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© 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
© 6 Prof. Ing. Josef Macháček, DrSc.
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
© 6 Prof. Ing. Josef Macháček, DrSc.
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
© 6 Prof. Ing. Josef Macháček, DrSc.
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
© 6 Prof. Ing. Josef Macháček, DrSc.
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
© 6 Prof. Ing. Josef Macháček, DrSc.
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)
© 6 Prof. Ing. Josef Macháček, DrSc.
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)
© 6 Prof. Ing. Josef Macháček, DrSc.
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
© 6 Prof. Ing. Josef Macháček, DrSc.
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).
© 6 Prof. Ing. Josef Macháček, DrSc.
OK3 10
Resistance of a concrete filled tube:
0
1000
2000
3000
4000
5000
6000
7000
Nsteel
Nconcrete
Nsteel+Nconcrete
Nwith confinementeffect
N
N
© 6 Prof. Ing. Josef Macháček, DrSc.
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 −+−=
© 6 Prof. Ing. Josef Macháček, DrSc.
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 )
© 6 Prof. Ing. Josef Macháček, DrSc.
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:
© 6 Prof. Ing. Josef Macháček, DrSc.
OK3 14
Possible details
© 6 Prof. Ing. Josef Macháček, DrSc.
OK3 15
Higher fire resistance due to concrete: