Page 50 - Mechanics Analysis Composite Materials

Chapter 2. Fundamentals of mechanics of solid? 35

where

(2.13)

are invariant characteristics (invariants) of the stressed state. This means that if we
refer the body to any Cartesian coordinate frame with directional cosines specified
by Eqs. (2.1), take the origin of this frame at some arbitrary point and change
stresses in Eqs. (2.13) with the aid of Eqs. (2.8) and (2.9), the values of I,, I2,13 at
this point will be the same for all such coordinate frames. Eq. (2.12) has three real
roots that specify three principal stresses 61.62, and 03. There is a rule according to
which 61 2 a. 2 63, Le., a[is the maximum principal stress and 03 is the minimum
one. If, for example, the roots of Eq. (2.12) are 100 MPa, -200 MPa, and 0, then
61 = 100 MPa, 02 = 0, and a3 = -200 MPa.
To demonstrate the procedure, consider a particular state of stress, important for
applications, namely pure shear in the xy-plane. Let a thin square plate referred to
coordinates x, y, z be loaded with shear stresses T uniformly distributed over the
plate thickness and along the edges (see Fig. 2.5).
One principal plane is evident -it is plane z = 0, which is free of shear stresses. To
find two other planes, we should take in Eqs. (2.13) a, = a,,= a, = 0, T~~ = T>= = 0,
and z,, = 5. Then, Eq. (2.12) acquires the form

c3- = 0 .
The first root of this equation gives = 0 and corresponds to plane z = 0. Two
other roots are 6 = h.Thus, we have three principal stresses, i.e., 61 = z,62 = 0,
03 = -7. To find the planes corresponding to (TI and 03 we should put Zr = 0,
substitute r~ = fzinto Eqs. (2.lo), write them for the state of stress under study, and

z
4

Fig. 2.5. Principal stresses under pure shear.