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 July 3, 2026 Exercise 1: Static aeroelasticity A model of a glider in a wind tunnel is suspended to two springs of stiffness K, attached at the trailing edge of the wing tips. The system is in
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 July 3, 2026 Exercise 1: Static aeroelasticity A model of a glider in a wind tunnel is suspended to two springs of stiffness K, attached at the trailing edge of the wing tips. The system is in
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STRUCTURAL DYNAMICS AND AEROELASTICITY Prof. Giuseppe Quaranta and Andrea Zanoni July 3, 2026 Exercise 1: Static aeroelasticity A model of a glider in a wind tunnel is suspended to two springs of stiffness K, attached at the trailing edge of the wing tips. The system is in equilibrium when there is no wind. The wing is flexible with tor- sional stiffness GJ, while the bending stiffness must be considered infinite. The total span is 2b, and the chord is constant and equal to c. The aerodynamic centre is located at 0.25c in front of the elastic axis, which is positioned at mid-chord. The wing uses a constant airfoil and is not twisted. The lift coefficient is CL/α and flow should be considered incompressible. 1.a) Compute the divergence pressure with the Ritz-Galerkin approach, using a single well-chosen shape function for the wing torsion. Discard the lift of the horizontal tail. 1.b) Consider ST to be the surface of the horizontal tail, M to be the total aircraft mass, d the distance between the aerodynamic centre of the tail and the centre of gravity, and CM AC0 , the moment coefficient of the wing with respect to the aerodynamic centre. Discard the aerodynamic properties of the fuselage and the moment coefficient of the tail plane. Write the equations to compute the lift of the horizontal tailplane and the angle of attack α0 to fly at a load factor n = 1. (Suggestion: read carefully the description to infer the position of the CG of the aircraft). 1.c) Consider now the effect of the ailerons placed at the wing tips of span ba ≪ b, of coefficients CL/β and CM AC/β : compute the dynamic pressure of control reversal. Exercise 2: Dynamic response Consider a wind tunnel model consisting of a straight, untapered, uniform wing of bending stiffness EJ, tor- sion stiffness…
Exercise 1. 1.a) For the analysis of static divergence, since the problem is symmetric, the only degrees of freedom that will influence the solution are the rigid heave displacement w and the torsion of the wing θ(y). Therefore it can be reduced to the problem of the divergence of wing of semi-span b contrained at its root to slide along z. The internal virtual work is: Z b T T δWint = − δθ/y GJθ/y dy − δwTE KwTE . (1) 0 Assume for the twist angle the following Ritz-like discretization (other choices are possible): 1 2y − b θ(y) = n(y)qθ = sin π +1 . (2) 2 2b The vertical displacement of the wing tip at the trailing edge is: c c wTE = w − θ(b) = w − qθ . (3) 2 2 Therefore the internal work is Z b T c T c δWint = −δqθ n(y)T/y GJn(y)/y dyqθ − δw − δqθ K w − qθ (4) 0 2 2 π 2 c T c = −δqθT GJ qθ − δw − δqθ K w − qθ (5) 8b 2 2 The structural stiffness matrix is, therefore c K − K 2 Ks = c π2 . (6) c2 − K K + GJ 2 4 8b The external work, assuming a symmetric airfoil, is Z b T δWext = δwAC L(y)dy, (7) 0 with the lift per unit span given by, according to steady strip theory L(y) = qcCL/α θ(y), (8) and the transverse displacement of the aerodynamic centre by wAC = w + eθ = w + e n(y)qθ , (9) therefore: Z b δWext = (δw + e n(y)δqθ ) qcCL/α n(y)qθ . (10) 0 Which results in two contributions to the aerodynamic stiffness matrix: Z b 2b δwT qcCL/α n(y)dyqθ = δwT qcCL/α qθ , (11) 0 π Z b b δwT qecCL/α n2 (y)dyqθ = δwT qecCL/α qθ . (12) 0 2
with GJπ 2 Kθθ = (24a) 8b b Ka,θ = cCL/α (24b) 2 2b fa,α0 = ec C (24c) π L/α 2b ma,0 = c2 CM AC0 (24d) π The solution of (23) is q (efa,α0 + ma,0 ) qθ = (25) (Kθθ − q Ka,θ ) Combining (16), (18), (19) and (25) it is possible to compute α0 for a generic flight condition at n = 1: qScCM AC0 − M gd q (efa,α0 + ma,0 ) α0 = − (26) (3/4c + d)qSCL/α 2(Kθθ − q Ka,θ ) 1.c) Considering the roll motion, the deformation now is anti-symmetric. One possible choice for the twist Ritz-like discretization is π y θ(y) = n̂(θ)qθ = sin qθ (27) 2b The virtual work in this case is Z b 0 = δW = δqθT n̂T GJ n̂dyqθ − (bδφ − eδqθ )T 2K(bφ − e qθ )T + (28) −b Z b + (yδφ + en̂δqθ )T qcCL/α n̂qθ dy + (29) −b Z −b+ba Z b − δqθT n̂T qc2 CmAC/β βdy + δqθT n̂T qc2 CmAC/β βdy (30) −b b−ba Z −b+ba Z b − (yφ + en̂δqθ )T qcCL/β βdy + (yφ + en̂δqθ )T qcCL/β βdy (31) −b b−ba The static equilibrium is thus held by: Ks,φφ Ks,φθ 0 Ka,φθ φ maφβ −q =q β (32) Ks,θφ Ks,θθ 0 Ka,θθ qθ maθβ with Ks,φφ = 2Kb2 (33) Ks,φθ = −2Kbe (34) Ks,θφ = −2Kbe (35) GJπ 2 Ks,θθ = 2Ke2 + (36) 8b 8b2 Ka,φθ = cCL/α (37) π2 Ka,θθ = eSCL/α (38) ma,φβ = ba (2b − ba )cCL/β (39) ba ba ma,θβ = (2b − ba )c2 CM AC/β + (2b − ba )ecCL/β (40) 2 2
with mb c mb M − −M 5 10 4 M= (52a) c2 mb c mb Jb − −M + +M 10 4 3 100 3 c 0 Cs = 16 (52b) 0 0 SCL/α 5 0 Ca = SC (52c) Lα 0 20 4EJ k 3 + 0 Ks = b 16 (52d) GJ 0 b S 0 CLα 4 Ka = 1 S (52e) 0 CLα 5 3 k ( ) Fz = 4 (52f) 0 (c) Fż = 4 (52g) 0 Setting zC (t) = 0 and moving the equations of motion in state space form, considering the state vector T u = q̇w q̇θ qw qθ : ü q −M −1 Cs + Ca −M (Ks + qKa ) u̇ −1 = U∞ . (53) u̇ u I 0 It is now possible to investigate flutter by computing the eigenvalues of the state matrix at increasing values of U∞ . The flutter conditions are met when at least one eigenvalue real part crosses the real axis changing sign from negative to positive. T 2.b) Calling x = u̇ u , A the state matrix, and reintroducing the forcing term: ẋ = Ax + Bz zC (t) + Bż żC (t) (54) with M−1 Fz M−1 Fż A = A(q̂), Bz = , Bż = , (55) 0 0 i.e. A is the same state matrix built in 2.a), evaluated at the given flight condition q = q̂. Since zC (t) = step(t), żC (t) = δ(t): the equations of motion contain an impulsive term that produces a jump in the generalized velocities at t = 0, while the generalized displacements, being continuous, remain zero. Assuming the system is at rest for t < 0: u(0) = 0, u̇(0) = M−1 Fż (56) (the impulsive velocity of the slider is transmitted to the wing through the damper r, contained in Fż ). For t > 0 the forcing is constant, zC (t) = 1, so the response is the superposition of a steady-state and a transient (homogeneous) contribution, u(t) = uss + uh (t), with the steady state given by (Ks + q̂ Ka ) uss = Fz . (57)
Remark: a Lyapunov-equation route confirms the divergence. Φ(ω) √ zC ,zC is rational, so zC (t) can be generated by filtering unit-intensity white noise w(t) through Hf (jω) = N (1 + jωa)/(ω02 − ω 2 + jξω0 ω), i.e. √ ẍf + ξω0 ẋf + ω02 xf = N w(t), zC = xf + aẋf . (68) Appending {xf , ẋf } to x gives an augmented, white-noise-driven system ẋaug = Aaug (q̂)xaug + Baug w(t), whose stationary covariance σx2 aug xaug solves Aaug σx2 aug xaug + σx2 aug xaug ATaug + Baug BTaug = 0. (69) Because a ̸= 0, żC = ẋf + aẍf pulls in w(t) undifferentiated, so Bż żC feeds w(t) directly into ü (the top block of ẋaug ). Since F = M eT ü, e = {1, −c/10}T , the output equation inherits this term: √ F = CF xaug + DF w(t), DF = M a N eT M−1 Fż ̸= 0 (generically). (70) So the augmented system has a genuine feedthrough term: computing σF2 would require adding DF2 Var[w(t)] to CF σx2 aug xaug CTF , and since white noise has infinite variance (Var[w(t)] → ∞), this is unbounded. The Lyapunov approach thus reproduces the divergence found above: the shear force does not have a finite variance under the stated assumptions (a quasi-steady model plus an input whose PSD does not decay fast enough at high frequency). A finite answer would require some high-frequency roll-off not included in the idealized model (e.g. unsteady aerodynamics).
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