Downhole tubular must have the capability to withstand maximum expected-
- Hook load
- Torque
- Bending stress
- Internal pressure
- External collapse pressure
General Stress-Strain curve of a material
Stress-Strain curve is the behavior of the material when subjected to load.
Stages are:
1. Proportional Limit: Hooke’s Law is applicable i.e. Stress ∝ Strain.
2. Elastic Limit: Material is elastic up to this point. On removal of stretching force material will return to its original dimension.
3.Yield Stress Point: It’s the stress after which permanent material elongation takes place.
4. Ultimate Stress Point: It’s the maximum stress that material has to bear before breaking. Ultimate stress corresponds to peak point on the stress-strain curve.
5. Breaking Point/Stress: It’s the maximum stress at which material finally breaks/rupture.
1. Proportional Limit: Hooke’s Law is applicable i.e. Stress ∝ Strain.
2. Elastic Limit: Material is elastic up to this point. On removal of stretching force material will return to its original dimension.
3.Yield Stress Point: It’s the stress after which permanent material elongation takes place.
4. Ultimate Stress Point: It’s the maximum stress that material has to bear before breaking. Ultimate stress corresponds to peak point on the stress-strain curve.
5. Breaking Point/Stress: It’s the maximum stress at which material finally breaks/rupture.
Yield Strength: When DP is stretched, it goes through a region of elastic deformation i.e. DP will return to its original dimensions if stretching force is removed. The upper limit of this elastic deformation is called as Yield Strength.
Tensile Strength: Beyond yield strength, there exists a region of plastic deformation. In this region, DP becomes permanently elongated, even when stretching force is removed. The upper limit of plastic deformation is called the Tensile Strength.
DP fails if tensile strength is exceeded.
Why tensile failure occurs on the top of the drill string while pulling on stuck drill pipe?
As pull exceeds Yield Strength of the metal distorts with thinning in the weakest area of DP (or smallest cross-sectional area). If pull then exceeds Tensile Strength, drill string will part. Such failure usually occurs on top of the drill string, as the top of the string is simultaneously subjected to upward pulling force and downward weight of the drill string.
Drill Pipe Grades
DP grades define the Yield Strength and Tensile Strength of the steel being used.
E
|
X-95
|
G-105
|
S-135
|
|
Minimum Yield Strength
|
75000
|
95000
|
105000
|
135000
|
Minimum Tensile Strength
|
100000
|
105000
|
115000
|
145000
|
Grade E is composed of a lower grade of steel. It has a lower Yield Strength in psi than high strength DP grades, however, once the Yield Strength is exceeded, it can withstand a greater percentage of stretch or strain prior to parting. Lower grades of steel such as Grade E are also more resistant to corrosion and cracking. Grades above E are referred to as high strength DP.
One drawback of higher grades of steel is that they are generally less resistant to corrosion, like that caused by Hydrogen Sulphide (H2S).
Yield strength and Tensile strength are in pounds per square inch of cross-sectional area of the DP. In order to calculate yield strength in pounds, this cross-sectional area must be known.
Drill Pipe Class defines the physical condition of the drill pipe in terms of dimension, surface damage, and corrosion.
CLASS
|
COLOR of BANDS
|
1 (New)
|
One White (No wear)
|
Premium
|
Two White (Uniform wear, 80% wall thickness of new DP)
|
2
|
One Yellow (65% wall thickness of new DP)
|
3
|
One Orange (55% wall thickness of new DP)
|
4
|
One Green
|
Scrap
|
One Red
|
Class 1 DP is new and therefore the strongest. With use, wall thickness will reduce consequently, reducing cross-sectional area which results in a lower Total Yield Strength (in pounds).
This Yield Strength in pounds can be calculated using the following formula:
This Yield Strength in pounds can be calculated using the following formula:
YIELD STRENGTH (in pounds) = Yield Strength (in psi) x π/4 (OD2 - ID2)
5" grade G-105, class 1 (new) drill pipe has a nominal weight of 19.5 Ib/ft and an ID of 4.276” ...therefore:
Minimum Yield Strength in pounds = 105,000 x π/4 x (52 - 4.2762)
= 553,833 lbs.
= 553,833 lbs.
Tool joints are short sections of pipe that are attached to the tubing portion of DP by means of using a friction welding process. The internally threaded tool joint is called a ‘Box’, while the externally threaded tool joint is the ‘Pin’.
API specifications also apply to tool joints:
• Minimum Yield Strength = 120,000 psi
• Minimum Tensile Strength = 140,000 psi
Nominal weight of DP is always less than the actual weight of DP + Tool Joint because of the extra weight added by tool joint and due to extra metal added to the pipe ends to increase the pipe thickness. Hence, in many calculations, Approximate Weight is used.
The tool joints on drill pipe may contain internal and/or external upsets. An upset is a decrease in the ID and/or an increase in the OD of the pipe which is used to strengthen the weld between the pipe and the tool joint. It is important to note that under tension, the tool joint is stronger than the tubular.
Archimedes Principle states that an upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged is equal to the weight of the fluid that the body displaces and it acts in the upward direction at the center of mass of the displaced fluid.
To determine the string weight in mud, it’s multiplied by the Buoyancy Factor.
Overpull
The pull applied to a drill string beyond the normal pick-up weight is called as Overpull.
The pull applied to a drill string beyond the normal pick-up weight is called as Overpull.
(1) Maximum Hook load When One Grade Of Drill Pipe Is Used
Eg: If you are using a 5” DP of Grade-S (Nominal Wt. = 19.5 ppf and Approximate Wt. = 22.6 ppf) and at a depth of 4415 m, pick-up weight is 410,000 lbs.
Eg: If you are using a 5” DP of Grade-S (Nominal Wt. = 19.5 ppf and Approximate Wt. = 22.6 ppf) and at a depth of 4415 m, pick-up weight is 410,000 lbs.
Minimum Yield Strength = Cross-sectional area x Yield Strength
Minimum Yield Strength in pounds = 135,000 x π/4 x (52 - 4.2762) = 711,709 lbs.
= 711,709 – 410,000 = 301,709 lbs
The operator can pull 301,709 lbs over the hook load before reaching the upper limit of elastic deformation (Yield Strength).
Or else you could do this way,
Yield strength of S-Grade 5” DP = 135000 psi
Yield strength of S-Grade 5” DP = 135000 psi
Minimum Yield Strength = Cross-sectional area x Yield Strength
Minimum Yield Strength in pounds = 135,000 x π/4 x (52 - 4.2762)
= 711,709 lbs.
= 711,709 lbs.
Now after including safety factor of 90%, Minimum Yield Strength in pounds
= 711,709 x 0.90
= 640,538 lbs
Hence, in this case of a stuck operator can pull up to 640klbs before reaching the upper limit of elastic deformation (Yield Strength).
Now, as previously told that there are different classes of DP corresponding on the wear of OD. Let’s see how the maximum pull is determined in such case.
Eg: If you are using a Premium Class (80% wall thickness) 5” DP of Grade-S (Nominal Wt. = 19.5 ppf and Approximate Wt. = 22.6 ppf) and at a depth of 4415 m, pick-up weight is 410,000 lbs.
New DP wall thickness = (5” – 4.276”)/2 = 0.362”
As said Premium Class DP has 80% wall thickness.
As said Premium Class DP has 80% wall thickness.
=> Wall Thickness = 80% of 0.362” = 0.286”
=> OD = Nominal ID + 2(wall thickness) = 4.276” + 2(0.286”) = 4.848”
Minimum Yield Strength in pounds = 135,000 x π/4 x (4.8482 - 4.2762) = 553,076 lbs.
Now after including safety factor of 90%, Minimum Yield Strength in pounds
= 553,076 x 0.90
= 497,768 lbs
Hence, in this case of a stuck operator can pull up to 640klbs before reaching the upper limit of elastic deformation (Yield Strength).
Obviously, as depth increases, hook load increases, at a certain depth the hook load will equal the yield strength (in lbs) for the drill pipe in use. Practically, an operator would never intend to reach this limit. For a directional well, the prediction of pick-up load is best obtained using a Torque and Drag program, as well as including the capacity for overpull. Some operators include an additional safety factor by basing their calculations on 80% - 90% of the yield strength values quoted in API RP7G.
(2) Maximum Hook load When Two Grades Of Drill Pipe Are Used
When two grades of drill pipe are used and a sufficient number of higher grade (i.e. the pipe with the higher load capacity) drill pipe stands is placed above the lower grade pipe. The maximum tension to which the top joint can be subjected is based on the yield strength of the higher grade of a pipe. Calculations similar to those already dealt with may be used to determine the maximum length of both grades of pipe.
(3) Maximum Hook load When Only a few Stands Of Higher Grade Drill Pipes Are Used
Provided the higher grade pipe is in the vertical section,
Maximum Hook load = Yield Strength of Lower Grade Pipe + Air Weight of Higher Grade Pipe
Maximum Hook load = Yield Strength of Lower Grade Pipe + Air Weight of Higher Grade Pipe
The air weight of the higher grade pipe is used because the buoy force acting on the drill string is acting on the bit and components of the BHA. The hydrostatic pressure which the mud exerts on the drill pipe in the upper (vertical) section of the hole does not create a resultant force acting upwards.
(4) Higher Grade Pipe In The Inclined Section Of The Well
If a higher grade pipe extends below the vertical part of the well, then an accurate analysis of the axial stresses requires the use of “Torque and Drag” programs.
Make-Up Torque
It is very important that the correct make-up torque is applied to the tool joints.
It is very important that the correct make-up torque is applied to the tool joints.
If a tool joint is not torqued enough,
1. Bending between the box and pin could cause premature failure.
2. Shoulder seal may not be properly seated, resulting in mud leaking through the tool joint, causing a washout.
Exceeding the torsional yield strength of the connection by applying too much torque to the tool joint could cause the shoulders to bevel outward or the pin to break off the box. Recommended make up torques for drill pipe and tool joints are listed in the API RP 7G.
Exceeding the torsional yield strength of the connection by applying too much torque to the tool joint could cause the shoulders to bevel outward or the pin to break off the box. Recommended make up torques for drill pipe and tool joints are listed in the API RP 7G.
BHA Weight & WOB (Weight-On-Bit)
1. When drilling vertical or low inclination directional wells, ordinary drill pipe must not run into compression (recommended by Lubinski in 1950). This is achieved by making sure that the buoyed weight of the drill collars and heavy-weight pipe exceed the maximum WOB with an acceptable safety margin.
In large hole sizes (16” or greater) drill pipe should not be run in compression.
In large hole sizes (16” or greater) drill pipe should not be run in compression.
2. In smaller hole sizes on high-angle wells (> 45°), drill pipe may be run in compression to contribute to WOB, provided the maximum compressive load do not exceed the critical buckling force. This critical buckling force is the minimum compressive force which will cause sinusoidal buckling of drill pipe.
Analysis of drill pipe buckling in inclined wells, by a number of researchers (most notably Dawson and Paslay), has shown that drill pipe can tolerate significant levels of compression in small diameter, high inclination boreholes. This is because of the support provided by the low side of the borehole.
3. Drill pipe is always run in compression in horizontal wells, without apparently causing damage to it.
When two contacting surfaces (i.e. drill pipe and the borehole wall) are in relative motion, the direction of the frictional sliding force on each surface will act along a line of relative motion and in the opposite direction to its motion. Therefore, when a BHA is rotated, most of the frictional forces will act circumferentially to oppose rotation (torque), with only a small component acting along the borehole (drag).
Buckling occurs due to too much compression of column (like a drill Pipe). It occurs when compressive loads in a tubular exceed a critical value, beyond which the tubular is no longer stable and deforms into a Sinusoidal or Helical shape.
1. SINUSOIDAL BUCKLING is the 1st mode of buckling, corresponds to a tube that snaps into a sine waveform. It’s also called as Lateral Buckling, Snaking, 2-D Buckling.
2. HELICAL BUCKLING is the 2nd mode of buckling, corresponds to a tube that snaps into a helical (spiral) shape.
Dawson and Paslay developed the following formula for Critical Buckling force in Drill Pipe:
Where, E is Young's modulus
I is axial moment of inertia for Drill Pipe
W is buoyed weight per unit length
Inc is borehole inclination.
r is radial clearance between the pipe tool joint and the borehole wall.
If compressive load reaches critical load Fcr, then sinusoidal buckling occurs.
The reason that pipe in an inclined hole is so resistant to buckling is that the hole is supporting and constraining the pipe throughout its length. The low side of the hole tends to form a trough that resists even a slight displacement of the pipe from its initial straight configuration.
Young’s Modulus for Steel, Esteel = 29 x 106 psi
Moment of Inertia for Drill Pipe,
Ihollow cylinder = π (OD4 – ID4)/64
Radial Clearance = (Hole Dia. – Tool Joint OD)/2
Radial Clearance = (Hole Dia. – Tool Joint OD)/2
Eg:
Calculate the critical buckling load for 5” grade S drill pipe with a nominal weight of 19.5 lb/ft (approximate weight 22.6 lb/ft; tool joint OD 6.625”; Pipe ID 4.276”) in an 8.5” hole at 8° inclination.
Sol’n:
Esteel = 29 x 106 psi
I = π (54 – 4.2764)/64 = 14.263 in4
W = 22.6 lb/ft = 1.88 lb/in
Sin 8 = 0.139
r = (8.5 – 6.625)/2 = 0.9375”
FCR = 2
Critical Buckling Force = 21,475 lbs
Calculating BHA Weight With Drill Pipe In Compression
It’s suggested that 90% of the critical buckling force be used as the maximum contribution to the weight on bit from ordinary drill pipe.
WBIT = {WBHA x SF x BF x Cos (Inc)} + 0.9FCR
Where, WBHA is total air weight of the BHA
WBIT is weight on bit
SF is Safety Factor
FCR is critical buckling load
FCR is critical buckling load
Eg:
Prior to drilling a 12.25” tangent section in a hard formation using
an insert bit, the directional driller estimates that they expect to use 50,000 lbs WOB. The hole inclination is 60° and the mud density is 11 ppg.
an insert bit, the directional driller estimates that they expect to use 50,000 lbs WOB. The hole inclination is 60° and the mud density is 11 ppg.
a) What air weight of BHA is required if we are to avoid running any drill pipe in compression? Use a 15% safety margin. (Assume that drill pipe is not run in compression)
Sol’n:
BF = 1-(11/65.5) = 0.832
Required WOB (in mud) = Available BHA Wt (in air) x BF x Cos (Inc)
Required WOB (in mud) = Available BHA Wt (in air) x BF x Cos (Inc)
=> 50000 = Available BHA Wt (in air) x 0.832 x Cos 60
=> Available BHA Wt (in air) = 120192.31 lbs
=> 120192.31 = Available BHA Wt (in air) x 85% (15% SF means only 85% BHA Wt. Is considered)
=> Available BHA Wt (in air) = 120192.31 x 100/85 = 141402.72 lbs = 63.63 T which is too large.
b) Recalculate assuming that grade-S 5” drill pipe is run in compression.
Sol’n:
Wbit = {WBHA x BF X Cos (Inc)} + 0.9FCR
=> 50000 = (WBHA X 0.832 X Cos 60) + 0.9(30,948)
=> WBHA = 53,237 lbs = Available BHA Wt (in air) x 85% (15% SF means only 85% BHA Wt. Is considered)
=> Available BHA Wt (in air) = 62,631.76 lbs = 28.2 T which is feasible.
BHA weight for steerable assemblies on typical directional wells
• The WOB is low, especially when a PDC bit is used.
• When the drill string is not rotated, the drill pipe is not subjected to the cyclical stresses which occur during rotary drilling. Therefore, sinusoidal buckling can be tolerated when there is no rotation of the drill string. Helical buckling, however, must be avoided.
Helical buckling occurs at 1.41 FCR,
Where, FCR is the compressive force at which sinusoidal buckling occurs.
Therefore, if BHA weight requirements are evaluated as for rotary
drilling, the results should be valid for steerable systems in the oriented mode except for unusual well paths which create exceptionally high values of axial drag.
drilling, the results should be valid for steerable systems in the oriented mode except for unusual well paths which create exceptionally high values of axial drag.
Neutral Point is the point in drill string where axial stress changes from compression to tension.
Location of the neutral point depends on WOB and buoyancy factor of the drilling fluid. Drillstring components located in the transition zone ( the section where axial stress changes from compression to tension) alternately experience compression and tension. These cyclical oscillations can damage downhole tools such as Drilling Jar or Drill pipe when running in vertical wells.
Bending Stress
Where, E = Young’s Modulus (psi)
D = Diameter of Tubular (in.)
R = Radius of curvature (in.)
R = Radius of curvature (in.)
Fatigue damage will ultimately lead to pipe failure.
Case 1
Sigma β < 1/8 Sigma Ultimate
· Fatigue damage should not occur.
· Life should be infinite.
Case 2
1/8 Sigma Ultimate < Sigma β < 1/4 Sigma Ultimate
· Fatigue damage may occur.
· Life may be as low as 1,000,000 cycles or be infinite.
Case 3
1/4 Sigma Ultimate < Sigma β < 0.42 Sigma Ultimate
· Fatigue damage will occur.
· Life will be less than 1,000,000 cycles.
Case 4
Sigma β > 0.42 Sigma Ultimate
· Fatigue damage will occur.
· Do not rotate.
