Volumetric and Linear Strain of the Austenite–to–Martensite Transformation: 21-Mat-B6 Ceramic Materials, December 2017, Q1

Question I: Volumetric and Linear Strain of the Austenite–to–Martensite Transformation (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Carbon concentration $x=1.0$ wt% C. Martensite (BCT) lattice parameters $c=0.2866+0.0166x$ nm, $a=0.2866-0.0013x$ nm. Austenite (FCC) lattice parameter $a_o=0.3555+0.0044x$ nm. Unit-cell volumes $V_{\gamma}=\tfrac{1}{2}a_o^3$ (austenite), $V_{\alpha'}=ca^2$ (martensite). Linear-strain relation $\Delta l/l=\Delta v/3v$.

Find. (i) The relative volume change $\Delta V/V$ on transforming austenite to martensite at $x=1.0$ wt% C. (ii) The corresponding linear (lattice-length) strain $\Delta l/l$.

Fig. 1.1 — parent FCC austenite unit cell (left) vs. product BCT martensite unit cell (right). Relative edge lengths are exaggerated for visibility — $a_o$, $a$ and $c$ actually differ by only a few percent.

Approach. Evaluate the three empirical lattice-parameter equations at $x=1.0$ wt% C, form each phase's unit-cell volume from the given geometric relations, take the volume change relative to the parent austenite cell, then convert that relative volume change to a linear strain with the supplied isotropic relation.

Evaluate the martensite lattice parameters at $x=1.0$. $$c = 0.2866 + 0.0166(1.0) = 0.3032\ \text{nm}$$ $$a = 0.2866 - 0.0013(1.0) = 0.2853\ \text{nm}$$
Evaluate the austenite lattice parameter at $x=1.0$. $$a_o = 0.3555 + 0.0044(1.0) = 0.3599\ \text{nm}$$
Compute the parent austenite unit-cell volume. $$V_{\gamma} = \tfrac{1}{2}a_o^3 = \tfrac{1}{2}(0.3599)^3 = 0.023309\ \text{nm}^3$$
Compute the product martensite unit-cell volume. $$V_{\alpha'} = ca^2 = (0.3032)(0.2853)^2 = 0.024679\ \text{nm}^3$$
Form the relative volume change, referenced to the parent (austenite) cell. Substituting the two volumes: $$\frac{\Delta V}{V} = \frac{V_{\alpha'}-V_{\gamma}}{V_{\gamma}} = \frac{0.024679-0.023309}{0.023309} = \boxed{+5.88\%}$$ The positive sign confirms the well-known result that the diffusionless shear from FCC austenite to the carbon-supersaturated BCT martensite EXPANDS the unit cell — it never contracts.
Convert the volume change to an equivalent linear strain. Using the given relation with $\Delta v/v$ from Step 5: $$\frac{\Delta l}{l} = \frac{\Delta v}{3v} = \frac{5.88\%}{3} = \boxed{+1.96\%}$$ so each lattice direction lengthens, on average, by about $1.96\%$ — the microscopic origin of the well-known dimensional growth and residual quenching stress that accompanies martensite formation in a hardened steel part.

Final results — Question I (at $x=1.0$ wt% C)
Quantity	Value
Martensite $c$	0.3032 nm
Martensite $a$	0.2853 nm
Austenite $a_o$	0.3599 nm
$V_{\gamma}$ (austenite)	0.023309 nm³
$V_{\alpha'}$ (martensite)	0.024679 nm³
Relative volume change $\Delta V/V$	+5.88%
Linear strain $\Delta l/l$	+1.96%

Concept & Theory

Topic. Crystallographic origin of the volume and linear-strain change on martensite formation, from empirical Fe–C lattice-parameter correlations.

Key relations. $V_{\text{FCC}}=\tfrac{1}{2}a_o^3$ and $V_{\text{BCT}}=ca^2$ as stated by the question; isotropic conversion $\Delta l/l=\Delta v/3v$ (exact for a small, quasi-isotropic dilation, since $v(1+\varepsilon_l)^3\approx v(1+3\varepsilon_l)$ to first order).

Why this works. The austenite-to-martensite shear is diffusionless — every carbon atom carried by the parent austenite is trapped in the product martensite — so the volume change is a pure CRYSTALLOGRAPHY effect: the same carbon atoms that occupied FCC octahedral interstices in austenite are forced into the more restrictive BCT interstitial sites of ferrite, tetragonally distorting the lattice ($c>a$) and net-expanding the unit cell. Because $c$, $a$ and $a_o$ are all linear in $x$, both the volume change and the derived strain grow with carbon content — a 1.0 wt% C steel swells measurably more on quenching than a low-carbon steel, which is exactly why high-carbon parts are more prone to quench cracking and distortion.

Common pitfall. Referencing the relative volume change to the WRONG denominator — the "relative change" must divide by the STARTING (austenite) volume, not the ending (martensite) volume or an averaged volume; using $V_{\alpha'}$ in the denominator instead of $V_{\gamma}$ changes the answer by a factor of $(1+\Delta V/V)$, about 6% here, which is not a rounding-level error.

Source. Reed-Hill & Abbaschian, Physical Metallurgy Principles, Ch. 6 (martensitic transformations, FCC–BCT crystallography); Krauss, Steels: Processing, Structure, and Performance, Ch. 7 (martensite crystallography, volume change and residual stress on quenching).

Question 1 of 7: Volumetric and Linear Strain of the Austenite–to–Martensite Transformation

Question I: Volumetric and Linear Strain of the Austenite–to–Martensite Transformation (10 marks)