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How can I find the change in diameter with time?

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Hi all, I modelled a 3D model of a pressure tube to find the effect of creep on it.

I defined the boundary load, internal pressure, and a creep law.

Now I want to find the change in diameter of the pressure tube with time due to creep at a particular location (along length) of the pressure tube.

How can I find this. Please suggest me.

Thanks


2 Replies Last Post 2023年9月28日 GMT+2 06:53

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Posted: 11 months ago 2023年9月27日 GMT+2 19:49
Updated: 11 months ago 2023年9月27日 GMT+2 19:38

It depends on the geometry, boundary conditions and material properties. If it's a simple cylinder where at a given axial location the displacements are axysymmetric, then you can just get the displacement normal to an edge, which will be equal to your change in radius and multiply it by 2 to obtain the change in diameter. Image 1. A more general approach will be to create a parametric 3D curve snaped to the boundary of your interest (Image 2) and then get a line average of the radial displacement (image 3). If the center of this pipe moves due to external loads or uneven pressure distribution, then first you would need to calculate the average center of the parametric curve, and then calculate the average radius by subtracting those coordinates from sqrt((x-x0)^2+(y-y0)^2).

It depends on the geometry, boundary conditions and material properties. If it's a simple cylinder where at a given axial location the displacements are axysymmetric, then you can just get the displacement normal to an edge, which will be equal to your change in radius and multiply it by 2 to obtain the change in diameter. Image 1. A more general approach will be to create a parametric 3D curve snaped to the boundary of your interest (Image 2) and then get a line average of the radial displacement (image 3). If the center of this pipe moves due to external loads or uneven pressure distribution, then first you would need to calculate the average center of the parametric curve, and then calculate the average radius by subtracting those coordinates from sqrt((x-x0)^2+(y-y0)^2).


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Posted: 11 months ago 2023年9月28日 GMT+2 06:53

Thanks Luis for your response. This will help me.

Thanks Luis for your response. This will help me.

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