1.4 ¿Qué es un modelo de Gestión?
1.5 Integración de la gestión del conocimiento con el cuadro de mando integral
In the concrete slab of a composite bridge, around an internal support, there is no tension in the transverse direction but the tensile stress is very high in the longitudinal direction (about -9 MPa for the design example). This gives thus:
σcp = max (σc,long/2 ; – 1.85) = - 1.85 MPa
The values for CRd,c and k1 can be provided by the National Annex to EN1994-2. The recommended
values are (EN1994-2, 6.2.2.5(3), note): o CRd,c = 0.15/γc = 0,10
o k1 = 0.12
These values are different from those recommended by EN1992-1-1 (0.18/γc and 0.1).It will be seen
that the note in EN1994-2, 6.2.2.5(3), only relates to concrete flanges in tension (σcp < 0) as part of a
steel/concrete composite structural beam, which is the case here in the longitudinal direction. In case of a concrete slab under bending moment or compression, the values for CRd,c and k1 would have
been provided by the National Annex to EN1992-1-1. See also paragraph 5.2.2.7. o vmin = 0.035 k 3/2 fck 1/2 5.2.4.2 Design example
The vertical load induced by the single wheel of the traffic load model LM2 is equal to:
VEd = γQ βQ Qak/2 = 1.35 x 1.00 x 400/2 = 270 kN
Its contact surface is a rectangular area of 0.35 x 0.6 m².
To calculate the depth d, the wheel of LM2 is put along the outside edge of the pavement on the cantilever part of the slab. The centre of gravity of the load surface is therefore at 0.5 + 0.6/2 = 0.8 m from the free edge of the slab. At this location, the slab thickness to consider is equal to 0.30 m. Hence:
d = 0.5.[(0.30-0.035-0.016/2) + (0.30-0.035-0.016-0.016/2)] = 0.249 m
The reference control perimeter is defined following the contact surface dimensions.
u1 = 2(0.35+0.6+4x0.11) + 4πd = 5.91 m is obtained.
The shear stress along this control perimeter is then equal to:
vEd = β
V
Edu
1d
= 0.18 MPa (with β = 1)
The design value of the resistance to punching shear is as follows:
l ly lz 0394%.052%045% k = 1 +
200
249
= 1.90 ≤ 2.0 σcp = - 1.85 MPa CRd,c = 0.10 k1 = 0.12 CRd,ck(100ρlfck) 1/3 = 0.48 MPa vmin = 0.035 x 1.90 3/2 x 351/2 = 0.54 MPa > 0.48 MPa vRd,c = vmin + k1σcp = 0.32 MPaThe punching shear is thus verified:
vEd = 0.18 MPa ≤ vRd,c = 0.32 MPa.
There is no need to add shear reinforcement in the concrete slab.
5.3 Second order effects in the high piers
5.3.1 MAIN FEATURES OF THE PIERS
Pier height : 40 m Pier shaft o external diameter : 4 m o wall thickness : 0.40 o longitudinal reinforcement: 1.5% o Ac = 4.52 m 2 o Ic = 7.42 m 4 o As = 678 cm 2 o Is = 0.110 m4 Pier head: o volume : 54 m 3 o Weight : 1.35 MN Concrete o C35/45 o fck = 35 MPa Ecm = 34000 MPa
Ic is the uncracked inertia of the pier shaft, Is is the second moment of area of the reinforcement about
the centre of area of the concrete cross-section.
5.3.2 FORCES AND MOMENTS ON TOP OF THE PIERS
Forces and moments on top of the piers are calculated assuming that the inertia of the piers is equal to 1/3 of the uncracked inertia.
Two ULS combinations are taken into account:
o Comb 1: 1.35G + 1.35(UDL + TS) + 1.5(0.6FwkT) (transverse direction)
o Comb 2: 1.35G + 1.35(0.4UDL+ 0.75TS + braking) + 1.5(0,.6Tk) (longitudinal direction)
Table 5.3 Forces and moments on top of piers
Fz (vertical) Fy (trans.) Fx (long.) Mx (trans.)
G 14.12 MN 0 0 0 UDL 3.51 MN 0 0 8.44 MN.m TS 1.21 MN 0 0 2.42 MN.m Braking 0 0 0.45 MN 0 FwkT(wind on traffic) 0 0.036 MN 0 0.11 MN.m Tk 0 0 0.06 MN 0 Comb 1 25.43 MN 0.032 MN 0 14.76 MN.m Comb 2 22.18 MN 0 0.66 MN 7.01 MN.m
5.3.3 SECOND ORDER EFFECTS
The second order effects are analysed by a simplified method: EN1992-1-1, 5.8.7 - method based on nominal stiffness. For the design example, the analysis is performed only in longitudinal direction.
Geometric imperfection (EN1992-2, 5.2(105)):
o θl = θ0αh where θ0 = 1/200 (recommended value) αh = 2/l 1/2 ; αh ≤ 1
l is the height of the pier = 40 m
o θl = 0.0016 resulting in a moment under permanent combination
M0Eqp = 1.12 MNm at the base of the pier
First order moment at the base of the pier:
o M0Ed = 1.35 M0Eqp + 1.35 Fz (0.4UDL + 0.75TS) l θl + 1.35 Fx(braking) l +1.5(0.6Fx(Tk) l
M0Ed = 28.2 MNm
ϕef = ϕ(∞,t0).M0Eqp / M0Ed
where:
ϕ(∞,t0) is the final creep coefficient according to 3.1.4
M0Eqp is the first order bending moment in quasi-permanent load combination (SLS)
M0Ed is the first order bending moment in design load combination (ULS)
ϕef = 2 (1.12/28.2) = 0.08
Nominal stiffness (EN1992-1-1, 5.8.7.2 (1)
EI = KcEcdIc + KsEsIs (Expression 5.21)
where:
Ecd is the design value of the modulus of elasticity of concrete, see 5.8.6 (3)
Ic is the moment of inertia of concrete cross section
Es is the design value of the modulus of elasticity of reinforcement, 5.8.6 (3)
Is is the second moment of area of reinforcement, about the centre of area of
the concrete
Kc is a factor for effects of cracking, creep etc, see 5.8.7.2 (2) or (3)
Ks is a factor for contribution of reinforcement, see 5.8.7.2 (2) or (3)
The expression (5.21) is the sum of the full stiffness of the reinforcement (Ks = 1) and of the reduced
stiffness of the concrete, function of the axial force and of the slenderness of the pier. In the design example:
Ecd= Ecm/γcE = 34000/1.2 = 28300 MPa Ic = 7.42 m 4 Es = 200000 MPa Is = 0.110 m4 Ks = 1
Kc is given by the following expression
Kc = k1k2 / (1 + ϕef)
where:
ρ is the geometric reinforcement ratio, As/Ac
As is the total area of reinforcement
Ac is the area of concrete section
ϕef is the effective creep ratio, see 5.8.4
k1 is a factor which depends on concrete strength class, Expression (5.23)
k2 is a factor which depends on axial force and slenderness, Expression (5.24)
k1 = fck/20(MPa)
k2 = n⋅
λ
170
≤ 0.20
where:
n is the relative axial force NEd/(Acfcd)
λ is the slenderness ratio, see 5.8.3 In the design example:
ρ = 0.015
NEd = 22.18 MN
n = 22.18/(4.52.19.8) = 0.25
λ = l0/i ; l0 = 1.43.l = 57.20 m (taking into account the rigidity of the second pier) ; i = (Ic/Ac) 0,5
= 1.28 m
l0 is the effective length of the elastic buckling mode. It is calculated taking into account the restraints at the end of the column. Here, the pier is assumed to have a full restraint at the bottom. Due to the presence of the other pier, there is an elastic restraint for the horizontal displacement at the top. This can be modeled by a spring which stiffness is equal to 3EI/l3(taking the same EI for both piers). In the longitudinal direction, the rotation is free at the top of the pier.
λ = 45
k2 = 0.25.(45/170) = 0.066
Kc= 1.32x0.066/1.08 = 0.081
EI = 39200 MN.m2 (≈ EIuncracked/6)
Moment magnification factor (EN 1992-1-1, 5.8.7.3)
(1) The total design moment, including second order moment, may be expressed as a magnification of the bending moments resulting from a linear analysis, namely:
MEd = M0Ed
!
1+β
NBNEd -1
"
(5.28) where:
M0Ed is the first order moment; see also 5.8.8.2 (2)
β is a factor which depends on distribution of 1st and 2nd order moments, see 5.8.7.3 (2)-(3)
NEd is the design value of axial load
NB is the buckling load based on nominal stiffness
(2) For isolated members with constant cross section and axial load, the second order moment may normally be assumed having a sine-shaped distribution. Then
β = π2 / c0 (5.29)
In the design example, c0 = 12, assuming a triangular distribution of the first order moment. Then:
M0Ed = 28.2 MN.m β β β β = 0.85 (c0 = 12) NB = π 2 EI/l0 2 = 118 MN
NEd = 26 MN (mean value on the height of the pier)
Finally, the moment magnification factor is equal to 1.23, and:
MEd = 1.23 M0Ed = 33.3 MN.m
This method gives a safe design of the piers, but it is possible that the longitudinal displacements are overestimated. For a better assessment of the displacements, a general method, based on moment- curvature relationship is necessary.