Question 2 (30 Marks) A space-satellite structure is made of a composite laminate tailored for the following thermal expansion properties: = ax*0.1 x 10-6 m/m/C ; = +10.5 x 10-6 m/m/C A certain percentage of the plies are oriented at +/-75. Given a maximum temperature change of -200C from the material's stress-free temperature, use the Tsai-Wu failure criterion to calculate the factor of safety in these plies. [30] Given: E = 204.2 GPa, E = 15.77 GPa, G12 = 7.31 GPa, V12 = 0.32 P.T.O. T C T X = 1300 MPa, X = 1080 MPa, X = 90 MPa, X = 130 MPa X6= 160 MPa 1 = +1.35 x 10 m/m/C ; = +8.96 x 10-6 m/m/C Page 3 of 8 Supplementary Equations (1) [Q] IN TERMS OF ENGINEERING CONSTANTS E1 V12E2 Q11= Q12 [1 - v12(E2/E1)] [1 v12(E2/E1)] E2 Q22= Q66 = G12 [1-v12(E2/E1)] STIFFNESS TRANSFORMATIONS FOR COMPOSITE PLIES Q11 Q12 15 16 17 18 = Q16 = Q22 Q11 m + 2mn (Q12 + 2Q66) + Q22n (Q11 Q224Q66) mn + Q12 (m + n) [(Q11 - Q12 - 2Q66)m - (Q22 - Q12 - 2Q66)n] mn 4 Q22m + 2(Q12 + 2Q66) mn + Qn Q26 = [(Q22 - Q12 - 2Q66)m - (Q11 - Q12 - 2Q66)n] m == Q66 = (Q11+ Q22 - 2Q12) mn + Q66 (m - n) P.T.O. BEAM BENDING EQUATIONS d4w bD dx4 = q(x) M = -bD dw dx Page 6 of 8 Supplementary Equations (2) TRANSFORMATION OF STRESS AND STRAIN Z, 3 P.T.O. 1 2 -y 79 T12 = F 8/12/2 66 = [T] Ex Y/2 m P Where [T] 2 n m mn mn m 2mn - 2mn - n 2 TSAI-WU FAILURE THEORY 2 Fo +Fo+F+F + F + F +2F =1 66 12 F = 0 where F and F = 1 1 T 1 X C F 1 = 1 2 T 1 C F66 = F = X X X X X 11 T and F12 C 22 T 1 T 2 2 X, X, X, X, 2 2 1 2 Page 7 of 8 Supplementary Equations (3) THERMAL STRESS RESULTANTS k + Q16 k XY AT.dz k + 026 k k AT.dz k k AT.dz k + 066 XY N. X Ny" = T Nxy" = NXX N k=1 N hx-1 hy hk-1 k=1 hk hk-1 11 k Q12 X k k 216 + * + k Q12 k 022 k k x + Q26 k THERMAL RESIDUAL STRESSES k Q11 k k + X 12 k (, - $ ) - Q 16 AT XY = Q12 ( x - x ) + Q * (, * -, *) - (a) AT k k (ax k k -dy 22 -226 k k XY Q16* (ax - ax") + Q26 * ( - a )-066* (*) AT k dy XY Page 8 of 8