d3=0.05m
Ma = 900 Kg, Mb = 700 Kg
ka = 0.85m, kb = 0.55m
C = 80 GN/m2
a) 3.37
b) 7.95
c) 6.95
d) 5.95

Explanation: We know that torsional equivalent shaft length is given by
l1/(d1)4 + l2/(d2)4 + l3/(d3)4
substituting the values we get
L = 8.95 m
now calculating J
we get polar moment of inertia J = 8×106 m4
Now we know that natural frequency is given by the formula
f = 12(C.Jl.I)−−−−−√
f = 3.37 Hz.

4 - Question

From the following data, calculate the location of node from the left end of shaft (l1).
l1=0.6m, l2=0.5m, l3=0.4m
d1=0.095m, d2=0.06m, d3=0.05m
Ma = 900 Kg, Mb = 700 Kg
ka = 0.85m, kb = 0.55m
a) 0.855m
b) 0.795m
c) 0.695m
d) 0.595m

Explanation: We know that for the system to be in vibration, the frequency should be same for all rotors
therefore,
LaIa = LbIb
Where La is the length of the left end shaft.
Substituting the values we get,
l = 0.855m.

5 - Question

For a vibration system having different shaft parameters, calculate which of the following cannot be the diameter of the equivalent shaft if the diameters of shafts in m are: 0.05, 0.06, 0.07.
a) 0.05
b) 0.06
c) 0.07
d) 0.08

Explanation: In a torsionally equivalent shaft, it is assumed that the equivalent shaft has the diameter which is equal to one of the diameters of the shaft and equivalent length is calculated.

6 - Question

A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a single node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2
a) 171
b) 181
c) 191
d) 201

Explanation: For a single node system, the node occurs at a distance 1.146m from left end.
The polar moment of inertia = 2.36×106 m4
La = 0.4356m (from same frequency relation)
Substituting these values into the frequency relation we get
f = 171 Hz.

7 - Question

A single cylinder oil engine works with a three rotor system, the shaft length is 2.5m and 70mm in diameter, the middle rotor is at a distance 1.5m from one end. Calculate the free torsional vibration frequency for a two node system in Hz if the mass moment of inertia of rotors in Kg-m2 are: 0.15, 0.3 and 0.09. C=84 kN/mm2
a) 257
b) 281
c) 197
d) 277

Explanation: For a two node system, the first node occurs at a distance 0.4356 m from left end.
The polar moment of inertia = 2.36×106 m4
La = 0.4356m ( from same frequency relation )
Lc = 0.726 m
Substituting these values into the frequency relation we get
f = 277 Hz.

8 - Question

Frequency is independent of the no. of nodes.
a) True
b) False