Question

A shaft supported freely at its ends has a mass of 100 kg placed 200 mm from one end. The shaft diameter is 50 mm. If the length of the shaft is 500 mm, then the frequency of natural transverse vibrations is

$\text{[Assume}E=200GN/m^2]$

• A. 40.25 Hz
• B. 80.5 Hz
• C. 161 Hz
• D. 232 Hz

Answer 1

The correct option is B 80.5 Hz

$m=100\text{kg}$
$l=0.5\text{m}$
$a=0.2\text{m}$
$b=0.5-0.2=0.3\text{m}$
$d=0.5\text{m}$
$E=200\times {10}^9{\text{N/m}}^2$

$I=\frac{\pi }{64}{d}^4$

$=\frac{\pi }{64}(0.05{)}^4=0.307\times {10}^{-6}{m}^4$

For shaft supported freely at both ends,

$\Delta =\frac{mg{a}^2{b}^2}{3EIl}$

$=\frac{100\times 9.81\times \left(0.2{)}^{\times }\right(0.3{)}^2}{3\times 200\times \times {10}^9\times 0.307\times {10}^{-6}\times 0.5}$

$=0.383\times {10}^{-4}m$

$f_n=\frac{1}{2\pi }\sqrt{\frac{g}{\Delta }}$

$=\frac{1}{2\pi }\sqrt{\frac{9.81}{0.383\times {10}^{-4}}}=80.5\text{Hz}$