The question of adding reinforcement areas is only raised for the shear plane transverse reinforcement layer is provided for both the
longitudinal shear flow.
For the longitudinal slab section above the steel main girder, the minimum required by the transverse bend
5.2.2.3). The minimum reinforcement area 14.9 cm²/m.
In general terms, it should be verif
Asup ≥ Aflex,sup
Ainf ≥ Aflex,inf
Ainf + Asup ≥ max { Ashear ;
For the design example, the criterion is satisfied
Aflex,sup = 18.1 cm 2
/m ; Afle
Ashear/2 + Aflex,sup = 14.9/2
Ashear/2 + Aflex,inf = 14.9/2
Ashear = 10.8 ≤ Ainf + Asup
If these conditions are not satisfied, a more refined method given in annex MM may be used.
5.2.2.10 ULS of fatigue under transverse bending
For this verification, the slow lane is assumed to be close to the parapet and the fatigue load is centred on this lane.
Fig.
Fatigue load model FLM3 is used. Verification method (EN1992-1-1, 6.8.5 and EN1992
085 = 1.29 MPa ≤ 6.02 MPa
The question of adding reinforcement areas is only raised for the shear plane a
transverse reinforcement layer is provided for both the transverse bending moment and the
For the longitudinal slab section above the steel main girder, the minimum reinforcement area required by the transverse bending assessment at ULS is equal to 18,1 cm²/m (see paragraph
reinforcement area Ashear required by the longitudinal shear flow is equal to
In general terms, it should be verified that:
; Ashear/2 + Aflex,sup ; Ashear/2 + Aflex,inf }
, the criterion is satisfied:
flex,inf = 0 ; Ashear = 14.9 cm 2 /m 2 + 18.1 = 25.6 ≤ Ainf + Asup = 30.3 cm 2 /m = 7.5 ≤ Ainf + Asup = 30.3 cm 2 /m = 30.3 cm2/m
If these conditions are not satisfied, a more refined method given in annex MM may be used.
ULS of fatigue under transverse bending
verification, the slow lane is assumed to be close to the parapet and the fatigue load is
Fig. 5.12 Position of the slow lane
Fatigue load model FLM3 is used. Verifications are performed by the damage equivalent stress range 1, 6.8.5 and EN1992-2, Annex NN).
a-a where the upper transverse bending moment and the
reinforcement area Aflex,sup
cm²/m (see paragraph required by the longitudinal shear flow is equal to
If these conditions are not satisfied, a more refined method given in annex MM may be used.
verification, the slow lane is assumed to be close to the parapet and the fatigue load is
Fig. 5.13 Load model FLM3 (axle loads 120 kN) – Variation of bending moment.
The variation of the transverse bending moment in the section above the main steel girder during the passage of FLM3 is calculated using a finite element model of the slab. The moment variation is equal to 39 kNm/m. The corresponding stress range in the reinforcement is calculated assuming a cracked cross section (even if under permanent loads, the section may be considered as uncracked):
∆σs(FLM3) = 63 MPa
The damage equivalent stress range method is defined in EN1992-1-1, 6.8.5 and the procedure for road traffic load on bridges is detailed in EN1992-2, Annex NN.
Adequate fatigue resistance of the reinforcing (or prestressing) steel should be assumed if the following expression is satisfied:
( )
σ
( )
γ
= ∆σ
∗≤∆
γ
∗ , , , Rsk F fat S eq F fatN
N
where:∆σRsk(N*) is the stress range at N* cycles from the appropriate S-N curve in Figure
6.30. For reinforcement made of straight or bent bars, ∆σRsk(N*) = 162.5 MPa
(EN1992-1-1, table 6.3N)
∆σS,eq(N*) is the damage equivalent stress range for the reinforcement and considering
the number of loading cycles N*.
γF,fat is the partial factor for fatigue load (EN1992-1-1, 2.4.2.3). The recommended value is 1.0
γs,fat is the partial factor for reinforcing steel (EN1992-1-1, 2.4.2.4). The
recommended value is 1.15.
The equivalent damage stress range is calculated according to EN1992, Annex NN, NN.2.1: ∆σs,equ = ∆σs,Ec.λs
where
o ∆σs,Ec = ∆σs(1.4 FLM3) is the stress range due to 1.4 times FLM3. In the case of
pure bending, it is equal to 1.4∆σs(FLM3). For a verification of fatigue on intermediate
supports of continuous bridges, the axle loads of FLM3 are multiplied by 1.75. o λs is the damage coefficient.
λs = ϕfat.λs,1. λs,2. λs,3. λs,4
where
o λs,1 takes account of the type of member and the length of the influence line or
surface
o λs,2 takes account of the volume of traffic
o λs,3 takes account of the design working life
o λs,4 takes account of the number of loaded lanes.
Fig. 5.14 λλλλs,1 value for fatigue verification (EN1992-2, Figure NN.2)
λs,1 is given by figure NN.2, curve 3c). In the design example, the length of the influence line is 2,5 m.
Therefore, λλλλs,1 ≈≈≈ 1.1 ≈ λs,2 = Nobs 2.0 k2 (expression NN.103) where
o Nobs is the number of lories per year according to EN1991-2, Table 4.5
o k2 is the slope of the appropriate S-N line to be taken from Tables 6.3N and
6.4N of EN1992-1-1
o is a factor for traffic type according to Table NN.1 of EN1992-2
Fig. 5.15 Factors for traffic types (Table NN.1 of EN1992-2)
For the design example: k2 = 9 (Table 6.3N); Nobs = 0.5x10 6
(EN1991-2, Table 4.5), assuming medium distance traffic: = 0.94
Finally: λλλλs,2 = 0.81
λ λ λ
λs,3 = 1 (design working life = 100 years)
λ λ λ
λs,4 = 1 (different from 1 if more than one lane are loaded)
ϕ ϕ ϕ
ϕfat = 1.0 except for the areas close to the expansion joints where ϕfat = 1.3
It comes:
λs = 0.89 (1.16 near the expansion joints)
∆σs,Ec = 1.4x63 = 88 MPa
Then:
∆σs,equ = 78 MPa (102 MPa near the expansion joints)
∆σRsk/γs,fat = 162.5/1.15 = 141 MPa > 102 MPa
The resistance of reinforcement to fatigue under transverse bending is verified. Generally, for transverse bending of road bridge slabs, ULS of resistance is more unfavourable than ULS of fatigue.
Note
In addition, EN1992-2, 6.8.7, requires fatigue verification for concrete under compression. The verification should be made using traffic data (6.8.7(101)) or by a simplified method (6.8.7(2)). In this case, the condition to satisfy is:
σc,max/fcd,fat ≤ 0.5 + 0.45 σc,min/fcd,fat (Expression 6.77) where σc,max and σc,min are the
maximum and minimum compressive stresses in a fibre under frequent combination of actions. fcd,fat is
the design fatigue strength of concrete, given by Expression 6.76:
fcd,fat = k1βcc(t0)fcd(1 – fck/250) where k1 = 0.85 (recommended value) and t0 is the
time at the beginning of the cyclic loading.
For the design example, depending on t0, βcc(t0) is lying between 1,1 and 1,2 and fcd,fat is between 16
and 17.5 MPa. The maximum and minimum compressive stresses on the lower fibre – calculated in the cracked section with a modular ratio equal to 5.9 – are 11.9 MPa and 3.5 MPa. The condition is not satisfied. However, this condition is very conservative and does not represent the effects of cyclic traffic load: the effects of the frequent traffic loads are much more aggressive than those of fatigue traffic loads. Moreover, the design fatigue strength is based on fcd, calculated with the recommended
value of αcc, equal to 0.85 in EN1992-2. It seems to be no reason to take this value – relevant for long
term loading – for fatigue verifications (with αcc = 1 and βcc(t0) = 1,2 the condition given by Expr.6.77 is
satisfied).
For concrete fatigue verification, there is no method of damage equivalent stress range as for reinforcement. Such a method, intermediate between the rough and conservative condition of Expr.6.77 and a more sophisticated method using traffic data, would be useful.
5.2.3 LONGITUDINAL REINFORCEMENT VERIFICATIONS
5.2.3.1 Resistance for local bending – Adding local and global bending effect
The local longitudinal bending moment at ULS in the middle of the concrete slab – halfway between the structural steel main girders – is equal to Mloc = 90 kN.m/m. It causes compression in the upper
longitudinal reinforcement layer (just below the contact surface of a wheel, for example).
The internal forces and moments from the longitudinal global analysis at ULS cause tensile stresses in the reinforcement for the composite cross-section at support P1 which are equal to σs,sup = 190
MPa in the upper layer and to σs,inf = 166 MPa in the lower layer (see Figure 6.6 in the chapter
“Composite bridge design”). The corresponding values for the internal forces and moments in the concrete slab are:
Nglob = As,supσs,sup + As,infσs,inf
= 24.2 cm2/m x 190 MPa + 15.5 cm2/m x 166 MPa = 715 kN/m
Mglob = – As,supσs,sup(h/2 – dsup) + As,infσs,inf(dinf – h/2)
= –24.2 cm²/m x 171 MPa x (308/2 – 60) mm + 15.5 cm²/m x 149 MPa x (240 – 308/2) mm = – 21 kNm/m
(dsup and dinf are the distance of the upper layer – resp. lower layer – to the top face of the
slab)
The longitudinal reinforcement around support P1 should be designed for these local and global effects. The local (Mloc) and global (Nglob and Mglob) effects should be combined according to Annex E
to EN1993-2. The following combinations should be taken into account: (Nglob + Mglob) + ψ Mloc and Mloc + ψ (Nglob + Mglob)
where ψ is a combination factor equal to 0.7 for spans longer than 40 m.