A unidirectional aerospace composite lamina is subjected to a state of plane stress along its principal material axes: $\sigma_1 = 600\text{ MPa}$, $\sigma_2 = 40\text{ MPa}$, and $\tau_{12} = 30\text{ MPa}$. The ultimate material strengths are: longitudinal tensile strength $X_T = 1{,}000\text{ MPa}$, transverse tensile strength $Y_T = 50\text{ MPa}$, and in-plane shear strength $S = 60\text{ MPa}$. According to the Maximum Stress failure criterion, the minimum safety factor against lamina failure is:

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### 1. Strategy & Theoretical Principles Under the Maximum Stress failure theory for orthotropic composite laminas, failure occurs when any of the principal stresses exceeds its respective allowable strength independently: $$\text{SF}_1 = \frac{X_T}{\sigma_1}, \quad \text{SF}_2 = \frac{Y_T}{\sigma_2}, \quad \text{SF}_{12} = \frac{S}{|\tau_{12}|}$$ The overall safety factor of the lamina is the minimum across all failure modes: $$\text{SF}_{\min} = \min(\text{SF}_1, \text{SF}_2, \text{SF}_{12})$$ ### 2. Analytical Formulation & Calculation Given: - Longitudinal mode: $\text{SF}_1 = \frac{1000}{600} = \frac{5}{3} \approx 1.667$ - Transverse mode: $\text{SF}_2 = \frac{50}{40} = 1.250$ - Shear mode: $\text{SF}_{12} = \frac{60}{30} = 2.000$ ### 3. Step-by-Step Calculation $$\text{SF}_{\min} = \min(1.667, 1.250, 2.000) = 1.250$$ Failure is governed by transverse matrix cracking with a safety factor of 1.25. ### 4. Distractor Forensics - Candidate checking only the fiber direction gets $1.67$, failing to recognize that transverse matrix strength is the critical bottleneck. ### 5. Exam Trap & Standard Reference - **Trap Alert:** In unidirectional composites, transverse tensile strength $Y_T$ is typically an order of magnitude weaker than longitudinal strength $X_T$ and frequently governs failure. - **Standard Reference:** Jones, R.M., *Mechanics of Composite Materials*, Chapter 2.

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