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sheared pins
- Thread starter luke81
- Start date
- Thread starter
- #21
Hi luke81
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
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
- Thread starter
- #23
For reference here is Holo-Krome catalog of properties. Their 1" pins have a shear capacity of 117,800 lbs., about 150,000 psi shear strength.
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
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
hi luke81
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
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
desertfox
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.
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.
Hi israelkk
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
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
hi israelkk
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
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
desertfox
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.
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.
Hi israelkk
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
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
desertfox
This was basically a custom made Oldham Coupling see:
This was basically a custom made Oldham Coupling see:
I used an Oldham coupling in a handheld grinder. With it I didn't have to be concerned about precise parallel alignment of motor shaft and output splindle. That meant allowing looser manufacturing tolerances for shaft alignment. It is also a self-captured coupling.
If redesign is an option, an Oldham coupling should be considered here.
Ted
If redesign is an option, an Oldham coupling should be considered here.
Ted
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