Question

A vibratory body of mass 150 kg supported on springs of total stiffness 1050 kN/m has rotating unbalance force of 525 N at speed of 6000 rpm. If the damping factor is 0.3.

The actual force transmitted is

• A. 53.5 N
• B. 43.5 N
• C. 67 N
• D. 435.7 N

Answer 1

The correct option is B 43.5 N

$\frac{F_T}{F}={\left[\frac{1+(2\xi r{)}^2}{(1-r{)}^2+(2\xi r{)}^2}\right]}^{\frac{1}{2}}$

$\frac{F_T}{F}=0.083$

$F_T=0.083\times 525=43.57N$