# Internal Forces Computation

Nodes are numbered from 1 to $n$ , and strands are numbered from 1 to n-1 (strand $k$ goes from node Nk to node Nk+1).

## Averaged Force

The averaged force in the multistrand is computed as:

Linear spring $F=\frac{K}{{L}^{0}}\delta +\frac{C}{{L}^{0}}\stackrel{˙}{\delta }$

Nonlinear spring $F=f\left(\epsilon \right)\cdot g\left(\stackrel{˙}{\epsilon }\right)+\frac{C}{{L}^{0}}\stackrel{˙}{\delta }$

or $F=f\left(\epsilon \right)+\frac{C}{{L}^{0}}\stackrel{˙}{\delta }$ if $g$ function identifier is 0

or $F=g\left(\stackrel{˙}{\epsilon }\right)+\frac{C}{{L}^{0}}\stackrel{˙}{\delta }$ if $f$ function identifier is 0

Where, $\epsilon$ is engineering strain: $\epsilon =\frac{L-{L}^{0}}{{L}^{0}}$

Where, ${L}^{0}$ is the reference length of element.

## Force Into Each Strand

The force into each strand $k$ is computed as:

${F}_{k}=F+\text{Δ}{F}_{k}$

Where, $\text{Δ}{F}_{k}$ is computed an incremental way:(1)
$\text{Δ}{F}_{k}\left(t\right)=\text{Δ}{F}_{k}\left(t-1\right)+\frac{K}{{l}_{k}^{0}}\delta {\epsilon }_{k}-\frac{K}{{L}^{0}}\delta \epsilon$

with ${l}_{k}^{0}$ the length of the unconstrained strand $k$ , $\delta \epsilon =\epsilon \left(t\right)-\epsilon \left(t-1\right)$ and $\delta {\epsilon }_{k}=\delta t{u}_{k}\cdot \left({v}_{k+1}-{v}_{k}\right)$ .

Where, ${u}_{k}$ is the unitary vector from node Nk to node Nk+1.

Assuming:(2)
$\frac{{l}_{k}}{{l}_{k}^{0}}=\frac{L}{{L}^{0}}$

Where, ${l}_{k}$ is the actual length of strand $k$ .

Therefore, 式 1 reduces to:(3)
$\text{Δ}{F}_{k}\left(t\right)=\text{Δ}{F}_{k}\left(t-1\right)+\frac{K}{{l}_{}^{0}}\left(\delta {\epsilon }_{k}\frac{L}{{l}_{k}}-\delta \epsilon \right)$

## Friction

Friction is expressed at the nodes: if $\mu$ is the friction coefficient at node $k$ , the pulley friction at node Nk is expressed as:(4)
$\text{Δ}{F}_{k}\left(t\right)=\text{Δ}{F}_{k}\left(t-1\right)+\frac{K}{{l}_{k}^{0}}\delta {\epsilon }_{k}-\frac{K}{{L}^{0}}\delta \epsilon$

When equation 式 4 is not satisfied, $|\text{Δ}{F}_{k-1}-\text{Δ}{F}_{k}|$ is reset to $\left(2F+\text{Δ}{F}_{k-1}+\text{Δ}{F}_{k}\right)\mathrm{tanh}\left(\frac{\beta \mu }{2}\right)$ .

All the $\text{Δ}{F}_{k}$ (k=1, n-1) are modified in order to satisfy all conditions upon $\text{Δ}{F}_{k-1}-\text{Δ}{F}_{k}$ (k=2, n-1), plus the following condition on the force integral along the multistrand element:(5)
$\sum _{k=1,n-1}{l}_{k}\left(F+\text{Δ}{F}_{k}\right)=LF$
This process could fail to satisfy 式 4 after the $\text{Δ}{F}_{k}\left(k=1,n-1\right)$ modification, since no iteration is made. However, in such a case one would expect the friction condition to be satisfied after a few time steps.

## Time Step

Stability of a multistrand element is expressed as:(6)
$\text{Δ}t\le \frac{\sqrt{{C}_{k}^{2}+\rho {l}_{k}{K}_{k}}-{C}_{k}}{{K}_{k}},\forall k$
with ${K}_{k}=\frac{Mass\text{\hspace{0.17em}}of\text{\hspace{0.17em}}the\text{\hspace{0.17em}}multistrand}{{L}^{0}}$ and (assuming 式 2):(7)
${K}_{k}=\mathrm{max}\left(\frac{K}{{l}_{k}^{0}},\frac{F}{{l}_{k}-{l}_{k}^{0}}\right)=\mathrm{max}\left(\frac{KL}{{l}_{k}{L}^{0}},\frac{FL}{{l}_{k}\left(L-{L}^{0}\right)}\right)$
(8)
${C}_{k}=\frac{\left(f\left(\epsilon \right)\frac{dg}{d\stackrel{˙}{\epsilon }}\left(\stackrel{˙}{\epsilon }\right)+C\right)}{{l}_{k}^{0}}=\left(f\left(\epsilon \right)\frac{dg}{d\epsilon }\left(\stackrel{˙}{\epsilon }\right)+C\right)\frac{L}{{l}_{k}{L}^{0}}$