Multi-cell box beams

Why it matters

Most real wings are multi-spar: a leading-edge D-nose cell, the main box between front and rear spars, sometimes a third spar or a trailing-edge cell. Each interior spar web splits the section into another closed cell. The cells share the torque and shear according to their stiffness, and finding that split is needed to size every skin panel and spar web, and to check fail-safety if one spar is damaged.

Key ideas

Indeterminacy. A section with N closed cells has N unknown constant shear flows (one per cell) under torsion or shear. Moment equilibrium gives one equation; the other N − 1 come from compatibility: every cell must twist by the same rate dθ/dz, because ribs or frames keep the cross-section shape.

Sign convention. Give each cell its own constant flow q_R, positive anticlockwise. An outer wall of cell R carries q_R. An interior web shared by cells R and R+1 carries the difference q_R − q_(R+1) in the sense of cell R.

Pure torsion.

• Equilibrium: T = Σ 2·A_R·q_R, the sum of the Bredt torques of all cells.
• Compatibility for cell R: dθ/dz = (1/(2·A_R·G))·∮_R q ds/t, where the integral runs round cell R only, using the actual (net) flow in each wall.
• For two cells this gives two equations in q₁ and q₂.

Physical results.

• Two identical cells under pure torque carry equal flows, so the common web carries no torsional shear flow; the section behaves as a single cell. Interior webs add little torsional stiffness: the outer contour does most of the work.
• In unequal cells the interior web carries the small difference q₁ − q₂.
• Interior webs matter for vertical shear (they carry part of it), for panel buckling, for crushing loads, and for fail-safety.

Shear (bending) in multi-cell sections. Cut one wall in every cell to make an open section, find the open-section flow q_b, add an unknown constant q_s,0,R to each cell, and use N compatibility conditions (zero twist if the load passes through the shear centre, or a common twist rate otherwise) plus the moment equation. The method is the single-cell method repeated, and with booms the open-section flow jumps at each boom.

Torsion constant. Once the flows for a torque T are known, J = T/(G·dθ/dz).

Assumptions. Thin walls, linear elastic, cross-sections held in shape by ribs, free warping, same G throughout (otherwise use G·t in place of t).

Formulas

T = Σ 2·A_R·q_R

• T: torque (N·m), A_R: mid-line area enclosed by cell R (m²), q_R: constant shear flow of cell R (N/m), anticlockwise positive.

dθ/dz = (1 / (2·A_R·G))·∮_R (q / t) ds

• Compatibility for each cell R. dθ/dz: rate of twist (rad/m), G: shear modulus (Pa), q: net flow in each wall of cell R, t: wall thickness (m).

dθ/dz = [q₁·(δ₁ + δ₁₂) − q₂·δ₁₂] / (2·A₁·G) and dθ/dz = [q₂·(δ₂ + δ₁₂) − q₁·δ₁₂] / (2·A₂·G)

• Two cells. δ₁, δ₂: Σ(length/t) of the outer walls of cells 1 and 2 (dimensionless); δ₁₂: length/t of the shared web.

q_web = q₁ − q₂

• Net shear flow in the shared web, in the sense of cell 1.

J = T / (G·dθ/dz)

• Torsion constant (m⁴).

Worked examples

Example 1 (standard): two equal cells. Given: box of mid-line width 600 mm and depth 200 mm, split by a central web into two cells 300 × 200 mm. Skins t = 1.5 mm, all three webs t = 2.0 mm, G = 27 GPa, T = 30 kN·m.

1. By symmetry q₁ = q₂ = q. The central web carries q₁ − q₂ = 0.
2. Equilibrium: T = 2·A₁·q + 2·A₂·q = 2 × 120 000 × q, so q = 30 × 10⁶/240 000 = 125 N/mm.
3. Skin stress: τ = 125/1.5 = 83.3 N/mm²; outer web stress τ = 125/2.0 = 62.5 N/mm².
4. Twist rate, cell 1: δ₁ = 2 × 300/1.5 + 200/2.0 = 500, δ₁₂ = 100; dθ/dz = [125 × 600 − 125 × 100]/(2 × 60 000 × 27 000) = 62 500/3.24 × 10⁹ = 1.93 × 10⁻⁵ rad/mm.
5. This equals the single-cell 600 × 200 box without the central web: q = 125 N/mm, ∮ds/t = 1000, dθ/dz = T·∮(ds/t)/(4A²G) = 1.93 × 10⁻⁵ rad/mm. Answer: q = 125 N/mm in all outer walls, zero in the central web; τ = 83.3 MPa (skins); dθ/dz = 1.93 × 10⁻⁵ rad/mm.

Example 2 (GATE level): two unequal cells. Given: same box but the interior web placed 200 mm from the front, giving cell 1 of 200 × 200 mm (A₁ = 40 000 mm²) and cell 2 of 400 × 200 mm (A₂ = 80 000 mm²). Same thicknesses, G = 27 GPa, T = 30 kN·m.

1. Wall terms: δ₁ = 2 × 200/1.5 + 200/2.0 = 366.7, δ₂ = 2 × 400/1.5 + 200/2.0 = 633.3, δ₁₂ = 200/2.0 = 100.
2. Equal twist: [q₁·466.7 − q₂·100]/(2 × 40 000) = [q₂·733.3 − q₁·100]/(2 × 80 000). Multiply by 160 000: 2·(466.7·q₁ − 100·q₂) = 733.3·q₂ − 100·q₁, so 1033.3·q₁ = 933.3·q₂ and q₁ = 0.9032·q₂.
3. Equilibrium: 2 × 40 000 × q₁ + 2 × 80 000 × q₂ = 30 × 10⁶, i.e. 80 000 × 0.9032·q₂ + 160 000·q₂ = 30 × 10⁶, so q₂ = 30 × 10⁶/232 258 = 129.2 N/mm and q₁ = 116.7 N/mm.
4. Interior web: q₁ − q₂ = −12.5 N/mm (12.5 N/mm in the sense of cell 2).
5. Twist: dθ/dz = [116.7 × 466.7 − 129.2 × 100]/(80 000 × 27 000) = 41 528/2.16 × 10⁹ = 1.92 × 10⁻⁵ rad/mm. Check with cell 2: [129.2 × 733.3 − 116.7 × 100]/(160 000 × 27 000) = 1.92 × 10⁻⁵ rad/mm.
6. Check equilibrium: 2 × (40 000 × 116.7 + 80 000 × 129.2) = 30.0 × 10⁶ N·mm. Answer: q₁ = 116.7 N/mm, q₂ = 129.2 N/mm, interior web 12.5 N/mm; dθ/dz = 1.92 × 10⁻⁵ rad/mm (1.10°/m).

Common mistakes

• Using the cell's own flow in the shared web instead of the difference q₁ − q₂.
• Including walls of the neighbouring cell in ∮_R, or leaving out the shared web.
• Using T = 2·A·q with the total area and one flow, as if the section were a single cell, when cells are unequal.
• Changing the positive direction of q between cells.
• Assuming interior webs greatly increase torsional stiffness; they mostly help shear, buckling and fail-safety.
• Forgetting that ribs are needed to keep the cross-section shape for the equal-twist assumption to hold.

For GATE AE

Expect two-cell torsion: shear flows in each cell and in the interior web, the rate of twist, and the torque from given flows. Conceptual questions test why the common web of two identical cells carries no torsional flow and how many unknowns a multi-cell section has. Practise setting up the two compatibility equations quickly with δ terms.

Quick check

1. How many unknown constant shear flows does a three-cell box have under torsion?
2. Two cells enclose 0.3 m² and 0.2 m² with flows 1500 N/m and 1200 N/m. What torque do they carry?
3. Why does the common web of two identical cells carry no torsional flow?
4. Which condition supplies the extra equations in multi-cell torsion? Answers: 1. three; 2. T = 2(0.3 × 1500 + 0.2 × 1200) = 1380 N·m; 3. by symmetry q₁ = q₂, so q₁ − q₂ = 0; 4. equal rate of twist of all cells (compatibility).