Yarnstrength: Difference between revisions

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Line 16: Line 16:
!Value
!Value
|-
|-
|alpha
|<math>\alpha</math>
|30deg
|<math>30^\circ</math>
|-
|-
|d
|d
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|-
|-
|Q
|Q
|30micron
|<math>30{\rm \mu m}</math>
|-
|-
|mu
|<math>\mu</math>
|0.13
|0.13
|-
|-
|L
|L
|30micron
|<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.
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.

Revision as of 10:56, 20 March 2009

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.