Given Q11=150 GPa, Q12=0.28, Q22=7 GPa, Q66=3 GPa, theta=45°, compute Qbar11.

• A. Approximately 42.4 GPa.
• B. Approximately 50.0 GPa.
• C. Approximately 35.0 GPa.
• D. Approximately 60.0 GPa.

Answer: (A)

When a ply is rotated, its in-plane stiffness along the new axis is given by the transformed reduced stiffness Qbar11. The expression is Qbar11 = Q11 cos^4 θ + Q22 sin^4 θ + 2(Q12 + 2Q66) sin^2 θ cos^2 θ. This formula combines the contributions from the material’s principal stiffnesses (Q11, Q12, Q22, Q66) as the orientation θ changes.

Here, θ = 45°. At this angle, sin^2 θ = cos^2 θ = 0.5, so sin^4 θ = cos^4 θ = 0.25 and sin^2 θ cos^2 θ = 0.25.

Compute each term:

• Q11 cos^4 θ = 150 × 0.25 = 37.5

• Q22 sin^4 θ = 7 × 0.25 = 1.75

• 2(Q12 + 2Q66) sin^2 θ cos^2 θ = 2(0.28 + 6) × 0.25 = 2(6.28) × 0.25 = 12.56 × 0.25 = 3.14

Add them: 37.5 + 1.75 + 3.14 = 42.39 GPa, which is approximately 42.4 GPa.

So the effective stiffness Qbar11 in the rotated orientation is about 42.4 GPa.