Investigation of quasi-periodic response of a buckled beam under harmonic base excitation with an “unexplained” sideband structure

Abstract

The aim of this paper is to investigate quasi-periodic response of a fixed–fixed buckled beam experimentally studied in detail by Kreider and Nayfeh (Nonlinear Dyn 15(2):155–177, 1998. https://doi.org/10.1023/A:1008231012968) with an “unexplained” sideband structure. After a description of the equation of motion of the buckled beam with quadratic and cubic nonlinearities, it is shown that the proposed approach, the incremental harmonic balance (IHB) method with two time scales, can be used to find the quasi-periodic response of the buckled beam with the “unexplained” sideband structure and its understanding is revealed. The traditional IHB method with a single time scale is first used to automatically track periodic and period-doubling solutions of nonlinear responses of the buckled beam, and stability and bifurcations of periodic solutions are investigated using the Floquet theory. In the case of 1:1 internal resonance between the first two modes of the buckled beam, it is found that anti-symmetrical modes cannot be excited due to small excitation amplitudes. However, the anti-symmetrical modes are excited via period-doubling bifurcation with an increased excitation amplitude. By continuously increasing the excitation amplitude, Hopf bifurcation occurs, which leads to quasi-periodic response whose spectrum contains uniformly spaced sidebands around integer multiples of a half of the excitation frequency, where the uniformly spaced sidebands were called an “unexplained” sideband structure in Kreider and Nayfeh (1998). The results obtained from the IHB method with two time scales are shown to well correlate with the experimental results in Kreider and Nayfeh (1998). Moreover, results obtained from the IHB method with two time scales are in excellent agreement with those from numerical integration using the fourth-order Runge–Kutta method.

Appendix A Derivation of the equation of motion of the nonlinear beam with fixed–fixed boundary conditions

Consider the nonlinear buckled beam with fixed–fixed boundaries. The bending moment $M\left( {\tilde{x}},{\tilde{t}}\right)$ of the beam at position ${\tilde{x}}$ and time ${\tilde{t}}$ is

$$\begin{aligned} M\left( {\tilde{x}},{\tilde{t}}\right) =EI{{\tilde{y}}}'', \end{aligned}$$

(A.1)

where E is the Young’s modulus, I is the area moment of inertia of a cross section, ${{\tilde{y}}({\tilde{x}},{\tilde{t}})}$ is the transverse displacement of the beam at position ${{\tilde{x}}}$ and time ${{\tilde{t}}}$, and a prime denotes a partial derivative with respect to x. The axial strain $\varepsilon$ is defined by

$$\begin{aligned} \varepsilon =\frac{\partial {U}}{\partial {{\tilde{x}}}}+\frac{1}{2}{\left( \frac{\partial {{\tilde{y}}}}{\partial {{\tilde{x}}}}\right) }^{2}, \end{aligned}$$

(A.2)

where $U\left( {\tilde{x}},{\tilde{t}}\right)$ is the longitudinal displacement of the beam at position ${\tilde{x}}$ and time ${\tilde{t}}$. Consider governing equations for nonlinear beam vibrations in the following form [28]:

$$\begin{aligned} \rho \frac{\partial ^{2}{{\tilde{y}}}}{\partial {{{\tilde{t}}}^2}}+\frac{\partial ^{2}{M}}{\partial {{{\tilde{x}}}^2}}-\frac{\partial }{\partial {\tilde{x}}}\left( T\frac{\partial {\tilde{y}}}{\partial {\tilde{x}}}\right) =0, \end{aligned}$$

(A.3)

$$\begin{aligned} \rho \frac{\partial ^{2}U}{\partial {{\tilde{t}}}^2}-\frac{\partial {T}}{\partial {{\tilde{x}}}}=0, \end{aligned}$$

(A.4)

where $T=EA\varepsilon$, in which A is the area of a cross section. The Kirchhoff hypothesis is that the axial inertial term in Eq. (A.4) can be neglected. One then obtains

$$\begin{aligned} \frac{\partial {T}}{\partial {{\tilde{x}}}}=0 \end{aligned}$$

(A.5)

and consequently

$$\begin{aligned} \varepsilon =\frac{\partial {U}}{\partial {{\tilde{x}}}}+\frac{1}{2}{\left( \frac{\partial {{\tilde{y}}}}{\partial {{\tilde{x}}}}\right) }^{2}=\text {const}. \end{aligned}$$

(A.6)

Integrating Eq. (A.6) over the spatial domain $\left[ 0,l\right]$ yields

$$\begin{aligned} \varDelta {l}= & {} U^{\left( l\right) }-U^{\left( 0\right) } +\frac{1}{2}\int _{0}^{l}{{{\left( {\frac{{\partial {\tilde{y}}}}{{\partial {\tilde{x}}}}}\right) }^{2}}\mathrm{d}{\tilde{x}}}\nonumber \\= & {} -U^b+\frac{1}{2}\int _{0}^{l}{{{\left( {\frac{{\partial {\tilde{y}}}}{{\partial {\tilde{x}}}}}\right) }^{2}}\mathrm{d}{\tilde{x}}}, \end{aligned}$$

(A.7)

where $U^{\left( 0\right) }$ and $U^{\left( l\right) }$ are the constant longitudinal displacements of the beam at ${\tilde{x}}=0$ and ${\tilde{x}}=l$, respectively. Therefore, the axial force in the beam is given by

$$\begin{aligned} T=\frac{EA}{l}\left[ -U^b+\frac{1}{2}\int _{0}^{l}{{{\left( {\frac{{\partial {\tilde{y}}}}{{\partial {\tilde{x}}}}}\right) }^{2}}\mathrm{d}{\tilde{x}}}\right] . \end{aligned}$$

(A.8)

Substituting Eq. (A.8) into Eq. (A.3) yields

$$\begin{aligned}&\rho \frac{\partial ^{2}{{\tilde{y}}}}{\partial {{{\tilde{t}}}^2}}\nonumber \\&\quad +EI\frac{\partial ^{4}{\tilde{y}}}{\partial {{\tilde{x}}^{4}}}\nonumber \\&\quad +\frac{EA}{l}\left[ U_b-\frac{1}{2}\int _{0}^{l}{\left( \frac{\partial {\tilde{y}}}{\partial {\tilde{x}}}\right) }^{2}\mathrm{d}{\tilde{x}}\right] \frac{\partial ^2{\tilde{y}}}{\partial {{\tilde{x}}}^2}=0. \end{aligned}$$

(A.9)