Question

A Brinell hardness tester is used to determine the hardness of 1/2" thick steel plate. The specifications for the steel require that the steel have a tensile strength between 125 ksi and 150 ksi. A 3000 kg load is used for the test.

A. Calculate the minimum and maximum Brinell hardness of the steel.

B. If a hardness test reveals that the Brinell hardness of the steel plate is 260 BHN what is the diameter of the indention in the surface of the plate from the Brinell hardness test?

Answer 1

Solution:-

Given- ₁ =125 Ksi = 125 x 6.89 x10⁶ pa

₂ = 150 ksi = 150 x 6.89 x 10⁶ pa

load P = 3000kg = 3000x9.81 N

so that we can find diameters by strength

= P/A

for 125 ksi,

125 x6.89 x 10⁶ = 3000 x9. 81/(D₁²/4)

D₁ = 0.0065 m =6.5mm

and for 150 ksi

150 x 6.89 x 10⁶ = 3000 x 9.81/( D₂²/4)

D₂ =0.0060m = 6.0mm

A) Now calculate Brinell Hardness

BHN = 2P/D(D - (D² - d²))

where, P = load in Kg

D = diameter of indentor/diameter of steel ball in mm

d =diameter of indentation in mm

generally a 10mm diameter is used for steel ball indentation.

D = 10 mm

now for indentation dia, 6.5mm

BHN = 2x 3000/ x10(10- (10² - 6.5²))

BHN = 79.55

Now for indentation dia, 6mm

BHN = 2X3000/ x 10 (10 - (10²-6²))

BHN = 95.49

therefore, maximum Brinell Hardness of steel = 95.49

and minimum Brinell Hardness of steel =79.55

B) for indentation dia

BHN = 2P /D(D - (D² - d²))

260 = 2x3000 / x10(10- (10² - d²))

So, d² = 14.15

therefore the indentation diameter, d =3.76mm =0.148 inch.