Video Training Courses in Offshore Structures Design

8/20/2019 Video Training Courses in Offshore Structures Design

1/102

Video Training Courses in Offshore Structures

Below

you

can

have

a

look

at

some

parts

of

video

course

notes

about

Onbottom

stability

of

jackets (Mudmat Design) , Design of Tubular members and Pile groups effects

Item

no.

Subject

of

Training

Course

Video

Duration

Remarks

1 Loads on offshore structures 2.5 hours These Video Courses are

collected

in

10

DVDs,

if

you

are

interested

in

provide

it

pls

contact

me

at

email

[email protected]

in

order

to

arrange

for

dispatching

DVDs

through

TNT

or

DHL

cash

on

delivery

Service.

Total

Cost

is

about

130

us$

(90

for

DVDs

+

40

for

Delivery)

Cash

on

delivery

Service

is

a

model

of

payment

under

which

you

pay

upon

you

received

the

order

2

Marine operation for jackets and topsides (loadout,

sailout,

installation)

2 hours

3 Design

of

Tublar

members

for

jackets 4 hours

4 Design of Tublar joints for jackets 8 hors

5 Inplace

analysis

of

jackets 1 hour

6 onbottom

stability

of

jackets

(mudmat

design) 1.5 horse

7 Basics of Soil Mechanics for Foundation of offshore

Structures

4.5 hours

8

Pile foundations for offshore structures (Design,

Analysis)

6 hours

9 Piles

installation

and

load

test 3 hors

10 Offshore

special

foundations 2 hours

11

Jackup

rig

analysis

and

design

3

hours

12 Sacs

modeling

for

offshore

Structures 3 hours

13

Sacs

analysis

of

offshore

structures

(inplace,

seismic,

Fatigue)

5 hours


2/102

30 May 2008 Dr. S. NallayarasuDepartment of Ocean Engineering

Indian Institute of Technology Madras-36

1

Mudmat Concepts and Design

OUTLINE FOR SESSION 10

Mudmat

Concepts Stability Requirements

Design

Special Foundations

Bucket Foundations

Gravity Foundations


Mudmat

Mudmats are temporary floor support for the

jacket immediately after the jacket has beenupended from floating horizontal position prior tosupported by piles.

Need to designed with adequate surface area andsufficient strength strength to avoid excessive

ttl t f th j k t


3/102



3


Advantages of

FRP and Timber Mudmat

FRP and Timber mudmats are used when liftweight is a concern. They will reduce the weightconsiderably.

The design requirement for Cathodic Protectionwill also be reduced


Large Timber Mudmat


4/102



5


FRP Mudmat


MUDMAT CONCEPTS


5/102



7


Jacket with Rectangular Mudmat


Triangular Mudmat


6/102



9


Rectangular Mudmat


Circular Mudmat


7/102



11


Triangular Mudmat


Mudmat Panels

Mudmat panels can be any one of the following.

Flate Plate (Steel)

Corrugated Plate (Steel)

Timber Plank


8/102



13


Flat Steel plate


Timber Plank


9/102



15


Corrugated Steel plate


FRP PANEL


10/102



17


Design Requirements

When the jacket is resting on seabed, it shallsatisfy following requirements

Stability against bearing

Stability against sliding

Stability against overturning

Structural members shall have adequate

strength


Design Loads

Dead loads

Bouyancy Loads

Wave and Current Loads

Wind Loads


11/102



19


Design Requirements

When the jacket is resting on seabed, it shall

satisfy following requirements (API RP 2A) Stability against bearing

Stability against sliding

Stability against overturning

Sometimes it is also called “Unpiled Stability” sincethis is prior to the piling of the jacket after which the

jacket is firmly fixed to the seabed by piles


Stability Against Bearing

As explained earlier, stability against bearing is to

have adequate bearing area to avoid excessivesettlement of jacket / failure of mudmat. This hastwo parts.

Geotechnical Requirement

Structural Requirement


12/102



21


Factor of Safety against Bearing

The Factor of Safety against bearing shall becalculated as below.

. . u

a

Q F O S

P

The minimum Factor of Safety shall be 2.0 forloads arising from dead weight of the jacket only

and 1.5 for dead weight + environmental loads.

Where Qu is the ultimate bearing capacity of soiland Pa is the applied pressure


Applied Mudmat Pressure (Dead Load)

The applied mudmat pressure can be calculated for dead

loads alone very easily.

2S x S

a

M yy

W e W H P

A I

Where WS is the total submerged weight of the jacket


13/102


14/102



25


Factor of Safety against Overturning

The Factor of Safety against Overturning shall becalculated as below (for each edge).

. . e

s

F h F O S

W x

Where x is the distance between the verticalload (jacket submerged weight) and the geometriccentre of mudmat system at mudline.

The minimum FOS of 1.5 shall be required.


Jacket Settlement

Most of Settlement will take place immediately after the

jacket has been placed on seabed. Hence the only immediate settlement using elastic theorywill suffice.


15/102



27


W Fe


Jacket Settlement

Settlement of jacket is an important criteria in designing

the mudmat system as excessive settlement woill leadsubmergence of bottom framing in to the soil. This will leadfollowing issues.

The mudline framing will be subjected to constantupward force on the members

The conductor guide if any will be submerged in to mud


16/102



29


Jacket Settlement

Elastic settlement of jacket on to the seabed canbe calculated as below.

2(1 ) s

qB I

E

Where q is the uniform applied pressure, B is thewidth of the mudmat, E is the Modulus of the soil, is the poissons ratio and Is is the influence

coefficient and shall be calculated depending on theshape of the mudmat.


Settlement of Circular Footing

Vertical settlement of circular footing is given by

QGR

41

QGR

uv

4

1 Where


17/102



31


bh Am 4

23

)2/2/(4

12

4b Bbh

hb I yy

yy xxm

sa

I

xM

I

yM

A

W P

)()(

23

)2/2/(412

4h H bh

bh I xx

Where x and y are co-ordinates of points at which the mudmat pressure is

required

Rectangular Mudmat system


22

4 44 H DDI

Circular Mudmat system

2

4

4 D Am

yy xxm

sa

I

xM

I

yM

A

W P

)()(


18/102



33


23

32

22

36

4b Bbh

bh I yy

23

32

22

36

4 H bh

bh I xx

Triangular Mudmat system

24

bh Am

yy xxm

sa

I

xM

I

yM

A

W P

)()(


23

3bh

Triangular Mudmat system

2

4bh

Am

yy xxm

sa

I

xM

I

yM

A

W P

)()(


19/102



35


BEARING CAPACITY OFMUDMATS


BEARING CAPACITY

The ultimate bearing capacity (qu) isdefined as the least pressure which

would cause shear failure of the

supporting soil immediately below


20/102



37


MODES OF FAILURE

a) General failure

b) Local shear

c) Punching failure

The mode of failure depends on thefollowing

- Foundation type and geometry- Soil compressibility


MODES OF FAILURE


21/102



39


THEORY OF PLASTICITY

A suitable failure mechanism shall befound by either inspection, trial or limit

theorems. Two bounds can be defined.

Lower Bound True failure load is large than the load

corresponding to an equilibrium system

Upper Bound The true failure load is smaller than the load

corresponding to a mechanism if that load isdetermined using the virtual work principle


EQUILIBRIUM SYSTEM

An equilibrium system, or a statically admissible field

of stresses is a distribution of stresses that satisfiesthe following conditions

a) it satisfies the conditions of equilibrium in each pointof the body

b) it satisfies the boundary conditions for the stresses

) h i ld di i i d d i i f h


22/102



41


Mechanism A mechanism, or a kinematically admissible field

of displacement is a distribution of displacementsand deformations that satisfies the followingconditions.

a) the displacement field is compatible, i.e. nogaps or overlaps are produced in the body(sliding of one part along another part isallowed)

b) it satisfies the boundary conditions for thedisplacements

c) wherever deformations occur the stressessatisfy the yield conditions


IDEALIZED STRESS-STRAIN RELATIONSHIP

s t r e s s

Y’


23/102



43


STATE OF PLASTIC EQUILIBRIUM


cos2)sin1()sin1(

)cot2(2

1

)(2

1

sin

13

31

31

c

c

)sin1(

2sin1

2


24/102



45


LOWER BOUND SOLUTION


cq

c

)1(2)1(

)1(2)1(

0for12/45tan2/45

245tan22/45tan

1.31.2

2

2

31


25/102



47


qcq

BqB B Bc

B Bq

ult

ult

2

022

UPPER BOUND SOLUTION


Simplified bearing capacity for a ø – c soil


26/102



49


yqcult

p

p p p

p

p

ult

p

ult

p p p p

H

O

H

O p

yBN N qcN q

K K yB K K

q K K

cq

P cA

H B y

Bq

K cH K H q K yH P

dz cq yz dz P

cos4coscos

2

0cossin

cos2

.22

.2..2

245tan2

245tan)()(

2

2

2

1


FAILURE UNDER A STRIP FOOTING


27/102



51


FOOTING AT DEPTH D BELOW THE

SURFACE


Width offoundation (B)not less than 1m. Water tableat least B belowbase offoundation

>600

200 – 600

300

100 – 300

Dense gravel or dense sand and gravel

Medium dense gravel or medium dense

sand and gravel

Loose gravel or loose sand and gravel

Compact sand

Medium dense sand

RemarksBearingvalue(kN/m²)

Soil type

PERSUMED BEARING VALUS (BS 8004: 1986)


28/102



53


factorscapacity bearingand,

Depth

Breadth

capacity bearingultimatethe2

1

qc

u

qcu

N N N

D

B

q

DN cN BN q


tan)1(801N

factorscapacityBearing,

cot1

/2)45(tan)tan(exp 2

N

N N

)(N N

N

cq

qc

o

q


29/102



55


qc f DN cN BN q 2.14.0

Length

Breadth

capacity bearingultimateThe

2.13.0

L

B

q

DN cN BN q

f

qc f

Circular footing

Square footing



30/102



57


Skempton’s values of Nc

for øu = 0 (Reproduced

from A.W.Skempton (1951)

Proceedings of the BuildingResearch Congress,

Division 1, p.181, by

permission of the Building

Research Establishment, ©

Crown copyright)


RECOMMENDED

BEARING

CAPACITY

FACTORS


31/102



59


ECCENTRICALLY-

LOADED FACTORS


AREA REDUCTION

FACTORS

ECCENTRICALLY-

LOADED


32/102



61


42 – 5885 – 100Very dense> 50

25 – 4265 – 85Dense30 – 50

8 – 2535 – 65Medium dense10 – 30

3 – 815 – 35Loose4 – 10

0 – 30 – 15Very loose0 – 4(N

I

)60

Id

(%)ClassificationN Value

DENSITY INDEX OF SANDS


Bearing capacity calculationsby Davis and Booker

The bearing capacity can be calculated when the soil profile

is varying linearly with depth

)1(4

ccuor u S B

N C F q

factor Shape

L

B

N

N S

c

c

NC= 5.14 for strip footing


33/102



63


Fr - Shear Strength Facto r

0.80

0.90

1.00

1.10

1.20

1.30

1.40

1.50

1.60

1.70

0.000 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000

Rh o


SPECIAL FOUNDATIONS


34/102



65


Special Foundations

Suction Anchor(Bucket Foundation)

Gravity Foundation


Suction Anchors (Piles)

A suction anchor is an inverted top capped hollow

cylinder of fairly large diameter with a length todiameter ratio (L/D) of 1.0 to 2.0 that is embeddedinto the sea bed. Self-weight and differential waterpressure can facilitate easy installation of this typeof anchor into the sea bed. This differential water

( ti ti ) b t d b


35/102



67


The main pile advantages of this anchor over tension piles aredue to the weight of the soil plug inside and the freely availablehigh ambient water pressure which offers two advantages; easyinstallation of the anchor with its active suction arrangement andmobilization of passive suction force at the anchor bottom during

uplift. Further, the large-diameter sealed top provides asubstantial space for additional ballast, which can increase thebreakout resistance


Suction Breakout Factors

From the equilibrium considerations (referring tofigure 1) the uplift pullout capacity of the suction

anchor is given by

Pu = Wa + Fext + Ws + Wb + R b

Where

W i th i ht f th h


36/102



69



Pu = Wa + Ws + Fext + Rb

Rb1 = Pu – (Wa + Ws + Fext)

From consideration of rupture in clay under tensile loading (Vesic, 1971) the

bottom breakwater resistance is expressed in a non-dimensional form as

Fext = Cu Ase

From the plug equilibrium (refer to figure 13) equations can be written as:

Rb2 + Ws - Ps + Fint

Rb2 = Ps + Fint - Ws

Design of Tubular Members Buckling


37/102

9/16/2015 Dr. S. NallayarasuDepartment of Ocean Engineering


1

CONTENTS Introduction

Necessity of tubular

Loading and Load types

Factors affecting strength

Method Tubular Fabrication

Steel Making process

Seam Less Pipes

Fabricated Pipes

Residual stresses

Material Properties

Yield and Tensile Strength Modulus of Elasticity

Imperfections

Out-of roundedness

Misalignment

Straightness deviation

Ultimate Strength

Factors affecting ultimate strength

Ultimate strength of sections and span

Buckling

Local Buckling

Global buckling (Euler)

Effective Length

Design Methods

Allowable Stress Design (ASD)

Load and Resistance Factor Design (LRFD)

API RP 2A - ASD

Applied stresses

Allowable stresses

Interaction

API RP 2A - LRFD Load and Resistance factors

Interaction

Hydrostatic Pressure

Hoop stresses

Interaction

Design examples

Tubular section

Ring stiffened cylinders

Design of Tubular MembersT b l M b


38/102



2

Tubular Members

Good Hydrodynamic Properties (Low Cd and Cm)

good buoyancy to weight ratio

Good resistance against hydrostatic pressure

Uniform property across the section

No torsional buckling Good Ultimate strength compared to others

Full moment connections possible

Tubulars or circular hollow sections (CHS) are

used for jacket structures commonly due totheir versatility in resisting various forces. Themajor reasons are listed below.

However, the tubular member connections aresusceptible to fatigue cracks and havefabrication difficulty due to non-linear surfacesat intersection !.

Design of Tubular Members


39/102



3

Load Categories

Gravity loads

Wind Loads

Wave and Current Loads

Seismic Loads

Drilling Loads

Following external forces are applied tothe structure which in turn induceinternal loads on the members.

The above forces shall be applied to thestructure in a three dimensional analysis.

The member internal loads shall beextracted from the analysis results.



40/102



4

Member internal loads



41/102



5

FREE BODY DIAGRAM

Following member internal loads mayneed to be considered

Following member internal loadsmay need to be considered

Axial (Compression or tension)

Bending (In-plane or Out-off plane)

Torsion

Shear (in-plane or Out-off plane)

External Pressure



42/102



6

Material properties (E, Fy, Ft )

Imperfections and residual stresses

Production method of tubular

Boundary conditions Loading

Geometric proportions: L/D, D/t

Stiffeners: circumferential or longitudinal

Factors Affecting Strength

Following factors affect the strength of the member.


i l i ( l)


43/102



7

Density

7850 kg/m3 or 78.5 kN/m3

Tensile stress (Ft)

Varies between 490 to 600 MPa

Yield stress (Fy )

Is in the range of 250 – 400 MPa

Modulus of Elasticity (E)

Normally taken as 200000 – 210000 MPa

Strain in elastic range is 0.2%. Poisson Ratio is in the range of 0.3 to 0.4

Friction coefficient is around 0.3 to 0.4

Material Properties (Steel)

The physical and mechanical properties of steel used in the design are listedbelow.

Design of Tubular MembersImperfections


44/102



8

Imperfections

Variation is cross section

Variation in thickness

Residual stresses

Out-off roundedness

Out-off straightness

Misalignment across thickness

Misalignment along length

Imperfections in fabrication and assembly can

cause the reduction in the strength of thestructure and must be minimized. Hencematerial and fabrication specifications shallinclude control parameters to limit the same.This is called “Tolerances”. Following are some

of the imperfections that need to be included.

Design of Tubular MembersTubular Production Methods


45/102



9

Tubular Production Methods

Tubular or Circular Hollow Sections (CHS) can be made using any one of thefollowing methods.

Seamless tube production by piercing of heated bars andextruding techniques

Hot forming steel plate and induction welding along thelongitudinal direction

Cold forming methods coils of plate and resistance welding alonglongitudinal direction

Cold forming of coils of plate and resistance welding along radialdirection

Cold forming of flat plates and assemble to make pipes

Each method has its own limitations, advantages and disadvantages. Hencedepending on the availability and technical requirement, production methodshall be selected.


Steel Making Process an outlook


46/102

01 August 13 Department of Ocean EngineeringIndian Institute of Technology Madras-36

10

BLAST FURNACE

STEEL MAKING

PROCESS

HEAT

TREATMENT

ROLLING

IRON ORE PIG IRON

PIG IRON INGOT, BILLETS

INGOT SLABS

SLABS PLATES & SHAPES

Steel Making Process – an outlook


Steel Making Process an outlook


47/102


11

Steel Making Process – an outlook

Source : Nippon Steel Corporation, Japan



48/102


12



49/102


13


Pilger and Piercing


50/102



14

Pilger and Piercing

The large size bars areused to produce pipes.

This has been in use forseveral decades in the

pipe producing mills.

Both thin and thick pipescan be made using this

method.

Limiting size for suchproduction depends onthe mill but generally

diameter larger than 20”is normally not availableby this method.

Design of Tubular MembersCold Forming Processes and Resistance welding


51/102



15

g g

In this method, sheet coilof plates is used to formcircular sections usingrollers.

The folded section is thenwelded by resistancewelding.

The application of thismethod is also limited bydiameter and generally to20”.

Design of Tubular MembersHot forming and induction welding


52/102



16

g g

This method is very similarto the forming and weldingmethod except that this isdone in hot condition.

The coils of plate is heatedfirst before it is bent androlled to the shape.

The folded section is thenwelded by inductionwelding.The application ofthis method is also limitedby diameter and generally to

20”.

Design of Tubular MembersCold Forming Processes


53/102



17

In this method, the plate

sections of specific lengthand width will be rolled toshapes either in semi-circular shape or in quarterarc of a circle.

The rolled sections of thecircular arc is then joined byarc welding to form a longpipe. This method is very

commonly used for makingpipes of any diameter usedin the steel fabricationindustry. Using this method,pipes of any diameter can be

made for use.As an alternative to the plates, rolls of plate can be used to form the pipe usingspiral form and then welded, and it is called “Spirally welded pipes”. Pipesmanufactured using this method is normally not used in the primary structure.

Design of Tubular MembersFabrication tubulars

T b la can be fab icated f om flat plates No mall flat


54/102



18

cold rolling a flat plate and weld at the seam toform a can (length up to 3m). The longitudinalseam may be one or more depending on the

width of the plate available. This one piece of pipemade from plates is called “Can”.

Several cans can be welded to form a long tube

The long seams shall be arranged such that theorientation in each can away by 90o.

Welding between Cans is called transverse seamor circumferential weld.

This method of fabrication introduces out-of-roundness, out of straightness imperfections andresidual stresses in both the longitudinal andcircumferential directions

Tubular can be fabricated from flat plates. Normally, flatplates are rolled to form circular arcs and welded toform circular section as shown in figure.

Design of Tubular MembersResidual Stresses


55/102



19

Residual stresses developed during welding of plates to form pipes and

welding of two pieces of pipes to form length may affect the final strengthunless these stresses are relieved.

Bending plates to form circular arcs induces bending strain andstresses depending on the radius of bend and D/t ratio. Larger the

bending radius, smaller the stresses. Larger the D/t ratio, strain willbe smaller.

Heat induced stresses during welding could be large due torestraint provided by the joining components.

Stresses induced during joining of pipe segments due to restrictionon the expansion during welding.


Consideration of Residual Stresses in design equations


56/102



20

Consideration shall be given to account for the residual stresses in

members in the design equation.

As these stresses exist even before the member is loaded, these

stresses shall be deducted from the allowable stresses. However,

it will not be practical to account for in each case.

Hence it is better to reduce the yield stress by certain percentage

to account for the residual stresses. DNV codes suggests a 5%

reduction in yield stresses to residual stresses of welded section

g q

Design of Tubular MembersEffective method of including Imperfections in design

The method to include the imperfections in fabrication is a difficult process as


57/102



21

The method to include the imperfections in fabrication is a difficult process asthe imperfections will not be known at the stage of design.

Hence certain assumptions has to be made during the design with limitations ondeviations that can be tolerated both with respect to design aspects andoperational aspects.

Design aspects will include change in cross sectional area, moment of inertia,center of gravity and other geometric properties. On the other hand, theoperational aspects include deviation from verticality, sagging of beams which

affects the daily operation for which the structures are built.

Hence restrictions on these imperfections which may happen during theconstruction stage may have to be imposed during the design stage.

These restrictions are called “Construction Tolerances” which shall beincorporated in the design equations so that the design need not be revised ifthese deviations are within the design tolerances.

Design of Tubular MembersOut-of Straightness


58/102



22

Out-of straightness tolerance o shall be

measured at all points along the length of themember and the maximum shall be taken forconsideration.

DNV (1982) specifies a maximumlimit of 0.0015L (L/666) as the limit

API Spec 2B specifies a maximumlimit of L/960 or 9.50mm in any12200mm length (L/1284) whichever

is lower

This tolerance is very important as thisdeviation will lead to eccentric load andcorresponding moment.

Design of Tubular MembersOut-of Roundedness


59/102



23

Out-of roundedness tolerance for fabrication of

tubular sections can be calculated as shown infigure using Dmean, Dmax and Dmin.

The Dmax and Dmin shall be measured across

diagonals at any angle and not necessarily at 90degrees. Out-of roundedness is normallyspecified as

max min %mean

D D D D Dδ −=

API Spec 2B specifies that the above tolerance

shall not exceed 2% and DNV specifies that thetolerance shall not exceed 1%.

Design of Tubular MembersEccentricity due to variation in Wall thickness


60/102



24

Maximum thickness variation = ∆t = tmax - tmin

Effective axial load eccentricity due to ∆t can be calculated and included inthe stress calculation.

Design of Tubular MembersMisalignment in Butt Joint


61/102



25

Misalignment in butt joint is veryimportant as it induces additionaleccentricity in axial loads andstresses.

API allows an eccentricity “e” of

• 0.2t1• e < 3.2mm for welding from

one side

• e < 6.4 mm for welding fromboth side.

DNV allows an eccentricity of 0.15t1 (minimum thickness) or 4mm whichever is less.

When the eccentricity in construction exceeds this limit, the design must be reviedadequate modifications shall be carried out to assure the d=safety of design.


Ulti t t th f ti

Ultimate Strength


62/102

9/16/2015 Dr. S. NallayarasuDepartment of Ocean EngineeringIndian Institute of Technology Madras-36

26

Ultimate strength of a section or

member depends on the efficientlyof the section to redistribute thestresses when the stresses exceedyield. Increase load carrying

capacity after reaching elastic limitis called “Ultimate Strength”.

Premature failure before reachingelastic limit is called “Buckling”.

Buckling strength of a member isfound to be considerably less thanthe theoretical elastic capacity.

Hence in order to determine the ultimate strength, first it is necessary to establish

that the section / member has sufficient buckling capacity to reach elastic capacity.The ultimate strength of the section / member can be computed based on thesection property and member boundary conditions.

Design of Tubular MembersBuckling Theory

B kling is a phenomenon that the bif ation of eq ilib i m to nstable state


63/102


27

Buckling is a phenomenon that the bifurcation of equilibrium to unstable state

under axial load when the slenderness exceeds 50. This was explained byLeonhard Euler in 1757 even if there is no axial load.

The column at its unstablebifurcation of equilibrium, fails dueto lateral displacement for aparticular load called “Critical Loador Buckling Load”.

The critical load differs if the endof the column is restrained inlateral direction. This is evidentfrom the photograph showing theexperiment.

Slenderness is the ratio of itslength to the radius of gyration of the section.

Design of Tubular MembersEffective Length Factors (K)


64/102


28

Effective length factor is

defines as the ratio ofbuckling strength of a columnwith simple pin-pin endconditions to that of a actual

column with any otherboundary conditions.

Buckling capacity of a columnwith pin-pin end conditions isgiven by

( )

2

2cr

EI P

KL

π =

In which K is called Effective length factor and is 1.0 for pin-pin endconditions of the column. For other cases, it is shown in the table above.

Design of Tubular MembersLocal and Global buckling

Buckling of thin walled tubes (D/t > 20) can be


65/102


29

Buckling of thin walled tubes (D/t > 20) can be

classified in to the following.

Local buckling – due to instability of local shell wall

Global buckling – due to slenderness

Local Global

In which the D is the diameter of the cylinder and tis the wall thickness.

Local buckling is governed by the D/T ratio and theglobal buckling is governed by the KL/r ratio. Local

buckling may also happen due to bending of largediameter tubular.

Design of Tubular MembersFactors influencing Ultimate strength

Following factors influences the ultimate strength of a column or beam


66/102


30

Following factors influences the ultimate strength of a column or beam

Cross section

Boundary condition at the ends

Load distribution

Stress strain characteristics of the material

Cross section influences the redistribution of stresses while the boundarycondition affects the redistribution of stresses across the length.

The stress strain relationship affects the ultimate load depending on the strainhardening range of the material. i.e. the gap between the yield point and theultimate point the stress strain curve.

All the factors put together, a beam or column can sustain larger load compared

to its load capacity at elastic range.

Design of Tubular MembersELASTIC AND PLASTIC MOMENT CAPACITY – RECTANGULAR SECTION


67/102

15th

April 2009 Dr. S. NallayarasuDepartment of Ocean EngineeringIndian Institute of Technology Madras-36

31

2 2

y F h P b=

22

2 2 3 6

y

y

F h h bhM b F

= =

2

22 4 4

p y y

h h bhM F b F

= =

2 p y

h F b=


ELASTIC AND PLASTIC MOMENT CAPACITY– CIRCULAR SECTION


68/102

15th


32

2

12 4

p y D P F π =

2 34

8 3 6 p y y

D D DM Pa F F π π

= = =

4

3

D

π

3

32 y

DM F

π =

Plastic moment capacity of solid cross section is give below.

Elastic moment capacity of solid crosssection is give below.

21

2 2 4

y F D P

π

=

Design of Tubular MembersPLASTIC MOMENT CAPACITY HOLLOW CIRCULAR SECTION


69/102

15th


33

2

Dds rd d φ φ = =a tds=

A hollow circular section of diameter Dand wall thickness t is divided in tofour symetric segments.

Consider a small arc of ds with area

of a in the first quadrant of the pipeas shown in figure.

The area of the segment can becalculated as tds where ds can be

calculated using small angleapproximation.

Using the symetry, the momentcapacity can be integrated for first

quadrant and multiplied by 4.

Design of Tubular MembersPLASTIC MOMENT CAPACITY– CIRCULAR HOLLOW SECTION


70/102

15th


34

p y P F dt π =

2

0

4 cos2

P y

DM AF

π

φ =

2

0

4 cos2 2

P y

D DM F t d

π

φ φ

=

22

0

cos P yM F D t d

π

φ φ =

2= P yM F D t

Design of Tubular MembersLoad category, Factors and combinations

Load category and the corresponding load factors are listed below


71/102


55

• D1 – Dead Load 1, e.g. Self weight• D2 – Dead Load 2, e.g. equipment weight• L1 – Live Load 1, e.g. weight of fluids• L2 – Live Load 2, e.g. operating forces

• We – Extreme wind, wave and current loads• Wo – Operating wind, wave and current loads• Dn – Inertial Load correspond to Wo

• Dead Load: 0.9 to 1.3• Variable Load: 1.3 – 1.5• Environmental load: 1.3 – 1.4

• Factored gravity loads•1.3D1 + 1.3D2 + 1.5L1 + 1.5L2

• Wind, wave and current loads• 1.1D1 + 1.1D2 + 1.1L1 + 1.35(We + 1.25Dn)• 0.9D1 + 0.9D2 + 0.8L1 + 1.35(We + 1.25Dn)

•1.3D1 + 1.3D2 + 1.5L1 + 1.5L2 + 1.2(Wo + 1.25Dn)• Earthquake

•1.1D1 + 1.1D2 + 1.1L1 + 0.9E•0.9D1 + 0.9D2 + 0.8L1 + 0.9E

Load combinations and the associated load factors required as per API RP 2A LRFD

Load category and the corresponding load factors are listed below

Design of Tubular MembersComparison of ASD and LRFD a beam column design with uniformly

distributed lateral load and axial load

Design lateral Load w kN/m


72/102


56

Design lateral Load = w kN/m

Axial Load = P kN Span = L m Self Wight = ρ kN/m Yield Strength = F y MPa

a

P L f

A

ρ +=

2

2b

wL f =

1 1 0.6a y F F φ φ = ≤

2 2 0.66b y F F φ φ = ≤

1.0a b

a b

f f

F F + ≤

Appliedstresses

Allowable Axial

stress AllowableBendingstress

Interaction

φ1 and φ2 are to be computed includingthe buckling and slenderness effects

1 2a

P L f

A

γ γ ρ +=

2

3

2b

wL f

γ =

0.85c c y c F F φ φ = =

0.95b b y b F F φ φ = =

1.0c b

c y b y

f f

F F φ φ + ≤

Applied

stresses

Allowable Axialstress

AllowableBendingstress

Interaction

φ1 and φ2 are to be computed including thebuckling and slenderness effects. γ 1, γ 2 and γ 3are load factors 1.5, 1.3 and 1.5 respectivelyfor live, dead and wind loads

Design of Tubular MembersASD DESIGN PROCEDURE FOR TUBULAR MEMBERSDivide the member in to sections and calculate the axial, bending and shear forces ineach section along the length At-least 3 sections shall be checked


73/102


57

each section along the length. At-least 3 sections shall be checked.

The variation in section propertysuch as diameter or wall thicknessshall also be taken in toconsideration for calculating thesection property along the memberlength in each section.

The axial buckling capacity shall be

calculated using the variable crosssection along the length.

Variation of internal forces shall

also be computed for varioussections along the length.

Free Body Diagram with member internal forces

Design of Tubular MembersASD DESIGN PROCEDURE FOR TUBULAR MEMBERS Divide the member in to sections and calculate the axial, bending and shear forces in

each section along the length At-least 3 sections shall be checked


74/102


58

each section along the length. At least 3 sections shall be checked.

Establish geometric properties such as sectional area, moment of inertia, effectivelength factors, radius of gyration for each section.

Calculate the applied axial(f a), bending(f bx , f by), hoop (f h) and shear stresses (f s)using the geometry of the section and the applied axial, bending, hydrostatic and

shear forces. Establish the slenderness ratio(kL/r) and calculate the allowable axial stress (Fa)

and calculate the elastic buckling stress (F xe) and inelastic buckling stress (F xc) Establish the D/t ratio and calculate the allowable bending stress (Fb) Compute the allowable stresses for hoop using Elastic Hoop buckling stress (F

he) and

critical hoop buckling stresses (Fhc). The combined effect of loads is obtained using interaction of these loads in an

appropriate manner using axial, bending, hoop and shear interaction formulae forthe following cases.

Axial Bending Shear Hoop

Axial and bending Axial and hoop Shear and bending


Following method shall be used in calculation of applied stresses in members.

Applied Stresses in Tubular members


75/102



59

a P f A

=

and y x

bx by

xx yy

M Y M Y f f

I I

= =

0.5 s

V f

A=

2

h

h

P D

f t =

Axial Stress

Bending Stresses

Shear Stress

Hoop Stress

Properties of Tubular section( )( )22 24

D D t A

π − −=

( )( )44 264

xx yy

D D t I I

π − −= =

Where P, V, M x , M y and Ph (= h) are the axial load, shear, in-plane and out-ofplane moments and hydrostatic pressure respectively. Y is the half diameter.


Following method shall be used in calculation of allowable stresses in members.

Allowable Stresses for Tubular members


76/102



60

Axial Stress – Allowable axial stress in compression shall include the effect of slenderness ratio (kL/r) to determine whether yielding or global buckling govern thedesign. This is applicable for compression where as in tension it is taken as 0.6F yThe effect of local buckling of tubular sections due to axial loads is taken in toconsideration by computing the limiting values of Fy using critical hoop bucklingstress (Fxc).

Bending Stresses – Allowable bending stress depends on the D/t ratio and the

maximum value is to be limited to 0.75F y.

Shear Stress – Allowable shear stress is to be taken as 0.4F y

Hoop Stress – The allowable hoop stress is computed based on local buckling

effects due to external hydrostatic pressure. This is done by computing criticalelastic buckling stress (Fhe) and inelastic buckling stress (Fhc).

Design of Tubular MembersAllowable Axial Stress(Compression)

Allowable AxialStress (Tension)


77/102



61

The allowable axial compressive stress, Fashould be determined from the followingformulae for members with a D/t ratioequal to or less than 60. Effect of localbuckling shall be considered by

substituting Fy with local buckling stress.2

2

3

3

2

2

122

( / )1

2 for /

3( / ) ( / )

5 / 3 8 8

12 for /

23( / )

2

y

c

a c

c c

a c

c

y

KL r F

C F KL r C

KL r KL r

C C

E F KL r C

KL r

where

E C

F

π

π

−

= <

+ −

= ≥

=

Fy = Yield stress (or min (Fxe, Fxc))

E = Young’s Modulus of elasticity

K = effective length factor

L = unbraced length

r = radius of gyration

The allowable tensile stress, Fafor cylindrical memberssubjected to axial tensile loads

should be determined from0.6a y F =

To account for local bucklingand imperfections, Fy shall bereplaced by minimum of Fxeand Fxc.

Design of Tubular MembersLocal Buckling Stress Due to Axial Load

The local buckling stress for use with axial stress limits


78/102

16 July 2007 Dr. S. NallayarasuDepartment of Ocean Engineering


62

Elastic Local Buckling StressThe elastic local buckling stress, Fxe for columns subjected to axial loadswhen D/t ratio greater than 60 and less than 300 should be determined

from:

Fxe = 2CE t/D

Where

C = Critical elastic buckling coefficient to be taken as 0.3 (instead of 0.6) to

account for imperfections as per API Spec 2B.D = outside diameter

t = wall thickness

Inelastic Local Buckling Stress

The inelastic local buckling stress, Fxc, should be determined from:Fxc = Fy x [1.64 – 0.23 (D/t)

¼] FxeFxc = Fy for (D/t) 60

g

shall be calculated in stages using elastic buckling stress

Design of Tubular MembersEffective length factor K as specified in API RP 2A

Deck Truss webmembers

Deck Trusschord members


79/102



72

members

SuperstructureLegs

chord members

Jacket Legs

Jacket Braces

Design of Tubular MembersAxial Tension and Hydrostatic Pressure

When member longitudinal tensile stress and hoop compressive stresses


80/102



81

(collapse) occur simultaneously, the following interaction equation should besatisfied.

2 2 2 1.0 A B A Bν + + ≤(0.5 )

( )a b h x y

f f f A SF

F

+ −= )(SF F

f B h

hc

h=

stressncompressiohoopof valueabsolute

stress bendingactingof valueabsolute

stressaxialactingof valueabsolute0.3,ratiosPoisson'

=

=

===

h

b

a

f

f

f v

Load case Axial

Tension(SFx)

Bending Axial

Comp.

Hoop Comp.

(SFh)

Operating 1.67 Fy /Fb 1.67 to 2.00 2.00

Storm 1.25 Fy /1.33Fb 1.25 to 1.50 1.50

ncompressiohoopfor factor safety

tensionaxialfor factor safety

stresshoopcritical

StrengthYield

=

=

=

=

h

x

hc

y

SF

SF

F

F

Factor of Safety against Hydrostatic collapse with other loads

Design of Tubular MembersAxial Compression and Hydrostatic Pressure

When longitudinal compressive stresses and hoop compressive stresses occur


81/102



82

simultaneously, the following equations should be satisfied.

0.1

0.1)()()5.0(

≤

≤++

hc

hh

h

y

b x

xc

ha

F f SF

SF F

f SF

F

f f

,

,

where

0.15.0

5.02

h

heha

x

xeaa

ha

h

haaa

ha x

SF

F F

SF

F F

F

f

F F

f f

=

=

≤

+

−

−

SF x = safety of factor for axial compression

SF b = safety of factor for bending

f x = f a+f b+(0.5 f h )

f x should reflect the maximum compressive stress combination

hafor f 0.5 x f >

Refer to Member Local Buckling stresses

F xe = Member elastic local buckling stress due

to axial compression

F xc = Member inelastic local buckling stress

due to axial compression


Circumferential stiffening ring size may be selected on the following

Ring Design


82/102



83

approximate basis.

hec F E

tLD I

8

2

=

Where

Ic = required moment of inertia

for ring composite sectionL = ring spacing

D = diameter of pipe

t = thickness of pipe

Fhe = Elastic buckling stress


The ring spacing is defines as the distancebetween supports or between the actual

Ring Spacing


83/102



84

ring location. Hence the following procedureshall be adopted in designing a ringstiffened cylinders against combined axialand hoop stress.

a) Compute the axial and bending stressesusing unstiffened cylinders

b) Assume the spacing of rings as initialmember length “L” between the supportsor nodal connection as shown in figure

c) Determine the critical elastic hoop stress(Fhe) and compute the inelastic hoopstress (Fhc).

d) Determine the interaction ratio usingappropriate factor of safety.

e) Repeat the above steps (b) to (d) using areduced spacing “S” and stop if the UC isless than 1.0


1.1b Dt =

Moment of inertia of Ring stiffeners

Effective


84/102



85

eff

( ) ( ) ( )

( )

0.5 0.5 0.5eff f w f f f na

eff w f

b t h t t t h t h bt t y

b t t h bt

+ + + + +=

+ +

( )

( )

( )

32

32

32

0.512

0.512

0.512

eff

xx eff f na

f na

f

f na f

b t

I b t h t y t

thth h t y

bt bt y t

= + + − +

+ + + −

+ + −

shell width

Neutral axis

Moment of inertia


Verify a jacket brace of diameter 762mm x 15.88mm against axial loads of 1200 kN, andin-plane and out-of-plane bending moment of 800 and 600 kNm respectively. The unbraced length of the member is 15m and yield strength is 345 Mpa.


85/102



86

DESIGN OF A TUBULAR MEMBER AS PER API RP 2A (WSD)INPUT DATA

Diameter of brace D 762 mm⋅:=

Wall thickness t 15.88 mm⋅:=

Yield Strength Fy 345 MPa⋅:=

Weight density ρ s 78.5kN

m3

⋅:=

Modulus of elasticity E 2.0 105

⋅MPa

⋅:=

Unbraced length Ls 15 m⋅:=

Effective length factors K y 0.9:= K z 0.9:=

Axial Load P 1200 kN⋅:=

Bending Moment about y axis My 800 kN⋅ m⋅:=

Bending Moment about z axis Mz 600 kN⋅ m⋅:=

Design of Tubular MembersGEOMETRIC PROPERTIES

Sectiona area Asπ

4D

2D 2 t⋅−( )

2−⋅:= As 3.7 10

4× mm

2⋅=


86/102



87

Moment of inertia about y axis Iyπ

64D

4D 2 t⋅−( )

4−⋅:= Iy 2.6 10

9× mm

4⋅=

Section Modulus for y axis bending Zy2 Iy⋅

D:= Zy 6.8 10

6× mm

3⋅=

Radius of gyration for y axis bending R yIy

As:= R y 263.9 mm⋅=

Due to symetry, z axis properties Iz Iy:= Zz Zy:= R z R y:=

Slenderness ratio for y axis bending KLRy K y Ls⋅R y

:= KLRy 51.165=

Slenderness ratio for z axis bendingKLRz

K z Ls⋅

R z:= KLRz 51.165=

Euler buckling stress Fe 12 π

2

⋅ E⋅23 KLRz

2⋅

:= Fe 393.4 MPa⋅=

Moment reduction factor Cm 1:=


ALLOWABLE BENDING STRESS AS PER API RP-2A SECTION 3.2.3

D


87/102

9/16/2015 Dr. S. Nallayarasu

Department of Ocean EngineeringIndian Institute of Technology Madras-36

88

Diametr to wall thickness ratio Ratio Dt

:= Ratio 47.985=

Allowable bending stress F b 0.75 Fy⋅ Ratio10340

Fy

≤if

0.841.74 Fy⋅ D⋅

E t⋅−

Fy⋅

10340

FyRatio<

20680

Fy≤if

0.72

0.58 Fy⋅ D⋅

E t⋅−

Fy⋅20680

Fy Ratio≤ 300≤if

:=

F b 240.1 MPa⋅=

Design of Tubular MembersALLOWABLE AXIAL STRESS AS PER API RP-2A SECTION 3.2.2

Critical elastic buckling coeficient Ceb 0.3:=

El ti l l b kli t F 2 C b E

t

F 2501 MP


88/102



89

Elastic local buckling stress Fxe 2 Ceb⋅ E⋅ D⋅:= Fxe 2501 MPa=

Inelastic local bukling stress Fxc FyD

t60≤if

min Fxe 1.64 0.23 Dt

1

4

⋅−

Fy⋅,

Dt

60>if

:=

Fxc 345 MPa=

Limiting Slenderness ratio Cc2 π

2⋅ E⋅

min Fy Fxc,( ):= Cc 107=

Allowable axial stress incompression

Fa

1KLRz

2

2 Cc2

⋅

−

min Fy Fxc,( )⋅

5

3

3 KLRz⋅

8 Cc⋅+

KLRz3

8 Cc

3

⋅

−

KLRz Cc


89/102



90

Computed Bending Stress f byMy

Zy:=

f by 117.6 MPa⋅=

f bzMz

Zy

:= f bz 88.2 MPa⋅=Computed Bending Stress

Unity Check ratio UCf a

Fa

f by2

f bz2

+

F b+

f a

Fa0.15≤if

UC1f a

Fa

Cm f by2

f bz2

+⋅

1f a

Fe−

F b⋅

+←

UC2f a

0.6 Fy⋅

f by2 f bz2+

F b+←

UC max UC1 UC2,( )←

f a

Fa0.15>if

:=

UC 0.86=


DESIGN OF A INTERNAL RING STIFFENER FOR BOUYANCY TANKS

Verify a buoyancy tank of diameter 2000mm x 15mm for a hydrostatic pressure of 100mdepth. The spacing of rings is 2m and yield strength is 250 Mpa.


90/102



91

DESIGN OF A INTERNAL RING STIFFENER FOR BOUYANCY TANKSInput

Water Depth Wd 100 m⋅:=

Outer Diameter D 2000 mm⋅:=

Thickness of shell t 15 mm⋅:=

Yield Strength of material Fy 250 MPa⋅:=

ρ s 78.5 kN

m3

⋅:= ρ w 10.25 kN

m3

⋅:=Density of steel and water

Young's Modulus E 2.0 105

⋅ MPa⋅:=

Assume Dia/Thickness ratio Dt

133.333=

Spacing of ring stiffeners Sp 2 m⋅:=

Design of Tubular MembersBuckling Coefficient

Maximum hydrostatic pressure ph ρ w Wd⋅:= ph 1.025 MPa⋅=


91/102



92

f h ph D⋅

2 t⋅:=

Maximum hoop stress f h 68.3 MPa⋅=

Geometric parameter MSp

D

2 D⋅

t

0.5

⋅:=

M 16.33=

Buckling Coefficient Ch 0.44t

D⋅ M 1.6

D

t⋅≥if

0.44t

D⋅ 0.21

D

t

3

M4

⋅+

0.825D

t⋅ M≤ 1.6

D

t


92/102



93

Fhe 2 Ch⋅ E⋅ D⋅:= Fhe 140.7 MPa⋅=Elastic Hoop BucklingStress

Critical HoopBuckling Stress

Fhc Fhe Fhe 0.55 Fy⋅≤if

0.45 Fy⋅ 0.18 Fhe⋅+ 0.55 Fy⋅ Fhe≤ 1.6 Fy⋅


93/102



94

Since the thickness of shell is given as 16mm, the thickness of the stiffener shall not exceed 16mmdue to welding limitations.

Assume a stiffener thicknessand dimension as

ts 15 mm⋅:= ds 150 mm⋅:=

ds

ts10= Less than 10, hence OK

Width of shell as part of ring Beff 1.1 t D⋅( )0.5

⋅:= Beff 190.5 mm⋅=

Nutral axis distance from bottomy

0.5 ts⋅ ds2

⋅ Beff t⋅ ds 0.5 t⋅+( )⋅+

ts ds⋅ Beff t⋅+:= y 121.2 mm⋅=

Moment of inertia of webIwp

ts ds3

⋅

12ts ds⋅ 0.5ds y−( )

2⋅+:=

Moment of inertia of flange IfpBeff t

3⋅

12Beff t⋅ ds 0.5 t⋅+ y−( )

2⋅+:=

Moment of inertia provided I p Iwp Ifp+:= I p 1.284 107

× mm4

⋅=

Irq < I p. Hence the provided stiffeners are adequate.


1 Check the axial load on the jacket leg of diameter 1524mm and wall thickness

Questions


94/102

16 July 2007 Dr. S. Nallayarasu


95

1. Check the axial load on the jacket leg of diameter 1524mm and wall thicknessof 50mm with yield strength of 345 MPa. The bending moment acting on the legis 200 Tonne.m. The unsupported length is 15m. The effective length factor Kand moment reduction factors Cm shall be taken as 1.0.

2. Calculate safe axial load that can be carried by the jacket leg of diameter1524mm and wall thickness of 50mm with yield strength of 345 MPa. Thebending moment acting on the leg is 200 Tonne.m. The unsupported length is15m. The effective length factor K and moment reduction factors Cm shall be

taken as 1.0.

3. Design a buoyancy tank of 2.2m diameter subjected to hydrostatic pressureat design water depth of 120m. The maximum thickness of the tank shall notexceed 16mm and the spacing of rings shall not be less than 1m. The materialof construction is ASTM A36. The initial unsupported length shall be taken as20m.


95/102


96/102


97/102


98/102


99/102


100/102


101/102


102/102

Documents

Video Training Courses in Offshore Structures Design