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faces, whose unit normals are
−
e
x
,
−
e
y
,
and
−
e
z
.
The fourth side of the tetrahedron has
the outward normal parallel to
a
If the area of this fourth side is
A
, then the areas of the
other three triangular faces are
a
x
A
,
a
y
A
and
a
z
A
.
.
The force equilibrium equation for the
isolated element of volume
V
is
p
V
V
+
σ
a
A
−
σ
x
a
x
A
−
σ
y
a
y
A
−
σ
z
a
z
A
=
0
where
p
V
is the body force. The negative signs are necessary because the stress vectors
acting on planes with unit normals
−
e
x
,
−
e
y
,
and
−
e
z
are
−
σ
x
,
−
σ
y
,
and
−
σ
z
.
The volume
V
is
Ah
3
where
h
is the height of the tetrahedron. In the limit as
h
approaches zero, the equilibrium
equation becomes
V
=
σ
a
=
a
x
σ
x
+
a
y
σ
y
+
a
z
σ
z
(1.79)
Thus, the stress vector on any arbitrarily oriented plane through
P
is a linear combination
of the stress vectors
σ
x
,
σ
y
,
and
σ
z
These three vectors, or equivalently the six stress com-
ponents in terms of which these vectors are written, define the state of stress at point
P
,in
the sense that the stresses on any cutting plane through
P
are expressible in terms of them.
The stress vector
σ
a
can be decomposed into normal and shear components. The normal
component
.
σ
is parallel to the unit vector
a
and is given by
=
σ
x
a
x
+
σ
y
a
y
+
σ
z
a
z
+
σ
=
σ
a
·
a
2
(τ
xy
a
x
a
y
+
τ
yz
a
y
a
z
+
τ
zx
a
z
a
x
)
(1.80)
If the orientation of the cutting plane is varied, by allowing the components of
a
to vary,
the normal stress
takes on different values. Of particular interest are the minimum and
maximum values attained by
σ
σ
and the corresponding orientations of the cutting plane.
Thus, extreme values of
σ
are sought under the constraint that
a
is a unit vector
a
x
+
a
y
+
a
z
=
1
(1.81)
This
extremum
problem
can
be
solved
by
introducing
the
Lagrange
multiplier
λ
(Appendix I) and finding the extreme points of the function
a
x
,a
y
,a
z
)
=
σ
−
λ
a
x
+
1
a
y
+
a
z
−
F
(
The conditions for
F
to assume its extreme values
∂
F
∂
F
∂
F
a
x
=
0
a
y
=
0
a
z
=
0
∂
∂
∂
are
(σ
−
λ)
a
x
+
τ
xy
a
y
+
τ
xz
a
z
=
0
x
τ
xy
a
x
+
(σ
−
λ)
a
y
+
τ
yz
a
z
=
0
y
τ
xz
a
x
+
τ
yz
a
y
+
(σ
−
λ)
a
z
=
0
z
where Eq. (1.80) has been introduced. A fourth condition for
F
to assume its extreme values,
∂
0, leads to
a
x
+
a
y
+
a
z
−
0 which assures that
a
is a unit vector. This set of
homogeneous linear equations for
a
x
,a
y
,a
z
has a nontrivial solution if and only if the
determinant of the coefficients is zero
F
/∂λ
=
1
=
σ
x
−
λτ
xy
τ
xz
τ
xy
σ
y
−
λτ
yz
=
0
(1.82)
τ
xz
τ
yz
σ
z
−
λ
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