Question

8.17 A [0/90] crossply antisymmetric carbon/epoxy lam- inate is cooled down during curing from 180 oC (356 °F) to 30 °C (86 °F) (Fig. 8.18). Compute the curvatures K, and k, for the lamina properties E 140 GPa (20.3 Msi) E2-10 GPa ( 1 .45 Msi) G12-6 GPa (0.87 Msi) V12 = 0.34 02 30 x 10-/C (16.7 x 10- F) β,-0.55 t = 1 mm (0.040 in) (layer thickness)

Answer 1

11E E1140 x 109 -1 = 7.142 × 10-12Pa-1,S22 = -1 = 1 × 10-10Pa-1,S12 = 10× 109Pa -2.42x 10-12Pa-1,S66 El - 140x109 Pa- Pa-1-1.66× 10-10Pa-1 - 92би 109 G12

S2 1SIS22 - Si2 12 S1192-ול2 Q12- 2 e11 Q2 Q66 So6

10-10 Q11-10-10×7.142× 10-12- (2.42 x 10 12)2 Pa = 0.1417 x 1012 Pa

7.142 x 10-12 10 Pa = 1.008×1010Pa 22 = x 7.142 x 10-12-(2.42 x 10-12)2

Q12 1010Pa 2.42 x 10-12 10-10 x 7.142x 10-12- (2.42x 10-12)2 P: 0.3416

Q66-G12-6 x 109Pa

For 0 and 90 degree,
Q11 Q12 0 2 2 2 2 0 0 66 3
,
Q22 Q12 0 Q90=Q12 Q11 0 2 0 0 66 3

Mid-plane strain = 0;

AT-180- 30 150°
, $\Delta C =$ weight of moisture absorption per unit weight of the lamina.

$[N] = [B] [\kappa]$......................................Eq(1)

where,

$N^T = [\alpha] \Delta T, N^C = [\beta] \Delta C$

where,
01 0
, $\beta= \begin{bmatrix} \beta_1\\ \beta_2 \\ 0 \end{bmatrix}$ ,

or,

Substituting the values of N , B in Eq. (1) we can get the value of vector k ,

$[\kappa] = \begin{bmatrix} \kappa_x\\ \kappa_y \\ 0 \end{bmatrix} = [B]^{-1}\times [N]$