 #### Oval Pipe a Non-Linear Problem ________________________

by Andrew C. Whyte, BSc, MSc, CEng, M.I.Mech.E.
Web site: https://acwhyte.droppages.com

1. Scope.

This paper describes one particular non-linear stress analysis problem, that of a long straight uniform thickness out-of-round pipe or tube subjected to internal pressure. The solution of this problem has been available from 1936 and can be readily analysed on a microcomputer.

2. Introduction.

Most structural engineering is based on linear analysis. The reasons for this are not difficult to understand:

• Linear solutions are relatively simple compared to non-linear ones.
• The load cases can be superimposed and scaled in a linear manner.
• Linear solutions are taught first in the colleges and universities.

One of the most common problems encountered by engineers is that of a thin walled cylinder (pipe or tube) subjected to internal pressure. The maximum stress induced in the cylinder can be obtained from thin shell theory and is given by: ƒm = pR/t                                                 eq(1). where   ƒm is the circumferential (hoop) membrane stress
p is the internal pressure
t is the wall thickness

By thin shell we mean geometries with ratio R/t >=10 although the theory is often used at ratios below 10. Because the shell is thin the radius is generally taken as the mean radius although strictly it should be the inside radius.

The circumferential stress distribution varies across the wall thickness, the stress being higher on the inside than the outside. Provided the cylindrical shell is thin the stress distribution is essentially uniform and hence the stress, given by eq(1), is considered a membrane stress and the non-uniform part of the stress distribution can be ignored.
For example if we consider an 8" pipe of standard wall thickness subject to an internal pressure of 3.5 N/mm2. The outside diameter is 219.1 mm and the mean wall thickness is 8.18 mm. If the usual 12.5% thinning allowance is applied and there is a 3 mm corrosion allowance then:

• The wall thickness, t = 8.18 * 0.875 - 3 = 4.16 mm minimum
• The mean radius, R = (219.1 - 4.16)/2 = 107.47 mm
• The geometry ratio R/t = 107.47/4.16 = 25.83
• The circumferential membrane stress, ƒm = 3.5 * 25.83 = 90.4 N/mm2

The actual circumferential stresses at the inside and outside radius calculated by the more accurate Lamè equations are: ƒi = 90.45 N/mm2 and ƒo = 86.95 N/mm2 respectively. Hence for a thin cylindrical shell the use of eq(1) is reasonable.

If however the cylindrical shell is not quite circular but is out-of-round then it is possible to generate large circumferential bending stresses across the wall thickness in addition to the membrane stress. A feature of these bending stresses is that they are non-linear with pressure, i.e. as the pressure is increased (or decreased) the bending stresses do not scale in a linear manner. The purpose of this note is to briefly discuss the problem and to show how the problem can best be analysed on a microcomputer by use of program OvalPipe-1. The program is public domain. You can download the program and documentation [0.75MB]. The README text file should be read. To run the program double click the file OVALPIPE (or OVALPIPE.EXE).

3. Analysis.

A solution to the problem of an out-of-round thin wall cylinder subjected to internal pressure was investigated by Professor B. P. Haigh in 1936 (reference 1). In this solution the out-of-round cylinder becomes more nearly circular as it expands under tension and bending stresses induced by the internal pressure loading.

The tensile stress is the circumferential membrane stress ƒm given by eq(1) above.

The bending stress is given by: ƒb = 6*ƒm*(Xo/t)*(cos(h*q))/(1 + parameter P)              eq(2). where   ƒb is the circumferential bending stress on the surface of the pipe
parameter P = 12*p*R3*(1 - s2)/(E*(h2 - 1)*t3)
p is the internal pressure as before
R is the mean radius as discussed above
t is the wall thickness as before
s is Poisson's ratio
E is Young's modulus
Xo is the out-of-round amplitude
q is angle theta
h is number of lobes

The amplitude Xo and number of lobes h, specify the out-of-roundness. The most basic case is an oval where the number of lobes = 2.

The bending stress will vary, +ve tensile, -ve compressive, around the inside and outside surfaces of the pipe as a function of angle theta. This can best be seen by example. If we take the geometry and loading of the previous example but for an oval pipe, i.e. lobes = 2 and out-of-round amplitude = 1.1 mm (approximately 2% ovality).
For a carbon steel pipe with Young's modulus = 210000 N/mm2 and Poisson's ratio = 0.3 the results, from OvalPipe-1, are as shown in Figures 1 to 3 below.

• Figure 1 shows the input data and stress results. The circumferential bending stress ƒb = 70.11 N/mm2, i.e. an ovality of only 2% has resulted in bending stresses of magnitude approaching that of the membrane stress.
• Figure 2 shows an X-Y plot of the stress results around the circumference of an oval pipe. The bending stress fluctuates + or - with the maximum bending stress at theta = 0, 90, 180 and 270°.
• Figure 3 shows a polar plot of the stress results shown in Figure 2. This plot is useful for showing the effect of the lobes.

By including the membrane stress the results shown in Figures 4 and 5 are obtained.

These results show the inside and outside surface stresses = ƒm + ƒb. These are the total stresses around the oval pipe. For the example given, this results in all tensile stresses in the pipe as ƒm > max.|ƒb|. At low pressures however, when ƒm < max.|ƒb|, it is possible to have some compressive stresses. This is discussed in more detail in the next section.

4. Non-Linear Effects.

At first glance eq(2) seems straightforward but on closer inspection it is apparent that the stresses are not linear with pressure. For example, if we double the pressure from say 3.5 to 7.0 N/mm2 the bending stresses given by eq(2) increase from 70.11 to 92.79 N/mm2. If the stresses increased in a linear manner we could have expected a stress = 140.22 N/mm2. The non-linear effect is best seen on a pressure - stress plot like that shown in Figures 6 and 7 below.

Figure 6 shows the results of a pressure - stress plot from OvalPipe-1 Help Menu. The Y-Axis plots the maximum bending stress as a function of pressure along the X-Axis. The plot clearly shows that the relationship between pressure and the bending stress is not linear.
If the pressure is allowed to increase to an arbitrary (unrealistic) large pressure the result shown in Figure 7 is obtained. The bending stress is not continuing to rise steeply with pressure, but is settling down to reach some maximum stress as the pipe becomes more nearly round.
It can be shown theoretically that as the pressure tends to infinity, the bending stress approaches an asymptotic. We proceed as follows: from eq(2) the maximum bending stress (at q = 0°) can be written in the form of eq(3).

ƒb = 6*(R*Xo/t2)*[1/{1/p + 12*R3*(1 - s2)/(E*(h2 - 1)*t3)}]                                             eq(3).

As the pressure approaches infinity, 1/p approaches zero, hence eq(3) reduces to eq(4).

The maximum possible ƒb = Xo*E*(h2 - 1)*t/(2*R2(1 - s2))                                           eq(4).

For the example given, the maximum possible ƒb = 137.14 N/mm2.

Figure 8 shows the results of the pressure - stress plot of Figure 6 but including the membrane stress. The plot shows the membrane stress increasing as a linear function with pressure but the bending stresses are quite clearly non-linear as expected.
At low pressures, when the membrane stress is < the bending stress, it is possible for surface stresses on the pipe to be compressive. As the rate of increase in the bending stress reduces with pressure, the total stresses = ƒm + ƒb eventually become all tensile. The pressure at which this occurs can be obtained by putting ƒb = ƒm in eq(3) and after rearranging, eq(5) is obtained.

pressure p' (to give ƒb = ƒm) = E*(h2 - 1)*t3*[6*Xo/t -1]/(12*R3*(1 - s2))                         eq(5).

For the example given, p' = 1.96 N/mm2. Pressures above this give all tensile stresses in the pipe.

The non-linear effects discussed above are not necessarily significant throughout the whole geometry range. This is best seen in a plot of geometry ratio R/t against maximum stress ratio shown in Figure 9 below.

Figure 9 shows a typical plot of maximum stress ratio = (ƒm + max.ƒb)/ƒm as a function of the geometry ratio R/t. This plot is obtained by keeping the pressure and mean radius constant and varying the wall thickness. For the example given, it is seen that the non-linear effects are more significant in a range of geometry ratio's around R/t = 20. At low R/t and high R/t ratios the non-linear effects are much less significant. This effect is related to the strain energy in the structure discussed in more detail in section 6.0 below.
The geometry ratio to give the maximum stress ratio can be calculated from eq(6) and the maximum stress ratio from eq(7).

R/t = [E*(h2 - 1)/(24*p*(1 - s2))]1/3                                                                             eq(6).

Max. Stress Ratio = 1 + 4*(Xo/R)*[E*(h2 - 1)/(24*p*(1 - s2))]1/3                                     eq(7).

5. Pressure Vessel Code Rules.

At least two well known pressure vessel codes: ASME Section III and BS5500 Appendix C Fatigue rules, make use of Haigh's analysis. The derivations will now be briefly discussed.

5.1 ASME code rules. A factor F1a is used in connection with piping stress indices. Where:

F1a = 1 + ((Dmax - Dmin)/t)*[1.5/(1 + 0.455(Do/t)3(p/E))]                                              eq(8).

Do = nominal outside diameter
Dmax = maximum outside diameter
Dmin = minimum outside diameter
t = nominal wall thickness
p = internal pressure
E = modulus of elasticity, i.e. Young's Modulus

Starting from eq(2) and substituting q = 0°, n = 2 (an oval pipe), Xo = (Dmax - Dmin)/4,
R = Do/2, ƒm = pDo/(2t) and Poisson's ratio = 0.3 then eq(2) becomes eq(9):

ƒb = (pDo/(2t))*((Dmax - Dmin)/t)*[1.5/(1 + 0.455(Do/t)3(p/E))]                                         eq(9).

Adding the membrane and bending stresses we have the total stress:

ƒm + ƒb = (pDo/(2t))*{1 + (Dmax - Dmin)/t)*[1.5/(1 + 0.455(Do/t)3(p/E))]}                          eq(10).

Therefore by inspection of eq(10) the term within the {} brackets is F1a of eq(8). We can conclude that the ASME rule is essentially Haigh's 1936 equation but based on the pipe outside diameter.

5.2 BS5500 code rules. A factor A2 is used in connection with fatigue rules. Where:

A2 = 1.5(Dmax - Dmin)/[e{1 + (0.5p(1 - n2)/E)*(D/e)3}]                                                   eq(11).

D = mean diameter
Dmax = maximum inside diameter
Dmin = minimum inside diameter
e = section thickness
p = design pressure
E = modulus of elasticity, i.e. Young's Modulus
n is Poisson's ratio

Starting from eq(2) and substituting q = 0°, n = 2 (an oval pipe), Xo = (Dmax - Dmin)/4,
R = D/2, s = n then eq(2) becomes eq(12):

ƒb = ƒm*1.5(Dmax - Dmin)/[e{1 + (0.5p(1 - n2)/E)*(D/e)3}]                                             eq(12).

Dividing both sides by the membrane stress gives:

ƒbm = 1.5(Dmax - Dmin)/[e{1 + (0.5p(1 - n2)/E)*(D/e)3}]                                             eq(13).

Therefore by inspection the right hand side of eq(13) is A2 of eq(11). We can conclude that the BS5500 rule is Haigh's 1936 equation with the out-of-round amplitude based on inside diameter.

5.3 Code Results. For the example given in section 3.0 above, OvalPipe-1 help menu gives the results shown in Figure 10 for information: ASME F1a = 1.75 and BS5500 A2 = 0.78

6. Strain Energy Considerations.

The help menu in program OvalPipe-1 includes options to allow a plot of the strain energy in the pipe as a result of the applied pressure loading. A plot of the strain energy can provide a useful picture of how a thin shell resists the applied loading by membrane (stretching) and bending actions.

The linear elastic strain energy of stretching and bending of the surface area of a thin shell, ignoring shear and twisting effects, is given by the following two equations (14) and (15).

Um = [(N1 + N2)2 + 2(1 + n)(-N1*N2)]/(2*E*t)                                                           eq(14).

Ub = 6[(M1 + M2)2 + 2(1 + n)(-M1*M2)]/(E*t3)                                                         eq(15).

Where:

Um is the strain energy per unit surface area of stretching in terms of the membrane forces
Ub is the strain energy per unit surface area of bending in terms of the bending moments

The forces and moments N1, N2, M1, and M2 are stress resultants defined as follows.
N1 = meridional (longitudinal) membrane force per unit circumference, +ve tensile
N2 = circumferential (hoop) membrane force per unit circumference, +ve tensile = ƒm*t
M1 = meridional bending moment per unit circumference
+ve if compression on the outside surface.
M2 = circumferential bending moment per unit circumference = ƒb*t2/6
+ve if compression on the outside surface.

t = wall thickness
E = modulus of elasticity, i.e. Young's Modulus
n is Poisson's ratio

For the present discussion there are no end discontinuity effects being considered, i.e. the cylinder ends are remote, therefore the meridional bending moment M1 = 0 and hence eq(15) can reduce to eq(16).

Ub = 6*M22/(E*t3)                                                                                               eq(16).

For the longitudinal membrane force N1 there is a choice of cases to consider. If only circumferential effects are being considered then N1 = 0 and eq(14) can reduce to eq(17).

Um = N22/(2*E*t)                                                                                                 eq(17).

Alternatively a common case is when the ends of the cylinder are capped. In this case the longitudinal membrane stress, from simple theory, is half the circumferential membrane stress, i.e. ƒL = 0.5*ƒm and hence N1 = N2/2. By substitution eq(14) can reduce to eq(18).

Um = [2.25 - (1 + n)]*N22/(2*E*t)                                                                             eq(18).

By inspection of eq(18) it can be seen that any material with a Poisson's ratio > 0.25 will result in a reduction of membrane strain energy compared to that calculated by eq(17), i.e. in general a two dimensional stress system does not necessarily lead to an increase in strain energy in the system. For many metallic materials with a Poisson's ratio = 0.3 a slight reduction in strain energy can be expected.

For the example given in section 3.0 above the following strain energies shown in Table 1 are obtained:

 Strain Energy Including N1 Excluding N1 Membrane Um 0.077 0.081 Bending Ub 0.016 0.016 Total Ut = Um+ Ub 0.093 0.097
Table 1 Strain Energy Results Nmm/mm2

A plot of the strain energy can provide a useful picture of how a thin shell resists the applied loading by membrane (stretching) and bending actions. For example Figures 11 and 12 below show linear and polar plots of the total strain energy Ut = Um + Ub around an oval pipe. Note that the energy is always positive.

Figure 13 shows a typical plot of maximum strain energy ratio = (Um + max.Ub)/Um as a function of the geometry ratio R/t. Like Figure 9 this plot is obtained by keeping the pressure and mean radius constant and varying the wall thickness. For the example given it is seen that the shape of the curve and its R/t position of maximum energy are similar to Figure 9 showing a clear relationship between the energy and stress levels induced in the structure.

The reasons for this shape of curve can be summarised:

1. At small R/t ratios the pipe is relatively thick and has a high bending stiffness, but the pressure loading is not high enough to induce large bending effects into the pipe.
2. At large R/t ratios the pipe is essentially a thin membrane shell with low bending stiffness, the pressure loading will deform the shell into a cylinder (or near cylindrical shape) with little bending effects.
3. For mid range R/t ratios there is sufficient bending stiffness and loading to induce significant bending effects.
7. References.
1. Haigh, B. P., "An Estimate of the Bending Stresses Induced by Pressure in a Tube that is not Initially Quite Circular", Welding Research Committee Second Report, Proceedings of the I.Mech.E., Volume 133, 1936.
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