Hardenability refers to the ability of steel to harden to a certain depth when quenched from its austenitizing temperature. It is typically measured by assessing the extent or depth of hardness in a standardized test specimen of fixed size and shape after a controlled quenching process. In the end-quench test, hardenability is determined by the distance from the quenched end of the specimen that corresponds to a specified hardness level.
The test involves quenching one end of a cylindrical specimen, 1.0 inch in diameter, in water and then measuring the hardness response as it varies with distance from the quenched end.
This method of Jominy Hardenability calculation from the chemical ideal diameter (DI) on a steel is based on the original work of M. A. Grossman [1] and described in ASTM A 255 standard [2]. The calculation method described here is applicable to the following range of chemical compositions:
| Element | Range, % |
|---|---|
| C | 0.10–0.70 |
| Si | 0.15–0.60 |
| Mn | 0.50–1.65 |
| Cr | ≤ 1.35 |
| Ni | ≤ 1.50 |
| Cu | ≤ 0.35 |
| Mo | ≤ 0.55 |
| V | ≤ 0.20 |
However, to facilitate melting process control for higher alloy steels, Hardenability Multiplying Factors have been included for calculating the DI within the following chemical composition ranges: ASTM A 255
| Element | Range, % |
|---|---|
| C | 0.01–0.90 |
| Si | 0.01–2.00 |
| Mn | 0.01–1.95 |
| Cr | 0.01-2.50 |
| Ni | 0.01-3.50 |
| Cu | 0.01-0.55 |
| Mo | 0.01-0.55 |
| V | 0.01-0.20 |
| Zr | 0.01-0.25 |
The procedure of hardennability curve calculation is the following.
(1.5, 3, 5, 7, 9, 11, 13, 15, 20, 25, 30, 40, 45, 50 mm).TODO: describe the method applicable to boron steels
This calculation uses a set of hardenability factors for each alloying element in the composition, which are multiplied together to determine the DI value in inches ($DI_{in}$). Only multiplying factors for DI in inch-pound units are provided for simplicity. The effects of phosphorus and sulfur are not considered, as they tend to offset each other. An austenitic grain size of No. 7 is assumed, as most steels with controlled hardenability are produced using fine-grain practices. Experience shows that a high percentage of heats conform to this grain size.
mf_C = \left\{
\begin{array}{ll}
0.54 \times C, & \text{if } 0.01 \leq C \leq 0.39 \\
0.171 + 0.001 \times C + 0.265 \times C^2, & \text{if } 0.39 < C \leq 0.55 \\
0.115 + 0.268 \times C - 0.038 \times C^2, & \text{if } 0.55 < C \leq 0.65 \\
0.143 + 0.2 \times C, & \text{if } 0.65 < C \leq 0.75 \\
0.062 + 0.409 \times C - 0.135 \times C^2, & \text{if } 0.75 < C \leq 0.90 \\
\end{array}
\right.
mf_{Si} = 0.7 \times Si + 1
mf_{Mn} = \left\{
\begin{array}{ll}
3.3333 \times Mn + 1, & \text{if } Mn \leq 1.2 \\
5.1 \times Mn - 1.12, & \text{if } Mn > 1.2 \leq 1.95 \\
\end{array}
\right.
mf_{Cr} = 2.16 \times Cr + 1
6. **Copper Factor**:
```math
mf_{Cu} = 0.365 \times Cu + 1
8. **Vanadium Factor**:
```math
mf_{V} = 1.73 \times V + 1
An example of a $DI_{in}$ calculation for an SAE 4118 modified steel is
provided below. The chemical composition of the steel along with the
corresponding multiplying factors for each element are listed in the table.
Chemical Composition and Multiplying Factors
| **Element** | **% Composition** | **Multiplying Factor** |
|----------------|-------------------|------------------------|
| Carbon (C) | 0.22 | 0.119 |
| Manganese (Mn) | 0.80 | 3.667 |
| Silicon (Si) | 0.18 | 1.126 |
| Nickel (Ni) | 0.10 | 1.036 |
| Chromium (Cr) | 0.43 | 1.929 |
| Molybdenum (Mo)| 0.25 | 1.75 |
| Copper (Cu) | 0.10 | 1.04 |
| Vanadium (V) | 0.05 | 1.09 |
To calculate the $DI_{in}$ value, the multiplying factors for each alloying
element are multiplied together:
```math
DI_{in} = 0.119 \times 3.667 \times 1.126 \times 1.036 \times 1.929 \times 1.75
\times 1.04 \times 1.09 = 1.95 \, \text{in.}
Converting inches to millimeteres.
DI = DI_{in} \times 25.4
The initial hardness is determined solely by the carbon content and is independent of hardenability. This value represents the martensite hardness.
HRC_M = 33.087 + 50.723 \times C + 33.662 \times C^2 - 2.7048 \times
C^3 - 107.02 \times C^4 + 43.523 \times C^5
The hardness at other positions along the end-quench specimen is found by dividing the initial hardness by the appropriate factor. The formulas for these factors are provided below.
To calculate the hardness at a specific distance from the quenched end use the following formula:
HRC_d = \frac{HRC_M}{df}
Here:
df_k = \sum_{i=0}^{5} \text{coeff}_i \times DI^i
df_3 = 0.170547 + 0.173925 \times DI - 0.0109281 \times DI^2
+ 3.13863 \times 10^{-4} \times DI^3 - 4.32086 \times 10^{-6} \times DI^4
+ 2.31674 \times 10^{-8} \times DI^5
df_5 = 3.03987 - 0.0855161 \times DI + 0.00138048 \times DI^2
- 9.98717 \times 10^{-6} \times DI^3 + 2.64963 \times 10^{-8} \times DI^4
+ 5.46044 \times 10^{-12} \times DI^5
df_7 = 4.32366 - 0.134451 \times DI + 0.00228151 \times DI^2
- 1.96250 \times 10^{-5} \times DI^3 + 8.35338 \times 10^{-8} \times DI^4
- 1.38456 \times 10^{-10} \times DI^5
df_9 = 4.46324 - 0.0992003 \times DI + 0.00119387 \times DI^2
- 7.40686 \times 10^{-6} \times DI^3 + 2.26087 \times 10^{-8} \times DI^4
- 2.46815 \times 10^{-11} \times DI^5
df_{11} = 4.40915 - 0.0792024 \times DI + 6.74319 \times 10^{-4} \times DI^2
- 1.97223 \times 10^{-6} \times DI^3 - 3.21758 \times 10^{-9} \times DI^4
+ 2.08025 \times 10^{-11} \times DI^5
df_{13} = 4.60261 - 0.0820023 \times DI + 7.18416 \times 10^{-4} \times DI^2
- 2.52800 \times 10^{-6} \times DI^3 + 2.30089 \times 10^{-10} \times DI^4
+ 1.25368 \times 10^{-11} \times DI^5
df_{15} = 5.01595 - 0.0957696 \times DI + 9.56240 \times 10^{-4} \times DI^2
- 4.62213 \times 10^{-6} \times DI^3 + 8.92787 \times 10^{-9} \times DI^4
- 8.74859 \times 10^{-13} \times DI^5
df_{20} = 5.51133 - 0.104310 \times DI + 1.15299 \times 10^{-3} \times DI^2
- 7.51801 \times 10^{-6} \times DI^3 + 2.75126 \times 10^{-8} \times DI^4
- 4.3110 \times 10^{-11} \times DI^5
df_{25} = 6.15369 - 0.127486 \times DI + 1.57885 \times 10^{-3} \times DI^2
- 1.12233 \times 10^{-5} \times DI^3 + 4.21359 \times 10^{-8} \times DI^4
- 6.42460 \times 10^{-11} \times DI^5
df_{30} = 7.16001 - 0.171328 \times DI + 2.42820 \times 10^{-3} \times DI^2
- 1.91259 \times 10^{-5} \times DI^3 + 7.67320 \times 10^{-8} \times DI^4
- 1.21571 \times 10^{-10} \times DI^5
df_{35} = 8.46964 - 0.229424 \times DI + 3.54915 \times 10^{-3} \times DI^2
- 2.97166 \times 10^{-5} \times DI^3 + 1.24831 \times 10^{-7} \times DI^4
- 2.0543 \times 10^{-10} \times DI^5
df_{40} = 9.13657 - 0.252296 \times DI + 3.94419 \times 10^{-3} \times DI^2
- 3.33383 \times 10^{-5} \times DI^3 + 1.41462 \times 10^{-7} \times DI^4
- 2.35541 \times 10^{-10} \times DI^5
df_{45} = 8.84696 - 0.223317 \times DI + 3.25787 \times 10^{-3} \times DI^2
- 2.62930 \times 10^{-5} \times DI^3 + 1.08190 \times 10^{-7} \times DI^4
- 1.76244 \times 10^{-10} \times DI^5
df_{50} = 8.10202 - 0.171039 \times DI + 2.12643 \times 10^{-3} \times DI^2
- 1.52754 \times 10^{-5} \times DI^3 + 5.78179 \times 10^{-8} \times DI^4
- 8.79890 \times 10^{-11} \times DI^5
TODO: describe SEP 1664 [3]