Exact solution of the Mindlin–Herrmann model for longitudinal vibration of an isotropic rod

Abstract

This paper presents a new approach to the problem of coupled longitudinal and transversal propagations of stress waves in an isotropic thick and elastic rod, based on the Mindlin–Herrmann theory. The novelty is that Hamilton’s variational principle is used not only for derivation of the governing equations and set of natural boundary conditions, but also for obtaining the exact solution in terms of Green’s functions directly from the Lagrangian. The success of this approach is based on the existence of multiple orthogonalities of the eigenfunctions. The proposed method is much easier than the standard approach of building Green’s functions. A numerical example illustrates the method of finding eigenfrequencies and eigenfunctions for isotropic Mindlin–Herrmann rod.

Appendices

Appendix 1: Derivation of the Mindlin–Herrmann model

The aim here is to show how to derive the equations of motion and the associated boundary conditions using the Hamilton variational principle.

Using Eq. (1) the strains are obtained as follows:

$$\begin{aligned}&\varepsilon _{xx}=\frac{\partial u}{\partial x}=u^{\prime }_{1},\quad \varepsilon _{yy}=\frac{\partial v}{\partial y}=u_{2},\quad \varepsilon _{zz}=\frac{\partial v}{\partial z}=u_{2}=\varepsilon _{yy}, \nonumber \\&\varepsilon _{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=yu^{\prime }_{2},\quad \varepsilon _{yz}=\frac{\partial v}{\partial z}+\frac{\partial w}{\partial y}=0,\quad \varepsilon _{zx}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=zu^{\prime }_{2}. \end{aligned}$$

(73)

The stresses of the rod are

$$\begin{aligned} \begin{array}{l} \sigma _{xx}=(\lambda +2\mu )\varepsilon _{xx}+\lambda (\varepsilon _{yy}+\varepsilon _{zz})=(\lambda +2\mu )u^{\prime }_{1}+2\lambda u_{2}, \\ \sigma _{yy}=(\lambda +2\mu )\varepsilon _{yy}+\lambda (\varepsilon _{xx}+\varepsilon _{zz})=2(\lambda +\mu )u_{2}+\lambda u^{\prime }_{1}, \\ \sigma _{zz}=(\lambda +2\mu )\varepsilon _{zz}+\lambda (\varepsilon _{xx}+\varepsilon _{yy})=2(\lambda +\mu )u_{2}+\lambda u^{\prime }_{1}=\sigma _{yy}, \\ \sigma _{xy}=\mu \varepsilon _{xy}=\mu yu^{\prime }_{2},\quad \sigma _{yz}=\mu \varepsilon _{yz}=0,\quad \sigma _{zx}=\mu \varepsilon _{zx}=\mu zu^{\prime }_{2}. \end{array} \end{aligned}$$

(74)

Kinetic energy is as follows:

$$\begin{aligned} K=\frac{\rho }{2}\int _{0}^{l}\int _{(s)}\left( \dot{u}^{2}+\dot{v}^{2}+\dot{w}^{2}\right) \mathrm{d}s\;\mathrm{d}x. \end{aligned}$$

(75)

Substituting expression (1) into Eq. (6) leads to

$$\begin{aligned} K=\frac{\rho }{2}\int _{0}^{l}\left( S\dot{u}^{2}_{1}+\dot{u}^{2}_{2}I_{p}\right) \mathrm{d}x. \end{aligned}$$

(76)

The arguments in all functions are sometimes omitted for simplicity.

Strain energy is as follows:

$$\begin{aligned} P=\frac{\rho }{2}\int _{0}^{l}\int _{(s)}\left( \sigma _{xx}\varepsilon _{xx}+\sigma _{yy}\varepsilon _{yy}+\sigma _{zz}\varepsilon _{zz}+\sigma _{xy}\varepsilon _{xy}+\sigma _{yz}\varepsilon _{yz}+\sigma _{zx}\varepsilon _{zx}\right) \mathrm{d}s\;\mathrm{d}x, \end{aligned}$$

(77)

where $\sigma _{ij}$, $\varepsilon _{ij}$ are given by Eqs. (73) and (74). Hence

$$\begin{aligned} P=\frac{1}{2}\int _{0}^{l}\left\{ S\left[ (\lambda +2\mu )u^{\prime 2}_{1}+4\lambda u_{2} u^{\prime }_{1}+4(\lambda +\mu )u^{2}_{2}\right] +\mu I_{p}u^{\prime 2}_{2}\right\} \mathrm{d}x. \end{aligned}$$

(78)

Let W be the work done by the distributed force $f=f(x,t)$

$$\begin{aligned} W=\int _{0}^{l}\int _{(s)}fu_{1}\mathrm{d}s\;\mathrm{d}x=\int _{0}^{l}fu_{1}S\;\mathrm{d}x. \end{aligned}$$

(79)

The Lagrangian is as follows:

$$\begin{aligned} \displaystyle L= & {} \displaystyle K-P+W \nonumber \\= & {} \displaystyle \frac{1}{2}\int _{0}^{l}\left\{ \rho S\dot{u}^{2}_{1}+\rho I_{p}\dot{u}^{2}_{2}-(\lambda +2\mu )Su^{\prime 2}_{1}-4\lambda Su^{\prime }_{1}u_{2} -4(\lambda +\mu )S u^{2}_{2}\right\} \mathrm{d}x \nonumber \\&+\,\frac{1}{2}\int _{0}^{l}\left\{ 2fu_{1}S-\mu I_{p}u^{\prime 2}_{2}\right\} \mathrm{d}x. \end{aligned}$$

(80)

Applying the Hamiltonian principle to the Lagrange functional Eq. (80), we obtain the system of equations of motion in general form:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{u}_1}\right) +\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial L}{\partial u^{\prime }_1}\right) -\frac{\partial L}{\partial u_1}=0, \quad \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{u}_2}\right) +\frac{\mathrm{d}}{\mathrm{d}x}\left( \frac{\partial L}{\partial u^{\prime }_2}\right) -\frac{\partial L}{\partial u_2}=0 \end{aligned}$$

(81)

and the associated boundary conditions in general form is as follows

$$\begin{aligned} u_{1}\bigg |_{x=0,l}=0,\; u_{2}\bigg |_{x=0,l}=0 \quad \text{ for } \text{ fixed } \text{ ends } \end{aligned}$$

(82)

or

(83)

We assume that the rod has a constant cross-section and that parameters such as $\lambda$, $\mu$, S and $I_{P}$ are constants. Hence, the explicit form of Eq. (81) is as follows:

$$\begin{aligned} \rho \ddot{u}_{1}-(\lambda +2\mu )u^{\prime \prime }_{1}-2\lambda u^{\prime }_{2}=f(x,t), \quad \rho I_{p}\ddot{u}_{2}+4(\lambda +\mu )Su_{2}-\mu I_{p}u^{\prime \prime }_{2}+2S\lambda u^{\prime }_{1}=0 \end{aligned}$$

(84)

with the following corresponding boundary conditions:

$$\begin{aligned} u_{1}|_{x=0,l}=0,\;u_{2}|_{x=0,l}=0 \quad \text{ for } \text{ fixed } \text{ ends } \end{aligned}$$

(85)

or

$$\begin{aligned} (\lambda +2\mu )u^{\prime }_{1}+2\lambda u_{2}|_{x=0,l}=0,\;\mu I_{p}u^{\prime }_{2}|_{x=0,l}=0 \quad \text{ for } \text{ free } \text{ ends. } \end{aligned}$$

(86)

Appendix 2: Analysis of the operator of the Sturm–Liouville problem

In this section, our goal is to determine the nature of the operator A of the Sturm–Liouville problem.

We firstly prove that A is self-adjoint on the class of functions $C^{2}(D)$ satisfying boundary condition (20). Let $\mathbf u =\left( \begin{array}{c} u_{1} \\ u_{2} \\ \end{array} \right)$ and $\mathbf v =\left( \begin{array}{c} v_{1} \\ v_{2} \\ \end{array} \right)$. We define the scalar product as follows:

$$\begin{aligned}&(\mathbf u ,\mathbf v )=\int _{0}^{l}\mathbf u \cdot \mathbf v \mathrm{d}x, \end{aligned}$$

(87)

$$\begin{aligned}&\displaystyle (A\mathbf u ,\mathbf v ) = \displaystyle \int _{0}^{l}A\mathbf u \cdot \mathbf v \mathrm{d}x \nonumber \\&\qquad \quad = \displaystyle \int _{0}^{l}\left( a_{11}\frac{\mathrm{d}^{2}u_{1}}{\mathrm{d}x^{2}}v_{1}+a_{12}\frac{\mathrm{d}u_{2}}{\mathrm{d}x}v_{1}\right) \mathrm{d}x \quad +\,\int _{0}^{l}\left( -a_{12}\frac{\mathrm{d}u_{1}}{\mathrm{d}x}v_{2}+a_{22}\frac{\mathrm{d}^{2}u_{2}}{\mathrm{d}x^{2}}v_{2}-c_{22}u_{2}v_{2}\right) \mathrm{d}x. \end{aligned}$$

(88)

Integrating twice and once by part the terms with the second and first derivatives, respectively, of the expression (88) and applying boundary conditions (20), we obtain

$$\begin{aligned} (A\mathbf u ,\mathbf v )= & {} \displaystyle \int _{0}^{l}\left( a_{11}u_{1}\frac{\mathrm{d}^{2}v_{1}}{\mathrm{d}x^{2}}-a_{12}u_{2}\frac{\mathrm{d}v_{1}}{\mathrm{d}x}\right) \mathrm{d}x \quad +\,\int _{0}^{l}\left( a_{12}u_{1}\frac{\mathrm{d}v_{2}}{\mathrm{d}x}+a_{22}u_{2}\frac{\mathrm{d}^{2}v_{2}}{\mathrm{d}x^{2}}-c_{22}u_{2}v_{2}\right) \mathrm{d}x \nonumber \\= & {} \int _{0}^{l}u_{1}\left( a_{11}\frac{\mathrm{d}^{2}v_{1}}{\mathrm{d}x^{2}}+a_{12}\frac{\mathrm{d}v_{2}}{\mathrm{d}x}\right) \mathrm{d}x \quad +\,\int _{0}^{l}u_{2}\left( -a_{12}\frac{\mathrm{d}v_{1}}{\mathrm{d}x}+a_{22}\frac{\mathrm{d}^{2}v_{2}}{\mathrm{d}x^{2}}-c_{22}v_{2}\right) \mathrm{d}x \nonumber \\= & {} \int _{0}^{l}\left( \begin{array}{c} u_{1} \\ u_{2} \\ \end{array} \right) \cdot A\left( \begin{array}{c} v_{1} \\ v_{2} \\ \end{array} \right) \mathrm{d}x = (\mathbf u ,A\mathbf v ). \end{aligned}$$

(89)

Equality (89) shows that the operator is self-adjoint which means that all the eigenvalues of the Sturm–Liouville problem (19)–(20) are real.