Casing design · MunaiHub

Quick refresher

The design goes in three steps: the sizes of the strings and the shoe depths (the depths where the strings end); the load cases for each string; the loads and the choice of pipes, with a check of tension.
A pipe is loaded not by a pressure but by the difference of the pressures inside and outside at the same depth. External excess pressure $p_{\text{out}} - p_{\text{in}}$ collapses the pipe, and internal excess pressure $p_{\text{in}} - p_{\text{out}}$ bursts it.
Collapse is most dangerous at the bottom of an emptied string: during lost circulation, when the well is brought in (started up), and in gas lift (lifting the fluid by injecting gas). Burst is most dangerous at the wellhead of a closed well with gas. For a production string, it is most dangerous at the packer (the seal between the tubing and the casing) if the tubing leaks.
The pressure inside a string in a closed well is limited by the fracture pressure of the rock under the shoe: it will not rise higher, because the rock will take the fluid. A column of gas is light, so at the wellhead the pressure is only a little lower than at the shoe.
The rating of a pipe must be not less than the load multiplied by the design factor. We take what is unknown on the cautious side: for collapse, the highest pressure outside; for burst, the lowest pressure outside.
The string is made of sections. At the bottom collapse decides the choice, and at the top internal pressure and tension decide it. Tension lowers the collapse resistance, so the places where the string changes to a weaker pipe are checked with a correction.

Key relationships

p_{\text{ne}} = p_{\text{out}} - p_{\text{in}}, \qquad p_{\text{ni}} = p_{\text{in}} - p_{\text{out}}

external and internal excess pressure at the same depth

p_{\text{col}} \ge n_{\text{col}}\,p_{\text{ne}}, \quad p_{\text{iy}} \ge n_{\text{bur}}\,p_{\text{ni}}, \quad F_y \ge n_{\text{ten}}\,F

strength conditions with design factors

p_{\text{wh}} = p_{\text{shoe}} - 0.00981\,\rho_{\text{g}}\,h_{\text{shoe}}

wellhead pressure of a closed well full of gas (MPa, g/cm³, m)

F = g\,k_{\text{b}}\sum m_i\,l_i, \qquad k_{\text{b}} = 1 - \dfrac{\rho_m}{\rho_{\text{steel}}}

weight of the sections below the cross-section, in mud

Y_{pa} = Y_p\left[\sqrt{1 - 0.75\,(\sigma_a/Y_p)^2} - 0.5\,\sigma_a/Y_p\right]

yield strength reduced for the axial tension $\sigma_a$ : the effective yield strength of a pipe that is also being pulled (for the collapse calculation)

Common pitfalls

Comparing with the rating of the pipe the pressure inside or outside, and not their difference.
Adding up incompatible conditions: pressures that do not exist at the same time in any operation.
Taking the margin “in the wrong direction”: lowering the external pressure in a collapse design, or the fracture pressure under the shoe in a burst design.
Assuming that the pressure at the wellhead of a closed well is much lower than at the shoe: a column of gas weighs little.
Checking collapse by the rating of the pipe where the pipe is in tension.

After this lesson you can

name the three steps of string design, and explain what a load case (a situation that loads the string) is and why the pressures inside and outside in a load case must be compatible
draw the pressure profiles inside and outside the string for collapse (emptying) and for burst (a closed well with gas, a leak in the tubing), and find the excess pressures along the depth
apply design factors and choose the sections of the string graphically, from the points where the load lines cross the ratings of the pipes
calculate the axial load from the weight of the string in mud and from additional forces, and check the strength of the pipe body
explain why tension lowers the collapse resistance, and apply the correction for biaxial loading (pull and pressure acting together)
list the conditions that need separate attention: salts, hydrogen sulfide, temperature, curvature of the hole, wear

The three steps of the design

We design a string when the casing programme of the well as a whole has been drafted. The design has three steps, and each rests on the previous one.

From the casing programme to the pipes

1
Sizes and depths
We find the number of strings and the depths of their shoes from the pore pressure, the pressure of the fluid in the rock, and the fracture pressure, the pressure at which the rock breaks (choosing shoes). We find the diameters from the bottom up (strings by purpose).
2
Load cases
For each string we list the situations in which it is loaded the most: while running and cementing, while drilling below it, and in production. For each case we state what is inside the string and what is outside.
3
Loads and pipes
We calculate the pressures inside and outside along the depth, their difference and the axial force. We multiply them by design factors and choose pipes whose ratings are not less. Then we check tension and its effect on the collapse resistance. If the check fails, we go back to the choice.

We choose the diameters starting from the smallest string, the production casing. Its size is set by the tubing, the completion equipment and the logging tools (Well completion: opening the reservoir and the tubing string). Each string above it must pass the bit for the hole below it. The smaller the diameters we can use, the cheaper the well: less steel, cement, mud and drilled rock.

We also check the shoe depth, found from the window of densities (the range of mud densities between the pore pressure and the fracture pressure), for a kick. When the crew circulates a pocket of gas out of a closed well, the pressure in the open hole rises, and it must not exceed the fracture pressure under the shoe (pressure at the shoe while circulating out gas). If it does, the shoe is set deeper, or the allowed influx volume is made smaller. This volume is the kick tolerance.

Load cases: what can happen to a string

A pipe is loaded not by the pressure itself but by the difference of the pressures on the two sides of the wall. At 3000 m, in mud of 1.2 g/cm³, the pressure is about 35 MPa. But if the same mud is both inside and outside the string, the wall is hardly loaded. A load appears when there are different fluids inside and outside, or when pressure is applied on one side only.

External and internal excess pressure

p_{\text{ne}}(z) = p_{\text{out}}(z) - p_{\text{in}}(z), \qquad p_{\text{ni}}(z) = p_{\text{in}}(z) - p_{\text{out}}(z)

where:

$p_{\text{ne}}$ external excess pressure, the collapse load, MPa
$p_{\text{ni}}$ internal excess pressure, the burst load, MPa
$p_{\text{out}},\ p_{\text{in}}$ pressure outside and inside the string at the depth $z$ , MPa
$z$ true vertical depth, m

Units: any consistent units of pressure

Holds when

Both pressures are taken at the same depth and at the same moment. The profiles inside and outside must be compatible, that is, they must exist together in one operation.
The pressures are gauge pressures: atmospheric pressure acts on both sides of the wall and cancels.
The pressure of a column of fluid is calculated from the vertical depth, and the length of the pipe sections from the length along the hole.

A positive $p_{\text{ne}}$ collapses the pipe, and a positive $p_{\text{ni}}$ bursts it. At one depth, in one load case, only one of the two is positive. A column of fluid with the density $\rho$ makes the pressure $0.00981\,\rho\,z$ at the depth $z$ (MPa, g/cm³, m). So the profiles are made of straight segments, and it is enough to calculate the load at the break points (where a line changes direction): at the wellhead, at the fluid level, at the top of cement, at the shoe.

A load case is a situation for which both sides are set: what is inside the string and what is outside. We go through all the operations in which the string takes part, and for each load we keep the heaviest of the realistic cases.

SchematicThree load cases. Left: collapse. The string is emptied by lost circulation, there is a column of mud outside, and the load grows with depth. Centre: burst. The well is closed after a kick and filled with gas, the pore pressure is outside, and the load is greatest at the wellhead. Right: a production string with a leaking tubing. The gas pressure from the wellhead passes to the fluid above the packer (the seal that closes the annulus), and the load grows towards the packer. The arrows show the direction and the relative size of the excess pressure. Not to scale.

Typical load cases

String: Conductor
Collapse: cementing; emptying during lost circulation
Burst: usually does not decide the choice: the well is not closed on a conductor, and the flow is diverted

String: Surface and intermediate casing
Collapse: full or partial emptying during lost circulation while drilling below; cementing
Burst: a closed well after a kick, filled with gas; pressure test

String: Production casing
Collapse: emptying while the well is brought in (made to flow), in gas lift (lifting oil with injected gas), or as the reservoir is depleted
Burst: a leak in the tubing at the wellhead; injection, hydraulic fracturing (cracking the rock with pumped fluid), killing the well

String	Collapse	Burst
Conductor	cementing; emptying during lost circulation	usually does not decide the choice: the well is not closed on a conductor, and the flow is diverted
Surface and intermediate casing	full or partial emptying during lost circulation while drilling below; cementing	a closed well after a kick, filled with gas; pressure test
Production casing	emptying while the well is brought in (made to flow), in gas lift (lifting oil with injected gas), or as the reservoir is depleted	a leak in the tubing at the wellhead; injection, hydraulic fracturing (cracking the rock with pumped fluid), killing the well

Which cases to consider, and what to assume inside and outside, is set by regulatory documents and company standards. For exploration wells, where the rocks are less well known, the requirements are stricter.

In each case we take what is unknown on the cautious side. For collapse, we take the highest possible pressure outside and the lowest inside. For burst it is the other way: the lowest outside and the highest inside. So the same quantity can enter two cases with different values. The pressure of the formations behind the string is taken at the upper end of the forecast for collapse, and at the lower end for burst.

Outside: the mud in which the string was run: It acts right after running and in the uncemented interval. Over the years the weighting material (the heavy solids that make the mud dense) settles, and the pressure of the column approaches that of the base liquid of the mud.
Outside: cement slurry: Until the cement sets, it presses like a heavy fluid. Set cement does not transmit hydrostatic pressure (the pressure of a standing column of fluid).
Outside: the pressure of the formations: Behind a cemented interval with a tight cement sheath, the pore pressure of the rocks acts on the string.
Outside: mix water: Mix water is the water used to mix the cement. A common assumption for a long service life: a column of fluid with a density of about 1.0 g/cm³ that stays after the cement sets and the weighting material settles.
Inside: mud, gas or an empty string: While drilling, mud up to the wellhead. In a closed well after a kick, in the worst case gas up to the wellhead. In lost circulation, an empty string down to the fluid level.
Inside: test and production pressure: The pressure at which the string is tested after cementing. For a production string, the pressures when the tubing leaks and during injection and treatments of the formation.

Collapse: an emptied string

The greatest external excess pressure appears when the fluid level in the string falls and a full column stays outside. While drilling, this happens with lost circulation. The mud goes into the formation below the shoe, and the level falls until the column of mud that is left balances the pressure in the loss zone (the formation that takes the mud). The most cautious assumption is full emptying: the string is empty down to the shoe. Then the load equals the external pressure, and it grows with depth in a straight line, from zero at the wellhead to the highest value at the shoe.

A less severe assumption is partial emptying: we find the level from the balance with the loss zone.

Depth of the mud level in the string during lost circulation

L = h_{\text{loss}}\left(1 - \frac{\rho_{\text{loss}}}{\rho_m}\right)

where:

$L$ depth of the mud level in the string, measured from the wellhead, m
$h_{\text{loss}}$ true vertical depth of the loss zone, m
$\rho_{\text{loss}}$ equivalent mud weight of the pressure in the loss zone (the density of mud that would give the same pressure there), g/cm³
$\rho_m$ density of the mud in the well, g/cm³

Units: any units of length; the densities in the same units

Holds when

The well is open at the wellhead, and the mud is at rest and has the same density throughout.
The formation takes mud while the pressure of the column above it is higher than the pressure in the formation: $\rho_m (h_{\text{loss}} - L) = \rho_{\text{loss}}\,h_{\text{loss}}$ .
While the level falls, the pressure on the formations above it decreases, and they may start to flow. Then the string will hold gas, not an empty space.

Above the level the string is empty, and the external excess pressure equals the full external pressure. Below the level, the column of mud inside reduces it. If the mud in the string is heavier than the fluid behind it, the greatest load falls at the depth of the level, not at the shoe.

For a production string, emptying is a normal working state, not an accident. When the well is brought in, the level in the well is lowered to start the inflow. In gas lift, there is gas in the annulus instead of liquid, and after its pressure is released the string stays almost empty. In a depleted gas well (one whose pressure has fallen after long production), the pressure inside falls almost to zero. So the string is designed for full emptying down to the packer or to the bottom.

During cementing, there is cement slurry outside, and it is usually much heavier than the fluid inside. The load is small compared with the emptying of a deep string. But for thin-walled pipes of large diameter, conductors and surface casing, it may be the deciding case (why thin-walled pipes collapse earlier).

Burst: a closed well with gas

The highest pressure inside a surface or intermediate string appears when the well is closed after a gas kick (Well control: primary barrier and kick indicators). In the worst case the gas has pushed all the mud out of it. A column of gas weighs little, so the formation pressure passes to the wellhead almost without loss: over three kilometres of gas only about 6 MPa is “lost”.

There is a natural limit. Below the shoe the hole is open, and the pressure in it cannot be higher than the fracture pressure of the weakest rock, usually the rock under the shoe (Fracture gradient and leak-off testing). If the formation is “stronger” than that rock, the rock under the shoe fractures and takes fluid, and the pressure in the string stays at the level of the fracture pressure. So the pressure inside the string at the shoe is the smaller of the two values.

Pressure in a closed well filled with gas

p_{\text{shoe}} = \min\!\left[\,p_p - \rho_{\text{g}}\,g\,(h - h_{\text{shoe}});\ p_f\,\right], \qquad p_{\text{wh}} = p_{\text{shoe}} - \rho_{\text{g}}\,g\,h_{\text{shoe}}

where:

$p_{\text{shoe}}$ pressure inside the string at the depth of the shoe, Pa
$p_p$ pore pressure of the highest-pressure formation of the next interval, at the depth $h$ , Pa
$p_f$ fracture pressure of the rock under the shoe, Pa
$\rho_{\text{g}}$ average density of the gas in the well, kg/m³
$h,\ h_{\text{shoe}}$ depths of the formation and of the shoe, vertical, m
$p_{\text{wh}}$ pressure inside the string at the wellhead, Pa
$g$ acceleration due to gravity, m/s²

Units: SI; the pressures are gauge pressures

Holds when

The well is closed and full of gas from the formation to the wellhead. This is the worst case. When a small pocket of gas is circulated out, the pressures are much lower (Well kill methods).
The gas density is taken as constant along the well, and taken with a margin on the low side: the lighter the gas, the higher the pressure at the wellhead.
For this calculation the fracture pressure is taken with a margin on the high side: an underestimated value here underestimates the load.
If there is a rock in the open hole that is weaker than the rock under the shoe, it sets the limit.

The burst load is the difference between this pressure and the external pressure. The pressure outside grows with depth faster than the pressure inside, where there is gas. So the internal excess pressure is greatest at the wellhead and falls towards the shoe.

In other unit systems (2)

Metric units

p_{\text{wh}}\,[\text{MPa}] = p_{\text{shoe}} - 0.00981\,\rho_{\text{g}}\,[\text{g/cm}^3]\;h_{\text{shoe}}\,[\text{m}]

pressures in MPa, density in g/cm³, depths in m.

0.00981 MPa is the pressure of a column of fluid with a density of 1 g/cm³ and a height of 1 m.

Oilfield units

p_{\text{wh}}\,[\text{psi}] = p_{\text{shoe}} - 0.052\,\rho_{\text{g}}\,[\text{ppg}]\;h_{\text{shoe}}\,[\text{ft}]

pressures in psi, density in ppg, depths in feet. The gas gradient in psi/ft is often given directly: a density of 0.20 g/cm³ corresponds to 0.087 psi/ft.

0.052 psi/ft per ppg comes from the oilfield formula for hydrostatic pressure.

For the burst design, we take the lowest realistic pressure outside: the pore pressure of formations of normal pressure, or a column of mix water. The lower the pressure outside, the greater the calculated load.

The string is part of the pressure shell, the whole chain that holds the pressure of a closed well, together with the casing head and the blowout preventer (BOP) equipment. We choose the working pressure of this equipment from the same wellhead pressure (Blowout prevention equipment): the pressure shell holds only as much as its weakest link. After cementing, the crew pressure-tests the string together with the wellhead equipment, and the test pressure itself becomes a load case: it is applied from inside along the whole length of the string.

Design factors and the choice of pipes

The pressures along the well are known from a forecast, the gas density and the level in lost circulation are assumed, and the pipe in the well wears and corrodes. We cover this uncertainty with the design factor: we multiply the design load by a number not less than one, and require that the rating of the pipe (strength ratings) is not lower.

Strength conditions of a casing pipe

p_{\text{col}} \ge n_{\text{col}}\,p_{\text{ne}}, \qquad p_{\text{iy}} \ge n_{\text{bur}}\,p_{\text{ni}}, \qquad F_y \ge n_{\text{ten}}\,F

where:

$p_{\text{col}}$ collapse pressure of the pipe, MPa
$p_{\text{iy}}$ internal yield pressure of the pipe, MPa
$F_y$ tensile strength of the pipe body; for a connection, its tensile strength (the pull-out load, at which the thread starts to pull out of the coupling), MN
$p_{\text{ne}},\ p_{\text{ni}},\ F$ external and internal excess pressure and axial force in the cross-section considered, MPa, MPa, MN
$n_{\text{col}},\ n_{\text{bur}},\ n_{\text{ten}}$ design factors for collapse, internal pressure and tension, —

Units: any consistent units for the rating and the load

Holds when

We check the conditions in every cross-section of the string, for every load case.
The ratings already include the minimum yield strength (the stress at which the steel starts to stretch permanently) and the tolerance on the wall thickness, but not wear, corrosion or axial load.
The values of the factors are set by regulatory documents and company standards.

Each of the three loads is checked separately, one at a time (a uniaxial design). Their effect on each other is taken into account by a correction for biaxial loading, and in difficult cases by a calculation with the equivalent stress.

Design factors: common ranges

Load: Collapse
Factor: 1.0–1.125

Load: Internal pressure
Factor: 1.0–1.33

Load: Tension
Factor: 1.0–2.0

Load: Equivalent stress (triaxial design)
Factor: 1.25

Load	Factor
Collapse	1.0–1.125
Internal pressure	1.0–1.33
Tension	1.0–2.0
Equivalent stress (triaxial design)	1.25

The ranges are from international practice. The less we know about the conditions, and the more serious the result of a failure, the larger the value we choose. The lesson's examples use 1.125 for collapse, 1.10 for internal pressure and 1.6 for tension.

It is convenient to choose pipes graphically. On a pressure–depth chart we draw the lines of the design loads, already multiplied by the design factors, and we draw the ratings of the pipes as vertical lines. A pipe is suitable at those depths where its line lies to the right of the load line. The crossing point shows to what depth we can use the pipe.

CalculatedChoosing pipes for the intermediate string of the example below. The solid lines are the design loads with design factors: collapse grows with depth, and burst falls. The dashed verticals are the collapse pressure and the internal yield pressure of 244.5 mm pipes of grade N-80 with a wall of 11.05 and 11.99 mm. The dots show the limits of use of the pipe with a wall of 11.05 mm: by collapse, not deeper than about 2180 m, and by burst, not shallower than about 270 m. The two load lines cross almost at the same point as the lower limit. This is a coincidence of the example data: the crossing point itself has no meaning.

Worked example244.5 mm intermediate string: loads and sections

Given (assumed data)

String: 244.5 mm (9⅝″), shoe at 2600 m; vertical well
Mud while running; pore pressure above the shoe: 1.09 g/cm³; 1.05 g/cm³
Next interval: to 4000 m, pore pressure at the bottom 1.60 g/cm³, mud 1.64 g/cm³
Fracture pressure under the shoe: 1.784 g/cm³; for the burst design, with a margin of 0.06 g/cm³ added
Gas density (with a margin on the low side): 0.20 g/cm³
Design factors: collapse 1.125; internal pressure 1.10
Pipes N-80, wall 11.05 mm: $p_{\text{iy}}$ = 43.66 MPa, $p_{\text{col}}$ = 26.27 MPa
Pipes N-80, wall 11.99 mm: $p_{\text{iy}}$ = 47.37 MPa, $p_{\text{col}}$ = 32.78 MPa

Solution

Collapse, full emptying during lost circulation. Outside is mud of 1.09 g/cm³: at the shoe $p_{\text{ne}} = 0.00981\cdot 1.09\cdot 2600 = 27.80$ MPa, and with the factor 1.125 it is 31.28 MPa. At the wellhead it is zero. The design load grows by 1.203 MPa every 100 m.
Burst, the well is closed and full of gas. Formation pressure at 4000 m: $0.00981\cdot 1.60\cdot 4000 = 62.78$ MPa. Under a column of gas down to the shoe it would be $62.78 - 0.00981\cdot 0.20\cdot 1400 = 60.04$ MPa. The limit by fracture is $0.00981\cdot(1.784 + 0.06)\cdot 2600 = 47.03$ MPa. This is lower than the pressure under the gas column, so $p_{\text{shoe}} = 47.03$ MPa.
Wellhead pressure: $p_{\text{wh}} = 47.03 - 0.00981\cdot 0.20\cdot 2600 = 41.93$ MPa. Outside is the pore pressure: zero at the wellhead, and at the shoe $0.00981\cdot 1.05\cdot 2600 = 26.78$ MPa. The internal excess pressure is 41.93 MPa at the wellhead and $47.03 - 26.78 = 20.25$ MPa at the shoe. With the factor 1.10 these are 46.13 and 22.28 MPa. The design load falls by 0.917 MPa every 100 m.
Check of the wellhead pressure in oilfield units: the shoe is at 8530 ft, the fracture pressure with the margin is $1.844\cdot 8.345 = 15.39$ ppg, and $p_{\text{shoe}} = 0.052\cdot 15.39\cdot 8530 = 6826$ psi. The gas gradient is 0.0868 psi/ft, and $p_{\text{wh}} = 6085$ psi, that is 41.96 MPa: the same, within the rounding of the constants.
Choice by burst. At the wellhead we need a pipe with $p_{\text{iy}} \ge 46.13$ MPa, and the wall of 11.99 mm (47.37 MPa) is suitable. The pipe with a wall of 11.05 mm (43.66 MPa) is suitable only below $(46.13 - 43.66)\cdot 100/0.917 \approx 270$ m.
Choice by collapse. The pipe with a wall of 11.05 mm (26.27 MPa) is suitable down to the depth $26.27\cdot 100/1.203 \approx 2180$ m. Below that we need the wall of 11.99 mm: 32.78 MPa is more than 31.28 MPa at the shoe.
Sections, rounded to the safe side: 0–300 m, wall 11.99 mm; 300–2100 m, 11.05 mm; 2100–2600 m, 11.99 mm.

Answer: The design loads: collapse from zero at the wellhead to 31.28 MPa at the shoe, burst from 46.13 MPa at the wellhead to 22.28 MPa at the shoe. The string has three sections: a wall of 11.99 mm at the top (to 300 m) and at the bottom (below 2100 m), and 11.05 mm in the middle.

What it means. Burst decided the top of the string, collapse decided the bottom, and in the middle a lighter pipe is enough: at 1800 m it saves about 9.4 t of steel. The drift diameter of the pipe with a wall of 11.99 mm is 216.55 mm, so the 215.9 mm bit for the next interval passes. The wellhead pressure of 41.93 MPa is above 35 MPa, so the casing head and the BOP equipment need the working pressure of the next step, 70 MPa. Full emptying is a severe assumption. Suppose the formation under the shoe, with a pressure of 1.05 g/cm³ at 2650 m, takes mud. Then mud of 1.64 g/cm³ will fall only to 953 m, and the collapse load will not exceed 10.2 MPa. The choice is not yet finished: the tension check and the correction of the collapse pressure for tension are still ahead. The data in this example are assumed for illustration.

Tension: the weight of the string and additional forces

Each cross-section of the string carries the weight of the pipes that hang below it. The tension is greatest at the wellhead, but we also check the top of each section: where a thinner wall or a weaker connection begins, the margin may be smaller than at the top of the string. In mud a string is lighter than in air, because the mud pushes up on it. We take this lifting force into account with the buoyancy factor.

Axial load from the weight of the string in mud

F(z) = g\,k_{\text{b}}\sum_i m_i\,l_i, \qquad k_{\text{b}} = 1 - \frac{\rho_m}{\rho_{\text{steel}}}

where:

$F(z)$ tensile force in the cross-section at the depth $z$ , N
$m_i$ nominal weight (mass per metre) of the pipes of the $i$ -th section, with couplings, kg/m
$l_i$ length of the part of the $i$ -th section that lies below the cross-section, m
$k_{\text{b}}$ buoyancy factor, —
$\rho_m,\ \rho_{\text{steel}}$ density of the mud and of steel (7.85 g/cm³), g/cm³
$g$ acceleration due to gravity, 9.81 m/s², m/s²

Units: SI; the force in newtons, 1 MN = 10⁶ N ≈ 102 tonnes-force ≈ 224.8 thousand pounds-force

Holds when

The string hangs freely, with mud of the same density inside and outside.
The hole is vertical. In an inclined hole, we take the projection of the length on the vertical instead of the length, and friction against the wall adds a load when the crew pulls the string or moves it up and down.
This is the weight of the string in the fluid. The true axial force in the pipe body below the wellhead is a little smaller (see the advanced section), so the estimate is on the safe side.

The heavier the mud, the lighter the string in it: in mud of 1.09 g/cm³ it loses 14% of its weight, and in mud of 1.90 g/cm³ almost a quarter.

Forces that arise in particular operations add to the weight. Three of them are easy to estimate.

Additional axial loads: bending, shock, pressure test

\sigma_{\text{bend}} = \frac{E\,D}{2R}, \qquad F_{\text{shock}} = \rho_{\text{steel}}\,c\,\Delta v\,A_s, \qquad F_{\text{test}} = p_{\text{test}}\,A_{\text{in}}

where:

$\sigma_{\text{bend}}$ the greatest bending stress in the pipe wall in a curved hole: tension on the convex (outer) side of the bend, Pa
$E$ Young's modulus of steel, about 207,000 MPa, Pa
$D$ outside diameter of the pipe, m
$R$ radius of curvature of the hole, m
$F_{\text{shock}}$ additional force when the speed of the string changes suddenly by $\Delta v$ : landing on the slips, a jerk, moving the string up and down, N
$c$ speed of the longitudinal wave along the pipe (the speed of sound along it, as in a rod), $\sqrt{E/\rho_{\text{steel}}}$ , about 5100 m/s; in massive steel the wave is faster, about 5900 m/s, m/s
$A_s,\ A_{\text{in}}$ area of the cross-section of the pipe wall and area of its inside cross-section, m²
$F_{\text{test}}$ additional tension when the string is pressure-tested with the pressure $p_{\text{test}}$ , when the top plug has landed on the float collar, N

Units: SI

Holds when

Bending: the pipe follows the axis of the hole. We add the force $\sigma_{\text{bend}} A_s$ to the axial load as the equivalent of bending, and we check the sections with sharp curvature.
Shock: an instant stop and an elastic wave in a long string. This is an upper estimate. With smooth work the force is much smaller.
Pressure test: the bottom of the string is closed with a plug, and the pressure acts on the whole area of the inside cross-section.

Bending and shock depend on how the hole is built and how the string is run. The pressure test depends on the test pressure we assume. In addition, we allow in advance for a margin of overpull (extra pull kept in reserve), in case the crew has to work the string up and down when the pipe sticks.

In other unit systems (2)

Metric units

\sigma_{\text{bend}}\,[\text{MPa}] = 60.2\;D\,[\text{m}]\;i\,[^\circ/30\ \text{m}], \qquad F_{\text{shock}}\,[\text{N}] = 40.3\;\Delta v\,[\text{m/s}]\;A_s\,[\text{mm}^2]

$i$ is the dogleg severity in degrees per 30 m; $D$ in m; $\Delta v$ in m/s; $A_s$ in mm².

$60.2 = 207\,000\,\pi/(2\cdot 30\cdot 180)$ : the radius of curvature is $30\cdot 180/(\pi i)$ m. $40.3$ MPa per 1 m/s is the product of the density of steel, 7850 kg/m³, and the speed of sound $\sqrt{E/\rho_{\text{steel}}} \approx 5135$ m/s.

Oilfield units

\sigma_{\text{bend}}\,[\text{psi}] = 218\;D\,[\text{in}]\;i\,[^\circ/100\ \text{ft}], \qquad F_{\text{shock}}\,[\text{lbf}] = 1780\;\Delta v\,[\text{ft/s}]\;A_s\,[\text{in}^2]

$i$ in degrees per 100 feet; $D$ in inches; $\Delta v$ in feet per second; $A_s$ in square inches.

218 comes from Young's modulus of $30\cdot 10^6$ psi. If we multiply the stress by the wall area $A_s \approx m/3.4$ ( $m$ is the nominal weight, lb/ft), we get the common form of the force, $64\,i\,D\,m$ pounds. 1780 psi per 1 ft/s is the same 40.3 MPa per 1 m/s.

Worked exampleTension in the chosen string

Given (assumed data)

Sections from the top down: 300 m with a wall of 11.99 mm; 1800 m of 11.05 mm; 500 m of 11.99 mm
Nominal weight with couplings: 69.9 kg/m (47 lb/ft) and 64.7 kg/m (43.5 lb/ft)
Mud: 1.09 g/cm³
Pressure-test pressure when the plug lands: 30 MPa
Design factor for tension: 1.6

Solution

Mass of the string: $300\cdot 69.9 + 1800\cdot 64.7 + 500\cdot 69.9 = 172{,}380$ kg. Weight in air: $172{,}380\cdot 9.81 = 1.691$ MN. Buoyancy factor: $1 - 1.09/7.85 = 0.861$ . Weight in mud, which is the tension at the wellhead: 1.456 MN (327 thousand lbf).
Top of the section with a wall of 11.05 mm (300 m): 151,410 kg hangs below, and the tension is 1.279 MN.
Strength of the N-80 pipe body: the wall area for 11.99 mm is $\pi\cdot(244.5 - 11.99)\cdot 11.99 = 8758$ mm², and $F_y = 552\cdot 8758 = 4.83$ MN (1087 thousand lbf). For the wall of 11.05 mm it is 8104 mm² and 4.47 MN.
Margin for the weight: at the wellhead $4.83/1.456 = 3.32$ , and at 300 m $4.47/1.279 = 3.50$ . Both are much more than 1.6.
Pressure test: the area of the inside cross-section is $\frac{\pi}{4}\cdot 220.52^2 = 38{,}193$ mm², and the additional force is $30\cdot 38{,}193 = 1.146$ MN. In total at the wellhead $1.456 + 1.146 = 2.602$ MN, and the margin is $4.83/2.602 = 1.86$ , which is more than 1.6.

Answer: The tension at the wellhead is 1.456 MN from the weight in mud, and up to 2.602 MN during the pressure test. The strength of the pipe body, 4.83 MN, gives a margin of at least 1.86.

What it means. The weight alone loads the pipe by less than a third of its strength, and the case that decided was the pressure test: the pressure on the area of the inside cross-section added a force comparable to the weight of the string. A sudden stop of the string from a speed of 0.5 m/s would add up to 0.18 MN, so the crew runs the string smoothly. The connection is checked in the same way: its tensile strength is taken from the catalogue, and for round threads, which have a triangular profile, it is noticeably lower than the strength of the body. The weight in mud is taken for a mud of 1.09 g/cm³. During cementing there is heavier cement slurry behind the string, and the tension is lower, so the calculation is on the safe side. The data in this example are assumed for illustration.

Biaxial loading: a pipe in tension is weaker in collapse

The rated pressures of a pipe are determined for a pipe with no axial load. In a well, the pipe is in tension from its own weight, and this changes its resistance to pressure. Steel starts to yield when a combination of stresses reaches the limit, not each stress separately. External pressure compresses the wall around the circumference, and tension pulls it along the axis. Stresses of opposite signs add their effects, and the reserve of the steel against hoop compression (squeezing around the circumference) gets smaller. Internal pressure stretches the wall around the circumference. This stress has the same sign as the axial one, and a pipe in tension holds the internal pressure a little better.

How axial load changes the resistance of a pipe to pressure

Axial load: Tension
Collapse resistance: falls
Resistance to internal pressure: rises

Axial load: Compression
Collapse resistance: rises
Resistance to internal pressure: falls

Axial load	Collapse resistance	Resistance to internal pressure
Tension	falls	rises
Compression	rises	falls

Only the unfavourable combinations are taken into the design: we do not take a rise of the rating as a margin. In a string the main combination is tension and collapse.

Yield strength reduced for axial tension

Y_{pa} = Y_p\left[\sqrt{1 - 0.75\left(\frac{\sigma_a}{Y_p}\right)^2} - 0.5\,\frac{\sigma_a}{Y_p}\right]

where:

$Y_{pa}$ reduced yield strength: the yield strength of a pipe that is also being pulled. It is lower than $Y_p$ , and we use it instead of $Y_p$ to calculate the collapse pressure of a pipe in tension, MPa
$Y_p$ minimum yield strength of the steel, MPa
$\sigma_a$ axial tensile stress: the axial force divided by the area of the pipe wall, MPa

Units: any consistent units of stress

Holds when

The von Mises yield condition (a rule for when steel yields under combined stresses) for the hoop and axial stresses. The radial stress is not included (the derivation is in the advanced section).
The formula is used for tension. The favourable effect of compression is not taken as a margin.
Bending is not included in $\sigma_a$ . Internal pressure, if there is any, is taken into account separately.

We recalculate the collapse pressure by the formulas of the standard with $Y_{pa}$ instead of $Y_p$ . It falls less than the yield strength does: for pipes that collapse by loss of stability, the strength of the steel matters little (collapse regimes).

CalculatedThe reduced yield strength as a fraction of the original, against axial tension. The solid line is for the collapse calculation: the more a pipe is stretched, the less is left for resisting external pressure. The dashed line is the same for internal pressure: moderate tension raises its resistance, so the value goes above 1. The dots are the string from the lesson's examples: the cross-section at 2100 m and the wellhead.

Worked exampleA change to a thinner wall: the correction for tension

Given (assumed data)

String: from the previous examples; below 2100 m there are 500 m of pipes with a wall of 11.99 mm (69.9 kg/m)
Pipe in the cross-section at 2100 m: wall 11.05 mm, N-80: wall area 8104 mm², $p_{\text{col}}$ = 26.27 MPa with no axial load
Design collapse load: 1.203 MPa per 100 m (with the factor 1.125)

Solution

Tension in the cross-section at 2100 m: $500\cdot 69.9\cdot 9.81\cdot 0.861 = 0.295$ MN; $\sigma_a = 0.295\cdot 10^6/8104 = 36.4$ MPa; $\sigma_a/Y_p = 0.066$ .
Reduced yield strength: $Y_{pa} = 552\cdot[\sqrt{1 - 0.75\cdot 0.066^2} - 0.5\cdot 0.066] = 552\cdot 0.9654 = 532.9$ MPa.
Collapse pressure with $Y_{pa}$ by the formulas of the standard (the advanced section): 25.98 MPa instead of 26.27 MPa, which is 1.1% less.
Design load at 2100 m: $1.203\cdot 21 = 25.26$ MPa. The condition is met: 25.98 MPa is 2.8% more.
If the change had been made at the limiting depth found with no correction (about 2180 m), the pipe would not have passed the check: with the correction the limit is about 2160 m.
At the wellhead the tension is greatest: $\sigma_a/Y_p = 0.30$ , the factor is 0.815, and the collapse pressure of the pipe with a wall of 11.99 mm falls from 32.8 to 29.6 MPa, by 10%. But there is no external excess pressure there.

Answer: In the cross-section at 2100 m, tension lowers the collapse pressure of the pipe from 26.27 to 25.98 MPa. The design load of 25.26 MPa is lower, so the sections were chosen correctly.

What it means. The correction is small here, because only 500 m of pipe hangs below the change. Tension weakens most the top of the string, where there is almost no collapse load. At the bottom, where the load is large, the pipe is hardly stretched. The middle of the string, and the points of change to a weaker pipe in long heavy strings, are the dangerous places: there the correction reaches several per cent or more, and the change must be moved up. We always check the lowest point of every section. The data in this example are assumed for illustration.

Production string: a leaking tubing and emptying

We design a production string for the conditions of production. In a producing well the formation fluid goes up through the tubing, and the annulus above the packer (a seal that closes the annulus above the formation) is filled with packer fluid and cut off from the formation (Well completion: opening the reservoir and the tubing string). The string is a backup barrier, in case the tubing or the packer starts to leak.

The worst case for burst is a leak in the tubing right at the wellhead of a closed gas well. The wellhead pressure of such a well equals the formation pressure less the light column of gas, and through the leak it passes into the annulus. Below, the weight of the packer fluid is added to it, so the highest pressure inside the string is at the packer, not at the wellhead. For collapse, the worst case is an empty annulus with the highest pressure outside.

Worked example177.8 mm production casing of a gas well

Given (assumed data)

String and equipment: 177.8 mm (7″) to 4000 m; packer at 3900 m; perforations at 3950 m
Formation pressure at 3950 m: equivalent mud weight 1.60 g/cm³
Gas; packer fluid: 0.20 g/cm³; 1.10 g/cm³
Outside, for burst: a column of mix water of 1.00 g/cm³
Outside, for collapse: the mud in which the string was run, 1.64 g/cm³
Design factors: collapse 1.125; internal pressure 1.10

Solution

Formation pressure: $0.00981\cdot 1.60\cdot 3950 = 62.00$ MPa. Pressure at the wellhead of the closed well: $62.00 - 0.00981\cdot 0.20\cdot 3950 = 54.25$ MPa.
A leak in the tubing at the wellhead: 54.25 MPa acts on the packer fluid. At the packer, inside the string, the pressure is $54.25 + 0.00981\cdot 1.10\cdot 3900 = 96.33$ MPa, and outside it is $0.00981\cdot 1.00\cdot 3900 = 38.26$ MPa.
Internal excess pressure: 54.25 MPa at the wellhead and 58.08 MPa at the packer. With the factor 1.10, these are 59.67 and 63.88 MPa.
Collapse: the annulus is empty down to the packer, and outside is mud of 1.64 g/cm³: $0.00981\cdot 1.64\cdot 3900 = 62.74$ MPa, and with the factor 1.125 it is 70.59 MPa.
The pipe with a wall of 11.51 mm of grade P-110: $p_{\text{iy}} = 0.875\cdot 2\cdot 758\cdot 11.51/177.8 = 85.9$ MPa, and $p_{\text{col}}$ = 74.4 MPa. Both conditions are met. The same wall in N-80 steel (62.5 and 59.4 MPa) fails both.

Answer: The design loads: internal pressure up to 63.88 MPa at the packer, and collapse 70.59 MPa. The 177.8 × 11.51 mm pipe of grade P-110 is suitable. The margin in collapse above the design factor is 5%.

What it means. The greatest burst load turned out to be at the packer, although the gas pressure is applied at the wellhead: the packer fluid is heavier than the mix water behind the string. The heavier the packer fluid, the higher this load. The wellhead pressure of 54.25 MPa needs wellhead equipment for 70 MPa. Suppose a liner had been run instead of a string to the wellhead. Then the production pressures above its top would fall on the 244.5 mm intermediate string of the first example, whose internal yield pressure is not more than 47.4 MPa. A string that becomes the production string after a liner is run is checked for these cases too. If the gas contains hydrogen sulfide, P-110 steel cannot be used at moderate temperature (grades), and the same strength is obtained with a thicker wall of a steel with limited hardness. The data in this example are assumed for illustration.

What else is taken into account, and what the whole design looks like

Flowing salts. Over time, salt puts a pressure on the string that is close to the overburden pressure. The collapse design uses a pressure gradient (the rise of pressure per metre of depth) of about 22–23 kPa/m (1 psi/ft) instead of the pressure of the mud, and the string in this interval is made thick-walled.
Hydrogen sulfide. In its presence high-strength steels crack under load. So the choice is limited to grades with controlled hardness, and the strength needed is obtained from the thickness of the wall (grades).
Temperature. As the temperature rises, the yield strength of steel falls: by several per cent at 100–150 °C. We take the correction from the manufacturer's data. Heating of the string during production causes compression, and it may lead to buckling (sideways bending) where the string is not cemented.
Curvature and wear. In a curved hole bending is added to tension, and over the time of drilling the tool joints of the drill string (the thick threaded ends of the pipes) wear the wall. The designer allows for the loss of thickness in advance: it lowers both the internal yield pressure and the collapse pressure.
Connections. A connection must be no weaker than the tension requires, and tight for the fluid that the string will carry. Where possible, one type of connection is used along the whole string.
Availability of pipes. A warehouse usually has only a few combinations of wall and grade. So, of several workable options, we choose the cheapest of those available, and limit the number of sections to two or three.

Order of string design

1
Input data
Depths and diameters of the strings, the forecast of the pore pressure and the fracture pressure, the densities of muds and cement, the method of production.
2
Load cases
For the string in question: what is inside and what is outside while running, cementing, drilling below, and producing.
3
Collapse and burst
Pressure profiles, excess pressures at the break points, multiplication by the design factors.
4
Choice of pipes
Pipes whose ratings are not less than the design loads at their depths; sections rounded to the safe side.
5
Tension
The weight in mud and additional forces; a check of the pipe body and the connection at the top of each section.
6
Biaxial correction
Recalculation of the collapse pressure for tension at the bottom of each section; a return to the choice if needed.
7
Documentation
All the assumptions made are written down: they let someone check the design when conditions change.

Terms and symbols

Load case: A situation in the life of the string for which the pressures inside and outside are set.
External excess pressure, $p_{\text{ne}}$: The difference of the external and internal pressures at the same depth; the collapse load.
Internal excess pressure, $p_{\text{ni}}$: The difference of the internal and external pressures at the same depth; the burst load.
Emptying of the string: A fall of the fluid level in the string during lost circulation, when the well is brought in, or during production. Full emptying is down to the shoe or the packer.
Design factor: The number by which the design load is multiplied before it is compared with the rating of the pipe.
Buoyancy factor, $k_{\text{b}}$: The ratio of the weight of the string in mud to its weight in air: $1 - \rho_m/\rho_{\text{steel}}$ .
Biaxial loading: The simultaneous action of pressure and axial force; tension lowers the collapse resistance.
Reduced yield strength, $Y_{pa}$: The yield strength lowered for axial tension; we use it to calculate the collapse pressure of a pipe in tension.
String section: A part of the string made of pipes of one wall thickness, one grade and one connection type.

Common mistakes and misconceptions

Wrong

“At 3000 m the pressure is 35 MPa, so the pipe must withstand 35 MPa.”

In fact

The pipe is loaded by the difference of the pressures inside and outside. With equal columns of fluid on both sides it is close to zero. The dangerous cases are those where one side is emptied or is under pressure.

Wrong

“The pressure at the wellhead of a closed well is much lower than at the shoe.”

In fact

If the well is full of gas, the difference is only the weight of the gas column: about 5 MPa over 2600 m at a density of 0.2 g/cm³. So the internal pressure decides the choice of pipes right at the wellhead.

Wrong

“It is better to take the fracture pressure under the shoe lower, for safety.”

In fact

It is safe only in the calculation of the allowable wellhead pressure when killing a well. For a string, an underestimated fracture pressure underestimates the burst load, so we take the upper estimate.

Wrong

“Tension is checked only at the wellhead.”

In fact

We check the margin at the top of every section and for every connection, and we check the effect of tension on collapse at the bottom of every section.

Wrong

“A design factor of 1.0 means no margin.”

In fact

The ratings already contain the minimum yield strength and the wall tolerance, and the load cases are chosen on the cautious side. But with such a factor there is indeed no margin left for wear, corrosion and an error in the pressure forecast.

In depth: advanced detail

The collapse pressure formulas of the standard

The standard (API TR 5C3, ISO 10400) calculates the collapse pressure by one of four formulas, depending on the ratio $D/t$ and the yield strength. Thick-walled pipes collapse when the steel on the inside surface reaches yield. The thinnest-walled pipes lose stability while the steel is still elastic (it springs back). Between them lie the “plastic” regime, described by an empirical formula (a formula fitted to test results), and the “transition” regime, which joins it smoothly to the elastic one.

Collapse pressure in the four regimes

\begin{aligned} p_Y &= 2\,Y_p\,\frac{D/t - 1}{(D/t)^2} && \text{— yield regime} \\ p_P &= Y_p\left(\frac{A}{D/t} - B\right) - C && \text{— plastic} \\ p_T &= Y_p\left(\frac{F}{D/t} - G\right) && \text{— transition} \\ p_E &= \frac{46.95\cdot 10^{6}}{(D/t)\,(D/t - 1)^2} && \text{— elastic} \end{aligned}

where:

$p_Y,\ p_P,\ p_T,\ p_E$ collapse pressure in the yield, plastic, transition and elastic regimes, psi
$Y_p$ minimum yield strength of the steel (or the reduced $Y_{pa}$ ), psi
$D/t$ ratio of the outside diameter to the wall thickness, —
$A,\ B,\ C,\ F,\ G$ coefficients that depend on the yield strength (see below), C in psi, the others dimensionless

Units: the formulas of the standard are written in psi: convert the yield strength to psi (1 MPa = 145.04 psi), and convert the result back to MPa

Holds when

We choose the regime by $D/t$ : yield up to $(D/t)_{YP}$ , plastic up to $(D/t)_{PT}$ , transition up to $(D/t)_{TE}$ , and elastic above. The boundaries are given below.
The plastic-regime formulas are empirical, from tests of pipes. The numerator of the elastic formula is the theoretical value reduced by about 29% to allow for imperfections of shape.
The pipe has no axial load and no internal pressure. We allow for tension by putting $Y_{pa}$ in place of $Y_p$ .

In the elastic regime the yield strength does not enter the formula at all: a stronger steel does not help a thin-walled pipe. In the plastic and transition regimes the collapse pressure grows with the yield strength, but more slowly than in proportion.

Coefficients of the collapse pressure formulas

\begin{aligned} A &= 2.8762 + 0.10679\cdot 10^{-5}\,Y_p + 0.21301\cdot 10^{-10}\,Y_p^2 - 0.53132\cdot 10^{-16}\,Y_p^3 \\ B &= 0.026233 + 0.50609\cdot 10^{-6}\,Y_p \\ C &= -465.93 + 0.030867\,Y_p - 0.10483\cdot 10^{-7}\,Y_p^2 + 0.36989\cdot 10^{-13}\,Y_p^3 \\ F &= \frac{46.95\cdot 10^{6}\,s^3}{Y_p\,(s - B/A)\,(1 - s)^2}, \qquad s = \frac{3\,B/A}{2 + B/A}, \qquad G = \frac{F\,B}{A} \end{aligned}

where:

$Y_p$ yield strength, psi
$s$ an auxiliary quantity, —

Units: yield strength in psi

Holds when

The coefficients are valid only with $Y_p$ in psi.
The boundaries of the regimes: $(D/t)_{PT} = \dfrac{Y_p\,(A - F)}{C + Y_p\,(B - G)}$ , $(D/t)_{TE} = \dfrac{2 + B/A}{3\,B/A}$ . The boundary $(D/t)_{YP}$ is the point where the yield formula and the plastic formula give the same value.

For N-80 steel ( $Y_p$ = 80,000 psi) we get $A$ = 3.071, $B$ = 0.0667, $C$ = 1955, $F$ = 1.998, $G$ = 0.0434, and the regime boundaries 13.38, 22.47 and 31.02.

Worked exampleCollapse pressure of a 244.5 × 11.05 mm pipe of two steels

Given (assumed data)

Pipe: $D$ = 244.5 mm, $t$ = 11.05 mm
Steels: N-80 (80,000 psi, 552 MPa) and P-110 (110,000 psi, 758 MPa)

Solution

Ratio $D/t = 244.5/11.05 = 22.13$ .
N-80: $A = 3.071$ , $B = 0.0667$ , $C = 1955$ psi; the regime boundaries are 13.38, 22.47 and 31.02. The value 22.13 lies between the first and the second, so the regime is plastic: $p_P = 80{,}000\cdot(3.071/22.13 - 0.0667) - 1955 = 3810$ psi, that is 26.27 MPa.
P-110: $F = 2.066$ , $G = 0.0532$ ; the boundaries are 12.44, 20.41 and 26.22. Now 22.13 lies between the second and the third, so the regime is transition: $p_T = 110{,}000\cdot(2.066/22.13 - 0.0532) = 4418$ psi, that is 30.46 MPa.

Answer: The collapse pressure is 26.27 MPa for N-80 and 30.46 MPa for P-110.

What it means. The steel is 37.5% stronger, but the collapse pressure grew by only 16%: a pipe with such a $D/t$ is already close to loss of stability, and the strength of the steel helps it little. The internal yield pressure would grow by the full 37.5%. So it is more effective to raise the collapse resistance with the wall thickness, and the resistance to internal pressure with both the wall and the steel. The values match the tables of the standard (3810 and 4420 psi). The sizes follow the standard series.

Buoyancy: the weight in fluid and the true axial force

The buoyancy factor gives the correct load on the hook, but it does not show how the axial force is spread along the length. A fluid presses on a pipe only at right angles to its surface. So it pushes the string up only where there is a downward-facing area, at the lower end face. The force on the end face equals the pressure at the shoe multiplied by the area of the wall cross-section, and it compresses the bottom of the string. Higher up, the axial force grows by the weight of the pipes in air.

True axial force in a string of one wall thickness

F_a(z) = g\,m\,(L - z) - p_L\,A_s

where:

$F_a(z)$ axial force in the pipe body at the depth $z$ ; positive is tension, N
$m$ mass of a metre of plain pipe (without couplings), kg/m
$L$ depth of the shoe, m
$p_L$ fluid pressure at the depth of the shoe, Pa
$A_s$ area of the wall cross-section, m²

Units: SI

Holds when

A vertical string of one wall thickness with an open bottom, and fluid of the same density inside and outside.
If the bottom is closed by a float valve and the pressures inside and outside differ, the force on the end face is $p_{\text{out}} A_{\text{out}} - p_{\text{in}} A_{\text{in}}$ , where $A_{\text{out}}$ and $A_{\text{in}}$ are the areas of the outer and inner cross-sections.
In a string of sections with different walls, the pressure also acts on the steps at the changes.

At the wellhead the formula gives the same result as the buoyancy factor. At the depth $z$ the true force is smaller than the weight in fluid of the pipes below by $p(z)\,A_s$ . So the biaxial correction made from the weight in fluid is an estimate on the safe side.

Worked exampleWhere the axial force changes sign

Given (assumed data)

String: 2600 m of pipes 244.5 × 11.99 mm; wall area 8758 mm², mass of plain pipe 68.75 kg/m
Mud: 1.09 g/cm³

Solution

Weight in air: $2600\cdot 68.75\cdot 9.81 = 1.754$ MN. Pressure at the shoe: 27.80 MPa. Force on the lower end face: $27.80\cdot 8758 = 0.243$ MN, compression.
At the wellhead: $1.754 - 0.243 = 1.510$ MN. By the buoyancy factor: $1.754\cdot 0.861 = 1.510$ MN, the same.
The axial force is zero where the weight of the pipes below the cross-section equals the force on the end face: at a height of $0.243\cdot 10^6/(68.75\cdot 9.81) = 361$ m above the shoe.

Answer: The lower 361 m of the string (14% of its length) is compressed axially, and above it the string is in tension; at the wellhead it is 1.510 MN.

What it means. The share of the compressed part equals the ratio of the densities of the mud and of steel. This compression does not threaten buckling: the fluid pressure also acts on the side surface of the pipe, and a string that hangs freely in fluid stays straight. For the collapse calculation at the bottom it is even favourable. The data in this example are assumed for illustration.

Equivalent stress: a triaxial calculation

Three stresses act in the wall of a pipe under pressure: axial $\sigma_a$ , hoop (circumferential) $\sigma_t$ and radial (across the wall) $\sigma_r$ . Steel starts to yield when a combination of them reaches the yield strength: the von Mises equivalent stress. The uniaxial checks and the biaxial correction are simplifications of this condition. The full calculation is made for the most important strings: at high pressures and temperatures, in wells with hydrogen sulfide, and with difficult combinations of loads.

Von Mises equivalent stress on the inside surface of a pipe

\sigma_{\text{eq}} = \sqrt{\sigma_a^2 + \sigma_t^2 + \sigma_r^2 - \sigma_a\sigma_t - \sigma_t\sigma_r - \sigma_r\sigma_a}, \qquad \sigma_r = -p_{\text{in}}, \qquad \sigma_t = \frac{p_{\text{in}}\,(r_{\text{in}}^2 + r_{\text{out}}^2) - 2\,p_{\text{out}}\,r_{\text{out}}^2}{r_{\text{out}}^2 - r_{\text{in}}^2}

where:

$\sigma_{\text{eq}}$ equivalent stress; we compare it with the yield strength divided by the design factor (usually 1.25), MPa
$\sigma_a$ axial stress: the true axial force divided by the wall area, plus the bending stress, MPa
$\sigma_t,\ \sigma_r$ hoop and radial stresses on the inside surface (the Lamé solution for a thick-walled pipe), MPa
$p_{\text{in}},\ p_{\text{out}}$ pressure inside and outside the pipe, MPa
$r_{\text{in}},\ r_{\text{out}}$ inside and outside radii, mm

Units: any consistent units of stress and pressure; radii in the same units

Holds when

An elastic pipe with a uniform wall. Under pressure, the highest equivalent stress is on the inside surface.
Tension is positive. Shear stresses from torsion are not included.
The condition describes the start of yield, not loss of stability: the collapse of thin-walled pipes is checked separately.

If we put $\sigma_r = 0$ and take the hoop stress as compressive, the condition $\sigma_{\text{eq}} = Y_p$ gives the formula of the reduced yield strength from the main text: under tension $\sigma_a$ , the hoop compression can only reach $Y_{pa}$ . With a tensile hoop stress, a plus sign appears in front of the second term. This is why the resistance to internal pressure rises.

Worked exampleThe wellhead of the intermediate string with the well closed

Given (assumed data)

Pipe: 244.5 × 11.99 mm, N-80 (552 MPa)
Pressures: 41.93 MPa inside, zero outside
Axial force: 1.456 MN, the weight of the string in mud

Solution

Radii: $r_{\text{out}} = 122.25$ mm, $r_{\text{in}} = 110.26$ mm. Hoop stress: $\sigma_t = 41.93\cdot\dfrac{110.26^2 + 122.25^2}{122.25^2 - 110.26^2} = 407.7$ MPa; radial stress $\sigma_r = -41.93$ MPa.
With no axial load: $\sigma_{\text{eq}} = 430.2$ MPa, and the margin is $552/430.2 = 1.28$ .
With tension $\sigma_a = 1.456\cdot 10^6/8758 = 166.3$ MPa: $\sigma_{\text{eq}} = 389.7$ MPa, and the margin is 1.42.
For comparison, the same pipe with the smallest allowed wall ( $0.875\,t$ ) and no axial load: $\sigma_{\text{eq}} = 490.9$ MPa, and the margin is 1.12.

Answer: The margin by the equivalent stress is 1.28 with no axial load and 1.42 with tension. Both are greater than 1.25.

What it means. Tension reduced the equivalent stress: hoop and axial stresses of the same sign partly “subtract” in the von Mises condition. With the smallest allowed wall, the margin with no axial load, 1.12, is almost the same as the rated internal yield pressure gives (1.13). The Barlow formula with the factor 0.875 is a cautious uniaxial approximation of the same condition. The data in this example are assumed for illustration.

After cementing: heating and changes of pressure

When the cement has set, the string is fixed at both ends: at the bottom by the cement, and at the top in the casing head. The axial force in it is “frozen” at the value that it had at that moment. Everything that afterwards tries to change the length of the free (uncemented) part changes the axial force, because the length cannot change.

Change of the axial force in a fixed string

\Delta F = -E\,\alpha\,\Delta T\,A_s + 2\nu\left(A_{\text{in}}\,\Delta p_{\text{in}} - A_{\text{out}}\,\Delta p_{\text{out}}\right)

where:

$\Delta F$ change of the axial force; positive is tension, N
$E,\ \alpha,\ \nu$ Young's modulus (about 207,000 MPa), the coefficient of linear expansion (about $12\cdot 10^{-6}$ 1/°C) and Poisson's ratio (0.3) of steel, Pa, 1/°C, —
$\Delta T$ average change of temperature of the free part, °C
$\Delta p_{\text{in}},\ \Delta p_{\text{out}}$ average change of pressure inside and outside along the free part, Pa
$A_s,\ A_{\text{in}},\ A_{\text{out}}$ area of the wall, and areas of the inside and outside cross-sections, m²

Units: SI

Holds when

The part is fixed at both ends and stays straight (no buckling).
The changes of temperature and pressures are averaged along the length of the free part.
The steel is elastic.

Heating compresses the string: the pipe wants to get longer but cannot. A rise of the internal pressure inflates it, and it tends to get shorter, so tension appears. So that heating during production does not lead to compression and buckling, the string is sometimes hung with extra tension after the cement has set.

In other unit systems (1)

Oilfield units

\Delta F\,[\text{lbf}] = -200\,\Delta T\,[^\circ\text{F}]\;A_s\,[\text{in}^2] + 0.6\left(A_{\text{in}}\,\Delta p_{\text{in}} - A_{\text{out}}\,\Delta p_{\text{out}}\right)

areas in square inches, pressures in psi, temperature in degrees Fahrenheit.

$200 = 30\cdot 10^6\cdot 6.67\cdot 10^{-6}$ psi per degree Fahrenheit: the same expansion coefficient $12\cdot 10^{-6}$ 1/°C is $6.67\cdot 10^{-6}$ 1/°F. Handbooks also give 207, with the coefficient $6.9\cdot 10^{-6}$ 1/°F. $0.6 = 2\nu$ .

Worked exampleHeating by 30 °C and a pressure rise of 20 MPa

Given (assumed data)

Pipe: 244.5 × 11.99 mm: $A_s$ = 8758 mm², $A_{\text{in}}$ = 38,193 mm²
Changes on the free part: temperature +30 °C; pressure inside +20 MPa, outside unchanged

Solution

Temperature: $E\alpha = 207{,}000\cdot 12\cdot 10^{-6} = 2.48$ MPa per degree; $\Delta F = -2.48\cdot 30\cdot 8758 = -0.65$ MN, compression.
Pressure: $\Delta F = 2\cdot 0.3\cdot 38{,}193\cdot 20 = 0.46$ MN, tension.

Answer: Heating by 30 °C removes 0.65 MN of tension, and a rise of the internal pressure by 20 MPa adds 0.46 MN.

What it means. Each of these forces is a noticeable share of the weight of the string in mud (1.46 MN). So for hot and high-pressure wells the axial load is calculated in stages: the state at the time the cement sets, and then the changes during drilling and production. The data in this example are assumed for illustration.

Connections to other topics

Casing: purpose, types, properties and standards — strings by purpose, grades and the rated characteristics of pipes.
Fracture gradient and leak-off testing — shoe depths and the fracture pressure that limits the pressure in the string.
Well kill methods — what pressures arise in the string and at the shoe while a well is killed.
Blowout prevention equipment — the working pressure of the wellhead and BOP equipment, the same pressure shell as the string.
Abnormal formation pressures: origin, prediction and recognition — the forecast of the pore pressure, on which all the load cases depend.
Well completion: opening the reservoir and the tubing string — the tubing, the packer and the packer fluid: the working conditions of the production string.
Primary cementing — pressures during cementing and the height of the cement.
Wellheads and running casing — hanging the strings, casing heads and running the string.

Check yourself

1
Which load case usually decides the wall thickness in the lower part of an intermediate string?
1. A.Emptying of the string during lost circulation
2. B.A closed well filled with gas
3. C.A pressure test of the string
4. D.The weight of the string in mud
Show the answer
Correct answer: A. Emptying of the string during lost circulation
The external excess pressure during emptying grows with depth and is greatest at the shoe. The burst load in a closed well with gas, on the contrary, is greatest at the wellhead, and tension is also greatest at the top. So we choose the bottom of the string by collapse.
2
A 339.7 mm surface casing is run to 800 m in mud of 1.12 g/cm³. Find the design external excess pressure at the shoe with full emptying and a design factor of 1.1.
Show the answer
There is a column of mud outside and nothing inside: $p_{\text{ne}} = 0.00981\cdot 1.12\cdot 800 \approx 8.79$ MPa. With the design factor: $8.79\cdot 1.1 \approx 9.7$ MPa. The collapse pressure of the pipe at the shoe must be not less than this.
3
The casing shoe is at 1900 m, and the fracture pressure under it, with a margin, is 1.70 g/cm³. Below, at 3000 m, there is a formation with a pressure of 1.35 g/cm³. The well is closed and full of gas of density 0.2 g/cm³. What will the wellhead pressure be?
Show the answer
Formation pressure: $0.00981\cdot 1.35\cdot 3000 \approx 39.7$ MPa. Under a column of gas the pressure at the shoe would be $39.7 - 0.00981\cdot 0.2\cdot 1100 \approx 37.6$ MPa. The limit by fracture is $0.00981\cdot 1.70\cdot 1900 \approx 31.7$ MPa. This is lower, so at the shoe it is 31.7 MPa. At the wellhead: $31.7 - 0.00981\cdot 0.2\cdot 1900 \approx 28.0$ MPa.
4
Why is the external pressure taken as the lowest possible in a burst design?
1. A.It is easier to calculate
2. B.External pressure does not affect burst
3. C.The burst load is the difference of the internal and external pressures: the lower the external pressure, the greater the load, so the calculation is on the safe side
4. D.Because cement does not transmit any pressure
Show the answer
Correct answer: C. The burst load is the difference of the internal and external pressures: the lower the external pressure, the greater the load, so the calculation is on the safe side
The external pressure supports the wall. If we take it higher than it turns out to be, the design load will be underestimated. For collapse the logic is the opposite: there we take the highest pressure outside and the lowest inside.
5
How does axial tension affect the ratings of a casing pipe?
1. A.It does not: the rated pressures do not depend on axial load
2. B.It raises the collapse resistance
3. C.It lowers the collapse resistance and slightly raises the resistance to internal pressure
4. D.It lowers the resistance to collapse and to internal pressure equally
Show the answer
Correct answer: C. It lowers the collapse resistance and slightly raises the resistance to internal pressure
Yield depends on a combination of stresses. Under external pressure the hoop stress is compressive and the axial stress is tensile: stresses of different signs strengthen each other, and we recalculate the collapse pressure with the reduced yield strength $Y_{pa}$ . Under internal pressure the signs are the same, and tension helps, but we do not take this as a margin.
6
A 339.7 mm string at 600 m is cemented to the wellhead: outside is cement slurry of 1.85 g/cm³, and inside is displacement fluid of 1.10 g/cm³. What is the external excess pressure at the shoe, and why must this case not be added to emptying?
Show the answer
$p_{\text{ne}} = 0.00981\cdot(1.85 - 1.10)\cdot 600 \approx 4.4$ MPa. Liquid cement behind the string exists only during cementing, when the string is full of displacement fluid. Emptying during lost circulation is possible later, when the cement has already set and does not transmit hydrostatic pressure. These are different load cases, and we take the larger of them, not the sum.

Updated: 19 October 2026