sheared pins
sheared pins
(OP)
This is a sheared drive pin. I am having a hard determining if it was pure shear that caused it, or combined with bending due to misalignment of the pins.
[img http://picasion.com/pic8/98fc7b5d5b0ab69a973a3935ea4aa2c8.gif]
[img http://picasion.com/pic8/98fc7b5d5b0ab69a973a3935ea4aa2c8.gif]





RE: sheared pins
RE: sheared pins
Is this one pin in two halfs or two seperate pin failures?
There appears to be what are termed river lines on the pictures you have posted and also a small smooth area almost like a chamfer on the edge.
Have you any more information on the loading of the pin and how long its been in service.
My initial thoughts is it looks like a brittle failure in shear however the smooth area and area with the river lines are usually seen in fatigue failures which is why I am requesting some more information.
For a fatigue failure however you require tensile stresses for the fatigue crack to grow now that might be the case if the pins were in torsional shear or as you say subject to bending.
Finally what function does the central hole play?
desertfox
RE: sheared pins
1. Pictures are two halves of one pin. (it is one of two drive pins and both pins failed)
2. Torque rating is about 120,000 in-lb. We have very little information on service; another reason i'm having difficulties.
3. Unfortunately, one of our technicians drilled holes pull out the pins.
I am attaching a simplified diagram of the pin arrangement.
RE: sheared pins
RE: sheared pins
As desertfox mentioned it seems that thereis a fatigue crack (the smooth area). When the crack is large enough the pin break/shear in static mode.
RE: sheared pins
It's a pity the holes had to be drilled in them to get them out; it may well have taken some evidence out with the drilling.
Okay I presume the two pins just engage against a plate or bar and drive them round a bit like a carrier and catch plate on a lathe.
What diameter and length are the pins and more importantly how far away from the base where the pins are fitted does the driven plate or bar engage on the pins?
Your left hand drawing of the file you have sent, I cannot understand it very well as it scanned correctly?
Also are the pins that have failed always in contact with the driven plate or do they ever impact on the driven plate due any clearance or lost motion between them.
Firstly why two pins is it so each pin engages separately to drive in opposite directions?
I think firstly we need to understand exactly what the pins are meant to do.
Then with the torque rating we can get the shearing force on the pins and hopefully knowing the pin dimensions we can verify the shear stress which is within the pin and make sure it's strong enough.
If both pins are meant to engage simultaneously on the driven plate then that might be part of the problem because it would be extremely difficult to guarantee two separate pins engaging the drive plate together which means that only one pin as engaged on the drive plate and the other pin as a slight clearance undetectable to the human eye.
So if the two pins are sized to take equal share of the load while driving in reality one pin is seeing all the load until it fails and then the other pin takes over until that fails too. Lets have as much info as possible and people on here might well be able to give you a better answer.
Regards
desertfox
RE: sheared pins
How hard are the pins? The failure does not appear to be very ductile.
Ted
RE: sheared pins
- The upper part drives the bottom part, which has a significant mass through the two pins. The forces on each pin act in opposite directions.
- Pins are 1" dia. X 2" (see picture for other dimensions)
- There is small clearance between pins and the driven plate, but prefitted with the driving side. Also there is no constance motion; max speed may be 1rpm.
I don't know if this information is enough.
RE: sheared pins
If you look at the deformation pattern, you see an elliptical shape of the brittle failure area, with the noted crescent of galling just before that. This is typical shear pin failure.
RE: sheared pins
Thanks for the information, a quick calculation on the average shear stress on a single pin is 101859lb/in^2 based on the torque you gave in an earlier post, this seems quite a high shear stress but without the material properties and its heat treated condition (if any)we have nothing to compare it with.
Now I based the shearing force on the radius of 1.5" based on the pin centre's of 3" from your last file, however if the driven plate is between those pins which from your last post that would appear to be the case then the true shearing force and subsequent
Shear stress will be a lot higher than my calculation because the radius arm will be a lot smaller than 1.5" so it looks like your pins might not be big enough to take the load. General consensus appears to be a shear failure which was my initial reaction, see what information you can get on the pin material. Also the pin that's failed if you view them from the side instead of on top what do the failed edges look like are they more or less at right angles to the central axis of pin or are the angled to the central axis of the pin. Might be easier to post a picture of the failed pins at 90 deg's to the ones in your original post.
desertfox
RE: sheared pins
I get half of the shear stress than you. You either need to use two pins shear at a distance of 1.5" from the center of the shaftas the pivot point or one pin shear at a distance of 3" from the other pin as the pivot point. This gives 50930 psi.
If we assume that the reaction on the pin shift to the edge of the pin the bending stress becomes 407436 psi almost 10 times the shear stress. Therefore, we can assume with confidense that this is is not a pure shear case and quite a large bending stress is involved even if the reaction doesn't apply on the pin edge.
As you said if we will know the pin alloy and tensile strength we can probably be sure if it is just a shear or a bending too.
RE: sheared pins
Looking at your application and the pictures of the pins, I'll assume these are off the shelf precision ground dowel pins which can vary quite a bit from supplier to supplier. These pins also look to be through hardened and not case hardened which may be a solution. A well heat-treated and tempered dowel pin doesn't have the brittle fracture you're seeing here.
Unless you can re-design and/or retro-fit the application, you should try a pin from different suppliers and vary degrees of hardness. Most pins of this type will take stresses in excess of 130-150Ksi if made to ANSI/ASME B18.8.2-1995 or better. Unbrako and Holo-Krome make some of the best pins available.
RE: sheared pins
The reason I have assumed one pin is in contact that it would be difficult in practice to make two pins contact a drive plate simultaneously on two opposite faces and the OP as said there is a clearance between the driven plate and the pins,in addition he also stated that the forces on the pins act in opposite directions which I have taken to mean that one pin drives the plate in one direction say clockwise and the other anti-clockwise, so to achieve that as I see it would mean the plate with the two pins mounted in it would have to rotate about its central axis thereby the radius arm would be in the order of 1.5".
Bear in mind that the stresses we have calculated are average and not maximum.
I agree if two pins were driving it the shear stress would be in the order of 50000 lb/in^2, anyway if my reasoning is wrong please shout up.
Perhaps luke81 could clarify whether my understanding is correct or not.
regards
desertfox
RE: sheared pins
Have a look at this link it may help you with the failure:-
ht
desertfox
RE: sheared pins
Note: I haven't uploaded images before so bear with me if they don't show up properly.
RE: sheared pins
http
Okay, this should work. Click on the links above to see the sketches
RE: sheared pins
Sorry luke several more questions: - the photograph shows the one pin in two half's which you told us had to have holes drilled in them to remove them, so the two faces I look at on the photograph presumably would have been joined as one before the failure? Now that being the case the hole in the pin on the right hand side of the photograph is blind as it doesn't appear to go through which puzzles me because the piece on the left hand side also looks as though its had a drill on it, not sure how that can be? Finally if the pin was already in two parts surely you would have only had to drill one part of the pin?
Back to the fracture face, the part that looks like a small crest or chamfer what is it on the actual pin is it flat or does it stand up like a little rib with an angle of about 45 deg?
RE: sheared pins
- the guys managed to take out the pins before drilling it all the way through.
- the crest is flat along the fracture surface.
RE: sheared pins
But why is there drill marks on both parts it still doesn't make sense to me?
Anyway can you get some material information on properties,
Also how old is this equipment was it a new piece of kit.
If you pick both parts of the pin up and try to push them together do they fit a bit like a jigsaw or because of deformation do they not want to go together, I get the impression that there is very little deformation on the parts which would support a brittle failure.
The material fail in shear if the shear strength of the material is less then the tensile strength irrespective of whether or not there was bending combined with the shear force but without more info we can't go any further.
regards
desertfox
RE: sheared pins
Going back to one of your previous posts; the pins are "meant" to engage simultaneously, but we don't know what happened when they failed. One might have failed before the other one, or both failed at once.
RE: sheared pins
If the driven arm is fixed on a rotation centre and the drive dog is also fixed on the rotation centre then it is unlikely you will get two pins to contact the drive plate at the same time the only way you can gurantee that two pins will engaged is if the driven plate can translate as well as rotate.
I think what as happened is that its only engaged on one pin and when the pin as failed the next pin takes over until that fails.
We need to fully understand how this thing works can you put some sketches up on file, secondly we need the material information ie strength etc and how long its been in use.
If the pins have been designed to share the load but in practice only one pin is doing the driving the pins are undersized and thats the reason for failure.
regards
desertfox
RE: sheared pins
- i'm attaching a sketch.
- we don't have material information, but the failed pins will be sent to a lab shortly.
luke
RE: sheared pins
htt
The shear failure may have been due to impact loading which could cause the apparent rapid failure.
In design the pins should probably be the weak link in order to save the driving and driven components from failing.
Ted
RE: sheared pins
Okay the pins fit into a round block then sit in holes located in the driven block.
What are the tolerances on the holes centres on both the driving and driven blocks?, what are the tolerances on the hole sizes and he pin diameters?
I am fairly certain if you do a tolerance check on those parts you will find that its possible to have only one pin doing all the driving in which case my original thoughts and my calculation of shear stress will be in the right ball park.
If you load the tolerances up on here I will check them also.
regards
desertfox
RE: sheared pins
The only way one pin touches the hole and other doesn't touch the other hole is if the driving shaft protrude snugly into the driven shaft or the driven shaft is protrude snugly into the driving shaft. Because then the outside of the protruding shaft takes the other reaction. If the arrangment is as in the last sketch then both pins participate. However, it is almost sure that one of the pins will fail first but, each pin still takes half of the torque until failure.
RE: sheared pins
Sorry I don't agree with you, if you have two perfect holes of 50mm dia on 300mm centres and two drive pins on 300mm centres, however one pin as a diameter of 49.95mm and the other 49.9mm dia, sit each pin perfectly central in each 50mm dia hole on a 300mm pitch circle, rotate the pins at the centre of the 300mm spacing as these are the drive pins, the pin with the largest dia contacts the 50mm hole first, the smaller pin cannot touch the edge of the 50mm hole as it is constrained geometrically with other pin and its fixed at a distance of 300mm, the only way in this case you would get the smaller pin to touch is by increasing its distance from the other pin which can't be done.
Now when you consider there is tolerances on the pin diameters, hole diameters, hole positions on both driver and driven it would take an awful lot to get two pins and hole diameters perfect to each other and then you also need to make sure all the hole centres are perfect too.
I drew this out in Autocad and all I changed was the pin diameters to the dimensions above but scaled everything up by factor so I could measure clearences before and after rotating.
As you can see I have asked luke81 for this information on tolerances and then maybe we can prove or disprove what I believe to be the problem.
desertfox
RE: sheared pins
Yes, you are right if the tow shafts are aligned and cn not move sudeways. I figured it just as I sent the post.
RE: sheared pins
no problem I am glad we concurr was worried for a minute I had missed something.
So I think thats the problem, I believe they designed it for each pin to take half the load but of course in practice I think only one pin takes it and gets overloaded fails and then the other pin takes over till that breaks too.
desertfox
RE: sheared pins
Few years ago I designed something similar but added one short shaft (cylinder) between the two shafts that could move sideways. I used integral key on the driving and driven shafts and the middle short shaft has two matching slots at 90 degrees that matched the keys on the driving and driven shafts. The short shaft allowed a mismatch between the driving and the driven shafts.
RE: sheared pins
Trying to visualize your design you describe it sounds quite clever any chance of a sketch?
luke81 does this device were looking at only drive in the one direction? Reason I ask is the flat crest shape appears on only one side of the pin, also what does the other pin failure look like? I would presume very similar with the flat crest also on one edge. What about the holes in the mating part are they showing any marks or suffering any distortion? I believe hydtools mentioned this in his earlier posts as it's a good point.
Anyway luke I believe that what's happened is that when this device as gone into service from day one, only one pin as been doing the work and because of this the pin as been over stressed, possibly to its limit and cracked at some point, every subsequent operation as made the crack worse assuming that the flat crest is the area where the pin and driven part are in contact, eventually the pin as failed rapidly leaving the area as you see it below the flat crest.
The same process would now follow on the second pin until it also failed.
The way the driver and driven are assembled its possible and from the fact that the pins had to be drilled out that rubbing of the two failed surfaces may have taken place again hydtools mentioned such an observation in his post.
It maybe worth looking at both sets of failed pins, as the pin that failed last would not necessarily have any marks from the fracture faces rubbing.
This being the case it would indicate that one pin failed before the other and point to which pin failed first.
desertfox
RE: sheared pins
This was basically a custom made Oldham Coupling see:
http://www.ruland.com/ps_couplings_oldham.asp
RE: sheared pins
Got it now thanks
desertfox
RE: sheared pins
If redesign is an option, an Oldham coupling should be considered here.
Ted