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Sausages universally split parallel to the length of the sausage. Why is that?

enter image description here

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The video explains it fairly well. There is greater tension release when the split propagates along the length. –  LDC3 Apr 1 at 4:49
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Being cooked too long probably. –  BigHomie Apr 1 at 14:43
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First use of the fracture tag. –  metacompactness Apr 1 at 18:40
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In the image in the question, two of the sausages appear to split in a spiral about the length. –  einnocent Apr 1 at 22:51

2 Answers 2

up vote 52 down vote accepted

This behaviour is well explained by Barlow's formula, even though the English Wikipedia article is incomplete in this context. The German version, on the other hand, gives the full picture (which I will quote in the following).

Stresses in a thin-walled cylinder; Source: Wikimedia commons

The walls of a pipe (or a similar cylindric container, say, a sausage) experience two types of stresses: Tangential ($\sigma_{\rm{t}}$) and axial ($\sigma_{\rm{a}}$). For given pressure $p$, diameter $d$ and wall-thickness $s$, the individual stresses can approximately be calculated from $$\sigma_{\rm{t}} = \frac { p \cdot d } { 2 \cdot s }$$ and $$\sigma_{\rm{a}} = \frac { p \cdot d } { 4 \cdot s }.$$ Here, you can directly see that the tangential stress will always be larger, which is why it is likely that cracks in the container/sausage will first form in this direction. In fact, this is why the first formula is often stated on its own, just as it is the case in the English Wikipedia article.

Fun fact: The sausage example is used by many German students as a mnemonic helping to remember which of the stresses is larger. As a result, the formulas are often called "Bockwurstformeln" (sausage formulas).

Edit: In response to the comments below, I will try to summarize some details about the above formulas

  • The formulas do not directly indicate how and where the container will split. Assuming that the tensile strength is identical in all directions, we can see that there will be a greater release of tension when the crack propagates length-wise (See the video posted by JoeHobbit and the comment by LDC3)
  • A real sausage will of course have various imperfections, which is why the crack path will not be straight in practice
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Cool ,you learn something new every day :-) A minor clarification, the cracks form normal to the tangential stress so that they reduce the tangential stress as they widen. –  John Rennie Apr 1 at 6:52
    
of course stating that the guaranteed higher tangential stress causes a break to (almost) always be lengthwise assumes that tensile strength is identical in all directions. –  jwenting Apr 1 at 10:52
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@jwenting: Indeed, there are quite a couple of assumptions behind these formulas... –  bigge Apr 1 at 11:01
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@SethBattin ideal sausage <3 –  n00b Apr 2 at 11:40
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The cylinder is not assumed infinite. Fluid pressure always acts orthogonally to a surface. A cylinder's side is parallel to its axis so doesn't contribute to axial stress. Only the ends can do that, and an infinite cylinder doesn't have ends. –  Hugh Allen Apr 2 at 15:33

I'll have to take a page from my EE background and say it's because of the path of least resistance. If you look at the end of a sausage, there is already tension along that plane, in multiple locations:

enter image description here

A chain is only as strong as its weakest link (see what I did there? :) ), so it's natural for a hot dog/sausage to split along a path that's already strained.

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Very punny. :) –  Brian S Apr 1 at 16:17
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You guys are going to make me split at the seams with laughter. –  WernerCD Apr 1 at 22:25
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That's the wurst joke that I've heard all day. –  Patrick Collins Apr 2 at 23:09
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This appears to be a link-only answer. –  Jason C Apr 3 at 5:53
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@JasonC nice one, I should have brat my A game. –  BigHomie Apr 3 at 12:39

protected by Qmechanic Apr 1 at 19:50

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