Question

A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg was used. Calculate the Brinell hardness (in HB) of this material.

Answer 1

Answer:

HB=131.69      

Explanation:

Given that

Diameter ,    D= 10 mm

Depth of indentation ,d= 1.55 mm

Load ,P =500 kg

Brinell hardness is given as

$HB=\dfrac{P}{\pi D(D-\sqrt{D^2-d^2})}$

Now by putting the values in the above equation we get

$HB=\dfrac{500}{\pi \times 10\times (10-\sqrt{10^2-1.55^2})}$

HB=131.69

Therefore the hardness of the material will be 131.69 HB.