Page 207 - Mechanics Analysis Composite Materials

192 Mechanics and analysis of composite materials

Repeating the derivation described in Section 2.5 we have

where

(4.138)

Using Eq. (4.137) we arrive at

(4.139)

where cos a = &Ids, and sin a = dy/ds,.
In a similar way, we can find the angle a’ after the deformation, i.e.,

. dv’
slna =-=-
I
ds, 1 +E,
(4.140)
cosa = -=-
dxr
I
ds, 1 + E,
Now return to the ply element in Fig. 4.60. Taking a = 4 in Eqs. (4.139) and (4.140)
we obtain

ax
sin$!=-[(1 +El1 1 +-2)sin++%cos4], (4.141)
aY 1
1
cos4) = -[(1 +%)cos4 +-sin4 .
1 +E1
Putting a = n/2 + 4 we have

sin(# + $) =-[(1 +2)cos 4 -%sin 41
1 +E2 (4.142)
cos(4’ + +) = -[-(I +z)cos4+aysing 1 .
1 +E2