Question

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured damping ratio of 0.2. The machine produces a harmonic force of 450 N at 13 rad>s during steady-state operating conditions. Determine (a) the amplitude of motion of the machine, (b) the phase shift of the motion (with respect to a zero phase exciting force), (c) the transmissibility ratio, (d) the maximum dynamic force transmitted to the floor, and (e) the maximum velocity of the machine.

Answer 1

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

• Mass of the machine, m=70 kg
• Stiffness of the system, k=30000 N/m
• Damping ratio=0.2
• Damping force, F=450 N
• Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}$

Where

• $\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

• $\omega=13\ \rm rad/s$
• $\rho=0.2$

• $F_0=450\ \rm N$
• $m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$

d)The magnitude of the force transmitted to the ground is

$F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N$

e)The maximum velocity is given by $V_{max}$

$V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s$