DPT1 - 02 - Drill String Design

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DPT1

Drill String Design

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References

• API RP 7G Drill Stem Design and Op Limits

• API SPEC 7 Specifications for Rotary Drilling Elements

• API SPEC 5D Specifications for Drill Pipe

• SLB Drill String Design manual

• TH Hill DS-1 Drill String Design

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At the end of this lecture YOU will be able to:

• Describe the functions of DPs, DCs and HWDPs

• Define and explain the use of upsets, tool-joints, rotary

shouldered connections, BSR, SR, Stress Relief features,

grades

• Explain Buoyancy and Define neutral points

• Design a BHA for a vertical hole

• Explain Margin Of Overpull, Slip Crushing Force

• Perform DP design calculations based on Tension, Torsion,

Burst and collapse

• Define buckling and Calculate Max WOB

Objectives

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I.I.I.I. Introduction to Drill String Design: OverviewIntroduction to Drill String Design: OverviewIntroduction to Drill String Design: OverviewIntroduction to Drill String Design: Overview

II.II.II.II. Drill String ComponentsDrill String ComponentsDrill String ComponentsDrill String Components

• Drill Collars Drill Collars Drill Collars Drill Collars ---- Drill Pipe Drill Pipe Drill Pipe Drill Pipe ---- HWDPHWDPHWDPHWDP

III.III.III.III. Drill String DesignDrill String DesignDrill String DesignDrill String Design

• Pipe Rating and Design FactorsPipe Rating and Design FactorsPipe Rating and Design FactorsPipe Rating and Design Factors

• Buoyancy, Axial Loads and BucklingBuoyancy, Axial Loads and BucklingBuoyancy, Axial Loads and BucklingBuoyancy, Axial Loads and Buckling

• Bottom Hole Assembly SelectionBottom Hole Assembly SelectionBottom Hole Assembly SelectionBottom Hole Assembly Selection

• Drill Pipe SelectionDrill Pipe SelectionDrill Pipe SelectionDrill Pipe Selection

Agenda

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The drill string is the mechanical linkage connecting the drill bit to the rotary drive

system on the surface.

The drillstring serves the three main following functions :

1.1.1.1. Transmit and support axial loads Transmit and support axial loads Transmit and support axial loads Transmit and support axial loads ---- WOBWOBWOBWOB

2.2.2.2. Transmit and support Transmit and support Transmit and support Transmit and support torsionaltorsionaltorsionaltorsional loads loads loads loads ---- rpmrpmrpmrpm

3.3.3.3. Transmit hydraulics to clean the hole and cool the bit. Transmit hydraulics to clean the hole and cool the bit. Transmit hydraulics to clean the hole and cool the bit. Transmit hydraulics to clean the hole and cool the bit.

Functions of the Drill String

The Drill String includes all tubular equipment between the Swivel and the bit:

Kelly, Surface Safety Valves, DP, HWDP, DC, Jars, Shock Subs, Bumper Subs, Junk

Baskets, Accelerators etc…

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Drill Pipe

FunctionFunctionFunctionFunctionTo serve as a conduit or conductor for drilling fluid

To transmit the rotation from surface to the bit on bottom

ComponentsComponentsComponentsComponentsA pierced, seamless tube of forged steel or extruded Aluminum

Tool joints attached to each end of the seamless tube

Tool JointsTool JointsTool JointsTool JointsProvide connections for the drill string

Separate pieces of metal welded to the seamless tube

Thick enough to have pin or box cut into them

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Mechanical Properties of Steel

Young ModulusYoung ModulusYoung ModulusYoung Modulus

E = Stress divided by Strain = 30,000,000 psi

Stress & StrengthStress & StrengthStress & StrengthStress & Strength

Stress = Load divided by Cross Section Area

Strain & stretchStrain & stretchStrain & stretchStrain & stretch

Strain = Stretch divided by original length

Elastic LimitElastic LimitElastic LimitElastic Limit

Limit of stress beyond which, when the stress is removed, the steel will have acquired a

permanent stretch.

Minimum Yield StressMinimum Yield StressMinimum Yield StressMinimum Yield Stress

The stress which gives a strain of 0.5% (0.005”). When the stress is removed, the steel

will have acquired 0.2% of permanent deformation.

Ultimate Tensile StressUltimate Tensile StressUltimate Tensile StressUltimate Tensile Stress

The max stress on the curve, very close to the stress which will break the steel

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Mechanical Properties of Steel

0

20

40

60

80

100

120

0 0.002 0.004 0.006 0.008 0.010

Stress

σ (ksi)

Strain

ε (in/in)

0.22 0.24

Yp (API method)Yp (ASTM method) Uy

Proportional Limit

Elastic Limit

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Drill Pipe Classification

1. Size 2 3/8” to 6 5/8” – refers to OD of pipe body

2. Length Range 1: 18-22 ft, Range 2: 27-30ft, Range 3: 38-45 ft

3. Grade E - 75, X – 95, G – 105, S – 135

the numbers denote 1000’s of psi minimum yield stress

each grade has a min, max and average yield stress

4. Weight Called nominal weight

5. Class API classification for used pipe

For example a drill pipe could be - 5”, Range 2, G-105, 19.5ppf, New

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Exercise DP-00

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New: No wear, has never been used

Premium: Remaining wall not less than 80%.

Class 2: Remaining wall not less than 70%.

Class 3: Remaining wall less than 70%.

Other details such as, dents and mashing, slip area mechanical damage, stress

induced diameter variations, corrosion cuts and gouges, specified on Table 24

(Classification of Used Drill Pipe ) of API RP 7G.

Unlike casing and tubing, which are normally run new, drill pipe is normally

used in a worn condition. It therefore has Classes:

Used Drill Pipe Classification

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Where the pipe joins the tooljoint, the pipe wall thickness is increased

or “upset”.

•This increased thickness is used to decrease the frequency of pipe

failure at the point where the pipe meets the tool-joint.

•The drill-pipe can have

• Internal upsets (IU), (OD stays the same )

• External upsets (EU)

• Internal and External Upsets (IEU).

Drillpipe Upsets

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Drill Pipe Weights

When referring to Drill Pipe Weights, there are four important ones:

Plain end Weight – Refers to the weight per foot of the pipe body.

Nominal Weight - Refers to an obsolete standard. ( Weight of Range I pipe

with connections ) Is used today to refer a class of Drill pipe.

Adjusted Weight – Refers to the weight per foot of pipe including the upset

but excluding the tool joint based on a length of 29.4 ft

Approximate Weight – The average weight per foot of pipe and tool joints

of Range II pipe. This approximate weight is the number to use in Design

calculations.

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lengthjttool29.4

jttoolwt.29.4DPwt.adj.Wt/ft

+

+×=

Calculating Approximate Weights

429 .

WtupsetWtTubeWt AdjDP +=

( ) ( ) ( )TETETJ DDd.DDx.dDL.Wt −××−−+−×= 23322 501016702220

( )ft

DD.LL TE

JtTool12

2532 −×+=

(1)

(2)

(3)

Data from Table 7 Sepc 5D, Fig 6 Table 7 Spec 7

L= combined length of pin and box (in),

D= outside diameter of pin (in)

d= inside diameter of pin (in)

DTE= diameter of box at elevator upset (in)

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Calculate the approximate weight of tool joint and drillpipe assembly

for 5” OD, 19.5 lb/ft DP having NC50 tool joints with 6.625” OD, 2.75” ID

and being IEU.

Compare the value against the one published on Table 9 of API RP7G.

Exercise DP-01 - Homework

Tables 7API 5D and Table 7 of the Specification can be found in handout # 1 of

tables.

Table 9 of API RP7G can be found on handout # 2 of tables.

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The API has set manufacturing tolerances on DP tubes:

-OD tolerance:

- For OD<= 4”, the tolerance is +/- 0.031”

- For OD>4”, the tolerance is +1% -0.5%

- Wall Thickness tolerance: -12.5%, +??

- Mass tolerance:

- For a single joint: +6.5% -3.5%

- For a string: +6.5% -1.75%

- There is no tolerance on the ID which is solely governed by the

OD and the weight tolerance

- Tolerances do not apply to used pipe

DP Manufacturing Tolerances

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Based on the API Tolerances, and considering a single joint of 5”

DP, 19.5# nominal weight, please calculate:

• How big or small can the 5”tube be?

• How heavy or light can the 5” tube be?? (plain end)

• What is the range of tolerated ID?

• What is the range of tolerated wall thickness?

• What would be the range of internal capacities of 10000’ of

that tube

Exercise DP-02 Tolerances

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A study on pipe OD has been

conducted by Hydril and 1800

tests were done on randomly

selected pipes of different

sizes, coming from 16

different mills and taken from

180 different orders covering

several years.

Tolerances & Reality Check

None of the pipes is actually of the nominal size or less!

They are all bigger, sometimes even exceeding the API tolerances.

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• All API tool joints have a minimum yield stress of 120,000 psi regardless of the

grade of the drill pipe they are used on (E, X, G, S).

• Tool joints are always stronger in tension than the tube to which they are

attached, and almost always weaker in torsion than the tubes to which they

are attached.

• API says that tool joint torsional strength should be at a minimum of 80% of the

tube torsional strength.

• Make up torque is determined by pin ID or box OD. The make up torque is 60%

of the tool joint torsional capacity. The equation for determining make up can

be obtained from the appendix of API RPG7.

• The API has developed a series of charts to find the recommended make up

torque to any connection given the tool jt OD of box and ID of pin.

Tool Joints

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Make-Up Torque Charts

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Exercise DP03

Using some tables (?) and some figures (?) of API RP7G what

should be the make up torque of NEW 5” - 19.5 ppf G105 and S135

drill pipe, NC50 ?

How do these values compare to the ones reported on Table 10 ?

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The most common thread style in DP is NC

The thread has a V-shaped form and is identified by the pitch diameter, measured at

a point 5/8 of an inch from the shoulder

Connection Number is Pitch dia*10 truncated to two digits

5/8”

GAUGE POINT PITCH DIAMETER

The size of a rotary shouldered connection is fixed

by its gauge point pitch diameter.

Drillstring Connections

Multiply 5.0417 by 10 → 50.417

Choose first two digits → 50

Hence NC 50

If the pitch diameter is 5.0417 in �

This is an NC50 connection

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• Typical sizes: NC 50 for tool joints with 6 1/2” OD for 5” pipe and

NC 38 for 4 3/4” tool joints and 3 1/2” pipe.

• Seal is provided by shoulder not threads. A clearance exists

between the crest of one thread and the root of the mating

thread

• Use of Zinc based dope (used to be Lead) vs Copper based

dope for DCs. Not for sealing but for lubrication, to help make-up

and prevent galling

• There are 17 NC’s in use : NC-10 (1 1/16”) through NC-77 (7

3/4”)

NC Drillstring Connections

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Connection Interchangeability

Ext Flush

Slim Hole

Dbl

Streamline

Extra Hole

Full Hole

Int Flush

4-1/2EF

4-1/243-1/22-7/8SH

5-1/24-1/23-1/2DSLDSLDSLDSL

54-1/23-1/22-7/8XH

4FH

4-1/243-1/22-7/82-3/8IF

NC50NC46NC 40NC 38NC 31NC 26

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Drill CollarsDescriptionDescriptionDescriptionDescription

They are heavy walled metal tubes

The ends are threaded (box and pin)

FunctionsFunctionsFunctionsFunctions

To put weight on bit (WOB)

To keep the drill string from buckling

TypesTypesTypesTypes

Comes in many OD and ID sizes

Typically 4 ¾” to 9 ½” OD

Most commonly in lengths of 30-31 feet

Square collars where the holes tend to be crooked

Spiral collars where there is chance of getting stuck

Collars with elevator and slip recesses

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1. Protect the Drill string from Bending

2. Control direction and inclination of wells

3. Drill straighter holes or vertical holes

4. Provide Pendulum effect

5. Reduce dog legs, key seats and ledges

6. Improve the probabilities of getting casing in the hole.

7. Increase bit performance

8. Reduce rough drilling, sticking and bouncing

9. As a tool in fishing, testing, completing

More functions of Drill Collars

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Drill Collars Strapping

length

Fish neckelevator

recess

slip

recess

OD

IDconnection

Well# TRG 1 Bit # 1

Date: 28-Jul-03 Sl # 1234

Rig: IDPT Type atm 234

BHA#: 1 Manuf Hughes

Hole Size 26" Jets 20-20-20

Item Sl # ID OD FN Pin Box Length Remarks

Bit 1234 26" 7 5/8" R 0.75 New

Bit Sub SL 235 3 1/8" 9 1/2" 7 5/8 R 1.01

9 1/2" Drill Collar 9546 3 1/8" 9 1/2" 0.67 7 5/8" R 7 5/8 R 8.96

Stab 237689 3 1/8" 9 1/2" 0.93 7 5/8" R 7 5/8 R 2.36

9 1/2" Drill Collar 9503 3 1/8" 9 1/2" 0.78 7 5/8" R 7 5/8 R 9.01

9 1/2" Drill Collar 9521 3 1/8" 9 1/2" 0.95 7 5/8" R 7 5/8 R 9.04

9 1/2" Drill Collar 9520 3 1/8" 9 1/2" 1.03 7 5/8" R 7 5/8 R 8.99

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• DC connections are rotary shouldered connections and can mate the

various DP connections

• The shoulder provide the only positive seal against fluid leakage

• The connection is the weakest part of the entire BHA

• Improper M/U torque, improper or insufficient lubricant, galling can all

lead to connection failure

• The DC connections go through cycles of tension-compression and are

subject to bending stresses

Drill Collar Connections

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• Stresses in DC connections are concentrated at the base of the pin and

in the bottom of the box

• DP body bends easily and takes up the majority of the applied bending

stress, DP connections are therefore subjected to less bending than the

DP body.

• DCs and other BHA components are however much stiffer than the DPs

and much of the bending stresses are transferred to the connections.

• These bending stresses can cause fatigue failure at the connections

Stress Relief Groove / Bore Back

Stress Relief Features

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Stress Relief Pin & Box Features

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Stress Relief Features

• The stress relief groove is to mitigate the fatigue cracks where the

face and threads would have otherwise joined

• The Bore Back serves the same purpose at the bottom of the box

• Pin stress relief grooves are not recommended on connections

smaller than NC-38 because they may weaken the connection’s

tensile and torsional strength.

• Bore Back boxes could be used on smaller connections.

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Lo-Torq Feature

• The low torque feature consists in

removing part of the shoulder area of

the pin and box.

• This allows for lower make-up torque

maintaining adequate shoulder loading.

• It is a common feature in large OD

connections.

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DC Make-Up Torque

M/U Torque as % of torsional yieldAPI recommended make-up

torque for connections is a

percentage of the total

torsional yield of the

connection 62.5%56.8%API NC

56.2%51.1%H-90

N/a79.5%PAC

DC>7 inDC< 7 in

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Heavy Weight Drill Pipe

DesignDesignDesignDesign

Heavier wall and longer tool joints

Center wall pad

Also available in spiral design

FunctionFunctionFunctionFunction

Used in transition zones between DC and DP

This prevents the DP from buckling

Can be used in compression (?)

Used for directional drilling

Used in place of DC sometimes (?)

Not to be used for Weight on Bit in vertical wells

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• Has the same OD as a standard DP but with much

reduced inside diameter (usually 3” ID for 5” DP)

and has an integral wear pad upset in the middle.

• It is used between standard DPs and DCs provide a

smooth transition between the different sections of

the drillstring components.

• Tool-Joint and Rotary shouldered connection, just

like DP

• HWDP, although stiffer than DP, can also buckle

CharacteristicsCharacteristicsCharacteristicsCharacteristics


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• HWDP can be run both in tension and in compression

BUT!!!BUT!!!BUT!!!BUT!!!

• They should not be buckled

• Manufacturers recommend not to run HWDP in compression in

hole sizes larger than 12 ¼”

• Experience shows that they should not be run in compression in

Vertical Holes

• If run in compression, rules of thumb are:

• TJOD + 6” > OH diameter

• 2 x TJOD > OH diameter

HWDP in Compression?HWDP in Compression?HWDP in Compression?HWDP in Compression?


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Agenda

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The objectives of Drill String Design are to:

1. Keep the maximum stress at any point in the drill stem less

than Yield Strength down-rated by a design factor

2. Select components and configure assemblies to retard fatigue

as much as economically possible

Why bother about DS Design?

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Design Factor

DF = Rating /Applied Load

- Used for casing and tubing design and for Drill String Design

- DF < 1.0 ⇒ Failure may or may not occur

Note: not to be confused with Safety Factor

- SF=Failure Load/Applied Load

- SF<1 ⇒ Failure is imminent

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DP tube Tensile Rating is given by:

YpIDODTYield *)(*4

22 −=π

Where Yp is Min Yield Strength and OD and ID are nominal diameters

Pipe Ratings - Tension

Tool Joints are always stronger in tension than the pipe they are attached to

The results can be found in API RP7G Table 2, 4 & 6

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Pipe Ratings - Torsion

DP tube Torsional Yield is given by:

Where J is the polar moment of inertia:

Tool joints Yield formula is much more complex.

Tool joints are almost always weaker than the tubes they are attached to.

OD

YJQ

p**096167.0=

)IDOD(J 44

32−=

π


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DP tube Burst Rating is given by Barlows formula:

Where Yp is Min Yield Strength and OD and ID are nominal diameters

t is the thickness

OD

tYpPBurst

**2*875.0=


Pipe Ratings - Burst

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DP tube Collapse Rating is given by 7 formulas (!):

The ratings will depend on the D/t ratio and one the Grade of the pipe

Derivations can be found in API RP7G Appendix A and yield the results found

in Table 3, 5 & 7

One major assumption to use the tables is that the pipe is under no axial load

Pipe Ratings - Collapse

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The strength of DP is determined by the strength of the weakest point, thus

the “worst case” has been assumed for calculating tensile, torsional,

burst and collapse resistance of DP

Pipe Ratings of Used Pipe

For calculating tensile and torsional ratings of used DP, it must be assumed

that the ID has its nominal value and that all the wear has taken place

uniformly on the outside of the pipe

For calculating burst and collapse ratings of used DP, it must be assumed

that the OD has its nominal value and that all the wear has taken place

uniformly on the inside of the pipe

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For a 5” nominal OD, 19.5# nominal weight, X-95, both New and Premium

Calculate:

- Torsional Rating (Torsional Yield Strength)

- Tensile Rating: (Min Tensile Yield Strength)

- Burst Rating: (Min Internal Yield Pressure)

Compare with Table 2,3,4,5 of API RP7G

Exercise DP04 - Ratings

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When the string is in tension, the Collapse rating is further de-rated:

1<=KP

P

CollapseNonimal

CollapseBiaxial

Pipe Ratings – Biaxial Collapse

Average

CollapsealNo

CollapseBiaxial

YpIDOD

AxialLoadZ

ZZ

P

P

*)(7854.0

2

34

22

2

min

−=

−−=

Notes: 1. the use of the Average Yield Stress not minimumnot minimumnot minimumnot minimum

2. For used pipe, ID has nominal value and wear is on the outside

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• For nominal Collapse

• Use D/t and correct formula Spec 7G Appendix A 3

• Use the results found in Table 3-6 RP-7G

• For OD and ID, use Table 1 RP-7G

• For Avg Yp Use Table in section 12.8 RP 7G

145,000S

120,000G

110,000X

85,000E

YpAvgGrade


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Simultaneous torsion reduces DP tube tensile capacity and vice versa

When backreaming, pulling on stuck pipe or fishing, high magnitude combined

tension and torsion on the drill pipe may occur

DP Torsional Yield Strength under tension is given by:

Where T is the total tension applied

This value may, or may not, be lower than the Tool-Joint Torsional Yield

strength

2

220961670

A

TY*

D

J*.Q p −=

Pipe Ratings – Biaxial Torsion

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Drill String Design Factors

Tension: 1.1

Burst:1.2

Collapse: 1.15

Torsion: No Design Factor Required.

Not DF but will be considered in Drill String Design:

Margin of OverPull: 50-100K

Excess BHA Wt: 1.15

Buckling: In highly deviated wells it is possible to use DP in compression, provided it is

not buckled

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Agenda

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• Buoyancy is the weight of the displaced fluid

• Buoyancy is usually accounted for via BF

• Buoyancy is creating a hydrostatic effect: the Pressure-Area

Force

• The forces acting on a drillstring are the self-weight and the

hydrostatic pressure of the drilling fluid

• Buoyancy is creating a force acting at the bottom of the drill

string and placing the lower portion of the drill string in

compression and reducing the hook load by HP x CSA

Buoyancy

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Exercise DP05 - Buoyancy• We are running open ended DCs

9” x 3” – 192ppf

• The fluid in the well is 14 ppg

• The depth is 10000 ft

• What is the hook load with BF?

• What is the hook load with Archimedes principle?

• What is the hook load with Pressure Area Force?

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Buoyancy• What happens if the fluids inside and outside are different?

• What happens if the well is deviated?

• How do you calculate the Axial Load at any point in the string?

• What happens at the junction from a DC to a DP?

The buoyancy factor can only be used to determine the Axial Load on surface provided the fluids inside and outside are the same.

The pressure-area approach is the preferred technique used to determine the axial tension or compressive load at any point in a tubular string

Free body diagrams (FBD) are constructed for that purpose

The FBD is constructed by simply drawing a sketch of the physical system and identifying all of the loads that are applied to the tubular. These loads are then summed and set equal to zero (since the system is static).

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Example – Axial Loads

We are running:

540’ of open ended DCs, 8” x 3” – 147ppf

5” DP, 19.5# and 4.276” ID to surface, G-105, NC-50

The fluid in the well is 14.8 ppg

The depth is 13500 ft

The string is hanging freely in the mud, no WOB is applied

Determine the Axial Load (Fa) at surface, at the bottom of the DP, at the top of the DCs and at the bottom of the string.

Plot Fa against depth

Where is the neutral point of Tension & Compression?

What happens if a WOB of 30000 lbs is applied?

Fa

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The only force acting on the bottom of the pipe is due to pressure exerting on the exposed cross-sectional area.

The force is negative, indicating a compressive force.

HP13500 x CSA

∑ −=⇒+== DC13500aDC13500ay APFAHPF0F

( ) ( )lbs.

in) psix.x.

800448F

384(135008140520F

a

222

a

−=

−−= π

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The forces acting on the top of the DCs are due to pressure on the exposed

cross-sectional area, and to the weight

HP13500 x CSAp

DCDC13500aDC13500DCay WAPFAHPWF0F +−=⇒+−==∑

( ) ( )lbs.

(lbs)xin) psix.x.

400369F

540147384(135008140520F

a

222

a

−=

+−−= π

The force is again negative, indicating the top of the DCs are in compression

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The forces acting on the bottom of the DP are due to pressure on the exposed

cross-sectional areas, and to the weight of the DCs

HP13500 x CSAp

∑ +−−−== DC13500DCDPDC12960ay AHPW)A(AHPF0F

( ) ( )lbs

(lbs)in.)( psi

850.8F

369400327645849974F

a

22222

a

=

−−+−+= π

The force is positive, indicating the bottom of the DPs is in tension.

There is a discontinuity in the axial force where the x-section of the pipe changes.

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The forces acting at the top of the DP (surface) are due to pressure on the

exposed cross-sectional areas, to the weight of the DCs and to the weight

of the DPs

∑ +−−−−== DC13500DCDPDC12960DPay AHPW)A(AHPWF0F

( )( )

lbs

(lbs)ft

293062F

885012960#93.21F

a

a

=

+=

Of course the force is positive, indicating a tension at surface.

Note: with the use of BF:

Fa

( )( )[ ]( )

lbs

(lbs)..ft)#)x((ft#.

281437F

5658141540147129609321F

a

a

=

−+=

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The plot of Fa against depth does

show the discontinuity at the

junction from DC to DP.

In this particular example, the Neutral

Point of tension compression

(tension=compression=0lbs) is

exactly at the junction.

If a WOB of 30klbs is applied, the

curve will shift to the left and the

Neutral Point will move up into the

DPs. The bottom of the DP will then

be in compression -21150 lbs

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Exercise DP06 – Axial Loads

Same exercise with the same string but with the following conditions:

- MW is 12 ppg, WOB is zero

- MW is 12 ppg, WOB is 30.000

- Depth of 13500’ and 15500’

Determine the Axial Load (Fa) at surface, at the bottom of the DP, at the top of the DCs and at the bottom of the string.

Plot Fa against depth

What happens to the neutral point of Tension & Compression as the depth increases?

Homework: construct a spreadsheet valid for any MW, any TD

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

11000

12000

13000

14000

15000

-600.0 -500.0 -400.0 -300.0 -200.0 -100.0 0.0 100.0 200.0 300.0 400.0

Load (kip)

TV

D R

T (

ft)

Fa

Feff

Stab Force

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• A tube subjected to a load will bend

• Bent is a condition in which the bending

increases proportionally with load

• When a little increase in load will result in

large displacements, the tube is said to be

buckling

• The tube may not necessarily be yielded as

buckling does not necessarily occur plastically

• The load which produces buckling is called

the Critical Buckling Load

Buckling

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Lubinski’s Effective Force

The two factors that promote buckling in pipes are axial force and pressure. In addition to the axial stress, the pressures (in and out) induce a radial and a tangetial stress: the pipe is subject to triaxial stresses

A convenient way of reducing a triaxial system to a pseudo-uniaxial system, has been developed by Lubinski who describes a fictitious force, relating pressure to buckling.

The fictitious force is sometimes called the stability force. It is not the buoyancy force.

It is combined with the axial force to form a so-called “effective force”.

iioofict PAPAF −=

iiooaeff PAPAFF −+=

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Buckling Criteria

If tension is a positive axial force, a positive Feff indicates no buckling.

Likewise, a negative Feff indicates that buckling may occur if the magnitude of the critical force (Fcrit) required to initiate buckling is exceeded.

To initiate buckling:

Once buckling is initiated:

Feff < 0 (at top and bottom of the pipe) -> totally buckled pipe

Feff > 0 (at top) and Feff < 0 (at bottom) -> partially buckled pipe

Feff = 0 -> neutral point (buckling stops)

criteff FF <

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Buckling: an instability failure

Buckling is a stability failure. In order to understand buckling, it is necessary to understand the nature of stability

Consider three balls at rest on three different surfaces.

What happens if each ball is disturbed slightly from its equilibrium and then released?

Ball A returns to its original position, Ball B remains in its new position, and Ball C moves away from its original position until it reaches a new equilibrium.

The equilibrium of the Ball A is stable, Ball B is neutral, and BaIl C is unstable.

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Buckling - Neutral Point

Klinkenberg has described the neutral point as the transition between the region of the tube where buckling may occur and the region where buckling is impossible.

- above the neutral point The tube is in a state of stable equilibrium

- below the neutral point it is in a state of unstable equilibrium

- at the neutral point it is in a state of neutral equilibrium.

Lubinski describes the neutral point as the point that divides the string into two portions: the weight of the upper portion being that suspended from the eIevators and the weight of the lower portion being equal to the applied force at the lower end.

Lubinski’s definition also can be stated as the point where the tube can be cut in two without changing the weight suspended at the surface. If if can be done, it must be neutral at the point of the cut.

There are numerous synonymous definitions of the neutral point.

It is called the Neutral Point of Bending or Neutral Point of Buckling.

It must not be confused with the Neutral Point of Tension & Compression.

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Neutral Point of Tension & Compression:

The point within a tube where the sum of the axial forces = 0

Useless in DS design

Neutral Point of Bending:

The point within a tube where the buoyed weight of the tube hanging below that point is equal to an applied force at its bottom end

The point within a tube where the sum of moments = 0

The point where the VME triaxial stress =0

0Fa =

0Feff =

Neutral Points

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• Buckling will occur if Feff < Fcrit

• Initially the pipe will buckle sinusoidally.

• As the compression increases, the pipe will eventually buckle helically

Sinusoidal and Helical Buckling

( )31

2

effcrit EIw x 3.5F −=

( )31

2

effcrit EIw x 5.55F −=

2

1

c

effcrit

r

sin4EIwF

−=

φ

2

1

c

effcrit

r

sinφEIw5.6F

−=

Sinusoidal Buckling for a vertical well: Sinusoidal Buckling for inclined wells

Helical Buckling for a vertical well: Helical Buckling for inclined wells

Pi,, Po = internal, external pressure at point of interest (psi)

rc

= radial clearance (in)f = inclination from vertical (deg)

w = nominal pipe weight (lbf/in)

I = moment of inertia (in4)

weff

= ooii AAw ρρ −+

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Beyond this set of equations, it must be noted that:

• In vertical wells, to simply avoid buckling of the DP or HWDP, it is just

necessary to keep the buoyed weight of the BHA exceeding the WOB

• In deviated wells, DP can be used to provide WOB because buckling of the

DP is delayed by the effort required to lift that DP to the high side of the hole.

Sinusoidal buckling will not occur as long as the (effective) compressive

force in the drillstring does not exceed the critical buckling load.

• Helical buckling load is usually very high in high inclination wells and it

would be easier to run out of pipe weight before helical buckling can occur.

Before you reach that load in the inclined part, you would have helical in the

vertical part of the well.

• Neutral point of bending can intuitively be found at the following distance

from the bit: H = WOB / buoyed weight per foot of string

Sinusoidal and Helical Buckling

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Exercise DP06bis – Continued

Calculate the Neutral Point of Buckling depth (or distance from the bit)

- MW is 12 ppg, WOB is zero

- MW is 12 ppg, WOB is 30.000

- Depth of 13500’ and 15500’

What happens to the neutral point of Bending distance from the bit as the depth increases?

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1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

11000

12000

13000

14000

15000

-600.0 -500.0 -400.0 -300.0 -200.0 -100.0 0.0 100.0 200.0 300.0 400.0

Load (kip)

TV

D R

T (

ft)

Fa

Feff

Stab Force

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Agenda

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Drill Collar Selection Principles

• Drill Collar selection is governed by two major factors:

Weight and Stiffness Weight and Stiffness Weight and Stiffness Weight and Stiffness ------------ Size!Size!Size!Size!

• Usually the largest OD collar that can be safely run is the best selection

• More weight available to account for WOB, buoyancy, hole inclination and a safety factor

• Greatest stiffness to resist buckling and smooth directional tendencies

• Cyclical movement is restricted due to tighter Clearances

• Usually Shortest BHA possible to • Reduce handling time at surface

• Minimize # of Connections in the hole

• Minimize total DC in contact with the wall for differential sticking exposure

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Sufficient Drill Collar Weight should be

available so that the Neutral Point is in

the DCs and buckling is avoided

For excess BHA weight, use 15%

15.1=WtWorkingMax

WtAvailableMax

BHA Design

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Procedure For Selecting Drill Collars:Procedure For Selecting Drill Collars:Procedure For Selecting Drill Collars:Procedure For Selecting Drill Collars:

1. Determine the buoyancy factor for the mud weight in use using the formula:

MW =Mud weight in use, ppg

65.5 =Weight of a gallon of steel, ppgBF = 1BF = 1BF = 1BF = 1---- (MW/65.5)(MW/65.5)(MW/65.5)(MW/65.5)

BHA Design

2. Calculate the required collar length to achieve the desired weight on bit:

WOB=Desired weight on bit , lbf (x 1000)

Wdc =Drill collar weight in air, lb/ft

1.15 =15% safety factor.

DC Length = 1.15* WOB / (BF*DC Length = 1.15* WOB / (BF*DC Length = 1.15* WOB / (BF*DC Length = 1.15* WOB / (BF*WWWWdcdcdcdc))))

DC Length = DC Length Vertical / DC Length = DC Length Vertical / DC Length = DC Length Vertical / DC Length = DC Length Vertical / CosCosCosCos IIII

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Determine the number of 9 inch OD by 3 in ID drill collars required

to provide a weight-on-bit of 55,000 lbf assuming

Hole deviation = 0°

Mud density = 12 ppg

Where does this place the Neutral Point of Bending?

What if the hole inclination is 30° ?

Number And Size Of Drill CollarsNumber And Size Of Drill CollarsNumber And Size Of Drill CollarsNumber And Size Of Drill Collars

Exercise DP-07

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Stiffness

• The BHA must have sufficient Stiffness to stabilize the BHA,

optimize ROP and prevent the formation of Key Seats, ledges and

doglegs

• The larger the DC, the stiffer the BHA

• Stiffness Coefficient :

= Moment of Inertia x Young’s Modulus of Elasticity

= л (OD4 – ID4) / 64 x 30.000.000

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Exercise DP-08

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Bending Strength RatioBSR is the relative stiffness of the box to the pin of a given connection.

Describes the Balance between two members of a connection and how they are likely to behave in a rotational cyclical environment

R

)dR(D

)bD(

Z

ZBSR

pin

box44

44

32

32−

−

==π

π

Where:

Zbox = box section modulus

Zpin = pin section modulus

D = Outside diameter of pin and box

b = thread root diameter of box threads at end of pin.

R = Thread root diameter of pin threads ¾“ from shoulder of pin.

. d= inside diameter or bore.

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BSR in DC Connections

• A Connection is said to be balanced if

the BSR is 2.5

• When BSR is higher tend to see pin

failures

• When BSR is lower tend to see

more box failures

• However, field experience has shown

that:

• 8” Dc having BSR’s of 2.5 usually

fail in the box

• 4-3/4” DC having BSR as low as 1.8

very rarely fail in the box.

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BSR in Connections

This table is from T.H. Hill & Associates Inc. Standard DS-1.

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• Fortunately for you API have

worked the problem!!!

• Pages 39-44 of RP 7G list the BSR of

Connections by OD and ID of the

collar

• TH-Hill DS1 gives tables directly

listing BSRs

Finding BSR

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Stiffness Ratio• The SR measures the stiffness of a connection in a transition between 2 types of

pipe

• Based on field experience, in a transition

from one collar or pipe to another the SR

should not exceed

• 5.5 for routine drilling

• 3.5 for severe or rough drilling

( )( )44

44

upruprlwr

lwrlwrupr

upr

lwr

IDODOD

IDODOD

Z

ZSR

−

−==

Note: Stiffness ratios are calculated using tube ODs & IDs, not connections.

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Torsion limits for DC

Torque is rarely limited by the DC connection because it is usually

higher in the DP at surface and lower in the DC.

• If DC make-up torque > DP make-up torque you have no routine

problems.

• BH Torque at any point should not exceed 80% of make-up

torque for the connections in the hole to avoid over tightening

connections which can lead to damage of seals.

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BHA Design Process

• Design the Collars• Max OD DC which can be handled, fished and drilled with

• Excess BHA wt

• WOB

• Buoyancy

• Excess factor

• Connection Selection

• BSR

• Torque capability

• Transition between collars and/or pipes

• SR

• Stabilization and other directional requirements

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Exercise DP-09On Seeyoulater land rig we find the following collars:

9” OD x 3” ID – 6 5/8” FH connection

8” OD x 3” ID – 6 5/8” REG connection

6 ¼” OD x 2 ¼” ID – NC46 connection

Given that we will drill a vertical 12 ¼” hole, with 9.5 ppg mud and 65000 pounds in a

relatively hard formations, what API collar would you recommend?

What would your recommendation on BSR be for the connection chosen?

Check your recommended DCs with your recommended BSR

What would be the SR between the DC and 5” DP be?

Is it acceptable?

If not what would you do?

What would be your final BHA? Length? Buoyed Weight?

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Agenda

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Drill Pipe Selection Principles

• Drill Pipe selection is governed by two major factors:

Size + Weight Size + Weight Size + Weight Size + Weight ------------ Strength!Strength!Strength!Strength!

• Usually the Drill Pipe with largest OD and ID is preferred

• Less pressure loss in the string

• More hydraulics available at the bit

• The Drill Pipe selection must address the following:

• Drill Pipe must allow to drill to TD

• Drill Pipe must support all weight below it (BHA+DP)

• Drill Pipe must provide Overpull capacity

• Drill Pipe must withstand slip crushing force

• Drill Pipe must resist burst and collapse loads

• Drill Pipe and tool joints must withstand torsion loads

• Drill Pipe might have to work in H2S environment

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The drillstring is not designed according to the minimum yield strength!!!

If DP reaches yield: it can have permanent deformation.

API recommends the use of maximum allowable design load

TTTTmaxmaxmaxmax = 0.9 x = 0.9 x = 0.9 x = 0.9 x TTTTyieldyieldyieldyield

IPM Defines a tension DF of 1.1 be applied to design loads. These accomplish the

same thing. Do not double dip!

Tmax = Max. allowable design load in tension , lb

Tyield = theoretical yield strength from API tables , lb

Tension Design

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1. Determine max design load (Tmax) :

Tmax = 0.9 x Minimum Yield Strength (or /1.1)..lb

Tension Design

2. Calculate total load at surface using

( )[ ] BFWLWLT dcdcdpdpsurf ××+×=

3. Margin Of Overpull: minimum margin of tension above expected working load.

Margin of overpull is nominally 50-100k. Choice of MOP should consider: Overall

drilling conditions, hole drag, likelihood of getting stuck, slip crushing force,

dynamic loading.

surfmax- TTMOP =

(1)(1)(1)(1)

(2)(2)(2)(2)

(3)(3)(3)(3)

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dc

dp

dc

dp

yield

dp LW

W

BFW

MOP.TL ×−

×

−×=

90

4. The maximum length of Drill Pipe that can be used is obtained by

combining equations 1,2 and 3 and solving for the length of DP

………….(4).(4).(4).(4)

Tension Design

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Exercise DP-10• Drill Collars length : 600’ and weight in air is 150 lb/ft.

• 5” / 19.5 lb/ft Premium G-105 DP with NC50 connections.

Calculate the maximum hole depth that can be drilled ?

Assume BF= 0.85

• Carry out calculations with MOP of 100,000 lb

• Use API - RP7G Tables for the values of Approximate Weight

(Wdp) and for Minimum Yield Strength

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Slip Crushing Force

• Slips because of the taper try to crush the Drill Pipe. This hoop

stress is resisted by the tube, and this increases the overall

stress in the steel

( )

( )dopeforFrictioncoeffArcTanz

TaperSlipyzyK

inlengthSlipLinODPipeD

L

DK

L

DK

S

S

StressTensile

StressHoop

s

sst

h

08.0;)(

)45279(;)tan(/1

;)(

221

'''

2

==

=+=

==

++=

µµ

ο

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• Generally expressed as a Factor

DP

TUBE 12 in 16 in2 3/8 1.25 1.18

2 7/8 1.31 1.22

3 1/2 1.39 1.28

4 1.45 1.32

4 1/2 1.52 1.37

5 1.59 1.42

5 1/2 1.66 1.47

6 5/8 1.82 1.59

SLIP LENGTH

Horz to Tang Stress Ratio

LoadAxialEquivalentStressTensile

StressHoop*loadWorking =

Axial

t

hLoad T

S

SHk =

Slip Crushing Force

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Tension Design

• You can only drill as far as you can set pipe in the slips.

• Different than overpull, this is based on working loads

dc

dp

dc

dp

T

h

yield

dp LW

W

BFW

SS

.T

L ×−×

×

=

90

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A drill string consists of 600 ft of 8 ¼ in x 2 13/16 in drill collars and the rest is

a 5 in, 19.5 lbm/ft Grade X95 drill pipe with NC50 connections. If the

required MOP is 100,000 lb and mud weight is 10 ppg, calculate:

1) The maximum depth of hole that can be drilled when using Premium

Drill Pipe. (MOP only)

2) What is the maximum depth that can be drilled taking into consideration

slip crushing force with 16” slips? To what hook-load does this

correspond? What is the MOP in this case?

3) What if we use 12” slips

Exercise DP-11

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• Step 2Step 2Step 2Step 2

• Drill collars and bottom drillpipe act as the weight carried by top

section…effectively the drill collar

• Apply the equation for top drill pipe last

• Step 1Step 1Step 1Step 1

• If we use different drill pipe, the weaker pipe goes on bottom and stronger on top

• Apply equation to bottom drill pipe first

dc

dp

dc

dp

yield

dp LW

W

W

MOP.TL ×−

−×=

90

Mixed String Design

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An exploration rig has the following grades of DP to be run in a 15,000 ft deep well :

• Grade E : New 5” OD –19.5 # NC 50

• Grade G : New 5” OD – 19.5# NC 50

It is desired to have an MOP of 50000 lbs on the grade E pipe. The total length and weight

of DCs plus HWDP are 984 ft and 101,000 lb respectively.

MW at 15,000’ = 13.4 ppg. The slips being used are 16” slips.

1. Calculate the Max. length of E pipe that can be used

2. What is the length of G pipe to use?

3. Check the MOP for the G pipe. Your conclusion?

4. Can the G pipe handle the slip crushing force at TD?

Exercise DP12 - Mixed DP

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Melun Oil Co wants to drill the following well:

TD = 17500 ft with 6 1/8" bit

8 1/2 hole section from 9200 to 14000 ft cased with 7"liner - TOL at 9000 ft

12 1/4"section from 3000 to 9200 ft cased with 9 5/8" casing

17 1/2" section from surface to 3000 ft cased with 13 3/8" casing

It is planned to use the following assembly: 650 ft of 4 3/4" DCs at 57.58 ppf

1200 ft of 3 1/2" HWDP at 25.34 ppf

3 1/2" DP 15.5# E-75 Premium and 5" DP 19.5# S-135 Premium to surface

MW is 9 ppg, and BF is 0.863

Calculate the Maximum Overpull if I get stuck at TDCalculate the Maximum Overpull if I get stuck at TDCalculate the Maximum Overpull if I get stuck at TDCalculate the Maximum Overpull if I get stuck at TD

Exercise DP13 - Mixed DP

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Other Loads

• Collapse & Collapse under Tension

• Burst

• Torsion & Torsion under Tension

• Other loads not covered here

• Bending Loads

• Fatigue estimations

• Vibrations Loads

• Shock Loads

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Collapse Design

• The DP will collapse if:

External Differential Pressure Load > Collapse pressure rating

• DF = Collapse Rating / C

• C = External Differential Pressure Load

• C = (Po-Pi)+2Pi /(D/t)

• A Design factor of 1.15 is used:

External Differential Pressure Load < Collapse rating / 1.15

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Exercise DP14 - Collapse

We will run a BHA in on 5” 19.5 #/ft Grade E premiumpremiumpremiumpremium grade DP.

There is a float valve in the BHA.

The annulus will have 12.0 ppg mud.

What is the collapse rating?

How much 5” DP can we run not to exceed a DF of 1.15?

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Exercise DP15 – Biaxial Collapse

• We are going to dry test a liner lap at 9,000 ft. We will run in with a packer

set in tension with 50,000 lb. We will run the packer in on 5 in 19.5 #/ft

Grade E premiumpremiumpremiumpremium grade DP. At the time of the test there will be nothing

inside the drill pipe. The annulus will have 12.0 ppg mud.

What is the collapse rating?

What is the collapse load on the bottom joint of DP?

What is the DF?

• For info: New 5” Gr E 5” OD, 4.276” ID, Avg Yp= 85,000 psi

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Burst Design

• The DP will Burst (yield by the action of internal pressure) if:

internal differential pressure load > burst pressure rating

• DF = Burst Rating /B

• B = Internal Differential Pressure Load

• B = (Pi-Po)

• A Design factor of 1.20 is used:

Internal Differential Pressure Load < Burst rating / 1.20

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Exercise DP16 - Burst

Worst load case happens during DST operations in a gas well. Pressure at

surface is BHP- gas gradient with no backup

• In the last example assume we are performing a DST test in the well at

9000 ft with BHP 200 psi less than the mud wt.

• What is the burst DF on the top of the Premium Grade E

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Torsion Design

Tool joints are normally weaker than the tubes to which they are attached.

To prevent downhole make-up and a possible resulting torsional failure, the operating torsion should never exceed tool joint make-up torque.

Provided this condition above is met, there is no need for a design factor

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ToolJoint Torsion Under Tension

1. Simultaneous tension reduces the torsional yield strength of pin-weak connections

2. Connection make-up torque beyond a given point reduces connection tensile capacity

It is commonly assumed that the string tensile capacity is limited by the tube.

Which is true.

But make-up imparts stresses to the tool-joint pin neck, which can be additive to the string

tension. As make-up torque is increased, a point will be reached when the pin neck and no

longer the tube, is the weak member in tension.

Conclusion: do not exceed Maximum make-up torque.

If the wellbore and drill string geometry were predicting a surface torque higher than the make-

up torque, consider the effect this may have on the string overall tensile capacity. Use API RP

7G, appendix A.

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Exercise DP17 – Torsion & Tension

A string with a hookload of 250klbs is stuck:

New 5” DP, 19.5#, G105.

What is DP Torsional Yield Strength?

What is TJ Torsional Yield Strength?

What is DP Make-Up Torque?

What is the maximum torque that can be applied to the pipe if 100klbs

Overpull is applied to work the string free?

Conclusion?

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Drill String Design Process-2

After the BHA Design is performed:

• Slip Crushing forces on DP

• Overpull tensile design at surface

• Lengths of DP Sections

• Burst Design Check

• Collapse under tension Design check

• Torsion under tension Design check

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Now you should be able to describe:Now you should be able to describe:Now you should be able to describe:Now you should be able to describe:

Drillstring Design

• Functions of Drill Pipe , Drill Collars and BHA selection

• Grades of Drill Pipe and strength properties

• Thread types and tool joints

• Drill collar weight and neutral point

• Critical Buckling force and Neutral Point of Bending

• Bending Stress Ratios and Stiffness Ratios

• Margin of overpull – Slip crushing force

• Basic design calculations based on depth to be drilled.

Documents

DPT1 - 02 - Drill String Design