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Up Topic Welding Industry / Inspection & Qualification / Australia Standard AS 2207 UT weld inspection
- - By Bill M (***) Date 08-20-2012 14:36
While I wait for my copy to arrive I was wondering if anyone has any experience using this UT standard for weld inspection?

thanks for any comments
Attachment: weldcriteria.doc (224k)
Parent - - By G.S.Crisi (****) Date 08-21-2012 21:30
SHANE!!!
A question for you to answer!!!
Giovanni S. Crisi
São Paulo - BRazil
Parent - - By Bill M (***) Date 08-23-2012 18:24 Edited 08-23-2012 18:27
finally got my copy... I added a page above for weld quality limits for review

So say I have one meter long weld in 25mm plate with a slag inclusion.  What is the Maximum permissible imperfection level allowed?
Parent - By jwright650 (*****) Date 08-23-2012 18:53 Edited 08-23-2012 18:56
dang Bill, that would take some using to get used to after using Table 6.2 in D1.1 all of these years.

your question as I read it,  L/2 or half a meter for the maximum.

edit: what I don't understand is what happens below the max permissible(using that weighted portion of the chart)
Parent - By G.S.Crisi (****) Date 08-24-2012 19:40
American standards (ASME VIII, API 650 etc.) include pictures showing the maximum allowable discontinuity level. Doesn't the Australian standard have a similar picture?
Giovanni S. Crisi
Parent - - By Flash Date 08-25-2012 22:00 Edited 08-25-2012 22:09
Hi Guys
this table is out of AS/NZS 1554.1 (D1.1 equivalent from Aust and Shanes favourite standard ;o) ), AS2207 is the method of tests for UT and it does not set acceptable criteria it covers calibrations, number of faces to scan from etc

with that being said Table 6.2.1 out of AS/NZS 1554.1-2011 is pretty straight forward read note 5 of the table

We need more information to answer your question definitively, like what is the height of the defect, I will make some assumptions to show how in my opinion it should be read

column 2 = 25mm plate
column 3 = weighting factor - if the height of the defect is 3mm you times the recorded length of the imperfection x by the weighting factor (in this case 2)
column 4 = gives you the maximum allowable imperfection (length/2)

worked example 1 = 100mm long inclusion, 3mm recorded height = gives you a weighted imperfection of 200mm, in 25mm plate you are allowed 500 mm in 1000mm so it complies (the way most people apply it)

applying it another way also works max imperfection level x by weighting factor = actual max imperfection level such as L/(2x2) = L/4 or 250mm allowable imperfection in 1000mm

both achieve the same outcome

worked example 2 = 101mm long inclusion, 4.5mm recorded height = gives you a weighted imperfection of 505mm, in 25mm plate you are allowed 500 mm in 1000mm so it does not complies

worked example 3 =  21mm long inclusion, 10.5mm recorded height = does not comply (anything with an X is not permissible)

worked example 4 = 2 of 200mm long adjacent inclusions separated by a distance of 150mm (becomes total imperfection length of 550mm), with a height of 3mm = does not comply as you treat them as one imperfection

There is a few things to watch
note 1 - deals with adjacent imperfects, if the distance between two imperfections is less than the length of one of those imperfections you treat them as one defect, this is true for imperfections in both the vertical and horizontal plane, have a look at figure 6.2.3 for assessing adjacent imperfections, as a picture tells a thousand words
note 6 - there shall be no imperfection with a height greater than 2 (or t/20) with a distance of T of the end of the weld
note 7 - if the length of weld exceeds 1m and you have imperfections in two adjacent metres you have to move the division to get the worst case
note 8 - where the length of the weld is greater than 1m, you must assess it in 1m lots (so you do not end up with defects greater than 500mm if weighting of one is used)


I hope this goes some one to answering your question
R
Flash
Parent - - By Bill M (***) Date 08-27-2012 19:01
Thanks Flash-
I did the same calculation but...

Say I have a 2mm tall lack of fusion indication in a 1 meter long weld in 25mm plate....the maximum length allowable for this indication is 1/2 a meter??   (19.7")

I just think I am missing something because the mag particle section says a maximum allowable dimension for lack of fusion or incomplete penetration is 2t/3, or 20mm max (t = plate thickness)

I read it the same- but accepting a 19-inch long UT defect...just doesn't sound right to me.
Attachment: 1554.1mtspecs.pdf (101k)
Parent - By Bill M (***) Date 08-27-2012 19:27
Sr. Crisi-  No defect pictures in the 1554.1 standard.  They do reference another specification for porosity charts.  (AS 4037)

They state that for routine UT or RT exam the defect height "H" shall be taken as 2mm.

Jwright:  I agree with you, I only use table 6.3 here....so give me D1.1 anytime!
Parent - - By Flash Date 08-28-2012 08:27
Thats right Bill
it works on the theory of loss of cross sectional area
You make an interesting comparison between internal and external allowances in your post
R
Flash
Parent - - By Bill M (***) Date 08-29-2012 19:55
Thanks-

Well around here if a welder puts down 15 inches of lack of fusion in 40 inches of weld -more that a couple times...he is down the road. 

Glad to know he can still get a job in Australia.
Parent - - By Flash Date 08-30-2012 20:55
Bill, nice reply
I didn't say the welder would still have a job, I just said that is what the standard allows
You have remember all our welders have to weld over head because we are down under ;o)
R
Flash
Parent - - By Bill M (***) Date 08-31-2012 12:37
No drama mate-  good point....been there a few times myself
I occasionally had trouble in keeping a pint on the bar sometimes
Parent - - By Shane Feder (****) Date 09-04-2012 12:17
Flash,
Firstly, this is not a personal shot at you because I consider you a good mate (be nice to catch up for a beer sometime)

I am pulling my hair out here and wondering what sort of drugs the Australian code committees are on ?
This "loss of cross sectional area" is an absolute joke.
The major players in the world welding community (America, Britain and Europe) have been judging defects on their length since god knows when ?
Now we have a minor player (who unfortunately some of us are forced to follow) basically saying they don't care what the big players are saying - height of a defect is the important thing - irrespective of whether the length would judge it a reject to all major welding codes.
I have reviewed a welder qualification radiograph today to AS 2980 and the welder had a gas pore of 2.5 mm length on a 6" Sch 160 (18.26 mm) coupon.
Based on the wall thickness this would be acceptable to all major welding codes - both pressure (ASME) and structural (AWS) but not the Australian code.

How can anyone understand this:
Table 4.3.5.3 from AS 2980 (Acceptance criteria for UT and RT)
Maximum allowable dimensions or number of imperfections (see Notes 1 and 2)
Note 2
For the purpose of radiographic and ultrasonic examination, the height of the detected imperfection is to
be taken as 2 mm.
Porosity - Internal—See loss of cross-sectional area
Loss of cross-sectional area (see Note 5) - 5% maximum

So the maximum "loss of cross sectional area" allowed based on the 5% maximum on an 18.26 mm thick coupon is 0.91 mm - obviously rejected based on Note 2.
If all RT indications are to be given a 2 mm depth (as per note 2) and you also have to comply with 5% maximum  of "loss of cross sectional areas" then any discontinuity on less than 100 mm thick is going to be a reject ?

To make it even more of a joke slag inclusions are acceptable up to a length of "t" (18.26 mm) but a 2.5 mm gas pore fails.

Hang on, what am I saying about a joke - I have to get up in the morning and try and administer these laughable codes.
And they wonder why we drink ??????
Cheers,
Shane
Parent - - By Flash Date 10-01-2012 03:39 Edited 10-01-2012 03:50
Hey Fed Express,
All comments are welcome, my apologies for not getting back to this earlier, I have been OS
as for what drugs they are on, I have submitted a TQ to the committee to find out ;o)

I tend to agree with some of your comments regarding AS but this is what we (I am) are stuck with
As for AS2980, my understanding this is largely based on ISO 9606 (with the acceptance criteria of 5817), this explains why are standards are messy
An example of what is in ISO5817, it uses a 'really smart' (in theory and said with a degree of sarcasim) allowance for weld reinforcement 1mm = 0.1m x w with w being the width of the weld
in theory this is a sound idea the wider the cap the greater the allowable reinforcement to control toe angle, but try inspecting this on a job when the cap vary from 6mm to 10mm wide, its fun, I have been there

The current trend is if we (no SAI Global) adopt an ISO rather than develop new or amend existing standards, which from my experience only confuses the situation

Anyway back to your current issue of LoCSA
To be perfectly honest very few people apply this as it is so cumbersome to apply

But this is my take

Porosity - (that is grouped not GP, elongated cavities, worm hole, piping porosity whatever you want to call it) refer to porosity charts out of AS4037 which are the same as ASME, as you point out it states 'Internal - see loss of cross sectional area', but it also states for RT use Porosity charts from AS4037, I would be opting for the later every time

As we are For GP, worm holes, elongated cavities, piping porosity  you have to use LoCSA, Do not refer to the porosity section as this is just for Porosity and not individual pores

I know, I know clear as mud

LoCSA (is listed in the AS2812 glossary of terms - effective loss of weld cross-sectional area due to imperfections) - so I can not categorically say if they mean transverse of longitudinal to the weld or both, to me it makes sense to do it on the shear area of the weld which is longitudinal but knowing the AS system, they are referring to transverse

but here is a worked example based on my humble opinion

a Single Vee butt on 20mm plate, 1 meter long weld the cross sectional area is

Transverse area = Area of triangle 1/2 b x h = 13 x 20 =260 mm2 therefore you are allowed 13mm2 of discontinuity (defect of say 6mm wide x 2mm height = 12mm2) (if you are working on transverse area only)

Longitudinal Area = 20mm x 1000mm = 20000mm2 therefore you are allowed 1000 mm2 of combined defects but none exceeding transverse allowance (defect of say 400mm long x 2mm height = 800mm2)

as I am typing this it is reinforcing that if anything it should be volume (mm3) not area (mm2) so it takes into account both length, width and height

back to your example

Using only the transverse area

Single Vee in 18.26WT
= A (of a triangle) mm2 =1/2bxh
= using Trig to calculate = 1/2b=10.5mm
= 10.5 x 18.26mm
=191.7mm2 cross section area of weld

allowance of 5% = 9.58mm2

a GP of 2.5mm diameter has an area of 4.9mm2 therefore it complies to this example

and yes I agree this is a long winded way of doing it but you did ask

I trust this is some help

R
Flash
Parent - By Shane Feder (****) Date 10-01-2012 11:03
Hey Flash,
Thanks for the response,
Cheers mate,
Shane
Up Topic Welding Industry / Inspection & Qualification / Australia Standard AS 2207 UT weld inspection

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