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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.

User Reuven Lax
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4 votes

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Step-by-step 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=((F_0)/(m))/(√(\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2))\\

Where


  • \omega_n=\sqrt{(k)/(m)}\\\\=\sqrt{(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=((450)/(70))/(√(20.7^2-13^2)^2 +(2* 0.2*20.7*13)^2))\\[tex]X=0.0229\ \rm m

b) The phase shift of the motion is given by


\tan\phi=(2\rho \omega_n \omega )/(\omega_n^2-\omega^2)\\\\(2*0.2*20.7*13 )/(20.7^2-\13^2)\\\\\phai=0.393\\

c)Transmissibility ratio is given by


T.R.=\sqrt{\frac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\frac{1+(2*0.2*0.628)^2}{(1-0.628^2)^2+{(2*0.2*0/628)^2}}}\\\\=1.57

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


F_T=(T.R)* F_0\\\\=450*1.57\\\\=707.6\ \rm N

e)The maximum velocity is given by
V_(max)


V_(max)=\omega A_0\\\\=13* 0.0229\\\\=0.298\ \rm m/s

User Aneroid
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