Hello,

I am doing the branch tee reinforcement pad requirement calculation according to EN 13480-3 section 8.4.3. I have some queries as below.

(1) In Equation 8.4.3-4 ( lpl <= ls). I have calculated the effective reinforcement of shell pipe (ls) value according to equation 8.4.1-2 which is coming around 190 mm. So According to the equation I can use the reinforcement pad width less than the term ls means less than 190 mm.

If I follow the above equation, It will never satisfy the final equation 8.4.3-6. It seems this equation is not correct.

If code restrict the reinforcement pad width how we can satisfy the final equation without increasing the pad width where reinforcement pad width is also limited to eas according to equation 8.4.3-5. Refer attached code page attachment (EN13480-3-Page-82.pdf). Please suggest.

(2) EN13480-3 equation 8.4.3-5 condition eapl<=eas shall be satisfy.
I am using the reinforcement pad thickness same as shell thickness* means pad will be cut from shell pipe during fabrication.

I am using the nominal pad thickness same as shell thickness 16 mm with assuming same corrosion allowances. Refer below calculation.
My nominal shell thickness en=16 mm. c0=3.4* c1=0.3 mm.so eas=16-3.4-0.3= 12.3 mm.
Plate thickness used enpl=16mm. c0=3.4* c1=0.3 mm. so eapl=12.3 mm.

Is my calculation & understanding is correct.

As in brief above code restrict the reinforcement pad thickness as well as width to compensate the cutting area for branch.

Please download my detail calculation pdf and code page at below mention path for your reference.

Reinforcement Pad calculation EN13480-3


EN13480-3, Code Page

Hi,

I am waiting for your comments.

A few additional question
What material You used to calculation ???
I can not find it on Your Sheet

How You calculated a design stress ??

Why You assumed joint coeficient Z=0.85

If You have a seam pipe the pipes in standard are calculated with Z=1

Hi Adrian,

Pipe material is S235JR (1.0038) welded plate material (Properties as per EN1025-2:2004).

Design stress is calculated as per EN13480-3 equation 5.2.1-1 as below.

f= min(Reht/1.5 or Rp0,2t/1.5; Rm/2.4) (For steel other than austenitic steel)

Where Rm=360 N/mm2
Reht=235 N/mm2
Rp0,2t=205 N/mm2 @20degc

f=min(235/1.5 or 205/1.5, 360/2.4)
=min(156.6 or 136.6, 150)= 136.6 N/mm2
Joint factor=0.85
So design stress =f*Z=136.6*0.85= 116.11 N/mm2

Weld joint coefficient (Z) is considered 0.85 according to EN13480-3 section 4.5 (Joint Coefficient) and EN13480-5 section 8.3 table.
______________________________________________________________________
4.5 Joint coefficient (EN13480-3)

The joint coefficient z shall be used in the calculation of the thicknesses of components which include one or
several butt welds, other than circumferential, and shall not exceed the following values:

 for equipment subject to destructive and non-destructive testing which confirms that the whole series of
joints show no significant imperfections: 1;

 for equipment subject to random non-destructive testing: 0,85;""

_______________________________________________________________________
8.3 Longitudinal welds (EN13480-5)

Longitudinal welds, except those in components made to material specifications acceptable to this European
Standard, shall be subject to the level of NDT appropriate to the joint coefficient required (see Table 8.3-1).
Table 8.3-1 — Extent of NDT for longitudinal welds

Joint coefficient 0,7 < z <= 0,85
VT 100%
MT or PT 10 %
RT or UT 10 %
_______________________________________________________________________

thanks for Your explanation.

I know aobut this what You wrote above.

I wonder why Yoiu choose z=0,85 is more appropriate for vessels but of course standard give Us this possibility


Ok Now I have all neccessery information.

I will try to calculate it today.

Regards

You know that this material can be used only in 20°C maks

I check this and You calculate it correct
in my opinion of course
But when I assumed a reinforcing pad eapl=16 mm and a diameter less than ls in my case 2382 mm it is not enough.

Thanks for your efforts. I assumed reinforcement pad nominal thickness same as run pipe thickness 16 mm. But in calculation I used analysis thickness eapl 12.3 mm same as run pipe thickness (eas) to satisfy equation 8.4.3-5 (En13480-3).

You are right dia of reinforcement pad approx. 2382 mm is not enough.
Now my doubt why reinforcement pad width (lpl) is limited to only ls.
I can not satisfy the code equation without Increasing the pad width.

width (ls) in our case is only 180 mm. however AWWA code recommend pad width maximum 2 times of the branch OD can be used.

This equation in EN 13480-3 is a requirement to first discontinuity of material. I don't know why but this is obligatory.

But I checked the calculation again and if You increase a WT of branch pipe also to 16 mm from 12 mm it will be ok for assumed like above reinforcment pad

It is simple solution and the results are sufficient

Should I ask to the EN code community, why this equation is obligatory, since other codes like ASME, AWWA permits more width of pad to compensate branch cutting area.

In our case branch pipe and header run pipe is already ordered for the thickness 12 and 16 mm respectively. However the connection between header and branch is stub type not welding tee.

However with Increasing thickness of the branch 12 mm to 16 mm, still equation is not satisfy with ls value 180 mm. Can you share your calculation.Thanks

sumitgoel
You have right I assumed to my caluclation joint coeficient z=1.0 not
0.85 as You. So in my case it is enough sorry for confusion.

Can You explain me what is stub type connection.
It sie typical flange connection ???
Or maybe something different ??

Yes sorry I checked my calculation with a joint coeficient Z=0.85 so You had right. In this case the WT 16 mm for nozzle doesn't change anything.
Still is not sufficient for this two big diameters.

Stub type means unreinforce fabricated tee.
If you see In caesar- SIF & Tees option Type-2 (Unreinforced).

SIF will be calculated according to EN13480-3, Table H.1 type-6.
Flexibility factor(h) =2en/dm
SIF=0.9/((h)^2/3)

A okey. For me is clear
I send You mail sumitgoel.
I don't receive any answer but I hope that You received it.

You have a problem with this big tee !!
In my opinion You should try to change joint coeficient from z=0.85
to z=1.0
I know that this change have a consequency in NDT Amount but You can have a smaller WT.
Maybe You should change also a corrosion allowance to 1 or 0.5 mm becasue the problem is that the Straight pipe has a 69 percentage of the allowed load so when You weaken it by opening it is simple that This caluclation will be failed.
Maybe You can use some special software like FEA.
I can't help you in this case becasue I don't have it.

Thanks. I received your calculation. I am checking your calculation and revert to you soon.