Full exam for STRUCTURAL DYNAMICS AND AEROELASTICITY in the Aerospace Engineering degree programme at Politecnico di Milano. The document covers: STRUCTURAL DYNAMICS AND AEROELASTICITY Prof. Giuseppe Quaranta and Andrea Zanoni September 4, 2026 Exercise 1: Static aeroelasticity A wind tunnel model is composed by two identical flexible wings of span b, chord c, torsion stiffness GJ and lift slope at α = 0 CL/α . Consider
Full exam for STRUCTURAL DYNAMICS AND AEROELASTICITY in the Aerospace Engineering degree programme at Politecnico di Milano. The document covers: STRUCTURAL DYNAMICS AND AEROELASTICITY Prof. Giuseppe Quaranta and Andrea Zanoni September 4, 2026 Exercise 1: Static aeroelasticity A wind tunnel model is composed by two identical flexible wings of span b, chord c, torsion stiffness GJ and lift slope at α = 0 CL/α . Consider
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STRUCTURAL DYNAMICS AND AEROELASTICITY Prof. Giuseppe Quaranta and Andrea Zanoni September 4, 2026 Exercise 1: Static aeroelasticity A wind tunnel model is composed by two identical flexible wings of span b, chord c, torsion stiffness GJ and lift slope at α = 0 CL/α . Consider symmetric airfoils, so that CM AC0 = 0. The two wings are con- nected at the elastic axis, located at mid-chord, by a bar of length ℓb , rigid in bending and torsion. A beam of length ℓ, bending stiffness EJ in the x-z plane, ap- proximately rigid axially and for bending in the y-z plane, is welded at the mispan section of the bar and at the wind tunnel floor. Each wing is fitted with a control surface, spanning the entire trailing edge, that can deflect by and angle βi . The lift and moment coef- ficients of both the control surfaces are, respectively, CL/β and CM AC/β . The aerodynamic centre of the wings is located at c/4 from the leading edge. 1.a) Compute the divergence pressure with the Ritz-Galerkin approach, using a single well-chosen shape function for the wings torsion. Consider β1 = β2 = 0. 1.b) Compute the tip displacement of the beam when β1 = β2 = β. 1.c) Compute the reversal pressure when β1 = −β2 = β. Exercise 2: Dynamic response A rigid wing of span b, chord c, uniform mass per unit sur- face m, uniform aerodynamic properties, is suspended in the wind tunnel by two identical cantilever beams of length ℓ, bending stiffness EJ. The beams are rigid in torsion and of negligible mass and clamped to the tunnel wall upstream and welded to the wing at the two ends of its trailing edge. A concentrated mass M , of negligible own moments of iner- tia, is rigidly attached to the mid-span section of the trailing edge. Each beam carries, in parallel, a viscous damper of co- efficient r acting…
Exercise 1. Consider ϑ (about y) as the pitch rotation, positive when it increases the incidence. Each wing is described by a local spanwise coordinate y ∈ [0, b], running from the root (junction with the bar) to the tip. The model is mounted at null rigid incidence: the whole angle of attack of a section is therefore produced by the elastic deformation of the model, and the only external command is the flap deflection βi . With strip theory, the lift and the pitching moment about the aerodynamic centre, per unit span of the wing i, are l(y) = q c CL/α ϑ(y) + CL/β βi , (1a) mAC (y) = q c2 CM AC/β βi , (1b) where ϑ(y) is the total elastic rotation of the section about y. Since the aerodynamic centre lies e = c/4 ahead of the elastic axis, its vertical displacement is zAC = e ϑ(y), so that the whole aerodynamic contribution can be collected in the moment per unit span about the elastic axis ma,EA (y) = e l(y) + mAC (y) = qc e CL/α ϑ(y) + e CL/β + c CM AC/β βi . (2) The support. The vertical beam is a cantilever of length ℓ, clamped at the tunnel floor and loaded at its tip, through the rigid bar, by the resultant of the loads of the two wings. That resultant consists of the total lift 2L, which is directed along z and applied on the elastic axis, i.e. along the beam axis, and of the resultant pitching moment Mt = 2Mr about y, Mr being the root torque of one wing. The lift is therefore purely axial and produces no bending; the beam is bent in the x–z plane by a tip couple only, so that its bending moment is constant. Using the Principle of Complementary Virtual Work: Mt z 2 Mt ℓ M t ℓ2 ℓ u(z) = =⇒ φ = u/z (ℓ) = , utip = u(ℓ) = = φ, (3) 2EJ EJ 2EJ 2 where u is the transverse displacement in the x direction and φ the rotation about y of the beam section. Because the…
1.c) With β1 = β = −β2 the aerodynamic loading is antisymmetric, and so is the response: the two wings twist in opposite directions. The kinematics is now much simpler than in the symmetric case, because the support does not participate at all: the resultant of the antisymmetric lift distribution is a pure moment Mx , which would bend the beam in the y–z plane: the beam being rigid in that plane, Mx is simply reacted by the clamp at the tunnel floor and produces no displacement. Roll is therefore not a degree of freedom, and control reversal has to be stated as the vanishing of the transmitted rolling moment. The antisymmetric problem is thus governed by the single degree of freedom qθ (taken with the sign of wing 2, whose flap deflection β2 is renamed β) and the second row of (11) with φ = 0 gives qA 2 2 qBβ Kθθ − qθ = q B β =⇒ qθ = . (17) 2 π π qA Kθθ − 2 Rolling moment. With yg = ℓb /2 + y the global spanwise station on wing 2, and using the antisymmetry of the load distribution, Z b ℓb Mx = 2 + y l(y)dy = 2qc CL/α I1 qθ + CL/β I0 β , (18) 0 2 with, using (7), Z b Z b b ℓb 4b2 ℓb b (ℓb + b) ℓb I0 = + y dy = , I1 = + y n(y)dy = + 2. (19) 0 2 2 0 2 π π The first term is the moment of the lift generated by the aeroelastic twist, the second the one directly generated by the flaps. Reversal. Control reversal occurs when the flaps produce no rolling moment, i.e. Mx = 0 for β ̸= 0. Substituting (17) into (18) and dividing by β ̸= 0, 2 CL/α I1 qR B + CL/β I0 = 0, (20) π qR A Kθθ − 2 which, being linear in qR once multiplied by the denominator, gives CL/β I0 Kθθ qR = (21) A 2 CL/β I0 − CL/α I1 B 2 π with A, B from (10) and Kθθ = GJπ 2 /(8b). Exercise 2. Because the beams are rigid in torsion the rotation of the wing about x is prevented at both attachments,…
u simply extracting the incidence, ϑ = uT q. The bending moment is obtained from the equilibrium of the free portion of the beam: T ℓ−x F Mb (x) = b (x) Q, b(x) = , Q= , (25) 1 Mt F being the tip force along z and Mt the tip couple, work-conjugate to wℓ and φℓ respectively, and Mb = EJ w/xx . Any variation δQ of the tip loads generates, through the same (25), a statically admissible virtual moment field δMb = bT δQ. The Principle of Complementary Virtual Work equates the complementary work of this virtual stress field on the true curvature χ = Mb /EJ to the complementary work of the virtual external loads on the true displacements: Z ℓ ∗ Mb T wℓ ∗ δWint = δMb dx = δQ = δWext . (26) 0 EJ φ Substituting (25) and requiring the equality to hold for every δQ gives the tip flexibility matrix directly, 3 2 Z ℓ ℓ ℓ b(x) bT (x) δ 1 3 2 , = Fb Q, Fb = dx = (27) φ 0 EJ EJ ℓ2 ℓ 2 Rℓ Rℓ Rℓ having used 0 (ℓ − x)2 dx = ℓ3 /3, 0 (ℓ − x)dx = ℓ2 /2 and 0 dx = ℓ. The flexibility can be inverted into the equivalent stiffness of a single beam, −1 EJ 12 −6ℓ K b = Fb = 3 , (28) ℓ −6ℓ 4ℓ2 and, mapping the tip state onto the generalised coordinates with (22) and summing the two beams, T 2EJ 12 6ℓ Ks = 2 T Kb T = 3 . (29) ℓ 6ℓ 4ℓ2 Aerodynamic loads. With quasi-steady strip theory the effective incidence is evaluated from the downwash at the three-quarter-chord point and the resulting lift is applied at the aerodynamic centre; by (23) and (24), ż 34 c nT3/4 q̇ VG T VG αe = ϑ − + =u q− + , L = q S CL/α αe , (30) U∞ U∞ U∞ U∞ VG being the (upwash-positive) gust velocity, and S = bc. The lift is: q q L = q S CL/α , uT q − S CL/α nT3/4 q̇ + S CL/α VG (31) U∞ U∞ . 2.a) The model has two degrees of freedom (22); a virtual displacement is δz(ξ) = nT (ξ) δq with…
2.b) Since the quasi-steady loads (40) are polynomial in s, the aeroelastic problem is exactly a linear, time-invariant system and no iteration on the reduced frequency is needed. Algebraic approach. Build the coefficients ai (U∞ ) of (41) — here in closed form, (42) — and find the smallest positive root of the scalar function ∆R (U∞ ) of (43) by bisection or Newton’s method, after bracketing the sign change on a coarse sweep; the condition a0 > 0 of (44) must be checked in parallel, since divergence may well precede flutter. The flutter frequency comes for free, ωF2 = a1 /a3 . It is the cheapest method, but it does not scale beyond a few degrees of freedom, where the symbolic expansion of the determinant becomes impractical. T Eigenvalue tracking. Cast (38) in first-order form, ẋ = A(U∞ )x with x = qT q̇T and 0 I A= , −M−1 (Ks + Ka ) −M−1 (Cs + Ca ) sweep U∞ , compute the eigenvalues at each step and interpolate the zero crossing of ℜ (s). The eigen- values must be tracked from one speed to the next (by eigenvector correlation, e.g. a MAC criterion, or by continuation from the previous solution) to avoid jumping between branches, and all branches must be followed, the critical speed being the lowest crossing; a crossing with ℑ (s) = 0 is a divergence, one with ℑ (s) ̸= 0 a flutter. This is the method of choice, because it also yields the damping trend and, being a standard eigenvalue problem, it scales to large models. 2.c) Load recovery. The bending moment at the root of a beam is recovered from the same statically admissible field (25), evaluated at x = 0 and with the tip loads expressed through (28): T T 2EJ 3 Mr = Mb (0) = b (0) Kb T q = r q, r= 2 , (45) ℓ ℓ Transfer function. From (38), with Z(s) = s2 M + s (Cs + Ca ) + Ks + Ka , Mr b2 s2 + b1 s +…
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