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Design of Bearings Bearings : Elastomer material properties properties : AASHTO 14.7.6.2
page 1474
Commonly used elastomers have a shear modulus between 0.6 and 1.75 MPa, and a nominal hardness between 50 and 70 on the Shore A Scale. The shear force on the structure induced by defonnation defonn ation of the elastomer in PEP, PEP, FGP and steelreinforced steelreinforced elastomeric bearings shall be based on a G value not less than that of the elastomer at23°C. Effects of relaxation shall be ignored.
Elastomer bearind hardness = 60 AASHTO
14.7 14.7.6 .6.2 .2-1 -1
page page 1474 1474
Therefore ; G. ≔ 1.14 ⋅ MPa
As an average of the range between 0.9 - 1.38
Design Forces on Bearings of Interior Girders at the Intermediate Pier :
Max Factored Reaction
Strength 1 Service 1
Max Reaction Due to LL
R DStrength ≔ 1453 ⋅ kN
R LLStrength ≔ 812 ⋅ kN
R DService ≔ 1133 ⋅ kN
R LLService ≔ 464 ⋅ kN
According to the Commentary of AASHTO C14.7.5.3.2 page 1467, the effect of the dynamic load allowance on the elastomeric bearing reaction may be ignored. The reason for this is that the dynamic load allowance effects are likely to be only a small proportion of the total load and bec ause the stress limits are based on fatigue damage, whose limits are not clearly defined.
The minimum bearing area :
The Bearing os fixed against shear deformations. Therefore; the maxmum compressive stress limit under service limit state according to AASHTO 4.7.5.3.2 page 1467 is :
σ s ≤ 2.0 ⋅ G. ⋅ S i ≤ 12.0 ⋅ MPa σ L ≤ 1.0 ⋅ G. ⋅ S i
Where : σ s
= service
L σ
= service average compressive stress due to live load (MPa)
average compressive stress due to the otal load (MPa)
G = shear Si = shape
modulus of elastomer (MPa) factor of the thickest layer ofthe bearing
To satisfy the 1.75 MPa limit, the minimum bearing area, Areq, should satisfy
R DService Areq ≔ ―― 12 ⋅ MPa
Areq = 9.442 ⋅ 10
4
mm
2
The corners of the bottom flanges of the girder are usually chamfered. The bearing should be slightly narrower than the flat part of the flange unless a stiff sole plate is used to insure uniform distribution of the compressive stress and strain over the bearing area. The bearing should be as short along the length of the girder as practical to permit rotation about the transverse axis. This requires the bearing to be as wide as possible which is desirable when stabilizing the girder during erection.
First Estimation for bearing dimensions :
Consider : Chamfer ≔ 25 ⋅ mm
b fBottom ≔ 710 mm
EdgeClearance ≔ 25 ⋅ mm W. ≔ b fBottom − 2 ⋅ Chamfer + EdgeClearance
W. = 610 mm
Consider the longitudinal dimension to ensure that the maximum compressive stress limit is satisfied : L. ≔ 190 ⋅ mm
Bearing area ≔ W. ⋅ L.
Bearing area = 1.159 ⋅ 10
5
mm
2
‖ Bearing Area ≔ if Bearing area ≥ Areq ‖ ‖ “OK” ‖ if Bearing area < Areq ‖ “N.G” ‖ Bearing Area = “OK”
The shape factor : AASHTO 14.7.5.1-1
page 1465
The shape factor of a layer of an elastomeric bearing, Si is taken as the plan area of the layer divided by the area of perimeter free to bulge. For rectangular bearings without holes, the shape factor of the layer may be taken as:
L ⋅ W S i ＝ ―――― 2 ⋅ hri ⋅ L + W
Where : L = length of a rectangular elastomeric bearing (parallel to longitudinal bridge axis) (mm) W = width of the bearing in the transverse direction (mm) hri = thickness of ith elastomeric layer in elastomeric bearing (mm)
The thickness of the elastomeric layer due to dead load is : L ⋅ W hri ＝ ―――― 2 ⋅ S i ⋅ L + W
The Compressive stress : AASHTO 14.7.5.3.2
page 1467
In any elastomeric bearing layer, the average compressive stress at the service limit statewill satisfy the following provisions. These provisions limit the shear stress and strain in the elastomer. The relationship between the shear stress and the applied compressive load depends directly on the shape factor, with higher shape factors leading to higher capacities.
The Shape factor under total load is : σ sTL
S TL ≥ ― 2 ⋅ G.
Where : P TL
σ sTL ＝ ―――
Bearing area
PTL = maximum bearing reaction under total load (kK) P TL ≔ R DService
P TL
σ sTL ≔ ―――
Bearing area
9.776 MPa σ sTL =
G. = 1.14 MPa
Shear Modulus
σ sTL
S TL ≔ ― 2 ⋅ G.
S TL = 4.288
The Shape factor under live load is : σ L
S LL ≥ ― 1 ⋅ G.
Where : P LL
σ s ＝ ―――
Bearing area
PLL = maximum bearing reaction under live load (kK) P LL ≔ R LLService
P LL
σ sLL ≔ ―――
Bearing area
4.003 MPa σ sLL =
G. = 1.14 MPa
σ sLL
S LL ≔ ― 1 ⋅ G.
S LL = 3.512
The Minimum Shape Factor is : S min ≔ max S TL , S LL
S min = 4.288
The elastomer thickness is : L. ⋅ W. hriTL ≔ ―――― 2 ⋅ S TL ⋅ L. + W.
hriTL = 16.895 mm
L. ⋅ W. hriLL ≔ ―――― 2 ⋅ S LL ⋅ L. + W.
hriLL = 20.627 mm
Use an interior elastomer layer thickness of : hri ≔ 15 ⋅ mm
The shape factor is : L. ⋅ W.
L. ⋅ W. S. ≔ ―――― 2 ⋅ hri ⋅ L. + W.
S. = 4.829
Compressive deflection :
AASHTO 14.7.5.3.3 page 1468 This provision need only be checked if deck joints are present on the bridge. Since this design example is a jointless bridge, commentary for this provision is provided below, but no design is investigated. Deflections of elastometric bearings due to total load and live load alone will be considered separately.
Instantaneous deflection : δ. ＝ Σ εi ⋅ hri
Where : εi = instantaneous compressive strain in ith elastomer layer of a laminated bearing. hri = thickness of ith elastomeric layer in a laminated bearing (mm)
Shear deformation :
AASHTO 14.7.5.3.4
page 1469
This provision need only be checked if the bearing is a movable bearing. Since the bearing under consideration is a fixed bearing, this provision does not apply.
Combined compression and rotation :
AASHTO 14.7.5.3.5
page 1469
Service limit state applies. Design rotations are taken as the maximum sum of the effects of initial lack of parallelism between the bottom of the girder and the top of the superstructure and subsequent girder end rotation due to imposed loads and movements. The goal of the following requirements is to prevent uplift of any corner of the bearing under any combination of loading and corresponding rotation.
For rectangular Bearings :
σ s > 1.0 ⋅ G. ⋅ S. ⋅
⎛ θ s ⎞ ⎛ B ⎞ 2 ⋅ ⎝ n ⎠ ⎝ hri ⎠
Where : n=
number of interior layers of elastomer, where interior layers are defined as those layers which are bonded on each face. Exterior layers are defined as those layers which are bonded only on one face. When the thickness of the exterior layer of elastomer is more than one-half the thickness of an interior layer, the parameter, n, may be increased by one- half for each such exterior layer.
hri = 15 mm
σ s ≔ max σ sTL , σ sLL
9.776 MPa σ s =
maximum compressive stress in elastomer (MPa)
B = length of pad if rotation is about its transverse axis or width of pad if rotation is about its longitudinal axis (mm.) B ≔ L.
s θ
=
maximum service rotation due to the total load (rads) For this example, θ s will include the rotations due to live load and construction load (assume 0.005 rads) θ s ≔ 0.005 ⋅ rad
⎛
⎛ B ⎞ 2 ⎞ 1.0 ⋅ G. ⋅ S. ⋅ θ s ⋅ ⎝ ⎝ hri ⎠ ⎠ nu ≔ ――――――― σ s
nu = 0.452
To prevent excessive stress on the edges of the elastomer, rectangular bearings fixed against shear deformation must also satisfy :
σ s < 2.25 ⋅ G. ⋅ S.
⎛ ⎝
1 − 0.167 ⋅
⎛ θ s ⎞ ⎛ B ⎞ 2 ⎞ ⋅ ⎝ n ⎠ ⎝ hri ⎠ ⎠
⎛
⎛ B ⎞ 2 ⎞ −0.167 ⋅ θ s ⋅ ⎝ ⎝ hri ⎠ ⎠ ncBearing ≔ ―――――― ⎛ ⎞ σ s ―――― − 1 ⎝ 2.25 ⋅ G. ⋅ S. ⎠
ncBearing = 0.636
Use 2 interior layers of 15 mm thickness an d 2 exterior layers of 5 mm thickness
nhri ≔ 2
t hri ≔ 15 ⋅ mm
nhre ≔ 2
t hre ≔ 5 ⋅ mm
Stability of elastomeric bearings :
AASHTO 14.7.5.3.6
page 1471
Bearings are investigated for instability at the service limit state load combinations specified in AASHTO Table 3.4.1-1. page 69
Bearings satisfying the below equations are considered stable, and no further investigation of stability is required. 2 ⋅ A. ≤ B.
2 ⋅ A. ≤ B.
For Which : For which : 1.92 ⋅
hri
L. A. ≔ ―――― ‾‾‾‾‾‾‾ 2.0 ⋅ L. 1 + ―― W.
A. = 0.119
2.67 B. ≔ ―――――― ⎛ L. ⎞ S. + 2.0 ⋅ 1 + ― ⎝ 4 ⋅ W. ⎠
B. = 0.363 Bearing ≔ ‖ if 2 ⋅ A. ≤ B.
‖ ‖ “Stable” ‖ if 2 ⋅ A. > B. ‖
“Not.Stable”
Bearing = “Stable”
Reinforcement :
AASHTO 14.7.5.3.7
page 1472
The reinforcement should sustain the tensile stresses induced by compression on the bearing. With the present load limitations, the minimum steel plate thickness practical for fabrication will usually provide adequate strength
At the service limit state :
3 ⋅ hri ⋅ σ s h sTL ≥ ―― F y
Where : Fy = yield strength of steel reinforcement ( MPa ) F y ≔ 420 ⋅ MPa
3 ⋅ hri ⋅ σ s h sTL ≔ ―― F y
h sTL = 1.047 mm
At the fatigue limit state : 2 ⋅ hri ⋅ σ L h sLL ≥ ―― Δ F TH
Where : Δ FTH
= constant amplitude fatigue threshold for Category A as specified in AASHTO Table 6.6.1.2.5-3 page 614 (MPa)
For Category A : Δ F TH ≔ 165 ⋅ MPa
2 ⋅ hri ⋅ σ sLL h sLL ≔ ――― Δ F TH
h sLL = 0.728 mm
Use h s ≔ 5 ⋅ mm
nhs ≔ 3
The total height of the bearing, h rt : h st ≔ nhre ⋅ t hre
+ nhri ⋅ t hri + nhs ⋅ h s
h st = 55 mm
Plan View :
Elevation View :
