Aero Structures 1: Governing Equations and Failure Analysis in Aircraft Design

Abstract

This article walks through the foundational framework of linear elasticity that underpins aircraft structural analysis, then connects it to practical failure prediction methods. We examine the three governing equation sets, principal stress concepts, and yield criteria commonly applied to aircraft components. A worked example demonstrates how these tools combine to assess structural safety in a thin-walled beam under combined loading.

Background

Aircraft structures must satisfy competing demands: they must be light, stiff, and strong enough to survive flight loads without permanent deformation or fracture. Achieving this requires a rigorous analytical foundation. The discipline of aero structures rests on three coupled mathematical frameworks that together form a complete boundary value problem ^{[governing-equations-of-linear-elasticity]}.

The first framework is equilibrium: internal stresses must balance applied loads and body forces at every point. The second is kinematics: deformations must be geometrically compatible—no gaps or overlaps in the material. The third is constitutive behavior: the material's stress-strain relationship (typically Hooke's Law for elastic materials) closes the system. When these three equation sets are solved together with appropriate boundary conditions, the engineer obtains the complete stress, strain, and displacement field throughout the structure ^{[governing-equations-of-linear-elasticity]}.

In practice, aircraft loads are rarely uniaxial. A wing spar experiences bending, torsion, and shear simultaneously. Stringers and skins carry multi-directional stresses. To predict failure in these complex states, engineers use failure criteria that reduce a multi-axial stress state to a single equivalent measure comparable against material strength.

Key Results

The Three Governing Equations

Linear elasticity is built on three coupled systems ^{[governing-equations-of-linear-elasticity]}:

1. Equilibrium equations ensure force and moment balance.
2. Kinematic equations relate strains to displacements, enforcing geometric compatibility.
3. Constitutive equations encode material response (e.g., Hooke's Law).

Together, these allow solution for stresses, strains, and displacements given loads and boundary conditions.

Principal Stresses and Failure Prediction

At any point in a structure, there exist three orthogonal directions—the principal axes—along which shear stresses vanish and only normal stresses remain. The principal stresses $\sigma_1 \geq \sigma_2 \geq \sigma_3$ represent the extreme normal stresses at that point ^{[principal-stresses-and-strains]}.

This is critical for failure analysis because materials typically fail along planes of maximum normal or shear stress. By computing principal stresses at critical locations (stress concentrations, attachment points, etc.), engineers identify the most damaging stress components and apply failure criteria to assess safety ^{[principal-stresses-and-strains]}.

Yield Criteria for Multi-Axial Loading

Under uniaxial loading, a material yields when stress reaches the yield strength $\sigma_y$. Under multi-axial loading, yield criteria extend this concept by combining principal stresses into an equivalent stress ^{[yield-failure-criteria]}.

The Von Mises criterion predicts yield when: $\sqrt{\frac{1}{2}[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2]} = \sigma_y$

The Tresca criterion predicts yield when: $\max(|\sigma_1-\sigma_2|, |\sigma_2-\sigma_3|, |\sigma_3-\sigma_1|) = \sigma_y$

Von Mises is more commonly used in aircraft design because it better matches experimental data for ductile metals under complex loading ^{[yield-failure-criteria]}.

Shear Center and Torsion in Thin-Walled Beams

Open thin-walled sections (channels, I-beams) have a unique point called the shear center. When a transverse load passes through the shear center, it produces pure bending without torsion. Loads applied elsewhere induce unwanted twisting ^{[shear-center-of-open-thin-walled-beams]}.

For symmetric sections, the shear center coincides with the centroid. For unsymmetric sections, they differ. In aircraft design, this distinction matters because torsion can trigger flutter, fatigue, or failure. Proper load path design ensures loads are applied near the shear center ^{[shear-center-of-open-thin-walled-beams]}.

Multi-Cell Beams in Aircraft Structures

Modern aircraft wings use closed multi-cell box beams (the wing box) rather than open sections. These structures consist of multiple thin-walled compartments that distribute shear flow and resist torsion far more efficiently than open sections ^{[thin-walled-multi-cell-beams]}.

A multi-cell beam under torsion or shear develops shear flow that circulates around and through each cell. The torsional rigidity depends on cell geometry, wall thickness, and material. Analyzing multi-cell beams requires solving for shear flow distribution and computing resulting stresses and deflections—more complex than single-cell analysis, but essential for modern aircraft ^{[thin-walled-multi-cell-beams]}.

Worked Example: Stress Analysis and Failure Check of a Loaded Beam Section

Problem: A thin-walled aluminum beam carries a transverse shear load $V = 50$ kN and a bending moment $M = 200$ kN·m. At a critical section, the stress state (in the material coordinate system) is:

$\sigma_x = 180 \text{ MPa}, \quad \sigma_y = 20 \text{ MPa}, \quad \tau_{xy} = 30 \text{ MPa}$

The aluminum alloy has a yield strength $\sigma_y = 350$ MPa. Determine whether the material yields using the Von Mises criterion.

Solution:

Step 1: Find principal stresses.

For a 2D stress state, the principal stresses are found from: $\sigma_{1,3} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

$\sigma_{1,3} = \frac{180 + 20}{2} \pm \sqrt{\left(\frac{180 - 20}{2}\right)^2 + 30^2}$

$\sigma_{1,3} = 100 \pm \sqrt{80^2 + 30^2} = 100 \pm \sqrt{6400 + 900} = 100 \pm \sqrt{7300}$

$\sigma_{1,3} = 100 \pm 85.4$

Thus: $\sigma_1 = 185.4$ MPa, $\sigma_3 = 14.6$ MPa, and $\sigma_2 = 0$ (out-of-plane).

Step 2: Apply Von Mises criterion ^{[yield-failure-criteria]}.

$\sigma_{\text{eq}} = \sqrt{\frac{1}{2}[(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2]}$

$\sigma_{\text{eq}} = \sqrt{\frac{1}{2}[(185.4-0)^2 + (0-14.6)^2 + (14.6-185.4)^2]}$

$\sigma_{\text{eq}} = \sqrt{\frac{1}{2}[34373 + 213 + 29180]} = \sqrt{\frac{63766}{2}} = \sqrt{31883} = 178.6 \text{ MPa}$

Step 3: Compare to yield strength.

Since $\sigma_{\text{eq}} = 178.6$ MPa $< \sigma_y = 350$ MPa, the material does not yield. The section is safe with a margin of approximately 1.96.

This example illustrates how the three governing equations ^{[governing-equations-of-linear-elasticity]} are applied: equilibrium and kinematics determine the stress state; constitutive relations (Hooke's Law) relate it to material properties; and failure criteria ^{[yield-failure-criteria]} assess safety.

References

• ^{[governing-equations-of-linear-elasticity]}
• ^{[principal-stresses-and-strains]}
• ^{[yield-failure-criteria]}
• ^{[shear-center-of-open-thin-walled-beams]}
• ^{[thin-walled-multi-cell-beams]}

AI Disclosure

This article was drafted with AI assistance from class notes (Zettelkasten). All factual and mathematical claims are cited to source notes. The worked example was generated from the framework described in the notes and verified for internal consistency. The author reviewed all content for technical accuracy and relevance to the course material.