Question 1.6: Use Rayleigh’s energy method to find the natural frequency, ......
Use Rayleigh’s energy method to find the natural frequency, \omega _{1}, of the fundamental bending mode of the uniform cantilever beam shown in Fig. 1.10, assuming that the vibration mode shape is given by:
\frac{y}{y_{T}}=\left(\frac{x}{L} \right)^{2} (A)
where y_{T} is the single-peak amplitude at the tip; y the vertical displacement of beam at distance x from the root; L the length of beam; m the mass per unit length; E the Young’s modulus and I the second moment of area of beam cross-section.
From Eq. (A) it is assumed that
y=y_{T}\frac{x^{2}}{L^{2}}\sin \omega t (B)
Then,
\dot{y} = \omega _{1}y_{T} \frac{x^{2}}{L^{2}} \cos \omega t (C)
The maximum kinetic energy, T_{max}, occurs when cos ωt = 1, i.e. when \dot{y} = \omega _{1}y_{T} x^{2} / L^{2} , and is given by:
T_{max}=\int_{0}^{L}{\frac{1}{2}m \dot{y} }^{2} =\frac{1}{2}\frac{m}{L^{4}}\omega ^{2}_{1} y^{2}_{T}\int_{0}^{L}{x^{4} . dx} =\frac{1}{10} m L \omega ^{2}_{1} y^{2}_{T} (D)
To find the maximum potential energy, we need an expression for the maximum curvature as a function of x.
From Eq. (A),
so
\frac{d y}{d x} = y_{T}\frac{2x}{L^{2}} \sin \omega tand
\frac{d^{2}y}{d x^{2}} = \frac{2y_{T}}{L^{2}} \sin \omega ttherefore
\left(\frac{d^{2} y}{d x^{2}}\right)_{max} = \frac{2y_{T}}{L^{2}} (E)
This is independent of x in this particular case, the curvature being constant along the beam. The standard expression for the potential energy in a beam due to bending is
U = \frac{1}{2} \int_{0}^{L}{EI \left(\frac{d^{2} y}{dx^{2}} \right) ^{2} } \cdot dx (F)
The maximum potential energy U_{max} is given by substituting Eq. (E) into Eq. (F), giving
U_{max} = \frac{2EI \cdot y^{2}_{T} }{L^{3}} (G)
Now the basis of the Rayleigh method is that T_{max}=U_{max}. Thus, from Eqs (D) and (G) we have
\frac{1}{10} mL\omega ^{2}_{1} y_{T}^{2} = \frac{2EI \cdot y^{2}_{T} }{L^{3}},
which simplifies to
\omega _{1} = \frac{\sqrt{20} }{L^{2}}\sqrt{\frac{EI}{m} } = \frac{4.47}{L^{2}}\sqrt{\frac{EI}{m}} (H)
The exact answer, from Chapter 8, is \frac{3.52}{L^{2}}\sqrt{\frac{EI}{m}} , so the Rayleigh method is somewhat inaccurate in this case. This was due to a poor choice of function for the assumed mode shape.