A10 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?

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 =    

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

                HB =                            =  

                       =                          

                       = 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 =

                         =

                          = 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.

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