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Design of Elastomeric Bearing
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Herman L. Guillermety AssociatesConsulting Engineers
document.xls 1
Project: Bridge over Gurabo River 6/20/2005
ELASTOMERIC BEARINGS
14.4.1 Method A - Design Procedure for Steel Reinforced Bearings.
14.4.1.1 Compressive Stress
300
by dead and live load excluding impact (See Appendix A)
Hardness = 60
G = 130 psi
L = 8 in
W = 26 in
0.50 in
No Layer = 4
S = 6.1
1 Average compressive stress
A = 208 in2 DL DL + LL (excluding impact)
89.6 kips 431 psi 663 psi
48.2 kips OK
795 psi
14.4.1.2 Compressive Deflection
2.00 in
3.2 % See Figure 14.4.1.2B
0.0640 in
Limit of initial compressive Deflection
Average compressive strees sc shall satify:
sc, TL < GS/b
sc, TL < 1,000 psi for steel reinforced bearings
sc, TL < 800 psi for plain pads
G, shear modulus of elastomer (psi) at 73oF (See Table 14.3.1)
S, shape factor: LW / 2hri (L+W))
b, Modifying factor
hri, Thickness of elastomer layer
sc, P/A = average compressive stress on the bearing caused
hri =
b =
PDL = sc = sc =
PLL =
GS/b =
Dc , Instantaneous compressive deflection of bearing (in.)
Dc = S ecihri eci , Intantaneous compressive strain in elastomer layer
S hri = hrt , Total elastomer thickness in an elastomeric bearing (in.)
S hri =
eci =
Dc =
Herman L. Guillermety AssociatesConsulting Engineers
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Initial thickness = 1.94 in 0.035 per layer OK
Final thickness after creep = 1.91 in Creep 35% see Table 14.3.1
14.4.1.3 Shear
Temperature Zone
15 Temperature Rise
25 Temperature Fall
Thermal Coefficient = 0.000006
Length (3 Span) = 86.4 m
0.816 in 1.63 OK
14.4.1.4 Rotation
Rotation Allowable Actual
0.01600 0.0075 OK
0.00492 0 OK
14.4.1.5 Stability
Total thickness of the bearing shall not exceed the smallest of:
for reinforced bearings Total thickness of bearing = 2.00 in OK
L/3 = 2.67 in
W/3 = 8.67 in.
14.4.1.6 Reinforcement
Reinforcement resistance in pounds per linear inch at working stress
in each direction shall not be less than:
for fabric Min. Reinf. Resistance Steel Plate Resistance
for steel OK 0.5 in 0.075 in
850 lb/in 20000 psi
f = 1500 lb/in
14.5 Anchorage
If the design shear force, H, due to bearing deformation exceeds 1/5 of the compressive force P
due to dead load alone, the bearing shall be secured against horizontal movement.
17.92 kips
0.07hri =
oFoF
per oF hrt > 2Ds
Ds = Min thickness of Elastometer = 2 Ds =
qTL,x < 2DC/L
qTL,z < 2DC/W qTL,x 2DC/L
qTL,z 2DC/W
1,400 hri
1,700 hri hri = tsteel =
1,700 hri = Fa =
PDL / 5 =
Herman L. Guillermety AssociatesConsulting Engineers
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11.035 kips OKH = GADh/hrt =
Keeper Plate Schedule (for zero rotation on bearing pad)
Min. Slope 0.016
PG t1 t2 t1 t2 Slope
Structure Elevation Span Slope in in in in Check
A1 7.698 0.75 1.06 3/4 1 1/16 0.03906
P1 BACK 8.835 29.1 0.039 0.75 1.06 3/4 1 1/16 0.03906
P1 FWD 8.835 0.75 0.95 3/4 15/16 0.02344
P2 BACK 9.553 29.4 0.024 0.75 0.95 3/4 15/16 0.02344
P2 FWD 9.553 0.75 0.75 3/4 3/4 0
P3 BACK 9.846 29.4 0.010 0.75 0.75 3/4 3/4 0
P3 FWD 9.846 0.75 0.75 3/4 3/4 0
P4 BACK 9.993 29.4 0.005 0.75 0.75 3/4 3/4 0
P4 FWD 9.993 0.75 0.75 3/4 3/4 0
P5 BACK 10.137 29.4 0.005 0.75 0.75 3/4 3/4 0
P5 FWD 10.137 0.75 0.75 3/4 3/4 0
P6 BACK 10.019 29.4 -0.004 0.75 0.75 3/4 3/4 0
P6 FWD 10.019 0.90 0.75 7/8 3/4 -0.0156
P7 BACK 9.47 29.4 -0.019 0.90 0.75 7/8 3/4 -0.0156
P7 FWD 9.47 1.02 0.75 1 3/4 -0.0313
P7 BACK 8.488 29.4 -0.033 1.02 0.75 1 3/4 -0.0313
P7 FWD 8.488 1.13 0.75 1 1/8 3/4 -0.0469
A2 7.09 29.1 -0.048 1.13 0.75 1 1/8 3/4 -0.0469
Appendix A
From Simple (Page 4)
Live Load Reactions
Distribution factor 1.157
Truck axle loads control / lane: 97.098 kips
Additional live load / beam 0 kips
Number of lanes 3
Number of girders 7
Total LL excluding Impact 337.03 kips
LL excluding Impact/beam 48.15 k/beam
Dead Loads
Total dead load at left support 627.246 k
DL / beam 89.607 k/beam
