The tail section is connected to the main body of the helicopter by an eccentric structure, The natural frequency of a tail section is observed as 135 rad/sec, During flight the rotor operates at 900 rpm, What is the vibration amplitude of the tail section if one of the blades falls off during rotation?

Assume a damping ratio of 0.05.

m = 28.5 kg

mo = 2.3 kg

e = 0.17 m

Wn1 = 135 r/sec when there are 4 blades.

w = 94.24 r/sec.

Find : X = SSA when one blade falls down at 900 rpm.

Case 1: When there are 4 blades.

$w_{n1} = \sqrt{ \frac{K_eq}{m + 4 mo}}$

$135 = \sqrt{ \frac{K_eq}{28.5 + 4 \times 2.3}}$

$K_{eq} = 687.08 \times 10^3$ N/m

Case 2: when one blade falls down.

$w_{n2} \sqrt{ \frac{keq}{m + 3mo}}$

$w_{n2} = \sqrt{ \frac{687.08 \times 10^3}{28.5 + 3 \times 2.3}} = 139.31$ r/sec.

$\frac{x}{\frac{mo.e}{m + 3mo}} = \frac{r^2}{\sqrt{ (1-r^2)^2 + (2 \zeta r)^2}}$

$\frac{X}{\frac{2.3 \times 0.17}{28.5 + 3 \times 2.3}} = \frac{ [ \frac{2 \pi (900)}{60}]^2/ 139.3}{\sqrt{(1- 0.6785)^2 + (2 \times 0.05 \times 0.676)^2}}$

X = 9.22 mm

Case 3: forced vibration with support excitation.

1] Analyze the problems of figure for steady state.

2] Response of the mass.

ANSLOM,

If = $\sum$ of all forces

$+ \ m\ddot{x} \ = \ - c(\dot{x}-\dot{y})- kx \ $

m$\ddot{x}$ + c$(\dot{x}-\dot{y})$ + kx = 0

m$\ddot{x}$ + c$\dot{x}$ - c$\dot{y}$ + kx = 0

m$\ddot{x}$ + c$\dot{x}$ + kx = c$\dot{y}$

m$\ddot{x}$ + c$\dot{x}$ + kx = c $y_0$ w cos wt

m$\ddot{x}$ + c$\dot{x}$ + kx = c $y_0 w [sin \ wt \ + \ \frac{\pi }{2}]$

Std form:

m$\ddot{x}$ + c$\dot{x}$ + kx = Fo sin wt

Fo = c$y_0$w

$\frac{X}{X_{st}} = \frac{1}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$X = \frac{\frac{c Y_o w}{k}}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

ANSLOM,

IF = $\sum$ all ext. forces.

m$\ddot{x}$ = - c$\dot{x}$ - k (x - y)

m$\ddot{x}$ + c$\dot{x}$ + k(x-y) = 0

m$\ddot{x}$ + c$\dot{x}$ + kx – ky = 0

m$\ddot{x}$ + c$\dot{x}$ + kx = ky

m$\ddot{x}$ + c$\dot{x}$ + kx = k (Yo sin wt)

m$\ddot{x}$ + c$\dot{x}$ + kx = kYo. Sin wt.

We know that

m$\ddot{x}$ + c$\dot{x}$ + kx = Fo sin wt.

Fo = k.Yo

$\frac{X}{Xst} - \frac{1}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$

$X = \frac{ \frac{k. \ Yo}{k}}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

$\frac{X}{Yo} = \frac{1}{\sqrt{ (1-r^2)^2 + (2 \xi r)^2}}$