[MW:26923] RE: 26921] Re: 26918] Weld metal thickness range QW-452.1 (b)

Hi,

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,

Lakshman Kumar B,

+91 9440031459

 

From: materials-welding@googlegroups.com [mailto:materials-welding@googlegroups.com] On Behalf Of venkkatesh krishnan
Sent: Tuesday, October 17, 2017 12:32 PM
To: Materials & Welding <materials-welding@googlegroups.com>
Subject: [MW:26921] Re: 26918] Weld metal thickness range QW-452.1 (b)

 

Dear Mr. Lakshman,

 

Thanks for your reply,

 

This is approved interpretation on ASME Sec IX No: IX-10-23 & File no: 10-1918.

 

For the Question no: 1, My answer was also No, but Interpretation answer is Yes i don't know why? that's what i need explanation from you and experts .

 

I am quite ambiguity about your reply: 

 

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)

 

could you explain me on above statement. 6060(F# is 3) am i correct?

 

Regards,

Venkkatesh Gopalakrishnan. 

On Tuesday, October 17, 2017 at 7:14:57 AM UTC+3, Lakshman Kumar wrote:

Hi,

 

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,

+91 9440031459

 

From: material...@googlegroups.com [mailto:material...@googlegroups.com] On Behalf Of venkkatesh krishnan
Sent: Tuesday, October 17, 2017 12:42 AM
To: Materials & Welding <material...@googlegroups.com>
Subject: [MW:26918] Weld metal thickness range QW-452.1 (b)

 

Dear Gents,

 

Would some one explain me the below interpretation:

 

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: yes

Question 2. Using 7018 is the welder qualified to deposit .644 in weld metal. 644 = 2t
Reply was Yes.

Question 3. Is the welder qualified to deposit .864 in. of weld metal using 6010 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)

 

why a welder gets 2T coupon thickness limit (for weld metal deposited) when he only deposited the root and hot-pass using 6010. Then same welder gets only 2t weld metal thickness for the 7018?

 

I will be thankful for a piece of advice on this issue.

 

Regards,

Venkkatesh Gopalakrishna/

 

 

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