Where does the pressure thrust of a bellows come from?
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When calculating the thrust on pipe supports, the thrust associated with bellows expansion joints includes the pressure thrust exerted by the bellows. I’ve looked up relevant information, but what’s available isn’t comprehensive. Some people on the internet say that this thrust is caused by the resistance of the pipe to the fluid flow; if that’s the case, then using formulas from fluid dynamics, the thrust should be calculated as F=ΔP*A. However, the information I found states that F=P*A, using only the operating pressure rather than the pressure difference. In some books, it seems that the formula is applicable only to pipes with closed ends, but it’s still used in cases where the ends aren’t closed. I hope experts can help me out; I’d be extremely grateful:handshake
Has this problem been solved? I have the same problem as well
It is the structure and stiffness of the expansion joint itself that prevent it from absorbing axial tension; the axial force is caused by pressure, not by a pressure difference. Think of an expansion joint as a balloon that is being inflated.
I don’t understand either; can someone who knows more explain it?
Let’s use a practical example to illustrate the force acting on a bellows blind plate: 1 Assume that a free-type bellows expansion joint has been manufactured and is now subjected to strength testing. First, both ends of the expansion joint are sealed, and then pressure is applied. During this pressurization process, it can be observed that the wavelength of the bellows gradually increases until the bellows become straight – clearly, this is not the desired outcome. 3 To ensure that the bellows maintains its basic shape during the testing process, supports are used to hold both ends of the expansion joint; in this way, the forces that cause deformation of the bellows are absorbed by the supports. The maximum force acting on the bellows is equal to the product of the internal pressure and the cross-sectional area of the bellows, which corresponds to F=P*A as described by the original poster. 4 The same principle applies to expansion joints in operation. 5 Please note one premise: for expansion joints subjected to axial loads by tie rods, hinges, etc., no external devices are required to handle the force from the blind flanges.
There are detailed explanations and examples on EJMA.