Geology Reference
In-Depth Information
or
∂
u
k
∂
x
j
+
dx
j
δ
x
k
=
∂
u
j
∂
x
k
0.
(1.237)
Hence, if the displacement is locally rigid-body,
e
ij
must vanish. We have already
shown that if
e
ij
vanishes, the displacement is rigid-body. Thus, a necessary and
su
cient condition for a rigid-body displacement field is that
e
ij
=
0. Therefore,
∂
u
j
1
2
∂
x
i
+
∂
u
i
e
ij
=
(1.238)
∂
x
j
measures the true deformation of the medium. It is called the
strain tensor
.
Now, let
ds
=|
d
r
|
, δ
s
=|
δ
r
|
,
d
r
=
μ
ds
, δ
r
=
ν
δ
s
. Then,
d
r
·
δ
r
=
dx
k
δ
x
k
=
ds
δ
s
μ
k
ν
k
=
ds
δ
s
(
μ
·
ν
)
=
ds
δ
s
cosθ
(1.239)
before deformation. Since the scalar product is not a
ff
ected by rigid-body rotation,
after deformation it is
dx
k
+
e
jk
dx
j
e
lk
δ
x
l
e
jk
+
e
kj
dx
j
δ
x
k
δ
x
k
+
≈
dx
k
δ
x
k
+
cosθ
+
2
e
jk
μ
j
ν
k
ds
δ
s
.
=
(1.240)
(
ds
)
2
. After deformation, it is
If
P
and
P
coincide, before deformation
dx
k
δ
x
k
=
1
2
e
jk
μ
j
μ
k
(
ds
)
2
(1.241)
The relative extension, or change in length per unit length, in the
PP
direction
is then
e
jk
μ
j
μ
k
.Inthe
x
i
-direction, the relative extension is
e
ii
. This is the
normal
component of the strain tensor
in the
x
i
-direction. The diagonal elements of the
strain tensor represent the relative extensions in the three co-ordinate directions
arising from the deformation.
If
PP
and
PP
are orthogonal before deformation, θ
=
π/2andcosθ
=
+
.
cosπ/2
=
0. After deformation, they are no longer orthogonal and cos (π/2
+Δ
θ)
=
cosπ/2
+
2
e
jk
μ
j
ν
k
. Then,
2
e
jk
μ
j
ν
k
. (1.242)
If, before deformation,
PP
is aligned with the
x
1
-axis and
PP
is aligned with
the
x
2
-axis, the situation is as illustrated in Figure 1.5. The change in the angle
P
PP
is
cos (π/2
+Δ
θ)
=−
sin
Δ
θ
≈−Δ
θ
=
Δ
θ
=−
2
e
jk
μ
j
ν
k
.
(1.243)
1
2
=
δ
=
δ
The unit vectors have components μ
j
j
and ν
k
k
. Thus,
∂
u
1
∂
x
2
Δ
θ
=−
=−
2
e
12
(1.244)
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