## Test Coupon

Consider a 10 cm diameter steel coupon used for a tensile strength test is subjected to a 500 N compressive force for 5 minutes. After completing the test, the specimen measured a length decrease of 0.5 cm and a new diameter of 10.01 cm. If the Poisson’s ratio is 0.3, what was the coupon’s original length in cm?

##
__
__**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 lateral direction:

$$$\varepsilon_{lateral}=\frac{\Delta D}{D_o}=\frac{(10.01cm-10cm)}{10cm}=\frac{0.01cm}{10cm}=0.001$$$

Because the Poisson’s ratio is known, let’s solve for the strain in the longitudinal direction next:

$$$0.3=\frac{0.001}{\varepsilon_{longitudinal}}\Rightarrow \varepsilon_{longitudinal}=\frac{0.001}{0.3}=0.0033$$$

Thus, the original length is:

$$$0.0033=\frac{0.5cm}{L_o}\Rightarrow L_o=\frac{0.5cm}{0.0033}=150\:cm$$$

150 cm