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The problem of finding the extreme values of
σ
is seen to be algebraically identical to the
eigenvalue problem for the 3
×
3 symmetric stress matrix (tensor)
T
σ
τ
τ
x
xy
xz
T
=
τ
σ
τ
(1.83)
xy
y
yz
τ
τ
σ
xz
yz
z
The eigenvectors of
T
are the normal vectors defining the planes on which
σ
assumes its
λ.
extreme values, and eigenvalues of
T
are the Lagrange multipliers
λ
λ
λ
3
are real numbers, and the eigenvec-
tors corresponding to distinct eigenvalues are perpendicular. If exactly two of the eigenval-
ues are identical, then there are two mutually perpendicular eigenvectors corresponding
to these two eigenvalues. In this case, any two mutually perpendicular vectors lying in
the plane defined by these two eigenvectors are also eigenvectors corresponding to the two
identical eigenvalues. If all three eigenvalues are identical, then any three mutually perpen-
dicular vectors are eigenvectors. In the following it will be shown that the eigenvalues are
the minimum and maximum values assumed by normal stresses as a function of cutting
plane orientation. These eigenvalues of the matrix
T
are called the
principal stresses.
It is
always possible to choose the eigenvectors
a
1
,
a
2
,
a
3
such that they form a right-handed
triad of unit vectors. The coordinate axes defined by such a triad are termed
principal axes
.
The columns of the matrix
T
are made up of the
x, y, z
components of the stress vectors
σ
x
,
σ
y
,
and
σ
z
Since
T
is symmetric, its three eigenvalues
1
,
2
,
.
Therefore, for any unit normal vector
a
,
Ta
=
[
σ
x
σ
y
σ
z
]
a
=
a
x
σ
x
+
a
y
σ
y
+
a
z
σ
z
=
σ
a
(1.84)
In this equation, all vectors represent vectors of coordinates with respect to the unit vectors
e
x
,
e
y
,
e
z
. Thus, if the
x, y, z
coordinates of a unit normal vector
a
are multiplied by the
matrix
T
,
the
x, y, z
coordinates of the stress vector on the plane defined by
a
are obtained.
In particular, the expression for the normal stress
σ
on the plane defined by
a
may be written
in terms of the matrix
T
as
σ
=
·
σ
a
=
·
a
a
Ta
(1.85)
Suppose that, given the state of stress by the matrix
T
, the three eigenvalues
3
and
the corresponding unit eigenvectors
a
1
,
a
2
,
a
3
of
T
, have been determined. Then the normal
stresses on the planes defined by the eigenvectors are
λ
1
,
λ
2
,
λ
σ
k
=
a
k
·
Ta
k
=
a
k
·
λ
k
a
k
=
λ
k
k
=
1
,
2
,
3
(1.86)
Thus, the three values of the Lagrange multiplier
λ
,
which are the eigenvalues of
T
, are the
extreme values of the normal stress.
If
a
is a unit eigenvector of
T
, with the corresponding eigenvalue
λ
, then the stress vector
σ
a
is given by
σ
a
=
Ta
=
λ
a
This equation shows that the stress vector
σ
a
consists only of a component along
a
In other
words, the shear stress on the plane defined by
a
is zero. Hence, principal stress planes are
free of shear stress. Conversely, if
a
is to be the normal to a plane that is free of shear stress,
then the condition for finding
a
is just the eigenvalue problem for the matrix
T
.
Ta
=
σ
a
=
λ
a
This establishes that a cutting plane at a point of the body is a principal stress plane if and
only if it is free of shear stress.
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