Nivel Estatal

Clause 6.2.2 Clause 6.2.2 is for beams without web encasement. The whole of the vertical shear is usually assumed to be resisted by the steel section, as in previous codes for composite beams. This enables the design rules of EN 1993-1-1, and EN 1993-1-530_{where necessary, to be used. The}

assumption can be conservative where the slab is in compression. Even where it is in tension and cracked in flexure, consideration of equilibrium shows that the slab must make some contribution to shear resistance, except where the reinforcement has yielded. For solid slabs, the effect is significant where the depth of the steel beam is only twice that of the slab,50_but

diminishes as this ratio increases.

Clause 6.2.2.3(2)

In composite plate girders with vertical stiffeners, the concrete slab can contribute to the anchorage of a tension field in the web,51_{but the shear connectors must then be designed for}

vertical forces (clause 6.2.2.3(2)). The simpler alternative is to follow Eurocode 3, ignoring both the interaction with the slab and vertical tension across the interface.

Bending and vertical shear

Clause 6.2.2.4 Clause 6.2.2.4(1)

The methods of clause 6.2.2.4 are summarized in Fig. 6.8. Shear stress does not significantly reduce bending resistance unless the shear is quite high. For this reason, the interaction may be neglected until the shear force exceeds half of the shear resistance (clause 6.2.2.4(1)).

Clause 6.2.2.4(2)

Both EN 1993-1-1 and EN 1994-1-1 use a parabolic interaction curve. In clause 6.2.2.4(2) the reduction factor for the design yield strength of the web is (1 – ρ), where

ρ = [(2VEd/VRd) – 1]2 (6.5)

and VRdis the resistance in shear. For a design shear force equal to VRd, the bending resistance

is that provided by the flanges alone, denoted Mf, Rd. This is calculated in Example 6.2.

The bending resistance at VEd= 0 may be the elastic or the plastic value, depending on

the class of the cross-section. Where it is reduced to Mb, Rdby lateral–torsional buckling,

interaction between bending and shear does not begin until a higher shear force than VRd/2 is

present, as shown in Fig. 6.8(a).

Where the shear resistance VRdis less than the plastic resistance to shear, Vpl, Rd, because of

shear buckling, clause 6.2.2.4(2) replaces Vpl, Rdby the shear buckling resistance Vb, Rd.

Where the design yield strength of the web is reduced to allow for vertical shear, the effect on a Class 3 section in hogging bending is to increase the depth of web in compression. If the change is small, the hole-in-web model can still be used, as shown in Example 6.2. For a higher shear force, the new plastic neutral axis may be within the top flange, and the hole-in-web method is inapplicable.

Clause 6.2.2.4(3)

The section is then treated as Class 3 or 4, and clause 6.2.2.4(3) applies. It refers to EN 1993-1-5. For beams, the rule given there is essentially

MEd/MRd+ (1 – Mf, Rd/Mpl, Rd)(2VEd/VRd– 1)2£ 1 (D6.8)

These symbols relate to the steel section only. For a composite section, longitudinal stresses are found by elastic theory. These lead to values of MEd,NEd and VEd acting on the steel

section, which is then checked to EN 1993-1-5. It is fairly easy to check if a given combination of these action effects can be resisted – but calculation of bending resistance for a given vertical shear is difficult.

Example 6.2: resistance to bending and vertical shear

Vertical shear is more likely to reduce resistance to bending in a continuous beam than in a simply-supported one, and it is instructive to consider its influence on a beam with bending resistance found by the hole-in-web method. Where the web is not susceptible to shear buckling, the application of clause 6.2.2.4 is straightforward. This example is therefore based on one of the few UB sections where web buckling can occur, if S355 steel

Bending resistance 0.5 VEd/VRd 0 Mf, Rd Mb, Rd Mel, Rd or Mpl, Rd (a) (b) 132 180 100 48 102 102 108 43 326 118 100 213 213 ₃₂₈ fyd Mpl, a, fl

VRd is the lesser of Vpl, Rd and Vb, Rd

1.0

is used. It is the section used in Example 6.1, 406 × 140 UB39, shown in Fig. 6.6, with longitudinal reinforcement of area As= 750 mm2.

The bending resistance will be calculated when the design vertical shear is VEd= 300 kN.

All other data are as in Example 6.1.

From EN 1993-1-1, resistance to shear buckling must be checked if hw/tw> 72ε/η, where

η is a factor for which EN 1993-1-5 recommends the value 1.2. These clauses are usually

applied to plate girders, for which hwis the clear depth between the flanges. Ignoring the

corner fillets of this rolled section, hw= 381 mm, and for S355 steel, ε = 0.81, so

hwη/twε = 381 × 1.2/(6.3 × 0.81) = 90.5

The resistance of this unstiffened web to shear buckling is found using clauses 5.2 and 5.3 of EN 1993-1-5, assuming a web of area hwtw, that there is no contribution from the

flanges, and that there are transverse stiffeners at the supports. The result is

Vb, Rd= 475 kN

which is 85% of Vpl, Rd, as found by the method of EN 1993-1-1 for a rolled I-section. From

clause 6.2.2.4(2),

ρ = [(2VEd/VRd) – 1]2= (600/475 – 1)2= 0.068

The reduced yield strength of the web is (1 – 0.068) × 355 = 331 N/mm2_{. For hogging}

bending, the tensile force in the reinforcement is

Fs= 750 × 0.5/1.15 = 326 kN (D6.9)

From plastic theory, the depth of web in compression that is above the neutral axis of the I-section is

326/(2 × 0.331 × 6.3) = 78 mm

This places the cross-section in Class 3, so the hole-in-web method is applied. From Fig. 6.6(a), the depth of the compressive stress-blocks in the web is 20tε. The value for ε should be based on the full yield strength of the web, not on the reduced yield strength, so each block is 102 mm deep, as in Fig. 6.6(c). The use of the reduced yield strength would increase ε and so reduce the depth of the ‘hole’ in the web, which would be unconservative. However, the force in each stress block should be found using the reduced yield strength, and so is now

228 × 331/355 = 213 kN

The tensile force in the web is therefore 2 × 213 – 326 = 100 kN

and the longitudinal forces are as shown in Fig. 6.8(b). From Example 6.1,

Mpl, a, flanges= 183.5 kN m

Taking moments about the bottom of the slab,

Mpl, Rd= 183.5 + 326 × 0.1 + 213(0.118 + 0.328) – 100 × 0.043

Mpl, Rd= 307 kN m (D6.10)

For this cross-section in bending only, the method of Example 6.1, with As= 750 mm2,

gives

Mpl, Rd= 314 kN m

The alternative to this method would be to use elastic theory. The result would then depend on the method of construction.

6.3. Resistance of cross-sections of beams for buildings with