1. The beam element with a constant circular cross section#

A beam is a solid generated by a surface of area \(S\), whose geometric center of inertia \(G\) describes a curve \(C\) called mean fiber or neutral fiber.

As part of this modeling, only straight, constant and circular beams are taken into account.

For the study of beams in general, the following hypotheses are formulated:

  • The straight section of the beam is undeformable,

  • The transverse displacement is uniform on the right section.

These hypotheses make it possible to express the displacements of any point in the section, as a function of an increase in displacement due to the rotation of the section around the transverse axes.

Discretization into « exact » beam elements is performed on a linear element with two nodes and six degrees of freedom per node. These degrees of freedom are divided into three translations \(u\), \(v\), \(w\) (movements in the directions \(x\), \(y\) and \(z\)) and three rotations \({\theta }_{x}\), \({\theta }_{y}\) and \({\theta }_{z}\). (around the axes \(x\), \(y\) and \(z\)).

_images/Shape1.gif

In the case of straight beams, the mean line is along the \(x\) axis of the local base, the transverse movements thus taking place in the \((y,z)\) plane.

To store quantities related to the degrees of freedom of an element in an elementary vector or matrix (therefore of dimension \(12\) or \({12}^{2}\)), we first order the variables for node 1 then those for node 2. For each node, we first store the quantities related to the three translations, then those related to the three rotations. For example, a displacement vector will be structured as follows:

\(\underset{\mathit{sommet}1}{\underset{\underbrace{}}{{u}_{1},{v}_{1},{w}_{1},{\theta }_{\mathit{x1}},{\theta }_{\mathit{y1}},{\theta }_{\mathit{z1}}}}\), \(\underset{\mathrm{sommet}2}{\underset{\underbrace{}}{{u}_{2},{v}_{2},{w}_{2},{\theta }_{\mathrm{x2}},{\theta }_{\mathrm{y2}},{\theta }_{\mathrm{z2}}}}\)

1.1. Defining landmarks#

We define:

  • \(x\) is the axis of the neutral fiber of the beam,

  • \(y\) and \(z\) are the main inertia axes of the section,

  • \({R}_{0}\) is the absolute coordinate system linked to a section in the initial configuration,

  • \(\mathrm{R}\) is the coordinate system linked to a section in the deformed configuration,

By not considering the twist, the transition from frame \({\mathrm{R}}_{0}\) to frame \(\mathrm{R}\) is carried out using 3 rotations, two following \(y\) and \(z\), and one rotation around \(x\), noted \(\varphi\), such that:

\(\dot{\varphi }\): natural rotation speed of the shaft

1.2. specifications#

Each element is an isoparametric beam element with a circular and constant cross section. Transverse shear is taken into account in the formulation of this element (Timoshenko straight beam).

Ratings:

  • \(x\) is the axis of the neutral fiber of the tree line,

  • density: \(\rho\)

  • element length: \(L\)

  • Young’s modulus: \(E\)

  • Poisson module: \(G\mathrm{=}\frac{E}{2(1+\nu )}\)

  • section:

  • inner radius: \({R}_{i}\)

  • outer radius: \({R}_{e}\)

  • Do: \(A\mathrm{=}\pi ({R}_{e}^{2}\mathrm{-}{R}_{i}^{2})\)

  • polar inertia: \({I}_{x}=\frac{\pi }{2}({R}_{e}^{4}-{R}_{i}^{4})\)

  • section inertia: \({I}_{\text{yz}}={I}_{y}={I}_{z}=\frac{\pi }{4}({R}_{e}^{4}-{R}_{i}^{4})\)

1.3. Calculation of the kinetic energy of the Timoshenko beam#

The kinetic energy of the Timoshenko beam element is calculated by considering membrane and bending deformations. The expression of kinetic energy is obtained by integrating over the length of the beam element:

\(T\mathrm{=}\frac{1}{2}\rho \text{.}A\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\left[{\dot{u}}^{2}+{\dot{v}}^{2}+{\dot{w}}^{2}\right]\text{dx}+\frac{1}{2}\rho \text{.}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{.}\left[J\right]\text{.}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{dx}\)

with: \(\left[J\right]=\left[\begin{array}{ccc}{I}_{x}& 0& 0\\ 0& {I}_{y}& 0\\ 0& 0& {I}_{z}\end{array}\right]\) with \(I\) (in \({m}^{4}\))

Given that it is a right beam with axis \(o\overrightarrow{x}\) for the undeformed configuration, it is necessary to define two intermediate bases to characterize the rotation speed vector \({\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\).

  • Passage from base B (O, \(\overrightarrow{x}\),, \(\overrightarrow{y}\), \(\overrightarrow{z}\)) to base \({B}_{1}\) (O, \(\overrightarrow{{x}_{1}}\), \(\overrightarrow{{y}_{1}}\), \(\overrightarrow{{z}_{1}}\)) by an axis rotation \(o\overrightarrow{y}\) of amplitude \({\theta }_{y}(x,t)\) such as:

\(\overrightarrow{{y}_{1}}\mathrm{=}\overrightarrow{y}\)

  • Passage from base \(B\) (\(O\), \(\overrightarrow{{x}_{1}}\), \(\overrightarrow{{y}_{1}}\), \(\overrightarrow{{z}_{1}}\)) to base \({B}_{2}\) (\(O\), \(\overrightarrow{{x}_{2}}\), \(\overrightarrow{{y}_{2}}\), \(\overrightarrow{{z}_{2}}\)) by rotating the e-axis with an amplitude-of-magnitude such as: \(o\overrightarrow{{z}_{1}}\) \({\theta }_{z}(x,t)\)

\(\overrightarrow{{z}_{2}}=\overrightarrow{{z}_{1}}\) and \(\overrightarrow{{y}_{1}}\mathrm{=}\mathrm{cos}{\theta }_{z}(x,t)\mathrm{.}\overrightarrow{{y}_{2}}+\mathrm{sin}{\theta }_{z}(x,t)\mathrm{.}\overrightarrow{{x}_{2}}\)

  • The rotation at angular speed \(\dot{\phi }(t)\) takes place along the \(o\overrightarrow{{x}_{2}}\) axis.

  • So the rotation vector is written as: \({\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\mathrm{=}\dot{\phi }(t)\mathrm{.}\overrightarrow{{x}_{2}}+\dot{{\theta }_{y}}(x,t)\mathrm{.}\overrightarrow{{y}_{1}}+\dot{{\theta }_{z}}(x,t)\mathrm{.}\overrightarrow{{z}_{1}}\)

  • Since the operator \([J]\) of the beam element is written in the base B2which corresponds to the deformed position, it is imperative, unless the inertia operator is based, to write the rotation speed vector \({\overrightarrow{\Omega }}_{R/\mathrm{R0}}\) in the base B2.

\({\overrightarrow{\Omega }}_{R/\mathrm{R0}}=\dot{\phi }(t)\mathrm{.}\overrightarrow{{x}_{2}}+\dot{{\theta }_{y}}(x,t)\mathrm{.}(\mathrm{cos}{\theta }_{z}(x,t)\mathrm{.}\overrightarrow{{y}_{2}}+\mathrm{sin}{\theta }_{z}(x,t)\mathrm{.}\overrightarrow{{x}_{2}})+\dot{{\theta }_{z}}(x,t)\mathrm{.}\overrightarrow{{z}_{2}}\)

  • Considering that the angles \({\theta }_{y}(x,t)\) and \({\theta }_{z}(x,t)\) are small, it is legitimate to carry out a development limited to order 1. The expression for speed vector \({\overrightarrow{\Omega }}_{R/\mathrm{R0}}\) then becomes:

\({\overrightarrow{\Omega }}_{R/\mathrm{R0}}=(\dot{\phi }(t)+\dot{{\theta }_{y}}(x,t)\mathrm{.}{\theta }_{z}(x,t))\mathrm{.}\overrightarrow{{x}_{2}}+\dot{{\theta }_{y}}(x,t)\mathrm{.}\overrightarrow{{y}_{2}}+\dot{{\theta }_{z}}(x,t)\mathrm{.}\overrightarrow{{z}_{2}}\)

  • It remains to develop the following dot product:

\(\frac{1}{2}\rho \underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{.}\left[J\right]\text{.}{\overrightarrow{\Omega }}_{R\mathrm{/}\mathit{R0}}\text{dx}\mathrm{=}\frac{1}{2}\rho {I}_{\mathit{yz}}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\left[{\dot{\theta }}_{y}^{2}+{\dot{\theta }}_{z}^{2}\right]\text{dx}+\frac{1}{2}\rho {I}_{x}\text{.}L\text{.}{\dot{\phi }}^{2}+\rho \dot{\phi }{I}_{x}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}{\dot{\theta }}_{y}\text{.}{\theta }_{z}\text{dx}\)

For a beam element with a constant cross section, the expression becomes:

\(T\mathrm{=}\frac{1}{2}\rho A\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\left[{\dot{u}}^{2}+{\dot{v}}^{2}+{\dot{w}}^{2}\right]\text{dx}+\frac{1}{2}\rho {I}_{\mathit{yz}}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\left[{\dot{\theta }}_{y}^{2}+{\dot{\theta }}_{z}^{2}\right]\text{dx}+\frac{1}{2}\rho {I}_{x}\text{.}L\text{.}{\dot{\phi }}^{2}+\rho \dot{\phi }{I}_{x}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}{\dot{\theta }}_{y}\text{.}{\theta }_{z}\text{dx}\)

With:

\({I}_{x}\mathrm{=}\frac{\pi }{2}\text{.}\left[{R}_{e}^{4}\mathrm{-}{R}_{e}^{4}\right]\)

\({I}_{\text{yz}}\mathrm{=}{I}_{y}\mathrm{=}{I}_{z}\mathrm{=}\frac{\pi }{4}\text{.}\left[{R}_{e}^{4}\mathrm{-}{R}_{e}^{4}\right]\)

The various kinetic energy terms represent:

  • for the first term, the kinetic energy of translation,

  • for the following two terms, the kinetic energy of rotation,

  • for the fourth term, the gyroscopic effect term.

1.4. Interpolation functions#

For membrane deformations (tension — compression), field \(u(x)\) is approximated by a linear function of the displacements of nodes 1 and 2 of the beam element:

\(u(x)=\langle {N}_{1}^{L}(x){N}_{2}^{L}(x)\rangle \left\{\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right\}\) with \(\{\begin{array}{c}{N}_{1}^{L}(x)=1-\frac{x}{L}\\ {N}_{2}^{L}(x)=\frac{x}{L}\end{array}\)

For flexural deformations, cubic functions of the modified Hermite type are used. The degrees of freedom \(v(x),{\theta }_{y}(x),w(x),{\theta }_{z}(x)\) are therefore interpolated as follows:

\(v(x)\mathrm{=}\mathrm{\langle }{\xi }_{1}(x)\mathrm{-}{\xi }_{2}(x){\xi }_{3}(x)\mathrm{-}{\xi }_{4}(x)\mathrm{\rangle }\left\{\begin{array}{c}{v}_{1}\\ {\theta }_{\mathit{z1}}\\ {v}_{2}\\ {\theta }_{\mathit{z2}}\end{array}\right\}\)

\({\theta }_{z}(x)\mathrm{=}\mathrm{\langle }\mathrm{-}{\xi }_{5}(x){\xi }_{6}(x)\mathrm{-}{\xi }_{7}(x){\xi }_{8}(x)\mathrm{\rangle }\left\{\begin{array}{c}{v}_{1}\\ {\theta }_{\mathit{z1}}\\ {v}_{2}\\ {\theta }_{\mathit{z2}}\end{array}\right\}\)

\(w(x)\mathrm{=}\mathrm{\langle }{\xi }_{1}(x){\xi }_{2}(x){\xi }_{3}(x){\xi }_{4}(x)\mathrm{\rangle }\left\{\begin{array}{c}{w}_{1}\\ {\theta }_{\mathit{y1}}\\ {w}_{2}\\ {\theta }_{\mathit{y2}}\end{array}\right\}\)

\({\theta }_{y}(x)\mathrm{=}\mathrm{\langle }{\xi }_{5}(x){\xi }_{6}(x){\xi }_{7}(x){\xi }_{8}(x)\mathrm{\rangle }\left\{\begin{array}{c}{w}_{1}\\ {\theta }_{\mathit{y1}}\\ {w}_{2}\\ {\theta }_{\mathit{y2}}\end{array}\right\}\)

We define \({K}_{\mathit{yz}}\) shear coefficient in the directions \(y\) and \(z\).

For a beam element with a constant cross section:

\({K}_{\text{yz}}\mathrm{=}\frac{7+\text{20}\text{.}{\alpha }^{2}}{6}\) with \(\alpha =\frac{{R}_{i}}{{R}_{e}\text{.}(1+\frac{{R}_{i}^{2}}{{R}_{e}^{2}})}\)

Noting \({\phi }_{\text{yz}}\mathrm{=}\frac{\text{12}\text{.}E\text{.}{I}_{\text{yz}}}{{K}_{\text{yz}}\text{.}A\text{.}G\text{.}{L}^{2}}\), functions \({\xi }_{i}\) are defined as follows:

\({\xi }_{1}(x)\mathrm{=}\frac{1}{1+{\phi }_{\text{yz}}}\left[2\text{.}{(\frac{x}{L})}^{3}\mathrm{-}3\text{.}{(\frac{x}{L})}^{2}\mathrm{-}{\phi }_{\text{yz}}\text{.}(\frac{x}{L})+(1+{\phi }_{\text{yz}})\right]\)

\({\xi }_{5}(x)\mathrm{=}\frac{6}{L\text{.}(1+{\phi }_{\text{yz}})}\text{.}(\frac{x}{L})\text{.}\left[1\mathrm{-}(\frac{x}{L})\right]\)

\({\xi }_{2}(x)\mathrm{=}\frac{L}{1+{\phi }_{\text{yz}}}\left[\mathrm{-}{(\frac{x}{L})}^{3}+\frac{4+{\phi }_{\text{yz}}}{2}{(\frac{x}{L})}^{2}\mathrm{-}\frac{2+{\phi }_{\text{yz}}}{2}\text{.}(\frac{x}{L})\right]\)

\({\xi }_{6}(x)\mathrm{=}\frac{1}{1+{\phi }_{\text{yz}}}\left[3\text{.}{(\frac{x}{L})}^{2}\mathrm{-}(4+{\phi }_{\text{yz}})\text{.}(\frac{x}{L})+(1+{\phi }_{\text{yz}})\right]\)

\({\xi }_{3}(x)\mathrm{=}\frac{1}{1+{\phi }_{\text{yz}}}\left[\mathrm{-}2\text{.}{(\frac{x}{L})}^{3}+3\text{.}{(\frac{x}{L})}^{2}+{\phi }_{\text{yz}}\text{.}(\frac{x}{L})\right]\)

\({\xi }_{7}(x)\mathrm{=}\frac{\mathrm{-}6}{L\text{.}(1+{\phi }_{\text{yz}})}\text{.}(\frac{x}{L})\text{.}\left[1\mathrm{-}(\frac{x}{L})\right]\)

\({\xi }_{4}(x)\mathrm{=}\frac{L}{1+{\phi }_{\text{yz}}}\left[\mathrm{-}{(\frac{x}{L})}^{3}+\frac{2\mathrm{-}{\phi }_{\text{yz}}}{2}{(\frac{x}{L})}^{2}+\frac{{\phi }_{\text{yz}}}{2}\text{.}(\frac{x}{L})\right]\)

\({\xi }_{8}(x)\mathrm{=}\frac{1}{1+{\phi }_{\text{yz}}}\left[3\text{.}{(\frac{x}{L})}^{2}+(\mathrm{-}2+{\phi }_{\text{yz}})\text{.}(\frac{x}{L})\right]\)

Note:

In the case of Euler beam elements (Elements POU_D_E) the term \({\phi }_{\text{yz}}\) is null.

The vector of the degrees of freedom of the beam element is defined by:

\(\mathrm{\langle }q\mathrm{\rangle }\mathrm{=}\mathrm{\langle }{u}_{1}{v}_{1}{w}_{1}{\theta }_{\mathit{x1}}{\theta }_{\mathit{y1}}{\theta }_{\mathit{z1}}{u}_{2}{v}_{2}{w}_{2}{\theta }_{\mathit{x2}}{\theta }_{\mathit{y2}}{\theta }_{\mathit{z2}}\mathrm{\rangle }\)

We ask:

\(\mathrm{\langle }\delta u\mathrm{\rangle }\mathrm{=}\mathrm{\langle }{u}_{1}{u}_{2}\mathrm{\rangle }\)

\(\langle \delta v\rangle =\langle {v}_{1}{\theta }_{\mathrm{z1}}{v}_{2}{\theta }_{\mathrm{z2}}\rangle\)

\(\mathrm{\langle }\delta w\mathrm{\rangle }\mathrm{=}\mathrm{\langle }{w}_{1}{\theta }_{\mathit{y1}}{w}_{2}{\theta }_{\mathit{y2}}\mathrm{\rangle }\)

Substituting the previous approximations in the expression for kinetic energy, we obtain:

\(\begin{array}{c}T\mathrm{=}\frac{1}{2}\mathrm{\langle }\delta \dot{u}\mathrm{\rangle }\left[{M}_{1}\right]\left\{\delta \dot{u}\right\}+\frac{1}{2}\mathrm{\langle }\delta \dot{w}\mathrm{\rangle }(\left[{M}_{2}\right]+\left[{M}_{4}\right])\left\{\delta \dot{w}\right\}+\frac{1}{2}\mathrm{\langle }\delta \dot{v}\mathrm{\rangle }(\left[{M}_{3}\right]+\left[{M}_{5}\right])\left\{\delta \dot{v}\right\}\\ +\dot{\phi }\text{.}\mathrm{\langle }\delta \dot{v}\mathrm{\rangle }\left[{M}_{6}\right]\left\{\delta w\right\}+\frac{1}{2}\text{.}\rho \text{.}{I}_{x}\text{.}{\dot{\phi }}^{2}\end{array}\)

With:

\(\left[{M}_{1}\right]=\underset{x=0}{\overset{L}{\int }}\rho \text{.}A\text{.}\left\{\begin{array}{c}{N}_{1}^{L}(x\\ {N}_{2}^{L}(x)\end{array}\right\}\text{.}\langle {N}_{1}^{L}(x){N}_{2}^{L}(x)\rangle \text{.}\text{dx}\)

\(\left[{M}_{2}\right]\mathrm{=}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\rho \text{.}A\text{.}\left\{\begin{array}{c}{\xi }_{1}(x)\\ {\xi }_{2}(x)\\ {\xi }_{3}(x)\\ {\xi }_{4}(x)\end{array}\right\}\text{.}\mathrm{\langle }{\xi }_{1}(x){\xi }_{2}(x){\xi }_{3}(x){\xi }_{4}(x)\mathrm{\rangle }\text{.}\text{dx}\)

\(\left[{M}_{3}\right]\mathrm{=}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\rho \text{.}A\text{.}\left\{\begin{array}{c}{\xi }_{1}(x)\\ \mathrm{-}{\xi }_{2}(x)\\ {\xi }_{3}(x)\\ \mathrm{-}{\xi }_{4}(x)\end{array}\right\}\text{.}\mathrm{\langle }{\xi }_{1}(x)\mathrm{-}{\xi }_{2}(x){\xi }_{3}(x)\mathrm{-}{\xi }_{4}(x)\mathrm{\rangle }\text{.}\text{dx}\)

\(\left[{M}_{4}\right]\mathrm{=}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\rho \text{.}{I}_{\text{yz}}\text{.}\left\{\begin{array}{c}{\xi }_{5}(x)\\ {\xi }_{6}(x)\\ {\xi }_{7}(x)\\ {\xi }_{8}(x)\end{array}\right\}\text{.}\mathrm{\langle }{\xi }_{5}(x){\xi }_{6}(x){\xi }_{7}(x){\xi }_{8}(x)\mathrm{\rangle }\text{.}\text{dx}\)

\(\left[{M}_{5}\right]\mathrm{=}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\rho \text{.}{I}_{\text{yz}}\text{.}\left\{\begin{array}{c}\mathrm{-}{\xi }_{5}(x)\\ {\xi }_{6}(x)\\ \mathrm{-}{\xi }_{7}(x)\\ {\xi }_{8}(x)\end{array}\right\}\text{.}\mathrm{\langle }\mathrm{-}{\xi }_{5}(x){\xi }_{6}(x)\mathrm{-}{\xi }_{7}(x){\xi }_{8}(x)\mathrm{\rangle }\text{.}\text{dx}\)

\(\left[{M}_{6}\right]\mathrm{=}\underset{x\mathrm{=}0}{\overset{L}{\mathrm{\int }}}\rho \text{.}{I}_{x}\text{.}\left\{\begin{array}{c}\mathrm{-}{\xi }_{5}(x)\\ {\xi }_{6}(x)\\ \mathrm{-}{\xi }_{7}(x)\\ {\xi }_{8}(x)\end{array}\right\}\text{.}\mathrm{\langle }{\xi }_{5}(x){\xi }_{6}(x){\xi }_{7}(x){\xi }_{8}(x)\mathrm{\rangle }\text{.}\text{dx}\)

1.5. Calculation of equilibrium equations#

The Lagrange equations for the kinetic energy of the beam are written in the following form:

\(\frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }{\dot{q}}_{i}})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }{q}_{i}}\mathrm{=}0\) with \(\mathrm{\langle }q\mathrm{\rangle }\mathrm{=}\mathrm{\langle }uvw\mathrm{\rangle }\)

That is: \(\mathrm{\{}\begin{array}{c}\frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }(\delta \dot{u})})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }(\mathit{\delta u})}\mathrm{=}\left[{M}_{1}\right]\left\{(\delta \ddot{u})\right\}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }(\delta \dot{v})})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }(\mathit{\delta v})}\mathrm{=}(\left[{M}_{3}\right]+\left[{M}_{5}\right])\left\{(\delta \ddot{v})\right\}\mathrm{-}\dot{\phi }\text{.}\left[{M}_{6}\right]\left\{(\delta \dot{w})\right\}\mathrm{-}\ddot{\phi }\text{.}\left[{M}_{6}\right]\left\{(\delta w)\right\}\\ \frac{d}{\text{dt}}(\frac{\mathrm{\partial }T}{\mathrm{\partial }(\delta \dot{w})})\mathrm{-}\frac{\mathrm{\partial }T}{\mathrm{\partial }(\mathit{\delta w})}\mathrm{=}(\left[{M}_{2}\right]+\left[{M}_{4}\right])\left\{(\delta \ddot{w})\right\}+\dot{\phi }\text{.}{\left[{M}_{6}\right]}^{T}\left\{(\delta \dot{v})\right\}\end{array}\)

These equations can be put in the form:

\(\left[M\right]\mathrm{\langle }\ddot{q}\mathrm{\rangle }+\left[{C}_{\mathit{gyro}}\right]\mathrm{\langle }\dot{q}\mathrm{\rangle }+(\left[K\right]+\left[{K}_{\mathit{gyro}}\right])\mathrm{\langle }q\mathrm{\rangle }\mathrm{=}\mathrm{\langle }0\mathrm{\rangle }\)

The gyroscopic damping matrix \(\left[{C}_{\mathit{gyro}}\right]\) of the system is formed from the matrix \([{M}_{6}]\) and its transpose. It is antisymmetric, and its contribution must be multiplied by angular speed \(\dot{\phi }\).

Noting: \(\phi \mathrm{=}{\phi }_{\text{yz}}\)

\(\left[{M}_{6}\right]\mathrm{=}\frac{\rho \text{.}{I}_{x}}{\text{30}L(1+\phi {)}^{2}}\left[\begin{array}{cccc}\mathrm{-}\text{36}& \mathrm{3L}(1\mathrm{-}5\phi )& \text{36}& \mathrm{3L}(1\mathrm{-}5\phi )\\ \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& {L}^{2}(4+5\phi +\text{10}{\phi }^{2})& \mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}{L}^{2}(1+5\phi \mathrm{-}5{\phi }^{2})\\ \text{36}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}\text{36}& \mathrm{3L}(\mathrm{-}1+5\phi )\\ \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}{L}^{2}(1+5\phi \mathrm{-}5{\phi }^{2})& \mathrm{-}\mathrm{3L}(\mathrm{-}1+5\phi )& {L}^{2}(4+5\phi +\text{10}{\phi }^{2})\end{array}\right]\)

\(\begin{array}{c}\left[{C}_{\mathit{gyro}}\right]\mathrm{=}\frac{\rho \text{.}{I}_{x}}{30L{(1+\phi )}^{2}}\mathrm{\times }\\ \left[\begin{array}{cccccccccccc}\mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}\\ & 0& \text{36}& \mathrm{-}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& 0& \mathrm{-}& 0& \mathrm{-}\text{36}& \mathrm{-}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& 0\\ & & 0& \mathrm{-}& 0& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& \text{36}& & \mathrm{-}& 0& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )\\ & & & \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& & \mathrm{-}& \mathrm{-}& \mathrm{-}\\ & & & & 0& {L}^{2}(4+5\phi +\text{10}{\phi }^{2})& \mathrm{-}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& & \mathrm{-}& 0& \mathrm{-}{L}^{2}(1+5\phi \mathrm{-}5{\phi }^{2})\\ & & & & & 0& \mathrm{-}& 0& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& {L}^{2}(1+5\phi \mathrm{-}5{\phi }^{2})& 0\\ & & & & & & \mathrm{-}& \mathrm{-}& & \mathrm{-}& \mathrm{-}& \mathrm{-}\\ & & & & & & & 0& \text{36}& \mathrm{-}& \mathrm{3L}(1\mathrm{-}5\phi )& 0\\ & & & & & & & & 0& \mathrm{-}& 0& \mathrm{3L}(1\mathrm{-}5\phi )\\ & & & & & & & & & \mathrm{-}& \mathrm{-}& \mathrm{-}\\ & & & & & & & & & & 0& {L}^{2}(4+5\phi +\text{10}{\phi }^{2})\\ & & & & & & & & & & & 0\end{array}\right]\end{array}\)

As matrix \(\left[{C}_{\mathit{gyro}}\right]\) is anti-symmetric, only the upper triangle is shown.

(—) means that the degree of freedom is not affected by gyroscopic matrices.

The gyroscopic stiffness matrix \(\left[{K}_{\mathit{gyro}}\right]\) of the system is formed from the matrix \(\mathrm{[}{M}_{6}\mathrm{]}\). Its contribution must be multiplied by angular acceleration \(\ddot{\varphi }\).

\(\begin{array}{c}\left[{K}_{\mathit{gyro}}\right]\mathrm{=}\frac{\rho \text{.}{I}_{x}}{30L{(1+\phi )}^{2}}\mathrm{\times }\\ \left[\begin{array}{cccccccccccc}\mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}\\ \mathrm{-}& 0& 36& \mathrm{-}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& 0& \mathrm{-}& 0& \mathrm{-}36& \mathrm{-}& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& 0\\ \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& & \mathrm{-}& 0& 0\\ \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& & \mathrm{-}& \mathrm{-}& \mathrm{-}\\ \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& & \mathrm{-}& 0& 0\\ \mathrm{-}& 0& \mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& \mathrm{-}{L}^{2}(4+5\phi +\text{10}{\phi }^{2})& 0& \mathrm{-}& 0& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& {L}^{2}(1+5\phi \mathrm{-}5{\phi }^{2})& 0\\ \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& & \mathrm{-}& \mathrm{-}& \mathrm{-}\\ \mathrm{-}& 0& \mathrm{-}36& \mathrm{-}& \mathrm{3L}(1\mathrm{-}5\phi )& 0& \mathrm{-}& 0& 36& \mathrm{-}& \mathrm{3L}(1\mathrm{-}5\phi )& 0\\ \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& 0\\ \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}& \mathrm{-}\\ \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& 0& \mathrm{-}& 0& 0\\ \mathrm{-}& 0& \mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& {L}^{2}(4+5\phi +\text{10}{\phi }^{2})& 0& \mathrm{-}& 0& \mathrm{-}\mathrm{3L}(1\mathrm{-}5\phi )& \mathrm{-}& \mathrm{-}{L}^{2}(4+5\phi +\text{10}{\phi }^{2})& 0\end{array}\right]\end{array}\)

The full matrix \(\left[{K}_{\mathit{gyro}}\right]\) is filled in full (upper and lower triangles).

Reminder:

    • with \(\langle q\rangle =\langle {u}_{1}{v}_{1}{w}_{1}{\theta }_{\mathrm{x1}}{\theta }_{\mathrm{y1}}{\theta }_{\mathrm{z1}}{u}_{2}{v}_{2}{w}_{2}{\theta }_{\mathrm{x2}}{\theta }_{\mathrm{y2}}{\theta }_{\mathrm{z2}}\rangle\)

    • in the case of Euler beam elements (Elements POU_D_E) the term \({\phi }_{\text{yz}}\) is null.