Test Specimen

Consider a titanium test specimen with a 2” diameter and a 10” length is subjected to a 150,000 lbs compressive force for 5 minutes. After completing the test, the specimen measured a length decrease of 0.25”. If the Poisson’s ratio was 0.2, how many inches did the diameter increase by?

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 the longitudinal direction:
$$$\varepsilon_{longitudinal}=\frac{\Delta L}{L_o}=\frac{0.25in}{10in}=0.025$$$
Because the Poisson’s ratio is known, let’s solve for the strain in the lateral direction next:
$$$0.2=\frac{\varepsilon_{lateral}}{0.025}\Rightarrow \varepsilon_{lateral}= (0.2)(0.025)=0.005$$$
Thus, the diameter increased by:
$$$0.005=\frac{\Delta D}{2in}\Rightarrow \Delta D=(0.005)(2in)=0.01\:in$$$
0.01 in