## 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**

**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**

**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