Question 3.10: A simple structure is found to vibrate as a single degree of...
A simple structure is found to vibrate as a single degree of freedom system. The spring constant is determined using static testing and is found to be 1500 N/m, and the equivalent mass of the structure is assumed to be 2 kg. By using a simple vibration test, the ratio of successive amplitudes is found to be 1.2. Determine the structural damping coefficient and the equivalent viscous damping coefficient. Determine also the energy loss per cycle for an amplitude of 0.05 m.
The logarithmic decrement δ is given by
δ = ln \frac{x_{i}}{x_{i}+1} = ln 1.2 = 0.18232
The structural damping coefficient c_{s} is given by
c_{s} = \frac{2δ}{\sqrt{(2π)² + δ²}} = \frac{2(0.18232)}{\sqrt{(2π)² + (0.18232)²}} = 0.058
The equivalent viscous damping coefficient can then be determined using Eq. 3.104 as
c_{e} = \frac{kc_{s}}{ω} = c_{s} \sqrt{km} = 3.1768 N · s/m
Equation 3.100 can be used to determine the energy loss per cycle as
ΔD = πc_{s}kX² = π(0.058)(1500)(0.05)² = 0.6833 N · m