Question

(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 270 HB when a 500-kg load is used

Answer 1

(a) We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.

Thus, the Brinell hardness is computed as

`HB=2P/\pi D{D-√(D^2-d^2)`

`=2*1000hg/\pi (10mm)[10mm-√((1000^2-(2.3mm)^2) ]`

(b) This part of the problem calls for us to determine the indentation diameter d which will yield a 270 HB when P= 500 kg.

`d=\sqrt{D^2-[D-(2P)/((HB)\pi D) } ]^2\\=\sqrt{(10mm)^2-[10mm-(2*500)/(450( \pi10mm)) } ]^2`