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Chapter 5
Analytical
techniques and solutions for linear elastic solids
5.2 Airy Function Solution to Plane
Stress and Strain Static Linear Elastic Problems
In this section we outline a general technique for solving
2D static linear elasticity problems.Â
The technique is known as the `Airy Stress Function’ method.
A typical plane elasticity problem is illustrated in the
picture. 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  Â plane.Â
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.Â
In this case, the plane strain
solution is applicable.
Some
additional basic assumptions and restrictions are:
 Â The Airy stress function is applicable only to
isotropic solids. We will assume that the solid has Young’s modulus E, Poisson’s ratio  and mass density
 The Airy Stress function can only be used if the
body force has a special form. Specifically,
the requirement is
where  is a scalar function of position. Fortunately, most practical body forces can
be expressed in this form, including gravity.
 The
Airy Stress Function approach works best for problems where a solid is
subjected to prescribed tractions on its boundary, rather than prescribed
displacements. Specifically, we
will assume that the solid is loaded by boundary tractions .
5.2.1 The Airy
solution in rectangular coordinates
The Airy
function procedure can then be summarized as follows:
1. Begin by finding a scalar function  (known as the Airy potential) which
satisfies:
where
In addition  must satisfy the following traction boundary
conditions on the surface of the solid
where
 are the components of a unit vector normal
to the boundary.
2. Given ,
the stress field within the region of interest can be calculated from the
formulas
3. If the strains are needed, they may be computed from
the stresses using the elastic stressstrain
relations.
4. If the displacement field is needed, it may be
computed by integrating the strains, following the procedure described in
Section 2.1.20. An example (in polar
coordinates) is given in Section 5.2.4 below.
Although
it is easier to solve for  than it is to solve for stress directly,
this is still not a trivial exercise.Â
Usually, one guesses a suitable form for ,
as illustrated below. This may seem
highly unsatisfactory, but remember that we are essentially integrating a
system of PDEs. The general procedure
to evaluate any integral is to guess a solution, differentiate it, and see if
the guess was correct.Â
5.2.2
Demonstration that the Airy solution satisfies the governing equations
Recall
that to solve a linear elasticity problem, we need to satisfy the following
equations:
 Displacementstrain
relation
 Stressstrain
relation
 Equilibrium
Equation Â
 where we have
neglected thermal expansion, for simplicity.
The Airy function is chosen so as to satisfy the
equilibrium equations automatically.Â
For plane stress or plane strain conditions, the equilibrium equations
reduce to
Substitute for the
stresses in terms of  to see that
so
that the equilibrium equations are satisfied automatically for any choice of . To ensure that the other two equations are
satisfied, we first compute the strains using the elastic stress-strain
equations. Recall that
with  for plane stress and  for plane strain. Hence
Next,
recall that the straindisplacement
relation is satisfied provided that the strains obey the compatibility conditions
    Â
All
but the first of these equations are satisfied automatically by any plane
strain or plane stress field. Substitute into the first equation in terms of
stress to see that
Finally,
substitute into this horrible looking equation for stress in terms of  and rearrange to see that
A few more weeks of
algebra reduces this to
which is the result we
were looking for.
This
proves that the Airy representation satisfies the governing equations. A second important question is  is it possible to find an Airy function for all 2D plane stress and plane strain
problems? If not, the method would be
useless, because you couldn’t tell ahead of time whether  existed for the problem you were trying to
solve. Fortunately it is possible to
prove that all properly posed 2D elasticity problems do have an Airy
representation.
5.2.3 The
Airy solution in cylindrical-polar coordinates
Boundary
value problems involving cylindrical regions are best solved using
Cylindrical-polar coordinates. It is
worth recording the Airy function equations for this coordinate system.
In
a 2D cylindrical-polar coordinate system, a point in the solid is specified
by its radial distance  from the origin and the angle . The solution is independent of z.Â
The Airy function is written as a function of the coordinates as . Vector quantities (displacement, body
force) and tensor quantities (strain, stress) are expressed as components in
the basis  shown in the picture.
The governing equation for
the Airy function in this coordinate system is
The state of stress is
related to the Airy function by
In polar coordinates the
strains are related to the stresses by
for plane strain, while
for
plane stress. The displacements must
be determined by integrating these strains following the procedure similar to
that outlined in Section 2.1.20. To
this end, let  denote the displacement vector. The strain-displacement relations in polar
coordinates are:
These
can be integrated using a procedure analogous to that outlined in Section
2.1.20. An example is given in Section
5.2.5.
In
the following sections, we give several examples of Airy function solutions
to boundary value problems.
5.2.4 Airy function solution to the
end loaded cantilever
Consider
a cantilever beam, with length L,
height 2a and out-of-plane
thickness b, as shown in the
figure. The beam is made from an isotropic linear elastic solid with Young’s
modulus  and Poisson ratio .
The top and bottom of the beam  are traction free, the left hand end is
subjected to a resultant force P,
and the right hand end is clamped.Â
Assume that b<<a, so
that a state of plane stress is developed in the beam. An approximate
solution to the stress in the beam can be calculated from the Airy function
You
can easily show that this function satisfies the governing equation for the
Airy function. The stresses follow as
To see that this solution
satisfies the boundary conditions, note that
1. The top and bottom surfaces of the beam  are traction free (  ).Â
Since the normal is in the  direction on these surfaces, this requires
that . The stress field clearly satisfies this
condition.
2. The plane stress assumption automatically satisfies
boundary conditions on .
3. The traction boundary condition on the left hand end
of the beam ( Â ) was not specified in detail: instead, we
only required that the resultant of the traction acting on the surface is . The normal to the surface at the left hand
end of the beam is in the  direction, so the traction vector is
The resultant force can be calculated by integrating
the traction over the end of the beam:
The stresses thus satisfy the boundary
condition. Note that by Saint-Venant’s
principle, other distributions of traction with the same resultant will
induce the same stresses sufficiently far ( Â ) from the end of the beam.
4. The boundary conditions on the right hand end of the
beam are not satisfied
exactly. The exact solution should
satisfy both  and  on . The displacement field corresponding to the
stress distribution was calculated in the example problem in Sect 2.1.20,
where we found that
where  are constants that may be selected to
satisfy the boundary condition as far as possible. We can satisfy  and  at some, but not all, points on . The choice is arbitrary. Usually the boundary condition is
approximated by requiring  at ,
. This gives ,
 and .  By Saint-Venant’s principle, applying
other boundary conditions (including the exact
boundary condition) will not influence the stresses and displacements
sufficiently far from the end.
5.2.5 2D Line load acting
perpendicular to the surface of an infinite solid
As
a second example, the stress fields due to a line load magnitude P per
unit out-of-plane length acting on the surface of a homogeneous, isotropic
half-space can be generated from the Airy function
The formulas in the
preceding section yield
The stresses in the  basis are
The
method outlined in section 5.2.3 can be used to calculate the displacements:
the procedure is described in detail below to provide a representative
example. For plane strain deformation,
we find
to
within an arbitrary rigid motion. Note
that the displacements vary as log(r) so they are unbounded both at the
origin and at infinity. Moreover, the
displacements due to any distribution of traction that exerts a
nonzero resultant force on the surface will also be unbounded at infinity.Â
It
is easy to see that this solution satisfies all the relevant boundary
conditions. The surface is traction
free (  ) except at r=0. To see that the
stresses are consistent with a vertical point force, note that the resultant
vertical force exerted by the tractions acting on the dashed curve shown in
the picture can be calculated as
The
expressions for displacement can be derived as follows. Substituting the expression for stress into
the stress-strain laws and using the strain-displacement relations yields
Integrating
where  is a function of  to be determined. Similarly, considering the hoop stresses
gives
Rearrange and integrate
with respect to
where
 is a function of  to be determined. Finally, substituting for stresses into the
expression for shear strain shows that
Inserting the expressions
for displacement and simplifying gives
The
two terms in parentheses are functions of  and r,
respectively, and so must both be separately equal to zero to satisfy this
expression for all possible values of  and r.
Therefore
This ODE has solution
The second equation gives
which
has solution . The constants A,B,C represent an arbitrary rigid displacement, and can be taken
to be zero. This gives the required
answer.
5.2.6 2D Line load acting parallel to
the surface of an infinite solid
Similarly,
the stress fields due to a line load magnitude P per unit out-of-plane
length acting tangent to the surface of a homogeneous, isotropic half-space
can be generated from the Airy function
The formulas in the
preceding section yield
The
method outlined in the preceding section can be used to calculate the
displacements. The procedure gives
to
within an arbitrary rigid motion.Â
The stresses and
displacements in the  basis are
5.2.7 Arbitrary pressure acting on a
flat surface
The
principle of superposition can be used to extend the point force solutions to
arbitrary pressures acting on a surface. For example, we can find the (plane
strain) solution for a uniform pressure acting on the strip of width 2a
on the surface of a half-space by distributing the point force solution
appropriately.
Distributing
point forces with magnitude  over the loaded region shows that
5.2.8 Uniform normal pressure acting
on a strip
For the particular case
of a uniform pressure, the integrals can be evaluated to show that
where  and Â
5.2.9
Stresses near the tip of a crack
Consider
an infinite solid, which contains a semi-infinite crack on the (x1,x3) plane. Suppose that the solid
deforms in plane strain and is subjected to bounded stress at infinity. The stress field near the tip of the crack can
be derived from the Airy function
Here,
 and  are two constants, known as mode I and mode II stress intensity
factors, respectively. They
quantify the magnitudes of the stresses near the crack tip, as shown below.
Their role will be discussed in more detail when we discuss fracture
mechanics. The stresses can be calculated as
Equivalent expressions in rectangular coordinates are
while the displacements can be calculated by integrating the strains,
with the result
Note that this displacement
field is valid for plane strain deformation only.
Observe that the stress intensity factor has the bizarre units of .
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