Strain analysis 423Example. Continuing from before, with e,, = 8000 p, eyy = 2000 ps,
eq = 4500 ,us:
B I
\
- B
Q at (2000,4500)I
\
4500)r = .\j3O0O2 + 4500' = 5410ys = maximum mathematical shear strainMaximum engineering shear strain = 2 x 5410
Principal strains are
= 10,820 ps.
(5000 + 5410) = 10,410p
(5000 - 5410) = -410,~~.Determination of principal struins from meusured
s trcr ins
This is the practical use of the two-dimensional strain transformation
equations.