Flexible actuator finite element applied to spatial mechanisms by a finite deformation dynamic formulation

Abstract

A flexible actuator finite element is developed and applied for the modelling of spatial mechanisms present in several industrial applications. A total Lagrangian framework is employed for the development of the finite deformation dynamic equilibrium using solid-like shell and 3D frame finite elements. Exploiting the total Lagrangian aspect of the formulation, the actuator motion is imposed by controlling the element reference configuration. It has the advantage of retaining the actuated bar flexibility, an important factor when simulating flexible mechanisms, and not requiring special treatments as constraint enforcement impositions. As the employed elements use alternative nodal parameters such as positions and generalized vectors to describe their kinematics, a treatment on the introduction of rotational connections—spherical, revolute and pinned joints—largely present in actuated mechanisms, is developed. The nonlinear equations of motion are solved by the Newton–Raphson method. Examples are presented to evaluate the proposed flexible actuator finite element regarding its dynamical behaviour in mechanisms where its use is of importance.

Appendix

The Stewart platform motion, as described in the example, is imposed by varying the actuators lengths calculated from the following displacements of the control points in Fig. 12a. The central point G displacements is:

$$\begin{aligned} {{\mathbf {d}}_{\text {G}}}=\left\{ \begin{matrix} -R(\psi )\sin \psi \\ R(\psi )\cos \psi \\ \frac{{{h}_{0}}}{2}\sin \left( \frac{2\pi }{5}t \right) \\ \end{matrix} \right\} , \end{aligned}$$

(52)

where the z-axis amplitude is ${{h}_{0}}=0.30\,\text {m}$ and the parametric radius evolution is written as:

$$\begin{aligned} R(\psi )=\left\{ \begin{array}{l} \frac{{{R}_{\max }}\psi }{2\pi }\quad \quad \text {if}\quad \quad \psi \le 2\pi \, \text {rad} \\ {{R}_{\max }}\quad \quad \,\,\,\, \text {if}\quad \quad \psi >2\pi \, \text {rad} \\ \end{array} \right. , \end{aligned}$$

(53)

with the maximum radius being ${{R}_{\max }}=0.30\,\text {m}$. The angle evolution function is adopted as:

$$\begin{aligned} \psi (t)=\frac{\pi }{5}t\quad \quad 0\,\text {s}\le t\le 20\,\text {s}. \end{aligned}$$

(54)

The displacements of the other control points are:

$$\begin{aligned} {{\mathbf {d}}_{\text {A}}}= & {} {{\mathbf {d}}_{\text {G}}}+\left\{ \begin{array}{c} 0 \\ \frac{2h}{3}(\cos \beta -1) \\ \frac{2h}{3}\sin \beta \\ \end{array} \right\} \nonumber \\&+\left\{ \begin{array}{c} -\frac{2h}{3}\sin \gamma \\ \frac{2h}{3}(\cos \gamma -1) \\ 0 \\ \end{array} \right\} , \end{aligned}$$

(55)

$$\begin{aligned} {{\mathbf {d}}_{\text {B}}}= & {} {{\mathbf {d}}_{\text {G}}}+\left\{ \begin{array}{c} 0 \\ \frac{h}{3}(1-\cos \beta ) \\ -\frac{h}{3}\sin \beta \\ \end{array} \right\} \nonumber \\&+\left\{ \begin{array}{c} -\frac{\ell }{2}\cos \gamma +\frac{h}{3}\sin \gamma +\frac{\ell }{2} \\ -\frac{h}{3}\cos \gamma +\frac{\ell }{2}\sin \gamma +\frac{h}{3} \\ 0 \\ \end{array} \right\} , \end{aligned}$$

(56)

and

$$\begin{aligned} {{\mathbf {d}}_{\text {C}}}= & {} {{\mathbf {d}}_{\text {G}}}+\left\{ \begin{array}{c} 0 \\ \frac{h}{3}(1-\cos \beta ) \\ -\frac{h}{3}\sin \beta \\ \end{array} \right\} \nonumber \\&+\left\{ \begin{array}{c} \frac{\ell }{2}\cos \gamma +\frac{h}{3}\sin \gamma -\frac{\ell }{2} \\ -\frac{h}{3}\cos \gamma +\frac{\ell }{2}\sin \gamma +\frac{h}{3} \\ 0 \\ \end{array} \right\} , \end{aligned}$$

(57)

in which $\ell =1.0\,\text {m}$ and $h=\sqrt{3}/2\,\text {m}$ are the A-B-C equilateral triangle side and height. Equations (55)–(57) are valid for the following rotation angles evolution around the z-axis, $\gamma$, and x-axis, $\beta$, adopted as:

$$\begin{aligned} \beta (t)=\left\{ \begin{matrix} 0\quad \quad \quad \quad \,\,\quad \quad \quad 0\,\text {s}\le t\le 10\,\text {s} \\ -\frac{\pi }{6}\sin \left( \frac{\pi }{10}t \right) \quad \quad 10\,\text {s}\le t\le 20\,\text {s} \\ \end{matrix} \right. \end{aligned}$$

(58)

and

$$\begin{aligned} \gamma (t)=\left\{ \begin{array}{l} \frac{\pi }{4}\sin \left( \frac{\pi }{10}t \right) \quad \quad 0\, \text {s}\le t\le 10\, \text {s} \\ 0\quad \quad \quad \quad \,\,\,\quad 10\, \text {s}\le t\le 20\, \text {s} \\ \end{array} \right. . \end{aligned}$$

(59)

Known the control points proposed displacements, the i-leg length variation $\varDelta {{L}_{i}}$ can be calculated as:

$$\begin{aligned} \varDelta {{L}_{i}}=\left\| {{{\mathbf {p}}}_{\text {X0}}}+{{{\mathbf {d}}}_{\text {X}}}-{{{\mathbf {p}}}_{i}} \right\| -{{L}_{i0}}, \end{aligned}$$

(60)

in which the initial length is ${{L}_{i0}}=1.0513\,\text {m}$, ${{\mathbf {p}}_{i}}$ is the leg base coordinates, ${{\mathbf {p}}_{\text {X0}}}$ the initial coordinates of the associated platform control point and ${{\mathbf {d}}_{\text {X}}}$ its displacement as calculated above.