D3. Interactions between rigid bodies

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As seen in unit D2, interactions between pairs of particles P and Q are described by a single force with direction [math]\displaystyle{ \overline{\Ps \Qs} }[/math]. When the interaction is between pairs of rigid bodies (which can be considered as two sets of infinite particles), the particle-by-particle description would lead to a system of infinite forces. In this case, we must move on to a compact description of this system of forces that nevertheless retains the information necessary to study the dynamics of rigid bodies: the system of forces is replaced by a resultant torsor.

This unit introduces the concept of torsor associated with a system of forces, and then applies it to the various interactions between solids (at a distance, in contact, and through intermediate elements).

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D3.1 Torsor associated with a system of forces

The reduction of a system of forces on a rigid body to a torsor is mandatory when the number of forces is very high (infinite). When it is a system of just a few forces, it is usually optional.

An effective mathematical operation to reduce the number of forces on a rigid body S is addition: however high the number of forces, the sum leads to a single resultant force. However, this drastic reduction implies a loss of essential information in many cases. As long as there is no interest in studying the deformation of objects (i.e., when only the dynamics of rigid bodies is studied), this is solved by adding a second vector to the compact description of the system: the resultant moment (or torque) about a point Q. The set of these two vectors (resultant force and resultant moment) is the torsor of the system of forces at point Q (Figure D3.1).

Figure D3.1 Torsor associated with a system of forces on a rigid body.

By way of an example, consider the case of a rigid bar initially at rest undergoing a system of forces with zero resultant force. Figure D3.2 shows three different situations corresponding to this situation: (a) free from forces, (b) forces parallel to the bar, (c) forces perpendicular to the bar. In the first two cases (a, b), the system of forces does not modify the state of rest. In the third case (c), the forces provoke a clockwise rotation of the bar. The resultant torsor at any point allows us to distinguish between (c) and (a, b).

Figure D3.2 Rigid bar undergoing a zero resultant force.

The resultant force never depends on the point where the torsor is calculated. However, the resultant moment generally does depend on that point (Figure D3.3).

[math]\displaystyle{ \hspace{1cm}\sum \overline{\Ms}(\Os) = (\Ls\Fs_\Ps+\Ls\Fs_\Qs) \otimes }[/math] [math]\displaystyle{ \hspace{1cm}\sum \overline{\Ms}(\Ps) = 2\Ls\Fs_\Qs \otimes }[/math] [math]\displaystyle{ \hspace{1cm}\sum \overline{\Ms}(\Qs) = 2\Ls\Fs_\Ps \otimes }[/math] [math]\displaystyle{ \hspace{1cm}\sum \overline{\Ms}(\Ss) = (3\Ls\Fs_\Ps-\Ls\Fs_\Qs) \otimes }[/math]

Figure D3.3 Rigid bar undergoing two forces with different value.

The resulting torsor is represented at the point about which the resultant moment has been calculated (Figure D3.4).

Figure D3.4 Torsors at diferent points for the exemple on the rigid bar in Figure D3.3.

The resultant moment about a point Q’ can be obtained from the torsor about a point Q:

[math]\displaystyle{ \overline{\mathbf{M}}_\mathrm{R}(\Qs ')=\overline{\mathbf{M}}_\mathrm{R}(\Qs) + \overline{\Qs '\Qs} \times \overline{\mathbf{F}}_\Rs }[/math]

💭 Proof ➕

[math]\displaystyle{ \overline{\mathbf{M}}_\Rs(\Qs)=\sum \overline{\Qs \Ps_\is} \times \overline{\mathbf{F}}_\is }[/math]

[math]\displaystyle{ \overline{\mathbf{M}}_\Rs(\Qs ')=\sum \overline{\Qs ' \Ps_\is} \times \overline{\mathbf{F}}_\is=\sum (\overline{\Qs ' \Qs}+\overline{\Qs \Ps_\is}) \times \overline{\mathbf{F}}_\is=\sum \overline{\Qs ' \Qs} \times \overline{\mathbf{F}}_\is + \sum \overline{\Qs \Ps_\is} \times \overline{\mathbf{F}}_\is=\overline{\Qs ' \Qs} \times (\sum \overline{\mathbf{F}}_\is) + \sum \overline{\Qs \Ps_\is} \times \overline{\mathbf{F}}_\is }[/math]

[math]\displaystyle{ \overline{\mathbf{M}}_\Rs(\Qs ')=\overline{\Qs ' \Qs} \times \overline{\mathbf{F}}_\Rs+\overline{\mathbf{M}}_\Rs(\Qs) }[/math]

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D3.2 Gravitational attraction

The calculation of the resultant gravitational torsor on a rigid body [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math] due to a rígid body [math]\displaystyle{ \mathrm{S}_\mathrm{Q} }[/math] is not simple. The resultant gravitational force on a mass differential dm([math]\displaystyle{ \Ps }[/math]) of [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math] ([math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}_{\mathrm{Q}} \rightarrow \Ps} }[/math]) is derived from the forces ([math]\displaystyle{ \overline{\mathbf{F}}_{\Qs \rightarrow \Ps} }[/math] ) that each mass differential exerts on P (Figure D3.5). The resultant torsor on [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math] is obtained from all these forces [math]\displaystyle{ \overline{\mathbf{F}}_{\Qs \rightarrow \Ps} }[/math] on all mass differentials of [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math].

[math]\displaystyle{ \mathbf{F}_{\mathrm{S}_{\mathrm{Q}} \rightarrow \Ps} }[/math] is the resultant force of all the [math]\displaystyle{ \mathbf{F}_{\Qs \rightarrow \Ps} }[/math] Figure D3.5 Calculation of the resultant gravitational force of rígid body [math]\displaystyle{ \mathrm{S}_\mathrm{Q} }[/math] ([math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}_{\mathrm{Q}} \rightarrow \Ps} }[/math]) on a mass differential dm([math]\displaystyle{ \Ps }[/math]) of rigid body [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math].

When it comes to the Earth gravitational attraction ([math]\displaystyle{ \mathrm{S}_\mathrm{Q} = \mathrm{Earth} }[/math]) on a rigid body of small dimensions compared to those of the Earth and close to the Earth's surface, we usually apply the uniform field approximation: the forces [math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}_{\mathrm{Q}} \rightarrow \Ps} }[/math] are practically parallel to each other and their value is [math]\displaystyle{ \mathbf{F}_{\mathrm{S}_{\mathrm{Q}} \rightarrow \Ps} = \mathrm{gdm}(\Ps) }[/math], with g constant and equal to the gravitational field at the Earth's surface: [math]\displaystyle{ \mathrm{g}=\mathrm{G}_0 \frac{\mathrm{M}_\Ts}{\Rs_\Ts^2} }[/math] (where [math]\displaystyle{ \mathrm{G}_0 }[/math] is the universal gravitational constant, and [math]\displaystyle{ \mathrm{M}_\Ts }[/math] and [math]\displaystyle{ \Rs_\Ts }[/math] are the mass and radius of the Earth, respectively. In this case, it can be shown that there exists a point in [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math] where the gravitational torsor is a resultant force [math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}_{\mathrm{T}} \rightarrow \mathrm{S}_{\mathrm{P}}} }[/math] with value mg (where m is the [math]\displaystyle{ \mathrm{S}_\mathrm{P} }[/math] mass) pointing towards the centre of the Earth, and a zero gravitational resultant moment. That point is the gravity centre of the rigid body, and will be represented by letter G (Figure D3.6).

Figure D3.6 Earth gravitational attraction on a rigid body under the approximation of uniform field.

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D3.3 Interaction through linear and torsion springs and dampers

Linear springs and dampers

When a linear spring or a linear damper connects two points P and Q of two different rigid bodies, we have to guarantee that the connection is made in such a way that the force transmitted between the points has the direction of the element, and that no moment (or torque) is transmitted.

In some cases, this can be achieved by inserting the element between two lengths of inextensible thread (Figure D3.7a). Then, the force between points P and Q can only be an attraction (due to the unilateral nature of the thread action).

In other cases, it is necessary to use revolute joints (if it is a planar problem, Figure D3.7b) or spherical joints (if it is a 3D problem).

Figure D3.7 Linear springs and dampers between two different rigid bodies.

✏️ EXAMPLE D3.1: repulsion force of a linear spring and a linear damper with linear behaviour

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The spring, which has linear behaviour, acts between the support fixed to the ground and a vertical axis that is in contact with the bar. For [math]\displaystyle{ \theta = 0 }[/math], the system is in equilibrium, and the force exerted by the spring between its endpoints is [math]\displaystyle{ \mathrm{F}_0 }[/math].

Without the spring, the bar, which is hinged to the support at point O, would fall (clockwise rotation). If there is equilibrium for [math]\displaystyle{ \theta = 0 }[/math], the spring must exert a repulsive force between its endpoints in that configuration. Therefore, the logical thing to do is to formulate the spring force for a general configuration as a repulsive force:

[math]\displaystyle{ \mathrm{F}_{\mathrm{rep}}^{\mathrm{spring}}= \mathrm{F}_0 - \mathrm{k}\Delta\rho=\mathrm{F}_0-\mathrm{k}[\rho(\theta) - \rho(\theta=0)] }[/math].

The length increase [math]\displaystyle{ \Delta \rho }[/math] of the spring from equilibrium configuration is proportional to the tangent of the angle: [math]\displaystyle{ \mathrm{tan}\theta=\Delta\rho/\mathrm{L} }[/math]. Therefore: [math]\displaystyle{ \mathrm{F}_{\mathrm{rep}}^{\mathrm{rping}}=\mathrm{F}_0-\mathrm{kL}\mathrm{tan}\theta }[/math].

The damper repulsion force can be obtained from the length increase through a time derivative:

[math]\displaystyle{ \rho=\mathrm{Ltan}\theta \Rightarrow \dot{\rho} = \frac{\mathrm{L}\dot{\theta}}{\mathrm{cos}^2\theta} \Rightarrow \mathrm{F}_{\mathrm{rep}}^{\mathrm{damper}}=\mathrm{c}\dot{\rho}=-\mathrm{c} \frac{\mathrm{L}\dot{\theta}}{\mathrm{cos}^2\theta} }[/math]

✏️ EXAMPLE D3.2: attraction force of a linear spring and a linear damper with linear behaviour

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The spring, with linear behaviour, has one endpoint fixed to the chassis of the vehicle, and the other one to an inextensible thread that wounds on a roller of radius r. That roller is fixed to the wheel of radius 2r, which does not slide on the ground. For [math]\displaystyle{ \xs = 0 }[/math], the spring is stretched and exerts a force [math]\displaystyle{ \mathrm{F}_0 }[/math] between its endpoints.

The x coordinate describes the position of the chassis relative to the ground, and therefore also that of the centre of the wheel of radius 2r. Since the spring has one endpoint attached to a thread wounded on the roller, the displacement of this endpoint relative to the ground can be obtained through the integration of its speed.

Given the velocities of the two spring endpoints, it is evident that its length is decreasing. The approaching speed between the endpoints is:

[math]\displaystyle{ \mathrm{v}_{\mathrm{approaching}}(=-\dot{\rho})=\frac{3}{2}\dot{\xs}-\dot{\xs}=\frac{1}{2}\dot{\xs} }[/math]. Therefore, the spring length decrease from the [math]\displaystyle{ \xs = 0 }[/math] configuration is: [math]\displaystyle{ \frac{1}{2}\xs(=-\Delta \rho) }[/math].

The spring attraction force (since [math]\displaystyle{ \mathrm{F}_0 }[/math] is an attraction force because the spring is stretched for [math]\displaystyle{ \xs = 0 }[/math]) is: [math]\displaystyle{ \mathrm{F}_{\mathrm{at}}^{\mathrm{spring}} + \mathrm{k}\Delta \rho = \mathrm{F}_0 - \frac{1}{2}\mathrm{k}\xs }[/math].

The damper attraction force can be obtained from the approaching velocity calculated previously:

[math]\displaystyle{ \mathrm{v}_{\mathrm{approaching}}(=-\dot{\rho})=\frac{1}{2}\dot{\xs} \Rightarrow \mathrm{F}_{\mathrm{at}}^{\mathrm{damper}}=\mathrm{c} \dot{\rho}=-\frac{1}{2} \mathrm{c} \dot{\xs} }[/math]

Torsion springs and dampers

Torsion springs and torsion dampers introduce moments (not forces) between the two rigid bodies they connect. As with linear elements, the connection to each rigid body must guarantee that no other moments but that of the spring or camper are transmitted. As with linear elements, there are several ways to achieve this.

When torsion springs and torsion dampers have a linear behaviour, the moment increment they introduce between the rigid bodies, when the relative orientation between them increases by an angle [math]\displaystyle{ \Delta \theta }[/math], is proportional to [math]\displaystyle{ \Delta \theta }[/math] and [math]\displaystyle{ \theta }[/math], respectively (Figure D3.8).

Figure D3.8 Formulation of torsion springs and torsion dampers with linear behaviour.

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D3.4 Direct constraint interactions

Direct constraint interactions between two rigid bodies [math]\displaystyle{ \mathrm{S}_1 }[/math] and [math]\displaystyle{ \mathrm{S}_2 }[/math] occur when they are in contact, and they come from small local deformations of the rigid bodies in the contact zone. From a macroscopic point of view, this results in impenetrability and roughness of the rigid bodies. As mentioned in section D2.7, as this course deals with the dynamics of rigid objects, those deformations are not formulated, and therefore the associated forces are unknowns of the dynamic problem.

When the constraints involve rigid bodies (not particles), it is necessary to specify whether they are smooth or rough. A smooth surface cannot prevent an element from sliding on it while in contact. However, a rough surface can prevent this. This has a direct consequence on the constraint characterization.

As an introduction to the characterization of constraints between rigid bodies, it is useful to consider the simplest case of a single-point contact.

Let us consider two rigid bodies S1 and S2 with a single-point contact. The contact points are [math]\displaystyle{ \Ps_1 }[/math] and [math]\displaystyle{ \Ps_2 }[/math], respectively. The characterization of the constraint exerted by S2 on [math]\displaystyle{ \Ps_1 }[/math] (or exerted by S1 on [math]\displaystyle{ \Ps_2 }[/math]) is obtained as that of theparticle-surface constraint (Figure D3.9).

Figure D3.9 Characterization of the single-point contact between S1 and S2.

When the constraint between S1 and S2 is associated with a multiple-point contact, the orthogonality condition between the constraint force and the allowed velocity can be applied to each point where contact occurs (which implies accepting that the múltiple-point constraint can be obtained as a superposition of independent single-point contacts). Sometimes, this leads to a high number of constraint force components (even infinite, if contact occurs along a continuous linear section or a continuous surface section), and the force system must be reduced to a constraint torsor. In some cases, this reduction can be obtained very easily from the point-to-point constraint description.

Although it is possible to calculate the torsor of a system of forces at any point (section D3.1), when dealing with constraint torsors it is convenient that that point belongs to the rigid body undergoing that system of forces, because an important property is derived from it.

✏️ EXAMPLE D3.3: constraint torsor in a two-point contact

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The block has a two-point contact with a smooth floor, and has a planar motion.

The description of the constraint as a superposition of two single-point contacts leads to two normal forces: if there were only contact at P, the constraint would introduce only a force at P orthogonal to the ground ([math]\displaystyle{ \Ns_\mathrm{P} }[/math]); if the contact were only at Q, it would introduce a force at Q orthogonal to the block ([math]\displaystyle{ \Ns_\mathrm{Q} }[/math]). The [math]\displaystyle{ \Ns_\mathrm{P} }[/math] and [math]\displaystyle{ \Ns_\mathrm{Q} }[/math] values are independent (knowing [math]\displaystyle{ \Ns_\mathrm{P} }[/math] does not imply knowing [math]\displaystyle{ \Ns_\mathrm{Q} }[/math]). The two-point contact, therefore, introduces two constraint unknowns.

Since there are only two forces, the reduction to a torsor is not necessary. On the other hand, this description is useful for studying the two limit conditions: [math]\displaystyle{ \Ns_\mathrm{P}=0 }[/math] indicates loss of contact at P (therefore, clockwise tipping), while [math]\displaystyle{ \Ns_\mathrm{Q}=0 }[/math] indicates loss of contact at Q (therefore, counterclockwise tipping).

In this case, the torsor associated with that system of forces does not reduce the number of unknowns, which is two. The resulting force has two independent components:

[math]\displaystyle{ \left\{\overline{\mathrm{F}}_{\mathrm{T} \rightarrow \text { block }}\right\}=\left\{\begin{array}{l} \mathrm{F}_1 \\ \mathrm{F}_2 \\ 0 \end{array}\right\} \text {, amb }\left\{\begin{array}{l} \mathrm{F}_1=-\mathrm{N}_{\mathrm{Q}} \sin \theta \\ \mathrm{F}_2=\mathrm{N}_{\mathrm{p}}+\mathrm{N}_{\mathrm{Q}} \cos \theta \end{array}\right. \text {. } }[/math]

The resulting moment depends on the point of the block where the torsor is characterized, and when it is not zero, it can be written in terms of the force components.

Moving from the point-to-point description to the torsor has no advantage in this case: it does not reduce the number of constraint unknowns, and it makes the study of boundary conditions more difficult.

✏️ EXAMPLE D3.4: constraint torsor associated with a continuous linear contact

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The roller has a sliding contact with a horizontal ground S. The description of the multiple-point constraint as a superposition of single-point contacts leads to a set of infinite forces [math]\displaystyle{ \Ns_{\rightarrow \mathrm{J}} }[/math] at the contact points J in the direction orthogonal to the ground (direction 3). The resultant constraint force on the roller is therefore also in direction 3, and must be strictly positive since it is a unilateral constraint:

[math]\displaystyle{ \int_{\mathrm{J} \in \text { roller }} \mathrm{N}_{\rightarrow \mathrm{J}}=\mathrm{N}\gt 0 . }[/math]

The resultant constraint moment depends on the point on the roller at which it is calculated. At points P or Q, the sign of resultant moment is given:

[math]\displaystyle{ \{\overline{\mathbf{M}}(\mathbf{P})\}=\left\{\int_{\text {J} \in \mathrm{roller }} \overline{\mathbf{P J}} \times \overline{\mathbf{N}}_{\rightarrow \mathrm{J}}\right\}=\left\{\begin{array}{c} 0 \\ \mathbf{M}_2\gt 0 \\ 0 \end{array}\right\},\{\overline{\mathbf{M}}(\mathbf{Q})\}=\left\{\int_{\text {J} \in \mathrm{roller }} \overline{\mathbf{QJ}} \times \overline{\mathbf{N}}_{\rightarrow \mathrm{J}}\right\}=\left\{\begin{array}{c} 0 \\ \mathbf{M}_2\lt 0 \\ 0 \end{array}\right\} . }[/math]

The description of the constraint through the torsor is very advantageous: it drastically reduces the number of constraint unknowns (we now have only two). If the torsor is characterized at P or Q, the study of the limit condition for overturning is easy: [math]\displaystyle{ \mathrm{M}_2=0 }[/math]. If it is characterized at any other point on the contact line, we must go to P or Q to investigate the overturning.

Analytical characterization of the constraint torsor between two rigid bodies S1 and S2

The characterization of the constraint torsor between two rigid bodies can be done without going through the point-to-point description of the constraint: if the characterization point belongs to one of the two rigid bodies, it is enough to combine the orthogonality condition between force and velocity at each contact point with the rigid solid kinematics, and add all the equations that result from it (Figure D3.10).

Figure D3.10 Analytical characterization of a multiple-point contact between two rigid bodies.

Vectors [math]\displaystyle{ \overline{\mathbf{v}}_\mathrm{S2}(\Ps) \text{ and }\overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} }[/math] can be factored out. Finally:

[math]\displaystyle{ \Biggl(\sum_\mathrm{J} \overline{\mathbf{F}}_{\rightarrow \mathrm{J}} \Biggl) \cdot \overline{\mathbf{v}}_\mathrm{S2}(\Ps) + \Biggr[ \sum_\mathrm{J} (\overline{\mathbf{PJ}} \times \mathbf{F}_{\rightarrow \mathrm{J}}) \Biggr] \cdot \overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} = 0 \Rightarrow \overline{\mathbf{F}}_{\mathrm{S2}\rightarrow \mathrm{S1}}\cdot \overline{\mathbf{v}}_\mathrm{S2}(\Ps) + \overline{\mathbf{M}}_{\mathrm{S2}\rightarrow \mathrm{S1}}(\Ps) \cdot \overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1}=0 }[/math]

This is the equation of the analytical characterization of the constraint torsor. It establishes the orthogonality between the constraint torsor [math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{S2}\rightarrow \mathrm{S1}} \quad \overline{\mathbf{M}}_{\mathrm{S2}\rightarrow \mathrm{S1}}(\Ps)\bigl\}^\top }[/math] and the kinematic torsor [math]\displaystyle{ \bigl\{\overline{\mathbf{v}}_\mathrm{S2}(\Ps) \quad \overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} \bigl\}^\top }[/math] of S1 relative to S2:

[math]\displaystyle{ \left\{\begin{array}{c} \overline{\mathbf{F}}_{\mathrm{S2}\rightarrow \mathrm{S1}}\\ \overline{\mathbf{M}}_{\mathrm{S2}\rightarrow \mathrm{S1}}(\Ps) \end{array}\right\} \cdot \left\{\begin{array}{c} \overline{\mathbf{v}}_\mathrm{S2}(\Ps)\\ \overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} \end{array}\right\} =0. }[/math]

This orthogonality does not imply orthogonality between force and velocity on the one hand, and between moment and angular velocity on the other. In principle, [math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S2}\rightarrow \mathrm{S1}}\cdot \overline{\mathbf{v}}_\mathrm{S2}(\Ps) \neq 0 \text{, } \overline{\mathbf{M}}_{\mathrm{S2}\rightarrow \mathrm{S1}}(\Ps) \cdot \overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} \neq 0 }[/math].

When using the analytical characterization equation, it is initially necessary to consider that both the resulting force and moment have three non-zero components. As for the kinematic torsor, its non-zero components must be written as a function of the DoF of S1 relative to S2.

Moreover, since the force and velocity of the point are multiplied scalarly on the one hand, and the moment and angular velocity on the other, different vector bases can be used for each of these scalar products, since the result does not depend on the basis:

[math]\displaystyle{ \bigl\{\overline{\mathbf{F}}_{\mathrm{S2}\rightarrow \mathrm{S1}} \bigl\}_\mathrm{B} \cdot \bigl\{\overline{\mathbf{v}}_\mathrm{S2}(\Ps) \bigl\}_\mathrm{B} + \bigl\{\overline{\mathbf{M}}_{\mathrm{S2}\rightarrow \mathrm{S1}}(\Ps) \bigl\}_\mathrm{B'} \cdot \bigl\{\overline{\boldsymbol{\Omega}}_\mathrm{S2}^\mathrm{S1} \bigl\}_\mathrm{B'} }[/math]

✏️ EXAMPLE D3.5: analytical characterization of the constraint torsor associated with a linear contact

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The roller moves without sliding on the horizontal ground S. The analytical characterization of the torsor of the direct constraint of the ground on the roller, at point P, is:

[math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}}\cdot \overline{\mathbf{v}}_\mathrm{S}(\Ps) + \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}}(\Ps) \cdot \overline{\boldsymbol{\Omega}}_\mathrm{S}^\mathrm{roller}=0 }[/math]

[math]\displaystyle{ \left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} 0\\ 0\\ 0 \end{array}\right\} + \left\{\begin{array}{c} \mathrm{M}_1\\ \mathrm{M}_2 \\ \mathrm{M}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} \Omega_1\\ 0 \\ 0 \end{array}\right\} =0 }[/math]

All the components (but [math]\displaystyle{ \mathrm{M}_1 }[/math]) of the constraint torsor can have any value, since they are multiplied by zero. However, since [math]\displaystyle{ \Omega_1 }[/math] is not zero in principle, [math]\displaystyle{ \mathrm{M}_1 }[/math] has to be zero for the orthogonality equation to be satisfied. Thus, the resulting constraint torsor is:

[math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}} \quad \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}}(\Ps)\bigl\}^\top = \bigl\{\mathrm{F}_1 \quad \mathrm{F}_2 \quad \mathrm{F}_3 \qquad 0 \quad \mathrm{M}_2 \quad \mathrm{M}_3 \bigl\}^\top . }[/math]

As discussed in example D3.4, has to be positive. A negative value would indicate overturning.

The five non-zero components are independent, and it is true that the sum of the number of independent components of the torsor and that of DoF of the roller with respect to S is 6.

The torsor of the same constraint at a different point can be obtained either by applying again the analytical characterization equation, or from the torsor at P (section D3.1). For example, for point C these two methods lead to:

• Analytical characterization: [math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}}\cdot \overline{\mathbf{v}}_\mathrm{S}(\mathbf{C}) + \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}}(\mathbf{C}) \cdot \overline{\boldsymbol{\Omega}}_\mathrm{S}^\mathrm{roller}=0 }[/math]

[math]\displaystyle{ \left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} 0\\ \mathrm{v}_2 = - \rs \Omega_1\\ 0 \end{array}\right\} + \left\{\begin{array}{c} \mathrm{M'}_1\\ \mathrm{M'}_2 \\ \mathrm{M'}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} \Omega_1\\ 0 \\ 0 \end{array}\right\} = 0 \quad \Rightarrow \quad \bigl\{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}} \bigl\}= \left\{\begin{array}{c} \mathrm{F}_1\\ ...\\ \mathrm{F}_3 \end{array}\right\} \quad \text{,} \quad \bigl\{ \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{C}) \bigl\} =\left\{\begin{array}{c} ...\\ \mathrm{M'}_2 \\ \mathrm{M'}_3 \end{array}\right\} }[/math]

[math]\displaystyle{ (-\rs \mathrm{F}_2 + \mathrm{M'}_1)\Omega_1 =0 \quad \Rightarrow \quad -\rs\mathrm{F}_2 + \mathrm{M'}_1=0 }[/math]

Finally: [math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}} \bigl\}=\left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\}\quad \text{,} \quad \bigl\{ \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{C}) \bigl\} =\left\{\begin{array}{c} \mathrm{M'}_1\\ \mathrm{M'}_2 \\ \mathrm{M'}_3 \end{array}\right\} }[/math], amb [math]\displaystyle{ \mathrm{M'}_1=\rs\mathrm{F}_2 }[/math]

• From the constraint torsor at P: [math]\displaystyle{ \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{C})=\overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{P})+ \overline{\mathbf{CP}} \times \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}} }[/math]
  

[math]\displaystyle{ \bigl\{ \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{C}) \bigl\} =\left\{\begin{array}{c} 0\\ \mathrm{M}_2 \\ \mathrm{M}_3 \end{array}\right\}+ \left\{\begin{array}{c} 0\\ 0 \\ -\rs \end{array}\right\} \times \left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\} = \left\{\begin{array}{c} \rs \mathrm{F}_2\\ \mathrm{M}_2 - \rs \mathrm{F}_1 \\ \mathrm{M}_3 \end{array}\right\} \equiv \left\{\begin{array}{c} \mathrm{M'}_1\\ \mathrm{M'}_2 \\ \mathrm{M'}_3 \end{array}\right\} }[/math]

Finally: [math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{S}\rightarrow \mathrm{roller}} \bigl\}=\left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\}\quad \text{,} \quad \bigl\{ \overline{\mathbf{M}}_{\mathrm{S}\rightarrow \mathrm{roller}} (\mathbf{C}) \bigl\} =\left\{\begin{array}{c} \mathrm{M'}_1\\ \mathrm{M'}_2 \\ \mathrm{M'}_3 \end{array}\right\} }[/math], amb [math]\displaystyle{ \mathrm{M'}_1=\rs\mathrm{F}_2 }[/math]

The number of non-zero components of the torsor in C is 6, but the number of independent components is 5, since there is a dependency relationship between [math]\displaystyle{ \mathrm{M}_1' }[/math] and [math]\displaystyle{ \mathrm{F}_2 }[/math]. Therefore, it is still true that the number of independent components of torsor plus that of DoF is 6.

✏️ EXAMPLE D3.6: analytical characterization of the constraint torsor associated with a nonsliding single-point contact

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The ball with radius r does not slide inside the spherical cavity. The analytical characterization of the torsor of the direct constraint of the cavity on the ball, at its center G, is:

[math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}}\cdot \overline{\mathbf{v}}_\mathrm{cav}(\mathbf{G}) + \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}}(\mathbf{G}) \cdot \overline{\boldsymbol{\Omega}}_\mathrm{cav}^\mathrm{ball}=0 }[/math]

[math]\displaystyle{ \left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} \mathrm{v}_1\\ 0\\ \mathrm{v}_3 \end{array}\right\} + \left\{\begin{array}{c} \mathrm{M}_1\\ \mathrm{M}_2 \\ \mathrm{M}_3 \end{array}\right\} \cdot \left\{\begin{array}{c} \Omega_1\\ \Omega_2\\ \Omega_3 \end{array}\right\} = 0 \quad \Rightarrow \quad \bigl\{ \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}} \bigl\}= \left\{\begin{array}{c} ...\\ \mathrm{F}_2\\ ... \end{array}\right\} \quad \text{,} \quad \bigl\{ \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}} (\mathbf{G}) \bigl\} =\left\{\begin{array}{c} ...\\ ... \\ ... \end{array}\right\} }[/math]

Since there is no sliding at the contact point, the velocity components and are proportional to [math]\displaystyle{ \mathrm{v}_1 }[/math] and [math]\displaystyle{ \mathrm{v}_3 }[/math], respectively: [math]\displaystyle{ \mathrm{v}_1=-\rs \Omega_3 }[/math], [math]\displaystyle{ \mathrm{v}_3=\rs \Omega_1 }[/math] (the negative sign of the first equality indicates that a positive [math]\displaystyle{ \Omega_3 }[/math] generates a negative [math]\displaystyle{ \mathrm{v}_1 }[/math]). Substituting in the characterization equation and developing the scalar product (without including the component [math]\displaystyle{ \mathrm{F}_2 }[/math], because it is multiplied by zero):

[math]\displaystyle{ (-\rs \mathrm{F}_1 + \mathrm{M}_3) \Omega_3 + (\rs \mathrm{F}_3 + \mathrm{M}_1) \Omega_1 + \mathrm{M}_2\Omega_2=0. }[/math]

Since the three rotations are independent, the coefficients that multiply them have to be zero for the equation to hold for any value of [math]\displaystyle{ \Omega_3 }[/math] and of [math]\displaystyle{ \Omega_1 }[/math]: [math]\displaystyle{ -\rs \mathrm{F}_1 + \mathrm{M}_3=0 }[/math] ,[math]\displaystyle{ \rs \mathrm{F}_3 + \mathrm{M}_1 }[/math] , [math]\displaystyle{ \mathrm{M}_2=0 }[/math]. Finally:

[math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}} \quad \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}}(\mathbf{G})\bigl\}^\top = \bigl\{\mathrm{F}_1 \quad \mathrm{F}_2 \quad \mathrm{F}_3 \qquad \mathrm{M}_1 \quad 0 \quad \mathrm{M}_3 \bigl\}^\top }[/math], amb [math]\displaystyle{ \mathrm{M}_3=\rs \mathrm{F}_1 }[/math] i [math]\displaystyle{ \mathrm{M}_1=-\rs \mathrm{F}_3. }[/math]

The torsor has five non-zero components, but only three are independent, and the sum of the number of independent components of the torsor and the DoF of the ball with respect to the cavity is 6.

This torsor can also be obtained from the characterization at J (section D3.1). Since it is a nonsliding contact point, there are three constraint force components and no moment at J:

[math]\displaystyle{ \bigl\{ \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}} \quad \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}}(\mathbf{J})\bigl\}^\top = \bigl\{\mathrm{F}_1 \quad \mathrm{F}_2 \quad \mathrm{F}_3 \qquad 0 \quad 0 \quad 0 \bigl\}^\top }[/math]

[math]\displaystyle{ \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}} (\mathbf{G})=\overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}} (\mathbf{J})+ \overline{\mathbf{GJ}} \times \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}} = \overline{\mathbf{GJ}} \times \overline{\mathbf{F}}_{\mathrm{cav}\rightarrow \mathrm{ball}} }[/math]

[math]\displaystyle{ \bigl\{ \overline{\mathbf{M}}_{\mathrm{cav}\rightarrow \mathrm{ball}} (\mathbf{G}) \bigl\} = \left\{\begin{array}{c} 0\\ -\rs\\ 0 \end{array}\right\} \times \left\{\begin{array}{c} \mathrm{F}_1\\ \mathrm{F}_2 \\ \mathrm{F}_3 \end{array}\right\} = \left\{\begin{array}{c} -\rs \mathrm{F}_3\\ 0 \\ \rs \mathrm{F}_1 \end{array}\right\} \equiv \left\{\begin{array}{c} \mathrm{M}_1\\ 0 \\ \mathrm{M}_3 \end{array}\right\} }[/math]

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As seen in the previous examples, the analytical characterization equation ensures that the sum of the number of independent components of the constraint torsor between two rigid bpdies and the number of relative DoF between them is always 6:

[math]\displaystyle{ \boxed{\text{indep. torsor comps. + relative DoF = 6}} }[/math]

Straightforward characterization of the constraint torsor

When we choose a characterization point P whose velocity, [math]\displaystyle{ \overline{\mathbf{v}}_{\mathrm{S1}}(\Ps) }[/math], is independent from [math]\displaystyle{ \overline{\mathbf{\Omega}}_{\mathrm{S1}}^{\mathrm{S2}} }[/math], and a vector basis such that the [math]\displaystyle{ \overline{\mathbf{v}}_{\mathrm{S1}}(\Ps) }[/math] components of are independent of each other and those of [math]\displaystyle{ \overline{\mathbf{\Omega}}_{\mathrm{S1}}^{\mathrm{S2}} }[/math] are also independent, the characterization is straightforward: each zero component of the kinematic torsor corresponds to a non-zero component of the dynamic torsor, and each non-zero component of the kinematic torsor corresponds to a zero component of the dynamic torsor.

✏️ EXAMPLE D3.7: straightforward characterization of the constraint torsor associated with a continuous multiple-point contact

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The conical millstone does not slide on the conical ground (E). The characterization of the direct constraint torque of the ground on the millstone at any of the contact points J can be straightforward, because [math]\displaystyle{ \vvec_\mathrm{E}(\mathbf{J}) }[/math] independently of the angular velocity [math]\displaystyle{ \velang{millstone}{E} }[/math].

For the straightforward characterization to be possible, we must choose a vector basis for the constraint moment such that the [math]\displaystyle{ \velang{millstone}{E} }[/math] components of are independent. Since the [math]\displaystyle{ \velang{millstone}{E} }[/math] direction is univocally determined (the wheel has 1 DoF relative to the ground), any vector basis having an axis parallel to the millstone ISA is suitable:

Finally: [math]\displaystyle{ \braq{\fvec{E}{millstone}}{} = \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\quad \text{,} \quad \braq{\mvec{E}{millstone}(\mathbf{J})}{} = \vector{\mathrm{M}_1}{0}{\mathrm{M}_3}. }[/math]

✏️ EXAMPLE D3.8: characterization of the constraint torsor of a helical joint

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The characterization of the direct constraint torsor of the male of the helical joint on the female cannot be straightforward.

In a helical joint, there is no point on the female whose speed relative to the male is independent of the rotation between the two pieces. Therefore, straightforward characterization is not possible.

If we choose any point O on axis 3, the analytical characterization leads to:

[math]\displaystyle{ \fvec{male}{female}\cdot \vvec_\mathrm{male}(\mathbf{O})+ \mvec{male}{female}(\mathbf{O}) \cdot \velang{female}{male}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{0}{0}{\mathrm{v}_3}+\vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3}\vector{0}{0}{\Omega_3}=0 \quad \Rightarrow \quad \braq{\fvec{male}{female}}{}=\vector{\mathrm{F}_1}{\mathrm{F}_2}{...}\quad \text{,} \quad \braq{\mvec{male}{female}(\mathbf{O})}{}=\vector{\mathrm{M}_1}{\mathrm{M}_2}{...}. }[/math]

The speed of O along the axis is directly proportional to the rotation through the thread pitch e. Since [math]\displaystyle{ \Omega_3 }[/math] is measured in rad/s, [math]\displaystyle{ \mathrm{v}_3 }[/math] in m/s, and e is given in mm/turn, a conversion of units must be made:

[math]\displaystyle{ \mathrm{v}_3\left[\frac{\ms}{\ss}\right] = \Omega_3\left[\frac{\mathrm{rad}}{\ss}\right]\es\left[\frac{\mathrm{mm}}{\mathrm{ volta}}\right]\frac{1 \ms}{10^3 \mathrm{ mm}}\frac{1 \mathrm{turn}}{2 \pi \mathrm{ rad}} \Rightarrow \mathrm{v}_3= \Omega_3 \frac{\es}{2 \pi \cdot 10^3} }[/math]

Finally: [math]\displaystyle{ \braq{\fvec{male}{female}}{}=\vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\quad \text{,} \quad \braq{\mvec{male}{female}(\mathbf{O})}{}=\vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} }[/math], amb [math]\displaystyle{ \mathrm{M}_3= -\frac{\es}{2 \pi \cdot 10^3} \mathrm{F}_3 }[/math]

Torsors associated with the usual constraints joints between rigid bodies

Usual constraint joints between rigid bodies have been analyzed from a kinematic point of view. From this description, the corresponding constraint torsors can be characterized (Figure D3.11).

Figure D3.11 Straightforward characterizaion of usual constraint joints.

In multibody systems formed only by rigid bodies with non-negligible mass connected by these usual joints, the dynamic description of the constraint is made by considering separately each pair of connected rigid bodies as if all the other elements of the system did not exist.

✏️ EXAMPLE D3.9: analysis of unknonws in a multibody system

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The multibody system contains 5 usual constraint joints: 3 revolute joints (or hinges) 1 cylindrical joint 1 spherical joint (or ball-and-socket joint) Taking into account that a hinge introduces 5 independent torsor components, a cylindrical joint introduces 4, and a spherical joint 3, there are a total of 22 constraint unknowns in the system.

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D3.5 Indirect constraint interactions: Constraint Auxiliary Elements (CAE)

In multibody systems, it is common for some rigid bodies to have negligible mass compared to the others, and to be undergoing exclusively constraint interactions with other rigid bodies. These elements are called Constraint Auxiliary Elements (CAE), and they can be treated in a particular way when characterizing constraints.

En els sistemes multisòlid, és freqüent que alguns sòlids tinguin massa negligible comparats amb els altres i que no reben altres interaccions que les d’enllaç amb altres sòlids. Aquests sòlids s’anomenen Sòlids Auxiliars d’Enllaç (SAE), i se’ls pot donar un tractament particular quan es tracta de caracteritzar enllaços.

Figure D3.12 shows a rigid body S1 in contact with a rigid body S of negligible mass, undergoing only constraint interactions, in two different situations: S just in contact with S1 (Figure D3.12a) and S in contact with two rigid bodie S1 and S2 (Figure D3.12b).

Figure D3.12 Interaction between rigid bodies S1 and S2 through a constraint auxiliary element (CAE).

Without S2, S (in contact with S1) is dynamically irrelevant: S is not an obstacle to moving S1. Consequently, S is not able to introduce any constraint force on S1.

But when S connects S1 and S2, it becomes a transmitter: S2 can be an obstacle when it comes to triggering certain movements of S1, and this translates into constraint forces on S1.

As a simple case to illustrate this, Figure D3.13 shows two examples (which we will consider 2D examples) where two rigid bodies S1 and S2 on a smooth plane are connected through a bar of negligible mass (compared to that of S1 and S2) hinged at both ends. We will assume that:

(a) the joints are perfect joints (without friction)

(b) one of the joints is associated with a torsion spring

Figure D3.13 Indirect constraint between two rigid bodies S1 and S2 through a massless bar S with hinges.

In case (a), the bar is a CAE (undergoing only constraint interactions). Moving point O of S1 in the direction of the bar implies moving S and S2. On the other hand, moving O in the direction orthogonal to the bar provokes motion of S but not of S2, while rotating S1 around O provokes motion of neither of them (neither of S nor of S2). Therefore, it only makes sense to associate a constraint force on S1 in the direction of the bar. The element responsible for that force is not S but S2: when it comes to characterising the constraint on S1, the kinematics of S1 must be assessed from S2. We say then that there is an indirect constraint through the CAE between S1 and S2.

In case (b), the bar is not a CAE because it undergoes a non-constraint interaction (that of the spring). As in case (a), moving O in the direction of the bar forces S2 to move. But now, moving O in a direction perpendicular to the bar forces the spring to deform, which acts on S2 and provokes its motion. Rotational motion of S1 about O is still possible without having to move S2 or deform the spring. In characterizing the constraint on S1, the kinematics of S1 must be assessed such that O has no velocity: it must be assessed from S.

✏️ EXAMPLE D3.10: characterization of an indirect constraint

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The chariot has 2 free DoF relative to the ground (longitudinal translation and vertical rotation). If the mass of the three elements (wheels and chassis) is comparable, the simplified representation of the system and the number of constraint unknowns it contains is as follows:

The total number of constraint unknowns is 17.

This number can be reduced if the mass of the wheels is negligible (compared to that of the chassis): the chassis has three constraints with the ground, one direct and two indirect through the wheels, which are CAE. The number of unknowns introduced by one of these indirect constraints is not always obvious a priori, and it is advisable to characterise it analytically. But it must be remembered that the kinematical description must correspond to that of the chassis with respect to the ground as if only the constraint under consideration acted on it. In this case, therefore, in the characterisation of the indirect constraint between the ground and the chassis through the wheels, the kinematics must be evaluated as if direct contact with the ground (at the front of the chassis) did not exist.

Characterization of the indirect constraint torsor between ground and chassis through one wheel, at point C and vector basis (1,2,3):

[math]\displaystyle{ \overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(wheel)} \rightarrow \mathrm{chassis}} \cdot \vvec_\Es (\Cs) + \overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(wheel)} \rightarrow \mathrm{chassis}}(\Cs) \cdot \velang{chassis}{E}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{\mathrm{v}_1}{\mathrm{v}_2}{0}+ \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{\Omega_1}{\Omega_2}{\Omega_3}=0 \Rightarrow \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(wheel)} \rightarrow \mathrm{chassis}}}{}=\vector{...}{...}{\mathrm{F}_3} }[/math]

The [math]\displaystyle{ \mathrm{v}_1 }[/math] speed is only possible if there is [math]\displaystyle{ \Omega_2 }[/math] rotation: [math]\displaystyle{ \mathrm{v}_1=\rs\Omega_2 }[/math]. Introducing that relationship and proceeding to the scalar product:

[math]\displaystyle{ (\rs\mathrm{F}_1+\mathrm{M}_2)\Omega_2 + \mathrm{F}_2\mathrm{v}_2+ \mathrm{M}_1 \Omega_1 +\mathrm{M}_3 \Omega_3=0 }[/math]

Since [math]\displaystyle{ (\mathrm{v}_2,\Omega_1,\Omega_2,\Omega_3) }[/math] are independent, the coefficient of each of the velocities components must be zero. Therefore, the indirect constraint introduces only 2 unknowns:

[math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(wheel)} \rightarrow \mathrm{chassis}}}{}=\vector{\mathrm{F}_1}{0}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(wheel)} \rightarrow \mathrm{chassis}}(\Os)}{}=\vector{0}{\mathrm{M}_2}{0} }[/math], amb [math]\displaystyle{ \mathrm{M}_2=-\rs \mathrm{F}_1 }[/math]

✏️ EXAMPLE D3.11: characterization of an indirect constraint

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The P vertex of the plate is forced to move within the guide, and point Q is connected to point O of the ground (E) through a bar with hinges at its endpoints. The system has only 1 DoF, described by [math]\displaystyle{ \dot{\theta} }[/math]. If we treat it as a 2De problem and no element has negligible mass, the total number of constraint unknowns is 5.

If the bar has a negligible mass and is treated as a CAE, the characterization of the indirect constraint torsor though the ground and the plate at point Q is: [math]\displaystyle{ \overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(bar)} \rightarrow \mathrm{plate}} \cdot \vvec_\Es (\Qs) + \overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(plate)} \rightarrow \mathrm{plate}}(\Qs) \cdot \velang{plate}{E}=0 }[/math] [math]\displaystyle{ \left\{\begin{array}{c} \mathrm{F}_2\\ \mathrm{F}_3 \end{array}\right\}\left\{\begin{array}{c} 0\\ \mathrm{v}_3 \end{array}\right\} + \left\{\begin{array}{c} \mathrm{M}_1 \end{array}\right\}\left\{\begin{array}{c} \Omega_1 \end{array}\right\}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(bar)} \rightarrow \mathrm{plate}}}{}= \left\{\begin{array}{c} \mathrm{F}_2\\ ... \end{array}\right\}\quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(bar)} \rightarrow \mathrm{plate}}(\Qs)}{}=\left\{\begin{array}{c} ... \end{array}\right\} }[/math] The [math]\displaystyle{ \mathrm{v}_3 }[/math] speed comes from the rotation of the bar, but not from that of the plate: since there is a revolute joint at Q, the plate may have a circular translational motion when the bar rotates. Therefore, it is a straightforward characterization: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(bar)} \rightarrow \mathrm{plate}}}{}= \left\{\begin{array}{c} \mathrm{F}_2\\ 0 \end{array}\right\}\quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(bar)} \rightarrow \mathrm{plate}}(\Qs)}{}=\left\{\begin{array}{c} 0 \end{array}\right\} }[/math]

✏️ EXAMPLE D3.12: characterization of an indirect constraint

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The ball has 4 free DoF relative to the ground (translational motions along the r-r’ axis and axis 2, rotations about the r-r’ axis and axis 2). If there are no elements of negligible mass, the total number of constraint unknowns is 8.

If the T-element has a negligible mass and is treated as CAE, the characterization of the indirect constraint torsor between the ground and the ball through this element at point G is:

[math]\displaystyle{ \overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}} \cdot \vvec_\Es (\Gs) + \overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}}(\Gs) \cdot \velang{ball}{E}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{\mathrm{v}_1}{\mathrm{v}_2}{\mathrm{v}_3} + \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{\Omega_1}{\Omega_2}{0}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}}}{}=\vector{...}{...}{...} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}}(\Gs)}{}=\vector{...}{...}{\mathrm{M}_3} }[/math]

The [math]\displaystyle{ \mathrm{v}_3 }[/math] speed is only possible if the T-element (and therefore the ball) rotate with [math]\displaystyle{ \Omega_2 }[/math]: [math]\displaystyle{ \mathrm{v}_3=-\xs\Omega_2 }[/math]. If this relationship is introduced in the equation and the scalar product is developed:

La velocitat [math]\displaystyle{ \mathrm{v}_3 }[/math] només és possible si l’element T (i per tant la bola) giren amb [math]\displaystyle{ \Omega_2 }[/math] : [math]\displaystyle{ \mathrm{v}_3=-\xs\Omega_2 }[/math]. Si s’introdueix aquesta relació i es desenvolupa el producte escalar:

[math]\displaystyle{ \mathrm{F}_1\mathrm{v}_1+\mathrm{F}_2\mathrm{v}_2+(-\xs\mathrm{F}_3+\mathrm{M}_2)\Omega_2+\mathrm{M}_1\Omega_1 }[/math]

[math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}}}{}=\vector{0}{0}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(T-elem)} \rightarrow \mathrm{ball}}(\Gs)}{}=\vector{0}{\mathrm{M}_2}{\mathrm{M}_3} }[/math], amb [math]\displaystyle{ \mathrm{M}_2=\xs\mathrm{F}_3 }[/math]

Although the torsor at G has three nonzero components, there are only two independent ones.

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A paradigmatic case where the dimension is drastically reduced is that of the ball bearing (Figure D3.14). If the problem is treated as a planar one and the N balls are not considered as CAE, the number of constraint unknowns in the system is 4N (each ball has a nonsliding single-point contact with the two rígid bodies S1 and S2, and in the planar case, a single-point contact introduces two constraint unknowns). If they are treated as CAE, the indirect constraint between S1 and S2 through the balls is reduced to two constraint unknowns.

Figure D3.14 Ball bearing.

✏️ EXAMPLE D3.13: analysis of constraints in a multibody system

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The multibody system has 2 DoF relative to the ground: the circular translational motion of the plate relative to the axis and the rotation of all elements around the vertical axis (allowed by the bearing). If no element has negligible mass, the total number of constraint unknowns is 25.

If the masses of the axis and the two bars are neglected and they are treated as CAE, the system constraints are reduced to an indirect one:

[math]\displaystyle{ \overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plae}} \cdot \vvec_\Es (\Gs) + \overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plate}}(\Gs) \cdot \velang{plate}{E}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{\mathrm{v}_1}{\mathrm{v}_2}{\mathrm{v}_3} + \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{0}{\Omega_2}{0}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plate}}}{}=\vector{...}{...}{...} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plate}}(\Gs)}{}=\vector{\mathrm{M}_1}{...}{\mathrm{M}_3} }[/math]

The [math]\displaystyle{ \mathrm{v}_3 }[/math] speed comes from the rotation of the system about the vertical axis: [math]\displaystyle{ \mathrm{v}_3=-(\mathrm{h} + \mathrm{Lcos}\theta)\Omega_2 }[/math]. The [math]\displaystyle{ \mathrm{v}_1 }[/math] and [math]\displaystyle{ \mathrm{v}_2 }[/math] speeds come from the circular translational motion of the plate, and therefore are not independent: [math]\displaystyle{ \frac{\mathrm{v}_1}{\mathrm{v}_2}=-\mathrm{tan}\theta }[/math]

If the scalar product is developed taking into account these relations:

[math]\displaystyle{ (-\mathrm{F}_1\mathrm{tan}\theta+\mathrm{F}_2)\mathrm{v}_2+ [-(\hs+\Ls\mathrm{cos}\theta)\mathrm{F}_3+\mathrm{M}_2]\Omega_2=0, }[/math]

[math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plate}}}{}=\vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\Es \rightarrow \mathrm{(CAE)} \rightarrow \mathrm{plate}}(\Gs)}{}=\vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} }[/math], amb [math]\displaystyle{ \mathrm{F}_2=\mathrm{F}_1\mathrm{tan}\theta }[/math] i [math]\displaystyle{ \mathrm{M}_2=(\hs+\Ls\mathrm{cos}\theta)\mathrm{F}_3 }[/math]

The number of constraint unknowns has been reduced to 4.

✏️ EXAMPLE D3.14: analysis of constraints in a multibody system

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If we add a torsional spring with one end connected to the axis to the system in example D3.13, the number of DoF is not modified, but in the description of the constraints, the axis cannot be treated as CAE even though its mass is negligible. The two bars, however, are still subjected only to constraint interactions, and therefore, if their mass is neglected, they are CAE:

[math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{axis} \rightarrow \mathrm{(bars)} \rightarrow \mathrm{plate}} \cdot \vvec_\mathrm{axis} (\Gs) + \overline{\mathbf{M}}_{\mathrm{axis} \rightarrow \mathrm{(bars)} \rightarrow \mathrm{plate}}(\Gs) \cdot \velang{plate}{axis}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{-\mathrm{v}_2\mathrm{tan}\theta}{\mathrm{v}_2}{0} + \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{0}{0}{0}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\mathrm{axis} \rightarrow \mathrm{(bars)} \rightarrow \mathrm{plate}}}{}=\vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\mathrm{axis} \rightarrow \mathrm{(bars)} \rightarrow \mathrm{plate}}(\Gs)}{}=\vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3}, \mathrm{F}_2=\mathrm{F}_1\mathrm{tan}\theta }[/math]

The total number of constraint unknowns is 10: 5 associated with the indirect constraint that has been characterized, plus 5 associated with the bearing between ground and axis.

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D3.6 Interactions through linear and rotatory actuators

Actuators are elements designed to control a degree of freedom (translation or rotation) between two rígid bodies and eliminate any other motion.

From the outside, an actuator has two elements (P1 and P2) with only the DoF to be controlled between them (in the case of rotary actuators, those elements are called stator and rotor). When inserted between two solids S1 and S2, the type of connection with the solids determines the GL eliminated between them. In this course, the mass of the actuators is always considered negligible.

The interaction between S1 and S2 is therefore described by means of a force [math]\displaystyle{ \mathrm{F}_\mathrm{ac} }[/math] or a torque [math]\displaystyle{ \Gamma }[/math] (or moment), depending on whether it is a linear or rotational actuator, and a constraint torsor.

There are two ways to describe the system consisting of S1, S2, P1 and P2:

• Option 1: P1 and P2 are treated as two rigid bodies of the system. Between them, the direct constraint torsor is characterized as if the actuator were deactivated (allowing the DoF that it controls between the two elements when activated). Whether the actuator is linear or rotatory, the constraint torsor contains 5 independent components. In addition, between P1 and P2 there is the [math]\displaystyle{ \mathrm{F}_\mathrm{ac} }[/math] force or the [math]\displaystyle{ \Gamma }[/math] torque. The interaction between S1 and P1, and between S2 and P2 is described by the corresponding constraint.
  

• Option 2: The actuator is considered deactivated, and P1 and P2 are treated as CAE. The interaction between S1 and S2 is then explained by the torque of the indirect constraint associated with this CAE chain, and the [math]\displaystyle{ \mathrm{F}_\mathrm{ac} }[/math] force or the [math]\displaystyle{ \Gamma }[/math] torque .

✏️ EXAMPLE D3.15: linear actuator between two rigid bodies

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The massless linear actuator acts between the ground and the pendulum, and controls the speed of point O in direction 1. We want to describe the interaction between the ground and the pendulum.

Option 1

Since the P1 element of the actuator is attached to the ground, they can be considered as a single element. Part 2, on the other hand, is not attached but hinged to the pendulum.

The characterization at point O of both constraints is straightforward:

• Prismatic constraint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{ground}\rightarrow\mathrm{P}2}}{}=\vector{0}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{ground}\rightarrow\mathrm{P}2}(\Os)}{}=\vector{\Ms_1}{\Ms_2}{\Ms_3} }[/math]
  

• Revolute joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{P}2\rightarrow\mathrm{pendulum}}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{P}2\rightarrow\mathrm{pendulum}}(\Os)}{}=\vector{\Ms_1}{\Ms_2}{0} }[/math]
  

Option 2

The indirect constraint torsor between ground and pendulum through the actuator at point O can be characterized analytically:

[math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{ground} \rightarrow \mathrm{(ac)} \rightarrow \mathrm{pendulum}} \cdot \vvec_\mathrm{ground} (\Os) + \overline{\mathbf{M}}_{\mathrm{ground} \rightarrow \mathrm{(ac)} \rightarrow \mathrm{pendulum}}(\Os) \cdot \velang{pendulum}{ground}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{\mathrm{v}_1}{0}{0} + \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{0}{0}{\Omega_3}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\mathrm{ground} \rightarrow \mathrm{(ac)} \rightarrow \mathrm{pendulum}}}{}=\vector{0}{\mathrm{F}_2}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\mathrm{ground} \rightarrow \mathrm{(ac)} \rightarrow \mathrm{pendulum}}(\Os)}{}=\vector{\mathrm{M}_1}{\mathrm{M}_2}{0} }[/math]

In this case, it is a straightforward characterization.

The interactions between ground and pendulum due to the actuator are summarized in the following diagram:

✏️ EXAMPLE D3.16: linear actuator between two rigid bodies

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The linear actuator acts between the support and the propeller, and controls the speed of point O in direction 2 (which causes the propeller to rotate as long as it does not go through singular configurations – for example, when the direction of the actuator is vertical and goes through O). We want to describe the interaction between the support and the propeller.

Opció 1
None of the elements of the actuator are fixed to the support or the propeller. Therefore:

The characterization of the three constraints is straightforward:

• Revolute joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{support}\rightarrow\mathrm{P}1}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{support}\rightarrow\mathrm{P}1}(\Os')}{}=\vector{\Ms_1}{\Ms_2}{0} }[/math]
  

• Prismatic joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{ground}\rightarrow\mathrm{P}2}}{}=\vector{0}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{ground}\rightarrow\mathrm{P}2}(\Os)}{}=\vector{\Ms_1}{\Ms_2}{\Ms_3} }[/math]
  

• Revolute joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{P}2\rightarrow\mathrm{pendulum}}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{P}2\rightarrow\mathrm{pendulum}}(\Os)}{}=\vector{\Ms_1}{\Ms_2}{0} }[/math]
  

Option 2

The torsor of the indirect constraint between support and propeller through the actuator is zero: as there are spherical joints at both ends of the actuator, point O may move in all directions with respect to the support.

The interaction between support and propeller through the actuator is reduced to a force (which is the [math]\displaystyle{ \Fs_{ac} }[/math] force of the actuator does when it is activated):

✏️ EXAMPLE D3.17: rotatory actuator between two rigid bodies

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The motor, of negligible mass, acts between the ground and the ring, and controls the angular velocity between the two. We want to describe the interaction between the ground and the ring. Option 1 In this case, the stator (P1) is fixed to the ground, and the rotor (P2) is fixed to the ring. Therefore, only option 1 makes sense. Les interaccions entre terra i anella per causa de l’actuador es resumeixen en el diagrama següent: The constraint torsor between the ground (+P1) and the ring (+P2) corresponds to a revolute joint, and its characterization is straightforward: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{ground}\rightarrow\mathrm{ring}}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{ground}\rightarrow\mathrm{ring}}(\Os)}{}=\vector{\Ms_1}{\Ms_2}{0} }[/math]

✏️ EXAMPLE D3.18: rotatory actuator between two rigid bodies

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The motor acts between the fork and the wheel, and controls the angular velocity between the two. We want to describe the interaction between the fork and the wheel.

Option 1

One of the elements of the the actuator (the rotor) is fixed to the wheel. Therefore:

The characterization of both constraints is straightforward:

• Revolute joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{fork}\rightarrow\mathrm{P}1}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{fork}\rightarrow\mathrm{P}1}(\Os)}{}=\vector{\Ms_1}{0}{\Ms_3} }[/math]
  

• Revolute joint: [math]\displaystyle{ \braq{\overline{\mathbf{F}}_{\mathrm{P}1\rightarrow\mathrm{wheel}}}{}=\vector{\Fs_1}{\Fs_2}{\Fs_3}, \braq{\overline{\mathbf{M}}_{\mathrm{P}1\rightarrow\mathrm{wheel}}(\Os)}{}=\vector{0}{\Ms'_2}{\Ms'_3} }[/math]
  

Option 2

The characterization of the indirect constraint between fork and wheel through the motor is:

[math]\displaystyle{ \overline{\mathbf{F}}_{\mathrm{fork} \rightarrow \mathrm{(motor)} \rightarrow \mathrm{wheel}} \cdot \vvec_\mathrm{fork} (\Os) + \overline{\mathbf{M}}_{\mathrm{fork} \rightarrow \mathrm{(motor)} \rightarrow \mathrm{wheel}}(\Os) \cdot \velang{wheel}{fork}=0 }[/math]

[math]\displaystyle{ \vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3}\vector{0}{0}{0} + \vector{\mathrm{M}_1}{\mathrm{M}_2}{\mathrm{M}_3} \vector{\Omega_1}{\Omega_2}{0}=0 \quad \Rightarrow \quad \braq{\overline{\mathbf{F}}_{\mathrm{forq} \rightarrow \mathrm{(motor)} \rightarrow \mathrm{wheel}}}{}=\vector{\mathrm{F}_1}{\mathrm{F}_2}{\mathrm{F}_3} \quad \text{,} \quad \braq{\overline{\mathbf{M}}_{\mathrm{fork} \rightarrow \mathrm{(motor)} \rightarrow \mathrm{wheel}}(\Os)}{}=\vector{0}{0}{\mathrm{M}_3} }[/math]

The interaction between fork and wheel through the motor is summarized in the following diagram:

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