94
3 Analysis of Strain
∴ At point (2, 2, 3),
ε x = 4K ,
ε y = 8K , ε z = 24K
γ xy = 4K ,
γ yz = K ,
γ zx = 2K
Now, the strain in any direction is given by
ε r = ε x l
2
+ ε y m
2
+ ε z n
2
+ γ xy lm + γ yz mn + γ zx nl (i)
Case (a)
Substituting the values in expression (i), we get
ε r = 4K (0) + 8K
1
√
2
2 + 24K
1
√
2
2 + 4K (0) + K
1
√
2
1
√
2
+ 2K (0)
∴ ε r = 4K + 12K +
1
2
K
∴ ε r = 16.5K
Case (b)
ε r = 4K (1)
2
+ 8K (0) + 24(0) + 4K (0) + K (0) + 2K (0)
∴ ε r = 4K
Case (c)
ε r = 4K (0.6)
2
+ 8K (0) + 24(0.8)
2
+ 4K (0) + K (0) + 2K (0.8) (0.6)
∴ ε r = 17.76K
3.18 Exercises
1. Explain strain tensor.
2. Derive the strain–displacement relation at any point in an elastic body.
3. The displacement at a point (x, y) is as given below
u = 5x
4
+ 3x
2 y
2
+ x + y
v = y
3
+ 2x y + 4
Compute the values of normal and shearing strains at a point (3, −2) and verify
whether compatibility exists or not?
4. Determine the strain components at point (1, 2, 3) for the following displacement
field.
u = 8x
2
+ 2y + 6z + 10