Abstract Generated abstract
The paper studies unsteady flexural oscillations of an unbounded free three-layer strip under assumptions that reduce the plate model to a single equation for a displacement function. For the special case in which the relative flexural stiffness parameter equals one, the governing equation is transformed by Fourier and Laplace methods, yielding an integral representation for an auxiliary function and its kernel. The analysis derives a convolution formula suitable for arbitrary loading and then specializes it to a unit impulse moving at constant velocity along the strip. The resulting expressions provide the auxiliary function and, through a further integral relation, the displacement function, while indicating the convergence and asymptotic character of the oscillatory integrals involved.
Full Text
Reports of the Academy of Sciences of the USSR
1967. Volume 172, No. 5
UDC 533.3:534.1
THEORY OF ELASTICITY
A. I. SMIRNOV
UNSTEADY OSCILLATIONS OF A FREE THREE-LAYER STRIP
(Presented by Academician L. I. Sedov on 6 April 1966)
§ 1. It is known \((^{1,2})\) that the problem of flexural vibrations of a three-layer plate reduces, under certain assumptions, to a single partial differential equation with respect to the displacement function \(\chi(x,t)\). In the case of an unbounded plate this equation may be written in the form
\[ (1-\vartheta k\nabla^2)\nabla^2\nabla^2\chi(x,\tau) +\chi_{\tau\tau}(x,\tau) -k\nabla^2\chi_{\tau\tau}(x,\tau) = \frac{l^4}{D}q_0(x,\tau), \tag{1,1} \]
where \(\nabla^2=\partial^2/\partial x^2\); \(\chi_{\tau\tau}=\partial^2\chi/\partial\tau^2\); \(q_0(x,\tau)\) is the external load; \(\tau\) is dimensionless time; \(\vartheta, k\) are dimensionless parameters characterizing the relative flexural stiffness of the load-bearing layers and the relative shear stiffness of the core; \(l\) is a linear quantity; \(D\) is the flexural stiffness of the three-layer beam (see \((^{1,2})\)).
The deflection \(w(x,\tau)\) is related to the displacement function by the relation
\[ w(x,\tau)=l(1-k\nabla^2)\chi(x,\tau). \tag{1,2} \]
For \(\vartheta=1\), equation (1,1) takes the form
\[ -k\left(\nabla^2\nabla^2\chi-\frac{1}{k}\chi\right) -k\left(\nabla^2\chi-\frac{1}{k}\chi\right)_{\tau\tau} = q(x,\tau), \tag{1,3} \]
where
\[ q(x,\tau)=\frac{l^4}{D}q_0(x,\tau). \tag{1,4} \]
Let us denote
\[ \Phi(x,\tau)=\nabla^2\chi(x,\tau)+a_0\chi(x,\tau), \tag{1,5} \]
where
\[ a_0=-1/k. \tag{1,6} \]
Substituting (1,4)—(1,6) into (1,3), we obtain
\[ \nabla^2\nabla^2\Phi(x,\tau)+\Phi_{\tau\tau}(x,\tau)=a_0q(x,\tau). \tag{1,7} \]
Applying successively to (1,7) the Fourier and Laplace transforms with respect to the variables \(x\) and \(\tau\), we obtain
\[ \Phi^{*}(u,s)=\frac{1}{\omega^2+s^2}\,\overline{q}^{\,*}(u,s) \tag{1,8} \]
(\(u,s\) are the transform parameters), where
\[ \omega^2=u^4. \tag{1,9} \]
According to the convolution rule,
\[ \overline{\Phi}(u,\tau)=\frac{a_0}{\omega}\int_0^\tau \overline{q}(u,\tau)\sin\omega(\tau-\tau_1)\,d\tau_1. \tag{1,10} \]
The original (1.10) has the form
\[ \Phi(x,\tau)=\frac{a_0}{2\pi}\int_{-\infty}^{\infty} e^{-ixu}\left[\frac{1}{\omega}\int_0^\tau \bar q(u,\tau_1)\sin \omega(\tau-\tau_1)\,d\tau_1\right]du. \tag{1.11} \]
Since
\[ \bar q(u,\tau)=\int_{-\infty}^{\infty} q(x,\tau)e^{iux}\,dx, \tag{1.12} \]
expression (1.11) becomes the following:
\[ \Phi(x,\tau)=\frac{a_0}{2\pi}\int_0^\tau \int_{-\infty}^{\infty} q(v,\tau_1)\,dv \int_{-\infty}^{\infty} e^{iu(v-x)} \frac{\sin u^2(\tau-\tau_1)}{u^2}\,du. \tag{1.13} \]
Transform the right-hand side of (1.13). Denote
\[ I(\mu)=\int_{-\infty}^{\infty} e^{i\xi u}\frac{\sin \mu u^2}{u^2}\,du. \tag{1.14} \]
Compute the partial derivative of \(I(\mu)\) with respect to \(\mu\):
\[ \frac{\partial I(\mu)}{\partial \mu} = \int_{-\infty}^{\infty} e^{i\xi u}\cos \mu u^2\,du = \frac{1}{2}\int_{-\infty}^{\infty} e^{i\xi u+i\mu u^2}\,du + \frac{1}{2}\int_{-\infty}^{\infty} e^{i\xi u-i\mu u^2}\,du. \tag{1.15} \]
Transform the integral
\[ \int_{-\infty}^{\infty} e^{i\xi u+i\mu u^2}\,du = \int_{-\infty}^{\infty} e^{i\mu(u+\xi/2\mu)^2-i\xi^2/4\mu}\,du = \sqrt{\frac{\pi}{|\mu|}}\,e^{i\cdot 1/4\pi\,\operatorname{sgn}\mu-i\xi^2/4\mu}. \tag{1.16} \]
Similarly, we find
\[ \int_{-\infty}^{\infty} e^{i\xi u-i\mu u^2}\,du = \sqrt{\frac{\pi}{|\mu|}}\,e^{-i\cdot 1/4\pi\,\operatorname{sgn}\mu+i\xi^2/4\mu}. \tag{1.17} \]
Taking (1.16) and (1.17) into account, we write (1.15) in the form
\[ \frac{\partial I}{\partial \mu} = \sqrt{\frac{\pi}{|\mu|}}\cos\left(\frac{\pi}{4}-\frac{\xi^2}{4|\mu|}\right). \tag{1.18} \]
Integrating (1.18), we obtain
\[ I(\mu)=I(0)+\int_0^\mu \frac{\partial I}{\partial \mu}\,d\mu. \tag{1.19} \]
Bearing in mind the zero initial conditions, set \(I(0)=0\). Then
\[ I(\mu)=\int_0^\mu \sqrt{\frac{\pi}{|\mu|}}\cos\left(\frac{\pi}{4}-\frac{\xi^2}{4|\mu|}\right)d\mu = \operatorname{sgn}\mu\int_0^{|\mu|}\sqrt{\frac{\pi}{u}}\cos\left(\frac{\pi}{4}-\frac{\xi^2}{4u}\right)du. \tag{1.20} \]
Putting \(\xi=v-x\), \(\mu=\tau-\tau_1\), we obtain
\[ \int_{-\infty}^{\infty} e^{iu(v-x)} \frac{\sin u^2(\tau-\tau_1)}{u^2}\,du = I(\tau-\tau_1) = \int_0^{\tau-\tau_1}\sqrt{\frac{\pi}{u}}\cos\left(\frac{\pi}{4}-\frac{v-x}{4u}\right)du. \tag{1.21} \]
Changing the order of integration in the right-hand side of (1.13) and using (1.20), we find
\[ \Phi(x,\tau)=-\frac{a_0}{2\sqrt{\pi}}\int_0^\tau \frac{d\xi}{\sqrt{\xi}} \int_0^{\tau-\xi} d\tau_1 \int_{-\infty}^{\infty} q(v,\tau_1)\cos\left(\frac{\pi}{4}-\frac{v-x}{4\xi}\right)\,dv. \tag{1.22} \]
It is hardly possible to evaluate the integrals on the right-hand side of (1.22) for an arbitrary function \(q(v,\tau)\) and an arbitrary range of variation of the variables. However, some asymptotic estimates are possible.
§ 2. Let the load \(q(v,\tau)\) be a unit impulse moving with velocity \(V\) in the positive direction of the \(x\)-axis. Then
\[ \int_{-\infty}^{\infty} q(v,\tau_1)\cos\left(\frac{\pi}{4}-\frac{v-x}{4\xi}\right)\,dv = \cos\left(\frac{\pi}{4}-\frac{V_*\tau_1-x}{4\xi}\right), \tag{2.1} \]
where \(V_* = V/{}^{[2]}\!\sqrt{\Omega/D}\) is the dimensionless velocity of motion of the load. Taking (2.1) into account, we write (1.22) in the form
\[ \Phi(x,\tau)=-\frac{a_0}{2\sqrt{\pi}}\int_0^\tau \frac{d\xi}{\sqrt{\xi}} \int_0^{\tau-\xi} \cos\left(\frac{\pi}{4}-\frac{V_*\tau_1-x}{4\xi}\right)\,d\tau_1, \tag{2.2} \]
or, evaluating the inner integral,
\[ \Phi(x,\tau)= -\frac{2a_0}{V_*\sqrt{\pi}}\int_0^\tau \sqrt{\xi}\, \sin\left[\frac{\pi}{4}-\frac{V_*(\tau-\xi)-x}{4\xi}\right]\,d\xi + \]
\[ +\frac{2a_0}{V_*\sqrt{\pi}}\int_0^\tau \sqrt{\xi}\, \sin\left(\frac{\pi}{4}+\frac{x}{4\xi}\right)\,d\xi. \tag{2.3} \]
The integrands on the right-hand side of (2.2) oscillate strongly near \(\xi=0\). It is obvious, however, that the integrals converge. If (1.22) has been evaluated, then, using (1.6), we find
\[ \overline{\Phi}(u,\tau)=(u^2-b^2)\chi(u,\tau), \tag{2.4} \]
where \(b=1/\sqrt{k}\).
Finally, the displacement function \(\chi(x,\tau)\) will be
\[ \chi(x,\tau)=\frac{1}{b}\int_0^\tau \Phi(x,\tau_1)\operatorname{sh} b(\tau-\tau_1)\,d\tau_1, \tag{2.5} \]
where \(\Phi(x,\tau)\) is determined according to (2.3).
The problem was proposed by E. I. Grigolyuk. The author expresses gratitude to him and to K. I. Babenko for remarks and advice.
All-Union Institute of Scientific
and Technical Information
Received
16 III 1966
REFERENCES
- E. I. Grigolyuk, P. P. Chulkov, DAN, 149, No. 1 (1963).
- E. I. Grigolyuk, P. P. Chulkov, Izv. AN SSSR, Mekh. i mashinostr., No. 1 (1964).