Aeroelasticity Master Notes

Cheatsheet Content

### Introduction to Aeroelasticity Aeroelasticity is the study of the interaction among inertial, elastic, and aerodynamic forces acting on an aerospace vehicle. This interaction can lead to phenomena such as divergence, flutter, and control reversal. #### Collar's Triangle The fundamental concept is visualized by Collar's Triangle, illustrating the interplay: - **Aerodynamic Forces:** Generated by airflow over the structure. - **Elastic Forces:** Restoring forces within the structure due to deformation. - **Inertial Forces:** Forces due to the mass and acceleration of the structure. ### Quasi-Steady Strip Theory #### Physical Intuition This theory simplifies aerodynamic calculations by assuming that each spanwise strip of the wing acts independently in 2D flow, and that the unsteady aerodynamic effects are negligible (quasi-steady). It's a fundamental tool for initial aeroelastic analysis. #### Assumptions & Boundaries - **2D Flow:** Assumes that the flow over each wing section is independent of adjacent sections. No spanwise flow. - **Independent Segments:** Each wing strip responds to local conditions without influencing others. - **Small Angles:** Valid for small angles of attack and small deformations. - **Thin Airfoil Theory:** Often used to determine aerodynamic coefficients. - **Steady Aerodynamics Locally:** Unsteady effects like wake lag are ignored or simplified. ### Module 1: Static Divergence #### Physical Intuition Divergence occurs when the aerodynamic twisting moment on a wing, due to elastic deformation, overcomes the wing's structural restoring moment. Beyond a critical speed, the wing twists uncontrollably, leading to structural failure. It's an interaction between aerodynamic and elastic forces. #### Continuum Wing Divergence: Derivation of 2nd Order BVP Consider an unswept isotropic wing of span $L$ with uniform torsional rigidity $GJ$ and a constant aerodynamic center (AC) eccentricity $e$ from the elastic axis (EA). Let $\theta(y)$ be the twist angle at spanwise position $y$. 1. **Aerodynamic Moment:** The elemental lift $dL$ on a strip of width $dy$ is given by: $$dL = q c a_0 \theta dy$$ where $q$ is the dynamic pressure ($q = \frac{1}{2} \rho U^2$), $c$ is the chord, $a_0$ is the 2D lift curve slope, and $\theta$ is the local angle of attack (due to twist). The elemental aerodynamic moment $dM_A$ about the elastic axis is: $$dM_A = dL \cdot e = q c a_0 \theta e dy$$ The distributed aerodynamic twisting moment per unit length, $M_A'(y)$, is: $$M_A'(y) = q c a_0 e \theta(y)$$ 2. **Elastic Moment:** For a beam in torsion, the internal elastic twisting moment $M_T(y)$ is related to the twist angle by: $$M_T(y) = GJ \frac{d\theta}{dy}$$ The rate of change of internal twisting moment is the distributed elastic moment per unit length, $M_E'(y)$: $$M_E'(y) = - \frac{dM_T}{dy} = - GJ \frac{d^2\theta}{dy^2}$$ The negative sign indicates that the elastic moment opposes the deformation. 3. **Moment Equilibrium:** For static equilibrium, the sum of aerodynamic and elastic moments must be zero: $$M_A'(y) + M_E'(y) = 0$$ $$q c a_0 e \theta(y) - GJ \frac{d^2\theta}{dy^2} = 0$$ Rearranging, we get the second-order boundary value problem (BVP) for static divergence: $$GJ \frac{d^2\theta}{dy^2} + q c a_0 e \theta = 0$$ This can be written as: $$\frac{d^2\theta}{dy^2} + \lambda^2 \theta = 0$$ where $\lambda^2 = \frac{q c a_0 e}{GJ}$. 4. **Boundary Conditions (Cantilever Wing):** For a cantilever wing of length $L$ (fixed at $y=0$, free at $y=L$): - **Fixed End (Root):** No twist at the root: $\theta(0) = 0$ - **Free End (Tip):** No internal twisting moment at the tip: $M_T(L) = 0 \implies GJ \frac{d\theta}{dy}\Big|_{y=L} = 0 \implies \frac{d\theta}{dy}\Big|_{y=L} = 0$ 5. **Solution of the BVP:** The general solution to $\frac{d^2\theta}{dy^2} + \lambda^2 \theta = 0$ is: $$\theta(y) = A \sin(\lambda y) + B \cos(\lambda y)$$ Applying boundary condition $\theta(0) = 0$: $$A \sin(0) + B \cos(0) = 0 \implies B = 0$$ So, $\theta(y) = A \sin(\lambda y)$. Now apply boundary condition $\frac{d\theta}{dy}\Big|_{y=L} = 0$: $$\frac{d\theta}{dy} = A \lambda \cos(\lambda y)$$ $$A \lambda \cos(\lambda L) = 0$$ For a non-trivial solution (i.e., $A \neq 0$), we must have: $$\cos(\lambda L) = 0$$ This implies $\lambda L = \frac{n\pi}{2}$ for $n = 1, 3, 5, \dots$. The smallest non-zero value occurs when $n=1$, so $\lambda L = \frac{\pi}{2}$. Therefore, the fundamental eigenvalue for a cantilever wing is $\lambda = \frac{\pi}{2L}$. 6. **Divergence Dynamic Pressure:** Substituting $\lambda = \frac{\pi}{2L}$ back into the definition of $\lambda^2$: $$\left(\frac{\pi}{2L}\right)^2 = \frac{q_D c a_0 e}{GJ}$$ Solving for the divergence dynamic pressure $q_D$: $$q_D = \frac{\pi^2 GJ}{4 L^2 c a_0 e}$$ The divergence speed $U_D$ is then $U_D = \sqrt{\frac{2 q_D}{\rho}}$. **Exam Tip:** - The eccentricity $e$ is positive if the AC is aft of the EA (destabilizing), and negative if AC is forward of EA (stabilizing). Divergence only occurs for positive $e$. - Always check units carefully. $GJ$ is typically in $N \cdot m^2$, $c$ in $m$, $a_0$ in $rad^{-1}$, $e$ in $m$, $L$ in $m$. #### Discrete Spring Divergence (Moment Equilibrium) For a rigid wing with a discrete torsional spring at a pivot, divergence is determined by balancing the aerodynamic moment and the spring restoring moment. Consider a rigid wing pivoted about an elastic axis, restrained by a torsional spring $K_T$ (e.g., at the root or an internal pivot). Let $\theta$ be the twist angle. 1. **Aerodynamic Moment:** The total aerodynamic moment $M_A$ about the pivot is: $$M_A = q S c_L e = q S (a_0 \theta) e$$ where $S$ is the wing area. For a uniform wing of span $L$ and chord $c$, $S=cL$. $$M_A = q (cL) (a_0 \theta) e = q c L a_0 e \theta$$ 2. **Elastic Moment:** The restoring moment from the torsional spring is $M_E = -K_T \theta$. 3. **Moment Equilibrium:** $$M_A + M_E = 0$$ $$q c L a_0 e \theta - K_T \theta = 0$$ $$(q c L a_0 e - K_T) \theta = 0$$ For a non-trivial solution ($\theta \neq 0$), the term in the parenthesis must be zero: $$q_D c L a_0 e - K_T = 0$$ $$q_D = \frac{K_T}{c L a_0 e}$$ **Exam Tip:** - If a spring is located at the trailing edge, its contribution to the moment must be calculated by multiplying the spring force by its moment arm relative to the pivot. - Always define your pivot point clearly and ensure all moments are taken about that point. ### Module 2: Aileron Reversal #### Physical Intuition Aileron reversal is a static aeroelastic phenomenon where, beyond a certain speed (reversal speed), the roll produced by aileron deflection becomes opposite to the intended direction, or the aileron loses its effectiveness entirely. This occurs due to the elastic deformation (twist) of the wing counteracting the aerodynamic forces generated by the aileron. It's primarily an interaction between aerodynamic and elastic forces. #### Definition of Reversal Aileron reversal is defined as the condition where the rolling moment produced by an aileron deflection becomes zero or reverses its sign. This implies that the effectiveness of the aileron, defined as the rolling moment per unit aileron deflection, becomes zero. #### Derivation for a Flexible Wing with an Aileron Consider a wing with an aileron deflected by $\delta_f$. This deflection creates an aerodynamic moment that twists the wing. This twist, in turn, modifies the lift distribution and thus the rolling moment. 1. **Aerodynamic Moment due to Aileron and Twist:** The local twist $\theta(y)$ is induced by the aileron deflection $\delta_f$. The total angle of attack for a local section is $\alpha_{eff} = \alpha + \theta(y)$. The local lift coefficient for a section with aileron deflection is: $$c_l = a_1 (\alpha + \theta(y)) + a_2 \delta_f$$ where $a_1$ is the wing lift curve slope (for the angle of attack) and $a_2$ is the aileron effectiveness parameter (lift change per unit aileron deflection). The elemental lift $dL$ is: $$dL = q c c_l dy = q c [a_1 (\alpha + \theta(y)) + a_2 \delta_f] dy$$ The elemental aerodynamic moment about the EA, $dM_A$, is: $$dM_A = dL \cdot e = q c e [a_1 (\alpha + \theta(y)) + a_2 \delta_f] dy$$ The distributed aerodynamic twisting moment per unit length is: $$M_A'(y) = q c e [a_1 (\alpha + \theta(y)) + a_2 \delta_f]$$ 2. **Elastic Moment and Equilibrium:** The elastic restoring moment per unit length is $M_E'(y) = -GJ \frac{d^2\theta}{dy^2}$. For equilibrium: $$GJ \frac{d^2\theta}{dy^2} + q c e [a_1 (\alpha + \theta(y)) + a_2 \delta_f] = 0$$ Assuming $\alpha=0$ (or absorbed into $\theta$ for simplicity in reversal analysis): $$GJ \frac{d^2\theta}{dy^2} + q c e a_1 \theta(y) + q c e a_2 \delta_f = 0$$ This is a non-homogeneous differential equation. 3. **Rolling Moment:** The total rolling moment $M_R$ is obtained by integrating the elemental lift multiplied by its moment arm $y$ along the span: $$M_R = \int_0^L y \cdot dL = \int_0^L y \cdot q c [a_1 (\alpha + \theta(y)) + a_2 \delta_f] dy$$ Assuming $\alpha=0$: $$M_R = q c \int_0^L y [a_1 \theta(y) + a_2 \delta_f] dy$$ Aileron reversal occurs when $\frac{dM_R}{d\delta_f} = 0$. This means the rolling moment effectiveness becomes zero. For a simplified 1-DOF rigid wing with a torsional spring, the twist $\theta$ is directly proportional to the applied aerodynamic moment. Let the aerodynamic moment due to aileron deflection be $M_{\delta_f} = q c_M \delta_f$, where $c_M$ is a moment coefficient. This moment causes a twist $\theta$. The total twist is: $$\theta = \frac{M_{\delta_f}}{K_T} = \frac{q c_M \delta_f}{K_T}$$ The total lift $L_{total}$ is composed of lift due to aileron deflection and lift due to twist: $$L_{total} = L_{\delta_f} + L_\theta = q S (a_2 \delta_f) + q S (a_1 \theta)$$ Substituting $\theta$: $$L_{total} = q S a_2 \delta_f + q S a_1 \left(\frac{q c_M \delta_f}{K_T}\right)$$ $$L_{total} = q S \delta_f \left(a_2 + \frac{q a_1 c_M}{K_T}\right)$$ Aileron reversal occurs when $L_{total}$ becomes zero for a non-zero $\delta_f$, which means the term in the parenthesis is zero: $$a_2 + \frac{q_R a_1 c_M}{K_T} = 0$$ $$q_R = -\frac{K_T a_2}{a_1 c_M}$$ Note that $c_M$ is typically negative for a trailing edge aileron producing a positive roll (it creates a nose-down moment, twisting the wing downwards). So $q_R$ will be positive. **Relationship between $a_1$ and $a_2$ in $q_R$:** The reversal dynamic pressure $q_R$ depends on the ratio of the aileron effectiveness $a_2$ to the wing lift curve slope $a_1$, and the structural stiffness $K_T$, as well as the aerodynamic moment coefficient $c_M$ that the aileron deflection produces. A higher $a_2$ (more effective aileron) or lower $a_1$ (less effective wing section) tends to increase $q_R$, making reversal less likely. A stronger spring $K_T$ also increases $q_R$. **Exam Tip:** - Understand that the aileron creates a local lift, which produces a moment about the EA. This moment twists the wing, which changes the local angle of attack and thus the lift distribution, ultimately affecting the rolling moment. Reversal happens when the twist-induced lift cancels the direct aileron-induced lift. ### Module 3: 1-DOF & 2-DOF Flutter #### Physical Intuition Flutter is a dynamic aeroelastic instability where a sustained oscillation of the structure occurs, extracting energy from the airflow. It involves the interaction of all three forces: aerodynamic, elastic, and inertial. Unlike divergence, flutter is dynamic and oscillatory. #### Mechanism of Energy Extraction Flutter occurs when the phase relationship between the structural motion (elastic and inertial forces) and the aerodynamic forces becomes such that the aerodynamic forces do positive work on the structure over a cycle of oscillation. This positive work feeds energy into the oscillation, leading to increasing amplitudes until structural failure. #### The Flutter Determinant For a system with $N$ degrees of freedom (DOF), the equations of motion in the frequency domain (assuming harmonic motion $h = h_0 e^{i\omega t}$, $\alpha = \alpha_0 e^{i\omega t}$) can be written in matrix form: $$[M] \ddot{\mathbf{q}} + [C] \dot{\mathbf{q}} + [K] \mathbf{q} = \mathbf{F}_A$$ where $\mathbf{q}$ is the vector of generalized coordinates, $[M]$ is the mass matrix, $[C]$ is the damping matrix, $[K]$ is the stiffness matrix, and $\mathbf{F}_A$ is the vector of generalized aerodynamic forces. For simple harmonic motion $\mathbf{q} = \mathbf{q}_0 e^{i\omega t}$, we have $\dot{\mathbf{q}} = i\omega \mathbf{q}$ and $\ddot{\mathbf{q}} = -\omega^2 \mathbf{q}$. The aerodynamic forces $\mathbf{F}_A$ are also dependent on $\mathbf{q}$, $\dot{\mathbf{q}}$, and $\omega$. They can be expressed as: $$\mathbf{F}_A = -\omega^2 [A_1] \mathbf{q} - i\omega [A_2] \mathbf{q} - [A_3] \mathbf{q}$$ where $[A_1]$, $[A_2]$, $[A_3]$ are matrices representing the added mass, aerodynamic damping, and aerodynamic stiffness, respectively. These matrices are generally complex and frequency-dependent. Substituting these into the equation of motion: $$-\omega^2 [M] \mathbf{q} + i\omega [C] \mathbf{q} + [K] \mathbf{q} = -\omega^2 [A_1] \mathbf{q} - i\omega [A_2] \mathbf{q} - [A_3] \mathbf{q}$$ Rearranging terms to get an eigenvalue problem: $$ \left( -\omega^2 ([M] + [A_1]) + i\omega ([C] + [A_2]) + ([K] + [A_3]) \right) \mathbf{q} = \mathbf{0} $$ For a non-trivial solution ($\mathbf{q} \neq \mathbf{0}$), the determinant of the coefficient matrix must be zero: $$\det \left( -\omega^2 ([M] + [A_1]) + i\omega ([C] + [A_2]) + ([K] + [A_3]) \right) = 0$$ This is the **Flutter Determinant**. The solution of this determinant yields complex eigenvalues, where the imaginary part relates to the damping and the real part to the frequency. Flutter occurs when the imaginary part (damping) becomes zero for a non-zero real part (frequency). #### 1-DOF Pitching System (Elastic Axis at Leading Edge) Consider a rigid wing section free to pitch about its leading edge, restrained by a torsional spring $K_\alpha$. Let the pitching DOF be $\alpha$. The EA is at the leading edge. The equation of motion for pitching about the leading edge is: $$I_{LE} \ddot{\alpha} + K_\alpha \alpha = M_{LE}$$ where $I_{LE}$ is the mass moment of inertia about the leading edge, and $M_{LE}$ is the aerodynamic moment about the leading edge. Using quasi-steady aerodynamics (for simplicity, though unsteady is needed for flutter): The aerodynamic moment about the leading edge (LE) is often expressed in terms of the moment about the AC ($M_{AC}$) and lift $L$: $$M_{LE} = M_{AC} - L \cdot x_{AC}$$ where $x_{AC}$ is the distance from LE to AC. For thin airfoils, the AC is typically at the quarter-chord, so $x_{AC} = c/4$. $$L = q S a_0 \alpha$$ $$M_{AC} = q S c m_0$$ (usually $m_0=0$ for symmetric airfoils at $\alpha=0$) So, $M_{LE} = - (q S a_0 \alpha) (c/4) = - q S a_0 (c/4) \alpha$. The equation of motion becomes: $$I_{LE} \ddot{\alpha} + K_\alpha \alpha = - q S a_0 (c/4) \alpha$$ $$I_{LE} \ddot{\alpha} + \left( K_\alpha + q S a_0 (c/4) \right) \alpha = 0$$ This is a simple harmonic oscillator equation. The effective stiffness is $K_{eff} = K_\alpha + q S a_0 (c/4)$. If $K_{eff}$ becomes negative, it implies divergence. For flutter, we need unsteady aerodynamics and at least two DOFs. A 1-DOF system cannot flutter in the classical sense because it lacks the necessary coupling between DOFs to extract energy. It can only diverge if the stiffness becomes negative. **Exam Tip:** - A true flutter analysis *requires* unsteady aerodynamics and typically at least two coupled degrees of freedom (e.g., plunge and pitch). A 1-DOF system can only exhibit divergence, not flutter. - For a 2-DOF system (plunge $h$ and pitch $\alpha$), the equations are coupled and involve complex aerodynamic terms. ### Module 4: Unsteady Aerodynamics (Theodorsen Theory) #### Physical Intuition Unsteady aerodynamics accounts for the time-dependent effects of airflow due to an oscillating body. Theodorsen's theory provides expressions for lift and moment on an oscillating airfoil, considering both the instantaneous motion and the shed wake. It's crucial for flutter analysis. #### Breaking Down the Unsteady Moment $M$ The unsteady aerodynamic moment $M$ on an oscillating airfoil (pitching $\alpha$ and plunging $h$) about the elastic axis can be conceptually broken into two main parts: 1. **Non-Circulatory (Added Mass) Part:** This part is associated with the inertia of the fluid that must be accelerated and decelerated as the airfoil moves. It's proportional to the acceleration of the airfoil and is independent of the wake. - **Physical Analogy:** Imagine pushing a plate through water; you not only accelerate the plate but also some surrounding water. - **Terms:** These terms typically involve $\ddot{h}$, $\ddot{\alpha}$, $\dot{\alpha}$ (if moment reference is not AC), and are proportional to $\rho b^2$ or $\rho b^3$. 2. **Circulatory (Wake-Influenced) Part:** This part arises from the circulation around the airfoil and the influence of the wake shed by the airfoil. As the airfoil oscillates, it sheds vorticity into the wake, which in turn affects the flow field around the airfoil. This effect is captured by the Theodorsen function $C(k)$. - **Physical Analogy:** The "memory" of the flow, where past motions influence current forces. - **Terms:** These terms are proportional to the velocity components and angles, and are multiplied by the Theodorsen function $C(k)$. Let's consider the general expression for the unsteady moment $M$ about the elastic axis (EA). For an airfoil oscillating in plunge $h$ and pitch $\alpha$ about the EA, with $x_{EA}$ being the distance from the leading edge to the EA (normalized by semi-chord $b$ as $a = (x_{EA}-b)/b$), the unsteady moment (positive nose-up) is given by: $$M = \pi \rho b^2 \left[ -U_\infty \dot{h} - U_\infty^2 \left( \frac{1}{2} - a \right) \dot{\alpha} + b \left( \frac{1}{8} - a \right) \ddot{\alpha} \right] $$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ U_\infty \dot{\alpha} + \dot{h} + U_\infty \alpha - b \left( \frac{1}{2} - a \right) \dot{\alpha} \right]$$ This is a complex expression. Let's simplify and identify terms. Assuming harmonic motion: $h = h_0 e^{i\omega t}$ $\alpha = \alpha_0 e^{i\omega t}$ $\dot{h} = i\omega h_0 e^{i\omega t} = i\omega h$ $\ddot{h} = -\omega^2 h_0 e^{i\omega t} = -\omega^2 h$ $\dot{\alpha} = i\omega \alpha_0 e^{i\omega t} = i\omega \alpha$ $\ddot{\alpha} = -\omega^2 \alpha_0 e^{i\omega t} = -\omega^2 \alpha$ Also, define reduced frequency $k = \frac{\omega b}{U_\infty}$. So $\omega = \frac{k U_\infty}{b}$. Substituting these into the moment equation (and grouping terms): $$M = \pi \rho b^2 \left[ (-\omega^2 h) \left( b \frac{1}{2} - a \right) - U_\infty^2 \left( \frac{1}{2} - a \right) \dot{\alpha} + b \left( \frac{1}{8} - a \right) \ddot{\alpha} \right] $$ This is a common representation: $$M = \pi \rho b^2 \left[ -U_\infty^2 \left( \left(\frac{1}{2}-a\right)\alpha + \left(\frac{1}{2}-a\right)\frac{b\dot{\alpha}}{U_\infty} - \frac{\ddot{h}}{U_\infty^2} - \left(\frac{1}{8}-a\right)\frac{b^2\ddot{\alpha}}{U_\infty^2} \right) \right.$$ $$\left. + 2 C(k) \left( U_\infty \alpha + \dot{h} + U_\infty \left(\frac{1}{2}-a\right) \frac{b\dot{\alpha}}{U_\infty} \right) \right]$$ A more standard form often used is: $$M = \pi \rho b^2 \left[ b \left( \frac{1}{8} - a \right) \ddot{\alpha} - U_\infty \left( b \left( \frac{1}{2} - a \right) + b \left( \frac{1}{2} - a \right) C(k) \right) \dot{\alpha} - \left( \frac{1}{2} - a \right) U_\infty C(k) \dot{h} \right] $$ $$+ \pi \rho b^2 \left[ U_\infty^2 \left( \left( \frac{1}{2} - a \right) C(k) \right) \alpha - \left( \frac{1}{2} - a \right) U_\infty C(k) h \right]$$ This equation is still quite dense. Let's refer to the problem statement's given form for $L$ and generalize to $M$. The provided expression for Lift $L$ is: $$L = \pi \rho b^2 [\ddot{h} - U_\infty \dot{\alpha} - b a \ddot{\alpha}] + 2 \pi \rho U_\infty b C(k) [\dot{h} + U_\infty \alpha + b (\frac{1}{2} - a) \dot{\alpha}]$$ Let's assume a similar structure for moment $M$ about EA. The unsteady moment $M$ about the elastic axis ($x_{EA}$) can be expressed as: $$M = M_{NC} + M_C$$ Where $M_{NC}$ is the non-circulatory (added mass) part and $M_C$ is the circulatory part. **Non-Circulatory Terms:** These are pure inertial terms, proportional to accelerations. For pitching motion about $x_{EA}$, the non-circulatory moment is: $$M_{NC} = \pi \rho b^4 \left( \frac{1}{8} + a^2 \right) \ddot{\alpha} - \pi \rho b^3 a \ddot{h}$$ (Note: this is for pitch about the mid-chord, then converted to EA. The constants like $1/8$ change based on the reference axis.) **Circulatory Terms:** These terms depend on $C(k)$ and are associated with the wake. $$M_C = -2 \pi \rho U_\infty b^2 C(k) \left[ U_\infty \alpha + \dot{h} + U_\infty b \left( \frac{1}{2} - a \right) \dot{\alpha} \right]$$ (The sign convention for moment might differ; this is a common form from Fung, "An Introduction to the Theory of Aeroelasticity"). #### Step-by-Step Non-Dimensionalization into $m_h$ and $m_\alpha$ The goal is to express the moment $M$ in a non-dimensional form. This typically involves dividing by $q S b$ and normalizing the kinematic variables by $b$ and $U_\infty$. Let's use the given form for the Lift $L$ to guide the non-dimensionalization for $M$. The problem asks to non-dimensionalize $M$ into $m_h$ and $m_\alpha$ from an expression similar to: $$M = \pi \rho b^2 [X \ddot{h} + Y \dot{h} + Z h + P \ddot{\alpha} + Q \dot{\alpha} + R \alpha] $$ where $X, Y, Z, P, Q, R$ are coefficients that may contain $U_\infty, b, a, C(k)$. Let's consider a generic moment equation in the form: $$M = M_h \frac{h}{b} + M_{\dot{h}} \frac{\dot{h}}{U_\infty} + M_{\ddot{h}} \frac{\ddot{h}b}{U_\infty^2} + M_\alpha \alpha + M_{\dot{\alpha}} \frac{\dot{\alpha}b}{U_\infty} + M_{\ddot{\alpha}} \frac{\ddot{\alpha}b^2}{U_\infty^2}$$ where $M_h, M_{\dot{h}}, \dots$ are coefficients. The problem asks for an expression like: $$M = \pi \rho b^3 \omega^2 \{ m_h \frac{h}{b} + m_\alpha \alpha \} e^{i\omega t}$$ This implies that $M$ is written in terms of non-dimensional coefficients $m_h$ and $m_\alpha$, which are themselves functions of $k$, $a$, etc. Let's assume the given moment expression (from the problem statement): $$M = \pi \rho b^2 \left[ b \left( \frac{1}{8} - a \right) \ddot{\alpha} - U_\infty b \left( \frac{1}{2} - a \right) \dot{\alpha} - U_\infty \dot{h} \right] $$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ U_\infty \alpha + \dot{h} + U_\infty b \left( \frac{1}{2} - a \right) \dot{\alpha} \right]$$ This is the moment about the EA. Let's rewrite it using harmonic motion and $k = \omega b / U_\infty$. Substitute: $\dot{h} = i\omega h$, $\ddot{h} = -\omega^2 h$, $\dot{\alpha} = i\omega \alpha$, $\ddot{\alpha} = -\omega^2 \alpha$. $$M = \pi \rho b^2 \left[ b \left( \frac{1}{8} - a \right) (-\omega^2 \alpha) - U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) - U_\infty (i\omega h) \right] $$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ U_\infty \alpha + (i\omega h) + U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) \right]$$ Group terms by $h$ and $\alpha$: **Terms with $h$:** $$M_h = \pi \rho b^2 [-i\omega U_\infty h] + 2 \pi \rho U_\infty b^2 C(k) [i\omega h]$$ $$M_h = \pi \rho b^2 i\omega h [-U_\infty + 2 U_\infty C(k)]$$ $$M_h = \pi \rho b^2 i\omega h U_\infty [-1 + 2 C(k)]$$ Recall $i\omega U_\infty = i \frac{k U_\infty}{b} U_\infty = i \frac{k U_\infty^2}{b}$. $$M_h = \pi \rho b^2 \left( i \frac{k U_\infty^2}{b} \right) h [-1 + 2 C(k)]$$ $$M_h = \pi \rho b k U_\infty^2 h i [-1 + 2 C(k)]$$ **Terms with $\alpha$:** $$M_\alpha = \pi \rho b^2 \left[ -b \omega^2 \left( \frac{1}{8} - a \right) \alpha - i\omega U_\infty b \left( \frac{1}{2} - a \right) \alpha \right] $$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ U_\infty \alpha + i\omega U_\infty b \left( \frac{1}{2} - a \right) \alpha \right]$$ Factor out $\alpha$: $$M_\alpha = \pi \rho b^2 \alpha \left[ -b \omega^2 \left( \frac{1}{8} - a \right) - i\omega U_\infty b \left( \frac{1}{2} - a \right) + 2 U_\infty b^2 C(k) \left( \frac{U_\infty}{b} + i\omega U_\infty \left( \frac{1}{2} - a \right) \right) \right]$$ This is getting very long. The key is to match the form: $M = \pi \rho b^3 \omega^2 \{ m_h \frac{h}{b} + m_\alpha \alpha \} e^{i\omega t}$. So we need to factor out $\pi \rho b^3 \omega^2$. Let's redefine the coefficients $m_h$ and $m_\alpha$ as given in the problem's example for $L$: $$L = -\pi \rho b^3 \omega^2 \left\{ L_h \frac{h}{b} + \left[ L_\alpha - L_h \left( \frac{1}{2} + a \right) a_0 \right] \alpha \right\} e^{i\omega t}$$ This implies $L_h$ and $L_\alpha$ are non-dimensional coefficients. The problem specifically asks to obtain $m_h$ and $m_\alpha$ from an expression similar to: $$M = \pi \rho b^3 \omega^2 \{m_h \frac{h}{b} + m_\alpha \alpha\}$$ Let's use the moment expression provided in the problem statement for M (Question 5a on the second OCR image): $$M = \pi \rho b^2 \left[ b a \ddot{h} - U_\infty b \left( \frac{1}{2} - a \right) \dot{\alpha} + b \left( \frac{1}{8} + a^2 \right) \ddot{\alpha} \right]$$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ \dot{h} + U_\infty \alpha + U_\infty b \left( \frac{1}{2} - a \right) \dot{\alpha} \right]$$ (Note: The expression for M in the prompt's OCR is slightly different from standard forms. I will use the one provided in the OCR image for consistency.) Substitute harmonic motion: $h = h_0 e^{i\omega t}$, $\dot{h} = i\omega h$, $\ddot{h} = -\omega^2 h$ $\alpha = \alpha_0 e^{i\omega t}$, $\dot{\alpha} = i\omega \alpha$, $\ddot{\alpha} = -\omega^2 \alpha$ $$M = \pi \rho b^2 \left[ b a (-\omega^2 h) - U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) + b \left( \frac{1}{8} + a^2 \right) (-\omega^2 \alpha) \right]$$ $$+ 2 \pi \rho U_\infty b^2 C(k) \left[ (i\omega h) + U_\infty \alpha + U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) \right]$$ Now, collect terms for $h$ and $\alpha$, and factor out $\pi \rho b^3 \omega^2$: **1. Terms involving $h$:** $$M_h = \pi \rho b^2 \left[ -b a \omega^2 h \right] + 2 \pi \rho U_\infty b^2 C(k) \left[ i\omega h \right]$$ $$M_h = -\pi \rho b^3 a \omega^2 h + 2 \pi \rho U_\infty b^2 C(k) i\omega h$$ To non-dimensionalize, we want to extract $\pi \rho b^3 \omega^2 \frac{h}{b}$: $$M_h = \pi \rho b^3 \omega^2 \frac{h}{b} \left[ -a + \frac{2 U_\infty b^2 C(k) i\omega}{\pi \rho b^3 \omega^2} \right]$$ $$M_h = \pi \rho b^3 \omega^2 \frac{h}{b} \left[ -a + \frac{2 U_\infty C(k) i}{\pi \rho b \omega} \right]$$ This doesn't seem to simplify nicely to the desired form. Let's recheck the question's format for $m_h$ and $m_\alpha$. The question is asking for $m_h$ and $m_\alpha$ from an expression like $M = \pi \rho b^3 \omega^2 \{m_h \frac{h}{b} + m_\alpha \alpha\}$. This means $m_h$ and $m_\alpha$ should be non-dimensional coefficients. Let's restart the non-dimensionalization carefully. We want to express $M$ as: $$M = (\pi \rho b^3 \omega^2) \left( m_h \frac{h}{b} + m_\alpha \alpha \right)$$ Divide the equation for $M$ by $\pi \rho b^3 \omega^2$: $$\frac{M}{\pi \rho b^3 \omega^2} = \frac{1}{\pi \rho b^3 \omega^2} \left( \pi \rho b^2 \left[ b a (-\omega^2 h) - U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) + b \left( \frac{1}{8} + a^2 \right) (-\omega^2 \alpha) \right] \right.$$ $$\left. + 2 \pi \rho U_\infty b^2 C(k) \left[ (i\omega h) + U_\infty \alpha + U_\infty b \left( \frac{1}{2} - a \right) (i\omega \alpha) \right] \right)$$ Simplify the common factor $\pi \rho b^2$: $$\frac{M}{\pi \rho b^3 \omega^2} = \frac{1}{b \omega^2} \left( \left[ -b a \omega^2 h - i\omega U_\infty b \left( \frac{1}{2} - a \right) \alpha - b \left( \frac{1}{8} + a^2 \right) \omega^2 \alpha \right] \right.$$ $$\left. + 2 U_\infty C(k) \left[ i\omega h + U_\infty \alpha + i\omega U_\infty b \left( \frac{1}{2} - a \right) \alpha \right] \right)$$ Now, collect terms for $h$ and $\alpha$: **For $m_h \frac{h}{b}$:** Coefficient of $h$: $$ \frac{1}{b \omega^2} \left[ -b a \omega^2 + 2 U_\infty C(k) i\omega \right] h $$ $$ = \left[ -a + \frac{2 U_\infty C(k) i\omega}{b \omega^2} \right] h $$ $$ = \left[ -a + \frac{2 U_\infty C(k) i}{b \omega} \right] h $$ Since $k = \frac{\omega b}{U_\infty} \implies \frac{U_\infty}{b\omega} = \frac{1}{k}$: $$ = \left[ -a + \frac{2 i C(k)}{k} \right] h $$ So, $m_h \frac{h}{b} = \left[ -a + \frac{2 i C(k)}{k} \right] h$. Therefore, $m_h = \left[ -a + \frac{2 i C(k)}{k} \right] \frac{b}{h}$... This is not right. $m_h$ should be dimensionless and independent of $h$. Let's re-examine the target form: $M = \pi \rho b^3 \omega^2 \{m_h \frac{h}{b} + m_\alpha \alpha\}$. This implies $m_h$ is the coefficient of $h/b$ and $m_\alpha$ is the coefficient of $\alpha$. Let's factor out $h/b$ and $\alpha$ from the original moment equation: **Coefficient of $h/b$ (to become $m_h$):** Terms containing $h$: $$- \pi \rho b^3 a \omega^2 h + 2 \pi \rho U_\infty b^2 C(k) i\omega h$$ Factor out $\pi \rho b^3 \omega^2 \frac{h}{b}$: $$ \pi \rho b^3 \omega^2 \frac{h}{b} \left[ -a + \frac{2 U_\infty b^2 C(k) i\omega}{\pi \rho b^3 \omega^2 \frac{h}{b}} h \right] $$ This approach is incorrect. The $h$ should be divided out. Let's directly find $m_h$ and $m_\alpha$ by dividing the total moment by $\pi \rho b^3 \omega^2$ and then separating terms. $$ \frac{M}{\pi \rho b^3 \omega^2} = \frac{1}{\pi \rho b^3 \omega^2} \left( \text{terms with } h + \text{terms with } \alpha \right) $$ **1. Collect terms with $h$:** $$ T_h = \pi \rho b^2 \left[ b a (-\omega^2 h) \right] + 2 \pi \rho U_\infty b^2 C(k) \left[ i\omega h \right] $$ $$ T_h = h \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right) $$ Now, divide $T_h$ by $\pi \rho b^3 \omega^2$ and then by $h/b$: $$ m_h = \frac{T_h / (\pi \rho b^3 \omega^2)}{h/b} $$ $$ m_h = \frac{b}{h} \frac{h \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right)}{\pi \rho b^3 \omega^2} $$ $$ m_h = \frac{b}{\pi \rho b^3 \omega^2} \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right) $$ $$ m_h = -a + \frac{2 U_\infty C(k) i\omega b}{\pi \rho b^3 \omega^2} \cdot \frac{\pi \rho b^3 \omega^2}{\pi \rho b^3 \omega^2} $$ This simplification is messy. Let's try again by factoring out $\pi \rho b^3 \omega^2$ from the beginning. Let $M = M_h + M_\alpha$. $M_h = h \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right)$ $M_\alpha = \alpha \left( -\pi \rho b^2 U_\infty b (\frac{1}{2}-a) i\omega - \pi \rho b^3 (\frac{1}{8}+a^2) \omega^2 + 2 \pi \rho U_\infty b^2 C(k) (U_\infty + U_\infty b (\frac{1}{2}-a) i\omega) \right)$ Now, we need to express this in the form $\pi \rho b^3 \omega^2 (m_h \frac{h}{b} + m_\alpha \alpha)$. **For $m_h$:** $$ m_h \frac{h}{b} \cdot \pi \rho b^3 \omega^2 = h \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right) $$ $$ m_h = \frac{b}{\pi \rho b^3 \omega^2} \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b^2 C(k) i\omega \right) $$ $$ m_h = -a + \frac{2 U_\infty C(k) i\omega b}{\omega^2 b^3} = -a + \frac{2 U_\infty C(k) i}{k U_\infty}$$ $$ m_h = -a + \frac{2 i C(k)}{k} $$ This looks like a standard form for $m_h$. **For $m_\alpha$:** $$ m_\alpha \alpha \cdot \pi \rho b^3 \omega^2 = \alpha \left( -\pi \rho b^3 (\frac{1}{2}-a) i\omega \frac{U_\infty}{b} - \pi \rho b^3 (\frac{1}{8}+a^2) \omega^2 + 2 \pi \rho U_\infty^2 b^2 C(k) + 2 \pi \rho U_\infty^2 b^3 C(k) (\frac{1}{2}-a) i\omega/b \right) $$ Divide by $\pi \rho b^3 \omega^2 \alpha$: $$ m_\alpha = \frac{1}{\pi \rho b^3 \omega^2} \left( -\pi \rho b^3 (\frac{1}{2}-a) i\omega \frac{U_\infty}{b} - \pi \rho b^3 (\frac{1}{8}+a^2) \omega^2 + 2 \pi \rho U_\infty^2 b^2 C(k) + 2 \pi \rho U_\infty^2 b^2 C(k) (\frac{1}{2}-a) i\omega b/U_\infty \right) $$ Simplify terms: $$ m_\alpha = -(\frac{1}{2}-a) \frac{i\omega U_\infty}{\omega^2 b} - (\frac{1}{8}+a^2) + \frac{2 U_\infty^2 C(k)}{\omega^2 b} + \frac{2 U_\infty^2 C(k) (\frac{1}{2}-a) i\omega b}{\omega^2 b U_\infty} $$ $$ m_\alpha = -(\frac{1}{2}-a) \frac{i U_\infty}{\omega b} - (\frac{1}{8}+a^2) + \frac{2 U_\infty^2 C(k)}{\omega^2 b^2} + \frac{2 U_\infty C(k) (\frac{1}{2}-a) i}{\omega} $$ Using $k = \frac{\omega b}{U_\infty} \implies \frac{U_\infty}{\omega b} = \frac{1}{k}$ and $\frac{U_\infty^2}{\omega^2 b^2} = \frac{1}{k^2}$: $$ m_\alpha = -(\frac{1}{2}-a) \frac{i}{k} - (\frac{1}{8}+a^2) + \frac{2 C(k)}{k^2} + \frac{2 U_\infty C(k) (\frac{1}{2}-a) i}{\omega} $$ The last term needs to be simplified further: $\frac{2 U_\infty C(k) (\frac{1}{2}-a) i}{\omega} = \frac{2 U_\infty C(k) (\frac{1}{2}-a) i}{\frac{k U_\infty}{b}} = \frac{2 b C(k) (\frac{1}{2}-a) i}{k}$. This is a bit different from the standard forms. Let me re-verify the initial moment formulation. The standard Theodorsen moment (positive nose-up) about the elastic axis at $x_{EA} = b(1+a)$ is: $$ M = -\pi \rho b^4 \ddot{\alpha} \left( \frac{1}{8} + a^2 \right) + \pi \rho b^3 a \ddot{h} + \frac{1}{2} \rho U_\infty^2 (2 \pi b) [ C(k) \alpha + C(k) \frac{\dot{h}}{U_\infty} + C(k) b (\frac{1}{2}-a) \frac{\dot{\alpha}}{U_\infty} ] $$ $$ M = -\pi \rho b^4 \ddot{\alpha} \left( \frac{1}{8} + a^2 \right) + \pi \rho b^3 a \ddot{h} + 2 \pi \rho U_\infty^2 b C(k) \left[ \alpha + \frac{\dot{h}}{U_\infty} + b (\frac{1}{2}-a) \frac{\dot{\alpha}}{U_\infty} \right] $$ This is a more commonly accepted form. Let's use this to derive $m_h$ and $m_\alpha$. Substitute harmonic motion: $$ M = -\pi \rho b^4 (-\omega^2 \alpha) \left( \frac{1}{8} + a^2 \right) + \pi \rho b^3 a (-\omega^2 h) + 2 \pi \rho U_\infty^2 b C(k) \left[ \alpha + \frac{i\omega h}{U_\infty} + b (\frac{1}{2}-a) \frac{i\omega \alpha}{U_\infty} \right] $$ $$ M = \pi \rho b^4 \omega^2 \alpha \left( \frac{1}{8} + a^2 \right) - \pi \rho b^3 a \omega^2 h + 2 \pi \rho U_\infty^2 b C(k) \alpha + 2 \pi \rho U_\infty b C(k) i\omega h + 2 \pi \rho U_\infty b^2 C(k) (\frac{1}{2}-a) i\omega \alpha $$ Collect terms for $h$ and $\alpha$: **Terms for $h$:** $$ T_h = -\pi \rho b^3 a \omega^2 h + 2 \pi \rho U_\infty b C(k) i\omega h $$ Factor out $\pi \rho b^3 \omega^2 \frac{h}{b}$: $$ m_h = \frac{b}{h} \frac{T_h}{\pi \rho b^3 \omega^2} = \frac{b}{\pi \rho b^3 \omega^2} \left( -\pi \rho b^3 a \omega^2 + 2 \pi \rho U_\infty b C(k) i\omega \right) $$ $$ m_h = -a + \frac{2 U_\infty b C(k) i\omega}{\omega^2 b^2} = -a + \frac{2 U_\infty C(k) i}{\omega b} $$ Using $k = \frac{\omega b}{U_\infty} \implies \frac{U_\infty}{\omega b} = \frac{1}{k}$: $$ m_h = -a + \frac{2 i C(k)}{k} $$ **Terms for $\alpha$:** $$ T_\alpha = \pi \rho b^4 \omega^2 \alpha \left( \frac{1}{8} + a^2 \right) + 2 \pi \rho U_\infty^2 b C(k) \alpha + 2 \pi \rho U_\infty b^2 C(k) (\frac{1}{2}-a) i\omega \alpha $$ Factor out $\pi \rho b^3 \omega^2 \alpha$: $$ m_\alpha = \frac{1}{\alpha} \frac{T_\alpha}{\pi \rho b^3 \omega^2} = \frac{1}{\pi \rho b^3 \omega^2} \left( \pi \rho b^4 \omega^2 \alpha \left( \frac{1}{8} + a^2 \right) + 2 \pi \rho U_\infty^2 b C(k) \alpha + 2 \pi \rho U_\infty b^2 C(k) (\frac{1}{2}-a) i\omega \alpha \right) $$ $$ m_\alpha = b \left( \frac{1}{8} + a^2 \right) + \frac{2 U_\infty^2 C(k)}{\omega^2 b^2} + \frac{2 U_\infty C(k) (\frac{1}{2}-a) i\omega}{\omega^2 b} $$ $$ m_\alpha = \left( \frac{1}{8} + a^2 \right) + \frac{2 U_\infty^2 C(k)}{\omega^2 b^2} + \frac{2 U_\infty C(k) (\frac{1}{2}-a) i}{\omega b} $$ Using $k = \frac{\omega b}{U_\infty} \implies \frac{U_\infty}{\omega b} = \frac{1}{k}$ and $\frac{U_\infty^2}{\omega^2 b^2} = \frac{1}{k^2}$: $$ m_\alpha = \left( \frac{1}{8} + a^2 \right) + \frac{2 C(k)}{k^2} + \frac{2 C(k) (\frac{1}{2}-a) i}{k} $$ This looks more consistent with standard results for unsteady aerodynamic moment coefficients. **Exam Tip:** - Be extremely careful with sign conventions and the reference axis for moments (leading edge, mid-chord, elastic axis). The parameter 'a' (distance from mid-chord to elastic axis, normalized by semi-chord $b$) is critical. - The Theodorsen function $C(k)$ is a complex function of reduced frequency $k$, representing the circulatory contribution. It accounts for the wake effect. - The non-circulatory terms are often called "added mass" terms because they are proportional to accelerations, similar to inertial forces.