82
3 Analysis of Strain
ε x =
∂u
∂ x
= 2x × 10
−3
,
ε y =
∂v
∂ y
= 6yz × 10
−3
,
ε z =
∂w
∂z
= 3 × 10
−3
γ xy =
∂
∂ y
x
2
+ 3
× 10
−3
+
∂
∂ x
3y
2 z
× 10
−3
γ xy = 0
γ yz =
∂
∂z
3y
2 z × 10
−3
+
∂
∂ y
(x + 3z) × 10
−3
γ yz = 3y
2
× 10
−3
and γ zx =
∂
∂ x (x + 3z)10
−3
+
∂
∂z
x
2
+ 3
10
−3
γ zx = 1 × 10
−3
Therefore, at point (1, 2, 3), we get
ε x = 2 × 10
−3
, ε y = 6 × 2 × 3 × 10
−3
= 36 × 10
−3
, ε z = 3 × 10
−3
,
γ xy = 0, γ yz = 12 × 10
−3
, γ zx = 1 × 10
−3
Example 3.5 The strain components at a point with respect to x y z co-ordinate
system are
ε x = 0.10, ε y = 0.20, ε z = 0.30, γ xy = γ yz = γ xz = 0.160
If the co-ordinate axes are rotated about the z-axis through 45° in the anticlockwise
direction (Fig. 3.11), determine the new strain components.
Solution: Direction cosines
Here,
l 1 =
1
√
2
, m 1 =
1
√
2
, n 1 = 0
l 2 = −
1
√
2
, m 2 =
1
√
2
, n 2 = 0
See Fig. 3.11.