Prove Poisson's Ratio is 0.5 [closed]

Poisson's ratio is the negative ratio of the transverse strain (_T) to the axial strain (_A). For an incompressible (density doesn't change), homogeneous (everything is the same molecule), isotropic (it doesn't matter which direction you pull or push on it, it will act the same way), linear elastic (Hookean) material, $v = 0.5$ for small strains. Using a first order approximation (sort of like a Taylor Series approximation), prove $v = .5$.

$$\mathrm{Strain} = (\Delta L)/L$$

$$v = - \frac{E_T}{E_A} = \frac{(\Delta L_T)/L_T}{(\Delta L_A)/L_A}$$

I have no idea where to go from here.

For the small strain setting, the trace of the strain tensor is a measure for the volume change, which should be zero for the incompressible material. That would be two times the lateral straining plus the longitudinal straining. Replacing the lateral straining by minus eta times longitudinal strain, you see immediately that eta must be 0.5.

Imagine a cube that is stretched in one direction and contracted equally in the transverse directions. The volume of the new cube is $$V=l_1 l_2 l_3$$

the old volume is

$$V_0=l_{01} l_{02} l_{03}$$

Now calculate $V/V_0$, and replace $l_i=l_{0i}(1+\varepsilon_i)$. The $l_{0i}$-product cancels out, so what is left is

$$V/V_0=(1+\varepsilon_1)*(1+\varepsilon_2)*(1+\varepsilon_3)\\ =1 + \varepsilon_1 + \varepsilon_2 + \varepsilon_3 + \varepsilon_1 \varepsilon_2 + \varepsilon_1 \varepsilon_3 + \varepsilon_2 \varepsilon_3 + \varepsilon_1 \varepsilon_2 \varepsilon_3$$

You want $V/V_0=1$ in case of incompressibility. Neglecting the quadratic and the cubic terms with the small strain assumption there remains only

$$0 = \varepsilon_1 + \varepsilon_2 + \varepsilon_3$$ which you replace by $$0 = \varepsilon_{lat} + \varepsilon_{lat} + \varepsilon_{long}.$$

• How does the change in volume equal two times the lateral strain plus the longitudinal strain? – MaryG Nov 8 '14 at 16:17
• ok will edit the answer... – Rainer Glüge Nov 8 '14 at 16:47