3.17 Numerical Examples
93
Solution: An octahedral plane is one which is inclined equally to the three principal
co-ordinates. Its direction cosines are
1
√
3
,
1
√
3
,
1
√
3
.
Now, the normal strain on the octahedral plane is
(ε n ) oct = ε x l
2
+ ε y m
2
+ ε z n
2
+ γ xy lm + γ yz mn + γ zx nl
=
1
3
[0.01 − 0.02 + 0.03 + 0.015 + 0.02 − 0.01]
∴ (ε n ) oct = 0.015
The strain tensor can be written as
⎛
⎝
ε x ε xy ε xz
ε xy ε y ε yz
ε xz ε yz ε z
⎞
⎠ =
⎛
⎝
0.01
0.015
2
−
0.01
2
0.015
2
−0.02
0.02
2
−
0.01
2
0.02
2
0.03
⎞
⎠ =
⎛
⎝
0.01 0.0075 −0.005
0.0075 −0.02 0.01
−0.005 0.01 0.03
⎞
⎠
Now, the resultant strain on the octahedral plane is given by
(ε R ) oct =
1
3
ε x + ε xy + ε xz
2 +
ε xy + ε y + ε yz
2 +
ε xz + ε yz + ε z
2
=
1
3
(0.01 + 0.0075 − 0.005) 2 + (0.0075 − 0.02 + 0.01) 2 + (−0.005 + 0.01 + 0.03) 2
=
√
0.0004625
∴ (ε R ) oct = 0.0215
and octahedral shearing strain is given by
(ε S ) oct = 2
(ε R )
2
− (ε n )
2
= 2
(0.0215)
2
− (0.015)
2
∴ (ε S ) oct = 0.031
Example 3.13 The displacement field is given by
u = K
x
2
+ 2z
, v = K
4x + 2y
2
+ z
, w = 4K z
2
where K is a very small constant. What are the strains at (2, 2, 3) in directions
(a) l = 0 m =
1
√
2
n =
1
√
2
, (b) l = 1 m = n = 0, (c) l = 0.6 m = 0 n = 0.8
Solution:
ε x =
∂u
∂ x
= 2K x, ε y =
∂v
∂ y
= 4K y, ε z =
∂w
∂z
= 8K z
γ xy =
∂v
∂ x
+
∂u
∂ y
= 4K + 0 = 4K
γ yz =
∂w
∂ y
+
∂v
∂z
= 0 + K = K
γ zx =
∂u
∂z
+
∂w
∂ x
= 2K + 0 = 2K
.