Re: Re: [MW:28936] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

Using E6010, is the welder qualified to deposit 0.864 in. maximum of weld metal? Reply (1): Yes. Which is 21.94mm using 6010? Deposition is 2.54mm during welder test; where as in actual job he can able to deposit 21.94 mm? Though the process is single,  will u add F3 without backing deposition + F4 with backing deposition?  When you are restricted on F3 and your case is individually F3, how will you go for F4, if it is F4 to F3 i agree with you but this is F3 to F4???

Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 file: 10-1918''. So during welder qualification obviously you have to dip into QW-452.1(b) and definitely also with QW-433, I mean when you come with QW-433, for this mentioned scenario your F4 gets knocked out and just F3 remains, without QW-433 you cannot make qualifications just with QW-452.1 (b)

Hence the answer is NO.

Thanks.

---

Regards,

Mohamed Muneeb Mahaboob

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IRCA-9001:2015 LA

ASNT-NDT-L2-RTFI-RT-UT-MT-PT.

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''If you know any art by any means then SCULPT IT - no matter how cheap or useless it is - JUST SCULPT IT, you will never know when experts will be on demand'' 

 

From: rajesh kaswala

Date: 2019-01-17 11:38

To: materials-welding

Subject: Re: Re: [MW:28925] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

yes because 7018 f4 so welder qualify f 1 to 4 and 6010 f no. 3 so qualify

your answer no , why ?

regard

rajesh

On Wed, Jan 16, 2019 at 3:17 PM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:

Hi Rajesh,

Greetings,

I hope its F3, so when you are restricted with F3, you cannot pull and trigger the F4 mode, which is what i have observed and you have observed, hence Question (1): Using E6010, is the welder qualified to deposit 0.864 in. maximum of weld metal? Reply (1): YES, but answer should be NO.

I want to see more experts view on this.

---

Regards,

Mohamed Muneeb Mahaboob

CSWIP 3.2-89247/2 

IRCA-9001:2015 LA

ASNT-NDT-L2-RTFI-RT-UT-MT-PT.

SAP#:70021650 APs:JAPID, QCS & QCI.

E muneebslsa@gmail.com M +91-636-955-7189

''If you know any art by any means then SCULPT IT - no matter how cheap or useless it is - JUST SCULPT IT, you will never know when experts will be on demand'' 

 

From: rajesh kaswala

Date: 2019-01-15 16:09

To: materials-welding

Subject: Re: Re: [MW:28909] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

your  all question and answer is correct 

1) total thickness 0.432 so qualified 2t = 0.864 not 1.528

2) 6010 f no. 2 and 7018 f no. 4 so qualified 6010, allow total thickness 0.864

regard

rajesh

On Tue, Jan 15, 2019 at 6:01 AM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:

Hi Rajesh,

Great catch to the point, lets talk in mm, why not 8.43 mm? why only 2.54 mm what is the reason? which is 5.08 mm only.

0.332 in - 8.43 mm

0.100 in - 2.54 mm

0.432 in - 10.97 mm

Thanks.

---

Regards,

Mohamed Muneeb Mahaboob

CSWIP 3.2-89247/2 

IRCA-9001:2015 LA

ASNT-NDT-L2-RTFI-RT-UT-MT-PT.

SAP#:70021650 APs:JAPID, QCS & QCI.

E muneebslsa@gmail.com M +91-636-955-7189

''If you know any art by any means then SCULPT IT - no matter how cheap or useless it is - JUST SCULPT IT, you will never know when experts will be on demand'' 

 

From: rajesh kaswala

Date: 2019-01-13 18:47

To: materials-welding

Subject: Re: [MW:28903] Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918

depositing 0.100 in. of E6010 and the balance  - 2t only as per 452.1 b , means 0.100 2t = 0.200 not 0.864 

regard

rajesh 

On Wed, Jan 9, 2019 at 7:09 PM muneebslsa@gmail.com <muneebslsa@gmail.com> wrote:

Greetings,

I have clarifications with below interpretaion for question 1 and agree with 2 and 3, does anyone has any observations on Q1?

Interpretation: IX-10-23 Subject: QW-433 and QW-452.1(b) Date Issued: March 14, 2011 File: 10-1918 

Background: A welder tests on an NPS 6 Sch. 80 (0.432 in. wall) coupon, depositing 0.100 in. of E6010 and the balance of 0.332 in. using E7018. Question (1): Using E6010, is the welder qualified to deposit 0.864 in. maximum of weld metal? Reply (1): Yes. Question (2): Using E7018, is the welder qualified to deposit 0.664 in. maximum of weld metal? Reply (2): Yes. Question (3): Is the welder qualified to deposit 0.864 in. of weld metal using E6010 plus 0.664 in. of E7018 weld metal deposit thickness for a total of 1.528 in. in the same groove? Reply (3): No. See QW-452.1(b).

Thanks.

---

Regards,

Mohamed Muneeb Mahaboob

CSWIP 3.2-89247/2 

IRCA-9001:2015 LA

ASNT-NDT-L2-RTFI-RT-UT-MT-PT.

SAP#:70021650 APs:JAPID, QCS & QCI.

E muneebslsa@gmail.com M +91-636-955-7189

''If you know any art by any means then SCULPT IT - no matter how cheap or useless it is - JUST SCULPT IT, you will never know when experts will be on demand'' 

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