Shear

.50 (MOA)

If a pin is inserted through two parts that are touching (or practically so), like a bolted lap joint but with only a pin, then loaded at 90 degress to the pin axis with single shear, what is the value of the material property to use for strength calculations- the pin should not experience any permanent deformation.

I found a reference on another public posting site (can't even remember where now) to it being .5 x UTS or .577 x UTS, but i cannot find any other refernces on the internet or in books (I'm in university so there is nearly too many books available)

Any values for shear that i found are (i think) for shear in torsion not direct shear (is this the correct term) the material i intend to use is EN8/ BS970 080M40 with a UTS of 600N.mm^2 and a yield strength of 340 N/mm^2

I know that the stress will be F/A, I need to know the limit for this.

It will either be the Yield strength or the .5 x UTS value just need to find a reference.

Thanks in advance for any help.

Turbulent Bill

The answer is - it depends! With no factor of safety and assuming no stress concentation at the pin/part interface (i.e., an unrealistic best case) you could use the elastic limit of the material to prevent plastic deformation.

In practice, I'd say a factor of safety would be used to account for overloading and local stress concentrations - this rule of thumb would probably be application-dependent. If you're not given this factor I'd make a reasonable one up and state it as an assumption in your answer.

.50 (MOA)

sounds good, I didn't mention the safety factor but I would be using one anyway. its a mater of which limit to use combined with the safety factor to make a calculated limit.

grateface

Is this a simple college question?
What i mean is, by the phrase "the pin should not experience any permanent deformation." i'm assuming you're just asking whether you should be using the UTS or the yield strength. For this part you should only be using the Yield strength, as permanent deformation does start to occur after this.
For your Factor of safety etc. Have a look at this page, it might help.
http://www.roymech.co.uk/Useful_Tables/Screws/Bolted_Joint.html

Sorry if none of that was any help to ya.

.50 (MOA)

http://img211.imageshack.us/img211/5880/pin1.jpg

This is the application. it is the cyan coloured part in the centre. this pin is through the ram head (where there is double shear) and then sticks into the black structure on the right, this is where it is in single shear. there will, i know be bending on this pin too, but from a shear perspective is where i wanted to know what to use. the yield strength is what would be used for a bending application but for some reason i feel unsure whether it applies to direct shear like this. this is for a final year project- being in final year i should definitely know this but its just a small void in my knowledge.

Turbulent Bill

.50 (MOA) wrote: »

http://img211.imageshack.us/img211/5880/pin1.jpg

This is the application. it is the cyan coloured part in the centre. this pin is through the ram head (where there is double shear) and then sticks into the black structure on the right, this is where it is in single shear. there will, i know be bending on this pin too, but from a shear perspective is where i wanted to know what to use. the yield strength is what would be used for a bending application but for some reason i feel unsure whether it applies to direct shear like this. this is for a final year project- being in final year i should definitely know this but its just a small void in my knowledge.

Remember that the yield strength is regarded as a property of the material rather than the application, though strictly speaking it's only applicable to tensile loading applications (i.e., those that mimic the material test setup). Therefore you can use it regardless of whether the stress is tensile, shear etc.

I'd be more concerned about calculating the stresses in the pin. These will definitely not be uniform in reality, so the pin could deform locally due to high local stresses while the average stress is still below the yield strength. I don't know how much detail you're going into for the project, just something to consider.

Offy

You have to calculate what force the ram and the chassis will exert on the pin and that will define the minimum shear strength you require. Thats your minimum yeild strenght required for that application. I assumed from the material you selected you want it to shear but then I re-read that you dont want any permanent deformation suggests that you actually want the pin not to shear. If this be the case dont use EN8/ BS970 080M40. In a product if you wanted the pin not to shear or suffer any permanent deformation you would calculate the yield strength required and select a material with a higher value. How high the value would depend on the application. From the values you give 340N/mm^ is enough force to cause perment deformation, this isnt a lot of force for a pin to withstand. EN8 is an unalloyed medium carbon stee l, Id pick a high carbon steel or an alloy depending on the application/enviorment. For college to demonstraigh a prototype or a concept it might be fine though. Its easy to work and its cheap but it would not be my choice for a pin holding a ram to a chassis.

.50 (MOA)

Offy wrote: »

You have to calculate what force the ram and the chassis will exert on the pin and that will define the minimum shear strength you require.

done 104313N

Offy wrote: »

Thats your minimum yeild strenght required for that application. I assumed from the material you selected you want it to shear but then I re-read that you dont want any permanent deformation suggests that you actually want the pin not to shear.

thats right the mechanism is lifting a large weight, and i would like it to stay lifted when it should.

Offy wrote: »

If this be the case dont use EN8/ BS970 080M40. In a product if you wanted the pin not to shear or suffer any permanent deformation you would calculate the yield strength required and select a material with a higher value.

instead of designing the size of the pin first and then applying the material to it such that it would be suitably strong enough, i left the pin size relatively open and calculated what size it needed to be to suit a specific material, in this case EN8. my understanding is that i could use any material (within reason) provided the pin was then made thick enough to resist the shear on it.

Offy wrote: »

How high the value would depend on the application. From the values you give 340N/mm^ is enough force to cause perment deformation, this isnt a lot of force for a pin to withstand.

340N/mm^2 is the yield strength of EN8 (think i'm right- lets assume it is for the minute) not the force in a non specific material that is to be specified

Offy wrote: »

EN8 is an unalloyed medium carbon stee l, Id pick a high carbon steel or an alloy depending on the application/enviorment. For college to demonstraigh a prototype or a concept it might be fine though. Its easy to work and its cheap but it would not be my choice for a pin holding a ram to a chassis.

if its yield strength resists shear and it is hard enough, then does this not suffice


not knowing this makes me look amateurish- which i kinda am, but i cannot remeber ever being told anything different

Offy

.50 (MOA) wrote: »

my understanding is that i could use any material (within reason) provided the pin was then made thick enough to resist the shear on it.

Yes and no. This is why the application is important. In a prototype it rarely matters but in a product things are different. You have to factor in life expectancy, environment the device will be used in (outdoor, home use or chemical plant?), will it be subject to impact? safety factor, etc. The application often defines the material, for example in a chemical plant silicon iron is often used in place of stainless steel as it resists corrosion better in that application. EN8 will corrode, if you want to use it in a product you have to treat it, plating or something similar. Stainless is harder and doesnt corrode as easily in general use. Again the environment is important, what ambient/working temp. is expected? EN8 is not designed to be used for shear pins, it can be used for them but other materials are more suitable. I dont know if this device is being designed as a proof of concept or a product and both are very different. Products that are being used be member of the public are generally designed to meet certain safety standards and shear pin material can be predefined.

.50 (MOA)

Offy wrote: »

Yes and no. This is why the application is important. In a prototype it rarely matters but in a product things are different.

its a product but this is the first so it will also be the prototype

Offy wrote: »

You have to factor in life expectancy, environment the device will be used in (outdoor, home use or chemical plant?),

outdoor use

Offy wrote: »

will it be subject to impact?

no impact loads

Offy wrote: »

safety factor,

2.5

Offy wrote: »

The application often defines the material, for example in a chemical plant silicon iron is often used in place of stainless steel as it resists corrosion better in that application. EN8 will corrode, if you want to use it in a product you have to treat it, plating or something similar.

Zinc electroplating

Offy wrote: »

Stainless is harder and doesnt corrode as easily in general use. Again the environment is important, what ambient/working temp. is expected?

average outdoor temperatures no extremes

Offy wrote: »

EN8 is not designed to be used for shear pins, it can be used for them but other materials are more suitable.

the pin must not shear- ever, the shear on it is due to it going from one part to another with relative movement between those parts possible

Offy wrote: »

I dont know if this device is being designed as a proof of concept or a product and both are very different. Products that are being used be member of the public are generally designed to meet certain safety standards and shear pin material can be predefined.

standards will be met


thanks for the replies, lots of useful info but i had already considered many of the things like safety factor, coating and the like. it was merely the material property that defined the materials resistance to a load perpendicular to the cross section, and it turns out it is the yield strength i need to use as this value.

right now the part will be a 40mm diameter pin made from EN8, with zinc electroplating

this will lift (as i have calculated) 104313N, with a 2.5 safety factor+ a bit more by rounding it up from 36,8mm to 40mm for standardisation.