Question

A mass m, of 20 kg is attached to the free end of a steel cantilever beam of length 1000 mm having a cross-section of $25\times 25$ mm. Assume the mass of the cantilever to be negligible and $E_{\text{steel}}=200$GPa. If the lateral vibration of this system is critically damped using a viscous damper, the damping constant of the damper is

• A. 1250 Ns/m
• B. 625 Ns/m
• C. 312.50 Ns/m
• D. 156.25 Ns/m

Answer 1

The correct option is A 1250 Ns/m

Moment of inertia of the shaft (I)
$I=\frac{1}{12}\times (25\times {10}^{-3}{)}^4$
$=3.25\times {10}^{-8}{m}^4$

Now, deflection at the free end of the cantilever beam

$\delta =\frac{W{L}^3}{3EI}$

$=\frac{20\times 9.81\times (1{)}^3}{3\times 200\times {10}^9\times 3.25\times {10}^{-8}}$

$=0.01m$

Now, we know that

${\omega }_n=\sqrt{\frac{g}{\delta }}=\sqrt{\frac{9.81}{0.01}}=31.32\text{rad/s}$

Critical damping constant,

$C_C=2m{\omega }_n=2\times 20\times 31.32$
$\approx 1250\text{Ns/m}$

POINTS TO REMEMBER
For a critically damped system, the damping factor is one. As we know,

$\text{Damping factor,}\xi =\frac{C}{2\sqrt{km}}$

From a critical damped system,

$\xi =1,C=C_C$

i.e. $1=\frac{C_c}{2\sqrt{Km}}$

$C_c=2\sqrt{Km}$

This problem is also based on deflection of the beam.