Question

The indentation on a steel sample has been taken using 10 mm tungsten carbide ball at 500 kgf load. If the average diameter of the indentation is 2.5 mm, the BHN will be nearly
1. 90
2. 100
3. 110
4. 120

Answer 1

Correct Answer - Option 2 : 100

Concept:

Brinell hardness number (BHN) \( = \frac{{2P}}{{\pi \;D\left\{ {D\;-\;\sqrt {{D^2} - {d^2}} } \right\}}}\)

Where P is applied load in kgf

D is the diameter of indenter ball in mm, d is the diameter of indentation in mm.

Calculation:

BHN \(= \frac{{2\; \times 500}}{{\pi \times 10\left\{ {10\; - \;\sqrt {{{10}^2} - {{2.5}^2}} } \right\}}}\)

∴ BHN = 100.242