Question

A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this material when a load of 1000 kg produced an indentation of 2.50|| mm in diameter. [1 Mark] b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used?

Answer 1

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

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

Now, putting the given values into the above formula as follows.

                HB = $\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}$                            = $\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}$  

                       = $\frac{2000}{9.98}$                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

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

                         = $\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}$

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.