Answer :
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 = [tex]\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}[/tex]
Now, putting the given values into the above formula as follows.
HB = [tex]\frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}[/tex] = [tex]\frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}[/tex]
= [tex]\frac{2000}{9.98}[/tex]
= 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 = [tex]\sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}[/tex]
= [tex]\sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}[/tex]
= 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.