**A reinforcement pad or RF pad is used to strengthen the fabricated or welded pipe branch connection joints. Standard branch fittings such as Tee and Olets Fittings are generally used to take out the branch from the header or runner pipe, but these standard fittings are not easily available and economical for a large size range.**

In such cases, we are required to make the fabricated branch connection (stub-in or stub-on). Fabricated branch connections are not as strong as the standard branch fittings, so to recover this strength gap between the standard branch fittings and fabricated branch joints, it is required to use the reinforcement pad or RF pad. But, it is not compulsory that all the fabricated branch connection joints are required a reinforcement pad.

**In this article, we will learn, how to decide whether the reinforcement pad is required or not for the size combination of the header and branch pipe? What should be the thickness and dimensions of the RF pad or reinforcement Pad?**

Table of Contents

**Fabricated Branch Connection Nomenclature**

A pipe having a fabricated or welded branch connection is weakened by cutting out some area from the header pipe to fit the branch pipe onto it.

Unless the wall thickness of the pipe is sufficient to sustain the pressure, it is necessary to provide added reinforcement pad. The amount of reinforcement or strength required to sustain the pressure is determined in accordance with **para. 304.3.3 or 304.3.4 of ASME B31.3.** Let’s see the nomenclature as per para 304.3.3 below-

**Where,**

**A _{1} =** Area removed from the header or run pipe

**A**Extrathickness available in the header pipe

_{2}=**A**Extrathickness available in the branch pipe

_{3}=**A**Area of RF Pad & area of welding

_{4}=**L _{4}** = Height of the reinforcement zone

**β**= Angle between axes of the branch and header or run pipe

**d _{1}**

**=**Effective length removed from the run or header pipe

**d**

_{2}**=**Half-width of the reinforcement zone

**D**Outside diameter of branch pipe

_{b}=**D**Outside diameter of run or header pipe

_{h}=**T _{h }= T_{b} =** Thickness after deducting mill tolerance from nominal pipe thickness or ordering thickness of run and branch pipe respectively

**t_{h} = t_{b} = **Calculated pipe thickness as per the

**equation of**

**para. 304.1 of ASME B31.3**for run and branch respectively.

**C =** Corrosion allowance**T _{r} = **Minimum thickness of reinforcement pad

**Formulas used for Reinforcement Pad Calculation**

To decide, whether the reinforcement pad or RF pad is required or not, the following formulas are used-

**Inputs Required for RF Pad Calculation**

**Run pipe material-** A672Gr.B**Branch pipe material-** A106 Gr.B**Run pipe type-** Welded**Branch pipe type- **Seamless**D _{h}–** 20″ = 508 mm

**D**14″ = 355.6 mm

_{b}–**β-**90°

**Design Pressure- **503 Psi**Design Temperature-** 400°C**Mill tolerance for run or header pipe- **±0.3 mm**Mill tolerance for branch pipe- **±12.5% mm**Corrosion allowances for run pipe (C)-** 3 mm**Corrosion allowances for branch pipe (C)-** 3 mm

**Reinforcement Pad Calculation Steps**

Follow the below steps to get the result:

**Step 1: Calculate the Run and Branch Pipe Thicknesses**

I have already written an article on Pipe Thickness Calculation. **Click here to see the calculation steps**

After following the steps mentioned in above articles, we have got the following value

**t _{h }=** 11.83 mm

**t**7.06 mm

_{b}=Now, we can get **T _{h }& T_{b}** from the above value

**T**16-0.3 = 15.70 mm

_{h }=**T**= 15.09×0.875 = 13.20 mm

_{b}**Step 2: Find d**_{1}

_{1}

Put the required values in **equation (2)** of formulas section

**d _{1 }** = [355.6 – 2(13.20 – 3)] / Sin 90° =

**335.20 mm**

**Step 3: Calculate A**_{1}

_{1}

Put the required values in **equation (1)**

**A _{1}**

_{ }= 11.83 × 335.20 (2 – Sin 90°) =

**3965.41**

**mm**

^{2}**Step 4: Find d**_{2}

_{2}

Put the required values in **equation (4)**

d_{2 } = Max[335.20 ; {(13.20-3)+(15.70-3)+(335.20/2)}]^{} = Max[335.20 ; 190.50]**d _{2 } = 335.20 mm**

**Step 5: Now, Calculate A**_{2}

_{2}

Put the required values in **equation (3)**

**A _{2}**

_{ }= [(2×335.20)-335.20)(15.70-11.83-3)]=

**291.62 mm**

^{2}**Step 6: Find L**_{4}

_{4}

Put the required values in **equation (6)**

L_{4 } = Min[{2.5(15.70-3)} ; {2.5(13.20-3)+0}] (Considering T_{r} = 0)^{} = Min[31.75 ; 25.5]**L _{4 } = 25.50 mm**

**Step 7: Now, Calculate A**_{3}

_{3}

Put the required values in **equation (5)**

**A _{3}**

_{ }= 2×25.5(13.20-7.06-3)/Sin90° =

**160.14 mm**

^{2}**Step 8: Find t**_{c}

_{c}

Put the required values in **equation (8)**

t_{c }= Min [0.7×15.09 ; 6]

= Min [10.56 ; 6]**t _{c }= 6 mm**

**Step 9: Now, Calculate A**_{4}

_{4}

Put the required values in **equation (7)**

**A _{4}** = 2×6

^{2 }=

**72 mm**

^{2}**Step 10: Check if A**_{1}≥A_{2}+A_{3}+A_{4}

_{1}≥A

_{2}+A

_{3}+A

_{4}

Add ** A_{2}+A_{3}+A_{4}** = 291.62+160.14+72 =

**523.76**

**mm**and we have A

^{2}**=**

_{1 }**3965.41**

**mm**

^{2}**Now, we can see A _{1}>A_{2}+A_{3}+A_{4}**

**Therefore, it is clear that we need to provide an RF pad to strengthen the joint.**

**Important Note:**

**If A _{1}<A_{2}+A_{3}+A_{4} then RF Pad or Reinforcement Pad is not needed to provide**

**Reinforcement Pad Thickness Calculation**

In most cases, RF Pad thickness is the same as the header or run pipe thickness, but we need to cross-check whether the reinforcement pad can be made from the run pipe.

So, Let’s consider RF Pad is taken from the header with the outer diameter (D** _{r}** ) = 2D

**=**

_{b}**711.20 mm**and thickness (T

**) = T**

_{r }**=**

_{h}**15.70 mm**

**Note: RF Pad diameter is taken as two times the outer diameter of the branch pipe.**

**Step 1: Calculate L**_{4} again considering RF Pad is already provided

_{4}again considering RF Pad is already provided

**L _{4 } = Min[{2.5(T_{h}-C)} ; {2.5(T_{b}-C)+T_{r}}]**

L

_{4 }= Min[{2.5(15.70-3)} ; {2.5(13.20-3)+15.70}] = 291.62 mm

^{2}= Min[31.75 ; 41.12]

**L**

_{4 }= 31.75 mm**Step 2: Now, Calculate A**_{3} considering RF Pad

_{3}considering RF Pad

**A _{3} = 2×L_{4} (T_{b}-t_{b}-C)**

= 2×31.75 (13.20-7.06-3)

**A**

_{3}= 199.39 mm^{2}**Step 3: Calculate A**_{4} considering RF Pad

_{4}considering RF Pad

**A _{4} = [{2×d_{1} – (D_{b} / Sinβ)}(T_{h})+2t_{c}^{2}]**

= [{2×335.20 – (355.6/sin90°)}(15.70)+(2×6

^{2})]

**A**

_{4}= 5014.36 mm^{2}**Step 4: Now Check if A**_{1}<A_{2}+A_{3}+A_{4}

_{1}<A

_{2}+A

_{3}+A

_{4}

**We have from the old calculation,****A _{1}_{ }**=

**3965.41**

**mm**

^{2}**A**=

_{2}**291.62 mm**

^{2}**We know from the latest calculation,****A _{3}= 199.39 mm^{2}**

**A**

_{4}= 5014.36 mm^{2}Now, let’s check the condition below

IF **A _{1}**<

**A**+

_{2}**A**+

_{3}**A**= 3965.41<291.62+199.39+5014.36

_{4}**3965.41< 5505.37**

**We can see from the above calculation A _{1}<A_{2}+A_{3}+A_{4}**

The above result ensures that a reinforcement pad can be made from the header or run pipe for our case.

If still **A _{1}**≥

**A**+

_{2}**A**+

_{3}**A**then we need to use a higher thickness pipe for reinforcement pad or RF pad.

_{4}We need to continue the calculation with the increased thickness of the pipe until the condition “**A _{1}**<

**A**+

_{2}**A**+

_{3}**A**” met.

_{4}I hope, now you can easily do the calculation for Reinforcement Pad. Do practice once or twice with other examples to clear all the doubts. Thanks for reading.

**You may also like to read**

Pipe Wall Thickness Calculation For External Pressure or Vacuum

A Presentation on Heat Exchanger

Some Important Piping Codes and Standards

Pipe Welding Positions: 1G, 2G, 5G, and 6G

Hydrotest Procedure for Piping system

Pipe Rack Design and Calculations

**Image Source**

ASME B31.3

http://www.copcosa.com/soft/mecheng07/rob3.htm

THIS RF PAD CALCULATION IS REALLY VERY HELPFUL FOR ME, IT HAS CLEARED MY DOUBT. THANK YOU.

Thanks, Ravi

Your article is very simple.

Thanks for the read Anthony

hello dear Rehan

Thank you for sharing this article.

Please help me about two tiny problems!

1- Is the parameters t_h and t_b used in above formulas include c (corrosion and mechanical allowances)? i.e. t_h=PD/(SEW+PY)+c? (the 4″ pipe in my case include a 2mm depth groove and 0mm corrosion allowance)

sincerely.

Hello Mr. Mehrshad,

Thanks for your reading.

No, t_h does not include corrosion and mechanical allowances. The below equation shall be used:

t_h=PD/(SEW+PY)

in above case 14″ th=6.82 with E,W=1and y=0.4 … But tb you have got 7.06mm

s=12912psi

Can we make RF pad in split form?

yes you can

Sir, calculated thickness of branch tb=7.06mm is not correct, kindly cross check it once

how to obtain tb = 7.06 mm? fromclause 304.1.1 t=PD/(2SEW+2PY) … E,W=1 and y=0.4 ? could you please clarify

Hello, Great article!

I have a question about calculating the RF pad thickness. In step 3, when calculating A4, do you use the Header thickness (Th) or is this the RF pad thickness (Tr)? I understand that in this example Th=Tr however, if I put the Tr value at 0 it still says A1<A2+A3+A4 which contradicts the theory that an RF pad is required per the previous equation?

Thanks in advance.

How we get Th and Tb values as 16 and 15.09mm respectively??

Please tell the stress and Y value in this case