On the grain boundary network characteristics in a martensitic Ti–6Al–4V alloy

Abstract

The characteristics of the intervariant boundary network that resulted from the $\beta \to \alpha^{\prime}$ martensitic phase transformation in a Ti–6Al–4V alloy were studied using the crystallographic theories of displacive transformations, five-parameter grain boundary analysis and triple junction analysis. The microstructure of Ti–6Al–4V martensite consisted of fine laths containing dislocations and fine twins. The misorientation angle distribution revealed four distinct peaks consistent with the intervariant boundaries expected from the Burgers orientation relationship. The phenomenological theory of martensite predicted four-variant clustering to have the lowest transformation strain among different variant clustering combinations. This configuration was consistent with the observed Ti–6Al–4V martensitic microstructure, where four-variant clusters consisted of two pairs of distinct V-shape variants. The $63.26^\circ /[\overline{10}\, 5\, 5\, \overline{3}]_{{\alpha^{\prime}}}$ and $60^\circ /[1\, 1\, \overline{2}\, 0]_{{\alpha^{\prime}}}$ intervariant boundaries accounted for ~ 38% and 33% of the total population, respectively. The five-parameter boundary analysis showed that the former had a twist character, being terminated on the $(\overline{3}\, 2\, 1\, 0)_{{\alpha^{\prime}}}$ plane, and the latter revealed a symmetric tilt $(1\, 0\, \overline{1}\, 1)_{{\alpha^{\prime}}}$ boundary plane. The $63.26^\circ /[\overline{10}\, 5\, 5\, \overline{3}]_{{\alpha^{\prime}}}$ and $60^\circ /[1\, 1\, \overline{2}\, 0]_{{\alpha^{\prime}}}$ had the highest connectivity at triple junctions among other intervariant boundaries. Interestingly, the boundary network in Ti–6Al–4V martensite was significantly different from the commercially pure Ti martensite, where only $60^\circ /[1\, 1\, \overline{2}\, 0]_{{\alpha^{\prime}}}$ intervariant boundaries largely were found at triple junctions due to the formation of three-variant clustering to minimize the transformation strain. This difference is thought to result from a change in the martensitic transformation mechanism (slip vs twinning) caused by the alloy composition.

Graphic abstract

Appendices

Appendix

Phenomenological theory of martensite transformation

The phenomenological theory of martensite transformation, which is based on the double shear mechanism introduced by Greninger-Troiano [84], has been developed by Bowles and Mackenzie (B–M theory [85, 86]) and by Wechsler, Lieberman, and Read (W–L–R theory [52, 53]). This implies that the crystallographic changes during the martensite transformation produce a linear strain resulting in existence of an interface plane with no rotation and distortion (invariant plane) at the vicinity of the martensite lath and the matrix. Thus, the strain (i.e. deformation) on this plane is called invariant plane strain. However, for an employed homogeneous deformation resulting from the crystallographic changes of the parent and product lattices, known as the Bain deformation, an invariant plane condition cannot be met. This can be generally observed through the imposing of the Bain distortion on sphere parent crystal. The deformation resulting in an ellipsoid shape shows that no lines or planes in the parent sphere crystal (shown by letters in Fig. 

Figure 11

(a) The effect of Bain strain (dashed ellipsoid) on the parent phase (solid circle), (b) after combination with a rigid body rotation. The invariant line “cd” can be observed in (b) [40, 87]

11a) can be unextended or undistorted, meaning that the Bain distortion is not an invariant plane strain in nature. In this regard, there should be an additional shear, which does not impose a macroscopic shape, although having a microscopically inhomogeneous character. This lattice invariant shear (LIS), when superimposed on the Bain strain, results in an invariant plane within the shear transformation. Since the LIS cannot produce any crystal changes, it should have an inhomogeneous character such as deformation by dislocation slip or twinning. This can be illustrated in Fig. 7 where the lattice strain changes the initial crystal into a deformed one (Fig. 7a), and thus, the magnitude of A′B′ vector can be brought back to AB through a lattice invariant shear either by slip (Fig. 7d) or twinning (Fig. 7c) [40, 87]. Moreover, there should be an additional rotation within the matrix to coincide the A′B′ vector with the original AB (Fig. 7b).

Therefore, the martensite transformation should consist of a lattice deformation (Bain distortion), a lattice invariant shear and a lattice rotation, which can be represented in a matrix form of B, P and R, respectively. The total shape deformation can be represented, as $P_{1} = RBP$. $P_{1}$ can be described by assuming the lattice invariant shear P. Based on the prior descriptions and the fact that the LIS can be ascertained by slip or twinning, the total shape deformation in martensite transformation is described through Wechsler, Lieberman and Read theories in the following.

Wechsler–Lieberman–Read theory

Here, the shape deformation is presented as the P₁, which can be described through the production of three matrices as follows:

$$P_{1} = RBP$$

(5)

where the R, B and P are known as the rigid body rotation, lattice deformation and a simple shear, respectively. It is now necessary to determine the plane and direction for the lattice invariant shear. The rigid body rotation does not change the length of any vectors; therefore, the vector v must remain unchanged in magnitude as a result of the combination of P and B. Meaning,

$$v^{\prime}P^{\prime}B^{\prime}BPz = v^{\prime}v$$

(6)

Here, the prime mark denotes the transpose of the matrix. Through this equation, the habit plane is determined as an undistorted plane.

If the combination matrix of BP can be defined as F, the relation changes into:

$$v^{\prime}F^{\prime}Fv = v^{\prime}v$$

(7)

Now, it is more convenient to express F as the product of an orthogonal matrix R₃ and a symmetric matrix F_s. Therefore, the F_s can be diagonalized by means of orthogonal transformation:

$$z^{\prime} = R_{4} z$$

(8)

In this regard, in the basis where the F_s is diagonal, the P₁ can be described as:

$$P_{1} = RR_{3} R_{4} F_{d} R^{\prime}_{4}$$

(9)

where F_d is a diagonal matrix described as:

$$F_{d} = \left[ {\begin{array}{*{20}c} {\lambda_{1} } & 0 & 0 \\ 0 & {\lambda_{2} } & 0 \\ 0 & 0 & {\lambda_{3} } \\ \end{array} } \right]$$

(10)

Now, based on the this theorem, where the condition for Eq. (6) is the presence of an undistorted plane, one of the eigenvalues of F_d must remain unity and the eigenvector must remain in the undistorted plane, which gives,

$$F^{\prime}F = R_{4} F_{d}^{2} R^{\prime}_{4}$$

(11)

Therefore, a characteristic equation:

$${\text{Det}} \left( {F^{\prime}F - \lambda^{2} I} \right) = 0$$

(12)

which can be solved. Here, I is the identity matrix.

Now, an orthonormal basis, which the lattice invariant shear takes from, needs to be introduced. This basis is defined by R₅ as a 3 by 3 matrix, where each column defines the unit shear direction (i.e. slip direction, d₂), the unit vector parallel to the shear plane normal (i.e. slip plane normal, p₂) and t through the cross product of d₂ and P₂.

In this regard, the P in the orthonormal basis (P₀) can be defined as:

$$P_{o} = \left[ {\begin{array}{*{20}c} 1 & g & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} } \right]$$

(13)

Now, Bain strain can be transformed into the orthonormal basis through,

$$B_{o} = R^{\prime}_{5} BR_{5}$$

(14)

Then, F can be defined easily in the orthonormal basis,

$$F_{o} = B_{o} P_{o}$$

(15)

In this regard, Eq. (12) takes a quadratic form and can be solved. Concurrently, the corresponding eigenvectors define the orthogonal matrix R₄, which as indicated diagonalizes the $F^{\prime}F$ (i.e. Eq. 11).

Now, the g value and the undistorted plane in Eqs. 6 and 13 can be solved, respectively. The R can be also obtained using the Euler theorem. Here, two vectors in the habit plane can be considered in the orthonormal basis, naming $\sigma$ and $\vartheta$. Then, the vectors should be defined in the orthogonal matrix R₄, as $\overline{\sigma }$ and $\overline{\vartheta }$. The magnitude of u (i.e. vector parallel to the desired axis of rotation) and the tangent half angle of the rotation can be obtained:

$$\frac{{\left[ {\overline{\vartheta } - \vartheta \left] \times \right[\overline{\sigma } - \sigma } \right]}}{{\left[ {\overline{\sigma } - \sigma } \right] \cdot \left[ {\overline{\vartheta } + \vartheta } \right]}} = u\tan \left( {\frac{\theta }{2}} \right)$$

(16)

Then, u and θ can be utilized to calculate the rigid body rotation (R). The R and P₁ values can be obtained using the following equation and Eq. 6, respectively:

$$R = \left[ {\begin{array}{ccc} {\alpha _{{11}}^{2} \beta + \cos \theta } & {\alpha _{{11}} \alpha _{{12}} \beta - \alpha _{{31}} \sin \theta } & {\alpha _{{11}} \alpha _{{31}} \beta + \alpha _{{21}} \sin \theta } \\ {\alpha _{{21}} \alpha _{{11}} \beta + \alpha _{{31}} \sin \theta } & {\alpha _{{21}}^{2} \beta + \cos \theta } & {\alpha _{{21}} \alpha _{{31}} \beta - \alpha _{{11}} \sin \theta } \\ {\alpha _{{31}} \alpha _{{11}} \beta + \alpha _{{21}} \sin \theta } & {\alpha _{{31}} \alpha _{{21}} \beta + \alpha _{{11}} \sin \theta } & {\alpha _{{31}}^{2} \beta + \cos \theta } \\ \end{array} } \right]$$

(17)

where the $\alpha_{ij}$ are the indices for the orthonormal matrix R₅.

For the accommodation of lattice invariant strain through twinning, the parent phase is evolved through two twin-related orientations in the martensitic phase. Therefore, two equivalent crystallographic strains need to be operated to produce two twin-related variants in the product phase. This is expected to form the symmetry configuration point of view, as the transformation process has a general tendency to restore the reduction in symmetry elements through creating a number of crystallographic variants. An illustration of such is depicted in Fig 12. It is observed that the parent lattice indicated by the notation ABCD produces two twin-related equivalent lattices. Therefore, the parent lattice loss of mirror symmetry is compensated through forming the twinned product crystals.

Figure 12

Schematic representation of the symmetry correspondence between two crystallographic variants [40]

For such transformation, the lattice deformation (Bain strain, B₁) for the two twin-related orientations can be represented in their principal axes systems, as follows:

$$\begin{gathered} B^{\prime}_{1} = \left[ {\begin{array}{*{20}c} {\eta_{1} } & 0 & 0 \\ 0 & {\eta_{2} } & 0 \\ 0 & 0 & {\eta_{3} } \\ \end{array} } \right] \hfill \\ B^{\prime}_{2} = \left[ {\begin{array}{*{20}c} {\eta_{1} } & 0 & 0 \\ 0 & {\eta_{3} } & 0 \\ 0 & 0 & {\eta_{2} } \\ \end{array} } \right] \hfill \\ \end{gathered}$$

(18)

However, it is more convenient to represent the lattice strains in the parent crystal coordinate system, meaning that the axis system of the martensite crystals should be rotated into the parent phase. Therefore, B₁ and B₂ can be obtained, as follows:

$$\begin{gathered} B_{1} = \left[ {\begin{array}{*{20}c} {\frac{{\eta_{1} + \eta_{2} }}{2}} & {\frac{{\eta_{2} - \eta_{1} }}{2}} & 0 \\ {\frac{{\eta_{2} - \eta_{1} }}{2}} & {\frac{{\eta_{1} + \eta_{2} }}{2}} & 0 \\ 0 & 0 & {\eta_{3} } \\ \end{array} } \right] \hfill \\ B_{2} = \left[ {\begin{array}{*{20}c} {\frac{{\eta_{1} + \eta_{2} }}{2}} & 0 & {\frac{{\eta_{2} - \eta_{1} }}{2}} \\ 0 & {\eta_{3} } & 0 \\ {\frac{{\eta_{2} - \eta_{1} }}{2}} & 0 & {\frac{{\eta_{1} + \eta_{2} }}{2}} \\ \end{array} } \right] \hfill \\ \end{gathered}$$

(19)

Similar to the Bowles and Mackenzie theory, B₁ and B₂ can be brought into twin-related orientations through a rigid body rotation of $\phi_{1}$ and $\phi_{2}$, respectively. These rotations are indicated by the 1 and 2 numbers, respectively, as clearly observed in Fig. 12. Therefore, $\phi_{1}$ and $\phi_{2}$ describe the rotations of the principal axes of the pure distortions in regions 1 and 2 relative to an axis system fixed in the untransformed parent phase.

It should be considered that the volume fraction of each orientation can be expressed in terms of their thickness ratio, which is determined based on the IPS condition required for the lattice invariant deformation. Therefore, an arbitrary vector such as r in the matrix (Fig. 7) becomes like a twinned martensite crystal shown as a zigzag line where each orientation consumes a thickness of (1 − x) and x, respectively. The transformation of vector r into the r′ can be expressed as the sum of the vectors $OV = OA + AB + BC + \cdots + UV$ or by the indicated lattice distortions and rigid body rotation, as follows:

$$r^{\prime} = \left[ {\left( {1 - x} \right)\phi_{1} B_{1} + x\phi_{2} B_{2} } \right]r$$

(20)

$$r^{\prime} = Er$$

(21)

$$E = \left[ {\left( {1 - x} \right)\phi_{1} B_{1} + x\phi_{2} B_{2} } \right]$$

(22)

In this regard, the total distortion matrix, E, can transform any vector in the parent matrix into an internally twinned martensite.

To define the rigid body rotations, $\phi_{1}$ and $\phi_{2}$ cannot be determined from the available data set. However, a rotation $\phi$ ($\phi_{2} = \phi_{1} \phi$), which gives the relative rotation between $\phi_{1}$ and $\phi_{2}$, can be defined and the total macroscopic shear can be expressed, as follows:

$$E = \phi_{1} \left[ {\left( {1 - x} \right)B_{1} + x\phi B_{2} } \right] = F\phi_{1}$$

(23)

where $F = \left( {1 - x} \right) B_{1} + xB_{2}$.

Therefore, the macroscopic shear tacking place during the martensitic transformation contains three components, naming $\phi_{1}$(i.e. rigid body rotation), F (i.e. the fraction of the shear by the twins) and $B_{1}$ (i.e. the Bain strain). For the certain values of x (magnitude of lattice invariant shear, LIS), the matrix produces an eigenvalue problem for the vectors in the habit plane.