6.3 Strain-Compatibility Equation
167
Subtracting Eq. (6.9h) from Eq. (6.9g) and using Eq. (6.9e), we get
∂
2
ε θ
∂r 2 −
1
r
∂
2
γ r θ
∂r ∂θ
=
1
r
∂ε r
∂r
−
ε r
r 2
−
1
r 2
∂
2
v
∂r ∂θ
−
1
r
∂ε θ
∂r
+
1
r 2
∂γ r θ
∂θ
−
1
r 2
∂
2
ε r
∂θ 2 +
ε θ
r 2
=
1
r
∂ε r
∂r
−
1
r
ε r
r
+
1
r
∂
2
v
∂r ∂θ
−
ε θ
r
−
1
r
∂ε θ
∂r
−
1
r
.
∂γ r θ
∂θ
+
1
r
∂
2
ε r
∂θ 2
=
1
r
∂ε r
∂r
−
1
r
∂ε θ
∂r
−
1
r
∂ε θ
∂r
+
1
r 2
∂γ r θ
∂θ
−
1
r 2
∂
2
ε r
∂θ 2
=
1
r
∂ε r
∂r
−
2
r
∂ε θ
∂r
+
1
r 2
∂γ r θ
∂θ
−
1
r 2
∂
2
ε r
∂θ 2
∴
1
r 2
∂γ r θ
∂θ
+
1
r
∂
2
γ r θ
∂r ∂θ
=
∂
2
ε θ
∂r 2 +
2
r
∂ε θ
∂r
−
1
r
∂ε r
∂r
+
1
r 2
∂
2
ε r
∂θ 2
6.4 Stress–Strain Relations
In terms of cylindrical coordinates, the stress–strain relations for three-dimensional
state of stress and strain are given by
ε r =
1
E
[σ r − ν(σ θ + σ z )]
ε θ =
1
E
[σ θ − ν(σ r + σ z )]
ε z =
1
E
[σ z − ν(σ r + σ θ )]
(6.10)
For two-dimensional state of stresses and strains, the above equations reduce to,
For Plane Stress Case
ε r =
1
E
(σ r − νσ θ )
ε θ =
1
E
(σ θ − νσ r )
γ r θ =
1
G
τ r θ
(6.11)