Question 8.2: (a) Use the classical Rayleigh–Ritz method to derive a dynam......
(a) Use the classical Rayleigh–Ritz method to derive a dynamic matrix enabling the approximate natural frequencies and normal modes of a uniform cantilever beam, in bending, to be found. Use the following two-term series to represent the displacement:
y = a_{1} x² + a_{2} x³ (A)where x is the distance along the beam from the fixed end. The length of the beam is L; Young’s modulus is E; the second moment of area of the cross-section is I and the mass per unit length is μ .
(b) Show that the same result can be obtained by the use of Lagrange’s equations, taking q_{1} \mathrm{and} q_{2} as generalized coordinates, defined by:
Part(a)
From Eq. (A):
From Eq. (8.26), using Eqs (C) and (E):
R = \int_{0}^{L}{EI \left(\frac{\partial^{2}y}{\partial x^{2}}\right) ^{2}}\;dx – \omega^{2} \int_{0}^{L}{\mu y^{2}}\;dx (8.26)
= EI (4a^{2}_{1}L + 12a_{1}a_{2}L^{2} + 12a^{2}_{2}L^{3}) – m_{x}\omega^{2} (\frac{a^{2}_{1} L^{5}}{5} + \frac{a_{1}a_{2}L^{6}}{3} + \frac{a^{2}_{2}L^{7}}{7}) (G)\\ \frac{\partial R}{\partial a_{1}} = EI (8a_{1}L + 12a_{2}L^{2}) – \mu \omega^{2} \left(\frac{2}{5}a_{1}L^{5} + \frac{1}{3}a_{2}L^{6} \right) = 0 (H)\\ \frac{\partial R}{\partial a_{2}} = EI (12a_{1}L^{2} + 24a_{2}L^{3}) – \mu \omega^{2} \left(\frac{1}{3}a_{1}L^{6} + \frac{2}{7}a_{2}L^{7} \right) = 0 (I)
or combining Eqs (H) and (I) in matrix form:
This can now be seen to be an eigenvalue problem, and it can be solved in any of the usual ways to obtain the two natural frequencies and the two ratios (a_{1} / a_{2})_{1} \mathrm{and} (a_{1} / a_{2})_{2}. The two normal mode shapes are then found from Eq. (A).
Part (b)
The equivalent modern approach is to use Lagrange’s equations to derive the equations of motion in terms of the generalized coordinates q_{1} \mathrm{and} q_{2} (which replace a_{1} \mathrm{and} a_{2}). With no damping and no external forces, Lagrange’s equations are
To find the kinetic energy, T:
From Eq. (8.19),
From Eq. (B),
y = q_{1}x² + q_{2}x³ \\ \mathrm{and} \\ \frac{\partial y}{\partial t} = \dot{q} _{1}x² + \dot{q} _{2}x³ (M)\\ \mathrm{so} \\ \left(\frac{\partial y}{\partial t}\right)^{2} = \dot{q}^{2}_{1}x^{4} + 2\dot{q}_{1}\dot{q}_{2}x^{5} + \dot{q}^{2}_{2}x^{6} (N)From Eqs (L) and (N):
T = \frac{1}{2} \mu \int_{0}^{L}{ \left(\dot{q}^{2}_{1}x^{4} + 2\dot{q}_{1}\dot{q}_{2}x^{5} + \dot{q}^{2}_{2}x^{6} \right) dx} = \frac{1}{2} \mu \left(\frac{L^{5}}{5}\dot{q}^{2}_{1} + \frac{L^{6}}{3}\dot{q}_{1}\dot{q}_{2} + \frac{L^{7}}{7}\dot{q}^{2}_{2} \right) (O)To find the potential energy, U:
From Eq. (8.18):
From Eq. (E), replacing a_{1} \mathrm{by} q_{1} \mathrm{and} a_{2} \mathrm{by} q_{2} in each case, we have
\left(\frac{\partial^{2}y}{\partial x^{2}}\right) ^{2} = 4q^{2}_{1} + 24q_{1}q_{2}x + 36q^{2}_{2}x^{2} (Q)From Eqs (P) and (Q):
U = \frac{1}{2} EI \int_{0}^{L}{\left(4q^{2}_{1} + 24q_{1}q_{2}x + 36q^{2}_{2}x^{2}\right) dx} = EI (2L^{2}q^{2}_{1} + 6L^{2}q_{1}q_{2} + 6L^{3}q^{2}_{2}) (R)Using Eqs (O) and (R), the partial derivatives required for Lagrange’s equations are
\frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}_{1}} \right) = \mu \left(\frac{L^{5}}{5}\ddot{q}_{1} + \frac{L^{6}}{6}\ddot{q}_{2} \right) (S_{1}) \\ \mathrm{and} \\ \frac{\partial}{\partial t} \left(\frac{\partial T}{\partial \dot{q}_{2}} \right) = \mu \left(\frac{L^{6}}{6}\ddot{q}_{1} + \frac{L^{7}}{7}\ddot{q}_{2} \right) (S_{2}) \\ \frac{\partial U}{\partial q_{1}} = EI (4Lq_{1} + 6L^{2}q_{2}) (T_{1})\\ \mathrm{and} \\ \frac{\partial U}{\partial q_{2}} = EI (6L^{2}q_{1} + 12L^{3}q_{2}) (T_{2})Substituting – \omega^{2}q_{1} \mathrm{for} \ddot{q}_{1} \mathrm{and} – \omega^{2}q_{2} \mathrm{for} \ddot{q}_{2} , and writing in matrix form:
\left(EI \begin{bmatrix} 4L & 6L^{2} \\ 6L^{2} & 12L^{3} \end{bmatrix} – \mu \omega^{2} \begin{bmatrix} \frac{1}{5}L^{5} & \frac{1}{6}L^{6} \\ \frac{1}{6}L^{6} & \frac{1}{7}L^{7} \end{bmatrix} \right) \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = 0 (U)Comparing Eq. (U) with Eq. (J), it can be seen that each term in Eq. (J) is twice as large as the corresponding term in Eq. (U). As pointed out above, this is due to the use of Eq. (8.26), the historic version of the Rayleigh–Ritz equation, in Part (a) of the solution, and Lagrange’s equations in Part (b). It is of no importance, since both sets of equations are equated to zero.