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where
I
is the unit tensor, I
pq
= 1 (0) for p = q (p
q), and tr is the trace operator. For a
matrix
A
¼ A
pq
;
tr
A
¼ A
xx
þ
A
yy
þ
A
zz
. In Eq. (
5.2
), the factor 1/3 comes from the
three-dimensionality; in two-dimensional theory the factor is ½. The eigenvalues of
A
are
the solutions (
≠
ʻ
) of the eigenvalue equation det
Ak
I
ð
Þ
¼0
, and the corresponding
eigenvectors
. The eigenvalues of the stress tensor are
called the principal stresses, and the eigenvectors give the directions of these stresses.
ʛ
are the solutions of
Ak
I
ð
ÞK
¼ 0
Example 5.1.
Hydrostatic pressure is a stress
r
¼
pI
, where p > 0 is scalar pressure. It is
spherical with
r
S
¼
p
, compressive (i.e.
σ
S
< 0), and in any direction,
r
n ¼
pn
.In
lake water,
, where p
a
is the surface atmospheric pressure and D is the
depth. At D =10m,p =2p
a
, i.e. each 10 m of water adds the pressure equivalent to one
atmosphere to the total.
p ¼ p
a
þ q
w
gD
5.1.2 Strain and Rotation
The techniques to describe deformation are given in topics on continuum mechanics
(e.g., Mase 1970; Hunter 1976). A brief summary is presented here. Let
x
stand for the
reference con
guration of a material body, and consider a change of the body to
X
,
represented by a mapping
x
→
X
. The displacement
d
¼
Xx
consists of translation,
rotation and strain. Translation and rotation correspond to rigid body motion, while strain
corresponds to physical deformation of the body. Strain and rotation are included in the
displacement gradient
∇d
. For small deformations (|
∇d
|
≪
1),
the linear theory is
employed. Then the strain
are given by the symmetric and anti-
symmetric parts, respectively, of the displacement gradient:
ʵ
and the rotation
ˉ
T
e
¼ 1
=
2½
rd þðrdÞ
ð
:
Þ
5
3a
T
x
¼ 1
=
2½
rdðrdÞ
ð
5
:
3b
Þ
where the superscript T stands for the transpose. Rotation has three independent com-
ponents, which give the rotations of the body around the co-ordinate axes (Fig.
5.3
). Strain
is symmetric and possesses six independent components, three representing the normal
strains and three representing the shear strains analogous with the stress components.
A rotation of 1 % corresponds to turning of the body by 0.01 rad
0.573°. There are
two modes of strain: normal strain and shear strain (Fig.
5.3
). Inversely, any physical
deformation of a body can be decomposed into normal and shear strains. Normal strain is
extensive (positive) or contractive (negative), while shear strain changes the shape of
particles. A tensile/compressive strain of 1 % means that the corresponding material
dimension has lengthened/shortened by 1 %, and a shear strain of 1 % means that a right
angle in the material con
≈
guration has changed by 0.01 rad
≈
0.573º.
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