Publication by Pan, S and He, BB and Huang, MX, Nature Materials, 2026 https://doi.org/10.1038/s41563-026-02588-5
This is a nice piece of original research worthy of study.
Research summary
A strong (2.4 GPa) but ductile (11% total elongation) steel (Fe58.96Co15.7Ni22Al2.96C0.38 wt%) is created by causing the untempered martensite within to be body-centred tetragonal. This in essence reduces twin boundary energy (Kajiwara, Mat. Sci. Engng A 273–275 [1999] 67-88) so a contribution to plastic deformation comes from mechanical twinning.
For comparison purposes, similar properties have been achieved in a weldable, low-alloy, untempered mass-produced martensitic steel by a completely different mechanism in which the scale of the martensite plates has been dramatically reduced. Toughness levels in the untempered state reach 75 MPa m1/2.
Study guide
Crystallography of twin variants: from BCC to BCT
In a cubic body-centred lattice, twins can form on any of twelve \(\{112\}\) planes. But in a body-centred tetragonal lattice, the number of twin variants with the lowest boundary energy is reduced because of the lower symmetry.
The reduction in crystal symmetry from cubic (\(m\bar{3}m\)) to tetragonal (\(4/mmm\)) causes the twelve \(\{112\}_\text{BCC}\) planes to become crystallographically non-equivalent, splitting them into distinct sets with different energies.
1. Splitting of the \(\{112\}\) family
In a BCC lattice, all twelve \(\{112\}\) planes are equivalent by symmetry. However, when the lattice is elongated or compressed along the \(c\)-axis to become BCT, the \(\{112\}\) family splits into two subsets based on their relationship to the unique \(c\)-axis:
- \(\{112\}\) Types (4 planes): These planes have indices where the '2' corresponds to the \(c\)-axis (e.g., \((112)\), \((1\bar{1}2)\), etc.).
- \(\{211\}\) and \(\{121\}\) Types (8 planes): These planes have the '2' on the \(a\) or \(b\) axes (e.g., \((211)\), \((\bar{2}11)\), \((121)\)).
2. Energetic differentiation
The twin boundary energy is closely linked to the magnitude of the twinning shear (\(s\)). In a BCT lattice, the shear required to restore the lattice symmetry across a twin boundary depends on the \(c/a\) ratio.
For a BCC lattice, the twinning shear is constant at \(s = 1/\sqrt{2} \approx 0.707\).
In a BCT lattice:
- The twinning shear for the \(\{112\}\) set decreases as the \(c/a\) ratio increases.
- The twinning shear for the \(\{211\}\) set typically increases or becomes more complex.
Because strain energy is proportional to the square of the twinning shear (\( \propto s^2\)), the \(\{112\}\) variants (where the shear is reduced by the tetragonality) become energetically "cheaper" than the others.
Note that Kajiwara's explanation for the reduction of energy at twin interfaces is different, arguing the the spacing between atoms at the interface becomes similar to that remote from the interface at BCT twin boundaries. However, a reduction in boundary energy per unit area is likely to be a small effect when compared with the reduction in strain energy per unit volume due to the reduction in shear strain.
3. Selection of variants
It follows that the twinning variants with a reduced shear strain would be favoured in the system studied - this would be worth of investigation. This can be done by studying using high resolution techniques, the alignment of BCT twins to the stress axis. Alternative, a more representative study would analyse the crystallographic texture expected if only the compliant BCT twin-variants form.
Secondly, it is important to measure the twin volume fraction if it is to be related to ductility via the shear strain. This is to permit differentiation between twinning-induced plasticity and an increase in work hardening due to the partitioning of the untwinned grains. It may turn out, as it does in TRIP-assisted steels, that the latter function is dominant. Given a volume fraction of twinning it is easy to calculate the contribution to elongation via the twinning shear.
1. The BCC baseline
In a perfect body-centred cubic (BCC) lattice, where the tetragonality ratio \(x = c/a = 1\), the twinning shear strain \(s\) on the \(\{112\}\langle11\bar{1}\rangle\) system is a constant geometric property:
\[s_\text{BCC} = \frac{1}{\sqrt{2}} \approx 0.7071\]
2. The BCT equation
For a body-centred tetragonal (BCT) lattice, the twinning shear depends on the tetragonality ratio \(x\). For the specific \(\{112\}\) variants where the tetragonal distortion reduces the shear magnitude, the relation is defined as (Christian and Mahajan, Progress in Materials Science 39, 1995, 1-158):
\[s_\text{BCT} = \frac{2 - x^2}{\sqrt{2}x}\]
3. Step-by-step calculation (\(x = 1.1\))
Using the specified ratio \(x = 1.1\), we substitute the values into the BCT shear formula:
\[1. \quad x^2 = (1.1)^2 = 1.21\]
\[2. \quad 2 - x^2 = 2 - 1.21 = 0.79\]
\[3. \quad \sqrt{2} \cdot x = 1.4142 \cdot 1.1 = 1.5556\]
Final calculation for BCT shear strain:
\[s_\text{BCT} = \frac{0.79}{1.5556} \approx 0.5078\]
4. Summary of results
Comparing the BCC state to the BCT state at \(c/a = 1.1\):
- Absolute Reduction: \(0.7071 - 0.5078 = 0.1993\)
- Percentage Reduction: \(\left( \frac{0.1993}{0.7071} \right) \times 100 \approx \mathbf{28.2\%}\)
- Energy implication: we note that the interface energy per unit area is a state function of the boundary's crystallography (the atomic mismatch across the plane), whereas the shear strain is a descriptor of the mechanical path taken to reach that state. On the basis of the shear strain alone, the strain energy density is \(\propto s^2\), which would lead to a reduction of approximately 48.5% for the {112}BCT twins.
In the study of martensitic transformations, it is vital to distinguish between the cost of the interface itself and the energy required to accommodate the lattice deformation. While tetragonality lowers both, their magnitudes differ by several orders of magnitude.
1. Volumetric strain energy density (\(U_\text{v}\))
The strain energy per unit volume is proportional to the square of the twinning shear strain (\(s\)). Assuming a typical shear modulus (\(G\)) of 80 GPa for steel:
\[U_\text{v} = \frac{1}{2} G s^2\]
Comparing the transition from BCC to BCT (\(c/a = 1.1\)):
- BCC (\(s \approx 0.707\)): \(U_\text{v} \approx 20,000 \text{ MJ/m}^3\)
- BCT (\(s \approx 0.508\)): \(U_\text{v} \approx 10,300 \text{ MJ/m}^3\)
The energetic "saving" from tetragonality is approximately 9,700 MJ/m³.
2. Interfacial energy per unit volume (\(U_{\gamma}\))
To convert the interfacial energy per unit area (\(\gamma_\text{tb}\)) into a volumetric term, we multiply by the amount of twin surface per unit volume (\(S_\text{V}\)). For fine twins with a thickness (\(t\)) of 10 nm:
\[S_\text{V} = \frac{2}{t} = 2 \times 10^8 \text{ m}^2/\text{m}^3\]
Even using a generous estimate for interfacial energy (\(\gamma_\text{tb} = 1 \text{ J/m}^2\)) and applying a 48% reduction for BCT:
- BCC Interfacial Cost: \(1 \times (2 \times 10^8) = 200 \text{ MJ/m}^3\)
- BCT Interfacial Cost: \(0.52 \times (2 \times 10^8) = 104 \text{ MJ/m}^3\)
3. Energy balance summary
| Energy Component |
BCC magnitude (approx.) |
BCT magnitude (approx.) |
| Volumetric strain energy (\(U_\text{v}\)) |
20,000 MJ/m³ |
10,300 MJ/m³ |
| Volumetric interfacial energy (\(U_{\gamma}\)) |
200 MJ/m³ |
104 MJ/m³ |
Key Conclusion: The interfacial energy per unit volume is roughly 100 times smaller than the strain energy density. While the reduction in symmetry does lower the specific energy of the boundary, the primary driver for twinning in BCT lattices is the massive reduction in the elastic "penalty" required to accommodate the shape change.
