For bridge projects, bearing pads come in various types, including bulging tolerances and specifications
For bridge projects, bearing pads come in various types, including bulging tolerances and specifications
Design Step 6.1.2-Steel-renforced elastomeric bearing – Method B (S14.7.5)
For bridges where the roadway has a positive or negative grade, the thickness of the may need to be varied along the length of the rafter. This is typically accomplished through the use of a tapered steel top plate. In this example, the bridge is assumed to be at zero grade; All internal layers of elastomer shall be of the same thickness. 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;
Si = LW/[2hri(L+W)] ………………… (S14.7.5.1-1)
Where: L= length of a rectangular elastomeric bearing (parallel to the longitudinal bridge axis) (in.)
W = Width of Elastomeric (in)
hri = Thickness of Elastomer Layers (in).
Now keeping in view Lahore Sialkot Motorway being shape factor is being calculated as followings.
(1)Bearing pad Size 500x450x77.5 mm .Steel plate (5 no) 1.5 mm each,4 no elastomer internal layers 15mm each,2 no cover layer 5 mm each (RD’s 21+755, 22+100, 30+236, 47+507, 56+310, and 61+910)
Si = LW/ [2hri (L + W)] ……………………………………….. (S14.7.5.1-1)
L = Length of Elastomeric bearing pads (500mm).
W = Width of Elastomeric (450mm).
hri= Thickness of Elastomer Layers (15mm).
Si = 500×450/ [2×15(500 + 450)]
Si =7.89 (Shape factor)
Elastomer Layer free to bulge is = 15mmx7.89%
Shape factor per layer free to bulge = 1.18mm
Total No of Layers = 4×1.18 = 4.72mm
Total Thickness Difference = 77.5-4.72 = 72.78mm
500x450x77.5 mm bearing pad Shape Factor free to bulge up till72.78 mm
(2) 500x500x77.5 mm.Steel plate (5 no) 1.5 mm each, 4 no elastomer internal layers 15mm each, 2 no cover layer 5 mm each (RD’s 31+966)
Si = LW/ [2hri (L + W)] ……………………………….. (S14.7.5.1-1)
L = Length of Elastomeric bearing pads (500mm).
W = Width of Elastomeric Bearing Pads (500mm).
hri= Thickness of Elastomer Layers (15mm).
Si = 500×500/ [2×15(500 + 500)]
Si =8.33(Shape factor)
Elastomer Layer free to bulge is = 15mmx8.33%
Per layer free to bulge = 1.24mm
Total No of Layers = 4×1.24= 4.96mm
Total Thickness Difference = 77.5-4.96 = 72.54mm
500x500x77.5mm Shape Factor free to bulge up till 72.54mm
(3) 500x400x77.5 mm.Steel plate (5 no) 1.5 mm each, 4 no elastomer internal layers 15mm each, 2 no cover layer 5 mm each (RD’s 37+777, 59+761)
Si = LW/ [2hri (L + W)] ……………………………… (S14.7.5.1-1)
L = Length of Elastomeric bearing pads (500mm).
W = Width of Elastomeric (400mm).
hri= Thickness of Elastomer Layers (15mm).
Si = 500×400/ [2×15(500 + 400)]
Si =7.40(Shape factor)
Elastomer Layer free to bulge is = 15mmx7.40%
Per layer free to bulge = 1.11mm
Total No of Layers = 4×1.11 = 4.44mm
Total Thickness Difference = 77.5-4.44 = 73.06mm
500x400x77.5mm bearing pads Shape Factor free to bulge up till 73.06mm
(4) 400x450x77.5 mm.Steel plate (5 no) 1.5 mm each, 4 no elastomer internal layers 15mm each, 2 no cover layer 5 mm each (RD’s 42+673, 43+089)
Si = LW/ [2hri (L + W)] ………………………..….. (S14.7.5.1-1)
L = Length of Elastomeric bearing pads (450mm).
W = Width of Elastomeric (400mm).
hri= Thickness of Elastomer Layers (15mm).
Si = 400×450/ [2×15(400 + 450)]
Si =7.05(Shape factor)
Elastomer Layer free to bulge is = 15mmx7.05%
Per layer free to bulge = 1.05mm
Total No of Layers = 4×1.05 = 4.20mm
Total Thickness Difference = 77.5-4.20 = 73.30mm
400x450x77.5mm Shape Factor free to bulge up till 73.30mm