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# Machine Dynamics MCQ – Torsionally Equivalent Shaft

From the following data, calculate the equivalent length of shaft in m.

l_{1}=0.6m, l_{2}=0.5m, l_{3}=0.4m

d_{1}=0.095m, d_{2}=0.06m, d_{3}=0.05m

a) 8.95

b) 7.95

c) 6.95

d) 5.95

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Answer: a

Explanation: We know that torsional equivalent shaft length is given by

l_{1}/(d_{1})^{4} + l_{2}/(d_{2})^{4} + l_{3}/(d_{3})^{4}

substituting the values we get

L = 8.95 m.

In a system with different shaft parameters, the longest shaft is taken for calculations.

a) True

b) False

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Answer: b

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

From the following data, calculate natural frequency of free torsional vibrations in Hz.

l_{1}=0.6m, l_{2}=0.5m, l_{3}=0.4m

d_{1}=0.095m, d_{2}=0.06m, d_{3}=0.05m

Ma = 900 Kg, Mb = 700 Kg

ka = 0.85m, kb = 0.55m

C = 80 GN/m^{2}

a) 3.37

b) 7.95

c) 6.95

d) 5.95

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Answer: a

Explanation: We know that torsional equivalent shaft length is given by

l_{1}/(d_{1})^{4} + l_{2}/(d_{2})^{4} + l_{3}/(d_{3})^{4}

substituting the values we get

L = 8.95 m

now calculating J

we get polar moment of inertia J = 8×10^{6} m^{4}

Now we know that natural frequency is given by the formula

f = 12(C.Jl.I)−−−−−√

f = 3.37 Hz.

From the following data, calculate the location of node from the left end of shaft (l_{1}).

l_{1}=0.6m, l_{2}=0.5m, l_{3}=0.4m

d_{1}=0.095m, d_{2}=0.06m, d_{3}=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

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Answer: a

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.

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

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Answer: d

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.

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-m^{2} are: 0.15, 0.3 and 0.09. C=84 kN/mm^{2}

a) 171

b) 181

c) 191

d) 201

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Answer: a

Explanation: For a single node system, the node occurs at a distance 1.146m from left end.

The polar moment of inertia = 2.36×10^{6} m^{4}

La = 0.4356m (from same frequency relation)

Substituting these values into the frequency relation we get

f = 171 Hz.

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-m^{2} are: 0.15, 0.3 and 0.09. C=84 kN/mm^{2}

a) 257

b) 281

c) 197

d) 277

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Answer: d

Explanation: For a two node system, the first node occurs at a distance 0.4356 m from left end.

The polar moment of inertia = 2.36×10^{6} m^{4}

La = 0.4356m ( from same frequency relation )

Lc = 0.726 m

Substituting these values into the frequency relation we get

f = 277 Hz.

Frequency is independent of the no. of nodes.

a) True

b) False

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Answer: b

Explanation: The frequency of the free torsional vibration depends on the number of nodes it forms, it also depends on the no. of rotors as the no. of rotors have an effect on the number of nodes.