Prove that the theoretical value of Poisson’s ratio is 0.5.

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To prove this, we shall consider a cube of a material of side length unity (say 1mm) as shown in Figure 2.21. One pair of the faces of cube be subjected to a tensile stress which produces strain ε along the direction of tensile stress.

Change in length along the stress

d l=\varepsilon L=\varepsilon \quad(\because \quad L=1 mm )

∴ The changed length along the stress = (1 + ε)

The strain along the remaining two sides is ε/m, that is, compressive.

∴ Changed length along other two sides = 1 – (ε/m)

Hence, volume of the cube = =(1+\varepsilon)\left\lgroup 1-\frac{\varepsilon}{m} \right\rgroup\left\lgroup 1-\frac{\varepsilon}{m} \right\rgroup

=(1+\varepsilon)\left\lgroup 1-\frac{2 \varepsilon}{m} \right\rgroup \text { after neglecting terms containing } \varepsilon^2

=1+\varepsilon-\frac{2 \varepsilon}{m} \text { again omitting term containing } \varepsilon^2

The original volume = 1 × 1 × 1 = 1 mm³

Since change in volume is negligible, the changed volume is almost equal to original one.

\begin{gathered} 1-\varepsilon-\frac{2 \varepsilon}{m}=1 \\ m=2 \quad \therefore \quad \frac{1}{m}=0.5 \end{gathered}

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