3.17 Numerical Examples
81
ε 0 = 750 × 10
−6
ε 45 = −110 × 10
−6
ε 90 = 210 × 10
−6
Now, for a rectangular rosette,
ε x = ε 0 = 750 × 10
−6
ε y = ε 90 = 210 × 10
−6
γ xy = 2ε 45 − (ε 0 + ε 90 )
= 2
−110 × 10
−6
−
750 × 10
−6
+ 210 × 10
−6
γ xy = −1180 × 10
−6
∴ The magnitudes of principal strains are
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 =
750 + 210
2
10
−6
±
1
2
(750 − 210)10 −6
2 +
(−1180)10 −6
2
= 480 × 10
−6
±
1
2
(1297.7)10
−6
= 480 × 10
−6
± 648.85 × 10
−6
∴ ε max = ε 1 = 1128.85 × 10
−6
ε min = ε 2 = −168.85 × 10
−6
The directions of the principal strains are given by the relation
tan 2θ =
γ xy
ε x − ε y
∴ tan 2θ =
−1180 × 10
−6
(750 − 210)10 −6 = −2.185
∴ 2θ = 114.6
0
∴ θ 1 = 57.3
0 and θ 2 = 147.3
0
Example 3.4 If the displacement field in a body is specified as u =
x
2
+ 3
10
−3
, v = 3y
2 z × 10
−3 and w = (x + 3z) × 10
−3 , determine the strain
components at a point whose co-ordinates are (1, 2, 3).
Solution: From Eq. (3.3), we have