78
3 Analysis of Strain
ε x =
400 × 10
−6
−
750 × 10
−6
cos
60
0
− 36.8
0
= −290 × 10
−6
ε y =
400 × 10
−6
+
750 × 10
−6
cos
60
0
− 36.8
0
= 1090 × 10
−6
Because point x
lies above the ε-axis and point y
below ε-axis, the shear strain
γ x y is negative.
Therefore,
γ x y
2
= −
750 × 10
−6
sin
60
0
− 36.8
0
= −295 × 10
−6
Hence, γ x y = −590 × 10
−6 .
Solution: (b)
From the Mohr’s circle of strain, the principal strains are (Fig. 3.9)
ε 1 = 1150 × 10
−6
ε 2 = −350 × 10
−6
The directions of the principal axes of strain are shown in Fig. 3.10.
Example 3.2 By means of strain rosette, the following strains were recorded during
the test on a structural member.
ε 0 = −13 × 10 −6 mm/mm, ε 45 = 7.5 × 10 −6 mm/mm, ε 90 = 13 × 10 −6 mm/mm
Determine (a) Magnitude of principal strains
(b) Orientation of principal planes.
Solution: (a) We have for a rectangular strain rosette the following:
ε x = ε 0
ε y = ε 90
γ xy = 2ε 45 − (ε 0 + ε 90 )
Substituting the values in the above relations, we get
ε x = −13 × 10
−6
ε y = 13 × 10
−6
γ xy = 2 × 7.5 × 10
−6
−
−12 × 10
−6
+ 13 × 10
−6
∴ γ xy = 15 × 10
−6
The principal strains can be determined from the following relation.