A nonlinear two-node superelement for use in flexible multibody systems

Abstract

A nonlinear two-node superelement is proposed for the modeling of flexible complex-shaped links for use in multibody simulations. Assuming that the elastic deformations with respect to a corotational reference frame remain small, substructuring methods may be used to obtain reduced mass and stiffness matrices from a linear finite element model. These matrices are used in the derivation of potential and kinetic energy expressions of the nonlinear two-node superelement. By evaluating Lagrange’s equations, expressions for the internal and external forces acting on the superelement can be obtained. The inertia forces of the superelement are derived in terms of absolute nodal velocities and accelerations, which greatly simplifies the dynamic formulation. Three examples are included. The first two examples are used to validate the method by comparing the results with those obtained from nonlinear beam element solutions. We consider a benchmark simulation of the spin-up motion of a flexible beam with uniform cross-section and a similar simulation in which the beam is simultaneously excited in the out-of-plane direction. Results from both examples show good agreement with simulation results obtained using nonlinear finite beam elements. In a third example, the method is applied to an unbalanced rotating shaft, illustrating the potential of the proposed methodology for a more complex geometry.

Appendices

Appendix A

For infinitesimal elastic deformations, the absolute velocities of the ith node of the FE model, $\dot {\bar {\boldsymbol {u}}}^{(i)}$, expressed in the reference frame attached to superelement node p, can be written as

(44)

where $\boldsymbol {V}_{n}^{(i)}$ is the vector with the x, y and z contributions of the nth constraint mode for the ith node of the FE model.

Rigid body translations and infinitesimal rigid body rotations can be expressed as linear combinations of the constraint modes,

(45a)

(45b)

(45c)

(45d)

where $\bar {\boldsymbol {r}}_{0}$ is the position vector of the nodes in the FE model in undeformed configuration. Substituting in (44) for V ₁ through V ₆ the expressions of (45a)–(45d) gives

$$ \dot {\bar {\boldsymbol {u}}}^{(i)} = \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {v}^{p} + \bigl( \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {\omega }^{p} \bigr) \times \bar {\boldsymbol {r}}_{0}^{(i)} + \dot {\bar {\boldsymbol {e}}}^{(i)}, $$

(46a)

with

$$ \dot {\bar {\boldsymbol {e}}}^{(i)} = \bigl[\boldsymbol {V}_{7}^{(i)}, \boldsymbol {V}_{8}^{(i)}, \boldsymbol {V}_{9}^{(i)}, \boldsymbol {V}_{10}^{(i)}, \boldsymbol {V}_{11}^{(i)}, \boldsymbol {V}_{12}^{(i)} \bigr] \left \{ \begin{array}{c} \bar {\boldsymbol {v}}^{q}_{x} - \bar {\boldsymbol {v}}^{p}_{x} \\ \bar {\boldsymbol {v}}^{q}_{y} - \bar {\boldsymbol {v}}^{p}_{y} - l_{\rm{0}} \bar{\omega }^{p}_{z} \\ \bar {\boldsymbol {v}}^{q}_{z} - \bar {\boldsymbol {v}}^{p}_{z} + l_{\rm{0}} \bar{\omega }^{p}_{y} \\ \bar{\omega}^{q}_{x} - \bar{\omega}^{p}_{x} \\ \bar{\omega}^{q}_{y} - \bar{\omega}^{p}_{y} \\ \bar{\omega}^{q}_{z} - \bar{\omega}^{p}_{z} \end{array} \right \}, $$

(46b)

and

$$ \begin{array}{l@{\qquad}l} \left \{ \begin{array}{c} \bar {\boldsymbol {v}}^{p}_{x} \\ \bar {\boldsymbol {v}}^{p}_{y} \\ \bar {\boldsymbol {v}}^{p}_{z} \\ \end{array} \right \} = \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {v}^{p} , & \left \{ \begin{array}{c} \bar {\boldsymbol {v}}^{q}_{x} \\ \bar {\boldsymbol {v}}^{q}_{y} \\ \bar {\boldsymbol {v}}^{q}_{z} \\ \end{array} \right \} = \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {v}^{q} , \\ \left \{ \begin{array}{c} \bar{\omega}^{p}_{x} \\ \bar{\omega}^{p}_{y} \\ \bar{\omega}^{p}_{z} \\ \end{array} \right \} = \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {\omega }^{p} , & \left \{ \begin{array}{c} \bar{\omega}^{q}_{x} \\ \bar{\omega}^{q}_{y} \\ \bar{\omega}^{q}_{z} \\ \end{array} \right \} = \boldsymbol {R}^{e \mathrm {T}} \boldsymbol {R}^{p \mathrm {T}} \boldsymbol {\omega }^{q}. \end{array} $$

(46c)

Appendix B

In this appendix the time derivatives of the velocity transformation matrices B ₁, B ₂ and B ₃ are given (see (38)).

2.1 B.1 Time derivative of B ₁

Taking the time derivative of matrix B ₁ in (24) gives

$$ \dot{\boldsymbol {B}}_{1} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{p\mathrm {T}} & & & \\& 2 \boldsymbol {R}^{e\mathrm {T}} \dot {\bar {\boldsymbol {\varLambda }}}^{p} & & \\& & \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{p\mathrm {T}} & \\& & & 2 \boldsymbol {R}^{e\mathrm {T}} (\dot{\boldsymbol {R}}^{p\mathrm {T}} \boldsymbol {\varLambda }^{q} + \boldsymbol {R}^{p\mathrm {T}} \dot {\boldsymbol {\varLambda }}^{q}) \\\end{array} \right ], $$

(47)

where the time derivative of R ^p can be computed by

$$ \dot{\boldsymbol {R}}^{p} = 2 \dot {\boldsymbol {\varLambda }}^{p} \bar {\boldsymbol {\varLambda }}^{p\mathrm {T}}, $$

(48)

where $\bar {\boldsymbol {\varLambda }}^{p}$ is defined in (33).

2.2 B.2 Time derivative of B ₂

Taking the time derivative of matrix B ₂ in (31) gives

$$ \dot{\boldsymbol {B}}_{2} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{r\mathrm {T}} & & & \\& 2 \boldsymbol {R}^{e\mathrm {T}} (\dot{\boldsymbol {R}}^{r\mathrm {T}} \boldsymbol {\varLambda }^{p} + \boldsymbol {R}^{r\mathrm {T}} \dot {\boldsymbol {\varLambda }}^{p}) & & \\& & \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{r\mathrm {T}} & \\& & & 2 \boldsymbol {R}^{e\mathrm {T}} (\dot{\boldsymbol {R}}^{r\mathrm {T}} \boldsymbol {\varLambda }^{q} + \boldsymbol {R}^{r\mathrm {T}} \dot {\boldsymbol {\varLambda }}^{q}) \\\end{array} \right ], $$

(49)

where the time derivative of R ^r is

$$ \dot{\boldsymbol {R}}^{r} = 2 \dot {\boldsymbol {\varLambda }}^{r} \bar {\boldsymbol {\varLambda }}^{r\mathrm {T}}. $$

(50)

Here the evaluation of (50) requires an expression for $\dot {\boldsymbol {\lambda }}^{r}$. Taking the derivative of λ ^r in (29) gives

$$ \dot {\boldsymbol {\lambda }}^{r} = \frac{\partial \boldsymbol {\lambda }^{r}}{\partial \boldsymbol {\lambda }^{p}} \dot {\boldsymbol {\lambda }}^{p} + \frac{\partial \boldsymbol {\lambda }^{r}}{\partial \boldsymbol {\lambda }^{q}} \dot {\boldsymbol {\lambda }}^{q}. $$

(51)

It can be shown that

$$ \frac{\partial \boldsymbol {\lambda }^{r}}{\partial \boldsymbol {\lambda }^{p}} = \frac{\partial \boldsymbol {\lambda }^{r}}{\partial \boldsymbol {\lambda }^{q}} = \frac{\boldsymbol{I} - \boldsymbol {\lambda }^{r}\boldsymbol {\lambda }^{r\mathrm {T}}}{\Vert \boldsymbol {\lambda }^{p} + \boldsymbol {\lambda }^{q}\Vert } = \frac{\boldsymbol {\varLambda }^{r\mathrm {T}}\boldsymbol {\varLambda }^{r}}{\Vert \boldsymbol {\lambda }^{p} + \boldsymbol {\lambda }^{q} \Vert }. $$

(52)

Substituting (52) in (51) then gives

$$ \dot {\boldsymbol {\lambda }}^{r} = \frac{\boldsymbol {\varLambda }^{r\mathrm {T}}\boldsymbol {\varLambda }^{r}}{\Vert \boldsymbol {\lambda }^{p} + \boldsymbol {\lambda }^{q} \Vert } \bigl(\dot {\boldsymbol {\lambda }}^{p} + \dot {\boldsymbol {\lambda }}^{q} \bigr). $$

(53)

2.3 B.3 Time derivative of B ₃

Taking the time derivative of matrix B ₃ in (32) gives

$$ \dot{\boldsymbol {B}}_{3} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{r\mathrm {T}} & & & \\& 2 \boldsymbol {R}^{e\mathrm {T}} \dot {\bar {\boldsymbol {\varLambda }}}^{p} & & \\& & \boldsymbol {R}^{e\mathrm {T}} \dot{\boldsymbol {R}}^{r\mathrm {T}} & \\& & & 2 \boldsymbol {R}^{e\mathrm {T}} \dot {\bar {\boldsymbol {\varLambda }}}^{q} \\\end{array} \right ]. $$

(54)

Appendix C

In this appendix expressions are given for $\boldsymbol {B}^{*}_{1}$, $\boldsymbol {B}^{*}_{2}$ and $\boldsymbol {B}^{*}_{3}$ (see (38)).

3.1 C.1 Expression for $\boldsymbol {B}^{*}_{1}$

For $\boldsymbol {B}^{*}_{1}$ we can write

$$ \boldsymbol {B}_{1}^{*} = \frac{\partial \bar {\boldsymbol {v}}}{\partial \boldsymbol {x}} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \boldsymbol{0} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \boldsymbol{0} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \boldsymbol{0} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{q}} \\ \end{array} \right ], $$

(55)

where

(56a)

(56b)

(56c)

(56d)

(56e)

The derivative of a rotation matrix to its Euler parameters is,

(57)

which can be used to evaluate $\displaystyle\frac{\partial \boldsymbol {R}^{p}}{\partial \boldsymbol {\lambda }^{p}}$ in (56a)–(56e).

3.2 C.2 Expression for $\boldsymbol {B}^{*}_{2}$

For $\boldsymbol {B}^{*}_{2}$ we can write

$$ \boldsymbol {B}_{2}^{*} = \frac{\partial \bar {\boldsymbol {v}}}{\partial \boldsymbol {x}} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{q}} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{q}} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{q}} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{q}} \\ \end{array} \right ], $$

(58)

where

(59a)

(59b)

(59c)

(59d)

(59e)

(59f)

(59g)

(59h)

The tensor $\displaystyle\frac{\partial R^{r}_{lk}}{\partial\lambda ^{p}_{j}}$ can be expressed as

$$ \frac{\partial R^{r}_{lk}}{\partial\lambda^{p}_{j}} = \frac{\partial R^{r}_{lk}}{\partial\lambda^{r}_{n}} \frac{\partial\lambda ^{r}_{n}}{\partial\lambda^{p}_{j}}, $$

(60)

and can be evaluated using (52) and (57).

3.3 C.3 Expression for $\boldsymbol {B}^{*}_{3}$

For $\boldsymbol {B}^{*}_{3}$ we can write

$$ \boldsymbol {B}_{3}^{*} = \frac{\partial \bar {\boldsymbol {v}}}{\partial \boldsymbol {x}} = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{q}} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \boldsymbol{0} \\ \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{p}} \quad& \boldsymbol{0} \quad& \displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{q}} \\ \boldsymbol{0} \quad& \boldsymbol{0} \quad& \boldsymbol{0} \quad& \displaystyle\frac {\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{q}} \\ \end{array} \right ], $$

(61)

where $\displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{p}}$, $\displaystyle\frac{\partial \bar {\boldsymbol {v}}^{p}}{\partial \boldsymbol {\lambda }^{q}}$, $\displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{p}}$ and $\displaystyle\frac{\partial \bar {\boldsymbol {v}}^{q}}{\partial \boldsymbol {\lambda }^{q}}$ are the same expressions as given by (59a)–(59h). Furthermore, $\displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{p}}$ and $\displaystyle\frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{q}}$ are given by

$$ \frac{\partial \bar {\boldsymbol {\omega }}^{p}}{\partial \boldsymbol {\lambda }^{p}} = -2 \boldsymbol {R}^{e\mathrm {T}} \dot {\bar {\boldsymbol {\varLambda }}}^{p} \quad \text{and} \quad \frac{\partial \bar {\boldsymbol {\omega }}^{q}}{\partial \boldsymbol {\lambda }^{q}} = -2 \boldsymbol {R}^{e\mathrm {T}} \dot {\bar {\boldsymbol {\varLambda }}}^{q}. $$

(62)