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Substitution of
x
,y
,z
from Eq. (1.9) into Eq. (1.8) gives
ds
2
dx
2
dy
2
dz
2
du
2
2
2
=
+
+
+
2
du dx
+
2
d
v
dy
+
2
d
w
dz
+
+
d
v
+
d
w
(1.11)
or, using Eq. (1.7),
ds
2
ds
2
du
2
2
2
−
=
(
+
v
+
w
)
+
+
v
+
w
2
du dx
d
dy
d
dz
d
d
(1.12)
Substitution of the total differentials of Eq. (1.10) into Eq. (1.12) leads to
2
2
]
dx
2
1
2
[
ds
2
ds
2
2
2
−
=
∂
x
u
+
(∂
x
u
)
+
(∂
v)
+
(∂
w)
x
x
2
2
]
dy
2
1
2
[
2
2
+
∂
v
+
(∂
y
u
)
+
(∂
v)
+
(∂
w)
y
y
y
2
2
]
dz
2
1
2
[
2
2
+
∂
w
+
(∂
z
u
)
+
(∂
v)
+
(∂
w)
z
z
z
+
2
(∂
x
v
+
∂
y
u
+
∂
x
u
∂
y
u
+
∂
x
v∂
y
v
+
∂
x
w∂
y
w)
dx dy
+
2
(∂
x
w
+
∂
z
u
+
∂
x
u
∂
z
u
+
∂
x
v∂
z
v
+
∂
x
w∂
z
w)
dx dz
+
2
(∂
y
w
+
∂
z
v
+
∂
y
u
∂
z
u
+
∂
y
v∂
z
v
+
∂
y
w∂
z
w)
dy dz
(1.13)
Note that
ds
2
ds
2
is zero if no relative displacement occurs between the points
A
and
B
as they move to
A
and
B
. This would correspond to a rigid body motion. For
ds
2
−
ds
2
not
equal to zero, the line element
ds
has changed in length, i.e., the solid is strained. Therefore
ds
2
−
ds
2
can be chosen as an appropriate measure of deformation of the solid. To define
the strain components we write Eq. (1.13) as
−
ds
2
ds
2
x
dx
2
y
dy
2
z
dz
2
−
=
2
+
2
+
2
+
4
xy
dx dy
+
4
xz
dx dz
+
4
yz
dy dz
(1.14)
where
∂
2
2
∂v
∂
2
∂w
∂
=
∂
u
1
2
u
x
+
+
+
(1.15a)
x
∂
∂
x
x
x
∂
2
2
∂v
∂
2
∂w
∂
=
∂v
∂
1
2
u
y
+
+
+
(1.15b)
y
∂
y
y
y
∂
2
2
∂v
∂
2
∂w
∂
z
=
∂w
∂
1
2
u
z
+
+
+
(1.15c)
∂
z
z
z
∂v
∂
1
2
x
+
∂
u
y
+
∂
u
x
∂
u
y
+
∂v
x
∂v
y
+
∂w
x
∂w
xy
=
(1.15d)
∂
∂
∂
∂
∂
∂
∂
y
∂w
∂
1
2
x
+
∂
u
z
+
∂
u
x
∂
u
z
+
∂v
x
∂v
z
+
∂w
x
∂w
xz
=
(1.15e)
∂
∂
∂
∂
∂
∂
∂
z
∂w
∂
1
2
y
+
∂v
z
+
∂
u
y
∂
u
z
+
∂v
y
∂v
z
+
∂w
y
∂w
=
(1.15f)
yz
∂
∂
∂
∂
∂
∂
∂
z
which can be abbreviated with the help of the notation of Section 1.1
1
2
(
=
u
i,k
+
u
k,i
+
u
l,i
u
l,k
)
(1.16)
ik
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