Critical velocities of anisotropic tubes under a moving pressure incorporating transverse shear and rotary inertia effects

Abstract

Closed-form expressions for critical velocities and middle-surface displacements of anisotropic tubes subjected to a uniform internal pressure moving at a constant speed are derived using a first-order shear deformation (FSD) model for axisymmetric orthotropic cylindrical shells. The FSD shell model is first formulated employing a variational method based on Hamilton’s principle, which incorporates both the transverse shear and rotary inertia effects. The general 3-D constitutive relations for orthotropic elastic materials are utilized to describe the tube anisotropy, which enables a unified treatment of orthotropic, transversely isotropic, cubic and isotropic tubes representing various composite and metallic cylindrical shells. Closed-form formulas for four critical velocities of the anisotropic tube are derived for the general case with the transverse shear, rotary inertia and radial stress effects, which divide the range of the pressure velocity into four segments. For each of these segments, closed-form expressions for the mid-surface radial displacement are obtained. By suppressing the transverse shear, rotary inertia or radial stress effect, the general formulas are reduced to those for special cases. In particular, when the transverse shear, rotary inertia and radial stress effects are all neglected, the newly derived critical velocity formulas for orthotropic and isotropic tubes recover the two existing ones for thin tubes. To quantitatively illustrate the new model, a numerical example is provided for an isotropic tube, where eight values of the lowest critical velocity are directly determined using newly derived formulas and compared with an existing value computationally obtained by others.

Appendices

Appendix A

The derivation of Eq. (23) from Eqs. (22b) and (22c) is provided here.

From Eq. (22b), it follows that

$$ \frac{{\partial \psi_{x} }}{\partial x} = \frac{2}{{hk_{s} C_{55} }}\left[ {\rho h\ddot{w} - f_{z} + \frac{h}{R}\left( {C_{12} \frac{\partial u}{{\partial x}} + \frac{{C_{22} }}{R}w} \right)} \right] - \frac{{\partial^{2} w}}{{\partial x^{2} }}. $$

(A.1)

Taking the derivative with respective to x on both sides of Eq. (22c) gives

$$ \frac{{h^{3} C_{11} }}{12}\frac{{\partial^{3} \psi_{x} }}{{\partial x^{3} }} - \frac{{hk_{s} C_{55} }}{2}\left( {\frac{{\partial^{2} w}}{{\partial x^{2} }} + \frac{{\partial \psi_{x} }}{\partial x}} \right) = \frac{{\rho h^{3} }}{12}\frac{{\partial^{3} \psi_{x} }}{{\partial x\partial t^{2} }}. $$

(A.2)

Substituting Eq. (A.1) into Eq. (A.2) yields

$$ \begin{gathered} \frac{{h^{3} C_{11} }}{12}\left\{ {\frac{2}{{hk_{s} C_{55} }}\left[ {\rho h\frac{{\partial^{4} w}}{{\partial x^{2} \partial t^{2} }} - \frac{{\partial^{2} f_{z} }}{{\partial x^{2} }} + \frac{h}{R}\left( {C_{12} \frac{{\partial^{3} u}}{{\partial x^{3} }} + \frac{{C_{22} }}{R}\frac{{\partial^{2} w}}{{\partial x^{2} }}} \right)} \right] - \frac{{\partial^{4} w}}{{\partial x^{4} }}} \right\} \hfill \\ \,\,\,\, - \left[ {\rho h\frac{{\partial^{2} w}}{{\partial t^{2} }} - f_{z} + \frac{h}{R}\left( {C_{12} \frac{\partial u}{{\partial x}} + \frac{{C_{22} }}{R}w} \right)} \right] \hfill \\ = \frac{{\rho h^{3} }}{12}\left\{ {\frac{2}{{hk_{s} C_{55} }}\left[ {\rho h\frac{{\partial^{4} w}}{{\partial t^{4} }} - \frac{{\partial^{2} f_{z} }}{{\partial t^{2} }} + \frac{h}{R}\left( {C_{12} \frac{{\partial^{3} u}}{{\partial x\partial t^{2} }} + \frac{{C_{22} }}{R}\frac{{\partial^{2} w}}{{\partial t^{2} }}} \right)} \right] - \frac{{\partial^{4} w}}{{\partial x^{2} \partial t^{2} }}} \right\}. \hfill \\ \end{gathered} $$

(A.3)

Rewriting Eq. (A.3) in descending order of spatial derivatives then leads to

$$ \begin{gathered} \frac{{h^{3} C_{11} }}{12}\frac{{\partial^{4} w}}{{\partial x^{4} }} - \frac{{C_{11} C_{22} }}{{6k_{s} C_{55} }}\frac{{h^{3} }}{{R^{2} }}\frac{{\partial^{2} w}}{{\partial x^{2} }} + \frac{{hC_{22} }}{{R^{2} }}w - \frac{{h^{3} C_{11} C_{12} }}{{6Rk_{s} C_{55} }}\frac{{\partial^{3} u}}{{\partial x^{3} }} + \frac{{hC_{12} }}{R}\frac{\partial u}{{\partial x}} + \frac{{h^{2} C_{11} }}{{6k_{s} C_{55} }}\frac{{\partial^{2} f_{z} }}{{\partial x^{2} }} - f_{z} \hfill \\ = \frac{{\rho h^{3} }}{12}\left( {1 + \frac{{2C_{11} }}{{k_{s} C_{55} }}} \right)\frac{{\partial^{4} w}}{{\partial x^{2} \partial t^{2} }} - \rho h\left( {1 + \frac{{h^{2} C_{22} }}{{6R^{2} k_{s} C_{55} }}} \right)\frac{{\partial^{2} w}}{{\partial t^{2} }} - \frac{{\rho^{2} h^{3} }}{{6k_{s} C_{55} }}\frac{{\partial^{4} w}}{{\partial t^{4} }} - \frac{{\rho h^{3} C_{12} }}{{6Rk_{s} C_{55} }}\frac{{\partial^{3} u}}{{\partial x\partial t^{2} }} + \frac{{\rho h^{2} }}{{6k_{s} C_{55} }}\frac{{\partial^{2} f_{z} }}{{\partial t^{2} }}, \hfill \\ \end{gathered} $$

(A.4)

which is what is given in Eq. (23), thereby completing the derivation.

Appendix B

It is shown here that after the determination of V_cr0 – V_cr3 and w in Sect. 3, all of the displacement, strain and stress components in the tube can be readily obtained.

With w determined, the middle-surface x-displacement component u can be found from Eq. (29) by direct integration.

After both w and u are obtained, the rotation angle ψ_x can be solved from Eq. (22b).

With u, w and ψ_x known, the displacement field in the axisymmetric cylindrical shell (tube) can be fully determined from Eqs. (1a–c) as

$$ u_{x} (x,z,t) = u(x,t) + z\psi_{x} (x,t),\,\,\,\,u_{\theta } (x,z,t) = 0,\,\,\,\,u_{z} (x,z,t) = w(x,t). $$

(B.1)

From Eqs. (4) and (B.1), the strain components can then be obtained as

$$ \begin{gathered} \varepsilon_{xx} = \frac{\partial u}{{\partial x}} + z\frac{{\partial \psi_{x} }}{\partial x},\,\,\,\,\varepsilon_{\theta \theta } = \frac{w}{r},\,\,\varepsilon_{zz} = 0, \hfill \\ \varepsilon_{zx} = \frac{1}{2}\left( {\frac{\partial w}{{\partial x}} + \psi_{x} } \right) = \varepsilon_{xz} ,\,\,\,\,\varepsilon_{\theta z} = 0 = \varepsilon_{z\theta } ,\,\,\,\,\varepsilon_{x\theta } = 0 = \varepsilon_{\theta x} . \hfill \\ \end{gathered} $$

(B.2)

Finally, using Eq. (B.2) in Eq. (5) yields the stress components in the tube as

$$ \begin{gathered} \sigma_{xx} = C_{11} \left( {\frac{\partial u}{{\partial x}} + z\frac{{\partial \psi_{x} }}{\partial x}} \right) + C_{12} \frac{w}{r},\,\,\,\,\sigma_{\theta \theta } = C_{12} \left( {\frac{\partial u}{{\partial x}} + z\frac{{\partial \psi_{x} }}{\partial x}} \right) + C_{22} \frac{w}{r}, \hfill \\ \sigma_{zz} = C_{13} \left( {\frac{\partial u}{{\partial x}} + z\frac{{\partial \psi_{x} }}{\partial x}} \right) + C_{23} \frac{w}{r},\,\,\,\,\sigma_{zx} = \frac{1}{2}C_{55} \left( {\frac{\partial w}{{\partial x}} + \psi_{x} } \right) = \sigma_{xz} , \hfill \\ \sigma_{\theta z} = 0 = \sigma_{z\theta } ,\,\,\,\,\sigma_{x\theta } = 0 = \sigma_{\theta x} . \hfill \\ \end{gathered} $$

(B.3)