120
4 Stress–Strain Relations
Now,
σ x = (2G + λ)ε x + λ
ε y + ε z
= (2 × 82.03 + 104.42)10
6
× 0.001 + 104.42 × 10
6
(−0.003 + 0)
∴ σ x = −44780 kN/m
2
or
σ x = −44.78M Pa
σ y = (2G + ν)ε y + λ(ε z + ε x )
= (2 × 82.03 + 104.42) × 10
6
× (−0.003) + 104.42 × 10
6
(0 + 0.001)
∴ σ y = −701020 kN/m
2
or
σ y = −701.02 MPa
σ z = (2G + λ)ε z + λ
ε x + ε y
= (2 × 82.03 + 104.42)10
6
(0) + 104.42 × 10
6
(0.001 − 0.003)
∴ σ z = −208840 kN/m
2
or
σ z = −208.84 MPa
τ xy = Gγ xy
= 82.03 × 10
6
× 0
∴ τ xy = 0
τ yz = Gγ yz = 82.03 × 10
6
× 0.001 = 82030 kN/m
2
or
τ yz = 82.03 MPa
τ xz = Gγ xz = 82.03 × 10
6
× (−0.004) = −328120 kN/m
2
or τ xz = −328.12 MPa
∴ The stress tensor is given by
σ i j =
⎛
⎝
σ x τ xy τ xz
τ xy σ y τ yz
τ xz τ yz σ z
⎞
⎠ =
⎛
⎝
−44.78
0
−328.12
0
−701.02 82.03
−328.12 82.03 −208.84
⎞
⎠