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Fatigue calculation problems for expansion joints

2021-01-26View Original

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As shown in the GB16749 expansion joint standard, when the allowable stress of the material at the design temperature is greater than 2 times the yield strength of the bellows material at the same temperature, a fatigue life calculation is required. How should this be understood? I don’t understand; at the same temperature, shouldn’t the yield strength generally be greater than the allowable stress? Why does the allowable stress exceed 2 times the yield strength? Solve
Reply #22021-01-26
This post was last edited by Shutong on 2021-1-26 at 22:40. The standards for calculating the fatigue life of expansion joints are as follows: EN13445, ASME EJMA, GBT 12777, GBT 16749. Among them, the EJMA in the United States is the most authoritative. Due to the different calculation methods specified in GBT 16749, for austenitic stainless steel expansion joints, the approach suggested by the original poster – that is, using an allowable stress greater than twice the yield strength – is based on extensive experimental data; as temperature rises, the fatigue life decreases. Don’t doubt it – the calculation of expansion joints is extremely complex; there are spherical elements and ring plates involved, and stress analysis requires high precision. But many of us don’t know how to use these tools or how to create models; it’s difficult! !
Reply #32021-01-27
Your reply doesn’t address the question I asked; I don’t understand why the allowable stress could be greater than the yield strength
Reply #42021-01-27
It’s fine if you don’t understand; you also lack even the basic respect.
Reply #52021-01-29
For pipeline stress analysis, 2 times the yield strength serves as a boundary for stability analysis. It is not the allowable stress that must be greater than the yield strength, but rather the calculated stress; when this value exceeds 2 times the yield strength, the system is theoretically not in a stable state during repeated cycles of heating and cooling. Additional plastic deformation occurs with each cycle, and over time this ultimately leads to failure. As for the allowable stress being greater than the yield stress, that does not exist. The allowable stress is obtained by dividing the material’s ultimate strength by the safety factor. I think you misread it.
Reply #62021-01-30
It’s indeed my fault for not looking carefully; that explanation referred to the range of normal stress in the meridional direction. What it should have described was a comparison between the total stress on the equipment and the allowable stress of the expansion joint material at its designed temperature
Reply #72021-02-01
This post was last edited by llu85 on 2021-2-1 at 18:29. Please refer to point 7.6.3: δt does not denote the allowable stress; it represents the total stress range in the meridional direction of the bellows (the actual stress). It refers to the actual maximum stress in the bellows when it is subjected to internal pressures and external loads such as displacement. δ3 is the meridional thinning caused by pressure; δ4 is the meridional bending caused by pressure; δ5 is the meridional thinning caused by displacement; δ6 is the meridional bending caused by displacement. All of these represent the forces acting in actual situations, and through certain mathematical calculations, the total stress value δt is obtained.
Reply #82021-04-06
Where are you looking? That symbol over there represents the allowable stress of the material at the design temperature

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