Chapter 5
Analytical techniques and solutions for
linear elastic solids
5.3 Complex Variable Solution to Plane
Strain Static Linear Elastic Problems
Airy
functions have been used to find many useful solutions to plane elastostatic
boundary value problems. The method does
have some limitations, however. The
biharmonic equation is not the easiest field equation to solve, for one thing.
Another limitation is that displacement components are difficult to determine
from Airy functions, so that the method is not well suited to displacement
boundary value problems.
In this section we outline a more versatile representation
for 2D static linear elasticity problems, based on complex potentials. The main goal is to provide you with enough
background to be able to interpret solutions that use the complex variable
formulation. The techniques to derive
the complex potentials are beyond the scope of this book, but can be found in most
linear elasticity texts.
A typical plane elasticity problem is illustrated in the
picture. Just as in the preceding
section, the solid is two dimensional, which means either that
1. The solid is a thin sheet, with small
thickness h, and is loaded only in
the plane.
In this case the plane stress
solution is applicable
2. The solid is very long in the direction, is prevented from stretching
parallel to the axis, and every cross section is loaded
identically and only in the plane.
In this case, the plane strain
solution is applicable.
Some
additional basic assumptions and restrictions are:
The complex variable method outlined below is
applicable only to isotropic solids. We will assume that the solid has Young’s modulus E, Poisson’s ratio and mass density
We will assume no body forces, and constant
temperature
5.3.1 Complex variable
solutions to elasticity problems
The
picture shows a 2D solid. In the complex
variable formalism,
The position of a point in the solid is
specified by a complex number
The position of a point can also be expressed
as where
You can show that these are
equivalent using Euler’s formula ,
which gives
The displacement of a point is specified using
a second complex number
The displacement and stress fields in
rectangular coordinates are generated from two complex potentials and ,
which are differentiable (also called `analytic’ or `holomorphic’) functions of
z (e.g. a polynomial), using the following
formulas
Here, denotes the derivative of with respect to z, and denotes the complex conjugate of . Recall that to calculate the complex
conjugate of a complex number, you simply change the sign of its imaginary
part, i.e. .
The displacement and stress in polar coordinates
can be derived as
The formulas given here for displacements and
stresses are the most general representation, but other special formulas are
sometimes used for particular problems.
For example, if the solid is a half-space in the region with a boundary at the solution can be generated from a single complex potential ,
using the formulas
For example, you can use these
formulas to calculate stresses from the potentials given in Sections
5.3.7-5.3.9. The conventional
representation gives the same results, of course.
5.3.2
Demonstration that the complex variable solution satisfies the governing
equations
We
need to show two things:
1. That the displacement field satisfies the equilibrium
equation (See sect 5.1.2)
2.
That the stresses
are related to the displacements by the elastic stress-strain equations
To
do this, we need to review some basic results from the theory of complex
variables. Recall that we have set ,
so that a differentiable function can be decomposed into real and imaginary
parts, each of which are functions of ,
as
This shows that
Next,
recall that if is differentiable with respect to z, its real and imaginary parts must
satisfy the Cauchy-Riemann equations
We
can then show that the derivative of with respect to is zero, and similarly, the derivative of with respect to z is zero. To see these, use
the definitions and the Cauchy-Riemann equations
We
can now proceed with the proof. The
equilibrium equations for plane deformation reduce to
These
equations can be written in a combined, complex, form as
It
is easy to show (simply substitute and use the definitions of differentiation
with respect to and ) that this can be re-written as
Finally, substituting
and
noting that and shows that this equation is indeed satisfied.
To
show that the stress-strain relations are satisfied, note that the
stress-strain relations for plane strain deformation (Section 3.1.4) can be
written as
Substituting
for D in terms of the complex
potentials and evaluating the derivatives gives the required results.
5.3.3
Complex variable solution for a line force in an infinite solid (plane strain
deformation)
The
displacements and stresses induced by a line load with force per unit out of plane
distance acting at the origin of a large (infinite)
solid are calculated from the complex potentials
The
displacements can be calculated from these potentials as
We
will work through the algebra required to calculate these formulae for
displacement and stress as a representative example. In practice a symbolic manipulation program
makes the calculations painless. To
begin, note that
and
The displacements are thus
Finally, using Euler’s formula and taking real and imaginary
parts gives the answer listed earlier.
Similarly, the formulas for stress give
Adding
the two formulas for stress shows that
Using
Euler’s formula and taking real and imaginary parts of this expression gives
the formulas for and

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Finally,
we need to verify that the stresses are consistent with a point force acting at
the origin. To do this, we can evaluate
the resultant force exerted by tractions acting on a circle enclosing the point
force. Since the solid is in static
equilibrium, the total force acting
on this circular region must sum to zero.
Recall that the resultant force exerted by stresses on an internal
surface can be calculated as
A
unit normal to the circle is ;
multiplying by the stress tensor (in the basis) gives
Evaluating the integrals
shows that ,
so as required.
5.3.4
Complex variable solution for an edge dislocation in an infinite solid

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A
dislocation is an atomic-scale defect in a crystal. The defect can be detected directly in
high-resolution transmission electron microscope pictures, which can show the
positions of individual atoms in a crystal.
The picture shows a typical example (a Lomer dislocation at the
interface between InGaAs and GaAs, from Tillmann
et al Microsc. Microanal. 10, 185–198, 2004). The
dislocation is not easy to see, but can be identified by describing a `burger’s
circuit’ around the dislocation, as shown by the yellow line. Each straight portion of the circuit connects
seven atoms. In a perfect crystal, the
circuit would start and end at the same atom.
(Try this for yourself for any path that does not encircle the
dislocation). Since the yellow curve
encircles the dislocation, it does not
start and end on the same atom. The
`Burger’s vector’ for the dislocation is the difference in position vector of
the start and end atom, as shown in the picture.
A
continuum model of a dislocation can be created using the procedure illustrated
in the picture. Take an elastic solid,
and cut part-way through it. The edge of
the cut defines a dislocation line . Next, displace the two material surfaces
created by the cut by the burger’s vector b,
and fill in the (infinitesimal) gap. Note that (by convention) the burger’s
vector specifies the displacement of a point at the end of the Burger’s circuit
as seen by an observer who sits on the start of the circuit, as shown in the
picture.
HEALTH WARNING: Some texts define the Burger’s vector to be the
negative of the vector defined here that is to say, the vector pointing from the
end of the circuit back to the start.
A
general Burger’s vector has three components the component parallel to the dislocation line is known as
the screw component of b, while the two remaining components are known as the edge components of b.
The stress field induced by the dislocation depends only on and b,
and is independent of the cut that created it.
The
displacement and stress field induced by a pure edge dislocation, with line
direction parallel to the axis and burgers vector at the origin of an infinite solid can be
derived from the complex potentials

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The displacement and
stresses (in polar coordinates) can be derived from these potentials as
The
displacement components are plotted in the picture below, for a dislocation
with . The contours show a sudden jump in at (This
is caused by the term involving in the formula for - we assumed that when plotting the displacement contours).
Physically, the plane corresponds to the `cut’ that created the
dislocation, and the jump in displacement across the cut is equal to the
Burger’s vector.

Contours
of stress are plotted in the figure below.
The radial and hoop stresses are compressive above the dislocation, and tensile
below it, as one would expect. Shear
stress is positive to the right of the dislocation and negative to the left,
again, in concord with our physical intuition.
The stresses are infinite at the dislocation itself, but of course in
this region linear elasticity does not accurately model material behavior,
because the atomic bonds are very severely distorted.

Radial or hoop stress
Shear stress
5.3.5 Cylindrical hole in an infinite
solid under remote loading
The
figure shows a circular cylindrical cavity with radius a in an infinite, isotropic linear elastic solid. Far from the
cavity, the solid is subjected to a tensile stress ,
with all other stress components zero.
The solution is generated
by complex potentials
The displacement and stress
state is easily calculated as
5.3.6 Crack in an infinite elastic solid
under remote loading
The figure shows a 2D crack with length 2a in an infinite solid, which is
subjected to a uniform state of stress at infinity. The solution can be generated by
complex potentials
Some care is required to evaluate the square
root in the complex potentials properly (square roots are multiple valued, and
you need to know which value, or `branch’ to use. Multiple valued functions are made single
valued by introducing a `branch cut’ where the function is discontinuous. In crack problems the branch cut is always
along the line of the crack). For this
purpose, it is helpful to note that the appropriate branch can be obtained by
setting
where the angles and distances and are shown in the figure, and the angles and must lie in the ranges ,
respectively to select the correct branch.
The solution is most conveniently expressed
in terms of the polar coordinates centered at the origin, together with the
auxiliary angles and .
If I got the algebra correct, (which is unlikely the algebra involved in getting these results
from the complex potentials is unbelievably tedious and unfortunately beyond
the capabilities of MAPLE) the displacement and stress fields are
5.3.7 Fields near the tip of a crack on bimaterial
interface

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The figure shows a semi-infinite crack, which lies in the plane, with crack tip aligned with the axis.
The material above the crack has shear modulus and Poisson’s ratio ;
the material below the crack has shear modulus and Poisson’s ratio . In this section we give the complex variable
solution that governs the variation of stress and displacement near the crack
tip. The solution is significant because
all interface cracks (regardless of
their geometry and the way the solid is loaded) have the same stress and
displacement distribution near the crack tip.
Additional elastic constants for bimaterial problems
To simplify the
solution, we define additional elastic constants as follows
1. Plane strain moduli ,
2. Bimaterial modulus
3. Dundur’s elastic constants
Evidently is a measure of the relative stiffness of the
two materials. It must lie in the range for all possible material combinations, with signifying that material 1 is rigid, while signifies that material 2 is rigid. The second parameter does not have such a
nice physical interpretation it is a rough measure of the relative
compressibilities of the two materials.
For Poisson’s ratios in the range ,
one can show that that .
4. Crack tip singularity parameter
For most material combinations the value of is very small typically of order 0.01 or so.
The full displacement and stress fields in
the two materials are calculated from two sets of complex potentials
where and are parameters that resemble the mode I and
mode II stress intensity factors that characterize the crack-tip stresses in a
homogeneous solid. In practice these
parameters are not usually used in fracture criteria for interface cracks instead, the crack tip loading is characterized
the magnitude of the stress intensity factor ,
a characteristic length L, and a phase angle ,
defined as
This means that .
Complete expressions for the displacement components and
stress components at a point in the solid can be calculated from these
potentials. To simplify the results, it
is helpful to note that
Then, in
material 1
while in material 2
The
individual stress components can be determined by adding/subtracting the last
two equations and taking real and imaginary parts. Note that .
Features of this solution are discussed in more detail in Section 9.6.1.
5.3.8
Frictionless rigid flat indenter in contact with a half-space

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A
rigid, flat punch with width 2a and
infinite length perpendicular to the plane of the figure is pushed into an
elastic half-space with a force per unit out of plane distance. The half-space
is a linear elastic solid with shear modulus and Poisson’s ratio .
The interface between the two solids is frictionless.
The
solution is generated from the following complex potentials
where
is
an arbitrary constant, representing an unknown rigid displacement. Note that
the solution is valid only for .
Stresses
and displacements can be determined by substituting for and into the general formulas, or alternatively,
by substituting into the simplified representation for
half-space problems given in 5.3.1. Some
care is required to evaluate the square root in the complex potentials,
particularly when calculating and .
The solution assumes that
where the angles and distances and are shown in the figure, and and must lie in the ranges .
The
full displacement and stress fields can be determined without difficulty, but
are too lengthy to write out in full.
However, important features of the solution can be extracted. In particular:
1. Contact
pressure: The pressure exerted by the
indenter on the elastic solid follows as
2.
Surface displacement: The displacement of the surface is
Note that there is no unambiguous way to determine the
value of . It is tempting, for example, to attempt to
calculate by assuming that the surface remains fixed at
some point far from the indenter.
However, in this case increases without limit as the distance of the
fixed point from the indenter increases.
3. Contact
stiffness: the stiffness of a contact
is defined as the ratio of the force acting on the indenter to its displacement
,
and is of considerable interest in practical applications. Unfortunately, the solution for an infinite
solid cannot be used to estimate the stiffness of a 2D contact (the stiffness
depends on ). Of
course, the stiffness of a contact between two finite sized elastic solids is
well defined but the stiffness depends on the overall
geometry of the two contacting solids, and varies as ,
where R is a characteristic length
comparable to the specimen size, and a
is the contact width.
5.3.9
Frictionless parabolic (cylindrical) indenter in contact with a half-space
A
rigid, parabolic punch with profile
(and
infinite length perpendicular to the plane of the figure) is pushed into an
elastic half-space by a force (this profile is often used to approximate a
cylinder with radius R). The interface between the two solids is
frictionless, and cannot withstand any tensile stress. The
indenter sinks into the elastic solid, so that the two solids make contact over
a finite region ,
where
The
solution is generated from the following complex potentials
where is
an arbitrary constant, representing an unknown rigid displacement. Note that
the solution is valid only for . You can use the formulas given at the end of
Section 5.3.1 to determine displacements and stress directly from the . In addition, the formulas in 5.3.7 should be
used to determine correct sign for the square root.
Important
features of the solution are:
1. Contact
pressure: The pressure exerted by the
indenter on the elastic solid follows as
2.
Surface displacement: The vertical displacement of the surface is
As discussed in 5.3.8, or the contact stiffness cannot be determined
uniquely.
3. Stress field

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4. Critical
load required to cause yield. The elastic limit is best calculated using
the Tresca yield criterion, which gives
where Y is
the tensile yield stress of the solid.
To derive this result, note that the stresses are proportional to . This means we can write
where is the stress induced at for a contact with a=1 subjected to load . The yield criterion can therefore be
expressed as
where denotes maximizing with respect to position in
the solid. The figure shows contours of
:
the maximum value is approximately 0.3823, and occurs on the symmetry axis at a
depth of about . Substituting this value back into the yield
criterion gives the result.
5.3.10 Line
contact between two non-conformal frictionless elastic solids
The
solution in the preceding section can be generalized to find stress and
displacement caused by contact between two elastic solids. The solution assumes:
1. The two contacting solids initially meet at along a
line perpendicular to the plane of the figure (the line of initial contact lies
on the line connecting the centers of curvature of the two solids)
2. The two contacting solids have radii of curvature and at the point of initial contact. A convex surface has a positive radius of
curvature; a concave surface (like the internal surface of a hole) has a
negative radius of curvature
3.
The two solids
have Young’s modulus and Poissons ratio and .
4.
The two solids
are pushed into contact by a force
The solution is expressed
in terms of an effective contact radius and an effective modulus, defined as
The
contact width and contact pressure can be determined by substituting these values
into the formulas given in the preceding section. The full stress and displacement field in
each solid can be calculated from the potential given in the preceding section,
by adopting a coordinate system that points into the solid of interest.
5.3.11 Sliding
contact between two rough elastic cylinders

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Two
elastic cylinders with elastic constants ,
radii ,
and infinite length perpendicular to the plane
of the figure, are pushed into contact by a forces acting perpendicular to the line of contact,
and acting parallel to the tangent plane. The interface between the two solids has a
coefficient of friction ,
and cannot withstand any tensile stress.
The tangential force is sufficient to cause the two solids to slide
against each other, so that .
We give the solution for solid (1) only: the solution for the second solid can
be found by exchanging the moduli appropriately.
The
coordinate system has origin at the initial point of contact between the two
solids. The two solids make contact over a finite region ,
where
and
Only
the derivatives of the complex potentials for this solution can be found
analytically: they are
Note that the solution is valid only for . You can use the formulas given at the end of
Section 5.3.1 to determine stresses directly from . In addition, the branch of must be selected so that
`
where the angles and distances and are shown in the figure, and and must lie in the ranges .
Important
features of the solution are:
1. Contact
pressure: The tractions exerted by
the indenter on the elastic solid follow as
In practice, the value of is very small (generally less than 0.05), and
you can approximate the solution by assuming that without significant error.
2. Approximate
expressions for stresses. For ,
the stresses can be written in a simple form.
The stresses due to the vertical force are given in Section 5.3.8. The stresses due to the friction force are
5.3.12
Dislocation near the surface of a half-space

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The
figure shows a dislocation with burgers vector located at a depth h below the surface of an isotropic linear elastic
half-space, with Young’s modulus and Poisson’s ratio . The surface of the half-space is traction
free.
The solution is given by the
sum of two potentials:
where
is the solution for a dislocation at position in an infinite solid, and
corrects
the solution to satisfy the traction free boundary condition at the surface.
The
displacement and stress fields can be computed by substituting and into the standard formulas given in Sect 5.3.1
(do not use the half-space representation).
A symbolic manipulation program makes the calculation painless. Most
symbolic manipulation programs will not be able to differentiate the complex
conjugate of a function, so the derivatives of and should be calculated by substituting
appropriate derivatives of and into the following formulas
As an example, the
variation of stress along the line is given by