Question

A refrigerator unit having a mass of 35 kg is to be supported on three springs, each having spring stiffness. The unit operates at 480 rpm. If only 10% of the shaking force is allowed to transmit to the supporting structure, the value of stiffness will be nearly
1. 2.7 N/mm
2. 3.2 N/mm
3. 3.7 N/mm
4. 4.2 N/mm

Answer 1

Correct Answer - Option 1 : 2.7 N/mm

Concept:

Transmissibility is defined as the ratio of force transmitted to the shaking force.

 \(\in = \frac{{{F_t}}}{{{F_0}}} = \frac{{\sqrt {1 + {{\left( {\frac{{2{\bf{\zeta \omega }}}}{{{{\bf{\omega }}_n}}}} \right)}^2}} }}{{\sqrt {{{\left[ {1 - {{\left( {\frac{{\bf{\omega }}}{{{{\bf{\omega }}_n}}}} \right)}^2}} \right]}^2} + {{\left( {\frac{{2{\bf{\zeta \omega }}}}{{{{\bf{\omega }}_n}}}} \right)}^2}} }}\)

where, \({F_t}\) = Transmitted force, \({F_0}\) = shaking force

Calculation:

Given:

m = 35 kg

N = 480 rpm

ω = \(\frac{{2\pi N}}{{60}}\) = 50.27 rad/s

As Damper is not used

ζ = 0

\({F_t}\) = (0.1) \({F_0}\)

Since, \({{\bf{\omega }}_n} = \sqrt {\frac{{{K_{eq}}}}{m}} \) 

The three springs will be in parallel, therefore K_eq = 3K

∴ \({{\bf{\omega }}_n} = \sqrt {\frac{{3K}}{{35}}}\)

For zero damping, the Transmissibility equation gets reduced to:

 

 \(\in = \frac{{\left( {0.1} \right){F_0}}}{{{F_0}}} = \frac{1}{{\sqrt {{{\left[ {1 - {{\left( {\frac{{\bf{\omega }}}{{{{\bf{\omega }}_n}}}} \right)}^2}} \right]}^2}} }}\)

0.1 = \(\frac{1}{{{{\left( {\frac{{\bf{\omega }}}{{{{\bf{\omega }}_n}}}} \right)}^2} - 1}}\)

\({\left( {\frac{{\rm{\omega }}}{{{{\rm{\omega }}_n}}}} \right)^2} = 10 + 1\)

\(\frac{{\rm{\omega }}}{{{{\rm{\omega }}_n}}}\) = 3.317

\(\therefore {{\rm{\omega }}_n} = \frac{{50.27}}{{3.317}}\) = 15.15

\(\frac{{3K}}{{35}} = {\left( {15.15} \right)^2}\)

K = 2677.76 N/m = 2.67 N/mm ≃ 2.7 N/mm