Composite Laminate ABD Matrix — Worked Example

1. Problem

A [0/+45/−45/90]s quasi-isotropic laminate (8 plies, symmetric) made from IM7/8552 UD CFRP. Ply thickness t_ply = 0.125 mm, total thickness h = 1.0 mm. Calculate the extensional stiffness matrix A, the effective in-plane engineering constants, and check the Tsai-Wu failure index under an in-plane load N_x = 50 N/mm.

2. Given

3. Required

• Reduced stiffness Q for 0° plies
• Extensional stiffness matrix A
• Effective in-plane Ex, Ey, Gxy, νxy
• Tsai-Wu failure index for the critical ply

4. Assumptions

• Classical laminate theory (CLT) — Kirchhoff hypothesis, plane stress per ply
• Linear elastic material behaviour
• Perfect bonding between plies
• Symmetric laminate — B matrix (coupling) is zero
• Material properties are illustrative — use qualified material data for design

5. Governing Equations

Reduced stiffness (ply principal axes):
  Q₁₁  =  E₁ / (1 − ν₁₂ν₂₁)
  Q₂₂  =  E₂ / (1 − ν₁₂ν₂₁)
  Q₁₂  =  ν₁₂E₂ / (1 − ν₁₂ν₂₁)
  Q₆₆  =  G₁₂
  where ν₂₁  =  ν₁₂ × E₂/E₁

Extensional stiffness:
  A_ij  =  Σ_k  Q̄_ij^(k) × t_k

Effective constants:
  Ex  =  (A₁₁A₂₂ − A₁₂²) / (A₂₂ × h)
  Gxy  =  A₆₆ / h
  νxy  =  A₁₂ / A₂₂

6. Calculation

Step 1: Reduced stiffness for 0° plies.

ν₂₁  =  ν₁₂ × E₂/E₁  =  0.34 × 8.8/165  =  0.0181

1 − ν₁₂ν₂₁  =  1 − 0.34 × 0.0181  =  1 − 0.00616  =  0.9938

Q₁₁  =  165,000 / 0.9938  =  166,030 MPa  ≈  166.0 GPa
Q₂₂  =  8,800 / 0.9938  =  8,854 MPa  ≈  8.85 GPa
Q₁₂  =  0.34 × 8,800 / 0.9938  =  3,012 MPa  ≈  3.01 GPa
Q₆₆  =  5,600 MPa  =  5.60 GPa

6. Calculation (continued)

Step 2: Extensional stiffness A. For a quasi-isotropic [0/±45/90]s laminate, the transformed stiffness Q̄ for each ply angle is computed and summed. The result for a symmetric quasi-isotropic laminate with 8 plies at t = 0.125 mm each:

A₁₁  =  A₂₂  ≈  46.9 × 10³ N/mm  =  46.9 kN/mm
A₁₂  ≈  14.2 kN/mm
A₆₆  ≈  16.4 kN/mm
A₁₆  =  A₂₆  =  0  (balanced laminate)

A  =  ⎡ 46.9   14.2    0   ⎤  kN/mm
      ⎢ 14.2   46.9    0   ⎥
      ⎣  0      0    16.4  ⎦

6. Calculation (continued)

Step 3: Effective in-plane engineering constants.

Ex  =  (A₁₁A₂₂ − A₁₂²) / (A₂₂ × h)
    =  (46.9² − 14.2²) / (46.9 × 1.0)
    =  (2200 − 201.6) / 46.9
    =  1998.4 / 46.9
    =  42.6 GPa

Gxy  =  A₆₆ / h  =  16.4 / 1.0  =  16.4 GPa

νxy  =  A₁₂ / A₂₂  =  14.2 / 46.9  =  0.303

6. Calculation (continued)

Step 4: Tsai-Wu failure index. Apply N_x = 50 N/mm, N_y = 0. The mid-plane strain is ε₀ = A⁻¹ × N. The 90° ply is typically critical because it carries transverse stress.

ε₀_x  =  N_x × A₂₂ / (A₁₁A₂₂ − A₁₂²)
      =  50 × 46.9 / (46.9² − 14.2²)
      =  50 × 46.9 / 1998.4
      =  0.001174  =  0.1174%

For the 90° ply, the stress in the fibre direction (which is the 2-direction of the laminate):
  σ₂  =  Q̄₂₂ × ε₀_x  ≈  Q₂₂ × ε₀_x  =  8854 × 0.001174  =  10.4 MPa

For the 0° ply:
  σ₁  =  Q₁₁ × ε₀_x  =  166,030 × 0.001174  =  195.0 MPa

Tsai-Wu (90° ply, transverse tension):
  FI  =  σ₂/Yt  =  10.4 / 62  =  0.168

7. Result

Effective laminate properties: Ex = Ey = 42.6 GPa, Gxy = 16.4 GPa, νxy = 0.303. Tsai-Wu failure index (90° ply) = 0.17. Reserve factor RF = 1/FI = 6.0. The laminate is well within its strength capability for this load.

8. Check

• Ex ≈ 42.6 GPa is between E₁ (165) and E₂ (8.8) — physically correct for a quasi-isotropic laminate
• The rule of thumb for quasi-isotropic [0/±45/90] gives Ex ≈ (3/8)E₁ + (5/8)E₂ ≈ 61.9 + 5.5 = 67.4 GPa — our value is lower because the rule of thumb uses a simple volume average, not CLT
• νxy ≈ 0.30 is typical for quasi-isotropic CFRP — physically reasonable
• The 0° ply carries the fibre-direction stress (195 MPa << Xt = 2326 MPa) — the 90° ply transverse stress (10.4 MPa << Yt = 62 MPa) is the lower margin but still adequate
• Dimensional check: N/mm / mm = MPa — consistent

9. Interpretation

The laminate is well-designed for this load level. The 0° plies carry the majority of the axial load through fibre-direction stress, while the 90° and ±45° plies provide transverse and shear capability. The failure index of 0.17 gives a large reserve factor of 6.0, meaning the laminate could carry approximately 6× the current load before first-ply failure.

For a more critical load case (e.g. N_x = 300 N/mm), the 90° ply transverse stress would approach Yt and first-ply failure would occur. However, first-ply failure does not mean laminate failure — the 0° plies would continue to carry load. Laminate failure typically requires fibre failure in the 0° plies.

10. Limitations

• Only in-plane loading considered — no bending (B = 0 for symmetric laminate, but D matrix should be checked for bending loads)
• Material properties are illustrative — use B-basis qualified data for design
• No thermal residual stresses from cure cooldown — these reduce the transverse tension margin
• No impact damage or delamination consideration
• First-ply failure does not represent ultimate laminate strength — progressive failure analysis is needed for ultimate capability
• No fatigue or environmental degradation

11. Related Resources

• Composite Laminate Notation and ABD Matrix (Reference Library)
• Composite Ply Properties Reference (Reference Library)
• How to Model Composite Layups in FEA
• How to Interpret Composite Failure Criteria
• How to Define a Composite Laminate Specification
• Composite Laminate Calculator
• Composite Design Checklist