Page 329 - Mechanics Analysis Composite Materials

314 Mechanics and analysis of composite materials
0 for the external skin:

A\:) = q cos44, + sin44c+2(J$Y2 + 2q,)sin2 4, cos24, = 99.05 GPa,
Ag) = sin44, + cos44c + 2(qv:, + 2q2)sin24, cos24c = 13.39 GPa,
A$) = @JC, + [g+e- 2(qvT2+2q2)]sin' +c cos24c = 13.96 GPa .

Using Eqs. (7.18) we find the thermal coefficients of the layers (the temperature is
uniformly distributed through the laminate thickness):

(AT,), = (AT2), = EfafAT= 1715 x AT GPa/"C,
=
(AT,), = = E-i(l + v~-~)c~,LAT32.08 x lop6AT GPa/"C,
6r
(AT,), = 2Er-gr cos2&AT = 4.46 x 10-'AT GPa/"C,
a,
6
(AT'), = 2Er'ar sin' &AT = 1.06 x AT GPa/"C,
ar
(AT,), = [q(aC +~~~a~)cos~~+l?!j(a~+v~,a~)sin~~]AT
= 132.43 x lo-' AT GPa/"C,
(AT2)4 = [q(aC + vF2a3sin24+q(aC,+ V;,IX:) cos' 4]AT
= 317.61 x 10-6AT GPa/"C .
Because the layers are orthotropic, AT, = 0 for all of them. Specifyingcoordinates of
the layers (see Fig. 5.10), i.e.
to = 0, tl = 0.02, t2 = 1.02, t3 = 10.02, t4 = 13.52 (mm)

and applying Eq. (7.27) we calculate parameters .I:; for the laminate

J!? = (AT,),(tI -to) + (AT,),(t2 - tl) + (AT,),(t3 - t2)
+ (,4TI),(t4 - t3) = 570 x lop6AT GPa mm/OC,
.I;:)= 1190 x IO-' AT GPa mm/OC,
(1) - 1 -6)+ + (ATl)4(f2 -ti)]
Jll -2 [(ATl)I(t: -$1 + (ATL)3(': -
= 5690 x lo-' AT GPa mm2/OC,
.I;;)= 13150 x 10-'AT GPa mm'/"C .

To determine we need to specify the reference surface of the laminate. Assume
that this surface coincides with the middle surface, i.e., that e = h/2 = 6.76 mm.
Then, Eqs. (7.25) yield

N: =J{p' = 570 x AT GPa mm/'C,
N& =J;:) = 1190 x AT GPa mrn/OC,