# Origins of the "threaded fastener pullout formula"

I was looking through NASA's document for "Installation Torque Tables for Noncritical Applications" and noticed their formula for the pullout load of a fastener on a certain material. It goes as follows:

$$P = \frac{\pi\cdot d_m \cdot F_s\cdot L}{3}$$

Where:

• P = pullout load, lbs.
• $$d_m$$ = pitch diameter of threads, in.
• $$F_s$$ = material ultimate or yield shear stress, psi
• $$L$$ = length of thread engagement, in.

Does anyone know what the derivation, or origins for this formula are? It's stated that the /3 is due to a factor of safety. So, the "real" formula would be:

$$P = \pi\cdot d_m \cdot F_s\cdot L$$

I assume the $$\pi\cdot d_m$$ looks like you get the circumference of a circle with the pitch diameter, then you're multiplying that by the length of the thread engagement, which gives us the area made by a theoretical cylinder made by the pitch diameter and length of thread engagement. Multiply that by the yield strength and we get a maximum force applied over an area which gives us a force. Since stress = force/area, force = stress × area, force = ultimate stress × cylinder. Anyway, that's just my thoughts, was hoping someone who knows more could lend some insight.

## general calculation

The calculation is indeed based on a cylinder surface/plane that transfer the force from one thread to the other.

Figure: Shear plane source: Analysis and design of threaded assemblies

As you can see the root of the thread is subjected to shear loads. The total load is the pull out force P, and the assumption is that the force is distributed over that plane/surface

However, the force that is transmitted through the threads is not uniform over the surface. It usually has an exponentially asymptotic form like in the following picture.

i.e. the 1st thread is loaded more. If you notice the first thread approximately take 1/3 of the total load (that varies with the application and the modulus of elasticity of the materials used, and also the tolerances of the thread/screw/bolt).

So, when you are designing the thread you need, to make sure that none of the threads will fail (either with significant plastic deformation or by shearing). Since the first thread is the most "tortured" one, the pullout load is estimated with the 1st thread as a constraint. And since the 1st thread roughly gets about 1/3 of the load, hence the 1/3 factor in the equation.

IMHO, the fact that the yield stress is used (instead of the ultimate tensile stress), is the actual safety factor, preventing the thread from shearing. The use of the yield stress makes sure that the thread will not enter the plastic range (hence it will be easy to unfasten the bold and reassemble). Also, because the ratio of UTS to the yield stress range in steels is usually over 1.4, the threads can carry at least over 40% of the estimated load before starting to pull out.

I think you're correct about the cylinder thing. It appears to be a crude equation though since it does not include thread percentage or thread pitch and does not even appear to be assuming a trapezoidal threadform.

I would think the correct way to do it is to unrolling the the trapezoidal thread form and calculating how much force it takes to shear that.

Maybe the equation does account for thread pitch because:

$$F_s\pi D_m (PL)T$$ where P is the thread pitch in terms of threads per length and $$T$$ is the thickness of the thread. $$(PL)$$ gives the number of engaged threads and if your threads are butted up right against each other (i.e. thick enough to take up as much space as possible next to each other) then $$T = (\frac{1}{P})$$ and everything cancels out to your result.

Not all screws have threads like that. Self-tapping wood screws for instance have a thin thread relative to the spacing between threads.

Still does not account for the trapezoidal thread form or thread percentage though.

You are right about the shear cylinder. With a SF of three, the thread's pitch or angle are irrelevant at the point of greatest strength. They assume that ultimately the threads will yield a stripped cylinder, and that ultimately the shear at the surface of the cylinder sets the limit of P.

If one tries to be more fussy, the load on the bolt surface decreases as we get near its end. because the end threads can expand a bit more freely.

In the figure I have shown the strech of the bolt by $$dL$$ and the shear stress by red line. the shear stress can readily be assumed uniform and arive at the formula!

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