"F" factor Chart Fig. UG-37 4

rww88,

It is a little confusing at first, but it refers to vessel longit. axis, where 0 deg in the figure is the long. axis.

It does not have to do with location of the nozzle as such, rather with the plane thru nozz centerline for which you are carrying out reinforcement calculations.

Calculations of plane parallel to long. axis are normal to hoop stress, F = 1. For plane in circumferential direction, normal to longit. stress which is half of hoop stress, angle = 90 deg., F can equal 0.5 if nozzle is "integrally reinforced" that is, has no added reinforcing element such as a pad added.

Calculation in plane other than longit. is of no practical interest for a typical radial, round nozzle. For offset (hillside) nozzles, angled nozzles, etc. depending on the size of the "finished opening" planes other than the longit. plane can govern the reinf. calculations and need to be checked, using appropriate F factor.

A little involved, hope you find this helpful.

Should have added:

The examples in Appx L illustrate the use of the F factor.

Thanks very much SnTMan. See if this sounds correct: If I interpret the diagram in Fig. UG-37 as a plan view of the nozzle, and not the vessel, I will be right?

rww88-

Yes, you can consider the figure as a plan view of the nozzle. Or, to put it another way, imagine that you are looking down the throat of a horizontal nozzle placed in a horizontal vessel.

Note that this can be of interest in ordinary radial nozzles where just a bit of weld buildup on the nozzle wall or shell will get you where you need to be as far as available reinforcement. Weld buildup on the nozzle wall or shell is considered to be integral reinforcemnt per UW(16)(c)(1).

jt

Thanks very much jt. Not to beat a dead horse but, is the angle formed by the long. plane of a circular hillside nozzle, with the chord subtended by the tangent to the shell radius, theta?

rww88-

Eeek! That's some vocabulary I haven't used in 15 or so years! I'll try to translate that to my level of understanding - let me know if I got it right:

I'm looking into a nozzle placed in a horizontal vessel. The nozzle is a hillside nozzle, ie it is just like a radial horizontal nozzle but the elevation is shifted up or down from the horizontal centerline of the vessel. Typical example would be a level gage/bridle connection. Now... if I take a horizontal plane and put it through the axis of the nozzle, theta is 0 for an F factor of 1. This plane could be (but, yes, this is a practice to be avoided) lined up with a longitudinal seam. If I take that plane and rotate it 45° about the axis of the nozzle, theta is 45° and F is 0.75. When I've rotated the plane to be vertical, potentially aligned with a circumferential seam, then theta is 90° and F is 0.5.

I find that the concept is easiest to deal with when I visualize whether I'm reinforcing a longitudinal seam (probably at around 20 ksi membrane stress) or a circumferential seam (probably at around 10 ksi membrane stress). Makes sense to me that I'd need less reinforcement along the circ seam which needs to redistribute half of the force imposed on the developed cut.

What I haven't figured out yet is the formula for the F factor. The best I can come up with is Theta = 0.75-sin(theta)/4 +cos(theta)/4 but that only works at 0°, 45°, and 90°. Maybe someone else out there more clever than me can provide a formula which works!

jt

Well, kind of a guess, but if you plot a Mohr's circle of the principal stresses, then the magnitude of the membrane tensile stress at any plane rotated from the longitudinal by Theta per Fig UG-37 is: (.75 + .25*cos(2*Theta))*Sigma, Sigma being the circumferential stress.

Or, F = .75 + .25*cos(2*Theta)

A simple semi-graphical interpretation, recalling (as I did by consulting a textbook!) that the angle considered in the Mohr's circle is twice the actual in-plane angle. Of course all this ingnores shear stresses.

Thanks again guys. jt, in return for your help, I derived the following formulas from a least squares polynominal regression analysis of Figure UG-37:

For 0 <= (theta) < 30,
F = 1 - 1.273E-4(theta)^2 - 1.645E-4(theta)

For 30 <= (theta) < 60,
F = 1.158 - 0.009(theta)

For (theta) >= 60,
F = 1.49 + 1.167E-4(theta)^2 - 0.0215(theta)

You will find these any many additional formulas and equations in a online course I am developing for PDHcenter.com