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

Answer:

(a). The HB of this material is 217.8

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

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

$HB=\dfrac{2P}{\pi D(D-\sqrt{D^2-d^2})}$

Put the value into the formula

$HB=\dfrac{2\times1000}{\pi\times10(10-\sqrt{10^2-2.4^2})}$

$HB=217.8$

(b). Given that,

Hardness = 300 HB

Load = 500 kg

We need to calculate the diameter of an indentation

Using formula of diameter

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

Put the value into the formula

$d=\sqrt{10^2-[10-\dfrac{2\times500}{300\pi\times10}]^2}$

$d=1.45\ mm$

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

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