as4100 unsw design of steel members

7/18/2019 AS4100 UNSW Design of Steel members

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CVEN3302 Structural DesignSemester 2 2009

CVEN3302 Structural DesignSemester 2 2009

School of Civil and Environmental Engineering, UNSWSchool of Civil and Environmental Engineering, UNSW

IntroductionbyZora Vrcelj


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LectureOutline

Introduction Housekeepingrules

Structuralsteelro erties fabrication

Limitstatesdesi ntoAS 100Strength,serviceability

Loads

Dead live wind earth uake


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tructura


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Thisiswhatyouendupwith

ifyoudontbuildwithsteel!

causeprematurefailure


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Steelisbyfarthemostusefulmaterialforbuildingstructureswithstren th ofa roximatel tentimesthatofconcrete,steelistheidealmaterialformodernconstruction.

Steelisalsoaveryecofriendlymaterialandsteelstructurescan

eeas y sman e an so asscrap.

Duetoitslargestrengthtoweight

ratio,steelstructurestendtobemoreeconomicalthanconcretestructuresfortallbuildingsandlargespanbuildingsandbridges.


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Togetthemostbenefitoutofsteel,steel

s ruc ures

s ou

e

es gne

an

protectedtoresistcorrosionandfire.

Steelstructuresareductileandrobustandcanwithstandsevereloadingssuchasearthquakes.

The effectsoftemperatureshouldbeconsideredindesign.Specialsteelsandprotect vemeasures orcorros onan reareavailableandthedesignershouldbefamiliarwiththeoptionsavailable.

Topreventdevelopmentofcracksunderfatigueandearthquakeloadstheconnectionsandin articulartheweldsshouldbedesignedanddetailedproperly.Theyshouldbedesignedanddetailedforeasyfabricationanderection.Goodqua ycon ro sessen a oensure

properfittingofthevariousstructuralelements.


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Thusthelifecyclecostofsteelstructures,whichincludesthecostofconstruction maintenance,repairanddismantling,canbelessthanthatforconcretestructures.Sincesteelisproducedinthefactoryunderbetterqualitycontrol,steelstructureshavehigherreliabilityandsafety.

Steelstructurescanbeconstructedveryfastandthisenablesthestructuretobeusedearly

thereb leadin tooveralleconom .Steelstructurescanbeeasilyrepairedandretrofittedtocarryhigherloads.


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StructuralSteel MildSteel

conom c, uc e

Hotrolled

into

standard

shapes Box sections

Easilyfabricatedbywelding

Cell Form BeamsStandard sections


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TypicalSteelFabricationShop

otro nga r cat onwor s op


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ductility.

Ductility allowsverylargestrainstodevelopwithlittleincreaseinstress,priortofailure.

Theadvantagesofductilityare:

Itallowsenergyabsorptionindynamicloading orinresistingbrittlefracture

Itallows

for

redistribution

of

actions

which

is

usuall

beni n

N.BAt resent in achievin a ductile stress-strain curve it re uires the ield stressfy

to be less than 450 MPa. The yield stress is also called the Grade of the mildsteel, i.e. Grade 350 steel has fy= 350 MPa.


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Pro ertiesofmildsteel

Not to scaleress

fuUpper yield stress

Not to scaleress

fuUpper yield stress

fyStrain hardening EstfyStrain hardening Est

Plastic Tensile rapturePlastic Tensile rapture

Strainy

st

as c

Strainy

st

as c

Idealisedstressstrainrelationshipforstructuralsteel

. . esames resss ra ncurve sassume ncompress on, u wes a see a

bucklingof

members

and

elements

in

compression

usually

prevents

high

strains

from

beingrealised


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IdealisedStressStrain(

) Diagram

,

,


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Yieldingunderbiaxialstresses

MisesYieldCriterion

1.0

Uniaxial

tension

MohrCircleConstruction

2/fy

Pure

shear

-1.0 1.0

ratio

r nc pa stress rat o 1 y

ip

alstress

Uniaxial

Princompression

Maximumdistortionenergy

-1.0f1, f2 normal stresses shear stress

f12

-f1f2+f22

+ 32

= fy3


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IntroductiontoStructural

SteelDesign


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DesignProcess

Problem! Includes: setting criteria, constraintsfunctional and structural requirements

Definition of problem

(Design brief)

Information search

,

information from other

consultants, loads

Preliminary designs and selection

Detailed designStructural systems

Drawings and specificationsIncludes: type of system,

spacing of major members,

onceptua es gn

Solution(completed job)

construction techniques

Advice on construction


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Whentheneedforanewstructurearises,anindividualoragency

hastoarrangethefundsrequiredforitsconstruction.

Theindividual

or

agency

henceforth

referred

to

as

the

owner

then

approachesanarchitect.

Thearchitectplansthelayoutsoastosatisfythefunctionalrequirementsandalso

ensuresthatthestructureisaestheticallypleasingandeconomicallyfeasible.

Inthisprocess,thearchitectoftendecidesthematerialandtypeofconstructionaswell.

Theplanisthengiventoastructuralengineerwhoisexpectedtolocatethestructuralelementssoastocauseleastinterferencetothefunctionandaestheticsofthestructure.

Hethenmakesthestrengthcalculationstoensure

safetyandserviceabilityofthestructure.Thisprocessisknownasstructuraldesign.

Finally,thestructuralelementsarefabricatedanderectedbythe

. , ,

aestheticandeconomicalstructureisconceived.


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Howeverinpractice,manystructuresfulfiltherequirementson ypar a y ecauseo na equa ecoor na on e ween epeopleinvolvedandtheirlackofknowledgeofthecapabilities

andlimitationsoftheirownandthatofothers.

Sinceastructuralengineeriscentraltothisteam,itisnecessaryforhim.

Itishisresponsibilitytoadviseboththearchitectandthecontractoraboutthepossibilitiesofachievinggoodstructureswitheconomy.

Ever

since

steel

began

to

be

used

in

the

construction

of

s ruc ures, asma eposs esomeo egran es structuresbothinthepastandalsointhepresentday.


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SuccessfulStructures

Functionalrequirements setbyclient

u ng e nc u ngcons ruc onper o STRUCTURALENGINEERS

Aestheticsatisfaction setby

architects

Economy Capitalcostisnotjustthestructuralcomponentbutalsofinancingandconstructionspeed

maintenancecostscaneffectlongtermlifecyclecosting


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LimitsStatesDesi ntoAS 100Forrest Centre,Perth WA

rs e ourne u ngto use concrete filledtubular steel columns

en ra concre e corewith steel beams andmetal formwork

Steel skeleton connectedto the central core

o umns were concre efilled composite steel

box columnsCasseldon Place,Melbourne


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Desi na roachofAS 100

Basedonlimitstatedesign

Principallimitstates

collapse: ieldin buckling

overturning Serviceabilitylimitstate,concernedwith

function: e ection vibration


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Aim Satisfactor erformanceunderavariet ofdifferentusesorloadscenarios

Strength Rarescenarios:

nofailure

Serviceability

Commonscenarios:

wantsatisfactoryperformanceinserviceundercommonloadingsnocracking,nobouncing,


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Determineappropriate

actions

Analyseusingappropriatemethodsand

accountin forvariabilit todetermine:

Designeffects{S*

},and

Ensure

no

limit

state

is

exceeded

*


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EffectofFactoredLoadsFactoredResistance

*

For load combinations, the effect of factored loads (S*

) is thestructural effect due to the specified loads multiplied by loadfactors.


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Variabilit ofactions

Precision

of

modelling

actions

varies:deadloadsrelatedtomaterialdensityand

thickness

imposedloads

based

on

type

of

occupancy

data

Probabilit ofloadcombinationsvaries


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ACTIONS

Weights

of

the

various

structural

members

and

the

weights

of

any

. .ofthestructure+superimposeddeadload)

LIVELOADS Buildingloads

r ge oa s

Windloads

Snow

loads Earthquakeloads

HydrostaticandSoilPressure

OtherNaturalLoads(theeffectofblast,temperaturechanges,different

settlementofthefoundation)


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DESIGNLOADS GENERAL

Forthedesignofstructuralsteelworkthefollowingloadsandinfluencesshallbeconsidered:

G

Dead

loads,

including

the

weight

of

steelwork

and

all

permanent

materials

of

, , ,concreteandfinishesresultingfromdeflectionsofsupportingmembers,andtheforcesduetoprestressing;

Q Liveloads,

including

load

due

to

intended

use

and

occupancy

of

structures;

movableequipment,snow,rain,soil,orhydrostaticpressure;impact;andanyotherliveloadsti ulatedb there ulator authorit

T Influencesresultingfromtemperaturechanges,shrinkage,orcreepof

com onentmaterials

or

from

different

settlements

W Liveloadduetowind;

E Liveload

due

to

earthquake


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Loadsactinverticaldirection.

The

specified

dead

load

for

a

structural

member

consists

of:

theweightofthememberitself,

ewe g o a ma er a so cons ruc on ncorpora e n o

thebuildingtobesupportedpermanentlybythemember,

theweightofpartitions,

theweightofpermanentequipment,and

theverticalloadduetoearth lantsandtrees.


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DeadLoad G

Services ventilation electricit ducts etc.

Superimposed dead load


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Load

path?!


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Variabilit ofmaterialandsectionproperties

Resistance{R}isrelatedprincipallytomaterial

Yieldstren thofsteelis uaranteed

Other

properties,

notably

Youngs

modulus

(E),

aremuchlessvariable


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Stability:overturning(equilibrium)

appropriate

Itmayalsobenecessarytoconsider:

Secondordereffects

Rupture(duetofatigue)


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StrengthLimitState

*

. . . .

=

Lefthand

side

is

factored

strength

load

effect,

S*

=capacityreductionfactor

Capacity

Factor

Capacity

Factor

Givesconsistentreliabilitytowholestructure

= . ors ee mem ers, u, u, u =0.8 (connectorsandconnections)


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Stren thLimitState

i.e.liveloadcombination. .. .. .. .

0.8G + 1.25Q0.8G + 1.25Q0.8G + 1.25Q0.8G + 1.25Q

i.e.windloadcombination

.. uu.. uu

++++

.. uu.. uu

(2) and(4) are

used

if

the

loads

act

in

opposite

directions.


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Stren thLimitState

Loadsaregenerally:UDLs(orpressures=Force/Area)andPointLoads

AxialforceN*

BendingmomentM*

Shear

force

V

*

Whentheseloadeffectsaredeterminedusingfactoredloads(*)theyarecalleddesignloads

Whentheloadeffectsaredeterminedwithoutusingfactoredloadstheyarecallednominalloads

G nominaldealload(giveninloadingcaseAS1170.1)

Q nominallive

load

iven

in

loadin

code

AS11 0.1 **

Wu nominalultimatewindload(giveninloadingcodeAS1170.2)

Mu, Nu, Vu the strengths are determined from

the steel code AS4100RR


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Stren thLimitState

Anexampleofthedesignequationmaybeestablishingthat:

*

where

M*

is

the

factored

bending

moment

in

a

beam

(determined

from

structuralanalysis)

b

buckling)and

=0.9


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Stren thLimitState

*c

where

N* isthefactoredaxialcompressioninacolumn(theloadeffector

Nc isitsstrengththataccountsfortheeffectsofcolumnbucklingifthecolumnisslender(thenominalcompressivestrength)and

=0.9


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Serviceabilit LimitState

Thefollowingconditionsmayneedtobeconsidered:

excessivevibrations

Both

conditions

are

associated

with

stiffness

rather

thanstren th

Formostbuildings,controllingdeflectionswillalsolimit

vi rations

S i bili f b


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Serviceabilit ofbeams

appearance(sagging)

fitness

for

purpose

(machinery,

pipe

grades)s ruc ura avo un n en e oa pa s

S i bili Li i


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Serviceabilit Limits

Code

ives

uidanceonl

(i.e. /L=1/250,1/500,etc.)

Mainmessageis THINKanddiscusswithclient

i f i bili


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Desi nforServiceabilit

1. greeon e ect on m ts lim w t c ent

2.Evaluateserviceabilityloadcombinationsthathavelimit lim

splitcombinationintoconstituentloadswi

estimateduration

of

each

constituent

load

D i f S i bili


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Desi nforServiceabilit

3. (udl,ss)

4

384

5

w

E

LI

Note:Designloadfactorsusedfor StrengthLimitStatedonotapplytoServiceabilit LimitState i.e.weusewnotw*

4. Select cross-section to ive I

5. ec en ng, s ear, ax a s reng

Mu

Vu

Nu

R f M i l


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ReferenceMaterial

: tan ar s ustra a, y ney.

AS1170.1&AS1170.2LoadingCodes:

StandardsAssociation

of

Australia,

Sydney.

NSTrahair&MABradford:TheBehaviorandDesi nofSteel

Structures

to

AS4100,

3rd

Australiaedition,

E&FN

Spon,

London,

1998.

, :SteelStructures,3rd edition,AISC,Sydney,1997.

STWoolock,SKitipornchai&MABradford:DesignofPortalFrameBuildings,3rd edition,AISC,Sydney,1999.


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BEAMS


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eam es gn:

Name overnin LimitStates?

______________________________

______________________________


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Modesof

failure

LocalBucklin andSectionClassification

Compact

Noncompact

Sectioncapacityinbending

S ti C it


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SectionCapacity

or

Design

Capacity

of

Fully

LaterallyRestrainedBeams

or

DesignCapacityofveryShortLaterally

Unrestrained

Beams


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Beams aremembersofstructureswhichcarryloads

transverse

to

their

length.

Thesemembersresistflexure (bending)andshear,andsometimestorsion,introducedbytransverseloads.

Purlins,rafters,joists,spandrels,lintels,floorbeams,stringersand

.

simultaneously

are

beam

columns.

Steel beam


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Steelbeam

Beam UDL -major axis loading) (couple)

Beam(torsion)

Beam (UDL -minor axis loading)

Beam-column

+ transverse loading)

Steel Beam where do we use it?


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SteelBeam wheredoweuseit?

Strengthlimitstatebending


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g g

Designequationforbendingstrength

*)(

es gn

capac y

>

ac ore

s reng

m

s a e

moment


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Theusualstrengthmodes offailurefor

structuralsteel

beams are:

Plastification

Flangelocal

buckling

Webcrippling

Weblocalbuckling undershear

Weshallconsidereachofthesestrengthlimitstatesinturn.


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pdevelops,orwhensufficientplastichingesdeveloptoforma

mechanism.

M = S fS plastic section

modulus

fy yield stress

MPa = N mm

S is

tabulated

for

most

rolled

sections

in

handbooks

(mm3

).

Ductile stressstrain curve


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Ductilestressstraincurve

p .

fy

ELong plastic

1

ductile

y

y= y e s ra n = y = y x

MPa

Maximum moment


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Maximummoment

Maximummomentthatcanbeattainedisthe PLASTICMOMENTM

h

Cequa areas

T

plastic neutral axis

NominalcapacityMmax=Ms =Mp

Ms iscalledtheSECTIONCAPACITY.Itisthemomenttocausefailureofthecrosssection.

Here,Mp =

Cxh

=Txh


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C1C

2

p

20020 C1

1h2105

00

2

T2

200

20 T1

yf=350 a


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*

beamshownbelow:

W*

Sx = 1190 x 103 mm3

3000 mm

yf= a

Nm3001011909.0 3* == yfxfSM

kNm3.321=

*** 3.321*

* M *

33.

Desi n bendin ca acit


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Desi nbendin ca acit

S

9.0=,

buckling usuallyresultin

ower ngmax

e owp

.

PS =

BeamsareusuallyunabletoreachMp becauseof

theoccurrenceofpremature

BUCKLING


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LocalBucklin

ec on ass ca on


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Rolledorweldedsectionsmaybeconsideredasanassemblyofindividualplateelements

OutstandSomeare outstand

Internal

Internal

langeso Ibeams

legs

of

angles

and

T

sections

WebWebSomeareinternal

FlangeFlange

websofopenbeams flangesofboxes

Rolled I-section Hollow section

Basis of section classification


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Basisofsectionclassification

Rolledorweldedsectionsma beconsideredasanassembl

ofindividualplateelements

Outstand

Someareoutstand

Internal

langeso Ibeams

legs

of

angles

and

T

sections

Someareinternal

Flange websofopenbeams flangesofboxes

Welded box section


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Astheplateelementsarerelativelythin,whenloadedincompressionthey

Thetendencyofanyplateelementwithinthecrosssectiontobucklemay

limit

the

axial

load

carrying

capacity,

or

the

bending

resistance

of

the

section,bypreventingtheattainmentofyield.

Avoidanceofprematurefailurearisingfromtheeffectsoflocalbuckling

elementswithinthecrosssection.

Internal

Internal Internal

u s an

n ernaWeb

Flange Flange

Rolled I-section Hollow section

ange

Welded box section

Flan elocalbucklin


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Buckled flange

Compression flange

Web

Buckled web

Flange andtopcompressiveregionofthewebDISTORT,buttheline

junctionbetween

the

flanges

and

web

remainsstraight.

Occursin

slender

COMPRESSION

FLANGES

Flan elocalbucklin


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Ifthecompressionflangeofabeamisslender,itmaybuckleLOCALLY andpreventthe

The stress to cause ELASTIC LOCAL BUCKLINGol is given

beamfromreachingitsmaximumbendingstrengthMp(PLASTICMOMENT).

2

2

112 = f

olb

tEk

FLANGE OUTSTANDS

where:tf

k= the local buckling coefficient that dependson edge and loading conditions (= 0.425 here)

3

bfbf

= Poissons ratio (0.3 for steel)


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forces along its short edges has an elastic critical buckling

stress iven b

= t

k is the plate buckling parameter which accounts for edge

,

plate

Platebucklin in com ression


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Bounded plate

in uniform

compression

For bounded

flanges kb = 4

Flangeplatebehaviourincompression


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Critical buckling coefficient k

therefore depends on:

BoundaryconditionsFlange in Compression

Stressdistribution

t

Aspectratio

(b)(a)

Simply supported onall four edges

b

5Buckling coefficient k

(width/thickness) Simply supportedlongitudinal edge b

3

4

b

L FreeExact

NOTE; for a web in pure

compression both longitudinal

ed es are sim l su orted and(c)L

1

2

k = 0.425 + (b/L)

k= 4.0.

(d)

Free

longitudinal edge1 2 3

04 5

Plate aspect ratio L / b

0.425

FREE FLANGE OUTSTAND

Exam le


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f f ensure yielding at fyfwill occur before elastic local buckling?

,

22 t

( )2

112

=f

ol

b

k

2

2

32 10200425.0

=> folyf

tf

. f

fb yf

ft

y


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4100 e nes ree ypeso crosssec on:

(a)COMPACTSECTION

(b)NON

COMPACT

SECTION

c

SLENDER

SECTION


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:

slenderness ofeachelement(definedby

aw t tot c nessrat o


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yf

efb =

f

y

Variations in

andep due to residualstress effects

Thesectionslenderness


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Fromthepreviousexamplewesawthatthesectionslenderness(b/t)is

Itwill

be

shown

how

this

can

be

extended

even

further.

importantinenforcingyieldingtooccurbeforeelasticbuckling.

The SLENDERNESS e

fymust be in units of MPa ( = N/mm

2)

the significance of the term isyf

apparent from Example 3.The normalising with respect to 250

yf

e

fb

=

had fy= 250 MPa. Yield stresses are

now higher.f obviously is more

transparent than

( ) 250yff ftb

( ) yff ftb

Sectionclassification


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a COMPACTSECTION

ThesesectionsallowtheFULLPLASTICMOMENTMp andfor

BUCKLINGoccurs.

SectionsmustbeCOMPACT ifplasticanalysis/designistobeutilised.

Thesectionslendernessis overnedb :

The limits on pare much tighter

epean w en ol= y n xamp e

because higher strains at fyareneeded to make local buckling occur

e p is constant

in the strain hardening region.

SectionclassificationThere are limits for flange and webclassification


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e m s are ere ore:

ep =10 [stressrelievedflanges]=9 o ro e

=8

welded

Thedifference

is

due

to initial

geometric

out

of

classification

ResidualStresses


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Com actsection


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-

PLASTIC SECTIONomen

MP Inelastic localbuckling well into the

strain-hardeningrange

curvature -

The design equation is then:

SMM *

9.0=

SfMM

yPS

==

Exam le


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530UB92.4

=3 3

x fyf= 300 MPa

209

10.253

209 N.B. Some UBs have e > 9.We have not considered the

will be done latter).



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(b)NONCOMPACTSECTION

ThesesectionsallowtheFIRSTYIELDMOMENTMytobereached,butbucklelocallybeforeMp canbeattained.

Thedesignequationisthen:

Theirmoment/curvatureresponseis:

M- of a NON-COMPACT SECTION

S

MM *

M

MP 9.0=

andfora

Inelastic local buckling

beforeMp is reached

NONCOMPACT

SECTION

curvature - eyS ZfM =

eye

Noncompactsection


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- eyeep

e ey m s are ooser an e ep m s an essen a ycorrespond to the coincidence of yielding and elastic local buckling,

but they are modified to include residual stresses and initial geometric

The limits are therefore

imperfections in the strength.

ey = 16 [most flange outstands]= 15 [welded flange outstands]

N.B. We saw in Example 3 that first yield [MY ] and elastic local bucklingcoincided when , or277=ftb

This is close to the above limits.

( ) 5.17250277250 === eyff ftb

Noncom actsection


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Ze istheeffectivesectionmodulus.

Z =S if M =Mo course,

Ze =

Z if

MS =

MY

[Z=elastic

section

modulus,

Y y

Noncom actsection


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Moment

ea e av or

MPM

MY Linearapproximation

ne as c oca

buckling

curvature -y

For non-compact sections we can interpolate linearly between

MYand MP, based on the value of e.

Noncom actsection


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Moment Linear

MP

approximation

MY

S

( )ZSZZepey

eey

e

+=

ep e ey e

Check:

Section strength ofsection with e

e ep e = - =

e = ey Ze =Z + 0(S-Z) =Z [non-compact]



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b SLENDERSECTION

Thesesectionsbucklelocallyevenbeforetheyieldstress(andMy)arereached.

Forslender

sections:Moment

Themoment/curvature

response

is:

eye >M- of a SLENDER SECTION

MY

MP 9.0=

Buckling failure prior to MY

SLENDERSECTION:

curvature -eyS

ZfM =

Slendersection


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Method1:An

effective

width

approach

omits

from

each

flange

the

width

in

excess

of

that

whichcorrespondstoey.

tf

be be

compressionflange(partiallyeffective,2be)

ineffective(i nore)

tensionflange(fullyeffective,b)

Slendersection


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Theeffectivewidthbe isdefinedsuchthat:

ey

ye fb

= or eyeb =

Althoughaccurate,themethodmaybecumbersomeforbeamcrosssectionsasthe

e

effectivesectionbecomesMONOSYMMETRIC,i.e.

CC y

IZ =T

IZ =yCCentroidoforiginalsection

[ ]TCe ZZZ ,min= and since CT yy


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:

Aneasierandsimplermethodtouse:

ZZ ey

e

= e

whereZistheelasticmoduluscalculatedforthefullsection.

Sectionclassificationbasedonweb

s en erness


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SofarwehaveconsideredthecompressionflangewhichmaybucklelocallyunderUNIFORMSTRESS.

Thewebissubjectedtobendingstress (compressionalongoneedge,

.

web

C C

Underbending,thecoefficientkinwebisapproximately23.9.

Weblocalbucklin


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Stocky flange

web Buckled web

Stocky flange occursin

slender

webs

withlargebending and/or

s ears ress


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Occursduetolocalisedyieldingofthewebnearconcentratedloads.

overasmallwebregion.

web

Sectionclassificationbasedonweb


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:

Thelimitsare:

epe


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boththeFLANGESandWEBmustbecompact.

ForaSECTION to

be

NON

COMPACT:

EITHERtheFLANGEorWEBorBOTHarenoncompact.

ForaSECTION TOBESLENDER:

EITHERthe

FLANGE

or

WEB

or

BOTH

are

slender.

Compactflange

Slenderweb

.e.t s

SECTION

s

classifiedasSLENDER

Exam le


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=

thecrosssectionshown.

2 0

8

10

240

Exam le


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ThecrosssectionisthereforeNONCOMPACT.

Exam le


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Boxcrosssections


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Forthecompressionflange,k=4.0Forthewebinbending,k=23.9

bf bcompress onflange

bf

tf

bending

[RHSorSHS]

Boxcrosssections

For the compression flange k 4 0


98/403

Forthecompressionflange,k=4.0

Theclassificationsarethesameasforflangeoutstands,butwith:

ep = 30ey -

= 40 [ lightly welded]

= 35 [ heavily welded]

COMPACT if ( ) epyffe ftb


99/403

DESIGN

OF

UNRESTRAINED

BEAMS

Lecture Outline

L t l t i l b kli


100/403

Lateral torsional buckling

Elastic lateral buckling

Twisting moment

warping

Moment gradient factor, m

Idealised end conditions

Slenderness reduction factor, s

full, lateral, partial and unrestrained

In-plane bending


101/403

X (u)

Y (v)

Out-of-plane buckling


102/403

X (u)

Y (v)or LateralTorsional Buckling

Lateral Torsional Buckling

or


103/403

orFlexural-Torsional Buckling

or

or

Lateral-Torsional buckling

or

Out-of-Plane buckling

Lateral buckling or flexural torsional buckling


104/403

u

Buckled web

u lateral displacementtwist

Original

configuration

Buckled configuration

IntroductionClamp atroot


105/403

root

Slender structural elementsloaded in a stiff plane tend to

fail by buckling in a moreflexible plane.

LateralLateral--torsionaltorsionalbucklingbuckling

Dead weightload appliedvertically

Buckledposition

Unloadedposition

about its major axis, failure

may occur by a form ofbuckling which involves bothlateral deflection and twisting.

Consider an I-beam ..


106/403

Perfectly elastic, initiallystraight, loaded by equaland opposite end moments

about its major axis.

M M

L

ElevationSection

an

y

z x

u

nres ra ne a ong s eng .

End Supports Twisting () and lateral

deflection (u)prevented.

Free to rotate both in theplane of the web and onplan.

Strength limit state bending momentcapacity


107/403

Design equation for bending strength

Design capacity factored strength limit

state moment

Lateral buckling

Lateral buckling is the most influential strength limit


108/403

Lateral buckling is the most influential strength limitstate in the design of steel beams.

Lateral buckling is also called flexural-torsional buckling

Beams with FULL LATERAL RESTRAINT do not bucklelaterally and their strength is the

CROSS-SECTION STRENGTH defined by:

instability of long slender beams.

SMM * 9.0=

yeS fZM =

Lateral buckling


109/403

Most commonly beams do not have full lateral restraint and thenominal strength Msmust be reduced to the MEMBER

BENDING STRENGTH Mb.

The design equation is:

Lateral buckling is catastrophic and sorepresents a STRENGTH LIMIT STATE.

bMM * 9.0=

Lateral buckling or flexural torsional buckling

Occurs in slender, laterally unrestrained beams.


110/403

Like columns, the beam reaches an energy configuration at which it prefers tosnap into an OUT-OF-PLANE buckled position rather than continuing to bendIN-PLANE.

Beam deflects laterally ( = sideways) by u and twist ....

{unstable if u, = 0 after buckling moment}

buckling moment or point of bifurcation

stable buckled position

u,

M

Lateral buckling


111/403

We must therefore calculate Mbbased onlateral buckling of the beam.

This is done by undertaking firstly an

ELASTIC BUCKLING ANALYSIS.

The elastic buckling resistance depends on the



112/403

The elastic buckling resistance depends on thefollowing cross-section properties:

Minor axis bending stiffness EIyTorsion resistance GJWarping resistance EIw

Iy= minor axis second moment of area (mm4)

J= torsional constant (mm4)Iw= warping constant (mm

6)

E = Youngs modulus 200,000 MPaG= Shear modulus 80,000 MPa = 0.3

forsteel ( )+

=12

EG


I J I are all tabulated in the One Steel Handbook


113/403

Iy, J, Iware all tabulated in the One Steel Handbook.

Alternatively for the doubly-symmetric I-section:

6

3

ffy

tbI =

( )wwff

n

iii

dtbttbJ 33

1

3 23

1

3

1+==

=

( )symmetricdoublyhII yw 42

=

(N.B. web ignored)

tw

bf

tf

dw

Twisting moment - torque

C id b ilt i til bj t d t


114/403

Consider a built-in cantilever subjected to atwisting moment ( torque)

ELEVATION

Mt

PLAN

move in move in

move out

Mt

An interpretation of warping:

Warping


115/403

An interpretation ofwarping:

A situation where plane sections do not remain.

This occurs with lateral buckling and its effect isreflected in the warping constant Iw.

Torsion

The equation of torsion is:


116/403

The equation of torsion is:

3dd

= angle of twist

3

dzdz wt

uniform torsion

resistance

warping torsion

resistance

Lateral buckling


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Basic model for lateral buckling is a simply supported I-beamsubjected to a uniform bending moment M.

mp y supporte n t e atera uc ng sense

means lateral deflection and twist are prevented atthe beam ends (u= 0, = 0), but the flanges arefree to rotate in their planes when the beam buckles

laterally.

Simply supported I-beam - model

M M free to rotate in plane during


118/403

L

buckling, but u = = 0

ELEVATION

M M

BMD

PLAN

buckled

top flange

Elastic buckling moment


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The ELASTIC BUCKLING MOMENT is:

(N.B. This is stated without proof. See Chapter 6 of Trahair & Bradford.)

2

2

2

2

LEIGJ

L

EIM wyo +=

Elastic buckling

Recall fo the elastic b ckling of a pin ended col mn


120/403

Recall for the elastic buckling ofa pin-ended column:

2 pin-ended

Euler buckling load(elastic critical buckling load pin-ended column)

2LN

y

oc =

co umn

deformedshape

Example 1

Calculate the elastic lateral buckling moment for a simply supported


121/403

460UB82.1 beam of lengthL = 3m subjected to uniform bending.

191

0

16

43

46106.18 =yI

From OneSteel Tables section properties handbook:

460UB82.1

9.9

191

4

16

6910919 =wI

N.B. this is the ELASTIC BUCKLING MOMENT and not the actual

BUCKLING STRENGTHMb which also depends on the yield stress fy.

Mb will be determined later.

Of b i li l l d d i if b di



122/403

Of course beams in reality are rarely loaded in uniform bending,nor are they pin-ended (or simply supported) so the formula for

Moneeds some modification.

o ar we ave cons ere un orm en ng on y.

This is very conservative as the elastic buckling moment(Mo) is increased by unequal moments, transverse loads etc.

This effect is reflected in the mvalues given inTable 5.6.1 of AS4100.

M Msingle curvature M Mdouble curvature



123/403

L L

M M

BMD+

BUCKLEDSHAPE

moment is highest

in this region

moment is

very small in

this region

M

M

+ = compression

- = tension+

-


The effect of the moment gradient is reflected in the mvalues given inTable 5.6.1 of AS4100 (see attachment).


124/403

( )Alternatively, the BMD is often given using analysis software (Microstran,Spacegas, Multiframe etc.). Therefore

*

M*m = maximumdesign moment within a

( ) ( ) ( )

5.2.

2

*4

2

*3

2

*2

++

=

MMM

mm segment

M*2, M*4 = designmoments at quarterpoints of a segment

M*3 = design momentat the mid-length of asegment

x

M*m M*3M*4

M*2

S/4 S/4 S/4 S/4 i.e. BMD segment

where

Moment modification factorsMoment modification factors


125/403


The SLENDERNESS REDUCTION FACTORsconverts the ELASTICREFERENCE BUCKLING MOMENT M i DESIGN STRENGTH


126/403

REFERENCE BUCKLING MOMENT Mo into a DESIGN STRENGTH.

2/12 where Mo - the elastic reference

=

ooS

MM. buckling moment determined

from Le

(effective length)

Ms the section bending

capacity depending onwhether the cross-section iscompact, non-compact or

slender = Zefy

2

2

2

2

e

w

e

y

oL

EIGJ

L

EIM

+=

s represents a transition betweenfull yielding (at Ms) and elastic

buckling (at Mo)

s Section strength at Ms



127/403

1.0

Elastic buckling at Mo

Le

0

Short beam does not bucklelaterally and

s= 1

Long beam is not influencedby yielding as its buckingmoment is very small

s

There are four types:



128/403

FULL RESTRAINT (F)

yp

PARTIAL RESTRAINT (P)

LATERAL RESTRAINT (L)

UNRESTRAINED (U)

FULL RESTRAINT (F)



129/403

Lateral deflection and twist are effectively prevented, i.e.

stiffeners

brace

compressionflange

concrete

slab

seat support shearconnector

PARTIAL RESTRAINT (P)



130/403

Lateral deflection prevented at some point other than at the

compression flange, and partial twist thus occurs during bucking.

P

Seat support restrains tension (T) flange fully at ends.

compress onflange

buckledconfiguration

tensionflange


LATERAL RESTRAINT (L)LATERAL RESTRAINT (L)


131/403

Compression flange is restrained against translation during

buckling, but the cross-section is free to twist during buckling.

brace

thin sheeting bendsduring buckling

thin roof sheeting quite stiff in-plane but

flexible in bending

screws


UNRESTRAINED (U)


132/403

Free to both displace and twist during buckling,i.e. cantilever tip.

W*

cantilever tip


For these idealised end conditions, AS4100 specifies a

TWIST RESTRAINT FACTOR k


133/403

TWIST RESTRAINT FACTOR kt

RESTRAINTS kt

F F F L L L F U

F P P L P U

P P

1.0

3

2

1

+

w

fw

t

t

L

d

3

221

+

w

fw

t

t

L

d

end 1end 2 1 2

Effects of load height

Load applied above the shear centre ( centroid for doublysymmetric I-section) causes an increased destabilising


134/403

torque that lowers the buckling load.

WW

load at top flange level

W

load at shear centre

W

Effects of load height

To account for the height of application of the load, AS4100 specifies a

LOAD HEIGHT FACTOR kl


135/403

RESTRAINTS AT SHEAR CENTRE AT TOP FLANGE

For load WITHIN THE BEAM SEGMENT

RESTRAINTS AT SHEAR CENTRE AT TOP FLANGEF F F P F L 1.0 1.0P P P L LL

F U PU 1.0 2.0

For load AT THE END OF THE BEAM SEGMENT

. .P P P L LL

F U PU 1.0 2.0

Load and rotation factors: AS4100


136/403

conservativelykr= 1

Lateral restrain classificationsLateral restrain classifications


137/403

Lateral restrain classificationsLateral restrain classifications


138/403

End restraints: examples


139/403



140/403



141/403

Twist factor: AS4100


142/403

Effective length Le

The reference buckling moment Mo is written in terms of the

EFFECTIVE LENGTH Lesimilarly as previously as:


143/403

22

wy EIEI =

22

ee

oLL lte

L = segment length or length of asub-segment between full and/orpartial restraints

kt= twist restraint factor

kl= load height factor

N.B.AS4100 also has a rotational

restraint factor krthat is difficult toquantify and which we shall take

equal to unity (conservatively):

Le= ktklkrL= ktxklx 1.0 x LLe= ktklL

Finally, the design equation for bending within a segment is:

*

Bending capacity,Mb


144/403

bMM *

9.0=

sssmb MMM =

mreflects the effect of the distribution of the bendingmoment along the beam.

sreflects the elastic lateral buckling (viaLe Mo) andyielding (Ms). It accounts for load height and restraint (via Le).

Bending capacity,Mb

Clearly ifms < 1.0, the full SECTION STRENGTH in bending is

not attained (Ms), and the beam will buckle laterally at Mb. Thisis very often the case


145/403

is very often the case.

mreflects the bending moment effects (moment gradient) and isreasonably difficult to control as the loading is fixed.

scan be increased by using a bigger section(larger Iy, Iw, J) or by bracing the beam to decrease Le.

bis the MEMBER STRENGTH in deference to swhich is

the SECTION STRENGTH.

Determine the maximum design momentM* of a 200UC52.2 .

The effective lengthLe = 3.5m and the end moments are as shown.

Example 2


146/403

20412.5

M * 0.4M*

200UC52.2fyf= 300 MPa

8.0

204

206

12.5

69

43

46

33

10166

10325

107.17

10570

=

=

=

=

w

y

x

I

J

I

S

1. Determine SECTION CAPACITY Ms

Example 2


147/403

2. Determine MEMBER CAPACITYMb

Example 2


148/403

Example 2


149/403

In this case the SECTION STRENGTH Msgoverned ratherthan the MEMBER STRENGTH Mb that is determined by

Example 2


150/403

than the MEMBER STRENGTH Mbthat is determined by

lateral buckling.

the low value of Le the high moment modification factor am that produced avery high elastic buckling momentMo . m.

This is NOT always the case and commonly:

Mb


151/403

A B C D

4W W

200UC52.2

3m 3m 3m

8.0

204

204

206

12.5

12.5

69

43

46

33

1016610325

107.1710570

=

=

=

=

w

y

x

IJ

I

S

P L U

In-plane analysis

Example 3


152/403

( ) *** 34360 WWRM AC +==

***

***

335.435.1

.

WWM

WWM

C

B

A

==

==

=

D

A B

C

4.5W*

3W*

+

-

BMD

Segment ABC moment gradient

Example 3


153/403

F

16

3 FLm

Segment ABC effective length

Example 3


154/403

Example 3

Segment ABC elastic buckling capacity


155/403

Segment ABC section capacity

Example 3

Segment ABC slenderness reduction factor,s


156/403

Example 3

Segment CD elastic buckling capacity


157/403

Example 3

Segment CD slenderness reduction factor,s


158/403

Example 4

A simply supported beam with a span of 15m has a nominal central concentratedlive load of 100 kN acting on the top flange. The beam is restrained againstlateral displacement and twist only at the ends, and is free to rotate in plan.Design a suitable WB in accordance with AS4100 of Grade 300 steel.


159/403

g

150kN

15m

Example 4

Assume fyf= 300 MPa, compact section

35.1=mGuess 25.0=


160/403

Guess 25.0s

( ) kNm9.185125.035.19.0/105.562M 6sx =

kNm108.6172300/109.1851S 36x =

Try a 800WB192

mm10t

mm816d

mm28t

mm300b

w

f

f

=

=

=

=

69

w

43

46y

33

x

2

g

mm1019600I

mm104420J

mm10126I

mm108060S

mm24400A

=

=

=

=

=

fyf= 280 MPa

Example 4


161/403

Example 4


162/403


163/403

CAPACITY

Stren th limit state

Local buckling


164/403

Intermediate transverse stiffeners

m s a e

BUCKLING limit stateCombined shear and bending

oa ear ng s eners

YIELD limit state

BUCKLING limit state

Strength limit state

Design equation for shear


165/403

s reng

*)( VVv

es gn capac y ac ore s reng mstate shear

e o a s ee mem er res s s eSHEAR STRESSES.


166/403

Local buckling in shear may restrict theSHEAR CAPACITYof a BEAM.

The design equation is:

*= Vv= nominal

v. s ear capac y

Consider firstl when the shear stress inthe web is APPROXIMATELY UNIFORM.


167/403

stressd dw tw

ara o c ut

approximately

uniform

V*

=ww td

For this case DEFINE Vv

= Vu

(u uniform)

Equation


168/403

22 wtE2112 w

old

with shear stresses.

Exam le 1

Unsitffened web yielding in shear before buckling locally.

If the web is to yield before buckling locally then:


169/403

ol

y

yf =

For a long web, k= 5.35 and using E= 200 x 103 MPa, = 0.3

roduces

2

32

3.0112

1020035.5

wyw

d

tf

yww fd

.250

wt

Exam le 1

The yield capacity is then ywwyw

w fAA 58.03


170/403

82wd

250yww ft

The web yields before buckling and wu VV =

whereywww .

For universal sectionsA =dt

For welded sectionsAw =dwtw

Exam le 1

when 250/82 ywww ftd > the capacity equals the

l l b kli i l


171/403

local buckling capacity wol82wd

250/yww ft

e we uc es e ore y e s an wvu =

where www fAV 6.0 and2

82

v

250

yw

w

w

t

Exam le 1

N.B.

When bucklin overns it is common to add

( )21 di l iff V d ff idl


172/403

( )21 ww tdvertical stiffeners as Vu drops off rapidly as

The provision of these vertical stiffeners will not

82

ncrease e capac y w en y e ng governs, .e.

when

250ywww

f

(they only increase the buckling capacity)

If a web has vertical stiffeners, its strength is increased markedly because:

the elastic local buckling coefficient is increased a benign tension field action develops similar to a truss action

Of hi l li h V V


173/403

Of course this only applies when Vu < Vw

flangewebstiffener stiffener

dw s

Verticall stiffened webs

wwdvu

2

82


174/403

( ) 1/75.082 2

+

= sdwv

250 t

d y

w

w

when

wds 2

( )[ ]75.0/822+= sdfd

wv

250 tw when

wds = 2.3

d = bolt diameter

tp = thickness of ply


352/403

tp c ess o p yfup = ultimate tensile strength of ply

Checking the capacity of a connection lap splice connection -

70

40

35703535 357070

Sp ice p ates, 2x10mm t ick

fy = 260 MPa steel, fu = 410 MPa

40

70

70 N*N*

10

20

PLAN9M24,8.8/S

ELEVATIONSpliced plate, 20mm thickf = 250 MPa steel, fu = 410 MPa

a. Bolt strength

M24 8.8 /S in dholes= 26mm

Design capacity of bolts in shear = Vfn + Vfx = 133 + 186 = 319 kN [TA2.2]

, .

.V = 3 2 x 24 x 20 x 410 = 630 kN 35x 20 x410 = 287 kN


353/403

Vb = 3.2 x 24 x 20 x 410 = 630 kN 35x 20 x410 = 287 kNSince 287 < 319, plate bearing capacity governs.


354/403

rvi ilit limit t t f r ltrvi ilit limit t t f r ltrvi ilit limit t t f r ltrvi ilit limit t t f r lt

For 8.8/TF where slip at service load is to be limited

Although a serviceability limit state still uses a capacity reduction factor

Shear:

sfsf VV *

9.0=

V*sf= nominal bolt shear capacity for friction-type connection

Vsf= neiNtikn

= slip factor (coefficient of friction) taken usually as 0.35

nei = number of effective interfaces kn = 1.0 for standard holes

= 0.85 for short slotted or oversize

t = m n mum o pre ens on ngenerating a friction-type connection

holes0 7 f l l tt d h l


355/403

generating a friction type connection = 0.7 for long slotted holes

++++2

*2

*

0.1+ tfsf

7.0=tis

*sf =

Ntf* = design strength shear force (bolt pre-tension)

N = bolt retension

MinimumNti VALUES (kN)

M20 145

M24 210

M30 335


356/403

M36 - 490

Bolt group subjected to inBolt group subjected to in--planeplaneBolt group subjected to inBolt group subjected to in--planeplane

loadingloadingloadingloading

boltweld

omen

connection

stiffenerend plate

column

stiffener

Considern bolts of equal area subjected to forceP eccentric e

to they-axis which passes through the bolt centroid.

Considern bolts of equal area subjected to forceP eccentric e

to they-axis which passes through the bolt centroid.

y eP

Pe

C P


357/403

C x P



Separate shear caused byP andPe are hard to calculate and sum vectorially.

Instead we calculate position of

INSTANTANEOUS CENTRE OF ROTATION.

Force Vfi* oni-th bolt and radiumri from centre of rotation.

y

Pe(xc, 0)

Vfi*ri

x


358/403



AkrV ifi =* k = constant

A = area of bolt zI

Pe

k =

n22 P

=z about centroids Akx nc =

N.B.1 force inx-direction as well its effect can be included in the same wa and

{ } 2/12*2** yVxVV fififi +=

integrals along the weld


359/403

Exam le 2Exam le 2Exam le 2Exam le 2

A typical web side plate connection is shown in the figure on

the next slide in which a single 10mm thick side plate is bolted

o e we o a eam an s we e o e ange o e co umn.

In designing the welds, the beam reaction is assumed to act

from the face of the column. In designing the bolts, the beam

reaction is assumed to act at the line of the weld at a distance of

90mm from the centroid of the bolt group.

Problem:

For a design beam reaction of 250 kN, determine the maximum

shear force in a bolt of the bolt group


360/403



361/403


The calculations are based on the instantaneous centre ofrotation a roach of Cha ter C9 of the Commentar AS4100 .

By inspection, the centroid of the bolt group is at its geometric

centre.


362/403


Determine the maximum shear in the bolt group in the beam

splice shown. 25Member desi n

65

actions at bolt group

centroid

70

70

Moment = +20kNm

8-M20 8.8/S bolts.

Threads in shear plane.

35

2x280mm E48XX fillet welds.

Single web plate.

This problem demonstrates the in-plane elastic analysis of a bolt group.


363/403



364/403

Exam le 4Exam le 4Exam le 4Exam le 4 .

plane for a design shear of Vf*

= 43.1kN

55 140 30 75 Member design


centroid

35

70

Shear = +160kN

Moment = +20kNm

8-M20 8.8/S bolts.


35


Single web plate.

This problem illustrates the determination of the shear capacity of a bolt.


365/403



366/403


transmitted in conjunction with a design shear force of Vf* =43.1kN by an M20 8.8/TF bolt whose threads intercept a single

shear plane.

25Member design


65

centroidShear = +160kN

Moment = +20kNm

8-M20 8.8/S bolts.

70

70 .


Single web plate.

35

This problem illustrates the checking of the strength of a bolt under combinedshear and tension.


367/403



368/403


be transmitted in conjunction with a serviceability shear forceof V * = 30.0kN by an M20 8.8/TF bolt in a standard hole.

This problem illustrates the checking of the serviceability of a bolt in a friction-

grip connection with a single interface.


369/403

e P

Pys = P/n

y

PTxBolt sheardue to

x

torsion

CG

y

CG of bolt group

= V Ty PTxbolt shear due

to torsion

From mechanics of solids PT = torsional constant x rTorque per bolt Ti = PT x r = C x r

2 = C x (xi2 + yi

2)

=

= 2 =

2 + 2

Torsional constant C = T / (xi2 +yi

2)


370/403

Bolt rou s in torsionBolt rou s in torsionBolt rou s in torsionBolt rou s in torsionIn bending = My/I

In torsion PT = Tr/IP

or torsional com onents P = T /I P = Tx/I

where IP = (xi2 +yi

2)

PTy

PTyPTy

2

max ySTyTx PPPP PTy PTy

PTy

VECTORIAL SUMMATION

P

PTy

PTy

PTyPTy

PTy PT


371/403

TyPTy PTy


372/403

Lecture OutlineLecture OutlineLecture OutlineLecture Outline

Welds, weld group ,

omen connec onsbeam moment splice

Force and moment connectionsseat for the beam-to-column connection, semi-rigid beam

to-column connection, the full strength beam splice


373/403

WeldsWeldsWeldsWeldsStructural connections between steel members are often made b arc-weldin

techniques, in which molten weld metal is fused with the parent metal of the

members or joint plates being connected.

Welding is often cheaper than bolting because of the great reduction in the

re aration re uired while reater stren th can be achieved the members or

plates no longer being weakened by bolt holes, and the strength of the weldmetal being superior to that of the material connected.

In addition, welds are more rigid than other types of load-transferring

connectors.

On the other hand, welding often produces distortion and high local residual

s resses, an resu s n re uce uc y, w e e we ng may e cu

and costly.


374/403

Force connection

Force connection


375/403


376/403

Butt weldsButt weldsButt weldsButt welds

Fillet welds

.

A full penetration weld enables the full strength of the member to be developed,

w e e u ng oge er o e mem ers avo s any o n eccen r c y.

Butt welds often require some machining of the elements to be joined.

Special welding procedures are usually needed for full strength welds between

c mem ers o con ro e we qua y an uc y, w e spec a nspec on

procedures may be required for critical welds to ensure their integrity.


377/403


378/403

We will onl consider e ual le fillet welds here:

t 2

t

tweld size

Design actions are calculated/unit length of weld on plane of throat:

Longitudinal shear, transverse shear, normal force all act on throat andare summed vectorially to produce:

vw* = design force per unit length of weld


379/403


380/403

The force er unit len th of fillet weld in the and

directions may be determined using the familiar expressions:

zxx

yMPv

***

ygeneral fillet

weld group*

wpw

xP**

*

centroid of filletweld group

y

M*

wpw

yIL

v

x

P x

*

yxzz

I

x

I

yM

L

Pv

***

M*zx

Weld in x-y

zplane, z = 0

General fillet weld group


381/403

Stren th desi n:

*

ww

2*2*2**

zyxw vvvv

es gn o any genera e we group su ec o a genera es gn ac on se

(P *x, P*y, P

*z, M

*x, M

*y, M

*z )

may be obtained by evaluating the property set

Lwx, Lwy, Lwz, Iwx, Iwy, Iwp (see Table on next slide)and substituting into the governing equation

- - - ,

checking that the governing inequality is satisfied, at each of the critical points.


382/403

Practical fillet weld rou sPractical fillet weld rou sPractical fillet weld rou sPractical fillet weld rou sMany fillet weld groups comprise lines of welds parallel to the x and y axes.

For such relatively regular fillet weld groups, the identification of possible critical

points is correspondingly more straightforward.

y The possible critical points fora fillet weld rou consistin of

3 8 lines of weld parallel to the xand y axes only are numbered 1

4 7

x

.

5 6

Possible critical oints in

particular weld group


383/403

where:

Lw

- the total length of the weld;

-wx wy (treated as a line element) about the x and y axes respectively;

I - the olar moment of area of the weld elements about thecentroid of the weld group (treated as a line element)

= Iwx + Iwy

The previous expressions can be

sightly modified in order to allow wp

z

wx

xx

I

yM

L

Pv

***

where:

L , L , Lthem to reflect realistic

distributions of the design force

set P* P* P* between wp

z

wy

y

yI

xM

L

Pv

**

*

the lengths of the weld

assumed to receive the

components of the total length ofthe weld group, as follows:

wy

y

wx

x

wz

zz

IxM

IyM

LPv

***

*

the individualx,y andzaxes respectively


384/403

Fillet Weld Group LoadedFillet Weld Group LoadedFillet Weld Group LoadedFillet Weld Group Loaded

nn--p anep anenn--p anep ane

Fillet weld group loaded in-plane by a common* * * x, y x :

**

wp

z

wx

xx

Iy

Lv*

*

zy

y

xMP

v

**

*

y

Pz*

*wpwy

0* =zv

z


385/403

Fillet Weld Group LoadedFillet Weld Group Loaded


ouou --oo --p anep aneouou --oo --p anep ane

Fillet weld group loaded out-of-plane by a common

desi n action set of forces F* , F* and desi n moment M* :

0* =v

yP*

*=

*

wy

yL

P ***

y

Pz*

*

wxwz

zIL

v x


386/403



-- ---- --

Line welds

unit thicknessP e

z Mx

Lw1 L

w1

Centroid ofPy

ycLw2

yweld groupy y

v*y =P*y/Lw = force per unit length acing iny-direction

Lw = total weld length = 2Lw1 + 2Lw2

v*

z =M*

xyc/Iwx = normal force per unit weld inx-direction at point AIwx = second moment of area of unit weld about centroid (mm3)


387/403

Weld group subjected to outWeld group subjected to out--ofof--Weld group subjected to outWeld group subjected to out--ofof--

vzvnthroat

vt

vy (perpendicular)

v = produces normal component and transverse component

force per unit length

on throat. z z

2/1222

Therefore at A (say)22

zy

zzyw

vv +=

For bolts calculateIx as n

yA 2and n

PyVy = etc.


388/403

Desi n e uationDesi n e uationDesi n e uationDesi n e uation

Stren th desi n:=

* .ww

GP6.0

vw = 0.6fuwttkr

vw = nom na capac y o e we per un eng

SP special purpose

kr = reduction factor for length of weldLw(m)= 1.0 (Lw < 1.7)

(high degree of inspection)GP general purpose

(low degree of inspection)

. . w . w .

= 0.62 (Lw > 8.0)

t = roa c ness =

= ultimate tensile stren th of weld

= 480 MPa for E48XX electrodes (most common)= 410 MPa for E41XX electrodes


389/403

Weld stress tra ectoriesWeld stress tra ectoriesWeld stress tra ectoriesWeld stress tra ectories


390/403


391/403


392/403

thickens weld group caused by a design shear force of 160kN

throu h the centroid of the bolt rou and a moment of 20kNm

about the centroid of the bolt group.

25 Member design

55 140 30 75 actions at bolt groupcentroid

Shear = +160kN

8-M20 8.8/S bolts.

35

70

Moment = +20kNm

rea s n s ear p ane.


Single web plate.

35

This problem demonstrates the in-plane elastic analysis of a fillet weldgroup under combined shear and bending.


393/403

=

the centre of the bolt group.

.. =

**= .., =

}2/75280212/2802/)(23

tII yx

mm5602802/

mm.

=

tA

**

560100.46/10446.410160 663

=

tt

yxc =cy

.

1406.272/75 22

max r /max

** yxw IItrMv

mm4.154kN/mm597.1

10446.44.154100.46

=


394/403

Determine the weld leg size required for the equal leg fillet weld

group, if the weld category is SP and the electrode is E48XX.

55 140 30 75

25 Member design


6535

centroidShear = +160kN

Moment = +20kNm

8-M20 8.8/S bolts.


70

70

35


Single web plate.

This problem illustrates the design of a fillet weld group


395/403

uw =

0.1=rk

8.0=

mmkNv /597.1* =

0.14806.08.010597.1 3 tt

mmtt .

mmt 8.9293.6 = Use 10x10 SP E48XX weld.

N.B. A smaller weld could be used if the weld group dimensions

.

group.


396/403

An 8mmx8mm SP fillet weld from E48XX electrodes has a

longitudinal design shear per unit length of vwL*

= 1.0kN/mm and* wx = .

vwy* = 0.4 kN/mm. Check the adequacy of the weld.

( ) mmkNvw /233.14.06.00.1 222* =++=

.

MPafuw 480=

mmt 66.52/8 ==

0.1=rk 8.0=*...... ww

Therefore OK.


397/403

(Fillet loaded out-of plane)

1

8

2

3

weld

rou

y

cetroid

305

x

450 kN

4

5 6

7

203

0,kN450,0

***

***=zyx PPP

,,m =zyx

Weld grou ro erties:

mm10162033052 =wL


398/403

s assume a e ver ca s ear s pr mar y a en y e we s

of the box section, then this vertical shear must be assumed to be

.

Hence,

mm305mm6103052

=

=

dLw

mm203b=

3623.wx

mm152.583,2,1,ointsat =

mm5.1527,6,5,4 y


399/403

106)/14.25.521((90000)v z*

-

450/610-v y*

=

=0v x*=

152.5)-(y76,5,4,pointsat0.967- =

65,2,1,pointsat0 =

esu tant orce per un t engt : 967.0738.0

22*wv

kN/mm22.1

weldfilletE48XXmm8MPa480f

=uw

66.528t =t8t=w

8.0

rtw kN/mm30.1ktf0.6v wuw v 1rk


400/403

180 kNy

275 175 mm300d

123

mm275=b

centroidweld

centroid x300

xmm0.89

2

2

=

=

db

bx

54 6

0kN1800

actionsdesign

***=PPP

0.89175275180

00

*

**

=yx

M

MM

kNmm64980


401/403

mm8503002752Lw =

assume mm850LLLL wwzwywx =

III

32 30022752753002756300 =

36

mm108.21

3002752312

wp

1502/300

1860.89275:6,1pointsat

x

1500.89:5,4,3,2pointsat yx


402/403

Global desi n actions er unit len th:

15064980 6

*

*

= yMv zx

esu an orce per un eng :

points 1, 6

)150(3,2,1pointsat447.0

.

y

wp

kN/mm888.0

767.0447.0

=

wv

150(6,5,4po ntsat447.0

y

**

ktf0.6v

weldfilletE48XXmm6

rtw= uw

108.21850 6*

=

wp

z

wy

y

yIL

v

OK

kN/mm978.0 *

> wv

0.8964980180

,.

6

=

MPa480f =uw

24.426t =6t=

8.0

)criticalnot(5,4,3,2pointsat054.0

.

w

1rk


403/403

Structures to AS4100, 3rd Australian edition, E&FN Spon,

London, 1998.

ST Woolcock, S Kitipornchai & MA Bradford: Design of Portal

rame u ngs, e on, , y ney, .

th

edition, AISC, Sydney, 1994.

as4100 unsw design of steel members

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