nus composite beam ii

8/12/2019 NUS Composite Beam II

1/29

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


2/29

3

Profiled steel deckings

4

Site welding of headedshear connectors


3/29

5

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.


4/29

7

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


5/29

9

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


6/29

11

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


7/29

13

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


8/29

15

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


Additional deflection Reduced moment capacity

Strain diagram Force diagram

Dp

Ds

Be

Summary of structural requirements


9/29

17

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)


14/29

27

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


15/29

29

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


16/29

31

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


17/29

33

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


18/29

35

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


19/29

37

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


20/29

39

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

+=


21/29

41

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


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


22/29

43

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


23/29

45

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


24/29

Transverse Reinforcement in

Concrete Slab

J Y Richard Liew

Department of Civil Engineering

National University of Singapore

48

Longitudinal Splitting


25/29

49

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 :


26/29

51

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


27/29

53

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


28/29

55

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.


29/29

57

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


Eurocode 4: Cl. 6.6.3.1


Eurocode 4: Cl. 6.7.4

nus composite beam ii

Documents