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  1. EN13480-3 Tee Branch Reinforcement Pad Calculation

    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 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 find attached my detail calculation. (Terna Reinforcement pad Calculation30052017.pdf)
  2. EN13480-3 Tee Branch Reinforcement Pad Calculation

    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 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 find attached my detail calculation. (Terna Reinforcement pad Calculation30052017.pdf)
    Attached Thumbnails Attached Files