The design example in the studied slab section above the steel main girder gives successively:
fck = 35 MPa
CRd,c = 0.12
k = 1 +
200360 = 1.75
Asl = 1848 mm2 (high bond bars with diameter of 20 mm and spacing of 170 mm).
bw = 1000 mm ρl =
1848
1000x360
= 0.51% CRd,ck(100ρlfck) 1/3 = 0.55 MPa σcp = 0 vmin = 0.035x1.75 3/2 x351/2 = 0.48 MPa < 0.55 MPa The shear resistance without shear reinforcement is:VRd,c = CRd,ck(100ρlfck) 1/3
bwd = 198 kN / m < VEd = 235 kN / m.
According to EN1992-1-1, shear reinforcement is needed in the slab, near the main girders. With vertical shear reinforcement, the shear design is based on a truss model (EN1992-1-1 and EN1992-2, 6.2.3, fig. 6.5) where α is the inclination of the shear reinforcement and θ the inclination of the struts:
Fig. 5.8 Truss model (fig. 6.5 of EN1992-1-1)
With vertical shear reinforcement (α =π/2), the shear resistance VRd is the smaller value of:
VRd,s = (Asw/s) z fywd cotθ and
VRd,max = αcw bw z ν1 fcd/(cotθ + tanθ)
where:
o z is the inner lever arm (z = 0.9d may normally be used for members without axial force) o θ is the angle of the compression strut with the horizontal, must be chosen such as
1 ≤ cotθ ≤ 2.5
o Asw is the cross-sectional area of the shear reinforcement o s is the spacing of the stirrups
o fywd is the design yield strength of the shear reinforcement
o ν1 is a strength reduction factor for concrete cracked in shear, the recommended value
o αcw is a coefficient taking account of the interaction of the stress in the compression and any applied axial compressive stress
prestressed members. In the design example, choosing 1-m-wide slab strip:
VRd,s = 0.00068x(0.9x0.36)x435x2
VRd,max = 1.0x1.0x(0.9x0.
Note
In EN1992-1-1, the shear resistance without shear reinforcement is the same for beams and for slabs. This does not take account of the 2
redistribution. The calibration of the expression of
For these reasons,vmin has been modified by the French National Annex to EN1992
vmin = 0.035 k 3/2
fck 1/2
for beam elements vmin = (0.34/γc) fck
1/2
for slab elements where transverse redistribution of loads is possible is based on French experience. Such a difference already existed in former French
expression: vmin = (0.34/1.5).351/2 = 1.34 MPa > 0.55 MPa
reinforcement in the slab.
5.2.2.8 Resistance to longitudinal shear stress
The longitudinal shear force per unit length at the steel/concrete int
analysis at characteristic SLS and at ULS. The number of shear connectors resist to this shear force per unit length and thus to ensure the longitudinal deck.
Fig. 5.9 Shear force per unit length resisted by the studs (MN/m)
At ULS this longitudinal shear stress should also be resisted to for any potential shear failure within the slab. This means that the
is a coefficient taking account of the interaction of the stress in the compression and any applied axial compressive stress; the recommended value of
prestressed members.
In the design example, choosing cotθ = 2.5, with a shear reinforcement area Asw
36)x435x2.5 = 240 kN/m > VEd
.36)x0.6x(1 – 35/250)x35/(2.5+0.4) = 2.02 MN/m >
1, the shear resistance without shear reinforcement is the same for beams and for slabs. This does not take account of the 2-dimensional behaviour of slabs and of the possibility of transverse redistribution. The calibration of the expression of vmin is based on tests made on beam elements only.
has been modified by the French National Annex to EN1992 r beam elements (it is the recommended value)
for slab elements where transverse redistribution of loads is possible is based on French experience. Such a difference already existed in former French
5).351/2 = 1.34 MPa > 0.55 MPa and there is no need to add shear
Resistance to longitudinal shear stress
The longitudinal shear force per unit length at the steel/concrete interface is determined by an elastic SLS and at ULS. The number of shear connectors is
resist to this shear force per unit length and thus to ensure the longitudinal composite behaviour of the
Shear force per unit length resisted by the studs (MN/m)
At ULS this longitudinal shear stress should also be resisted to for any potential surface of longitudinal shear failure within the slab. This means that the reinforcing steel bars holing such kind
is a coefficient taking account of the interaction of the stress in the compression chord ; the recommended value of αcw is 1 for non
sw/s = 6.8 cm 2
/m for a
02 MN/m > VEd
1, the shear resistance without shear reinforcement is the same for beams and for slabs. onal behaviour of slabs and of the possibility of transverse is based on tests made on beam elements only. has been modified by the French National Annex to EN1992-1-1 :
for slab elements where transverse redistribution of loads is possible. This is based on French experience. Such a difference already existed in former French code. With this and there is no need to add shear
determined by an elastic designed thereof, to composite behaviour of the
Shear force per unit length resisted by the studs (MN/m)
surface of longitudinal reinforcing steel bars holing such kind of surface
should be designed to prevent any shear failure of the concrete or any longitudinal splitting within the slab.
Two potential surfaces of shear failure are defined in EN1994-2, 6.6.6.1 (see Fig. 5.10(a)):
o surface a-a holing only once by the two transverse reinforcement layers, As = Asup + Ainf o surface b-b holing twice by the lower transverse reinforcement layer, As = 2.Ainf
(a) potential surfaces of shear failure (b) shear resistance for the shear plane a-a
Fig. 5.10 Potential surfaces of shear failure in the concrete slab
According to Fig. 5.9, the maximum longitudinal shear force per unit length resisted to by the shear connectors is equal to 1.4 MN/m. This value is used here for verifying shear failure within the slab. The shear force on each potential failure surface is proportional to the first moment of area, about the centre of gravity of the composite section, of the part of the slab outside the failure surface. For a nearly horizontal concrete slab, it can then be considered that the shear force applied on a potential failure surface is proportional to the area of the part of the concrete section situated outside this surface. The shear force on each potential failure surface (see fig. 5.10(a)) is as follows:
o surface a-a, on the cantilever side : 0.59 MN/m o surface a-a, on the central slab side : 0.81 MN/m o surface b-b : 1.4 MN/m