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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
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
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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