2.23 Numerical Examples
35
τ x y = −4 ×
1
2
+ 6 ×
1
2
+ 0 + 1
1
2
−
1
2
+ 0 + 0
= 1 MPa
τ y z = 0 + 0 + 0 + 0 + 0 + 2
−
1
√
2
= −
√
2 MPa
τ x z = 0 + 0 + 0 + 0 + 0 + 2
1
√
2
=
√
2 MPa
Hence, the new stress tensor becomes
⎡
⎣
6
1
√
2
1
4 −
√
2
√
2 −
√
2 8
⎤
⎦ MPa
Now, the new invariants are
I
1 = 6 + 4 + 8 = 18
I
2 = 6 × 4 + 4 × 8 + 6 × 8 − 1 − 2 − 2 = 99
I
3 = 6 × 30 − 1 × 10 +
√
2
−
5
√
2
= 160
which remains unchanged. Hence proved.
Example 2.2 The state of stress at a point is given by the following array of terms
⎡
⎣
9 6 3
6 5 2
3 2 4
⎤
⎦ MPa
Determine the principal stresses and principal directions.
Solution The principal stresses are the roots of the cubic equation
σ
3
− I 1 σ
2
+ I 2 σ − I 3 = 0
Here,
I 1 = 9 + 5 + 4 = 18
I 2 = 9 × 5 + 5 × 4 + 9 × 4 − (6)
2
− (2)
2
− (3)
2
= 52
I 3 = 9 × 5 × 4 − 9 × 4 − 5 × 9 − 4 × 36 + 2 × 6 × 2 × 3 = 27