D1.4 WPS Flare-bevel L1 AGAIN...

Date 08-18-2010 01:32

I've asked this before,

Does this look right to you, for solving to find weld length ‘L1’ in D1.4 Figure 6.5C?
Fu=80,000 (A706)
Fxx= 80,000 (E8018)
D= .75 (3/4”)
n=4

I get
5.23x80000(.75)/80000(4) =.980625
Is .98 measured in inches?

And what about 6.2.4.3 (2) "length of connecting welds"?
is the above equation supposed to provide ".6x the min. spec. tensile strength x the nominal area of the solid bar"???
Tensile strength of what??? filler metal?

I got a wps from Xxxxx Xx at XXXX Steel, XXXXX told me that the PQR welds were 4+” long, per XXXX’s calcs
I believe the PQR welds should be much shorter
the weld length surprised me the first time I ran calc, but it makes sense, controlled length (weld volume) on test piece; if weld length is too long, its like ‘cheating’ on tensile test…
Xxxx told me that he reduced tensile strength Fu in equation to reflect test load of actual area of bar (#6 = .44") My opinion is that we aren't testing rebar, we're testing weld strength.
My colleague also told me that he was told that "L1"= feet, i dont see that anywhere...

Bottom line... how to calculate L1 for Figure 6.5(C) test assembly (W/2 plates)
I don't like my opinion any more, now that sparks are flying... I need to know how this is done.

By pax23

Date 08-24-2010 15:17

Your calculations look right to me.

The 6.2.4.3(2) issue is a little strange. I see that as the actually code requirement for the length of the welds on the specimens. If you derive a formula based on that requirement what you end up with is the formula in Figure 6.5 note 2, but that is only assuming you are using a flare-V-groove weld. If you derive it for a flare bevel groove weld specimen you get nearly the same equation except the 5.23 constant becomes 7.9 which would give you a longer weld (about 50% longer) for a flare-bevel specimen. I think technically you could use the alternate formula for flare-bevels based on the 'shall' in 6.2.4.3(2) and if you could show how you derived the formula to anyone who challenges you.

So why does the alternate formula not show up in Figure 6.5, note 2. I would guess (and its only a guess) that the code is trying to keep things simple by given you one one formula to deal with, the one with the '5.23' constant would be the more conservative of the two formulas.

The weld lengths will get longer as the bar diameters go up, if there is any undermatching filler chosen, or if you use an angle which will give you 2 for n. I suppose you could arrange a test in D1.4 where the weld lengths approached 4", but your calculations for your specific example look good to me.

Agreed on the 'cheating' comment, but at the same time if the welds are nominally longer than the calculated value I think it would be OK. They don't provide any tolerances, and the acceptance criteria for the test is 125% of the yield of the bar. I see the calculated value as a ballpark number and no one should get their nose out of joint if any of the welds are slightly longer.

Don't reduce Fu in the equation. The cross sectional area of the bar is already accounted for with 'D' and the constant provided.

The dimensions of L1 will be a product of whatever values you use, just be consistent. If you use ksi for Fu and Fxxx, then make sure D is in inches -- L1 will be in inches. If you use MPa for Fu and Fxx, then make sure you use meters for D -- L1 will be in meters. If you use MPa for the forces and mm for the rod diameter, then your L1 will be off by a factor of 1000.

By Duke

Date 08-26-2010 14:09

Thanks, a very clear argument... my colleague called a D1.4 engineer at AWS, and he said much the same thing.

By 68suzuki Date 06-06-2013 19:52

pax23,
hoping you are still out there. where does the 5.23 constant vs 7.9 come from. I assume it is from an engineering standard or formula? In my circumstance, I need a little more length to get an acceptable weld profile and this could be a big help if I can document my use of it in my WPS.