Question

a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 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 / Explanation

Let us first understand what the is for us to be able to properly answer this question:

The Brinell hardness equation or number as known by some can be refereed to as the load divided by the surface area of the indentation.

The Brinell Hardness Number can be calculated using the equation:

HB = 2P / π D ( D - √D² - d² )

Now, according to the definition above, if we recall the equation above:

Recalling the Equation for Brinell Hardness:

We have:

HB = 2P / π D ( D - √D² - d² )

Now, if we insert the values of the parameter, we have:

2 ( 1000) / 10 π ( 10 - √ 10² - 2.5² )

2000 / 10π ( 10 - √ 100 - 6.25)

2000 / 10π ( 10 - √93.75 )

2000 / 10π ( 10 - 9.682)

2000 / 10π ( 0.318)

2000 / 31.42 (0.381)

= 2000 / 11.97102

Finally, we have: 167.070

And for part B,

We should know that the Brinell hardness equation cann be rearranged by making d the subject of formula to be able to solve fro d:

Therefore:

d = √ D² - (D - 2P / HBπD )²

Now, if we move forward to insert the values as given in the question:

We have:

d = √ 10² - ( 10 - 2 ( 500) / 300 π (10) ²

Solving further, we derive at a final answer of:

1.45mm.

Therefore, we can conclude that The HB of the material is 167.070 while the diameter of an indention to yield a hardness of 300HB when 500kg load is used is: 1.45mm

Answer 2

a)
`HB = 200.484`, b)
`d \approx 1.453\,mm`

Step-by-step explanation:

The Brinell hardness can be determined by using this expression:

`HB = (2\cdot P)/(\pi\cdot D^(2))\cdot \left(\frac{1}{1-\sqrt{1-(d^(2))/(D^(2)) } } \right)`

Where
`D` and
`d` are the indenter diameter and the indentation diameter, respectively.

a) The Brinell hardness is:

`HB = (2\cdot (1000\,kgf))/(\pi\cdot (10\,mm)^(2)) \cdot \left[\frac{1}{1-\sqrt{1-((2.50\,mm)^(2))/((10\,mm)^(2)) } } \right]`

`HB = 200.484`

b) The diameter of the indentation is obtained by clearing the corresponding variable in the Brinell formula:

`d = D\cdot \sqrt{1-(1-(2\cdot P)/(\pi\cdot HB \cdot D^(2)) )^(2)}`

`d = (10\,mm)\cdot\sqrt{1-\left[1-(2\cdot (500\,kgf))/(\pi\cdot (300)\cdot (10\,mm)^(2)) \right]^(2)}`

`d \approx 1.453\,mm`