bridge bearings - caltrans · bridge bearings . introduction . aashto defines a bearing as "a...

I bull MEMO TO DESIGNERS 7-1 bull JUNE 1994 ----c=~~~

11ztrans

Bridge Bearings

Introduction

AASHTO defines a bearing as a structural device that transmits loads while facilitating translation andor rotation 1 In the past Caltrans has used a variety of bearings with varying degrees of success These include rockers rollers pins pots steel girder hangers PTFEelastomeric and elastomeric pads Of all the bearings mentioned the reinforced elastomeric bearing (introduced in 19 55) has been the most widely used the past four decades

As design trends have shifted toward designs that favor structures with longer frames and fewer joints for seismic reasons not to mention the widespread use ofcurved and skewed bridges the demands on bearings have increased Provisions must be made for large longitudinal displacements due to temperature prestress shortening shrinkshyage creep and seismic activities as well as rotations produced by changes in camber live load and misalignment of bearing seats due to construction tolerances In shon Designers need a selection of bearing types to handle varying demands

Increased demands on bearings have led to the development of new bearings (post World War II) Improvements in engineering materials particularly plastics and elastomers are largely responsible for the innovative designs and refinements made in the past three decades The three new bearing types most widely used today in the United States are pot spherical and disk Collectively these bearings are known as High Load Multi-Rotational bearings Of the three bearing systems mentioned spherical bearings have the greatest rotation capacity and most trouble-free mainteshynance record Pot bearings have been troublesome in the past and are still not considered trouble-free Disk bearings on the other hand have fewer documented failures than pot bearings however up until 1992 they were a patented system made by a single manufacturer In addition to the three bearings mentioned above CaltralS has used PIFEJelastomeric bearings on several structures with large longitudinal displacements

Supersedes Memo to Designers 7-1 dated November 1989

BRIDGE BEARINGS PAGE 1

4lf llllr --------------------------~M~E~M~O_T~0~0=E~SI~G~N~E~R~S~7-middot~1_bull~J~U~N=E~1~9~9~4 Gzltrans

Bearing Selection

Bearing selection is influenced by many factors such as loads geometry mainteshynance available clearance displacement rotation deflection availability policy designer preference construction tolerances and cost The designer must consider all the applicable variables early in the design stage and design the structure and bearing as a unified system Too often bearings are selected at the last minute when forces and available space are fixed Such an approach increases the chances of future mainteshynance problems

The official policy of the Division of Structures is to avoid using an alternative bearing system where a conventional reinforced elastomeric pad can provide the required characteristics through shear deformation When the practical limits of elastomeric bearing pads are exceeded designers should consider using PTFF1 spherical or PTFEelastomeric bearings The three bearing systems mentioned should provide enough versatility to satisfy the design requirements of most strucshytures designed by the Division of Structures Other bearing systems may be approshypriate for special circumstances designers should consult with the Bearing Technical Specialist for unique applications

On widenings designers are cautioned against mismatching bearing types It has been common practice to use elastomeric bearing pads a yielding bearing to widen structures supported on steel rocker bearings an unyielding bearing While this practice has worked satisfactorily on short to moderate length structures it has created problems when thick elastomeric bearing pads have been used on structures with long spans 6

PAGE2 BRIDGE BEARINGS

4lf llllr --------------------------~ ~ ~9~9~4M~EM~O~T~O~D~E~S~IG~N~E~A~S~7~-~1_bull~JU~N~E~1lbltrans

Reinforced Elastomeric Bearing Pads

General

Reinforced elastomeric bearing pads designed in accordance with Bridge Design Specifications Section 14 should be considered the preferred type of bearing for all structures The typical hinge or abutmentconfigurationusing many small elastomeric bearing pads has proved highJy reliable and redundant In addition these bearings are extremely forgiving of loads and translations exceeding those considered in design

Designing and Detailing Notes

The following data illustrates our current practice and provides practical information about elastomeric bearings In addition design examples and various charts in this memo provide background for the Bridge Design Specification

Pad thickness is determined in increments of 1h inch Minimum laminated pad thickness is one inch Maximurn laminated pad thickness is 6 inches at abutments and 4 inches at hinges Plan dimensions (length width) are determined in two inch increments Maximum dimensions should not exceed 30 inches

The minimum shape factor (S) for any reinforced bearing shall be 50 Unless shear deformation is prevented the average compressive stress Oc in any layer of any reinforced bearing with an S ~ 50 shall not exceed 800 psi The minimum average compressive stress due to dead load will not be less than 200 psi The Transportation Laboratory has found that as overall pad thickness increases the compressive stiffness of the pad decreases although the shape factor is held constant 3

Rotational stresses may be minimized by specifying the smallest pad width possible within the limits of the application Orient rectangular bearing pads so that the long side is parallel to the axis about which the largest rotation occurs (see Figure 1 page 9)

All pads ata hingeorabutment should be thesamesizeand oriented similarly Ensure that the orientation is clearly detailed

Bearing pads onskewedstructures should beoriented parallel to the principal rotation axis (see Figure 1 ) When insufficient seat width exists the bearing pads may be oriented normal to the support The effects of skew andor curvature must be considered This may result in varying the pad spacing to accommodate the increased load at the obtuse comer Minimum loads must be maintained to ensure that slippage (movement) of the bearing does not occur


4lf llllr --------------------------~M~EM~O~T~0~0~E~S~IG~N~E~R~S~7-~1_bull ~~J~U~N~E~1~9~94 lizatl16

Slippage will be prevented by maintaining a minimum compressive force five times greater than the largest possible shear force under all service load conditions including live load plus impact

The effects ofprestress shortening creep shrinkage and thermal movements will be included in bearing pad designs The Bridge Design Specifications (Article 1426) state that the shear deformation shall be taken as the maximum possible deformation caused by creep shrinkage post tensioning and thermal effects unless a positive slip apparatus is installed

Testing at the Transportation Laboratory with positive slip apparatuses have shown that prestress shortening may be partially accommodated by placing a greased galvanized sheet metal plate (sliding bearing) above the pad The plate should extend a minim urn ofone inch in all directions beyond the calculated movement (See Figure 2 page 1 0) Long term tests have demonstrated that 50 percent of the total anticipated prestress shortening may be relieved by this sliding bearing without any significant shear deformation of the elastomeric bearing pad The remaining prestress shortenshying creep and shrinkage must be included in the bearing pad design Note that prestress shortening may continue beyond the calculated long term shrinkage panicularly in the case of shallow structures with depthspan ratios less than 004

The prestress shortening percentage (50 percent) used to design the bearing pad may be reduced at hinges that have delayed hinge closure pours when the sliding bearing detail is utilized The reduction may be calculated by adding 20 weeks to the duration of the closure pour waiting period and determining a new shortening value from the prestress shortening curve (Attachment 1)

The designer needs to specify silicone grease on the plans when using the sliding bearing to differentiate it from the previously used multipurpose automotive and industrial greases Testing at the Transportation Laboratory _has demonstrated that multipurpose petroleum base greases do not provide the desired sliding effect and may damage the elastomer because they are absorbed by the pad in a very short period of time

Minimum edge distance to any vertical face (backwall face of abutment or hinge seat) should be equal to t (pad design thickness) or 3 inches whichever is greater For cast-in-place structures surround the bearing pads with polystyrene of the same thickness as the actual pad thickness (See Figure 3 page 10)

Plain pads are acceptable during stage construction of precast prestressed girder superstructures that are continuous for live load where in the final condition the bent cap becomes monolithic with the girders and slab


fit llllr --------------------------~M~EM~O~T~O~D~E~S~IG~N~E~R~S~7~-~1_bull~J~U~N~E~1~9~9~4

liztltz1$

Steel reinforced and fabric reinforced pads have different design criteria Where possible the designer should prepare both designs (with one set of details) and allow the contractor the choice as specified in Section 51-112H of the Standard Specificashytions The following is an example of a note that should be shown on the plans

Fabric reinforced elastomeric bearing pads 22 x 28 x 2 orsteel reinforced elastomeric bearing pads 20 x 26 x 2 (elastomer only)

In most cases the designer need not be aware of the increase in thickness due to the steel reinforcing since the design thickness relates to the thickness ofelastomer and changes due to the actual thickness are taken care of in the specifications and during construction Exceptions may include retrofit projects where the actual thickness should be shown on the plans In these cases a subtitution would not be allowed and the plans should state this clearly

Properties of Elastomer

Durometer Hardness 55 plusmn5

Creep in Compression 25 of initial vertical deflection

Shear Modulus (Adjusted)

70degF = 135 psi 20degF 110 x 135 = 149 psi 0degF 125 x 135 = 169 psi -20degF 190 x 135 = 257 psi

Shear modulus (G) is the most important material property for design and it is therefore the preferred means ofspecifying the elastomer Hardness has been widely used in the past because the test for it is quick and simple However the resul~ obtained from it are variable and correlate only loosely with shear modulus 1

The shear modulus of elastomer obtained from testing is approximately 100 psi at 70degF 3 The design value was increased to 135 psi at 70degF to include a safety factor of 35 percent against horizontal overloads For design calculations use the modulus at 0degF (169 psi) unless temperatures will be substantially lower

Railroads require 60plusmn5 hardness on their structures Specifications handles the change and there is no change in design


4lf Jlllr --------------------------~M~E~M~O~T~0~0~E~S~I~G~N~ER~S~7--1~middot-J~U~N~E~19~9~4 liztrans

Elastomer properties are in accordance with the Research Report A Laboratory Evaluation of Full Size Elastomeric Bridge Bearing Pads dated June 1974 by the Transportation Laboratory of the State of California except as noted

Design Criteria

1 Temperature movement shall be calculated as per the examples in this memoranshydum In calculating movement use 11h times the coefticients as it is not possible to always place the pad at a mean temperature Rise and fall temperature values are given in the Bridge Design Specifications Article 316

2 Long term prestress shortening and shrinkage shall be included in the bearing movement calculation Prestress shortening per 100 feet of contributory length will equal 010 feet for post tensioned concrete structures and a minimum of001 feet for pretensioned concrete structures Shortening (shrinkage) ofconventional reinforced concrete structures will equal a minimum of0005 feet per 100 feet of contributory length Fifty percent of the prestress shortening may be discounted when the sliding bearing is used (See Figure 2 page 10)

3 Pad thickness shall not exceed 113 of the length or width or be less than twice the calculated horizontal movement Maximum thickness for plain pads is 12 inch Maximum thickness for laminated pads is 6 inches at abutments and 4 inches at hinges Minimum thickness for laminated pads is -1 inch two lh-inch layers) When design procedures require a pad thickness greater than the maximum recommended thickness investigate the use of PTFE bearings

4 Average pressure on the pad shall not exceed 800 psi under a service load combination of dead load plus live load not including impact For steel reinshyforced bearing pads with a Shape Factor 7 5 the average pressure shall not exceed 1000 psi Minimum pressure on any pad due to dead load shall not be less than 200 psi

5 Initial vertical deflection (compressive strain) shall not exceed 7 percent (excludshying the effect of rotation) of the uncompressed thickness of the pad

Determine the initial compressive strain from Figure 4A or 4B (page 11 and 12) Compressive Strain Percentage using the compressive stress and shape factor

PAGE 6 BRIDGE BEARINGS

~I

4 jllr ------------------~----~M~E~M~O_T~0~0~ES~I~G~N~ER~S~7- _~A~R~CH~~9~96~1_bull M 1lizltlt111S

Note

one loaded surface area Shape Factor= -------- shy

total free- to - bulge area

WxL = ----------------------2(W+ L) x(thickness per layer)

Where

L = length of bearing pad in direction of horizontal movement W = width of bearing pad normal to direction of horizontal movement

For laminated pads since the thickness per layer is always 112 the formula reduces to

WLShape Factor = - shy

W+L

6 If some combination of service loads (including live load plus impact) exists which causes a shear force greater than 1A of the simultaneously occurring compressive force the bearing should be secured against horizontal movement

No experimental detennination has been made of the starting friction between bearing pad and concrete or steel But it has been observed in laboratory tests on full size pads that slippage does not occur so long as the shear stress does not exceed 1A the compressive stress acting on the interfaces Therefore the shear force in bearing pad designs should be limited to 1A the minimum vertical load (usually Dead Load)

ElastomericBearing Pad Coefficient of Friction for Seismic Analysis pound6 ~ 0

Sf- ~ ~~~ ~~~ f 0

t ) ~

Seismic analysis ofeXkfuig structures with elastomeric bearing pads often equirei a determination of1M friction middotcoefficient between the bearing pad and concrete

~ ~

substructure -~ ~ middot 0 ~ ~ ~K ~~ _~ ~ middot~ ~

l ~ -middotmiddot~~ ~ if

0Bearings onnew structures are normally designed to Tesist slippage by linliting the shear force to 15 the ininimorh compressive stress acting on the neopreneconcrete interface ie~ coeffiCJentofJnction equals 020 This value is conservative and - enSures that ~gs~o hot~p out of position under service load conditions However tlie value ismiddotunconservatively low for seismic analysis and should not be used to determine subsrrucrureforces


4f illr ------------------------~M~E~M~O~~~O~D~E~S~IG~N~E~R~S~7~-~1-bull~M~A~R~C~H~1=9==96

lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy

- middot Friction aeterminarl~anin~actscience dePeident on many vari~Ieswhicbmiddotare- bull -- -- iJ1 il ~(~--~ bull ~ ~ ~ ~

-~~~-1 ~_~oot easQf~~~~-~ljd~~~r~~~~foremiddota~~~~~~~~t~ middotmiddot - middotvalue must be ~tt ensure thaf-substroctureltforces are-middot not underestiJilatcuAmiddot

T ( ltr irl~middot ~gt ) middot review ofseveral tistlepOrts indicates that fiietion coefficients equal to 040 for

o

concrete to neoprene31ld035 forsteel to neoprene interfaces aremore realistic values for seismic analysis middot-~ middot

middot _ - AL~-~ middot - ~o bullbull --middot middotxamp~ _ ~middot middot -~ ~ 4R bull bull~ke bullbull middot IDfu middot ~R 5 -~ -~-w t~ -~ bull ~ middot$

Desi~shltgtUld alsO mvestigare~~ maxim~force exerted by the bearing pad throughmiddot shear translation priorto middotslippage anadetermine which Caslaquo oo1_1trols It is

_~ --=estimateltitbat elastomericbearing~ will reSist a maximum shear straW ofplusmn~Q middot -7~ middot ~~rcentbpnor tQ fa95middot ~g~~tests~~~~~ed sho~ negligible~d~age~o

middotela5torriericbeiuiOgtffiristatetlplusmn1OO~rcentJttlieifdesign thickness plusmn100 perceiit - ltjillyen ~

shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force

Shear Modulus = _S_rre_s_s = __ ___Pa_d_Area Strain Deflection

Pad Thickness

Fs G = A = Fs T

bs Ab - s

T

FsT GMsRearrangmg tenns d s =- and Fs = --GA T

Substituting G =169 psi 0degF and Fs max= Vs DL

PAGE e BRIDGE B EARINGS

4lt llllr --------------------------~M~E~M~O~T~0~0~E~S~IG=N~E=R~S~7~middot~1-middot~J~U~N~E~1~9~9~4

tizltran6

Jr---1__ ~ Bearing and principle rotation axis

Plan Pad at Centerline of Girder

lttAbutment

Section 8 -8

Figure 1 Bearing Pad Orientation


4lt llllr --------------------------~M~EM~O~T~O~D~E~S~IG~NE=R~S~7- ~JUN~E~19~9~4~1~middot ~~liztrll15

trExtend galvanized sheet metal plate beyond pad a sufficient distance to provide for long-term prestress shortening + 1h temperature movement (1 minimum in all directions beyond calculated movement)

14 ga (0079 in) galvanized sheet metal Coat top of pad with silicone grease

Figure 2 Sliding Bearing Detail

Expanded polystyrene same thickness as bearing pad

Figure 3 Detail

Note Abutment shown - hinges similar


------------------------------------------------------- ---------------

illlr --------------------------~M~EM~O~T~0~0~E~S~I~G~N=E~RS~7--1~middot~J~U~N~E~1~9~9~4

littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300

Rberglass Reinforced Bearing Pad 55 Durometer Hardness

200

100

0 0 1 2 3 4 5 6 7 8

Compressive Strain percentage

Figure 4A Recommended Compressive Stress vs Strain Curve for Fiberglass Reinforced Bearing Pads

PAGE 11

I MEMO TO DESIGNERS 7-1 bull JUNE 1994 --~~~~

tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4


Recommended Compressive Stress vs Strain Curves for Steel Reinforced Bearing Pads


4lf illlr --------------------------~M~EM~O~T~O~D~E~S~IG=N~E~R~S~7~middot~1-middot~J~U~N=E~1~9~9~4

liztrans

Design Procedures

Example 1 Fabric Reinforced Pad Steel Girder Structure

Given

Span Length - one end flxed =112 feet Contributory Length -structural steel = 112 feet DL Reaction - Service Load = 69 kips LL Reaction - Service Load = 56 kips Bottom Flange Width = 16 inches Moderate Temperature Zone- Rise and Fall = 50degF

Step 1 Temperature Movement= 15 x 00000065 x 50deg x 112 x 12 = 066

Step 2 Concrete Shonening-Not Applicable (Steel Girder)

Step 3 Minimum Thickness ofPad= 2 x horizontal movement= 2 x 066 = 132 Try T= 15

W =flange width= 16

Maximum Pressure= DL+ LL = 800 psi WxL

DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16

Trial Pad 10 x 16 x 15

10 Maximum Thickness of Pad 13 of length=-= 333 gt 15

3 T= 15 Okay

Step 4 Compressive Stress

DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10

Compressive Stress (Dead Load Only)

DL 69000 = =431 pSIgt 200

WxL 16x10


I bull MEMO TO DESIGNERS 7-1 bull JUNE 1994 --~~~~

lbltftlllS

Step 5 Initial Vertical Deflection

W x L 16x10 Shape Factor = = ( ) = 615 gt 50

W+L 16+10

From Figure 4A for compressive stress= 781 psi and Shape Factor= 61 initial compressive strain is 51 percent of pad thicknesslt 7 percent (by extrapolation)

Initial Thickness= 150- (15 x 0051) = 142

Final Thickness after creep= 142- (15 x 0051) x 025 = 140

Step 6 Calculate the maximum shear force at slippage (minimum compressive force= DL in this example)

DL 69 Fsmu =-=-= 138 kips

5 5

S F F GA~sActual Des1gn hear orce s = = T

Modulus x Area x Movement= 169 x 160 x 066 = 1 kips1 9

Pad Thickness x 1000 15 x 1 000

11 9 kips lt 138 kips

Therefore T= H-2 Okay

Specify Fabric Reinforced Elastomeric Bearing Pads 10 x 16 x 15

To reduce rotational (compressive) stresses orient rectangular bearing pads so that the long side is parallel to the axis about which the largest rotation occurs

To complete the design repeat steps 3 through 6 using the design criteria for steel reinforced pads (if applicable) and include this design on the contract plans


- 4lf llllr --------------------------~M~EM~O~T~O~D~E~S~IG~N~E~R~S~7-~1-bull~J~U~N~E~1~9~9~4

tizltlans

Steel Laminated Elastomeric Bearing Pads

Background

Our policy has been to standardize on lh layers ofelastomer Until recently we used very thin steel plates and a minimal elastomer cover at the top and bottom for the steel reinforced pads The minimal thickness of cover and of steel was ignored and the bearing thickness shown on the plans was the sum of the lh -inch thick layers This resulted in a simple standard Cal trans procedure for the design and manufacture of both the fabric reinforced and steel reinforced bearing pads

Steel reinforcement option was removed frorrt the 1981 Standard Specifications because the bearing manufactures could not properly mold the bearing with the thin steel plates

Current Policy

The current specifications for elastomeric bearings permit the use of the steel reinforced bearing as an option However the proper design of the steel reinforced bearings requires 14 gauge (0075 inch) steel plates full lh-inch elastomer layers between the plates and a lA-inch cover top and bottom These two 14-inch layers are considered one lh-inch layer for design purposes Therefore because ofthe steel plate thickness the steel reinforced bearing will always be slightly thicker than the corresponding fabric reinforced (fiberglass) bearing pad

Design

In permitting the use of the steel reinforced bearing as an option the specifications require that the contractor notify the Resident Engineer of their choice If the steel reinforced bearing is selected the bearing seat elevation will be adjusted (lowered) by the Resident Engineer to allow for the increased thickness The minor increase in compression on the steel plates due to the 1h side cover may be ignored

For most cast-in-place concrete precast concrete and steel superstructures there should be no difficulty in adjusting the bearing seat elevation at the time the contractor selects the bearing type In general there is no need for the designer to be concerned with the choice

For some applications the designer may want to limit the bearings to only one of the two types If this is the case the designer should send a memo to the specification writer who will denote the specific type of pad in the Standard Special Provisions


I bull MEMO TO DESIGNERS 7-1 bullJUNE 1994------=~~~

1zltrans

The maximum size ofsteel reinforced bearings is governed by the fabricators ability to vulcanize a large volume of elastomer uniformly and completely 4 Since elastomers are poor conductors of heat achieving a full cure in the center of the bearing without overcuring the outside becomes increasingly difficult as the bearing size increases 4 Steel reinforced elastomeric bearings should be limited in size to approximately 500 kips based on an allowable stress of 1000 psi to ensure proper vulcanization of the elastomer

The new LRFD Bridge Design Specifications will allow a maximum allowable compressive stress of up to I 600 psi in the absence of rotation (service limit) To utilize the higher stress limits the designer will have to use a more complex design procedure and specify more rigorous testing The current Bridge Design Specificashytions allow alternative design procedures as outlined in NCHRP Report 298

The various modes offailure for steel reinforced pads are debonding fracture ofsteel plates and instability 4

Steel reinforced bearings have a greater overload capacity before failure than fabric reinforced pads The Transportation Laboratory reponed that the ultimate compresshysive stress of steel reinforced pads was approximately 6000 psi before the 14 gage steel yielded In comparison the ultimate compressive stress offabric reinforced pads was approximately 1800 psi 2 Therefore steel reinforced elastomeric bearings provide a greater factor of safety against overloads than do fabric reinforced

Holes are strongly discouraged in steel reinforced bearings However ifholes are used their effect should be accounted for when calculating the shape factor because they reduce the loaded area and increase the area free to bulge 1 Suitable shape formulae are

LW -ltr d2

for rectangular bearings Smiddot = 4 hn[2L+ 2W + lmi)

for circular bearings D2 -UJ2

S=----shy4hn(D+Ld)

where

L = length of a rectangular bearing (parallel to the longitudinal bridge axis) (in) w = width of bearing in transverse direction (in) hri = thickness of the elastomeric layer (in) d = diameter of the hole or holes in the bearing (in)


41 illir --------------------------~ I ~1_bull JU~N~E~~9~9~4M~EM~O~T~O~D~E~S~G~N~E~R~S~7~- ~~ 11zltIUI$

D = diameter of the projection of the loaded surface of the bearing in the horizontal plane (in)

To assist the designer the total thickness for steel reinforced bearings is tabulated below

Design Number of Number of Acrual Thickness

Thickness 12 Layersbull Steel Plates Minimum Maximumbullbull

10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724

Plan Dimension Length Width or Diameter 18 Min

V4 Max1

Internal elastomer layers 0075 Plates

Steel Laminated Elastomeric Bearing Pad 15 3-Layer Pad Shown

The 14 layer top and bottom is considered one lh layer for design purposes Includes consideration for allowable tolerances for steel plates and ela~tomer thickness


illlr --------------------------~M~EM~O~T~0~0~E~S~IG~N~E=R~S~7-~1~middot~J~U~N~E~1~9~9~4 11zlaans

Example 2 Steel Reinforced Pad - CIP PIS Structure

Given

Span 5 Hinge (See Attachment 2)

Structure Length = 7 40 feet Contributory Length- 100 (CIP PS) 156 (PCC) = 256 feet DL ReactionGirder- Service Load = 148 kips LL ReactionGirder- Service Load No Impact = 41 kips Moderate Temperature Zone- Rise and Fall = 35degF Sliding Bearing Used

Step 1 Temperature Movement= 15 x 00000060x 35 x 256 x 12 = 097

Step 2 Concrete Shortening (sliding bearing accounts for 50 percent ofshortening)

PIS Shortening 05 X (010100) X 100 X 12= 060

Concrete Shrinkage (0005100) x 156 x 12 =009

Step 3 Minimum Thickness of Pad = 2 x horizontal movement= 2 x (097 + 060 + 009) = 332 Try T= 35 (elastomer only)

p DL+IL Maxnnum ressure = = 1000 pst

WxL

148 41wXL= + = 189 in2 = 137 X 137 10


Note 14 X 16 pad was selected because 14 X 14 pad has a shape factor of only 7 0 which results in a maximum compressive stress of 800 psi

Maximum Thickness of Pad 13 of length or width= ~ = 47 3

T= 35 Okay


jllr --------------------------~ ~1~middot~J~U~N=E~1~9~9~4M~EM~O~T~0~0~E~S~IG=N~E~R~S~7-11ztrtzn$


DL+ LL = 148+ 41 = si lt l OOO843 Okay (ShapeFactor~7 5W x L 14 x 16 P required)


148DL = = 660 si gt 200 Okay

WxL 14xl6 P


W xL 14 x16 Shape Factor= = = 747- 75

W+ L 14+16

From Figure 4B for compressive stress = 843 and shape factor= 75 initial compressivestrain is45 percent of pad thicknesslt percent (by extrapolation)

Initial Thickness= 35 - (35 x 0045) = 334

Final Thickness after creep = 334 - (35 x 0045) x 025 = 330


DL 148 Fsmax = - =- = 296 kips (Allowable) 5 5

Actual Design Shear Force =

Modulus x Area x Movement= 169 x 16 x 14) x 166 = _ kips18 0

Pad Thickness x 1 000 35 x 1 000

296 kipsgt 180 kips Okay

Therefore T= 35 Okay

Specify Steel Reinforced Elastomeric Bearing Pads 14 x 16 x 35 (elastomer only)


To complete the design repeat steps 3 through 6 using the design criteria for fabric reinforced pads (if applicable) and include this design on the contract plans


AI( illlr --------------------------~M~EM~O~T~0~0~E~S~JG~N~E~R~S~7-~1~middot~J~U~N~E~1~9~9~4

tizltrDnS


Given

Span 5 Hinge (See Attachment 2) Due to construction staging spans 5 and 6 will not be completed for 1 year

Span Length = 740 feet Contributory Length- 100 CIP PS (Stage 1) 156 PCC (Stage 2) = 256 feet DL ReactionGirder- Service Load = 148 kips LL ReactionGirder- Service Load No Impact = 41 kips Moderate Temperature Zone- Rise and Fall = 35degF Sliding Bearing Used

Step 1 Temperature Movement= 15 x 00000060 x 35 x 256 x 12 = 097

Step 2 Concrete Shortening

PS Shortening (refer to Attachment 1) Enter graph with 72 weeks after stressing (52 week waiting period + 20 weeks for sliding bearing)

Shonening100 = 00251100 (Stage 1 after 72 weeks)

(0025100) X 100 X 11 = 030

Concrete Shrinkage (0005100) x 156 x 12 = 009

Step 3 Minimum Thickness of Pad = 2 x (097 + 030 + 009) = 272 Try T = 30 (elastomer only)

p DL+ LL 1 OOO Max1mum ressure = = ps1WxL

148 41 w XL= + =189 in2 =137 X 137 10


Note 14 x 16 pad was selected because 14 x 14 pad has a shape fac tor of only 7 0 which results in a maximum compressive stress of 800 psi


bull lr 1zltrans

MEMO TO D ESIGNERS 7 middot1 bull JUNE 1994 --~~~~

14Maximum Thickness of Pad 1J of lenth or width= = 47

- 3 7=30 Okay


DL+ LL = 148+41 = si lt l OOO843 Okay (Shape Factor~75 WxL 14 xl6 p required)

Compressive Stress (Dead Load only)

148DL = = 660 si gt 200 Okay

WxL 14x16 P


WxL 14xl6Shape Factor= = = 747- 75

W+L 14+16

From Figure4B for compressive stress =843 and Shape Factor= 75 initial compressive strain is 45 percent of pad thickness lt 7 percent (by extraposhylation)

Initial Thickness = 30- (30 x 0045) = 287

Final Thickness after creep= 287 - (30 x 0045) x 025 = 284


bull bull MEMO TO DESIGNERS 7middot1 bull JUNE 1994-------=~~~

Gzatl16

Step 6 Calculate the maximum shear force at slippage (minimum compressive force equals DL in this example)

DL 148 Fsmu = - =- = 296 kips (Allowable)

5 5


Modulus x Area x Movement= 169 x16 x 14) x 136 = kips17

2


172 kips lt 296 kips Okay

Therefore T = 30 Okay





41[ jlllr --------------------------~M~EM~O~T~O~D~E~S~IG~N~E~R~S~7~-~1_bull~J~U~N~E~1~9~9~4

11ztrans

PTFE Sliding Surfaces

General

Polytetrafluoroethylene (PTFE) was first used in bridge bearings in the early 1960s because of its low frictional characteristics chemical inenness and resistance to weathering 4 Many modern bearing systems such as pot spherical disk etc use PTFE in contact with stainless steel as a sliding surface

P1FE is usually used in the form ofsheet resin (filled or unfilled) and woven fabric made fromPTFE fibers 4 Although the acrual chemical formulation ofP1FE is oflittle interest to the designer the physical properties and performance characteristics are

The two most imponant design aspects of PTFE are the coefficient offriction and the wear rate 4 The coefficient of friction controls the forces transmitted to other parts of the bearing device and the substructure The wear rate affects the design life and maintenance of the bearing

Extensive research and testing by Stanton Roeder and Campbell has demonstrated that the coefficient of friction and the wear rate are affected by several variables in addition to the type of PTFE and mating surface

PTFE Types

Unfilled PTFE Sheet Resin (Dimpled Lubricated)

Dimpled lubricated PTFE manufactured from unfilled sheet resin has the lowest coefficient of friction under all load conditions 4 The dimples function as reservoirs that store the grease for continuous lubrication Testing by Stanton Roeder and Campbell has shown that the absence of dimples will result in a higher coefficient of friction as the grease is dissipated While dimpled lubricated PTFE has the lowest coefficient of friction it is also the most sensitive to problems and deficiencies 4 middot

Woven Fabric PTFE

Woven fabric PTFE made from PTFE fibers has a higher coefficient of friction than dimpled lubricated PTFE however it has a greater resistance to creep Woven fabric can take up to 30 times the compressive stress without cold flow as compared to PTFE resin 4


lllr --------------------------~M~EM~O~T~0~0~E~S~IG=N~E=R~S~7-~1-middot~J~U~N~E~1~9~9~4

tizltrtiiiS

Fil1ed PTFE Sheet Resin

Filled PTFE sheet resin has the highest coefficient of friction as compared to unfilled PTFE sheet resin and woven fabric PTFE Fillers such as glass fiber graphite and ceramics etc are incorporated in PTFE resins to alter propenies such as cold flow compressive strength etc 4

Factors that affect PTFE Performance 4

Friction

bull Friction increases with decreasing contact pressure decreasing temperature increasing sliding speed and increasing number of cycles

bull Static friction is usually higher than dynamic friction

bull Maximum coefficient of friction (spike) occurs at the onset of movement

bull Type of surface fmish on mating surface affects friction

bull Direction ofsurface finish on PTFE mating surface affects coefficient offriction Surface fmishes parallel to the direction of sliding give lower friction values

bull Friction increases to near static value ifPTFEsurface remains loaded for a period of time without movement This generally is nota problem onhighway structures

bull Lubrication reduces friction

bull Friction values could be 5 to 10 times higher at seismic speeds as compared to thermal expansion speeds

Wear and Creep (Cold Flow)

bull Sliding speed is the most significant factor affecting wear Increased sliding speed increases wear

bull Decreased temperature increases wear

bull Lubricant reduces the wear rate of sheet resin

bull Wear and creep increase with increasing loads

bull Sheet resin disks should be recessed half their thickness into their backing plate to reduce creep


illlr --------------------------~M~E~M~O~T~O~D~E~S~I~G~N~E~R~S~7- middot ~J~U~N~E~1~9~9~4- 1~lilltrans

Suggested Values for Coefficient of Friction

The minimum coefficient of friction used for design shall be as specified by the bearing manufacturer or as given in Anicle 1526 of the Bridge Design Specificashytions Designers are cautioned against using design friction values that are too low Low friction values are sometimes difficult to obtain in manufactured bearings and may result in transfer oflarger forces to the bearing and substructure than calculated

Suggested Design Bearing Pressures for PTFE Surfaces

The average bearing pressure on PTFE surfaces due to all loads shall not exceed

Filled PTFE 3500 psi

Unfilled PTFE (Recessed) 3500 psi

Woven Fabric PTFpound 3500 psi

Additional values may be found in Article 152 7 ofthe Bridge Design Specifications Bearing pressures below 2000 psi are not recommended as they produce high coefficients of friction and poor bearing performance Unfilled PTFE (not recessed) should not be used on Division of Structures designs


4lf illlr --------------------------~M~E~M~O~T~O~D==ES~I~G~N~E~R~S~7_-~1-middot~J~U~N=E_1~9~9~4

liztrans

PTFESpherical Bearings

General

PTFESpherical bearings designed inaccordance with Bridge Design Specifications Section 15 should be considered for use only when the practical limits of reinforced elastomeric bearing pads have been exceeded

The basic spherical bearing design is comprised of a convex base with a mating concave element for rotation On expansion bearings an upper sliding plate is added for translation All contact surfaces are polytetrafluoroethylene (PTFE) to stainless steel

PTFESpherical Bearing Types

PTFESpherical bearings are available in three different forms each designed to meet different functional requirements The three standard types ofPTFEJSpherical bearings in use today are (1) expansion (non-guided) (2) expansion (guided) and (3) fixed

Expansion non-guided bearings allow horizontal movement and rotation in all directions (see Figure 5 page 29) Expansion guided bearings allow horizontal movement along only one axis and rotation in all directions (see Figure 6 page 30) Fixed bearings are restrained from horizontal movement in all directions while allowing rotation in all directions (see Figure 7 page 31 ) Due to previous problems experienced with guided bearings they will not be considered for use by the Division of Structures

PTFESpherical expansion bearings (non-guided) are comprised of four basic comshyponents (1) sole plate (2) concave plate (3) convex plate and (4) masonry plate PTFESpherical fixed bearings are similar except that a nonsliding sole plate may or may not be required The function of the four components mentioned above and the typical manufacturing materials are as follows (see Figure 8 pages 32 and 33)

bull SolePlate-Transfers superstructure loads to the bearing and provides a stainless steel sliding surface for superstructure translation The sole plate is fabricated from A36 A36M steel and has a stainless steel surfacing

bull Concave Plate- Provides PTFE sliding surface for sole plate and PTFE concave surface for rotation The concave plate is fabricated from A36N36M steel A woven PTFE pad is epoxy bonded and mechanically fastened to the flat and concave surfaces Dimpled lubricated PTFE has been used by some manufac-


4lf llllr --------------------------~M~E~M~O~T~O~D~E~S~I=G~N~ER~S~7~-1~middot~J~U~N~E~1~9~9~4

tizltrtZnS

tures However woven PTFE fabric is preferred for this type of bearing and should be used for Division of Structures designs

bull Convex Plate - Provides stainless steel mating surface for rotation of concave plate and transfers load to masonry plate The convex plate is usually made from solid stainless steel or A36A36M with a stainless steel weld overlay

bull Masonry Plate- Transfers load from convex plate to bearing seat The masonry plate is fabricated from A36 A36M steel

Design Requirements

PTFESpherical bearings are designed in accordance with Section 15 of the Bridge Design Specifications All loads are service loads Minimum vertical loads are for dead loads and superimposed dead loads Maximum vertical loads are for dead loads superimposed dead loads and live loads plus impact PTFE fabric stresses are limited to 3500 psi maximum The coefficient of friction for fabric containing P1FE fibers varies from 008 to 004 at bearing pressures of 500 psi and 3500 psi respectively A design coefficient of friction of 006 is recommended for designs with bearing pressures from 2000 psi to 3500 psi Bearing pressures below 2000 psi (DL only) should not be used

Design Guidelines

The nucleus of all spherical bearings is the concaveconvex plate interface (spherical surface) All loads vertical and horizontal are transmitted through the interface Since the spherical interface slides on low friction materials (PTFE to stainless steel) all stresses that pass through the interface are assumed to be radially transmitted through the geometric center of the sphere (see Figure 9 page 34) The low friction interface is assumed to provide no frictional resistance to horizontal loads Due to the complexity of the analysis required to accurately determine the stresses at the concaveconvex plate interface simplified design guidelines were developed by bearing committees and adopted by AASHTO The procedure to design the interface complete with formula derivations and other PTFEspherical bearing componentS Is outlined below Refer to Figure 9 for bearing geometries


I 16 __ MEMO TO DESIGNERS 7-1 bull JUNE 1994___~~~

11ztrt111S

PTFESpherical Bearing Notations

Ah = total assembly height (inches)

APTFE =P1FE area of flat sliding surface (in2)

c = minimum vertical clearance between rotating and non-rotating bearing parts

Cm = minimum convex chord length (inches)

DBact = minimum concave bearing pad diameter (arc length)

Dm = diameter ofminimum allowable projected bearing area concave plate (inches)

H = height of convex spherical surface (inches)

Haa = overall height of convex plate (inches)

Lcp = length and width of concave plate (inches)

Lmu = maximum longitudinal movement (inches)

L0 = sole plate safety overhang (inches)

Lrp = longitudinal length of sole plate (inches)

Mm = minimum metal depth of concave surface (inches)

PHrnax = maximum horizontal load (kips)

PLrnax = maximum longitudinal load (kips)

Prrnax = maximum transverse load (kips)

Pvrnax = maximum vertical load (kips)

Pvmin = minimum vertical load (kips)

Ract = actual radius of concave bearing sliding surface (inches)

Rrnax = maximum allowable radius of concave bearing sliding surface (inches)

Tm = maximum transverse movement (inches)

Umu = maximum allowable unit pressure for PTFE (35 ksi)

W rp = transverse width of sole plate (inches)

arrun = minimum angle to prevent uplift (degrees)

f3 = minimum design rotation capacity ofbearing (3=f3s + f3c) 2 degrees minimum

f3c = maximum rotation res ulting from construction tolerances (002 radians)

f3s = minimum design rotation capacity of structure from DL U camber changes and constructionerection sequences

Ymin = minimum angle of convex surface (degrees)

Imin = minimum angle of concave bearing surface (degrees)


4f llllr --------------------------~M~EM~O~T~O~D~E~S~IG=N~E=R~S~7-~_~~~ ~9~9~41 bull JUN~E~1t1zlatln6

1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11

I I II I I I I I I I I 1 I I I

I I

_ -- l----- - --==

Plan

Anchorage Typical

Stainless Steel Sliding Surface Woven PTFE Pad

bonded to flat and ---- Concave Platemiddot concave surfaces

~~~~~~~~~~~~~~~~~~~~===----convex Plate _____ Masonry Plate Stainless ~~~~~~--~~~~--~~~~~~~~~~

Steel Convex Surface

Section X-X

Figure 5 PTFEJSpherical Expansion Bearing (Non-Guided)


- ---------

4lt lllr --------------------------~M~E~M~O~T~0~0~E~S~IG~NE=R~S~7- -J~U~N~E~19~9~4~1~middot

lizluans

I I I I lr------ -~----- ------------- ~ ~ ------- I II II II 11 I I 1l II I I 11 II I I I 1 1t

1111 Ill+111 I

1dr--+-shy 11 I II I I II I I II I I

II I II ---III----- - --== I I I I

I I I I I I

I Ill --t---~ I Ill I I II

I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan

Woven PTFE Pad bonded to flat and concave surfaces

Stainless Steel Convex Surface

Section X-X

Figure 6 PTFESpherical Expansion Bearing (Guided)


--

I bull MEMO TO DESIGNERS 7-1 bull JUNE 1994--------=~~~

tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan

Woven PTFE Pad bonded to concave surface


Section X-X

Figure 7 PTFESpherical Fixed Bearing

Sole Plate


-----

41 jlllr --------------------------~M~E=M~O~T~0~0~E=S=I~G~N=E~RS~7--1~middot~J~U~N=E~1~9~9~4

tbltrans

0 0 0

0

0

Stainless Steel Sliding Surface

Elevation

Transfers superstructure load to the bearing and provides stainless steel sliding surface for superstructure translation

Plan

Sole Plate

I I I I

Woven PTFE Pad I I

Epoxy bonded and mechanically Elevation fastened to flat and concave surfaces

Plan Provides PTFE sliding surface for sole plate and PTFE concave surface for rotation

Concave Plate

Figure 8 PTFESpherical Expansion Bearing Components


I bull MEMO TO DESIGNERS 7-1 bull JUNE 1994 -------=~~~

azJaans

Stainless ---- Steel Convex ~ Surface

r777J7h Elevation

Plan

Provides stainless steel mating surface for rotation of concave plate and transfers load to masonry plate

Convex Plate

middot

Elevation

Transfers bearing load to bearing seat

Plan

Masonry Plate

Figure 8 PTFESpherical Expansion Bearing Components (continued)

BRIDGE B EARINGS PAGE 33

llllr --------------------------~MEM~O~T~O~D~E~S~IG~N~E~R~S~7~_~~~ ~9~9~4~ ~-1 bull JUN~E~1tizallll$

c 1-------m12---= 1 D ~~--D-m12--~

Concave Plate---~middot~lt

Convex Plate (_

Center of Sphere

Figure 9 Spherical Bearin_g Geometries


__ ______~~~bull bull MEMO TO DESIGNERS 7middot1 bull JUNE 1994

lizltriZIIS

Design Procedure

Concave Plate

bull Diameter (Dm) of Minimum Allowable Projected Bearing Area

The minimum diameter (Dm) of the concave spherical plate must be large enough to ensure that the maximum bearing stress ( CTu) on the horizontal projected area of the plate does not exceed the maximum allowable stress on the PTFE fiber (3500 psi) Therefore the minimum diameter (Dm) may be determined from the maximum venicalload (Pvmu ) and the PTFE maximum allowable unit pressure (Umv)

Concave Concave Sliding Surface Plate Projected Area

Pvmv = Maximum Venical Load - - ~ ---+--------- shyI

Umax =Maximum allowable unit I I

pressure (PTFE = 3500 psi)

Derivation

I

Pvmu ]]2 1 Pvmu =Area= nDr Dm = 4 Umax Dm = 2[ Pvmax ]2 Umax 4 1C 1r Unux[ [

bull PTFE Area (APTFE) ofRat Sliding Surface

The flat PTFE sliding surface area on the concave plate should be size to the nearest 025 inch using the maximum allowable stress on the P1FE fiber (3500 psi)

Concave Plate

Pvmu Flat PTFE Ap1FE =--

UmaxSliding Surface


lllir --------------------------~M~EM~O~T~O~D~E~S=IG~N~E~R~S~7~-~1_bull~J~U=N~E~1~9~9~4

tizluans

bull Minimum Angle (lXnun) Required to Prevent Uplift

The minimum angle (amin) is used to calculate the spherical radius required to resist the greatest ratio ofhorizontal to vertical load without unseating of the concave plate On expansion (non-guided) bearings the maximum horizontal load cannot be transferred from the sole plate to the concave plate because the sole plate is free to slide horizontally Therefore on expansion non-guided bearings it is recommended that a horizontal load equal to 10 percent of the maximum vertical load and a minimum dead load of 50 percent of the maximum vertical load or the actual minimum dead load whichever is smaller be used to determine amiD

For fixed bearings the horizontal to vertical ~oad ratio should not exceed 40 percent when using simplified design procedures Using the simplified procedure on load cases above the 40 percent level will result in over stressing the PTFE fabric at the spherical interface Hence external shear devices are required to fix bearings during seismic activity exceeding the 40 percent ratio

PHmax= Max horizontal load

Pvmin= Min vertical load (including uplift)

PH maxamin =tan- _ _ _

Pvrruo

It is evident from the formula above that spherical bearings virtually have no resistance to horizontal loads unless a vertical load is present Without the presence of a vertical load the concave plate will ride up and off the convex plate


I bull MEMO TO DESIGNERS 7-1 bull JUNE 1994--------=~~__

litltrtln6

bull Minimum Angle Cvrmm) of Concave Bearing Surface

The minimum angle (Vmio) of the concave bearing surface determines the combined rotation and horizontal load capacity of the bearing The minimum design rotation capacity (3) for spherical bearings is usually 2 degrees and should include rotations from DL U camber changes construction tolerances and erection sequences

Vmio =asruo + 3 3 = minimum design rotation capacity of bearing (3 = f3s +3c) 2 degrees minimum

bull Maximum Allowable Radius (Rmu) of Concave Bearing Sliding Surface

The maximum allowable radius (Rmu) defmes the spherical curvature which resists the applied horizontal forces and provides rotation Rmu is calculated from the minimum projected diameter (Dm) and the minimum angle (1fmi0 ) Due to manufacshyturing limitations Rrrw should not exceed 36 inches Hence the actual radius (Ratt) of the concave bearing surface may be less than Rmu This limitation is usually not a problem because the smaller radius will increase the allowable rotation andor increase the lateral load capacity

Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)

If RIJW gt 36 use Rmu =36


4f illlr --------------------------~M~EM=O~T~0~0~E=S~G=N~E~R~S~7~-~1_bull~J~U~N~E~1~9~9~4I l1ziJam

bull Minimum Concave Bearing Pad Diameter (DBacJ

The minimum concave bearing pad diamter (DBact) is used to calculate the minimum metal depth (Mm) ofthecancavesurface and the minimum angle(Ymin) of the convex surface

DBact is the arc length along the concave bearing sliding surface

Derivation

Arc length =radius x angle in radians

(Dm2)VIm= sm-1 T (degrees)

7tl180 converts to radians multiply by 2 to get diameter

bull Minimum Metal Depth (Mm) of Concave Surface

Derivation

Aic length =radius x angle in radians

DBact ( )( n ) DBact-2- = ~ Vlmin 180 - 2shy

_ DBact (180)VIuun - 2Ract n

y cos VImin = - - Y = Ra cosVImin

Ra


4lt illlr --------------------------~M~EM~O~T~O~D~E~S~IG~N~E~R~S~7~-~1-middot~J~U~N~E~1~9~9~4

lilltrtln6

Use 009375 thick PTFE for design Actual values vary form 1A2 to Ys as specified in the Standard Special Provisions

Note Some manufacturers attach the PTFE fabric to a substratum which is attached to the concave surface

bull Minimum Metal Thickness at Center Line (Tmin)

Tmio = 075 minimum

bull Maximum Metal Thickness (Tmax)

Tmu =Trruo + Mm + 0125 (Ys allows for maximum PTFE thickness and and substratum

bull Length and Width (Lcp) of Concave Plate

Lcp =Projected diameter (Dm) + 1125

Note 1125 allows for edge distance substratum and PIFE

Convex Spherical Plate

bull Minimum Angle ( Ymin) of Convex Surface

The minimum angle ( Yrruo) of the convex surface allows the maxim urn rotation of the concave plate without loss of contact area

Ymin = lrrun + fJ


______~_______~ bull MEMO TO DESIGNERS 7-1 bull JUNE 1994

Dzltrans

bull Minimum Convex Chord (Cm) Length

Derviation

sinrmin = Cm 2

Cm =2[ Ract(sin(rmin))]Ract

bull Height of Convex Spherical Surface (H)

Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm

H=Ract- Raebull - ( 2 )

bull Overall Height of Convex Plate (HacJ

Hacl =H + 075

Bact includes thickness from stainless steel surfacing and Y4 recessed into masonry plate The 075 inch vertical sides may be increased to provide minimum clearance or to provide minimum fillet weld height


4f llllr --------------------------~M~EM~O~T~0~0~E~S~I~G~NE~R~S~7-~1 ~J~U~N~E ~~middot ~1~9~94

Ozltrtul6

bull Minimum Vertical Clearance (c)

The minimum vertical clearance (c) ensures that the concave plate does middotnot contact the base plate during maximum rotation

Spherical bearings square in plan

c= 07 Lcp 3+0125

Spherical bearings round in plan

C = 05 Dm3 + 0125

V ~_j_c ~----------~lr

L--_______~----Y 3]_j_ e= 1s minimum at 1---------_JIT maximum rotation

Sole Plate

The sole plate must be sized so that it remains in full contact with the concave plate under all loading conditions

The safety overhang (L 0 ) provides a minimum edge distance and allows for additional sliding surface beyond the calculated movement The value reflects the certainty or uncertainty of the total movement calculation

Lcp = length and width of concave plate Lrrw = maximum longitudinal movement (including creep shrinkage postshy

Ltensioning thermal effects and seismic)

0 = safety overhang Tm =maximum transverse movement

bull Longitudinal Length of Plate (Lsp)

Lsp =Lcp +~(total)+ LoL


fit Ill[ --------------------------~M~E~M~O~T~0~0~ES~I~G~N~E~R~S~7_-~1_bull~J~U~N=E~1~9~9~4 tizUIUI$

bull Tranverse Width of Sole Plate (Wsp)

bull Plate Thickness (Tsp)

Design in accordance with AISC design procedure for column base plates when mounted on concrete

Recommended minimum thickness 075 inch

bull Anchorage

Anchorage may be accomplished with shear studs bolts or welding depending on the structure type Studs smaller in diameter than 34 inch are not recommended

bull Bevel

Bevel sole plate to provide a level bearing plate on steel and precast concrete girder structures

Masonry Plate


Convert convex plate area to equivalent square area to design plate thickness


Length and width to accomodate the seating of the convex plate

Anchorage may be accomplished with shear studs sleeved anchor bolts or welding depending on the structure type Studs smaller in diameter than 3J4 inch are not recommended


4lt lllr --------------------------~M~EM~O~T~0~0~E~S~IG~NE~R~S~7-~1_bull~J~U~N~E~1~9~9~4

liztrans

Example 4 PTFESpherical Expansion Bearing (Non-Guided) CIP PIS Structure

Given


SLructure Length = 785 feet Contributory Length - 176 + 255 (CIP PIS) = 431 feet DL ReactionGirder- Service Load = 271 kips LL + I ReactionGirder - Service Load = 50 kips Moderate Temperature Zone- Rise and Fall = 35degF fc = 4 ksi Fy = 36 ksi

Concave Plate


Pvmu (maximum vertical load)= 271 +50= 321 kips

Umu (maximum allowable PTFE pressure)= 3500 psi

I I

321 000Dm = 2[Pvmv I 1r p = 2[ bull I 1rp = 1081 Say 1100 (Okay to round) Umax J 3500 J

bull PTFE Area (APTFE) ofFlat Sliding Surface

ApTfE = Pvnux = 321000 = 917 in2 (min) Umu 3500

(917)111 = 9582 use 9752 APTFE = 95 in2

Pv

bull Minimum Angle ( lXmin) Required to Prevent Uplift

PHmv (maximum horizontal load)= (010)(321) = 321 kips

011n (minimum verticalload)=(050)(321)= 1605kipslt271 kips use 160Skips

a n = tan-l pHm=gtx ) = tan-l ( - 321 ) - = 1131deg 011 ( A nun 1605


~ 1111 --------------------------~M~E~M~O~T~O~D~E~S~IG~N~E~R~S~7-_1~middot-J~U~N~E~1~9~9~4

liztrans

bull Minimum Angle ( Vmin) of Concave Bearing Surface

ltXuuo = angle required to resist uplift = 1131 deg

f3s = Structure rotation 0003 radians = 0 17deg

f3c=Construction rotation assume 002 radians= 115deg

f3 = f3s+ f3c= 132deg use 2deg min = 20deg

Note Bridge Design Specifications Anicle 1522 requires a minimum rotation capacity of 0015 radians (086deg)

bull

ljmjo =ltluun + 3 = 1131deg + 2deg =1331deg

Maximum Allowable Radius (Rmu) of Concave Surface

Rmax = Dm = 11 middot 00 =2389

2sm (ljoun) 2sin (1331 deg) (Okay to round)

bull

Use Rad = 2375 lt 36 Okay

Minimum Concave Bearing Pad Diameter (DBacJ

1 0012 = 2[2375[__(sin-bull 1 )]] = 1110 (Do not round)

180 2375


M =Ra - M =Ra - [ Ra cos((~rO))]+PTFE Thickness

M = 2375 -[2375cos(( 2~1~5))C0) )]+009375= 0739 (Do not round)


4lt ------------------------~M~EM~O~T~0~0~E~S~IG=N~E~R~S~7_ 1 bull~J~U~NE~1~99~4-~_lbltran6

bull Minimum Metal Thickness at Centerline (Tmin)

Tmin= 075


Tmax= Tmin+ Mm + 0125 = 075 + 0739 + 0125 = 161- 1625 (Round up)


Lcp = Dm + 1125 = 1100 + 1125 = 12125 (Okay to round)

Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r


Cm = 2[2375(sin(1539))] = 1261 Say 12625 (Okay to round)


H=2375- 23752 -c2-~25r =0854-085


Jlllr --------------------------~M~E~M~O~T~0~0==ES~I~G~NE=R~S~7-~ ~J~U~N~E~1~9~9~41~middot

lizluans


Hact = H + 075

Hact = 085 + 075 = 160


c = 07Lp3 + 0 125 (3 is in radians)

c =07(121252degCo)) +0125 =042 minimum required

Actual vertical clearance provided is greater than 050 (075 inch vertical sides on the convex plate are recessed 025 inch into masonry plate)

Sole Plate

L


Lsp = Lcp + lmu (total) + LoL

Lcp = 12125 (Concave Plate)

Temperature movement= (15)(2)(00000060)(35degF)(431 ) (12 inft) = plusmn326

PIS shortening= (070) (010 ft100 ft) (431 )(12 inlft) = 362

Seismic Movement = plusmn30

Lrnax = 326 + 362 + 2 (30) = 1288

0L = 2 (10) = 20 (Edge distance)

Lsp = 12125 + 1288 + 20 = 2700


illlr --------------------------~M~EM~O~T~0~0~E~S~IG~N~E~R~S~7~-~1_bull~J~U~N~E~1~9~9~4

lbltrans

Notes 1) Thermal movement was multiplied by 2 because 35degF is rise or fall temperature

2) Thermal movement was multiplied by 15 because it is not always possible to place the sole plate at a mean temperature

3) Position sole plate to account for one directional movement of PS shortening

4) Approximately 70 of PS shortening remains at time of hinge closure pour (see Attachment 1)

bull Transverse Width of Sole Plate (Wsp)

L cp =12125

Tm = plusmn10 (Seismic)

Lor= 2 (10) =20 (Edge distance)

Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600

bull Plate Thickness (T sp)

Design in accordance with AISC design procedure for column base plates

Lsp = 2700

Wsp =1600

J =4 ksi Fy = 36 ksi

(Ref BDS Article 815213)

Assume for this example that A2A 1 = 15

Maximum bearing pressure (fb) on loaded area

b = 030 X 4 000 X Jl3 = 1 470 psi


4lt ijlllr --------------------------~M~E~M~O~T~0~0~E~S~I~G~NE~R~S~7- 1~middot~J~U~N~E~1994~ ~~Drlfr1111S

Detennine Required Plate Area

321 000bull = 21837 in2

1470

Since the length of the sole plate was detennined for sliding purposes detennine the required length to distribute the load to the concrete

21837 = 1365 2 125 160

PTFE t~~~~=r--rSliding Ibull Use plate size of 16 x 14 to design Surface io

io C1 lttthickness

0) Ci

Actual bearing pressure (b) on loaded area

12125 321 000fb = bull = 1433 psilt 1470 Okay

1614 x 16

Design ModelNote Model shows 1 transverse movment (Tm)

Detennine Plate Thickness

(AISC 3-106 9th Edition)

1 433 Tsp=2(4125) =165 use 175 gt 075 minimum

36000

bull Anchorage

Since structure is cast-in-place PIS use shear studs

bull Bevel

Bevel is not required for cast-in-place concrete


lltL --------------------------~M~E~M~O~T~0~0~E=S~IG~N~E~R~S~7~-~1_bull~J~U~N=E_1~9~9~4

liztrans

Masonry Plate


Cm = 12625

fb =1470 psi (A2A1 middot= 15)

1 = 4000 psi

bull Determine Required Area

321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15

Actual Bearing Pressure (fb) on Loaded Area

321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152

Equivalent 15 Square Area

bull Determine Plate Thickness

Convert convex plate area to equivalent square to determine (n)

2625Area Convex Plate= n(l 2middot ) = 12519 in2

4


4f 1111 --------------------------~M~E~M~O~T~0~0~E~S~I~G~N~ER~S~7~-1~middot~J~U~N~E~1~9~9~4

lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000

Tmp = 076 + 025 = 101 use lOO plate

bull Anchorage

Since structure is cast-in-place use shear studs

bull Determine Total Bearing Height (Ah)

Ah = Tmp + Hac1 - 025 + Trrun + Tsp + 2(PTFE thickness)+ (sole plate stainless thickness)

= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510

Notes Actual bearing thickness may vary slightly depending on the thickness of the PTFE substratum and stainless steel used Minimum and maximum values are given in the Standard Special Provisions

middotUse Bridge Standard Detail Sheet XS 12-80 ifpossible (see Attachrnent4)



Uztrans

PTFEElastomeric Bearings

General

P1FEelastomeric bearings designed in accordance with Bridge Design Specificashytions Sections 14 and 15 should be considered for use only when the practical limits of reinforced elastomeric bearing pads have been exceeded

The PTFEielastomeric bearing concept and design procedure covered in this section was adopted with few exceptions from Ted Jensens paper titled ElastomericrrFE Bearings (Ocrober 1987)

The basic PTFEelastomeric bearing design is simply comprised of PIFE disks sliding on stainless steel surfaces tO accommodate the longitudinal movements and elastomeric bearing pads to accomodate the rotational movements (see Figure 10 page 58)

P1FEelastomeric bearings are suitable for structures with moderate to large longishytudinal translations and relatively small rotations This non proprie tary bearing is simple to design and fabricate Good performance can be attained with careful attention to loading rotation and the physical propertieslimitations of the manufacshyturing materials

PTFEElastomeric Bearing Components

PTFEelastomeric bearings are comprised of five basic components (1) sole plate (2) PTFE disk (3) intermediate plate (4) elastomeric bearing pad and (5) masonry plate The function of the five components mentioned above and the typical manufacturing materials are as follows (see Figure 10 page 58)

bull Sole Plate- Transfers superstructure loads to the bearing and provides a stainless steel sliding surface for super structure translation The sole plate is fabricated from A36 A36M steel and has a stainless steel surfacing The stainless steel surface is bonded to the sole plate with epoxy resin and stainless steel cap screws or by perimeter welding

bull PTFE Disk - Provides a low fric tion sliding surface for the sole plate The PTFE disk is manufactured from I 00 percent pure virgin unfilled dimpled sheet resin The PTFE disk must be recessed one-half its thickness to control cold flow

bull Intermediate Plate- Transfers loads from PTFE disk to elastomeric pad The intermediate plate is manufactured from A36 A36M steel


~ M~E~M~O~T~O~D~E~S~~G~NE~R~S~7~-~~middot 1 9~9~4--------------------------~ I 1 ~J~U~N~E~~tizlaans

bull Elastomeric Bearing Pad- Allows rotation of the superstructure while maintainshying 100 percent contact between the PTFE disk and the sole plate The steel reinforced elastomeric bearing is fully vulcanized to the steel plates Fabric reinforced pads are not allowed

bull Masonry Plate- The masonry plate transfers load from the elastomeric bearing pad and anchors the bearing to the seat The masonry plate is fabricated from A3 A 36M steel

Design Requirements

PTFEelastomeric bearings are designed in accordance with Bridge Design Specifishycations Sections 14 and 15 All loads are service loads Minimum vertical loads are for dead loads and superimposed dead loads Maximum vertical loads are for dead loads superimposed dead loads and live loads (no impact)

Unfilled PTFE sheet resin stresses are limited to 3500 psi maximum The design coefficient of friction varies from 008 to 004 at bearing pressures of 500 psi and 3500 psi respectively

Steel reinforced elastomeric bearing pads with shape factors 275 may be loaded to a maximum stress of 1000 psi The shear modulus (G) used for design is 100 psi

Design Guidelines

PTFE surfaces should be loaded to a minimum of 2000 psi (DL only) for optimum performance A design coefficient of friction of 006 is recommended for designs with bearing pressures from 2000 psi to 3500 psi Actual lubricated friction values are lower however they should not be used for design because the long term effects of the grease are unknown

A minimum and maximum PTFE thickness of 3J6 inch and 114 inch respectively should be shown on the contract plans as the limits are not specified in the Standarfl Special Provisions

To reduce rotational stresses orient rectangular bearings so the long side is parallel to the axis about which the largest rotation occurs

A bearing pad with a low shape factor accomodates rotation most readily and a bearing pad with a high shape factor is best for resisting compresssion 1 Therefore the best choice represents a compromise between the two A minimum shape factor of 75 is recommended


4f llllr --------------------------~M~E~M~O~T~O~D~E~S~IG=N~E~R~S~7-~1~middot~J~U~N~E~1~9~9~4

lbtltul$

Design Procedure

Elastomeric Bearing Pad

1 Determine the width (W) and length (L) of the elastomeric pad for the applied vertical load (DL + LL not including impact) using an allowable unit stress of 1000 psi sTo use 1000 psi the shape factor (S) ~ 75

DL+U =WxL 1000 psi

2 Check the compressive strain of the elastomer due to dead load and live load from the stressstrain curves for various shape factors shown in Figure 11 (page 59) Thesecurves developed by the California Transportation Laboratory2 are based on tests ofpads constructed with 1h inch layers ofelastomers between steel plates meeting California specifications To account for compressive creep of the elastomer under sustained dead load the initial deflection from dead load is increased by 25 percent The total deflection from dead load (DL) and live load (LL) shall not exceed 007 times the thickness of the elastomeric bearing

3 Determine the initial thickness of the elastomer required for structure rotation The structure rotation should include rotations from DL U camber changes construction tolerances and erection sequences

The relative rotation between top and bottom surfaces of the bearing shall be limited by

LaL + Waw ~ 2~c for rectangular bearings

D~al + av ~ 2~c for circular bearings

~c =Lecit ~c =eotT

Therefore the elastomer thickness (7) may be determined from

T~ LaL + Waw (rectangular bearings) 2eTot

T ~ D~al + av (circular bearings) 2poundTor


bull MEMO TO DESIGNERS 7-1 bull JUNE 1994 --~~~~

DzIYIUI$

L = gross length of rectangular bearing parallel to the longitudinal axis of the bridge (in)

W =gross width of rectangular bearing perpendicular to longitudinal axis of the bridge (in)

D =gross diameter of circular bearing (in)

a Lbull (aw) = relative rotation oftop and bottom surfaces ofbearing about an axis perpendicular (parallel) to the longitudinal axis of the bridge (radishyans)

lJc = instantaneous compressive deflection of bearing (in)

Ec = compressive strain of ith elastomer layer (change in thickness divided by unstressed thickness)

poundToe = total compressive strain of elastomer

t = thickness of ith elastomer layer (in)

T = total elastomer thickness of bearing (in) = It

4 Determine the Maximum Allowable Shear Force (Fs) in the Elastomer

AFs = G-Js

T

lJs = Shear deflection of bearing (in)

G =Shear modulus of elastomer (psi) at 73degF

A = plan area of bearing (in2)

The maximum allowable shear force in the elastomer must be greater than the maximum lateral force required to slip the P1FE disk under dead load (see Figure 12 page 60)

Note that the shear modulus (G) decreases with increasing temperature and increases with decreasing temperature A value of 100 psi is recommended for this calculation

The maximum shear deflection (6s) in the elastomer shall be limited by


4lt Jill --------------------------~ ~N~E~R~S~7~- 1_ 4M~EM~O~T~0~0~E~S~IG ~ bull~J~U~N~E~1~9~9~l1ztrans

If the maxim urn allowable shear force is exceeded the area of the elastomeric pad may be increased to provide greater shear capacity It is evident from the above formulas that the elastomer design is sensitive to both the shear modulus and the friction force transmitted through the stainless steel sliding surface

PTFE Disk

1 Determine the area of the PTFE disk required to support vertical loads (DL+U no impact) using a 3500 psi maximum compressive stress Note that the allowable compressive stress for the PTFE is 35 times the allowable stress for the elastomer To minimize the thickness of the intermediate plate in which the PTFE is recessed the length width or radius of the PTFE should be such that the edge distance is held to a minimum A 2000 psi to 2500 psi (DL only) design compressive stress on the PTFE should provide a reasonable intermediate top plate thickness

PTFE disks are recommended to facilitate fabrication of the recess in the steel intermediate plate

2 Calculate the Lateral Force (F1) required to slip the PTFE disk under Dead Load

F1 =J1N

J1 = friction coefficient

N =Dead Load

The friction values given in the PTFE Sliding Surfaces section and Article 1527 of the Bridge Design Specifications should be used for this calculation Note that the actual coefficient of friction will probably be less because the stainless steel slider plate will be coated with silicone grease Initial coefficients of friction as low as 1 to 2 percent were observed by the Transportation Laboratory on greased samples loaded to 3170 psi However these low friction values should not be used for design because the long term affects of the silicone grease are not known

3 Compare the Maximum Allowable Shear Force (Fs) in the elastomer with the Lateral Force (F1) required to slip the PTFE under Dead Load


41 lllir --------------------------~M~EM~O~T~0~0~ES~I~G~N~E~R~S~7--1~middot ~J~U~N~E~1~9~9~4 l1ztrans

Intermediate Plate

1 Size the intermediate plate length and width to match the dimensions of the elastomeric bearing pad

2 Determine required plate thickness Design in accordance with AISC design procedure for column base plates Suggestion-conven PTFE disk area to equivalent square area to design plate thickness

Sole Plate

The sole plate must be sized so that it remains in full contact with the PTFE disk under all loading conditions

Thesafety overhang (L0 ) provides a minimum edgedistance and allows foradditional sliding surface beyond the calculated movement The value reflects the cenainty or uncertainty of the total movement calculation

Do = Diameter of PTFE disk ~ = Maximum longitudinal movement (including creep shrinkage post

L

tensioning thermal effects and seismic) Tm = maximum transverse movement

0 = safety overhang

1 Longitudinal Length (Lsp) and Transverse Width (Wsp) of Sole Plate

Single Disk

2 Plate Thickness (Tsp)

Design in accordance with AISC design procedure for column base plates when mounted on concrete Recommended minimum thickness 075 inch

3 Anchorage may be accomplished with shear studs bolts orwelding depending on the structure type Studs smaller in diameter than 34 inch are not recommended


jlllr --------------------------~M~E~M~O~T~0~0~E~SI~G~N~E~R~S~7_-~1_bull~J~U~N=E~1~9~9~4

liztrans

Masonry Plate

1 Size the masonry plate length and width to match the dimensions of the elastomeric bearing pad unless a larger plate is required for anchorage

2 A plate thickness of075 inches is recommended

3 Anchorage may be accomplished with shear studs sleeved anchor bolts or welding depending on the structure type Studs smaller in diameter than inch are not recommended

Testing

Until recently our policy was to test scale bearings fabricated in the same manner as the full size bearings The test bearings were detailed in the contract plans and tested at the Transponation Laboratory This practice was abandoned June 1994 after it was determined that some test bearings were not representative of the actual bearings delivered to the job site

The current specifications required that full sized PTFE bearings be proof tested and evaluated for compression and coefficient of initial static friction in the presence of the Engineer The specifications also require that the manufacturer furnish one sample of elastomeric bearing to the Transportation ~boratory for testing Test bearings should not be detailed in the contract plans


IllEr --------------------------~M~EM~O~T~O~ ~1~middotD~E~S~IG~N~E~R~S~7- ~J~U~N~E~1~9~9~4 11ztrans

Longitudinal Movement Sole II

Intermediate II +~~~~~--fo_~~-~~~---J---~ Rotation vc Movement ~(j~Elastomeric cOBearing Pad -~o middot

)) (j~

~~ ~Q

vt

Figure 10 PTFEEiastomeric Bearing


1111 --------------------------~M~E~M~O~T~O~D~E~S~IG~N~E=R~S~7-~1~middot-J~U~N~E~19~9~4

liztrans

AFs = G-11s

T

T ~ 211s

Ft =JIN Ft $ Fslll3X

Fs = shear force in elastomer

G =shear modulus (100 psi 70degF)

T = total elastomer thickness

ls = shear deflection of bearing

F1 = force required to slip P1FE disk under DL

N =DL

J1 = coefficient of friction

Figure 12 Shear Force in Elastomer


illlr ---------------------------M~E=M~O~T~0~0~ES~I~G~N=ER~S~7--1~middot-J~U~N~E~19~9~4 tiztram

Example 5 PTFEEiastomeric Bearing CIP PIS Structure

Given


Structure Length = 785 feet Contributory Length- 176 + 255 (CIP PIS) = 431 feet DL ReactionGirder - Service Load = 271 kips U ReactionGirder- Service Load No Impact = 43 kips Moderate Temperature Zone - Rise and Fall = 35degF Jc = 4 ksi Fy = 36 ksi


1 Determine Width (W) and Length (L)

DL + U =WxL 271 kips+ 43 kips = 314 in2

1000 psi 1000 psi

Try 12 x 28 Area = 336 in2 gt 314 Okay

Note Slender bearing pad selected to maximize rotation capacity

12 x 28 Shape Factor (S) = = 84 gt 75 1000 ps1 Okay

12+28

Actual load on elastomer

271 kips+ 43 kips = si934 (DL+ LL) 12x 28 P

271 kips = 807 si (DL)12 x28 P


bull bull MEMO TO DESIGNERS 7-1 bull JUNE 1994 --~~~~

l1zltrans

2 Check Compressive Strain

S =84 obtain strain values from the curves shown on Figure 11 (page 59)

Pn =934 psi en =44

PDL = 807 psi EDL=40oo

ETotal = 44 + ( 40) (025) = 54 lt 70 Okay

3 Determine Initial Thickness (7) for Rotation

assume Waw negligible

T ~ LaL =(l2)(00IS) =167 say 2 (elastomer only) 2ET()( 2(0054)

Structure Rotation f3s = 0003 radians

Construction Rotation f3c = 001 radians

Bridge Design Specifications require 0015 radians (minimum)


A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2


lllr --------------------------~M~EM~O~T~0~0~E~S~IG~N~E~R~S~7-~1_bull~J~U~N~E~1~9~9~4

11zaans

PTFE Disk

l Determine Area of PTFE Disks

Use Two Disk Design

271 kips 1r D2 0DL Area Required = =-- Do =83 0

2 x 2500 psi 4

Try Do = 850

DL Stress on PTFE = 271 ki~~2 =2388 psi gt 2000 psi Okay 2xnxshy

4

271 2DL + LL on P1FE = +~5 = 2 766 psilt 3 500 psi Okay 2xnxshy

4

Use two Y4 x 8lh0 PTFE disks

2 Calculate the Lateral Force (F1) Required to Slip PTFE

J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips

3 Compare Allowable Shear Force (Fs) with Slip Force (F F)

Fs ~ 168 gt 162 kips OkayF1



11ztrans

Intern1ediate Plate

1 Size Intennediate Plate Length (L) and Width (W) to match Elastomeric Pad Dimensions

L =12 W=28

2 Determine Plate Thickness (Tp)

28 14

- n 7 smiddot n

- r------ I I I

75 I I I

I_-- -1 middot PTFE Disks Intermediate

Plate

Actual Bearing Design Model

T =2nftP fp = Actual bearing pressure on elastomer p F

y Fy =Steel yield strength Fb=075 Fy

Model -Convert Disk to Equivalent Square to Determine (n)

t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2

Total Plate Thickness (Tp) including 1h Recess

TP =105 + 0125 = 117 say 125 thick


llllr --------------------------~M~E~M~O~T~O~D~E~S~~G~N=ER~S~7--~middot U~N~E~19~9~4I 1 -J~tizluans

Sole Plate

1 Longitudinal Length (Lsp) and Width (Wsp)

Temperature movement = ( 15)(2)(00000060)(35degF)( 431)(12 inft) = plusmn 326 PIS Shortening= (070)(010 ft100 ft)(431)(12 inft) = 362 Seismic Movement = plusmn30

L4nv = 326 + 362 + 2 (30) = 1288

0L = 2(10) (Edge distance) =20 DD = 85 Tm =(Seismic) = plusmn10 Lsp =DD + 4rw (total) + LoL Lsp = 85 + 1288 + 2 = 2338 say 240 Wsp =DD + Tm + L0

LOT = 2(10) (Edge distance) =20 Wsp = [(2)(85) + 55) + 2 (10) + 20 = 265



3) Position sole plate to account for one directional movement of PIS shortening

4) 70 of PIS shortening remains at time of hinge closure pour (see Attachment 1)


Design in accordance with AISC design procedure for column base plates mounted on concrete

Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi

b = 030j-)A2 A1 $ 060J (Ref BDS Article 815213)

Assume for this example thatA21A 1 = 15

Maximum bearingpressure(fb)onloadedarea b = (030)( 4000)Ji5 = 1470 psi


4lt illir --------------------------~M~E~M~O~T~O~D~E~S~IG=N~E~R~S~7~-~1-bull~J~U~N~E~1~9~9~4

liztrans

3 Determine Required Plate Area

271 + 43 = 2136 in2 1470

Since the length of the sole plate was determined for sliding purposes determine the required length to distribute the load to the concrete

2213 6 middot in =81 usediskdiameter + edgedistance(85+2)=105

265

Design Thickness for 265 x 105 Plate

~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24

Design Model Sole Plate

Actual Bearing

Model Information

bull Convert disks to equivalent square bull 1 transverse movement (Tm) shown

b =Actual bearing pressure Fy = Steel yield strength (36 ksi) Fb =075 Fy


4lt lllr --------------------------~M~EM~O~T~O~D~ES~I~G~N=E~R~S~7_ J~U~N~E~1~9~9~4-~1_bull~

liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate

1 Size Plate Area to Match Elastomer Area

Use 12 x 28 plate

2 Plate Thickness

A plate thickness of is adequate since the masonry plate has the same area and load as the elastomeric pad

934 psi lt 1470 psi

3 Anchorage

Since structure is cast in place use shear studs


4lt illr --------------------------~M~E~M~O~T~O~D==ES~I~G~N~E~R=S~7_-~_1 bull~J=U~N=E~1~9~9~4 l1ztrtll1$

Summary of Calculations

Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate

Total Bearing Height

Sole Plate Stainless Steel PTFE Disk (lA thick recessed 1h) Intermediate Plate Elastomer 2 + (4)(0075) middot Masonry Plate

Total

cY~ crmvKJjfe~ A McKee

2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments


I amp MEMO TO DESIGNERS 7-1 bull JUNE 1994----------=-~~~

11dtrtzn6

References

1 AASHTO LRFD Design Specification (Version 402) Section 14 Joints and Bearings

2 An Evaluation of Fiberglass and Steel Reinforced Elastomeric Bridge Bearing Pads dated January 1982 by the Transportation Laboratory of the State of California

3 A Laboratory Evaluation of Full Size Elastomeric Bridge Bearing Pads dated June 1974 by the Transportation Laboratory of the State of California

4 NCHRP Report 10-20A High Load Multi-Rotational Bridge Bearings Final Report- John F Stanton Charles W Roeder T Ivan Campbell

5 Caltrans ElastomericrrFE Bearings by Ted Jensen PE dated October 1987

6 Cal trans QAI Newsletter No3 Suggestions to Avoid Future Bridge Problems August 3 1992


4lf llllr --------------------------~M~E~M~O_T~0~0=E~SI~G~N~E~R~S~7-middot~1_bull~J~U~N=E~1~9~9~4 Gzltrans

Bearing Selection

Bearing selection is influenced by many factors such as loads geometry mainteshynance available clearance displacement rotation deflection availability policy designer preference construction tolerances and cost The designer must consider all the applicable variables early in the design stage and design the structure and bearing as a unified system Too often bearings are selected at the last minute when forces and available space are fixed Such an approach increases the chances of future mainteshynance problems

The official policy of the Division of Structures is to avoid using an alternative bearing system where a conventional reinforced elastomeric pad can provide the required characteristics through shear deformation When the practical limits of elastomeric bearing pads are exceeded designers should consider using PTFF1 spherical or PTFEelastomeric bearings The three bearing systems mentioned should provide enough versatility to satisfy the design requirements of most strucshytures designed by the Division of Structures Other bearing systems may be approshypriate for special circumstances designers should consult with the Bearing Technical Specialist for unique applications

On widenings designers are cautioned against mismatching bearing types It has been common practice to use elastomeric bearing pads a yielding bearing to widen structures supported on steel rocker bearings an unyielding bearing While this practice has worked satisfactorily on short to moderate length structures it has created problems when thick elastomeric bearing pads have been used on structures with long spans 6




General




















liztltz1$















Design Criteria








~I


Note




Where




W+L




Sf- ~ ~~~ ~~~ f 0

t ) ~


~ ~






lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










General




















liztltz1$















Design Criteria








~I


Note




Where




W+L




Sf- ~ ~~~ ~~~ f 0

t ) ~


~ ~






lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References


















liztltz1$















Design Criteria








~I


Note




Where




W+L




Sf- ~ ~~~ ~~~ f 0

t ) ~


~ ~






lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










Design Criteria








~I


Note




Where




W+L




Sf- ~ ~~~ ~~~ f 0

t ) ~


~ ~






lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








~I


Note




Where




W+L




Sf- ~ ~~~ ~~~ f 0

t ) ~


~ ~






lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lbltrtllt6

--~- ~ -~- --yen~~ bull 1(~ - ~11-- 1 jib shy




o






shear strain)_

Equations

Pad Area A =LX W

By Definition

Shear Force


Pad Thickness

Fs G = A = Fs T

bs Ab - s

T





tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizltran6



lttAbutment

Section 8 -8








Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References













Figure 3 Detail



------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








------------------------------------------------------- ---------------


littrans

iii Q

vi VI Q)

U5 Q) gtiii VI Q) Q

E 0 ()

BRIDGE BEARINGS

700 Shape Factor

12

600

500

400

300


200

100

0 0 1 2 3 4 5 6 7 8



PAGE 11


tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizlaans

Shape Factor 12

1000

Shape Factor 6

200

1 2

Figure 48

3 4





liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans

Design Procedures


Given





W =flange width= 16


DL+ LL 125 UL = = = 9 8 mmtmum se 10

08x W 08x16



3 T= 15 Okay


DL+ IL 125000 ------ = = 781 pst lt 800 WxL 16x10


DL 69000 = =431 pSIgt 200

WxL 16x10



lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lbltftlllS



W+L 16+10






5 5











tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizltlans


Background



Current Policy


Design






1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









1zltrans






LW -ltr d2



S=----shy4hn(D+Ld)

where








10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References













10 2 2 115 129

Jj 3 3 173 189

20 4 4 230 248

25 5 5 288 308

30 6 6 345 367

35 7 7 403 427

40 8 8 460 486

45 9 9 518 546

50 10 10 575 605

55 11 11 633 665

60 12 12 690 724


V4 Max1







Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










Given









WxL

148 41wXL= + = 189 in2 = 137 X 137 10




T= 35 Okay






148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References












148DL = = 660 si gt 200 Okay

WxL 14xl6 P



W+ L 14+16
















tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizltrDnS


Given







(0025100) X 100 X 11 = 030




148 41 w XL= + =189 in2 =137 X 137 10




bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








bull lr 1zltrans



- 3 7=30 Okay




148DL = = 660 si gt 200 Okay

WxL 14x16 P



W+L 14+16






Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Gzatl16



5 5



2









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









11ztrans


General





PTFE Types



Woven Fabric PTFE




tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizltrtiiiS




Friction



























liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References



















liztrans


General











tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizltrtZnS




Design Requirements


Design Guidelines




11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









11ztrt111S

































1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









1 r------------ -- --- --------------

J

r--------------- -- --- shy--- -- I

I ---- I _ I

I I 1

I I 1 I I I 1

I 1 I I 1 I

II Ill+1

1 I I It 11


I I

_ -- l----- - --==

Plan

Anchorage Typical





Section X-X



- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








- ---------


lizluans


1111 Ill+111 I


II I II ---III----- - --== I I I I

I I I I I I


I I II I I 1

1I I I

1 I I - I I I

- -- -- ~

Plan



Section X-X



--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








--


tizltran6

I--------------------------~

I ------ -I shy

shyf I

I I I

I I I I I I

Illll I

----+-----~~--------11 I II I I I I I I

I I I -- shy--------==

+

I

I I

1

IllI I I

--------L~----~~---I II I I I

I I I I I I

I I I I I

I I

- ---

Plan



Section X-X


Sole Plate


-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References








-----


tbltrans

0 0 0

0

0


Elevation


Plan

Sole Plate

I I I I

Woven PTFE Pad I I



Concave Plate




azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









azJaans


r777J7h Elevation

Plan


Convex Plate

middot

Elevation


Plan

Masonry Plate




c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









c 1-------m12---= 1 D ~~--D-m12--~


Convex Plate (_

Center of Sphere




lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lizltriZIIS

Design Procedure

Concave Plate







Derivation

I




Concave Plate


UmaxSliding Surface



tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









tizluans







Pvrruo




litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









litltrtln6






Derivation

Dm2smVImin =-Raw--

Rmu = Dm 2sm(Vmio)







Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References












Derivation





Derivation





Ra



lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lilltrtln6




Tmio = 075 minimum









Ymin = lrrun + fJ



Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Dzltrans


Derviation

sinrmin = Cm 2



Derivation

H= Rad-z

Cm 2 2 2 (Cm )

2

( 2 ) +z~ =Racl z= Rad - 2

2 2 Cm



Hacl =H + 075




Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Ozltrtul6




c= 07 Lcp 3+0125


C = 05 Dm3 + 0125

V ~_j_c ~----------~lr


Sole Plate














bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References













bull Anchorage


bull Bevel


Masonry Plate








liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans


Given



Concave Plate




I I




(917)111 = 9582 use 9752 APTFE = 95 in2

Pv







liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans







bull





bull




180 2375







Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










Tmin= 075





Convex Plate


0 = ( 1110 )(180)+ 20 = 15390 Ynuo 2(2375) 1r




H=2375- 23752 -c2-~25r =0854-085



lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lizluans


Hact = H + 075

Hact = 085 + 075 = 160





Sole Plate

L







Lrnax = 326 + 362 + 2 (30) = 1288


Lsp = 12125 + 1288 + 20 = 2700



lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lbltrans






L cp =12125



Wsp = 12125 + (2)(10) + 20 = 16125 Say 1600



Lsp = 2700

Wsp =1600





b = 030 X 4 000 X Jl3 = 1 470 psi




321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










321 000bull = 21837 in2

1470


21837 = 1365 2 125 160


io C1 lttthickness

0) Ci



1614 x 16





36000

bull Anchorage


bull Bevel




liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans

Masonry Plate


Cm = 12625


1 = 4000 psi


321 000bull = 21837 in2 = 1478 X 1478

1470

Use 15 x 15 plate

Lmp =15


321000 2 0 0 15 fb = =14 7 ps1 lt 147 kay

152





4



lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lilltrans

12519 in2 = 1119 x 1119

n = (15 -1 119)2 = 191

1427 Tmp = (2)(191) = 076+ recess depth

36000


bull Anchorage




= 100 + 160-025 + 075 + 175 + 2(009375) + 0060

=510





Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Uztrans


General














Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References











Design Requirements




Design Guidelines







lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









lbtltul$

Design Procedure



DL+U =WxL 1000 psi






~c =Lecit ~c =eotT






DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









DzIYIUI$











AFs = G-Js

T










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










PTFE Disk




F1 =J1N


N =Dead Load





Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Intermediate Plate



Sole Plate




L


0 = safety overhang


Single Disk






liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans

Masonry Plate




Testing







)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References











)) (j~

~~ ~Q

vt




liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans

AFs = G-11s

T

T ~ 211s







N =DL






Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










Given






1000 psi 1000 psi




12+28



271 kips = 807 si (DL)12 x28 P



l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









l1zltrans



Pn =934 psi en =44


ETotal = 44 + ( 40) (025) = 54 lt 70 Okay








A 12 x28 Fs =G-As= 100 X X 1= 168 1dps

T 2

T 2 lll =- =- = smu 2 2



11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









11zaans

PTFE Disk


Use Two Disk Design


2 x 2500 psi 4

Try Do = 850


4


4



J1 =006 N = 271 kips (DL)

F1 =006 X 271000 =162 kips





11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









11ztrans

Intern1ediate Plate


L =12 W=28


28 14

- n 7 smiddot n

- r------ I I I

75 I I I


Plate





t - 271 k + 43 k- 934 J P - 12 X 28 - pSI

n = 14-75 =325 2


TP =105 + 0125 = 117 say 125 thick



Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









Sole Plate



L4nv = 326 + 362 + 2 (30) = 1288









Lsp = 240 Wsp = 265

J =4 ksi F1 =36 ksi






liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans


271 + 43 = 2136 in2 1470



265


~ I I 15middot

---~ 105 I I

JW I I I

--- -1 middot-- -_ I I

I I I 24


Actual Bearing

Model Information





liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









liztrans

fJ = 271 k + 43 k si= 1 128 b 2(105 X 1325) p

1 128Tsp =2 X 425 X bull =150

36000

Masonry Plate


Use 12 x 28 plate

2 Plate Thickness


934 psi lt 1470 psi

3 Anchorage





Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References










Plan Bearings


PTFE Disk (2)

Intermediate plate

Sole Plate

Masonry Plate



Total


2 X 1-0 X 2-4

A X 8Yz0

111 X 1 -0 X 2-4

1112 X 2-0 X 2-212

JA X 1 -0 X 2-4

150 0060 0125 125 230 075

599- 60

RLjgf Attachments



11dtrtzn6

References









11dtrtzn6

References








bridge bearings - caltrans · bridge bearings . introduction . aashto defines a bearing as "a...

Documents