Question

A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Calculate the HB of this material.

Answer 1

Answer:

$HB \approx 200.484$

Explanation:

The Brinell Hardness is obtained from the following formula:

$HB = \frac{2\cdot P}{\pi \cdot D^{2}}\cdot \left(\frac{1}{1 - \sqrt{1-\frac{d^{2}}{D^{2}} } } \right )$

Where $P$ is the load measured in kilograms-force, $D$ is the diameter of indenter measured in millimeters and $d$ is the diameter of indentation measured in millimeters.

$HB=\frac{2\cdot (1000\,kgf)}{\pi\cdot (10\,mm)^{2}}\cdot \left[ \frac{1}{1-\sqrt{1-\frac{(2.50\,mm)^{2}}{(10\,mm)^{2}} } } \right ]$

$HB \approx 200.484$