Structural idealisation: booms and skin

Why it matters

A real wing or fuselage section has dozens of stringers, spar caps and skin panels of different thicknesses. Calculating its bending stresses and shear flows exactly is tedious, so structural engineers idealise it: the direct-stress-carrying material is lumped into concentrated areas called booms, and the skin between them is assumed to carry only shear. The idealised section gives quick, accurate hand calculations and is the basis of the classic wing-box and fuselage analyses.

Key ideas

The idealisation.

• Booms: concentrated areas, located at the centroids of stringers and spar caps, that carry all the direct (axial) stress from bending. The direct stress is taken as constant over each boom.
• Skin and webs: carry shear flow only. In the simplest idealisation they carry no direct stress, so the shear flow between two adjacent booms is constant.
• Skin that does carry direct stress is not lost: its direct-stress capacity is added to the adjacent booms as an equivalent area.

Equivalent boom area from a skin panel. Replace a panel of thickness t_D and width b between booms 1 and 2 by additional areas at booms 1 and 2 that produce the same moment about the neutral axis, for a linear variation of direct stress across the panel. This gives the contribution to boom 1: t_D·b/6·(2 + σ₂/σ₁). Under pure bending σ₂/σ₁ = y₂/y₁ (distances from the neutral axis), so the boom area depends on the loading case. For a panel with both edges at the same distance from the neutral axis (σ₂/σ₁ = 1) the contribution is t_D·b/2 to each boom; for a web symmetric about the neutral axis (σ₂/σ₁ = −1) it is t_D·b/6.

Section properties. With booms, I_xx = Σ B_r·y_r², I_yy = Σ B_r·x_r², I_xy = Σ B_r·x_r·y_r, the booms' own second moments about their centroids being neglected.

Bending stress. The general (unsymmetrical) bending formula applies unchanged, evaluated at each boom.

Shear flow. Because the walls carry no direct stress, the open-section flow is constant between booms and jumps at each boom r by Δq = −(S_x·I_xx − S_y·I_xy)/(I_xx·I_yy − I_xy²)·B_r·x_r − (S_y·I_yy − S_x·I_xy)/(I_xx·I_yy − I_xy²)·B_r·y_r. For a symmetric section with only S_y: Δq = −(S_y/I_xx)·B_r·y_r. Closed sections then need the constant q_s,0 exactly as before, and the moment of a constant flow between two booms about a point O is 2·A_enclosed·q, where A_enclosed is the area swept between the wall and O.

When the idealisation is good. Closely spaced stringers, thin skins, bending-dominated loading. It is poor for thick skins with few stiffeners (use the direct-stress-carrying skin formula above) and for local load introduction.

Formulas

B₁ = t_D·b / 6·(2 + σ₂/σ₁)

• Boom area contribution (m²) at boom 1 from a panel of thickness t_D (m) and width b (m) between booms 1 and 2; σ₁, σ₂: direct stresses at booms 1 and 2. Add the contributions of every panel meeting the boom and the stringer or cap area itself.

σ₂/σ₁ = y₂/y₁

• Pure bending about the x-axis; y measured from the neutral axis.

I_xx = Σ B_r·y_r²

• Second moment of the idealised section (m⁴).

σ_r = M_x·y_r / I_xx

• Boom stress (Pa) for a symmetric section under M_x; use the general formula otherwise.

Δq = −(S_y / I_xx)·B_r·y_r

• Jump in shear flow (N/m) on passing boom r, symmetric section under S_y (N).

P_r = σ_r·B_r

• Axial load carried by boom r (N).

Worked examples

Example 1 (standard): idealising a spar. Given: a spar with a web 300 mm deep and 4 mm thick, and two caps of 400 mm² each at the top and bottom edges of the web. Pure bending about the horizontal axis through mid-depth.

1. Web contribution to each boom with σ₂/σ₁ = −1: B = t_D·b/6·(2 − 1) = 4 × 300/6 = 200 mm².
2. Boom areas: B₁ = B₂ = 400 + 200 = 600 mm².
3. Idealised second moment: I_xx = 2 × 600 × 150² = 2.70 × 10⁷ mm⁴.
4. Exact second moment for comparison: caps 2 × 400 × 150² = 1.80 × 10⁷ plus web 4 × 300³/12 = 0.90 × 10⁷, total 2.70 × 10⁷ mm⁴. Answer: B = 600 mm² per boom; I_xx = 2.70 × 10⁷ mm⁴, identical to the exact value for a linear stress distribution.

Example 2 (GATE level): four-boom wing box. Given: rectangular box with booms at the corners of a 600 mm × 200 mm mid-line rectangle. Each spar cap is 500 mm². Top and bottom skins are 2 mm thick, spar webs 3 mm. Loads: M_x = 100 kN·m (pure bending case) and, separately, S_y = 50 kN along the vertical axis of symmetry.

1. Skin contribution (both edges at the same y, σ₂/σ₁ = 1): 2 × 600/6 × (2 + 1) = 600 mm².
2. Web contribution (σ₂/σ₁ = −1): 3 × 200/6 × (2 − 1) = 100 mm².
3. Boom area: B = 500 + 600 + 100 = 1200 mm².
4. I_xx = 4 × 1200 × 100² = 4.8 × 10⁷ mm⁴.
5. Boom stress: σ = M_x·y/I_xx = 100 × 10⁶ × 100/4.8 × 10⁷ = 208.3 N/mm² (tension at the bottom booms for sagging, compression at the top).
6. Shear: by symmetry q = 0 in the top and bottom skins (the cut is at the top-skin mid-point and there is no boom between it and the corner). Passing a top boom: Δq = (S_y/I_xx)·B·y = 50 000 × 1200 × 100/4.8 × 10⁷ = 125 N/mm in magnitude.
7. Each web: q = 125 N/mm, constant; τ = 125/3 = 41.7 N/mm². Check: 2 × 125 × 200 = 50 000 N = S_y. Answer: B = 1200 mm²; σ = ±208.3 MPa; web q = 125 N/mm, τ = 41.7 MPa; skins carry no shear for this load.

Common mistakes

• Forgetting that the boom area depends on the load case through σ₂/σ₁; the same skin gives different booms for bending about x and about y.
• Adding a panel's contribution to only one of its two booms.
• Using σ₂/σ₁ = +1 for a web that crosses the neutral axis; it is −1 for a symmetric web.
• Letting the shear flow vary between booms in an idealised section; it is constant there.
• Losing the sign of y_r in Δq, so that flows fail to add up to the applied shear.
• Treating booms as carrying shear, or skins (in the basic idealisation) as carrying bending stress.

For GATE AE

Common items: boom area from t_D·b/6·(2 + σ₂/σ₁), I_xx of a boom section, boom stresses, and shear flows in two-, four- or six-boom boxes, including the jump Δq at each boom and the closed-section constant. Practise the four-boom box both for symmetric shear and for offset shear (add the Bredt flow T/(2A)).

Quick check

1. What do booms carry and what do skins carry in the basic idealisation?
2. A 2 mm skin, 120 mm wide, joins two booms at the same distance from the neutral axis. What area does it add to each boom?
3. How does the shear flow vary between two booms of an idealised section?
4. In Example 2, what axial force does each boom carry? Answers: 1. booms carry direct stress, skins carry shear flow; 2. 2 × 120/6 × 3 = 120 mm²; 3. it is constant; 4. 208.3 × 1200 = 250 kN.