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During normal operation, the maximum tensile force occurs at the top of the drill pipe section during pick–ups. Since this point has (normally) the smallest section area, it is also the point of maximum stress. In addition to static load (the buoyed weight of the drillstring), inertial effects (the force to accelerate the drillstring), friction effects between the drillstring and the borehole wall, and vis-cous effects must be considered. Also, in the event of stuck pipe¹, the drillstring must be able to support the overpull applied during pipe freeing operations.

Due to several uncertainties involved in the calculation of the various load-ings, relative large safety factors must be used. It is practice to use 125% of the static load as the design parameter (25% of overpull.) In addition, it is im-portant to note that the API defines yield stress as the stress that will cause a certain permanent (plastic) deformation. Based on this, the minimum yield strength of a pipe is defined. The minimum yield strength is the minimum axial tensile load that will cause the yield of the material. Normally we want to avoid any plastic deformation of the drill pipes. Therefore, only a fraction (normally 90%) of the minimum yield strength is allowed during drilling operations. This figure is called tensile strength of the pipe.

The margin of overpull (MOP) is defined as the excess of the tensile load ca-pacity of the drillstring to the normal tensile load for normal operations. Knowing the MOP is important in case of stuck pipe. In practice, the determined MOP must not be exceeded since the drillpipe would fail. Typical values of MOP re-quirements for drillpipe selections are in the range from 50,000 to 100,000 lbf.

Example 16: Calculate the minimum yield strength required for the drill pipe of the previous example.

1A stuck pipe is the situation in which the force required to move, or the torque required to rotate the drillstring is larger that its strength.

Solution: The drillstring buoyed weight is 291000 lbf, and 125% of this value (25% of overpull) is 1.25 x 291000 lbf = 363750 lbf. The minimum tensile strength required is

F_y = 363750

0.9 = 404167 lbf

If a new 5 in, 19.5 lb/ft (nominal) is used, the minimum yield stress is:

Y_min = F_y

A_s = 404167

π

4 (5²− 4.276²) = 76625 psi Therefore, a X–95 grade pipe is required (Fy = 501087 lbf).

The minimum yield strength in the example is obtained from drill pipe tables, or calculated from the pipe parameters:

F_y = π

4 OD²− ID²
Y_s

where Ys is the yield strength of the material. For the example above we have:

F_y = π

4 × 5²− 4.276²
× 95000 = 501087 psi

This figure is for new pipe dimensions. For premium pipes (every new pipe is re–classified to premium in the first time it goes to operation), it should be considered that the wall thickness is reduced to 80% of the original wall thick-ness. Normally, the worst scenario is used and, in this case, we assume that the thickness reduction occurred in the outside diameter (external wear). There-fore, the new dimensions are:

ID = 4.276 in

OD = ID + 80% (OD − ID) = 4.276 + 80% × (5 − 4.276) = 4.8552 in Therefore, for premium DP, the minimum yield strength is:

F_y = π

4 4.8552² − 4.276²
× 95000 = 394611 lbf .

Note that if a premium DP is considered in the previous example, the X-95, a premium pipe can not be used in that operation.

Situations like this suggest the use of tapered drill pipe section. In tapered drill pipe sections, the segment with the lowest capacity is placed right above the drill collar section (or HWDP when used), and the maximum length for that pipe is calculated. Then a higher capacity drill pipe is used and the maximum length for this drill pipe is calculated. The process is repeated until the expected total length (the maximum depth for that drillstring) is reached. For a multi di-ameter, multi weight, or multi grade drillstring, the topmost joint of each section must be checked for tension.

Curtin University of Technology Department of Petroleum Engineering

Master of Petroleum Well Engineering Drilling Engineering Fundamentals

Example 17: Calculate the drillstring for the previous example data using pre-mium drill pipes of 5 in–19.5 lb/ft and grades E-75, X-95, and G-105.

Solution:

This problem is simplified by the fact that the total weight of the drillstring is the same, no matter the grade and length of the drillpipe. Using 25% of the buoyed weight, the MOP is

M OP = 25% × 291000 = 72750 lbf .

It is important to realize that any overpull applied to the top of the drillstring will manifest in every element of the drillstring (assuming that the stuck point is in the bit).

The minimum yield strength and maximum tensile load for 5 in–19.5 lbf/ft (actual 22.28 lbf/ft) drill pipes for the various grades, and using a maximum of 90% as operational limits are:

E–75: F_y = π

Starting with E–75 (on top of the drill collars), the maximum length for this grade is

The maximum length is 933 + 4705 = 5638 ft, and the partial weight is 933 × 147 + 4705 × 22.28 = 241978 lbf .

Continuing with the next grade (X–95):

(241978 + LX–95× 22.28) ×

The maximum length is 5638 + 3912 = 9550 ft, and the partial weight is 241978 + 3912 × 22.28 = 329137 lbf .

Continuing with the next grade (G–105):

(329137 + L_G–105× 22.28) ×

LG–105 = 1956 ft

The maximum length is 9550 + 1956 = 11506 ft, enough to reach the depth of 10000 ft.

Note that if different DP diameters or different linear weights are used, the final weight of the drillstring is not known before it has been designed, and the MOP cannot be imposed directly (and an iterative process will be needed).

In this case, we can set the MOP absolutely, based, for example in the worst scenario (heaviest DP) or using field experience.