© Copyright August 2001, January 2021

Web site: https://acwhyte.droppages.com

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.

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: ƒ_{m} = pR/t **eq(1)**.

p is the internal pressure

R is the radius

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/mm^{2}. 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, ƒ
_{m}= 3.5 * 25.83 = 90.4 N/mm^{2}

The actual circumferential stresses at the inside and outside radius
calculated by the more accurate Lamè equations are: ƒ_{i} = 90.45 N/mm^{2}
and ƒ_{o} = 86.95 N/mm^{2} 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).

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 ƒ_{m} given by **eq(1)** above.

The bending stress is given by: ƒ_{b} = 6*ƒ_{m}*(X_{o}/t)*(cos(h*q))/(1 + parameter P) **eq(2)**.

parameter P = 12*p*R

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

X

q is angle theta

h is number of lobes

The amplitude X_{o} 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/mm^{2} and
Poisson's ratio = 0.3 the results, from **OvalPipe-1**, are as shown in **Figures 1** to **3** below.

Figure 1 |
Figure 2 |
Figure 3 |

**Figure 1**shows the input data and stress results. The circumferential bending stress ƒ_{b}= 70.11 N/mm^{2}, 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.

Figure 4 |
Figure 5 |

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

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/mm^{2} the
bending stresses given by **eq(2)** increase from 70.11 to 92.79 N/mm^{2}. If the
stresses increased in a linear manner we could have expected a stress = 140.22 N/mm^{2}. The non-linear
effect is best seen on a pressure - stress plot like that shown in **Figures 6** and **7** below.

Figure 6 |
Figure 7 |
Figure 8 |

**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)**.

ƒ_{b} = 6*(R*X_{o}/t^{2})*[1/{1/p + 12*R^{3}*(1 - s^{2})/(E*(h^{2} - 1)*t^{3})}] **eq(3)**.

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

The maximum possible ƒ_{b} = X_{o}*E*(h^{2} - 1)*t/(2*R^{2}(1 - s^{2})) **eq(4)**.

For the example given, the maximum possible ƒ_{b} = 137.14 N/mm^{2}.

**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 = ƒ_{m} + ƒ_{b} eventually become all tensile. The pressure at which this occurs can be obtained by
putting ƒ_{b} = ƒ_{m} in **eq(3)** and after rearranging, **eq(5)** is obtained.

pressure p' (to give ƒ_{b} = ƒ_{m}) = E*(h^{2} - 1)*t^{3}*[6*X_{o}/t -1]/(12*R^{3}*(1 - s^{2})) **eq(5)**.

For the example given, p' = 1.96 N/mm^{2}. 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 |

**Figure 9** shows a typical plot of maximum stress ratio = (ƒ_{m} + max.ƒ_{b})/ƒ_{m} 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*(h^{2} - 1)/(24*p*(1 - s^{2}))]^{1/3} **eq(6)**.

Max. Stress Ratio = 1 + 4*(X_{o}/R)*[E*(h^{2} - 1)/(24*p*(1 - s^{2}))]^{1/3} **eq(7)**.

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 *F*_{1a} is used in connection with piping stress indices. Where:

*F*_{1a} = 1 + ((D_{max} - D_{min})/t)*[1.5/(1 + 0.455(D_{o}/t)^{3}(p/E))] **eq(8)**.

D_{o} = nominal outside diameter

D_{max} = maximum outside diameter

D_{min} = 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),
X_{o} = (D_{max} - D_{min})/4,

R = D_{o}/2, ƒ_{m} = pD_{o}/(2t) and Poisson's ratio = 0.3 then **eq(2)** becomes **eq(9)**:

ƒ_{b} = (pD_{o}/(2t))*((D_{max} - D_{min})/t)*[1.5/(1 + 0.455(D_{o}/t)^{3}(p/E))] **eq(9)**.

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

ƒ_{m} + ƒ_{b} = (pD_{o}/(2t))*{1 + (D_{max} - D_{min})/t)*[1.5/(1 + 0.455(D_{o}/t)^{3}(p/E))]} **eq(10)**.

Therefore by inspection of **eq(10)** the term within the {} brackets is *F*_{1a} 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 *A*_{2} is used in connection with fatigue rules. Where:

*A*_{2} = 1.5(D_{max} - D_{min})/[e{1 + (0.5p(1 - n^{2})/E)*(D/e)^{3}}] **eq(11)**.

D = mean diameter

D_{max} = maximum inside diameter

D_{min} = 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),
X_{o} = (D_{max} - D_{min})/4,

R = D/2, s = n then **eq(2)** becomes **eq(12)**:

ƒ_{b} = ƒ_{m}*1.5(D_{max} - D_{min})/[e{1 + (0.5p(1 - n^{2})/E)*(D/e)^{3}}] **eq(12)**.

Dividing both sides by the membrane stress gives:

ƒ_{b}/ƒ_{m} = 1.5(D_{max} - D_{min})/[e{1 + (0.5p(1 - n^{2})/E)*(D/e)^{3}}] **eq(13)**.

Therefore by inspection the right hand side of **eq(13)** is *A*_{2} 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** *F*_{1a} = 1.75 and **BS5500** *A*_{2} = 0.78

Figure 10 |

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)**.

U_{m} = [(N1 + N2)^{2} + 2(1 + n)(-N1*N2)]/(2*E*t) **eq(14)**.

U_{b} = 6[(M1 + M2)^{2} + 2(1 + n)(-M1*M2)]/(E*t^{3}) **eq(15)**.

Where:

U_{m} is the strain energy per unit surface area of stretching in terms of the membrane forces

U_{b} 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 = ƒ_{m}*t

M1 = meridional bending moment per unit circumference

+ve if compression on the outside surface.

M2 = circumferential bending moment per unit circumference = ƒ_{b}*t^{2}/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)**.

U_{b} = 6*M2^{2}/(E*t^{3}) **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)**.

U_{m} = N2^{2}/(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. ƒ_{L} = 0.5*ƒ_{m} and hence N1 = N2/2. By substitution **eq(14)** can reduce to **eq(18)**.

U_{m} = [2.25 - (1 + n)]*N2^{2}/(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 U_{m} |
0.077 | 0.081 |

Bending U_{b} |
0.016 | 0.016 |

Total U_{t} = U_{m}+ U_{b} |
0.093 | 0.097 |

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 U_{t} = U_{m} + U_{b} around an oval pipe.
Note that the energy is always positive.

Figure 11 |
Figure 12 |

**Figure 13** shows a typical plot of maximum strain energy ratio = (U_{m} + max.U_{b})/U_{m} 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.

Figure 13 |

The reasons for this shape of curve can be summarised:

- 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.

- 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.

- For mid range R/t ratios there is sufficient bending stiffness and loading to induce significant bending effects.

- 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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