134
5 Two-Dimensional Problems in Cartesian Co-ordinate System
=
1
E
1 − v
2
σ x − v(1 + v)σ y
=
1
E
(1 + v)
(1 − v)σ x − vσ y
and
ε x =
1
E
σ y − υ(σ x + σ z )
=
1
E
σ y − v
σ x + v
σ x + σ y
=
1 + v
E
(1 − v)σ y − vσ x
Now, rewriting the compatibility Eqs. (5.8) in terms of stress with the help of the
above relations, we get
2(1 + v)
E
.
∂
2
γ xy
∂ x∂ y
=
1 + v
E
.
∂
2
∂ y 2
(1 − v)σ x − vσ y
+
1 + v
E
.
∂
2
∂ x 2 [(1 − v)σ x − vσ x ]
or
2∂
2
γ xy
∂ x∂ y
= (1 − v)
∂
2
σ x
∂ y 2 − v
∂
2
σ y
∂ y 2 + (1 − v)
∂
2
σ y
∂ x 2 − v
∂
2
σ x
∂ x 2
or
2∂
2
γ xy
∂ x∂ y
= (1 − v)
∂
2
σ x
∂ y 2 +
∂
2
σ y
∂ x 2
− v
∂
2
σ y
∂ y 2 +
∂
2
σ x
∂ x 2
(5.36)
Now, differentiating Equation of (5.28) with respect to x and (5.29) with respect
to y, we get
∂
2
σ x
∂ x 2 −
∂
2
∂ x 2 +
∂
2
τ xy
∂ x∂ y
= 0
∂
2
σ y
∂ y 2 −
∂
2
∂ y 2 +
∂
2
τ xy
∂ x∂ y
= 0
Substituting the above in Eq. (5.36), we get
2
∂
2
τ xy
∂ x∂ y
= (1 − v)
∂
2
σ x
∂ y 2 +
∂
2
σ y
∂ x 2
− v
∂
2
∂ x 2 +
∂
2
∂ y 2 −
2∂
2
τ xy
∂ x∂ y
= 0
or