Engineering & Science
Tsai-Hill Orthotropic Composite Ply Failure Index & Strength Ratio
Evaluate biaxial + shear lamina failure under combined in-plane stresses (sigma_1, sigma_2, tau_12) using the quadratic Tsai-Hill failure index FI and reserve strength factor SF = 1 / SQRT(FI).
Account for stress interaction when checking composite plies!
Checking maximum fiber stress sigma_1 < X, transverse stress sigma_2 < Y, and shear stress tau_12 < S independently ignores multiaxial stress interaction—where simultaneous transverse tension and in-plane shear trigger matrix cracking well below either standalone strength! Azzi and Tsai adapted Hill's anisotropic yield theory to a thin orthotropic lamina under plane stress: FI = (sigma_1 / X)^2 - (sigma_1 * sigma_2) / X^2 + (sigma_2 / Y)^2 + (tau_12 / S)^2!
First-ply failure occurs whenever FI >= 1.0! Because Tsai-Hill is homogeneous of degree 2 in stress, scaling all applied loads proportionally by factor SF multiplies FI by SF^2—meaning the linear reserve strength factor (margin to failure) is SF = 1 / SQRT(FI), NOT 1 / FI! At sigma_1 = 600 MPa, sigma_2 = 30 MPa, tau_12 = 40 MPa on a carbon-epoxy ply (X = 1500, Y = 50, S = 70 MPa), FI = 0.8385 (< 1 Safe, SF = 1.092), whereas raising transverse stress to sigma_2 = 45 MPa pushes FI to 1.2845 (Failure!).
| Load Case (X=1500, Y=50, S=70 MPa) | Applied (s1, s2, t12) | Tsai-Hill Index FI | Strength Factor SF = 1/SQRT(FI) |
|---|---|---|---|
| Design Flight Load | (600, 30, 40) MPa | 0.8385 (< 1.0 Safe) | SF = 1.092 (+9.2% Margin) |
| Thermal + Transverse Overload | (600, 45, 40) MPa | 1.2845 (>= 1.0 Failure!) | SF = 0.882 (First-Ply Matrix Crack) |
| Invalid Zero/Negative Strength | Y <= 0 | Guarded Input Error | "Invalid Strength Allowables" |
Tsai-Hill Quadratic Failure Index FI (Safe < 1.0, Failure >= 1.0)
Computes (s1/X)^2 - (s1*s2)/X^2 + (s2/Y)^2 + (t12/S)^2 after verifying positive material strengths X, Y, and S.
=IF(OR(B5 <= 0, B6 <= 0, B7 <= 0), "Invalid Strength Allowables", ROUND(((B2 / B5) ^ 2) - ((B2 * B3) / (B5 ^ 2)) + ((B3 / B6) ^ 2) + ((B4 / B7) ^ 2), 4))Linear Reserve Strength Factor SF = 1 / SQRT(FI)
Returns the proportional load multiplier required to reach FI = 1.0.
=IF(OR(B5 <= 0, B6 <= 0, B7 <= 0), "Invalid Strength Allowables", ROUND(1 / SQRT(((B2 / B5) ^ 2) - ((B2 * B3) / (B5 ^ 2)) + ((B3 / B6) ^ 2) + ((B4 / B7) ^ 2)), 3))Try the example
Use the original values or change them to test your understanding.