3.8 Change in Length of a Linear Element
63
Similarly, S
be the length P
Q
with its components
x
= x + u, ,y
= y + v, ,z
= z + w
∴
S
2 =
P
Q
2 = (x + u)
2
+ (y + v)
2
+ (z + w)
2
From Eqs. (3.17)–(3.19),
x
=
1 +
∂u
∂ x
x +
∂u
∂ y
y +
∂u
∂z
z
y
=
∂v
∂ x
x +
1 +
∂v
∂ y
y +
∂v
∂z
z
z
=
∂w
∂ x
x +
∂w
∂ y
y +
1 +
∂w
∂z
z
Taking the difference between
S
2 and (S)
2 , we get
P
Q
2 − (P Q)
2
=
S
2 − (S)
2
x
2 +
y
2 +
z
2
−
(x)
2
+ (y)
2
+ (z)
2
= 2
ε x x
2
+ ε y y
2
+ ε z z
2
+ ε xy xy + ε yz yz + ε zx xz
(3.20)
where
ε x =
∂u
∂ x
+
1
2
∂u
∂ x
2
+
∂v
∂ x
2
+
∂w
∂ x
2
(3.20a)
ε y =
∂v
∂ y
+
1
2
∂u
∂ y
2
+
∂v
∂ y
2
+
∂w
∂ y
2
(3.20b)
ε z =
∂w
∂z
+
1
2
∂u
∂z
2
+
∂v
∂z
2
+
∂w
∂z
2
(3.20c)
ε xy = ε yx =
∂v
∂ x
+
∂u
∂ y
+
∂u
∂ x
∂u
∂ y
+
∂v
∂ x
∂v
∂ y
+
∂w
∂ x
∂w
∂ y
(3.20d)
ε yz = ε zy =
∂w
∂ y
+
∂v
∂z
+
∂u
∂ y
∂u
∂z
+
∂v
∂ y
∂v
∂z
+
∂w
∂ y
∂w
∂z
(3.20e)
ε zx = ε xz =
∂u
∂z
+
∂w
∂ x
+
∂u
∂z
∂u
∂ x
+
∂v
∂z
∂v
∂ x
+
∂w
∂z
∂w
∂ x
(3.20f)
Now, introducing the notation
ε P Q =
S
− S
S