Vibration: find the stiffness of the spring

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SUMMARY

The discussion focuses on calculating the required stiffness of helical springs to achieve a transmissibility ratio (Tr) of 0.1 for a 100-tonne machine operating at 200 RPM. The key equations involved are the natural frequency formula ωn = sqrt(k/m) and the transmissibility ratio formula. The user initially assumed zero damping (ζ=0) and derived a natural frequency of approximately 20, leading to a stiffness value of k = ωn² × m, which resulted in an unexpectedly high stiffness value of 3.9 x 10^8 N/m. Participants in the discussion emphasized the importance of understanding the implications of the calculated stiffness in relation to the machine's weight and the expected spring compression.

PREREQUISITES
  • Understanding of simple harmonic motion and its equations
  • Familiarity with the concept of transmissibility ratio in vibration analysis
  • Knowledge of natural frequency calculations and their significance
  • Basic principles of mechanical spring design and material properties
NEXT STEPS
  • Study the derivation and implications of the transmissibility ratio formula in vibration isolation
  • Learn about the design and selection criteria for industrial helical springs
  • Explore the effects of damping on vibration systems and how to incorporate it into calculations
  • Investigate the relationship between mass, stiffness, and natural frequency in mechanical systems
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Mechanical engineers, vibration analysts, and students studying dynamics and vibration control who are involved in designing systems for vibration isolation and spring selection.

bigbang42
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Homework Statement


a machine of mass 100 tonne generates a simple harmonic disturbing force when operating at a speed of 200rpm. to protect the floor and surrounding machinery it is proposed tp mount the machine on helical springs so that the transmissibility ratio is reduced to 0.1

Q1 determine the total stiffness of the springs required assuming the damping to be negligible ??

I'm having with this question because apparently certain sections you have to make "assumptions" which to me is more like guess work


Homework Equations



ωn = sqrt ( k/m)

tr = \sqrt{(1 + ( 2ζ ω/ωn) / ((1 - (ω/ωn)^2)^2 + (2ζ ω/ωn)^2) }





The Attempt at a Solution



right what I've done is I've said that ζ=0 which gets rid of anything in the tr formula with ζ in as obviously anything multiplied by 0 is 0

so that leaves sqrt [ 1 / ((1-(ω/ωn)^2)^2) ]

i found ω is 200/60 x 2pi so you get something like ω=(20/3)pi

i then rearranged the tr formula two lines above to get ωn = 20

then used the formula ωn= sqrt (k/m)

rearranged to get k = ωn^2 x m and got a stupidly high answer so i imagine i done all the above wrong

can anyone give me any steps on what to do or where I'm going wrong, i have been tearing my hair out over it :(
 
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What makes you think the result is "stupidly high"?
It is a 100T machine, after all. So what are you using as your benchmark for "high"?
 
i got something silly like 3.9x10^8 for my k value in the end, is that not too high ?
 
"too high" compared with what?

what sort of value were you expecting and why?
(get into the habit of justifying your hunches.)

i.e.
How far would that 100T machine compress the springs under gravity alone?
What would happen to that compression under the kind of value you were expecting?
i.e. how tall would the springs have to be just to support the machine off the floor?
i.e. how long are industrial springs ... normally?
 
Letting you off the hook:
With no damping, the transmissibility ratio is $$Tr=\frac{1}{\left | 1-\frac{\omega^2}{\omega_0^2}\right |}$$ ... This is the same as yours, but written more clearly so you can see what is going on better.
Bearing this in mind, you told us:
i found ω is 200/60 x 2pi so you get something like ω=(20/3)pi
i then rearranged the tr formula two lines above to get ωn = 20
Notice that ω = 20π/3 ≈ 20 also ... which suggests a frequency ratio close to 1 - or resonant transmissibility (Tr → ∞): the opposite of what is wanted. The question calls for Tr=1/10 ...
 

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