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# Poissons ratio question...3

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## Poissons ratio question...

(OP)
Hi everyone. I have what I'm sure is a very simple question.

Incompressible materials are supposed to have a poissons ratio of 0.5. I am wondering if there is a mathematical proof for this because I'm not getting on so swell at understanding this concept. What prompted this question is uniaxial testing of soft biological tissues which are assumed to be almost incompressible due to their high water content.

Anyway here's whats bugging me.

Suppose you've a cube of dimension l*w*t=v (i.e length*width*thickness=volume).

For the sake of a numerical example lets say this cube has a volume of 1 (and hence l,w,t = 1) and is filled with an incompressible liquid.

Ok. So lets stretch this cube of material uniaxially by stretching its length. In fact lets stretch it so that its length is now x2 the original length. i.e. l=2

To satisfy the incompressibility of the material its new dimensions must now be 2*w*t=1. Suppose w=t... I think its safe to say this.

So now w = 1/SQRT(2) = 0.707 and Delta w = -0.2929

So obviously strain in the transverse direction is -0.2929 and strain in the londitudinal axis is 1. So this gives a poissons ratio of 0.2929.

So what gives? Maybe when its said that poissons ratio for an incompressible material is 0.5 it only applies to small deformations. This indeed is true when small strains are applied to my simple numerical example. Perhaps an 'incompressable' material is not supposed to deform like I have done above.

Anyway anyone who will take the time to tweak my understanding of this is much appreciated.

Yours,

Brian MacG

### RE: Poissons ratio question...

bmgri....by subjecting your model to a length increase of 100 percent, you took away the premise of incompressibility (or in this case the opposite of such).  That's why your relationship doesn't hold.

### RE: Poissons ratio question...

(OP)
Yeah I see what your saying. I thought that incompressibility just ment no net change in volume though. Water is incompressable but it also can deform. So does incompressible also mean not deformable?

### RE: Poissons ratio question...

"....... incompressibility just ment no net change in volume though. Water is incompressable......."

Compared to gases water is incompressible. Compared to steel water is very compressible indeed.

### RE: Poissons ratio question...

Bmgri,

Yes, you are correct that small deformations (small strains) are assumed in developing the theory leading up to Poisson's approximation.  For strains much over a few percent, you would need to consider second-order and possibly higher terms.

### RE: Poissons ratio question...

0.5 is the upper limit of poisson's ratio

### RE: Poissons ratio question...

2
Materials respond to an applied stress by straining. For applied stresses within the elastic range, most materials deform linearly; i.e. the stress is proportional to the strain. For a state of simple tension we know this relationship can be expressed as

S = Ee

where S is the nominal stress, e is the nominal strain, and E is the modulus of elasticity or Young’s modulus. Consider a bar of material that is loaded in uniaxial tension. If the deformation is elastic and the strains are relatively small, then for what value of Poisson’s ratio will there be no volume change during this process? Consider a material that has an initial length of Lo, and a square cross section with initial lateral dimension of lo. From the definition of strain e,

e = (L - Lo)/Lo

where L is the final length and Lo is the original length of the material. Carrying out the division,

e =  L/Lo - 1
So,

L/Lo = 1 + e

From the definition of the Poisson ratio v,

v = - lateral strain/tensile strain

= - e(lateral)/e(tensile)

= - (l-lo)/e*lo

where l is the final lateral dimension of the component, and lo is the initial lateral dimension. Multiplying both sides of the equation by e,

v*e = l - l/lo
So,
l/lo = 1 - v*e

If there is no volume change, then

L*l^2 = Lo*lo^2

(l/lo)^2 = Lo/L

We find,

(1 - v*e)^2 = 1/(1 + e)

(1 + e)(1 - v*e)^2 = 1

(1 + e)[1 - 2v*e + (v^2)*(e^2)] = 1

Since the strains are small, the strain squared is even smaller, and can be ignored. So,

(1 + e)(1 - 2v*e) = 1

1 + e - 2v*e - 2v*(e^2) = 1

Again ignoring the term involving the strain squared, we find

e - 2v*e = 0

which reduces to

e = 2v*e

1 = 2v

v = 1/2

Maui

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