Page 266 - Intro to Tensor Calculus

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Finally, from the equations (2.4.47) and (2.4.48) we obtain the desired result that

1 %F i
u i,kk + u j,ji + =0.
1 − 2ν µ
Consequently, the equation (2.4.39) is a solution of equation (2.4.38).
As a special case of the above theorem, note that when the body forces are zero, the equations (2.4.40)
become
2 ~
2
~
∇ φ =0 and ∇ ψ = 0.
In this case, we find that equation (2.4.39) is a solution of equation (2.4.38) provided φ and each component of
~
ψ are harmonic functions. The Papkovich-Neuber potentials are used together with complex variable theory
to solve various two-dimensional elastostatic problems of elasticity. Note also that the Papkovich-Neuber
~
potentials are not unique as different combinations of φ and ψ can produce the same value for ~u.
Compatibility Equations

If we know or can derive the displacement field u i ,i =1, 2, 3 we can then calculate the components of
the strain tensor
1
e ij = (u i,j + u j,i ). (2.4.49)
2
Knowing the strain components, the stress is found using the constitutive relations.
Consider the converse problem where the strain tensor is given or implied due to the assigned stress
field and we are asked to determine the displacement field u i ,i =1, 2, 3. Is this a realistic request? Is it even
possible to solve for three displacements given six strain components? It turns out that certain mathematical
restrictions must be placed upon the strain components in order that the inverse problem have a solution.
These mathematical restrictions are known as compatibility equations. That is, we cannot arbitrarily assign
six strain components e ij and expect to find a displacement field u i ,i =1, 2, 3 with three components which
satisfies the strain relation as given in equation (2.4.49).

EXAMPLE 2.4-3. Suppose we are given the two partial differential equations,
∂u ∂u 3
= x + y and = x .
∂x ∂y
Can we solve for u = u(x, y)? The answer to this question is “no”, because the given equations are inconsis-
tent. The inconsistency is illustrated if we calculate the mixed second derivatives from each equation. We
2
2
∂ u ∂ u 2
find from the first equation that = 1 and from the second equation we calculate =3x . These
∂x∂y ∂y∂x
√
mixed second partial derivatives are unequal for all x different from 3/3. In general, if we have two first
∂u ∂u
order partial differential equations = f(x, y)and = g(x, y), then for consistency (integrability of
∂x ∂y
the equations) we require that the mixed partial derivatives
2
2
∂ u ∂f ∂ u ∂g
= = =
∂x∂y ∂y ∂y∂x ∂x
be equal to one another for all x and y values over the domain for which the solution is desired. This is an
example of a compatibility equation.