Biomedical Engineering Reference
In-Depth Information
3
10
10
3
3
y
3
x
Figure 8.12
Stress components.
If the normal to the plane of interest equals
n
=
e
x
,
then, the stress vector on this plane follows from
s
=
σ
·
n
=
(10
e
x
e
x
+
3(
e
x
e
y
+
e
y
e
x
))
·
e
x
=
10
e
x
+
3
e
y
.
Clearly, as expected, the operation
σ
·
e
x
extracts the stress components acting
on the right face of the rectangle shown in Fig.
8.12
. The stress vector
s
may be
decomposed into a component normal to this face and a component tangent to
this face. Clearly, the normal component should be
s
n
=
10
e
x
, while the tangent
component should be
s
t
=
3
e
y
. This also follows from
s
n
=
((
σ
·
n
)
·
n
)
n
=
((10
e
x
+
3
e
y
)
·
e
x
)
e
x
=
10
e
x
.
Generalization to three dimensions
As noted before, there are six independent
stress components in the three-dimensional case. These may be stored in the three-
dimensional stress tensor using the sum of nine dyads:
σ
=
σ
xx
e
x
e
x
+
σ
yy
e
y
e
y
+
σ
zz
e
z
e
z
+
σ
xy
(
e
x
e
y
+
e
y
e
x
)
+
σ
xz
(
e
x
e
z
+
e
z
e
x
)
+
σ
yz
(
e
y
e
z
+
e
z
e
y
),
(8.54)
and also in the symmetric stress matrix:
⎡
⎣
⎤
⎦
σ
xx
σ
xy
σ
xz
σ
=
σ
yx
σ
yy
σ
yz
.
(8.55)
σ
zx
σ
zy
σ
zz
