Can you clarify whether this is a thread and screw design you need to verify or if it is a failed screw you wish to analyse for failure mode?
Generally, a screw will strip its thread if the shear force in the thread helix exceeds the yield strength of the material. If you wish to calculate this you need good data on the thread geometry and the applied torque.
Turn the torque into a linear force by using the thread pitch / circumference as a lever (of sorts), ignore friction.
Calculate the area of material experiencing the shear force.
Calculate the shear force.
Compare this to the yield strength. You want to be below 1/2 of that for safety.
Then you can work back and specify a torque setting for the driving device to avoid problems.
The component of the force which acts along the centre line of the screw is the shear force - it is this force which is trying to shear the thread material off of the body of the screw / nut. Say I have a thread which has a 1mm pitch and a 10mm nominal radius at the thread. If the torque on the screw is 1Nm (one Newton at a radius of 1m) then the torque at 10mm radius is 100N. (As one gets smaller, the other gets bigger to give the same product)
The circumference of the thread is 2*PI*10mm = 62.83mm
So, for every 1mm we go down, we go around by 62.83mm. This ratio (62.83:1) is applied to the torque to give us shear.
We had 100N, multiply that by 62.83 to get 6283N shear force.
Now we need to know the area which joins the thread to the rest of the material. Say on the thread we are using the root width is 0.8mm. We now need to know how long our thread helix is. We have a circumference of 62.83mm which we can use for simplicity. Multiply that by the number of turns of engagement (say 5) to get 5*62.83=314.15 length of thread. Multiply this by the root width giving 314.15*0.8=251.32mm^2
So now we know we have 6283N acting to shear 251.32mm^2
We need to convert the area into square metres so :
251.32/10^6 = 2.5132*10^-4 m^2
So our shear force is 6283N acting on 2.5132*10^-4 m^2
Stress = Force/Area
Stress = 6283/2.5132*10^-4 = 25*10^6Nm^-2 (exactly, as it happens)
Which is 25MNm^-2
As we only calculated the shear force, this is shear stress.
Although this seems to have been talked around a number of times, a lot of the topics and discussions are quite technical and very interesting but not that helpful when you want to do a simple calculation.
If this equation has been run through in another thread then I couldn't find it. As it is, I hope I have been of some help.
I understand your desire for a simple equation, but I don't think this is sufficient. There are three thread failure modes - internal thread shear, external thread shear, and external thread tensile fracture. All three need to be computed. Equations for these are available in FED-STD-H28 (free downloadable .pdf file) or VDI 2230 or Handbook of Bolts and Bolted Joints referred to in the other threads.
CoryPad,
I'm sure the link you have posted is the ultimate solution. In my experience, engineers don't usually start worrying about thread stripping unless they have a hard screw going into a soft material without a great deal of thread engagement. It is that situation which rings alarm bells and in that situation I believe the maths I have supplied are correct.
I think there is also a bigger point to be made- if aleovol uses your computational software I'm sure he'll(assuming gender) get the right answer but he won't have learned a great deal and next time the problem arises it will be back to the internet. I know the screw problem is complex (a mixture of torsion, tension and shear) and material failure modes even more so but I do find the readiness to trust somebodies software over a real understanding a bit worrying.
Could I suggest that aleavol tries both methods and compares the results. That way you'll get the best of both worlds.
I agree that most thread stripping problems are due to insufficient thread engagement with weak internal thread material (plastic, cast Al, etc.). I have doubts that your "maths" are accurate. Your statement "So, for every 1mm we go down, we go around by 62.83mm. This ratio (62.83:1) is applied to the torque to give us shear." is one that I have never seen. Even if it is correct, I believe your analysis is off by at least a factor of 10. According to the equations shown in the references I previously listed, the force generated by 1 Nm on an M10 x 1.0 hexagonal head screw with friction coefficient between 0.1 and 0.2 produces a force of only 400-750 N.
You refer to my "computational software", but what I showed were published guidelines for performing calculations. These guidelines have references, equations, and illustrations to educate users. Therefore, a first-time user should be able to avoid using the Internet the next time a similar problem occurs.
I'd like to think that the maths were accurate in what they were trying to be: a way of translating the torque into shear force without taking anything else into account. The purpose of this was to give a feel for what was going on. I made up the screw dimensions to illustrate the point so they might not correspond well to actual screws. By ignoring the friction I have added to the safety margin.
Ultimately none of this really matters. The main point is that I was suggesting, and still do suggest, that as a first order calculation to determine whether a screw thread would strip (which was the question) then calculating the shear force on the threads is a good way forward. Yes, the whole situation is complex and yes, there are probably more accurate ways of determining the exact stripping force / failure mode but its nice to start with some maths you understand.
I apologise if I misunderstood the nature of the link you provided. I hope that Engineers out there are wary of using equations they don't fully understand. Your link didn't work for me when I tried it. I'll try again later.
Just for the record, I checked out Biggadike's unique approach for calculating preload on a frictionless bolt and found it to be correct under the stated assumption of zero friction. P = T/(K D), but if friction is neglected, Eq. 3 of
becomes K = p/(2 pi D). Substituting, gives P = T/(K D) = (2 pi T)/p. Using this expression for Biggadike's example, P = (2 pi T)/p = 2(pi)(1000 N mm)/(1 mm) = 6283 N. Whenever possible, keep units in N, mm, and MPa, because, very conveniently, 1 MPa = 1 N/mm^2. For more info., see
. And stress and material strengths are normally reported in MPa.
Thus, if area A = 251.32 mm^2 and you retain units in N and mm, that would simply be stated as shear stress tau = P/A = 6283/251.32 = 25.0 MPa, whereas shear yield strength of, say, ASTM A36 is about 145 MPa.
Neglecting friction gives you a built-in factor of safety of about 7 (if you use no other factor of safety; 9 or more if you do), so the resultant engagement length (tap depth) could end up, say, 3.5 times longer than necessary. Admittedly an ultra conservative first-order check. Nonetheless your post is very interesting for looking at fundamental concepts, to help the user begin to learn the more complex situation by degrees, with the above links also provided by CoryPad.
Thank you so much Vonlueke.
The maths may be unique in that they aren't pulled out of a book but they are just a geometric translation of torque into shear force to give a worst case for thread stripping. Pascalls are indeed the SI unit for pressure but I was using their equivalent Nm^-2 which perhaps makes more sense when calulating in Newtons and metres.
What is interesting is just how much difference friction makes: As the screw bites down, friction on the threads turns the torque into torsion in the screw body (with and equal and opposite torsion in the 'nut' ) This torsion is at its greatest at the head of the screw (as it has the most opposing friction below it) which is why the screw itself will tend to fail first as soon as the engagement length exceeds about 2 x the nominal diameter.
That the stripping safety factor is so high just goes to show how inefficient screw threads are at converting torque into linear movement under load.
T. SOFRONAS, P. E., Consultant Engineer, Houston, Texas
Reference to Page 95, Hydrocarbon Processing February 2003.
Damaged threads on larger 1- to 4 in. - diameter blind tapped hole, such as found on compressor casings and vessel flanges, usually required a downtime to repair. Anyone can ask for a downtime. However, if a repair is not immediately required and the plant can be kept operating safely, shutting down could be a very expensive decision. These are the tough decisions engineers help make. This article demonstrates how a simple analysis can aid in this decision-making process.
Twenty 3-in. UNC-8 TPI stud were threaded into a ductile steel vessel flange. Some studs were accidentally removed during an inspection and eight of the 31 internal threads were damaged in several threaded holes. Could the vessel be put back into service or was an extended downtime required?
Consider a stud in a threaded hole (Fig. 1). The stud is steel and the part is ductile steel and has a shear strength of ó shearpart , which is less than the stud, L is the engaged length at which “strip-out” of the threads in the part could be expected, D is the stud nominal diameter, P is the load in the stud and k = 0.9 converts the nominal diameter to the stud stressed diameter.
Now at complete yield conditions of the thread, the load is pretty evenly distributed along the threads. This is different form normal elastic conditions, where the first few threads can take up the shear strength, ó shearpart, is about one half of the yield strength, The failure stress, ó fail, can be approximated as:
ó fail = ó shearpart P/L share = P/ðkDL
The thread length that will strip-out is:
L = P/( ðkD ó shearpart)
Nominal stress, ó stud, in the stud due to assembly torque or stretch and any other loading produces the load
P = [ð (kD)²/4] ó stud
This simplifies to:
L/D = (0.225) ó stud /ó shearpart)
When ó shearpart = 17,300 lb/in², and ó stud 45,000 lb/in².
L/D = (0.225)(45,000 / 17,300) = 0.585
L = 0.585 x 3 = 1.76 in.
The threads could be expected to start to strip out when the engagement was 1.76 in. the remaining engagement is 23 good threads divided by 8 tpi or 2.88 in. there is adequate margin (2.88/1.76 = 1.64) and stripping-out isn’t expected.
Analyzing thread stripping is a complex subject and has to be blended with experience and common sense. While this analysis was verified with a nonlinear finite-element model, performing a risk assessment that determine the consequence of a thread strip-out is always prudent, Will a stripped threads result in a small, nonhazardous leak allowing an orderly shutdown? Or could a major leak occur, whit the potential for a catastrophic event? An analysis such as this provides quantitative input that can be used to help make the final decision.
1. Sopwith, D.G.., “The Distribution of Load in Screw-Thread,” Institution of Mechanical Engineers, Proceedig, V159, 1948
