Question

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively.

Answer 1

L = 75.25 mm

Step-by-step explanation:

First we need to find the lateral strain:

Lateral Strain = Change in Diameter/Original Diameter

Lateral Strain = (20.025 mm - 20 mm)/20 mm

Lateral Strain = 1.25 x 10⁻³

Now, we will find the Poisson's Ratio:

Poisson's Ratio = (E/2G) - 1

where,

E = Elastic Modulus = 105 GPa

G = Shear Modulus = 39.7 GPa

Therefore,

Poisson's Ratio = [(105 GPa)/(2)(39.7 GPa)] - 1

Poisson's Ratio = 0.322

Now, we find longitudinal strain by following formula:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

Longitudinal Strain = - Lateral Strain/Poisson's Ratio

Longitudinal Strain = - (1.25 x 10⁻³)/0.322

Longitudinal Strain = - 3.87 x 10⁻³

Now, we can fin the original length:

Longitudinal Strain = Change in Length/L

where,

L = Original Length = ?

Therefore,

- 3.87 x 10⁻³ = (74.96 mm - L)/L

(- 3.87 x 10⁻³)(L) + L = 74.96 mm

0.99612 L = 74.96 mm

L = 74.96 mm/0.99612

L = 75.25 mm