Thickness Qualified For 1

[Spindraft]

I was reading some old posts as I am a new CWI and I am trying to clarify various definitions as given by the code books. One that I have been studying is QW 452.1(b). I teach an advanced pipe welding and fabrication class and I have been introducing various qualification requirements to the class as well as using a WQR mock up to record their grades. I want this to be accurate as to min/max thickness qualified for and the QW 452.1(b) is a bit tricky to comprehend at first. I have went through all of the referenced material and these mitigating factors are not the issue, it is the wording of the table itself that could be much improved.

QUOTE FROM OLD POST
"The problem i am having now is this ; Ordinarily when ASME uses a number as a cutoff point (as in QW 452.1(a))it is written explicitly (eg less than 1/2 , greater than 3/4 etc.). This makes me wonder if the conditions listed in QW 452.1(b) should be read together , not individually . For example ; if a test coupon is .625"(5/8)and welded with F3 to 1/8 then filled with F4 to 1/2" ,then the welder is qualified up to 1 1/4 . However if the coupon is .750 and the ticket then gets written for unlimited thickness the (2t) requirement for ALL thicknesses in QW 452.1(b) has been busted ."

I get what this person is presenting here but taken in context of the table it would seem to me that "all" dictates that the welder has welded the entire joint and is therefor qualified to 2t the thickness of that weld. In order to be qualified to unlimited thickness the welder would be required to complete a weld at least 1/2" thick with at least three layers of weld metal.

I don't believe that nulls the 2t rule. The welder can qualify to 2t the thickness of any weld (in accordance with other restrictions such as F#'s)meaning that the welder can technically qualify to 7/8" if welding on a 7/16" joint, if the weld is 1/2" they are qualified to unlimited which satisfies the 2t rule by default. I have also heard arguments that F4 covers or blankets F3 electrodes which would then make the weld described in the quote qualify to unlimited. If F4 does not qualify F3 then the correct determination in the above quote would be the welder is qualified to 1/4" using F3 electrodes and qualified to unlimited thickness using F4 electrodes.

I don't have QW 432 in front of me at the moment so cannot make a clear determination on the F# consideration as to thickness qualified in this argument. Given the definition of the table as I propose here it would be possible to qualify for weld thickness' over 1" without qualifying for unlimited, ex- 3/8" welded with F3 and 3/8" welded with F4 in the same joint if the F3 and F4 must qualify separately would technically qualify the welder to weld up to a 1 and 1/2" weld as long as the F# maximums were adhered to. Either way both rules are always satisfied but I must agree with a portion of the original post that this table is muddy at first glance.

 

[Spindraft]

The original post I was reading

http://www.eng-tips.com/viewthread.cfm?qid=227369

 

[Duwe6]

The cited previous discussion was for QW-451 PROCEDURE Qual's. Sticking metal together using a designated process & filler(s). A demonstration of the WPS to be able to generate sound welds, *IF* the welder does his part.

QW-432 and "test" put this discussion into WELDER quals. Totally different; welder quals are all about the demonstration of SKILL to be able to adequately implement the WPS. And yes, if you put a 5/8" thick Supercoupon in front of a welder, and you allowed that welder to put in 5/16" of F3 - E6010, and 5/16" of F4 - E7018; then that welder would only have a 1¼" qualification range - 2T = 2*(t3 + t4). Review NOTE (1) of table QW-452.1(b). That's why the instructions to the welder are "No more than root & hot-pass using 6010".

The secret to Sect IX is to keep Procedure and Performance separate in your mind; and not to mux up big T with little t. Then you can use IX like a cookbook, as it was intended to be used.

 

[Spindraft]

I do appreciate the destiction between Prodcedure Qual and Welder Qual but the opening argument in the thread from which the quoted post is taken concerns itself with Welder Performance Qualification and only talks about the WPS as the mitigating factor that ultimately defines the qualification. Not sure where you got QW 451 as they are clearly discussing QW 452 Performance Qualification Thickness Limits and Test Specimens. If we are to hack terminology then I would surely make the assumption that Performance is concerned with the welder's skill in this instance and not the ability of the weld itself to perform under any given conditions at all.

 

[DekDee]

spindraft,
Not sure where the confusion is coming from - seems pretty clear to me.
"All" covers exactly that - all weld deposit thicknesses (less than 1/2" 13 mm and/or less than 3 layers).
Put in a 10 mm weld with 3 x layers - qualified to 2 x t
Put in a 20 mm weld with 2 x layers - qualified to 2 x t
Put in a 13 mm weld with 3 x layers - qualified to "unlimited"

Now if you have F3 & F4 you just apply the same factors to each F #

"For example ; if a test coupon is .625"(5/8)and welded with F3 to 1/8 then filled with F4 to 1/2" ,then the welder is qualified up to 1 1/4 ."

This statement in the original post is incorrect (if the F4 was welded with 3 x layers)
It should be F3 less than or equal to 1/4" / F4 unlimited

Hope that helps,

Cheers,
DD

 

[kukes2]

read QW-452.1(b)(note 1),QW-404.15,QW-433 AND QW-353

 

[Spindraft]

True enough DekDee, when i wrote that statement in my mind I was not taking into consideration the number of passes required to fill the rest of the joint with F4.

 

[kukes2]

If a welder qualifies on a .625"(5/8") thick coupon with F3 to 1/8" and filled the rest of the way with F4 with a minimum of 3 layers his qulifications are as follows for production welding.

F4- Unlimited thickness with backing
F3-unlimited thickness with backing
F3-1/4" without backing

Refer to QW-433