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dynamic response spring mass system 3

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zekeman

Mechanical
Aug 30, 2004
1,311
I have a design problem that I can do numerically but I'm looking for a more elegant method.

Here goes:

A spring, S is interposed between 2 masses (M-S-m aligned vertically) and compressed by the gravity of the much larger mass, M which sits on top. The smaller mass, m which is much smaller than the spring mass is at the bottom and is initially restrained. Now the restraint is suddenly removed. The system is now moving vertically under gravity.
What is the response?

I am after the dynamic equations that include the spring mass.


Thanks for any ideas.

 
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"Mr. Ruby's work assumes the spring expands in a uniform manner."

I'm very skeptical about that assumption. This contradicts the wave equation which more accurately describes the motion.

I could accept it for a quasi static motion, but certainly not here-- at least without some proof.

As far as I can remember, and I have used it, the 1/3 assumption has been used where m was larger than the spring mass but I have never seen it for the problem at hand,

m<<spring mass

This looks more like a compressed spring suddenly released.
 
The 1/3 assumption does not rely on m being smaller than the mass of the spring.

If you have it, take a look at Harris' Shock and Vib Handbook 6th ed pages 242-248

=====================================
(2B)+(2B)' ?
 
I should clarify for people with other editions of that book, it is chapter 7 of systems with distributed mass and elasticity.

=====================================
(2B)+(2B)' ?
 
Well maybe there are some assumptions built into that 1/3 about relative masses.

At any rate table 7.10 gives solution (resonant frequency) for various cases similar to yours.

=====================================
(2B)+(2B)' ?
 
Actually, not in the table. One would have to study the text to extract a solution for this specific case.

=====================================
(2B)+(2B)' ?
 
"
The derivation of m/3 can be found here:

Good reference.fter the derivation there was a footnote caveat that said

'Real spring

However, the above calculations are only suitable for small values of \frac{M}{m}. Since Jun-ichi Ueda and Yoshiro Sadamoto have done the experiment[citation needed]. It is found that, as the ratio \frac{M}{m} increases beyond 7, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value \frac{m}{3} and eventually reaches negative values. This unexpected behavior of the effective mass can be explainable in terms of the elastic'

m is the spring mass in this case and I assume the M is the fixed mass which is << m [ the reverse notation I have used]

Pretty much rules out the arbitrary 1/3 rule for general application. However, for the problem proposed, it looks like it might be valid. I will check further and post results if possible.

Thanks for your input as well as everybody else who responded. Your comments are most appreciated.

zeke
 
Correction

I think the reverse is true. I must have gotten it backwards.i.e.

As the ratio of spring mass to fixed mass increases beyond 7 it looks like the effective mass of the spring decreases below 1/3 and eventually goes negative.

I think I'll assume the 1/3 factor and test it . Analytic work is too time consuming for this. In fact a solution to the free spring with similar boundaries I found* indicated that 0.4 is more appropriate

*
Mechanics Applied to Vibration and balancing, D laugharne Thornton, Chapmen & Hall, 1951
 
Way out of my expertise but here goes. The mention of hydraulic cylinders above post hit neuron in my small brain.


Isn't there a very similar problem with mass dampening system, like earthquake mitigation and situations the infamous foot bridge bridge in England? Trying to recall a paper I saw about the swinging bridge in England.
 
Is this school work or home work? This sounds an awful lot like our senior lab project in ME Design. We used analog computers to solve it and compose the equations.

rmw
 
"Is this school work or home work? This sounds an awful lot like our senior lab project in ME Design. We used analog computers to solve it and compose the equations."

Yeah, mine too, in the 60's but not with analog computers. You must be talking the 50's.
 
Interesting discussion. When m is less than spring mass the problem is one of vibration of distributed or continuous systems. These are governed by a wave equation of the form

diff(diff(u,t),t) = E/Rho x diff(diff(u,x),x).

where u is the displacement of a point on the spring. In addition to this wave equation you need the appropriate boundary values and initial conditions (the two masses at the ends affect the boundary conditions). You can find more details in many Mechanical Vibration textbooks in the advanced chapters. I found a solution on page 239 of Vibration of continuous systems By Singiresu S. Rao (from Google books), which I'll summarize below:

The natural frequencies expression is

wn_i = alpha_i x c / L

where alpha_i are the multiple solutions of

alpha_i x tan(alpha_i) = beta

and
beta = spring mass / end mass
and
c = sqrt (E/Rho) of spring

In this problem we should be primarily interested in the first of these natural frequencies (i=1).

This equation actually results in the 1/3 factor for large values of end mass/spring mass (greater than 7). The other limit of end mass very small gives a factor of 4/Pi^2 which is approximate 0.4 as Zekeman mentioned.

It's important to gain that insight. However, in practice I would consider using finite elements or some other numerical tool to model this however.

I checked these equations and the final expression, but not thoroughly enough!

Nagi Elabbasi
Veryst Engineering
 
Nagi,

Thank for your detailed post.

I believe Timoshenko did this problem first around 1910 and a detailed analysis can be found in his book,"vibration Problems in Engineering". I have a copy of the second edition where it appears in pp 317-321.

He derives the 1/3 factor based on assuming a small ratio of spring to lumped mass; however, this holds pretty good over a a larger range of ratios and tends toward 4/Pi^2 or about 0.4, my case.

Thanks again for the discussion and valuable comments.


zeke

 
The behaviour of a spring having a mass Ms and a mass m suspended, being Ms>>m is similar to that of a slinky coil spring. So the assumption of a spring with uniform elongation that leads to an inertial equivalent mass Ms/3, is not valid anymore (my bad). In this case it is not possible to neglect the influence of waves and impulses which propagate over the spring.

E.E. Galloni and M. Kohen, in their “Influence of the mass of the spring on its static and dynamic effects,” Am. J. Phys. 47, 1076–1078 (Dec. 1979), described the behaviour of a slinky coil spring and found a value for the inertial equivalent mass equal to [Ms* 4/(PI^2)]. This value has been already noticed above (Zekeman and Ellabasi).
 
 http://files.engineering.com/getfile.aspx?folder=ad1fffa8-9f02-4b14-903e-bab9921e9e42&file=PTE45_0100_1648.pdf
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