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Pin Design 1

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Badri07

Badri07

Mechanical
Mar 17, 2008
36
GB
Hi All,
i have designed a pin (which has got circular cross section),thats going to be in shear,i have assumed the pin to be short coloumn and considered only shear.

I did try to do some basic bending stress calculation ,but the bending stress is way above the allowable,while shear stress is below the allowable.

I am bit confused about this reponse.i thought short coloumns fail more due to shear than baending.
i appreciate your comments on this.

Thanks
Badri07.
 
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Dont you mean a beam? A column usually decribes an axially loaded member. The pin will only go into bending if it spans between supports or has a moment applied. A pin usually describes an element in a connection loaded by adjacent plates or ply, which will induce single or double shear only. Can you give some further information? what exactly is this element? how was the bending analysis performed?
 
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Thank you for your post Parrot77,

yes,what you are saying is right,infact its a beam connected between adjacent plates,this has got double shear.on bending calculation i just did a basic hand calc to check the bending stress --->sigma(b)=M/Z.This calculation was performed assuming it to be a simply supported beam with point load in the center.

infact i did an FEA on the pin using beam elemnts,this is giving an bending stress of sigma(b)=133mpa.

Thanks
Badri07
 
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For significant bending to be induced there will need to be a substantial gap between the outer plates and the middle plate where the load is applied. The pin is then forced to span between the two outer plates. If however the plates are side by side and the pin is threaded through as per most common connections then only shear will be critical.. much like a regular beam loaded adjacent a support.
 
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Hi Badri07

Just post your pin size, distance between supports and load applied and we can maybe help further.

Regards

desertfox
 
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Desertfox,

The thickness of the connecting plates will also be required as bearing stresses may be critical.
 
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Hi All,
Thank you all for your time,the distance between the two plates is around 150mm and the 2.22x10exp6 is applied on to the pin,the dia of the pin is 126.5mm(5").

Thanks
Badri07.
 
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that's some pin ... shear and bending margins are dependent on different aspects of the geometry, so it's not surprising that it's critical in bending (particularly if its a solid pin).

E looks funny, units ?

what end conditions did you assume (in calculating the moment in the pin) ? ... i'd suspect that fixed would be appropriate.

where did you apply the shear forces ? mid-thickness would be conservative, closer to the mating plane is more realistic.

did you allow plastic bending ?
 
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How is this load applied to the pin? By a third middle plate? if so what are the thickness of all 3? The thickness would be governed by bearing stresses. What is the load? I didn't underdstand your annotation
 
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Hi rb1957,
E=2.0e+005mpa,and the load applied in the pin 2.5sh tons,thats 2.22e+006N(as in my previous post),Boundary condition for Mmax is simply supported,i applied the shear force in the center of the beam,and the moment i calculated applies to elastic bending.
Thanks
Badri07
 
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Hi Parrot77,
The thickness of the end plates is 50 mm and the thickness of the middle plate is 150mm.

Thanks
Badri07.
 
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Badri 07,

I would be inclined to ignore bending if the middle plate takes up the full 150mm gap between the two edge plates. If the pin tends to span and act like a beam in bending then it will deflect in the middle slightly relative to the stiff midddle plate shedding load towards the edges/ plate interpace where it is back to a pure shear situation. I would consider shear and bearing only.
 
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modelling the pin as a simply supported beam, with a concentrated load at the mid-span and reactions at the mid-thickness of the outer plates is highly conservative. the pin is a pretty neat fit into the bore, yes? so it really can't bend like a beam (before it bears up against the bore and starts to react bending differently).

the ends of the pin are reasonably fixed in the outer plates. the shear loads can be reasonably applied close to the mating planes between the inner and outer plates (this'll reduce the moment, lots).

the applied load is 2.5 tons ? ... that's about 6,000 lbs, about 25,000 N, no? ... not 2E6N, no??
 
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I recently reviewed a pin design for a lifting beam with similar size of plates and pin. The design engineer designed the pin exactly as parrot77 describes. I questioned it at first, but the engineer convinced me that his design method was correct.
 
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rb1957,
yes,it was by oversight i had an typo in my load,
 
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parrot77 is correct. From your descriptions, your pin spanning in between two 50mm end plates, which spaced 150mm apart. Then the 150mm thick middle plate (which introduces the load) perfectly fit into the space. The effect of bending can be ignored, or you may check bending by distributing the load uniformly accross the 150mm space/span, now your moment will be much smaller. If the resulting stress is still higher than the allowable, then you have to increase the pin size, or using higher grade materials.
 
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Unless I'm missing something there's no need to design a 5" dia pin for 2.5 tons, just use it.
A half inch bolt will almost support that in double shear.
 
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Hi,
just that the middle plate is inducing 250sh tons,2.22e+006N,not 2.5sh tons
Thanks
Badri07
 
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