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

In summary, Hooke's Law, which states that F=kx, applies to linear springs where the force applied is directly proportional to the displacement. However, not all springs behave linearly and may have a non-linear force-displacement relationship. In this case, the spring constant k is not constant and instead becomes a function of x. Various mathematical equations, such as power series, can be used to determine the relationship between force and displacement in a non-linear spring. This concept can also be seen in other fields, such as non-linear optics.
  • #1
chandran
139
1
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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  • #2
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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  • #3
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
 
  • #4
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

[tex] \sigma = E \epsilon [/tex]

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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  • #5
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.
 
  • #6
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;

[tex] \vec{P}=\epsilon_0\chi\vec{E} [/tex]

The polarisation is described by;

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

Where [tex] \chi_n [/tex] 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.
 
  • #7
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.
 
  • #8
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.
 

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

1. What is a non-linear spring?

A non-linear spring is a type of spring that does not follow Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. Non-linear springs have a variable stiffness, meaning that the force exerted by the spring changes as it is compressed or stretched.

2. How is the force-displacement relationship of a non-linear spring derived?

The force-displacement relationship of a non-linear spring can be derived using mathematical equations and principles such as differential calculus and the principles of equilibrium. This involves considering the spring's geometry, material properties, and any external forces acting on it.

3. What factors affect the stiffness of a non-linear spring?

The stiffness of a non-linear spring can be affected by several factors, including the material properties of the spring (such as its Young's modulus and yield strength), its geometry (such as the number of coils and their diameter), and any external forces acting on it (such as gravity or other applied loads).

4. Can non-linear springs be used in real-life applications?

Yes, non-linear springs are commonly used in various applications, such as shock absorbers, suspension systems, and mechanical watches. They are useful in situations where a constant force is needed over a specific range of displacement.

5. How does a non-linear spring compare to a linear spring?

A non-linear spring differs from a linear spring in that its stiffness is not constant. A linear spring follows Hooke's Law and has a constant stiffness, while a non-linear spring has a variable stiffness. This means that the force-displacement relationship of a linear spring is a straight line, while that of a non-linear spring is a curve.

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