Why is Poisson's ratio necessary, when we know that volume stays constant?

Why is the Poisson ratio necessary, when volume is conserved? I read that volume is conserved when a body is subjected to longitudinal (compressive or tensile) stress or shear stress, so given that volume is conserved, can we not simply find the change in diameter (and hence the lateral stress) without the Poisson ratio? Is either the Poisson ratio or the conservation of volume only applicable in certain limits? If so which ones? Thanks!

• Who says volume is conserved when a body is subjected to longitudinal compressive or tensile stress? This is the case only if the Poisson ratio is 1/2. – Chet Miller Jun 15 at 12:07

We need Poisson's ratio $$\sigma$$ precisely because the volume is usually not conserved when we stretch, squash or twist something. An exception is ordinary rubber which is, to a reasonable approximation, incompressible, so for rubber $$\sigma=1/2$$. For steel it is about $$.3$$.

Notice, only normal stresses are responsible for causing volumetric strain or change in volume of a material. The shear stress causes distortion of shape of the object but doesn't change its volume.

Engineering stress-strain curve can be divided into two regions

a) Elastic region: Where material obeys Hook's law & poisson's ratio comes into picture to find elastic strains.

In general, the volumetric straight $$\large \epsilon_v$$ of any object or component (having Young's modulus $$E$$ & Poisson's ratio $$\nu$$ ) subjected to three normal stresses $$\sigma_x, \ \sigma_y \$$ $$\sigma_z$$ along X, Y & Z axes respectively within the elastic limit, is given by $$\large \epsilon_v=\frac{(\sigma_x+\sigma_y+\sigma_z)(1-2\nu)}{E}$$ The above volumetric strain $$\epsilon_v$$ of a material will be zero i.e. volume of material will remain conserved in following two cases

1.) if $$\sigma_x+\sigma_y+\sigma_z=0\$$ i.e. when a material is subjected to three normal stresses such that their (algebraic) sum is zero for example $$\sigma_x=50\ kPa$$, $$\sigma_y=20\ kPa$$ & $$\sigma_z=-70\ kPa$$, the volumetric strain $$\epsilon_v=0$$ i.e. volume of the object or component remains conserved withing elastic limit irrespective of material or Poisson's ratio $$\nu$$.

2.) if Poisson's ratio $$\nu=0.5$$ for example rubber, the volumetric strain $$\epsilon_v=0$$ i.e. the volume of such material remains conserved withing elastic limit irrespective of types of stresses which the material is subjected to.

Besides above two cases, the volume of a component subjected to various stresses within elastic limit is not conserved i.e. the volume changes which requires Poisson's ratio $$\nu$$ to be known to find normal (axial) strains $$\epsilon_x, \epsilon_y, \epsilon_z$$, or volumetric strain $$\epsilon_v$$ within elastic limit.

b) Plastic region: Where material fails to obey Hook's law & poisson's ratio doesn't come into picture. Material follows power law of strain-hardening

In plastic region (beyond elastic limit), the volume of a component subjected to external forces or loads remains ideally conserved. The material obeys power law of strain-hardening i.e. $$\sigma=\sigma_o+K\epsilon^n$$ Where, $$\sigma$$ is flow-stress, $$\sigma_0$$ is yield flow-stress beyond which material deforms plastically, $$K$$ is strength coefficient, $$\epsilon$$ is plastic strain, $$n$$ is strain-hardening exponent.

• When can stress have a negative value? – Meripadhai Jun 17 at 16:04
• @Meripadhai Compressive stress is taken as negative stress. While tensile stress is taken as positive stress – Harish Chandra Rajpoot Jun 17 at 16:51

Take a (circular) rod. Stretch it along its axis in the reversible (elastic) region of behavior. Assume the material is isotropic. To first order, the following expression can be derived:

$$-\frac{dV}{V_o} = \nu^2\epsilon_a^3 + \nu(\nu - 2)\epsilon_a^2 + (1 - 2 \nu)\epsilon_a$$

where $$dV/V_o$$ is the relative change in volume referenced to the initial volume, $$\nu$$ is the Poisson ratio, and $$\epsilon_a$$ is the axial strain $$dl/l_o$$.

Eliminate all terms that are higher order of magnitude than $$\epsilon_a$$ to obtain

$$-\frac{dV}{V_o} \approx (1 - 2\nu)\epsilon_a$$

From this you can see that the only time we will have anything approaching a true conservation of volume is for $$\nu = 0.5$$ for an isotropic material under exceptionally small axial strain. The latter restriction is so that the higher order terms do fall off to zero.

Volume is conserved in plastic deformation, so the density of a broken sample after a tensile test should be the same as the initial one.

But during the test, while stressed, there is a (very small) change in the volume. For no change at all the poisson ratio should be 0.5.

• I cannot simply trust that volume is conserved in plastic deformation. I would have to see an explicit proof. In any case, the Poisson ratio does not apply in plastic deformation. – Jeffrey J Weimer Jun 13 at 19:58
• When an ingot comes from the melt shop, it is possible to have some reduction of volume in the first rolling passes, due to micro (or sometimes macro) voids from foundry. But after that it is constant. – Claudio Saspinski Jun 13 at 20:52