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

(a) HB = 241

(b) diameter = 1.19 mm

Step-by-step 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 -
`√(D^2-d^2)` ) ................1

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

HB = (2× 500 ) /
`\pi` 10 ( 10 -
`√(10^2-1.62^2)` )

HB = 241

In second part, we calculate d from given equation

d =
`\sqrt{D^2-  (D- (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- (2×500)/(450 \pi  10))^2}^(2)`

so diameter = 1.19 mm