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8/12/2019 NUS Composite Beam II
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Shear Connection
in
Composite Structures
J Y Richard Liew
Department of Civil Engineering
National University of Singapore
2
Composite beam with composite slab using profiled steel deckings
Composite beam with solid concrete slab
D
B
Beam span perpendicular to slab span
D
B
DpDs
Beam span parallel to slab span
DsDp
D
B
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Profiled steel deckings
4
Site welding of headedshear connectors
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Shear connection in composite structures
Interfacial shear resistance is important between
concrete and steel sections to ensure composite
sections.
Shear connectors are used to transmit forces
between concrete and steel sections.
Shear connectors should be strong enough to resist
high shear forces, stiff enough to limit relative
slippage without fracture, i.e. ductility.
6
Outline of Presentation
Basic considerations and material specifications.
Degree of shear connection vs. strength utilization in composite
beams.
Full shear connection versus partial shear connection
Possible arrangement of headed shear connectors.
Other shear connectors.
Push-out tests.
Shear resistance / initial stiffness / deformation capacity
Characteristic resistance of headed shear connectors.
Shape factor, k.
Dimensional detailing.
Different degrees of shear connection.
Shear connection in composite columns.
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Headed Studs
Welded to the steel section, either directly or
through profiled steel sheets. The purpose of the
head of the studs is to resist any uplift component
of the forces applied to the studs.
Shear connectors
1.5 d
0.4 d
d
Typical dimensions
d ranges from 16 to 25 mm
headed shear studs with d = 19 mm are most commonly used.
minimum diameter and the minimum depth of the head of a headedstud shall be 1.5 d and 0.4 d respectively.
8
Structural performance
Strength calculation:
Full shear connection vs. partial shear connection
Deflection calculation:
Rigid shear connectors vs. flexible shear connectors
Shear connectors
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Basic resistancesHow to mobilize full moment resistance of a composite section
Rc = Resistance of concrete flange
= 0.45fcuBe (Ds Dp)
Rs = Resistance of steel beam
= A py
Rq = Resistance of shear connection
= N P
Dp
Ds
Be
h
Rc
Rs
Rq
10
Full shear connection
1R
Rand1
R
Rk
s
q
c
q
sc =
ksc = 1.0
i.e. full shear connection
Lot of shear connectors provided
Rq
Rc or Rq
Rs
Dp
Ds
Be Full shear connection
hRc
Rs
Rq
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Compressive
force
Tensile
force
Lot of shear connectors provided
Moment resistance envelope
with ful l shear connection
Moment resistance
Appl ied moment (UDL)
12
Lot of shear connectors provided:
Rq Rc or Rs ; full resistance of concrete slab or steel section to bemobilized.
Moment equilibrium: Mc = Rs x h or Rc x h whichever is
smaller
Moment resistance at full shear connection
h
Rc
Rs
Dp
Ds
Be Rigid shear connector
Strain diagram Force diagram
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Partial shear connection
1R
Rand1
R
Rk
s
q
c
q
sc =
ksc 1.0
i.e. partial shear connection
Few shear connectors provided
Rq Rc or Rq Rs
Dp
Ds
Be Partial shear connection
hRc
Rs
Rq
14
Compressive
force
Tensile
force
Few shear connectors provided
Moment resistance envelope withpartial shear connection
Moment resistance
Appl ied moment (UDL)
Reduced moment resistance
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Few shear connectors provided:
Rq < Rc or Rs ; neither full resistances of concrete slab and steel section to
be mobilized.
Moment equilibrium: Mc = Rq x h ;Resistance of shear connectors controls
Moment resistance at partial shear
connection
DpDs
BeRigid shear connector
Strain diagram
h
Rq
Rq
Force diagram
16
hRc
Rs
Rigid shear connector Full shear connection
Deformation
requirement
Strength
requirement
Slippage due to
flexible shear
connectors
Flexible shear connector
Rq
Rs
h
Partial shear connection
Additional deflection Reduced moment capacity
Strain diagram Force diagram
Dp
Ds
Be
Summary of structural requirements
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Degree of shear connectionThe degree of shear connection, ksc, is a measure of the strength utilization
of a composite beam, and is defined as
1.0
smalleriswhicheverR
Ror
R
Rk
s
q
c
q
sc
4
cmcu Efd 8.029.02
48.0
2dfu
+ 12.0
d
h
d
h
d
h
Characteristic resistance of headed
shear studs
26
Design of Shear Connection BS 5950:Part3:3.1
Design capacity
Rq= 0.8 Qk Cl 5.4.3
For light weight concrete
Rq = 0.9 (0.8Qk)
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In a solid concrete slab, the design resistance of headed shearstuds against longitudinal shear is given by:
For positive moments, Rq = 0.8 Qk
Forces acting in a headed shear stud
embedded in a solid concrete slab
Design resistance of headed shear studs
in sol id concrete slab
28
Design resistance of headed shear studsin composite slabs
In composite slabs, the design resistance of headed shear studs
against longitudinal shear is given by:
For positive moments, Rp = 0.8 x k x Qk
where
k is the shape factor to allow for strength reduction due to
the presence of profiled steel decking.
Forces acting in a headed shear stud
embedded in a composite slabusing profiled steel decking
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Ribs perpendicular to beam1 stud per rib: k = 1.0
2 studs or more per rib: k = 0.8
Ribs parallel to beam
All cases: k = 1.0
0.85 1r
p p
b hD D
0.6 1r
p p
b h
D D
16.0
pp
r
D
h
D
b
In most modern deckings with wide troughs, k is equal to 1.0.
Decking Shape factor, k
30
Beneficial side for shear connectors
depends on the direction of force relative to
the position of central stiffeners in the
trough, and such effect is reflected in the
value of the shape factor, k.
Presence of large concrete
block to resist force effectively
Insufficient concrete to
resist force effectively
Effect of centralstiffeners in prof iled
steel decking
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Maximum longitudinal spacing lesser of 600 mm or 4Ds Minimum spacing
Spacing
5d along the beam
4d between adjacent studs
3d between staggered studs
Unless located directly over the web, nominal diameter of a
stud 2.5 times the thickness of the flange to which it is welded
Diameter
Edge distance
20 mm
min.
50 mm min.
45
Dimensional details of headed shear studs
32
Use d = 19 mm for illustration
max. spacing = 600 mm or 4Ds (slab depth) typically = 4 x 125 = 500 mm edge distance > 20 mm min. spacing = 5d = 95 mm along the beam
= 4d = 76 mm between adjacent studs
Staggered arrangement may be used.
B
> 4d
> 5d
> 20
> 20
Dimensional details of headed shear studs
> 4d
> 20> 20
B
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Provided that individual connectorspossess sufficient ductility, the shear
connection as a whole may be
designed assuming all those in a
shear span fail as a group
R Smaller of R or R q s c>
Rq = n kQs k( )
Deck reduction factor
No of connector
No. of Shear Connectors for Full
Composite
34
Summary on Shear Stud Design
Characteristic strength Qk of stud from Table 5
Design capacity:Normal concrete Rq = 0.8 Qk x k
Lightweight concrete Rq = 0.9 x 0.8 Qk x k
k = deck reduction factor
For concrete slab k = 1.0
Number of studs n = min (Rs or Rc) / Rq,
to be distributed between max moment to zeromoment
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Example 2:From Example 1, determine the number of shearstuds required for full composite.
Design Data:
UB457x191x74
280kN
Span = 12.0m
Concrete slab depth = 125mm
Concrete Grade = 30
Shear Studs: 19mm, 95mm long
Loading
Dead Load = 15.0kN/mImposed Load = 16.0KN/m
Design moment = 839kNm
Design Shear force = 280kN
M = 839kNm
280kN
280kN
36
For UB 457 x 191 x 74
Smaller of Rc and Rs is 2615kN.
Capacity of shear connector (19mm diameter and 95mm long)
Qk = 100kN Table 5
Design capacity R = 0.8Qk = 80kN
No. of connectors per half span = 2615/(80) = 32.7
Use 34 connectors with two connectors per trough in pairs with spacing
as shown
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34
R A kNs y= = 2615
Centre Line
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Moment capacity of a composite beam
with different degrees of shear connection
Ms
0 0.4 1.0 ksc
a) b)
Mc
Design methods
a) Linear interpolation method (LIM) conservative approach
b) Plastic stress block method (PSBM)
38
Design Methods
b) Plastic stress block method (PSBM)
+=
2
DD
R
RD
2
DRM
ps
c
sssc Full shear connection
( )4
T
R
RR
2
DD
R
RDR
2
DRM
f
2
qsps
c
q
sqsc
+= Partial shear connection
a) Linear interpolation method (LIM)
( )scscsco MMkMM +=
where ksc = degree of shear connection
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Plastic moment capacity of partial composite
Case 3a: R Rq w< (PNA in web)
Rq
y
y
y
x
Ms
2y=
yBe
D
PNARq
Rq
Dp
Ds
xt R and R tdy q v y2 = =
xR
t
d R
R
q
y
q
v
= =2 2
R R yD D
q cs p
=
= yR
RD D
q
cs p( )
Take moment about the centre of steel
section
Mc = Ms +Rq[D/2+Ds-y/2]-Rqx/2
Substituting for x & y:
M M R D D RR
D D RR
dc s q s
q
c
s p q
v= + +
2 2 4
2
Rw =Dty Rv =dty
40
Case 4: R Rq w (PNA in steel flange)
Ds- Dp
y
Rq
y
y
2yx
D
PNA
Rs
Rq
Ds
Be
Rs-Rq
Dp
xR R
B
R R
R T
s q
y
s q
f
=
=
2 2 /
in which Rf= 2BTy
yR
R
D Dq
c
s p= ( )
Moment about top of steel flange
2
x)RR(
2
yDR
2
DRM qssqsc
+=
4
T
R
)RR(
2
)DD(
R
RDR
2
DRM
f
2
qsps
c
q
sqsc
+=
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Worked example
633.7PSBM
PSBM
LIM
LIM
Design
methods
704.4
673.10.78
2 studs per
trough
567.80.49
1 stud per
trough
Mco(kNm)
ksc
Arrangement
of
shear studs
Partial shear connection
Notes: Linear interpolation method (LIM)
Plastic stress block method (PSBM)
0 0.49 0.78 1.0 ksc
Mco
Ms
Mc
Ms = 389.1 kNm Mcp = 753.2 kNm Full shear connection
Dp = 50 mm
2.8 m width
125 mm thick
UB 457 x 152 x 52 Grade 50
Grade 30 concrete
19 mm dia. Headed shear studs,
95 mm as welded height
42
hecks Connectors at Other Locations
N1N3 =0 N2 N1 N1
N2 N1
N2
Ms
N1 N2 N1
M1 M2
Ni = Np (Mi-Ms) / (Mc-Ms)Ni = number of shear connectors between the intermediate load point and the
adjacent support.
Np = number of shear connectors provided.
Mi = moment at the intermediate point i.
Ms = moment capacity of the steel section.
Mc = moment at the composite section
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Example: Moment capacity of partial composite
beam
Moment capacity of steel section Ms = 278 kNm
Moment capacity of full composite section Mc = 612 kNm
No. of connectors from zero to max. moment, Np = 38
15 38-15 = 23 23 15
M1 = 410 M2 = 565N1
N2
Ni = Np (Mi-Ms) / (Mc-Ms)N1 = 38(410-278)/(612-278)
= 15
N2 = 38
No. of connectors
44
Partially-encased Composite
Beams
Steel section is encased to enhance fire resistance
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Partially-encased Composite
Beams 0.45fcu
0.45fcu
0.45fcu
PNA
PNA
PNA
46
HomeworkDesign data:
S275 Partially Encased Universal Beams
Span = 12.0m
Spacing of steel beams = 3.0m
Concrete slab depth = 125mm
Concrete Grade = 30
Loading
Dead Load = 15.0kN/m
Imposed Load = 16.0KN/m
Determine the beam size and check
moment capacity
Beam3.0m
3.0m
12m
Ds
Be = 3m
Partially encased by Grade 30 concrete
2 T20 bars
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Transverse Reinforcement in
Concrete Slab
J Y Richard Liew
Department of Civil Engineering
National University of Singapore
48
Longitudinal Splitting
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Transverse reinforcement refers to the
reinforcement in the concrete slab runningtransversely to the span of the beam.
Sufficient transverse reinforcement should be
used to enable the concrete slab to resist the
longitudinal shear transmitted by the shear
connectors, both immediately adjacent to the
shear connectors and elsewhere within its
effective breadth (Be
).
When profiled steel sheets are used, they may
also act as transverse reinforcement.
Transverse reinforcement
50
The total longitudinal shear force per unit length (v) to be resisted at any
point in the span of the beam should be determined from the spacing of
the shear connectors by the following equation:
N = Number of shear connectors in a group
s = Longitudinal spacing of shear connectors
Q = Smaller of Qp and Qn
v = N Q / s
v vr
Check longitudinal shear force
For structural adequacy, the longitudinal shear force, v, should not
be larger than the local shear resistance in the concrete slab, vr :
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The local shear resistance of the reinforced concrete slab is
fcu = characteristic cube strength of concrete in N/mm2
but 40 N/mm2
= 1.0 for normal weight concrete and 0.8 forlightweight concrete
Acv = mean cross-sectional area, per unit length of the
beam, of the concrete shear surface under
consideration = (Ds + Dp )/2
Asv = mean cross-sectional area, per unit length of thebeam, of the combined top and bottom
reinforcement crossing the shear surface
vp = contribution of the profiled steel sheeting
vr= 0.7 Asv fy + 0.03 Acv fcu + vp
but vr 0.8 Acvfcu + vp
Local shear resistance
Transverse
Reinforceme
nt
52
d) Composite slab.Profiled decking spanning perpendicular to the beam
e) Composite slab.Profiled decking spanning parallel to the beam
Transverse shearsurfaces, Asv
Surface Asv1-1 (Ab +At)
2-2 2Ab3-3 At
a) Solid slabs
Ab
At
2
1
1 2
3 Lap joint inprofiled decking
3
At 3
3
At3
3
Profiled
decking
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Profiled sheeting may be assumed to contribute to the transverse
reinforcement provided that it is either continuous across the top flangeof the steel beam or that it is welded to the steel beam by stud shear
connectors.
d = nominal shank diameter of the stud
n = 4
vp = tp pyp
a) Continuous + Ribs perpendicular to beam span
vp = (N/s)(n d tp pyp) but vp tppyp
b) Discontinuous + Studs welded to steel beam
Contribution of profiled decking
54
Example:
Dp = 50mm Ds = 130mm
Light-weight concrete Grade 30 to be used
K = 0.8 = reduction factor due to metal decking
Design shear force
V = NQ/S
N = 2 studs per rib
Q = 0.8 (0.9 x 0.8Qk) = 58 kN
(0.8 is the reduction factor for decking perpendicular to thebeam; 0.9 is reduction fator for light-weight concrete;
Qk = 100 kN for 19mm stud)Spacing of stud = 375 mm
V = 2 x 58/.375 = 309kN
For intermediate beam, there are two shear planes
For each shear plane 1-1 as shown in the figure
V = 309/2 = 155 N/mm
Dp = 50mm
Ds = 130mm
Metal Decking
1
1
1
1
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Shear Resistance
Acv = (80 + 50/2) x 1mm = 105 mm2 / mm
= 0.8fcu = 30 N/mm2For sheeting continuous across the beam:
Vp = tp x py = 1 x 280 N/mm2 = 280 N/mm
Assume A142 mesh: Asv = 142 mm2/m or 0.142 mm2/mm,
fy = 460 N/mm2
0.7Asvfy = 0.7 x 0.142 x 460 = 46 N/mm
0.03Acvfcu = 0.03 x 0.8 x 105 x 30 = 76 N/mm
= 46 + 76 + 280 = 402 N/mm
Vr= 0.8 x 0.8 x 105 (30)0.5 + 280 = 648 N/mm
Therefore Vr= 402 N/mm > V = 155 N/mm OK
r sv y cv cu p0.7A f 0.03 A f V = + +
cucvysvr fA03.0fA7.0 +=
r sv y cv cu p0.7A f 0.03 A f V = + + r cv cu p0.8 A f V +
Max. value of Vr
56
However, for edge beam, there is only one
shear plane. In this case
V = 155 x 2 = 310 N/mm, hence same
reinforcement can be used for the edge beam.
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Shear connection in composite structures
Composite Slab
Composite Beam
Composite Column
BS5950: Part 4: Cl.6.4.1Eurocode 4: Cl. 9.7.3
BS5950: Part 3: Table 5
BS5400: Part 5: Table 7
Eurocode 4: Cl. 6.6.3.1
BS5400: Part 5: Table 7
Eurocode 4: Cl. 6.7.4