Question

(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.4 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 300 HB when a 500-kg load is used?

Answer 1

(a). The HB of this material is 217.8

(b). The diameter of an indentation is 1.45 mm.

Step-by-step explanation:

Given that,

Diameter of Brinell hardness D= 10.0 m

Diameter of steel alloy d= 2.4 mm

Load = 1000 kg

(a). We need to calculate the HB of this material

Using formula of Brinell hardness

`HB=(2P)/(\pi D(D-√(D^2-d^2)))`

Put the value into the formula

`HB=(2*1000)/(\pi*10(10-√(10^2-2.4^2)))`

`HB=217.8`

(b). Given that,

Hardness = 300 HB

Load = 500 kg

We need to calculate the diameter of an indentation

Using formula of diameter

`d=\sqrt{D^2-[D-(2P)/((HB)\pi D)]^2}`

Put the value into the formula

`d=\sqrt{10^2-[10-(2*500)/(300\pi*10)]^2}`

`d=1.45\ mm`

Hence,(a). The HB of this material is 217.8

(b). The diameter of an indentation is 1.45 mm.