Page 281 - Mechanics Analysis Composite Materials

266 Mechanics and analysis of composite materials

= -*[I+ e-m(o. 1 I sin tx - 0.052 cos &)I,
2m

E =-pB21 [I +e-'"(0.51 sin tx - 0.24costx)], (5.108)
." 2nRB
PB21 -*
0 -- e (6.3 cos tx -2.3 sin a) ,
- 2nRB
where r= 7.9/R and t = 8.75/R. As can be seen, solution in Eqs. (5.79) is
supplemented with a boundary-layer solution that vanishes with a distance from the
cylinder end.
To determine transverse shear stress T,,, we integrate the first equation in
Eqs. (5.73) under the condition T=(Z = 0) = 0. As a result, the shear stress acting in
the angle-ply layer is specified by the following expression:

where

Substitution of Eqs. (5.108) and calculation yield
-
T::) =*e-* [(23.75cos tx - 9.75sin &)z + (6.3cos tx +24.9 sin tx)z2] .
2nR2B
(5.109)

Transverse normal stress can be found from the following equation similar to
Eq. (5.86):

For a thin cylinder, we can neglect z/R in comparison with unity. Using Eqs. (5.108)
and (5.109) for the angle-ply layer, we get
P
0;') = -O.O68-{z+ e-"[(O.l8costx - 0.0725sintx)z
R2h
-(0.12cos&+ 0.059sintx)2 + (0.05costx - 0.076sintx)2]} .