ResearchForge / Calculators
← all articles
aero-structureselasticitybeam-theorythin-walled-structuresstructural-analysisFri Apr 24
Narrated explainer (90s)
3Blue1Brown-style animation reel

Aero Structures 1: Step-by-Step Derivations

Abstract

This article presents the foundational framework of linear elasticity and its application to aircraft structural components. We derive the three governing equation sets—equilibrium, kinematic, and constitutive—and show how they combine to form a complete boundary value problem. We then examine failure prediction through principal stresses and yield criteria, and conclude with specialized topics in thin-walled beam theory essential to modern aircraft design.

Background

Aircraft structures must support complex, multi-axial loading while remaining as light as possible. This demands precise prediction of internal stresses, strains, and displacements throughout the structure. The mathematical framework that enables this prediction is linear elasticity [governing-equations-of-linear-elasticity].

Linear elasticity rests on three assumptions: (1) material behavior is linear (stress proportional to strain), (2) displacements are small relative to structural dimensions, and (3) the structure does not undergo large rotations. Under these assumptions, the superposition principle holds, and we can solve complex loading cases by combining simpler ones.

Key Results

The Three Governing Equations

Linear elastic structural analysis requires three coupled equation systems [governing-equations-of-linear-elasticity]:

Equilibrium equations enforce force and moment balance. At every point in the structure, the divergence of the stress tensor must equal the body force:

σ+b=0\nabla \cdot \boldsymbol{\sigma} + \mathbf{b} = 0

where σ\boldsymbol{\sigma} is the stress tensor and b\mathbf{b} is the body force per unit volume. This ensures that internal stresses can support applied loads without acceleration.

Kinematic equations relate strains to displacements. The strain tensor is the symmetric part of the displacement gradient:

ε=12(u+(u)T)\boldsymbol{\varepsilon} = \frac{1}{2}(\nabla \mathbf{u} + (\nabla \mathbf{u})^T)

where u\mathbf{u} is the displacement field. These equations ensure geometric compatibility—deformations must fit together without gaps or overlaps.

Constitutive equations encode material behavior. For a linear elastic isotropic material, Hooke's Law relates stress to strain:

σ=C:ε\boldsymbol{\sigma} = \mathbf{C} : \boldsymbol{\varepsilon}

where C\mathbf{C} is the stiffness tensor. For an isotropic material, this simplifies to:

σij=2μεij+λδijεkk\sigma_{ij} = 2\mu \varepsilon_{ij} + \lambda \delta_{ij} \varepsilon_{kk}

where μ\mu and λ\lambda are Lamé parameters, and δij\delta_{ij} is the Kronecker delta.

Together, these three systems form a complete boundary value problem. Given boundary conditions (prescribed displacements or tractions) and applied loads, we can solve for the stress, strain, and displacement fields everywhere in the structure.

Principal Stresses and Failure Prediction

At any point in a loaded structure, there exist three orthogonal directions—the principal axes—along which the stress tensor contains only normal stresses and zero shear stresses [principal-stresses-and-strains]. The stresses along these directions are the principal stresses, ordered as σ1σ2σ3\sigma_1 \geq \sigma_2 \geq \sigma_3.

Principal stresses are critical for failure analysis because materials typically fail along planes of maximum normal or shear stress. By identifying principal stresses at critical locations (such as stress concentrations in aircraft wings), engineers apply failure criteria to assess whether the structure will yield or fracture.

Yield Failure Criteria

Two widely used yield criteria extend uniaxial yield strength to multi-axial stress states [yield-failure-criteria]:

The Von Mises criterion predicts yield when the equivalent stress reaches the material's yield strength:

σeq=12[(σ1σ2)2+(σ2σ3)2+(σ3σ1)2]=σy\sigma_{\text{eq}} = \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 the maximum shear stress reaches half the yield strength:

τmax=12max(σ1σ2,σ2σ3,σ3σ1)=σy2\tau_{\max} = \frac{1}{2}\max(|\sigma_1-\sigma_2|, |\sigma_2-\sigma_3|, |\sigma_3-\sigma_1|) = \frac{\sigma_y}{2}

The Von Mises criterion is more commonly used in aircraft structural analysis because it better matches experimental data for ductile materials under complex loading.

Thin-Walled Beam Theory

Aircraft wings and fuselages are often modeled as thin-walled beams. For open sections (such as channels or I-beams), the shear center is the point through which transverse loads must pass to produce pure bending without torsion [shear-center-of-open-thin-walled-beams]. In symmetric beams, the shear center coincides with the centroid; in unsymmetric sections, they differ. Loads applied away from the shear center induce unwanted torsion, which can trigger flutter or fatigue failure.

For closed sections, multi-cell beams are superior to open sections for resisting torsion [thin-walled-multi-cell-beams]. A multi-cell beam consists of multiple closed compartments formed by thin walls. Under torsional or transverse shear loading, shear flow circulates around and through each cell. The main spar of a modern aircraft wing is typically a multi-cell box beam that carries bending, torsion, and shear loads simultaneously. Analyzing these structures requires determining shear flow distribution across cells and calculating the resulting stresses and deflections.

Worked Example: Von Mises Stress in a Wing Spar

Consider a point in an aircraft wing spar where the principal stresses are σ1=200\sigma_1 = 200 MPa, σ2=50\sigma_2 = 50 MPa, and σ3=30\sigma_3 = -30 MPa. The material is aluminum 7075-T6 with yield strength σy=505\sigma_y = 505 MPa.

The Von Mises equivalent stress is:

σeq=12[(20050)2+(50(30))2+((30)200)2]\sigma_{\text{eq}} = \sqrt{\frac{1}{2}[(200-50)^2 + (50-(-30))^2 + ((-30)-200)^2]}

=12[1502+802+2302]= \sqrt{\frac{1}{2}[150^2 + 80^2 + 230^2]}

=12[22500+6400+52900]= \sqrt{\frac{1}{2}[22500 + 6400 + 52900]}

=40900202 MPa= \sqrt{40900} \approx 202 \text{ MPa}

The safety factor is σy/σeq=505/2022.5\sigma_y / \sigma_{\text{eq}} = 505 / 202 \approx 2.5, indicating the structure is safe against yield at this location.

References

AI Disclosure

This article was drafted with AI assistance from class notes (Zettelkasten). All mathematical statements and technical claims are cited to source notes. The worked example is original. The author reviewed all content for technical accuracy and relevance to the course material.

References

AI disclosure: Generated from personal class notes with AI assistance. Every factual claim cites a note. Model: claude-haiku-4-5-20251001.