Question 13.2.1: A system having a rotating unbalance, like that shown in Fig......
A system having a rotating unbalance, like that shown in Figure 13.2.1, has a total mass of m = 20 kg, an unbalanced mass of m_u = 0.05 kg, and an eccentricity of ϵ = 0.01 m. The machine rotates at 1150 rpm. Its vibration isolator has a stiffness of k = 2 × 10^4 N/m. Compute the force transmitted to the foundation if the isolator’s damping ratio is (a) ζ = 0.1 and (b) ζ = 0.5.
First convert the machine’s speed to radians per second.
\omega=1150 rpm =\frac{1150(2 \pi)}{60}=120 rad / sThen
m_u \epsilon \omega^2=0.05(0.01)(120)^2=7.25 Nand the frequency ratio is
r=\frac{\omega}{\omega_n}=\frac{\omega}{\sqrt{k / m}}=\frac{120}{\sqrt{2 \times 10^4 / 20}}=3.81We can calculate the transmissibility ratio from (13.2.10).
T_r=\frac{F_t}{F_r}=\sqrt{\frac{1+4 \zeta^2 r^2}{\left(1-r^2\right)^2+4 \zeta^2 r^2}} (13.2.10)
T_r=\sqrt{\frac{1 + 58.1 \zeta^2}{182.7 + 58.1 \zeta^2}}The force transmitted to the foundation is
F_t=m_u \epsilon \omega^2 T_r=7.25 T_ra. If ζ = 0.1, T_r = 0.093 and F_t = 7.25(0.093) = 0.67 N.
b. If ζ = 0.5, T_r = 0.28 and F_t = 7.25(0.28) = 2 N. The more highly damped isolator in this case transmits more force to the foundation because r > \sqrt{2}.