SCI AD266

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Advisory Desk Note

AD 266: Shear Connection in composite beams Page 1

SCI ADVISORY DESK

AD 266: Shear Connection in composite beams

Following some recent questions on the requirements given in BS 5950-3: 1990 for the

design of the shear connection in composite beams, a clarification of three major issues isgiven in this advisory desk article. These issues are:

Effective breadth of the concrete flange (BS 5950-3: 1990 Clause 4.6)

Partial shear connection (BS 5950-3: 1990 Clause 5.5)

Transverse reinforcement (BS 5950-3: 1990 Clause 5.6)

Effective breadth of the concrete flange (BS 5950-3: 1990 Clause 4.6)

In normal composite construction, a relatively thin concrete floor slab acts as thecompression flange of the composite beam. The longitudinal compressive bending stresses

in the slab cause shear stresses in the plane of the slab as shown in Figure 1.

b b

B

Figure 1 Shear stresses in a composite beam

The shear stresses cause shear strains in the plane of the slab. One effect of these shear

strains is that the areas of slab further from the steel beams are not as effective at resisting

longitudinal bending stresses as the areas close to the steel beams. This effect is called

shear lag. As a result, the longitudinal bending stress across the width of the slab is notconstant, see Figure 2. The longitudinal stress tends to be a maximum over the web of the

steel section, and reduces non-uniformly away from the centre-line of the beam.

In order that simple engineers bending theory may be applied (i.e., plane sections remain

plain in bending), the effective width concept is introduced. The section properties are

calculated using the effective width, Be, which is assumed to carry a uniform stress across

the width Be. The value of the stress in the concrete calculated using these effective section

properties is equal to the maximum stress resulting from the effects of shear lag in the actual

slab.

Advisory Desk Notes - AD 266: Shear Connection in composite beams

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Advisory Desk Note


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A G K F

H J

DC E

O

Mean bending stressin concrete flange,

x

Be

b b

B

xmax

Figure 2 Use of effective width to allow for shear lag

The effective width is defined mathematically by the following equation:

dyBb

x

xe

=

0max

2 (1)

The above equation allows the actual flange width Bto be replaced by an effective width Be,

such that the area GHJK equals the area ACDEF. Previous research, based on elastic theory,

has shown that the ratio of Be/Bdepends in a complex way on:

the ratio of Bto the span L

the type of loading

the boundary conditions at the supports

other variables.

The results of this work have been simplified and incorporated in BS5950-3:1990

Clause 4.6. According to this code of practice, the total effective breadth of the concreteflange Be should be taken as the sum of the effective breadths of the portions of flange be

each side of the centre-line of the steel beam. In the absence of any more accurate

determination, the effective breadth of each portion may be taken to be:

a) for a slab spanning perpendicular to the beam, b/Lb ze >/= 8

b) for a slab spanning parallel to the beam b./Lb ze 808>/=

where Lz

is the distance between points of zero moment (taken as the span L for simply-

supported beams) and b is the actual breadth of each portion of the concrete flange (taken as

the lesser of half the beam spacing and, when the beam is adjacent to a free edge, thedistance between the centre-line of the web and the free edge).


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Advisory Desk Note


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It should be noted that unless a reduction in composite action has been justified by tests or

numerical analyses, it is not appropriate to consider a smaller effective width than given by

the expressions shown above. This is because an underestimate of the effective width of the

concrete flange will result in unsafe designs for the shear connectors. For this particular

reason, the values of the effective breadth given in Eurocode 4 are generally higher than

those in Eurocode 2 for reinforced concrete T-beams.

Partial shear connection (BS 5950-3: 1990 Clause 5.5)

The basic requirement for shear connectors is that they are capable of maintaining their

design resistance to shear at large slips, to enable the composite beam to have sufficient

rotation capacity to develop its full design bending resistance. The ductility of a shear

connector is defined by its slip capacity, which is established from a standard push test, andis defined by the maximum slip that the connector can resist while still maintaining its design

resistance. Provided that studs have a slip capacity greater than required in a beam design,

they are considered to be ductile, allowing a plastic distribution of force to be assumed at

the shear connection. This means that the studs may be spaced equally along the beam andassumed to be equally loaded under flexural failure of the beam. Note that the deformations

of the connectors will not be equal, but will be much greater at the ends of the beam than at

midspan.

From extensive numerical analyses and full-scale beam tests, it has been shown that, to

enable a beam to develop its full bending resistance, the slip required increases with the

beam span and the degree of shear connection. In addition, for steel sections that have a

bottom flange area greater than the top flange (i.e., asymmetric sections), the slip required

increases further, due to the neutral axis lying further down within the steel section.

Rather than stating slip capacities directly, the current codes of practice allow designers toassume a plastic distribution of force at the shear connection by specifying minimum degrees

of shear connection in terms of the beam span and, in the case of Eurocode 4, the degree of

asymmetry of the steel section. These code rules are based on numerical studies of

composite beams that considered the slip capacity of the shear connection explicitly.

According to BS5950-3:1990, for a steel beam with equal flanges, the following relationship

for the degree of shear connection should be satisfied:

For spans up to 10 m 40.N/N pa (2)

For spans between 10 and 16 m 10/)6(/ LNN pa but 40.N/N pa (3)

where Nais the actual number of shear connectors provided,

Np is the number of shear connectors required for full shear connection and

L is the beam span in metres.

The partial shear connection rules given in Eurocode 4 are applicable for much larger spans

and additional guidance is given for steel sections with unequal flanges (provided that the

bottom flange area does not exceed 3-times the upper flange area). In these cases, ductile

connectors are defined as those with a characteristic slip capacity of 6 mm. Although

BS5950-3: 1990 and Eurocode 4 give different expressions for partial shear connection, they


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Advisory Desk Note


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are identical in one respect viz. the degree of shear connection provided in a composite beamshould not be less than 0.4.

Transverse reinforcement (BS 5950-3:1990 Clause 5.6)

In composite beams, the longitudinal shear force that has to be transferred between the steelbeam and the concrete flange is dependent on whether full shear connection or partial shear

connection is provided. If the shear connectors are ductile (see requirements for partial

shear connection above), a plastic distribution of force may be assumed at the shear

connection. For full shear connection, the magnitude of this longitudinal shear force V isequal to lesser of either the cross-sectional resistance of the concrete flange or the cross-

sectional resistance steel section. For partial shear connection, the longitudinal shear force is

equal to the resistance of the shear connection (i.e., the design resistance of the stud

multiplied by the number of connectors provided between the support and the critical section

under consideration).

For composite beams using composite or solid slabs, BS5950-3:1990 requires that sufficienttransverse reinforcement should be provided to resist the longitudinal shear force V, toprevent longitudinal splitting of the concrete flange along the potential shear planes shown in

Figure 3.

A b

a

a b b

A t

a) Solid slab

te

eSheeting

A

b) Composite slab with the sheetings annin er endicular to the beam

c) Composite slab with the sheetings annin arallel to the beam

tA e

e

e

eLap jointin sheeting

Figure 3 Potential shear planes according to BS5950-3:1990

For typical internal composite beams that are equally spaced, the effective breadth of each

portion of the concrete flange be is equal. In this case, the longitudinal shear force in each

portion of the concrete flange is equal, and the longitudinal shear force that has to be

transferred along potential shear planes such as a-a and e-e has a value of V/2. Due to thefact that it is normal to assume a plastic distribution of force at the shear connection, it is

often more convenient to work in terms of a longitudinal shear force per unit length v, inwhich case the following equation may be used:


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Advisory Desk Note


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s/Vv 2= (4)

where Vis the longitudinal shear force, and is:

for full shear connection, taken as either the lesser of resistance of the concreteflange or the steel section,

for partial shear connection, taken as the resistance of the shear connection

and s is the spacing of the shear connectors.

However, for cases when composite beams are not spaced equally, or when a beam is

adjacent to a free edge (such as at a hole in the slab), the longitudinal shear force per unit

length along potential shear planes either side of the beam is no longer equal. This design

case is illustrated in Figure 4.

f

f

g

g

bb1 2

Be

Figure 4 Composite beam with an unsymmetrical concrete flange

For the special case shown in Figure 4, the longitudinal shear force per unit length along

shear planes f-f and g-g may be calculated from the following expressions:

sB/Vbv eff 2= (5)

sB/Vbv egg 1= (6)

For further information contact:

Dr Stephen Hicks, SCI.

Tel.: 01344 623345

E-mail: [email protected]


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SCI AD266