Given. A 0.6 wt%C steel is quenched from austenite to 100% martensite. FCC austenite and BCT martensite lattice parameters vary with carbon content by the standard empirical (Roberts) relations.
Given data
Quantity
Value
Carbon content
0.6 wt%C
Austenite (FCC) lattice parameter
$a_\gamma=3.548+0.044(\text{wt\%C})$ Å
Martensite (BCT) $a$-axis
$a_{\alpha'}=2.861-0.013(\text{wt\%C})$ Å
Martensite (BCT) $c$-axis
$c_{\alpha'}=2.861+0.116(\text{wt\%C})$ Å
Young's modulus, $E$
206 GPa
Find. (i) the relative volume change $\Delta V/V$ on transformation to 100% martensite; (ii) the equivalent linear dimension change $\Delta L/L$; (iii) the tensile stress that would produce the same elastic strain.
Approach. Compute the volume per atom in each lattice from its own unit cell (FCC has 4 atoms/cell, BCT has 2 atoms/cell), since no atoms are gained or lost in a diffusionless transformation; convert the volumetric strain to an isotropic linear strain, then apply Hooke's law.
Austenite lattice parameter and volume per atom.
$$a_\gamma=3.548+0.044(0.6)=3.5744\ \text{\AA}\qquad V_\gamma^{atom}=\frac{a_\gamma^3}{4}=\frac{(3.5744)^3}{4}=11.417\ \text{\AA}^3$$
Martensite lattice parameters and volume per atom.
$$a_{\alpha'}=2.861-0.013(0.6)=2.8532\ \text{\AA}\qquad c_{\alpha'}=2.861+0.116(0.6)=2.9306\ \text{\AA}$$
$$V_{\alpha'}^{atom}=\frac{a_{\alpha'}^2\,c_{\alpha'}}{2}=\frac{(2.8532)^2(2.9306)}{2}=11.929\ \text{\AA}^3$$
Relative volume change. Since the atom count per unit volume is conserved (no diffusion, same number of Fe and C atoms carried straight across the transformation):
$$\frac{\Delta V}{V}=\frac{V_{\alpha'}^{atom}-V_\gamma^{atom}}{V_\gamma^{atom}}=\frac{11.929-11.417}{11.417}=\boxed{+4.48\%}$$
Martensite is LESS densely packed than austenite — the transformation is an EXPANSION.
Equivalent linear dimension change. For a small, approximately isotropic volumetric strain, $\Delta V/V\approx3(\Delta L/L)$:
$$\frac{\Delta L}{L}=\frac{1}{3}\cdot\frac{\Delta V}{V}=\frac{4.48\%}{3}=\boxed{+1.49\%}$$
Equivalent tensile stress via Hooke's law. Treating the linear transformation strain as if it were purely elastic:
$$\sigma=E\cdot\varepsilon=206\times10^9\ \text{Pa}\times0.0149=3.08\times10^9\ \text{Pa}=\boxed{3078\ \text{MPa}\ (\approx3.1\ \text{GPa})}$$
Check: this 3.1 GPa "equivalent stress" is roughly 2–3 times the tensile strength of even the strongest commercial steels (typically 1–2 GPa), so it can never actually develop as a real elastic stress. The physical meaning of this result is the teaching point of the question: the volumetric mismatch between austenite and martensite generates internal strains far beyond what the surrounding, still-elastic material can accommodate, so it is instead relieved by local PLASTIC deformation and/or MICROCRACKING at the transformation front — this is the direct microstructural origin of quench cracking and quench distortion in hardened steel components, and is precisely why quenching a large or geometrically complex section to full martensite is a practical risk, not just a metallurgical curiosity.
