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5 Two-Dimensional Problems in Cartesian Co-ordinate System
σ x = 2G ε x + λ ε x + λ
ε y + ε z
or
σ x = λ
ε x + ε y + ε z
+ 2G ε x
Similarly,
σ y = λ
ε x + ε y + ε z
+ 2G ε y
σ z = λ
ε x + ε y + ε z
+ 2G ε z = 0
τ xy = G γ xy τ yz = G γ yz = 0 τ zx = G γ zx = 0
Denoting
ε x + ε y + ε z
= J 1 = First invariant of strain, then
σ x = λ J 1 + 2 Gε x , σ y = λ J 1 + 2 G ε y , σ z = λ J 1 + 2 G ε z = 0
(a)
From, σ z = 0, we get
ε z = −
λ
(λ + 2G)
ε x + ε y
Using the above value of ε z , the strain invariant J 1 becomes
J 1 =
2G
λ + 2G
ε x + ε y
(b)
Substituting the value of J 1 in equation (a), we get
σ x =
2Gλ
λ + 2G
ε x + ε y
+ 2Gε x
σ y =
2Gλ
λ + 2G
ε x + ε y
+ 2Gε y
5.2.2 Plane Strain Case
Here ε z = 0
∴ σ x = λJ 1 + 2Gε x = λ
ε x + ε y
+ 2Gε x
σ y = λJ 1 + 2Gε y = λ
ε x + ε y
+ 2Gε y
σ z = λ J 1 = λ
ε x + ε y