# Yarnstrength

### Tensile Strength Formula from M. Zhang et al. (2004)

The formula for the tensile strength <math>\sigma_y</math> of the yarn relative to the tensile strength <math>\sigma_f</math> of the fiber given in M. Zhang et al. (2004) is:

<math>\sigma_y/\sigma_f = \cos^2 \alpha (1 - k/\sin\alpha)</math>,

with

<math>k=\sqrt(dQ/\mu)/3L</math>,

where <math>\alpha</math> is the tilt of the fibers to the yarn axis, d the fiber diameter, Q the 'migration length', that is the length scale on which a fiber goes from the yarn's surface to the interior, L is the length of the fibers, and <math>\mu</math> the friction coefficient between the fibers.

If we enter the following values:

Quantity | Value |
---|---|

<math>\alpha</math> | <math>30^\circ</math> |

d | 1nm |

Q | <math>30{\rm \mu m}</math> |

<math>\mu</math> | 0.13 |

L | <math>30{ \rm \mu m}</math> |

we get <math>\sigma_y/\sigma_f=0.75</math>. For <math>\sigma_f=150{\rm GPa}</math>, the yarns would be 112GPa strong, strong enough for the SE! If you read more closely the M. Zhang et al. (2004) paper, you'll see that the <math>\cos^2</math>-term in the formula is due to the force component supported by the fibers in the direction of the yarn. The second term in brackets is supposed to represent the transmission of forces between the fibers.

The formula has some problems and I do not believe it is applicable to our problem at all. For example, what happens for <math>\alpha \rightarrow 0</math>? The ratio would become <math>-\infty</math>. If you use the numbers from the table on the left, the second term is very close to 1, which would mean that there is no reduction of the yarn strength due to the limited force exchange between the individual fibers. This is however our main problem.

**It would be good if someone could come up with a correct formula describing <math>\sigma_{\rm yarn}/\sigma_{\rm fiber}</math> for yarns made of CNTs.**

Return to the Review of M. Zhang et al. (2004).