Hi guys,
I am in the process of designing an air fill valve and pressure regulator that are to be screwed to either end of a cylinder that holds 250bar of compressed air. The fill valve is brass and the regulator chassis will possibly be BDMS. Both screw into the cylinder using an M25 x 1.0 thread.
How would I work out what length the thread would have to be before the pressure strips the thread and blows them out of the ends of the cylinder?
All help greatly appreciated.
May Thanks.
Last edited by Kelly_Bramble; 04-25-2015 at 01:29 PM.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Hi,Originally Posted by Kelly Bramble
I did see the calculators however they didnt make sense to me.
When calculating the tensile stress area for 100 ksi and less, a major diameter of .980 and 28 threads per inch gives a tensile stress area of 0.70168.
When calculating the minimum thread engagement length I enter a tensile stress area of 0.70168, a major diameter of 0.980 and threads per inch of 28, this gives a minimum thread engagement length of 0.93284.
The original part I am trying to re-design only has a thread engagement length of 0.393 so I dont get how the minimum could be calculated to higher than that.
Also wouldn't you have to take the material into account? for example wouldnt a part made of nylon need a bigger thread engagement length than something made of steel? The calculators dont seem to take that into account.
Regards.
The equations and calculator give the minimum full thread engagement to obtain full thread strength based on shear area of the material.
Regardless of the material, the calculator has shown you that need 0,93284 inches of thread engagement to get full strength. So, if you only get .5 inches of thread then your not getting full advantage of the base material and could get a thread pullout at some loading less than the base materials strength.
Also, if you know the tensile area of the thread you can calculate the yield based on the material characteristics. see:
http://www.engineersedge.com/calcula...eload-calc.htm
and
http://www.engineersedge.com/materia...alculation.htm
Last edited by Kelly_Bramble; 04-26-2015 at 07:59 AM.
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Thanks but I still cant get it, iv spent all afternoon running over the calculations. Im trying to find the minimum thread length i can get away with before the thread strips, the original part has a thread length of 0.393", lower than the calculator gives, and is able to handle the pressure, im looking for a thread length smaller than this.
This is the first time i have been working with thread design so im clueless.
Regards
The equations given are more for bolting design and not appropriate for your application. As Kelly states, they are matching the thread shearing strength to the tesile load strength of the fastener.
For your application, you should start by calculating the pressure load applied to valve thread which is equal to 250 bar x the x-sectional area of the pressure sealing area of your fitting's threaded connection to the tank.
The max shear loading capability of the fitting thread is equal to the fitting thread max allowed shear strength (as a percentage of the material yield strength for safety) x the fitting thread I.D. x pi x the length of thread engagement.
As a result, by setting up an equation equating the calculated thread load to the thread max shear equation with the thread length as the unknown you can then determine your required minimum thread length to resist the fittings pressure loading.
Assuming the max allowed shear strength of the mating tank is the same or greater than that of your fitting, using this minimum length x an appropriate safety factor will be the safe minimum thread length for your design.
On the other hand, if the max shear strength of the mating tank I.D. thread is less than that of the fitting material, then you need to check the minimum allowed thread length for the tank I.D. thread; and, this can be achieved by the same above equation by substituting the fitting thread O.D. for the fitting thread I.D. and the max allowable shear strength of the mating tank material in the equation. Then the minimum required thread length for the threaded joint is the maximum length of the two equations x the appropriate safety factor.
As a final step, you should then verify that the tensile strength of your designed fitting threaded connection is sufficient for the applied pressure loading by dividing the applied pressure load by the wall thickness of the bored section of the thread, which is the thread I.D. area minus thru bore max O.D.
If you have any questions or concerns about the above, don't hesitate to post them and I will be glad to respond.
Here's another resource that should help...
Thread Yield and Tensile Strength Equation and Calculator
Tell me and I forget. Teach me and I remember. Involve me and I learn.
Thats for the info guys you have been most helpful, it is greatly appreciated.
Jalberts I tried to run a few calculations based on your formula and came up with the following:
pressure load = pressure x cross sectional area
pressure load = 3625.94 psi (250 bar) x 0.75 inches squared
pressure load = 2719.455 (is it psi?)
max shear loading = shear strength x fitting thread (should it be OD instead of ID? the fitting has a male thread that screws into the pressure cylinder) x pi x thread engagement length
max shear loading = 34100 psi x 0.980 inches (assuming it is OD) x 3.1415 x 0.393 inches
max shear loading = 41258.180271
Assuming what i have done so far is correct, I kind of lost what you said after that, please could you clarify the part about setting up an equation?
Kelly thank you for the link.
By inputting this on the imperial side of the calculator:
Thread pitch size - 0.039 in
Thread shear stress area - 0.6985 in^2
Thread engagement - 0.393 in (this is the length of the standard part's thread engagement)
Material yield strength - 19580 psi
I get the following:
Shear strength of nut thread - 11356.4 psi
Tensile force to thread yield - 410.115 psi
Axial force applied to yield - 112.581 Lbs
Would I be right in saying the tensile force to thread yield is the max pressure the thread would be able to take before it goes?
Again thanks to both of you.
To answer your question about the load units of measure, the results of both of my equations are in lbs, not psi, multiplying the area in lbs/sq in x the area in sq inches cancels out the sq in units and you are left with lbs.
I will discuss the relationship between the equations more later; but for now, I am making a general review of all elements of your calculation as well as the results of the referenced site of that calculation.
In order to help me do that, can you tell me the manufactures specified maximum operating pressure for the sample fitting you are using for your example?
Also, your .393 engagement length is in inches, correct? This just seems to be a very short engagement for such a large diameter thread although I have seen similar fitting for low pressure applications.
Additionally, what is the yield strength of your material; and, are you using exactly the same specification material as the sample fitting?
Last edited by JAlberts; 04-30-2015 at 01:58 AM.
Ahh ok thank you.To answer your question about the load units of measure, the results of both of my equations are in lbs, not psi, multiplying the area in lbs/sq in x the area in sq inches cancels out the sq in units and you are left with lbs.
The maximum operating pressure the manufacturer states is 232 bar which is also the normal working pressure. I have heard that a 250 bar fill is possible.In order to help me do that, can you tell me the manufactures specified maximum operating pressure for the sample fitting you are using for your example?
Yes that is in inches, in the UK i have been brought up with mm however as most of the calculators on engineers edge work in inches I have simply been converting mm to inches. Yes after research I thought the same about the length, I read somewhere that a general rule of thumb is 1.5 x diameter, however that is the engagement of the original part from the manufacturer and it works perfectly fine.Also, your .393 engagement length is in inches, correct? This just seems to be a very short engagement for such a large diameter thread although I have seen similar fitting for low pressure applications.
The original fitting is made from brass, there is no way I can know for sure what type of brass the part is, I simply ordered a brass bar from ebay and im using that, however I will consider using a stronger material if it means I can make the thread engagement length smaller. According to Elgin Fastneners the yield strength of brass is 19600 psi.Additionally, what is the yield strength of your material; and, are you using exactly the same specification material as the sample fitting?
Many Thanks
Last edited by Kelly_Bramble; 04-30-2015 at 06:22 AM.
perky416
Thanks for the quick response and input.
The material and yield strength definitely help clear up an issue I was seeing in my calculation formulas' results.
As I stated above, the units for both of my formula results are lbs. and since you chose to enter a value for the thread length in the second formula then it is possible to determine the SF (Safety Factor) of the sample part's thread by dividing the thread's allowable shear load from the second formula by the applied load from the first formula. Using your values this results in a Safety Factor = 41258.180271 / 2719.455 = 15.1714, which very high even for a pressure fitting.
However, now seeing that you are using brass with an 19600 psi yield clears up that that issue.
The maximum allowable shear strength of 34100 psi you used in the second formula is an error because it exceeds the maximum yield strength of the material. The generally accepted value for the maximum shear strength of a material is .58 x the material's allowable yield strength. For your brass material the resulting maximum allowable shear strength = .58 x 19600 = 11368 psi; and, using this value in the second formula:
max shear loading = 10440 psi x 0.980 inches (using the tread OD) x 3.1415 x 0.393 inches
max shear loading = 13754.746 lbs
Which indicates that the SF (Safety Factor) for the sample fitting thread shear strength = 12631.91 lb / 2719.455 lb = 5.058; and, this is a much more reasonable and understandable SF value.
With that issue resolved, with respect to your additional questions that I did not previously answer:
1. I specified that you use the male thread I.D. for the shear area calculation because this results in the minimum shear area for the thread joint and it is where a male thread shear failure commonly occurs.
In that respect, there are a number of methods of selecting “correct” male and female thread shear stresses that are used by different engineering standards and I am not claiming that the method using the minimum male thread diameter and maximum female thread diameter and basic equation I have offered is in any way the best of these but the results obtained with this method should be very close to that of any alternatives for this purpose and I would suggest that any values obtained by online calculators that vary substantially from these results should be inspected carefully for formulations, assumptions and intended applications.
2. With regard to your confusion about the equations, by telling you to set the two equations equal to each other with the thread engagement length (L) as an unknown I was replying to your request for a method of determining the minimum thread length you would need. In simple terms (using your thread shear diameter value):
L min = 2719.455 / (shear strength x effective thread shear diameter x pi)
= 2719.455 / (11368 x 3.1415) = .078 in. (without any SF)
3. With regard to my discussion about determining whether the male thread or the female thread will determine the required minimum thread length, as long as the female thread part material yield strength is equal to, or greater than that of the male thread component then the male component will always fail first; otherwise, if the material of the female is less than that of the male, then you must calculate the required thread length of the female thread using the female thread O.D. as well to insure that this required thread length is equal to or greater than that calculated for the male thread.
I hope this helps clear up your issues and clearly explains the issue of the maximum allowable shear strength for a material.
One important note about material selections and ordering, especially for gas pressure applications, it is very important that you request material certifications with your material order to insure the material you order and receive meets the strength requirements you expect and use in your design calculations.
Again, if you have further questions or want to discuss any of the above issues please post them and I will reply.
PS, Starting on Sunday I will be intermittently offline for awhile so for quick responses please post any questions as soon as possible.
Brilliant mate that has cleared it up a hell of a lot for me.
There are just a few more things that are confusing me in the calculations however.
Where you state:
Where does 10440 psi come from, as right before that the shear strength was calculated to be 11368 psi?max shear loading = 10440 psi x 0.980 inches
Also when determining the safety factor of the sample thread where did 12631.91 lb come from?
Finally just to clear something up...when you state the ID of the thread do you mean the minor diameter? Similar to the following...
You help is greatly appreciated, you have taken more time than anybody on several forums to explain it to me. Love it when someone is willing to teach.
Thanks again
OK, first, I apologize for the errors in the equation values. What happened is that I somehow initially got it in my mind that the Sy value for the brass was 18000 psi and used that value for the calculations and results in my initial draft post; but then I realized my error, recalculated everything for the Sy 19600 psi value and revised (I thought) all of my draft post calculation values accordingly. Unfortunately, in the two instances you caught, I correctly updated the equation result values but (clearly) failed to update the values within those two formulas.
Immediately below is a corrected version of the two paragraphs with the value corrections highlighted for clarity.
"The maximum allowable shear strength of 34100 psi you used in the second formula is an error because it exceeds the maximum yield strength of the material. The generally accepted value for the maximum shear strength of a material is .58 x the material's allowable yield strength. For your brass material the resulting maximum allowable shear strength = .58 x 19600 = 11368 psi; and, using this value in the second formula:
max shear loading = 11368 psi x 0.980 inches (using the tread OD) x 3.1415 x 0.393 inches
max shear loading = 13754.746 lbs
Which indicates that the SF (Safety Factor) for the sample fitting thread shear strength = 13754.746 lb / 2719.455 lb = 5.058; and, this is a much more reasonable and understandable SF value."
Bottom line, fortunately, only the two equation values in the original post were in error and all of the results of the calculations are correct and valid for an Sy = 19600 psi.
With regard to your question about my choice of an effective diameter for the thread shear area calculation. Now that I have seen your reference thread diagram, I understand your confusion all along. I should have been using “Minor Diameter” instead of "min diameter" all along in my original posts. I guess maybe my old age memory of correct dimension terminologies is not what it used to be.
I am glad to have been assistance to you in problem resolution and apologize for my glitches along the way in getting there. In fact, there really isn't any short cut method for discussing this type of issue; and, one advantage I have by being a retired engineer is that I am able to devote time to projects like this without compromising other priorities. As usual, “if you have any questions just post them and I will respond”.
Hi JAlbert ,
I saw the your thread length calculation formula, the way you have explained in details was really helped me a lot.
But just want to the know, Do the formula you mentioned in the forum for thread length calculation refers any text book. If 'yes' please let me know the book for my future reference.
thanking you in advance.
with regards,
Mohanraj
There are probably more recent references that cover this but unfortunately the only one I have is my college textbook "Machine Design" by Joseph Edward Shigley, McGraw-Hill Book Co, 1950 Edition, Chapter 6 "The Design of Screws, Fasteners, and Joints", pg 185 that specifically presents the thread stress calculation equations as I described them in my post.
There is also references in Machinery's Handbook - Torque and Tension in Fasteners. It may not be spot on to your particular needs but its not a bad starting point. I love that book BTW.
Hi,
I'm struggling with the equation for length of thread engagement. The shear area of the nut should be greater than the tensile area of the bolt so the bolt fails first. The minimum condition is when they are equal.
At = tensile area of the screw (bolt)
The shear area of the minimum nut is
1/4n (this is the length of the nut thread at the minor diameter for a 60deg V thread)
+ 2*tan(30deg)*(Esmin-Knmax) (the additional length moving from the minor diameter at a distance of the screw pitch minor and the nut minor maximum)
The circumference at Esmin is Esmin*pi so the shear area for one thread is Esmin*pi*[1/4n+2tan(30deg)*(Emin-Knmax)]
The number of engaged threads is Le*n so
An (shear area of the nut)= Esmin*pi*Le*n*[1/4n+2tan(30deg)*(Esmin-Knmax)]
Equating this to At =Esmin*pi*Le*n*[1/4n+2tan(30deg)*(Esmin-Knmax)] rearranging and solving for Le
Le = 2*At/{Esmin*pi*[1/2+0.57735*n*(Esmin-Knmax)]}
But the spec (and they are smarter than me or I missed something) rearranges to
Le=2*At/{Knmax*pi*[1/2+0.57735*(Esmin-Knmax)]}
So now I'm lost.
The only reason I went down this road was that I wanted to come up with similar equations for other thread forms.
Corrections and comments welcome.
Ed
Last edited by eckotapish; 06-19-2018 at 10:28 AM. Reason: correction
In your first paragraph you state: "1/4n (this is the length of the nut thread at the minor diameter for a 60deg V thread)" but then for calculating the circumference you use Esmin (the diameter of the external thread) in calculating the circumference when I believe you should be using Knmax because that is the basis for the 1/4n length value.
Here's how I thought it through. At Knmax the shear width of the internal thread is 1/4*pitch or 1/4n. The shear area would then be 1/4n*Knmax*pi. At Esmin the shear width is [1/4n+2*tan(30deg)*(Esmin-Knmax)] so the shear area is [1/4n+2*tan(30deg)*(Esmin-Knmax)]*Esmin*pi.
Last edited by eckotapish; 06-24-2018 at 06:06 PM.
Sorry for the double post but I forgot to mention that Esmin is where the theory states the internal thread will fail.
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