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

Answer:

(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

$HB=2P/\pi D{D-\sqrt{D^2-d^2}$

$=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]$

(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.  

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