Page 173 - Mechanics Analysis Composite Materials

I58 Mechanics and analysis of composite materials
The inverse form of these equations is

(4.83)

where

(4.84)

Now, decompose the strip displacements, strains, and stresses into two components
corresponding to
(1) free tension (see Figure 4.23), and
(2) bending.
For the free tension we have z,, = 0 and u = 0. So, Eqs. (4.81) and (4.82) yield

(4.85)

where E] = u{ and dl'= cr =F/ah, where F is the axial force applied to the strip, a
the strip width and h is its thickness. Because dl)= constant, Eqs. (4.85) give
F
O
d
el = qx.vx- = constant, e: = 81 = -= - . (4.86)
EX Ex ah
Adding components corresponding to bending (with index 2) we can write the total
displacements and strains as

The total stresses can be expressed with the aid of Eqs. (4.83), i.e.,

Transforming these equations with the aid of Eqs. (4.84) and (4.86) we arrive at

(4.87)

These stresses are statically equivalent to the axial force P,bending moment M, and
transverse force V, which can be introduced as