Bronze Bearing Design

'file:///tt/file_convert/5489032db4795921448b45af/document.xls'#$BRG DESIGN

SPAN 1 & 3 PIGGYBACK BEARING (USE INFORMATION FOR SPAN 1 OR 3, WHICHEVER CONTROLS)

MAX UNFACTORED DC1 FROM EXT. OR INT. REACTION = 123.7 KMAX UNFACTORED DC2 FROM EXT. OR INT. REACTION = 30.10 KMAX UNFACTORED FWS FROM EXT. OR INT. REACTION = 18.8 K

MAX UNFACT. LL FROM EXT. OR INT. REACT. (INCL IMPACT, DIST. FACT.) = 157.7651 K

STRENGTH I LIMIT STATEDC1 LOAD FACTOR (ALRFD 3.4.1) = 1.25DC2 LOAD FACTOR (ALRFD 3.4.1) = 1.25FWS LOAD FACTOR (ALRFD 3.4.1) = 1.50

LL LOAD FACTOR (ALRFD 3.4.1) = 1.75

496.54 K

FIND THE MAXIMUM REQUIRED LENGTH OF THE BRONZE PLATE:

MAXIMUM LENGTH BASED ON BEARING RESISTANCE

3.0 KSI1.00

MAXIMUM UPPER OR LOWER FLANGE WIDTH = 24 INW (LOWER FLANGE WIDTH + 2 * 4.5 IN MIN FROM BC-755M EXP BRG IIIE) = 32 IN

5.17 IN

MAX LENGTH BASED ON BC-755M EXP BRGS IIIE DETAIL & MOVEMENT REQ.ANCHOR BOLT DIAMETER = 1.25 IN

MIN SLOT LENGTH IN BRONZE PLATE FROM BC-755M EXP BRG IIIE = 7.63 IN

EXPANSION LENGTH (SPAN 1 + SPAN 2) = 208.50 FT

TEMPERATURE RANGE FOR STEEL BEARINGS (DM-4 3.12.2.1) = 156

COEFFICIENT OF THERMAL EXPANSION (ALRFD 6.4.1) = 6.5E-06

MOVEMENT DUE TO THERMAL CONTRACTION/EXPANSION = 2.54 INDC1 ROT. MOVEMENT (FROM STAAD) = 0.277 INDC2 ROT. MOVEMENT (FROM STAAD) = 0.069 IN

LL ROT. MOVEMENT (FROM STAAD) = 0.323 IN

MAXIMUM FACTORED LOAD AT STRENGTH LIMIT STATE, Pu =

CYLINDRICAL FACTORED COMPRESSIVE RESISTANCE, Pr = fDWsss (ALRFD 14.7.3.2)

SET Pu = Pr AND SOLVE FOR D

BEARING STRESS AT STRENGTH LIMIT STATE, sss (ALRFD 14.7.7.3) =f (ALRFD 14.7.3.2) =

D = Pu / ( f W sss ) =

oF

IN/IN/oF

SO No. 104906

Subject : SR 219 SB OVER SR 281

BRONZE BEARING DESIGN FOR PIGGYBACK BRGS

Computed By: KEW Checked By: Date: 4/29/2005


SO No. 104906




TOTAL MOVEMENT = 3.21 IND = MIN SLOT LENGTH + MAX OF [ (2 * 1.5 IN ) OR TOT. MOV. ] = 10.83 IN

USE D (b) = 10.875 IN

CALCULATE THE LENGTH & WIDTH OF THE TOP PLATE

LENGTH (FROM BC-755M, S = b + 1.5 IN) = 12.375 INWIDTH OF TOP PLATE = WIDTH OF BRONZE PLATE = 32 IN

CALCULATE THE REQUIRED MINIMUM TOP PLATE THICKNESS

ASSUME A TOP PLATE THICKNESS (H) = 1.0000 IN

NOTE: TOP PLATE IS DESIGNED ACCORDING TO BD-613M SHT 1 OF 15 "SOLE PLATE DESIGN."

UPPER FLANGE WIDTH, W = 24 INEFFECTIVE TOP PLATE WIDTH USING 1V:1.5H SLOPE = 27 IN

PRESSURE BASED ON BD-613M, w = 1.691 KSI

CANTILEVER LENGTH, L = 1.5* H = 1.5000 IN

1.90 K-IN/IN

0.167

1.00

36 KSI

6.01 K-IN/IN

6.01 > 1.90 OK

BEVELED SOLE PLATE LONGITUDINAL SLOPE = 0.0365BEVELED SOLE PLATE LENGTH (FROM ABOVE), L = 12.375 IN

RADIUS (BC-755M SHT 3 OF 4) = 18.00 IN

1.0000 IN

DESIGN MOMENT, Mu = w * L2 / 2 =

SECTION MODULUS, S = H2 / 6 = IN3 / IN

f, BENDING (ALRFD 6.5.4.2) =

Fy =

RESISTING MOMENT, Mr = f * Fy * S =

Hmin = T =


SO No. 104906




1.8711 IN

2.3227 IN

CALCULATE THE LENGTH & WIDTH OF THE BOTTOM PLATE

FROM BC-755M, THE LENGTH OF THE BOTTOM PLATE IS THE MAX OF:BRONZE PLATE LENGTH (D OR b) + 2 IN = 12.875 IN

OR14.25 IN

T1 = [ R - 1/2 ( 4 R2 - L2 )0.5 ] + T - ( L / 2 * LONGIT SLOPE ) =

T2 = [ R - 1/2 ( 4 R2 - L2 )0.5 ] + T + ( L / 2 * LONGIT SLOPE ) =

BRONZE PLATE LENGTH (D OR b) + DL =

L

T1

T2

T

STA AHDR 18.00in.


SO No. 104906




USE L = 14.25 IN

WIDTH OF BOTTOM PLATE = WIDTH OF BRONZE PLATE, W = 32 IN

CALCULATE THE REQUIRED MINIMUM BOTTOM PLATE THICKNESS

1.0000 IN

NOTE: TOP PLATE IS DESIGNED ACCORDING TO BD-613M SHT 1 OF 15 "SOLE PLATE DESIGN." ASSUME THAT BOTTOM PLATE IS DESIGNED SIMILARLY.

FLANGE WIDTH OF LOWER GIRDER = 33 INEFFECTIVE BOTTOM PLATE WIDTH USING 1V:1.5H SLOPE = 36 IN

PRESSURE BASED ON BD-613M, w = 1.268 KSI

CANTILEVER LENGTH, L = 1.5* H = 1.5000 IN

1.43 K-IN/IN

0.167

1.00

36 KSI

6.01 K-IN/IN

6.01 > 1.43 OK

BEVELED BOTTOM PLATE LONGITUDINAL SLOPE = 0BEVELED PLATE LENGTH (FROM ABOVE) = 14.25 IN

1.00 INBOTTOM PLATE THICKNESS @ C.L. BRG = 1.00 IN

1.00 IN

ASSUME A BOTTOM PLATE THICKNESS (Hmin) =

DESIGN MOMENT, Mu = w * L2 / 2 =

SECTION MODULUS, S = H2 / 6 = IN3 / IN

f, BENDING (ALRFD 6.5.4.2) =

Fy =

RESISTING MOMENT, Mr = f * Fy * S =

Hmin =

BOTTOM PLATE THICKNESS @ Hmax =


SO No. 104906




CHECK THAT HORIZONTAL LOAD IS LESS THAN HORIZONTAL RESISTANCE (ALRFD 14.7.3.3)

COEFFICIENT OF FRICTION FOR BRONZE (ALRFD 14.7.7.2)= 0.1

22.05 K

172.6 K

R = 18 INL = 10.875 IN

W = 32 INTOTAL DC1 ROTATION = 0.006 RADTOTAL DC2 ROTATION = 0.002000 RAD

TOTAL LL ROTATION (INCL. IMPACT & DIST FACTORS) = 0.004831 RADTOLERANCE (DM-4 14.7.7.5P) = 0.02 RAD

0.0328 RAD

3.0 KSI

7.28 DEG

17.58 DEG

64.12 K

22.05 < 64.12 OK

CHECK THE MINIMUM THICKNESS OF THE BRONZE PLATE

MIN THICKNESS AT C.L. BRG DUE TO BC-755M OR AS REQ'D BY DESIGN = 1.5 IN

0.659 IN

MIN EDGE THICKNESS FROM DM-4 14.7.3.1 = MAX OF:

OD X 0.045 = 0.489 IN

Hu = MAX FACTORED LONGITUDINAL HORIZONTAL FORCE (USE m * DL) =

PD =

qu TOTAL SERVICE ROTATION (DM-4 14.4.2) =

sss =

b = tan-1 ( Hu / PD ) =

Y = sin-1 ( L / 2 R ) =

2 R W sss sin (Y - b - qu ) sin b =

MIN EDGE THICK. BASED ON BR. PLATE LENGTH = T - [R - 1/2 ( 4 R2 - L2 )0.5 ] =


SO No. 104906




OR

0.5 IN

SINCE THE ACTUAL EDGE THICK. > MIN THICK. REQ'D - OK

CHECK MIN VERTICAL CLEAR DIST. BETW. ROTATING & NON-ROTATING PARTS (DM-4 14.7.3.1)

TOP PLATE LENGTH = 12.375 INMIN VERT. CLEAR DIST. BETW. TOP PL & BOT PL NOT INCLUDING ROTATION:

0.403 IN

SUBTRACT OUT VERT. DIST. TOP PLATE MOVES DOWN DUE TO ROTATIONMAXIMUM ROTATION FROM ABOVE = 0.0328 RAD

0.203 IN

FINAL MINIMUM VERTICAL CLEARANCE BETW. TOP & BOT PL = 0.2 IN

MIN ALLOWED VERT. CLEAR. BETW. ROTATING & NONROTATING PARTS (DM-4 14.7.3.1)

D = 10.875 IN

0.0328 RAD

0.375 IN

THE ACTUAL VERT. CLEAR IS LESS THAN THE ALLOWABLE - NG

T - [R - 1/2 ( 4 R2 - L2 )0.5 ] =

AMT TOP PLATE LOWERS FROM ORIGINAL INITIAL SET POSITION DUE TO ROTATION = (L / 2 ) * TAN (ROTATION) =

qu FACTORED DESIGN ROTATION (DM-4 14.4.2) =

c = 0.7Dqu + 0.125 =

LROTATION ANGLE – DEG

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Filename: 'file:///tt/file_convert/5489032db4795921448b45af/document.xls'#$Bolt Check

Calculate the maximum horizontal force for bolt checks:

SEISMIC FORCE (EQ) AASHTO 3.10

Acceleration Coefficient, A = 0.05 DM-4 Figure 3.10.2-1Seismic Zone = 1 AASHTO Table 3.10.4-1Factor EQ/DL = 0.2 See below

For transverse force, the tributary load is the dead load reaction at this bearing.Total Brg DL Reaction = 220.45 kips

Transverse seismic force EQ(T) = 44.09 kips/bearing

Longitudinal force goes to the fixed bearings.Number of Long. Fixed Bearings = 0Longitudinal seismic force EQ(L) = 0.00 kips/bearing

WIND ON STRUCTURE FORCE (WS) AASHTO 3.8.1.2.2

Structure depth = 11.41667 ft. (use max grdr dpth + 4"hnch + 8"dck + 42"barrier )Length = 64.250 ft. (1/2 span 1)

Arm, trans. force to bearing = 5.708333 ft. (from mid-height of superstructure)Girder spacing = 29.66667 ft.

I = 440.1 ft.^2 (pile group analogy)Vertical force due to

Wind Pressure, KSF Force, kips overturning effect, kipsAngle Trans. Long. Trans. Long. Ext. gdrs.

0 0.050 0.000 36.68 0.00 42.3415 0.044 0.006 32.27 0.00 37.2630 0.041 0.012 30.07 0.00 34.7245 0.033 0.016 24.21 0.00 27.9460 0.017 0.019 12.47 0.00 14.39

FRICTION FORCE FROM BEARING DESIGN (FR)

22.05 kipsLONGITUDINAL HORIZONTAL FORCE (USE m * DL) =

SO No.: 104906

Subject : SR 31 OVER SR 219

BRONZE BEARING ANCHOR BOLT CHECK

Computed By: JWB Checked By: Date: 4/6/2005

AASHTO 3.10.9.2 - The horizontal design connection force in the restrained directions shall not be taken to be less than 0.2 times the vertical reaction due to the tributary permanent load. (No live load is included per DM-4 3.4.1.)

Each bearing is to be designed to resist all the wind force on the pier, in accordance with DM-4 14.6.1.2P. Therefore the forces in the table below are assumed to go to one bearing.

SO No. 104906




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SO No.: 104906




SO No. 104906




By inspection, Strength III controls the maximum horizontal force for bolt and weld design:

Total Strength III transverse force:

WS load factor = 1.4WS force = 36.68 kips

Factored transverse WS force = 51.35 kips

Total Strength III longitudinal force:

FR load factor = 1.0FR force = 22.05 kips

Factored longitudinal FR force = 22.05 kips

Resultant = 55.88 kips

Shear Check

Number of bolts 2

Resistance factor for shear 0.8 (AASHTO 6.5.4.2)

Diameter of bolt 1.25 in

1.227 in^2

Min. tensile strength of bolt 120

Number of shear planes 1

Strength horizontal reaction 55.88 kips STR-III

2 * 0.8 * 0.38 * 1.227 * 120 * 1 = 89.54 kips

Check:Is 89.54 > 55.88 ?? YES

Nb =

fs =

db =

Area of bolt = p * db^2 / 4 = Ab =

Fub = ksi (ASTM A325) (AASHTO 6.4.3)

Ns =

STR Rh =

Rr = fs * Rn (AASHTO 6.13.2.2-2)

Rn = 0.38 * Ab * Fub * Ns (AASHTO 6.13.2.7-2)

Rr = Nb * fs * 0.38 * Ab * Fub * Ns =

Rr =

Is Rr > STR Rh ??

9 OF 12


SO No.: 104906




SO No. 104906




Bearing Check

Number of bolts 2

Diameter of bolt 1.25 in

Thickness of top plate 1 in

Thickness of bottom plate 1 in

t = 1 in

Tensile strength of connected material 58Clear distance between holes or between the hole

and the edge of the material 1.1875 in

Resistance factor for bolts in bearing 0.80 (AASHTO 6.5.4)

2 * 0.8 * 1.0 * 1.1875 * 1 * 58 = 110.20 kips

2 * 0.8 * 2.0 * 1.25 * 1 * 58 = 232.00 kips

Minimum of 110.20 or 232.00 = 110.20 kips

110.20 kips

Check:Is 110.20 > 55.88 ?? YES

Nb =

db =

ttp =

tbp =

Thickness of connected material = t = MIN( ttp or tbp)

Fu = ksi (Gr 36)

Lc =

fbb =

fbb * Rn = Nb * fbb * 1.0 * Lc * t * Fu

fbb * Rn =

fbb * Rn = Nb * fbb * 2.0 * db * t * Fu

fbb * Rn =

Rr =

Is Rr > STR Rh ??

10 OF 12

Filename: 'file:///tt/file_convert/5489032db4795921448b45af/document.xls'#$Welds

Top and Bottom Plate Welds

Input

Resistance factor for fillet welds in shear 0.8 (AASHTO 6.5.4.2)Horizontal load P = 55.88 kips

Rolled beam bottom flange width (minimum) 24 in

Supporting girder top flange width (minimum) 24 in

Minimum weld size W = 0.3125

Classification of weld metal 58 ksi (for EXX70 Electrode)

Minimum tensile strength of structural steel 58 ksi (for Fy = 36 ksi)

Minimum edge distance ED = 1

Calculations

Strength of weld metal:

(AASHTO Eq. 6.13.3.2.4b - 1)

Strength of connected material:

Therefore,

0.6 * 0.8 * Minimum( 58 or 58 )

27.84 ksi

Length of weld = L =L = 2 * ( Minimum( 24 or 24 ) - 2 * 1 )L = 44 in

55.88 / ( 27.84 * 44 * 0.707 )

0.065 in

Maximum( 0.065 or 0.3125 ) Required weld size = 0.3125 inch

fe2 =

Wcp =

Lsp =

in (AASHTO Table 6.13.3.4-1)

Fexx =

Fu =

in (DM-4 Table 6.13.3.10P)

Rr = 0.6 * fe2 * Fexx

Rr = 0.6 * fe2 * Fu

Rr = 0.6 * fe2 * Minimum( Fu or Fexx )

Rr =

Rr =

2 * ( Minimum( Wcp or Lsp ) - 2 * ED )

Weld size req'd = Wr = P / ( Rr * L * 0.707 )

Wr =

Wr =

Therefore, Required weld size = Maximum( Wr or W ) =

SO No.: 104906


BRONZE BEARING WELD DESIGN


Per AASHTO 6.13.3.2.4b, "Fillet-welded connections subjected to shear on the effective area shall be taken as the lesser of either the factored resistance of the connected material specified in Article 6.13.5 or the factored resistance of the weld metal."

SO No. 104906




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Filename: 'file:///tt/file_convert/5489032db4795921448b45af/document.xls'#$Welds

SO No.: 104906


BRONZE BEARING WELD DESIGN


SO No. 104906




= 5 / 16 inch

Note: Based on above weld calculation, use 5/16" weld for all bearing welds.

EXPANSION BRONZE BEARINGS IIIE

LOCATIONDEAD LIVE TOTAL

MARKDIMENSIONS AT CL BRG

WEIGHT (LB)LOAD (K) LOAD (K) LOAD (K) B J L T1 T2

PIGGYBACK SPAN 1 173 158 331 BR01 9 1/8" 12 3/4" 21" 1" 1" 405

PIGGYBACK SPAN 3 37 71 108 BR02 9 1/8" 12 3/4" 21" 1" 1" 405

NOTE: THE LOADING VALUES ARE UNFACTORED.

Documents

Bronze Bearing Design