3.17 Numerical Examples
89
Determine (a) octahedral normal and shearing strains. (b) Deviator and spherical
strain tensors.
Solution: For the octahedral plane, the direction cosines are l = m = n =
1
√
3
.
(a) Octahedral normal strain is given by
(ε n ) oct = ε x l
2
+ ε y m
2
+ ε z n
2
+ 2
ε xy lm + ε yz mn + ε zx nl
Here, ε xy =
1
2
γ xy , ε yz =
1
2
γ yz and ε zx =
1
2
γ zx
∴ (ε n ) oct = 0.0001
1
√
3
2 + 0.0003
1
√
3
2 + 0.0005
1
√
3
2
+2
0.0002
1
3
+ 0.0004
1
3
+ 0.0005
1
3
∴ (ε n ) oct = 0.001
Octahedral shearing strain is given by
γ oct = 2
(ε R )
2
oct − (ε n )
2
oct
where (ε R ) oct = Resultant strain on octahedral plane
∴ (ε R ) oct =
1
3
ε x + ε xy + ε xz
2 +
ε xy + ε y + ε yz
2 +
ε xz + ε yz + ε y
2
=
1
3
(0.0001 + 0.0002 + 0.0005) 2 + (0.0002 + 0.0003 + 0.0004) 2 (0.0005 + 0.0004 + 0.0005) 2
∴ (ε R ) oct = 0.001066
∴ γ oct = 2
(0.00106)
2
− (0.001)
2
∴ γ oct = 0.000739
(b) Deviator and spherical strain tensors.
Here,
mean strain = ε m =
ε x + ε y + ε z
3
=
0.0001 + 0.0003 + 0.0005
3
∴ ε m = 0.0003