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Basic Tubular Buckling, release 5, issued 16 May 2008. JIT version.
This worksheet exists in two versions. They are identical apart from the way they are formatted. The Work version hides intermediate calculations and allows the user to see the results just below the inputs. This is useful for quick “what-if” games, changing various inputs to see what works best. The JIT version displays all intermediate calculations, plus adds tutorial text to explain the methodology.
This “Just In Time Learning” tutorial version of the worksheet explains the theories behind the calculations. It shows the formulae and intermediate results. It assumes that you are familiar with the meanings and units of hydrostatic pressure, stress and force. You should also understand that in a deviated hole, axial force arising from the weight of the tubular is calculated using the average weight per foot (including connections) multiplied by the true vertical depth and not the measured depth.
Tutorial text is shown in blue. It should take 35-40 minutes to work through this tutorial. Leave the user input as it is for the tutorial.
At the end of this tutorial, YOU will be able to;
1. State the necessary condition for buckling to occur.
2. If buckling is possible, give possible solutions to prevent buckling taking place.
3. State why the casing might buckle later, as the well heats up when on production.
This worksheet takes a bunch of inputs for a tubular and it’s parameters. It uses the accepted engineering formulae to calculate the Axial Force Fa and the Stabilising Force Fs which arise from the axial (tensile/compressive) force and the internal and external pressures. These two forces added together make the Effective Force, Feff. Either of the two component forces can be negative and where the Effective Force becomes negative, it is possible for the casing to buckle. With thin walled tubes like casings and tubings, it is assumed that the tubular has no mechanical resistance to buckling.
If both ends of the tubular are fixed (such as in a cemented casing), a temperature or internal pressure increase will decrease Fa and if this decreases Feff to below zero, then buckling can occur. The temperature increase can be entered in this worksheet and the results calculated, allowing options to be explored to avoid buckling due to changes in temperature.
4.5″ 9.5 ppf4.5″ 10.5 ppf4.5″ 11.6 ppf4.5″ 13.5 ppf4.5″ 15.1 ppf5″ 11.5 ppf5″ 13 ppf5″ 15 ppf5″ 18 ppf5″ 21.4 ppf5″ 23.2 ppf5″ 24.1 ppf5.5″ 14 ppf5.5″ 15.5 ppf5.5″ 17 ppf5.5″ 20 ppf5.5″ 23 ppf5.5″ 26.8 ppf5.5″ 29.7 ppf5.5″ 32.6 ppf5.5″ 35.3 ppf5.5″ 38 ppf5.5″ 40.5 ppf5.5″ 43.1 ppf6.625″ 20 ppf6.625″ 24 ppf6.625″ 28 ppf6.625″ 32 ppf7″ 17 ppf7″ 20 ppf7″ 23 ppf7″ 26 ppf7″ 29 ppf7″ 32 ppf7″ 35 ppf7″ 38 ppf7″ 42.7 ppf7″ 46.4 ppf7″ 50.1 ppf7″ 53.6 ppf7″ 57.1 ppf7.625″ 24 ppf7.625″ 26.4 ppf7.625″ 29.7 ppf7.625″ 33.7 ppf7.625″ 39 ppf7.625″ 42.8 ppf7.625″ 45.3 ppf7.625″ 47.1 ppf7.625″ 51.2 ppf7.625″ 55.3 ppf7.75″ 46.1 ppf8.625″ 24 ppf8.625″ 28 ppf8.625″ 32 ppf8.625″ 36 ppf8.625″ 40 ppf8.625″ 49 ppf9.625″ 32 ppf9.625″ 36 ppf9.625″ 40 ppf9.625″ 43.5 ppf9.625″ 47 ppf9.625″ 53.5 ppf9.625″ 58.4 ppf9.625″ 59.4 ppf9.625″ 64.9 ppf9.625″ 70.3 ppf9.625″ 75.6 ppf10.75″ 32.75 ppf10.75″ 40.5 ppf10.75″ 45.5 ppf10.75″ 51 ppf10.75″ 55.5 ppf10.75″ 60.7 ppf10.75″ 65.7 ppf10.75″ 73.2 ppf10.75″ 79.2 ppf10.75″ 85.3 ppf11.75″ 42 ppf11.75″ 47 ppf11.75″ 54 ppf11.75″ 60 ppf11.75″ 65 ppf11.75″ 71 ppf13.375″ 48 ppf13.375″ 54.5 ppf13.375″ 61 ppf13.375″ 68 ppf13.375″ 72 ppf16″ 65 ppf16″ 75 ppf16″ 84 ppf16″ 109 ppf18.625″ 87.5 ppf20″ 94 ppf20″ 106.5 ppf20″ 133 ppf
Short RoundLong RoundButtress NormalButtress Special ClearanceOther
Select a casing
Select a connection type.
For the OD and connection selected,
additional weight per
If no data exists, you can enter a figure here for the additional weight per connection. If no value is entered in this field, then the worksheet will instead use the API nominal weight per foot to calculate downhole stresses.
Manually enter the additional weight per connection. Wtconn =
Enter the average joint length. L =
Enter the depth of the highest fixed point (eg top of cement or tubing anchor), TVD. TVDfixed =
lb per ftkg per m
Enter the pipe weight average with connections. wij =
This value is given in the worksheet “Casing Strengths”. If mixed string above TVDfixed, enter highest weight.
Enter the axial force at the surface (tensile is +ve). Fa =
Enter the internal pressure at the surface. pi =
Enter the external pressure (annular pressure) at the surface. po =
psi/ftSG or kg/lppglb/ft3ppbkPa/m
Enter the fluid gradient inside the tube. ρi =
psi/ftSG or kg/lppglb/ft3ppbkPa/m
Enter the fluid gradient in the annulus outside the tube. ρo =
Enter the expected average temperature increase with the well on production. TΔ =
(Enter -ve if well cools down instead of heats up).
Click here when any values are modified to update the result.
To calculate whether or not a thin walled tube will buckle, two values have to be assessed along the length of pipe between the surface and the highest fixed point in the tube (such as the top of cement).
1. The axial force Fa acting on the tubular. In the absence of any force opposing the tendancy to buckle, a thin walled tubular that is not supported against sideways movement would buckle as soon as it goes into axial compression (where Fa is negative).
2. The Stabilising Force Fs arises as a result of internal and external pressures. These pressures give rise to stresses within the wall of the tube which can oppose the negative Fa (if Fs is positive) or can add to the tendancy to buckle (if negative).
These two forces are added together to arrive at the Effective Force, Feff. Bear in mind that both the Stabilising and the Axial forces can be positive or negative, so it is possible for a tubular under axial tension to buckle, if the stabilising force is negative by an amount that overcomes the tensile force.
Where Feff is negative, if the surface pull on the tubing can be increased by an amount greater than the resulting negative force (to make the sum positive), then buckling can be avoided.
The Stabilising Force Fs is easy to calculate. It is the external pressure acting on a cross sectional area of a circle of diameter equal to the tube OD,
the internal pressure acting on a cross sectional area of a circle of diameter equal to the tube ID.
Fs = (0.7854 x OD2 x Po) – (0.7854 x ID2 x Pi)
where Po is the outside or external pressure and Pi is the internal pressure.
Note that increasing the internal pressure decreases Fs. Thus increasing the internal pressure makes buckling more likely, as long as this does not change the axial force in the tubular (both ends are fixed).
The internal and external pressures at the depth of interest arise from 1) surface pressures on the annulus or tube ID plus 2) hydrostatic pressures from fluids in the annulus or tube inside.
A casing is in axial compression of 45,000 lbs at the top of cement. OD = 9.625”, ID = 8.615”, Po = 2000 psi, Pi = 1500 psi. Can the casing buckle?
Fa = -45,000 lbs
Fs = (0.7854 x 9.6252 x 2000) – (0.7854 x 8.6152 x 1500) = 58,083 lbs
Feff = Fs + Fa = 13,083 lbs, > 0 therefore the casing cannot buckle.
The effective force Feff can be plotted on a graph against depth. The depth where this line crosses zero is the neutral point for buckling (NPbuckling). Above this point, the casing cannot buckle.
If this neutral point is below the top of cement or below the tubing anchor depth, then buckling will not occur.
The surface axial force is entered by the user as part of the worksheet inputs. It is calculated for the situation at the end of cement displacement. If necessary, once the cement has set, extra pull can be applied to the casing before setting the hanger (if the hanger type allows this), to prevent buckling. Alternatively the top of cement can be adjusted to bring it above the neutral point for buckling.
Starting at the surface, the axial force Fa will arise due to the weight of the casing in the hole, reduced by buoyant effects. Once cement is displaced, Fa will decrease (neglecting friction) because the heavy cement replaces light mud outside the casing and this increases buoyancy. At the end of cement displacment, the axial force at the surface Fa can be calculated.
The axial force decreases with depth (because there is less weight hanging under the depth of interest). The stabilising force will normally increase with depth, unless the fluid density inside the pipe is much greater than the fluid density outside the pipe. Remember that the internal pressure acts over a smaller area than the external pressure.
To plot the graph of effective force vs depth, the worksheet will add Fa and Fs at the surface. If internal and annular pressures are zero then Fs at the surface will also be zero. Then the slope of the graph below the surface will be the addition of the gradients of Fa and Fs. The first calculations are shown below, using the values in the input fields above. Fa at surface was input above and does not need to be calculated. Note that metric units (forces in Newtons) are used for these calculations.
Calculate the stabilising force at surface (arising from internal and external pressures).
Calculate the change in stabilising force with depth arising from hydrostatic fluid gradients (assumes same pipe dimensions).
If this is -ve, internal fluid gradient > external fluid gradient and Fs decreases with depth.
Calculate the reduction in axial force with depth (due to decreasing axial force).
Combining the effects of decreasing Fa and changing (increasing or decreasing) Fs with depth;
Plot the effective force at depths between the surface and the highest supported depth.
Buckling will not occur above NPbuckling depth (this is TVD, not MD).
If after fixing the tubular in the hole (cement sets or tubing anchor set) the temperature changes, this has an effect on the axial force in the tubular due to thermal expansion or contraction.
If a steel tube is free to move, increasing temperature will cause it to expand. If a steel tube is fixed at both ends and the temperature changes, then the length cannot change – instead, the stress in the steel changes (by a known amount). This stress change can be multiplied by the cross sectional area to find out the change in force due to the temperature. If the pipe cools down (eg by pumping water into a reservoir to increase pressure) then Fa will increase. If the pipe heats up then Fa will decrease.
Change in force due to temperature; stress x area
New axial force at surface due to temperature change.
Buckling will not occur above NPbuckling_hot depth after a temperature change (this is TVD, not MD).
Axial force at the top of the pipe
Top of cement or tubing anchor depth
Neutral point for buckling at the initial temperature (depth above which the tubular cannot buckle)
If the neutral point for buckling occurs within the free pipe (the pipe above the upper fixed point eg top of cement) then the NPbuckling is plotted as a blue dotted line, below which buckling can occur.
The axial force at surface was entered as or . The minimum axial force required at the surface to avoid .
If the minimum axial force required < Fa, then the options include;
1) Picking up an additional surface force after the cement has set to make Fa exceed the minimum required surface force.
2) Raise the top of cement above , or
3) Support the casing laterally, such as by using rigid centralisers up to the NPbuckling depth.
The situation shown in the
Neutral point for buckling after the temperature change
This replots the graph above after the temperature change. The same advice as above applies, if the Effective Force goes below zero at a depth above the fixed point in the tubing.
After the temperature change, the
Benham PP and Warnock FV. Mechanics of Solids and Structures. Pitman 1976.
Version 1 of this worksheet released on 5 February 2008.
Version 4 released after testing against API 5C2 found an error in one of the 8.625″ casing wall thicknesses.
Version 5 released to correct an error in the conversion factor for kPa/m.