Basic Theory of TMD(2)

Observations of TMD Parameters

Since we have already know the equation governing the structure response:
$$
|X| = \underbrace{\frac{F_0}{mw_1^2}} _ {static} × \underbrace{\frac{|P|}{|Q|}} _ {Dyn.Amp\ D(w)}
\\ P = \alpha^2-\beta^2+2\zeta_d\alpha\beta i
\\ Q = [1-(1+\mu)\beta^2](\alpha^2-\beta^2+2\zeta_d\alpha\beta i)-\mu\beta^4
$$
Then how do we select the parameters $\alpha,\zeta_d,\mu$ to control the response?

Intuition tells us that we can iterate through every parameter value and find the smallest response. True, it sounds reasonable, but I have to say that it is not elegant.😂😂😂

So, no rush, let`s consider the plot of dyn. amp. vs excitation freq. ratio (D vs β) at first.

First check the extreme cases to see if it makes sense:

1.$\zeta_d = 0$

A 2- DOF structure with no damping, so two clear peaks in dyn. amp.

2.$\zeta_d = \infty$

TMD stuck with structure (almost zero relative disp. u(t)), so effectively an SDOF structure with mass $m(1+\mu)$ having only one peak.

In fact, we have found that when the damping ratio is close to 1, the double peaks are no longer obvious, and when it is above 1.5, it converges to only one peak.

So notice that the damping ratio of TMD is not the bigger the better, which is a little counter-intuitive. In fact, as we will see later in the discussion, we are instead looking for the maximum response and optimal damping ratio of the structure.

Classical TMD Design

Now we fix $\mu$ and consider typical values of $\zeta_d$ and $\alpha$ to try to find an ingenious observation. What a hard luck, we truly observe something:

For given $\mu$ and $\alpha$, there are two points invariant to $\zeta_d$.❗❗❗

Derivation:

$D=\frac{|P|}{|Q|}\Rightarrow D^2 =\frac{A^2 }{B^2 }×\frac{1+(2\zeta_d\alpha\beta)^ 2/A^ 2}{1+C^ 2(2\zeta_d\alpha\beta)^2 /B^2}$
where

$A=\alpha^2 -\beta^2;$

$B = [1-(1+\mu)\beta^2 ](\alpha^2-\beta^2)-\mu\beta^4;$

$C=1-(1+\mu)\beta^2$

Since A, B and C do not depend on $\zeta_d$, D is invariant to $\zeta_d$ if $\zeta_d→\infty$ or $1/A^2 =C^2 /B^2$.

Algebra gives $AC+B = \mu\beta^4≠0$ and so the only possibility is $AC-B=0$, which yields the following equation:

$(2+\mu)\beta^4-2[1+(1+\mu)\alpha^2]\beta^2+2\alpha^2=0…(1)$ at fixed point;
$D^2=\frac{A^2}{B^2}=\frac{1}{C^2}=\frac{1}{[1-(1+\mu)\beta^2]^2}…(2)$ at fixed point.

This is a really powerful clue: with this property, engineers have the chance to make these two points the control points of the structural response. To be specific:

1.The two fixed points have the same height

2.The peaks pass through the fixed points

In this way, we can say with certainty that the response of the structure must be lower than these two high fixed points. Such approach is called "classical TMD design".

So, what are the mathematics expressions?

1.The two fixed points have the same height

$$
\alpha = \frac{1}{1+\mu}
$$

From (1):

$\beta_ 1^ 2+ \beta_ 2 ^2 = \frac{2[1+(1+\mu)\alpha^2]}{2+\mu}…(3)$

From (2):

$\frac{1}{1-(1+\mu)\beta_ 1 ^ 2}=-\frac{1}{1-(1+\mu) \beta _2 ^2}
\Rightarrow \beta _1 ^2+ \beta_2 ^2 = \frac{2}{1 + \mu}…(4)$

Combining (3) and (4):

$\alpha = \frac{1}{1+\mu}…(5)$

2.The peaks pass through the fixed points

$$
\zeta_d=\sqrt{\frac{3\mu}{8(1+\mu)^3}}
$$

Subs. (5) into D equation,

$\zeta_d^2 = \frac{\mu[3 ± \sqrt{\mu / (\mu + 2)} ] }{8(1+\mu)^3}$

For small $\mu$, the two values of $\zeta_d^2$ differ by a negligible fraction of the order of $\sqrt{\mu}$. The average value may be empirically taken, giving:

$\zeta_d=\sqrt{\frac{3\mu}{8(1+\mu)^3}}$

Response When Optimally Tuned

$$
u(t) = U e^ {iwt}; |U| = \frac{F_0}{mw_1^2}D_u
\\ D_u = \frac{\beta^2 D}{\sqrt{(\alpha^2 - \beta^2)^2 + (2\zeta_d \alpha\beta)^2}}
\\ D=\sqrt{1+\frac{2}{\mu}}
$$

*Could be derived from the two fixed points when optimally tuned.

We can observe that when optimally tuned:

The response of TMD is much larger than that of structure.

So, if you were engineer, make sure the space is enough to install and allow movement of TMD.

$D$ and $D_u$ will both decrease with increase of TMD mass ratio and TMD damping ratio.

In this case, heavier TMD is better.