Page 175 - Mechanics Analysis Composite Materials

160 Mechanics and analysis of composite materials

The total angle of rotation 8 = 81 + 02, where 8, is specified by Eqs. (4.86),
should be zero at the ends of the strip, i.e., 8(x = f1/2) = 0. Satisfying these
conditions we get

(4.94)

Substitution into Eq. (4.93) and integration allow us to find the deflection

(4.95)

This expression includes two constants, C, and C4, which should be determined
from the boundary conditions U~(X= 1/2) = 0. The final result following from
Eqs. (4.86), (4.94), and (4.95) is

(4.96)

where i= l/a and X =x/l. Deflection of a carbon-epoxy strip having 4 = 45” and
10
I=is shown in Fig. 4.28.
Now, we can write the relationship between modulus E, corresponding to the
ideal test shown in Fig. 4.23 and apparent modulus E.: that can be found from the
real test shown in Figs. 4.22 and 4.26. Using Eqs. (4.84), (4.86), (4.92), and (4.96) we
finally get

c=Exa~ ,

-0.1 1

Fig. 4.28. Normalized deflection of a carbon-epoxy strip (Q= 45”, T = IO).