# Pipes Under Internal Pressure Load

### Technical Article

001513 04/04/2018

Pipe systems are exposed to a large number of loadings. Internal pressure is one of the most governing loads. Therefore, this article describes the stresses and deformations resulting from a pure internal pressure load in the pipe's wall and for the pipe respectively.

For better understanding, the following example is used.

The pipe is considered to be closed on both sides. On the one hand, the pressure is applied perpendicular to the internal "surface area of the cover plate".

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

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

It becomes obvious that the stresses depend on the considered radius r. The reverse conclusion is that those are non-uniformly distributed over the cross-section. However, for thin-walled pipes (outside/inside diameter < 1.2) a uniform stress distribution can be assumed. So, the mean tangential and radial stress are the following:
$\begin{array}{l}{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;-\;{\mathrm r}_\mathrm i}\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{\mathrm P\;\cdot\;{\mathrm r}_\mathrm i}{{\mathrm r}_\mathrm e\;+\;{\mathrm r}_\mathrm i}\end{array}$

Setting the initial values into the formulas, we get the following stresses:
$\begin{array}{l}{\mathrm\sigma}_\mathrm l\;=\;\frac{2\;\cdot\;103.25^2}{109.55^2\;-\;103.25^2}\;=\;15.9\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm t\;=\;\frac{2\;\cdot\;103.25}{109.55\;-\;103.25}\;=\;32.8\;\mathrm N/\mathrm{mm}²\\{\overline{\mathrm\sigma}}_\mathrm r\;=\;\frac{-2\;\cdot\;103.25}{109.55\;+\;103.25}=\;-1.0\;\mathrm N/\mathrm{mm}²\end{array}$

A change in the pipe's length also results from the internal pressure. Generally speaking, the change in length is equal to the product of the length and the epsilon strain:
ΔL = L ∙ ε

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

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

For the deformation it is important to activate the Bourdon effect in the general calculation parameters of RFEM. If you don't want to consider only the axial strain of pipes but also the stretching of pipe bends, use the add-on module RF-PIPING.

#### Reference

 [1] Franke, W.; Platzer, B.: Rohrleitungen Grundlagen - Planung - Montage. München: Carl Hanser, 2014 [2] Wikipedia: Barlow's formula

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