Aluminum bolt calculation request 3

The quick answer is yes, the M14 fastener should withstand 2000 N. The method is:

stress = force / area

so,

force = stress * area
force = 180 MPa * 115 mm^2
force = 20.7 kN

The stress I used is a typical yield stress for 356 T6 material. The area I used is the fastener stress area. I recommend reviewing the contents of this website to get more information. Try these:

Thread288-20937, Thread725-22078, Thread727-22951, and Thread288-21503

CoryPad:

Your answer (based on the external thread tensile stress area) appears to assume that the length of thread engagement is sufficient to fail the bolt thread in tension (that is, that the length and strength of the external and internal threads is sufficient that they will not "strip" or "shear" prior to bolt tensile failure, but will support the full tensile capability of the bolt). There doesn't appear (to me) to be enough information in the original post to support that assumption.


I think additional information is needed:

We are told the "bolt length" is 25mm. But is this the length of the engaged internal and external threads? If not, what is the length of engaged threads?

Are both the external and internal threads the same material/strength (A356-T6)?

Is the pitch 2mm, as you apparently assumed based on your tensile stress area (needed for thread shear area calculations)?

While the externally threaded meter of the coupling has been described as a “bolt”, on the off chance does this “bolt” have a hole through the middle of it? The use of “coupling” in the post makes me wonder if this is a “fluid” coupling – which might be hollow...

What is the thread class/fit (this can have and effect of thread stripping/shear calculations)?


See thread404-11447 for a similar discussion regarding stripping of internal aluminum threads. JohnF_Al, you can calculate the length of engagement required to prevent shear or "stripping" of the aluminum thread at the anticipated design load. This calculation is based on the geometric shear area of the thread at minimum material condition and shear strength of the parent thread material. An excellent source of formulas, information and examples for performing these (and other) thread strength calculations are found in Appendix B of FED-STD-H28/2B -- available for free at These formulas are relatively simple, accurate and readily lend themselves to entry and reuse in a spreadsheet. If you don’t know the shear strength of your materials, two possible sources of free online information are MatWeb (usually “typical” values) at or in MIL-HDBK-5 (usually statistically adjusted values to a given confidence level) at (Warning: HUGE download, but the smaller, separately downloadable chapters available at euler9 managed to somehow lose the cast aluminum alloys at the end of Chapter 3). Alternately, you wish to use an applicable material specification “minimum guaranteed” value.

CoryPad, don’t take this personally, I am (and remain) a fan...

Kenneth,

I didn't take anything you said as negative criticism. You are correct that I assumed a 2 mm pitch, and axial and diametral thread engagements that would cause tensile failure. Considering the large margin between typical force (2000 N) and calculated break (20,000 N), I would wager that any small differences (like pitch, thread class, etc.) will not materially affect the results.

My philosophy on this forum is to try and move people in the right direction, not necessarily do the complete analysis that may be required. That is why I did a simple calculation, and encouraged JohnF_Al to seek the detailed information available at this and other sites.

Thanks to both of you for the info. The Fed std H28/2B document was very helpful.

John.