Division of Engineering
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.
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:
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
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
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.
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
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
Consider where d is a real constant. Evidently this solution consists of two dislocations located at and at
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
is helpful for future reference.
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
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
The spatial coordinates corresponding to each value of can be calculated with the mappings
Crack in an infinite solid under remote loading
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
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.