Torsion of thin-walled open and closed sections

Torsion of thin-walled sections: Bredt–Batho shear flow and torsion constant for closed cells, thin-strip theory for open sections, twist rate, and why closed boxes are so much stiffer.

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Why it matters

Wings, tails and control surfaces are twisted by aerodynamic and inertia torques, and the fuselage is twisted by fin and asymmetric tail loads. Torsional stiffness controls aileron reversal, divergence and flutter speed, while torsional strength sets skin and web thickness. The difference between open and closed thin-walled sections is enormous, often three orders of magnitude in stiffness, and it explains why every primary aircraft structure is built as a closed box.

Key ideas

Closed sections (Bredt–Batho). A torque T on a closed single cell is carried by a shear flow that is constant round the cell, q = T/(2A), A being the area enclosed by the wall mid-line. Constant q means the shear stress τ = q/t is highest where the wall is thinnest. The lever arm of this flow is large (it acts round the whole perimeter), so closed sections are stiff and strong in torsion. The torsion constant is J = 4A²/∮(ds/t), which for constant thickness is 4A²t/S, S being the perimeter.

Open sections (Saint-Venant thin strip). In an open section (channel, I, Z, slit tube) the torque is resisted by shear stress that varies linearly through the wall thickness, zero at the mid-line and maximum at the surfaces. Each wall acts like a thin rectangular strip, so J = Σ s·t³/3 (s = wall length) and τ_max = G·t·dθ/dz = T·t/J at the thickest wall. The internal lever arm is only of the order of t, so J is tiny.

Comparison. For a thin circular tube of radius r and thickness t, closed J = 2πr³t, slit (open) J = 2πrt³/3; the ratio is 3(r/t)². For r/t = 25 that is 1875. Cutting a closed cell (for example a large access hole without reinforcement) destroys most of its torsional stiffness.

Rate of twist. dθ/dz = T/(G·J) for both cases with the appropriate J. The total twist over a length L with uniform section and torque is θ = T·L/(G·J).

Effect of thickness. Doubling t roughly doubles J for a closed section (J ∝ t) but multiplies it by 8 for an open section (J ∝ t³).

Effect of shape. For a given perimeter and thickness, a closed section is stiffest when its enclosed area is largest, so a circle beats a square beats a flat rectangle. A deep, wide wing box is therefore better than a shallow one.

Warping. Non-circular sections warp (cross-sections do not stay plane) under torsion. If warping is free, the formulas above apply. If it is restrained, for example at a wing root fixed to a stiff fuselage, extra direct stresses arise and the effective stiffness of open sections rises sharply; this is beyond the scope of these formulas.

Assumptions. Thin walls, linear elastic, cross-section shape maintained (by ribs or frames), free warping, pure torque.

Formulas

q = T / (2·A) and τ = q / t

  • Closed single cell. T: torque (N·m), A: mid-line enclosed area (m²), q: shear flow (N/m), t: local wall thickness (m), τ: shear stress (Pa).

J = 4·A² / ∮(ds/t)

  • Torsion constant of a closed cell (m⁴); ∮(ds/t) is the sum of (wall length / thickness) round the cell (dimensionless).

dθ/dz = T / (G·J) and θ = T·L / (G·J)

  • dθ/dz: rate of twist (rad/m), G: shear modulus (Pa), θ: total twist (rad) over length L (m).

J = Σ s·t³ / 3

  • Open thin-walled section; s: length of each wall (m), t: its thickness (m).

τ_max = G·t·(dθ/dz) = T·t / J

  • Open section: maximum shear stress at the surface of a wall of thickness t.

J = π·(D⁴ − d⁴) / 32

  • Exact polar second moment of a circular tube (m⁴), D and d outer and inner diameters (m).

Worked examples

Example 1 (standard): closed tube versus slit tube. Given: thin circular tube, mean radius r = 50 mm, t = 2 mm, G = 27 GPa, T = 50 N·m. Compare the closed tube with the same tube slit along its length.

  1. Closed: A = π·r² = π × 50² = 7854 mm², q = T/(2·A) = 50 000/(2 × 7854) = 3.183 N/mm, τ = q/t = 1.59 N/mm².
  2. Closed: J = 2·π·r³·t = 2π × 125 000 × 2 = 1.571 × 10⁶ mm⁴; dθ/dz = T/(G·J) = 50 000/(27 000 × 1.571 × 10⁶) = 1.18 × 10⁻⁶ rad/mm.
  3. Slit: J = 2·π·r·t³/3 = 2π × 50 × 8/3 = 837.8 mm⁴; τ_max = T·t/J = 50 000 × 2/837.8 = 119.4 N/mm²; dθ/dz = 50 000/(27 000 × 837.8) = 2.21 × 10⁻³ rad/mm.
  4. Stiffness ratio: J_closed/J_open = 3·(r/t)² = 3 × 25² = 1875. Answer: closed: τ = 1.59 MPa, twist 1.18 × 10⁻⁶ rad/mm; slit: τ = 119 MPa, twist 2.21 × 10⁻³ rad/mm, 1875 times more flexible.

Example 2 (GATE level): wing box with different wall thicknesses. Given: rectangular box, mid-line width 400 mm and depth 150 mm, top and bottom skins t = 1.2 mm, spar webs t = 2.5 mm, T = 20 kN·m, G = 27 GPa.

  1. A = 400 × 150 = 60 000 mm²; q = T/(2·A) = 20 × 10⁶/120 000 = 166.7 N/mm.
  2. Stresses: skins τ = 166.7/1.2 = 138.9 N/mm²; webs τ = 166.7/2.5 = 66.7 N/mm².
  3. ∮ds/t = 2 × 400/1.2 + 2 × 150/2.5 = 666.7 + 120 = 786.7.
  4. J = 4·A²/∮(ds/t) = 4 × 60 000²/786.7 = 1.831 × 10⁷ mm⁴.
  5. dθ/dz = T/(G·J) = 20 × 10⁶/(27 000 × 1.831 × 10⁷) = 4.05 × 10⁻⁵ rad/mm = 2.32°/m. Answer: q = 166.7 N/mm; τ = 138.9 MPa in skins, 66.7 MPa in webs; twist rate 4.05 × 10⁻⁵ rad/mm (2.32°/m).

Common mistakes

  • Using the outer-contour area or the solid cross-section area for A in Bredt's formula instead of the mid-line enclosed area.
  • Assuming shear stress is constant round a closed cell of varying thickness. Shear flow is constant; stress is not.
  • Using J = π·D⁴/32 or other solid-section results for thin-walled open sections.
  • Forgetting the cube in J = Σst³/3, or applying it to closed sections.
  • Taking the closed-section result for a section with an unreinforced cut or slit.
  • Using degrees in T·L/(GJ); the result is in radians.

For GATE AE

Expect Bredt–Batho shear flow and stress, twist rate of a closed box or tube with varying thickness, J and maximum stress of open sections, and the closed-versus-open stiffness ratio. Practise ∮ds/t for multi-thickness boxes and quick J = Σst³/3 for channels and I-sections.

Quick check

  1. A closed box has enclosed area 0.04 m² and carries 2 kN·m. What is q?
  2. A channel has two 50 mm flanges and a 100 mm web, all 2 mm thick. What is J?
  3. Doubling the wall thickness of an open section multiplies its torsional stiffness by what factor?
  4. In a closed cell with walls of 1 mm and 2 mm, which has the higher shear stress under torque? Answers: 1. q = 2000/0.08 = 25 kN/m; 2. J = (50 + 50 + 100) × 2³/3 = 533 mm⁴; 3. 8; 4. the 1 mm wall (twice the stress).

Interview questions

All Aircraft Structures interview questions

Try answering each one aloud before you open it.

  1. 1.What is torsion in the context of aircraft structures?Concept

    Torsion is twisting of a structural member by a torque about its longitudinal axis. In aircraft it arises when the resultant lift or inertia load does not pass through the flexural axis of the wing, from control-surface and fin loads, and from engine torque. It produces shear flow in skins and spar webs and a twist dθ/dz = T/(GJ), and wing twist directly affects aeroelastic behaviour such as divergence, aileron reversal and flutter.

  2. 2.Explain the difference between open and closed thin-walled sections in aircraft structures.Concept

    Open thin-walled sections have a cross-section that is not fully enclosed, like an I-beam, while closed thin-walled sections are fully enclosed, like a tube. Closed sections generally have higher torsional stiffness and strength compared to open sections, making them more effective in resisting torsional loads.

  3. 3.Why are closed sections preferred over open sections in aircraft structures for torsion resistance?Application

    Closed sections are preferred because they provide greater torsional stiffness and strength. This is due to the enclosed geometry, which allows the section to better resist twisting and distribute the stresses more evenly across the structure, enhancing the overall structural integrity.

  4. 4.What happens to the torsional stiffness of a thin-walled section if the wall thickness is doubled?Application

    For a closed section J = 4A²/∮(ds/t), so doubling t everywhere roughly doubles J and the torsional stiffness GJ (A changes only slightly). For an open section J = Σs·t³/3, so doubling t multiplies J by eight. Even so, an open section remains far less stiff than a closed one of similar size, because its internal lever arm is only of the order of t.

  5. 5.How does the shape of a cross-section affect its torsional properties?Concept

    The shape of a cross-section significantly affects its torsional properties. For example, circular sections are very efficient in resisting torsion due to their symmetry, while non-circular sections like rectangles or I-beams may have lower torsional stiffness and strength. The distribution of material around the axis of twist is crucial in determining the torsional characteristics.

  6. 6.Explain how shear flow is distributed in a closed thin-walled section under torsion.Concept

    In a closed thin-walled section under torsion, shear flow is distributed uniformly around the section. This uniform distribution helps in resisting the applied torque effectively, as the shear stresses are balanced throughout the section, minimizing the risk of structural failure.

  7. 7.What is the effect of a cut or opening in a closed section on its torsional strength and stiffness?Application

    A longitudinal cut turns the closed cell into an open section: torque can no longer be carried by a constant shear flow round the cell, only by through-thickness stress gradients in each wall. For a thin tube the stiffness drops by a factor of 3(r/t)², often more than a thousand, and the stresses for the same torque rise by a similar order. Access holes in wing boxes are therefore framed with reinforced edges or covered by load-carrying panels so the shear path round the cell stays continuous.

  8. 8.Calculate the torsional stiffness of a circular tube with an outer diameter of 0.1 m, an inner diameter of 0.08 m and a shear modulus of 25 GPa.Numerical

    The polar second moment of a circular tube is J = π(D⁴ − d⁴)/32 = π × (0.1⁴ − 0.08⁴)/32 = π × 5.904 × 10⁻⁵/32 = 5.796 × 10⁻⁶ m⁴. The torsional stiffness per unit length is GJ = 25 × 10⁹ × 5.796 × 10⁻⁶ = 1.45 × 10⁵ N·m², so a 1 m length twists 1 rad per 145 kN·m of torque.

  9. 9.For a closed thin-walled rectangular section of fixed perimeter and thickness, how does increasing the aspect ratio affect torsional stiffness?Application

    With constant thickness J = 4A²t/S, where S is the perimeter. At a fixed perimeter the enclosed area A is largest for a square and falls as the rectangle becomes more elongated, so J falls with A². For example, at the same perimeter a 2:1 rectangle has 8/9 of the square's area and about 79% of its stiffness, and a very flat box approaches the poor stiffness of an open strip.

  10. 10.Determine the angle of twist per unit length for a thin-walled circular tube with a torque of 500 Nm, a shear modulus of 30 GPa, and a polar moment of inertia of 0.0002 m⁴.Numerical

    The angle of twist per unit length (θ') is given by θ' = T / (GJ), where T is the torque, G is the shear modulus, and J is the polar moment of inertia. Substituting the given values: T = 500 Nm, G = 30 GPa = 30 × 10^9 Pa, J = 0.0002 m⁴. θ' = 500 / (30 × 10^9 × 0.0002) = 500 / 6 × 10^6 = 8.33 × 10^-5 rad/m.

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