Question

An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplitude 120 N is applied to the compressor, the maximum steady-state displacement of 5 mm occurred at a frequency of 320 r/min. Determine the equivalent stiffness and damping constant of the foundation.

Answer 1

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$)

put here value

0.005 = 120/k   ×  (  $\frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}}$)  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

$\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}$

$\varepsilon^{2} - \varepsilon^{4} = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}$

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m