## Critical lengths and cross-sections

It will be show in Section 3.7 that the bending moment at which yielding of steel first occurs in a simply-supported composite beam can be below 70% of the ultimate moment. If the bending-moment diagram is parabolic, then at ultimate load partial yielding of the steel beam can extend over more than half of the span.

At the interface between steel and concrete, the distribution of longitudinal shear is influenced by yielding, and also by the spacing of the connectors, their load/slip properties, and shrinkage and creep of the concrete slab. For these reasons, no attempt is made in design to calculate this distribution. Wherever possible, connectors are uniformly spaced along the span.

It was shown is Section 3.5.3 that this cannot always be done. For beams with all critical sections in Class 1 or 2, uniform spacing is allowed by Eurocode 4 along each critical length, which is a length of the interface between two adjacent critical cross-sections. These are defined as sections of maximum bending moment. supports.

sections subjected to heavy concentrated loads.

Places where there is a sudden change of cross-section of the beam, and free ends of cantilevers.

There is also a definition for tapering members.

Where the loading is uniformly-distributed, a typical design procedure, for half the span of a beam, whether simply-supported or continuous, would be as follows.

(1) Determine the compressive force required in the concrete slab at the section of maximum sagging moment, as explained in Section 3.5.3. Let this be Fc.

(2) Determine the tensile force in the concrete slab at the support that is assumed to contribute to the bending resistance at that section (i.e. zero for a simple support, even if crack-control reinforcement is present; and the yield force in the longitudinal reinforcement, if the span is designed as continuous). Let this force be F1.

(3) If there is a critical cross-section between these two sections, determine the force in the slab at that section. The bending moment will usually be below the yield moment, so elastic analysis of the section can be used.

(4) Choose the type of connector to be used, and determine its design resistance to shear, PRd, as explained in Section 2.5.

(5)The number of connectors required for the half span is

The number required within a critical length where the change in longitudinal force is Rd ΔF / P .

An alternative to the method of step (3) would be to use the shear force diagram for the half span considered. Such a diagram is shown if Fig.3.18 for the length ABC of a span AD, which is continuous at A and has a heavy point load at B. The critical sections are A, B, and C. The total number of connectors is shared between lengths AB and BC in proportion to the area of the shear force diagram, OEFH and GJH.

In practice it might be necessary to provide a few extra connectors along BC, because codes limit the maximum spacing of connectors, to prevent uplift of the slab relative to the steel beam, and to ensure that the steel top flange is sufficiently restrained from local and lateral bucking.

## Dactile and non-ductile connectors

The use of uniform spacing is possible because all connectors have some ductility, of slip capacity.

This term has no standard definition, but is typically assumed to be the maximum slip at which a connector can resist 90% of its characteristic shear resistance, as defined by the falling branch of a load slip curve obtained in a standard push test.

Slip enables longitudinal shear to be redistributed between the connectors in a critical length, before any of them fail. The slip required for this purpose increases at low degrees of shear connection, and as the critical length increases (a scale effect). A connector that is ‘ductile’ (has sufficient slip capacity) for a short span becomes ‘non-ductile’ in a long span, for which a more conservative design method must be used.

The definitions of ‘ductile’ connectors given in Eurocode 4 for headed studs welded to a steel beam with equal flanges are shown in Fig.3.19. The more liberal definition. The push test specified in British codes since 1965 is, in any case, unsuitable for this purpose, because the reinforcement in the slabs is insufficient to prevent longitudinal splitting (Section 2.5).

Where non-ductile connectors are used, design for full shear connection is the same (in Eurocode 4) as for ductile connectors, but for N/Nf<1, the two methods shown in Fig.3.16 are replaced by methods that require higher ratios N/Nf for a given value of Mpl Rd . . These are approximations to elastic behaviour, and so rely less on slip capacity. They are explained in Reforence 15. There is also a rule in Eurocode 4 that limits the use of uniform spacing of non-ductile connectors.

## Transverse reinforcement

The reinforcing bars shown in Fig.3.20 are longitudinal reinforcement for the concrete slab, to enable it to span between the beam shown and those either side of it. They also enhance the resistance to longitudinal shear of vertical cross-sections such as B-B. Bars provided for that purpose are known as ‘transverse reinforcement’, as their direction is transverse to the axis of the composite beam. Like stirrups in the web of a reinforced concrete T-beam, they supplement the shear strength of the concrete, and their behaviour can be represented by a truss analogy.

The design rules for these bars are extensive; as account has to be taken of many types and arrangements of shear connectors, of haunches, of the use of precast or composite slabs, and of interaction between the longitudinal shear per unit length on the section considered, vSd, and the transverse bending moment, show as Ms in Fig.3.20. The loading on the slab also causes vertical shear stress on planes such as B-B; but this is usually so much less than the longitudinal shear stress vsd/Acv on the plane, that it can be neglected.

The notation here is that of Eurocode 4; Part 1.1; Acv is the cross-sectional area per unit length of beam of the concrete shear surface being considered. The word ‘surface’ us used here because EFGH in Fig.3.20, although not a plane, is another potential surface of shear failure. In practice, the rules for minimum height of shear connectors ensure that in slabs of uniform thickness, planes such as B-B are more critical; but this may not be so for haunched slabs, considered later.

The design longitudinal shear per unit length for surface EFGH is the same as that for the shear connection, and in a symmetrical T-beam half of that value is assumed to be transferred through each of planes B-B and D-D. For an L-beam or where the flange of the steel beam is wide (Fig.3.21), the more accurate expressions should be used:

For planes such as B-B in Fig.3.20, the effective area of transverse reinforcement per unit length of beam, Ac, is the whole of the reinforcement that is fully anchored on both sides of the plane (i.e. able to develop its yield strength in tension). This is so even where the top bars are fully stressed by the bending moment Ms, because this tension is balanced by transverse compression, which transverse compression, which enhances the shear resistance in the region CJ by an amount at least equivalent to the contribution the reinforcement would make, in absence of transverse bending.

Effective areas are treated in more detail in Volume 2.

## 3.6.3.1 Design rules for transverse reinforcement in solid slabs

Part of a composite beam is shown is plan in Fig.3.22. The truss model for transverse reinforcement is illustrated by triangle ACE, in which CE represents the reinforcement for a unit length of the beam, Ac, and v is the design shear force per unit length. The force v, applied at some point A, is transferred by concrete struts AC and AE, at 45° to the axis of the beam. The strut force at C is balanced by compression in the slab and tension in the reinforcement. The model fails when the reinforcement yields. The tensile force in it is equal to the shear on a plane such as B-B caused by the force v, so the model gives a design equation of the from