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Non linear bend stiffener

Non linear material laws for the stiffener

In a bending stiffener, the relationship between stress and strain is considered to be non linear and function of the temperature. If stress is noted \(\sigma\) and strain \(\varepsilon\), the formula is given by:

\[ \sigma = B \varepsilon + A(1-exp(-C\varepsilon)) \]

Furthermore, the coefficients A , B , and C are temperature dependant. A typical range for the coefficients is:

\(4.3MPa \leq A \leq 10.8MPa\),

\(16.9MPa \leq B \leq 36.5MPa\),

\(23 \leq C \leq 37\).

The non linear laws presented in the above formula has been modified to avoid unrealistic high stress for compression. An anti-symmetric law is therefore used for compression:

\[ \begin{cases} \sigma = B \varepsilon + A(1-exp(-C\varepsilon)) &\text{if } \varepsilon>0\\ \sigma = B \varepsilon - A(1-exp(C\varepsilon)) &\text{if } \varepsilon<0 \end{cases} \]

The change is illustrated on Figure 1 for \(A=8MPa\), \(B =26MPa\), \(C = 30\).

Derivation of internal forces

The resulting internal efforts are the efforts \(\vec {N_{x}}^\xi = A^\xi (R_x^\xi \vec x_0^t - \vec e_3)\) and the moments \(\vec {M_{x}}^\xi = J_I^\xi \vec k_x^\xi\)

\(A^\xi\) and \(J^\xi\) are matrices giving the material stiffness of the riser.

The circumference needs to be divided in sub areas where the strain and stress are evaluated. The graph on Figure 2 presents the frames of reference and the discretisation of the pipe section for the default values ( nrp=1, ntp=8 ).

Discretization of the pipe section for stress/strain calculations.

The section is divided into nrp*ntp sub section, and the strain is evaluated at each point, the strain is considered constant over each subsection. The location of the center of the section with respect to the center of the section is defined by

\[ \vec x^* = \vec x_p - \vec x_M = \xi_1 \vec d_1 + \xi_2 \vec d_2 \]

The strain e can be expressed as: \(\vec \varepsilon = \vec x' (M) = \vec x'(P) + \vec k (P) \wedge \vec x^*\)

where \(\vec x '\) and \(\vec k\) are respectively the axial strain and the curvature at a given point. The stress can then be deduced.

This expression is used in DeepLines to compute the force and moment at P as well as the tangent matrix.

Note that in the following calculations, only the component of the strain and stress along \(d_3\) is considered. Therefore only the force along \(d_3\) and the moment in the ( \(d_1\),\(d_2\)) planes are evaluated with this law. (see section 3.3 of Ref. 5). The torsion moment and shear force are evaluated with the usual coefficients.

All integrals on the circumference are computed numerically, by assuming that the function to be integrated is constant on each cell. There are \(nrp \times ntp\) cells. The value of the function \(g\) on a cell is defined by \(g_{jk} = g(r_j, \theta_k), j \in [1, nrp];k \in [1, ntp]\), with:

\[ r_j = \frac {D_{int}} 2 + \left(j- \frac 1 2 \right) \frac {D_{ext}-D_{int}}{2 \times nrp} \]

\[ \theta_x = \frac {2\pi}{ntp} k \]

And \(D_{ext}\) and \(D_{int}\) the external and internal diameter of the pipe.

The axial strain at a given point of the beam section can be expressed as :

\[ \varepsilon_3 = R_x^t x_0 -1+ k_x^\xi(1)x^t(2) - k_x^\xi(2)x^t(1) = dex(3) + xkx(1)*rsin\theta-xkx(2)*rcos\theta \]

The force at P is given by the following expression in cylindrical coordinates:

\[ \vec N(P) = \int_S \vec \sigma(r, \theta)rdrd\theta \]

Introducing the nonlinear function \(f(\varepsilon_3)=B \varepsilon_3 - A(1-exp(-C \varepsilon_3))\) (for \(e_3>0\) \(\varepsilon_3\), anti- symmetric function otherwise), and considering that the strain is constant on each sub area, the axial force can be computed as:

\[ N_3(P) = \frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] f(\varepsilon _ {jk}) \]

For the moment, the expression in cylindrical coordinate is :

\[ \vec M(P) = \int_S \vec x * \wedge \vec \sigma(r,\theta) rdrd\theta \]

or

\[ M_1(P) = \frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] f(\varepsilon _ {jk}) r_j sin \theta_k \]

\[ M_2(P) = -\frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] f(\varepsilon _ {jk}) r_j cos \theta_k \]

In DeepLines, five coefficients are defined to describe the material stiffness of a linear law: \(afe\) (for axial force), \(afmu\) (for shear force in both directions) , \(x_{j11}\) (for bending moment in direction 1) , \(x_{j22}\) (for bending moment in direction 2) , \(x_{j33}\) (for torsion). Since shear force and torsion are not modified, \(afmu\) and \(x_{j33}\) are unchanged. Therefore, the expression for the shear force and torsion is:

\(N_1 = afmu \times x'_1(P)\) , \(N_2 = afmu \times x'_2(P)\) , \(M_3 = x_{j33} \times k_3(P)\) .

The derivative of forces and moments has to be computed to evaluate the tangent matrix. For the forces:

\[ \Delta N(P) = A\Delta \vec x'(P)+J_N^+ \Delta \vec k(P) \]

The matrix \(A\) is null except \(A(3,3)\) :

\[ A(3,3) = \frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] \frac {\partial f}{\partial \varepsilon} ( \varepsilon*) \]

For the other matrix related to the derivative of the force:

\[ J_N^* = \frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] \frac {\partial f}{\partial \varepsilon} ( \varepsilon_{jk}) \begin{pmatrix} 0 & 0 & -r_jsin \theta_k \\ 0 & 0 & r_jcosn \theta_k \\ r_jsin \theta_k & -r_jcos\theta_k & 0 \end{pmatrix} \]

Since we kept the shear force contribution, we have to add \(A(1,1) = A(2,2) = afmu\).

For the moment derivative, it can be written as:

\[ \Delta \vec M(P) = J_M^* \Delta \vec x ' (P) + J \Delta \vec k (P) \]

with

\[ J_M^* = -J_N^* \]

and

\[ J = \frac \pi {4ntp} \sum_{j=1}^{nrp} \sum_{k=1}^{ntp} \left [ \left ( D_{int} + \frac j {nrp} (D_{ext} - D_{int}) \right) ^2 - \left ( D_{int} + \frac {j-1} {nrp} (D_{ext} - D_{int}) \right) ^2 \right] \frac {\partial f}{\partial \varepsilon} ( \varepsilon_{jk}) \begin{pmatrix} (r_jcos \theta_k)^2 & -r_j^2cos \theta_k sin \theta_k & 0 \\ -r_j^2cos \theta_k sin \theta_k & (r_j sin \theta_k)^2 & 0 \\ 0 & 0 & (r_jcos \theta_k)^2 + (r_j sin \theta_k)^2 \end{pmatrix} \]

To account for torsion we have to add \(x_{j33}\) to \(J(3,3)\).