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sheared pins
- Thread starter luke81
- Start date
MiketheEngineer
Structural
Call a metalurgist - He/she can tell you exactly what happened..
hi luke81
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
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
- Thread starter
- #4
desertfox,
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.
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.
- Thread starter
- #5
To my experience there is no doubt that bending stresses are involved here because there is always a gap between the hole and the pin and it is growing with time. Thats shift the reaction force on the pin to the edge of the pin and the result is an increase in the bending stresses on the base of the pin. To avoid such a case the pin should be as short as possible compared to it's diameter.
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.
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.
Hi Luke81
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
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
It looks like shear failure. The slight cresent shape at the edge looks like initial shearing damage. The lines across the face look like marks resulting form one face dragging across the other, not fatigue beach marks. Are the holes into which the pins fit distorted/elogated? If not the pins most likely failed only in shear.
How hard are the pins? The failure does not appear to be very ductile.
Ted
How hard are the pins? The failure does not appear to be very ductile.
Ted
- Thread starter
- #9
desertfox,
- 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.
- 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.
It looks like a shear failure only. No "beach" marks to indicate bending fatigue.
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.
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.
Hi luke81
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
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
desertfox
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.
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.
Hi everybody, I just signed up for this forum and found this thread which is within my small realm of expertise. I worked with shearing many types of materials (and pins) as a Die Engineer and I'll have to agree with hydtools that this is a pure shear failure in a hard material. There is very little shear zone and quite a bit of fracture zone which is what I would see when shearing hard materials with a punch and die at close tolerance.
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.
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.
Hi iraelkk
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
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
Hi luke81
Have a look at this link it may help you with the failure:-
desertfox
Have a look at this link it may help you with the failure:-
desertfox
If you can re-design the shaft joint, the attached sketches show a better arrangement. One pin takes the same load as two pins, there is better support from the shafts themselves and you get an overall stronger joint. Just thought I would throw this out as an option without knowing the application or how many units you have.
Note: I haven't uploaded images before so bear with me if they don't show up properly.
Note: I haven't uploaded images before so bear with me if they don't show up properly.
Okay, this should work. Click on the links above to see the sketches
Hi luke81
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?
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?
- Thread starter
- #19
Ok luke
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
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
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