104
4 Stress–Strain Relations
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
σ x
σ y
σ z
τ xy
τ yz
τ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
D 11 D 12 D 12
0
0
0
D 11 D 12
0
0
0
D 11
0
0
0
1
2 (D 11 − D 12 )
0
0
1
2 (D 11 − D 12 )
0
1
2 (D 11 − D 12 )
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
ε x
ε y
ε z
γ xy
γ yz
γ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(4.19)
Thus, we get only 2 independent elastic constants.
Replacing D 12 and
1
2 (D 11 − D 12 ), respectively, by λ and G which are called
“Lame’s constants”, where G is also called shear modulus of elasticity, Eq. (4.19)
can be written as:
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
σ x
σ y
σ z
τ xy
τ yz
τ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
2G + λ λ
λ
0 0 0
2G + λ λ
0 0 0
2G + λ 0 0 0
G 0 0
Symmetry
G 0
G
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
ε x
ε y
ε z
γ xy
γ yz
γ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(4.20)
Therefore, the stress–strain relationships may be expressed as
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
σ x
σ y
σ z
τ xy
τ yz
τ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
=
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
2G + λ λ
λ
0 0 0
λ
2G + λ λ
0 0 0
λ
λ
2G + λ 0 0 0
0
0
0
G 0 0
0
0
0
0 G 0
0
0
0
0 0 G
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
ε x
ε y
ε z
γ xy
γ yz
γ zx
⎫
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭
(4.21)
Therefore,
σ x = (2G + λ)ε x + λ
ε y + ε z
σ y = (2G + λ)ε y + λ(ε z + ε x )
σ z = (2G + λ)ε z + λ
ε x + ε y
(4.22)
Also,
τ xy = Gγ xy
τ yz = Gγ yz
τ zx = Gγ zx
Now, expressing strains in terms of stresses, we get