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3 Analysis of Strain
Example 3.7 The displacement components in a strained body are as follows.
Determine the strain matrix at the point P (3, 2, −5).
u = 0.01x y + 0.02y
2
, v = 0.02x
2
+ 0.01z
3 y, w = 0.01x y
2
+ 0.05z
2
Solution:
ε x =
∂u
∂ x
= 0.01y
ε y =
∂v
∂ y
= 0.01z
3
ε z =
∂w
∂z
= 0.1z
γ xy =
∂v
∂ x
+
∂u
∂ y
= 0.04x + 0.01x + 0.04y
γ yz =
∂w
∂ y
+
∂v
∂z
= 0.02x y + 0.03z
2 y
γ zx =
∂u
∂z
+
∂w
∂ x
= 0 + 0.01y
2
At point P (3, 2, −5), the strain components are
ε x = 0.02, ε y = −1.25, ε z = −0.5
γ xy = 0.23, γ yz = 1.62, γ zx = 0.04
Now, the strain tensor is given by
ε i j =
⎡
⎣
ε x
1
2
γ xy
1
2
γ xz
1
2
γ yx ε y
1
2
γ yz
1
2
γ zx
1
2
γ zy ε z
⎤
⎦
∴ Strain matrix becomes
ε i j =
⎡
⎣
0.02 0.115 0.02
0.115 −1.25 0.81
0.02 0.81 −0.50
⎤
⎦
Example 3.8 The strain tensor at a point in a body is given by
ε i j =
⎡
⎣
0.0001 0.0002 0.0005
0.0002 0.0003 0.0004
0.0005 0.0004 0.0005
⎤
⎦