How to calculate the failure point of a pin

[pete]

Hi, I have a device I'm suspending from a chain and the chain attaches to the body with a pin. I'm curious how you would calculate how much weight a pin like that could take before failing. There is a block on the body and a bar on the end of the chain that slots into the block and the pin goes through both and out the other side so it would have to shear the pin in two, in three actually so it seems unlikely but I would like to learn how to work it out myself so I know for sure.
I tried googling it but I got so many different answers that I got a bit confused so I thought I'd come and ask here?

 

[.Scott]

They can be modeled and calculated, but mostly, the manufacturer will test his bolt or pin and then publish the result. Here is an example:
https://www.portlandbolt.com/technical/faqs/bolt-shear-strength-considerations/

First, let me better described what I think you have. There is a vertical channel through the block where the bar is inserted. Then the bar is held in place with a pin that crosses horizontally through the block and the bar.

It is best to determine the shear failure load by experimentation. It will be dependent on the size, fit, and materials used for the block and the bar. Also, there are other failure modes besides shear. If the block is made of a softer material, it could deformed and allow the pin to pulled through.

 

[Spinnor]

I like a Google image search for your terms in question. Your mind is very good at picking out pictures of your problem of interest and the pictures may lead to your answer.

https://www.google.com/search?q=SHE...Bs0KHRstCQIQ_AUIECgB&biw=931&bih=570&dpr=1.38

 

[pete]

Thanks for the response. Yes, that's it more or less. The block is just over 40mm wide with a 30mm diameter hole through it and the bar which is also 30mm diameter slides into the hole then a 6mm pin goes through the block passing through the center of the bar. The block and bar are aluminum and the pin stainless steel. The aluminum could deform and loosen up the connection but the block or bar would have to tear themselves apart to actually fail so I'm sure the pin would go first.

I was thinking the block being slightly softer may even help prevent sudden failure as it will be harder for it to behave like a guillotine on the pin.

The link you provided has some information on how to calculate it. I'll haven't had time to read it properly yet but will later.

 

[pete]

I'm afraid I've never done much maths so I'm not too good a reading stuff like that. To be clear what it means is that the stress on the bolt is equivalent to the force divided by the area. The area of the pin being the radius squared times Pi. I suppose the only difference is that there is no possibility for the pin to bend so when it fails it shears in half, although not in my case with the aluminum. I'm guessing it will all just get twisted up. But really I want it to be well within its capacity so it does not deform.

Maybe the best thing is to just make one up and load it till it fails, then I'll know for sure.

These equations are very neat and make sense but I never understand what units I would put in there to end up with a useful number. Like how much weight can the pin carry.

 

[Spinnor]

This might not be too hard to figure out,

https://www.engineersedge.com/material_science/bolt_double_shear_calcs.htm
From, https://www.google.com/search?q=SHE...hUKEwjz36GBkJnjAhVMWN8KHdxoCMAQ_AUoAHoECAAQAw