6. composite steel and concrete structures - fsv...

© 6 Prof. Ing. Josef Macháček, DrSc.

OK3 1

6. Composite steel and concrete structuresFull and partial shear connection, plastic and elastic shear connection, continuous beams, composite columns, Eurocode design.

SLSAlways elastic approach – with ideal cross-section: 1. Check elastic behaviour (all calculations with characteristic values).2. Determine deflection and vibration (or limit deflection instead) for individual

phases of assembly.

ULSPlastic approach Elastic (influenced by method of assembly)

beff

z

Ns

fyd = fy/γa

xAs

pl. n. a.

concrete in tensionneglected

Nc

fsd

0,85 fck/γc = 0,85 fcd

Npl,a

"ideal cross section"(equivalent effective steel section in steel units)

Asn x lower

or

beff/n

cd1 f≤σ1

2

el. n.a.

yd2 f≤σ

ratio n = Ea/(Ecm/2)

sds f≤σ

1σsσ

2σ


OK3 2

Shear connection

Resistance of shear connectors follows from "push-out tests". E.g.:

Headed studs: from shear

from bearing of concrete

480

2

v

uRd

df,P πγ

=

cmckv

2

Rd 290 Efd,Pγα

=

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: relevant references.≤ 1


OK3 3

Shear connection according to elastic theoryNecessary for:

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

Example:

VEdDistance of studs:

1

Rd

VPie ≤ < 800 mm

< 6 dp

Distribution according to VEd, but less than emax.

dpbeff/n

zShear flow at connection:

i

iEd1 I

SVV =

First moment of connected area: zdn

bS ⎟⎠⎞

⎜⎝⎛= p

effi

V1


OK3 4

Rd

cff P

Fn =

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

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)

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


OK3 5

Partial shear connection with ductile connectorsFrequently the required number of connectors 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)

Requirement:• 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)

• can be only 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


OK3 6

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+ adequately higher)


OK3 7

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 + Asfsdsíla ve spřaženíforce 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

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


OK3 8

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


OK3 9

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

td

Aa

Ac

Simple plasticresistance:

concretesteel

'fAfAN cdcydaRdpl, +=

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

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

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

5.0≤λ

Buckling resistance:

L( )

2eff

2

cr LIE

Nπ

= where effective elastic flexural stiffness:

( ) ceffc,aaeff 60 IE,IEIE +=

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


OK3 10

Resistance of a concrete filled tube:

0

1000

2000

3000

4000

5000

6000

7000

Nsteel

Nconcrete

Nsteel+Nconcrete

Nwith confinementeffect

N

N


OK3 11

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

thence Mmax and from design tables:Mpl,Rd= κ Mmax

(κ depends on the parameter ) Rdpl,

yda

NfA

=δ3'cd

2ydmax )2(

121)( tdftdtfM −+−=


OK3 12

Interaction of compression and bending (NEd + MEd)

← Example.

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

Mmax

N

M

Npl,Rd

Mpl,Rd

conc

rete

steel

Rdpl,

yda

NfA

=δ

Interaction curveconstruction: using various positionsof neutral axis determine N, M.

(depends on parameter )


OK3 13

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 MEd should be modified by coefficient k:

011 effcr,Ed

,,NN

k ≥−

=β

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:


OK3 14

Possible details


OK3 15

Higher fire resistance due to concrete:

6. composite steel and concrete structures - fsv...

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