Numerical prediction of the modal response of flexible cylinders in cross-flow with a current dependent form of damping

Abstract

A quadratic eigenvalue problem (QEP) was posed in order to study the dynamics of flexible cylinders in cross-flow, simulating slender offshore structures such as risers, catenaries or tendons. The Euler–Bernoulli equation was used to model the structure assuming a fluid loading model, and yielding a quadratic eigenvalue problem that included a form of damping dependent not only on the structural damping itself, but also on the free stream velocity and the fluid force coefficients. We solved the QEP using the finite element method. We also derived a simplified analytical solution in this work for comparison with the QEP, however this solution does not consider changes in tension along the length of the cylinder as the QEP does. In our study, the QEP solutions were first validated against the simplified analytical solution, and also against a well-known experimental dataset obtained in 2003, in which a flexible circular cylinder model was used to model the dynamics of a riser undergoing multi-mode vortex-induced vibrations.

Appendices

Appendix A: Details of the implementation using the finite element method

The flexible cylinder was idealized as a one-dimensional domain coinciding with the neutral axis of the structure. We used a mesh formed by one-dimensional elements $\Upomega^e$ consisting of two nodes, $\Upomega^e=[z^e_1,z^e_2],$ with the length of the element defined as h ^e = z ^e₂  − z ^e₁ , being z ^e₁ and z ^e₂ the global coordinates of the finite element at the first and the second node, respectively. The number of elements is n _e = L/h ^e and the number of nodes results in n = n _e + 1. We approximated the solution of Eq. 1 in each element with

$$ \begin{aligned} u^e(z,t)\simeq {\bf U}^e(t){\bf\Uppsi}^e(z)&=\sum_{j=1}^{N} u_j^e(t)\psi_j^e(z) \\ v^e(z,t)\simeq {\bf U}^e(t){\bf\Uppsi}^e(z)&=\sum_{j=1}^{N} v_j^e(t)\psi_j^e(z), \end{aligned} $$

(12)

where N is the number of degrees of freedom considered at each node of the element multiplied by the number of nodes of each element and ψ ^e _j (z) are the shape functions. These can be written in a matrix form as follows:

$$ u^e(z,t) \simeq ({\bf P^e_x R}^e){\bf\Uppsi}^e\\ $$

(13)

$$ v^e(z,t) \simeq ({\bf P^e_y R}^e){\bf\Uppsi}^e $$

(14)

with the displacement vector being ${\bf R^e}=[ u^e_1 u^e_2 v^e_1 v^e_2 | u^e_3 u^e_4 v^e_3 v^e_4]^T,$ including U ^e and V ^e. u ₁ is the displacement at the first node (z ^e₁ ), $u_2=-\frac{\partial u}{\partial z}$ is the rotation in the first node, u ₃ is the displacement at the second node (z ^e₂ ) and $u_4=-\frac{\partial u}{\partial z}$ is the rotation at the second node; the same applies to the components of V ^e. The P ^e matrices map the degrees of freedom with the corresponding displacements or rotations in the elemental displacement vector r ^e. The only non-zero elements in P ^e _x are the ones occupying positions [1,1], [2,2], [6,6] and [7,7], and those are 1. In P ^e _y , the non-zero elements are located at the [3,3], [4,4], [8,8] and [9,9] positions.

1.1 A.1 Weak form of the governing equations

A way to obtain the correct number of linearly independent equations when deriving the integral form of Eqs. 6 and 7 is to find its weak form [19]. We used a weight function to multiply the governing equations and, after integration by parts once in the second order term and twice in the fourth order term, the weak form of the differential equation was obtained. In the finite element method (FEM), the shape functions are used as the weight functions; therefore, by substituting the approximations in Eq. 12 into the governing Eq. 1, we obtained

$$ \begin{aligned} \int\limits_{z_1^e}^{z_2^e} &\Biggl[-EI\frac{\partial^2 \psi^e_i}{\partial z^2}\left(\sum_{j=1}^{N}u_j^e\frac{\partial^2 \psi^e_j}{\partial z^2}\right) + T(z)\frac{\partial \psi_i^e}{\partial z}\left(\sum_{j=1}^{N}u^e_j\frac{\partial \psi^e_j}{\partial z}\right) + c_{xx} \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2u^e_j}{\partial t^2}\right) \\ &+ c_{xy} \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2 v^e_j}{\partial t^2}\right) + m(z) \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2u^e_j}{\partial t^2}\right) \Biggr] {\rm d}z + Q^x_1\psi^e_i(z_1^e)+ Q^x_2\psi^e_i(z_2^e)\\ &+ Q^x_{\bar{1}}\left(\frac{\partial \psi^e_i}{\partial z}(z_1^e)\right) + Q^x_{\bar{2}}\left(-\frac{\partial \psi^e_i}{\partial z}(z_2^e)\right) = 0, \end{aligned} $$

(15)

with the secondary variables being

$$ Q^x_1(t)=\left[\frac{\partial}{\partial z}\left(EI\sum_{j=1}^{N_t}u_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\right) - T(z)\sum_{j=1}^{N_t}u_j^e\frac{\partial \psi^e_j}{\partial z}\right]_{z=z_1^e}, $$

(16)

$$ Q^x_2(t)=-\left[\frac{\partial}{\partial z}\left(EI\sum_{j=1}^{N_t}u_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\right) - T(z)\sum_{j=1}^{N_t}u_j^e\frac{\partial \psi^e_j}{\partial z}\right]_{z=z_2^e}, $$

(17)

$$ Q^x_{\bar{1}}(t)= - EI\sum_{j=1}^{N_t}u_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\Big|_{z=z_1^e}, $$

(18)

and

$$ Q^x_{\bar{2}}(t)=EI\sum_{j=1}^{N_t}u_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\Big|_{z=z_2^e}.$$

(19)

Doing the same for the y direction, we obtained

$$ \begin{aligned} \int\limits_{z_1^e}^{z_2^e}& \Biggl[-EI\frac{\partial^2 \psi^e_i}{\partial z^2}\left(\sum_{j=1}^{N}v_j^e\frac{\partial^2 \psi^e_j}{\partial z^2}\right) + T(z)\frac{\partial \psi_i^e}{\partial z}\left(\sum_{j=1}^{N}v^e_j\frac{\partial \psi^e_j}{\partial z}\right) + c_{yy} \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2 v^e_j}{\partial t^2}\right) \\ &+ c_{yx} \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2 u^e_j}{\partial t^2}\right) + m(z) \psi_i^e\left(\sum_{j=1}^{N}\psi^e_j\frac{\partial^2 v^e_j}{\partial t^2}\right) \Biggr]{\rm d}z \\ &+ Q^y_1\psi^e_i(z_1^e) + Q^y_2\psi^e_i(z_2^e) + Q^y_{\bar{1}}\left(\frac{\partial \psi^e_i}{\partial z}(z_1^e)\right) + Q^y_{\bar{2}}\left(-\frac{\partial \psi^e_i}{\partial z}(z_2^e)\right) = 0, \end{aligned} $$

(20)

with the secondary variables being

$$ Q^y_1(t)=\left[\frac{\partial}{\partial z}\left(EI\sum_{j=1}^{N_t}v_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\right) - T(z)\sum_{j=1}^{N_t}v_j^e\frac{\partial \psi^e_j}{\partial z}\right]_{z=z_1^e}, $$

(21)

$$ Q^y_2(t)=-\left[\frac{\partial}{\partial z}\left(EI\sum_{j=1}^{N_t}v_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\right) - T(z)\sum_{j=1}^{N_t}v_j^e\frac{\partial \psi^e_j}{\partial z}\right]_{z=z_2^e}, $$

(22)

$$ Q^y_{\bar{1}}(t) = - EI\sum_{j=1}^{N_t}v_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\Big|_{z=z_1^e}, $$

(23)

and

$$ Q^y_{\bar{2}}(t) = EI\sum_{j=1}^{N_t}v_j^e\frac{\partial ^2\psi^e_j}{\partial z^2}\Big|_{z=z_2^e}.$$

(24)

The above equations may be compacted in matrix form

$$ {\bf{K}}^{e}{\bf R^e} + \text{C}^{e}\dot{\bf R}^e + \text{M}^e{\ddot{\bf R}^e} = 0 $$

(25)

with K ^e, C ^e and M ^e as the elemental stiffness, damping and mass matrix, respectively. The stiffness matrix is

$$ {\mathbf{K^e}}= \left[ \begin{array}{ll} {\mathbf{K^e_x}} & 0 \\ 0 & {\mathbf{K^e_y}} \\ \end{array} \right] $$

where $\mathbf{K^e_x}=\mathbf{K^e_y}$ because EI and T(z) are the same in planes xz and yz. The stiffness matrix can be split into a bending and a tension matrix

$$ \bf{K^e_x=K^{e^{T}}_x+K^{e^{EI}}_x} $$

and from Eqs. 15 and 20, the components of the matrices are

$$ K{^e}^{T}_{x_{ij}} = K{^e}^{T}_{x_{ji}}=\int\limits_{z_1^e}^{z^B}T(z)\frac{d\psi_i(z)}{dz}\frac{d\psi_j(z)}{dz}{\rm d}z, $$

(26)

and

$$ K{^e}^{EI}_{x_{ij}} = K{^e}^{EI}_{x_{ji}}=\int\limits_{z_1^e}^{z^B}EI\frac{d^2\psi_i(z)}{dz^2}\frac{d^2\psi_j(z)}{dz^2}{\rm d}z. $$

(27)

The damping matrix can be written as

$$ {\mathbf{C^e}}= \left[ \begin{array}{ll} {\mathbf{C^e_{xx}}} & {\mathbf{C^e_{xy}}} \\ {\mathbf{C^e_{yy}}} & {\mathbf{C^e_{yx}}} \\ \end{array} \right] $$

with components

$$ C^e_{xx_{ij}}=C^e_{xx_{ji}}=\int\limits_{z_1^e}^{z^B}c_{xx}\psi_i(z)\psi_j(z){\rm d}z, $$

(28)

$$ C^e_{xy_{ij}}=C^e_{xy_{ji}}=\int\limits_{z_1^e}^{z^B}c_{xy}\psi_i(z)\psi_j(z){\rm d}z, $$

(29)

$$ C^e_{yy_{ij}}=C^e_{yy_{ji}}=\int\limits_{z_1^e}^{z^B}c_{yy}\psi_i(z)\psi_j(z){\rm d}z, $$

(30)

and

$$ C^e_{yx_{ij}}=C^e_{yx_{ji}}=\int\limits_{z_1^e}^{z^B}c_{yx}\psi_i(z)\psi_j(z){\rm d}z, $$

(31)

and finally the mass matrix is

$$ {\mathbf{M^e}}= \left[ \begin{array}{ll} {\mathbf{M^e_x}} & 0 \\ 0 & {\mathbf{M^e_y}} \\ \end{array} \right] $$

with $\mathbf{M^e_x}=\mathbf{M^e_y}$ and components

$$ M^e_{x_{ij}}=M^e_{x_{ji}}=\int\limits_{z_1^e}^{z^B}m\psi_i(z)\psi_j(z){\rm d}z. $$

(32)

For computer implementation purposes, a change of variable yielding a local generic element $\Upomega^R=[-1,1]$ was used. Integration of the equations was done using the Gauss-Legendre Quadrature formula.

Appendix B: Analytical solution of a simplified version of the problem

We sought to derive an analytical solution of a simplified version of Eqs. 6 and 7, where the damping matrix was diagonal and therefore the system was decoupled, so comparisons and an extra validation of the code for these simplified cases could be carried out. First, we dealt with the problem of a cylinder without tension applied on it (T = 0), and later on we introduced a constant tension along the length of the cylinder. The full problem with varying tension along the length of the cylinder and with the possibility of introducing a form of damping depending on C _l was too complicated to be treated analytically, which is why we solved it numerically, as explained in the previous section. The results produced with the equations derived in this section appear together with the numerical results for comparison, in Tables 2, 4 and in Fig. 3.

1.1 B.1 No tension is applied to the cylinder

Consider the one-dimensional Euler–Bernoulli eigenvalue equation,

$$ EI\frac{d^{4}\widehat{A}}{dz^{4}}+(\overline{c}\lambda+m\lambda^{2})\widehat{A} = 0, $$

(33)

with the same boundary conditions as in 3,

$$ \widehat{A}(z=0)=\widehat{A}(z=L) = 0 $$

(34)

$$ \frac{d^{2}\widehat{A}}{dz^{2}}\Bigg|_{z=0} = \frac{d^{2}\widehat{A}}{dz^{2}}\Bigg|_{z=L} = 0, $$

(35)

where $\widehat{A}$ is the generic deflection that can be $\widehat{u}$ or $\widehat{v},$ and $\overline{c}$ is a generic damping coefficient that can also be c _xx or c _yy depending on the direction considered. Looking for example to the x direction, in order to obtain a solution different to the trivial solution one ($\widehat{u}=0$), an eigenvalue condition must be satisfied. For the system in Eqs. 33–35 the condition is [25]

$$ \root{4}\of{\frac{-(m\lambda^{2}+\overline{c}\lambda)}{EI}}L=n\pi \qquad n=1,2,3,\ldots $$

(36)

and the eigenvalues s that satisfy Eq. 36 are

$$ \lambda=-\frac{\overline{c}}{2m}\pm\sqrt{\left(\frac{\overline{c}}{2m}\right)^{2}-\frac{EI}{m}\frac{n^{4}\pi^{4}}{L^{4}} } \qquad n=1,2,3,\ldots. $$

(37)

For small values of n, pure real eigenvalues can be obtained in some situations. The imaginary part in Eq. 37 gives the natural pulsations of the system.

1.2 B.2 Constant tension is applied to the cylinder

Let us consider now a more complicated case, where the cylinder has a constant tension T. The equation becomes:

$$ EI\frac{d^{4}\widehat{A}}{dz^{4}}+T\frac{d^{2}\widehat{A}}{dz^{2}}+(\overline{c}\lambda+m\lambda^{2})\widehat{A} = 0 $$

(38)

with the same boundary conditions as in Eq. 35. In this case an approximate eigenvalue equation can also be derived. We call σ _n

$$ \sigma_{n}=EI\frac{n^{4}\pi^{4}}{L^{4}}+T\frac{n^{2}\pi^{2}}{L^{2}} \qquad n=1,2,3,\ldots $$

(39)

and the eigenvalue equation is:

$$ \lambda=-\frac{\overline{c}}{2m}\pm\sqrt{\left(\frac{\overline{c}}{2m}\right)^{2}-\frac{\sigma_{n}}{m}}\qquad n=1,2,3,\ldots. $$

(40)