combined stresses on fasteners 5

1) its "principal"

i've never seen your interaction equation ... it looks like a von mises equation.

i've seen Rt^2+Rs^3 = 1, but i prefer Rt^2+Rs^2 = 1 'cause it's way easier to use (you don't have to solve a cubic to get your RF) and it's slightly conservative.

OK

rb1957, can you give me more info about your equations?

A principal stress is a stress on a plane with zero shear, which is not your condition. Your equation is based on an equivalent stress concept (known by the names Huber, von-Mises, distortion energy, etc.) and is used frequently to calculate the stress during tightening on a fastener (applied torsion producing shear in the shank, inclined threads producing tension in the shank). The interaction of tension and shear stresses on fasteners due to applied joint forces is usually done with interaction curves. NASA RP1228 and MIL-HDBK-5 provide various interaction equations, including the ones listed by rb1957.

hi bsmet95

If your tension and shear on the bolts is created by external forces and not by tightening then they can be calculated by the methods shown on this site:-


If your interested in the stresses during tightening then on the same site:-


regards

desertfox

Current US codes require interaction to be:

[(actual shear/allowable shear)+ (actual tension/allowable tension)]< 1.0

There's some good information on the Von Mises Yield Criterion on Wikipedia:

In ductile metals, this is arguably the most accurate way to predict yielding due to combined loading. Now, just about everything is in some sort of combined loading, so it turns out that it's pretty useful.

Your bolts are likely seeing tensile loading (mostly from preload) and shear loading from external loads. So, in a sense the formula given to you is correct. However, it's a good idea to understand where it's coming from before you use it.

If the shear loading on your joint is significant (more than you can handle with friction), consider a rabbet fit. Due to tolerances, etc, bolted joints in shear have to slip & yield a bit before all of the bolts become loaded. A rabbet or close-fit pin in your joint will predictably carry all of the applied shear load in excess of the friction force.

Hi bsmet95

I found the formula you are quoting or at least in a similar form to what you posted:-

Failure criteria: Refer to page Failure Modes


The notes on this page In order to estimate the design factors of safety it is necessary to consider the failure modes. The preferred failure criteria for ductile metals is the "Shear Strain Energy Theory" (Von Mises-Hencky theory). For a stress regime associated with a bolt i.e pure tensile stress sx combined with shear stress ? xy. The Factor of safety relative to the material tensile strength Sy..is calculated as follows

Factor of Safety = Sy / ( sx2 + 3 .? xy2 ) 1/2


the above can be found at this site:-

Scroll down to the title "failure critera"

desertfox

wouldn't code definitions (re bsmet95's post above) trump an equation in a book ? unless of course there are cavets in the code or it doesn't apply ...