Bearing Design Method A

7/27/2019 Bearing Design Method A

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102 kips

122 kips

1.0 in.

0.004 Radians

0.005 Radians (14.4.2.1)

0.009 Radians

Rectangular

yes

18 in.

17 in.

> W

> 17 in.

224 kips

224.00 in.2

N/A in. *

mi n = 13.18 in.

19 in.

L > L min

>

> N/A

> N/A N/A

323.0 in.2

*Note - Stress l imit increased 10 percent due to prevention of shear deformation per Section 14.7.6.3.2.

1.000 in.

2.000 in.

10.000 in.

= hri = 0.625 in.

Thickness of Top and Bottom Cover Layers (each) 0.250 in.

hcover < 0.7hri (14.7.6.1)

0.250 < in. OK

5

3.625 in. *

2.000 in.

12.00 in.

hrt > 2s (14.7.6.3.4-1)

3.625 > in. OK

hrt > 10s (14.7.6.3.4-2)

N/A > N/A

Calculated Rotation =

Total Unfactored Compressive Load = PT =

(n int = 0 for PEP and CDP)

2 s =

0.438

10 s =

III. SHEAR DEFORMA TION (AASHTO LRFD 14.7.6.3.4 )

Number of Interior Elastomeric Layers (Excluding Exterior Layer Allowance) = nin t =

Total Elastomer Thickness = hrt = 2hcover+ nin thri =

N/A

Maximum Total Shear Deformation of Elastomer at Service Limit = s= 0=

2.00

Elastomeric Layer Thickness

= hcover =

For PEP, FGP, and Steel-Reinforced Elastomeric Bearings:

N S B A E L A S T O M E R I C B E A R I N G D E S I G N ( E N G L I S H U N I T S )

M E T H O D A - E L A S T O M E R I C P A D S AND S T E E L - R E I N F O R C E D E L A S T O M E R I C B E A R I N G S - S E C T I O N 1 4 . 7 . 6

I. INITIAL DESIGN INPUTS

Dead Load = PD =

Steel-Reinforced BearingBearing Type:

A A S H T O L R F D , 3 R D E D . , 2 0 0 4 W I T H 2 0 0 5 I N T E R I M S

The following design program was developed based upon the above-referenced AASHTO LRFD code. The program is applicable to the design of

reinforced pads, the program assumes that interior elastomeric layers are of equal thickness, as are the two exterior elastomeric layers. For fiber-

outlined in articles 14.7.6.3.3 and 14.7.6.4 respectively.

elastomeric pads and steel-reinforced elastomeric bearings, both rectangular and circular in shape. For steel-reinforced elastomeric bearings and fiber-

layer per article 14.7.6.3.1. Compressive deflections and anchorage requirements for bearings are to be verified by the user in accordance with provisions

Live Load = PLL =

Flange Width

Rotation Construction Tolerance =

Design Rotation = s =

Bearing Shape:

Bearing Subject to Shear Deformation?

reinforced pads, the fiberglass reinforcement comprising h s shall be assumed to consist of a double fiberglass layer separated by a 0.125 in. bonding

Horizontal Movement of Bridge Superstructure =0

=

18 OK

II. BEARING GEOMETRY

Flange Width =

Bearing Width = W =

Based on service limit (14.7.6.3.2)

Minimum Bearing Length = L

Bearing Length = L =

Minimum Required Area of Bearing (shear deformation prevented) = A mi n =

Minimum Required Area of Bearing (shear deformation permitted) = A mi n =

N/A

Bearing Area = A =

19.0 13.18 OK

N/A

N/A

Total Elastomer Thickness of PEP = hrt =

Total Elastomer Thickness of CDP = hrt = tp =

For CDP Elastomeric Bearings:

*Note - Applies to FGP and steel-reinforced elastomeric bearings only.

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0.100 ksi

0.08 < < 0.250 ksi (14.7.6.2)

0.08 < < 0.250 ksi N/A

0.08 < < 0.175 ksi

0.08 < < 0.175 ksi OK

0.69 ksi

0.38 ksi

7.18 (14.7.5.1-1)

N/A (14.7.5.1-2)

< 0.80 ksi (14.7.6.3.2-1)

N/A < N/A ksi N/A

s


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0.004 radians

0.009 radians

0.003 radians

0.008 radians

(14.7.6.3.5b-1)

> ksi N/A

(14.7.6.3.5b-2)

> ksi N/A

(14.7.6.3.5b-3)

> ksi N/A

(14.7.6.3.5c-1)

> psi N/A

(14.7.6.3.5c-2)

< psi N/A

0 (14.7.6.3.5d)

5

(14.7.6.3.5d-1)

> ksi OK

(14.7.6.3.5d-2)

> ksi OK

(14.7.6.3.5d-3)

> ksi N/A

N/A

N/A

Total Service Rotation About Transverse Axis (with Construction Tolerance) = s,x =

Total Service Rotation About Longitudinal Axis (with Construction Tolerance) = s,z =

Exterior Layer Allowance = next =

N/A

PEP:

N/A

Rectangular Pads:

0.69

Circular Pads:

N/A

N/A

N/A

Equivalent Number of Interior Elastomeric Layers = n= nin t + next =

Circular Pads:

N/A

CDP:

V. ROTATION (AASHTO LRFD 14.7.6.3.5)

Service Rotation Due to Total Load About Transverse Axis =

Service Rotation Due to Total Load About Longitudinal Axis =

N/A

Rectangular Pads:

0.69

0.42

0.60

N/A

N/A

FGP and Steel-Reinforced:

N/A

xsrt

h

LGS

s ,

2

5.0

zsrt

h

WGS

s ,

2

5.0

pt

L

ss 000,6500,1

nh

DGS s

ri

s

2

375.0

nh

WGS

zs

ri

s

,

2

5.0

nh

LGS

xs

ri

s

,

2

5.0

spt

Ls

000,12

srt

h

DGS

s

2

375.0

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FGP: Service Limit State

36 ksi

N/A in.

N/A in. (14.7.6.3.1)

Steel-Reinforced: Service Limit State

36 ksi

0.036 in. Controls (14.7.5.3.7-1)

24 ksi (Table 6.6.1.2.5-3)

0.020 in. (14.7.5.3.7-2)

0.036 in.

0.1250 in.

> hs min> 0.036 in. OK

6.33 in.

5.67 in. Controls

N/A in.

< W/3

< 5.67 in. OK

17.0 in.

19.0 in.

0.625 in.

0.250 in.

5

3.625 in.

0.1250 in.

4.3750 in.

Reinforcement Thickness = hs =

Total Elastomer Thickness = hrt =

Number of Interior Elastomeric Layers (Excluding Exterior Layer Allowance) = nin t =

VII. STABILITY (AASHTO L RFD 14.7.6.3.6)

L/3

D/4

W/3

=

=

Minimum Yield Strength of Fiber Reinforcement = FyFIBER=

Total Bearing Thickness = hrt+ hs(nin t+1) =

Bearing Length = L =

=

VI. REINFORCEMENT (AASHTO LRFD 14.7.6.3.7)

Minimum Yield Strength of Steel Reinforcement = Fy =

Steel-Reinforced: Fatigue Limit State

Constant Amplitude Fatigue Threshold =FTH =

0.125

Double Fiberglass Layer with 0.125 in. Separation = hs min total= 2hs min+ 0.125 =

Elastomeric Layer Thickness = hri =

Thickness of top and Bottom Cover Layers (each) = hcover=

Required Minimum Reinforcement Thickness = hs min=

hs

Reinforcement Thickness = hs =

Total Bearing Thickness = hrt+ hs(nin t+1) =

Total Thickness of Bearing Shall Not Exceed the Least of:

4.38

VIII. FINAL DESIGN SUMMARY

Bearing Width = W =

y

ss

F

h.h max

min

03

TH

L

s

F

h.h

max

min

02

yFIBERF

rih.

sh

22

min

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Documents

Bearing Design Method A