# Machine Dynamics Questions and Answers – Torsionally Equivalent Shaft

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This set of Machine Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Torsionally Equivalent Shaft”.

1. 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. 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. 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 = $$\frac{1}{2}\sqrt{(\frac{C.J}{l.I})}$$
f = 3.37 Hz.

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