Vibration Isolation.

The inertia force developed in a reciprocating engine or unbalanced forces produced in any other rotating machinery should be isolated from the foundation so that the adjoining structure is not set into heavy vibrations. Another example may be the isolation of delicate instruments from their supports which may be subjected to certain vibrations.

In either case, the effectiveness of isolation is measured in term of the force or motion transmitted to that in existence. The first type is said to be free isolation and the second type as motion isolation.

For isolation, we place properly chosen isolating materials between the vibrating and the supporting structure. The isolating materials may be of rubbers, or metallic springs.

Force transmissibility:

The term transmissibility in the case of force excited system is defined as the ratio of the force transmitted to the foundation to that impressed upon the system. Imagine a mass m supported on the foundation by means of an isolator having an equivalent softness and damping coefficient k and c respectively and excited by a force f sin wc. Under steady state conditions, the forces acting on the mass are shown in vector diagram 14(b).

Out of the four forces acting on the mass, the spring force kx and the dash pot force x are the two common forces acting on the mass and also on the foundation. Therefore, the force transmitted to the foundation is the vector sum of these two forces acting in a direction opposite to that on the mass, as shown in figure 14(c).

These forces are 90˚ out of phase with each other and their vector sum $F_tr$ is the force transmitted to the foundation.

$\therefore$ $F_{tr} = \sqrt{(kx)^2 + (cwX)^2}$

Or $F_{tr} = X \sqrt{k^2 + (cw)^2}$

Now we know that

$X = \frac{F_0}{\sqrt{ (k – mw^2)^2 + (cw)^2}}$

Put this value of X in equation of $F_{tr}$

$F_{tr} = \frac{ F_0 \sqrt{k^2 + (cw)^2}}{\sqrt{(k – mw^2)^2 + (cw)^2}}$ - - - - (20)

Which, dimensionless form, can be written as

$T_r = \frac{F_{tr}}{F_0} = \frac{ \sqrt{ 1 + ( 2\xi \frac{w}{w_n})^2}}{\sqrt{ [ 1 – (\frac{w}{w_n})^2]^2 + [ 2 \xi \frac{w}{w_n}]^2}}$

$T_r = \frac{ \sqrt{ 1 + (2 \xi r)^2}}{\sqrt{ (1- r^2)^2 + (2 \xi r)^2}}$ - - - - (21)

$T_r$ being the transmissibility.

The angle through which the transmitted force lags the impressed force can be seen from figure 14(c )

$= tan^{-1} [ \frac{cwX}{Kx}]$

$tan^{-1} [\frac{cw}{k}]$

$ tan^{-1} [ 2 \xi \frac{w}{w_n}]$

$tan^{-1} (2 \xi r)$

And angle $\phi$ is known $\rightarrow$

$\phi = tan^{-1} [ \frac{2 \xi \frac{w}{w_n}}{ 1 – (\frac{w}{w_n})^2}] = tan^{-1} [ \frac{2 \xi r}{1 – r^2}]$

$\therefore$ $\phi - \alpha = tan^{-1} [ \frac{ 2 \xi \frac{w}{w_n}}{1 – (\frac{w}{w_n})^2}] – tan^{-1} [ 2 \xi \frac{w}{w_n}]$ - - - - (22)

Equations (21) and (22) give the transmissibility and the phase lag of transmitted force from the impressed force. All the curves start from unity value of transmissibility pass through the unit transmissibility at $w/w_n = \sqrt{2}$ and after that they tend to zero and as $(w/w_n) \rightarrow \infty$