The Burgers vector and the stress field of an edge dislocation

Sandbox Physics · 2026-10-30

Slide the top half of a perfect crystal over the bottom half in one go and every bond across the slip plane must break at once. That takes a shear stress of order the shear modulus divided by

2\pi

, far above the stress at which metals actually yield. Crystals deform instead by moving a line defect, the dislocation, which breaks one row of bonds at a time. This chapter defines that defect and computes the stress it carries.

One extra half-plane of atoms

Cut a crystal along a plane, push one side of the cut by one atomic spacing, and glue it back. Where the cut ends inside the crystal, the lattice cannot match up: above that line sits one more half-plane of atoms than below it. That line is an edge dislocation. Far from it every atom has its usual neighbours; near it the lattice is bent and stretched.

When the dislocation moves one spacing along the cut, only the bonds next to the line rearrange. After it has crossed the whole crystal, the top has slipped over the bottom by one spacing, which is the same result as the rigid slide but at a tiny fraction of the stress. The plane in which the line moves is its glide plane.

Source: A. Kauffmann and D. Schliephake

The circuit that fails to close

Draw a loop of lattice steps around the line, say four steps right, four up, four left and four down. In a perfect crystal the loop closes. Around a dislocation it misses its start by one lattice vector. That closure failure is the Burgers vector $\mathbf b$ . For an edge dislocation it is perpendicular to the line; for a screw dislocation it is parallel.

In elasticity the same statement reads as an integral of the displacement gradient around any closed curve. The displacement itself jumps by $\mathbf b$ across the cut, but its gradient, the strain, is continuous away from the core, so the integral does not depend on where the loop runs as long as it encloses the line once.

\oint_C \frac{\partial\mathbf u}{\partial s}\,ds=\mathbf b

The Burgers circuit in continuum form.

C

is any closed curve around the dislocation line and

s

is arclength along it; a loop that misses the line gives zero.

Source: A. Kauffmann and D. Schliephake

The Volterra field of a straight edge dislocation

Put the line along $z$ and the Burgers vector along $x$ in an isotropic crystal with shear modulus $\mu$ and Poisson ratio $\nu$ . The displacement that satisfies equilibrium and jumps by $b$ across the half-plane $x<0$ is given below; $\theta$ is the polar angle, and its jump of $2\pi$ produces the closure failure.

Differentiating and applying Hooke’s law gives the stress. Every component falls off as one over distance, and the shear stress on planes parallel to the glide plane changes sign on the lines $x=0$ and $x=\pm y$ , so it has four lobes. Above the glide plane, where the extra half-plane sits, $\sigma_{xx}$ is compressive; below it is tensile.

The formulas fail within a few Burgers vectors of the line, where strains are no longer small. The KIT lecture estimates the cut-off for an edge dislocation at about $1.5\,b$ . The Lab cuts the field off within one Burgers vector of each core and draws the atoms at their displaced positions, so you see the extra half-plane directly.

u_x=\frac{b}{2\pi}\Bigl[\theta+\frac{xy}{2(1-\nu)(x^2+y^2)}\Bigr],\qquad u_y=-\frac{b}{8\pi(1-\nu)}\Bigl[(1-2\nu)\ln(x^2+y^2)+\frac{x^2-y^2}{x^2+y^2}\Bigr]

Displacement of a straight edge dislocation in an isotropic crystal (plane strain).

\sigma_{xy}=\frac{\mu b}{2\pi(1-\nu)}\,\frac{x(x^2-y^2)}{(x^2+y^2)^2},\qquad \sigma_{xx}=-\frac{\mu b}{2\pi(1-\nu)}\,\frac{y(3x^2+y^2)}{(x^2+y^2)^2}

Its glide-plane shear stress and the normal stress along the Burgers vector.

Source: A. Kauffmann and D. Schliephake · X. Niu

Worked example: the shear stress at two nearby points

Problem. Take $\nu=0.3$ . What shear stress $\sigma_{xy}$ does an edge dislocation at the origin produce at $(4b,2b)$ and at $(2b,4b)$ ?

Prefactor. $1/[2\pi(1-\nu)]=1/(2\pi\times0.7)=0.2274$ , so $\sigma_{xy}=0.2274\,\mu b\,x(x^2-y^2)/(x^2+y^2)^2$ .

First point. $x(x^2-y^2)/(x^2+y^2)^2=4\times12/20^2=0.12/b$ , so $\sigma_{xy}=0.0273\,\mu$ .

Second point. $2\times(4-16)/20^2=-0.06/b$ , so $\sigma_{xy}=-0.0136\,\mu$ . Both points are the same distance from the core, yet the stress has opposite signs and different sizes, because the second lies past the line $x=y$ .

Check. The Lab colours this stress from $-0.04\,\mu$ to $+0.04\,\mu$ , and both points sit inside that range. The route benchmark integrates the displacement gradient numerically around $24$ seeded circles: the twelve that enclose the core close with a mismatch of $b$ to within $2.5\times10^{-11}\,b$ , and the twelve that miss it close to $1.3\times10^{-11}\,b$ . Hooke’s law applied to the numerical strain reproduces the stress formula to $1.5\times10^{-8}$ relative error.

Answer. $0.0273\,\mu$ at $(4b,2b)$ and $-0.0136\,\mu$ at $(2b,4b)$ .

Source: A. Kauffmann and D. Schliephake · X. Niu

Try it in the Lab

Find the four lobes of shear stress

Choose A bound dipole rocks and pause. Before looking closely, predict which of the four lobes around the upper dislocation the colour map will show as positive shear.
Drag the upper dislocation sideways along its dashed glide plane. Watch whether the lobes move with it or stay fixed in the crystal.
Drag it onto the glide plane of its partner and let go. Explain, using the Burgers circuit, why the lattice around the pair becomes perfect once they meet.

Exercises with explained answers

With $\nu=0.3$ , what shear stress does an edge dislocation produce on its own glide plane at $x=10\,b$ ?

Show the answer

On $y=0$ the formula reduces to $\sigma_{xy}=\mu b/[2\pi(1-\nu)x]$ . With $x=10\,b$ this is $0.2274\,\mu/10=0.0227\,\mu$ .

A Burgers circuit encloses two edge dislocations with opposite Burgers vectors. What closure failure does it find, and what does that say about the strain far from the pair?

Show the answer

The closure failures add, $\mathbf b+(-\mathbf b)=\mathbf 0$ , so the loop closes. Far away the two fields cancel to leading order and the strain falls faster than one over distance.

The elastic energy per unit length of an edge dislocation is $\mu b^2\ln(R/r_0)/[4\pi(1-\nu)]$ . Evaluate it for $R=1000\,b$ , $r_0=1.5\,b$ and $\nu=0.3$ , and say why the result depends on $R$ .

Show the answer

$\ln(1000/1.5)=6.50$ , so the energy is $6.50/(4\pi\times0.7)\,\mu b^2=0.739\,\mu b^2$ per unit length. A field that falls as one over distance has an energy density falling as one over distance squared, and its integral over area grows logarithmically, so the outer cut-off $R$ , set by the spacing to other dislocations, matters.

Sources and what we recomputed

A. Kauffmann and D. Schliephake, Plasticity, chapter 4c: Elastic theory of dislocations, lecture slides, Karlsruhe Institute of Technology, version 24-05-02 (2024)

Version 24-05-02, all slides. Slides 9 and 10: displacement and stress field of a straight edge dislocation with line along $z$ and Burgers vector along $x$ , after Hirth and Lothe. Slide 11: maps of the stress components. Slide 13: interactions follow from superposing the fields. Slide 15: the field is cut off near the core, at about $1.5\,b$ for an edge dislocation. Slide 23: the glide force per unit length is $b\tau$ .

X. Niu, Y. Zhu, S. Dai and Y. Xiang, A continuum model for distributions of dislocations incorporating short-range interactions, arXiv:1707.05161v2 (2018)

Version 2, Section 2, the discrete dislocation model. Eq. (2.1): the Peach–Koehler force $\mathbf f=(\boldsymbol\sigma\cdot\mathbf b)\times\boldsymbol\tau$ , with glide component $b\sigma_{xy}$ and climb component $-b\sigma_{xx}$ . Eqs. (2.2)–(2.3): the isotropic stress field of an edge dislocation. Eqs. (2.4)–(2.7): forces summed over all other dislocations plus the applied stress. Eq. (2.8): the mobility law $v_{\mathrm g}=m_{\mathrm g}f_{\mathrm g}$ . Section 4 discusses dipolar walls; its continuum results are not used here.