Bearing Design Method B PDF

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Summary

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 SM E T H O D B - 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. 5The following design program was developed based upon the above-referenced AASHTO LRFD code. The program is applica...

Description

NSBA

E LAS T OM E RI C AASHTO

METHOD

B

-

LR FD ,

BEARING 3RD

STEEL-REINFORCED

ED.,

2004

DESIGN

(ENGLISH

WITH

IN TER IMS

ELA STO MER IC

2005

BEARINGS

-

SECTION

UNITS) 1 4 .7 .5

The following design program was developed based upon the above-referenced AASHTO LRFD code. The program is applicable to the design of steelreinforced elastomeric bearings, both rectangular and circular in shape. The program is not applicable to design of rectangular bearings subject to combined rotation about the transverse and longitudinal axes. The program assumes that interior elastomeric layers are of equal thickness, as are the two exterior elastomeric layers. I. INITIAL DESIGN INPUTS Dead Load = PD = Live Load = PLL = Horizontal Movement of Bridge Superstructure = Δ0 = Axis of Pad Rotation:

102

kips

122

kips

1.0

in.

Transverse

Calculated Rotation =

0.004

Radians

Rotation Construction Tolerance = Design Rotation = θs =

0.005

Radians

0.009

Radians

Bearing Shape:

(14.4.2.1)

Rectangular

Bearing Subject to Shear Deformation?

yes

Bridge Deck Fixed Against Horizontal Translation?

yes

II. BEARING GEOMETRY Flange Width =

18

in.

Bearing Width = W =

15

in.

Flange Width

>

W

18 > 15 Total Unfactored Compressive Load = PT = Minimum Required Area of Bearing = Amin = Minimum Bearing Length = L

=

min

Bearing Length = L = L

>

13.5 >

L

in. 224 140.0

in.2

9.33

in.

13.5

in.

9.33

N/A

>

N/A

N/A

>

N/A 202.5

III. SHEAR DEFORMATION (AASHTO LRFD 14.7.5.3.4 ) Maximum Total Shear Deformation of Elastomer at Service Limit = Δs = Δ0 = 1.000

in.

OK

N/A

N/A

in.2 in.

2Δs =

2.000

in.

Elastomeric Layer Thickness = hri =

0.375

in.

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

0.250

in.

<

0.7hri

0.250 < 0.263 Number of Interior Elastomeric Layers (Excluding Exterior Layer Allowance) = nint = Total Elastomer Thickness = hrt = 2hcover + ninthri = hrt

Based on service limit (14.7.5.3.2)

min

Bearing Area = A =

hcover

OK

kips

(14.7.5.1) in.

OK

10 4.250

in.

> 2Δs

(14.7.5.3.4-1)

4.250> 2.000

in.

OK

IV. COMPRESSIVE STRESS (AASHTO LRFD 14.7.5.3.2) Service Average Compressive Stress (Total Load) =

Service Average Compressive Stress (Live Load) =

Rectangular Shape Factor = Si

Circular Shape Factor = Si

σ = s

PT = A

P σ L= LL= A LW = = 2 hri ( L+W ) D = = 4 hri

Shear Modulus of Elastomer = G = 0.080 <

1.11

ksi

0.60

ksi

9.47

(14.7.5.1-1)

N/A

(14.7.5.1-2)

0.100

ksi

< 0.175

ksi

0.080 < 0.100 < 0.175

ksi

OK

1.11 < 1.57 σs < 1.6 ksi

ksi

OK

1.11 < 1.6

ksi

OK

ksi

OK

G

(14.7.5.2)

For Bearings Subject to Shear Deformation: σs < 1.66GS

σL

(14.7.5.3.2-1)

(14.7.5.3.2-1)

< 0.66GS 0.60 < 0.63

(14.7.5.3.2-2)

For Bearings Fixed Against Shear Deformation: σs < 2.00GS

σL

(14.7.5.3.2-3)

N/A < N/A σs < 1.75 ksi

N/A

N/A < N/A

N/A

< 1.00GS

1 of 4

(14.7.5.3.2-3) (14.7.5.3.2-4)

NSBA

E LAS T O M E R I C AA S H T O

LR FD ,

BEARI NG 3RD

ED.,

2004

DESIGN

(ENGLISH

WITH

IN TER IMS

N/A < N/A

2 of 4

2005

N/A

UNITS)

NSBA

E LAS T O M E R I C AA S H T O

LR FD ,

BEARI NG 3RD

ED.,

2004

DESIGN

(ENGLISH

WITH

IN TER IMS

2005

UNITS)

V. COMBINED COMPRESSION AND ROTATION (AASHTO LRFD 14.7.5.3.5) RECTANGULAR BEARINGS: B = Length of Pad = 13.50

in.

Exterior Layer Allowance = next = 1.0 Equivalent Number of Interior Elastomeric Layers = n = nint + next = 11

(14.7.5.3.5)

σ s>1.0GS 1.11

>

( )( ) θs B n h ri

2 (14.7.5.3.5-1)

1.00

ksi

( )( ) ]

[

σ s

1.11

( )( ) θs D n h ri

N/A

[

1.11

θs

D hri

n

<

2

(14.7.5.3.5-5)

N/A

[

1.11

( )( ) ]

N/A

<

( )( ) ]

σ s

hs min

0.125

>

0.035

in.

VIII. FINAL DESIGN SUMMARY Bearing Width = W =

15

in.

Bearing Length = L =

13.5

in.

Elastomeric Layer Thickness = hri =

0.375

in.

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

0.250

in.

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

10

Total Elastomer Thickness = hrt =

4.250

in.

Reinforcement Thickness = hs =

0.1250

in.

Total Bearing Thickness = hrt + hs(nint+1) =

5.6250

in.

4 of 4

(Table 6.6.1.2.5-3)

(14.7.5.3.7-2)

OK...