## Edge dislocation displacement fields — Part II

### March 5, 2007

To calculate the displacement fields of a wedge dislocation:

The components of displacements: in the rectangular coordinates — $u$ and $v$, and in polar coordinates — $u_r$ and $u_{\theta}$.

The stress field of a wedge dislocation, in addition to being cylindrically symmetric, is also unchanged for any reflection along any radius. Thus, the two principal stresses at any point are everywhere radial, and transverse to the radius; and, their magnitude is independent of $\theta$. So are the principal strains.

Consider a polar grid scribed on the system (in Figure (a) of the previous post), with equally spaced radii, and equally spaced circles. After the introduction of the wedge dislocation, while the radii are still equally spaced, the circles need not be; in other words, $u_r$ is independent of $\theta$, while $u_{\theta}$ is proportional to $r$ and linear in $\theta$. From Figures (a) and (b), $u_{\theta} = \frac{\Omega}{2 \pi} \left(\theta + \frac{1}{2} \pi\right) r$. Since, as the wedge is created, the material will tend to shrink radially, $u_r$ is expected to be non-zero, and let $u_r = \frac{\Omega}{2 \pi} r f(r)$.

The rectangular components of the displacement are, $u = \frac{\Omega}{2 \pi} \left[ f x - (\theta + \frac{1}{2} \pi) y\right]$ and $v = \frac{\Omega}{2 \pi} \left[ (\theta + \frac{1}{2} \pi) x + f y \right]$.

Hence, for the edge dislocation, $u_e = \frac{b}{2 \pi} \left( \theta + (1 - r f^{\prime} \right) \frac{xy}{r^{2}} + \frac{b}{4}$, and $v_e = \frac{b}{2 \pi} \left \{ -f + (1 - r f^{\prime}) \right\} \frac{y^{2}}{r^{2}} - \frac{b}{2 \pi}$.

To find $f$:

Using the condition that the displacements satisfy the equilibrium conditions, we can obtain the function $f$ in the expressions above.

Dilatation: $e = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}$;

Rotation: $\omega = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right)$

Equations of mechanical equilibrium: $(1 - 2 \sigma) \nabla^{2} u + \frac{\partial e}{\partial x} = 0$, and, $(1 - 2 \sigma) \nabla^{2} v + \frac{\partial e}{\partial y}$

Or, introducing $\alpha = \frac{1 - \sigma}{1 - 2 \sigma}$ where $\sigma$ is the Poisson’s ratio, $\frac{\partial (\alpha e)}{\partial r} = \frac{\partial \omega}{\partial y}$ and $\frac{\partial (\alpha e)}{\partial y} = - \frac{\partial \omega}{\partial x}$, and these two equations hold for any orientation of the $x$ and $y$ axes.

Taking $x$ and $y$ to be the directions of increasing $r$ and $\theta$ at $(r, \theta)$, we obtain, $\frac{\partial (\alpha e)}{\partial r} = \frac{1}{r}\frac{\partial \omega}{\partial \theta}$ and $\frac{1}{r}\frac{\partial (\alpha e)}{\partial \theta} = - \frac{\partial \omega}{\partial r}$. However, for a wedge dislocation, the rotation is $\omega = \frac{\Omega}{2 \pi} (\theta + \frac{1}{2} \pi)$. This result also follows from the definition of $u_r$, $u_{\theta}$ and the equilibrium equations — the unknown $f$ contributes nothing to the rotation.

Using the expression for rotation, from the equilibrium equations above we obtain (a) $e$ is independent of $\theta$ and (b) $\alpha e = \frac{\Omega}{2 \pi} (\ln r + \mathrm{const})$.

To relate $u_r$ or $f$ to $e$:

Consider the change in volume of a disk of unit thickness and radius $r$ due to the introduction of a wedge dislocation: $2 \pi r u_r(r) = \int_0^{r} 2 \pi e r dr$. Since $u_r = \frac{\Omega}{2 \pi} r f(r)$ and $\alpha e = \frac{\Omega}{2 \pi} (\ln r + \mathrm{const})$, we obtain $\Omega r^{2} f(r) = \int_0^{r} \frac{2 \pi \Omega}{2 \pi \alpha} (r \ln r + (\mathrm{const}) r )$. Hence, $\Omega r^{2} f(r) = \frac{\Omega}{\alpha} \left[ \frac{r^2}{2}(\ln r - \frac{1}{2}) + \frac{r^2}{2} \mathrm{const} \right]$. Thus, neglecting the term $(\mathrm{const}) r$ which represents uniform expansion in the above, we obtain $f(r) = (\frac{1}{2 \alpha}) \ln r$.

The displacement fields obtained using $f = (\frac{1}{2 \alpha}) \ln r$ for the wedge dislocation — Field of a terminating tilt boundary made up of edge dislocations with Burgers vectors $b$ and spacing $h = b / \Omega$ at distances a few $h$ from the boundary.

Thus, substituting for $f$ in the edge dislocation expressions of displacements (and, neglecting rigid-body translations from the resulting expression), we obtain the usual displacement expressions.

For a wedge dislocation, the traction across any circle $r = \mathrm{const}$ is purely radial and is independent of $\theta$; thus, the total force and the couple on the circuit are zero. Thus, there are no couples or forces at the singularity $r = 0$; nor are there any induced by differentiation to obtain the edge dislocation fields.

Stresses and Airy stress function:

Stresses, as usual, are found by the differentiation of the displacements. Further, using the relationship between the dilatation and Airy stress function, by integration, we can obtain the Airy stress function (as we note below):

$P = - K e_e$; $P$ — Hydrostatic pressure; $K$ — Bulk modulus.

$(1 + \sigma) \nabla^{2} \chi = 3 K e$

While obtaining $\chi$ by integration from the above expression, terms of the type $A r^{2} + B \ln r$ are neglected since the first term represents uniform expansion, while terms of the $\ln r$ type in the Airy stress function will lead to singular dilatation at the origin.