Natural Frequency in Series? 2

[Twoballcane]

Hello All,

I know that if springs are in series I can sum them like this 1/K = 1/K1 + 1/K2 … to get the equivalent spring constant. Do you know if I can do the same with natural frequencies? If I had the following (made up for this example): fn1=10hz, fn2=50hz, and fn3=30hz in series, can I calculate the equivalent natural frequency? Thus 1/fn = 1/fn1 + 1/fn2 + 1/fn3, which will give me equivalent fn = 6.5hz

Does this make sense? Somebody is asking me to do this, but not sure if correct.

Thanks in advance for your input!


Tobalcane
"If you avoid failure, you also avoid success."

 

[electricpete]

what does "in series" mean.

If you have N single DOF discrete systems, then in general when you connect them you have N natural frequencies.

Dunkerley is one approach to break the system into pieces by examining the resonant frequency of one mass attached to the spring system at a time. Gives a bound for the lowest frequency

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[GregLocock]

"Does this make sense?"

No, not in general.

If we skip the maths, adding a tiny system to a large system has scarcely any effect on the large system's fundamental mode.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Twoballcane]

Thanks for responding,

electricpete, in series is when you have one spring connected to another spring which in turn is connected to another spring in a straight line oppose of them being in parallel which is side by side. And I agree with you that if I wanted to analysis Fn per DOF that this is the way to do it. However, if you have more than one spring you can sum them up and then calculate the Fn or the overall system (pg 24 Vibration Analysis for Electronic Equipment by Dave Steinberg). But, in my case I do not know the spring constants only the Fn of each system.

GregLocock, I see what you are saying, thanks for your input.


Tobalcane
"If you avoid failure, you also avoid success."

 

[izax1]

Natural frequencies is not only dependent on stiffness but also dependent on mass (f=Sqrt k/m)So you also have to take care of the mass in the total system.
So, I dont think your suggestion makes sense.

 

[Twoballcane]

Thanks izax1, so far I think this is not going to work and I'll push back on this.

Thanks everybody!

Tobalcane
"If you avoid failure, you also avoid success."

 

[Denial]

Elaborating upon what ElectricPete said above.

To get an approximate estimate of the lowest natural frequency for a multi-dof system of lumped masses, you can use Dunkerley's formula (also sometimes known as the Southwell-Dunkerley formula, or the "inverted squares" formula). This formula is

(1/F)^2 = (1/F1)^2 + (1/F2)^2 + (1/F3)^2 + ...

where F1 is defined as the frequency of a modified form of the structure in which all the springs remain present, but all masses except M1 are set to zero.

Points to note with this formula:
(1) It applies only to lumped mass systems. Thus the structure consists of a heap of lumped masses, interconnected by a tangle of linear springs.
(2) It always gives an underestimate of the true structural frequency. (This is not to say it is necessarily inaccurate, just to alert you to the direction any error will take.)
(3) You need to be careful in how you calculate your "component frequencies" Fi, and adhere to the above definition.
(4) As per GregLocock's post, there are situations where the result of the formula can be way off.

 

[Twoballcane]

Thank you Denial! I was just coming back from my meeting with exactly with what you have.

Tobalcane
"If you avoid failure, you also avoid success."

 

[Twoballcane]

Oh Thank you too electricpete!

Tobalcane
"If you avoid failure, you also avoid success."