Drill string design

Quick refresher

In mud the string is lighter. The weight of a metre is $w = m\,g\,k_{\text{b}}$ , with $k_{\text{b}} = 1 - \rho_m/\rho_{\text{steel}}$ . This gives the effective axial force. The weight indicator shows it, and it decides whether a pipe will buckle (bend). The true force in the steel below the wellhead is smaller, by the pressure multiplied by the wall area. But compression by pressure does not bend a pipe.
The pipes at the bottom of the string provide the weight on bit $G$ . The neutral point is the section where the effective force is zero. It lies at a height $G/(w\cos\theta)$ above the bit. In a vertical hole it must stay within the drill collars. So we take their length with a design factor of 1.10–1.20 (most often 1.15). This means the weight on bit is no more than 83–91% of the weight of the collars in mud.
In an inclined hole the wall supports the pipe. The pipe stays straight as long as the compression is less than the critical load $2\sqrt{EIw\sin\theta/r}$ . So in a straight inclined section HWDP can provide the weight on bit, and the crew runs only a few drill collars.
We design the string for tension from the bottom up. The allowable load $P_a$ is 90% of the strength of the pipe body (for used pipe, of the premium class). We leave a margin of overpull above the weight of the string, usually 0.22–0.44 MN (50–100 thousand lbf). We also check crushing in the slips. The weight carried by the slips, multiplied by $S_h/S_t$ (1.42 for a 127 mm pipe in 406 mm slips), must not exceed $P_a$ .
Less strong pipes go at the bottom and stronger ones at the top. The margin of overpull of the whole string is the smallest of the margins at the tops of the sections.
In an inclined hole, friction against the walls increases the hook load when pulling out and reduces it when running in. When rotating, it goes into torque. In the build section (where the hole curves and the angle grows), the tension presses the pipe against the high side (the upper wall), so drag there grows with the tension.

Key relationships

w = m\,g\,k_{\text{b}}, \qquad k_{\text{b}} = 1 - \dfrac{\rho_m}{\rho_{\text{steel}}}

weight of a metre of string in mud

L_{\text{np}} = \dfrac{G}{w\cos\theta}, \qquad L_{\text{dc}} = \dfrac{k_{\text{df}}\,G}{w\cos\theta}

height of the neutral point above the bit and the length of drill collars with a design factor

P_a = 0.9\,F_y, \qquad \text{MOP} = P_a - P, \qquad P\,\dfrac{S_h}{S_t} \le P_a

allowable load, margin of overpull and the check for crushing in the slips

F_{\text{cr}} = 2\sqrt{\dfrac{E\,I\,w\sin\theta}{r}}

the compression at which a pipe lying on the low side of an inclined hole begins to bend in a wave ( $r$ is the radial clearance between the pipe and the wall)

F_{\text{top}} = F_{\text{bot}} + w L\,(\cos\theta \pm \mu\sin\theta), \qquad M = \mu\,w L\sin\theta\;r_{\text{cont}}

straight inclined section by the soft-string model: plus for pulling out, minus for running in; torque when rotating ( $r_{\text{cont}}$ is the contact radius with the wall)

Common pitfalls

Calculating the neutral point from the true axial force, and adding drill collars “to compensate for the buoyant force”. The effective force decides buckling, and extra drill collars only make the string heavier.
Taking the rated strength of a new pipe for a string of used pipe.
Checking the margin of overpull only at the wellhead. In a tapered string the smallest margin is at the top of the lower, weaker section.
Forgetting about crushing in the slips in deep wells. It can limit the length of a section more than the margin of overpull does.
Carrying the rules of an inclined hole over to a vertical one. There the wall does not support the pipe, and compressed drill pipe buckles almost at once.
Taking the friction factor for a property of the rock. It is a tuning parameter of the model, refined from measurements of hook load and torque.

After this lesson you can

allow for buoyancy: calculate the weight of the string in mud, and explain how the effective axial force differs from the true one and why the effective one decides buckling (the bending of a pipe into a wave)
find the neutral point and choose the length of drill collars and HWDP for a given weight on bit, with a design factor, in a vertical and in a deviated hole
check compressed pipe in a straight inclined section against the critical buckling load
design the string for tension from the bottom up: find the allowable load and the margin of overpull, check crushing in the slips, and build a tapered string of pipes of different strength grades
use the soft-string model (the string treated as a flexible rope) to estimate the hook load when pulling out, running in and rotating, and the torque in a directional well
carry out a complete string design for a vertical and for a directional well

The loads on the string and where the design starts

The string works in several modes. In each mode it is loaded in its own way.

Drilling. Weight on bit compresses the bottom of the string, and the weight of the string stretches the top. Torque at the bit and friction against the walls twist the string. In curved intervals, rotating pipes bend back and forth.
Pulling out. Drag (friction) in the inclined hole adds to the weight. If the pipe is stuck, one more load is added: the overpull, which is the pull the driller applies above the weight to free the string.
Running in. Drag reduces the hook load. In holes with a large inclination, the pipe may not have enough weight to reach the bottom.
In the slips. During connections (adding a joint of pipe) and trips (running or pulling the whole string), the string hangs in the slips of the rotary table. The slips squeeze the pipe from outside. Together with the tension from the weight, this can crush the pipe body.
Special operations. These include a drill stem test (testing a formation through the pipe), when the string is run partly empty and outside pressure can collapse it. They also include pressure testing and working with the jar (a tool that strikes a hammer blow to free stuck pipe).

We do the design from the bottom up, because each section carries what hangs below it. First we choose the bottom assembly. It must give the weight on bit and meet the requirements for the direction of the hole. Then we choose the sections of drill pipe. They must carry the tension with a margin, and the slips must not crush them. Last, we check the torque, the pressure and the fatigue (damage from repeated bending). The lesson on the drill string covers the strength of the elements themselves: the pipe body in tension and torsion, torsion under tension, and the wear classes. Here we take ready-made numbers from it.

The order of the design

1
Input data
Gather the profile of the hole, the bit diameter, the mud density, the maximum weight on bit, the required margin of overpull, the length of the slips, and the pipe that is available with its class (the wear grade of used pipe).
2
Bottom assembly
Find the length of the drill collars and HWDP that gives the weight on bit with a design factor. Find the position of the neutral point. Make the change in stiffness gradual (the rule of steps).
3
Tension
Find the allowable load of each grade of pipe. Find the greatest length of the lower section that the margin of overpull and the crushing in the slips allow. Then do the next section, and so on up to the wellhead.
4
Inclined hole
Find the critical load for the compressed sections. Find the hook load when pulling out, running in and rotating, and the torque, by the soft-string model. Count the margin of overpull from the weight when pulling out.
5
Checks
Make sure the torque is less than the make-up torque (the torque that screws the joint together) of the weakest tool joint, and less than the allowable torque of the pipe body under tension (torsion under tension). Check collapse (crushing by outside pressure) when the string is run empty. Check fatigue in intervals with a high dogleg severity, where the hole bends sharply (bending in a curved hole).

The weight of the string in mud

In mud the string is lighter than in air, because a buoyant force (the upward push of the fluid) acts on it. If the mud inside and outside the string has the same density, we allow for this with the buoyancy factor. We do the same for casing strings (axial load on a casing string).

Weight of a metre of string in mud

w = m\,g\,k_{\text{b}}, \qquad k_{\text{b}} = 1 - \frac{\rho_m}{\rho_{\text{steel}}}

where:

$w$ weight of a metre of pipe in mud, N/m
$m$ mass per metre. For drill collars, use the formula for a plain pipe. For drill pipe and HWDP, use the adjusted mass from the catalogue (“adjusted weight”), which includes the tool joints and the upsets (the thickened pipe ends), kg/m
$g$ acceleration due to gravity, 9.81, m/s²
$k_{\text{b}}$ buoyancy factor, —
$\rho_m,\ \rho_{\text{steel}}$ density of the mud and of steel (7.85 g/cm³), g/cm³

Units: SI. The densities can be in any units, as long as both use the same ones.

Holds when

The mud inside and outside the pipe has the same density. If the densities differ (the string is run empty, or there is a weighted slug, a small batch of heavy mud, inside), we calculate the weight by the general formula in the advanced section.
The adjusted mass of a drill pipe is noticeably greater than the nominal one. Take a 127 mm 19.50 lb/ft pipe. Its nominal mass is 29.0 kg/m, the mass of the plain body is 26.7 kg/m, and with NC50 tool joints it is about 34 kg/m.
Weight in air differs from mass by the factor $g$ . One tonne-force (tf) is 9.807 kN.

The heavier the mud, the lighter the string in it. In a 1.25 g/cm³ mud the string loses 16% of its weight. In a 1.40 g/cm³ mud it loses almost 18%. The sum $w\,L$ over all the pipes below a section is the force that the weight indicator (the gauge of the hook load) reads when the string hangs off bottom in a vertical well. This leaves out the weight of the travelling block and the top drive (the hoisting equipment that hangs above the string).

In other unit systems (1)

Oilfield units

w\,[\text{lb/ft}] = W\,[\text{lb/ft}]\left(1 - \frac{\rho_m\,[\text{ppg}]}{65.5}\right)

$W$ is the adjusted mass in pounds per foot. The mud density is in pounds per gallon (ppg). The result is the weight in mud in pounds force per foot.

65.5 ppg is the density of steel (7.85 g/cm³). The conversion is 1 g/cm³ = 8.345 ppg.

Masses per metre and weights of the string elements in the examples of the lesson

Element: Drill collar 165.1 × 71.4 mm
Mass, kg/m: 136.6
Weight in 1.40 g/cm³ mud, N/m: 1101
Weight in 1.25 g/cm³ mud, N/m: 1127

Element: HWDP 127 × 76.2 mm with tool joints
Mass, kg/m: 75.0
Weight in 1.40 g/cm³ mud, N/m: 605
Weight in 1.25 g/cm³ mud, N/m: 619

Element: Drill pipe 127 mm G-105, NC50
Mass, kg/m: 33.9
Weight in 1.40 g/cm³ mud, N/m: 273
Weight in 1.25 g/cm³ mud, N/m: 280

Element: Drill pipe 127 mm S-135, NC50
Mass, kg/m: 34.5
Weight in 1.40 g/cm³ mud, N/m: 278
Weight in 1.25 g/cm³ mud, N/m: 285

Element	Mass, kg/m	Weight in 1.40 g/cm³ mud, N/m	Weight in 1.25 g/cm³ mud, N/m
Drill collar 165.1 × 71.4 mm	136.6	1101	1127
HWDP 127 × 76.2 mm with tool joints	75.0	605	619
Drill pipe 127 mm G-105, NC50	33.9	273	280
Drill pipe 127 mm S-135, NC50	34.5	278	285

The mass of a drill collar comes from the formula for a plain pipe (drill collars and heavy-weight pipe). The masses of HWDP and drill pipe are adjusted masses, which include the tool joints. They come from catalogues (assumed values): 50.4, 22.8 and 23.2 lb/ft. The 1.40 g/cm³ mud is used in the vertical well example, and the 1.25 g/cm³ mud in the directional one.

The weight in mud multiplied by the length of the pipes below gives a force. We call it the effective axial force (effective tension). The true force that stretches the steel is smaller below the wellhead. The fluid presses on the bottom end of the string from below. It also presses on the shoulders (the steps where the pipe size changes), from below or from above. So the bottom of even a freely hanging string is in compression. But compression by pressure does not bend a pipe, because pressure acts on it from all sides. So the effective force decides stability, and with it the neutral point. You will find more detail in the advanced section and in the lesson on casing design.

Weight on bit, the neutral point and the length of the drill collars

When the bit stands on the bottom under a weight $G$ (weight on bit, WOB), the effective axial force at the bit is $-G$ . The minus sign means compression. Higher up, the force grows by the weight of each metre in mud, or more exactly by the component of that weight along the hole. Somewhere it becomes zero. This section is called the neutral point, or the neutral section. Below it the pipes are compressed and may buckle. Above it they are in tension.

Height of the neutral point above the bit

L_{\text{np}} = \frac{G}{w\cos\theta}

where:

$L_{\text{np}}$ distance along the hole from the bit to the neutral point, m
$G$ weight on bit, N
$w$ weight of a metre of drill collar in mud, N/m
$\theta$ inclination of the hole in the interval of the assembly, degrees

Units: SI

Holds when

The compressed part is in a straight section of the hole. The assembly is of one size. For several elements, subtract the weight of the lower element first, then of the next, and so on.
Friction against the walls is not taken into account. With rotation, it hardly changes the axial force. Without rotation, part of the weight on bit goes into friction.
The calculation uses the effective force, so the pressure of the fluid does not enter it (advanced section).

The neutral point depends only on the weight on bit and the weight of the bottom of the string, not on its length. If you increase the weight on bit, the neutral point rises. If you make the mud heavier, it rises too, because the drill collars are lighter in it.

In a vertical hole the wall does not support the pipe. Compressed drill pipe buckles at a compression of the order of 10 kN, which is a small fraction of the usual weight on bit. So all the compression must fall on the drill collars, with a margin. The weight on bit fluctuates during drilling (the bit bounces, and the feed is uneven), and the neutral point must not enter the HWDP or the drill pipe. We choose the length of the drill collars like this:

Length of drill collars for a given weight on bit

L_{\text{dc}} = \frac{k_{\text{df}}\,G}{w\cos\theta}

where:

$L_{\text{dc}}$ length of the drill collars, rounded up to a whole number of joints, m
$k_{\text{df}}$ design factor: usually 1.10–1.20, most often 1.15, —
$G,\ w,\ \theta$ maximum weight on bit, weight of a metre of drill collar in mud, maximum inclination in the interval of the assembly, N, N/m, degrees

Units: SI

Holds when

Drill collars of one size provide the weight on bit. If there are several sizes, find the length of the upper ones from the weight that remains after the lower ones.
In a vertical hole $\cos\theta = 1$ . For an inclined hole, see the next section, where HWDP can also provide the weight.

With $k_{\text{df}} = 1.15$ the weight on bit is $1/1.15 = 87\%$ of the weight of the drill collars. The neutral point lies in the lower 87% of their length. At 1.10 and 1.20 it lies in the lower 91% and 83%.

In other unit systems (1)

Oilfield units

L_{\text{dc}}\,[\text{ft}] = \frac{k_{\text{df}}\,G\,[\text{lbf}]}{W_{\text{dc}}\,[\text{lb/ft}]\;k_{\text{b}}\cos\theta}

The weight on bit is in pounds force, and the mass of the drill collars is in pounds per foot.

Worked exampleBHA for a vertical well 4800 m deep

Given (assumed data)

Well: vertical, 4800 m, 215.9 mm (8½″) bit
Mud: 1.40 g/cm³
Maximum weight on bit $G$: 120 kN (about 12.2 tf)
Drill collars: 165.1 × 71.4 mm (6½ × 2 13/16″), joints of 9.45 m (31 ft)
HWDP: 127 × 76.2 mm, 12 joints of 9.45 m, adjusted mass 75.0 kg/m
Design factor $k_{\text{df}}$: 1.15

Solution

The buoyancy factor is $k_{\text{b}} = 1 - 1.40/7.85 = 0.8217$ . The mass of a metre of drill collar is $0.006165\,(165.1^2 - 71.4^2) = 136.6$ kg/m. Its weight in mud is $w = 136.6\cdot 9.81\cdot 0.8217 = 1101$ N/m.
The length is $L_{\text{dc}} = 1.15\cdot 120{,}000/1101 = 125.3$ m, which is 13.3 joints. We take 14 drill collars: $14\cdot 9.45 = 132.3$ m.
At the maximum weight on bit, the neutral point is $L_{\text{np}} = 120{,}000/1101 = 109.0$ m above the bit. That is 82% of the length of the drill collars. The actual design factor is $132.3\cdot1101/120{,}000 = 1.21$ .
Now the HWDP above the drill collars. The weight of a metre is $75.0\cdot 9.81\cdot 0.8217 = 604.6$ N/m, and the length is $12\cdot 9.45 = 113.4$ m. The weights in mud are: drill collars $132.3\cdot1101/1000 = 145.7$ kN, HWDP $113.4\cdot604.6/1000 = 68.6$ kN, and the whole BHA 214.3 kN.

Answer: The design needs 14 drill collars (132.3 m) and 12 HWDP joints above them. At a weight on bit of 120 kN, the neutral point is 109.0 m above the bit, in the lower 82% of the drill collars. The BHA weighs 214.3 kN in mud.

What it means. Rounding up to a whole joint has raised the design factor from 1.15 to 1.21. The limit is also easy to see. At a weight on bit of 145.7 kN the neutral point would reach the top of the drill collars. That weight on bit has no margin at all. If the bit needs more, the crew adds drill collars. The HWDP between the drill collars and the drill pipe split the jump in stiffness into two steps. The ratios of the section moduli (a measure of bending strength) are about 2.4 and 1.9 (the example in the lesson on the drill string). The jar sits among the HWDP, in the tensioned part, far above the neutral point. The data in this example are assumed for illustration.

Compressed pipe in an inclined hole

In an inclined hole the pipe lies on the low side (the lower wall), and the wall stops it from bending. To buckle, the pipe has to lift itself up the wall against its own weight. So compressed pipe withstands a much greater force here than in a vertical hole. First it bends in a wave along the low side. This is the sinusoidal (planar) mode of buckling. As the compression grows further, the pipe coils into a spiral, a helix (helical buckling). It presses against the walls along its whole length, friction grows, and the weight no longer reaches the bit.

Critical buckling load of a pipe in an inclined hole (the Dawson–Paslay formula)

F_{\text{cr}} = 2\sqrt{\frac{E\,I\,w\sin\theta}{r}}, \qquad I = \frac{\pi\left(D^4 - d^4\right)}{64}

where:

$F_{\text{cr}}$ compressive effective force at which the pipe begins to bend in a wave, N
$E$ Young's modulus of steel (its stiffness), $2.07\cdot10^{11}$ , Pa
$I$ second moment of area of the cross-section of the pipe body (a measure of its resistance to bending), m⁴
$w$ weight of a metre of pipe in mud, N/m
$\theta$ inclination, degrees
$r$ radial clearance between the pipe body and the wall of the hole, $(D_{\text{hole}} - D)/2$ , m
$D,\ d$ outside and inside diameters of the pipe body, m

Units: SI

Holds when

The section is straight, with constant inclination. The pipe lies on the low side under its own weight. The formula is used for inclinations of a few degrees and more. In a vertical hole there is no support, and the formula does not apply.
The clearance is taken for the pipe body. The tool joints and the central upset of HWDP reduce the clearance and raise the critical load, so the estimate is on the safe side.
The helical mode begins at a force roughly 1.4–2.8 times greater, depending on the solution. Rotation, irregularities and curvature of the hole change the picture. So we keep the compression below $F_{\text{cr}}$ with a margin.

The critical load grows with the stiffness of the pipe, its weight and the inclination. It decreases as the clearance grows, so in a narrow hole a pipe holds more. Each factor enters through a square root: twice the stiffness gives only 1.41 times the force.

Critical load

F_{\text{cr}}

in a 215.9 mm hole, 1.25 g/cm³ mud, kN

Inclination: 5°
127 mm drill pipe: 52
HWDP 127 × 76.2 mm: 106

Inclination: 15°
127 mm drill pipe: 89
HWDP 127 × 76.2 mm: 182

Inclination: 30°
127 mm drill pipe: 124
HWDP 127 × 76.2 mm: 253

Inclination: 45°
127 mm drill pipe: 148
HWDP 127 × 76.2 mm: 301

Inclination: 60°
127 mm drill pipe: 164
HWDP 127 × 76.2 mm: 333

Inclination: 90°
127 mm drill pipe: 176
HWDP 127 × 76.2 mm: 358

Inclination	127 mm drill pipe	HWDP 127 × 76.2 mm
5°	52	106
15°	89	182
30°	124	253
45°	148	301
60°	164	333
90°	176	358

We calculated the values by the formula above, with a clearance of 44.45 mm for the pipe body. $EI$ is 1.23 and 2.30 MN·m². For comparison, a compressed 127 mm drill pipe in a vertical hole buckles at a force of the order of 10 kN. The solution for a heavy vertical pipe resting on the bottom gives $1.94\,(EIw^2)^{1/3} \approx 9$ kN.

The table shows the main thing. In a 30° section, drill pipe withstands about 125 kN of compression and HWDP about 250 kN. So in directional wells HWDP often provides the weight on bit, and the crew keeps only a few drill collars. They give stiffness above the bit and a place for the measuring instruments. Fewer thick, heavy pipes on the low side mean less friction and torque, and a lower risk of differential sticking (the pipe being held against the wall by the pressure difference). We calculate the length of the compressed part by the same neutral point formula, but with $\cos\theta$ . We keep all the compression in the straight inclined section below the end of the build.

Design for tension from the bottom up

The weight of everything that hangs below stretches each section of the string. The greatest tension is at the wellhead. But in a string of several sections, we check the top of each section too. Less strong pipes go at the bottom, so the top of the lower section may turn out to be the most heavily loaded relative to its strength.

Allowable load and the margin of overpull

P_a = 0.9\,F_y, \qquad \text{MOP} = P_a - P

where:

$P_a$ allowable tensile load on the pipe body, N
$F_y$ tensile strength of the pipe body at the minimum yield strength (the load at which the steel starts to stretch for good). For used pipe, use the strength of the premium class (the example in the lesson on the drill string), N
$0.9$ a factor that keeps the load below the proportional limit of the steel (the load up to which the steel stretches in proportion to the force), —
$P$ weight of the string in mud below the section considered. In a deviated well, use the force in this section when pulling out, N
$\text{MOP}$ margin of overpull: the force that can be applied in excess of the weight, N

Units: Any units of force, as long as all quantities use the same units.

Holds when

The tension is static. The margin covers jerks and shocks.
Torsion (twisting) is not taken into account. With a torque acting at the same time, the allowable tension is smaller (torsion under tension).
The tensile strength of the tool joints is usually greater than that of the pipe body. Check it in the catalogue.

The margin of overpull is the force, in excess of the weight of the string, that you can apply to a stuck string without risking permanent stretching of the pipe. It is usually set at 50 to 100 thousand lbf (0.22–0.44 MN, about 23–45 tf). The value depends on the risk of sticking, on whether there is a jar, and on what the rig can do. Sometimes a design factor $P_a/P$ is set instead of a margin.

Slip crushing. During connections and trips, the string hangs in the slips of the rotary table. The slips hold the pipe by friction. Their taper converts the vertical load into a radial squeeze (a squeeze from the sides) roughly four times as great. The hoop stress (the stress around the circumference of the pipe) from the squeeze, together with the tension, can crush the pipe body even if the weight is less than the allowable load.

Ratio of the hoop stress from the slips to the tensile stress

\frac{S_h}{S_t} = \sqrt{1 + x + x^2}, \qquad x = \frac{D\,K}{2\,L_s}, \qquad K = \frac{1}{\tan(y + z)}

where:

$S_h/S_t$ how many times “heavier” a load in the slips is than the same load in free tension, —
$D$ outside diameter of the pipe body, mm or inches
$L_s$ length of the slips: 305 mm (12″) or 406 mm (16″), same units
$K$ lateral load factor (the ratio of the sideways squeeze to the vertical load), —
$y$ taper angle of the slips, 9°27′45″, degrees
$z$ friction angle between the slips and the bushing (the bowl that holds the slips), $\arctan\mu$ . For lubricated slips we take $\mu = 0.08$ , and then $K \approx 4.0$ , degrees

Units: $D$ and $L_s$ must be in the same units.

Holds when

The slips and bushings are in good order and grip the pipe evenly along its whole length. Worn slips, an uneven grip, and a shock when the pipe is set in the slips cause crushing sooner.
$\mu = 0.08$ is an assumed value. With greater friction $K$ is smaller, so a calculation with this value is on the safe side.
The condition of the check: the weight of the string under the slips $P$ , multiplied by $S_h/S_t$ , must be no more than the same allowable load $P_a$ .

Long slips spread the squeeze out and reduce the ratio. For pipes of larger diameter the ratio is greater.

The ratio

S_h/S_t

\mu = 0.08

Pipe: 88.9 mm (3½″)
305 mm (12″) slips: 1.39
406 mm (16″) slips: 1.28

Pipe: 114.3 mm (4½″)
305 mm (12″) slips: 1.52
406 mm (16″) slips: 1.37

Pipe: 127.0 mm (5″)
305 mm (12″) slips: 1.59
406 mm (16″) slips: 1.42

Pipe: 139.7 mm (5½″)
305 mm (12″) slips: 1.66
406 mm (16″) slips: 1.47

Pipe: 168.3 mm (6⅝″)
305 mm (12″) slips: 1.82
406 mm (16″) slips: 1.59

Pipe	305 mm (12″) slips	406 mm (16″) slips
88.9 mm (3½″)	1.39	1.28
114.3 mm (4½″)	1.52	1.37
127.0 mm (5″)	1.59	1.42
139.7 mm (5½″)	1.66	1.47
168.3 mm (6⅝″)	1.82	1.59

We calculated the ratios by the formula above, for the nominal outside diameter. For a worn pipe the diameter is smaller, and the ratio is slightly smaller (1.41 instead of 1.42 for a 127 mm premium-class pipe). So a calculation for the nominal diameter is on the safe side.

Greatest length of a drill pipe section

L_i = \min\left\{\frac{P_a - \text{MOP} - P_{\text{below}}}{w_i};\ \ \frac{P_a\,S_t/S_h - P_{\text{below}}}{w_i}\right\}

where:

$L_i$ greatest length of the $i$ -th section, m
$P_a$ allowable load of the pipes of this section, N
$P_{\text{below}}$ weight in mud of everything that hangs below the section (the BHA and the lower sections), N
$w_i$ weight of a metre of the pipes of the section in mud, N/m

Units: SI

Holds when

The well is vertical. In a deviated well, use the force when pulling out, found by the soft-string model, instead of the weight.
The first condition is the margin of overpull, and the second is crushing in the slips.

The stricter of the two conditions limits the length of a section. Crushing in the slips becomes the main limit when $P_a(1 - S_t/S_h)$ is greater than the required margin of overpull. At $S_h/S_t = 1.42$ this means that $P_a$ is greater than about 3.4 times the margin. This happens with strong pipes in deep wells.

If one strength grade is not enough up to the wellhead, we make the string tapered. It has pipes of lower strength at the bottom and stronger ones above them. We find the length of the lower section from the weight of the BHA. We find the length of the next section from the weight of the BHA and the lower section, and so on until we reach the wellhead. The margin of overpull of the whole string is the smallest of the margins at the tops of the sections, because the overpull is applied from the top and adds equally to all sections. If the string is run empty (a drill stem test), we check collapse by outside pressure too. We use the same formulas as for casing (rated properties of pipe), with a design factor.

Drag and torque in a deviated well

In an inclined hole the pipes lie on the wall and rub against it. The most widespread calculation scheme is the soft-string model. It treats the string as a flexible cable lying on the wall, and it ignores the bending stiffness. We find the contact force (the force with which the pipe presses on the wall) from the weight of the pipe and, in curved sections, from the tension. Friction equals this force multiplied by the friction factor $\mu$ . When pulling out, friction acts downward, against the motion, and increases the tension. When running in, it acts upward and reduces the tension. When rotating without axial motion, the friction force is directed around the circumference. It does not change the axial force, but it creates torque.

Axial force and torque in a straight inclined section

F_{\text{top}} = F_{\text{bot}} + w L\left(\cos\theta \pm \mu\sin\theta\right), \qquad M = \mu\,w L\sin\theta\;r_{\text{cont}}

where:

$F_{\text{top}},\ F_{\text{bot}}$ effective axial force at the top and at the bottom of the section. Tension is positive, N
$w$ weight of a metre of pipe in mud, N/m
$L$ length of the section along the hole, m
$\theta$ inclination of the section, degrees
$\mu$ friction factor. The plus sign is for pulling out and the minus sign for running in. When rotating, there is no term with $\mu$ , —
$M$ friction torque on the section when rotating, N·m
$r_{\text{cont}}$ contact radius with the wall (the radius of the tool joint or the drill collar), m

Units: SI

Holds when

The section is straight. The pipe lies on the low side, and only its own weight presses it, $w\sin\theta$ per metre.
The friction is Coulomb friction: the friction force is proportional to the contact force. When the string is rotated and moved at the same time, friction is shared between the axial and circumferential directions. The model gives only the extreme cases.
Bending stiffness is not accounted for. For a stiff BHA in a curved hole, the model underestimates the contact forces.

On a straight section, drag is proportional to the weight of the pipe on it and to the sine of the inclination. At 30° and $\mu = 0.25$ , drag adds about 14% to the component of weight along the hole.

On a curved section, the tension adds to the weight. A stretched cable on a bend presses against the inner side of the arc, like a rope against a pulley. The force is $F\kappa$ per metre, where $\kappa$ is the curvature of the hole. In a build section the centre of the arc lies on the high side. So the tension presses the pipe against the high side, and the weight presses it against the low side. The contact force is their difference, $F\kappa - w\sin\theta$ . At usual tensions the first term is much greater than the second, and drag grows in proportion to the tension. In a drop section (where the angle decreases), both forces press the pipe against the low side and add up.

Axial force in a build section (the trapezoid method)

\begin{aligned} F_{\text{top}} &= \frac{F_{\text{bot}}\,(1 + a) + w\,(H - \mu A)}{1 - a}\ \ \text{(pulling out)}, \qquad F_{\text{top}} = \frac{F_{\text{bot}}\,(1 - a) + w\,(H + \mu A)}{1 + a}\ \ \text{(running in)},\\ F_{\text{top}} &= F_{\text{bot}} + w H\ \ \text{(rotating)}, \qquad M = \mu\,r_{\text{cont}}\left(\bar F\,\Delta\theta - w A\right), \qquad a = \frac{\mu\,\Delta\theta}{2} \end{aligned}

where:

$\Delta\theta$ change of inclination in the section, rad
$H,\ A$ projections of the section on the vertical and the horizontal. For a build from the vertical to an inclination $\theta$ with radius $R$ , they are $R\sin\theta$ and $R(1 - \cos\theta)$ , m
$\bar F$ mean axial force in the section, $(F_{\text{bot}} + F_{\text{top}})/2$ , N
$F_{\text{top}},\ F_{\text{bot}},\ w,\ \mu,\ r_{\text{cont}},\ M$ as in the formula for a straight section, N, N/m, —, m, N·m

Units: SI. The angle is in radians.

Holds when

The build section is planar (it lies in one plane) and has constant dogleg severity. The azimuth (the compass direction of the hole) does not change.
The tension presses the pipe against the high side harder than the weight presses it against the low side, so $F\kappa > w\sin\theta$ along the whole section.
The integral of the contact force, $\bar F\Delta\theta - wA$ , is taken with the mean force. Long sections are divided into parts. The derivation is in the advanced section.

Friction on the arc is proportional to the axial force in it. If the tension lower in the string is large, a short build section gives drag and torque comparable to those of a long straight section.

The friction factor is not a tabulated property of the rock. It is a tuning parameter of the model, and it “hides” several things: cuttings on the low side, small irregularities of the hole, stabilisers (the tools that keep the string centred in the hole) and the lubricity (slipperiness) of the mud. Usually it lies within 0.15–0.35. In oil-based muds it is nearer the lower limit. We choose it from the hook load and torque measured in neighbouring wells or in the upper intervals of the same well.

On the rig, the crew plots calculated lines of hook load against depth in advance. There are lines for pulling out, running in and rotating, for several friction factors. The crew then plots the measurements for every stand (a few joints of pipe screwed together) on them. If the measurements when pulling out go above the calculated lines, and when running in go below them, drag is growing. Cuttings accumulate on the low side, the hole narrows, and ledges form. This is a reason to circulate the mud to clean the hole, or to ream it (open it up again), before the pipe sticks.

Complete design: a vertical well

Let us continue with the 4800 m well. The bottom hole assembly was chosen in the example above: 14 drill collars and 12 HWDP joints weigh 214.3 kN in mud. It remains to choose the drill pipe. We have 127 mm (5″) 19.50 lb/ft pipes with NC50 tool joints. They are of the premium class (used pipe with 80% of the wall left) and come in two strength grades, G-105 and S-135. The rig works with 406 mm slips. The required margin of overpull is 0.40 MN.

Worked exampleA tapered string of drill pipe down to 4800 m

Given (assumed data)

BHA: 14 drill collars and 12 HWDP joints, weight in mud 214.3 kN; 1.40 g/cm³ mud, $k_{\text{b}} = 0.8217$
Pipe: 127 mm, 19.50 lb/ft, premium class; G-105: 33.9 kg/m with tool joints, S-135: 34.5 kg/m
Margin of overpull: not less than 400 kN (0.40 MN, about 41 tf)
Slips: 406 mm (16″), $S_h/S_t = 1.42$

Solution

First we find the strength of the body of premium-class pipe, with uniform wear down to 80% of the wall remaining (the example in the lesson on the drill string). For G-105 it is 1939.3 kN, and for S-135 it is 2493.8 kN. The allowable loads are $0.9F_y$ , so $P_a$ = 1745.4 and 2244.4 kN.
The weight of a metre in mud is $33.9\cdot 9.81\cdot 0.8217/1000 = 0.2733$ kN/m for G-105 and 0.2781 kN/m for S-135.
The greatest length of G-105 pipe that the margin of overpull allows is $(1745.4 - 400 - 214.3)/0.2733 = 4139$ m.
Crushing in the slips allows a weight under them of no more than $1745.4/1.42 = 1229.2$ kN. The length is then $(1229.2 - 214.3)/0.2733 = 3714$ m, which is less. We take 3700 m of G-105 pipe.
The top of the G-105 section is at a depth of $4800 - 132.3 - 113.4 - 3700 = 854.3$ m. The tension there is $214.3 + 3700\cdot0.2733 = 1225.5$ kN. The margin of overpull is $1745.4 - 1225.5 = 519.9$ kN. In the slips the load is $1225.5\cdot 1.42 = 1740.2$ kN, which is no more than 1745.4.
Above that goes 854.3 m of S-135 pipe. The weight of the string at the wellhead is $1225.5 + 854.3\cdot0.2781 = 1463.1$ kN (149.2 tf). The margin of overpull is $2244.4 - 1463.1 = 781.3$ kN. In the slips the load is $1463.1\cdot1.42 = 2077.6$ kN, which is less than 2244.4.

Answer: The string from the bottom up is 14 drill collars, 12 HWDP joints, 3700 m of G-105 pipe and 854.3 m of S-135 pipe. The hook load is 1463.1 kN. The margin of overpull of the whole string is 519.9 kN. The top of the G-105 section sets it, not the wellhead.

What it means. Crushing in the slips limited the length of the G-105 section, not the margin of overpull. For these pipes $P_a$ = 1745.4 kN, which is more than 3.4 times the margin of 400 kN. With 305 mm slips ( $S_h/S_t = 1.59$ ), the G-105 section would have to be shortened to $(1097.7 - 214.3)/0.2733 = 3232$ m. At the wellhead the S-135 pipe would not pass the check either: $1463.1\cdot 1.59 = 2326.3$ kN, which is more than 2244.4. In that case the crew uses long slips, or stronger pipe in the upper part, for example new S-135 pipe with $F_y$ of about 3167 kN. The data in this example are assumed for illustration.

The string of the example from the bottom up, 1.40 g/cm³ mud

Section: Drill collars, 14 joints
Length, m: 132.3
Weight in mud, kN: 145.7
Tension at the top, kN: 145.7
Margin of overpull, kN: —

Section: HWDP with a jar, 12 joints
Length, m: 113.4
Weight in mud, kN: 68.6
Tension at the top, kN: 214.3
Margin of overpull, kN: —

Section: Drill pipe G-105
Length, m: 3700
Weight in mud, kN: 1011.2
Tension at the top, kN: 1225.5
Margin of overpull, kN: 519.9

Section: Drill pipe S-135
Length, m: 854.3
Weight in mud, kN: 237.6
Tension at the top, kN: 1463.1
Margin of overpull, kN: 781.3

Section	Length, m	Weight in mud, kN	Tension at the top, kN	Margin of overpull, kN
Drill collars, 14 joints	132.3	145.7	145.7	—
HWDP with a jar, 12 joints	113.4	68.6	214.3	—
Drill pipe G-105	3700	1011.2	1225.5	519.9
Drill pipe S-135	854.3	237.6	1463.1	781.3

The tension is for a string hanging off bottom. When drilling with a weight on bit of 120 kN, it is 120 kN less all along the string. The margin of overpull of the string is the smallest of the margins: 519.9 kN at the top of the G-105 section.

CalculatedTension design of the string of the example. The solid line is the weight in mud of everything that hangs below the section. For each section of drill pipe there are three vertical lines: the allowable load

P_a

(grey dashed), the limit by the margin of overpull

P_a - 0.40

MN (orange dashed) and the limit by crushing in the slips

P_a/1.42

(orange solid). The string passes the check as long as the weight line is to the left of both limits. For the G-105 section the slips are stricter, and the weight line almost touches their limit at the top of the section.

Complete design: a directional well

Take a build-and-hold well (the angle builds up and then stays constant). Its profile is calculated in the lesson on profile design. The well has a vertical section down to 500 m. Then it builds angle at 3° per 30 m to an inclination of 32.4° at 824 m along the hole. After that comes a tangent section (a straight inclined section) 1886.7 m long, to a target at 2711 m along the hole (2400 m vertically). The bit is 215.9 mm, the mud is 1.25 g/cm³, the maximum weight on bit is 100 kN, and the pipe is G-105 premium class up to the wellhead.

Worked exampleBHA for the tangent section

Given (assumed data)

Tangent section: inclination 32.4°: $\cos\theta = 0.844$ , $\sin\theta = 0.536$
Weight on bit $G$: 100 kN; design factor 1.15
BHA: 3 drill collars 165.1 × 71.4 mm (28.35 m; one is non-magnetic, so that it does not disturb the compass readings, and carries an MWD (measurement while drilling) tool), with HWDP 127 × 76.2 mm above them
Hole: 215.9 mm; clearance for the 127 mm pipe body 44.45 mm

Solution

The buoyancy factor is $k_{\text{b}} = 1 - 1.25/7.85 = 0.8408$ . The weight of a metre in mud is 1127 N/m for drill collars, 618.6 N/m for HWDP and 279.6 N/m for drill pipe.
Suppose drill collars provided all the weight on bit. Then $L_{\text{dc}} = 1.15\cdot 100{,}000/(1127\cdot 0.844) = 120.9$ m, which is 13 joints.
Three drill collars give $28.35\cdot1127\cdot 0.844/1000 = 27.0$ kN along the hole. The remaining $100 - 27.0 = 73.0$ kN must come from the HWDP. That needs $73.0\cdot 1000/(618.6\cdot 0.844) = 139.8$ m. With the factor of 1.15, it needs $(115 - 27.0)\cdot 1000/(618.6\cdot 0.844) = 168.6$ m, which is 18 HWDP joints, or 170.1 m.
The neutral point is at $28.35 + 139.8 = 168.2$ m above the bit, in the HWDP. The whole assembly (198.45 m) is in the tangent section.
Now the stability of the HWDP. The greatest compression is at their bottom, 73.0 kN. The second moment of area is $I = \pi(0.127^4 - 0.0762^4)/64 = 1.111\cdot10^{-5}$ m⁴, and $EI = 2.07\cdot10^{11}\cdot I = 2.30$ MN·m². The critical load is $F_{\text{cr}} = 2\sqrt{2.30\cdot10^6\cdot 618.6\cdot 0.536/0.04445} = 262.0$ kN, which is 3.6 times the compression.

Answer: The assembly is the bit, 3 drill collars and 18 HWDP joints with a jar, instead of 13 drill collars. The neutral point is 168.2 m above the bit. The greatest compression of the HWDP is 3.6 times less than the critical load.

What it means. The compressed HWDP work with a large margin. At this inclination even drill pipe ( $EI$ = 1.23 MN·m²) would withstand up to 128.8 kN. The limit is elsewhere. The compressed part must stay in the straight inclined section, below the end of the build, and here the end of the build is more than 1600 m away. The jar goes where the manufacturer's calculation says, and not at the neutral point. The data in this example are assumed for illustration.

Worked exampleHook load and torque in the same well

Given (assumed data)

String: 3 drill collars, 18 HWDP joints (170.1 m), G-105 drill pipe above them to the wellhead
Weight in mud, kN/m: drill collars 1.127; HWDP 0.6186; pipe 0.2796
Friction factor $\mu$: 0.25 along the whole hole (assumed)
Contact radius: tool joints (diameter 168.3 mm): 0.08415 m; drill collars (diameter 165.1 mm): 0.08255 m
Build section: $\Delta\theta = 32.4^\circ = 0.565$ rad, $H = 307.0$ m, $A = 89.2$ m
Torque at the bit while drilling: 4 kN·m (assumed)

Solution

Start with the tangent section. The drill pipe on it is $1886.7 - 28.35 - 170.1 = 1688.3$ m long. The weight in mud of everything lying on it is $28.35\cdot1.127 + 170.1\cdot0.6186 + 1688.3\cdot0.2796 = 609.2$ kN. The force at the end of the build is as follows. When pulling out, it is $609.2\cdot(0.844 + 0.25\cdot0.536) = 609.2\cdot0.978 = 595.8$ kN. When running in, it is $609.2\cdot0.710 = 432.5$ kN. When rotating, it is $609.2\cdot0.844 = 514.2$ kN.
Now the build section. Here $a = 0.25\cdot0.565/2 = 0.0706$ . When pulling out, $F_{\text{top}} = [595.8\cdot(1 + 0.0706) + 0.2796\cdot(307.0 - 0.25\cdot89.2)]/(1 - 0.0706) = 772.0$ kN. When running in, it is 461.5 kN. When rotating, it is $514.2 + 0.2796\cdot307.0 = 600.0$ kN.
The vertical section adds $0.2796\cdot500 = 139.8$ kN. The hook load off bottom is 911.8 kN when pulling out (93.0 tf), 739.8 kN when rotating, and 601.3 kN when running in. The drag is 172.0 kN when pulling out and 138.5 kN when running in.
Where does the drag when pulling out come from? Friction in the build section gives $772.0 - 595.8 - 0.2796\cdot307.0 = 90.4$ kN. Friction in the whole tangent section gives $595.8 - 514.2 = 81.6$ kN.
Now the torque when rotating off bottom. In the tangent section it is $0.25\cdot0.536\cdot(0.08255\cdot28.35\cdot1.127 + 0.08415\cdot(170.1\cdot0.6186 + 1688.3\cdot0.2796)) = 6.86$ kN·m. In the build section, $\bar F = (514.2 + 600.0)/2 = 557.1$ kN and $M = 0.25\cdot0.08415\cdot(557.1\cdot0.565 - 0.2796\cdot89.2) = 6.10$ kN·m. In total it is 12.96 kN·m.
While drilling, the tension in the build section is 100 kN less: $0.25\cdot0.08415\cdot((557.1 - 100)\cdot0.565 - 0.2796\cdot89.2) = 4.91$ kN·m. The friction torque is 11.77 kN·m, and with the torque at the bit the total is 15.8 kN·m.
Last, the tension. The allowable load of G-105 premium-class pipe is $P_a$ = 1745.4 kN. When pulling out, the load at the wellhead is 911.8 kN. This leaves a margin of overpull of 834 kN, which is more than the required 400 kN.

Answer: Off bottom, the hook load is about 911.8 kN when pulling out, 739.8 kN when rotating and 601.3 kN when running in. The torque while drilling is about 15.8 kN·m. The margin of overpull is 834 kN.

What it means. The 324 m build section gave more drag when pulling out than the 1887 m tangent section, and almost as much torque. The reason is that the tension presses the pipe against its high side. So designers make a build in the upper part of the hole, where the tension is large, gentle if possible. A calculation in small steps (about 8 m over the build section) gives the same values to within 0.1%. Suppose 13 drill collars, with six HWDP joints above them, provided the weight on bit. Then the hook load when pulling out would be 47 kN greater, and the torque would be 0.9 kN·m greater. But the light assembly has a more noticeable advantage: fewer thick pipes on the low side and a lower risk of sticking. The data in this example are assumed for illustration.

CalculatedEffective axial force along the hole of the example well by the soft-string model (

\mu = 0.25

). The lines show pulling out, rotating and running in off bottom, and drilling with a weight on bit of 100 kN. The black dashed line on the left is the critical buckling load with a minus sign. It is shown for the drill pipe in the tangent section and for the HWDP. The compression while drilling stays far from it.

Terms and symbols

Neutral point: The section of the string where the effective axial force is zero. Below it the pipes are compressed. Above it they are in tension.
Effective axial force: The axial force that allows for buoyancy, which is the weight in mud of the pipes below. It decides the stability of the string and the readings of the weight indicator.
True axial force: The force in the steel of the pipe. Below the wellhead it is smaller than the effective force by the pressure multiplied by the wall area.
Buoyancy factor $k_{\text{b}}$: The fraction of the weight in air that remains in mud. It equals $1 - \rho_m/\rho_{\text{steel}}$ .
Design factor $k_{\text{df}}$: The number of times that the weight of the drill collars (along the hole) exceeds the maximum weight on bit. It is usually 1.10–1.20.
Allowable load $P_a$: 90% of the tensile strength of the pipe body. For used pipe, the strength is taken for its wear class.
Margin of overpull (MOP): The force, in excess of the weight of the string, that can be applied when the string is stuck. It is usually 0.22–0.44 MN.
Slip crushing, $S_h/S_t$: The squeezing of the pipe by the slips. It is measured by the ratio of the hoop stress to the tensile stress, and we multiply the weight under the slips by this ratio.
Critical buckling load: The compression at which a pipe lying on the low side of an inclined hole begins to bend in a wave. It equals $2\sqrt{EIw\sin\theta/r}$ .
Soft-string model: A calculation of drag and torque in which the string is treated as a flexible cable lying on the wall of the hole.
Hook load: The force shown by the weight indicator. It is the weight of the string in mud, less the share carried by the wall in a deviated well, plus or minus friction. It also includes the weight of the travelling equipment, if that has not been deducted.
Tapered string: A string made of sections of pipe of different strength or mass. The less strong sections are at the bottom.

Common mistakes and misconceptions

Wrong

“Let us add drill collars so that their weight exceeds the buoyant force on the bottom end. Then the drill pipe will certainly be in tension.”

In fact

The effective force, which is the weight in mud, decides bending. If the drill collars are long enough for the weight on bit, with a margin, the drill pipe is already in tension in this sense. Extra drill collars increase weight, friction and the risk of sticking, and they eat into the margin of overpull.

Wrong

“It is enough to check the margin of overpull at the wellhead. The tension is greatest there.”

In fact

In a tapered string the less strong pipes are at the bottom, and the smallest margin is often at the top of the lower section. The margin of the string is the smallest of the margins at the tops of the sections.

Wrong

“If the weight under the slips is less than the allowable load, the pipe in the slips is in no danger.”

In fact

The slips squeeze the pipe. In 406 mm slips the load on a 127 mm pipe acts as 1.42 times greater. In deep wells this limit can be stricter than the margin of overpull.

Wrong

“The hook load when pulling out is the weight of the string in mud.”

In fact

In a deviated well drag is added to it, and when running in drag is subtracted. Even when rotating, when there is no axial drag, the hook load is less than the weight of the string in mud, because the wall carries part of the weight of the inclined sections. We count the margin of overpull from the weight when pulling out.

Wrong

“Since drill pipe can work in compression in an inclined hole, it can be compressed in the vertical part as well.”

In fact

The critical load is proportional to the square root of the sine of the inclination. In a vertical hole the wall does not support the pipe, and it buckles at a compression of the order of 10 kN. The compressed part must stand in the straight inclined section.

In depth: advanced detail

True and effective axial force

Fluid presses on a pipe only at right angles to its surface. On a vertical wall it presses sideways and creates no axial force. The axial force comes only from the end faces and shoulders: the bottom end of the string (more exactly, the ring of steel, if the bore, the channel inside the string, is open), the transitions from drill collars to HWDP and from HWDP to drill pipe, the upsets and the tool joints. So the true force in the pipe body at a given depth is the weight of the pipes below in air. From it we subtract the pressure forces on the downward-facing areas, because the fluid pushes these up. We add the pressure forces on the upward-facing areas, because the fluid pushes these down. We define the effective force so that pressure does not enter it:

Effective and true axial force

F_{\text{eff}} = F_{\text{true}} + p_{\text{out}}A_{\text{out}} - p_{\text{in}}A_{\text{in}}, \qquad w = g\left(m - \rho_{\text{out}}A_{\text{out}} + \rho_{\text{in}}A_{\text{in}}\right)

where:

$F_{\text{eff}},\ F_{\text{true}}$ effective and true axial force in the section. Tension is positive, N
$p_{\text{out}},\ p_{\text{in}}$ pressure outside and inside the pipe at this depth, Pa
$A_{\text{out}},\ A_{\text{in}}$ areas bounded by the outside and inside diameters of the pipe body, $\pi D^2/4$ and $\pi d^2/4$ , m²
$w$ effective weight of a metre of pipe, N/m
$\rho_{\text{out}},\ \rho_{\text{in}}$ density of the fluid outside and inside, kg/m³
$m$ mass of a metre of pipe, kg/m

Units: SI

Holds when

The pipe is vertical, in a fluid at rest. With equal pressures inside and outside, $F_{\text{eff}} = F_{\text{true}} + pA_s$ , where $A_s$ is the wall area. Then the weight of a metre is the usual weight in mud, $m g k_{\text{b}}$ .
The weight formula is written for a pipe of constant cross-section. For a pipe with tool joints and upsets, we take the difference $A_{\text{out}} - A_{\text{in}}$ in it as the mean along the length, that is $m/\rho_{\text{steel}}$ .
If the fluid inside is lighter (the string is run empty) or heavier (a weighted slug), the string on this section is lighter or heavier. The hook load shows this.

We arrange the effective force so that hydrostatics (the pressure of the fluid at rest) is “already accounted for” in it. It changes only with the weight in mud and with external forces such as the weight on bit and friction. The effective force decides whether a pipe will bend. A pipe immersed in a fluid and loaded only by pressure stays straight, although the true force in its lower part is compressive.

Worked exampleWhere the true force changes sign in a vertical well

Given (assumed data)

String: from the complete design of the vertical well: 14 drill collars (132.3 m), 12 HWDP joints (113.4 m), G-105 and S-135 pipe
Mode: drilling with a weight on bit of 120 kN; 1.40 g/cm³ mud inside and outside
Wall areas: drill collar 165.1 × 71.4 mm: 17,404 mm²; drill pipe 127 × 108.62 mm: 3401 mm²

Solution

At the bit the pressure is $0.00981\cdot1.40\cdot4800 = 65.92$ MPa. The true force is $-120 - 65.92\cdot17{,}404/1000 = -1267$ kN. This is a compression more than ten times the weight on bit.
The effective force becomes zero at the neutral point, 109.0 m above the bit.
At the top of the HWDP (depth 4554.3 m) the effective force is $-120 + 214.3 = 94.3$ kN. In the drill pipe just above the HWDP the pressure is 62.55 MPa, and the true force is $94.3 - 62.55\cdot3401/1000 = -118.4$ kN. This is still compression.
Higher up, the true force grows by the weight of a metre in mud plus the decrease in the pressure force. That is $273.3 + 0.00981\cdot1.40\cdot3401 = 320.0$ N/m. The true force becomes zero after $118.4\cdot1000/320.0 = 370$ m, at a depth of 4184 m.

Answer: The true force compresses the string for 616 m above the bit, up to a depth of 4184 m in the drill pipe. The effective force does so only for 109.0 m, within the drill collars.

What it means. If the true force decided stability, hundreds of metres of extra drill collars would be needed. In fact the drill pipe between the neutral point and a depth of 4184 m is straight. The pressure of the mud compresses it, and that pressure acts on the pipe from all sides. To check strength, on the contrary, we need the true force, because it is what stretches or compresses the steel. At the wellhead, where the pressure is zero, the two forces coincide. The data in this example are assumed for illustration.

CalculatedThe lower 700 m of the string of the vertical well while drilling with a weight on bit of 120 kN. The solid line is the effective axial force, and the dashed line is the true force in the pipe body. The effective force becomes zero at the neutral point inside the drill collars. The true force does so 500 m higher, in the drill pipe. The jumps of the dashed line are the shoulders where the wall area changes.

Where the formula for the build section comes from

Take a short piece of string of length $\dd s$ on a build arc with curvature $\kappa = \dd\theta/\dd s$ . The axial forces at its ends, $F$ and $F + \dd F$ , point along tangents that diverge by the angle $\dd\theta$ . Together they give a force $F\,\dd\theta$ towards the centre of the arc, which is towards the high side. The weight of the piece, $w\,\dd s$ , gives a component $w\sin\theta\,\dd s$ towards the low side and a component $w\cos\theta\,\dd s$ along the hole. The wall responds with a reaction $\dd N = F\,\dd\theta - w\sin\theta\,\dd s$ . The friction $\mu\,\dd N$ points against the motion. When pulling out:

Equilibrium of a piece of string on the build arc when pulling out

\frac{\dd F}{\dd s} = w\cos\theta + \mu\left(F\kappa - w\sin\theta\right)

where:

$F$ effective axial force, N
$s$ length along the hole, measured upward, m
$\kappa$ curvature of the hole, rad/m
$w,\ \theta,\ \mu$ weight of a metre in mud, inclination, friction factor, N/m, degrees, —

Units: SI

Holds when

The arc is planar. $F\kappa > w\sin\theta$ , so the pipe lies on the high side. Bending stiffness is not accounted for.

Integrate along the arc. Here $\int w\cos\theta\,\dd s = wH$ and $\int w\sin\theta\,\dd s = wA$ are the projections of the arc, and $\int F\kappa\,\dd s = \int F\,\dd\theta \approx \bar F\Delta\theta$ . This gives $F_{\text{top}} - F_{\text{bot}} = wH + \mu(\bar F\Delta\theta - wA)$ . Substitute $\bar F = (F_{\text{bot}} + F_{\text{top}})/2$ and solve for $F_{\text{top}}$ , and you arrive at the formula of the lesson. Without weight, the equation gives the well-known “capstan formula” $F_{\text{top}} = F_{\text{bot}}\,e^{\mu\Delta\theta}$ (the formula for a rope on a drum). At $\mu\Delta\theta = 0.1414$ the factor is $e^{0.1414} = 1.152$ . The trapezoid method gives $(1 + 0.0707)/(1 - 0.0707) = 1.152$ , which is the same.

When running in, friction changes sign. When rotating, the axial term with $\mu$ disappears, and $\mu\,\dd N$ multiplied by the contact radius gives the increment of torque. In a drop section the tension and the weight press the pipe against the same side, the low side. So the formulas have the sum $F\kappa + w\sin\theta$ instead of the difference. In a three-dimensional hole, where the azimuth also changes, the contact force is the geometric sum of the components in the vertical and horizontal planes. This is how the programs that calculate drag and torque from directional survey data work (Directional survey calculations).

Connections to other topics

Drill string: pipe, tool joints, drill collars and heavy-weight drill pipe — the elements of the string, their strength in tension and torsion, torsion under tension, the rule of steps in stiffness and fatigue in a curved hole.
Casing design — the buoyancy factor, the true axial force and the design of a casing string for tension.
Directional wells: applications and profile — the build-and-hold profile for which the directional well is designed.
Deflection tools and downhole motors — assemblies for building, holding and dropping angle, and drilling with a motor in the sliding mode.
Horizontal wells — horizontal wells: drilling them needs a rig with sufficient hoisting capacity and torque.
Drill bits: types and design — the bits to which the string transmits weight and rotation.
Bit performance and cost per metre — how the weight on bit and the rotary speed affect the rate of penetration.
Drilling rig: power and hoisting systems — the block and tackle, the weight indicator and the hoisting capacity of the rig.

Check yourself

1
A vertical hole, weight on bit 150 kN, weight of a metre of drill collar in mud 1.20 kN/m. Where is the neutral point, and what length of drill collars is needed with a design factor of 1.15?
1. A.104 m above the bit; 120 m of drill collars
2. B.125 m above the bit; 109 m of drill collars
3. C.150 m above the bit; 173 m of drill collars
4. D.125 m above the bit; 144 m of drill collars
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Correct answer: D. 125 m above the bit; 144 m of drill collars
The neutral point is $L_{\text{np}} = 150/1.20 = 125$ m. The length of the drill collars is $1.15\cdot150/1.20 \approx 144$ m, rounded up to a whole number of joints (16 drill collars of 9.45 m, which is 151.2 m). The 109 m option mixed up the factor. Dividing by 1.15 would leave the neutral point above the drill collars.
2
Why is the weight of the drill collars in mud used to calculate the neutral point, although the true axial force at the bit compresses the string much harder than the weight on bit?
1. A.Because friction against the walls compensates for the pressure
2. B.Because the pressure of the mud compresses the pipe from all sides and does not bend it. The effective force decides stability
3. C.Because the true force cannot be calculated
4. D.Because the pressure of the mud on the bottom end of the string is always negligible
Show the answer
Correct answer: B. Because the pressure of the mud compresses the pipe from all sides and does not bend it. The effective force decides stability
The force of pressure on the bottom end really is large: more than 1.1 MN in the example of the lesson. But a pipe loaded only by the pressure of the fluid stays straight, because the pressure also acts on its side surface. Forces in excess of hydrostatics cause bending. These are the weight on bit and the weight in mud, that is, the effective force.
3
In the vertical well of the lesson the rig can work only with 305 mm slips. What changes in the design of the string?
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The ratio $S_h/S_t$ for a 127 mm pipe rises from 1.42 to 1.59. The G-105 section will have to be shortened. The weight hanging below its top can be no more than $1745.4/1.59 = 1097.7$ kN. That is about 3230 m of pipe instead of 3700. Worse, at the wellhead the full string of about 1463 kN gives $1463\cdot1.59 \approx 2326$ kN in the slips. This is more than the allowable 2244 kN for premium-class S-135 pipe. The way out is long slips or stronger pipe at the top (for example, new S-135 pipe: $0.9\cdot3167 \approx 2850$ kN).
4
How will the critical buckling load of HWDP at the same inclination change if the 215.9 mm hole is replaced by a 311.1 mm (12¼″) hole?
1. A.It will not change: it depends only on the pipe
2. B.It will increase by about 30%
3. C.It will halve
4. D.It will decrease by about 30%
Show the answer
Correct answer: D. It will decrease by about 30%
The radial clearance will grow from 44.45 to 92.1 mm. The critical load is inversely proportional to the square root of the clearance, so it changes by the factor $\sqrt{44.45/92.1} = 0.69$ , which is about 30% less. In a wide hole a pipe has room to bend. So in large diameters HWDP provides less weight, and more drill collars are needed.
5
Why did the 324 m build section in the directional well example give more drag when pulling out than the almost 1900 m tangent section?
Show the answer
On a straight section, only the weight of the pipe presses it against the wall, $w\sin\theta$ per metre. On the build arc the tension adds to this. The stretched string presses on the high side with a force $F\kappa$ per metre. The tension in the build section when pulling out is from 596 to 772 kN. At a dogleg severity of 3° per 30 m ( $\kappa = 0.001745$ rad/m), this gives 1.0–1.35 kN per metre. That is 7–9 times the transverse component of the weight of the pipe (not more than 0.15 kN/m). So drag and torque grow with the tension, and a sharp build in the upper part of the hole is especially “expensive”.
6
At what allowable load $P_a$ does crushing in 406 mm slips limit the length of a 127 mm pipe section in the same way as a margin of overpull of 400 kN?
Show the answer
The limits are equal when $P_a - 400 = P_a/1.42$ , that is $P_a(1 - 1/1.42) = 400$ , hence $P_a \approx 1352$ kN. For strong pipes with a greater allowable load, the slips are stricter. In the example of the lesson, premium-class G-105 has 1745 kN. For less strong pipes, the margin of overpull is stricter.

Updated: 19 October 2026