What is the derivation for a non-linear spring and its effects on deflection?

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SUMMARY

The discussion centers on the derivation and implications of non-linear springs, contrasting them with linear springs governed by Hooke's Law (F = kx). Non-linear springs exhibit force-displacement relationships that may involve higher powers of displacement, such as F = kx² or F = g(x)f(x), where k varies with displacement. The conversation highlights the necessity of experimental determination of these relationships, particularly for materials that do not behave elastically. Additionally, the discussion draws parallels to non-linear optics, emphasizing the complexity of restoring forces in non-linear systems.

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  • Knowledge of material properties and stress-strain relationships
  • Basic principles of non-linear optics
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chandran
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we all know the linear hookes law as f=kx where f is the force applied on a spring
and x is the displacement and k is the spring stiffness.

All springs need not behave linearly. suppose a spring behaves non linearly
What is the theory that calculates the deflection for a non linear spring when
a force f is applied.

Basically i believe non linear finite element analysis is based on this non linear
spring theory.

in a non linear spring f will not be proportional to x but may be proportional to
square of x or cube of x or some power of x.
 
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Just a guess, but F=kx.

if it is non linear then if you say x^2, then F=kx^2, or in general, F=k*f(x).
Even more generally, K may also vary as a function of x, so then you will have F=g(x)f(x), where g(x) is the spring constant at various positions.

But idealy, the whole POINT of a GOOD spring is that it IS LINEAR. if it is not linear then the spring probably isint worth using in the first place.

Edit: I assumed that F=k*f(x); however, if it is nonlinear, then the relationship between force and displacement would have to be determined experimentally. It may work out that it is directly proportional to x, or it may work out to be inversely proportional. But I suppose by generalizing with f(x) it takes care of that problem, because if it were inversely proportional f(x) might be 1/x, etc.
 
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I think he means physically non-linear spring, I don't know, its not making much sense to me. If a spring doesn't deform linearly then however it deforms would be its force equation like cyrus said
 
Materials go under different phases until they break, they start in the proportional face when there is a proportional limit which could be equal to its elastic limit, but not necessarily as can be seen in materials such as rubber, after than the materials go to the fluency phase (Plasticity), when they deform easily without mush stress applied, then finally the material recombines to make itself stronger against the stress and goes into the estriction phase where a stress of a bigger magnitude will end up in the material being broken. These depends on the characteristics of the materials.

Hooke's Law

\sigma = E \epsilon

Stress = Elasticity Module * Axial Unitary Deformation, which can be rewriten as F = kx. This law only applies for linear elastic materials. This means the proportional limit is the same as the elastic limit in the materials. This is very important in engineering because most materials are linear elastic.

Materials that have nonlinear elasticity can bend in more than one way under stress or can take different forms before they get back to their original way. Mostly this require equations which can give more than one answer.

Here is a paper about NonLinear Elasticity

http://www.math.siu.edu/spector/fracture.pdf
 
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but may be proportional to
square of x

I would like to point out that we will not get oscillatory behavior if we leave off minus signs or use even powers of x. Only odd powers of x, and odd functions in general, will produce an oscillatory solution.
 
In the most general case, a non-linear restoring force will be a power series of x. The restoring force will depend on the physical characteristics of the spring. To find the restoring force, you would need to determine the coefficients of the power series. (Experimentally, I would imagine).

An interesting parallel to this sceanrio is that of non-linear optics. In non-linear optics, the polarisation is no longer proportional to the applied electric field, i.e. instead of;

\vec{P}=\epsilon_0\chi\vec{E}

The polarisation is described by;

\vec{P}=\epsilon_0\chi_1\vec{E}+\chi_2\vec{E}^2+\chi_3\vec{E}^3+...

Where \chi_n denotes the polarisability tensors of the nth order. The inclusion of these terms in the wave equation gives rise to new frequencies being generated, including new harmonics. For materials with a symmetric atomic restoring force, all the even orders are zero.

Crosson said:
would like to point out that we will not get oscillatory behavior if we leave off minus signs or use even powers of x. Only odd powers of x, and odd functions in general, will produce an oscillatory solution.

Odd functions give rise to symmetric restoring potentials, asymmetric restoring potentials, such as those obtained when even powers of x are included. Asymmetric potentials can still yield oscillatory solutions.



Claude.
 
A member that behaves as a nonlinear spring would have a stiffness that varies with x. The formula relating force to displacement is pretty much the same: F=k(x)x. you will of course need to know how k varies with x.
 
Cyrus said:
Just a guess, but F=kx.

if it is non linear then if you say x^2, then F=kx^2, or in general, F=k*f(x).
Even more generally, K may also vary as a function of x, so then you will have F=g(x)f(x), where g(x) is the spring constant at various positions.QUOTE]

For nonlinear springs, the spring constant k is no longer a constant. it becomes the function of x. it would be more correct to say F=k(x)*x. The units on k must always be in units of force over length in order for it to bear any significance in other formulas.
 

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