90
3 Analysis of Strain
∴ Deviator strain tensor =
⎡
⎣
(0.0001 − 0.0003)
0.0002
0.0005
0.0002
(0.0003 − 0.0003)
0.0004
0.0005
0.0004
(0.0005 − 0.0003)
⎤
⎦ .
i.e., E
=
⎡
⎣
−0.0002 0.0002 0.0005
0.0002
0 0.0004
0.0005 0.0004 0.0002
⎤
⎦ .
and spherical strain tensor = E
=
⎡
⎣
ε m 0 0
0 ε m 0
0 0 ε m
⎤
⎦ .
i.e., E
=
⎡
⎣
0.0003 0
0
0 0.0003 0
0
0 0.0003
⎤
⎦ .
Example 3.9 The components of strain at a point in a body are as follows:
ε x = c 1 z
x
2
+ y
2
ε y = x
2 z
γ xy = 2c 2 x yz
where c 1 and c 2 are constants. Check whether the strain field is compatible one?
Solution: For the compatibility condition of the strain field, the system of strains
must satisfy the compatibility equations.
i.e.,
∂
2 ε x
∂ y 2 +
∂
2 ε y
∂ x 2 =
∂
2 γ xy
∂ x∂ y
Now, using the given strain field,
∂ε x
∂ y
= 2c 1 yz,
∂
2
ε x
∂ y 2 = 2c 1 z
∂ε y
∂ x
= 2xz,
∂
2
ε y
∂ x 2 = 2z
∂γ xy
∂ x
= 2c 2 yz,
∂
2
γ xy
∂ x∂ y
= 2c 2 z
∴
∂
2
ε x
∂ y 2 +
∂
2
ε y
∂ x 2 = 2c 1 z + 2z = 2z(1 + c 1 ) and
∂
2
γ xy
∂ x∂ y
= 2c 2 z
Since
∂
2 ε x
∂ y 2 +
∂
2 ε y
∂ x 2 =
∂
2 γ xy
∂ x∂ y
, the strain field is not compatible.