Tension field beams

Wagner tension field beams: post-buckling diagonal tension in thin webs, stiffener and flange loads, the tension-field angle, and failure modes.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

A thin spar web buckles in shear at a stress that is a small fraction of its strength. Instead of thickening the web to prevent buckling, aircraft designers often let it buckle and carry the shear as diagonal tension, with vertical stiffeners and the flanges providing the compression struts. This tension field (Wagner) beam is much lighter than a shear-resistant web, and it is used in spars, ribs and fuselage floor beams. Its design changes the loads in the flanges and stiffeners, which must be sized for them.

Key ideas

Shear-resistant versus tension field webs. Pure shear τ on a web is equivalent to equal tension and compression of magnitude τ on planes at 45°. A thick (shear-resistant) web carries both. A thin web buckles when τ reaches τ_cr; beyond that the compressive diagonal can carry hardly any more, and additional load is carried by the tension diagonal alone. The buckles form inclined folds running along the tension direction.

Complete (ideal) tension field (Wagner). Assume the web carries no compression at all, only a uniform diagonal tension σ_t at angle α to the flanges. Equilibrium of a web element then gives σ_t = 2·τ/sin 2α, where τ = S/(t·d) is the nominal web shear stress. At α = 45°, σ_t = 2τ: the web carries twice the tension of a shear-resistant web, but no compression.

Load on stiffeners. The vertical component of the diagonal tension pulls the flanges together; vertical stiffeners at spacing b hold them apart and carry a compressive load P = S·b·tan α/d. They must be checked for column buckling (often with an effective length of about d/2 for double stiffeners, take from data sheets).

Load on flanges. The horizontal component of the tension field adds a compressive force S·cot α in total, shared equally between the two flanges (S/(2·tan α) each), on top of the bending loads ±M/d. The pull of the tension field between stiffeners also bends the flanges as continuous beams on the stiffeners, with a maximum secondary bending moment S·tan α·b²/(12·d) at the stiffeners.

Angle of the tension field. With rigid flanges and stiffeners α = 45°. Flexible members let the field rotate; neglecting flange bending, energy minimisation gives tan⁴α = (1 + t·d/(2·A_F))/(1 + t·b/A_S), where A_F is the area of each flange and A_S of each stiffener. In practice α is usually between about 38° and 45°.

Incomplete tension field. Real webs are between the two extremes: part of the shear is carried as pure shear (up to τ_cr) and part as tension field. Semi-empirical methods (NACA/Kuhn, in design data sheets) blend them through a diagonal-tension factor. Use them for real design; the complete tension field is the conservative bound for stiffener and flange loads.

Consequences and failure modes. Permanent buckles are acceptable only in certain areas (not in aerodynamic surfaces at cruise, not where fatigue under repeated buckling is critical). Failure can be by web rupture in tension, stiffener column buckling or forced crippling, flange failure under combined compression and secondary bending, or rivet failure along the flange and stiffener lines, which carry the tension-field pull as well as shear.

Formulas

τ = S / (t·d)

  • τ: nominal web shear stress (Pa), S: shear force (N), t: web thickness (m), d: depth between flange centroids (m).

σ_t = 2·τ / sin 2α = 2·S / (t·d·sin 2α)

  • σ_t: diagonal tension stress (Pa), α: angle of the tension field to the flanges. Complete tension field.

P = S·b·tan α / d

  • P: compressive load in each vertical stiffener (N), b: stiffener spacing (m).

F_T, F_B = ±M/d − S / (2·tan α)

  • Axial loads in the top and bottom flanges (N), tension positive; M: bending moment at the section (N·m).

M_max = S·tan α·b² / (12·d)

  • Maximum secondary bending moment in each flange (N·m), at the stiffeners.

tan⁴α = (1 + t·d / (2·A_F)) / (1 + t·b / A_S)

  • A_F: area of each flange (m²), A_S: area of each stiffener (m²); flange bending stiffness neglected.

Worked examples

Example 1 (standard): complete tension field at 45°. Given: web depth d = 400 mm, t = 1.5 mm, stiffener spacing b = 300 mm, shear S = 20 kN, rigid flanges and stiffeners (α = 45°).

  1. Nominal shear stress: τ = S/(t·d) = 20 000/(1.5 × 400) = 33.3 N/mm².
  2. Diagonal tension: σ_t = 2·τ/sin 90° = 66.7 N/mm².
  3. Stiffener load: P = S·b·tan 45°/d = 20 000 × 300/400 = 15 000 N (compression).
  4. Extra flange compression: S/(2·tan 45°) = 10 000 N in each flange, added to ±M/d. Answer: τ = 33.3 MPa, σ_t = 66.7 MPa, stiffener load 15.0 kN, extra flange compression 10.0 kN each.

Example 2 (GATE level): finding α and the member loads. Given: the same beam, but flexible members: each flange A_F = 400 mm², each stiffener A_S = 150 mm².

  1. t·d/(2·A_F) = 1.5 × 400/800 = 0.75; t·b/A_S = 1.5 × 300/150 = 3.0.
  2. tan⁴α = (1 + 0.75)/(1 + 3.0) = 0.4375, so tan α = 0.8133 and α = 39.1°.
  3. sin 2α = sin 78.2° = 0.9789; σ_t = 2 × 20 000/(1.5 × 400 × 0.9789) = 68.1 N/mm².
  4. Stiffener load: P = 20 000 × 300 × 0.8133/400 = 12 200 N.
  5. Extra flange compression: S/(2·tan α) = 20 000/(2 × 0.8133) = 12 300 N each.
  6. Flange secondary bending: M_max = S·tan α·b²/(12·d) = 20 000 × 0.8133 × 300²/(12 × 400) = 3.05 × 10⁵ N·mm. Answer: α = 39.1°, σ_t = 68.1 MPa, stiffener 12.2 kN, flange compression 12.3 kN each, flange M_max = 305 N·m. The flexible stiffeners let the field flatten, which lowers the stiffener load and raises the flange load.

Common mistakes

  • Using σ_t = τ; in a complete tension field at 45° it is 2τ.
  • Forgetting to multiply by the stiffener spacing b in the stiffener load.
  • Treating the tension-field flange load as tension; it is compressive in both flanges.
  • Assuming α = 45° when stiffeners and flanges are flexible.
  • Comparing τ with the tensile yield stress to judge buckling; buckling is set by τ_cr from plate theory, while web rupture is set by σ_t against the tensile allowable.
  • Ignoring secondary bending of the flanges between stiffeners.

For GATE AE

Expect conceptual questions on why thin webs are allowed to buckle and what carries the load afterwards, and short numericals on σ_t = 2S/(t·d·sin 2α), stiffener load S·b·tan α/d and the extra flange compression. Practise the tan⁴α formula and checking stiffeners as columns.

Quick check

  1. In a complete tension field at 45°, how does the diagonal tension compare with the nominal shear stress?
  2. Which members are put into compression by the tension field?
  3. If stiffener spacing doubles, what happens to each stiffener's load?
  4. Why are tension field webs avoided on some external surfaces? Answers: 1. σ_t = 2τ; 2. the vertical stiffeners and both flanges; 3. it doubles; 4. permanent buckles spoil the aerodynamic surface and repeated buckling causes fatigue.

Interview questions

All Aircraft Structures interview questions

Try answering each one aloud before you open it.

  1. 1.What is a tension field beam in the context of aircraft structures?Concept

    A tension field beam is a structural component used in aircraft that relies on tension fields to carry shear loads. It consists of a thin web that can buckle under compressive loads, allowing the tension field to develop and carry the shear loads efficiently. This design helps in reducing the weight of the structure while maintaining its strength.

  2. 2.Explain how tension field action works in a beam.Concept

    Tension field action occurs when the web of a beam buckles under compressive stress, allowing diagonal tension fields to form. These tension fields carry the shear loads across the web, effectively redistributing the loads to the flanges or stiffeners. This mechanism allows the beam to maintain its load-carrying capacity even after the web has buckled.

  3. 3.Why are tension field beams used in aircraft structures?Application

    Tension field beams are used in aircraft structures because they provide an efficient way to carry shear loads while minimizing weight. The ability to allow the web to buckle and still carry loads through tension fields means that less material is needed, which is crucial in aircraft design where weight savings are essential for performance and fuel efficiency.

  4. 4.What happens if the web of a tension field beam does not buckle?Application

    Then it simply works as a shear-resistant web, carrying pure shear τ = S/(t·d) with equal diagonal tension and compression, and the stiffeners and flanges carry no tension-field loads. That is not unsafe: the web is merely heavier than it needed to be. Tension field design is a weight-saving choice that accepts buckling, not a mechanism that requires it.

  5. 5.How does the presence of stiffeners affect the performance of a tension field beam?Application

    Stiffeners in a tension field beam help to control the buckling of the web and provide additional paths for load transfer. They ensure that the web buckles in a predictable manner, allowing the tension fields to develop effectively. Stiffeners also help to distribute loads more evenly and can increase the overall strength and stability of the beam.

  6. 6.Describe the role of flanges in a tension field beam.Concept

    Flanges in a tension field beam provide the primary paths for axial load transfer and help to stabilize the web. They work in conjunction with the tension fields to carry shear loads and prevent excessive deformation. The flanges also help to anchor the tension fields, ensuring that the loads are effectively transferred across the beam.

  7. 7.What are the potential failure modes of a tension field beam?Application

    Web rupture when the diagonal tension σ_t = 2S/(t·d·sin 2α) reaches the tensile allowable; column buckling or forced crippling of the vertical stiffeners under P = S·b·tan α/d; flange failure under the combined bending load, the extra compression S/(2 tan α) and secondary bending between stiffeners; and rivet failure along flanges and stiffeners. Repeated buckling can also cause fatigue cracking at the fold lines.

  8. 8.Estimate the shear capacity of a complete tension field web 2 mm thick and 1 m deep if the tensile allowable of the web material is 250 MPa and the field is at 45°.Numerical

    In a complete tension field σ_t = 2τ/sin 2α, which at 45° gives σ_t = 2τ. Web rupture occurs when σ_t = 250 MPa, so τ = 125 MPa and S = τ·t·d = 125 N/mm² × 2 mm × 1000 mm = 250 kN. The stiffeners and flanges must also be checked for the loads this shear produces.

  9. 9.A web 1 m deep and 3 mm thick carries 600 kN shear. Will it buckle, and can it carry the load as a complete tension field if the tensile allowable is 300 MPa?Numerical

    Nominal shear τ = 600 000/(3 × 1000) = 200 MPa. With stiffeners at, say, 300 mm, the elastic shear buckling stress k_s·π²E/(12(1 − ν²))·(t/b)² is only of the order of 30 MPa for aluminium, so the web buckles long before 200 MPa. As a tension field at 45°, σ_t = 2τ = 400 MPa, above the 300 MPa allowable, so the web is overstressed and must be thickened to at least 4 mm.

  10. 10.What design considerations must be taken into account when using tension field beams in aircraft?Application

    Design considerations for tension field beams in aircraft include ensuring that the web can buckle predictably, selecting appropriate materials for weight and strength, and incorporating stiffeners and flanges to stabilize the structure. Engineers must also consider load paths, potential failure modes, and the overall integration of the beam within the aircraft's structural framework to ensure safety and performance.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousBuckling of thin plates and stiffened panelsNext Composite laminae, laminates and classical lamination theory