Question

An aluminum specimen subjected to the Brinell hardness test, the indenter and indentation diameters are 12 mm&4.49 mm respectively. Determine the surface area of indentation, Also, determine the force applied on the aluminum specimen, if Brinell hardness value is 684 .

1. The surface area of Indentation in mm² is ___________________
2. The force applied in N is ________________

Answer 1

The surface area of indentation when performing the Brinell hardness test on an aluminum specimen with a 4.49 mm indentation diameter is 15.86 mm². The force applied, given a Brinell hardness value of 684, is approximately 132,648 N.

Step-by-step explanation:

To determine the surface area of indentation for the Brinell hardness test on an aluminum specimen:

1. The indentation diameter is 4.49 mm.
2. To find the surface area of a circle, use the formula A = π * (d/2)^2, where d is the diameter.
3. The calculated area A = π * (4.49 mm / 2)^2 ≈ 15.86 mm².

To find the force applied in the Brinell hardness test:

1. The Brinell hardness value (HB) is 684.
2. Brinell Hardness is defined as HB = 2P / (πD( D - √(D^2 - d^2) )), where P is the force, D is the diameter of the indenter, and d is the diameter of the indentation.
3. Rearrange to solve for P: P = HB * (πD( D - √(D^2 - d^2) )) / 2.
4. Put in the respective values: P = 684 * (π * 12 mm * (12 mm - √(12^2 mm² - 4.49^2 mm²))) / 2
5. The calculated force P is approximately 132,648 N.

Therefore, the surface area of indentation in mm² is 15.86 and the force applied in N is 132,648.