Page 273 - Mechanics Analysis Composite Materials

258 Mechanics and analysis of composite materials
My,associated with the change of the curvature of the cross-sectional contour in
Eq. (5.75) is very small.
For numerical analysis, we first use Eqs. (4.72) to calculate stiffness coefficients
for the angle-ply layer, i.e.,

Ai:) = 25 GPa, Ai:) = 10 GPa, A$) = 14.1 GPa, A:) = 11.5 GPa ,
(5.80)

and for the hoop layer

Ai:) = 9.5 GPa, Ai;) = 2.5 GPa, A$) = 44.7 GPa, A:) = 4 GPa ,
(5.81)

Then, we apply Eqs. (5.41) to find the I-coefficients that are necessary for the cases
(axial compression and torsion) under study:

1:;)= 21.2 GPa mm, 1;;)= 7.7 GPa mm,
1:;) = 35.6 GPa mm, 1$) = 9.5 GPa mm,
1;;)= 10.1 GPa mm’, Z::) = 3.3 GPa mm2,
1;;)= 27.4 GPa mm2, 1:) = 4.4 GPa mm’,
1;;)= 21.7 GPa mm3, Zii)= 5.9 GPa mm3,
1;;’ = 94 GPa mm3 .

To determine stiffness coefficients of the laminate, we should pre-assign the
coordinate of the reference plane (a cylindrical surface for the cylinder). Let us put
e = 0 for simplicity, i.e., we take the inner surface of the cylinder as the reference
surface (see Fig. 5.20). Then, Eqs. (5.28) yield

BII=I/? = 21.2 GPa mm, B12 =1:;)= 7.7 GPamm,
B22 =1;;)= 35.6 GPamm,
CII=Zi:) = 10.1 GPamm2, C12 =Z[i)= 3.3 GPamm’,
C22 =Z;:) = 27.4 GPamm2,
012 =1;;’= 5.9 GPamm3, 022 =I;;) = 94 GPamm3 ,

and in accordance with Eqs. (5.78) for R = 100 mm,

BI~= 7.7 GPa mm, 822 = 35.3 GPa mm,
E12 = 3.2 GPa mm2, c22 = 26.5 GPa mm2 .