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Calculating limit stress on a solid pin loaded in shear is straightforward, and there are several handy calculators which will do single shear and even double shear. However, everything I have found assumes that the pin is solid.

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For weight savings reasons, one might want to replace a solid pin with a hollow pin. However, stress flow in a hollow object is different from in a solid object, so it seems dangerous to assume that the yield stress is purely inversely proportional to area without consideration of the nature of the cross-section.

How to take into account the cross-section when calculating failure stress in single-shear and double-shear loading for a pin with arbitrary ID? Is it sane/safe to use the existing shear eqn. above with the hollow cross-section area and then bump the required safety factor by, say, 2x?

UPDATE: The best answer includes design tables and/or guidelines. Pure theoretical approaches are interesting but ultimately unsatisfactory, as 1) they can be easily simulated with FEA as well and 2) they might not include real-world experience.

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  • $\begingroup$ This has all been done before, do some researxh on wrist pins or gudgeon puns used in ice for pistons to con rods. $\endgroup$
    – Solar Mike
    Commented Jan 22, 2021 at 5:55
  • $\begingroup$ @SolarMike yes, I agree this has all been done before. I am having trouble finding any references to it, though. I can't imagine this hasn't been thoroughly studied and documented somewhere. $\endgroup$
    – Kenn Sebesta
    Commented Jan 22, 2021 at 15:32
  • $\begingroup$ Also, please if you downvote, mention why. I have done enough research to know that this question isn't easily answered on Google. Perhaps a downvote is because of question clarity, but then that should be explained. $\endgroup$
    – Kenn Sebesta
    Commented Jan 22, 2021 at 15:46
  • $\begingroup$ Thank you for your advice on how to use the site, much appreciated. $\endgroup$
    – Solar Mike
    Commented Jan 22, 2021 at 16:02

2 Answers 2

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You are right in thinking that a thin walled pin will fail primarily by compression/wall buckling (I tend to think about more as crushing), however that does not change the calculation of the shear stress.

Shear stress is defined in a very specific way: by dividing Forces acting on a direction parallel to a surface, divided by the surface area.

Of course that increases when considering shear stress which are induced due to bending depending on the cross-sections, or because of stress concentrations (see here https://www.fracturemechanics.org/hole.html). Despite these cases (and maybe some other I am forgetting) where the nominal shear stress is factored, the baseline calculation of shear stress remains the same.

If you want to understand failure, you'd consider all types of stress (normal and shear stresses) and the use some failure theory (Tresca, Von Mises etc) to determine if failure occurs. If you are worried about buckling you'd need to carry out a different type of analysis.

UPDATE:

From an additional look I've taken, my belief that hollow pins:

  • made out of common engineering materials
  • and $R_{in} <\frac{2}{3}R_{out}$, are not much in danger of a transverse collapse has been reinforced.

The closest I've found where references with respect to bending loading and thin walled structures. To be honest, this subject feels more of a hot topic in solid mechanics science, rather than the typical rule-of-thumb engineering. Nevertheless, just a couple of references I've found that seem relevant:

You can also have a look at this book:

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  • $\begingroup$ I agree with everything you say, and yet this isn't helping beyond pointing me back toward FEA. What I'd like is a canonical answer for the design, that is to say most fasteners are chosen by the book and so the answer I'm looking for is "The Book". $\endgroup$
    – Kenn Sebesta
    Commented Jan 22, 2021 at 15:29
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    $\begingroup$ I understand better now what you are looking for. I've never come across "The Book", (however that does not mean that it does not exist). I would hazard a guess, that for common engineering materials and for hollow pins with an $R_{in}<\frac{2}{3}R_{out}$, there is not much danger of a transverse collapse. Nevertheless, I'll have another look for any ref to "The Book". $\endgroup$
    – NMech
    Commented Jan 22, 2021 at 16:13
  • $\begingroup$ Two years on I missed that you had updated your answer. I think this comes close enough to "The Book" to count. The hollow pin at <2/3 R_out is 45% lighter, which is more than enough to accomplish a decent lightening exercise. $\endgroup$
    – Kenn Sebesta
    Commented Sep 19, 2023 at 21:46
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Maximum shear in a solid cylinder is

$$\tau_{max}= 4/3 V_{average}$$

Maximum shear in a hollow cylinder is intuitively zero at the top and bottom of the cylinder and a maximum of 2V/A on the vertical sides.

here is a link to analytical calculations: shear in a hollow cylinder.

$$\tau_{max}=2V/A$$

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  • $\begingroup$ I think this somewhat misses the thrust of the question, which is how to determine appropriate sizing for a pin loaded in a shear joint. For instance, below a certain wall thickness the pin will fail in compression before it fails in shear. I have updated the title to draw attention to this. $\endgroup$
    – Kenn Sebesta
    Commented Jan 22, 2021 at 3:59

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