Question 4.7.3: Consider the wing vibration model of Figure 4.20. Using the ......

Let m denote the mass of the wing section and J denote the rotational inertia about point G. The kinetic energy is

T=\frac{1}{2} m \dot{x}_G^2+\frac{1}{2} J \dot{\theta}^2

where x_G is the displacement of the point G. This displacement is related to the coordinate, x(t), of the point of attachment of the springs by

x_G(t)=x(t)-e \sin \theta(t)

which is obtained by examining the geometry of Figure 4.20. Thus \dot{x}_G(t) becomes

\dot{x}_G(t)=\dot{x}(t)-e \cos \theta(t) \frac{d \theta}{d t}=\dot{x}(t)-e \dot{\theta} \cos \theta

The expression for kinetic energy in terms of the generalized coordinates q_1=x and q_2=\theta then becomes

T=\frac{1}{2} m[\dot{x}-e \dot{\theta} \cos \theta]^2+\frac{1}{2} J \dot{\theta}^2

The expression for potential energy is

U=\frac{1}{2} k_1 x^2+\frac{1}{2} k_2 \theta^2

which is already in terms of the generalized coordinates. The Lagrangian, L, becomes

L=T-U=\frac{1}{2} m[\dot{x}-e \dot{\theta} \cos \theta]^2+\frac{1}{2} J \dot{\theta}^2-\frac{1}{2} k_1 x^2-\frac{1}{2} k_2 \theta^2

Calculating the derivatives required by equation (4.146) for i=1 yields

(4.146)

so that equation (4.146) becomes

m \ddot{x}-m {e} \ddot{\theta} \cos \theta+e m \dot{\theta}^2 \sin \theta+k_1 x=0

Assuming small motions so that the approximations \cos \theta \rightarrow 1, and \sin \theta \rightarrow \theta hold, and assuming that the term \dot{\theta}^2 \theta is small enough to ignore, results in a linear equation in x(t) given by

m \ddot{x}-m \ddot{\theta}+k_1 x=0

Calculating the derivatives of the Lagrangian required by equation (4.146) for i=2 yields

so that equation (4.146) becomes

\ddot{J \theta}-m e \cos \theta \ddot{x}+m e^2 \cos ^2 \theta \ddot{\theta}-m e^2 \dot{\theta}^2 \sin \theta \cos \theta+k_2 \theta=0

Again if the small-angle, small-motion approximation (i.e., \sin \theta \rightarrow \theta, \cos \theta \rightarrow 1, \left.\dot{\theta}^2 \theta \rightarrow 0\right) is used, a linear equation in \theta(t) results given by

\left(J+m e^2\right) \ddot{\theta}-m \ddot{x}+k_2 \theta=0

Combining the expression for i=1 and i=2 into one vector equation in the generalized vector x =\left[\begin{array}{ll}q_1(t) & q_2(t)\end{array}\right]^T=\left[\begin{array}{ll}x(t) & \theta(t)\end{array}\right]^T yields

\left[\begin{array}{cc} m & -m e \\ -m e & m e^2+J \end{array}\right]\left[\begin{array}{c} \ddot{x}(t) \\ \ddot{\theta}(t) \end{array}\right]+\left[\begin{array}{cc} k_1 & 0 \\ 0 & k_2 \end{array}\right]\left[\begin{array}{c} x(t) \\ \theta(t) \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \end{array}\right]

Note here that the two equations of motion are coupled, not through stiffness terms, as in Example 4.7.2, but rather through the inertia terms. Such systems are called dynamically coupled, meaning that the terms that couple the equation in \theta(t) to the equation in x(t) are in the mass matrix (i.e., meaning that the mass matrix is not diagonal). In all previous examples, the mass matrix is diagonal and the stiffness matrix is not diagonal. Such systems are called statically coupled. Dynamically coupled systems have nondiagonal mass matrices, and hence require the use of the Cholesky decomposition for factoring the mass matrix \left(M=L^T L\right) in the modal analysis steps of Window 4.5 (replacing M^{-1 / 2} with L ). This is discussed in Section 4.9. The programs in the Toolbox and the various codes given in Section 4.9 are capable of solving dynamically coupled systems as easily as those that have a diagonal mass matrix.