Civil Engineering Reference
In-Depth Information
C
11
=
λ
+
2
µ
C
12
=
λ
(1.24)
C
44
=
2
µ
=
C
11
−
C
12
The strain elements are related to the stress elements by
λδ
ij
2
µ(
3
λ
1
2
µ
σ
ij
e
ij
=−
+
2
µ)
(σ
11
+
σ
22
+
σ
33
)
+
(1.25)
e
11
e
22
e
33
e
13
e
23
e
12
1
/E
−
ν/E
−
ν/E
0
0
0
σ
11
σ
22
σ
33
σ
13
σ
23
σ
12
−
ν/E
1
/E
−
ν/E
0
0
0
−
ν/E
−
ν/E
1
/E
0
0
0
=
(1.26)
0
0
0
1
/
2
µ
0
0
0
0
0
0
1
/
2
µ
0
0
0
0
0
0
1
/
2
µ
where
E
is the Young's modulus and
ν
is the Poisson ratio. They are related to the Lame
coefficients by
µ(
3
λ
+
2
µ)
λ
E
=
+
µ
(1.27)
λ
ν
=
2
(λ
+
µ)
The shear modulus
G
is related to
E
and
ν
via
E
G
=
µ
=
(1.28)
2
(
1
+
ν)
Examples
Antiplane shear
The displacement field is represented in Figure 1.4. For this case, the two components
e
ij
which differ from zero are
1
2
∂u
2
∂x
3
e
32
=
e
23
=
(1.29)
The angle
α
is equal to
∂u
2
∂x
3
α
=
(1.30)
Using Equation (1.21), one obtains two components
σ
ij
which differ from zero:
σ
32
=
σ
23
=
µα
(1.31)
The coefficient
µ
is the shear modulus of the medium, which relates the angle of
deformation and the tangential force per unit area. The three components of the rotation
