Lateral Buckling of Axially Constrained Pipelines
Authors: Palmer A.C. , Baldry J.A.S. 
Published: December 2007

Introduction
Imagine a length of straight uniform pipeline, rigidly attached to a rigid anchor block at each end, and initially unstressed. If an internal pressure is applied, can this pressure cause the pipe to buckle sideways?Although a correct answer to this question follows from a simple argument, it is nevertheless the subject of dispute among pipeline engineers. A simple but misleading argument goes like this: pressure induces a hoop tensile stress around the pipe, and if it were unconstrained it would contract along its length (because Poisson's ratio is positive). Since the anchor blocks resist this contraction, a longitudinal tensile stress is induced. Since the pipe wall is in longitudinal tension, the pipe cannot buckle laterally.It is the final step in this argument that is incorrect. This article sets down reasoning that leads to the correct conclusionthat the pipe can buckleand compares the predicted buckling pressure with experiment.

The Critical Pressure
An elementary argument shows that the longitudinal stress set up in a straight thin-walled axially constrained pipe by an internal pressure, p, is vpD/2T, where v is Poisson's ratio, D is diameter, and T is wall thickness. The resultant force over a complete cross-section perpendicular to the tube axis is (pi/4)D2p (1 - 2v), compressive since v less than 1/2, the difference between the compressive force (pi/4)D2p carried by the fluid within the pipe and the tensile force (vpD/ 2T) (piDT) carried by the wall of the pipe. The pipe can then be thought of as a column with clamped ends carrying this resultant compressive force, and buckling can be expected to occur when the force reaches the Euler buckling load. If Young's modulus is E, the flexural rigidity is phi ED3T/8; by this argument, buckling will occur when
where L is the length of the pipe; that is, when
The above argument leads to a correct value for the buckling pressure, but leaves a certain uneasiness in one's mind, because an instinctive feeling that the axial compressive force must "follow" the buckling pipe seems to conflict with the fact that the ends are pipe seems to conflict with the fact that the ends are fixed. An alternative argument avoids this difficulty, and makes clearer what the force pushing the pipe sideways is.In the above argument the pipe wall itself and the contained fluid within it were considered together. Instead, consider them separately. Fig. 1a shows the forces and moments acting on an element ds of the pipe wall, and Fig. 1b shows the forces acting on the pipe wall, and Fig. 1b shows the forces acting on the fluid contained within the element. The resultant force exerted by the contained fluid pressure on the pipe wall is Rds, and psi denotes the inclination of the deflected pipe to its original line. Since the fluid element must be in equilibrium, resolving forces perpendicular to the pipe axis in Fig. 1b, so that
In words, the contained fluid exerts a lateral force on a deflected pipe; the magnitude of the force per unit length is the pressure multiplied by the cross-section multiplied by the curvature, and it acts towards the outside of the curve. If one analyzes buckling by examining the pipe deflected from its initial position, one must not forget the existence of this force (which is of course well known in other contexts).
JPT
P. 1283