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A tail rotor section of the helicopter consistent of four blades, each of mass 2.3 kg and an engine box of mass 28.5 kg. The centre of gravity of each blade is 170 mm from the rotational axis.
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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

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

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

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