I have ASTM F 606 and the 25th edition of Machinery's Handbook.  I am not sure what the second equation is - can you include it here?

The equation to calculate the thread stress area in ASTM F606-95b if found in paragraph 3.4.2, equation (1):
A= 0.7854[D- (0.9743)/n]squared
In the 26th Edition of Machinery Handbook, Page 1490, a topic: "Formulas for stress areas and lengths of engagement of screw threads"  The above equation is given as Eq. 2a, for steels of up to 100,000 psi ultimate tensile strength, and Eq. 2b:

For steels of over 100,000 psi ultimate tensile strength  

A = 3.1416[(E sub s min/2) - (0.16238/n)]squared

where E sub s min = Minimum pitch diameter of external thread for the class of thread specified, and n = number of threads per inch.

Equation 2b, is a more recent addition.  I believe that it has validity for high strength alloys.  Your comments?  Do you know the researcher or group that contributed this new equation?  Thanks for your interest.

Thanks for including the equations.  I have not seen the second equation before.  Is it possible that it is only a regression fit rather than based upon first principles?

I have derived the first equation, but have not had time to derive the second one.  A real issue in determining the area to be used in calculating tensile stresses in fasteners is whether to include any of the thread area in the cross-sectional area.  The first equation (used in ASTM F606, eq. 1) and commonly used in stress calculations is the accepted equation but as stated in Kent's Mechanical Engineering Handbook, 12th Ed., 1958 [which tells my age], Chapter 10, page 10-08, Section 2, entitled "Screw Thread Standards for Fasteners," it states, and I quote "... Stress Area (A sub S) is the ASSUMED AREA OF AN EXTERNALLY THREADED PART, used for the purpose of computing tensile strength.  It is given by the formula

     A sub s  =  3.1416[ (E sub m + K sum m)/4 ]squared

where E sub m = the mean pitch diameter and
      K sub m = the mean minor diameter.

The cross-sectional area used for calculating the yield and tensile stresses is important and needs to be reconciled in ASTM Committe F-16 on Fasteners.  

 

I know that the tensile stress area is usually defined as the average of the minor and pitch diameters.

A_t=(PI/4)*((d_p+d_r)/2)²

In your first equation, PI/4=.7854,  d_p=D-.649519P,  d_r=D-1.299038P, and P=Pitch=1/N.

Substitution and simplification yields your first equation.


The second equation looks more like the area of a circle
A=PI*r²
I can't quite figure out where the .16238 value comes from.



The page references for the 25th ed. are 1407, 1416 and 1640,  The 26th ed. are 1482, 1490, and 1740, respectively.


--Scott

Scott,

Thanks for including the page numbers for the equations.  On page 1406 of the 25th edition, Cornell University is mentioned, so perhaps they are the originators of the equation in question.

nialloyman,

Regarding reconciliation, I suppose there is one, and it uses the ASTM equation to which you refer.  This equation is also used in ASME B1.1 and the equivalent is used in ASME B1.13M and ISO 898-1 for metric threads.  This equation is used for screws threads in materials with ultimate tensile strengths much greater than 100 ksi.  I would ignore the other equation.

CoryPad and Scott,
If both of you would be so kind to supply your e-mail addresses I will be able to provide more data than is conviently possible with these messages.  A colleague plotted data and ran calculations that shows the ASTM equation overstates the effective area of applied stress.  An optical comparator was used to determine the minor diameter where the minor diameter is the outer diameter minus twice the thread depth.  This diameter best represents the actual area under stress since the threads appear to be free surfaces and thus carry no load.  The constant in the ASTM equation (0.9743) represents a ratio of pitch length to twice pitch depth, but the ratio should be higher.  The tooth depth  as a function of the pitch (in.) for various sized bolts is a linear function that has a straight line equation of Depth, inches = 0.6019(Pitch, in.) + 0.0013 (I can supply the plotted data). It would be more easily understood if I send you the study.
There are more interesting considerations in this study.  The results impact the pass/fail criteria for supplying fasteners to many customers, including the US Navy.  I hope we can continue to unravel this puzzle and your input as well as other interested parties is sought.  Your contributions have been insightful and other opinions are welcome.  Thanks,  Bill Mankins.

My address is:

cpadfield@omnimetalslab.com

I am happy to review your information.  I have spent some time studying this subject, and I think you may be underestimating the work that went into the thread equations.  Careful studies have been done in the past to compare the calculations with actual failures.  I can provide more info, but I will review yours first.

Best regards.

Cory, et. al.,

Both the 12th and 20th editions of Machinery's Handbook have both equations...interesting that the one for steels >100 ksi ult. appears to have been removed from the 25th edition and then replaced in the 26th edition. A simple error perhaps? Also interesting is that FED-STD-H28-2B also shows two equations, but instead of using minimum pitch diameter like eqn. 2b above (nialloyman), it uses basic pitch diameter. The result for me was that both equations in the FED-STD ended up calculating the same tensile stress area for 1.25 in bolts.

Andy

ASTM specification committee F16.96 (Bolting Technology) is considering sponsoring research to re-confirm/refine this equation.  As I hear more about this, I will post it in this thread.

Checkout lecture 28

http://www.utm.edu/departments/engin/lemaster/machine_design.htm