# A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original...

## Question:

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 22.529 mm and 22.557 mm, respectively, and its final length is 79.4 mm, calculate its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 102 GPa and 42.4 GPa, respectively.

## Elastic modulus, shear modulus and poisson's ratio

Elastic modulus is the ratio of stress to strain for a point below proportionality limit. Similarly, shear modulus is the ratio of shear stress to shear strain. Poisons ratio is the ratio of the lateral strain to longitudinal strain.

Given:

Original diameter of specimen {eq}d_o = 22.529\ mm {/eq}

Final diameter of specimen {eq}d_f = 22.557\ mm {/eq}

Final length of specimen {eq}l_f=79.4\ mm {/eq}

Elastic modulus E = 102 GPa

Shear modulus G = 42.4 GPa

Calculating the lateral strain :

{eq}\\\epsilon_x=\frac{\Delta d}{do}=\frac{22.557-22.529}{22.529}=1.24\times10^{-3} {/eq}

Relation between elastic constants :

{eq}\\E=2G(1+\mu) \\102\times10^{3}=2\times42.4\times10^{3}(1+\mu) \\\mu=0.203 \\\mu=\frac{-\epsilon_x }{\epsilon _z} {/eq}

Calculating the longitudinal strain using Poisson's ratio :

{eq}\\\epsilon _z=\frac{-\epsilon _x}{\mu}=\frac{-1.24\times10^{-3}}{0.203}=-6.11\times10^{-3} {/eq}

Now,

{eq}\\\epsilon _z=\frac{l_f-l_0}{l_0}=\frac{l_f}{l_0}-1 \\l_0=\frac{l_f}{1+\epsilon _z}=\frac{79.4}{1-6.11\times10^{-3}}=79.9\ mm {/eq}