Stress analysis of wing spars and fuselage frames

Why it matters

The wing spars and the fuselage shell carry the largest loads in the airframe: the spars take the lift from the whole wing to the root, and the fuselage carries tail loads, payload inertia and cabin pressure. Sizing a spar cap, a spar web or a fuselage skin is the everyday job of a stress engineer, and it uses the bending, shear and pressure results of the earlier topics in combination.

Key ideas

Wing spar loads. Treat each wing as a cantilever from the root (or from the side-of-body rib). The spanwise lift distribution minus the wing's own weight and fuel inertia gives the net load per unit span w(z). Integrating from the tip gives the shear force S(z), and integrating again gives the bending moment M(z). Both are largest at the root. Concentrated items (engines, landing gear, fuel tanks) add steps in S and kinks in M, and their inertia relieves the lift load.

Two-flange spar. In the simplest model the caps carry all the bending as axial forces P = M/h (h between cap centroids) and the web carries the shear as a constant shear flow q = S/h. A more refined model idealises the web's own bending contribution into the caps (structural idealisation topic).

Tapered spars. Real spars get shallower towards the tip. When the caps are inclined, their axial loads have components normal to the beam axis. When the depth decreases in the direction in which the bending moment decreases (the normal wing case), these components carry part of the shear, so the web shear is reduced: S_web = S − P₁·tan α₁ − P₂·tan α₂, with α the cap inclinations to the beam axis. If depth increased in that direction, the web shear would increase instead.

Fuselage as a beam. Between the wing attachment and the tail the fuselage is a thin-walled tube under bending (hogging from the weight of the rear fuselage and down-loads on the tailplane in most cases), vertical shear and torsion (from fin side loads). For a circular section of radius r and uniform skin t: I = π·r³·t, maximum bending stress σ = M·r/I, and under a shear S through the centre the skin shear flow varies as q = (S/(π·r))·sin θ (θ from the top), maximum S/(π·r) at the neutral axis and zero at the top and bottom.

Pressurisation. Cabin pressure differential p adds hoop stress p·r/t and longitudinal stress p·r/(2t) in the skin. The longitudinal pressure stress adds to the bending stress, so the top of a hogging fuselage is the critical tension region (with its fatigue sensitivity at joints and windows).

Frames. Frames keep the circular shape, limit the buckling length of stringers, and spread concentrated loads (wing and tail attachments, floor beams, landing gear) into the skin as shear flow. A frame carries its own bending moment, shear and normal force, found by treating it as a closed ring loaded by the applied point loads and reacted by the skin shear flow; it is statically indeterminate and is usually analysed with the energy methods of an earlier topic or by FE.

Combined stresses. A skin element generally sees direct stress (bending + pressure) and shear stress (vertical shear + torsion) together; check it with a yield criterion or a buckling interaction, and check the spar web for shear buckling.

Formulas

S(z) = ∫_z^tip w dz and M(z) = ∫_z^tip S dz

• w: net load per unit span (N/m), S: shear force (N), M: bending moment (N·m), z: spanwise station (m).

P = M / h and q = S / h

• Two-flange spar: P cap axial load (N), h distance between cap centroids (m), q web shear flow (N/m).

S_web = S − P₁·tan α₁ − P₂·tan α₂

• Tapered spar, depth decreasing towards the tip: P₁, P₂ cap loads (N), α₁, α₂ cap inclinations to the spar axis.

I = π·r³·t, σ = M·r / I

• Thin circular fuselage: r mean radius (m), t skin thickness (m), σ maximum bending stress (Pa).

q = (S / (π·r))·sin θ

• Shear flow (N/m) in a thin circular tube under shear S through the centre; θ measured from the top.

σ_h = p·r / t and σ_L = p·r / (2·t)

• Pressure stresses (Pa); p: cabin pressure differential (Pa).

Worked examples

Example 1 (standard): root loads and spar sizing. Given: semi-span L = 8 m, net upward load w = 5 kN/m uniform. Spar depth h = 0.40 m (between cap centroids), allowable cap stress 400 MPa, web t = 3 mm.

1. Root shear: S = w·L = 5 × 8 = 40 kN.
2. Root moment: M = w·L²/2 = 5 × 64/2 = 160 kN·m.
3. Cap load: P = M/h = 160/0.40 = 400 kN, tension in the lower cap, compression in the upper cap.
4. Cap area: A = 400 000/400 = 1000 mm².
5. Web: q = S/h = 40 000/400 = 100 N/mm, τ = 100/3 = 33.3 N/mm². Answer: S = 40 kN, M = 160 kN·m, cap area 1000 mm², web τ = 33.3 MPa (the compression cap must also be checked for buckling).

Example 2 (GATE level): tapered spar. Given: at a station, S = 100 kN, M = 200 kN·m, depth h = 0.50 m; the depth decreases towards the tip and each cap is inclined to the spar axis with tan α = 0.05.

1. Cap loads: P₁ = P₂ = M/h = 200/0.50 = 400 kN.
2. Vertical components: P·tan α = 400 × 0.05 = 20 kN per cap, both opposing the applied shear.
3. Web shear: S_web = 100 − 20 − 20 = 60 kN.
4. Web shear flow: q = 60 000/500 = 120 N/mm, compared with 200 N/mm for a parallel-flanged spar. Answer: S_web = 60 kN, q = 120 N/mm.

Example 3: pressurised fuselage in bending and shear. Given: r = 1.5 m, skin t = 1.2 mm, hogging M = 500 kN·m, vertical shear S = 100 kN, p = 60 kPa.

1. I = π·r³·t = π × 1500³ × 1.2 = 1.272 × 10¹⁰ mm⁴.
2. Bending: σ = M·r/I = 500 × 10⁶ × 1500/1.272 × 10¹⁰ = 58.9 N/mm² (tension at the top).
3. Pressure: σ_L = p·r/(2t) = 0.06 × 1500/2.4 = 37.5 N/mm², σ_h = 75.0 N/mm².
4. Top skin longitudinal stress: 58.9 + 37.5 = 96.4 N/mm².
5. Shear at the neutral axis: q = S/(π·r) = 100 000/(π × 1500) = 21.2 N/mm, τ = 21.2/1.2 = 17.7 N/mm². Answer: top skin σ_L = 96.4 MPa with σ_h = 75 MPa; side skin τ_max = 17.7 MPa.

Common mistakes

• Integrating the load from the root instead of from the free tip, or forgetting inertia relief from wing-mounted masses.
• Using the overall spar depth instead of the distance between cap centroids.
• Ignoring the shear carried by inclined caps in a tapered spar, or subtracting it when the taper is the other way.
• Using I = π·r⁴/4 (solid) instead of π·r³·t for a thin fuselage shell.
• Forgetting that pressure adds longitudinal tension to the bending stress.
• Putting maximum fuselage shear flow at the top and bottom; it is at the neutral axis.

For GATE AE

Expect root shear and moment of a cantilever wing under a given load, cap and web sizing of a two-flange spar, the tapered-spar web shear, thin-tube fuselage bending stress and shear flow, and combined bending plus pressure. Practise drawing S and M diagrams for a wing with a point mass and remember P = M/h and q = S/h.

Quick check

1. Where are wing shear force and bending moment largest?
2. A spar at a station has S = 50 kN, M = 100 kN·m, h = 0.4 m and parallel caps. Find the cap load and web shear flow.
3. What is the maximum shear flow in a thin circular fuselage of radius 2 m under 80 kN shear?
4. Does inertia of an engine on the wing increase or reduce root bending in flight? Answers: 1. at the root; 2. P = 250 kN, q = 125 kN/m; 3. q = 80 000/(π × 2) = 12.7 kN/m; 4. reduce.