6. Composite steel and concrete members. - FSv...

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© Prof. Ing. Josef Macháček, DrSc. 2C08 - 6 1 6. Composite steel and concrete members. Composite steel and concrete beams, full and partial shear connection, plastic and elastic shear connection, continuous beams, composite columns. Composite beams ε concrete (lightweight concrete is preferred) steel non- composite composite elastic plastic σ = E ε strains stresses Shear connectors headed studs Hilti bracket Ribcon Stripcon perforated block connector fired nails

Transcript of 6. Composite steel and concrete members. - FSv...

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© Prof. Ing. Josef Macháček, DrSc.

2C08 - 6 1

6. Composite steel and concrete members.Composite steel and concrete beams, full and partial shear connection, plastic

and elastic shear connection, continuous beams, composite columns.

Composite beams εconcrete (lightweight concrete is preferred)

steel non-composite

compositeelastic plastic

σ = E εstrains stresses

Shear connectors

headed studs Hilti bracket Ribcon Stripcon perforated blockconnector

fired nails

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2C08 - 6 2

be beσmax

b

Due to warping of cross-section (Bernoulli-Navier hypothesis „plane sections remain plane“is not more valid).

∫=2

0e d

/b

max sb σσ Eurocode: 28ebLb ≤=

Common calculation:beff

bLbb ≤==4

2 eeff

Effective width of concrete flange (shear lag effect)

Shear check (conservatively, only steel section is considered):

3yd

vRdc,Edf

AVV =<

For effect of shear on the moment resistancemay be neglected.

Rdc,Ed 21VV <

τ

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Classification of cross sections:

In accordance with steel part, depends on:

• distribution of compression stresses,

• slenderness b/t,

• yield strength of steel.

Example:

99 % casesClass 1 or 2

Class 1 (steel flange)

either if c/tf ≤ 10εor anchored by more studs

web in compression may also buckletw

tf α d

c

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Ultimate limit statePlastic check (Class 1, 2)

Example: Concrete slab without reinforcement (As = 0), plastic neutral axis in slab:

position of pl.n.a.:

plastic resistance:

Check: Mpl,Rd ≥ MEd

zfAzNM a ydaRdpl, ==

Important note: there is no influenceof construction method in plastic check !

plastic n.a.

concrete in tensionneglected

beff 0,85 fcd = 0.85 fck/γc

z

Na

Nc

Nsx

1.0

As

fsd

fyd = fy/γM0

1.5

design strength of reinforcement (γs = 1.15)

design (cylindrical) concrete strength in compression

cdeff

ydaca 85,0 fb

fAxNN =⇒=

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Elastic check (necessary for Class 3, 4) (conservatively also for Class 1, 2).

Modular ratio (for replacing composite section by all-steel cross-section):

in general: „effective“ secant modulus of elasticity of the concrete (incl. effects of creep and shrinkage)

For simplicity and for building structures (except storages):

'c

a

EEn =

2cm

a

/EEn =

Follows common calculation for: area elastic neutral axis y-y z1, z2second moment of area Ii

( )peffsai1 dbn

AAA ++=

„ ideal cross-section “ (in steel units)

beff As

dp

As

stress in concreten x lower

or

beff/n

yd2 f≤σ

cd1 f≤σ1

2

y y

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Check of stresses in ideal cross section:

yd2i

Ed2 fzM

≤=I

σcd1i

Ed1

1 fzMn

≤=I

σconcrete: steel:

Note: from equality may be derived Mel,Rdconversion to concrete

Influence of construction method (consider at elastic calculations only !!)

„Unpropped construction“:

„Propped constructions":All loads resisted by composite section.

phase a) is missing

wet concrete

a) steel girder b) composite cross-section total: a) + b)(own weight + wet concrete) („rest of dead weight" + variable loading)

++++ =

ideal cross-section

σa σi σa+σi

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Serviceability limit states (always elastic, for all Classes 1, 2, 3, 4)

1. Use characteristic loading: γG = γQ = 1

2. In case of plastic ULS design it is necessary to check elastic state during SLS (icl. assembly phases):

σ1 < fck

σ2 < fy

i.e. with factor γc = 1

i.e. with factor γa = 1

3. Determination of elastic deflection with respect to erection method:

4. Cracking of concrete in tension areas (e.g. in continuous beams, minimum reinforcement is required (see Eurocode 4).

variable loading δ2

Ia ... deflection δa

a) Steel section: b) Composite section: c) In total:

secondary beams: δ2 ≤ L/ 250primary beams: δ2 ≤ L/ 400vibration: f1 ≥ 3 Hz

(i.e. δmax ≤ 28 mm)

Ii ... deflection δi = δ1+δ2

δ = δa+δi ≤ δmax

dead weight δ1

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Shear connection

Resistance of shear connectors follows from „push-out tests":

Headed studs: from shear

from bearing of concrete

480

2

v

uRd

df,P πγ

=

When using trapezoidal sheeting: Rd'

Rd PkP =

(α = 1 for h/d > 4)

Shear connectors

headed studs Hilti bracket Ribcon Stripcon perforated connector block

Other connectors: see relevant references

≤ 1

min. 4d 20

dmin. 5d

h

cmckv

2

Rd 290 Efd,Pγα

=

PRk

δu

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Design of shear connection according to elastic theory:Elastic calculation is needed for:

• cross sections of class 3 and 4,• for "non ductile connectors" (if characteristic slip during push test is δuk < 6 mm)

dpbeff/n

zShear flow at connection:

i

iEd1 I

SVV =

First moment of connected area: zdn

bS ⎟⎠⎞

⎜⎝⎛= p

effi

V1

Example:

VEdSpacing of studs:

1

Rd

VPie ≤ emax ≤ 800 mm

≤ 6 dp

Distribution according to VEd, but less than emax.

e

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Rd

cff P

Fn =

The number of ductile shear connectors nf for full shear connection results from equilibrium:

where force in connected flange: sccf NNF +=

)85,0( cdeff fbx=Example:

Shear flow is redistributed, shear connectors(e.g. studs) may be distributed uniformly:

Mmax

nf nf

Mmax

nf nf

dp

e

Shear connection according to plastic theory: (for ductile connectors, e.g.welded studs)

Fcf

Ns

Npl,a

plast. n. o. Nc

Mmax = Mpl,Rd (the most stressed cross-section)

• full - transfers Mpl,Rd

• partial - transfers only MRd < Mpl,Rd

and determines resistance

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Partial shear connection with ductile connectorsFrequently the required number of connectors nf can not be placed (e.g. due to limited space in trapezoidal sheeting):

( )cf

cRda,pl,Rdpl,Rda,pl,Rd F

FMMMM −+=

n n < nf (nf is number of connectors for full connection)

Requirements:• The following is valid for ductile connectors only.

Eurocode guarantees ductile behaviour for studs ø 16÷25 mm and span Le < 25 m, depending on degree of shear connection η = n/nf: η ≥ 1 – (355/fy)(0,75 – 0,03Le)

• It can only be used in buildings and when Mpl,Rd ≤ 2,5 Mpl,a,Rd

linear approach

plastic theoryRdpl,

Rd

MM

fc,

c

f FF

nn

==η

1

1

requires shear connection

steel cross section

Rdpl,

Rda,pl,

MM

e.g.

For number of connectors n < nfthe resistance of the cross section:

or number of connectors for given MEd:

⎟⎟⎠

⎞⎜⎜⎝

−== cf

Rda,pl,Rdpl,

Rda,pl,Ed

RdRd

c 1 FMM

MMPP

Fn

MEd,max

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Transverse reinforcement in concrete slab(to ensure transfer of shear from connectors to slab)

Failure is protected by shear resistance of concrete, transverse reinforcement,possibly also steel sheeting.

Detailed relations are given in Eurocode 4.

critical planes for shear failure

anchorlength

As,min = 0,002 Ac1

2

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Continuous composite beamsGlobal analysis (determination of internal forces):

• plastic (rigid-plastic or elastic-plastic) – necessary to fulfil a number of conditions;• elastic (approximate, with redistribution or iterative).

Approximate elastic analysis with redistribution of moments:

a) „Uncracked analysis"

b) „Cracked analysis"

0,15 L1 0,15 L2 EaI1EaI2

acc. class

Uniform equivalent effective steel cross sectionassuming that concrete in tension is uncracked.

Reduction of moments:class 1: -40 % class 2: -30 %class 3: -20 % class 4: -10 %

Above supports equivalent eff. steel cross section neglecting concrete in tension (EaI2).

Reduction of moments:class 1: -25 % class 2: -15 %class 3: -10 % class 4: 0 %

EaI1

reducedacc. class

(M+ is adequately higher)

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běžnýAs (reinforcement

only considered)

common cross section

Effective widths of concrete flange:

beff bLbb ≤==4

2 eeeff

ULS

crosssections:

shearconnection:

Stability of compression flange above support:for IPE < 600 (S235) or 400 (S355)

HE < 800 (S235) or 650 (S355)need not be checked

Fcf Fcf + Asfsdshear in connection:force in the reinforcement

12

Ductile connectors should be distributed uniformly in sections 1 and 2, e.g.:

Rd

sdscf2 P

fAFn +=

Le = 0,8 L1 0,7 L2

L1 L2

differently for cantileverLe = 0,25(L1 + L2)

Rd

cf1 P

Fn =

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Composite columnsTypes and requirements to exclude local buckling:

Concrete filled sections:

Partially encased sections:

Concrete encased sections:

t

t

h

d ε90≤td

ε52≤th y

235f

tb

ε44≤tb

hc

bcmax. 0,4 bc

min. 40 mmmax. 0,3 hc

max. 0,06 Ac

NOTE:Round holes (ø 20÷30 mm) are required in each 5 m for release of vapour under fire.

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Concrete filled tubes without reinforcementOther sections similarly – more simple calculation, but usually to consider reinforcement is necessary.

td

Aa, Ia

Ac, Ic

Simple plasticresistance:

concretesteel

'fAfAN cdcydaRdpl, +=

'cdf ... commonly = 0,85 fcd, but for concrete filled sections increased:

• design strength without reduction 0,85;• on top of it another increase for circular cross sections due to

„confinement effect“ (but only for „short columns“ with and small eccentricities with e/d ≤ 0,1).

5.0≤λ

Buckling resistance:

L

( )2

eff2

cr LEN Iπ

= where effective elastic flexural stiffness:( ) ceffc,aaeff 60 III E,EE +=

reduced (effective) secant modulus of concrete taking account of long term effects (Ecm/2).

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Slenderness:

cr

Rkpl,

NN

=λcharacteristic plastic resistance

Check: 01Rdpl,

Ed ,NN

≤χ

Reduction coefficient χ for hollow section from buckling curve a.

Bending'fcd ydf

ydfUsually more suitable procedure:

From equilibrium:

plastic neutral axis

Mpl,Rd

i.e. calculate Mmax and from design tables:Mpl,Rd= κ Mmax

(κ depends on the parameter )see complementary notes. Rdpl,

yda

NfA

=δ3'cd

2ydmax )2(

121)( tdftdtfM −+−=

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© Prof. Ing. Josef Macháček, DrSc.

2C08 - 6

Mmax

N

M

Npl,Rd

Mpl,Rd

conc

rete

steel

18

Interaction of compression and bending (NEd + MEd)

← Example for tube.

Similar curves are available in literaturefor various cross sections, e.g.:

Rdpl,

Ed

MM

Rdpl,

Ed

NN

0 0,4 0,8 1,2 1,6

1,0

0,2

0,4

0,6

0,8 δ = 0,20,225

0,300,90

0,600,40

Interaction curve:construction - using various positionsof neutral axis determine N, M.

(depends on parameter ) Rdpl,

yda

NfA

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2C08 - 6 19

Check

90Rdpl,d

Ed

RdN,pl,

Ed ,M

MM

M≤=

μ

coefficient of uncertainty of the

model

Note: For members of sway frames second order effects shall be taken into account. Moment MEdshould therefore be modified by coefficient k:

011 effcr,Ed

,,NN

k ≥−

Ncr for effective rigidity

0,66+0,44ψ ≥ 0,44(for lateral loading β = 1)

Npl,Rd

N

M

Mpl,Rd

NEd

(moment resistance)

Mpl,N,Rd

1) In buckling (see above):

2) Interaction of bending + compression:

01Rdpl,

Ed ,NN

≤χ

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2C08 - 6 20

Complementary notes

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© Prof. Ing. Josef Macháček, DrSc.

2C08 - 6 21

Factor κ for determination of plastic moment of concrete filled tubeMpl,Rd= κ Mmax

Rdpl,

yda

NfA

δ , 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

0,20,30,40,50,60,70,80,9

0,5680,7150,8150,8850,9330,9660,9860,997

0,5860,7270,8230,8910,9370,9680,988

0,6030,7380,8310,8960,9410,9700,989

0,6190,7490,8390,9010,9450,9720,990

0,6340,7600,8460,9060,9480,9740,991

0,6480,7700,8530,9110,9510,9760,992

0,6620,7800,8600,9160,9540,9780,993

0,6760,7890,8670,9200,9570,9800,994

0,6890,7980,8730,9250,9600,9820,995

0,7020,8070,8790,9290,9630,9840,996

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Example of a realization

round hole (ø 20÷30 mm) for release of vapour under fire

More common connections of composite girders to filled tube sections: - bolt connection on fin-plates passing through tube (and welded to tube),- hinge bedding on blocks welded to the column.

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Behaviour under propped and unpropped construction:

Composite slabs (Possibilities acc. to Eurocode 4)

M

deflection

non-composite

composite propped

composite unpropped

ε σ

sheetingwith indents

dovetail sheeting HOLORIB(self-locking)

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Welded headed studs Thin-walled steel strip Hilti STRIPCON

Hilti brackets Thin-walled steel strip Hilti RIBCON

Welded perforated shear connector

fired nails

fired nails

fired nails

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Experiments at Faculty of Civil Engineering CTU in Prague

Composite truss girder