Estimation of impact forces during multi-point collisions involving small deformations

Abstract

Collision between hard objects causes abrupt changes in the velocities of the system, which are characterized by very large contact forces over very small time durations. A common approach in the analysis of such collisions is to describe the system velocities using an impulse–momentum based relationship. The time duration of impact and the deformations at the contact points are usually assumed to be negligible for such an impact models, and the system velocities are evolved in terms of the impulses on the system. This type of impact models are usually relevant for hard (rigid) impacts, where deformations at the contact points can be considered negligible. However, these models cannot determine the forces during the impact process. The main objective of this work is to reformulate the impulse–momentum based model to determine the forces during an impact event, by relaxing the rigidity assumption to allow small deformations at the contact points.

Appendix A: Frictional and rigid body constraints

This section discusses the derivation of ${\mathbf{C}}$ in (5) as well as the energetic termination condition for the impact events. During impact events, points may stick, slip or slip-reverse. The slip-state of any given impact point is characterized in terms of the sliding speed along the impact plane,

$$ \left \{ \textstyle\begin{array}{l@{\ }c@{\ }c@{\quad }l} s_{i} =0 & \text{and} & \dot{s}_{i} =0 & {\text{sticking},} \\ s_{i} =0 & \text{and} & \dot{s}_{i} \neq 0 & \text{stick--slip transition,} \\ s_{i} \neq 0 & & & {\text{slipping},} \end{array}\displaystyle \right . $$

(52)

where the subscript $i$ refers to the impact point. In the literature, stick–slip behavior is represented as a set of complementarity conditions [7], which establish frictional force relationships for different slip states. These complementarity conditions are defined based on Coulomb friction [4, 7, 28], and form the basis for the LCP, which are solved using optimization techniques [28, 61]. This work uses a similar set of complementarity conditions based on Coulomb friction law,

$$ \left \{ \textstyle\begin{array}{l@{\quad }l} {\text{sticking}} & dp_{t_{i}} \leq \mu _{s} \ dp_{z_{i}} , \\ {\text{stick-slip transition}} & dp_{t_{i}} =\mu _{s} \ dp_{z_{i}} , \\ {\text{slipping}} & dp_{t_{i}} =\mu _{d} \ dp_{z_{i}} , \end{array}\displaystyle \right . $$

(53)

where $dp_{t_{i}} = \sqrt{{dp_{x_{i}}}^{2} + {dp_{y_{i}}}^{2}}$, $\mu _{s}$ and $\mu _{d}$ are the values of static and dynamic coefficients of friction, and $i= \{ 1,\ldots,4 \}$. Here the complementarity conditions are expressed in terms of the forces in (5), where $dt$ is canceled from both sides, leaving the differential impulses.

1.1 A.1 Frictional constraint for slipping

If impact point $i$ slips, then based on (53), the tangential and normal differential impulses are related as

$$ \displaystyle \frac{dp_{x_{i}}}{dt} = -\mu _{i} \cos (\phi _{i}) \ \frac{dp_{z_{i}}}{dt}, \qquad \displaystyle \frac{dp_{y_{i}}}{dt} = -\mu _{i} \sin (\phi _{i}) \ \frac{dp_{z_{i}}}{dt}, $$

(54)

where $\mu _{i}$ is the coefficient of friction for point $i$. Here this value is selected as $\mu _{i} = \mu _{d}$. The equations in (54) also govern the force relations during slip reversal. The frictional constraints in (54) can be expressed as

$$ \left \lbrack \textstyle\begin{array}{c} dp_{x_{i}} + \mu _{i} \ \cos (\phi _{i}) \ dp_{z_{i}} \\ dp_{y_{i}} + \mu _{i} \ \sin (\phi _{i}) \ dp_{z_{i}} \end{array}\displaystyle \right \rbrack =U_{i} \ d{\mathbf{p}} ={\mathbf{0}}. $$

(55)

1.2 A.2 Frictional constraint for sticking

When impact point $i$ sticks, the tangential components of its velocity must remain equal to zero, $s_{i} = 0$, and therefore $v_{x_{i}} = v_{y_{i}} = 0$. The differential velocities of the sticking points are obtained from (3) as

$$ d{\boldsymbol{\vartheta }} =M \ d{\mathbf{p}} =\left [ {\mathbf{m}}_{x_{1}}^{T} \ \ {\mathbf{m}}_{y_{1}}^{T} \ \ {\mathbf{m}}_{z_{1}}^{T} \ \ \ldots \right ]^{T} d {\mathbf{p}} .$$

(56)

When an impact point sticks [75],

$$ \left \lbrack \textstyle\begin{array}{c} dv_{x_{i}} \\ dv_{y_{i}} \end{array}\displaystyle \right \rbrack =\left \lbrack \textstyle\begin{array}{c} {\mathbf{m}}_{x_{i}} \\ {\mathbf{m}}_{y_{i}} \end{array}\displaystyle \right \rbrack d{\mathbf{p}} =U_{i} \ d{\mathbf{p}} ={\mathbf{0}} .$$

(57)

1.3 A.3 Rigid-body constraints

The frictional constraints are sufficient for resolving single-point impacts, but multi-point analysis requires additional constraints. Here such constraints are derived using the rigidity properties of the impacting body. These constraints are expressed in terms of the line connecting two points on a rigid body:

$$ \hat{\boldsymbol{\eta }} =\frac{({\mathbf{P}}_{Oi} - {\mathbf{P}}_{Oj})}{||{\mathbf{P}}_{Oi} - {\mathbf{P}}_{Oj}||} =\left [ \eta _{x} \ \ \eta _{y} \ \ \eta _{z} \right ]^{T} $$

(58)

where ${\mathbf{P}}_{Oi}$ and ${\mathbf{P}}_{Oj}$ are the position vectors from the body-attached reference point $O$ to the impact points $i,j= \{ 1,\ldots,4 \}$, $i \neq j$. The relative motion of two impact points on the same rigid body must be equal to zero when observed from the body-attached reference frame and point, therefore

$$ \begin{gathered} ({\mathbf{v}}_{i} - {\mathbf{v}}_{j}) \cdot \hat{\boldsymbol{\eta }} =0, \\ (v_{x_{i}}-v_{x_{j}})\eta _{x} + (v_{y_{i}}-v_{y_{j}})\eta _{y} + (v_{z_{i}}-v_{z_{j}}) \eta _{z} ={\mathbf{\mathbf{w}}}_{ij} \ {\boldsymbol{\vartheta }} =0. \end{gathered} $$

(59)

Equation (59) provides a constraint equation which relates the velocities of points $i$ and $j$. Now, it is possible to transform this velocity based rigid-body constraint into a constraint on the differential impulses (or forces) based on a work–energy argument, as shown in [10, 11, 65, 66]. Let us consider one way constraining the velocities using (59), such that $v_{x_{i}}$ depends upon all other velocity components,

$$ {\boldsymbol{\vartheta }} = \left \lbrack \textstyle\begin{array}{c} v_{x_{j}} - (v_{y_{i}}-v_{y_{j}})\frac{\eta _{y}}{\eta _{x}} - (v_{z_{i}}-v_{z_{j}}) \frac{\eta _{z}}{\eta _{x}} \\ v_{y_{i}} \\ v_{z_{i}} \\ v_{x_{j}} \\ v_{y_{j}} \\ v_{z_{j}} \end{array}\displaystyle \right \rbrack = \underbrace{ \left \lbrack \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} -\alpha & -\beta & 1 & \alpha & \beta \\ 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 \end{array}\displaystyle \right \rbrack }_{P} {\boldsymbol{\vartheta }}_{s} $$

(60)

where ${\boldsymbol{\vartheta }}_{s}$ are independent velocities, $P$ is a matrix representing the rigid-body constraint, and $\alpha =\frac{\eta _{y}}{\eta _{x}}$ and $\beta = \frac{\eta _{z}}{\eta _{x}}$. This may be also expressed as

$$ {\boldsymbol{\vartheta }} = J \dot{\mathbf{q}} = P {\boldsymbol{\vartheta }}_{s} .$$

(61)

This yields

$$ {\boldsymbol{\vartheta }}_{s} = P^{+} J \dot{\mathbf{q}} $$

(62)

where $P^{+} = {(P^{T} P)}^{-1} P^{T}$ is the generalized inverse of the matrix $P$. As shown in [12], the virtual work due to the contact forces yields the relation,

$$ {\mathbf{F}}^{T} {\boldsymbol{\vartheta }} = {\boldsymbol{\Gamma }}^{T} \dot{\mathbf{q}} .$$

(63)

Let ${\mathbf{F}}_{s}$ be a vector containing the constrained force terms corresponding to the independent velocities ${\boldsymbol{\vartheta }}_{s}$. Enforcement of the velocity-level rigid-body constraints should not cause any change in the virtual work of the system. Hence,

$$ {{\mathbf{F}}_{s}}^{T} {\boldsymbol{\vartheta }}_{s} = {\boldsymbol{\Gamma }}^{T} \dot{\mathbf{q}} .$$

(64)

Now substituting the relation ${\boldsymbol{\Gamma }} = J^{T} {\mathbf{F}}$ and (62) into (64) yields

$$ {{\mathbf{F}}_{s}}^{T} P^{+} J \dot{\mathbf{q}} = {\left ( J^{T} {\mathbf{F}} \right )}^{T} \dot{\mathbf{q}} .$$

(65)

This yields the constrained force relationships,

$$ {\mathbf{F}} = (P^{+})^{T} {\mathbf{F}}_{s}, \qquad {\mathbf{F}}_{s} = P^{T} {\mathbf{F}} .$$

(66)

Therefore, from (66), a projection of rigid-body constraints onto the force space can be obtained as

$$ \left ( I - {\left ( P^{+} \right )}^{T} P^{T} \right ) {\mathbf{F}} = { \mathbf{0}} .$$

(67)

Note that in (67), ${\left ( P^{+} \right )}^{T} P^{T} \neq I$. The matrix ${\left ( P^{+} \right )}^{T} P^{T}$ projects the forces ${\mathbf{F}}$ to a space orthogonal to the velocity-level constraint such that it is equal to ${\mathbf{F}}$. Now using the coefficients of $P$ from (60), the matrix $I - {\left ( P^{+} \right )}^{T} P^{T}$ can be symbolically obtained as

$$ I - {\left ( P^{+} \right )}^{T} P^{T} = \frac{1}{2 \left ( \alpha ^{2}+ \beta ^{2} +1 \right )} \left \lbrack \textstyle\begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 1 & \alpha & \beta & -1 & - \alpha & - \beta \\ \alpha & \alpha ^{2} & \alpha \beta & -\alpha & -\alpha ^{2} & - \alpha \beta \\ \beta & \alpha \beta & \beta ^{2} & - \beta & - \alpha \beta & - \beta ^{2} \\ -1 & - \alpha & - \beta & 1 & \alpha & \beta \\ - \alpha & - \alpha ^{2} & - \alpha \beta & \alpha & \alpha ^{2} & \alpha \beta \\ - \beta & - \alpha \beta & -\beta ^{2} & \beta & \alpha \beta & \beta ^{2} \\ \end{array}\displaystyle \right \rbrack .$$

(68)

It can be seen that all equations obtained by substituting (68) into (67) yield the same force constraint:

$$ f_{x_{i}} + \alpha f_{y_{i}} + \beta f_{z_{i}} - f_{x_{j}} - \alpha f_{y_{j}} - \beta f_{z_{j}} = 0 .$$

(69)

Now substituting back $\alpha = \frac{\eta _{y}}{\eta _{x}}$ and $\beta = \frac{\eta _{z}}{\eta _{x}}$, and rearranging gives the rigid-body constraint,

$$ (f_{x_{i}}-f_{x_{j}})\eta _{x} + (f_{y_{i}}-f_{y_{j}})\eta _{y} + (f_{z_{i}}-f_{z_{j}}) \eta _{z} = {\mathbf{\mathbf{w}}}_{ij}{\mathbf{F}} = 0 $$

(70)

or in terms of differential impulses

$$ {\mathbf{\mathbf{w}}}_{ij} d{\mathbf{p}} = 0 .$$

(71)

The rigid-body constraint in (71) is valid for any two arbitrary contact points $i$ and $j$. Hence to resolve the differential impulses for an $n$-point impact in terms of a single independent parameter, $n-1$ such rigid-body constraints would need to be solved along with $2n$ frictional constraint equations.

1.4 A.4 Final frictional and rigid body constraints

The frictional and rigid-body constraints from (55), (57) and (71) for the example block in Fig. 1 can be expressed as

$$ H \ d{\mathbf{p}} =\left \lbrack \textstyle\begin{array}{c} U_{1} \\ \vdots \\ U_{4} \\ {\mathbf{w}}_{14} \\ {\mathbf{w}}_{24} \\ {\mathbf{w}}_{34} \end{array}\displaystyle \right \rbrack d{\mathbf{p}} =0 .$$

(72)

Note that the ${\mathbf{w}}_{ij}$ will always be linearly independent as long as $i \neq j$. The matrix $H \in \mathbb{R}^{11 \times 12}$ is rank deficient by one row. The linearly dependent and independent columns of $H$ can be separated as $H =\left [ H_{s} \ \ H_{r} \right ]$. Here $H_{s} \in \mathbb{R}^{11 \times 11}$ is a full rank matrix, $H_{r} \in \mathbb{R}^{11 \times 1}$ is a linearly dependent column in $H$, such that

$$ d{\mathbf{p}} =\left \lbrack \textstyle\begin{array}{c} -H_{s}^{-1} H_{r} \\ 1 \end{array}\displaystyle \right \rbrack dp_{n} ={\mathbf{C}} \ dp_{n} $$

(73)

where ${\mathbf{C}}$ relates all of the differential impulses to the single independent impulse parameter $dp_{n}$.

If $s_{i} = 0$, the differential impulses must satisfy the no-slip condition,

$$ \sqrt{ dp_{x_{i}}^{2} + dp_{y_{i}}^{2} } \leq \mu _{s} \ dp_{z_{i}} \quad \implies \quad \sqrt{ C_{x_{i}}^{2} + C_{y_{i}}^{2} } \leq \mu _{s} \ C_{z_{i}} $$

(74)

where $C_{x_{i}}$, $C_{y_{i}}$ and $C_{z_{i}}$ are the element of ${\mathbf{C}}$ associated with $dp_{x_{i}}$, $dp_{y_{i}}$, and $dp_{z_{i}}$ respectively. Thus (74) implies that the slip state of an impact point is dependent on the friction and rigid-body constraints, rather than the value of $p_{n}$. If the no-slip condition is not satisfied, then point $i$ slips along a new direction $\hat{\phi }_{i}$. This is known as slip reversal. The following section presents how this new slip direction $\hat{\phi }_{i}$ is calculated. Since this work considers rigid bodies, if two or more impact points stick, then sticking is enforced in all other contact points. Hence there can be only three possibility during stick–slip transition: 1) all points enter sticking, 2) a single point sticks, while the rest slip-reverse or 3) all points slip-reverse in some new directions. If the slip states of any of the impact points must be altered, $H$ must be re-evaluated and all conclusions of sticking and slipping must be rechecked.