Elastic forces on a rubber band


Posted by Diego Assencio on 2015.10.14 under Physics (Mechanics)

Consider a uniform elastic band of rest length $L$ and mass $M$ as shown in figure 1a. The rest length $L$ is such that if the elastic band is shaped as a circle of perimeter $L$ (i.e., a circle with radius $R = L/2\pi$), the elastic force over its entire extension is identically zero. In this rest configuration, we parameterize each point of the band using its arclength $s$: for that, we choose a fixed point to be the one with $s = 0$ and all other points of the band are assigned $s$ values which grow in a counterclockwise manner, so $s$ values are in the range $[0,L)$ as shown in figure 1b.

Elastic band in rest configuration
(a)
Elastic band at arbitrary configuration
(b)
Fig. 1: An elastic band. Figure (a) shows the elastic band in an arbitrary configuration described by a function ${\bf X}(s,t)$ and the unit vector $\pmb\tau(s,t)$ which is tangent to the band at the point ${\bf X}(s,t)$. Figure (b) shows how the arclength value $s$ is defined for each point of the interface in the rest configuration.

As the elastic band is allowed to move, we can track the motion of each of its points using the arclength value $s$ assigned to each point in the rest (circular) configuration, i.e., we can define the position of each point of the band through a function ${\bf X}(s,t)$ which defines the location of the point with assigned arclength value $s$ at time $t$ (see figure 1a).

Our goal is to compute the force per unit length ${\bf f}(s,t)$ at the point ${\bf X}(s,t)$ for an arbitrary configuration of the elastic band. For that, let us first model the band as a set of $n$ small masses (each one with mass $M/n$) connected by identical springs of elastic coefficient $k$ and rest length $\Delta{s} = L/n$ as shown in figure 2. In what follows, we will refer to this model as the "discrete elastic band".

Fig. 2: A discrete elastic band consisting of masses connected by springs.

The force at the mass located at ${\bf x}_i$ is the sum of the elastic forces produced by the two springs connected to it. Let $\Delta{\bf x}_i = {\bf x}_{i+1} - {\bf x}_i$ and let $\pmb\eta_{i}$ be a unit vector parallel to $\Delta{\bf x}_i$, i.e.: $$ \pmb\eta_{i} = \displaystyle\frac{{\bf x}_{i+1} - {\bf x}_i}{\|{\bf x}_{i+1} - {\bf x}_i\|} = \displaystyle\frac{\Delta{\bf x}_i}{\|\Delta{\bf x}_i\|} $$ With these definitions, the force at the mass located at ${\bf x}_i$ is given by: $$ \Delta{\bf F}_i = k \left(\|\Delta{\bf x}_i\| - \Delta{s}\right)\pmb\eta_{i} + k\left(\|\Delta{\bf x}_{i-1}\| - \Delta{s}\right)(-\pmb\eta_{i-1}) \label{post_be83d9c57ef2d0c138c880ce222471eb_force_discrete_band} $$ The reason for the minus sign before $\pmb\eta_{i-1}$ on the second term comes from the fact that if the spring which connects ${\bf x}_i$ and ${\bf x}_{i-1}$ is stretched beyond its rest length $\Delta{s}$, the force on the mass at ${\bf x}_i$ points towards the mass at ${\bf x}_{i-1}$, i.e., it is parallel to $(-\pmb\eta_{i-1})$. Since we are interested in computing the force per unit length ${\bf f}_i$ on ${\bf x}_i$, we can divide equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_discrete_band} by the band length $\Delta{L}_i$ associated with the mass at ${\bf x}_i$: $$ \Delta{L}_i = \displaystyle\frac{\|\Delta{\bf x}_i\| + \|\Delta{\bf x}_{i-1}\|}{2} \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_delta_L} $$ We then get: $$ \begin{eqnarray} \displaystyle {\bf f}_i &=& \frac{\Delta{\bf F}_i}{\Delta{L}_i} \nonumber\\[5pt] &=& \frac{\Delta{s}}{\Delta{L}_i} \frac{\Delta{\bf F}_i}{\Delta{s}}\nonumber\\[5pt] &=& \frac{\Delta{s}}{\Delta{L}_i} k \left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right] \nonumber\\[5pt] &=& (k\Delta{s}) \frac{\Delta{s}}{\Delta{L}_i}\frac{1}{\Delta{s}}\left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right] \label{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} \end{eqnarray} $$ As we increase the number of masses $n$ in the the discrete band while keeping the total mass equal to $M$, the discrete force density ${\bf f}_i$ will converge to the force density ${\bf f}(s,t)$ we are trying to compute. To obtain ${\bf f}(s,t)$, we must analyze each term on the last line of equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} and see how it behaves as $n \rightarrow \infty$. Since the elastic band is parameterized by a function ${\bf X}(s,t)$ as described earlier, and since $\Delta{s} \rightarrow ds$ as $n \rightarrow \infty$, we have that: $$ \displaystyle \frac{\|\Delta{\bf x}_i\|}{\Delta{s}} = \left\| \frac{\Delta{\bf x}_i}{\Delta{s}} \right\| \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1} $$ Similarly: $$ \displaystyle \frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} = \left\| \frac{\Delta{\bf x}_{i-1}}{\Delta{s}} \right\| \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2} $$ With the definition of $\Delta{L}_i$ given in equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_delta_L} as well as equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2}, we have that: $$ \displaystyle \frac{\Delta{L}_i}{\Delta{s}} = \displaystyle \frac{1}{2}\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} + \frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}}\right) \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_dL_ds} $$ Additionally: $$ \pmb\eta_i = \displaystyle\frac{\Delta{\bf x}_i}{\|\Delta{\bf x}_i\|} = \left(\displaystyle\frac{\Delta{\bf x}_i}{\Delta{s}}\right) \Big/ \left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}}\right) $$ and therefore (the same result is true for $\pmb\eta_{i-1}$): $$ \pmb\eta_i \longrightarrow \frac{\partial{\bf X}}{\partial s} \Big/ \left\|\frac{\partial{\bf X}}{\partial s}\right\| = \pmb\tau(s,t) \label{post_be83d9c57ef2d0c138c880ce222471eb_tau_converges} $$ where $\pmb\tau(s,t)$ is the unit vector which is tangent to the elastic band at the point ${\bf X}(s,t)$ as shown in figure 1a. From equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1}, \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_tau_converges}, we see that the term: $$ \frac{1}{\Delta{s}}\left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right] $$ converges to: $$ \frac{\partial}{\partial{s}}\left[ \left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) \frac{\partial{\bf X}}{\partial s} \Big/ \left\|\frac{\partial{\bf X}}{\partial s}\right\|\right] = \frac{\partial}{\partial{s}}\left[ \left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) \pmb\tau\right] \label{post_be83d9c57ef2d0c138c880ce222471eb_partial_term} $$ The last component we have to analyze is the elastic constant $k$. What does it converge to when $n \rightarrow \infty$? The answer to this question lies on equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} itself. Since all other terms converge to fixed quantities, $k\Delta{s}$ must converge to a fixed quantity as well. Given that $k\Delta{s}$ is physically associated with the elasticity of the band, we must have $k\Delta{s} \rightarrow \kappa$ for a fixed constant $\kappa$, meaning $k$ must grow proportionally to $1/\Delta{s}$ as $\Delta{s} \rightarrow 0$ in order for the assumption ${\bf f}_i \rightarrow {\bf f}(s,t)$ to hold. The constant $\kappa$ represents the elastic coefficient of the continuous elastic band. Using this fact as well as equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_dL_ds} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_partial_term}, we have that the force density ${\bf f}(s,t)$ on the elastic band at the point ${\bf X}(s,t)$ is given by: $$ \boxed{ \displaystyle {\bf f}(s,t) = \frac{\partial}{\partial{s}}(T\pmb\tau) \Big/ \left\|\frac{\partial{\bf X}}{\partial{s}}\right\| } $$ where $T(s,t)$ is the tension on the elastic band at the point ${\bf X}(s,t)$: $$ \boxed{ \displaystyle T(s,t) = \kappa\left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) } $$ Finally, notice that our derivation did not require the elastic band to be really closed, i.e., the results above apply to elastic fibers in general. Elastic fibers are commonly used in numerical simulations of fluid-structure interactions. For instance, see the immersed boundary method developed by Charles S. Peskin to simulate blood flow inside the human heart.

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