 # Pipes under internal pressure load

### Technical Article

001513

4 April 2018

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

For the sake of traceability, the following example is used.

The tube is considered to be closed on both sides. As a result, on the one hand, a pressure is exerted perpendicular to the inner "lid surface".

The resulting force must in turn be absorbed by the pipe wall. This results in a longitudinal stress that can be calculated as follows:
${\mathrm\sigma}_\mathrm l\;=\;\frac{\mathrm P\;\cdot\;\mathrm r_\mathrm i^2}{\mathrm r_\mathrm e^2\;-\;\mathrm r_\mathrm i^2}$
With
r i , r e = 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 that can be determined by the following formulas:

${\mathrm\sigma}_\mathrm t\;=\;\mathrm P\;\cdot\;\frac{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{\mathrm r}}\right)^2\;+\;1}{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{{\mathrm r}_\mathrm i}}\right)^2\;-\;1}$

${\mathrm\sigma}_\mathrm r\;=\;-\mathrm P\;\cdot\;\frac{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{\mathrm r}}\right)^2\;-\;1}{\left({\displaystyle\frac{{\mathrm r}_\mathrm e}{{\mathrm r}_\mathrm i}}\right)^2\;-\;1}$

With
r = radius in the limits r i ≤ r ≤ r e

It can be seen that the stresses depend on the considered radius r. This implies, conversely, that these run unevenly across the cross section. For thin-walled tubes (outer diameter / inner diameter <1.2), however, a uniform stress distribution can be assumed. This results in the mean tangential or radial stress to:
$\begin{array}{l}{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;-\;{\mathrm r}_\mathrm i}\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;+\;{\mathrm r}_\mathrm i}\end{array}$

Inserting the input values into the formulas yields the following voltages:
$\begin{array}{l}{\mathrm\sigma}_\mathrm l\;=\;\frac{2\;\cdot\;103,25^2}{109,55^2\;-\;103,25^2}\;=\;15,9\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{2\;\cdot\;103,25}{109,55\;-\;103,25}\;=\;32,8\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{-2\;\cdot\;103,25}{109,55\;+\;103,25}=\;-1,0\;\mathrm N/\mathrm{mm}²\end{array}$

From the internal pressure also follows a change in length of the tube. Generally speaking, the change in length is equal to the product of the length with the elongation epsilon:
ΔL = L ∙ ε

The elongation of the tube results from the three voltages 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 manually calculated results can also be reproduced in RFEM (deformation) or the steel design modules (stresses).

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 elongation of pipes, but also the expansion of pipe bends, you have to use the additional module RF-PIPING.

#### literature

  Franke, W .; Platzer, B .: Pipelines Basics - Planning - Assembly. Munich: Carl Hanser, 2014  Wikipedia: kettle formula 