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2
1
+ ν
2 p Vj, j =
where the relationship
0 has been employed. The solution of
these equations (for simply connected domains) provides stresses that satisfy the equi-
librium conditions and can lead to strains that can be derived from unique, continuous
displacements.
Equation (1.68) may be transformed into equations for unknown stress functions,
which are defined in terms of stresses so that the equilibrium conditions are identically
satisfied. To accomplish this place σ from Eq. (1.56) in Eq. (1.68) and solve
σ kk =−
ν
1
D 1 E 1 D 1 ψ
D 1 E 1 σ 0
=
0
(1.72)
which are equations often used in the force formulation. These differential equations
represent the conditions of compatibility.
Equation (1.72) is much less complicated for some cases that are frequently of interest.
In the two-dimensional stress state, only one stress function
ψ
is required to replace the
three unknown stresses
σ x ,
σ y ,
τ xy . If the body forces are zero, a very simple expression
for
can be derived from Eq. (1.72) or by following the process employed in deriving
Eq. (1.72). In the latter case, substitution of the material law of Eq. (1.40a) for plane stress
into the compatibility equation of Eq. (1.27) gives
ψ
+ ν)
2
τ xy
y =
y 2 x νσ y ) +
2
2
2
(
1
x 2 y νσ x )
(1.73)
x
Replace the stresses here by the Airy stress function relations of Eq. (1.55), giving
4
ψ
2
4
ψ
y 2 +
4
ψ
2
2
4
x 4 +
y 4 =
0or
ψ =∇
ψ =
0
(1.74)
x 2
2
y . The exact solution of plane stress problem must satisfy Eq. (1.74), a
biharmonic equation, as well as the boundary conditions.
2
2
x
where
=
+
1.6.3 Mixed Formulation
The differential governing equations considered in the two previous sections were written
in terms of either displacements or forces (stresses). Equations formulated in terms of both
displacement and force variables are referred to as mixed equations.
As might be expected, there are a variety of mixed equations that can be derived. Choose
as the fundamental unknowns the three components of the displacements u and the six
components of the stress vector σ . The constitutive relations of Eq. (1.32) or Eq. (1.43) can be
rewritten in terms of displacements u rather than strains by using the strain-displacement
relations of Eq. (1.21). Thus, =
E 1 σ + 0 or Du
E 1 σ = 0 . Rewrite the equations
Du
=
of equilibrium of Eq. (1.54) as D T σ =−
p V . Place these relations together in the form
.
u
σ
D T
0
p V
0
... .
...
=
(1.75)
.
E 1
D
The operators in this matrix are partitioned as
.
0
Equilibrium
Equations
..................... .
............
(1.76)
.
Strain-Displacement
Material
Relations
Law
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