UDC 593.3:534.1

THEORY OF ELASTICITY

B. I. KOGAN, P. Ya. BERSHTEIN

VIBRATIONS OF A ROTATING DISK UNDER PARTIAL BRAKING

(Presented by Academician I. N. Vekua on 9 X 1968)

Vibrations of an elastic disk under complete braking were considered by us in [1]. In the case of partial braking, the radial and tangential elastic displacements are coupled with one another through the rotational inertial force; therefore, the system of differential equations for the elastic displacements does not split into two independent equations, as in the case [1]. We have succeeded in reducing the solution of our system to Bessel’s equation. The stresses are represented by convergent series of trigonometric and Bessel functions; moreover, for the shear stress we were able to obtain a simple, readily surveyable formula.

Let a circular, isotropic, homogeneous, elastic disk of radius \(R\) and density \(\rho\) rotate about the central axis perpendicular to its plane with angular velocity \(\omega\). At some instant of time \(t = 0\), a torque is applied to an absolutely rigid rim mounted on the disk, such that the angular velocity becomes equal to \(\omega_0 < \omega\). In this case the material of the disk will be in a generalized plane stress state, independent of the polar angle \(\varphi\), which it is convenient to consider in a moving polar coordinate system rigidly connected with the rim, so that its pole coincides with the center of the disk. Then the differential equations of motion will have the form [2, 3]

\[ \frac{1}{c_2^2}\left[\frac{\partial^2 u_r}{\partial t^2} - \omega_0^2 r - 2\omega_0 \frac{\partial u_\varphi}{\partial t}\right] = \frac{\partial}{\partial r} \left\{ \frac{1}{r}\frac{\partial}{\partial r}(u_r r) \right\}; \]

\[ \frac{1}{c_2^2}\left[\frac{\partial^2 u_\varphi}{\partial t^2} + 2\omega_0 \frac{\partial u_r}{\partial t}\right] = \frac{\partial}{\partial r} \left\{ \frac{1}{r}\frac{\partial}{\partial r}(u_\varphi r) \right\}, \tag{1} \]

where \(c_1^2 = (\lambda^* + 2\mu)/\rho\), \(c_2^2 = \mu/\rho\), \(\lambda^* = 2\lambda\mu/(\lambda + 2\mu)\), and \(\lambda\) and \(\mu\) are the Lamé coefficients.

Thus, in contrast to complete braking (\(\omega_0 = 0\)), the radial and tangential displacements are coupled with one another. This creates additional difficulties in solving system (1). The indicated system must be solved under the following initial and boundary conditions:

\[ u_r(r,0) = f(r) = \omega^2 r(R^2 - r^2)/8c_1^2 \]

(\(f(r)\) is the value of \(u_r\) in a disk rotating uniformly with angular velocity \(\omega\)):

\[ \frac{\partial u_\varphi(r,0)}{\partial t} = \begin{cases} (\omega-\omega_0)r = \omega_1 r, & 0 \leqslant r < R,\\ 0, & r = R, \end{cases} \quad u_\varphi\big|_{t=0} = \partial u_\varphi/\partial t\big|_{t=0} = 0 \tag{2} \]

\[ u_r(R,t) = u_\varphi(R,t) = 0. \tag{3} \]

Applying to (1) the one-sided Laplace transform, and taking (2) into account, we obtain a system of equations for the \(L\)-images of the sought functions [4]:

\[ \frac{1}{c_1^2} \left[ -p f(r) + p^2 \bar{u}_r - \frac{\omega_0^2 r}{p} - 2\omega_0 p \bar{u}_\varphi \right] = \frac{d}{dr} \left\{ \frac{1}{r}\frac{d}{dr}(\bar{u}_r r) \right\}, \]

\[ \frac{1}{c_2^2}\left[p^2\bar u_\varphi-\omega_1 r-2\omega_0 f'(r)+2\omega_0 p\bar u_r\right] = \frac{d}{dr}\left\{\frac{1}{r}\frac{d}{dr}(\bar u_\varphi r)\right\}. \tag{4} \]

Expressing \(\bar u_\varphi\) in terms of \(\bar u_r\) from the first equation (4) and substituting it into the second, we obtain a differential equation with variable coefficients for \(\bar u_r\),

\[ D^2\bar u_r-\frac{p^2(c_1^2+c_2^2)}{c_1^2c_2^2}D\bar u_r + \frac{p^2(4\omega_0^2+p^2)}{c_1^2c_2^2}\bar u_r = \frac{p\omega^2 r}{c_1^4} + \frac{p(p^2+4\omega_0^2)\omega^2 r(R^2-r^2)}{8c_1^4c_2^2} + \frac{p(\omega_0^2+2\omega_1\omega_0)}{c_1^2c_2^2}\,r, \tag{5} \]

where \(D\) is the differential operator

\[ D\bar u_r=\frac{d}{dr}\left\{\frac{1}{r}\frac{d}{dr}(\bar u_r r)\right\}. \]

Consider the homogeneous equation

\[ D^2\bar u_r-aD\bar u_r+b\bar u_r=0, \tag{6} \]

\[ a=p^2(c_1^2+c_2^2)/c_1^2c_2^2,\qquad b=p^2(4\omega_0^2+p^2)/c_1^2c_2^2. \]

We write the equation in the form

\[ D\bar u_r+\varkappa \bar u_r=0, \tag{7} \]

where \(\varkappa\) is a coefficient to be determined. Applying the operation \(D\) to equation (7), we obtain

\[ D^2u_r+\varkappa D\bar u_r=0. \tag{8} \]

Eliminating \(D u_r\) from (7) and (8), we obtain:

\[ D^2\bar u_r-\varkappa^2\bar u_r=0. \tag{9} \]

Multiplying now (7) by \(a\) and subtracting from (9), we have

\[ D^2\bar u_r-aD\bar u_r-(\varkappa^2+a\varkappa)\bar u_r=0. \tag{10} \]

Comparing (10) with (6), we find \(a\varkappa+\varkappa^2=-b\), whence

\[ \varkappa_{1,2}=-a/2\pm\sqrt{a^2/4-b}. \]

Thus it is sufficient for us to find the solution of the equation

\[ D\bar u_r+\varkappa_i\bar u_r=0\qquad (i=1,2), \tag{11} \]

which reduces to Bessel’s equation. A particular solution of the nonhomogeneous equation is found without difficulty.

Knowing \(\bar u_r\) and using the first equation (4), we find \(\bar u_\varphi\). The expressions for the originals follow from the inversion formula for the \(L\)-transform, using the residue theorem \((^4)\).

\[ u_r= \sum_{k=1}^{\infty}2\left\{ \frac{a_k^2(\omega-\omega_0)^2+c_2^2(\omega_0^2-\omega^2)\alpha_k^2} {a_k\alpha_k^2\,[\alpha_k^2(c_1^2+c_2^2)-2a_k^2+4\omega_0^2]}\cos a_k t - \right. \]

\[ \left. - \frac{(\omega_0^2-\omega^2)\alpha_k^2c_2^2+(\omega-\omega_0)^2b_k^2} {\alpha_k b_k^2\,[2b_k^2-\alpha_k^2(c_1^2+c_2^2)-4\omega_0^2]}\cos b_k t \right\} \frac{J_1(\alpha_k r)}{J_0(\alpha_k R)} + \frac{\omega_0^2 r(R^2-r^2)}{8c_1^2}, \]

\[ u_\varphi= \frac{1}{2\omega_0}\frac{\partial u_r}{\partial t} + \frac{c_1^2}{\omega_0} \sum_{k=1}^{\infty}\left\{ \frac{a_k\,[a_k^2(\omega-\omega_0)^2+c_2^2(\omega_0^2-\omega^2)\alpha_k^2]} {a_k^3[\alpha_k^2(c_1^2+c_2^2)-2a_k^2+4\omega_0^2]}\sin a_k t - \right. \]

\[ \left. - \frac{\alpha_k[(\omega_0^2-\omega^2)\alpha_k^2c_2^2+(\omega-\omega_0)^2b_k^2]} {b_k^3[2b_k^2-\alpha_k^2(c_1^2+c_2^2)-4\omega_0^2]} \sin b_k t \right\} \frac{J_1(\alpha_k r)}{J_0(\alpha_k R)}. \tag{12} \]

where \(\alpha_k\) are the roots of the equation

\[ J_1(\alpha_k R)=0, \]

\(J_1(\alpha_k r)\) and \(J_0(\alpha_k R)\) are Bessel functions of the first kind,

\[ a_k=\sqrt{ \frac{ \alpha_k^2(c_1^2+c_2^2)+4\omega_0^2 -\sqrt{\alpha_k^4(c_1^2-c_2^2)^2+8\omega_0^2\alpha_k^2(c_1^2+c_2^2)+16\omega_0^4} }{2} }, \]

\[ b_k=\sqrt{ \frac{ \alpha_k^2(c_1^2+c_2^2)+4\omega_0^2 +\sqrt{\alpha_k^4(c_1^2-c_2^2)^2+8\omega_0^2\alpha_k^2(c_1^2+c_2^2)+16\omega_0^4} }{2} } \]

are the natural frequencies.

Let us note that the expression for \(u_r\) consists of two terms: the first describes the elastic vibrations of the disk as a result of braking, and the second describes the elastic displacements of points of the disk under uniform rotation with angular velocity \(\omega\). The presence of the displacement \(u_\varphi\) is entirely caused by partial braking. Under uniform rotation it is absent.

We now investigate the stressed state of the disk. The expressions for \(\sigma_{rr}\) and \(\sigma_{\varphi\varphi}\) are represented by series converging as \(1/\alpha_k^2\) and \(1/\alpha_k^3\).

As for the series for \(\tau_{r\varphi}\), it consists of two parts, one of which converges as \(1/\alpha_k^2\), and the other as \(1/\alpha_k\).

Using the asymptotic expressions for the Bessel functions and the formulas for trigonometric series (5), we write the expression for \(\tau_{r\varphi}\)

\[ \tau_{r\varphi}= \frac{\mu(\omega-\omega_0)R\sqrt{R}}{\pi c_2\sqrt{r}} \left\{ \frac{\pi(c_2t+r)}{2R} \cos\frac{\pi}{4R}(c_2t+r-R) \right. \]

\[ \left. +\sin\frac{\pi}{4R}(c_2t+r-R)\, \ln 2\cos\frac{\pi(c_2t+r)}{2R} +\frac{\pi}{2R}(c_2t-r)\sin\frac{\pi}{4R}(R+r-c_2t) \right. \]

\[ \left. +\cos\frac{\pi}{4R}(c_2t-r-R)\, \ln 2\cos\frac{\pi(c_2t-r)}{2R} +S(r,t) \right\} \]

\[ \left(0\leq t<\frac{R-r}{c_2}\right); \]

\[ S(r,t)\text{ is a series converging as }\sum_k O\!\left(\frac{1}{k^2}\right) \]

\[ \tau_{r\varphi}= \frac{\mu(\omega-\omega_0)R\sqrt{R}}{\pi c_2\sqrt{r}} \left\{ \frac{\pi[c_2t+r-2nR]}{2R} \cos\frac{\pi(c_2t+r-R)}{4R} -\right. \]

\[ \left. -\sin\frac{\pi}{4R}(c_2t+r-R)\cdot \frac12\ln \frac{1}{2\left[1+\cos\frac{\pi}{R}(c_2t+r)\right]} +\right. \]

\[ \left. +\frac{\pi[c_2t-r-2nR]}{2R} \sin\frac{\pi(R+r-c_2t)}{4R} -\right. \]

\[ \left. -\cos\frac{\pi}{4R}(c_2t-r-R)\cdot \frac12\ln \frac{1}{2\left[1+\cos\frac{\pi}{R}(c_2t-r)\right]} +S(r,t) \right\}; \]

\[ (2n-1)/c_2<t<[(2n+1)R-r]/c_2\quad (n=1,2,\ldots). \]

For

\[ \frac{R-r}{c_2}<t<\frac{R+r}{c_2} \]

\[ \frac{\pi[c_2t-r-2nR]}{2R} \]

must be replaced by

\[ \frac{\pi}{2R}(c_2t-r), \]

and

\[ -\frac12\ln \frac{1}{2\left[1+\cos\frac{\pi}{R}(c_2t-r)\right]} \]

by

\[ \ln\left[2\cos\frac{\pi(c_2t-r)}{2R}\right]. \]

At each point of the disk, at certain instants of time

\[ t_k=\frac{(2k-1)R+r}{c_2} \]

\((k=1,2,\ldots)\), the stress \(\tau_{r\varphi}\) becomes infinite. Thus there occurs

the so-called “phase impact.” The time interval between two successive “phase impacts” at any point of the disk is

\[ t_{k+1}-t_k=2R/c_2. \]

Thus, in the case of partial braking an infinitely large tangential stress arises. This indicates that, under real conditions, tangential stresses may be of an order higher than \(\sigma_{rr}\) and \(\sigma_{\varphi\varphi}\).

All-Union Scientific-Research Institute
of Hydraulic Engineering and Land Reclamation
named after N. Kostyakov

Received
30 IX 1968

References

1. P. Ya. Bershtein, B. I. Kogan, Abstracts of Reports, Third All-Union Congress on Theoretical and Applied Mechanics, Moscow, 1968.
2. I. N. Sneddon, D. S. Berry, Classical Theory of Elasticity, Moscow, 1961.
3. L. G. Loitsyanskii, A. I. Lur’e, Course of Theoretical Mechanics, Moscow–Leningrad, 1948.
4. H. Karslow, D. Eger, Operational Methods in Applied Mathematics, IL, 1948.
5. I. S. Gradshteyn, I. M. Ryzhik, Tables of Integrals, Sums, Series, and Products, Moscow, 1962.