Deflection in 3-D

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Hi

Sorry if this is the wrong place to post this as it is probably more of an engineering question.

I'm trying to derive the extension of a cylindrical coiled spring in terms of E (YOUNG's modulus, K(bulk modulus), No of turns per metre, spring diameter, (D), spring length (l), wire diameter (d).

Torsional deflection is important but I'm at a loss to how it could be extended to 3-D.

Anyone knows, the derivation or could point me in the right direction - Thanks.
 
2,013
483
What sort of load do you have on this spring?
 
Oh yes forgot to mention that.

The spring is hung vertically from a fixed point and a point load is applied to the end - as in a weight with a hook hung on the spring

so the parameters are
Tension, wire diameter, spring diamter, the 2 moduli of elasticity, spring length and possibly number of turns per metre
 
2,013
483
This analysis is givben in most mechanical engineering design books such as the book by Shigley & Mishkey or the book by M.F. Spotts. I suggest you look in one of these under the topic of spring design.
 
Great found it! Thank you.
For those interested, it is

Deflection = [tex]\frac{8FD3N}{Gd4}[/tex]

Don't know what happened there - here it is again

8FD3N/Gd4

Deflection is independent of length! (although N is related to length in a way)
 
2,013
483
N is the number of turns, so it is directly proportional to the length. Thus this says that the deflection is actually directly proportional to the length.
 
Yes and no. N is directly proportional to length if every turn is coiled in the same way - i.e. at the same angle to the previous turn.

A spring could be 'tightly' wound (small angle) or loosely wound (big angle) and the formula will not distinguish this but works if the way in which it is wound is uniform even though the spring length is different

However N is not affected if a spring is coiled in several different ways but, spring length is.

E.g. If a spring is wound 'tightly' and 'loose' and then tight again randomly, the formula will fail to show that deflection is proportional to spring length because N can no longer be directly proportional to spring length in this case.

Looking at the derivation, it does assume a uniformly wound coil, hence coil length is absent from the deflection equation.
 
2,013
483
Yes and no. N is directly proportional to length if every turn is coiled in the same way - i.e. at the same angle to the previous turn.

A spring could be 'tightly' wound (small angle) or loosely wound (big angle) and the formula will not distinguish this but works if the way in which it is wound is uniform even though the spring length is different

However N is not affected if a spring is coiled in several different ways but, spring length is.

E.g. If a spring is wound 'tightly' and 'loose' and then tight again randomly, the formula will fail to show that deflection is proportional to spring length because N can no longer be directly proportional to spring length in this case.

Looking at the derivation, it does assume a uniformly wound coil, hence coil length is absent from the deflection equation.
As you have said, the derivation is based on the assumption that the spring is uniformly wound. That in itself implies that N is proportional the the length in this equation.

While it is possible to make a spring with varying pitch, it is impractical in most cases. To do this in a controlled manner, to achieve a particular purpose, would require very careful and expensive special purpose spring winding machinery. This is not going to happen without a lot of cost justification.
 

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