4.9 Numerical Examples
121
Example 4.3 The stress tensor at a point is given as
⎛
⎝
200 160 −120
160 −240 100
−120 100 160
⎞
⎠ kN/m
2
Determine the strain tensor at this point. Take E = 210 × 10
6 kN/m
2 and ν = 0.3
Solution
ε x =
1
E
σ x − ν
σ y + σ z
=
1
210 × 10 6 [200 − 0.3(−240 + 160)]
∴ ε x = 1.067 × 10
−6
ε y =
1
E
σ y − ν(σ z + σ x )
=
1
210 × 10 6 [−240 − 0.3(160 + 200)]
∴ ε y = −1.657 × 10
−6
ε z =
1
E
σ z − ν
σ x + σ y
=
1
210 × 10 6 [160 − 0.3(200 − 240)]
∴ ε z = 0.82 × 10
−6
Now,
G =
E
2(1 + ν)
=
210 × 10
6
2(1 + 0.3)
= 80.77 × 10
6 kN/m
2
τ xy = Gγ xy = 80.77 × 10
6
× γ xy
∴ γ xy =
τ xy
G
=
160
80.77 × 10 6 = 1.981 × 10
−6
γ yz =
τ yz
G
=
100
80.77 × 10 6 = 1.24 × 10
−6
γ zx =
τ zx
G
=
−120
80.77 × 10 6 = −1.486 × 10
−6
Therefore, the strain tensor at that point is given by
ε i j =
⎛
⎝
ε x ε xy ε xz
ε xy ε y ε yz
ε zx ε zy ε z
⎞
⎠ =
⎛
⎝
ε x
γ xy
2
γ xz
2
γ xy
2
ε y
γ yz
2
γ zx
2
γ zy
2
ε z
⎞
⎠