4 . 4 . 2 . 1 – Ch a ra c t e r i st i c s o f t h e s t r u c t u r e.
This is a four-span pre-stressed structure, built as balanced cantilever method, with the main dimensions as shown on figure 4.4.
Figure 4.4: the main dimensions of the studied structure The load paths are summarized in the following table:
For a pier (2 bearings)
C0 and C4 P1 and P3 P2
Permanent loads V max (MN) 2.98 17.88 18.34
V min (MN) 2.87 17.76 18.32
max rot (10-3 rad) 1.1 -0.5 -0.2
min rot 0.8 -0.8 -0.2
Traffic loads V max (MN) 1.57 3.74 3.78
V min (MN) -0.56 -0.66 -0.54
concom rot Vmax (10-3 rad) -2.0 -0.5 0 concom rot Vmin (10-3 rad) 1.3 0.2 0
Thermal gradient V (MN) 0.29 -0.40 0.22
Rot (10-3 rad) 0.6 0.3 0
We show here the linear compressions taken into account for each span, as with the preceding case:
End spans (1) and (4) Central spans (2 and 3)
Shrinkage/creep 10.4 mm 15.6 mm
Uniform temperature 20.8 mm 31.2 mm
Total 31.2 mm 46.8 mm
The displacements are calculated as in paragraph 4.3. If there are sliding bearings, the difficulty lies in the fact that these bearings work up to a certain threshold. Below the sliding force, their flexibility is the same as if they were not sliding.
Above, their flexibility is "infinite". An initial calculation thus needs to be made, taking them as non-sliding bearings and then, if the force corresponding to the displacement goes beyond the threshold, their flexibility needs to be replaced by a corresponding limit force and the balance of horizontal force needs to be recalculated.
Laminated elastomeric bearings –Use on bridges, viaducts and similar structures 4V
4 .4 .2 . 2 – Pr e - d i m e n s io n in g b ea ri n g s
To do this, an initial dimensioning of the bearing needs to be carried out, mainly from vertical force and rotations (that are not dependent on horizontal force).
The following force is obtained for a bearing:
C0 and C4 P1 and P3 P2
V max (MN) 2.35 10.71 11.12
V min (MN) 1.16 8.39 8.87
| max rot | (10-3 rd) 2.4 1.2 2.1
| min rot | (10-3 rd) 1.0 0.6 1.5
We choose (thereby verifying the maximum stress of 30 MPa on the concrete):
C0 and C4 P1 - P2 - P3
1 bearing of 400 x 500 2 bearings of 700 x 600*
* the values calculated for maximum and minimum reactions on piers takes into account a difference in stiffness of ± 15 % for each twinned bearing (cf. note of § 5.3.3.7 of NF EN 1337-3).
Figure 4.5: the lay-out of pier bearings.
For horizontal displacements, calculations are greatly simplified if we consider a point 0 in the geometric centre of the structure. In this case, the maximum displacements to take into account will be:
C0 and C4 P1 and P3 P2 unit
vi max 78 47 0 mm
This calculation immediately shows that abutment bearings –of small size – must be sliding as it would be impossible to stack enough layers of elastomer to absorb the displacement.
With the principles of paragraph 4.2, we arrive at:
C0 and C4 P1 - P2 -P3
1 bearing of 400 x 500; 3 (12 + 3); 2 x 6 2 bearings of 700 x 600; 6 (16 + 4); 2 x 8
5M================Laminated elastomeric bearings – Use on bridges, viaducts and similar structures
4 . 4 . 2 . 3 – Ho r i zo n t a l f o r c e f o r sli d i n g b ea ri n g s
We calculate the flexibility of these bearings, all considered as non-sliding.
• Static
• For dynamic (breaking), flexibility is divided by 2.
Sliding bearings on abutments have a friction coefficient of around 5.3 % on maximum load. Taking into account a maximum vertical force of 2.33 MN (for one bearing), the maximum limit sliding force – per abutment – is:
Hlim = 0.053 x 2 x 2.33 = 0.25 MN 4 . 4 . 2 . 4 - Ho r i zo n t a l f o r c e f o r n o n - sl i d i n g b e a r i n g s
4 . 4 . 2 . 4 . 1 – F o r c e d u e t o l i n e a r v a r i a t i o n s o f t h e d e c k 1s t i t e r a t i o n
An initial calculation is carried out in a similar way to that of paragraph 4.3.
Abutments 0 and
4 Pier 1 Piers 2 and 3
Δstat Δstat Δstat
Bearing 0.1333 0.0741 0.0741
Foundation + shaft - 0.0091 0.0431
Total 0.1333 0.0832 0.1172
Ri = 1/Δ 7.502 12.019 8.532
For maximum deformation, taken as equal to 6 x 10-4 (CP and uniform temperature), the relative displacements di of the bearings in relation to the left-hand support are:
Displacement of support 1 = - 0.031 = - 0.031 m
Laminated elastomeric bearings –Use on bridges, viaducts and similar structures 5N Point 0 is therefore located at the distance x0 = 0 074
0 156, 260 00
, × , = 123.83 m The corresponding horizontal force can be deduced:
C0 P1 P2 P3 C4
displacement M 0.074 0.043 -0.004 -0.051 -0.082
H MN 0.56 0.52 -0.03 -0.43 -0.61
Hlim 0.25 - - - 0.25
We can confirm that the bearings C0 and C4 slide.
2n d i t e r a t i o n
Sliding bearings have zero stiffness. We replace the values of Ri for C0 and C4 by 0. The calculations gets more complicated, however, as we have to successively study three case scenarios, namely:
• The sliding bearings all have the same friction.
• Those of the supports situated on the left of the point 0 have a value equal to the minimum value (§ 4.4.1.3), and maximum for thus situated on the right.
• And the opposite, namely the maximum values for the supports on the left and the minimum for those on the right.
There are 4 sliding bearings on the full number of bearings.
The friction coefficients are therefore (§ 4.4.1.3):
• minimum μa = μmax = 5.3 %
• maximum μr = 0
C a s e 1 – t h e sa m e f ri c t i o n c o e f f i c ie n t v a l u e o n t h e l e f t a n d th e r i g h t
In this case, we replace the value of the product Ri x di by the limit value of H for the supports C0 and C4. The calculation becomes:
C0 P1 P2 P3 C4
Ri 0 12.019 8.532 8.532 0
Sum Ri 29.084
Ri x di or Hlim -0.235 -0.375 -0.666 -1.065 0.235
Sum Hi -2.105
point 0 120.65
displacement 0.072 0.041 -0.006 -0.052 -0.084
H 0.25 0.50 -0.05 -0.45 0.25
Hlim 0.25 - - - 0.25
5O================Laminated elastomeric bearings – Use on bridges, viaducts and similar structures
C a s e 2 - f ri c t i o n c o e ff i c i e n t o n t h e l e f t 0 % a n d o n t h e r i g h t 5 . 3 %
We obtain the following table:
C0 P1 P2 P3 C4
Ri 0 12.019 8.532 8.532 0
Sum Ri 29.084
Ri x di or Hlim 0 -0.375 -0.666 -1.065 0.235
Sum Hi -1.870
point 0 107.18
displacement 0.064 0.033 -0.014 -0.060 -0.092
H 0 0.40 -0.12 -0.52 -0.25
Hlim 0.25 - - - 0.25
C a s e 3 - f ri c t i o n c o e ff i c i e n t o n t h e l e f t 5 . 3 % a n d o n t h e r i g h t 0 %
We obtain the following table:
C0 P1 P2 P3 C4
Ri 0 12.019 8.532 8.532 0
Sum Ri 29.084
Ri x di or Hlim -0.235 -0.375 -0.666 -1.065 0
Sum Hi -2.340
point 0 134.11
displacement 0.080 0.049 0.002 -0.044 -0.076
H 0.24 0.59 -0.02 -0.38 0
Hlim 0.24 - - - 0.24
We conclude that the maximum displacement in P 1 can be 0.049 m
in P 2 0.014 m
in P 3 0.060 m
Under only permanent loads, the displacements are: in P 1 0.014 m in P 2 0.001 m in P 3 0.017 m
N.B: it must be understood that the presence of sliding bearings transforms the structure into a non-linear system. Strictly speaking, it is not therefore possible to superimpose the forces and displacements of each of these actions to combine them.
However, that would lead to complicated calculations that are quite useless in relation to the differences in the values to be calculated. We could therefore simply consider that the displacement value due to the uniform temperature is the difference between the calculation with the maximum linear compression (here 6.10-4) and the linear compression due to permanent loads (2.10-4).
Laminated elastomeric bearings –Use on bridges, viaducts and similar structures 5P 4 . 4 . 2 . 4 . 2 – C a l c u l a t i n g t h e d i s t r i b u t i o n o f a b r a k i n g f o r c e
As regards breaking, it is presumed to be distributed only on non-sliding bearings. The value of the force assumed by the support is directly proportional to the stiffness of the bearing.
We obtain:
Pier 1 Piers 2 and 3
v1 v1
Bearing 0.0370 0.0370
Foundation + shaft 0.003 0.0143
Total 0.040 0.0514
Ri = 1
v 24.969 19.474
The sum of the stiffnesses is equal to 63.92.
For a breaking force of 0.36 MN, the distribution is thus:
- pier 1 H1 = 24
When verifying bearings, the preceding calculations give rise to a slight over-sizing of bearings on the piers, with those on the abutments having been correctly dimensioned from the start.
When all the calculations have been completes, so as to adapt as precisely as possible to the forces and deformations applied to bearings, we could reduce the number of laminations.
C0 and C4 P1 P2 P3
1 bearing of 400 x 500 2 bearings of 700 x 600
3 x (12 + 3); 2 x 6 4 x (16 + 4); 2 x 8 3 x (16 + 4); 2 x 8 6 x (16 + 4); 2 x 8
(unchanged) (unchanged)
As the bearings on the piers have been modified, we need to recalculate the distribution of forces. This calculation proves that the bearings on the piers are suitable.