Shear flow in closed single-cell sections

Shear flow in closed single-cell thin-walled sections: the cut-and-close method, the redundant constant flow, Bredt–Batho torsion, rate of twist and offset loads.

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Why it matters

The wing box, the tailplane box and the fuselage shell are closed thin-walled cells. They carry the vertical shear from lift and the torque from the aerodynamic centre being off the flexural axis, and their skins and spar webs are sized by the resulting shear flow. Closed sections differ from open ones in one key way: the shear flow cannot be found from statics alone, and handling that one redundant is the whole of this topic.

Key ideas

Why a closed cell is different. In an open section q = 0 at free edges, so integrating from a free edge gives q everywhere. A closed cell has no free edge, so the shear flow at any chosen starting point is unknown. A closed single cell is therefore once statically indeterminate in shear.

The cut-and-close method.

  1. Make an imaginary cut in the wall to create an open section. Calculate the "basic" open-section shear flow q_b from the open-section formula, with q_b = 0 at the cut.
  2. Add an unknown constant shear flow q_s,0 all round the cell (a constant flow round a closed loop is self-equilibrating in force; it only produces a torque 2A·q_s,0).
  3. The total is q_s = q_b + q_s,0.
  4. Find q_s,0 from moment equilibrium: the moment of q_s about any point must equal the moment of the applied shear loads about the same point.
  5. If the shear centre is wanted (load position unknown), use instead the condition of zero twist: ∮ q_s/(G·t) ds = 0.

Symmetry shortcut. If the section is symmetric about the axis along which the shear acts and the load acts along that axis (through the shear centre), the shear flow is zero where the axis of symmetry cuts the walls. Cutting there gives q_s,0 = 0 directly.

Pure torsion (Bredt–Batho). A torque T on a closed single cell produces a constant shear flow q = T/(2A), where A is the area enclosed by the wall mid-line, whatever the wall thickness distribution. The shear stress τ = q/t is largest where the wall is thinnest. The rate of twist is dθ/dz = (T/(4A²G))·∮ds/t.

Superposition. A shear load not acting through the shear centre = the same load through the shear centre (bending shear flow, no twist) + a torque equal to load × offset (constant Bredt flow). This is the most efficient way to solve many closed-cell problems.

Booms. In idealised sections (topic on structural idealisation), the open-section flow jumps by −(S_y/I_xx)·B_r·y_r at each boom of area B_r; the method is otherwise identical.

Assumptions. Thin walls, linear elastic, cross-section shape maintained (by ribs or frames), direct stresses from the bending theory, no restraint of warping.

Formulas

q_s = q_b + q_s,0

  • q_s: total shear flow (N/m), q_b: open-section shear flow from the cut (N/m), q_s,0: constant shear flow at the cut (N/m).

S_x·η₀ − S_y·ξ₀ = ∮ p·q_b ds + 2·A·q_s,0

  • Moment equilibrium about a point O. S_x, S_y: applied shear loads (N) at (ξ₀, η₀) measured from O (m); p: perpendicular distance from O to the tangent of the wall (m); A: enclosed mid-line area (m²). Take the anticlockwise sense as positive consistently for moments and q.

q = T / (2·A)

  • T: torque (N·m), A: enclosed area (m²). Pure torsion of a closed single cell.

τ = q / t

  • Shear stress (Pa) in a wall of thickness t (m).

dθ/dz = (1 / (2·A·G))·∮ q_s/t ds

  • Rate of twist (rad/m), G: shear modulus (Pa). For pure torsion it becomes dθ/dz = T·∮(ds/t) / (4·A²·G).

∮ q_s / t ds = 0

  • Condition for a shear load through the shear centre (no twist).

Worked examples

Example 1 (standard): symmetric box under shear through the shear centre. Given: rectangular box, width b = 300 mm, depth h = 150 mm (mid-line), uniform t = 2 mm, vertical shear S_y = 30 kN on the vertical axis of symmetry.

  1. I_xx = 2·t·h³/12 + 2·b·t·(h/2)² = 2 × 2 × 150³/12 + 2 × 300 × 2 × 75² = 1.125 × 10⁶ + 6.75 × 10⁶ = 7.875 × 10⁶ mm⁴.
  2. By symmetry q = 0 at the middle of the top and bottom walls; cut there so q_s,0 = 0. S_y/I_xx = 30 000/7.875 × 10⁶ = 3.810 × 10⁻³ N/mm⁴.
  3. Top wall, from the middle to a corner: q = (S_y/I_xx)·t·s·(h/2). At the corner (s = 150 mm): q = 3.810 × 10⁻³ × 2 × 150 × 75 = 85.7 N/mm.
  4. Web, at the neutral axis: q = 85.7 + 3.810 × 10⁻³ × 2 × 75²/2 = 85.7 + 21.4 = 107.1 N/mm.
  5. Check: each web carries 85.7 × 150 + 3.810 × 10⁻³ × 2 × (75² × 150 − 150³/12)/2 = 12 857 + 2143 = 15 000 N; two webs give 30 kN. Answer: q = 85.7 N/mm at the corners, 107.1 N/mm at mid-depth of each web; τ_max = 53.6 MPa.

Example 2 (GATE level): same box, load along the left web. Given: as Example 1, but S_y = 30 kN acts upward along the left web, i.e. 150 mm left of the shear centre. G = 27 GPa.

  1. Equivalent system: S_y through the shear centre (Example 1) plus a torque T = S_y × 150 = 4.5 × 10⁶ N·mm, clockwise.
  2. Enclosed area: A = 300 × 150 = 45 000 mm². Bredt flow: q_T = T/(2·A) = 4.5 × 10⁶/90 000 = 50.0 N/mm, clockwise: up the left web, down the right web.
  3. Left web at mid-depth: q = 107.1 + 50.0 = 157.1 N/mm, τ = 157.1/2 = 78.6 N/mm². Right web: q = 107.1 − 50.0 = 57.1 N/mm.
  4. Check vertical equilibrium: left web carries 15 000 + 50 × 150 = 22 500 N up, right web 15 000 − 7500 = 7500 N up; total 30 000 N.
  5. Rate of twist from the torque: ∮ds/t = 2 × (300 + 150)/2 = 450, so dθ/dz = T·∮(ds/t)/(4·A²·G) = 4.5 × 10⁶ × 450/(4 × 45 000² × 27 000) = 9.26 × 10⁻⁶ rad/mm. Answer: left web q_max = 157.1 N/mm (τ = 78.6 MPa), right web 57.1 N/mm; twist 9.26 × 10⁻⁶ rad/mm (0.53°/m).

Common mistakes

  • Using q = SQ/I on a closed section without the constant q_s,0, unless symmetry justifies the cut position.
  • Using the enclosed area of the outer surface or the material area instead of the mid-line enclosed area in 2A·q_s,0 and q = T/(2A).
  • Confusing a shear force with a torque in Bredt's formula; q = T/(2A) only for torque.
  • Mixing up the sense of q_b, q_s,0 and moments; fix anticlockwise positive throughout.
  • Forgetting that a thinner wall carries the same Bredt flow but higher stress.
  • Ignoring the extra torque when the shear does not pass through the shear centre.

For GATE AE

Common items: Bredt–Batho shear flow and stress, rate of twist of a closed tube or box, shear flow in a symmetric box under vertical shear, and superposition of bending and torsion flows for an offset load. Practise the cut-and-close method on a two-boom or four-boom box and the moment equation about a corner.

Quick check

  1. Why is a closed single-cell section statically indeterminate in shear?
  2. A box with enclosed area 0.05 m² carries 10 kN·m torque. What is q?
  3. Where is the shear flow zero in a doubly symmetric box under vertical shear?
  4. Which wall has the highest Bredt shear stress? Answers: 1. there is no free edge, so the flow at any starting point is unknown; 2. q = 10 000/(2 × 0.05) = 100 kN/m; 3. at the mid-points of the top and bottom walls; 4. the thinnest one.

Interview questions

All Aircraft Structures interview questions

Try answering each one aloud before you open it.

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

    Shear flow q is the shear force per unit length of a thin wall's mid-line, q = τ·t, in N/m or N/mm. It is used for thin-walled skins and webs because it is continuous round the wall and balances at junctions, and its line integral over a wall gives that wall's force. In wing and fuselage cells it comes from both transverse shear (bending) and torque.

  2. 2.Explain the significance of shear flow in closed single-cell sections of an aircraft.Concept

    In a closed cell such as a wing box or fuselage, shear flow carries both the vertical shear from lift and the torque, with torque carried very efficiently as a constant flow q = T/(2A). Because there is no free edge, the flow at a cut is an unknown constant, so the cell is once statically indeterminate and needs a moment equation (or the zero-twist condition) to solve. The resulting skin and web flows set the skin thickness, rivet pitch and shear buckling checks.

  3. 3.How is shear flow calculated in a closed single-cell section?Concept

    Cut the wall to make it an open section and find the open-section flow q_b, which is zero at the cut. Add an unknown constant flow q_s,0 round the cell, so q_s = q_b + q_s,0, and find q_s,0 by equating the moment of the shear flows about a convenient point to the moment of the applied load: S_x·η₀ − S_y·ξ₀ = ∮p·q_b ds + 2A·q_s,0. If the load passes along an axis of symmetry, cutting on that axis gives q_s,0 = 0.

  4. 4.Why are closed single-cell sections preferred in aircraft structures?Application

    A closed cell carries torque through a constant shear flow q = T/(2A) with a lever arm set by the enclosed area, so its torsional stiffness is orders of magnitude higher than that of an open section of the same material. That keeps wing twist small, which matters for aileron effectiveness, divergence and flutter. The same closed shell also carries bending and pressure efficiently, and it fits naturally inside an aerofoil or fuselage contour.

  5. 5.What happens if there is an opening in a closed single-cell section?Application

    An opening in a closed single-cell section can significantly reduce its torsional rigidity and strength. This is because the closed path for shear flow is interrupted, leading to stress concentrations around the opening. It may require additional reinforcement or design modifications to maintain structural integrity.

  6. 6.How does shear flow affect the design of aircraft wings?Application

    Shear flow affects the design of aircraft wings by dictating how shear forces are distributed along the wing's cross-section. Proper calculation and distribution of shear flow ensure that the wing can withstand aerodynamic loads without excessive deformation or failure. It influences the choice of materials, thickness, and structural reinforcements used in the wing design.

  7. 7.How is shear flow related to buckling of skin panels and spar webs?Application

    The shear flow in a panel divided by its thickness is the shear stress that must be compared with the panel's critical shear buckling stress, τ_cr = k_s·π²E/(12(1 − ν²))·(t/b)². High shear flow in thin skins can cause diagonal shear buckling; this is accepted in some webs (tension field beams), while elsewhere stiffeners or thicker skin are added to keep τ below τ_cr. Shear flow does not prevent buckling; it is the load that causes it.

  8. 8.In a closed box section loaded by vertical shear along its axis of symmetry, the shear force is 5000 N, the first moment about the neutral axis of the wall area from the symmetry cut to the point is 0.02 m³ and I = 0.005 m⁴. What is the shear flow at the point?Numerical

    Because the load acts along the axis of symmetry, the shear flow is zero where that axis cuts the walls, so the open-section formula started there needs no constant correction. q = S·Q/I = 5000 × 0.02/0.005 = 20 000 N/m. Without symmetry, a constant q_s,0 from the moment equation would have to be added.

  9. 9.What design considerations are important when dealing with shear flow in composite materials used in aircraft?Application

    When dealing with shear flow in composite materials, it is important to consider the anisotropic nature of composites, which means their strength and stiffness vary with direction. The layup sequence, fiber orientation, and matrix material must be carefully chosen to ensure that the shear flow is effectively managed. Additionally, the bonding between layers and the potential for delamination under shear loads must be addressed.

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