should I be using pure shear or bending?

[pk123]

Hi
I have a design where I have a 2.5mm diameter pin that is 28mm long. It is inserted into a hole where the middle of the hole is open (ie. approx 7mm at each of the pin is situated in a hole (note: interference fit).BTW, the whole thing can be assumed to be symmetric.
The exposed middle (14mm) of the pin is subjected to a UDL (uniformly distributed load) across about 13.6mm of its length.
My question: In such a case, should I just use pure shear calc (F/A) to check the design or is beam theory applicable?
My feeling, based on some limited experimenting, calculations and feeling are that pure shear is the way to go but I would like to hear from other experienced proffessionals on the matter.
Thanks
Paul

 

[ruth44]

pk123

You should check both shear and bending.Draw the moment and shear diagram and check the section with the max. shear load and bending moment.
For a hand calc I would assume a triangular distribution of the reacting load.

Ruth44

 

[ishvaaag]

This looks as just a shear pin with two shearing surfaces. However, since you say a uniform load is in the middle section instead of being loaded by a fitting perforated plate, what would complete the 2 faces in shear case as per usual code provisions, the recommendation of ruth44 is sound.

 

[ishvaaag]

Also to add:

The calculation of devices in shear assumes for bolts the shaft can't or likely won't become dragged unto the hole where the pulling party is. The head of the bolt and its nut are there for this (and avoid lateral displacement), and in other cases the constraint geometries also ensure the wanted shearing faces' behaviour.

 

[pk123]

Hi Ruth44 and ishvaag
Thanks for your replies. I have a question though. I do not understand exactly what you mean by using a triangular distribution for the reaction load? Do you mean a right-angled triangle (meaning that the highest load is closest to the UDL)? If not, then what?
Also, how exactly does one calculate by hand a distributed reactionload? I have never even seen an example of such a thing?
Thanks for help so far.
Paul

 

[stel8]

Your case is similar with case 18, page 240, Machinery's Handbook, ed.25. Both ends are fixed (you said interference) and the 14 mm are UDL. Stress is max at ends = WL/12Z, where W is the equivalent concentrated load for UDL, L is 14 mm, and Z is the section modulus for round bar = pi X D^3/32 = 0.098 D^3
To be on the safe side check calculations also for case 19 (next page) for concentrated load, which is more severe, ie max stress is WL/8Z.

 

[ishvaaag]

I referred to the fact of that if the central part is not truly forcing "pure" shear action to the interfaces, it was sound to take unto account moment coming from bending as well. The recommendation of using a linear distribution within the embedments use to be sound when what is embedded is quite stiff when compared to the material surrounding it, say a relatively deep (but not exceedingly so) beam embedded in a thick wall brick. A moment generated by some bending central part like in your pin will be taken by a pair at two contact zones in opposite sides. Any convenient idealization may serve to study the behaviour there.

But if you simply have shearing at the end of the central part once plastified, you may analyze, or better, design or check for just shear with two faces in shear, provided all other requirements to ensure such behaviour are met.