Making Sense of Wedge Curvature According to ASME Sec. V Art. 4
Introduction
As of 2017, ASME Sec. V Art. 4 specifies the use of contoured wedges for ultrasonic weld examinations performed on curved surfaces with diameters less than 14 in. (350 mm). The language regarding contouring changed from “may” to “shall" in 2017, and is now a requirement rather than a suggestion. However, the way paragraph T-432.2 is written can be a little confusing. (Note: skip to the end if you just want the fancy table!)
Do I Need a Contoured Wedge?
The first sentence in T-432.2 uses the term “shall” regarding contouring, but this isn’t always the case. If the pipe diameter is sufficiently large compared the wedge geometry, a flat wedge may suffice.
Part (a) of T-432.2 states:
(a) Search units shall be contoured as required by the following equation:
How those equations are supposed to be used isn't well explained. It's in there, it's just that it's implied rather than stated outright. The equation really just tells you the minimum pipe size on which a flat wedge is allowable. A more user-friendly version might look like this (in metric, because it's the friendliest):
How About an Example?
Okee dokee. You are about to inspect a 6 in. nominal (6-5/8 in. OD, or 168.3 mm) circumferential butt weld. To do this, you will be scanning in the axial direction using an Olympus SA10-N55S wedge. Since you are scanning axially, we set A in the equation above equal to the wedge width of 23 mm.
Update: I was politely reminded by a couple sharp-eyed readers that it is the "footprint dimension" that should be used, which will be less than the catalogue width if the outside edges are chamfered. The width of a chamfered SA10 wedge would be 13 mm, and not 23 mm. I agree wholeheartedly. I have added a small disclaimer in the table to take this into consideration. Most Olympus PAUT wedges have chamfers on the sides to permit transit along the length of a weld, but we will continue the example assuming a wedge without chamfers.
Now, plug D (168.3 mm) and A (23 mm) into the left and right sides of the equation:
left side: 168.3 right side: [ (23 x 23) / 2.87 ]
168.3 < 184.3
The left hand side is less than the right hand side, so we definitely need to contour our wedge.
How Much Do I Contour?
This part is even easier. Simply refer to the tables in T-432.2 (b). Here's one below for examination from the OD:
On a 6-5/8 in. OD pipe, we land on the 2nd row in the table. There can be only an increase in contour diameter, so the wedge must be contoured greater than or equal to the pipe OD (6-5/8 in.), but less than 8-5/8 in. Note: a contoured wedge might be suitable for a smaller diameter pipe than what it matches, but can never be used on a larger pipe because of the potential gap at the sound entry point.
Can You Just Make Me a Table?
Here you go, just follow these two steps:
1) Find your wedge dimension (some common Olympus wedges are already waiting for you). If you’re scanning a circumferential weld, start on the left side of the table and find your wedge width. If you’re scanning circumferentially, look on the right side and find your wedge length. If your wedge is chamfered, be sure to use the footprint dimension (flat part on the bottom, not counting the chamfers).
2) Find your pipe diameter at the top, then follow it down and match up with your wedge. If you land in the grey, you need to contour. If you land in the white, you can use the wedge flat.
Click here to download the full PDF
Conclusion
This is my interpretation of T-432.2. I hope you find it useful. I find pictorial things like this much more helpful than paragraphs upon paragraphs of obtuse and legal-sounding mumbo jumbo. Just imagine how many trees we saved!
Disclaimer
Follow your procedure. If this isn’t in your procedure, consult your Level 3.
Manager
5 年I have some thoughts on this. There are wedges that are irrigated making them much wider than an non-irrigated wedge. So, if A is purely determined by the width of the wedge then irrigated wedges need will require more contouring then non irrigated wedges regardless the width of the transducer. Now we have other wedges that have Irrigation additions, are these considered part of the wedge as they just connect to the sides but are not integral. If not then what is the difference between integral irrigation and non-integral irrigation and to what extent does adding to the wedge for irrigation and cooling affect the need for curvature? As to the chamfer on wedges how does that affect ID curves? Now that I have asked all these questions, I would like to say that my answer is as follows For OD curvature: that “A” should be the the transducer itself if the wedge has curvature stabilization feet as that will stop the rocking of the transducer. If the wedge plus anything attached to it that is rigidly affixed without a means of stabilizing the rocking, then the whole mechanism should be taken into account. For ID curvature: anything extending the pane of contact with the scanning surface of the pipe must be contoured or the transducer and wedge will not contact the scanning surface. I have added this into my procedure at it makes things a little more practical. I realize that the code was written to be more straightforward, but I believe this is the intent of the section. To account for proper contact based on a transducer and reduce the loss of signal from rocking.
Freelancer QC INSPECTOR for Samsung, KBR, CB&I, TR, MHI BOILER, ECA FLOUR, Local companies like MTC, KAPADIA CONSULTANCY, GLOBAL INSPECTION SERVICES
6 年good
Freelancer QC INSPECTOR for Samsung, KBR, CB&I, TR, MHI BOILER, ECA FLOUR, Local companies like MTC, KAPADIA CONSULTANCY, GLOBAL INSPECTION SERVICES
6 年giid
Physics Eng. and CMO at Medscint | Innovative Dosimetry Solutions
6 年Nice document!