A VEGETACIÓN LIBRE

critical length might then have insufficient available slip. Use of an additional critical section would lead to a more suitable distribution.

Example 6.4: arrangement of shear connectors

As an example of the use of these rules, a simply-supported beam of span 10 m is considered. It has distributed loading, equal steel flanges, a uniform cross-section in Class 2, S355 steel, and stud connectors. At mid-span, MEd is much less than Mpl, Rd. The

cross-section is such that the required resistance to bending can be provided using 40% of full shear connection (n/nf= 0.4).

Clause 6.6.1.2(1) gives n/nf≥ 0.55. However, if the slab is composite, and the other

conditions of clause 6.6.1.2(3) are satisfied, n/nf= 0.4 may be used.

Suppose now that the span of the beam is 12 m. The preceding limits to n/nf are

increased to 0.61 and 0.48, respectively. One can either design using these limits, or go to

clause 6.6.1.3(5), which refers to ‘longitudinal shear calculated by elastic theory’. This

presumably means using vL= VEdA /I, where VEdis the vertical shear on the composite

section. This gives a triangular distribution of longitudinal shear, or separate distributions, to be superimposed, if creep is allowed for by using several modular ratios. The force in the slab at mid-span now depends on the proportion of MEdthat is applied

to the composite member, and on the proportions of the cross-section. Connectors corresponding to this force are then spaced accordingly, possibly with extra ones near mid-span to satisfy the rule of clause 6.6.5.5(3) on maximum spacing of studs.

Strictly, the envelope of vertical shear should be used for VEd, which gives non-zero

shear at mid-span. For distributed loading this increases the shear connection needed by only 2%, but the envelope should certainly be used for more complex variable loading.

For continuous beams where MEd at mid-span is much less than Mpl, Rd, the method

is more complex, as partial shear connection is permitted only where the slab is in compression. The envelope of design vertical shear from the global analysis should be used. For simplicity, longitudinal shear can be found using properties of the uncracked cross-section throughout, because this gives an overestimate in cracked regions. Examples 6.7 and 6.8 (see below) are relevant.

It does not help, in the present example, to define additional critical sections within the 6 m shear span of the beam, because the limits of clause 6.6.1.2 are given in terms of effective span, not critical length.

6.6.2. Longitudinal shear force in beams for buildings

Clause 6.6.2.1

Clause 6.6.2.2 Clauses 6.6.2.1 and 6.6.2.2 say, in effect, that the design longitudinal shear force should beconsistent with the bending resistances of the cross-sections at the ends of the critical length considered, not with the design vertical shear forces (the action effects). This is done for two reasons:

• simplicity – for the design bending moments often lie between the elastic and plastic resistances, and calculation of longitudinal shear becomes complex

• robustness – for otherwise longitudinal shear failure, which may be more brittle than flexural failure, could occur first.

Beam with Class 3 sections at supports and a Class 1 or 2 section at mid-span

Clause 6.6.2.1 applies because non-linear or elastic theory will have been ‘applied to

cross-sections’. The longitudinal forces in the slab at the Class 3 sections are then calculated by elastic theory, based on the bending moments in the composite section. At mid-span, it is not clear whether clause 6.6.2.2 applies, because its heading does not say ‘resistance of all cross-sections’. The simpler and recommended method is to assume that it does apply, and to calculate the longitudinal force at mid-span based on MRd at that section, as that is

consistent with the model used for bending. The total shear flow between a support and mid-span is the sum of the longitudinal forces at those points. The alternative would be to find the longitudinal force at mid-span by elastic theory for the moments applied to the composite section, even though the bending stresses could exceed the specified limits.

Clause 6.6.2.2(3) This absence of ‘all’ from the heading is also relevant to the use of clause 6.6.2.2(3), on the use of partial shear connection. The design for a beam with Class 3 sections at internal supports limits the curvature of those regions, so the ultimate-load curvature at mid-span will be too low for the full-interaction bending resistance to be reached. The use of partial shear connection is then appropriate, with MRdless than Mpl, Rd.

6.6.3. Headed stud connectors in solid slabs and concrete encasement

Resistance to longitudinal shear

In BS 5950-3-131_{and in earlier UK codes, the characteristic shear resistances of studs are}

given in a table, applicable only when the stud material has particular properties. There was no theoretical model for the shear resistance.

Clause 6.6.3.1(1) essential. Those given in clause 6.6.3.1(1) are based on the model that a stud with a shankThe Eurocodes must be applicable to a wider range of products, so design equations are diameter d and an ultimate strength fu, set in concrete with a characteristic strength fckand a

mean secant modulus Ecm, fails either in the steel alone or in the concrete alone.

The concrete failure is found in tests to be influenced by both the stiffness and the strength of the concrete.

This led to equations (6.18)–(6.21), in which the numerical constants and partial safety factor γVhave been deduced from analyses of test data. In situations where the resistances

from equations (6.18) and (6.19) are similar, tests show that interaction occurs between the two assumed modes of failure. An equation based on analyses of test data, but not on a defined model,70

PRd= k(πd2/4)fu(Ecm/Ea)0.4(fck/fu)0.35 (D6.16)

gives a curve with a shape that approximates better both to test data and to values tabulated in BS 5950.

In the statistical analyses done for EN 1994-1-171,72_{both of these methods were studied.}

Equation (D6.16) gave results with slightly less scatter, but the equations of clause

6.6.3.1(1) were preferred because of their clear basis and experience of their use in some

countries. Here, and elsewhere in Section 6, coefficients from such analyses were modified slightly, to enable a single partial factor, denoted γV(‘V’ for shear), 1.25, to be recommended

for all types of shear connection. This value has been used in draft Eurocodes for over 20 years.

It was concluded from this study72_{that the coefficient in equation (6.19) should be 0.26.}

This result was based on push tests, where the mean number of studs per specimen was only six, and where lateral restraint from the narrow test slabs was usually less stiff than in the concrete flange of a composite beam. Strength of studs in many beams is also increased by the presence of hogging transverse bending of the slab. For these reasons the coefficient was increased from 0.26 to 0.29, a value that is supported by a subsequent calibration study15

based on beams with partial shear connection.

Design resistances of 19 mm stud connectors in solid slabs, given by clause 6.6.3.1, are shown in Fig. 6.12. It is assumed that the penalty for short studs, equation (6.20), does not apply. For any given values of fuand fck,the figure shows which failure mode governs. It can

be used for this purpose for studs of other diameters, provided that h/d ≥ 4.

The ‘overall nominal height’ of a stud, used in equations (6.20) and (6.21), is about 5 mm greater than the ‘length after welding’, a term which is also in use.

100 80 60 20 30 40 50 PRd (kN) Normal-density concrete Density class 1.8 fu = 500 N/mm 2 450 400 fck (N/mm 2 ) fcu (N/mm 2₎ 25 37 50 60

Weld collars