PE Pipe Technical Catalogue (PDF) - Pipelife Norge AS

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Pipelife Norge ASPE CATALOGUE-SUBMARINE APPLICATIONS, PIPELIFE NORGE AS, December 2002.A.3 Static designIn this chapter we will present formulas to decide the wall thickness of the pipe taking into accountinternal and external forces acting on the pipeline.The internal diameter of the pipe is decided by the formulas in chapter A.2.We will underline that quite a few of these calculations are necessary to carry out in a realproject. It is important to sort out the significant factors with respect to the pipe’s service life.A.3.1Internal pressureThe internal pressure will create stress in the pipe wall both in the hoop direction and the longitudinaldirection. The stress in the longitudinal direction is dependent on the way the pipeline is able tomove (fixed or free movement).A.3.1.1 Hoop directionFig. A.3.1.1.1 indicate the static system.Fig. A.3.1.1.1 Static system forinternal pressure, cut pipe.Nσ rp PD mNσ rSNo shear stress will occur due to the internal pressure. There will only be a tensile force (N) inthe ring direction.If we integrate the pressure components we find the following result based on equilibrium offorces :2 ⋅ N = p ⋅ D m A.3-1)N = tensile force (N)p = pressure (N/m 2 = Pa)D m = mean diameter (m)Introducing the ring stress (σ r ) and the wall thickness (s), we can develop the following formulas :N=σ r ⋅SA.3-2)p ⋅ D mσ r =A.3-3)2 ⋅ Sp ⋅ DmS = A.3-4)2 ⋅ σrSince D m = D – s s = p ⋅ D(2 ⋅ σ r + p)A.3-5)σ r = design stress (see table A.1.2)D = external diameterSide 28 av 84
Pipelife Norge ASPE CATALOGUE-SUBMARINE APPLICATIONS, PIPELIFE NORGE AS, December 2002.Example 1Find the wall thickness for a Ø 200 mm PE80 pipe exposed to a design pressure of 1Mpa (10 bar)Design safety factor = 1.6.Solution :The wall thickness (s) is found from formula A.3-5). σ is taken from table A.1.2s =1⋅0.2m(2 ⋅ 5 + 1)=0.0182 m= 18.2mmThe stress (σ) for a given pipe in the hoop direction exposed to a pressure (p) can be calculated fromthe formula :pDσ r = (SDR −1)where SDR=A.3-6)2sExample 2Given a PE100 pipe SDR 17.6 exposed for a pressure of 0.8Mpa (8bar). Calculate the stress in thepipe wall and the safety factor against burst after 50 years of loading ?Solution :Formula A.3-6) gives the hoop stress :0.8σ r = (17.6 −1)MPa = 6.64 MPa210Safety factor confer A.1-1) : C = = 1. 506.64A pipe is always exposed to additional forces besides the internal pressure, for instance temperatureforces, forces in bends and reducers, depth of backfill in trenches, water hammer, forces fromcurrent and waves, installation forces etc.You have to consider the factor of safety (design factor) taking into account these other forces.The method process is to calculate all acting forces and find the maximum combined stress.This is the method adopted in the consecutive chapters.A.3.1.2 Longitudinal directionFig. A.3.1.2.1 below shows the stress and strains for a pipe exposed to internal pressure.ε lσ l∆Lε rσ rQpQLFig. A.3.1.2.1 Pipe exposed to internal pressureThe internal pressure will give a deformation in the longitudinal direction if the pipe is free to move.The tube will try to be shorter due to contraction :Side 29 av 84
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