Poisson’s Ratio

Consider an aluminum test specimen with a 3” diameter and a 12” length is subjected to a 100,000 lbs compressive force for 5 minutes. After completing the test, the specimen measured a length decrease of 0.2” and a diameter increase of 0.01”. Calculate the Poisson’s ratio.

Hint
Poisson’s ratio measures a material’s deformation in directions perpendicular to the loading/pressure direction:
$$$\upsilon=-\frac{\varepsilon_{lateral}}{\varepsilon_{longitudinal}}$$$
where $$\varepsilon_{lateral}$$ is the lateral strain, and $$\varepsilon_{longitudinal}$$ is the longitudinal strain.
Hint 2
Strain is the change in length per unit length.
$$$\varepsilon=\frac{\Delta L}{L_o} $$$
where $$\varepsilon$$ is the engineering strain, $$\Delta L$$ is the change in length, and $$L_o$$ is the original length.
Poisson’s ratio measures a material’s deformation in directions perpendicular to the loading/pressure direction:
$$$\upsilon=-\frac{\varepsilon_{lateral}}{\varepsilon_{longitudinal}}$$$
where $$\varepsilon_{lateral}$$ is the lateral strain, and $$\varepsilon_{longitudinal}$$ is the longitudinal strain.

Strain is the change in length per unit length.
$$$\varepsilon=\frac{\Delta L}{L_o} $$$
where $$\varepsilon$$ is the engineering strain, $$\Delta L$$ is the change in length, and $$L_o$$ is the original length.

First, let’s analyze the strain in both the lateral and longitudinal direction:
$$$\varepsilon_{lateral}=\frac{\Delta D}{D_o}=\frac{0.01in}{3in}=0.0033$$$
$$$\varepsilon_{longitudinal}=\frac{\Delta L}{L_o}=\frac{0.2in}{12in}=0.0167$$$
Therefore, Poisson’s ratio is:
$$$\upsilon=\frac{0.0033}{0.0167}=0.199\approx 0.2$$$
0.2