Page 42 - Aircraft Stuctures for Engineering Student

1.1 5 Stress-strain relationships 27

stresses. Find also the principal strains, the maximum shear stress, the maximum
shear strain and their directions at the point. Take E = 200000N/mm2 and v = 0.3.

If we assume that a, = 83 N/mm2 and ay = 65 N/mm2 then from Eqs (1.47)
1
E, = ~ (83 - 0.3 x 65) = 3.175 x lop4
200 000

E -- (65 - 0.3 x 83) = 2.005 x
- 200000
-0.3
e, = ~ (83 + 65) = -2.220 x lop4
200 000
In this case, since there are no shear stresses on the given planes, a, and av are
principal stresses so that E, and are the principal strains and are in the directions
of a, and cy. It follows from Eq. (1.15) that the maximum shear stress (in the plane of
the stresses) is
83 - 65
rmax = = ~N/IYUII'
2
acting on planes at 45" to the principal planes.
Further, using Eq. (1.45), the maximum shear strain is
2 x (1 +0.3) x 9
'Ymax =
200 000
so that -ymax = 1.17 x on the planes of maximum shear stress.

Example 1.3
At a particular point in a structural member a two-dimensional stress system exists
2
where a, = 60 N/mm , ay = -40 N/mm' and rxy = 50 N/mm2. If Young's modulus
E = 200000N/mm2 and Poisson's ratio v = 0.3 calculate the direct strain in the x
and y directions and the shear strain at the point. Also calculate the principal strains
at the point and their inclination to the plane on which a, acts; verify these answers
using a graphical method.

From Eqs (1.47)
+
E, = - 0.3 x 40) = 360 x lop6
(60
200 000
1
EY = -
(-40 - 0.3 x 60) = -290 x lop6
200 000
From Eq. (1.45) the shear modulus, G, is given by
E
G=- - 2oo Oo0 = 76 923 N/mm2
2(1+v)-2(1+0.3)
Hence, from Eqs (1.47)