Test Coupon

#452 / Mechanics of Materials Medium

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?

Expand 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 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


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