# Thread: Help required for determining for Code Stress in Caesar ii : Code: ASME 31.1

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## Help required for determining for Code Stress in Caesar ii : Code: ASME 31.1

My issue:

1. Pls refer the attached file "1.Node" so that you may get the idea of the system. - It is a plain piping system (OD=168.3) anchored at Node 10. with no temp./pressure. Only self weight acts causing bending and torsion.

2. Pls refer the stress "2. Stress table".
I have doubt in Code stress

**At node 10 & 18:
Node18:
Bending: 1552.8kPA, torsion: 78.8kPa, Maximum Stress Intensity = ((1552.8^2)+4*(78.8^2))^0.5=1560.8kPA
Code stress = same as Maximum Stress Intensity since because SIF =1 ? Am i right.

**At node 18:
Maximum stress intensity = (435^2+4*(78.8)^2)^0.5 = 462.7 kPA
SIF = 2.473
Therfore as per ASME B 31.1,
Stress = 0.75*i*(M/Z)
M/Z = Stress
Net Bending at node 18= 435-175.9=259.1kPA
Net torsion at node 18= 48.8-48.8 = 0

Therefore, Code stress = 0.75*2.473*259.1 = 480.56kPA
- But the Code stress in report is 438.2 kPA
- Pls explain how. I am mistaken somewhere.?

**Similary At node 19 = Unable to determine how did 205.8 come?

**But again at node 20, where torsion = 0, SIF=2.473
Code = 0.75*bending = 0.75*64.2 = 48.1 k PA
why SIF value is not used here?

I am also attaching the report file (ELB1.TXT) for your ref.

Regards,
Prabhjyot Singh

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