## EN224: Linear Elasticity

Division of Engineering

Brown University

9.2 Solution Anti-Plane Shear problems for anisotropic materials

We focus first on the simple case of anti-plane shear deformation.  In this case, there is only one field quantity  to be computed, and the stresses and strains are related by the two-dimensional expressions

Only three material constants appear in the constitutive relation.  The material is isotropic if .  Anisotropy has two effects: applying a shear strain parallel to one axis may induce a component of shear strain parallel to the orthogonal axis (regulated by  ).  In addition, the shear stiffness parallel to the two coordinate axes may differ.

The stiffnesses obviously depend on the choice of coordinate axes.  For future reference, it is worth recalling that the material constants transform under an axis rotation as follows

It is always possible to find a particular choice of axes for which .  For this special choice, only two material constants appear in the constitutive relation.

Method of solution for anti-plane shear deformations

A simple complex variable formulation can be used to solve anti-plane shear problems.  The procedure is very similar to the technique we used for isotropic materials.  The procedure is as follows.

Let

where f(z) is an analytic function.  However, instead of setting , we now put  where p is a complex number given by

The stresses can be computed from

The boundary conditions on f(z) are:

Displacement BVP:

Traction BVP:

The latter expression can also be used to calculate the resultant force on an arc in a solid, using exactly the same procedure we used for isotropic solids.

Note that the governing equations and boundary conditions are identical to those we obtained for isotropic solids.  Consequently, all the techniques we developed to solve problems involving isotropic materials can be applied to anisotropic solids.  In addition, many solutions for isotropic solids can immediately be extended to anisotropic materials.

It is helpful to visualize the mapping from the  plane to the complex plane, which is illustrated below.  If p contains both real and imaginary parts, the mapping both stretches and shears the solid.

Note that the transformation is not a conformal map, since it does not preserve angles. Note also that one can always define a pair of coordinate axes for which p is purely imaginary.  To see this, note that for the coordinate transformation  shown below

the material constants transform as follows

whence , ;

Choosing  gives .  For this particular basis, the mapping simply stretches parallel to the coordinate axes:

For this special choice of axes, the mapping has the form , and the expression for stresses simplifies to

Proof

The governing equation for the out-of-plane displacement is

In addition, the stresses are related to the displacement by

For , note that

and substitute into the governing equation to see that

This is satisfied by

as stated. Note that  is real, and that  for an isotropic solid, so we recover our original results.

The stresses can be computed from f(z) as follows.  Using the expressions for stress, we have

Note that

giving

The boundary conditions for a displacement BVP follow trivially.  For the traction BVP, note that the boundary condition may be expressed as

Using  and the expressions for stress components above, this may be rewritten as

and integrating both sides gives the required answer.

Solutions to selected anti-plane shear problems

All the procedures we used to solve anti-plane shear problems for isotropic solids can immediately be applied to anisotropic materials.

Uniform stress.

Without loss of generality, we can choose a coordinate system that ensures , as discussed earlier.  In this coordinate system we have that .  Choosing  and matching real and imaginary parts, we see that

For a general orientation, we have  and a similar calculation gives

Point force in an infinite solid

We need an analytic function f(z) with single valued real part, but with multiple-valued imaginary part such that

so that the tractions acting on any curve C enclosing the origin are in equilibrium with the point force F.

is clearly a suitable candidate.  Substituting into the force balance equation shows that

Screw Dislocation

The same approach can be used to obtain stresses due to a screw dislocation – evidently

produces a displacement field with a jump of magnitude b across a branch cut that coincides with the positive real axis.

Screw dislocation near a free surface

Without loss of generality, we place the stress free boundary along .  Note that the traction free boundary condition may be expressed as

along .

Consider  where d is a real constant.  Evidently this solution consists of two dislocations located at  and at

Then

is clearly zero along  and so f(z) satisfies our boundary condition.  This gives the solution for a dislocation at a distance d below the surface of a half-space.

Hole of radius a at the origin of an infinite solid subjected to uniform remote stress

Again, without loss of generality we can choose a coordinate system such that .  We can write the solution as the sum of a uniform stress and a correction g(z) that will free the boundary of the hole from traction:

The solution must satisfy  on , and  as .  Let , and note that

The inverse mapping

On the boundary of the hole , so that

This evidently describes an ellipse, with semiaxes a and .  We can map this boundary to the unit circle with the conformal transformation

The boundary condition now reduces to  on .  We can satisfy this boundary condition in the usual way – expanding the function  as a Laurent series, and substituting into the boundary condition gives

where

Comparing coefficients of powers of , and using the condition that , we see that

with all other .  We therefore obtain

The displacements follow as the real part of . The stresses can be calculated from the relation

where

The spatial coordinates corresponding to each value of  can be calculated with the mappings

Motivated by the similarity between isotropic and anisotropic solutions we try a solution of the form

where the branch cut for the square root is taken to lie on the real axis along .  The constants c and b are to be determined.  Evidently, to give the correct stress at  we must take

The constant b must be chosen to ensure that  on .  Recall that

.  Thus

For  the right hand side must be purely real.  For the last term to be real, we must ensure that  (b is pure imaginary), and in addition , so that .  The solution is therefore

The asymptotic crack tip fields can be computed by setting  and expanding for small , which gives

We see the usual square root singularity at the crack tip – with no oscillatory singularity.  Stress intensity factors can be defined in the usual way.