Page 53 - Mechanics Analysis Composite Materials

38 Mechanics and analysis of’ composite materials
where

(2.22)

can be treated as linear strain-displacement equations. Taking 1, = I, I, = I, = 0 in
Eqs. (2.22), Le., directing element LM in Fig. 2.6 along the x-axis we can readily see
that cX is the strain along the same x-axis. Similar reasoning shows that E,, and E= in
Eqs. (2.22) are strains in the directions of axes y and z. To find out the physical
meaning of strains y in Eqs. (2.22), consider two orthogonal line elements LM and
LN and find angle CI that they make with each other after deformation (see Fig. 2.6),
i.e.,

(2.23)

Here, dxl, dyl, and dzl are specified with Eqs. (2.17), dsl can be found from
Eq. (2.18) and

(2.24)
ds’,= ds’(1 +E’) .

Introduce directional cosines of element LN as

(2.25)

Because elements LM and LN are orthogonal, we have
lxl: + IJ;, + lzZ; = 0 .

Using Eqs. (2.14)’ (2.18), (2.24X2.26) and introducing shear strain y as the
difference between angles MIL1 NI and MLN, i.e., as
x
y=--m
2
we can write Eq. (2.23) in the following form

(2.26)