Pipes subjected to internal pressure

Technical Article

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Piping systems are exposed to a variety of loads. One of the most decisive is the internal pressure. This article will therefore deal with the stresses and deformations resulting from a pure internal compression load in the pipe wall or for the pipe.

The following example is used for traceability.

Figure 01 - System

The pipe is considered as closed on both sides. Thus, on the one hand, a pressure is applied perpendicularly to the inner "cover surface".

Figure 02 - Longitudinal stress

The resulting force must in turn be absorbed by the pipe wall. This results in a longitudinal stress, which can be calculated as follows:
${\mathrm\sigma}_\mathrm l\;=\;\frac{\mathrm P\;\cdot\;\mathrm r_\mathrm i^2}{\mathrm r_\mathrm e^2\;-\;\mathrm r_\mathrm i^2}$
where
ri , re = inner and outer radius

On the other hand, the internal pressure acts perpendicular to the pipe inner wall. This results in a tangential and radial stress, which can be determined using the following formulas:
${\mathrm\sigma}_\mathrm t\;=\;\mathrm P\;\cdot\;\frac{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{\mathrm r}}\right)^2\;+\;1}{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{{\mathrm r}_\mathrm i}}\right)^2\;-\;1}$
${\mathrm\sigma}_\mathrm r\;=\;-\mathrm P\;\cdot\;\frac{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{\mathrm r}}\right)^2\;-\;1}{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{{\mathrm r}_\mathrm i}}\right)^2\;-\;1}$
where
r = radius in the limits ri ≤ r ≤ re

It becomes apparent that the stresses depend on the considered radius r. Conversely, this means that they run unevenly over the cross-section. For thin-walled pipes (outer diameter/inner diameter <1.2), however, a uniform stress distribution can be assumed. Thus, the mean tangential or radial stress is obtained as follows:
$\begin{array}{l}{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;-\;{\mathrm r}_\mathrm i}\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;+\;{\mathrm r}_\mathrm i}\end{array}$

Inserting the input values into the formulas results in the following stresses:
$\begin{array}{l}{\mathrm\sigma}_\mathrm l\;=\;\frac{2\;\cdot\;103.25^2}{109.55^2\;-\;103.25^2}\;=\;15.9\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{2\;\cdot\;103.25}{109.55\;-\;103.25}\;=\;32.8\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{-2\;\cdot\;103.25}{109.55\;+\;103.25}=\;-1,0\;\mathrm N/\mathrm{mm}²\end{array}$

The internal pressure also results in a change in length of the pipe. In general terms, the change in length is equal to the product of the length with the strain epsilon:
ΔL = L ∙ ε

The strain of the pipe results from the three stresses just calculated:
$\mathrm\varepsilon\;=\;\frac{\left({\mathrm\sigma}_\mathrm l\;-\;\mathrm v\;\cdot\;\left({\mathrm\sigma}_\mathrm t\;+\;{\mathrm\sigma}_\mathrm r\right)\right)}{\mathrm E}\;=\;\frac{\left(15.9\;-\;0.3\;\cdot\;\left(32.8\;-\;1\right)\right)}{212,000}\;=\;3\;\cdot\;10^{-5}$
The change in length is therefore:
ΔL = 10,000 mm ∙ 3 ∙ 10 -5 = 0.3 mm

The results calculated manually can also be reproduced in RFEM (deformation) or in the steel design modules (stresses).

Figure 03 - Result

Important for the deformation is the activation of the Bourdon effect in the global calculation parameters of RFEM. If you want to consider not only the axial strain of pipes but also the expansion of pipe bends, you can use the RF-PIPING add-on module.

Literature

[1] Franke, W .; Platzer, B .: Piping Basics - Planning - Assembly. Munich: Carl Hanser, 2014
[2] Wikipedia: Boiler formula

 

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