0
1.9kviews
A thin disc of mass 0.8kg and radius 60 mm is attached to the end of 1.2 m steel shaft of dia. 20mm. The disc is subjected to harmonic torque of amplitude 12.5 Nm at a frequency 700 rad/sec.
0
141views

Find the steady state amplitude of angular oscillation. $(G = 80 \times 10^9 N/m^2$ and $= 7500 kg/m^3$)

Mass of disc, md = 0.8 kg

$= 60 \times 10^{-3m}$

Diagram of shaft = $d_s = 20 \times 10^{-3} \ m$

To or Mo = 12.5 N-m

G shaft = $80 \times 10^9 N/m^2$

S shaft = 7500 $kg/m^3$

Since, S shaft is given.

$\therefore$ we have to consider I shaft also in $w_n$

$w_n = \sqrt{ \frac{kt}{I disc \ + \ \frac{ Ishaft}{3}}}$

$kt = \frac{GJ}{L} = \frac{G}{L} \times \frac{\pi }{32} \times d^4$

$kt = \frac{80 \times 10^9}{1.2} \times \frac{\pi }{32} \times [20 \times 10^{-3}]^4$

$kt = 6.66 \times 10^{10} \times 1.57 \times 10^{-8}$

Kt = 1046.66 N-m / rad

$Idisc = \frac{Mdisc \ \times \ r^2 \ disc}{2}$

$= \frac{0.8 \times (60 \times 10^{-3})^2}{2}$

$\frac{0.8 \times (60 \times 10^{-3})^2}{2}$

$= 1.44 \times 10^{-3} kg-m^2$

$I shaft = \frac{M shaft \ \times \ (r shaft)^2}{2}$

$M shaft = S shaft \times A \times L$

$= 7500 \times \frac{\pi }{4} \times (20 \times 10^{-3})^2 \times 12$

= 2.827 kg

I shaft = $1.41 \times 10^{-4} kg-m^2$

$W_n = \sqrt{ \frac{1046.66}{1.44 \times 10^{-3} + \frac{1.41 \times 10^{-4}}{3}}}$

$w_n = 838.93$ rad/sec

$r = \frac{w}{w_n} = \frac{700}{838.93} = 0.834$

$\frac{\theta }{\theta st} = \frac{1}{\sqrt{ (1- r^2) + (2 \zeta r)^2}} = \frac{1}{1 – r^2}$

$\frac{\theta }{ \theta st} = \frac{1}{1 - r^2}$

$\frac{\theta}{0.0119} = \frac{1}{1 – (0.834)^2}$

$\theta = 0.039$ rad.