I've got a 35mm diameter pin holding a roller between two plates. I want to figure out the max force the pin can withstand in shear. I've been working on a formula but I'm getting answers of the order of 130,000+ N. The pin will be made from mild steel. Does anyone know the formula and information I should be using? Thanks for helping me out with this.
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How to Determine max shear force on a pin 1
- Thread starter Kwylie2
- Start date
racookpe1978
Nuclear
(Ultimate) Shear resistance (pin tear-through) is cross-section area x strength.
Area is pi*r^2 of course. The results are why shear is seldom a limiting load in design. There really is a lot of strength available even in small pins.
Area is pi*r^2 of course. The results are why shear is seldom a limiting load in design. There really is a lot of strength available even in small pins.
DesignerGuy16
Mechanical
If you find the pin manufacturer most give these as a clear minimum value as single and double shear. Or are you using a custom pin?
I just went through this excercise at my work with a project to replace a retaining ring holding a gear in place with a pin. With a roll pin I was able to get more thrust load with the pin, and removed any possibility of assembly error. What I noticed was a goodly (on the order of 20-30%+) increase in shear load on spring pins than dowel pins.
James Spisich
Design Engineer, CSWP
I just went through this excercise at my work with a project to replace a retaining ring holding a gear in place with a pin. With a roll pin I was able to get more thrust load with the pin, and removed any possibility of assembly error. What I noticed was a goodly (on the order of 20-30%+) increase in shear load on spring pins than dowel pins.
James Spisich
Design Engineer, CSWP
- Thread starter
- #6
The formula I've been using is
Shear Force = (pi * d^2 * Shear Strength)/ 4. Is this correct?
It's most likely to see no more 110,000N, when I worked this out it put my answers into a bit more perspective.
It's a custom pin, nothing complex. I think the most confusing part is finding an accurate figure for the shear strength. I don't know the exact mild steel composition but is there anywhere where I can find accurate material properites?
Thanks for your help.
Shear Force = (pi * d^2 * Shear Strength)/ 4. Is this correct?
It's most likely to see no more 110,000N, when I worked this out it put my answers into a bit more perspective.
It's a custom pin, nothing complex. I think the most confusing part is finding an accurate figure for the shear strength. I don't know the exact mild steel composition but is there anywhere where I can find accurate material properites?
Thanks for your help.
Google search usually turns things up, Matweb is one site.
In fact, some schools have their class notes etc. on the web so a case like yours is quite likely on the web somewhere.
KENAT,
Have you reminded yourself of faq731-376 recently, or taken a look at posting policies:
In fact, some schools have their class notes etc. on the web so a case like yours is quite likely on the web somewhere.
KENAT,
Have you reminded yourself of faq731-376 recently, or taken a look at posting policies:
Hi Kwylie
If your pin is taking the load between the two plates then what you have is double shear and not single shear which is what you have calculated.
The fomula should over 2 and not over 4 as you have it in your original post.
Shear Force = (pi * d^2 * Shear Strength)/ 4 should be:-
Shear Force = (pi * d^2 * Shear Strength)/ 2
regards
desertfox
If your pin is taking the load between the two plates then what you have is double shear and not single shear which is what you have calculated.
The fomula should over 2 and not over 4 as you have it in your original post.
Shear Force = (pi * d^2 * Shear Strength)/ 4 should be:-
Shear Force = (pi * d^2 * Shear Strength)/ 2
regards
desertfox
racookpe1978
Nuclear
A 1" dia shear pin (25 mm) is very large.
Thinking about this last night - one extra problem sprung up. We (everybody) have been considering this from the "what strength can I calculate for the shear pin?" viewpoint - then trying to figure out if the shear pin is one-sided (one shear surface), or two-sided (has two shear surfaces.) and what is the proper formula for the area.
But look also at whether you actually want the shear pin to fail (and protect the rotor and shaft from being torn up and stripped) or do you want the shear pin to be strong enough to carry any load that the rotor will get? If the second case, you will want the shear pin to be stronger than than the shaft, but under overload conditions, the pin hole and shaft will need to be replaced or re-machined after failure/overload.
in the first case, you would most likely only have to shutdown the machine and replace the shear pin.
Thinking about this last night - one extra problem sprung up. We (everybody) have been considering this from the "what strength can I calculate for the shear pin?" viewpoint - then trying to figure out if the shear pin is one-sided (one shear surface), or two-sided (has two shear surfaces.) and what is the proper formula for the area.
But look also at whether you actually want the shear pin to fail (and protect the rotor and shaft from being torn up and stripped) or do you want the shear pin to be strong enough to carry any load that the rotor will get? If the second case, you will want the shear pin to be stronger than than the shaft, but under overload conditions, the pin hole and shaft will need to be replaced or re-machined after failure/overload.
in the first case, you would most likely only have to shutdown the machine and replace the shear pin.
Hi racookpe
My view is thats its not a shear pin that needs to fail under certain conditions as the OP stated that he has a roller supported on the pin between two plates.
I read this to mean its like a truck wheel or similiar and the fact that the roller is between two plates suggests the pin is supported through both plates,which led me to believe that the pins in double shear.
Maybe the OP can confirm my observations or deny them.
regards
desertfox
My view is thats its not a shear pin that needs to fail under certain conditions as the OP stated that he has a roller supported on the pin between two plates.
I read this to mean its like a truck wheel or similiar and the fact that the roller is between two plates suggests the pin is supported through both plates,which led me to believe that the pins in double shear.
Maybe the OP can confirm my observations or deny them.
regards
desertfox
racookpe1978
Nuclear
Good point! If OP is looking at an "axle" then his (her ?) problem isn't just shear though. Axles (whether supported on one side, or on both sides of the roller) will almost always fail by under bending load - never by shear.
The load won't be a point source though - unless it's from a ball bearing ring. If a bushing or roller bearing is used, it would need to be approximated as a distributed load, right?
I'm guessing here - but the resistance to that load (the force holding the axle in place) could be either a point load (if it were short compared to the axle diameter) or distributed (if a thin axle were threaded into a deep hole, or a thin axle was held up by two thick plates on both sides of the load.)
The load won't be a point source though - unless it's from a ball bearing ring. If a bushing or roller bearing is used, it would need to be approximated as a distributed load, right?
I'm guessing here - but the resistance to that load (the force holding the axle in place) could be either a point load (if it were short compared to the axle diameter) or distributed (if a thin axle were threaded into a deep hole, or a thin axle was held up by two thick plates on both sides of the load.)
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1
- #14
GuyFromDenver
Mechanical
Most of what I've read in this thread is sound, but there is something else that you must consider: bending. If you are in double shear then it isn't much of an issue, but in single shear it is. Draw a free body diagram and you will see that there is a bending moment induced on the pin for two reasons: tolerance on fit and load distribution. Unless it is a tight press fit then there is some misalignment of the pin to the hole. this may be small, but the the bending moment should be considered in your calculation. Now assume that it is a press fit into both plates. You FBD will show that the net force is theoretically applied at the center of the length of the pin inserted into the plate. Consider that when sizing your pin as well.
Hi GuyFromDenver
The only problem with considering bending is that the normal bending formula is valid only if the length to depth ratio of the pin is 20:1. In my earlier post I stated that I assumed that the OP knew he had a shearing situation otherwise bending would have been mentioned then.
regards
desertfox
The only problem with considering bending is that the normal bending formula is valid only if the length to depth ratio of the pin is 20:1. In my earlier post I stated that I assumed that the OP knew he had a shearing situation otherwise bending would have been mentioned then.
regards
desertfox
I agree with including bending, also. I am curious about the 20:1 length/depth ratio, desertfox. Can you say where this comes from? I understand that the bending moment dominates over transverse shear for large length/depth ratios, but to say that "normal bending formula is valid only if the length to depth ratio of the pin is 20:1" puzzles me.
kwylie2:
Please consult with material guys on the proper allowable shear stress, which is dependent upon the characteristics of the tensile capacity of the subject material. (A shear failure is occured when the applied force causing a stress to exceed the breaking point of the material)
Please consult with material guys on the proper allowable shear stress, which is dependent upon the characteristics of the tensile capacity of the subject material. (A shear failure is occured when the applied force causing a stress to exceed the breaking point of the material)
Hi dvd
Yes I was over the top it appears with my length to depth ratio, Roarks says about a ratio of 8:1 for beams of compact metal section and about 15 or more for beams with relatively thin webs.
Other sites say a length to depth ratio of 10:1 like this one here:-
But either way the original poster I assume would know whether he had a bending situation or not.
Untill the OP confirms what dimensions, or how its actually supported we can only guess.
As the GuyFromDenver says if its in double shear its not much of an issue, but if you read the OP's post it suggests that the situation is that of double shear.
I know somewhere I did read a length to depth ratio in the order of 20:1 but I cannot find it unless of course I dreamt it.
regards
desertfox
Yes I was over the top it appears with my length to depth ratio, Roarks says about a ratio of 8:1 for beams of compact metal section and about 15 or more for beams with relatively thin webs.
Other sites say a length to depth ratio of 10:1 like this one here:-
But either way the original poster I assume would know whether he had a bending situation or not.
Untill the OP confirms what dimensions, or how its actually supported we can only guess.
As the GuyFromDenver says if its in double shear its not much of an issue, but if you read the OP's post it suggests that the situation is that of double shear.
I know somewhere I did read a length to depth ratio in the order of 20:1 but I cannot find it unless of course I dreamt it.
regards
desertfox
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