Question

A 10-mm-diameter Brinell hardness indenter produced an indentation 1.62 mm in diameter in a steel alloy when a load of 500 kg was used. (a) Compute the HB of this material. (b) What will be the diameter of an indentation to yield a hardness of 450 HB when a 500 kg load is used?

Answer 1

Answer:

(a) HB = 241

(b) diameter  = 1.19 mm

Explanation:

Given data

load (P) = 500 kg

diameter  (d) = 1.62 mm

Diameter (D) = 10 mm

hardness = 450 HB

To find out

HB of this material and diameter

Solution

First part we use equation to find brinell hardness

HB = (2× load ) / $\pi$ D ( D - $\sqrt{D^2-d^2}$ )    ................1

Now put the value load, D and d in equation 1

HB = (2× 500 ) / $\pi$ 10 ( 10 - $\sqrt{10^2-1.62^2}$ )

HB = 241

In second part, we calculate d from given equation

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

Now put the value HB = 450, load and D value in equation 2

d =$\sqrt{10^2- (10- \frac{2×500}{450 \pi 10})^2}^{2}$

so diameter  = 1.19 mm