122
4 Stress–Strain Relations
∴ ε i j =
⎛
⎝
1.067 0.9905 −0.743
0.9905 −1.657 0.62
−0.743 0.62
0.82
⎞
⎠ × 10
−6
Example 4.4 A rectangular strain rosette gives the data as below.
ε 0 = 670 micrometres/m
ε 45 = 330 micrometres/m
ε 90 = 150 micrometres/m
Find the principal stresses σ 1 and σ 2 if E = 2 × 10
5 MPa, ν = 0.3
Solution We have
ε x = ε 0 = 670 × 10
−6
ε y = ε 90 = 150 × 10
−6
γ xy = 2ε 45 − (ε 0 + ε 90 ) = 2 × 330 × 10
−6
−
670 × 10
−6
+ 150 × 10
−6
∴ γ xy = −160 × 10
−6
Now, the principal strains are given by
ε max or ε min =
ε x +ε y
2
±
1
2
ε x − ε y
2 + γ 2
xy
i.e. ε max or ε min =
670+150
2
10
−6
±
1
2
(670 − 150)10 −6
2 +
−160 × 10 −6
2
∴ ε max or ε min = 410 × 10
−6
± 272.03 × 10
−6
∴ ε max = ε 1 = 682.3 × 10
−6
ε min = ε 2 = 137.97 × 10
−6
The principal stresses are determined by the following relations
σ 1 =
(ε 1 + νε 2 )
1 − ν 2 .E =
(682.03 + 0.3 × 137.97)10
−6
1 − (0.3)
2
× 2 × 10
5
∴ σ 1 = 159 MPa
Similarly,
σ 2 =
(ε 2 + νε 1 )
1 − ν 2 .E =
(137.97 + 0.3 × 682.03)10
−6
1 − (0.3)
2
× 2 × 10
5
∴ σ 2 = 75.3 MPa