Background: A welder tests on a NPS 6 Sch80 .432 wall coupon, depositing .100 in of 6010 and the balance of .332 using 7018.
Question 1, Using 6010, is the welder qualified to deposit .864 in Maximum of weld metal? 864 = 2T
Reply was: NO (Reason:- as per QW 452.1 b, welder is qualified for 2t only ) for the maximum weld deposited thickness.
Question 2. Using 7018 is the welder qualified to deposit .644 in weld metal. 644 = 2t
Reply was YES - agreed
Question 3. Is the welder qualified to deposit .864 in. of weld metal using 6010
Yes qualified (because 7018 F # is 4, and when a welder qualified in F#4 he is qualified for F#3 also, refer QW 433)
plus .664 in. of 7018 weld metal for a deposit of 1.528 in the same groove.
Reply was: no See QW 452.1 (b) –and you are correct that the total thickness qualified is 2t
Thanks & Regards, only
Lakshman Kumar B,
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Yes qualified (because 7018 F # is 4, and when a welder qualified in F#4 he is qualified for F#3 also, refer QW 433)
Question 3. Is the welder qualified to deposit .864 in. of weld metal using 6010
Yes qualified (because 7018 F # is 4, and when a welder qualified in F#4 he is qualified for F#3 also, refer QW 433)
If the same welder tested for 6010 & 7018 for 2 different thickness (0.1+0.332=0.432 thick)
Then he is qualified for 0.864 thick weld metal.
Reason – 7018 F Number is 4, and 6010 F Number is 3, when a welder is tested for F Number 4 he will be qualified for F number 3 also and hence thickness range can be consider for 0.332*2 (2t) = 0.664In.
could you explain me on above statement. 6060(F# is 3) am i correct?
Thanks & Regards,