SUBSTRUCTURE EXAMPLE. Full Height Abutment on Spread Footing


 Noel Cunningham
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1 SUBSTRUCTURE EXAMPLE Full Height Abutment on Spread Footing This example illustrates the design of a full height abutment on spread footings for a single span astinplae posttensioned onrete box girder bridge. The bridge has a 160 feet span with a 15 degree skew. Standard ADOT 32inh fshape barriers will be used resulting in a typial dek setion onsisting of 15 barrier, 120 outside shoulder, two 120 lanes, a 60 inside shoulder and 15 barrier. The overall outtoout width of the bridge is A plan view and typial setion of the bridge are shown in Figures 1 and 2. The following legend is used for the referenes shown in the lefthand olumn: [2.2.2] LRFD Speifiation Artile Number [ ] LRFD Speifiation Table or Equation Number [C2.2.2] LRFD Speifiation Commentary [A2.2.2] LRFD Speifiation Appendix [BDG] ADOT Bridge Design Guideline Superstruture Design Example 1 demonstrates design of the superstruture and bearings for a single span astinplae posttensioned onrete box girder bridge using LRFD. Critial dimensions and loads are repeated here for ease of referene. Bridge Geometry Bridge span length Bridge width Roadway width Loads DC Superstruture DW Superstruture ft ft ft kips kips Substruture This example demonstrates basi design features for design of a full height abutment supported on a spread footing. The substruture has been analyzed in aordane with the AASHTO LRFD Bridge Design Speifiations, 4 th Edition, 2007 and the 2008 Interim Revisions. Geotehnial The soil profile used in this example is the one used for the Geotehnial Poliy Memo Number 1: Development of Fatored Bearing Resistane Chart by a Geotehnial Engineer for Use by a Bridge Engineer to Size Spread Footings on Soils based on Servie and Strength Limit States. 1
2 Figure 1 Figure 2 2
3 Material Properties [ ] Reinforing Steel Yield Strength f y = 60 ksi Modulus of Elastiity E s = 29,000 ksi Conrete f = 3.5 ksi [ ] [C3.5.1] Unit weight for normal weight onrete is listed below. The unit weight for reinfored onrete is inreased kf greater than that for plain onrete. Unit weight for omputing E = kf Unit weight for DL alulation = kf [C ] The modulus of elastiity for normal weight onrete where the unit weight is kf may be taken as shown below: E = 1820 f ' = = 3405 ksi [5.7.1] The modular ratio of reinforing to onrete should be rounded to the nearest whole number n = = 8.52 Use n = [ ] Modulus of Rupture [ ] β 1 = the ratio of the depth of the equivalent uniformly stressed ompression zone assumed in the strength limit state to the depth of the atual ompression zone stress blok. For onrete strengths not exeeding 4.0 ksi, β 1 = The modulus of rupture for normal weight onrete has several values. When used to alulate servie level raking, as speified in Artile for side reinforing or in Artile for determination of defletions, the following equation should be used: f r = 0.24 f ' = = ksi When the modulus of rupture is used to alulate the raking moment of a member for determination of the minimum reinforing requirement as speified in Artile , the following equation should be used: f r = 0.37 f ' = = ksi 3
4 Existing Soil The existing soil has the following properties: Depth ft Soil Type Total unit weight, γ s pf φ degrees 025 Fine to oarse sands Gravelly sands Fine to oarse sands Gravels The following assumptions have been made. No groundwater is present. The soils will not experiene any longterm (onsolidation or reep) settlement. The Fatored Net Bearing Resistane Chart plots the fatored net bearing resistane versus effetive footing width for a range of immediate settlements as shown in Figure 3. Figure 3 Bakfill Soil The soil used for bakfill has the following properties: γ s = kf k a =
5 Limit States [1.3.2] [ ] In the LRFD Speifiation, the general equation for design is shown below: η iγ iqi ϕrn = R r For loads for whih a maximum value of γ i is appropriate: [ ] η η η η 0.95 i = D R I For loads for whih a minimum value of γ i is appropriate: [ ] 1 η i = η η η D R I 1.0 [1.3.3] Dutility For strength limit state for onventional design and details omplying with the LRFD Speifiations and for all other limit states: η D = 1.0 [1.3.4] Redundany For the strength limit state for onventional levels of redundany and for all other limit states: η R = 1.0 [1.3.5] Operational Importane For the strength limit state for typial bridges and for all other limit states: η I = 1.0 For an ordinary struture with onventional design and details and onventional levels of dutility, redundany, and operational importane, it an be seen that η i = 1.0 for all ases. Sine multiplying by 1.0 will not hange any answers, the load modifier η i has not been inluded in this example. [BDG] For atual designs, the importane fator may be a value other than one. The importane fator should be seleted in aordane with the ADOT Bridge Design Guidelines. 5
6 SUBSTRUCTURE Loads Setion 3 Loads There are several major hanges and some minor hanges onerning the determination of loads. The DC loads must be kept separate from the DW loads sine different load fators apply. The live load is different as seen in the superstruture design. The dynami load allowane is a onstant rather than a funtion of the span and only applies to members above the ground. The Longitudinal Fore in the Standard Speifiations has been modified and replaed by the Braking Fore. A vehile ollision fore relating to protetion of piers or abutments has been added. The wind and wind on live load is similar but has a modifiation fator for elevations above 30 feet. The vertial wind pressure is the same but the speifiation larifies how to apply the fore to the proper load group. The lateral earth pressure inludes better larifiation of the following items: when to use the Rankine or Coulomb Method, when to use ative or at rest pressure, and when to use the equivalent fluid pressure method. The disussion of dead and live load surharges is enhaned. Figure 4 Abutment 1 is pinned while Abutment 2 is expansion. The pinned abutment will resist externally applied longitudinal fores. The expansion abutment will resist the frition and internal fores from the deformation of the bearing pads. Sine determining whih abutment is ritial is not obvious, the fores at eah abutment will be determined. [10.5.2] [10.5.3] Limit States For substruture design, foundation design at the servie limit state inludes settlement, lateral displaement and overall stability. Foundation design at the strength limit state inludes bearing resistane, limiting eentriity (exessive loss of ontat), sliding at the base of the footing, and strutural resistane. Three strength limit states require investigation. Strength I is the basi load ombination without wind. Strength III is the load ombination inluding wind exeeding 55 mph. Strength V is the load ombination ombining normal vehiular use with a wind of 55 mph. 6
7 [3.6.5] For substruture design, Extreme Event I load ombination inludes seismi events while Extreme Event II load ombination inludes ollision of substruture units by vehiles. These limit states are not onsidered in this example. A diagram showing the general dimensions (feet) for the abutment follows: Figure 5 7
8 [3.5] [3.5.1] PERMANENT LOADS DC Dead Load Strutural Components DC superstruture dead load inludes selfweight inluding intermediate and abutment diaphragms and barriers. DC Superstruture = k e long = (16.00 / ) = 1.25 ft M long = ( )(1.25) = 1537 ftk DC substruture dead load inludes the weight of the abutment inluding end bloks, wingwalls and footing. Item N H W L Weight X A M A Bakwall Seat / Stem Footing End Blk / Wing Total DC Substruture = k.g. =7631 / = ft e long = / = ft M long = (947.85)(0.051) = 48 ftk DW Dead Load Wearing Surfae and Utilities The DW superstruture load inludes the future wearing surfae and utility loads. This bridge has no utilities. DW = k M long = (85.63)(1.25) = 107 ftk EV  Vertial Earth Pressure The LRFD Speifiation does not provide data on unit weights of well ompated soils. For this example use a vertial earth pressure based on a unit weight of kf. In atual design use the values speified in the Geotehnial Report. 8
9 Design for the full height of the wall even though the soil only extends to the top of the seat. To avoid the omplexities of how to deal with the weight of the approah slab it is simpler to design for the taller height of soil. Item N H W L Weight X A M A Toe Heel Seat / Total Note: Some numbers may not add up due to rounding. EV = 850 kips.g. = 9879 / = ft e long = / = ft M long = (849.89)(3.624) = ftk [3.11.5] EH  Horizontal Earth Pressure Two deisions must be made before analysis begins: (1) whether to use atrest or ative lateral earth pressure and (2) whether to use the Rankine or Coulomb theory. Typial abutments supported on ohensionless soils with elastomeri bearings supporting the superstruture with strutural grade bakfill will deflet adequately to mobilize ative soil pressure. Therefore, ative pressure will be used in the design. [C ] The LRFD Speifiation states that the Coulomb method is neessary for design of retaining walls where the bak fae of the wall interferes with the development of the full sliding surfaes in the bakfill soil assumed in the Rankine theory. Abutment onrete antilever walls with short heels will require the use of the Coulomb method. Abutment onrete antilever walls with long heels may be designed with either the Rankine or Coulomb method. The LRFD Speifiation indiates that the Rankine method of determining lateral earth pressure is not appropriate when the heel is determined to be a short heel. However, the use of the Coulomb Method is a major departure from ADOT past pratie. In addition, a value of frition must be onsidered in the Coulomb Method yet the reommended value varies widely. 9
10 Foundation Analysis and Design by Bowles partially agrees with the LRFD Speifiation but omes to the onlusion that neither method in its pure form an be used. However, either method an be used if the following modifiation is made: the soil loads are applied to a vertial line extending from the end of the heel and the soil on top of the heel is treated as a stati load. [BDG] The Rankine formula provides more onservative designs, is allowed per ADOT Bridge Design Guidelines and will be used in this example. The soil data will be provided in the Geotehnial Report. For this problem assume that the soil extends to the full height of the abutment with the following properties: γ s = kf k a = The lateral earth pressure is assumed to be linearly proportional to the depth of earth and taken as: [ ] p = k a γ s z = (0.295)(0.120)(24.50) = ksf/ft EH = 0.5(0.867)(24.50)(46.41) = k/ft The resultant ats at a height of H/3 above the base of the wall. M long = (493.08)(24.50) / 3 = 4027 ftk/ft 10
11 TRANSIENT LOADS [ ] [ ] [ ] LL Vehiular Live Load The number of design lanes is the integer part of the ratio w/12 = 42.00/12 = 3 where w is the lear roadway width. The ritial live load reation is the ombination of the design lane (52.19 kips) and design truk (67.80 kips). Refer to the Superstruture Example 1 for alulation of the live load reations. Apply the multiple presene fator, m, for the reation. Critial values are underlined. One Vehile Two Vehiles Three Vehiles P = ( )(1.20)(1) = k e L = / os(15) = ft M trans = (143.99)(16.56) = 2384 ftk M long = (143.99)(1.25) = 180 ftk P = ( )(1.00)(2) = k e L = / os(15) = ft M trans = (239.98)(10.35) = 2484 ftk M long = (239.98)(1.25) = 300 ftk P = ( )(0.85)(3) = k e L = 4.00 / os(15) = 4.14 ft M trans = (305.97)(4.14) = 1267 ftk M long = (305.97)(1.25) = 382 ftk Figure 6 11
12 To simplify the problem, the maximum reation and moments will be used even though they do not our simultaneously. This will redue the number of load ases without substantially simplifying the design. [3.6.2] IM Dynami Load Allowane Dynami load allowane need not apply for foundation omponents that are entirely below ground suh as footings. For the portion of the abutment above the ground, the dynami load allowane is only a design load for the stem. [3.6.4] BR Vehiular Braking Fore The braking fore shall be taken as the greater of: 25 perent of the axle weights of the design truk or design tandem V = (0.25)( ) = k <= Critial V = (0.25)( ) = k 5 perent of the design truk plus lane load or 5 perent of the design tandem plus lane load V = (0.05)[ (160.00)(0.640)] = 8.72 k V = (0.05)[ (160.00)(0.640)] = 7.62 k It should be noted that the truk load will always ontrol and the tandem fore need not be alulated. The braking fore shall be plaed in all design lanes that are onsidered to be loaded whih arry traffi in the same diretion. For this bridge the number of lanes equals the lear roadway width of 42 feet divided by 12 foot lanes = 3.5. Sine only full lanes are used, use 3 lanes. The bridge is a one diretional struture with all lanes headed in the same diretion. Therefore, all design lanes shall be simultaneously loaded and the multiple presene fator shall apply. BR = (18.00)(3)(0.85) = k This load is applied 6 feet above the dek surfae. However, due to the pinned restraint the longitudinal fore will be applied at the seat level. V long = os(15) = k V trans = sin(15) = k M long = (44.34)(16.83) = 746 ftk M trans = (11.88)(16.83) = 200 ftk 12
13 [ ] LS  Live Load Surharge A live load surharge shall be applied where a vehiular load is expeted to at on the surfae of the bakfill within a distane equal to onehalf the wall height behind the bak fae of the wall. The inrease in horizontal pressure due to live load surharge may be estimated as: [ ] p = kγ s h eq where: [ ] γ s = total unit weight of soil = kf k = oeffiient of lateral earth pressure, k a, for walls that deflet h eq = equivalent height of soil for vehiular load from Table 1 = 2.0 ft for abutment height > 20.0 feet p = (0.295)(0.120)(2.0) = ksf P = (0.120)(2.0)(6.5)(46.41) = k V long = (0.0708)(24.50)(46.41) = k M long = (80.50)(24.50) / 2 = 986 ftk [C3.4.1] [ ] The vertial weight of the soil surharge is to be inluded for foundation designs where the load inreases the load effet but ignored where the load inreases the resistane. For bearing resistane the vertial soil weight on the heel will inrease the total load and therefore the load effet and should be inluded. For sliding resistane and overturning the vertial soil weight will inrease the resistane and therefore should be ignored. If the vehiular loading is transmitted through a strutural slab, whih is also supported by means other than earth, a orresponding redution in the surharge loads may be permitted. The standard ADOT approah slab satisfies this riterion. However, the abutment is tall ompared to the slab length and no method is provided to determine the amount of the redution, so the full live load surharge will be used. In addition, onstrution vehiles ould produe a live load surharge before the approah slab is onstruted. 13
14 [3.8] [ ] WS Wind Load on Struture Wind pressures are based on a base design wind veloity of 100 mph. For strutures with heights over 30 feet above the groundline, a formula is available to adjust the wind veloity. The wind is assumed to at uniformly on the area exposed to the wind. The exposed area is the sum of the areas of all omponents as seen in elevation taken perpendiular to the assumed wind diretion. Height = (44.83) = ft Area = (11.07)(160.00) = 1770 ft 2 Wind on Superstruture [ ] The base pressure for girder bridges orresponding to the 100 mph wind is psf. The minimum wind loading shall not be less than 0.30 klf. Sine the girder bridge has spans greater than 125 feet, the wind must be evaluated for various angles of attak. The enter of gravity of the loads is loated (11.07) / 2 = feet above the bottom of the footing. Wind fore in the diretion of the span will be applied at the top of the seat due to the pinned ondition. Wind pressures for various angles of attak are taken from Table Refer to Figure 4 for proper inlusion of the skew affet for the load ombinations. Critial values are underlined. Pinned Abutment 0 Degree Skew Angle V long = [(1770)(0.050)sin(15)]/2 + (1770)(0.000)os(15) = k V trans = [(1770)(0.050)os(15)]/2 + (1770)(0.000)sin(15) = k M long = [(1770)(0.050)(22.36)sin(15)]/2 + (1770)(0.000)(16.83)os(15) = 256 ftk M trans = [(1770)(0.050)(22.36)os(15)]/2 + (1770)(0.000)(16.83)sin(15) = 956 ftk 15 Degree Skew Angle V long = [(1770)(0.044)sin(15)]/2 + (1770)(0.006)os(15) = k V trans = [(1770)(0.044)os(15)]/2 + (1770)(0.006)sin(15) = k M long = [(1770)(0.044)(22.36)sin(15)]/2 + (1770)(0.006)(16.83)os(15) = 398 ftk M trans = [(1770)(0.044)(22.36)os(15)]/2 + (1770)(0.006)(16.83)sin(15) = 887 ftk 14
15 30 Degree Skew Angle V long = [(1770)(0.041)sin(15)]/2 + (1770)(0.012)os(15) = k V trans = [(1770)(0.041)os(15)]/2 + (1770)(0.012)sin(15) = k M long = [(1770)(0.041)(22.36)sin(15)]/2 + (1770)(0.012)(16.83)os(15) = 555 ftk M trans = [(1770)(0.041)(22.36)os(15)]/2 + (1770)(0.012)(16.83)sin(15) = 876 ftk 45 Degree Skew Angle V long = [(1770)(0.033)sin(15)]/2 + (1770)(0.016)os(15) = k V trans = [(1770)(0.033)os(15)]/2 + (1770)(0.016)sin(15) = k M long = [(1770)(0.033)(22.36)sin(15)]/2 + (1770)(0.016)(16.83)os(15) = 629 ftk M trans = [(1770)(0.033)(22.36)os(15)]/2 + (1770)(0.016)(16.83)sin(15) = 754 ftk 60 Degree Skew Angle V long = [(1770)(0.017)sin(15)]/2 + (1770)(0.019)os(15) = k V trans = [(1770)(0.017)os(15)]/2 + (1770)(0.019)sin(15) = k M long = [(1770)(0.017)(22.36)sin(15)]/2 + (1770)(0.019)(16.83)os(15) = 634 ftk M trans = [(1770)(0.017)(22.36)os(15)]/2 + (1770)(0.019)(16.83)sin(15) = 471 ftk Expansion Abutment 0 Degree Skew Angle V long = [(1770)(0.050)sin(15)]/2 = k V trans = [(1770)(0.050)os(15)]/2 = k M long = [(1770)(0.050)(22.36)sin(15)]/2 = 256 ftk M trans = [(1770)(0.050)(22.36)os(15)]/2 = 956 ftk 15 Degree Skew Angle V long = [(1770)(0.044)sin(15)]/2 = k V trans = [(1770)(0.044)os(15)]/2 = k M long = [(1770)(0.044)(22.36)sin(15)]/2 = 225 ftk M trans = [(1770)(0.044)(22.36)os(15)]/2 = 841 ftk 15
16 30 Degree Skew Angle V long = [(1770)(0.041)sin(15)]/2 = 9.39 k V trans = [(1770)(0.041)os(15)]/2 = k M long = [(1770)(0.041)(22.36)sin(15)]/2 = 210 ftk M trans = [(1770)(0.041)(22.36)os(15)]/2 = 784 ftk 45 Degree Skew Angle V long = [(1770)(0.033)sin(15)]/2 = 7.56 k V trans = [(1770)(0.033)os(15)]/2 = k M long = [(1770)(0.033)(22.36)sin(15)]/2 = 169 ftk M trans = [(1770)(0.033)(22.36)os(15)]/2 = 631 ftk 60 Degree Skew Angle V long = [(1770)(0.017)sin(15)]/2 = 3.89 k V trans = [(1770)(0.017)os(15)]/2 = k M long = [(1770)(0.017)(22.36)sin(15)]/2 = 87 ftk M trans = [(1770)(0.017)(22.36)os(15)]/2 = 325 ftk A onservative answer an be ahieved by simplifying the problem by using the maximum values in eah diretion ating simultaneously. If wind ontrols the design, the omplexities of ombining 5 wind ombinations should be performed. A summary of wind fores used in the design follows: Pinned V long = k V trans = k M long = 634 ftk M trans = 956 ftk Expansion k k 256 ftk 956 ftk Wind on Substruture [ ] The transverse and longitudinal fores to be applied diretly to the substruture are alulated from an assumed base wind pressure of ksf. Beause the longitudinal wind blows opposite the earth pressure, the ritial wind on substruture load in the longitudinal diretion will be zero. V long = 0 k V trans = 0.040[(10.50)(18.00)] = 7.56 k M long = 0 ftk M trans = 0.040(10.50)(18.00)(15.50) = 117 ftk 16
17 [ ] WL Wind Pressure on Vehiles Wind pressure on vehiles is represented by a moving fore of 0.10 klf ating normal to and 6.0 feet above the roadway. Loads normal to the span should be applied at a height of = ft Pinned Abutment 0 Degree Skew Angle V long = [(0.100)sin(15)/2 + (0.000)os(15)] = 2.07 k V trans = [(0.100)os(15)/2 + (0.000)sin(15)] = 7.73 k M long = [(0.100)(30.50)sin(15)/2 + (0.000)(16.83)os(15)] = 63 ftk M trans = [(0.100)(30.50)os(15)/2 + (0.000)(16.83)sin(15)] = 236 ftk 15 Degree Skew Angle V long = [(0.088)sin(15)/2 + (0.012)os(15)] = 3.68 k V trans = [(0.088)os(15)/2 + (0.012)sin(15)] = 7.30 k M long = [(0.088)(30.50)sin(15)/2 + (0.012)(16.83)os(15)] = 87 ftk M trans = [(0.088)(30.50)os(15)/2 + (0.012)(16.83)sin(15)] = 216 ftk 30 Degree Skew Angle V long = [(0.082)sin(15)/2 + (0.024)os(15)] = 5.41 k V trans = [(0.082)os(15)/2 + (0.024)sin(15)] = 7.33 k M long = [(0.082)(30.50)sin(15)/2 + (0.024)(16.83)os(15)] = 114 ftk M trans = [(0.082)(30.50)os(15)/2 + (0.024)(16.83)sin(15)] = 210 ftk 45 Degree Skew Angle V long = [(0.066)sin(15)/2 + (0.032)os(15)] = 6.31 k V trans = [(0.066)os(15)/2 + (0.032)sin(15)] = 6.43 k M long = [(0.066)(30.50)sin(15)/2 + (0.032)(16.83)os(15)] = 125 ftk M trans = [(0.066)(30.50)os(15)/2 + (0.032)(16.83)sin(15)] = 178 ftk 17
18 60 Degree Skew Angle V long = [(0.034)sin(15)/2 + (0.038)os(15)] = 6.58 k V trans = [(0.034)os(15)/2 + (0.038)sin(15)] = 4.20 k M long = [(0.034)(30.50)sin(15)/2 + (0.038)(16.83)os(15)] = 120 ftk M trans = [(0.034)(30.50)os(15)/2 + (0.038)(16.83)sin(15)] = 107 ftk Expansion Abutment 0 Degree Skew Angle V long = (0.100)sin(15)/2 = 2.07 k V trans = (0.100)os(15)/2 = 7.73 k M long = (0.100)(30.50)sin(15)/2 = 63 ftk M trans = (0.100)(30.50)os(15)/2 = 236 ftk 15 Degree Skew Angle V long = (0.088)sin(15)/2 = 1.82 k V trans = (0.088)os(15)/2 = 6.80 k M long = (0.088)(30.50)sin(15)/2 = 56 ftk M trans = (0.088)(30.50)os(15)/2 = 207 ftk 30 Degree Skew Angle V long = (0.082)sin(15)/2 = 1.70 k V trans = (0.082)os(15)/2 = 6.34 k M long = (0.082)(30.50)sin(15)/2 = 52 ftk M trans = (0.082)(30.50)os(15)/2 = 193 ftk 45 Degree Skew Angle V long = (0.066)sin(15)/2 = 1.37 k V trans = (0.066)os(15)/2 = 5.10 k M long = (0.066)(30.50)sin(15)/2 = 42 ftk M trans = (0.066)(30.50)os(15)/2 = 156 ftk 60 Degree Skew Angle V long = (0.034)sin(15)/2 = 0.70 k V trans = (0.034)os(15)/2 = 2.63 k M long = (0.034)(30.50)sin(15)/2 = 21 ftk M trans = (0.034)(30.50)os(15)/2 = 80 ftk 18
19 A onservative answer for wind on live load an be ahieved by using the maximum values in eah diretion ating simultaneously. If wind ontrols the design, the omplexities of ombining 5 wind diretions should be performed. Pinned V long = 6.58 k V trans = 7.73 k M long = 125 ftk M trans = 236 ftk Expansion 2.07 k 7.73 k 63 ftk 236 ftk [3.8.2] Vertial Wind Pressure A vertial upward wind fore of ksf times the width of the dek shall be applied at the windward quarter point of the dek. This load is only applied for limit states whih inlude wind but not wind on live load (Strength III Limit State) and only when the diretion of wind is taken to be perpendiular to the longitudinal axis of the bridge. When appliable the wind loads are as shown: P = (0.020)(44.83)(160) / 2 = upward M trans = [71.73(44.83) / 4]os(15) = 777 ftk M long = [71.73(44.83) / 4]sin(15) = 208 ftk [3.13] [ ] [ ] FR Frition Fores Frition fores from the greased bearings aused by superstruture movement will be transmitted to the substruture for Abutment 2, the expansion abutment. These fores will our during the stressing operation and for a short period of time afterwards while the bridge undergoes long term prestress shortening. This fore was alulated for the elastomeri bearing for the Superstruture Example 1 as repeated below: H bu = µp u P u = 1.25DC DW P u = 1.25(245.86) (17.13) = k H bu = (0.10)(333.02) = k per bearing pad It is important to note that this fore is already fatored and only applies to strength and extreme event limit states. Table does list FR as a load with a load fator of 1.0 for all limit states inluding servie. However, this FR does not apply to servie limit states. 19
20 V trans = (33.30)(5 bearings)sin(15) = k V long = (33.30)(5 bearings)os(15) = k M trans = (43.09)(16.83) = 725 ftk M long = (160.83)(16.83) = 2707 ftk Bearing Translation The elastomeri bearing pad will also transmit fores to the substruture due to horizontal displaements aused by temperature, shrinkage, reep and prestress shortening. For the greased pad the shrinkage, reep and prestress shortening are resisted as a frition load so there is no diret load for SH and CR. The fore due to deformation of an elastomeri bearing pad due to TU shall be taken as: [ ] H bu = GA h u rt u = shear deformation from appliable strength and extreme event load ombinations in Table h rt = ( )(3 interior) + ( )(2 exterior) = 1.70 [BDG] [BDG] For a posttensioned box girder with greased sliding pads the elasti shortening and reep are assumed to be taken by the greased pad in a sliding mode. Afterwards the grease hardens and the pad resists temperature movement by deformation of the pad. The strength limit state load fator for TU deformations is The 0.65 fator reflets the fat that the pads are not always onstruted at the mean temperature. The temperature range for elevations less than 3000 feet is 90 degrees. u = (0.50)(0.65)( )(90)(160)(12) = in H bu = ( 0.130) (28) (14) = kips 1.70 The fore from frition (33.30 k/pad) is higher than the fore resulting from the internal deformation of the elastomeri bearing (10.10 k/pad). This bearing translation fore only applies to the strength and extreme event load ombinations and is also already fatored. Sine the FR fores are greater than the TU fores and only one fore an our at a time, only the FR fores will be onsidered further. 20
21 Bearing Rotation Rotations in the elastomeri bearing pads will ause bending moments that will be transmitted to the substruture. For unonfined elastomeri bearings the moment shall be taken as: [ ] M u = 1.60 ( 0.5E I ) θ h s rt where: I = moment of inertia of plan shape of bearing I = WL 3 /12 = (28)(14) 3 / 12 = 6403 in 3 [C ] E = effetive modulus of elastomeri bearing in ompression E = 6GS 2 = 6(0.130)(11) 2 = ksi Refer to Example 1 Superstruture Bearing alulations. θ S = radians Strength Limit States [ ] M u = ( 1.60) (0.5) (94.38) (6403) 12 = 199 ftk 1.70 M long = (199)(5 bearings)os(15) = 961 ftk M trans = (199)(5 bearings)sin(15) = 258 ftk Again this moment is already fatored. The 1.60 fator in the formula is the load fator that allows for use of servie limit rotations. This load only applies to the strength and extreme limit states. 21
22 A summary of unfatored axial loads, shears and moments exept as otherwise noted follows: Abutment 1 (Pinned) Load P max P min V long V trans M long M trans kip kip kip kip ftk ftk DC super DC sub DC DW EV EH LL BR LS WS super WS sub WS WS vertial WL FR * Bearing Rotation* Abutment 2 (Expansion) Load P max P min V long V trans M long M trans kip kip kip kip ftk ftk DC super DC sub DC DW EV EH LL BR LS WS super WS sub WS WS vertial WL FR* Bearing Rotation* * Loads are fatored and only apply to strength limit states 22
23 [ ] LOAD COMBINATIONS STRENGTH I Max = 1.25DC DW EH EV (LL + BR +LS) + FR + Bearing Min = 0.90DC DW EH EV (LL + BR + LS) + FR + Bearing STRENGTH III Max = 1.25DC DW EH EV (WS + WS vert ) + FR + Bearing Min = 0.90DC DW EH EV (WS + WS vert ) + FR + Bearing STRENGTH IV Max = 1.50DC DW EH EV + FR + Bearing STRENGTH V Max = 1.25DC DW EH EV (LL + BR + LS) WS WL + FR + Bearing Min = 0.90DC DW EH EV (LL + BR + LS) WS WL + FR + Bearing SERVICE I Max = 1.00(DC + DW + EH + EV) (LL + BR + LS) WS WL The moment due to bearing rotation and frition fores from the bearing only apply to strength limit states and are already fatored. As previously disussed, CR and SH fores are not ritial for this bridge and are not inluded in the load ombinations above. 23
24 General The methods used to estimate loads for the design of foundations using LRFD are fundamentally the same as the proedures used in the past for ASD. What has hanged is the way the loads are onsidered for evaluation of foundation stability (bearing and sliding resistane of spread footing foundations) and foundation deformation. The design of foundations supporting bridge abutments should onsider all limit states loading onditions appliable to the struture being designed. The following Strength Limit States may ontrol the design and should be investigated: Strength I Limit State will ontrol for high live to dead load ratios. Strength III or V will ontrol for strutures subjeted to high wind loads Strength IV Limit State will ontrol for high dead to live load ratios A spread footing foundation will be evaluated for the following failure onditions: 1. Bearing Resistane Strength Limit States 2. Settlement Servie I Limit State 3. Sliding Resistane Strength Limit States 4. Load Eentriity (Overturning) Strength Limit States 5. Overall Stability Servie I Limit State 6. Strutural Resistane Servie I and Strength Limit States 24
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