Prove Poisson's Ratio is .5 for Small Strains

[_Anonymous_]

Homework Statement

Poisson's ratio v is the negative ratio of the transverse strain to the axial strain. 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 = .5 for small strains. Using a first order approximation (sort of like a Taylor Series approximation, prove v = .5.

Homework Equations

strain = ε = ΔL/L₀

Poisson's ratio = - ε_T / ε_A= - (ΔL_T/L_T) / (ΔL_A/L_A) = -(ΔL_TL_A)/(ΔL_AL_T)

ΔVolume = 0

The Attempt at a Solution

I drew a picture of a cube with original dimensions lwh = L_A³. Then I applied a transverse strain to the box and drew a new box inside and extending from the original box. The new box's dimensions are (ΔL_A + L_A)(L_A - ΔL_T)(L_A - ΔL_T) assuming that ΔL_T and ΔL_A are positive values. Since the change in volume is zero, the amount the box "shrinks" and the amount the box "extends" should be equal, so I tried summing the new volumes and setting them equal to zero:

0 = -(Volume that the cube shrunk) + (volume that the cube elongated)

0 = - (2(ΔL_T)(L_A) + (ΔL_T)²L_A) + ΔL_A(L_T - ΔL_T)

2(ΔL_T)(L_A) + (ΔL_T)²L_A = ΔL_A(L_T - ΔL_T)

I tried to factor out the terms that I wanted to be the numerator and denominator:

(ΔL_TL_A)(2 + L_T) = ΔL_AL_T - ΔL_AΔL_T

(ΔL_TL_A)(2 + L_T) + ΔL_AΔL_T = ΔL_AL_T

Divide both sides by ΔL_AL_T

v(2 + L_T + ΔL_T/L_T = 1

v = (1 - ΔL_T/L_T)/(2 + L_T)

Simplifying...

v = (L_T - ΔL_T) / (L_T(2 + L_T)

What should I be doing differently?

 

[OldEngr63]

This problem makes an interesting assertion about Poisson's ratio. It is particularly interesting since, for common metals like steel, Poisson's ratio is about 0.3.

Poisson's ratio = 0.5 applies to a material such as rubber, but not to most metals.

 

[haruspex]

0 = -(Volume that the cube shrunk) + (volume that the cube elongated)

0 = - (2(ΔL_T)(L_A) + (ΔL_T)²L_A) + ΔL_A(L_T - ΔL_T)

That last equation is dimensionally wrong. It mixes quadratic and cubic powers of length.
Once you have the correct equation, you need to make an approximation for small deformations. If L >> ΔL, what can you say about L ΔL compared with ΔL²?