A circular rod undergoing torsional vibration Is held fixed at one end and the e

Question

A circular rod undergoing torsional vibration Is held fixed at one end and the end has a flywheel. The rod and the flywheel are steel. The diameter and length of the rod are 2cm and 60cm. The flywheels thickness and diameter are 4cm and 10cm.

A) Find the torsional stiffness of the steel rod and the mass moment of inertia of the flywheel
b) find the natural frequency of the vibrating system by including the rods own mass moment of inertia.
C) If the rods flywheel assembly is placed in an oil housing and a torsional damper will be used for the vibration control purpose, what should the dampers damping coefficient, ct, be so that the system is critically damped. may need to google properties of steel, such as density, youngs modulus and shear modulus

Explanation / Answer

Given:

Dia of rod, d = 2 cm = 2r

Length odf the rod , l = 60 cm

Flywheel thickness, w = 4 cm

Flywheel dia, D = 10 cm = 2R

Shear modulus , C = 79.3 GPa

Solution:

Rod stiffness, k1 = CJ1/l = 79.3 x 103 x (/32) x 204/600 = 2.076 x 106 = 2076.1 N/m

Flywheel stiffness, k2 = CJ2/w =  79.3 x 103 x (/32) x 1004/40 = 1.94 x 1010 = 1.94 x 107 N/m

The stiffness of the assembly is given by

1/k = 1/k1 +1/k2 = 0.0004816 + 0.0000000515

=> k = 2076.41 N/m

Moment of inertia of the rod and flywheel, J =(/2)[r4l + R4w) = 3.176 x 10-3 kgm2

A) The stiffness of the assembly is k = 2076.41 N/m

Moment of inertia of the rod and flywheel, J  = 3.176 x 10-3 kgm2

B) Natural frequency n = (k/J) = 808.57 rad/s

C) Ccr = 2km = 181.57 Ns/m

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