5.4 Airy’s Stress Function
133
Substituting the values of σ x =
∂
2 φ
∂ y 2 −
, σ y =
∂
2 φ
∂ x 2 −
and τ xy = −
∂
2 φ
∂ x∂ y
in Eq. (5.31) and simplifying, one can get,
∂
4
φ
∂ x 4 +
∂
4
φ
∂ y 4 + 2
∂
4
φ
∂ x 2 ∂ y 2 + (1 − ν)
∂
2
∂ x 2 +
∂
2
∂ y 2
= 0
(5.32)
Equation (5.32) is the governing differential equation for plane stress problems.
If any function φ (x, y) can satisfy Eq. (5.32) and appropriate boundary conditions,
then a plane stress problem can be solved. If the body forces are constant or zero,
then Eq. (5.32) reduces to
∂
4
φ
∂ x 4 + 2
∂
4
φ
∂ x 2 ∂ y 2 +
∂
4
φ
∂ y 4 = 0
(5.33)
In general, ∇
4
φ = 0
where ∇
4
=
∂
4
∂ x 4 +
2∂
4
∂ x 2 ∂ y 2 +
∂
4
∂ y 4 .
Equation (5.32) is known as “Biharmonic equation” for plane stress case.
5.4.2 Stress Function for Plane Strain Case
For plane-strain problem,
ε z = 0 =
1
E
σ z − v
σ x + σ y
σ z = v
σ x + σ y
]
(5.34)
The stress-strain relations are
ε x =
1
E
σ x − v
σ y + σ z
ε y =
1
E
σ y − v(σ x + σ z )
ε z =
1
E
σ z − v
σ x + σ y
γ xy =
τ xy
G
=
2(1 + v)
E
τ xy
(5.35)
Substituting the value of σ z in Eq. 5.35, we get
ε x =
1
E
σ x − v
σ y + v
σ x + σ y