4.9 Numerical Examples
119
Volumetric strain = ε v =
ε x + ε y + ε z
= 5.67 × 10
−4
+ 2.57 × 10
−4
+ 7.14 × 10
−5
∴ ε v = 8.954 × 10
−3
To find Lame’s constants
We have,
G =
E
2(1 + ν)
G =
210 × 10
3
2(1 + 0.3)
∴ G = 80.77 × 10
3 N/mm
2
λ =
G(2G − E)
(E − 3G)
=
80.77 × 10
3
2 × 80.77 × 10
3
− 210 × 10
3
210 × 10 3 − 3 × 80.77 × 10 3
∴ λ = 121.14 × 10
3 N/mm
2
Example 4.2 The state of strain at a point is given by
ε x = 0.001, ε y = −0.003, ε z = γ xy = 0, γ xz = −0.004, γ yz = 0.001
Determine the stress tensor at this point. Take E = 210 × 10
6 kN/m
2 , Poisson’s
ratio = 0.28. Also find Lame’s constant.
Solution We have
G =
E
2(1 + ν)
=
210 × 10
6
2(1 + 0.28)
∴ G = 82.03 × 10
6 kN/m
2
But
λ =
G(2G − E)
(E − 3G)
=
82.03 × 10
6
2 × 82.03 × 16
6
− 210 × 10
6
210 × 10 6 − 3 × 82.03 × 10 6
∴ λ = 104.42 × 10
6 kN/m
2