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

Step-by-step 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 * 1000 kg}{3.14 * 10 mm [D - \sqrt{((10 mm)^(2) - (2.50)^(2))}]}`

=
`(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 - (2P)/(HB) \pi D]^(2))}`

=
`\sqrt{(10 mm)^(2) - [10 mm - (2 * 500 kg)/(300 * 3.14 * 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.