I'm with ya jon.
I can see kix's confusion, and will refrain from arguing that since I don't have a Seciton IX in my hands to review the note (though I should in a couple of days) but I think its being overthought. The 1/2" 3 pass requirement is it. One coupon.
This particular requirement was simply a long overdue adjustment (some 5 years ago) of the old 3/4" thickness qual for unlimited with no pass count considered.
Some on Section IX had decided (of which I agreed) there really was no special skill difference between a 1/2" deposit and 3/4" deposit. So the deposit thickness was reduced. But to make sure you were actually in that stable multi pass regime the 3 pass min was added.
Someone needs to check this, but I think you have bigger fish to fry than the diameter.
QW452.1(b) Note 1 also applies.
Summarizing, by my read it states that if there are two or more F numbers, each individual F number T is
considered to be for each F number singular, not the aggregate.
"shall be determined and used individually in the thickness t of weld metal in the coupon column to determine
thickness of weld metal qualified.".
Note 2 Two or more pipe test coupons with different weld metal thickness may be used to determine the
weld metal thickness qualified.
With that read, 6010 is an F3 7018 is an F4 and because of note 1 the thickness deposited is the thickness of
each F number, and not the combined thickness. So the F3 needs it's own individual thickness, and the F4 the same.
""I have 2 WPQR's from a welder..SMAW,
1 test conducted with 6010(F#3)/7018(F#4) on 2".218" thk.,
and the other test is 6" .432" thk.""
Based on note 1, you can't use the combination to qualify as a whole, so you have to use the individual layers.
To qualify this, I believe you have to have a minimum of 13mm F4 three pass, and will only qualify 2T for the F3 layers.
Judgements of diameter as listed in 452.3 are determined by O.D. without respect to thickness which gives you two
criteria to satisfy. That being the diameter criteria and thickness criteria.
For the OD less than 1" it's size welded up, 1 to 2 7/8 1" unlimited, and so on.
So for diameter you can go up, but not down in diameter.
I'd be interested in hearing others take on this. I wished ASME would not use the and/or business.
Regards,
Gerald
Gerald,
I understand what you ar4e saying. I am away of the different F# requirement, and I don't think I included any mention of the 6010 (F3) in the thicknesses I was talking about. I am only questioning the range of the 7018 since most 6010's will not be used to fill up a joint. Having said that, my question would be - one WPQR states the welded thickness of the 7018 is .125". The other WPQR states that the 7018 welded thickness is .375", thus giving the total thickness in two coupons of .500", and of course there are at least 3 layers between the two test - according to note 2, is the weld thickness range qualified for "max to be welded" per the table?
It really looks to me like the answer would be yes, and the welder is in fact qualified for unlimited thickness with the 7018. I respect the opinions of the more intelligent people here, and was looking to see if I am correct in my interpretation. And if not, tell me what note 2 is trying to say. What is the purpose of it?
I've just dropped an e-mail to the Vice Chair of IX to get an explanation. Likely won't get a response to post until Tuesday... I stand by my earlier statement but admit having been wrong in the past!!! ;-)
Given that Jon has sent an email to ASME I'll defer to the answer he comes up with. However; it is my opinion they are not in fact qualified as I believe the three layers are meant to be contiguous and not broken up.
Its too late in the day for my brain to study this too close and peruse all the finer points of this thread, so I'm not sure who said exaclty what in all the details, but I will respond to one thing, I suspect Walts answer is going to concur with Geralds that the layers have to be contiguous. It doesn't make sense any other way. Contiguous was the reason for the allowance of dropping the thickness from 3/4" to 1/2" in the first place.
1 test conducted with 6010/7018 on 2" .218" thk., and the other test is 6" .432" thk.
I took the mention of 6010 to mean in both.
As many of you have already mentioned, it's late and we're tired, but I did notice there was no mention of the thickness of the F3 deposit and the thickness of the F4 deposit. That being the case, as the third party inspector, I would reject the performance test records (and the PQR if I had access to it).
I usually refer to the welding documentation as a house of cards. If the PQR fails, the WPS collapses and so does the welder's performance test. It is a common problem that the "witness" (that isn't required) fails to record the thickness of the weld deposited by each process or for each F-number. Then the contractor tries to claim his qualification range is based on the thickness of the test piece, not considering each process or F-number separately.
Suppers must be ready, I heard a bell and I can't control the saliva.
Best regards - Al
Jon, thanks for taking the time to e-mail the vice chair in regards to this question. I eagerly await the response he gives.
Al - On the certs, it does give the deposited thickness of both F#'s. I may have left out the details of the 6010 to simplify my question. However, I'm only questioning the range qualified for the 7018 (F4), as that will be the rod that "fills up" the joint and would be the only one that would need to be unlimited thickness. Both certs have the total deposited thickness for the 7018 as equaling 1/2". The remaining thickness would be the 6010.
Alright, here's my final question (I hope) - suppose all variables as discussed remained the same - 2 coupons, 2 diameters, 2 thicknesses, etc - Let's say the 6" coupon that has .375" of 7018 was in fact deposited with 3 layers, so now we have the 1/2" thickness required, and now we also have the 3 layer minimum....in at least 1 coupon. Does that change anyone's stance on the subject or ASME's answer?
new tito... here's my question posed to him, with his response:
"You are not missing the point. Not sure how they get that interpretation."
On Aug 31, 2007, at 3:17 PM, Jon Lambert wrote:
Reading a thread on the AWS Message Board and am at a loss to explain... Note 2 of the above "seems" to indicate a welder can weld TWO 1/4" thick coupons and take credit for 1/2" of deposited weld metal thus having unlimited thickness quals... doesn't seem kosher somehow... am I missing the point????
Thanks Jon for your inquiry to the ASME guru.
I'm still scratching my head though. What the heck does note 2 mean????
Here is exactly what he means (example):
A welder tests with a process on NPS 2 Sch 40 (0.154 wall) is qualified down to 1 in. OD up to 0.308 maximum deposit thickness. If he also tests on NPS 6 XXS (0.864 wall) and puts in at least 1/2 inch of weld metal with the same process, that qualifies him down to 2-7/8 in OD and unlimited thickness. Those qualifications can be combined to qualify him down to 1 in OD and for unlimited thickness. What is prohibited is that if the second test was on plate, you cannot combine the tests and get unlimited thickness down to 1 in OD.
Forum
