Module 3 - Tubular Member Design

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Design of Tubular Members

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

Indian Institute of Technology Madras-36

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






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) areused 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 !.






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.






4

Member internal loads






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

f b l b






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.

D i f T b l M b






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.

D i f T b l M b






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 someof the imperfections that need to be included.

D i f T b l M b






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.

De ign of T b l Membe




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






11

Steel Making Process – an outlook

Source : Nippon Steel Corporation, Japan






12






13







14

Pilger and Piercing

The large size bars areused to produce pipes.

This has been in use for

several decades in thepipe producing mills.

Both thin and thick pipescan be made using this

method.

Limiting size for suchproduction depends on

the mill but generallydiameter larger than 20”is normally not availableby this method.







15

Cold Forming Processes and Resistance welding

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”.







16

Hot forming and induction welding

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”.







17

Cold Forming ProcessesIn 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 long

pipe. This method is verycommonly used for makingpipes of any diameter usedin the steel fabricationindustry. Using this method,

pipes of any diameter can bemade 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.







18

cold rolling a flat plate and weld at the seam toform a can (length up to 3m). The longitudinal

seam may be one or more depending on thewidth 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

Fabrication tubulars

Tubular can be fabricated from flat plates. Normally, flat

plates are rolled to form circular arcs and welded toform circular section as shown in figure.







19

Residual Stresses

Residual stresses developed during welding of plates to form pipes andwelding 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.







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

Consideration of Residual Stresses in design equations







21

Effective method of including Imperfections in designThe method to include the imperfections in fabrication is a difficult process as

the 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.




g



22

Out-of straightness toleranceo

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

Out-of Straightness

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 any

12200mm length (L/1284) whicheveris lower

This tolerance is very important as thisdeviation will lead to eccentric load and

corresponding moment.




g



23

Out-of Roundedness

Out-of roundedness tolerance for fabrication oftubular 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 toleranceshall not exceed 2% and DNV specifies that thetolerance shall not exceed 1%.




g



24

Eccentricity due to variation in Wall thickness

Maximum thickness variation = ∆t = tmax - tmin

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




g



25

Misalignment in Butt Joint

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 from

both 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.




g



26

Ultimate strength of a section ormember 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.

Ultimate Strength

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.






27

Buckling Theory

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

bifurcation 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 the

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






28

Effective Length Factors (K)

Effective length factor isdefines 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 column

with 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.






29

Local and Global buckling

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

global buckling is governed by the KL/r ratio. Localbuckling may also happen due to bending of largediameter tubular.






30

Factors influencing Ultimate strength

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 comparedto its load capacity at elastic range.




15th April 2009 Dr. S. NallayarasuDepartment of Ocean Engineering


31

ELASTIC AND PLASTIC MOMENT CAPACITY – RECTANGULAR SECTION

2 2

y F h P b=

22

2 2 3 6

y

y

F h h bh M b F

= =

2

22 4 4

p y y

h h bh M F b F

= =

2

p y

h F b=






32

ELASTIC AND PLASTIC MOMENT CAPACITY– CIRCULAR SECTION

21

2 4 p y

D P F π =

2 34

8 3 6 p y y

D D D M Pa F F

π

π

= = =

4

3

D

π

3

32 y

D

M 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 =






33

PLASTIC MOMENT CAPACITY HOLLOW CIRCULAR SECTION

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.






34

PLASTIC MOMENT CAPACITY– CIRCULAR HOLLOW SECTION

p y P F dt π =

2

0

4 cos2

P y

D M AF

π

φ =

2

0

4 cos2 2

P y

D D M F t d

π

φ φ

=

22

0

cos P y F D t d

π

φ φ =

2= P y F D t






35

ELASTIC MOMENT CAPACITY OF HOLLOW CIRCULAR SECTION

( )4 4

( 2 )64 D D t π

= − −

ELASTIC MOMENT

CAPACITY

ELASTIC MOMENTOF INERTIA

Y F Z =

( ) ( )4 2 2 2 4 2 2( 4 4 ) ( 4 )

64 64

D D t dt D D dt π π

= − + − = − +

( )3

4 4 2 2 3( 16 8 )64 8

t I D D D t D t

π π = − + − =

Expand and ignore t2

terms since t << D

( )32

8

/ 2 4

D t D t

Z D

π π = =






36

SHAPE FACTOR

3

3

8 1.70

32

y

y

D

F S

D F

π

π = =

2

2

4 1.5

6

y

y

bh F S

bh F

= =

Solid Circular Section

Rectangular section

Hollow circular section

2

21.27

4

y

y

F D t S

D t F

π = =

Shape factor of a section is defined as the ratio of plastic section modulus tothe elastic section modulus as defined below.






37

SHAPE FACTORS FOR VARIOUS SECTIONS




18 Oct 2011 Dr. S. Nallayarasu Department of OceanEngineering Indian Institute of

Technology Madras

38

Simply supported beam with central concentrated load

3

48

PL

EI δ =

4

C

PL M =

Elastic

2

Lδ θ =

4 p

C

P L

=

Plastic





Technology Madras

39

2

Lδ θ =

e C W P δ =2i pW M θ =

2

2

C p P L M θ

θ =

Equating and external and internal work done

Central displacement

External work done

Internal work done

Collapse Load

4 p

C P L

=

PLASTIC COLLAPSE LOAD


Fi d t d b ith t l t t d l d




Technology Madras

40

Fixed supported beam with central concentrated load

3

192

PL

I δ =

8C

PL M =

Elastic

2

Lδ θ =

8 p

C P

L

=

Plastic





Technology Madras

41

2

Lδ θ =

e C W P δ =2i p p pW M M M θ θ θ = + +

4

2C p P L M

θ θ =

Equating and external and internal work done

Central displacement

External work done

Internal work done

Collapse Load8 p

C P L

=

PLASTIC COLLAPSE LOAD

Design of Tubular MembersDesign of structural members





42

Design of structural members

Design of structural members or system means to select a suitable material

with sufficient strength satisfy the functional requirements with adequatesafety margin.

Strength (Capacity) of structure is determinedfrom structural mechanics principles of bending,

axial and shear due to load effects (Demand).

Design is deemed to be satisfied when

Capacity Demand≥

Capacity of the structure shall be evaluated for each individual element aswell as the overall system such that the functional requirements aresatisfied throughout the life of the structure.

Similarly, the load effects or demand shall be evaluated such that themaximum effects occurring during its functional life.

Design of Tubular MembersDesign Methods





43

Allowable Stress Design (ASD)

ASD method is based on principle that the load are estimated based

on historical data and associated effects are obtained by mechanicswhile the strength of the structure or system is evaluated usingselected material and their characteristics. A suitable factor of safetyis assigned against the overall strength.

Limit State Design (LSD) or Load and Resistance Factor Design (LRFD)

LRFD method is developed based on probabilistic approach using thecharacteristic variation of loads and resistance. The uncertainty andvariation of the loads is assessed using probability theory to estimate

the load factors while the resistance resistance factors. Theadvantage of this method lies in the approach of variable load andresistance factors depending on their variability and uncertainty.

Design methods can be classified in to following two categories.

Design Methods

Design of Tubular MembersAllowable Stress Method





44

Demand (Applied Stresses)

Applied stresses are calculated for category of load on a system or partof the structure using basic structural mechanics principle of axial,bending and shear or combination of all. Elastic theory is normally usedas principle with linear stress distribution across the section.

Applied stress = f (geometry, section and stiffness)

Capacity (Allowable Stresses)Allowable stresses are obtained as a fraction of yield strength of thematerial (yield strength divided by a suitable factor of safety) withdue consideration for the behavior such as buckling and slenderness

effects. Allowable stress = Yield Stress/Factor of Safety

Following procedure is adopted in the ASD method.

Allowable Stress Method

Design is deemed to be safe when Allowable stress > Applied Stress !

Design of Tubular MembersEstimation of load and its effects





45

Following procedure is adopted to estimate the loads and to compute the effectsloads on the structure.

Loads are estimated based on historical information and maximum value is takenwhich may occur during the design life. Design loads are adjusted from

estimated loads based on past experience. Variation of loads or the probability of exceedence of the loads during the design

life is not taken into consideration explicitly. Response of the structure is evaluated using linear elastic theory assuming worst

possible combination of loads including P-∆ effects and dynamic consideration ifrequired. For example;

1.00 * Dead Load + 1.00 * Live Load1.00 * Dead Load + 1.00 * Live Load + 1.00 * Environmental Load

1.00 * Dead Load + 1.00 * Seismic Load

Applied stresses are calculated for axial, bending, shear, torsion, hydrostatic andbuoyancy forces.

Estimation of load and its effects

Design of Tubular MembersEstimation of Strength





46

Following procedure is adopted to estimate the resistance of the structure.

Estimation of Strength

Design yield strength of the material istaken as target constant value assured byquality control during construction using

random sampling method. No consideration is given to variability of

the yield strength or probability of strengthoccurring below the target strength.

Allowable stresses are taken as fraction of

yield stress as shown in figure.

Design allowable stresses for each component is considered individually such asaxial, bending, shear, torsion, and hoop with a suitable factor of safety for each.

Due consideration is given to local and global buckling, torsional buckling ofmembers in the calculation of allowable stresses.

No global factor of safety is applied to the structure and no probability of failureis obtained.

Design of Tubular MembersASD DESIGN PROCEDURE





47

Design using ASD method is adopted as follows.

ASD DESIGN PROCEDURE

Design loads are taken as the maximum load that may occur during the life of thestructure using historical data. Effect of these loads on structures are calculatedusing the basic elastic principles of mechanics.

Structure resistance is calculated using the design strength of materials (target)

as capacity of structural elements. Allowable stresses for each category ofresistance such as axial tension, axial compression, bending inplane, bending out-of plane and shear are obtained as the fraction of yield stresses. These fractionsare proportional to the factor of safety (FOS) for each case.

Design is deemed to be satisfied when design resistance is greater than the loadeffects.

The combined effect of loads is obtained using interaction of these loads in anappropriate manner using axial, bending, hoop and shear interaction formulae.

(axial,bending,shear and hoop) y

F f

FOS >

Design of Tubular MembersLoad and Resistance Factor Design





48

Design Loads (Demand)

Loads are estimated based on probability of exceedence usingcharacteristic distribution of each type of load. The design loads areestimated by multiplying nominal loads by the load factors

corresponding to the probability of exceedence. The method of calculation of load effects are very similar to the ASD

Design Load= f (nominal load, load factor)=nominal load*load factor

Resistance (Capacity)The resistance of the structure is evaluated using the characteristicvalues of strength parameters such as yield strength, fabricationtolerances using probability based approach. The design resistance

factors are obtained using probability distribution.Design Resistance = f (Strength, Resistance factor)=strength/resistance factor

Following procedure is adopted in the LRFD method.

g

Design is deemed safe when Resistance > Design Load effects !

Design of Tubular MembersProbability distribution of Load and Resistance





49

o y o o o

Overlap

The probability distribution of Load effects (Q) and Resistance (R) is shown in

figure. The overlapped portion is indicating exceedence of load effects withresistance. This is note as Probability of failure. µQ and µR mean values of loadeffects and resistance. Even though µQ is less than µR, the failure is due to spread ofload and resistance effects. More the spread (deviation), the probability of failure is

higher. This is indicated by the standard deviation values of σQ and σ

R .

This can be compared to thedeterministic values of load effects (Q)and resistance (R ) in ASD method of

design.

The design is considered to be safewhen R is greater than Q. The Factor

of safe is defined as the gap betweenthe R and Q.

Design Margin = R-Q

Design of Tubular MembersRELIABILITY INDEX (β)





50

( )

β depends on the load effect “Q” and the resistance “R” and their probability

distributions. It represents how confident we are in our decision that the resistance of the material is higher than the load effects. For a normal distribution, the designmargin can be written as

2 2 2

( ) Probablity of failure

Z R Q

Z R R

Z

Z

Z R Q µ µ µ

σ σ σ

µ β

σ

φ β

= −= −

= −

=

=

Design of Tubular MembersLIMIT STATES





51

Limit state is defined as a state of the structure that cease to perform its intended

function for which it is designed. In offshore structures, limit states can be classifiedinto following categories. Ultimate Limit State (ULS) – This limit state defines the strength requirement

for the successful performance of structure to satisfy the functionalrequirements.

Fatigue Limit State (FLS) – This limit state defines the fatigue requirementfor the performance of the structure for the design life without deterioration.

Serviceability Limit State (SLS) – This limit state defines the serviceabilityrequirements such as deflection, vibration etc. for the successful performance of

the structure. Accidental Limit State (ALS) – This limit state defines the accidental cases of

loads arising from fire, blast and impact from vessels.

Each limit state is provided with loads and resistance factors to account for thevariability and corresponding partial safety factors and load factors are used todetermine the safety of the structure.

Design of Tubular MembersEstimation of Design load and its effects





52

Following procedure is adopted to estimate the design loads and to compute the

effects loads on the structure. Nominal Loads are estimated based on historical information and data

collected. These data is then used to derive the characteristic probability densityand exceedance values are then estimated using probability theory. Load factoris computed for the given probability of exceedance.

Design loads are then calculated by multiplying the nominal load with the loadfactor.

Each load category is assigned with individual load factor thus allowingdifferent variation for each load type. For example, dead loads are given lower

load factor when compared to environmental and live loads. Response of the structure is evaluated using the same principle as that of the

ASD method except that the load combinations are based on load factors. Forexample;

1.3 * Dead Load + 1.5 * Live Load1.3 * Dead Load + 1.5 * Live Load + 1.2 * Environmental Load1.1 * Dead Load + 0.9 * Seismic Load

Design Load Nominal Load *γ =

Design of Tubular MembersEstimation of Resistance





53

Following procedure is adopted to estimate the resistance of the structure.

Design yield strength of the material is obtained by probability theory ofexceedance below the target value using characteristic distribution. Resistancefactors for each load category is obtained

These resistance factors are multiplied by the nominal resistance to obtain the

design resistance. Following resistance factor are recommended by API RP 2A LRFD

Loading Type Resistance Factor (φ)

Axial Tension 0.95

Axial Compression 0.85

Bending 0.95

Shear 0.95

Hoop Buckling 0.80

Connections 0.9 - 0.95

Design Resistance Nominal Resistance *φ =

Design of Tubular MembersLRFD DESIGN PROCEDURE





54

Design using LRFD method is adopted as follows.

Load factors (γ) and resistance factors ( ) are selected for each type of load andstrength parameters.

Nominal loads and resistance for each set of load combination is obtained fromdata and design requirements.

Design loads and resistance is obtained by multiplying the nominal loads andresistance by load and resistance factors respectively.

Design is deemed to be satisfied when design resistance is greater than the loadeffects.

The load and resistance factors for each type of load is applied depending on thecombination of loads.

The above expression is evaluated for both structure component and system sothat a probability of failure is obtained with respect to the selected load andresistance factors.

n nQφ γ >

1 1 2 2 3 3 1 1 1 1 1 1( , , ,...) ( , , ,...) f R R R f Q Q Qφ φ φ γ γ γ >

Design of Tubular MembersLoad category, Factors and combinations





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





56

distributed lateral load and axial load

Design lateral Load = w kN/m Axial Load = P kN Span = L m Self Wight =

ρ

kN/m Yield Strength = F y MPa

a

L f

+=

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 in





57

Divide 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.

The variation in section propertysuch as diameter or wall thicknessshall also be taken in to

consideration 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





58

Divide 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.

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 (Fhe) andcritical 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

Design of Tubular MembersApplied Stresses in Tubular members





59

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

a P f =

and y x

bx by

xx yy

Y M Y f f

I I

= =

0.5 s

V f =

2hh

P D f t =

Axial Stress

Bending Stresses

Shear Stress

Hoop Stress

Properties of Tubular section( )( )22 2

4

D D t A

π − −=

( )( )44 2

64 xx yy

D D t I I

π − −= =

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

Design of Tubular MembersAllowable Stresses for Tubular members





60

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

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 y

The 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 bucklingeffects 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 AxialS ( i )





61

(Compression)

The allowable axial compressive stress, Fa

should be determined from the followingformulae for members with a D/t ratioequal to or less than 60. Effect of localbuckling shall be considered bysubstituting 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 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

Stress (Tension)

The allowable tensile stress, Fa

for cylindrical memberssubjected to axial tensile loads

should be determined from

0.6a y F =

To account for local bucklingand imperfections, Fy shall bereplaced by minimum of Fxe

and Fxc.

Design of Tubular MembersLocal Buckling Stress Due to Axial Load



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 determinedfrom:

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] ≤ Fxe

Fxc = Fy for (D/t) ≤ 60

The local buckling stress for use with axial stress limitsshall be calculated in stages using elastic buckling stress




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

63





64

Elastic coefficients for local buckling of steel cylindersunder axial compression

(Extracted from API RP 2A – Figure C3.2.2-1)

200 2000 20000300





65

Comparison of test data with design equation forfabricated steel cylinders under axial compression






66

Beyond KL/r = 120, no effectof yield strength on Fa

For KL/r = 0, Fa = 0.6 Fy

Design of Tubular MembersAllowable Bending Stress

The allowable bending stress Fb should be determined from:




67

The allowable bending stress, Fb should be determined from:

300680,20

for58.072.0

680,20340,10for74.184.0

340,10 for75.0

≤<

−=

≤<

−=

≤=

t

D

F

F

Et

D F F

F t D

F F

Et

D F F

F t

D F F

y

y

y

b

y y

y y

b

y

yb

Allowable Shear Stress

The allowable shear stress, Fs should be taken as:

0.4 s y F F =





68

Design equation for fabricated steel cylinders under bending






69

Design of Tubular MembersInteraction of Axial Compression and Bending

Cylindrical members subjected to combined compression and bending should be




70

Cylindrical members subjected to combined compression and bending should be

proportioned to satisfy following requirements at all points along their length.

0.1

1

22

≤

′−

++

b

e

a

bybxm

a

a

F F

f

f f C

F f 0.1

22

≤+

+b

bybx

a

a

F

f f

F f

0.16.0

22

≤

+

+b

bybx

y

a

F

f f

F

f

22

' '1 11.0

m bym bx

a a

ex eya

a b

C f C f

f f

F F f

F F

+

− − + ≤

f a/Fa ≤ 0.15f a/Fa > 0.15

General case

For asymetric sections

Design of Tubular MembersTerms used in axial and bending interaction formulae

F All bl i l t




71

( )

2'

2/e

E

F KL r

π

=

Euler buckling stress value Fe’ can be calculated using the following formula

This is similar for x and y axes also.

Fa

= Allowable axial stress

Fb = Allowable bending stress

F’ ex = Euler buckling stress in x axis

F

’

ey = Euler buckling stress in y axisCm = Moment reduction factor = 0.85 for uniform moment at ends

f a = Applied axial stress

Fb = Applied bending stress

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

Deck Truss web

Deck Truss




72

Deck Truss webmembers

SuperstructureLegs

chord members

Jacket Legs

Jacket Braces

Design of Tubular MembersMoment reduction factor Cm

When the members are subjected to unequal end moments (MA or MB) as shown in




73

A B

figure, the maximum moment may occur at the ends or anywhere in the span. Thelocation and magnitude of Mmax needs to be calculated for design purpose. Thecalculation can be eliminated by introducing the equivalent moment concept withoutloosing the magnitude.

eq m Bc M =

Values of the reduction factor Cm

referred to in the above table asfollows (with terms as defined byAISC)

a. 0.85

b. 0.6 – 0.4 , but not less

than 0.4, nor more than 0.85

c. 1-0.4 , or 0.85 whichever is less

1

2

a

e

f

F





74

Design of Tubular MembersElastic Hoop Buckling Stress:The elastic hoop buckling stress determination is based on a linear stress-strain relationship from




75

strain relationship fromF he = 2 C h E t/D

WhereThe critical hoop buckling coefficient Ch includes the effect of initialgeometric imperfections within API Specification 2B tolerance limits.

Ch = 0.44 t/D for M≥1.6 D/tCh = 0.44 t/D + 0.21 (D/t)2 / M4 for 0.825 D/t ≤M<1.6 D/tCh = 0.736/(M-0.636) for 3.5 ≤M<0.825 D/tCh = 0.755/(M-0.559) for 1.5 ≤M<3.5

Ch = 0.8 for M < 1.5

The geometric parameter, M, is defined as:

L = length of cylinder between stiffening rings, diaphragms, or endconnections.

21

)/2( t D D

M =






76

Comparison of test data with design equation for ring buckling andinelastic local buckling of steel cylinders under hydrostatic pressure







77

Comparison of test data with elastic design equations for local buckling ofsteel cylinders under hydrostatic pressure for M > 0.825 D/t







78

Comparison of test data with elastic design equations for local bucklingof steel cylinders under hydrostatic pressure (M < 0.825D/t)







79

Design of Tubular MembersCritical Hoop Buckling Stress

Elastic Buckling





80

F hc = 0.45 F y +0.18F he for 0.55 F y = F he ≤ 1.6 F y

for 1.6Fy < F he < 6.2 F y

F hc = F y for F he > 6.2 F y

)/(15.1

131

he y

y

hc F F

F F

+=

F hc = F he for F he ≤ 0.55 F y

Inelastic Buckling

Design of Tubular MembersAxial Tension and Hydrostatic Pressure

When member longitudinal tensile stress and hoop compressive stresses





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 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

ncompressiohoopforfactorsafety

tensionaxialforfactorsafety

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





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





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





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





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

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.





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⋅=





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 Zy

2 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





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:=

Elastic local buckling stress Fxe 2 Ceb E

t

D Fxe 2501 MPa



9/16/2015 Dr. S. Nallayarasu

Department of Ocean EngineeringIndian Institute of Technology Madras-36

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<if

12 π2

⋅ E⋅

23 KLRz2

⋅KLRz Cc≥if

:=

Fa 166.7 MPa⋅=

Design of Tubular MembersAPPLIED STRESSES AND COMBINED AXIAL AND BENDING INTERACTION RATIO

Computed Axial Stress f aP

As:= f a 32.2 MPa⋅=





90

Computed Bending Stress f by

My

Zy:=

f by 117.6 MPa⋅=

f bz

Mz

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

Fa

0.15>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.





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⋅=





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<if

0.736

M 0.636−

3.5 M≤ 0.825D

t<if

0.755M 0.559−

1.5 M≤ 3.5<if

0.8 M 1.5<if

:=

Ch 0.0469=


Hoop Stress Check

Fhe 2 Ch⋅ E⋅

t

D⋅:= Fhe 140 7 MPa⋅=Elastic Hoop Buckling





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⋅<if

1.31 Fy⋅

1.15Fy

Fhe+

1.6 Fy⋅ Fhe≤ 6.2 Fy⋅<if

Fy Fhe 6.2 Fy⋅>if

:=

Fhc 137.8 MPa⋅=

Factor of Safety againsthydrostatic collapse

SFh 2.0:=

Unity Check UC2

f h

Fhc SFh⋅:= UC2 0.992=

Design of Tubular MembersStiffener Design

Moment of inertia of rings required Irqt Sp⋅ D

2⋅

8 E⋅Fhe⋅:=

Irq 1.055 107

× mm4

=

Since the thickness of shell is given as 16mm, the thickness of the stiffener shall not exceed 16mmd t ldi li it ti





94

g ,due 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 10

7× mm

4⋅=

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



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.


4. Check the safety of a jacket horizontal brace supported between jacket legsspaced at 12m against wave slamming loads. The diameter of the brace is 508mm

and wall thickness is 16mm with yield strength of 345 MPa The platform is



16 July 2007 Dr. S. Nallayarasu


96

and wall thickness is 16mm with yield strength of 345 MPa. The platform issubjected to waves with a period of 6 sec and height of 5m at a water depth of60m. Assume the wave slamming force is uniform over the entire length of themember. Use linear wave theory to calculate the vertical water particle velocity.

5. Design a skirt pile to carry an axial load of 2000 Tonnes and lateral load of 100Tonnes. The skirt pile has been driven into ground for a penetration of 100m andthe ultimate soil capacity in axial direction varies linearly with diameter from10MN to 60MN for piles of diameter 1.2m to 2.2m respectively. The pile shall be

assumed to be fixed at 25m from seabed. The yeild strength of material of pileshall be 345 MPa. A minimum factor of safety against axial load of 2.0 in soilcapacity shall be maintained.

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Module 3 - Tubular Member Design

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