Caltrans_Bridge Design Practice.pdf

Contents

Section

I Bridge Loads ................................................................... February 1993

2 Rein forced Concrere Design ........................................... February 1994

3 Prestressed Concrete Design ............................................... March 199 3 Contents ................................................................................................. June 1994

3.5 Design of a 4Span Continuous Cast-in-Place Box Girder ............. June 1994

4 Design of Welded Steel Plate Girders ........................... December 1995

. 5 Substructures and Remining S!nctures ................. Jan 1982 Dec 1983

6 Underground Structures ................................................ December 1992

7 Bridge Design Aesthetics ................................................ February 1993

8 Metric Seismic Analysis of Bridge Stmcrures ............................. October 1995

r* - Brldge Design Practice . February 1993 - Section 1 . Bridge Loads

Contents

1.0.1 General .......................................................................................................... 1-1

1.0.2 Design Methods .......................................................................................... 1-1

1.1 Load Definitions ........................................................................................... 1-2

Dead Loads ................................. .... .................................................................. 1-2

Earth Pressure on Culverts ..................................................................................... 1-2

Live h a d s ........................... .. ................................................................................ 1-3 H Loads ............................................................................................................... 1-3

Alternative Loads ................................................................................................ 1 - 3

P Loads ................................................................................................................... 1-4

Highway Vehicle T&c h e s .......................................................................... 1 4

Highway Vehicle Load Applimtion ............ ,. .................................................... 1 4

Impact ........................................................................................................................ 1 -5

Longitudinal Force ................... .... ....................................................................... 1 4

.................................................. ........................... Wind Load on Structures ... 14

Wind on Live Loads ................................................................................................ 1-7

T h e 4 Forces, Shnnkage and Prestressing ....................................................... 1-7 Uplift Forces ...................................................................................................... 1 - 7

Forces of Sbeam Current, Floating Ice and Drift .............................................. 1-8

Buoyancy ................................................................................................................. 1-6

Earkh Pressure on Abutments and Retaining Wails ........................................... 1-8

Seismic Force ......................................................................................................... 1-9

Centrifugal Force .................................................................................................... 1-10

Section 7 . Contenrs Page 7-i

=* - Bridge Design Practice . February 1993 =

1.15.1 Load Distribution ......................................................................................... 7-11

1.15.2 Dead Loads .......................................................................................................... 1-11

1.15.3 Highway Vehicles .................................................................................................. 1-11

1.16.1 Load Factors ................................................................................................ 1-14

1.16.2 Gamrna Factor ........................................................................................................ 1-14

1.16.3 Beta Factor ............................................................................................................... 1-15

1.36.4 Phi Factor ............................................................................................................. 1-15

..................................................... ........................ 1.17 Loadcombinations ,. 1-16

1.18 References ................................................................................................. 1-17

Appendix .................................................................................................................... 1-13

A-1 H Loads and Al temative Load ............................................................................ 1.18

A-2 PLoads .................................................................................................................... 1-19

A-3 Simple Span Moment and Shear Comparison for P Loads and H Loads ...... 1-20

A 4 Example of Moment Envelope Calculations .................................................. .... 1-21

A-5 Example of Shear Envelope Caiculations .............................. ..... ................ 1-24

Page I - i i Section I . Contents

c* - Bridge Design Practice - February 1993 =

Bridge Loads

1.0.1 General

Loads are fundamental to bridge design, having evolved with experience and study over many pears. They have been codified in the United States since the mid-3920s in the Standard Specifica- tions for Highway Bridges of the American Association of State Highway and Transportation O f f i d s ( M H T O ) . In Calbans, these design specifications are contained in the Bridge Design Spedcations (BDS). BDS is indexed to correspond with PLASHTO, Division 1, Design. It includes some non-AASHTO material selected in anticipation of future AASHTO adoption or because of its local importance.

This section, Bridge Loads, deals with general aspects of the loads spedfied in the Thirteenth Edition of BDS, Sect ion 3, Loads. The mated covers load definitions; requirements and practices regarding distribution of load effects; adjustment of raw loads by specified load factors; and combination of loads with one another into specified groups acting together. In addition, it is prefaced by a brief introduction to Load Factor Design. This section may be conside~ed a commentary on BDS Section 3.

The specific uses of loads in the context of design processes are illustrated in the text sections dealing with structure elements and construction materials.

1.0.2 Design Methods

AASHTO provides specifications suitable for two distinct design methods, Load Factor Design (LFD) and Service Load Design (SLD). Caltrans' policy is to use LFD to the greatest extent possible. Therefore it receives the bulk of attention in BDS and in this text.

SLD, until recent years, was the primary design method upon which AASHTO was based I t hns k e n h o r n hstorically also as workmg stress design or eiastic design Its main objective is to equate load effects w ithallowable stress, a spedfied fraction of the yield strength of steel or uI timate strength of concrete. Its factor of safety against failure is implicit in that fraction. It draws on elastic theory for its fundamental concepts. SLD does not consider structural p e r f o m c e beyond the elastic range.

LFD implementation in Caltrans has been in progress since 1974. There are sti l l a few Lingering exceptions to its use, but it is now f i d y established as the basic Caltrans design method.

Themain objective in LFD is to ecluatedtimate load-carrymg capacity wikhapplied loads,after both have been modified by safety factors. Nominal or theoretical ultimate capacity, with stresses at the verge of failure (yield point of steel), is reduced by a materiais confidence factor. Applied loads are adjusted by multipIiers of both the individual loads and the combinations of loadsacting together.

The net effect is calculated to maintain stresses u s d i y within the elastic range. LFD follows ultimate strength theory as well as elastic theory for its fundamentals.

Section I - Bridge Loads Page 7 - 1 I

c* - Bridge Design Practice - February 1993

LFD was adopted by Caltrans as much for itsconsequences after construction as for its refinements in logic and precision during design. Because its safety factors relate directly to loads as well as materials, rather than just to materials as in SLD, it is possible to design bridges w i h consistently uruform usable live loadsallowed on the highway system by special permit. In LFD, usablelive load is dearly and duectly represented by design live load.

In SLD, design capacity is based on computed loads with a prescribed safety factor which is the same for dead and live loads. Capacity for pemit live loads isbased onhgher stresses (lower safety factor) than those used for design.

Ths sMt in stress levels provides usable live load capacity from two sources; from capacity oripally provided for live load, inueased now by higher allowable stresses; and from a s& increase in dead load capacity, which is not needed to support added dead load. The ratio of dead to live load in a structure varies markedly from one struche type to another. I t is relatively high for concrete; low for steel structures.

The use of excess dead load capacity as a source of live load capacity in bridges designed by SLD has resulted in a &orderly variation of permit load capacity from bridge to bridge along stretches of highway with mixed structure types. The bridge with the least permit capacity controls, thus preventing the use of available additional capacity in the rest of the group. The use of LFD avoids this problem.

1 .I Load Definitions

1.2 Dead Loads

Dead loads consist of the weight of permanent portions of the structure, including the effects of anticipated future additions.

Designs must provide for an additional 35 pounds per square foot dead load for future deck overlay. Long ramp comectors and s p e d major shctures in regions of d d dimate are exempt from this requirement.

The effectsof future utibtiesandplanned future widenings needspecial attention to assure they are accommodated in the design.

Research has indicated that the earth weight to be used indesign of culverts shallbe as modified in BDS 6.2 in order to provide suffiaent strength.

More information and references are contained in BDP Section 6, STRUCTURES UNDER ROADWAY EMBANKMENTS; BDS Section 6, CULVERTS.

Page 1-2 Section 7 - Bridge Loads

r-t. - Bridge Design Practice - February 1993 =

1.4. I Live Loads Bridges on the State Highway System are subjected to a vaiety of live loadings, including vehicular, equestrian, pedestrik and others.

Thts dixussion h limited tohighway vehicle loads, which are divided into thee loadsystem: H loads, Alternative loads and P loads. These loads are shah- in Appendixes A.l and A.2 and the e fkts of moment and shear are compared in Appendix A.3-

The example problems that follow in Appendixes A.4 and A.5 demonstrate the application oi H loads to a continuous bridge superstructure in order to obtain control.ling conditions for design, I t is assumed that the reader has knowledge of structural mechanics which will enab te him to make the necessan; compu tahons. Solution by compu ter greatly expedi tes h e work. The Office of Structure ~ e s i b makes extensive use o i the computer p r o w , Bridge Design System because it is tailor-made for the purpose.

1.4.2 H Loads

The H and HS trucks are live loads used in bridge design to enswe a minimum load carrying capacity. These loads represent a vast number of actual truck types and loadings to which the bridgemight besubjected under actual trafficconditions. They are theoriginal APSHTOdesign five load sjrstem, dating from t he 1920s. Thev have been revised and expanded periodically since hen;but mu retain intheir original chamiler.

The lane load is a simp lilied loading whch approx.ima tes a 20-ton truck preceded and followed by 15-ton tnrch.

For simple spans, one truck is the governing H load for moment in spans less that 145 feet, and the lane load governs for longer spans.

In conttnuous spans, the lane loading governs the maximum negative moment, except for spans less than about 45 feet inlength where the HStruck loading with its32 kip axles, variably spaced from 14 feet to 30 feet, may govern The exact point of change of controlhg load depends on ratios of adjacent span lengths. The positive moment of canhnuous spans is usually controUed by the Lane loadmg for spans of more than about 110 feet

1.4.3 AItemative Loads The basic alternative loading consists of kwo axles spaced four feet apart with each axle weighing 24 kips. Tkis load produces slightly greater Live load moments t . h H loads in spans under 40 feet.

The Alternative Load origma ted as a Federal Highway Admini.stration requirement for bridges on the kite-rstate hghway System in 1956. I t provides capacity for certain heary military vehicles. For convenience and uniformity, it isapplicable to the design of all bridges m thestate Highway System.

Se~rion 1 - Brjdge Loads Page 1-3


In addition, a single 32 kip axle with the weight equally chvided between two wheels centered six feet apart is used in design of transversely reinforced bridge decks.

For drscussion purposes, alternative loads are frequently combined with H loads and referred to jointly as H loads, in contrast to P loads.

1.4.4 P Loads

P loads are special designvehcles in addition to the H loads and Al temative loads specified by PLASHTO. The P loads were developed in California to ensure sufficient live load capacity to carry exhalegal live loads allowed by permit.

Permit design loads (P loads) consist of a family of idealized vehicles (see Appendix A.2) used by Office of Structure Maintenance and Investigations in rating bridge capaeties.The steenng axle and any number fiom two to six pairs of tandems (assumed as single concentrated loads) constitute a valid design v h d e . The combination that produces the maximum effect is used.

These loads were adopted for design in Caltrans because without them the AASHTO provhions for LFD would, in many cases, result in structures incapable of carrying permit loads in actual use or anticipated on California highways.

1.4.5 High way Vehicle Traffic Lanes The basic highway vehicle load width is 10 feet, which applies to all design bucks, lane loads and axles.

Virtua.Uy all design lanes are 12 feet wide.

The 10-foot wide loads are allowed to move within the 12-foot wide lanes which, in tum, may movebetween the curbs. The number of loads, their positions within the lanes, and the location of the lanes themselves are as required to produce maximum effects in the member under consideration.When applying h c k s to determine maximum effects in a member, only one truck per h e is utilized.

Fractional parts of lanes are not allowed for bent caps and substructure and members.

h v e load reduction factors are applicable to substnrcture members and some supersbcture members. These factors represent the probabdity that several lanes of full design load will not occur simultaneously on the bridge.

1.4.6 High way Vehicle Load Application

Thebasic live load design objective is to satisfy both H load and P load requirements. Structural components are proportioned for these loads at either the factored level or service level of magnitude as specified for the structural material or system under consideration.

Correction for sidesway is not normaUy made for live load because the duration of the loading is not long enough for sidesway to occur.

Page 1-4 Sedion I - Bridge Loads I

r4 - Bridge Design Practice - February 1993 =

1.5 Impact Impact is added to live loads lor most structural members which are above ground to account for hedynamic effect of theseloads. However,irnpactisnotadded toloads on timbermembers. Following are some illushations of the loaded length, 'It" for use in the impact formula for highway vehicles.

Positive Moments in Continuom Spans

I T TRUCK

t

Negative Mamenls in Continuous Spans

TRUCK

'L' = LI + + 2

Concentrated load I 1 Uniform load

LANE

'L' = L1 + L2

Moment in Carnilever A m 2

.

Use distance from moment center lo far end of truck. Max impact = 30%

Impact Examples Figure 1

Section I - Bridge Loads Page 1-5

r-t: - 8ridge Design Pradice - February 1993

1.6 Longitudinal Force

Provision is made for the longitudinal traction and braking effects of vekucular traffic headed in the same direction in Load Combination Groups Ilt and VI only (BDS 3.22). T h e longhdina l force of P loads is not considered.

The 1onl;itudhal force, when combined with the other forces, may affect the design of bents. Occasionally, in rigid frame structures where the bents are very stiff, longitudinal force, when added to other forces, may affect superstructure design.

The application of the farce six feet above the road way does not change grder moments much. We u e more concerned with longitudinal force as a shear on the column tops.

T h e spedications describe friction effects due to various types of expansion bearings. The friction forces are transmilted to the substruchre as reactions from horizontal design forces on the superstruchue. However, friction is not an independent primary force that requires consideration for group action in BDS 3.22.

7.7.1 Wind Load On Structures Wind loads are applied to all structures and hghway vehide live loads except P loads.

The bastc wind loads result horn a lugh wind of 100 mph and a moderate wind of 30 m p k In general, the high wind is assumed to act on the structure when live load is not present. Moderate wind acts on the structure when ljve load is present, for some load combinations.

The basic kigh wind of 100 mph produces 75 psf on arches and trusses, 50 psf on pders and beams, and 40 ps f on substructures. When Load Combinat ion Groups I I 1 and VI are considered, a moderate wind of 30% of the high wind pressure is used.

This force is applied in a variety of ways depending on whether one is designing supershucbre or subsmctu~, and whether the sbucture is usual or unusual.

Horizontal wind loads on the superstruhre are always based on the area seen in elevation view. They act both longitudinally and transversely. Loads on the substructure can be applied to elevation or hansverse views, or skew angles in between

When calculating the forces tending to overturn a structure, the upward high wind pressure of 20 psi (based on p h view area) is used for Groups 11, V and IX, while the moderate wind pressure of 6 psf is used for Groups 111 and VI.

The specifications provide for the use of judgement concerning wind velocities to be used in structure design. Permanent terrain features or predse data horn local weather service records may indicate that the basic 100 mph design wind should be rnoddied. Ilsuch is the case, the speded wind pressure is chmged in the ratio of the square of the design wind velodty to the square of 100. Whenever this is done, h e revised design wind must be stated in the General Notes of the bridge plans.

For high structures the wind effects on the bents and footings need to be thoroughly investigated, bolh laterally and longitudinally.

Page 1-6 Section I - Bndge Loads


The limiting height of column where wind m a y control varies with the span length, physical makeup of the shucture, and the magnitude of other lateral Ioads such as thak due tu earthquakes.

When applying lateral loads in continuous structures, consideration should be given to the rigidity oi the deck and its ability to transfer wind loads to a b u h e n t s which might be considerably stiffer than the bents. In these cases, the abutments must be designed to support t h e w lateral Ioads.

1.7.2 Wind On Live Loads

In addition to moderate wind pressure on the bridge structure when live load is present, a moderate wind force is exerted on the live load itself. This force is expressed asa line load acting both transversely and longitudinally 6 feet above the roadway surface. This offset locationisnot important except perhaps for the design of high piers.

1.8 Themal Forces, Shrinkage and Prestressing S t r u c d members are investigated to satisfy the Design Range of temperahues givenin BDS 3.1 6 or the Bridge Preliminary Report. The design range provides for movement corresponding to abouk one-half or less of the full air temperature range. Forces due to temperahue movement can become large on short stiff multicolumn bents but are usually reduced by distribution through the bent frame.

Expansion joint movement ratings are calculated to provide for tIw full air temperature range with dowance for m p and shnkage. Special insbctions are induded in Memo to Desiprs 7-10.

Shrinkage is the volume deaease which occurs when fresh concrete hardens and for a period of time thereafter. 1k is important in arches, where rib shxinbge produces rib and column moments, and preshessed girders, where shrinkage produces loss of stress% force.

Provision for innuence of movement and bending effects caused by prestresing is desaibed in Section 3 of this manual, PMSIRESED CONCRETE. Hinge location and subshcture design are sometimes determined by prestressing effects.

1.9 Uplift Forces Certain combinations of loading tend to Lift the bridge superskructure from the substructure. Elements of the bridge must be tied together to resist these uplift forces. T ~ J S can be accom- pkhed by either providing tension ties or by providing &cient mass in the superstructure to resist the uplift force. UpWt can become important with unusual span mnhprations. For instance, a very short end span adjacent to a long span will tend to lift at the abuhnent.

In LFD we must provide a resisting force sufficient to balance uphft caused by any load combination in 8DS 3.17. Forsenice load checks. calculated upMt force is m p u e d by factors to ensure safety.

Secrion 7 - Bridge Loads Page 1-7


1.10 Forces of Stream Current, Floating /ce and Driff C o l w and piers in streams are designed to resist the forces of wa ter, ice and dnft. The Bridge Preliminary Report wiU describe requirements for these items, whennecessary. Piersshould be located and skewed to afford minimum restriction to the waterway as recommended in BBDS Section 7, SLTBSTRUCWUS.

&ox girders or slabs are recommended superstructure types where less than 6 feet of clearance is provided over a stream crvryirtg MI.

1. I I Buoyancy Whenever a portion of a structure will be submerged, the effects of buoyancy should be considered in Ihe design. In small structures, its effects are unimportant and no economical advantage can be realized in the footing design. In large s t r u m e s , however, its effects should be taken into account in the design of footings, piles and piers.

1.12 Eaflh Pressure on Abutments and Retaining Walls

Abubnents and retaining walls should be designed so that any hydrostatic pressure is minimized by pruviding adequate drainage forthe backlill. References a t the end of this section and BDP Section 5, SUBSTtlUClVRES AND RETAINING WALLS, include more iniomtation on h e application of soil mechanics to abubnent and retaining wall design.

Symbols used in BDS 3.20 are: K, = active earth pressure: coeffjcient w = unit weight of soil (pounds per cubic foot) h = height (feet) S = l ive load surcharge height (feet)

For level backtill, the minimum active earth pressure is usually taken as an equivalent fluid pressure of 36 poundspe.r square foot per foot of height for abutments and retaining walls. This is based on an earth pressure coeffient [q) of -30 and a unit weight (w) of compacted earth of 120 pounds per cubic foot. This is used in design of folIowing elements:

1 Toe pressure or toe piles in retaking walls and abutments. (2) Bendmg and shear in retaining walls and abutments. (3) Sliding of spread footings or lateral Ioads in piles.

For t.hc design of rear piles in retaining walls or abutments, checks should be made using an equivalent fluid pressure of 27 pounds per square foot per foot of height. This corresponds to a 6 of ,225.

A trapezoidal pressure distribution is used where the top of wall is restrained. This provides a more realistic solution than the trimguhr pressure distribution which applies to typical retaining w a k without restraint.

Page 7-8 Section 7 - Bridge Loads


1.13 Seismic Force

Wqwkesand the response of shctures toearthquakes,are dynamiceven&-events that go into many cydes of shahg. An earthquake of magnitude &, such as that which o c m d m %n Francisco in '1906 and in Alaska in 19M, may have strong motions lasting for as long as 40 to 60 seconds. The San Fernando earthquake of magnitude 6.6 had about 12 seconds of sbong motion.

During thkperiod of strong motion, the structure passes though mimy cydes of deflection in response to the motions applied at the base of the structure. The strains resulting horn these deflections are the cause of the st-ructural d m g e .

Structures subjected to earthquake forces skaU he designed to survive theskaimresulMg horn the earthquake motion. Factors that are considered when desigrung to resist earthqua.ke motions are:

(1) The proximity of the site to known active Iaults. (2) The seismic response of the soil ar the site. (3) The dynamic response characterisiics of the total structure.

The foundation report prepmd by the C a l m Division of New Technology Materials and Research, Office of Engineering Geology, con t a im the seismic information necessary for design.

Three methods of analysis are available to d e s i g n s t r u m to resist earthquake motions. They are the Equivalent Sta tic Force Method; Response Specbum Modal Analysis; mdTime History Method.

Equivalent Static Force Method Column and member forces may be calculated using the equi\ia.lent static force method of analysis. It was developed as a simple way to design for the strains associated with earthquake motions and is suitable for hand calculations. This approach is effective when the mode shape (deflected shape under vibration) can be approximated in each direction being analyzed and when one mode domjna tes in each direction

The method assumes a predomiriant deflected shape and location of maximum displacement when vibrated in the direction under consideration.

Curves are used which consider this period, the depth of alluvium under the stnictute, and theexpected maximum acceleration of bedrock based on geology of the site. A value far the seismic coefficient is determined from these r u n e s whchrepresent the elastic response of the bridge to the earthquake. Tkis value is hen used to determine the maximum displacement in the s m c t u r e . Design forces in ndividual members are then computed for the displacement.

Structures with no more: Ihitn one intermediate hinge and having the following character- ist.ics may be analyzed using the equivalent static force method:

A. Tangent or nearly tangent alignment. B. Total deck length to width ration less than 15. C. Skew angles of abutments and bents less than twenty degrees. D. S a h c e d spans and supporting bents or piers of approximately equal stiffness.

Section 1 - Bridge Loads Page 1-9


Response Spectrum Modal Analysis me response specbum teduuque of modal analysis should be considered for determining earthquake loads when the bridge does not fall into the categories listed above. In this case, several modes of vibration will probably be sigruficant contributors to the overall seismic response of the structure.

T h i s method of analysis is computer oriented. The amount of calculation necessary makes it impractical, if not impossible, to do by hand. The computer first determines aU modes of vibration that a three dimensional mathematical model of the structure can have. It then applies a response spectrum loading ushg the same curves used for the equivalent static force method. These loadings are applied to the structure for each mode of vibration. The computer reports the deflections and forces thus induced both for each mode and for the root mean square summation of all modes.

Time History Method Time history analyses should be utilized for unusual structures. Structures for consideration have sites adjacent to active faults, sites with unusual geologic conditions, unique features, or a fundamental period greater than3.0 seconds. They are usually large, complex and important structures.

This computer analysis is the most complex (and expensive) of the three methods. The computer actually subjecks a a t h e m t i c a l model of the structure to an idealized earthquake. It does this by subjecting the computer model to earthquake irnpulses at predetermined t ime intends. These intervals are in small hactions of a second representing the ground accelerations varying with time. The forces in the various parts of the sbcture can either build or cancel under these impulses as time passes depending on the vibration characteristics of the structure. The computer reports these forces at their maximum and for any desiredpoint in time. l h s method gives the designer the best understanding of the true dynamic characteristics of earthquake loading.

All of these methods are based on elastic theory. None of them is capable of modehg shctural behavior when material strains are in the inelastic range. in a major earthquake, structural movement will almost surely be in the inelastic range. Recognizing this fact and realizing that it is impractical of design bridges to behave elas tically under attack from a large earthquake, a ductility and risk factor, 2, is introduced. The seismic forces that are produced by any of the described methods of analysis are divided by h factor before design. Thevalue of Z varies depending on anassessment of the ability of a particular bridge member to withstand strain in the inelastic range as weU as the member's importance in preventjng collapse. See Figure 3.21.1.2 in BDS 3.21.

I . I4 Centrifugal Force Centrifugal forces are included in all groups whch contain vehicular live load, includmg P loads. They act 6 feet above the centerline of roadway surface.

Page 1-10 Secfion I - Bridge Loads F


Centrifugal forces are significant in the design of bridges having small w e radii or w e d bridges with long columns.

These forces act as shears in girder end hames and as loads at tops of colurruts. Again, the 6-foot vertical dimension between the point of load application and the floor is seldom important.

1 .I 5.1 Load Distribution

Load dzshbution is the process by which the effects (forces, moment, shear, reaction, torsion) of a load flows horn the point of application ko dl other locations within a structure and into t he foundation.

Distribution of most loads is accomplished through rational analysis by some accepted mathema tical method. Statistics, moment distribution and stifhess analysis are common Corms of rational analysis.

The most important exception to Lkis approach is contained in the provisions of BDS 3.23 for vehcular live load and certain aspects of dead load dsMbution These simplifying assumptions and empirical formulae are specified for convenience and uniformity. They reflect the results of research and the state of the art current a t the times they were adopted by AASHTO.

Dead load is u d y disttibuted to supporting members by an appropriate rational analysis. However, forsimplicity, BDS 3.232.2 specifies that the weight of nubs , sidewalks, d i n g s and wearing surface may generally be distributed uniformly to all stringers and beams,

1.15.3 Highway Vehicles

Slabs are loaded by individual wheels. Our design specifications are based on plate theory to find the resulting maximum design effects. Standard designs are avarlabk for transverse deck slabs on Qirdersand for Iongi tudinally reinforced slab bridges. See Bridge D a i p Detnils &30and Bn'dge Design Aids 410 through 4-19.

Bridge gmders, stringers, and some floor beams are loaded by Lines of wheel loads that roll along the deck. A wheel line is half of a truck or lane load. The number of h e s assigned to each girder depends on the girder spacing and type of pder.

The live loads are moved longitudinally along the bridge, andas they move, they generate changurg effects in the bridge members. Refer to Appendices A.4 and A.5 for an example showing the application of H loads to a 3-span continuous structure.

The maximum moments or shears resulting from these moving loads are evaluated at the various locations by the computer program, Bridge Design System.

SpeoifIcaLly, in superstructure design we are concerned with the maximum Live load effect that any one member can experience regardless of the number of live load lanes the bridge can accommodate. This effect depends on the transverse stiffness of the superstructure--its abjli ty

Seclion I - Bndge Loads Page 7 - 1 1

c* - Bridge Design Practice - February 1993 m

to distribute loads IateraUy. For instance, a concrete box grder dishibu tes live Ioad much more completely than a steel shinger bridge.

BDS 3.23 gives the mechanics for determiningdisbibution to girders by empirical formulae that consider structure type and girder spadng. The r e d & are generally in fractions of wheel h e s per girder.

On box girder structures, the "5-over" distribution of BDS 3.23 is applied to the entire width of bridge as a unit.On structures other h n box girders the "Sover" distribution apphes to single girders only, where S is the girder spacing. For ex teriot grrders, the same dtstribution applies but with a factor to account for length of deck overhang.

As gvder spacing increases, a point is reached where the " h v e r " distribution no longer applies. Beyond this point che wheel loads are distributed to the girders assuming the deck slab acts as a simple beam between girders. Lii.iting girder spacing for this condition depends on type of superstructure.

Because of the wo aiteria for wheel load distribution, depending on whether girders are "closely spaced" or "widely spaced", P load dishbution for superstructure design is divided into two procedures.

(1) For closely spaced girders, P loads adyare applied. l2i.stri.b~ tionis by the "S-over" formula.

Tkis procedure effectively applies a major P fraction to every girder in a system. The total design load on a girder system two or more lanes wide exceeds the in tended single P load and adjacent H load as required in BDS3.11. Theexcesscapacity provlded isenough to aUow bonuses that exceed the P Ioad sometimes granted with permits.

(2 ) For widely spaced girders, a single P Load and an adjacent H load, bolh positioned for maximum effect are applied. Distributionis by static reactions on the guders, as specified in BDS Table 3.23.1, Footnote I.

Figure 2 shows examples of live load distribution to superskctures and substructures.

One other point to be kept in mind is the discontinuity in procedures for applying live Ioads to various elements of a bridge structure. The deck slab js designed according to one loading ai terion, the p d e r s by another. The live load reachons from these loads are then discarded,and we start anew as design proceeds to Ihe substructure. Different live loads may control design at different locations. Also, the effect of ljve load &tributes, and the effect of impact dissipates as they move down through the stmcture. Reductions are taken to allow for improbability that several heavy vehdes will uoss a structure shultaneousl y. For these reasons, it is difficult to trace logically a pven design live load from deck level to foundation. The t o h l live load considered td be taken by the foundationis almost always less than that which was applied to the su pershuctu re.

Page 1-12 Section I - Bridge Loads

Live

Loa

d D

istr

ibut

ion

to S

uper

stru

ctur

es

Clo

sely

Spa

ced

Glr

ders

W

idel

y S

pace

d G

irde

rs

Co

ncr

ete

~:e

Bea

m

Con

cret

e B

OX

Gir

der

C

on

cret

e b

eck

on S

teel

Str

inge

rs

Use

sam

e nu

ntbc

r or

lane

s fo

r gr

oups

I,, a

nd I,

wiih

no lane re

ducl

ion

facl

or.

No

frac

tia~r

d lane

s al

low

ed in

dct

erm

inal

ion

nf

1 la

ne

sim

plc

heam

deck

rcac

~io

~is.

D

is~

rib

~il

c bv S

- ovc

r Ibr

mul

ac m

ulllp

licd

by r

atio

: 2

whe

el li

ttcs

G

roup

Ill us

e m

a..

num

ber o

f la

nes

wil

h c

lnc

13

11

~

Iivc

load

la~

~es

S

R17

live

load

lane

s ov

er-a

ll dc

ck w

idrli

- - - -=

0.

68

- - in

~eri

or ai

rder

12

12

br

idge

14

re

duc!

ion

racr

ors.

DII

S 3J

2

56.7

5 z-

= 0.66~ 0.68

14

use

O! 68

=

4.05

live

load

sIsn

cs=

1m52

1-l (3

lanes of

!I loads

wirh

ia

leri

ur g

irdcr

ln

nc rc

ducl

ion

faci

or)

live

load

lane

s - 1.64

1 1 (2

lar

les

of I

l loa

ds)

L

J

live

load

Isne

s 4.

05

cxle

rior g

irdcr

lo

la1

live

load

larlc

s -

--

- 1 -

0.58

* =

2f0.

68) +

5(0.68)

= 4.

76 *

gird

er

(old

live

load

Ianc

s br

idgc

=

1.52

11 +

2(1.

fi4)I

I =

4.

8; 11

4 br

idgc

liv

e L

oad

Dis

trib

utio

n 10

Sub

stru

ctur

e G

roup

1,. us

c 2

lnnc

s or

I la

l-lc

lane

s

Exa

mpl

e of

Liv

e Lo

ad L

ocat

ion

Den

t cap

s ar

e co

nsid

ered

par

1 of

subs

uucc

urc .

U

seG

roup

Illo

r I,,

v

Num

ber n

f lan

cs a

nd p

osili

ons

acro

ss ro

adw

ay

shal

l be

as re

quire

d lo

pro

ducc

max

imum

cWcc

1

No

frac

tiona

l lan

es a

llow

ed.

Lane

redu

crio

n Fa

c~or

a sha

ll ap

ply.

For G

roup

lpw

.

only

one

P lo

ad, o

r one

P lo

ad a

nd

one

11 to

ad m

ay b

c ap

plie

d.

Figu

re 2

Hve

load

s la

ncs

(1 l

x~

e I1 lo

ad

= .6311+,84P

ilire

rior g

irde

r 1

lane

P Iu

ad)

live

load

lane

s (1

lanc

1-1

load

=

.511l+.1.141'

cxle

rior g

irder

1

tane

I' to

ad)

lola

l liv

e lo

ad lanes =

1.6511 +

3.12

P *

brid

ge

Crn

trm

l Nolrr:

.Cro

ups

I I,

1 k Ipw

ar

c cv

alu

slc~

l sepa

rste

ly,

nrll s

rrrr

ulrm

cnus

ly.

AN

I

IO

~~

~I

\~

P

Kofio

ua ar

c rl

rcfr

clar

rcrt

.*

Dcn

um

vah

cs g

ivul

Iw co

rnpa

rlvl

lr p

urp

osa

. U~

ert

gird

ers

.rrc

btd

tn~

rily

dcbi

glle

d rc

puhl

cly.

E-t: - Bridge Design Practice - February 1993 m

1.16.1 Load Factors

An essential feature of LFD, as stated in 1.0.2, Design Methods, requires raw design loads or related internal moments and forces to be modified by specified load factors (y, gamma and p, beta), and computed material strengths to be reduced by specified reduction factor (q, phi ).

These are safety factors which ensure certain margins for va.riation. The three different kinds of factors are each set up for a distinct purpose, each independent of the other two. In ths way, any one of them may be refined in the future without disturbing the other hvo.

1.1 6.2 y (Gamma) Factor

The y (gamma) factor is the most basic of the three. It varies in magnitude from one load combination to mother, but it always applies to all the loads in a combination. Its main eifect is stress control that says we do not want to use more than about 0.8 of the ultimate capacity. Its most common maptude, 1.3. lets US use 77%. Earthquake loads are not factored above 1.0 because we recognize that stresses in the phstic range are dowed. as long as collapse does not occur.

An example may be pven to justrfy the use of gamma of 1.3 for dead load.* Assuming the jive load being absen t, the probable uppe-r value of the dead load could be a minimum of 30% greater than c a l d a ted. For a simple structure thrs percentage may. be as follows:

1 0 due to excess weight. 5% due to misplaced reinforcement. 5% approximation in behavior of structure. 39"/. increase in stress, actual compared with calculated. 30% Tolal variation assumed to occur c o n m t l y at the section most heavily

stressed.

"Nola MI Load Factor Design for Reinforcert Cnncrw Bridge Su-uclms wilb k s i g n App1ic.ations" by Portland Cemenr Association, Page AB-9-

Page 1-74 Secrioh I - Bridge Loads

Em - Bridge Design Practice - February 1993

1.16.3 P(Beta) Factor The second factor, P (beta),is ameasure of the accuracy with which we can predict various kinds of loads. I t also reflects the probability of one load's simultaneous application with others in a combination. It applies separately, with different maptudes, to different loads in a combination, For example, it is usually 1.0 for dead load. It varies from 1.0 to 1.67 for live loads and impact.

Due regard has been given to sign in assigning values to beta factors, as one type of loading may produce effects of opposite sense to h a t produced by another g.pe.The loadcombinations with 6, = 0.75 are spedically included for the case where a higher dead load reduces the effects of other loads.'

The beta factors for prestressing force effects are set so that when multiplied by the respective gamma factor, tlw product is unity. Beta of 1.67 for live load plus impact from H Ioads reflects A ASHTO'S way of handling permit loads; the 1.00 and 1.15 for P loads on widely spaced guders accounts for bonuses sometimes granted in Cnltrans' permits.

1.16.4 #(Phi) Factor $ (phi), the third factor, relates to materials and is called either a capadty reduction factor or a strength reduction factor. Its purpose is to account for small adverse variations in material strength, worlmmhip, and dimensions. I t applies separately to different maptudes for various load effects in reinforced concrete, and various manuiactuing processes in presmssed concrete. Since $ relates to materials rather than loads, its values are given in the various material specifications. For structural steel it isalmost alwaysl.O. For conaete it varies from 0.7 to 1.0.

* Commenuq on Building Cde for Reinford Coucrere (ACI 3 f 8-77), Page 33.

Section 7 - Bridge Loads Page 1-15 a

r-t. - Bridge Design Practice - February 1993

f .I 7 Lo-ad Combinations

The various load combinations to which a bridge m a y be subjected as well as the appropriate load factors are p e n in EDS 3.22, Table 3.22,lA and Table 3.22.1B.Different groups control the design of different parts of the s h c t u r e , and it is often necessary to tabulate loadsand effects to determine the cont rohg loads on members such as abutment or bent columns. It is, of course, not necessary to investigate all the loadings for a given bridge. It is often evident by inspection that only a few loadings are Likely to control the design of any single type of structure.

Group 1, contains no P loads and applies to superstructures as well as substructures. Group I, is used only for P load application to superstructures with closely spaced gvders where the "Sover" famuhe apply.

Group I , is used for P load application to substructures and superstructures with widely spaced guders. Only one P load or one P load with one H load may be applied to thestntcture at a time and placed for maximum effect.

Loads as combined and factored for Service Load Design in Table 3.22.1B are for use in the service level considerations of LFD,and the mre occasions when SLD isappropriate.Stresses for the various groups arelimited to the specikied auowable stress for a materia1,adjusled by the percent overstress factors in t he table.

Page 7-76 Sect~on 1 - Bridge Loads ,


1.18 References

Bridge Compufer Manual loose leaf binder issued by Office of Struchre Design, Caltrans.

B d g e Dcsigrr Aids and B r i d g ~ Design Details, loose leaf binder issued by Office of Structure Design, caltrans.

Bridge Dmgn Spccifimtions, loose leaf binder issued by Office of Skucture Desjgn, Caltrans, containing Standard SpebJcations for Hicghzony B r i d p , 13 t h Ed ition, 19&3, with In terini SpeCiFcations, Bnlges, thm 1984, pubkhed by American Association of State Highway and Transportation Official5 with California modifications.

Ezrilding Code Requirements for Reinforced Concrcte lAC1 31 8-77) with Commentary, American Concrete Institute. Detroit, December 1977.

Memos to D ~ i p e r s , loose leaf binder issued by the Office of S b u c w Design, Caltrans.

Notes on had Fador Design /or Reinforced Concrete Bridge Structures with Desipt Applications, Portland Cement Association, 7974.

Steel Sheet Piling Design Munw1, U.S. Steel

Tay tor, Donald W ., F~lndarnm tals of Soil Mecimn ics, John Wiley and Sons, hc., New York, 1948

Tenaghi and Peck, Soil Mdurnics in Engineering Practice, John Wiley and Sons, Inc., New York, 1967.

Section r - Bridge Loads Page 1-17


Appendix A-1

Note: Oniy one truck per lane is ta be used far a maximum moment or shear determination for eirher simple or continu~us spans.

Clearance and

W = Combined weigh1 on the fist two axles which is the same as lor the ALTERNATIVE LOADING corresponding H truck.

V = Variable spacin-14 feet to 30 feet inclusive, Spacing to be used is that which produces maximum stresses.

18.000 Ibs. for Moment *** I- Concenlrared Load - 26.000 Ibs. for Shear

t Uniform Load 640 ibs per linear fool of load lane

* Width 01 tires shall be the same as the Standard H Truck

** For slab design the centerfine of wheel shall be assumed lo be one foot from face of curb.

* * * For continuous spans another concentrated load of equal weight shall be placed in one other span in the series. in such position as 10 produce maximum negative moment.

FIGURE 3

Page 1-18 Seclion I - Bridge Loads


Appendix A-2

P5 2Bk 48k 48k - - - - Minimum Vechicle

P7 26k 48k 48k 48k - - - P9 26k 48-8k 4gk 4ak - - PI1 26k 48K 4 8 " ~ ~ 48k 4ak - P I 3 26k 48k 48k 48-4k 48k 48'1 MaximumVechicle

10'4" clearance

Section A-A

P Loads Permit Design Vechicles

Figure 4

Section I - Bridge Loads Page 1-19

+PI I 8 I , 8 18' Appendlx A-3

!dBk / d B k 1 4 l k 4Bk 148' 1 l S k 1 PB* 18,000

. 4 8 k t48k f48k 148' t 4 8 k t26k PI1 Permit Deslgn Llve Loads

- 4ok +4ak t 48k t 26k P7

.64 /tl. ol toad lane H Loads P Load Moment

a 07 w -

-- Slmpla Span (One Lane no Impact) 40 -

Scale Horlz: 1"=20', Vart: lm=2009;1! Morn.* I (before reduction) Vert: 1 "= 40 , Shear, 10

50 100 150 200 25 0 290 P Simple Span Lengrh (it) 4

r-it - Bridge Design Practice - February 1993

Appendix A-4 Example of Moment Envelope Calculations

A-4. I Example of Moment Envelope Calculations For Conk inuous Structure

On continuous structures jt is not always obvious by inspection how the loads should be placed to produce maximum conditions. A great deal of guesswork can be eliminated in the placing of live 1,oads for maximum moment, shear, or reactions by the use of influence lines. The Bridge Design System t vd l automatically generate influence lines and determine girder moments and shears along with top of column moments and reactions. I tis difficult lo visualize the way buck loadings actually generate a moment or shear envelope, so we will go through a step-bystep development of -these envelopes.

This example treats only application of H loads. The vehide is moved auoss the structure horn left to right and from right to left. The moments and shears a! the various tenth points of the spans are noted as this movement occurs. The envelope is determined when the moment or shear at each tenth point is maximum. Table A 4 shows a summary of this work for maximum moment Table A-5 shows maximum conditions for shear. The maximum moments or shears are shown at the tenth points along with the position of the vehicle or lane load that produces the maximum condition.

Bridge Design System automatically moves the applicable Eve loads across the structure in both directions, notes the values of moment and shear and reports the maximums in tabula or graphcal form at each tenth paint.

We will use a simplified thee span continuous smcture with 50-foot spans simply supported (Figure 6).

AblR 1 Bent 2 Bent 3 Abtrl4

50' 50' 50' j

Figure 6

-

Page 1-21 Section Y - Bridge Loads -

-

E= constant

Girder dead load= 2klfl I= constant

c* - Bridge Design Practice - February 1993 D

For praposes of this example, assume h a t HS 20 live loading consists of 1.5 wheel line or 0.75 lane per girder. Impact is 28.6%

The (LL+I)H values given in T a b k A 4 and A-5 result from applying HS 20 loads converted as follows:

Truck Load

Jm paa 8 kip 1.286 1.5 wheel Line

x- x x lane = 7.7 Kjgirder lane I girder 2 wheel line 32 kip 1.286 1.5 wheel line

x- x x lane = 30.9 K/girder Ime I girder 2 wheel line

h e Load

0.640 IGp L.286 1.5 wheel line 1 lane x- x x = 0.6 17 K/girder/ft-

fr. lane 1 girder 2 wheel line

I8 Eirp 1.286 L 5 wheel line X

1 lane x- x = 17.4Ktgirder

lane 1 @r 2 wheel line

Maximum moment and shear envelopes are determined similarly when P loads are applied.

The values of (LL+l)P moments for 0.75 lane of P loads are p e n for comparison purposes on Table A+, line 21. Sirmlar values for shear are grven in Table A-5, h e 13.

1t should be noted that even though these raw values for P loads are lugher than the H load results, the application ofdifferent load factors for each brings the results closer together during design

A-4.2 Dead Load Moments (7kble A-4, Lines 2 and 3) Dead load momentsare givenby Bridge Design System using the uniform load of 2 kips per foot. A plot of these values dosely follows parabolic curves.

A -4.3. Envelope Curve For Positive Girder Moment in Span 1 (Table A-4, Lines 4- 1 1)

For the %foot span, the HS ~ckproducesmardmum positive moment in the span. Each horizontal Line of value indimtes dw moments at the tenth points of the first two spans of the three span shcture. moments are mused by ltbe 0.75 lane plus impact for the HS truck located as shown by the circles indatmg axles. T k mak M e r e p m t s the front axle of the truck

*

Page 1-22 Section I - Bridge Laads

- Bridge Design Practice - February 1993 - \ W e not always true, the maximum moment due to two or more moving concentrated loads generally occurs when the heaviest load is at the section.

From the tabulation it can be seen that the maximum positive moment- is at the 0.4 poizt of the span (Table A 4 , line 7). For prac t id purposes the h c k could be placed anywhere from the 0.35 to the 0.45 point to obtain the maximum positive moment, without appreciable error-

For the condit.ion of dead load and live load indicated, the tabulation of underlined (LL+I)H moments when added to the dead load moments gives the maximum positive moment thatcan be obta.ined at each of the points shown (line 19). A curve passed through these points constitutes an envelope curve pf positive moment (see Figure 7).

A-4.4 Envelope Curie For Negative Girder Moment (Table A-4, Lines 72-15)

Maximum negative &L+T)H momen1 over the suppon (Bent 2) is produced by the l a~e load io Spans 1 and 2 wirh concenmted loads for moment (line 12). Momenrs produced by h e lane loading are slightly greater rhan hose produced by the RS 20 uuck.

Except near Bent 2, maximum live loadnegative rnomentinSpan 1 is produced by loading Span 2. It is noted that a plot of Ihe Span I moments will produce a shight line. The load position is shown on line 13. T h e negative moment in unloaded spans is frequently overlooked by beginners.

The envelope curve for negative Live load moment in Span 2 is obtained by loading the adjacent span as shown o n h e 14.

An ad& tional re6nement to h e negative moment envelope is obtained by loading Spans 1 a i d 3 w i k the lane loading shown on line 15. Th.ts loading slightly modi6es the envelope curve between the 0.4 and 0.5 points on Span Z However, the modification is of little practical value in determining cutoff of negative bars in a concrete span, because the value Fa& below the resisting moment of long tudinal bars which are normally carried through

The envelope curve for nega tive moment is obtained by combing maximum values of dead load and &L+l)H. The valuis are tabulated on Line 20 and the envelope is plotted in Figure A 4 . I t is seen that the positive and ntptive moment envelopes overlap along the base line. 'Ibis is characteristic of continuous structures.

A-4.5 Envelope Curve For Posirive Girder Moment In The Interior Span Of Three Continuous Spans (Table A-4, Lines 7 6- 17)

The truck load again produces maximum positive center span live Ioad moments. The moments due to truck loadings at the 0.2 and 0.5 points in the span are shown. These points are control points which determine the envelope cure which approximates a 2nd degree parabola. The combined values of dead and l ive load m0me.n ts are shown on line 19.

Section 1 - Bridge Loads Page 1-23

r* - Bridge Design Practice - February 1993 =

Appendix A-5 Example of Shear Envelope Calculations

A-5. I Envelope Cuwe For Girder Shear In End Span Of Three Continuous Spans (Table A -5, Lines 4 -7)

The same threespan contjnuous structure is u&ed to illustrate h e placing of load in continuousspans lor determination oflive load shear. Dead loadshearsare tabulated onTable A-5 h e 3 and plotted on Figure A-5.

It can be found that the maximum live load shear at Abutment 1 is governed by the buck loading.

For practical utilization of the curve of maximum shear i t is necessary to determine the LL+l shear at only three points; at Abutment 1, at the 0.4 point of Span I , and left of Bent 2. These points are connected w i h straight h e s as indicated in the diagram on Table A-5.

The variation between the actual shear envelope wh& has a slight curve and the straight line method desaibed above is of no practical concern, as other empirical assumptions may introduce a greater ddference.

The shear diagram for use indeterminingstirrup spacing inconcrete beams,or stiffenerspacing in plate girders is constructed by combining the dead load shears with the maximum (LL+I)H shears as shown on Figure A-5.

A-5.2 Envelope Curve For Girder Shear In Center Span Of Three Continuous Spans (Table A-5, Lines 8-9)

In this structure the DL+(LL+I)H shear curve for Span 2 is symmebical about the mterline of span; therefore it is necessary to compute shear at right of Bent 2 and at the centerline only. The load positions and values of live loadshear for Span2 we given on lines 8and9. Negativeshear in Span 2 has the same value as pos tiveshear and isdetemined by placing the buck on the other end of the span.

The (LL-+1)H values are added to the dead load shears to give the DL+(LL+I)H w e shown in Figure A-5.

Subsequent sections of the course w ill cover the utilization of shear and moment envelopes for the design of various types of structures.

Page 1-24 Secfioion I - Bridge Loads

Moment, Table A-4

wheel line)( lane ) aK"anB (1'268'(1'5 girder 2 wheel line = 7 . f /girder 0 32K 0.286) = 30.gK/girder 0

Envelope Curve of Posltive Mornenl In

+075

S

ymm

atrl

cal

Co

nlln

uo

~s S

truc

lure

ab

out 'l S

pan

Wilh

Thr

ee SO

loo1

spa

ns

800

I N

oles

:

DL +

LL

I Inc

lude

eH

ecls

600

of 0.75

lane

of H L

oad.

For S

pan

3 va

lues

, see

S

pan

1

400

See

Tab

le A

-4 lo

r la

bula

led

valu

es o

f m

omen

t at

tent

h po

lnts

ol S

pans

.

200

I / /

\

/

0 -200

-400

-600

-800

Q;

Q; Q;

Ab

ut I

Ben

1 2

Ban

13

Mo

men

t E

nve

lop

e Fi

gure

7

Shea

r, Ta

ble

A-5

Figu

re A

-5

She

ar E

nvel

ope

-5

55

55

55

S

hear

al

Lelt

of 8

enl 2

7

0

6 26

-4

-4

-12

-12

-12

-12

-12

She

ar a

t R

lght

of

Ben

t 2

0 0 P

osili

ve S

hear

at .5 pt

. -3

-3

-3

-3

-3

-3

-3

-3

-3

-3

-3

31

21

21

21

21

-1

0 -

1 0 91

-41

-4

1 S

pan2

ur a

DL+

(LL+

I)H

--

3-50 lo

ot s

pan

z-

conl

lnuo

us s

truc

ture

0

No

te:F

ors

pan

3 JZ

VJ

valu

es s

ee

-50

span

1.

-54

45

37 30

23

16

11

7 4

2

2 -5 .S

.g

.I3

-20

-27

.34

-42

-49

-55

-60

94

75

57

40

23

6

-20 -37

-54

-72 -

89

-1 05

-1 2

0 -3

81

66 52

40 30

21

13 6

2 2

2 -7 -7 -12

-17

-25 -39

-52

-63 -77 -90

-103

--IZO

abou

t I$ sp

an 2

'

57

51

43

36

28

21

14

11

8

8 8

-0

-8 .8,-1, -1

4 -2, -28

-36

-43

-51

-57

107 91

73 56

30

21

4 -4

-2

1 -38 -

56 -73 -

91

-107

98

85

73

59

47

34

21

11

11

11

11

-11

-11

-11

-11

-21

-34

-47

-59

-73

.B5

-98

ILL4

I)HE

nvdo

e01

1nax

.She

~11

01

.7~

~a

ne

alR

1o

ad

s

Posi

live

Shea

r DL

+(LL

4 I)H

'1 1

N

egal

lve

Shea

r D

L+(L

L+I)H

12

.L

oc 7

Can

eolP

load

s 13

r* - Bridge Design Practice . February 1994 =

Section 2 . Reinforced Concrete

Notations and Abbreviations ................................................................................... 2-1

2.0.0 Introduction ..................... .. ............................................................................ 2-4

Part A . Design Example ........................................................................ 2-5

............................................................................... 2.1.0 Structure Requirements 2-5

.................... .............................................. 2.2.0 Typical Section Geometry .... 2-5

2.3.0 Superstructure Loads ..................... ... .......................................................... 2-8

2.3.1 Dead Loads, D .......................................................................................................... -2-8

2.3.2 Live Loads, L ............................................................................................................... 2-8

2.4.0 Effective Depth ......................-.-.... .... ............................................... 2-9

2.41 W u m Bar Cover ....................... ... ..................................................................... 2-9

2.4.2 Transverse Bars ........................................................................................................... 2.9

2.5.0 Factored Design Shears .............................................................................. 2-10

........................................................ ....................... 2.6.0 Girder Web Flares - 2-11

...................................... 2.7.0 FactoredDesign Moments .............................. 2-14

......................................................................... 2.8.0 Maximum Design Moments 2-15

2.9.0 Steel Requirements at Maximum Moment Section ....................... .. ......... 2-15

2.9.1 Positive Moment Section Parameters ............................................................ 2-15

2.9.4 Span 3 0.6 point. Mu = 14529 k-ft (Solution Method 3) ........................................ 2-17

2.9.5 Negative Moment Section Parametws .......................................................... 2 8

P

Section 2 . Contents Page 2-i

=* - Bridge Design Practice . February 1994

2.9.7 Bent 3. M u = -23577 k-ft ..................... .. ............................................................... 2-18

2.1 0.0 Maximum AlIowed Tension Steel .,.- ..----............. .... ---- .. ......................... 2 9

2 10.1 Maximum Tension Steel in the Soffit Slab ............................................................ 2-19

2.10.2 MaKimumTensionSteelintheDeckSlab ............................................................. 2-19

2.1 1.0 Effective Tension FIange Width - ...................--~-----.....-.-.--................. - 2-20

........................... . 2 11 . 1 Span 2 Positive Moment Tension Flange Width (soffit slab) 2-20

........................... . 211.2 Bent 3 Negative Moment Tension Flange Width (deck slab) 2-21

2.1 2.0 Positive Moment Bar Size Limitation .......................................................... 2-23

2.12.1 Span 2 inflection Points ........................................................................................... 2-23

2.122 Span 3 Inflection Point ....................................~..................................................... 2 - 2 3

2.12.3 Span 3 Abutment .................................................................................................... 2-24

2.1 3.0 Crack Control (Pre-Design) .......-......................... ....... ....................... 2-24

2.13.1 Span 2 0.5 point - #9 bars only ................................................................................ 2-27

2.14.0 Bar Spacing Limits ........................................................................................ 2-29

2.1 4.1 Minimum Bar Spacing Limits ................................................................................ -2-29

214.2 MaximumEkSpacjng Limits ................................................................................ 2-29

214.3 MinimumNumberofBarsRequired ..................................................................... 2-29

2.1 5.0 Minimum Reinforcement Requirements ......................... .... ................. 2-30 2.16.0 Bar Layout. Span 2 ..................................................................................... 2-31

216.1 ChooseBarGroups ................................... .... ............................................................ 2-31

2-17-0 Bar Layout, Span 3 - Positive Moment ...................................... ........... 2-34

2.17.1 Choose Bar Groups ............................................................................................ 2-34

2.1 8.0 Bar Layout, Bent 3, Negative Moment ....................... , .............................. 2-37

2.18.1 Choose Bar Groups ................................................................................................... 2-37

2.1 9.0 Fatigue Check ..--.....-......... ., .............~.~~~~~.......................................... 2-39

Page 2-ii Section 2 . Contenls

r* - Bridge Design Practice . February 1994 m

2.20.0 Final Bar Layouts .......................................................................................... 2 4 2

2.20.1 Span 2 . Bottom Slab Reinforcement ......................... .... ................................. 2 4

2.20.2 Span3-BottomSlabRemforcement ...................................................................... 2-43

2.20.3 Bent 3 - Top Slab Reinforcement ......................... .. ................... ............... 2 - 4 4

2.21.0 Longitudinal Web Reinforcement ............................ ... ....................... 2 4 5

2.22.0 Shear Reinforcement ................................................................................... 2 4 6

2.221 SlirmpDesignWithinTheFlares .......................................................................... 2-47

2.22.2 Stirrup Spacing Limits ..................... ...... .................................................... 2 - 4 8

2.22.3 Shear Capacity for Different Stirrup Spadngs ..................... .... .............. 2 8

2.22.4 Graphical Procedure ................................. .... ....................................................... 2-49

2.23.0 Bent 3 Model ....................... .. .................................................................... 2-50

2.24.0 Bent Loads .................................................................................................. 2-50

2.24.3 Dead Loads .................... .... ................................................................................. 2-50

2.24.2 Live Loads .................................................................................................................. 2-50

2.25.0 Bent Cap Geometry .................................. ... .................................................. 2-51

2.26.0 Face of Bent Support .................... .. ........................ ............................. 2-52

2.27-0 Factored Cap Design Moments ..................,................................................ 2-53

2.28.0 Maximum Design Moments ....................... - ................................................. 2-53

2.29.0 Bent Cap Minimum Fleinf arcernent Requirements , ................................. 2-54

2.29.1 Fo r Positive Moments .............................................................................................. 2.54

2.29.2 For Negative Moments ........................................... ...-

2.30.0 Cap Effective Depth ..................................................................................... 2-54

2-31 -0 Cap Steel Requirements ........................... ... .............................................. 2-55

2.31 -1 Positive Moment Sections ........................................................................................ 2.55

2.31.2 Negative Moment Sections ................................... ..... ........ 2-55

2.32.0 Crack Control ....................... ... ................................................................... 2-56

b

Section 2 . Contents Page 2-iii

r* - Bridge Design Practice . February 1994 m

2.33.0 Construction Reinforcement .............................~........................................ 2-57

2.34.0 Cap Side Face Reinforcement .................................................................... 2-59

............................... ........................................... 235.0 Cap Shear Reinforcement - 2-59

2.35.1 Span 1 Shear Design at Support Face ..................................................................... 260

2.35.2 Span 1 Shear Design at Cap End ........................... .. ....... .. ........................... 241

2.36.0 Final Cap Design ......,.................................................................................. . 2-63

2.37.0 Computer Output .......................-.......... .-.. . ............................................... 2-64

237.1 Factored Loads - BDS .............................................................................................. 2 4 5

2.37.2 Service Loads - BDS ................................................................................................. 2-78

237.3 Bent 3 Loads - BENT .................................. ... .......................................................... 2-86

Part B: Design Notes ........................................................................... 2-92

2.38.0 Service Load Design = Overview .. .....................,........................................ 2-92

2.39.0 Strength Design Method . Overview ....,.................................................... 2-93

........ 2.40.0 Face Support . Negative Moment Design .................................... - 2 - 9 5

2.40.1 Example ...................................................................................................................... 2-96

2.41.0 Cross Sections Experiencing Positive Bending ........................................ 2-97

2.42.0 Cross Sections Experiencing Negative Bending ...................... ,., ....... -2-98

2.43.0 T-Girder Compression Flange Width Positive Moment Case ....,............. 2-99

243.1 Example .................................................................................................................... 2-1 00

2.44.0 Box Girder Compression Flange Width ....,...... , ................-..-....-.~.......... 2-1 00

2.45.0 Box Girder Effective Tension Flange Positive Moment Case ....,........... 2-101

2.46.0 Box Girdern-Girder Effective Tension Flange Negative Moment Case ...................... .... ......................................... 2-1 03

Page 2-2 Section 2 . Contents

c* - Bridge Design Practice . February 1 994 =

2.47.0 Rectangular Sectlon With Tension Bars Only ...................... .. ............. 2-1 04

2.45.1 Example ......................... ... ................................................................................ -2-105

2.48.0 Flanged Section With Tension Bars Only ................... .. ..... .............. . . 2 0 6

2.48.1 Example .................................................................................................................... 2-108

2.48.2 Example ................................................................................................................... 2-108

2-49.0 Bar Spacing Limits for Girders .......-----......................,.....-..........--.-.....-... 2-1 09

2.50.0 Development of Reinforcement .............................................................. 2 - 1 0

..................................................... 2.51.0 Positive Moment Bar Size Limitation 2-1 I 0

2.51 -1 Example ................................... ..... ................ ...:... ..................................................... 2-1 10

2.520 Minimum Reinforcement Requirements .................................................. 2-1 I I

2.52.1 Example ................................................................................................................. 2 - 1 2

2.53.0 Moment Capacity Diagram ........................ ... ........................................ 2-1 13

............... 2.54.0 Moment Capacity Diagram Versus Design Moment Envelope 2-1 14

2.55.0 Bar Layout-Graphical Procedure ............................................................... 2-1 15

2.56.0 Matching Bar Ends ...................................................................................... 2-1 17

2.57.0 Working Stress Analysis Calculations ..................... ... .......................... 2-1 19

257.1 Example .................................................................................................................. 2-120

2.58.0 Crack Control Serviceability .......................... .. .............................. 2-1 21

2.59.0 Crack Control Check . Post Design Rectangular Sections ..................... 21122

259.1 Example .................................................................................................................... 2-123

2.60-0 Crack Control Check Post Design f Box Girder with Single Layer of Steel ........................................................................ . 2-1 24

2.60.1 Example ........................ .... ................................................................................. 2-125

2.61.0 Crack Control - Pre Design ......................... .. ......... ............................ 2-126

2.62.0 Pre Design Crack Control Derivation ........................................................ 2-128

Seclton 2 . Contenls Page 2-v

E-t: - Bridge Design Practice . February t994

2.63.0 Fatigue Serviceability ..................................................................-............ 2-130

2.63.1 Derivation for Procedure Outhed ...................... ... ....................................... 2-131 2.63.2 Example ................................. ... ............................................................................. 2-132

2.64.0 Shear Design .......................................... - ................................................... 2-133 2.65.0 Shear Design and Girder Webs .................................................................. 2-133

2.66.0 Shear Design of Flared Girder Webs = Example ....................................... 2 3 4

2.67.0 Shear Modification Due To Bar Cutoffs ....................... ,. ...................... 2 3 7

2.67.1 Modification Method 1 .......................................................................................... 2-137

...................................... 2.67.2 Modihca tion Method 2 2 3 7

2.67.3 Modification Method 1 . Derivation ................................................................ 2-139

267.4 Modification Method 2 . Derivation .................................. ... ......................... 2-139

2.68.0 Shear Friction Design ............................................................ . . . . . . 2-140

268.1 Basic Shear Friction Requirements ....................................................................... 2-1 42

2.682 Example-Shear Key ............................................................................................... 2-144

2.69.0 Compression Members ......................................................................... 2-1 45

269.1 Example ................................................................................................................... 2-146

2 69.2 Example ............................................. -. ................................................................... -2-1 SO

Page 2-vi Sedion 2 . Contents


Reinforced Cc

Notations

a = depth of eqwvalent rectangular stress block (BDS Artide 8.16.2.7)

ab = depth of equivalent rectangular stress block for balanced strain conditions, inches (BDS Artrcle 8.16.4.23)

A = effective tension area of concrete surrounding the flexural tension reinforcement and having the same centroid as that reinforcement, divided by the number of bars or wires, square inches; when the f l e d reinforcement consists of several bar sizes or wires the number of bars or wires shaU be computed as the total area of redorcement divided by the area of the largest bar or wire used (BDS Artide 8.16.8.4)

A, = area of an individual bar, square inches (BDS Article 8.25.1)

A, = area of concrete section resisting shear transfer, square inches (BE Artide 8.16.6.4.5)

A, = gross area of section, square inches

& = area of reinforcement in bracket or corbel resisting tensile force, N, (N,), square inches ( B E Artides 8.15.5.8 and 8.16.6.8)

= area of tension reinforcement, square inches

A', = area of compression reinforcement, square inches

ASf = area of reinforcement to develop compressive strength of overhanging flanges of I - and T-sections (BDS Article 8.16.3.3.2)

~ncrete Design

. = area of shear-friction reinforcement, square inches (BDS Article 8.15.5.4.3)

l b = width of compression face of member

= effective tension flange width (not a code variable)

bw = web width, or diameter of circular section. For tapered webs, the average width or 1.2 times theminimum width, whichever is smaller, inches (BDS Artide 8.15.5.1.1)

= dstance from extreme compression fiber to neutral axis (I3DS A.rticle 8.16.27)

d = distance from exbeme compression fiber to centroid of tension reinforcement, inches. For compumg shear shength of circular sections, d n e d not h less than the distance from extreme compression fiber to centroid of tension reinforcement in opposite half of member. For computing horizontal shear strength of composite members, d shall be the distance from extreme compression fiber to centsoid of tension reinforcement for entire composite section

d ' = distance from exbeme compression fiber to centroid of compression reinforcement, inches

d, = nominal diameter of bar or wire, inches

4 = thickmess of concrete cover measured from extreme tension fiber to center of bar or wire located closest there to (BD5 Article 8.16.8.4)

Ec = modulus of elasticity of concrete, psi (BDS Artide 8.7.1)

Es = modulus of elastidy of reinforcement, psi (BDS Artide 8.7.2)

Section 2 - Reinforced Concrete Page 2- I a


= extreme fiber compressive stress jn concrete at service loads ( B E Artide 8.15.21.1)

Mu = factored moment at section

n = modular ratio of elasticity = E,/E, @DS Article 8.15.3.4)

= specified compressive strength of conaete, psi

n = number of bars

N = effective number of bars = square root of specified comprmive

strength of concrete, psi Nu = factored axial load normal to the cross section cccurring simultaneously with v u

= fatigue stress range in reinforcement, ksi (BDS Artide 8.16.8.3)

P, = nomind axial load strength of a m o n at balanced sfrain conchions (BDS Artide 8.16.4.2.3)

= algebraic minimum stress level in remforcement (BDS Arhde 8.1 6.8.3)

= modulus of rupture of concrete, psi (BDS Artide 6.15.21.1)

P n = nominal axial load strength at given eccentricity

= tensile stress in reinforcement at service loads, psi (BE Article 8.15.22)

P, = factored axial load at given emenbicity

s = spacing of shear reinforcement in direction parallel to the longitudinal reinforcement, inches

= stress in compression reinforcement (different than defined in code)

= extreme fiber tensile sbess in concrete at service loads (BE Artide 8.15.21.1)

t = tension flange thicEoness (not a code variable)

= specified yield strength of reinforcement, psi

V, = nominal shear strength provided by conaete (BDS Article 8.16.6.1)

= overall, thickmess of member, inches V, = nominal shear strength (BDS Article 8.16.6.1) = compression flange thihess of I- and

T- sections V, = nominal shear sbength provided by shear reinforcement (BDS Article 8.16.6.1)

= tension flange thichess (not a code variable)

V, = factored shear force at section @DS m d e 8.16.6.1)

= moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement

Yt = distance from centroidal axis of gross section, neglecting reinforcement, to extreme h r in tension (BE Article 8.13.3)

= span length (not a code variable)

= additional embedment length at support or at p o d of inflection, inches (BDS Article 8.24.23) z = quantity limiting distribution of

f l e d reinforcement (BDS Article 8.16.8.4) = dear span length (not a code variable)

= development length, inches a+ = angle between shear-friction reinforcement and shear plane (BDS Artides 8.15.5.4 and 8.16.6.4)

= mcking moment (BDS Artide 6.13.3)

= n o d moment strength of a section = ratio of area of reinforcement cut off

Section 2 - Reinforced Concrefe Page 2-2 8


to total area of reinforcement at the (beta) section (BDS Article 8.24.1.4.2)

p, = ratio of depth of equivalent compression zone to depth from fiber of maxjmum compressive strain to the neutral axis (BDS Astide 8.16.2.7)

h = correction factor related to unit weight for concrete (BDS Articles 8.15.5.4 and 8.16.6.4)

p (mu)= coefficient of friction (BDS Article 8.15.5.4.3)

p (rho)= tension reinforcement ratio = A,/bd

P' = compression reinforcement ratio =

A',/ bd

p, = rehforcement ratio producing balanced s& conditions (BDS Article 8.16.3.1 . I )

9 (phi) = strength reduction factor (BDS Article 8.16.1.2)

Abbreviations

BDS = Bridge Design Spm)5ations

Secrion 2 - Reinforced Concrete Page 2-3

c* - Bridge Design Practim - Februav 1994

2.0.0 Introduction

The purpose of this section of the Bridge Design Pradice Manual is to assist design engineers with reinforced conaete design.

There are two parts to tkis chapkr:

Part A - Design Example

This section contains an example design solution for a reinforced connete box gvder superstructwe using Strength Design methods. The computer programs "Bridge Design Systemw and "Bent" were used to do the required structurd analysis.

If should be noted that the example does not constitute a complete bridge design. Only enough work has been done to demonstrate design methods. For example, tension steel has not been designed for every span of the shucture as would be done for an actual bridge design. Additionally, there are other design considerations not considered in the example. For instance, seismic design has not been addressed. It is hoped, however, that the example will provide a good foundation for the design of reinforced conaete bridge structures.

Also, note that the example does not completely meet current CALTRANS design standards. For example, current CA LTRANS standards require continuous small diameter f ension bars h box grrder bridgs in addition to large diameter bars. However, for simplicity, the small diameter bars were not utilized in tlus example.

It i.s also important to note that h e example design is only an example. It is the work of only one enpeer. 7% methods used should not be ziewed as Caltrans standards! There are often several different ways to solve a design problem.

Part B - Design Notes This section contains helpful fomulas, interprets tions of the specifics tions, derivations and exampies. It does not cover all sections of the spe&cations and it is not intended to be a commentary on h e specifications. It does, however, offer guidance on how to handle frequently encountered bridge design problems.

A j n d word ojcautim &appropriate at this point! The i n f o m tion contmed in h s section should not be used as a design p d e in place of reading the spedications. There may be certain instances where the methods desaibed in this section are not appropriate for use. Therefore, it is recommended that prior to applymg arty formula or procedure contained within this section, the designer should read the appropriate s e c h n of the spedfimtions to be certain that the described formula or procedure is appropriate for use.

Arty errors found in khis chapter, either technical or typographical in nature, should be reported to the chairman of the CaItrm Reinforced Concrete Committee.

Page 2-4 Section 2 - Reinforced Concrere

r* - Bridge Design Practice - February 1994

Part A - Design Example

2.1.0 Structure Requirements

Design a reinforced concrete box girder structure with the span configuration shown below. Provide for 44 feet of clearance between Type 25 barrier rails. Assume the use of two Type 2R columns with base h e t e r s of 4 feet. Assume f', = 3.25 ksi and fy = 60 hi.

2.2.0 Typical Section Geometry

- -

Deck width = (curb to curb clearance) + (two Type 25 barrier rails)

= 40' + Z(1.75') = 43.5'

AbUl 1

From the Bridge Design Aids manual, for a reinforced conuete box girder with continuous spans, an economical design will result when

S h c t u r e Depth = (0.055)(110') = 6.05'

Use Depth = 6'

Bent 2 Bern 3

20'

Slope the exte.rior prder web for aesthetic reasons.

AbOr 4 2 w

I ' Horizontal Exterior web slope =

2' Vertical

Assume exterior web width = 10"

Assume interior web width = 8"

The exterior webs are wider to allow for easier conae te placement which is difficult due to the webs slope.

Secrion 2 - Reinforced Concrere Page 2-5

c* 1- Bridge Design Practice - February I994 =

Fromthe Memo to Desipms manual, Memo 15 -2, the spacing between girder webs for a reinforced concrete box girder should be appro& tely 1% times the smctures deplh.

girder spacing = (1 .5)(6') = 9'

deck width - 43.5' - 4.83 --- girderspacing 9'

Assume 4 bays 8 9' and two deck overhangs.

43.5'4(9') -1 1.2/ 12 Overhang length = 'I = 3.283' =. 3'-3 %$

Use overhangs = 3 '- 3"

Assume overhangs to be 7 inches deep at outside edge of deck and 12 inches at the intersectjon of the overhangs with the exterior girder web.

Effective dear span between girder webs (interior bay) = S = (9 ') - (8") = 6 '- 4"

From the Bridge Dtzip Details manual, Page 8-30, when S = 8 '- 6",

deckslabdepth = 8%"

soffit slab depth = 6% "

Soffit slab width = deck width - 2Iaverhang) - 2(grrder slope)(girder depth )

= 43.5' - Z(3.25') - 2('h)(6' - 1') = 32'

Assume the use of two Type 2R c o I ~ at each bent.

See the Bridge Design Detaik manual, starting with Chapter 7, page 31, for standard architectural columns used by Calbans.

Page 2-6 Section 2 - Reinfored Concrete

c* - 8ridge Design Practice - February 1994 M

4 pa-91 Structure symmetrical

4 ' 4 " - 2-0'

SECTION A-A

@ Bent < SECTION 6-B

Section 2 - Reinforced Concrete Page 2- 7

c-t: - Bridge Design Practice - Februav 1994

2.3.0 Superstructure Loads

CaItrans currently uses a program titled "Bridge Design System" to perform shctural analysis of standard conaete box p d e r sbuchres. The user is required to input the number of ljve load Ianes wluchare tobe loaded o n the stnlcture. The program will analyze for different truckpositions along each span and also for lane loadings as desaibed in Chapter 3 of the Bridge Speci jht io t ts .

A simple way of obtaining factored results is to input more lanes into the program than actually exists. Factored results will then be output.

Design Loads to Consider (See 8DS table 3.22.1 A)

Load 1 = Service Load = 1 .O [D+ (L+I) H]

Load 2 = Group IA - - 1.3 [D+f .67 (L+l) HI

Load 3 = Group I, - - 1.3 ID+ (L+I) PI

2.3.1 Dead Loads, D

Superstruchre (box) weight = 0.15 kcf

Future AC wearing surface = (43.5'-3.5')(0.035 ksf) = 1.4 klf

Type 25 barrier rail = (2rails>T261cf/fi)(O.I 5 kcf) = 0.783 klf

Future AC plus barrier rails = 2183 kIf

2.3.2 Live Loads, L

BDS Art. 3.23.2.2 says:

The live load bendmg moment for each interior stnnger shallbe determined by applying to the stringer the fraction of a wheel load (both hont and rear wheels) d e t a e d in Table 3.23.1.

BDS Table 3.23.1 says:

Concretebox girders are designed as whole width units. The number of wheel lines applied to a box girder structure is:

overall deck width, feet 7.0

Number of design live load lanes

= (g wheel h e s ) ( 1 five load lane = 3.1 07 live load lanes 7 2 wheel lines

Page 2-8 . Section 2 - Reinforced Concrere

c-t: - Bridge Design Practice - February 1994

2.4.0 Effective Depth

Input Data To BDS Analysis Program

2.4.1 Minimum bar cover (BDS Art, 8.22.1)

top deck steel = 2"

bottom slab steel = 1.5"

Note: The above cover requirements assume n o d enviromental conditions.

2.4.2 Transverse Bars

Load 3 Group I,

P -Truck Factored 1.3(0.15) = 0.195

1.3(2.183) = 2.838

0

1.3(3.107) = 4.039

Superstructure DL. k d AC and Barrier DL. klf #+IS20 live bad lanes

#P -truck live load bnes

From C a l m Bridge Desigrr Details manual, Page 8 -30, dated June 1 986, and an effec live span lengh (dear span between girder webs) of 8' - 4" find:

top transverse bars = #6

Load 1

HS20 Service

0.15

2.1 83

3.1 07

0

bottom transverse bars = #4

Asswne main long~tudinal bars to be #I 1's. This is probably conservative.

Load 2

Group lH HS20 Factored

1.3(0.15) = 0.196

1.3(2.183) = 2.838

1.3(1.67)(3.107) = 6.745

0

Note: Bridge D&gn Details manual, page 8-30 has been updated. Fuhue designs should be based on the current standard.

Section 2 - Rein farced Concrete , Page 2-9

-1 Bridge Design Practice - February 1994

2.5.0 Factored Design Shears (D + L + I ) in kips

Location

See pagw 2-77 and 2-76

Page 2- 10 Section 2 - Bein forced Concrete

Span 2 S ~ a n 3

c* - Bridge Design Practice - February f 994

2.6.0 Girder Web Flares

a s s w e V, = ~ & b , , d

m h ~ m usable V,= 8&b,d

(BDS kt. 8.16.6.21)

@DS Art. 8.16.6.3.9)

maximum allowable Vu = 100 fib,d

bw. 2 v" is required. 104fid

Q = 0.85

f', = 3250 psi

d = 68.54 "

b, = 2(10") -+ 3(8") = 44" (initial assumption)

maximum allowable V, = lO(0.85)-(44)(68.54) - = 1461 k (I%,,)

M m u m design shears, V,, may be assumed to be the shears which occur at a distance d kom the face of abument and bent cap supports. ( B E Art 8.16.6.1.2)

At the abutments, h s point occurs at;

68 54" 1.25' + = 6.96' from abutment center lines. 12

At bents, thk, point occurs ak;

225' + 68 = 7-96' horn bent center lines. 12

Secfion 2 - Reinforced Concrele Page 2-1 1


Referring to Lhe table of factored design shears, it is seen that the maxium design shears for spans 1 and 3 will never exceed the m.aximum aUowable design shear of V, = 1461 k, For Span 2, design shears must be calcnla tcd at 7.96 feet from the bent center lines to determi.ne if web flares will be required.

7 96 Bent 2: V, = 1701 - ( 1 7 0 1 - 1382) = 1470 k > 1461 k I1 '

7 96 k t 3: V, = -2717 + ( 1 7 1 7 - 1398) = -1486 k > 1461 k 11

Web flares are required at both ends of Span 2.

Determine total web width required at the face of the bent caps. By observation it is recognized that it will be appropriate to calculate flare requirements at Benr 3 md apply the requirements to both bents.

225 k t 3 cap face: V, = -1717 + ( 1 7 1 7 - 1398) = - 1652 k 11

require that b vu = 1652(1000Ib/k)

" - lo$fld 10(0.89d%%(66.54)

require b, 1 49.7"

Use b, = 10" for each web at the cap face.

Total b, = 50" for the whole box girder.

Page 2- 12 Section 2 - Reinforced Concrefe

Ed - Bridge Design Practice - February 1994

Detamine the required length of flare. The webs must begin to flare at the point in the span where:

V, = maximum allowable V, = 1441 k

Let x = minimum distance from support center line to start of web flare.

x =- 8.83'

Required flare length = 8.83' - 2.25' = 6.58'

Minimum required flare length = 12ldtfference in web thichess) (EDS k t . 8.11.3)

= 12(10"-g"' = 2'

Use flare length = 7'

5 Bent Cap

Plan

Bent 2 - Typical Interior Girder (Bent 3 is similar)

8'

225'

Section 2 - Reinforced Concrefe Page 2- 13

1

I

7'

flare


2.7.0 Factored Design Moments (D .t L + 1) in k-ft

Location Positive Negative

Page 2-14 Section 2

0.6

0.7

0.8

0.9

1.0

Reinforced Concrete

14529 P

13814 P

11069 P

6831 P 0

See pages 2-70 and 2-75

ca - Bridge Design Practice - February 1994 - 2.8.0 Maximum Design Moments

Moments at faces of support may be used for negative moment design. (BDS Art. 8.8.2)

Bent 2

Span 1 side of cap: M.=-22476+?(22476-14690) =-20252k-H 7

L Span 2 side of cap: M, = - 25319 + - (25319 - 9835) = - 22504 k-ft

11

2 Span 2 side of cap: Mu = - 26386 + - (26386 - 10936) = - 23577 k-A

11

Span 3 side of cap: Mu = - 24568 c 2 (24568 - 14598) = -22076 k-A 8

2.9.0 Steel Requirements at Maximum Moment Sections

Lwalion

Span 1 0.4

Span 2 0.5

Span 3 0.6

8ent 2

Benl 3

2.9.1 Positive Moment Sectlon Parameters b = 43.5' = 522"

b, = 10+.8+8+8+10 = 44"

h, = 6.325"

d = 69.3"

= 0.9

Gesign Moment

1 0777 k-fi

19126

1 4529

- 22504

- 2357?

Secfion 2 - Reinforced Concrete Page 2- 15


2.92Span f 0.4point, Mu= 10777k-ft (Solution Method f )

assume ash, (i-e., section is rectangular in nature)

require $ M , I M ,

set

1.123%' - 3 7 m + (1W77)(12) = 0

This is a quadratic equation which cart be solved for 4. A, = 34.93 in2

check that a 5 hf

(BDS Art. 8.16.3.2.1)

okay

Required A, = 34.93 in2

2.9.3 Span2 0-Spoint, Mu = 19126k-fi (Solution Method 2)

assume a l h,

The above equation can be solved algebraically to yield a direct solukion for 4:

Uni ts for aU variables must be consistent.

Page 2- 76 Section 2 - Reinfotced Concrete


check that a 5 h,

Required 4 = 6250 in2

2.9.4Span 3 O.Gpoint, Mu = 14529k-ft (Solution Method 3)

assume a 5 h,

As fy = Mu where a = -

.St,%

Restate the above equations in a different form:

For the above equations, a = inches, = k-ft

Star! with an assumed value of a. Then iterate between equations.

a = 4" (initialassumption) A, = U.97 in2

a = 1.996" Pq = 47.27 in2

a = 1.967 A, = 47.26in2

a = 1.967" A, = 47-26 in2

a = 1.967" < 8.125" okay

Required fc, = 47.26 in2

Section 2 - Reinforced Concrete Page 2- 1 7


2.9.5 Negative Moment Section Parameters b = 32'= 3&4"

b, = 44" (web flares have been neglected)

h, = 6.375"

d = 68.54"

$ = 0.9

2-9.6 Bent 2, Mu = -22504 k-ft

Assume a <hf

Require $M,5M,

Use one of previous methods to solve for A,.

A, = 75.30 in2 a = 4.26" < 6.375"

Required 4 = 75.30 id

2.9.7 Bent 3, Mu = -23577 k- ft

Solve the same as for Bent 2.

Required 4 = 79.02 in2 a = 4.47 " < hf

Page 2- 18 Section 2 - Reinforced Concrete


2.1 0.0 Maximum Allowed Tension Steel (B DS Art. 8.1 6.3.1 .I)

p = !k and p, = balanced reinforcement ratio. bd

Maximum allowed p = 0.75 p,

From the above equations &d BDS Art. 6.16.3.3, it can be found that for a flanged section with the neutral axis below the flange,

maximum allowed A, - 87000 + fy

2.10.1 Maximum Tension Steel in the So flit Slab

f', = 3250 psi f, = 60,000 psi p, = 0.85 b = 522" b, = 44"

h, = 8.125"

d = 69.3" maximum &owed A, = 187.08 in2

2.10.2 Maximum Tension S tee1 in the Deck Slab

f', = 3250 psi f, = 60,000 psi j3, = 0.85 b = 384"

b, = 43" (Note: Do not include web Oares here) $ = 6.375" d = 68.54" Maximum allowed 4 = 127.23 in2

Seclion 2 - Reinlorced Concrete Page 2- 19


2.1 1.0 Effective Tension Flange Width (BDS Art. 8.1 72.1 )

2.1 I. I Span 2 - Positive Moment Tension Flange Width (soffit slab)

Exterior Girder:

First Interior Girder:

Second Interior Girder:

Total effective tension flange width.

= 45" + 81.25" + 84.5" + 81.25" + 45" = 337

Note: CaIdations for spans 1 and 3 are similar.

Page 2-20 Section 2 - Reinforced Concrete

EM - Bridge Design Practice - February 1994 H

2.11-2 Bent 3 - Negative Moment Tension Flange Width (deck slab)

Exterior Girder:

Fixst hterior Girder:

61 = 6 (8.125') = 48.75' overhang = 39'

L d, = 1/2 (997 = 49.5'

Second hterior Girder:

Total effective tension flange width

= 97.75" + 107.5" + 207.5" + 107.5" + 97.75" = 518"

'40 L = ( l O =132-

39" + 10' + 48.75" = 97.75'

Note: According to BDS Art. 8.17.21 -3, an effective tension flange width shall be calculated on each side of the bent cap. The larger of the two effective widths shall be used. Upon mspection it can be seen that the 51 8 inch width calculated above will control.

97.75"

Section 2 - Reinforced Concrete Page 2-27

Bridge Design Practice - February 1994 m

All main tension steel bars shall be distributed within the effective tension flange areas. -

L-bn

1

Span 1 0.4

Span 2 0.5

Span 3 0.6

Bent 2

Bent 3

Page2-22 - Section 2 - Reinforced Concrete

Design I Moment, M,

10777 k-fi

19126

14529

- 22504

- 235n

4 Requirement

34.93 inZ

62.50

47.26

75.30

79.02

Effective Tension Flange

336.5'

337

337

51 0

I

51 8

c* - 8ridge Design Practice - February 1994 - 2.12.0 Positive Moment Bar Size Limitation (BDS Art 8.24.2.3)

Requuement at simple supports and points of inflection:

2.12.1 Span 2 Inflection P o h ts

Mection points occur at 0.15 and 0.85 points of Span 2. (See moment envelope, page 2-33)

at 0.15 point, V, = 1362 - '/,(I382 - 1067) = 1225 k

at 0.65 point, V, = - 1398 + '/, (1398 - 1082) = - 1240 k

at 0.5 point, Mu = 19126 k-ft 5 $M,

At Ieast % of the steel present at the 0.5 point of Span 2 must be extended into the bent caps. (BDS Art. 8.24.21)

Therefore, it is safe to assume that the moment capacity at the inflection points is at least:

M, = 'A (21251) = 5323 k-ft.

t , = greater of d or 12 d, at points of inflection

e, = [69.3" or 12 (1.41 ) = 16.9"] for #ll bars = 69.3"

td = 66" for #I1 bars

Any bar size #I 1 or smaller may be used for positive moment steel in Span 2.

2.122 Span 3 Inflection Point

The inflection point occurs at 0.2 point of Span 3. (see page 2-36)

at 0.2 point, V, -1114 k

at 0.6 point, Mu = 14529 k-ft 5 @M,,

'I 4529 M, Z - = 16143 k-ft 0.9

At least 'A of the steel present at Lhe 0.6 point of Span 3 must be exkended to the bent cap.

(BDS Art. 8.24.2 1 )



Assume at 0.2 point, M,, = '1, (16143) = 4036 k-ft.

E, = 69.3"

t , = 66"

2.12-3 Span 3 Abutment

at the abutment, V, = - 1036 k a t 0.6 point, Mu = 14529 k-ft S $M,

14529 M,= - = 16143 k-ft 0.9

At Ieast I/, of !he steel present at the 0.6 point of Span 3 must be extended to the abuhnent.

(BDS Art. 8.24.2.1)

Assme at the abutment, M, = 1/3 (16143) = 5381 k-ft

fa = embedment Iengh beyond support center h e .

fx = (1'-3")- 3" = 12"

f, = 66"

Any bar size #11 or smaller may be used for positive moment steel in Span 3.

2.1 3.0 Crack Control (Pre-Design) (BDS Art. 8.1 6.8.4)

The following procedure can be used to find out how many tension bars should be used to satisfy uack control requirements. If an existing design is to be checked for ma& control, do not use this procedure. Also, please note that this pracedrtre is only valid if all of the tension bars are the same size (see pages 2-121 ttuu 2-129).

d, = distance from extreme conmte tension fiber to center of the closest tension bar.

P, = area of tension steel required to meet sbength design requirements.

A, = area of one tension bar.

A, = effective area of conaete in tension which surrounds the tension steel and has the same centroid as the tension steel.

Page 2-24 , Section 2 - Reinforced Concrete

c* - Bridge Design Practice - February 1 994

z = aadz control factor (see specifications).

f, = working stress in tension steel at service loads.

Q= number of bars required to satis9 sb-ength design,

I

= number of bars required to satisfy crack conbol allowable stress formula, fs = z / (dfi)J.

n,= number of bars required to create stresses in the tension steel of 24 ksi.

n,= number of bars required to create sbesses in the tension steel of 36 hi.

n = minimum number of bars required

f, = 60 ksi is assumed.

1. C a l d te required A, for the factored moment, Mu. 2 Calculate f, assuming A, = m o u n t of tension steel present. Use working sbess analysis and

service load moments, D+(L+I)H.

3. C a l d t e n, = LL *b

4. I f f s 1 2 4 k s i , u s e n = ~ -

5. Calculate 4, A,, and T=A&

A, = (b,) x lesser of I::' where = Wcbess of tension flange

b, = effective tension Range width

This definition of A, is only good if all tension bars are in a shgle layer.

7. If n, > nz, u s e n = l a r g e r o i n , , o r ~

I f nl, n, > n3& u w n = larger of n, or Q

If n, < n 3 ~ use n = larger of n,, or n,


Ed I Bridge Design Practice - February 1994 m

-

Service Load Moments (D + L + I) H in k-ft

See page 2-83

L m l i o n

Span 2 0.0

col. face'

0.1

0.5

0.9

col. face'

1 .O

Span 3 0.0

#[.face*

0.1

0.6

*Calculate moments at coEmu-, support faces.

Bent 2: M = - 14408 + % I (14408 - 6132) = - 12903 k-ft

Eknt3: M=-15547+2/i~(15547-7069)=-14006

M =- 14670 + 2/3 (14670 - 8782) = -13198

Calculate f, at each sectionusing required A, from strength design calculations (see page 2-1 19).

Positive

10669

8359

Negative

- 14408

- 12903

- 61 32

- 7069

- 14006

- 15547

- 14670

- 131 98

- 8782


Bent 2 Rt

384

50

6.375

9

75.30

68.54

-1 2903

31.58

0.6 Span 3

522

44

8.125

9

47.26

69.3

8359

32.09

b

k,

h n

4 d

M

f,

0.5 Span 2

522"

44 '

8.125'

9

62.5 in2

69.3 in2

1 M 9 k-ft

31.08 ksi

Bent 3 Lt

384

50

6.375

9

79.02

68.54

-1 40E

32.69

Bent 3 Rt

384

44

6.375

9

79-02

68.54

-1 31 98

30.78

EN - Bridge Design Practice - February 1994

2.13.1 Span 2 0.5 point - #9 bars only

A, = 625 in2 6 = 31.08 ksi T= 4;6 = (625)(31.08) = 1943 k

b, = effective tension flange width = 337'

d, = 1.5 + 0.5 + 1.128/2= 257"

2d, = 5.14''

h, = 6.375"

Pq. = (bJ(Jesser o12d, and h, ) = (337)(5.14) = 1732 in2

z = 170 for normal enviromental conditions

n, -= n36 Therefore, n =larger of n, or n, = 63

Section 2 - Reinforced Concrefe Page 2-27

Em - 8ridge Design Practice - February 1994

Bar

A,

4 2 4

Ae

"d

nm

b 4

"36

R

Bar

k

dc

2 4

A,

rb

b

h

n3s

n


Span 3 0.6 point

4 = 47.26 f, = 32.09

T = 1517 z = 170

h, = 6.375 b, = 337

Span 2 0.5 point

A, = 62.5 f, = 31.08

T =I943 Z =I70

h, = 6.375 b, =337

#9

1.0

2.57

5.14

1 732

48

43

64

43

48

#9

1 .O

2.57

5.14

1732

63

51

81

54

63

Bent 2 Right

A, = 75.30 f, = 31.58

T = 2378 X = 170

h, =8.125 b,= 518

#9

1.0

3.31

6.62

3429

76

75

99

66

76

# 10

1.27

2.M

5.28

1779

38

36

50

34

38

#I 0

I .27

2.64

5.28

1779

50

43

&I

43

50

# I 1

1.66

2.70

5.40

1820

31

31

41

27

31

#11

1.56

2.70

5.40

1820

40

38

52

35

40

#I 0

1.27

3.39

6.78

351 2

60

64

78

52

64

# t l

1.56

3.46

6.92

35 85

49

55

&4

43

55

c* - Bridge Design Practice - Februav 1994

2.14.0 Bat Spacing Limits

2.14.f Minimum barspacing (BDS Art. 8.21.1)

Mirumurn bar spadng = 212" assuming use of #ll bars.

2.14.2 Maximum bar spacing (BDS Art. 8.21.6)

11 -5 (slab thihess) = 1.5 (8.125") = 122" far top slab

lesser Of 118"

= 1.5 (6.375") = 9.56" for bottom slab

Maximum spacing = 122" for the top slab - 9.56" for the bottom slab

2.14.3 Minimum Number of Bars Required

I

10 0 1 0 C I Non-&&e lension

flange area

D 0 0 0

44" -2" -2" Top slab, 44 inch section = 3.3 spaces required betwen bars shown.

12,2"

5 bars required



109"- 2"- 2" Tou sbb, 7 09" section = 8.6 spaces

10 bars required

Bottom slab, 84" section 84"- 8'- 2" = 7.7 spaces

9.56" 10 bars required

Bottom slab, 42.25" section 42.25"- 2"- 2"

= 4 spaces 9.56"

5 bars required

Note: The 2 inch, 4 inch, and 8 inch dimensions on the above figure are only approximations.

2.1 5.0 Minimum Reinforcement Requirements (BDS Art, 8.17.1)

A minimum design sbength is required at any section where tension reidorcement is required,

@DS Art. 8.1 7.1.1)

From the BDS frame analysis output:

- I, = 363.38 ft ' is = 1- -

y, = 3.49' for positive moments.

M o w the design moment envelope as follows;

For positive moments

minimum Mu = 9 3 z O 363.38h4 144in /fi = 7693 k-ft '-1 349k )[ IWO;,:) For negat5ve moments

The minimum design moment requirements above may be waived if the steel provided at a section is one third greater than that required (BDS Art. 8.17.1.2) due to the applied factored moment, Mu

For example, if M, = 90 k-ft (factored D + L + I), then it is acceptable to design for an adjusted Mu value of Mu = 90 + '/S (90) = 120 k-ft

Page 2-30 Section 2 - Reinforced Concre 1 e I


2.16.0 Bar Layout, Span 2 - Positive Moment Try#IObars n=50

Effective tension flange = 337"

Extend at least 'A (54) = 13.5 bars into bent caps. (BDS Art. 8-24.2.1)

2.16.1 Choose Bar Groups Bars have been tentatively layed out as shown in the above diagram. It is assumed that the A bars will extend into the bent caps. The A bars within the girder webs will be continuous.

Bar Type No. Groups A, 4 M,

Draw the factored design moment envelope. Modify the envelope to meet minimum 4 requirements of BDS Art. 8.17.1. Draw lines represen- 4M, for each bar g~oup.

Mark off bar extensions in accordance with BDS Art. 8.24 1 2 7 .

Check Lhat all bars extend past the moment envetope at least a &stance equal to development length, P,, in accordance with BDS Art. 8.24.122.


=* - Bridge Design Practice - February 1994

Measure, in feet, the distance from the span center h e to the ends of each bar group.

Match bar ends to reduce the number of different bar lengths required in the field. Keep in mind that 60 feet is the longest practical bar length available. Anything longer will require splidng. Try to keep splicing to a minimum.

S + 5 =10 4bars 5 + 5 = 10' F 20 + 37 = 57 S bars 21 + 37 = 58' C 27 + 31 = 58 + Use 8 bars 27 + 31 = 58' E 31 + 26 = 57 8bars 31 + 27 = 58' D 38 + 20 = 58 8 bars 38 + 20 = 58' B continuous IS bars continuous A

Provide redorcement in noneffective tension flange areas. (BDS &t. 8.17.2.1 .I)

required area = (0.4%)(6.375")(23.5") = 0.60 in2

Place one #7 bar at khe center of the noneffective areas.



Section 2 - Reinforced Cancrele Page 2-33


2.17.0 Bar Layout, Span 3-PasitiveMoment

Try#lObars n=38

Effective tension flange = 337'

84 - (38) = 9.5 + hy 10 bars in 84" flange seckions. 337

42.25 ( 3 8 ) = 4.8 + t ry 5 bars in 4225" flange sections. 337

Total number of bars = 2(10 + 5 + 5) = 40

Extend at least ]A (40) = 10 bars into bent cap.

Extend at least '/5 (40) = 13.3 bars into abutment

(BDS Art. 8.24.21)

(BDS Art. 8.24.2.1)

2.17.1 Choose bar groups

BarType No. Groups 4 @Mn

A 12 12 15-24 in2 4735 k-ft

8 8 20 25.40 7867

c 8 28 35.56 10980

D 8 36 45.72 14074

E 4 40 50.80 15614



Perform graphical procedures as was done for Span 2 Match bar lengths.

3 + 2 0 =23' 4bars 3 + 20 = 23'

13 + 41 = 54' 8bars 15 + 41 = 56'

19 + 37 = 56' + Use 8bars 19 + 37 = 56'

26 + 30 = 56' Bbars 26 + 30 = 56'

con t:inuous I2 bars continuous

Place one 47 bar at the center of the noneffective area.

Section 2 - Reiniorced Concrete Page 2-35 I


Seelion 2 - Reinforced Concrete Page 2-36 ,


2.18.0 Bar Layout, Bent 3 - Negative Moment

Try #TO bars n = 68

Effective tension flange = 518"

The noneffective tension flange regions a.re very small. Consider the full 522 inch flange width for distribution of the tension steel.

44 - (68) = 5.7 + try 6 bars in 44" overhangs. 522

109 - (68) = 14.2 4 try 14 bars in each bay. 522

Total number of bars = 2 (6 + 14 + 14) = 68

Extend at least 'h (68) = 227 bars beyond inflection point. (BDS Art. 8.24.3.3)

2.18.1 Choose bar groups

&Type No. Groups As 4%

A 72 12 1524 in2 4675 k-ft

Perform graphical procedures similar to those done for Spans I and 2. Match bar lengths.

35 + 10 = 45 18 bars 35 + 10 = 45'

cont.huous 12 bars continuous


Span

2

0.7

0.8

Enve

lope

mod

llcal

lon

As

requ

lrem

enls

I

a! B

DS

Art

. 8.1

7.1

BD

S A

rt. 8

.24.

1 -2

.1

d =

68.6

1" =

5.7

' 15

db =

15

(1.2

7")

= 1.

6'

V20 hir =

VZo

(1 1

0' -

4.5'

) =

5.3'

U

se 6

' bar

ext

ensl

ons

BDS A

rt. 8

.24.

3.3

d =

5.73

' 12

db

= 12

(1.2

7")

= 1

.27'

Lc

lr =

Ys8

(1 10

' - 4.5

') =

6.0'

Use

7' b

ar e

xten

slon

s pas

t the

po

ints

ol I

nfle

ctio

n lo

r on

e thl

td

of b

ars

BD

S A

rt.

8.24

.1.2

.2

Id =

4.5'

us

e 5'

mln

lrnur

n

Spa

n 3

I 0.

1 0.2

0.3

0.4

0.5

I I

I I

/

I 12

Bar

s I

Con

t. D

M,

= 4

675

k-h

#

/

I /

1

# 1

0 B

ars

Vel

tlcal

Sca

le

, 50

00 k

-(t

Hor

lmnl

al S

cale

I

10 tl

I

Neg

ativ

e M

om

ent E

nvel

ope

- Ben

t 3


2.19.0 Fatigue Check (BDS Art. 8.1 6.8.3)

Requirement: 23.4 - 0.33fmi, 2 f,, - f,,,,, equivalent expressions

f,, - 0.67fh, 5 23.4 ksi

f,, = maximum stress in reinforcement from (I3 + L + r) HS service loads in ksi (dculatk using working stress analysis)

I,,, = minimum stress in reinfo~ement from (D + L e I) HS service loads in ksi (calculate using working stress analysis)

Sign convention: tensile stresses are positive;

compressive stresses are negative-

See page 2-83


(M,, - 0.67 Mmin) I N

11 9

120

139

- +- check I00

1 06

99

94

11 3 +- check - f- check

130

151 + check

141

146

11 9

127

Locat ion

Span 2

0.5

0.6

0.7

0.8

0.9

col. face

Span 3

ca!. face

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

M,

6323 k-ft

5 4 9

2953

- 1017

- 7069

- 14005

- 13198

- 8782

- 4333 - 1075

1407

31 13

4043

4196

35 74

2175

M,

1 W 9 k-fi

9656

6690

1 848

- 4605

- 101 90

- 9697

- 6120

- 1401

2620

5626

7527

8359

8080

6672

3994

NMI

54

50

34

18

18

18

12

12

12

20

36

36

40

36

36

20

NI,

12

12

12

30

40

68

68

50

30

30

30

12

12

12

12

12

c* - Bridge Design Practice - Februaw 1994 m

= Dead plus positive live load moment envelope,

M,% = Dead plus negative live load moment envelope.

M,, = Moment which causes maximum stresses in the tension steel.

Mmi, = Moment whichcausesminimum stressesin thetensionsteel.

N,, = Number of fully developed bars in the bottom slab.

N,, = Number of fully developed bars in the top sbb.

M,, - 0.67 M,, = internal member moment whch will result in a steel stress of La- 0.67 fmi, This is only h e when M,, and hi, have the same sign (ie. no moment reversal).

Do a fatigue check at the member Ioca tion yielding the largest value of (M, - 0.67 &--,I/ N. Do this check separately for positive moment locations and negative moment locations. Also do a fa t i p e check at loations where moment reversal W s place.

*M = appljed moment at locations where moment reversal occurs.

*M = M, - 0.67 Kin at locations where moment reversal does not occur.

Working Stress Anatysis


Nbo,

N ~ w

n

b

b w

h d d'

A, A', M '

f , top bars

f s

bot bars

Span 2 0.8

Span 3 0.2

12

30

9

384

44

6.375

68.61

2.64

38.10

15-24

- 3395

16.25

- 4.32

18

30

9

522'

44'

8.125'

69.36'

3.39'

22.86 in2 38.10 i#

1848 k-fl

-1.42 ksi

14.47 ksi

18

30

9

384

44

6.375

68.67

2.64

38.10

22.86

-101 7

4.86

-1.24

Span 3 0.5

36

12

9

522

44

8.725

69.36

3.39

45.72

15.24 5441

- 4.39

21.55

Span 3 0.3

20

30

9

522 44

8.125

69.36

3.39

25.40

38.10

2620

- 2-02

18.50

20

30

9

384 44

6.375

68.61

2.W 38.10

25.40

-1075

5.14

- 1.29


+an 2: 0.8 pt bottom steel 14.47 - 0.67 (- 1.24) = 15.3 < 23.4 ksi okay

0.3 pt top steel 4.66 - 0.67 (- 1.42) = 5.6 < 23.4

Span 3: 0.2 pt top steel 16.25 < 23.4

0.3 pt bottom steel 1850 - 0.67( - 1.29) = 19.4 < 23.4

0.3 pt top steel 5.14 - 0.67 ( - 202) = 6.5 < 23.4

0.5 pt bottom steel 21 -55 < 23.4

Fatigue requirements have been satisfied.

aka r okay

okay

okay

Section 2 - Reinforced Concrete Page 2 4 1


2.20.0 Final Bar Layouts

2.20.1 Span 2 - Bottom Slab Reinforcement

7 P Exterior Girder

jl " I ---- ---------------- --------------------- Cool

27 Ifr B Interior Girders 31 31

27 t

All bars are #10 except where noted. All bars shall be evenly spaced withln limits shown. Numbers at ends of bars represent distances fmm span center line.

Conl ---

* Extend al least 6" in10 bent caps.

- - - - - - - - - - - - - - - - . - - - - - - - - - - - - - - - - - - - - - -

Page 2-42 Section 2 - Reinforced Concrete 1

LC Bar layout syrnrnetriml about Q of structure.

c* - Bridge Design Practice - February I 994 =

2.20.2 Span 3 - Bottom Slab Reinforcement

Span 3 E Abut 1

15 Con?

---------------I--------'

15

26 30 19 37

3 20 Cant #7

C Interior 3 20 Girders 37

30

15

F̂B U e T r Girder

,.& n- - - - - - - - - - - - - -.I I

- Bar layout symmetrical aboul Q of structure.

- - - - - - - - - -

All bars are $1 0 except where noted. All bars shall be evenly spaced within limits shown. Numbers at ends of bars represent distances from span center line.

Extend at least 6' into bent cap.



+

2.20.3 Ben f 3 - Top Slab Reinforcement

Q Bent 3

5 Exlerior / Girder

1 onr r---------------'--------------------

Typiml 35 interior Bay 11

All bars are #10. G All bars shall be evenly spaced. Numbers at ends of bars represent distances from Ihe bent center line.

10 42

% Interior 11 / Girders 35

Page 244 Section 2 - Reinforced Concrete

42

I0


2.21.0 Longitudinal Web Reinforcement (BDS Art. 8.17.2.1.4) (B DS Att. 8.1 7.2.1.5)

The maximum amount of flexural reinforcement occurs at Bent 3, A, = 86.36 in2

2 10% of A, = 8,6411 = 1.73 in2/girder web.

5 guders

Minimum bar size to be used is a No. 4 bar (A, = 0.20 in2).

2 Number of No. 4 bars required = 1-73in = 9 ban = 5 bars/web face.

0.20jn2

Check maximum spacing requirement:

{;:b width = 8" + Maximum bar spadng = lesser of

Based on maximum spacing requirements, the number of 8 " spaces between bars is:

and therefore the minimum number of bars required = 8 bars along each girder web face.

Maximum spacing requirement controIs the design.

The top side face bar on each face of the guder web shall lx a No. 8 bar.

'-]gp~Lr 1 #4 tot 12 per girder girder

Section 2 - Reinforced Concrefe Page 2-45 I

ca - Bridge Design Practice - February 1994

2.22.0 Shear Reinforcement

See page 2-10

Location

Span 2 0.0

0.7

0.2 0.3

0.4

0.5

0.6

0.7

0.8

0.9

1 .O

Sections located less than d horn Lhe face of the bent caps may be designed (BDS Art. 8-16.6.1.2) for the factored shear, V,, which occurs at d from the face of the caps.

d = 68.61" = 5.71'

d from cap face = 225' + 5.71 ' = 7.96' horn support

The three interior girder webs are Oared from 8 inches to 10 inches over a 7 foot flare length.

Shear

1707 k 1 382

1067

749 470

- 243

- 488

- 765 - 1082 - 1398

- 1717

B Bent

Page 2-46 Seclion 2 - Reinforced Concrete

c* - 8ridge Design Practice - February 1 994

At the end of the 7 foot flare

b , = 1 O W + 6 + 8 + S+10=44"

At the cap face

b, = 5(10") = 50"

At d = 5.71 ' from the cap face

2.22.1 Stirrup Design Within the flares Ik was assumed when calculating the flare geometry that the stirrup steel would be utilized to the full extent allowed by BDS Art. 8.16.6.3.9

maximum V, = 8 P w d = BJ3250(45.1)(68.61) - = 1411 k (1 A)

Assuming #5 s m s a e used, A, = (5 girders)(2 'Wgirder)(0.31 tn2/leg) = 3.1 m2

when V? 4&bwd, S shall not exceed d or 12". (BDS Art. 8.16.6.3.8) 4

Use S = 9" within the flared sections.

Section 2 - Reinforced Concrete Page 247

c* - Bridge Design Practice - February 1994 - 2.22.2 Stirrup Spacing Limits (BDS Art. 8.1 6.6.3.8 and 8.19.3)

or 24" when V, 5 4$3,d maximum allowed S =

or 12" when 4@$,d < V, L 8T&,,d

Assume V, = 2 C b , d

6$Rb,d whenVs=4Eb,d 9v, = qw, + V,) =

10$Eb,d when V, = G b , d

b, = 44" at web sections where no flare is present.

$ = 0.85 for shear

(BDS Art. 8.16.6.2.1)

(HDS Art. 8.16.7.22)

10$&b,d = 1463 k

2 4 when $V, 1 878k maximum aliowed S =

12" when878k<$Vn11463k

2.22- 3 Shear Capacity for Different Stirrup Spacings

A, = 3.1 id

'1463 k is horn the limit, maximum $V,, = lo$ fib,d



2.22.4 Graphical Procedure (Steps are shown circled on the following graph) Step 1: Plot the V, design envelope to scale. Note that the maximum value of V, occurs at d

from the bent cap faces.

Step 2: For different S values plot $V, as a horizontal line. Only do this in the nonflaring web lengths.

Step 3: Plot $V, values which correspond to maximum S values.

Step 4: Choose reasonable stirrup s p a ~ g s . Graphically measuse distances along the span for each value of 5 chosen.

Step 5: Stipulate final design stirrup spacing.

Span 2

Seclion 2 . Reinforced Concrete Page 2-49


2.23.0 Bent 3 Model

2.24.0 Bent Loads

2.24.1 Dead Loads

From the analysis of the longttudinal model:

Dead load on h t = 1300.6 k (see page 2-80)

Assume this dead load is applied to the cap equally through each of the five girder webs.

dead load - 1300.6 k = 2M).I - girder 5

The 1300.6 k dead load did not take into account the existence of a solid cap section.

Extra cap dead load = (575")(54") (0.15 kcf) = 3.24 klf (see page 2-52)

Apply t h ~ ~ eextra dead load uniformly along the cap. Do not appiy it on the deck overhangs however.

2.24.2 Live Loads The following data is born the longitudinal model arialysis. All loads represent one unfactored mck/ lane.

See pages 2-72 and 2 - 7


Bent 3

(member #5)

HSZO truck P - t ~ k

Maximum Axial Load Case

P M b ~ Mht

119 k 57k-ft 0

31 0 142 0

Maximum Moment Case

P Mbp Mbol

64 294 0

200 76 1 0

r-t: - Bridge Design Practice - February 1994

Calhans currentjy utilizes a program namedr73W' to analyze bents. The program will appl y the above m c k loads dtrectly to the k t in the form of wheel lines. Jt moves the truck aaoss the cap to obtain maximum design forces for the cap and supporting coIumns. It will also put hu&s in more than one lane if necessary. 1 t should be noted that the program considers the bent to be fully supported against sidesway when computing forces due to live loads.

The bent loading shown below will result in a maximum negative moment in the lefk cantilever member. I t consists of a single P-buck, dead load due to the solid cap section, and dead load b-ansfmed to the cap though the guder stems.

The 4.9 foot distance shown above is from the edge of deck to h e approximate center of gravity of the exterior guder web.

The 3.75 foot distance = 1.75 feet barrier rail pl.us 2 feet from lane h e to wheel line-

2.25.0 Bent Cap Geometry

The bent coIumns have a very definite georneby and stif£rtess- The cap consists of a rectangular section with overhanging deck and soffit slabs. A reasonable assumption must be made to determine how much of the deck and soffit slabs can be induded as part of the cap.

U s e of BDS Art. 8.1 0.1.4 seems reasonable:

6 t = 6(6.375") = 38.27

' / io L = 'A0 (22') = 26.4" lho L = Ihd, 2 x 10.75') = 25.8" for cantilevers

A s m e cap web width = column diameter + 6" = 4.5' = 54" Total cap width = 25.8 + 54 -c 25.8 = 105.6" = 8.8'

Section 2 - Reinforced Conerere Page 2-5 1

EM - 6 ridge Design Practice - February 1994 -

2.26.0 Face of Bent Support (BDS Art. 8.8.2)

The face of the c o l m support sshall be considered to be at a section on the column face which is 1.5 times the structure depth below the deck surface.

Column

, Page 2-52 Section 2 - Reinforced Concrere


Equation of parabolic flare of column face: y = ax2

for x = 12, y = 2'

a = y/x2= 2/12?

y = (? / '44 )~2

for x = 9', ): = %.u (9)' = 1.125'

Face of support = 1.125 + 2 = 3.125' from the column centerhe

2.27.0 Factored Cap Design Moments (D + L + I) in k-ft

Location LeR 0.3 Canlilever 0.4 Span 1 0.5

0.6 0.7 0.8 0.9 1 .o

Middle 0.0 Span 2 0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Positive Negative 0

- I32 - 552

- 1172 - 1797 - 2427 - 3062 - 3945 - 3528 - 2082 - 1307 - 552

See page 2-91

All of the above moments are factored group IH or group IF loads.

2.28.0 Maximum Design Moments

Moments at face of column supports m a y be used for negative moment design. (BUS Art. 8.8.2)

3.125 Span I: support face is - = 02907 of span horn column centerhe

10.75

Section 2 - Reinforced Concrele Page 2-53

r* Bridge Design Practice - February 1994

3.125 Span 2 support face is - =

22 0.142 of span born column cen terhe

Mu = -2082 + 0.42(2082-1307) = -1757 k-ft

2.29.0 Bent Cap Minimum Reinforcement Requirements (BDS Art. 8.1 7.1 )

From the Bent analysis output (see page 2 8 7 ) :

&= 119.3 P

y, = 3.05' for positive moments

2.29.7 For Positive Moments

minimum Mu =

2.29.2 Foc Negafive Moments

m = 9 '''13 )(H)= 2988 k-ft 6 - 3.05 loo0

Noke that h e minimum design moments are larger than the factored moments produced by the h c k Loadings. Obviously, the columns are overdesigned such that the c;ap has a relatively low load applied to i t . A more practical design would either downsize the columns or change to a single column design.

2.30.0 Cap Effective Depth

structure depth 72 72 - clearance 1.5 2 - transverse bars 0.5 0.75 - longitudinal bars 1.27 1.27 - #I1 cap bars 1.4112 1.41/2

d,, = 68.02" d,, = 67.27'


r-t. - Bridge Design Practlce - February 1994 =

2.31.0 Cap Steel Requirements

For rectangular sections (a I h):

A S f Y a t - .65f',b

2.31.1 Positive Moment Sections

4 = 2890 k-ft b = 106" b, = 54" h, = 8.125" d = 68.02"

z =663.9h2

A, = 9.58 in2 a = 1.96"

maximum allowed 4 = 78.40 in2

2.31.2 Negative Moment Secthns

M, = -2988 k-ft b =106"

b, = 54" h, = 6375" d = 67.27'

z = 656.6 A, = 10.02in2 a = 2.05"

maximum allowed 4 = 74.55 in2

10'02 - 6.4 Number of # I 1 bars required = - - 1.56

Try using 7 #I 1 bars for both top and bottom steel in the bent cap.

& = (q(1.56) = 10.92 in2


c-t; - Bridge Design Practice - February 1994

2.32.0 Crack Control (BDS Art. 8.1 6.8.4)

Service Load Moments (I3 + L + I)H (see page 7-W)

Span 2 0.5pt M = 1204 k-ft

Span 1 col. face M = - 976 k-ft

I t should be sufficient to check only at the 1204 k-ft section.

Assume the use of 7 #11 bars.

Calculate the working stress in the steel:

b = 106"

bw = 54''

h, = 8.125"

n = 9

4 = 10.92 in2

d = 68.02" (for positive moment)

M =: 1204 k-ft

f = 20.46 ksi

Since f, = 20.46 < 24 ksi. serviceability is sarisfid for borb crack control and fatigue. (BDS h 8.14.1.6)

Therefore. use 7 # 1 1 bars for both top and bomm s w l



2.33.0 Construction Reinforcement (8 DS Art. 8.1 7-2.1.6)

Face of support C Column

Construction p i n t ---, ,

Seclion 2 - Reinforced Concrete Page 2-57

c* - 8ridge Design Practice - February 1994

Redorcement shall be placed appro* tely 3 inches below the construction job-it.

dead load negative moment of shaded portion of cap Design for M, = 1.3

and supershcture as shown in the above fig-wes 1 Dead load of cap and soffit slab:

Dead load of exterior girder web:

Dead load moment at the cantilevered face of support:

M = (2725')(11-Zk) + (swr)(znsft]'

C) = 49.1 k-ft

M, ;: 1.3(49.1)= 63.8 k-ft

Assume f', = 2500 psi at the time when the cap is required to resist construction loads.

b = 106" =: effective compression h g e width (BDS Art. 8.10.1.4) b, = 54" h, = 6375" d = 72- 8.125- 4 - 3 = 56.87' M, = 63.S k-ft required A, = 0.25 in2

The dimensions of the bent model are such that a very smallmoment was calculated for use in the design of the construction reinforcement. This has resulted in a very small steel requirement. It seems reasonable that some other method sh-ou,ld be considered for design of theconskuction steel. One possibility would be to use minimum reinforcement criteria of BDS Art. 8.77.1.

Design for minimum M, = 12 M,

For simplidty of the example calculations, the overhangs will be neglected. 1

1, = -(4.5')(4.99'))= 46.6 it' 12

Now find required 4 = 5.45 in2

Use 4 #ll bars (4= 6.24 in2)

Some designers will initially assume the use of 4 #I1 bars. They use # I1 bars because the main cap bars are #Z 1. They WLU then check their steel requirements by the procedure shown above. If 4 #I I barsareinsufficient, they willaddsteel. If 4#11 barsare toomuch, h e y will still use the4 #I1 bars.



2.34.0 Cap Side Face Reinforcement (BDS Art. 8.1 7.2.1.4)

Flexural 4 = 7(1.56) = 10.92 in2

10% of A,= 1.1 in'

Place this s tee1 within a distance of approximately 53 inches along the side faces of the cap.

Maximum bar spacing = 12"

Mwumum bar size = M

1.1 Number of #4 bars required = - = 5.5 or 6 bars

02

53 in - 4.5 spaces Number of 12" spaces between bars = - -

12 in

Therefore, 6 bars are required along each face of the cap. There is atready a #I1 bar at the bottom of the cap and just below the construction joint. Therefore, place 4 #4 bars along each face of the cap.

2.35.0 Cap Shear Reinforcement

See page 2-91

Location Span 1 0.3

0.4 0.5 0.6 0.7 0.8 0.9 1.0

Span 2 0.0 0.1 0.2 (3.3 0.4 0.5

Usually, themaximum designshear, V,, canbe takenas the shear that occursat a distance d from the face of the support. However, major conmnhated Ioads may occur on the c;ap between the face of the support and a &stance d from the support face. Therefore, it is reasonable to design for V, which occurs at the face of the support column.

Shear O k

-232 - 572 - 577 -581 - 586 - 590 - 826 1026 61 7 545 473 400 - 71

Seclion 2 - Rein!orced Concrete Page 2-59


2.35.1 Span I Shear Design at Support Face

3-1 25 Face of support = - = 0.29 of span horn the column centerline

10.75

V, = 586k approximately

required $V, = V, - $V,= 586 - 352 = 234 k

try using #6 stirrups u A, = (2 legs)(0.44 in2/leg) = 0.88 in2

Therefore, m i m u m s = lesser of (r"" 1- - = 33 inches

Use #6 63 12" at column face

Page 2-60 Section 2 - Reinfotred Concrete

r* - Bridge Design Practice - February 1994 m

2.35.2 Span 1 Shear Design at Cap End

To first whsd

To application of girder dead bad 4.73'

2 vertial 1 horizontal

3.75 feet from Edge of Deck

V"= 232 k

required +V, = 232 - 101= 132k

By inspection OV, < *& b,d A

I 24 inches 1 maximum s .; lesser of 1% ]=9-be8

R 6 1 6'' will be acceptable.

Section 2 - Reintofced Concrete Page 2-61

c-t: - Bridge Design Practice - February 1994

4.9 feet From Edge of Deck

V, = 572 k a p p r o b l e l y

@V, = 24 b,dB = 245 k

required $V, = 572 - 245 = 327 k

by inspection, $V. < f i b,ds

m u m S = lesser of = 2 . d 1 *

#6 u 43 6" will be acceptable

Remainder of Span 1

Use the same spadng as that calculated to be used at the face of the column support.

2.35.3 Span 2 Shear Design

3.125 Face of support = - -

22 - 0.142 of span korn the column centerline.

V, = 617 - 0.42(617 - 545) = 587 k

$V, = a & b,d = 352 k (using d= 67.T)

required $V, = 587 - 352 = 235 k

24 inches """"'=""'"'jt )="al

use #6 u 1 12" at column fire.



When S = 24 inches

Use #6 u 3 2 4 between the 0.3 arid 0.7 points of Span ?

2.36.0 Final Cap Design

Note: Bar configuration based on Bridge Design Details manual page 8-30, Dated June 1986.

wri rot 7 con! A[ At

Deck I

#6 @ 6 ' @ i Z - @ 12'- 8 12' @ 24' stirrup '

- e l l

Y,,'.". 'V

1 See Bridge Design Delails page


c* - Bridge Design Practice - Februav 1994

Page 2-64 Section 2 - Reintorced Concrel e

EM - Bridge Design Practice - February 1994



w m m . - - * d y l 4 m m VL bl rl

l l t

Secrion 2 - Reinforced Concrete Page 2-67

8OP

BR

STR

UC

TU

M

LX

W LOAD D

IAC

lNO

STIC

9 - LL UO. 1.

HS

30

-44

M

8H

FO

MM

IN

O WITHOUT ALTBIWATXVB

IPA

CT

O-D

)

mE

R OF

LIV

B LOAD

UN

X9

---------------------+

----------+

MnU

BU

P8R

ST

AU

CT

UII

B

SUB

STR

UC

TO

RII

N

O.

LT.EHD

RT. END

LT

RT. 8t4D

----

----

.-

"---

----

-- -*

----- --

----

----

-- --

- N

o. oC dsm

im live load Ismam

. (4

3.5

feet

f 7

) wheel line.

x I1

llvr load Ibarlfll uherl

Iln

ar

l - 3,1

07

live l

oad laam*

For fectorod results, factor.

may ba applied to

the

number of laumwr

(1.3)

x

(1.6

71

x

(3.1

07

1 - 6.

74

5 factored livw

loa

d lane.

MB

BT

O IMPACT F

AC

TO

RB

C

UC

UL

RT

BD

BY

PR

OU

RhM

K

HM

IN

PA

C'F

N

O

%

HS

20 - F

acto

red

c* - 8ridge Design Practice - February 1994

--

Section 2 - Reinfo~ed Concrete Page 2-69 I

LL N

O.

1.

DB

AB

b

OM

PL

US

WO

AT

IVB

L

IVE

L

OA

D

MO

ME

NT

EN

YE

LO

PI

(PA

CT

OR

BD

I

HE

M

LEQT

.1 PT

.3 PT

.3 PT

4 T

.5 PT

.6 PT

.7 PT

. 8 PT

.9 PT

RIG

HT

NO 1

0.

1719.

26

65

. 2

63

9.

22

41

. 8

70

. -1

17

4.

-41

90

- -7

87

8.

-12

81

4.

-19

86

4.

1 -1

18

23

. -9

53

0.

-10

79

. 3919.

68

95

. 7

63

9.

66

76

. 3

05

0.

-21

89

. -1

06

70

. -2

34

36

.

3 -13114.

-13

35

0.

-70

29

. -2

61

9.

78

2.

31

76

. 4

55

7.

49

31

. 4

29

6.

26

53

. 0

,

LL NO.

1.

DlUb

LO

m FLUS P

OS

ITIV

E

LIV

E W

N,

MO

PIBN

T B

PN

BW

PB

(P

ICT

OR

ED

)

WH

LIFT

-1 PT

-2 PT

-3

PT

-4 PT

.5 PT

.6 PT

-7 PT

.El PT

.9 PT

RIGHT

NO 1

0.

5

11

9.

86

69

. 1

03

94

, 1

07

17

. 9

65

8.

73

02

- 3

52

3.

-14

87

. -7

06

6.

-12036.

2 -1

19

8s

. -1

43

9.

ISA

S.

11

55

0.

15

79

1.

17

07

9.

15

59

8.

11J.61.

39

31

. -5

51

9.

-14501.

3 -1

39

09

. -7

51

1.

-66

s.

51

00

. 9

94

1.

12

75

7.

13

92

7.

13

36

3.

11

02

2.

66

02

. 0

.

HS

20 - F

acto

red

-


- n ; . . . a . . . . k . .

r w a & - -

W N W G W

. l Y L I

g:? w

P r - w - m 1

. . P W r l Q I

m m + m 0 w w w

3 my'- I 3 p - m . . - E'?? n m 0 IA l m 0 w

n o - n m . r r a m - . "l - 0 -

a n - - 4 e. - k a - E Y * p-l*

r( b-0 1 0 0 d m - 4 - f l d o - . w I * W


KAX.

MI

AL

L

OM

I H131 . L

ON

OIT

UD

XN

AL

M

OM8E

FF

WIT

H ASSOCIATED HOMEKTS

WITH

MS

OC

IIF

BD

UIA

L W

AD

S

--------*-*----"-----*"--*-

-*---*--------------------*-

Mh

L

----

-- M

Qm

m- - -

- - -

MIU

--

----

HO

mm

----

--

LU

hD

TO

P

BO

T ,

LO

MI

TOP

BO

T ,

SOPPORT JT

. 1

POSXTf VS

75

.1

0,

0

,

0.0

0

. 0

. H

gOhT

Im

-10.0

0,

0

. 0

.0

0.

0.

BUPPORT JT.

4 P

OS

ITIM

7

6.3

0

. 0

. 0

.0

0.

0

MO

hT

IVB

-0.1

0,

0.

0.0

0

.

0

AL

I aupport ramultm

xmpr

mm

ent

Inte

rn11

mupport

join

t raactionr

durn

to

tbo

application

to

thr

mupermtructura of

on

ly ana

tru

ck or

tru

ck lano

loa

din

g.


4

4 1 : " Z o c a - o d L I 8 I .

P r l U U


c* - Bridge Design Practice - February t 994

Page 2- 74 Seetian 2 - Reinforced Concrele

WKBBR 1 LSPT POS. V 5 3 7 . 6 MOM. 0 . NgO. V - 1 0 4 . 0 MOM. 0 .

WHBER 7 LEFT POS. V 0 4 5 . 2 MOM. -10151. NgO, V - 4 9 . 0 HOM. 1401.

LL NO. 4 D I M MAD PLUS L I V E LOAD aUMll

HgMBER 1 W P T P 0 S . V 9 0 5 . 8 HBO. V 163.1

RIOHT - 7 2 1 . 0

- 1 6 5 6 . 1

HBMBSR 3 LBPT POS. V 1 5 7 0 . 6 HaO. V 8 0 3 . 5

P - Factored

-

LZ

VB

LOAD SUPPORT R

ESU

LT

B

(BB

RV

TC

E

I UNPXCTOKBD)

mu

XXIIL

MiU

I M

AX

. LOHOLTUOXNAL HOHgKP

HITA ASBOCIhTEX) W

OW

EN

TS

WITH

AS

SO

CX

hT

gfI

hXXhL

LO

AD

S -----------------*---.-

----

----

----

----

----

----

----

----

N

(I

At

----

-- MO)Q!m------

IJ(X

AL

----

--M

QM

gm

----

--

LO19

TO

P

BO

T ,

LO N

l T

OP

DOT.

SUPPORT JT. 1

POSITIVE

13

3.1

0

. 0

, 0

.0

0.

0

. M

BO

AT

IM

-16

.0

0.

0

. 0

.0

0.

0.

SUP

PO

RT

JT.

& P

OS

ITIW

1

45

.9

0.

0.

mO

AT

IVE

-2

0.9

0

. 0

,

Al

l oupport

rmau

ltm

raprsmanc

intr

rna

l mupport

join

t raactlonr due to

tha

application

to t

he

rupsrmtructurm of

on

ly o

ne

tru

ck or

tru

ck

Ia

n.

loa

&in

g.


m o o 0 0 0 0 0 rl

0 0 0

- 4 A - . I . W, * *

d o o r * nu, . * 0 0 0 0 0 0 0 0 . I. m

3: - - - - - m o o 0 ..) 0 0 0 0 0 0 0 0 0 0 0 0

8: Cleo 0 rl - - -

% . 0 0 0 0 0 m & * 0 0 0 0 0 3 P 3 0 8 ~ ~ ~ 0 n ~ 0 0 0 w 0 w w 0 0 0 0 n m n m m m .+ n n n m m

Page 2-78 Section 2 - Reintorced Concrete

r* 1- Bridge Deslgn Practice - February 1994

Secfion 2 - Reinforced Concrete Page 2- 79


. . m w w . " I?. In

. . In R w w w * I

e. r- * - - - o m - rl I?. U) m m I 1 I


r* - Bridge Design Practice - February 1994 =

--

Section 2 - Reinforced Concrete Page 2-8 1

r-Jt - Bridge Design Practice - February 1994 R

w rl . . . . l-a o m

5 2 z s .l I

Page 2-82 Seclion 2 - Reinforced Concrete

r-t: - Bridge Design Practice - February t 994 - E Z Z O n u t -


LIVB LOAD # H E M EHVBLOPES lrPm ASSOCIATBD HOKBNTS (IBRVICEI

m N B B R 1 LEFT FOB. V 231.3 MOM. 0. W O . V -31.1 MOM. 0.

RIOHT

5 . 3 3 7 3 .

- 2 5 6 . 5 - 9 0 3 .

K B W 8 R 1 LBPT POS. V 3 5 5 . 3 MOM. - 1 1 8 1 . H80. V - 1 8 . 5 MOM. 511 .

MBMBER 3 LEFT P08. V 1 5 1 . 4 non. - 1 0 1 2 . ma. v - 3 . 5 MOM. 1 8 3 .

u m I?. m 3

LL NO. 1. Lorn PLUS

RIOHT - 5 6 0 . 3 - 6 2 1 . 1

KBMSIR 3 LBPT P 0 B . V 6 0 3 . 7 H80. V 6 1 1 . 1

HS20 - S

ervi

ce

-

tt

HO

. 1

tfM

W

M BUPPORT RBSULTS

(SERVICE)

HhX.

UX

AL

WAD

MA

X.

LO

N~

ITV

DIN

AL

MO

MB

NT

KTTB

h3R

OC

Lk

TII

D HOKKNTS

WIT

H

AS

SO

CIk

TE

D M

II

L W

AD

B

-----+

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

....

....

....

....

....

....

...

mu

- -

- - - -

MOMgm- - -

- - -

- - - -

- -H

Om

NT

- - -

- - -

LO-

TOP

ROT.

WA

D

TOP

BO

T .

SUP

PO

RT

J

T.

1

POBITIVB

75

.4

0.

0.

0.0

0

. 0

.

HBaF

CF1

VB

-10

.0

0.

0.

0.0

0

. 0

.

MBP

IBER

4 P

OSITI~

115.8

-82

. 0.

64

.7

20

~.

0

. t4

EO

hTT

VX

-7

.7

48

. 0

. 6

6.3

-280.

0.

HgW

BE

I S

PO

SIT

IM

119.1

57

. 0

.

63

.7

29

4.

0.

NBOITZVX

-5 .o

-4

2.

0.

65

.3

-as

k.

o .

SUPPORT JT.

1

POSIPXVB

76

.3

0.

0.

0

.0

0.

0.

Ng

OA

Tf V

IS

-B.I

0.

0.

0.0

0

. 0

.

***.

* AIL wupport rorultm rmpr.ment

imt+ruml ruppott joint

rss

ctl

on

m

du

e to

th

e

application to

tb

r m

upormtruccuro oL

on

ly one

tru

ck

or truck lane

loa

dtn

g.

TB

B

MT

IO OP BUBSTRVCTURB /

SUP

ER

STR

UC

TU

RE

L

OhD

INO

IS

0.3

22

2.37.3

Ben

t 3

Loa

ds -

Ben

t

,,...,

,..., l.

ll.l..l....,,,,

IHPU

T FILE FOR BENT

PR

OO

W

,,,,,,

,....,

.... ..

.I......*

BE

NT1

1

1 2C

H

1

08

0

03

25

0

00

0

0

0 0

00

s

BE

NT1

2

2 3

H

22

0

0 0

32

50

0

0

0 0

0

00

0

BENT1

3 3

4 C

H

10

8

0 0

32

50

0

0 0

0

0 0

00

B

EP

IT1

4 5

2P

ZOO

0

62

32

50

0

15

0

0 0

0

0

0 0

B E

NT

I 563P

20

0

0 62

32

50

0

15

0

0 0

0 0

0 0

BE

NT1

1

0 0

0

0 88

60

0 81

2 6

37

0

00

27

0

02

7

0 0

0 0 0

0 0 1

BE

NT

1 2

01

0

0 0

0

0 0

00

00

0

00

O

OO

OO

OO

O

BE

NT1

3

01

0

0 0

0 0

00

00

0

00

0

0~

00

0~

00

B

EN

T1

4 0

0 2

6

0 0

0 0

12

57

1

25

7

0 4

BE

NT

1 4

80

40

0

0

0 R

0 0

0 0

BE

NT

1 4

14

0

0 26

0

O 0

0 1

65

7

26

70

0

5 B

EN

T1

4 2

00

0

26

0

0 0

0 2

85

7

12

60

0

0 6

BENT1

SO

40

0

0

0 0

0

0 0

0 B

EN

T1

58

04

0

0

0 0

0 0

0

00

B

EN

T1

51

40

50

0

0

0 0

0

0 0

0 B

EN

T1

52

00

6

0 0

0

0 0

0 0

0 0

BENT1

0 1

3240

U

49

10

0 0

0 0

s

AD

DE

D C

AP

W

EIG

HT

BENT1

0

2 3

21

0U

0

22

0

0

0 0

AD

DE

D

CA

P

VlE

IGIlT

B

MT

L

0 3

32

40

U

0 5

9

0 0

0

AD

DEI

, C

AP

W

EIG

HT

BEM

T 1

0

0 0

00

0

17

54

00

0

00

0

0 0

11

0 B

EN

T l

L 1 2

60

1

49

0

BEb1

T1

2 1

26

01

7B

5 B

EN

T1

3 2

26

01

90

0

8EN

T1

4 1

26

01

785

B

EN

T1

11

9

57

0

66

294

0310

14

2

02

00

761

01

B

EN

T1

0

0 0

80 0

39

03

60

0

00

0

00

EEID SY P PORT C m Y O M R DISTRIBUTZON HEM JT JT CCKD OR PACTORS PACTORS NO LT RT LT RT D I A BPlrN B I HINOE Y t LT RT LT RT P ' C

I*.** Zp KEHBEll 113 HQIIIZOUTAL SUPPORT OR RINOU PIBLD BQUALB ZIOCATZON OP KINOE PROM bEPT IIUD OF KBKBLR *'*'* .am** ZP MBMBgR IS VERTICAL SUPPORT OR HINOB YIELD EQUALS SUPPORT WIDTH VSBD FOR MOWNT WDUCTXOIP . * m m * Y t r Umtmacs from bottom of mupmrotruotur~ moffit to tbm amnttoid of tho concrete mupsrmtructurr.

BENT 3

D X M LOAD W A L Y B I B * * * SIDBSWAY INCLUDED. "*

HORIZONTAL MgMBBR MOMENTS TRIAL 0 (SERVICE) MBM

NO LEFT . 1 P T . 2 P T . 3 P T . 4 P T . 5 P T 1 0 . 0 . 0 . 0 . 0 . - 1 3 0 .

2 - 1 5 0 6 . - 6 4 5 . - 2 9 3 . 4 3 . 3 6 3 . 6 4 1 . (face of wupport - 5 0 1 . )

VXRTICAL MEMBER MOHBNTB TRIAL 0 (SBRVICE) HEM

NO LEFT . l P T . 2 P T . 3 P T . O P T . 5 P T 4 0 . 9 . 17. 26 . 3 4 . 4 3 . 5 0 . -9. - 1 7 . - 2 6 . -34 . - 4 3 .

HORIZONTAL KBMBIIR SHEARB TRIAL 0 (BERVICE) mu NO LEFT . l P T . t P T . 3 P T . 4 P T . 5 P T 1 0 . 0 0 . 0 0 . 0 0 . 0 0 . 0 - 1 6 1 . 7 1 4 3 0 . 5 1 6 3 . 3 1 5 6 . 2 1 4 9 . 0 1 4 1 . 9 - 1 2 5 . 3 3 2 7 9 . 2 2 7 5 . 7 271 .2 268 .7 265 .2 .261.7

VBRTICAL KEMBBR SHEARl TRIAL 0 (SERVICE) HBM

NO LBPT . 1 P T . 1 P T . 3 P T . 4 P T . 5 P P 4 4 . 3 4 . 3 4 .3 4 . 3 4 . 3 4 . 3 5 - 4 . 3 -1 .3 - 4 . 3 - 4 . 3 - 4 . 3 - 4 . 3

. 7 PT . 0 PT . 9 PT RXOHT - 7 0 3 . - 9 9 5 . -1291 . - 1 5 9 1 .

( face of support - 7 4 1 . )

5 8 . - 2 5 7 . - 5 8 7 . - 1 4 8 0 . (face of support - 4 5 2 . )


Secrion 2 - Reiniorced Concrete Page 2-89

- Bridge Design Practice - February 1994

Page 2-90 Secrion 2 - Reinforced Concrele

THE LIVE LOADINOS USED TO OBNaRATB THE POLLOHINO PACTOMD INMMPBS CONSISTED OP ( A QROUP OR Hal0 TRUCKS ) hNDfOR ( A P TRUCK WITH OR UXTHOUT AH HS2O TRUCK ) .

DEN) WllD PLVB POSITIVE L'IYB L O M MOMKNT BNvgWPX FOR C I P DRSXON IPRCTOMD)

LEFT .1 PT. . a PT. , 3 PT. . 4 PT. . 5 PT. . 6 PT. . 7 PT. . B PT. .I PT. RIOHT

DBllD LOAD PLUS NEQATIM LIVE LQhD MONBNT BNVEZXIPB POR CAP DlClXON (FACTORBD)

LEFT .L PT. . a PT. . 3 PT. . 1 PT. . 5 PT. . 6 PT. - 7 PT. - 8 PT. . 9 PT. RIGHT

- 3 5 2 8 . 1 - 1 0 8 2 . 3 -1306.9 - 5 5 1 . 9 117 -1 0 . 0 3 4 -500.1 -1232.0 - 1976 .4 - 3 6 5 8 . 8 (faca of mupport -1765.1 ( face of mupport - 1671 .1

DBhD MAD PLUS POSITIVB / UlSOIITIM L I V B LO- S H B M B W 8 L O P l FOR CAP DBSION LFACTOMD)

LSPT . I PT. . 2 PT. . 3 PT, . 4 PT. , 5 PT. . 6 ST. . 7 PT. . 8 PT. , 9 PT. RIOHT

BENT 3

m 2. a m

TC1 a m I?. m 3

? nl 3. n m

rlt - Bridge Design Practice - February 1994

Part B - Design Notes

2.38.0 Service Load Design - Overview (BDS 8.15)

(Also known as Allowable or Working Stress Design)

In service load design, members are designed for the maximum load whch is actually expected to occur in the member. Stressesare calculated horn theloadjngcondition These appliedstresses are then compared to allowable stresses.

C = 1/2 x f, = total compressive fan%

T = Asfs = total tensile force

In the above figure:

Assume that conuete m o t resist tensile forces.

x = distanse from extreme concrete compression fiber to the neutralaxis. Note that x depends on section geometry and not on the applied load.

d = effectivedepth

f = compressive stress in the extreme concrete fiber

f, = tensile sbess in the steel

M = kf, d , - = internal resiswg moment ( ~ 3 ) f, and f, can be calculated using the familiar formula, f = - MY

1 Both conmete and steel are assumed to stay weU within the elastic range for the given loadings.

Page 2-92 Section 2 - Reinforced Concrele

r-t. - Bridge Design Practice - February 1994

2.39.0 Strength Design Method Overview - (BDS 8.16)

(Also known as Load Factor Design)

This is the p~edominant design method used by Caltrans (see BDS 8.14). Strength des jp diifers radically from service d e s w Factors are applied to the actual maximum loads which are expected to occur on a smcture. Members are then designed for these factored loads which should never occur. The concrete and steel are assumed to behave inelastidty as the factored loads are approached (this is not entirely true for aU parts of strength design, however it is the underlying basis for this philosophy of design). In general it isassumed that a structure designed this way will not have a catastrophic failure unless an actual factored load is applied to it.

For example:

Mu = factored moment at a section

M, = nominal moment capacity of a section

$M, = design moment capacity of a section

= strength reduction factor (safety factor)

Design Giteria is $M, 1 Mu

The stress distribution for flexure in girders changes as the lea- is increased from service loads to the nominal capadty af a section. The following figure shows the progression in the sbess dLstribution &gram as loads are inmeased horn service levels to the nonivlal capacity of a section.

Elastic Stress at Service Loads

Inelastic Stress Inelastic Stress at Factored Loads



The basic criteria far design by the Strength Design Method is:

Design Strength 2 RequiredSkength @DS 8.16.1.1)

(Strength Reduction Factor) (Nominal Strength) 2 (Load Factors) (Service Load Forces)

The following terms are very important if one wishes to understand the Strength Design Method.

Service Loads - These are the actual design loads. They are described in detail in BDS Section 3. From a designers point of view, these are the actual loads which a structure may be subjected to.

Factored toads - Service loads inaeased by factors. The appropriate factors to use axe covered -m BDS 3.22.

Required Strength - Strength necessary to resist the factored loads and forces applied to a s h m e in the combinations stipdated in BDS 3 .22 in determining the required shength of a section, the factored loads must be placed in such combinations and locations as to produce the maximum forces on the aoss section under considera tion.

N o d Strength - Strength of a cross section c a I d ted in accordance with the provisions and assumptions of the BDS Code. For flexure and axial loads, the assumptions are covered in BD5 8.16.2.

Design Strength - Nominal strength multiplied by a strength reduction factor, $. See BDS 8.16.1.22 for appropriate I$ factors.

The subscript "u" is used to denote required strengths or factored forces. The subscript "n" is used to dmote nominal sh-enghs. For example:

M, =. factored moment = required moment strength

M, = nominal moment strength = theoretical moment s m g t h

$4 = design moment strength = usable moment shengh

It should be emphasized that M, and M, are totally independent of each other.

Mu is determined horn an elastic analysis of the structure with the factored loads applied to it.

M, is a function of the geometry and materials present at a given cross section of a sb~~ctural element. It is in no way related to the loads applied to the structure.

For moment, shear and axial loads, the basic criteria for design is:



2.40.0 Face of Support - Negative Moment Design (BDS 8.8.2)

For continuous members, instead of designing for the negative moment which occurs at the center tine of the support, the maximum negative design moment may be taken as the moment which occurs at the face of the support (member and support must be monolithic).

4w continuous member O 1.5D M-

7 A /

\ / - face ol suppan a s 45"

Section 2 - Reinfoced Concrete Page 2-95


2.40.1 Example

Equation of the parabolic flare (based on X-Y axis shown)

y = ax2 where y = 2' when x = 10'

a = y/x2 = 2/102 = 0.02

Face of support occurs at x = 10' - 2' = 6'

y = (0.02)(8)2 = 1.28'

Face of support occurs at 2' + 1.28' = 3.28' from the center line of the column. Use the factored negative moment at this location for the flexuraI steel design.

Page 2-96 Section 2 - Reinfoxed Comere

r-it - Bridge Design Practice - February 1994 =

2.41.0 Cross Sections Experiencing Positive Bending

Effective compression flange (BDS 8.10. I )

&

Eff wive tension flange

Each 7" is designed as a single girder.

bl + b2 + b3 = Effective tension flange (BDS 8.17.2.1.1)

- _ - - - - - _ _- - - - - - - - - _ - - - - - - - - - - - _ - - - - - Positive bending

region



2.42.0 Cross Sections Experiencing Negative Bending

Effective tension flange (BDS 8.1 7.2.1 . I )

&

Effective compression flange

bl + b2 + b3 = Effective lension fbnge (BDS 8.1 7.2.1 -1) - - - - 3

b 1 b2 b3

t- I Effective compression flange (BDS 8.1 0.2)


I - - - _ _ - - - - - - J - - - --------7 Negative bending regions


2.43.0 T-Girder Compression Flange Width (BDS 8.10.1 .I)

Positive Moment Case

L = grrder span length

For a typical exterior prder, left:

1 = lesser of 6hf, overhang length

2 = lesser of 6b, ' 1 2 LCk1

3 = 1 + b , + 2

4 = ' / r L

b = lesser of 3 and 4

For a typical interior girder:

2 = Iesser of 6hf, /2 LCI,,

2 = lesser of 6hf, / 2 LElr 2

3 = 1 + b w + 2

4 = l /4L

b = lesser of 3 and 4

Also see BDS 6.10.1.2 - 8.10.1.4 for special considerations.


c* - Bridge Design IPractica - February 1994 m

girder span length .= 60'

Interior Girder:

1 = lesser of 6h, = 6(8) = 48" and ' /zLCI, = /2 (96) = 48"

2 = lesser of 6hf = 6(8) = 48" and I/z LC,, = l/2 (96) = 48"

3 = I +b,+2=48+15-t-48=1111'

4 = I / 4 L = l/4 (60') = 180"

b =lesserof3and4= 111"

Exterior Girder (right):

I = lesser of Sh, = 6(8) = 48" and 3/2L,1, = (96) = 48"

2 = lesser of 61% = 6(8) 4 8 " and overhang = 36"

3 = I + b,,+2 = 48 -t15 + 36 =99"

4 = I / q L = l / 4 (60') = 180"

b = lesser of 3 and 4 = 99"

2.44.0 Box Girder Compression Flange Width (BDS 8.10.2.1)

For box girders, the entire slab width shall be assumed effective for compression.



2.45.0 Box Girder Effective Tension Flange (BDS 8.17.2.1 .I) Positive Moment Case

Tension reinforcement shalt be distributed entirely within the effective tension flange areas.

L =girder span length

t =tension slab thickness

bl, 'b2, b3, etc = effec tjve tension flange widths for each grrder web.

Ldr1,2,3, dc = clear spans for each bay.

For a typical exterior girder, bl:

1 = Lesser of 6t, l / z L,,,,'/12L

2 = b , t 1

3 ='/10L

bl = lesser of 2 and 3

For a typical interior girder, b2:

1 = lesser of 6t, ' / 2 LCI,,

2 = lesser of 6t, /z LclrZ

3 = I + b , + 2

4 ='/10L

b2=lesser of 3 and 4

Whole box effective lension flange width = bl + b2 + b3 +


r* - Bridge Design Practice - February 1994 W

2.457 Example:

girder span length = 100'

\P 15' wide non-otlectiv.e

area

4 3' 2' 8' I 10' I 8'

I I I I I

Calculate the positive moment effective tension flange width. (ie. soffit)

Exterior Girder

Interior Girder

Total effective tensjon flange width, b,

=51 +go+ 90 c 51 =282"

I 6t = e(6.5) = 39' and

2 LI~ = 2 ~ 1 8 ) = 39'

j&rngh~ = X(108) = 54.

Page 2- 102 Section 2 - ReinTorcd Concrete

3

39 + 12 +39 = 90"

bt = 90'

c* - Bridge Design Practice - February 1 994 M

2.46.0 Box Girdern-Girder Effective Tension Flange (BDS 8.1 7.2.1.1) Negative Moment Case

Tension redorcement sbI1 be d.istributed entirely within the effective tension fla~ge ateas.

L = girder span length

t = tension slab thickness

bl , 5 2 b3, etc = efle&ve tension flange widths for each girder web.

LC,,, 23Atc = c ICR r spans for each bay.

For a typical kxterior girder, bf:

1 = Iesser of 6t, ovexhng length L 2 =lesser of&, z L,,,,

3 = l + b , - E ?

B = ' / lo L

b l= lesser of 3 and 4

For a typical h-~ terio~ prdcr, b2:

2 = I s e r of Gt, 4

2 = ~ ~ S S C T O ~ bt, LcbZ

3 =Z+bw+2

4 ='/ToL

b2= lesser of 3 and 4

Whole box effective tension flange width = bl -I- b2 -I- 53 -F . . .

Seclion 2 - Reinfored Concrele Page 2- 103

rit - Bridge Design Practice - February 1994

2.47.0 Rectangular Section with Tension Bars Only (BDS 8.16.3.2)

where a = Asfy

0.85 C b

Solving for A, algebraicafiy :

1.7f: bd where z =

2 f~

Solving for A, by iteration:

Assumean initial value fox a. Use that value to calculate As from the second equation, Use that As value to calculate a anew value of a from the first equation. Continue iterating behveen equations until a and As values converge to a h l solulion.

Page 2- 7 04 Section 2 - Reinforced Concrete


2.4 7.7 Example

f', = 3.25 ksi

1, - 60 ksi

b = 60"

d = 65"

Mu = 3000 k-ft

Solve algebraically:

rnax allowed PLj= 0.6375(0 .SS)(3.25 )(60)(65) 87

60 (-) = 67.751'

Solve by itera tion:

assume an i n i t i a l value of a - 2 in.

Note: The A, calculation above represents the minimum amounl o[ tension reinforcement required far k e above section with an %= 3000 k-ft.

Section 2 - Reinforced Concrele Page 2- 105


2.48.0 Flanged Section with Tension Bars Only (BDS 8,16.3.3)

Always start the analysis or design of a flanged section by assuming that a i hF l'his mc~ms, calculate the depth of the compressive stress block, a, using the equations for a rectangular section.

"sfy a =

0.85f b

I f a 5 h, thm

The above calculation of A, and a are correct.

1 f a > h, then

T l~e above calculation of 4 and a are incorrect.

The following equations apply.

maximum allowed A, = [ 870M)+ 87m0 fy )+(b-b,v)hr]

Note: The above equation for rnaxjmum allowed A, usually holds truc even if a 5 h,.

Page 2- 1 0 6 Section 2 - Reinforced Concrere

EM 7 Bridge Design Practice - February 1994 1

Solving for A,:

SpIit the fJanged section into 2 rectangular sect ions as shown. Perform calculations for each rectangle. Superimpose results.

web 2

1.7f :b,d where z =

fy



2.48.1 Example:

t= 60"

a = (1 9 -1 1)(60)

= 4.02 in < hf (rectangular compressjon area) 0.85(3.25)(60)

f', = 3.25 ksi

max allowed. A, = 0G375(325) 60 [0.85(15)(42)(~) 87 -t 60 + (60 - 15)(5)]

1 b.

5'

7

2.48.2 Example:

For the flanged section in the prior example, cdculak As required when EjI, = 3000 k. lt.

Assume a I h,

a = 17.14(60)

= 6.20 in 11~ (non--rectangular compression area) 0.85(3.25)(60)

42'

f = 6 0 k i -

Above ca lda t ions are incorrect.

Mu = 2000 k.A

A s m e a I h, (reckangular section)

Page 2- 1 08 Section 2 - Reinforced Concrete

4 a- i

c* - Bridge Design Practice - February 1 994 =

A,, = 0.85(3.25)(GO - 15)(5)

= 10.36 in2 60

Note: Suppose required A, > maximurn allowed A,. What options might be considered?

lncrease f', Provide compression reinforcement. Revise the geurnetry-of the concrete section.

Usually, the easiest remedy is to increase the thickness of t-hhe connet; compression flange wid^ the critically loaded regjons.

2.49.0 Bar Spacing Limits For Girders

Minimum dear har spndng @DS 8.21.1)

1.5 dL 1.5 (maximum aggregate size) 1.5 in

Maximum bar spacing in slabs (BDS 8.21.6)

1.5 (slab ttuchess) lesser of

18 in

For bundIed bars, treat the bund1.e as a single bar of a diameter such that the area of the single bar is rcpivalent to the total area of the bundled bars (BDS 8.21 5).

Section 2 - Reinforced Concrefe Page 2- 1 09


2.50.0 Development of Reinforcement

There is very Little to say here about calculating bar development lengths. BDS 8.25 tllrough BDS 8.30 covers the subject quite sufficiently. Numerous charks are available in Caltran.~ and other publications which Lisrs development lengths for various barsizesasused in various design details.

However, one very impol-lant thing to keep h mind is:

Development lengths of bars with standard hooks, as covwed in BDS 8.29, apply mly to bars in tension. To develop a hooked bar in compression, the formulas in BDS 8.26 must be used.

2.51.0 Positive Moment Bar Size Limitation (BDS 8.24.2.3)

Requirement a l simple supports and points of inflection*

M e, 5 1.3 -2 + P, if bar ends are confined by a compressive reaction. v,

%, at a support = bar embedment length beyond c e n t e ~ of the support.

e, at infleckion points = greater of {&lib

"Note: This requirement does not apply to bars tenninwting beyond tlw center line of simpje supports by a skmlard hook.

2.5 1.1 Example: End diaphragm abutment

20 - #I0 positive moment bars extend jnto the abuhnent.

M,, = 5300 k-ft for 20 bars

V, = 11GU k at the abutment

Abubnent wid01 = 2.5' = 30"

!, = j z (abubnent width) -clearance = /z (30") - 3" - 12"

Page 2- r 10 Section 2 - Rein/ormd Cuncrete

c* - Bridge Design Ptactice - February 1994 1

The direclion of the shear and reaction at the abutmenka re such that

I the bar ends are confined by a compressive reaction.

1 For #10 bars, P, = 54 inches

M t, < 1.3 J + t, #lo bars are acceptable. vu

2.52.0 Minimum Reinforcement Requirements (BDS 8.1 7.1)

A minimum design strength is required at any section where tension reinforcemenk is required.

minimum required $Mn=1.2M, (BDS 8.17.1.1)

where M,, = cracking moment = moment which will cause tensile cracks in a concrete section which has no stee! reinforcement.

since f, = 7.5 K= modulus of rupture for normal weight conaete

The above minimum @bin requirement: may be waived if the area of reinforcement provided at a section is a t least one third greater than that required by analysis (BDS 8.17.1.2).

The above two minimum design criteria can be satisfied by modifying the factored moment envelope, M, as fonows:

1. Draw to scale the factored moment envelope, M,.

4 3. Plot MI,= -Mu

3

4. Darken in the fiml modfied factored moment envelope as shown in the following example. Use the modified envelope to design for flexure.

Section 2 - Reinforced Concrete Page2-117


2.52.1 Example:

f ', = 3250 psi

1, = 18.52 fP

For positive bending:

gross section y, = 50 -16.8 = 33.2" = 2.77'

= 494 k- f

------------ @ M ; = 9 a lg/y, = 494 k-ft

elope horn loading analysis

Modified Positive Moment Envelope

Page 2- 1 12 Section 2 - Reinkrced Concrete

c* - 8ridge Design Practice - February 1994 =

2.53.0 Moment Capacity Diagram

I I I I t I I I 1 I

1 capacity envelope doe to 2 full length bars

1 and 2 parl ial (ength bars o h , I I

I 4 bars ,, I

,/,. -

/- 2 lull length bars

/-- 2 parliai length bars

I

a = flexural capacity increases from zero to OM, lor 2 tuUy developed bars b = flexural capacity is due to 2 fully developed bars

c = flexural capacity increases from OM, for 2 fully developed bars to OM, for 4 fully developed bars d = flexural capacity Is due to 4 fully developed bars

--

Section 2 - Reinforced Concrefe Page 2- 7 73

I I I I I I I

I I

4M" I I

1 I t

I capacity envelope due to 2 full length bars I

2 bars ,, I

I I

r7t. - Bridge Design Practice - February 1994 =

2-54.0 Moment Capacity Diagram Versus Design Moment Envelope

I I I

I I QIM, = Moment Capacity Oiagmm I I I I I I I 1 I

I I I

I I I I I M, = Design Moment Envelope

I 1 ' \ = Factored Moment Envelope I I 1 I theoretical cut ofl point for the last two bars I I I

safety faclor separating moment capacity diagram

I from the factored moment envelope (BDS 8.24.1.2.1)

I

B = point where bars are theoretically no longer required to rasist flexure (theoretical cut off poinl.)

A 0 = required bar extension (safety factor, BDS 8.24.1.2.1)

AB = greater of d, 15 db and 1/20 I+ AC 2. & is required (BDS 8.24.1.2.2)

Page 2- 1 1 4 Section 2 - Reinforced Concrete


2-55-0 Bar Layout - Graphical Procedure (BDS 8.24)

1. Draw the factored moment envelope, M,, to scale.

2. Modify the envelope to meet minimum reinforcement requirements of 0DS 8.17.1.

3. Choose bar groups. Several items should be considered in doing this:

a Bars within girder webs (inside stirrup bends) should be continuous.

At least one third of the positive moment steel must extend in to simple supports such as abutmen Is.

At least one tourth of tl-te posj tivem omen t steelmusk ex tend in to continuous supports such as bent caps.

A11 bars usedin calculating the strength of t h ~ section mustbe evenly distributed within the effective tension flange areas.

Bar layout should be made symmehjcal about girder web center lines if a t all possible.

Maximum and minimum bar spacing requirements must be met.

4. Calculate $M, values for eadi bar group. Draw horizontal lines representing $M, values for each group 0.n top of the factored moment envelope.

5. Mark off all points B and C as shown.

6 . Calculate required bar extensions. (BDS 8.24.1.2.2)

bar extension = greater of d, 15db, j / 2 o tClr

Draw extensions horn point B to A.

7. Cdcula te the required development length, !, (BDS 8.25). Check that t l ~ e clistanas froin poult A to point C is a t least P,. If i t j s not, extend point A ouhvard until it is.

8. Fox negative moment steel, calcdate the foljowing embedment length:

embedment length = greater of d, 12db, /16 Lei,

At least one this-d of the negative moment tension steel must extend beyond [he points of inflection by an amount not less than the above embedment length.

9. Measure the distances from the span center line to the ends of ea.ch bar group for positive momenl steel. Measure the distances from the support center line to the ends of eachbargroup for negative moment steel.

10. Match lmglhs of bar group ends to provide an efficient and simple h a 1 bar layout. Try to provide symmetry in the la you t and stagger bar cutoffs. Try to keep bar lengths less than or equal to 60 feet so that spliclslg will not be required.



L$ bent

Page 2-1 16 Section 2 - Reinforced Concrete

c* - Bridge Design Practice - February 1 994 m

2.56.0 Matching Bar Ends

For illustrati,on purposes o d y, suppose here exist a simple span rectangular girder whjch requ.ires a maximum of six bars at center span. The moment envelope and bar groupings are graphed below. Design an efficient bar Iayou t.

Tehica l ly , the following bar lengths can be used for the construction of this girder:

2 bars 24+24 = 48'

2 bars 361-36 = 72'

2 bars 41+41= 82'

Now, match bar ends to come up -4th the following preferred bar lengths:

2 bars 24+36 = 60'

2bars 36+24=601

2 bars 411-41 = 82'

Sechn 2 - Reinforced Concreie Page 2-1 17

c* - Bridge De-sign Practice - Febroaiy 1994

The two different bar layouts are shown below.

Numbers at bar ends represent distances from span center Line. Bars over 60 feet long will need to be spliced.

Note that both bar layouts are technica Uy the same. At any Ioca Lion dong the span, each design conkajm the same number of steel bars.

However, in the first layout, fow of the six bars will need to be spliced, and severa.1 different bar le.ngths wiU need to be used. ln the second layout, only Wci bars will need splicing and the other fou~barsareall60fcet long. Itisgc8neraUypreferable to usealayout witl~asfewsplicesaspossible. Inaddition, a Iayou t in whichrnostbarsare the same lmg-thjs easier toconstruct since workers don't have as m y different bar lengths to keep track of.

~sexamg1eillustrationmayno(.emphasize greatly enough just how much better a bar layout can be when barendsare matched. The bar layouts done in the example design a t .the beginning of this chapter show clearly h e advantage of matchkg bar mds.



2.57.0 Working Stress Analysis Calculations

The followingprocedure is valid for both rectangular and flanged sections. The t e n 2n seen in the equations is a direct result of BDS 8.15.3.5- Toanalyze a section based strictly an rnechrutical theory every 2n tern should be replaced by n.

E, Given: b, b,, h , d, d', A, A', n =- , M = applied moment Ec

d' b

4 L

Compression Flange X

- Neutral Axis 7 d

7

2 If hf*O and b 1 -[n(d-hZ)A,-(a-1)h-da)A',] henset b,=b

h: 1 set B = - b& - b,,) + d , - t . (& 7) A'J

b,,.

2 set c = --[hi (b-b,) /2 + ndA, + (h - 1) d' A',]

b,

I 1. 1 = -bx3 - - (b - b,)(x - hJ3 -I- nA,(d - x)' + (2n - I ) A', ( X -

3 3

Mx f, = - = stress in rap fiber of compression flange.

1

f', 2nM(x - d') = mC ( I - 5) = st rcss in compression steel. 1

fR = nh3(d- X)

=stress in tension steel. 1

Stxiion 2 - Reinforced Concrete Page 2-1 19


2.57.1 Example

f ', = 3.25 ksi Service Load, M = 15,000 k-ft

fu = 60 kst b =32 '=3&"

b, = (3)(8) = 24" A, = 76in2

d = 62" A', = 20inz

d' = 3.5"

x = d1627~ + 4593 -162.7 = 13.55 inches

fc = (1 SO00 x 12)(13.55)

1,937,890 = 1.26 ksi

fns = (18)(15000 x 12)(13.55 - 3.5)

1,937,890 = 16.80 ksi

(9)(15000 x 12)(62 - 13.55) fs = 1,937,890

= 40.50 ksi



2.58.0 Crack Control Serviceability (BDS 8.1 6.8.4)

To conbol cracking of concrete, the code requires tensionsteel to be well distributed within zones of maximun~ flexure.

Laboratory test have s l ~ o w n that crack width is generally proportional to steel stress. To b i t the size of cracks which may form, the tensile stress in steel at service levels is Lmi ted to an allowable skress wl~ich is a fuxlctjon of the geometry of the rejdorced concrete section under invest-iga tion.

allowable f, = lesser of and 0.6 1, 3@

f, = steel tensile stress due to D -1- (L -t I) HS

The variables in the allowable fs formula are described in the code and on following pages.

Note that f, is tile tensile stress due to applied service loads. P-loads are not considered service loads. Factors are not apphed at the service level

The variables d,md A are both dependent on the size of the tension bars. Thisleads to the fact that for a given amount of steel, A, as the sjze of the temio.nbars decreases, both d, and A decrease, thus resulting ina larger allowable ff {assuming Umt 0.6fy does not control for allowable 6).

Wmce, the conclusim m7f be drnzun flai srnnller h r s at a closer spaciltg are better t h r I ~ r g ~ r bars sraced farther apart, ~t Imstfron~ n m c k crnl trol point of vim.

For n @uew steel requirement, A, mzce a bar size isfound which meets crack cant rol criteria, r'f holds tlzn t any s w l l e r bclr size will nlso nrtct mck conl-rol criferin.

* In members such as bent caps, i t is of ten found tlu t crack control criteria cannot be met if only the main longitudinal bars are considered. In th i s case it may be advantageous to consider the transverse deck steeI over the cap. IY the transverse steel is at an angle wi.th the cap centex h e , an effective cross seclional area of s tee1 should be calculated. The service level steel stress, f, should becalculated at thecentroid of the bar layer located closest to theextreme tension fiber of the section.

Crack control can be viewed from two different angles:

2- Post Design Crack Control Check (easiest):

Cl~oose size and number of bars to use based on A, from strength designrequirements.

Calculate f, a t service hvels. Be sure to use service or working stress analysis to calculate f,.

CaLdate aUowa ble f,.

Compare fs to allowable f,.


r-t: - Bridge Design Practlce - February 1994 =

2. PreDesign G a c k Control Check:

Tlirs method is useful Lf the designer wishes to choose a bar size prior to performing additional de5i .p steps.

Choose a bar size to investigate.

Follow the predesignprocedure oullined on pages 2-126 and 2-127.

Keep in mind that f, is calculated using service loads cmd working stress analysis.

T h e benefits of using the Pre-Design Crack Control Check are questionable. However, it bas been used in Calkram for many years, and for that reason i t is included in h i s version of khe Bridge Dfs@ Prnchke Mnnz~d. The procedure has been simplhed to make i t easier to use than i t used to be and, therefor^, mm.y designers may not recognize i t a t first. I t should benoted that a Post Design Crack Control Check is easier to perform and understand, and is generally the recommended procedure to follow.

2.59.0 Crack Control Check - Post Design Rectangular Sections (BDS 8.1 6.8.4)

b, = effect-ive tension flange width (BDS 8.27.2.1)

d, =distance fromtheextreme connete tension fiber to thecenkrofthecloscst tension bar (inches).

N = number of bars = total effective tension steel area area of the largest bar


c* - Bridge Des'ign Practice - February 1994

- effectivetensiclnconcreteareawhichhasthesamece.ntmidasthetensionsfeel A - number of tension bars

f, = steeI tensile stress due to udactored D + (L + T) HS. Calculate& using working stress analysis (ksi)

allowable fs = lesser of 3& 0-6 4

If fs I 24 ksi or f, l allowable f, then crack control requirements h.ave been mek.

" 24 ksi is based on the use of grade 60 steel f B E 8.7 4.1.6)

2.59.1 Example: fa, =3.25 ksi

$ - 60ksi

M = 600 k-ft a service levels

5.83. N =- =4.57 bars 1.27

2" clear 10.6 f, =0.6 (60) = 36 ksi t

allowable f, = smaller of 170 I & = m 4 ) ( 2 0 . 4 ~ ) = 42.4 ksj

calculate service load stress jn the steel bars.

f, = 31.9 ksi

f, < allowable fs.

This section meets crack control criteria.



2.60.0 Crack Control Check - Post Design B Box Girder With Single Layer Of Steel (BDS 8.16.8.4)

b, = effective tensionflangewidl-h=bl +b2+ b3(BDS8.17.2.1)

d, = distance from the extreme concrete tension fiber to the center of the closest tension bar (in).

N = numberofbars = total effective tension steel area area of the Iarges t ,bar

A = effective tension concrete area which has the same centroid as the t m i o n steel number of tension bars

2dCbt A = lesser of - htb, N

and - N

I, = steel tensile s h e s due to unfact-0redD-1- (L+I) HS. Calculate f, using working stressmalysis (ksi).

aUowabIe f, = lesser of 3& and 0 . 6 f ~

Xf f, I 24 ksi or f, < allowable f, then crack control requirements have been met.

*24 ksi i s based on the use of grade 60 steel (BDS 8.3 4.1.6)

Page 2- 3 24 Section 2 - Reinfored Concrete


2.60.1 Example

t- 26'= 312' I 1

f', = 4 ksi

fy = 60 bi

b = 312"

b,= 3(1OU)=30"

h.r = 8"

d -57"

b, = effective tension flange widtl1=40 + 80 + 4Cl= 160"

tension flange thickness = 6.25"

required A, = 35 inZ M = 5200 k-ft at service levels

n =E, /E ,=8

35 Try using #11 bars. N = -=23 ba.rs with A, = 35.88 in2

1.56

Perform a service/workmg streas analysis to h d

f, = 32.23

Calculate allowable f5

d, = 3"

z =I70 k/in

z allowable If,=

170

3 ~ = 3~3)(41.74) = 33.98 ksi > f,

This section meets crack conk01 criteria.


c* Bridge Design Practice - February 1994 =

2.61.0 Crack Control - Pre Design (BDS 8.16.8.4)

The foilowing procedure can be used to find out how many tension bars should be used to satisfy crack control requirements. I t is only applicable when aU of tlze tension bars are the same size. I t should be noted that a post design crack control check wiU ofken be easier toperform. If an existing design is to be checked, use the post design crack controt procedure.

d, = &stance horn extreme cono-eke tension fiber to center of the closest k-ion bar.

A, = area of tension sf eel required to meet strength design .wqu iremen ts.

A,, = area of one tension bar-.

A = effective area of concrete jo tension wlljch surrounds the tension steel and has the same centroid as the tension steel.

z = crack con fro1 factor (see specifications)

f, = working stress in tension steel a1 service loads.

n, = number of bars required to satisfy strength design 1

n,, = number of bars required to satisfy ccra.& control allowable stress fomuta, f, = z/ (d ,~)5

n2, = nwnber of bars required to create stresses in the tension steel of 24 ksi .

= number of bars required to create stresses in the tension steel of 36 ksi.

n = minimumnumberofbarsrequired.

fY = 60 ksi is assumed.

See the design example in part A ol this chapter.

Page 2-126 Ssction 2 - Reinfarced Concrele


1. Calculaterequired &for the factored moment, Mu

2. Calculate 4 assuming A, = amount of tension steel present. Use working slress analysis and service load moments, D + (I, + OH.

3. Calculate - AL *b

4. 11 f, 524ksi,use n=rgd

5. Calculate d , A,, and T = A,f,

A, = (b,) x lesser of {:c

Where h, = thickness of tension flange

b, = effective tension flange width

This definjtion of A, is only g a ~ d if all tension bars are in a single layer

7. If n, > nz4 use n = larger of nZ4 or nsd

Ef nz4 >nee> n36 use n = larger of n,, or Q

L€ G < n36 use n = larger of nj, or n,

Section 2 - Rein forcad Concrete Page 2- 727

c* - Bridge Design Practice - February 1994 - 2.62.0 Pre Design Crack Control Derivation

Starr with an assumed A, value from strength design.

Assume T= &fs = constmt for relatively small adjustments in A,.

T Is = - E working stress in steel a! service loads. As

equate f,,,,,and fs.

solve for n:

1 - 4

= numbel- of bars to satisfy the empirical allowable formula

designate n, = n from above formula.

since T - constant is assumed arid T = Asf9

T = nA,f, = constant

T 1124 = - = number o.f bars to use for fs = 24 ksi 2 U b

= number ofbars to use €or f* = 36 ksi n36 = - 36 Ab

Page 2- 128 Secrion 2 - Reinforced Concrele

r* 1- Bridge Design Practice - February 1994

2.62.1 Rnal Logic:

1. If n,, > nz4, then

z For n, bars present, f, , - < 24 ksi 1

(dc A F

Therefore, u.se n = n,, (BDS 8.14.1 4)

2. If nz4 > n,, > n36 then,

For n,, bars present, 24 ksi < I, < 36 ksi 1

(dc*)?

Jherefore use n = n,

Z For bars present, f, , - > 36 ksi I

(d , A)?

But, maximum allowable I, - 36 ksi

Therefore, more bars are needed to bring stresses down to 36 ksi.

Use n = n,,


EN - Bridge Design Practice - February 1994

2.63.0 Fatigue Serviceability (BDS 8.1 6.8.3)

Fatigue is a result of stress fluctuations in tension steel. Fatigue serviwabilily is iiddressed by corn-paring the stress range which the steel expe.1-iences to m allotvable stress range.

Requirement:

f,, = maximum stress in redarcement from (D+L+I) WS xrvice loads in ksi (calculate using working stress d y s i s )

f i n = minimum stress in reinforcement from (D+L-I-I)HS sr!rvice loads in ksi (calculate using working stress analysis)

sign convention: te-iie skreses are positive compressive stresses are negative

M,,, = Maximum dead plus positive live load moment

M,% = Maximum dead pltrsnegakive liveloadmoment

M,* = Moment which causes maximum stress, f,,, in tl~e steel

M,,, = Moment wluch causes minimum stress, fmi,, in the stee t

N = Number of fully develop~d tension bass.

When checking bottom steel, M,, = M,, %in = M,,

When checking top skel, Mm,, = Mnw b i n = Mpos

1. rJ themember is prismatic and the only section properky which varies is A, then

A. At all sections where M,,, and M,,, are posiljve , calculate

or

Mn,, - 0'67 M ~ n if all bars are the same size N

Do a fatigue ch,eck on the section which yields the highest value.

Page 2-130 Secfion 2 - Rein forced Concrete

r* - Bridge Design Practice - February 1994 m

B. At all sections whew M,, and M,, are negative, calm late

1-1 il all bars are the same size

Do a fil(.igue check on the section which yields the highesl value.

C. At dl sections where M,,, and M,,, are o€ different signs, perform a fatigue check.

lf the member is non-prisrnalic or if some section properties other than A, differ from section to section, then a fatigue check must be performed at aU moss sect ions.

2.63.1 Derivation For Procedure Outlined In I A and IS

lf dll section properties except A, are held conslant, then it is found that fs is approximately proporkional to M/A,. Tl~erefore, f, h~crt?;~ses as h 4 / 4 increases.

.&,o,applyingarnornent. M = M,,-0.67 M,;, toa seclion will yield a steel strcss ecpiifalent to fs = fm, -0.67 fmi,.

M M,,, - 0.67 M,, Therefore f5=fm,-0.67fm, will increase as thevatueof - = increases.

As As

Therefore, for allpfismatic members wheremoment reversal doesnot occur, the section which M ,,, - 0.67 M

wiU be the most critical in fatigue is the one where is a rnaximun~. As

:lion 2 - Reinforced Concrels Page 2- 131


2.632 Example

2.7" top bars. A, = 4.0 irP bottom bars, A, = 5.0 i r@

M = 300 k-ft.= maximurn service D + (L+I) HS

M = - 4Ok-ft.= minimum service D t. (L+T) J-IS

Working Stress Analysis

Nobe: At sections where moment reversal occurs, compression steel must be included in the working stress analysis.

Check top steel: 3.35 - 0.67(- 3.02) = 5.37 ksi < 23.4 ksi 0 . K .

Checkbottomsteel: 19.46-O.67(-1.48)= 20.45ksi<23.4kscsi O.K.

Page 2- f 32 Secli'on 2 - Reinforced Concrete


2.64.0 Shear Design (BDS 8.16.6.1 - 8.16.6.3 and 8.19.1 - 8.19.2)

Require $V, 2 V,

Sections located less than a distance d h o n ~ the face of support may designed for the same shear, V,, as that computed at a distance d from the face of support. See the specifications for the exceptions to flus (BDS 8.16.6.1.2).

V, = V, + V, = nornin,d sl~ear capacity of a secljan.

V, = 2 fib,d may be assunred.

V, = wlwn shear bars are perpendicular to the member. 5

d s shall not exceed - or 24 in when V, I4 f b d 2 J;;w

d s shall not exceed - or 12 in when 4 K b , d ' Vr 5 8&t;,d 4

V, shall not be taken greater than 8 Kb,d

1 Shear reidorremen t is requ bed any lime V,, 2 .;-$V,

1 Anywhere that Vu -9Vc t l ~ e area of shear reinforcement provided sl-rall not be less than:

2

2.65.0 Shear Design and Girder Webs

Since V, shall not be taken greater than 8 fibivd, the maximum shear capaoly ol a section is:

QV,, = $(V, + v,) = t$(dcb,d + 8Kh,d) = 10~fib,d

Therefore, it is required ihak V. 5 10$ E b , d

required minimum b,, = vu 10qfl d

It iscommonpractice to flare girder websnear span ends when necessary to meet the above cri teria.

Secliorr 2 - Reinforced Concrele Page 2- 7 33

c* - Bridge Design Practice - February -1994 m

2.66.0 Shear Design of Flared Girder Webs - example

Assumed cross sect ion:

cl of symmetrical section

f', =3.25 ksi fy =60 ksi d =68in b, = 4(8) = 32 in

Analysis gives the following design shear envelope.

V, (kips)

I I

2' 7.67' Span Location



At d from support face, V, = 11 00 K

( ) = 1054K Is Vu <lo 9 fib,,d =(10)(0.85) -&%(32)(66) NO.

Therefore girder web flares are requjred.

Calculate h e wjdth of web required at the face of support. (Width of web required could be calculated at d horn the supPo& face, but this would probably make calculations of other flare djmer.lsions more complicated.)

required minimum b, = V~ - (1315)(1000) - (10)(0.85)d3250(68)

= 39.9 in.

Use b,, = 10 in. per girder = 40 in. for the whole box

Determine the required flare length.

Locate the span location where v,-104 E b , d = 1054k

By straight h e approximation of the V, envelope:

x = 8.88 ft. from support centerline

= 6.88 ft. from face of support

Note that x could have been found graphically also.

Based on x = 6.88 ft, a 7 f t flare would be adequate.

However, an 8 ft flare length would probably be better for construction purposes.

Minimum required flare length = 12 (difference in web width) (BDS 8.11.3)

= 12 (2in) =24 in.

Section 2 - Reinforced Concrete Pege 2- 135

c* Bridge Design Practice - February 1994 =

Plan View of Typical Inferior Girder Web

The flare width was calculated assuming that stirrup steel would be utilized to the Cull extent allowed by BDS 8.16.6.3.9.

At d = 5.67' from face of support

assumed V, = 8 b,d = 8 4 m (3.3)(68) - = 1064 k (1Bo)

Assum.ing use of 45 st.i.rrups,

A, = (4 girders)(2 legs/stirrup)(0.31 in2/leg) = 2.48 in2

A,fyd - (2.48j(60)(68) max allowed s = - - = 9.51 in.

Vs 1064

Since V, > 4 f i b,,d

max allowed s = 12 in.

d 68 rnax allowed s = - = - = 17 in.

4 4

Use #5 stirrups spaced a t 9 in. centers within th.e 6 R. long flared girder web sections.

Page 2- 136 Secrion 2 - Reinforced Concrete


2.67.0 Shear Modifications Due to Bar Cutoffs (BDS 8.24.1.4)

Any time flcxu.ralsree1 is t e m w t e d in a tensionzone the factored shearer~vcIopc must be n~odified ( a c i u a ~ ~ there is one ot.her option not discussed here).

Ttusmcldfication isonlyrequired if thetensionsteel i s terminated wi1h.h [he portion of themember used to calculate shear strength.

The shaded regions of I-he above members are used to calcdake shear strength. Bars are likely to be tcm~inated in a tension zone when using the T-Section. Thus, shear modifications are required. 1 1 1 ~ web bars in the box girder section shorlfd always be made continuous, thus no shear modifications are needed.

The spedfications allow h r h ~ o relatively shiplc ways to modify theshear design. Either OF boLh ways may be utilized.

2.67.1 Modification Method 3 (BDS 8.24.1.4. I )

Design for a modified factored shear force

V ', = 1.5 V,,

Design for thisn~odified valuewithin the regon bounded by the end of the terminated tension bar and a point lotsated at 0.75d born Ihe end of the terminated bar.

2.672 Modification Method 2 (BDS 8.24.1.4,2)

Design for a modified factored shear force

V', = V, e 60 $ b,d whenunits are Ibs and inches

\?'" = Vu 4 0.06 @ b,d when units are kips and inches

area of steel culoCf where p,, =

(area of steel cukoff + area of steel continuing)

Design for this modified value w i t h the region bounded by tJw end of the terminated \ension bar nnd a point located 0.75d from bhe end of the terminated bar.

Section 2 - Reinforced Goncrefe Page 2- 137

r* - Bridge Design Practice - February 1994 5

Shear Modification Example

+An, - 6 bars I I Bar cutofl point

I I -

I

I I I I i V j = VU + 0.06 +bwd

I

This section modifled for convenlencc only

The solid line represents the modified shear desi n envelope. Note that V,' = V, + 0.06 $b,d will be tlie tnost R ! ficient modification for this span.

Page 2- 138 Section 2 - ReinforCed Concrete

c4 I Bridge Design Practice - February 1994 ,

2.67.3 Modification Method I - Derivation (BDS8.24.1.4.1)

code requires V, 1 2/3 QV,,

$V, 1 1.5 V,

Therefore, design for V', = 1.5V,

2.67.4 Modification Method 2 - Derivation (BDS 9.24.9.4.1)

Code requires shear steel in excess of that which is normally required.

SOb,s Excess required A, 2 -

v

vS= - = shear capad ty of s tee1 S

(excessrequired~,)f~d excess required V, =

S

= 60 b,d

require $V, 1 V,, + $(excess reauired V,) = V, t. 60 $b,d

therefore, design for V', = V, + 60 9 b,,d

use metl~od 1 when V, I 120 b,d

use method 2 when V, > 120 $ b,d

Section 2 - Reinforced Concrele Page 2- 139


2.68.0 Shear Friction Design (BDS 8.1 6.6.4)

Shear friction concepts shall be applied when it is appropriate to consider shear transfer across a pven plane, such as an existing or potential crack, an interface between dissimilar materials, or an interface between two concretes cast a t different times (BDS 8.3 6.6.4.1)

As slipping begins to occur aIong a cracked surface, the two faces of the cracked surface must separate a minimum amount in order to allow further slippage to occur.

As the two facesseparate, a clamping force is developed in the bars crossing the interface. The shear force is then resisted by friction which develops between Ehe faces (other forces also help to resist slippage, but are not discussed here).

The rest of this section conveys only enough information for shear friction design of simple components such as shear keys and beam supporbs. Components such as brackets, corbels and Imge seats are much more complex The BDS and ACl codes should be stud-ied thoroughly before attempting design of one of these items. The PCA publication, Notes 071 ACl328-89 isa goodsource of infomalion for shear friction design.


r-t. 7 Bridge Design Practice - February 1994

Potential Crack Locations

dapped ends

support bearing

V//I

corbel

column base

L

Section 2 - Reinforced Concreie Page 2- 14 1

=* - Bridge Design Practice - February 1994

2- 68.1 Basic shear Friction Requirements (BDS 8.6 6.6.4.4)

Use units of p o d s and inches.

When shear-h-iction reinforcement is peipendicular to the assumed crack location

When shear-friction reinfo.rcemen t is at an angle to the assumed crack location:

Avf 2 VI,

I$ f y ( ~ sin af + cos af)

Net fmsik fmces across the assu-med mckshaU be resisted by additional terision reinforcement.

Petrtlanentnefcmpressizr)e f ~ ~ c c s aaoss the assumed crackmay be utibzed, jncaldating the shear strength of the section.

Page 2- 142 .. Section 2 - Reinforced Concrete

C* Bridge Design Practice - February 1994

Applied Forces

V, = R, sin a,: + T,cos g

Nu = T, sin af- k c o s a,

A,,, fy cos at

Resisting Forces

V, =Avf$ (p sin uf + cos cq) if Nu = tensile force

-r5

Vn = A& (p sin a, .i. cos aj) + !flu if Nu = compressive form

Tu - rf/,rT r reinforcement i

Nn = Anfy sin af = nominal tensile strength

A, = factored support reaction T, = factored force due lo shrinkage and temperature

effects and other loads. T, = 0.2 R, is generally considered a minimum design load to consider.

crack loca!ion

TU cos ur T, sin a1

R, sin cq k Tu

Total required steel, A, = AVf + A,

Distribute steel unifornzly along potentjal crack plane.

Section 2 - Reinfarced Concrete , Page 2-743


2.68.2 Example - Shear Key

f', = 3.25 h i

f, = 50 ksi

V, = 200 k potenlia t crack

Ij'shear key concrete is phced monolithica.lly:

p = 1.4 A = 1.4 (1.0) =1.4

200,000 Ibs = 362 inZ c

rn.inMun7 A,,= 200,000 Ibs

= 294 in2

v,, - required A,., = -- 230

= 2.80 inz M,P (0.85)(60)(1.4)

Shear reinforcementn~ust be anchored to develop Ihe steels yield strength on both sides of the potential crack plane.

Often, the height of the key isnot sufficient to develop straight bars, thus hooked bars areoften used. It is common to use "U" b a s for this purpose. This will require that an even number o t legs cross the potential crack.

2.8 For #5 bars, No. of legs required = .- = 9

0.31

2 8 - 6.4 For #6 bas , No. of legs required = - -

0.44

Choose 5 - #5 "U" b m . The legs of the bars sluU extend beneakh the poteniial crack plane mf ficiently to develop the sped tied yield strength.

Page 2- 7 44 Seclion 2 - Reinforced Concrete


2.69.0 Compression Members (BDS 8.1 6.4)

Compression members can fail in three ways:

1. Compression failure - concrete crushes prior to tension steel yielding.

2 Balanced failwe - concrete crushes as tension steel yields.

3. Tension failure - steel yields prior ro concrete crushing (actual1 y, the failure is sti l l defined as the point at which the concrete crushes).

As the axial strength, P,, of a member changes, so does the flexural sl~englh, M,. An interaction curve relates P,, to M,.

To check the adequacy of a section for a set of required sl~engtl-ts, M, and P,, ala~uys enter the djagramwithkhevalue of P, first. Project horizontally ko h e curve, and then read what then~ornent strength is.

Given: required P, = 1300k required M, = 500k-ft

Find: far P, = 1300k Mn = 5SOk-ft

M, > required M,

Section is adequate

Note: If P,>Pb a compression controls condit.ion exist

If P,< P, a tension controls rondi tion exist.



2.69.1 Example For the section shown, draw t l ~ e interaction diagram for bending about the x - x axis.

d' = 3"

f', = 3.25 ksi

fy = 6 0 k s i X -

4 - #8 bars at each face

A, = 3.16 in2 for 4 bars 0 0 0 0

Far pure axial compression:

P, = 0.85 f', (As-&,) + ht 4 = 0.85 (3.25)(2g2 - 6.32) + (6.32)(60) = 1953 k

M, = o

For pure flexure:

P, = 0

a = *sfy -

- (3'16X60' = 2.86" (Neglecting A;) 0.8Sf: b (0.85)(3.25)(24)

= (3.16)(60) ( 21-- 22G)(A) = 309 k-h.



Balanced failure condition:

Tension steel yields just as the concrete crushes

87000(~ - d') f's = S 60,000

C

- - 87000(12.429 - 3) = 66 ksi + 60 ksi

12.429

C, = A', (f', - 0.85 f',) = (3.16)[60 - (0.85)(3.25)] = 180.9 k 1.~ T = A,fy y y (3.16)(60) = 189.6 k

P, = C,+ C, - T = 700.4 + 180.9 - 189.6 = 692 k

Compression failure condition:

Concrete u u s t ~ e s before tension skeeI can yield.

For th i s condition, c > c,,

Try c = 20" a = 0.85~ = 17"

C, = 0.85f',ab = 1127 k

f', = $7(c - d')

= 74 + 60 ksi C

c, = A', (f', - 0.85 f',) = 180.9 k

f, = 4.35 ksi

T =&fs=13.7k

P, =C,+C,-T=1294k

M, = C, (12-2) + C, (12 - dS) + T(d -12) =475 k-ft



Tension failure condition:

Tension steel yields prior to concrete ouslMg.

For this condition, c c c ,.

Try c = 6 " a--0.85c=5.1"

C, = 0.85 f', ab = 338 k

compression mnlrols

I --------------------

I tension cmlrois

(0.309)

Page 2- 1 4 8 Seclion 2 - Reinforced Concrele


Suppose a certain loading produces the following forces on the member.

P , = W 6 k

with r) = 0.70 for tied menihers

P,/$ = 1294k

Mu/$ 477 k-ft

From the interaction diagram at point A,

When P,, 1294, M, = 475 >Mu/$

Therefare, the section is adequate.

Suppose P, = 0

M, = 330 k- ft

P,/$ = 0

Mu/$ = 471 k-ft

From the interaction diagram at point B,

Therefore, the section is not adequate.

Note that tk section was adequate when a higher axial load was applied. Therefore, it would have been erroneoqJs to assume that reducing the axial load while leaving t l ~ e moment constmt would still result in an adequate section.

Section 2 - Reinforced Concreis Page 2- 149


2.69.2 Example:

End diaphragm abutment P"

h = 30" = thickness of the abukrnent

b =12"

d = 27.5" P, = 20"

f', = 3.25 ksi

fy = 60 ksi Mu= SOk

Q = 0.70

For equilibrium:

P, = C - T = 0.85 f',ba - A,f,

In the above equations, note t h e following points. P, is assumed to act through the sections pk3stic centroid which is estimated a t h/2 fmln the ccompression lace. In calculating M,, mornen ts must be taken about the plastic ce.nkroid. The compression reinforcement has been iflored for simplicily.

Solvc the above two equations:

Set P, = P,/$ and M, = Mu/$ (be careful of units)

Find a = I .366" 4 = 0,279 in2/ ft

#S bars at [ z ) ( 0 . 3 1 ) = 0.279 13.33"

Try using #5 bars at 12".


E* Bridge Design Practice . June 1994 1

Section 3 . Prestressed Concrete Design

Contents

Notations and Abbreviafions .................*.m..... .. .................................................... *...3-1

3.0.0 Preface ............................................................................................................ 3-4

3.1.0 Introduction . Basic Theory and Principies ................................................... 3 4 3.1.1 Defrnitionc ...........................................................-................................................................ 3-5

3.1 2 Advantages of Restressed Concrete . .................................................................-..+..-...... 3á

.......................................................... 3.1.3 Dicadvantages of Prestressed Concrete 3 - 6

................................ 3.1.4 Prestressing Force .......................+...........-........................................... 3-7

.................................................................................................... .... 3.1.5 Formulas fm Design ., 3-7

3.1.6 AUowable Stresses ......................*....................++.......-.......................................................... 3-9

3.1.7 Decign Theory .....................................+..-...............*......................................................... 3 1 1

.................................. 3.1 -8 Design Paxameters .............................................--..-.................... 3-1 1 3.1.9 Assumptions ...................................................-.......................................................... 3-12

r) 3-1-10 De~ign ............&..............+..--...-. *....... .............-......................... .................. .........-.....-...... 3-1 2

.................................................. ......................... 3.1 -11 Load Factms ., ..................-...-.+.-+.........- 3-12

........................................... 3.1.12 Strmgth Reduction Factors ...................... .- ....-...-.......... .... 3-12

3.2.0 Design of a Simple Span Prestressed Precast 66 39 1 Girder Bridge .......................................................................................... 3-13

3.2.0.1 Design example ...........-....-.................-...................................-....-.-...--+.-..+..-..d........ 3-14

................................................................................... 3.2.1 Design - Cross Section, Pretension 3-17

3 -22 Design - Net flransfonned Sedon, Pretension ..........................-...................+.....+..... 3-21

....................................... 3.2.3 Design - Net JTransformed Section, Post-Tension ................- 3-27

................................ ....... 3.24 Ultimate Moment ...........en.en..en........ . ........... .......................-...-... 3-27

3.2.5 Path of Prestressing Steel .....-..............................+.......-..............................d..................... 3-29

3.2.6 Shear ......................................-.....................................................+.........m.................-.....d.. .. 3-30 ..................................................**...... .*..........................*...*.*..........*............ 3.2.7 Deflections .....- 3-30


E* I Bridge Design Practice . J une 1 994 m

.................................... 3.3.0 Design of an "1" Girder Continuous For Live Load 3-31

3.3.1 Negative Reinforcement ................................. ...-.......-..... ,.. .......... ....................*.... ........ 3-32 '1 .................................. 3.3.2 Sumniary ........................ ....-..........-.............. . .......................-....... 2-36

3.4.0 Design of a 2-Span Continuous Cast-ln-Place Prestressed Box Girder ........................................

3.4.1 Introduction .............................. .............-......................-..... . ......-........ ....................... 3-36

3.4.2 Crms Cectional Geometry ................. .....-........ ..................... ................, .................. ..-..... 3-36

3.4.3 Loads .........-...................-............................................. ..........................,........................... 3-39

3.4.4 Longitudind Section - .............................................................................. ......... ........... -1 3.4.5 Losses .................................................................................................-........-...................... 3 4 3

3.4.6 Force Cmffiaen-ts ..........................................................................................................- ... 3 4

3.4.7 Cecondary Moments .-. .... - ....................... .. ..........-..++... .. .......... - .. .- ................................. 3-50 - 3.4.8 Prestress Force ................-......................................................................-.................--..*+.. 3-39

3-43 S t r e s ~ e ~ ........-......... - ................... ........-........ ........=.h..... -...-.... ....................................*....... 3-63

3.4.10 Concrete Shength Required ............-........+..... - -.. ..-.....-...... ...-......--......... ................. 3-67

3.4.11 Ultimate Moment ...........-......,...............~.........................................~-.~........,....,......-.. 3-67'

3.412 Shear ..............................-......-........-.......................... ......... .......-..........................-.......... 3-74

........ 3.4.13 BDS Computer Output .....-............. ......,.. .................................... .........................-.. 3-78

...................... 3.5.0 Design of a 4Span Continuous Cast-in-Place Box Girdsr -3-1 03

Introductim ......................................... .......................,............... ........... ........,.............. 3103

Bridge Configuration and Compment Properíies ...................................... &lo3

........ h a d s .. ........+......-*.+....... .... .... ...........................................................................-..-.... 3-105

Longitudinal Section ........................................................................................................ 3-1 05

LOS% .................... ...-.............- ................................... ................-.......*...... .................... -3-108 Force Cwfficient ....................-................-.............................-...................-................... 3-110

Cecondary Moments ..................................................................................................... 3-11 1

Pres tress Shortening ......................................................................................................... 3-1 14

... Prestress Forre .......-................ ................................................................................... .... 3115

-.................. ...................... BDC Cornputef Output ...............- .......................... .......-....... 3- 121

Page 3-N Secfien 3 . Contents

E* -1 Bridge Design Practice - March 1 993 M

Prestressed Concrete Desig n

Notations

"D"

= bearing area of anchor pIate of p ost-tensioníng steel.

= maximum arpa of the portion of the anchorage surface that is geometrically similar to and cancentric &ith the ama of the bearing phte of post-tensioning steeL

= area of nonprestressed tension reinfoxcement .

= area of compression reinforcement. = area of prestressing steel. = steel area sequired to develop the

ultimate compressíve ctrength of the overhangíng portions of the flange.

= steel area requircd to develop the compress si ve strength of the web ~f a flanged section,

= area ef web rwiforcement. = width of flange of h g e d member

or width of rectangular member. = width of web of a M g e d member. = loss of prestress due to creep of

concrete. = Ioss ef prestress due to relaxatim

of prestressing steel. = los5 of prestress due to relocation of

post -tensioning steel. = center of gravity crf entire concrete

section, = center of gravity of ttansfomed

~ c t i o n . = center of gravity of steel area. = effect of dead Ioad. = nominal diameter of prestressing

steel. = dictance to centroid of prestress

ducts as per Memo fe Desig~ters 13- 28.

= distance from extreme compressive fiber to centraid of the prestressing force, or to centroid of negative moment reinforcing for precast girder bridges made contuiuous.

= loss of prestress due tci elastic shortenúig.

= flexura] modulus of elasticity af concrete.

= moduius of elasticity oi psestressing steel.

= base of Napesian logarithms. = distance from center of gravity of

section to the centroid of the cable path.

= resulting stress in bottom fiber - final.

= resulting stress in bottam fibet - initial.

= resulting stress in top fiber - final. = resdting stress in top fiber - iscitial. = concrete stress in bottom fiber due

to W,. = concrete stress in top fiber due to

wa- = concrete stress in bottom fiber due

to w,. = conueke stress in top fiber due to

Wd- = average conmete compressive

stress at the c.g. cif the prestressíng steel under full dead load.

= average concrete stress at &e c.g. of the prestressing steel at time of release.

= compressive strength of concrete a t 28 dayc.

= compressive ctrength of concrete at time of initial prestress.

= Tnitial concrete stress after transfer of prestress force.

Section 3 - Prestressed Concrete Page 3-7

E* 1 Bridge Oesign Practice - March 1993 1

i r = permissible horizontal shear stress. V, = nominñl sheax strength provided

by concrete.

va = nomina1 shear strength provided by concrete when diagonal cracking results from combined shear and rnorned.

V, = nominal shear strength provided by concrete when diagonal cracking results from excessive principal temile stress in web.

Vd = shear force at section due to unfactared dead load (not including added dead load shear).

V, = factored shear fnrce at section due to extemally appljed loadc occwing simulkaneously with wm,

V, = vertical component of effective prestress forre at section.

Irg = nominal shear strength provided by shear reinforcement.

Y,, = shear applied due to prestress secandary rnoments.

V, = factored shear forre at sectim.

WJ = live load or superimposed Ioad applied after prestressing comple ted.

W, = wejght of concrete, lb. per cu. ft. W, = dead Ioad acting at trme of

prestressing. Y , = distame from cmtroidal axis of

gross sec tion, neglec ting reinforcement, to extreme fiber in ~ompression.

Y, = distance from centroidal axis of gross section, neglecting reinfarcement, to extreme fiber in tension.

a = total angular change of pestressing steel profila in radians fram jachng end of point x.

P = hction curvature coeffiuent. Y, Bnf B,, PPS = load facters. $ = strength reduction factor. CI = rotation angle.

Abbreviations

BDA Bridge Design Aids EDD Bridge D~sigri Detuils BDF Bridge Desigrí Practice Mmnol BDS Brfdge Desip Sysfem Manzinl MTD Memos fo ~wi$ers Spec. Bridge D ~ s I - ~ Specj(imtions

Section 3 - Prestressed Concrefe Page 3-3

E* Bridge Design Practice - March 1993 m

3.0.0 Preface

The subject of presb-es5 concrek is very important withjn the Division of Stmchires by reason of tfie vast n m b e r of ctructures in the state whch are designed using this techo1 ogy.

Yet, the subject is not dways a part of ournew employees undergraduate studies. It is for that reason that "BasicTheo~andPrinciples" along with definitions and acsumptionsareintroduced early in this chapter. To iilustrate design pracedure, several example problems are worked through. Further study Lnto prestress tehology ís encouraged.

Prestsess technology hasbeena part of the Bridge Design PracticeManualsince the 1960 edition, Over the years, revisions and updates have beenmade. Again, there is aneed for updating. Ths editisn reflects revised friction coefficients and detailed computer outpu t. In all of the example problems, m efffort has been made to include thti "longhand" solution where appropriate, with the feeling that the designer must know "what '5 going on" before he can intelligentlv use the many computer propams available.

3.1 .O lntroduction - Basic Theory and Principies

The bacic principie of prestressed concrete canbe described as fallows: Strecses are introduced inthe concrete oppesitein sign te thase resultrng from Ioadsacting on the structure, andin such a manner that ailowabJe wwcrrking stresses will not be exceeded. Compressive stresses, that are induced across the section under consideratim, counter tensile skressec €hat develop due to loads. These compressive stresses r e d t freman axial force and a bendingmoment transmi t ted to the concrete by the prestressing steel after the concrete has attained sufficient strength.

Prestressed concrete makes f d l use of the compressive strength of the concrete and the tensile strengfh of the prestressing steel. Ordinary reidorced conmete does not use the concrete to its full advantage- This is shown in the following comparison using the same allowable concrete stress (see Figure 3-11> In reinforced concrete, the resulhg stresses are compressive on the top and tensile at the bottom. This leadc to csadcs developing in the tensile regionmd renderingthe concrete stress zero while the steel takec all temion stresses. In prestressed concrete, large compressive stresses whichaxe the result of the applica tion of an eccentricdlylocated prestress force, reduces the tension sbess to near z r o .

With b e m c of the same depth, ayrestressed sectian can resist over twice the moment that the reinforced concrete cection can resist. Furthemore, the allowable workirig stsess can be doubled for the prestressed section, thus rnaking the resisting moment over four times that of the reinforced concrete section. The prestressed section rnakes use of the entire concrete area; whereas, khe reinforced section uses abou t of t he area and ?4 iC uced to hold the reinforcing cteel away fiom the workuig secaon, recht shewing S tressec, md develop the bond betureen the concrete and reínforang striel.

Page 3-4 Section 3 - Presfressed Can~rere


Remforced Concrete

f c

PIS Efiects Prestressed Concrete -

fc

Figure 3 - 1

Refer to Spec. Article 9.1.3 - Definitions for an extensive list of terms used in prestress technologv* A few addítional tem used in this text are defined below:

GzbTeShea7-Verticai component ofa prectress force whidi is ínclined to thehorizontal. Used in sheas calculations and generally reduces the effects of applied shear forces.

Cmtinum~sfOt. Live hads-Mdti-spanstruchet~tiüzing pr~cas t -pr~qtr~ss units with a rast- in-place de&. The top slab is reinforced auass the bents thereby making the stmchire, '~continuous for live loa&." See Article 3.3.0, Design of an "1" Girder Continuous for Live Load.

Crmtinuouc Sfmcfure - M d ti-span structure constnicted cantinu ous aver sweral cpans withou t expansian j oints.

Tlinge Curl - Unwanted deflection of the cantilever portion of a hinged span caused by application of the fuU prestress ferce to the member in its unloaded state.

Lightu~eighf Concrete - Unit weight o£ concrete that is 5 1% Ib per cu ft.

N o m l weight Concrete- Unit weight af concrete that is 1145 lb. per cu ft.

P ~ r t i a I Presfrescing - Tedinoloay whereh the strucke is designed with a combina tion of prestrecshg steel and r d d steel rejnforcing. See "A Design Procedure for Partial Prestressing of Concrete Box Girder Bridges" by Steven B. McBride dated January 30,1987.

Prestress Fmme - That porbion af a centinuous stnrcture between expansian joints to be prestressed for i t 'S total length.

P r ~ t r s s Path - Trace of prestressing tendon throughout the length of the membes.

Prestress Sho~tening - Elastic m d inelastic shortening of a member due to the application of the prestressuig force.

Ptimrinj P r ~ t r e s s híornaz t - Mommt resulting from the eccentricit y of the prestress tcndon.

X s u l t ~ l z t Momenf - ?'he resultíng moment due ta prestressing 3s the algebraic surn of the primary and secondary moments.

Sectiun 3 - Presfresced Concrete Page 3-5

Bridge Pesign PracrEce - March 7 993 m

Secondarr~Mornmt - Moment resultingírom induced reactions at the supports of aprestressed continu~us rnember. The term "secondary" is somewhat rnisleading cince the moments are not always se con dar^ in magnrtude m d therefore play an important part irt the stresses along the member.

Ulf imateMoment Clreck-Comparicon of amernber 'S flexura1 strength tofactored loads given in Table 3.22.1A of the Spec, Article 3.22.

Worklng Sfress Design -0therwise temed "Service Load Design"; the design of memberc by application of loads without load factors.

3.1.2 A dvan tages of Prestressed Concrete

f . Reduction of concrete and steel quantities.

2. Considerable reduction in depbh of section, not only relative to reinforced concrete, but alco relative to sttuctural steel.

3. Crackless contirete within a h w n range of load. Thlc results in greater durabdity under severe conditions of exposure.

4. Possesses m-um rigidity under working loads and m a x h u m flexibility under excessive overlaads.

5. Provides capacityto cupport a loadin excess of the'designloadínwhichcratks appear but disappear completely on removal of the excesc loa d.

6. Provides resistance to repeating and altemtíng loads even whm exceeding the design load.

7. Produce definite reductien in diagonal tensimwhich leads fo fewer stimps needed for shear.

8. During the prestressing operations, the steelis tested to a stress that will never agUn be reached undes designloads. The m e applies to the concrete, inmany caces. Thus, it can be said tha t the materiaIs 2n the stmc ture are tested bef o ~ e being subjected to h e working loads. This "in place" testing is impocsible in ordinary reinforced concrete sínchrres .

9. Makes it possible to control and/m reduce long t em deflections.

10. Added flexibility for constructien o f continuous mullí-spm h e .

3.7.3 Disadvantages of Prestressed Concrete

1. The prectress stmcture ic more sensitive to quality of workrnanchrp and materials.

2. Creep and shrínkage of the concrete and relaxation of the prestressing stee3 are important censiderations which need to be considered by h e desigrier. ThiC is especially so if Lighhveight concrete being used.

Page 3-6 Section 3 - Presiressed Concrete


3.l.4 Prestressing Force

There are two mefhods of applying prestressing force."Pretensíoning" indicates that tensioning of the steel is done before the concrete is cast in the forms. "Post-tensioning" means that t h e steel is twsioned after h e concrete fiasbeen cast and attained the required strength. In the forrner, the fwce is tsancmjtted by bond between the ste,el and concrete. T1ie initial prestress is immediately reduced due to the deformatian and shinkage of the concrete. Gradually &ese Iosses are increased by further s h r h b g e and creep of the concrete. Iri "post-tensioning" there are no irnmediate losces but there is a gradual losc due the shrínkage md creep of the concrete and the relaxation of the steel. Consequently, for equivaient members the 'pretmioning" me thod requires a greater initial prestressing force to compensa te for the larger losses.

"Pretensionuig" is prachcal o d y with factory or mass pr oductionfacilities, sisice pemanent externa] ancharages are xequhred to take the reaction of the stressed strands until the concrete a ttains the requiwd strength.

Severalmethods of strecsing and anchnring "post-tensj oned" steel are ín use. The methods used most c o m o d y inthe Unrted States at the present time ase illustrated in publications ismed by the Uidustry. The Office of Stmcture Construction publishes its '.Califonltn Prestress M m i m T " which ako contains illuctrations of current hardware.

3.1.5 Formulas for Design Cince the theory of prestressed concrete design amountc to superimposing stresses caused bv various stages and conditiom of loading, a general equation cam be written as follows:

P Pev F = - f - 5 f (General Prestressing Equation)

A I

F = resulting stress in the concrete.

y = distance from the cenboidd axis ata desired stage and candition of loading.

P = the prestressing force acüng at this stage.

A = asea of concrete or equivalent area of concrete used iri a particularmekhod of desip.

e = díctame from the center of gravity of the section to the centroid of the cable path.

1 = section moment of inmtia.

5 = the stress y distance fromthe centroiddaxis causedby stage and conditien of loading on the member. This stress is cornputed as if the rnember ulere a hamogeneous material.

MY That is, f = - (Le., -DL + LL + 1 Stresses) 1

Section 3 - Prestressed Concrele - - Page 3-7

E* I Bridge Design Practice - March 1993

For example, consider Figure 3-2. me gjrder of span length "k'%as just been prectressed. A t this s tage the oniy externa1 load acting is the dead Ioad of thegirder (the effect of prestressing is to lift the girder frorn fts soffitji. Therefere, the stresses in the top and bottom fibes at this time are:

fdtr fdb = Girder DL stress top anci bottom

F,, = resulting stress in tep fiber - initial

F,, = resulting stress in botlorn fiber - hitial

C.G. of section -

m * P

1

Section At Midspan

f- C.G. of prectrecsing force

Elevat ion

Figure 3-2

Page 3-8 Section 3 - Prestressed Concrete


Accumenext that,intimerthegi.rderhas lostpart of itsp~es.trescingforce. Thisloss is due ta the reduction of s ~ s s jn &e prestr~shg stel caused by theshrinkage and aeep of theconcre te,and the creep of the steel. Tlx Pi in Equations 1 and 2 intime become P,.. Now any laad superimposed upan thP girder changes the stress dictribution inthe sedion &o, the fmce P,is increased due to the fact tha t the girder deflects under the superimposed or Líve load. The inuease iK P, is negiected in the discucsion at this time cince, in the design example, t he effect of the increase in steel stress will be considered. The shesses in tfie top and bottom f i k r at the stage of Ioading comisting of the dead lmd of the girder and the cuperimposed or live load are as foliows:

where fa,, fa, = stresses due to added DL and LL + 9, top and bottrim.

Note: Tn equations 1 through 4, use only the numerical values fdt, fat, fdb, and f,,. The equations will give the correct sip. Aplus sign úidicates compsecsion and a m h u s cign indicates tensiun.

3.1.6 AIIowable Stresses (Spec. AriicIe 9.15)

A. Prestresaíng Steel

Stress at iinchorage after seatíng:

Pretencioned rnembers: far stress-relieved strand:

for low relaxatian strand:

Post tensioned members:

Stress at service load after losses:

where: for stress-relieved strand: f = 0.85 f ',

for low-relaxation strand: f = 0.90 f ' , (Jacking stress is 0.75 f 2 max at any t ime)

B. Concrete

l. Ternp orary stresses before locses due ta creep md shrinkage:

Cornpression

Pret-ioned rnembers:

Post-tensioned membexs:


r-É ' Bridge Design Practice - March 1993

Tension

nonprestressed reinforcement in the tension zone: zero

Olher areas

In iension areas with no bonded reinforcement: 2W psi or 3&

Where the calculated tensile stress exceedc t h i s vdue, bondcd reinforcement shall be provided to resist the total temion force in the conmete computed on the assurnption of

an uncracked cection. The rnruámum tensile stress shall not exceed: %5&

2. Stress a t service laad af ter losses have oc~utred:

Compression: 0.40 f',

Tension in the precompressed tensile zone

(la) For members with bonded reiniorcement: 6 & (2a) For Em7irmaital Area III and marine enviroment: 3&

(b) Dead Ioad tension: zero

(c) For memberc without bonded reinforcement: zero

To determine the working stress in prestress, the following Ioscec must be considered: (1) anchorage slip, (2) friaion losses due to curvature (p$) and wohble (KL), (3) elastic shortening of concrete (ES), (4) creep of concrete (CR,), (5) shrinkage of concrete (SH), (6) relaxation of presliesshg steel (C$). The table below gives total losses for ait causes except (1) and (2):

C. Assumed Losses (Spec., Article 9.1 6.2.2)(Tabie 9.1 6.2.2)

Pretemioning strand: low relaxationr 55,OCíI psi

n o d relaxation: 45,000 psi

Post-tensiening (friction losces not jncluded)

Wire or strand: low relaxation: 20,000 psi

normal relaxation: 3Zj000 psi

Bars: 72,000 psi

Page 3- 1 U Secfion 3 - Prestressed Concrete

r-Jt -1 Bridge Design Practice -- March 1993

3.1.7 Design Theory

l'restressed concrete members shali t-ie designed as follows:

;%e prcstressing force s M I be detemiined by ALlowable Stress Decign usirig elacfic tl-irory tíir loads a t the sewice level considering 115 Loads.

The ultimate rnornent capacity chali be d~eclíed by Load Factor Design using ultimatc strength theory for ioads at the factored level considering HS Load or 1' Load, whi&ever is greatcr.

Shear design shall be based an strength (Load Factor Design) using ~dtirr ia te strength theorv wi th factored HC Lciad or J? Load, \vhr.chever i~ great~r.

3.1.8 Design Parameters

A. Minimum Deptb to Span Ratios (BDA 10-26 thru 10-28)

For Cast-in-Place Prestressed Box Girder

Smplc Span D/S = 0.945

Mulli -Span Contínuous D/S = 0.040

Ha unched Stnicture

A t the bents DIS = 0.048

At midspan and abutments D/S = 0.028

For Precast Prestressed 1 Girder

Simple Cpm D/S = 0.055

Multi-Span Continuous D/S = 0.050

B. Minirnum f ', and f Ici

For post-tencioned members fMTD 11-3)

Use f '= = 4000 psi, f ",j = 3500 psi

For pre-tensioned members (Spec. Articie 9.15) Use f '; = 4000 psi, f ', = 4000 psi

C. Frame Length

F r m c s with coniinuous tendons may be economical up to 750 f t . cir morc h-t length.

Section 3 - Prestressed Concrete Page 3- 11


D. Single End Stressing

Desiper should consider single-end stressing for Srames of k o , three, os four spans or where the Tncseace of prestressing forre is less than 3 percent.

3. J . 9. Assumptians The follewing assumptions are made:

1. Ctrain varies iinearly over the depth of the rnernber throughout the entire load range.

2. Before uackmg, stress is linearly proportional to straín.

3. After cracking, tension in the concrete is neglected.

3-7-10 Design

Bridges shall be investigated far stresses and deformations at ea& loadíng ctage that may be critica1 during constniction, handling, transportatl on, erection and &e service Me.

Load Factors g and b are mdtiples of the design load thus allowing for variation in loading of the struc ture and assuring its safety , The Required Str@ngth U is then ob taimed:

Mu, V,, LOAD,, STRESS, = y [Ib,)DL + l3d-L + I)3 + (pm)ps

where: y = load factor (Spec., Tables 3 . 2 1 A and 3 -22.1B)

p = coefficient (Spec., Table 3.22.114)

Savice Load Design; y = 1-00, = 1 .O, PG+ IyH = 1-00

Load Factor Design:~ = 1.30, P, = 1.0, P, + vH = 1-67, P,, + ,,, = 1-0

3.1.12 Strength Reduc tion Factors

Reductim factors (4) are related to material and are called Strength Capacity Reduction Fartorc in Spec. Article 9.14, or Strength Reduction Factors in Spec. Article 8.16.1.2. Reduction factors assures that a conservative estimate of the actual strength is used. Calcubted capacity of the rnember ic usualiy called Nominal C t r e n , ~ S,.

The produa of Nominal Ctrength (S,) and of the Strength Capacity Reduction Factor (1) is called Design Streq$h (%)of the member and in general t e rm may be expressed by the following equation:

Page 3- 12 Section 3 - Prestressed Concrete

E* -4 Bridge Design Practice - March 1993 1

In tems of moment, shear, or laad we obtain:

M, 5 &M,

v, 5 u, P, 5 4 P n

where: 0 = strength reducticin factor (Spec., Artide 8.16.1.2 and Article 9.14)

= 1.00 fer precast presfressed

= 0.95 for cast-in-place post-tensioned

= 0.90 for shear, torsion

S, = general t e m for nominal ctrength

V, = nominal shear force

P, = nominal axiaI force etc.

3.2.0 Design of a Simple Span Prestressed Precast " 1 " Girder Bridge

In designing prestressed strucbnec there are vario= factors wkiich influence the type of member, concrete stresses, and prestressing force to be used. In some caces where the mtumurn depth is the important factor, the limiting concrete stresses ceuld very well determíne the design. Hawever, where the depth is not critical, the detemiriing factors may be the qmntity of concrete compared to the quantity of prestressing s t e d It is quite apparent that a shallow depth, while requiring less concrete, will require more prestressing steel than a deeper section. Jn other words, for any site there arevarious combinatiom to be d-iecked in arder to obtain the most economical structure.

It is not the purpose in this section fo discuss the economics of the various types or desi,-. In the design ta follow, it is assumed &at the stnicture selected 1s the one best suited by site requüements, economics, and other factors.

On the following pages, calculations for the Standard "1" Girder are used to demonstrate the design of a precast prestressed girder with a composite slab. The calculations for any precast prestressed @der wodd be similar. The Standard "I" Section has been f m d to be best suited for arninimum depth to span ratio of0.055 for simple spanc and 0.050 for rontinuous structures, and 4th girder spacing frem 7 ft. to 8 ft.

The gross section properties are used first in the calcuhtions, followed by solutions using net/ transfomed sections. Most of our prestressed concrete designs allow the use of pretensioníng or thevarious systemsof post-tenciothg. However,pretensi~ning isthed-ioice most contractors will elecf d e s s other constraints east su& as girders too long to be transported or too large a prestressing force for the casting bed.


E* Bridge Design Practice - March 1 993

The net/ transfomed section will vary depending on fhe nurnber, size, and location of tendon holes for post-tensioned units; and the area and location of preslrescing steel for pretensioned units.

These variations complicate the netStransformed design of a @der.

It shodd be noted that computer programc forprestíessed concrete *des5 make it practica1 te design using net/trmsformed section properties.

At this time, bethpost-tension and pretensionsteels aremost ~amonlysuppIied wíthnaminat ultimate strengbh vdue of 270 ki. Thic value of f', is used in subsequent examples.

3.2-0.1 Design Example

32-0' - -

Type 25 Concrete Barrier

I

Trpical Section

Design Span = 68'

Depth / Span = 4.17' = 0.061 z 0.055 OK 68'

Figure 3-3

Note: We will designa SpicaI interior girder. For section properties of standard 1 gjrders see BDA 6-1.



Deck Slab

D e s i ~ Cpm = (7'4'') - (1 '-7") = S-9" @DD 8-30)

Top Slab Thícknesc, t = 7lhin

Loads and midcpan maments - lnteriar

Weight of concrete = 150 lb/ft3

Area of Type 25 conmte barrier = 2.61 ftl

Weight of Asphalt Concrete (AC) surfaring = 35 lb/@

= 494 lb/ft Use 0.50 kips/ft

DL Slab= l(0.594 x 7.33) + (0.08 x 1,38)] x 150 = 672 1b/ft Use 0.67 kips/ft

= 224 lb Jft Use 0.22 kips/ft

Total DL - 1.55 kipc/ft

wL2 For simple cpan, M,,, = - For live load moments in simple spans see BD A 9-1 and 9-81 -2.

8

MDL (slab) = =387kipft=4.65x106inIbs 8

(MLL+ jHs = MaximumMoment HS Loading, one h e x Impac t* x Number o f h e s per girdef"

(MLL+~P = 1560 x 1.259 x0.666 = 1309 kip f t = 1 5 . 7 8 ~ lo6 in lbs

Section 3 - P restressed Concrete Page 3-15


The sequence of impocition of stresses on the girder is as follows:

a. Tmdons are sfressed within the form far f d dead and live loads, the girder is poured and after the concrete sets the cables are cut, tmmfesring stresses to the girder (Figure 3-4,).

b. The gkders are erected on the struchire supparkc and the top slab is poured. Additjanal stresces are introduced into the girder due to the slab weightL prior to the setting of concrete, the shb daes not contribute to the section modulus (Figure 3-4b).

c. Afier the slab sets, í5 ~ c t s with the girder as a cornposite member and helps to resist stresses by adding to the sectionrnoment of inertia stresses are due to fulll dead Ioads and live load c (Figure 3-44.

Barrier &AC DC Slab DL

f a t f b t ? C t Loads: PIS i DL (Girder) PIS t DL (Girder + Slab) PIS + DL (Gird + Slab + Barrier. etc) \ : 1 gird 1 gifd 1 comp

Figure 3-4

*lmpact, 1 = 50 -- 50 = 0.259% (Cpec. Article 3.8.2.1)

Lt125 í;8+125

Girder Spacing 1 Lane - **Nmber of Lanes per girder = x Wheel Lines x -- =0.i;&

5.5 2 Wheel Limes 11 (Spec. Table 3.23.1)

Page 3-16 Sedion 3 - Prestressed Concrete


3.2-1 Deslgn - Cross Sectíon, Pretension

Properties of Girder only - Gross Section

Stresses at this stage and after t h ~ pouring of the slab, at the top and bottom od girder arc

MY computed from and f =- are as foll~ws: 1

Tap Fibet I+i) Bottom Fiber (psi)

DL Slah

Total DL = 1881 psi (Compression) = 1710 psi (Tencion)

(Note: Carry these stresses forward to the summation of DL + LL stresses.)

Properties of CornpociteSection (Girder & Stab)

Assume fdet thichess = ?4 .

Effective Flange Width: (Spec. Article 8.10.1-1)

6 x t, = 6 x 0.594 = 3.564 fi

%! S = M (5.75) = 2.875ft < 3.5ó4ft use 2.875ft

68 Total width = 2 ~ 2 . 8 7 5 + 1.38 = 7.35ft c 1 / 4 C = - = 17.0ft

4

Therefore, 7.33 f? contrak - use this as the effec tive flange width. Assume t he dimension of the bottom of the girder to the C.G. of the prestressing steel (cgs) to be 3% in. at midspan.

Then,

Section 3 - Prestressed Concrete Page 3- 1 7


Properties of Composite Section

Girder 474 20.00 9,480 189,600 93,000

T about the bottorn of girder = 1,541,834 + 97,653 = 1,639,487 in4

so:

I,, = 1639,487 - 1353953 = 286,134 iri.'

For hvestigation of fiber stresses at ttie top of the girder:

Y, = 42.0 - 35.06 = 6.94 in.

and, similarly, at the bottom of the girder:

Y, = 35.06 in.

Top of Girder (pd) Bottom of Girder (psi)

Total DL + LL = 2177 psi (Cempression) = 3204 psi (Tension)

+Previously calculated sbesces, see page 3-17

Page 3- 1 B Section 3 - Pres tressed Concrete


Determine Prestressing Force, P.

Decign stress, DL only = 2033 psi (no tension allowed under DL's Design for the larger

Design shecc, DL + LL = 3204 - allow ten = 3204 - 6& - 1 of the two stresses.

Thereforer f,, = 2825 psi

Tkis stressis contered by that induced crn the girdarby prectressing. At first, the girder will resist the prestress withaut the contribution of t h e top slab ts the sectionmodulus and:

P = 505,965 lbs., u ~ e P = 510 kips. (MTD 11-8)

It should be noted ihat in most pretension girder designs, the total stress (1 DL + LL) controls the final force. However, i t can bepredícked that jn certaln shctures, cuchas a pedesrrian overmossing, where live loads are c m U , dead load stress will be higher fhan the total stress less allowable tension and therefore control the pretensioning f orce.

We will complete the designforthepre-tensioned @rderonly,nating that theprocedure for a post-tensionedbeam is similar and that o d y the losses inthe psestressing steel are handled differently.

Effective Sbess = 0.75 f', - Assumed Losses (using low lax strands)

Effective Stress = 270 x 0.75 - 35 = 167.5 ksi

From above: P = P, = 510 kipc

At fhe t ime the girder is stressed a pertion of the total loss cif 35 ksi w U have occurred. We ascume h t h s amount is 13 ksi. The remainder, 22 kci, will be lost after initial prectressing force, pi, has been applied.

Therefore, for the purpoce of cornputirig f ';

Pi = P, -t A', (losses after transfer) = 510 + 3.04 x 22 = 577 kips

Sectian 3 - Presfressed Concrete Page 3-1 9

E* ' Bridge Design Practice - March 1993 m

x1000+8(34=-183 psi (Tensien)

x1 0Da+2177=1304 psi (Compression)

x100&731=2491 psi (Compression)

Table 3-3

Concrete Strength Determination (f ',} (Spec. Article 9.1 5.2.1)

T e m p o r q Stress Conditionc: 149 1

Comprecsion; f',, = - - - 4151 psi, use 4200 psi minimum. O. 60

Tension; allowable stress = 3 or 200 psi maximum.

3 44700 = 194 psi.

183 psi e 194 psi - OK.

Design Load Stress Conditions:

1304 Compression; f', = - = 3260 psi < 4200 psi, therefore use

0.40

Y,= f',¡ = 4200 psi mhhum.

Page 3-20 Section 3 - Presfrecsed Gancref e

E* + Bridge Design Practice - MarcR 1993 m

3.2.2 Design - NeVTrans formed Section, Pre fension (Holes f or Prestressing Steel)

It is logical to assume that Ft could be less in a netltransfomed cection desigm than ui khe preceding example. Using the gross sectirin design as a guideline, estimate that P, will be appraximately480 k. Since this is apretensioned member, the controlhgvahe fox working stress in the steel is 0.75 x 270 -35 = 167.5 ksi and for the purpose of design, A*, = 480/167.5 =2.87in.2.Use anaveragevalueofn= 7,andagainatmidspmassumethatcgs=3'h~.above the bottom of the girder-

Rroperties of the Girdes Only - Net Transformed Sectien

Girder 474 20.00 9,480-0 169,600 95,000

Bquiv. Conc. Area

&(n - 1) = 2.87 x 6 17 3.5Q 59.5 20 8 - - - - Z= 491 Ui? 9,539.5in3 189,808 in? 93,000 inn4

1 about #e bottom of @des = 189,808 + 95,000 = 284,808 in.'

y, = 42.0 - 19.43 = 2257 in.

e = 19.43 - 3 3 = 15.93 In,

Stresses Net Transformed Section

Top Fiber (psi) Bottom Fibet (psi)

DL Girder 3.47 x l O h x2257

= 788 99,443

Total DL = 18-43 psi (Compressian) = 1587 psi (Tensjon)

SectEon 3 - Prestressed Concrete Page 3-21

r-t: Bridge Design Practice - March 1993

Properties of Camposite Sedion

-.

Girder (Net/Trans.) 491 19.43 9540 185,365 99,443

Slab (88 in. x in.) 627 46 -4.11 29,118 1,35 2,234 2,633 - S = 1,118 h.' 38,658 1,537,599 in? 95,000

I about the bottom of girder = 1,537,599 + 102,096 = 1,639,695

(EA) y 2 = 1,118 x M.5B2 = 1236,878 in.4

and

Ir,. = 1,639,695 - 1,336,878 = 302,817 in!

For investigation of fiber stresses at the top of the girder:

yk = 42.0 - 34.58 = 7.42 h.

Stresses - Composite Section

Top of Girder Ipsi) Bottom of Girder (psi)

Previously calculated DL = 1843 * =f567+

Total DL + LL = 2141 psi (Compression} = 2981 psi (Tension)

*Stress carried over from girder only calcula tion.


rZ Bridge Design Practice - March 1993 1

Allowable tension at design load after losses = 6 8

Assume f', = 4000 psi

= 6- = 379 psi

Then, try 379-psi tension and using the net/transfarmed girder &ea:

f,, = 2981 - 379 = 2602 psi > 1889 pci (DL + added DL only)

Therefore, use fd, = 2602 psi.

2602

'£=( 1 . 15.93~19.43) = 505,146 Ibs. Use 510 kips.

Since this is about the same Pf as the psevious soIution, we wilInot redesign againas the answer wdi nof change signhcantlv.

fs = 0.75 f; - 35 = 167.5 kci

and for the purpose of computing f',;

P i = P f + A , ~ L o s ~ ~ = 5 1 O + 3 . 0 4 ~ S 2 = 5 ~ k i p s

FQT e = 19.43 - 3.5 = 15.93 in.

Section 3 - Psestressed Concrete - Page 3-23

c* Bridge Design Ptactice - March 1993 m

xl000-678=2293 psi (Compression}

Table 3-2 2293 f ' . = - = 3622 psi

" 0.60

Use f', = 4000 psi

Cment practice allows B a t pretencioned beams may be comtructed with as few as twa holddown positiom. The harping of these tendons near the girder ends help lo control deflectionc and tensile stresses in the upper flange. me path 05 the prestressing steel might then be:

By inspectlon we can sep that at the L J3 points, the stresses (from the prestrecs effect alarte) are unchanged from their values at midspan. Yhe stresses in the composite sec tion, however, are reduced in the ratio of reduced rnornent-

Page 3-24 Section 3 - Presfressed Concrete


L W x(L J 3 ) x C L - L / 3 ) MIL? Eor the wufarmly djshlbuted dead loads, morngnts at ;- = or -

3 2 9

wL2 wL2 anci the ratio of L/3 point stresses to midspan stresses is -+- or 0.89.

9 8

For a 68ft span it can be shown (infiuence h e s ) that LL + 1 moments @ L /S oqual approximately0.91 of the rnidspan values. For spanlengths in gmeral, i t is sufíiciently accurate to use the ratia 0.89 for all L/3 point stresses.

Then:

Table 3-3

Concrete Strength Requirernents

Temporary stress conditiuns:

2368 Compressicin; f', = - = 3947 psi > f', at midspan, hawever less than 4000 psi.

0.60

Tension; allawable stress = 3x or 344000 = 190 psi ar 200 psi rna-

Maxirnum temporary tensik stress = 21 0 > 190. It rna y be assumed 2 4 4 mi Id reinforcement will resist 20 pci tension (see belaw).

E* ! Bridge Design Practice - March 1993 m

1101 - 2753 psi @ L/3 point < f', (3340 psi) at midspan. Comprecsion; f', = - -

O. 40

Use f, =yn = 4000 psi.

The fibet stresses at the L/3 points were -210 psi and +2368 psi at the top and bottom of the girder, respectively. With these sbesses, fICi would appear to be govemed by the temporary temile stress in the top fibers of h e girder, with a required value of 4900 psi (f',i = (210 / JI2= 4900psi). ( 1 ) Speufications, however, allowthat awuliay nonprestressed rehforcement may be provided ta resist the total tension force (computed on the assmption of anuncracked section) provided the tensiie s w s s does not exceed 7 . 5 8 .

If fhe other controlhg stzesses permitted theuse of 40OOpci c~ncrete, it is more sensible to provide this reinforcement tfian to raise the vaIue of cuncrete strength above 4000 psi.

Locate neutral axic, x (cee Figure 3-5).

Condition of Stress:

2368 psi

figure 3-5


E* 1 Bridge Design Practice - March 1993

The upper portion of the girder cross section is:

We codd be slightly more accurate, but it is conservative and sufficientl y correct to say that the total tension force F, = ('h x 21Q) x 3.42 x 19 = 6823 lbs. Using nonprestrecsed reinforcement jmild stel) at a working stress of 24 ksi, the required steel asea = 6823 / 24,000 = 028 sq. in. whidi we would supply by using tw o (2) Ien@ #4 bars. (A, = 2 x O. 20 = 0.40 id). With the addition of th i s reinforcemerit, the design to thic point will be adequate using f', and f', = 4000 psi.

3.2.3 Design - NetfTrans formed Sec fjon, Pos f-Tension

The design of the girder for W e conditions would folEow the pattern of the preceding example and will not be repeated here.

Listed below are some differences Y1 the design approach whi& shodd be considered:

1. Girder praperties wauld be aosc section properties less the effects of the assumed number, s i z , and location of the tendon holes.

2. Itkn~ces~to~vestigatefiberstressesonlyatmidspmsincethetendmw~lldrape in an approxirna tely parabolic shape from anchorage to anchwage.

3. Post temion losses due to cablerelaxationare diffesent and there are additional losses due to anchorage slippage,

The girder des@ of Section 3.2.2 will be used in a tp ica l exampIe of the required check for ultimate flexud capacity of the girder. Refer ta Spec. Articles 9.1 7 and 9.16.

Sectiori 3 - Prestressed Concrete Page 3-27

E* - Bridge Design Practice - March 1993

A*, = 3.04 inZ

b = 7-4"

d =50"-3.50"=46.50m. or3.88ft

f, = 3250 psi = 3-25 ksi (deck concrete)

f, = 270 ksi

then

{Spec. Artide 9.17.4.1)

= 262 ksi

The neutral axis (at uItímate load) is assumed to be in the web ifthe flange thickness is Iess than

1.4 dp'f* $11 - 1 . 4 ~ 3 . 8 ~ ~ ( 7 . 4 4 x l 0 ~ ) ~ 2 6 ? ~ 1 2 - (Spec. Article 9.17.3) f 'C 3.25

= 3.90" < t = 7%" and the N.A. at u1 timate load is in the flange, therefore the design is treated as a rectangular section.

Mure of the steel rather than of the conmete will occur at dtrmate load and &e u l t ima te flexura1 strength of the section is taken to be

where L$ = 1 .O0 for precast members (see page 3-13)

(Spec. Article 9.17.2)

Page 3-28 Seclion 3 - Prestressed Concrete


= 2960 kip fi

Load factor rnoments applied (Spec Artirle 3.32)

= 29 80 > M,,(H) = 2890 kip ft OK

Had the value for M,, beein significantly smaUer thanload factor moment applied, i t would be possible to use nonpresbessed reinforcement ínthe mmner dowedby rhe Spec. Article 9.19. The contrlbution of tlus reinfarcing to the tensile stsength of the beam (at ultimate) can be computed from the relationship:

1 6 ~ b 1 (1) M',, = i) &f,d 1 -? (Spec. Artlcle 8.16.3-2.1)

where f, is taken as 6íI ksi

The ideal path for the prestressing steel iri a simple beam is parabolic with as much eccentncity from cgcat midcpan as is pessible. The reason for this should be evídent in that the moment envelope for dead and live loads is also approxiniately parabolic. In a pretensioned gjrder, the parabolic path is approximated withharping points located at two ar more positions irt the span length.

At the anchorage ends, eccentricitv from cgc may either be zma or come value above or below cgc which rmains in the kem h&t, so as to h u r e comprescion over the entire section. Kem Iimits for pxestrecsed t>eams are: I /Ay, as the limit below cgc, m d T/Ay, as the limit above cgc.

(1) Thic must use an equivaicnt comprescion blo& that is a combina tion of p/s and d d

s teel

a= A *' +As'y or equivalent equation for flansed sections. 0.65f' b

c

Section 3 - Presfressed Concrefe Page 3-29

E* 1- Bridge Design Practice - March 1993 m

Changing the eccentricity of cgs at the ends @ut remairung in the kern) is an e ffective rneans of controlhg the deflection pattern of the be- and is a method oftenused to establlsh the condition that DL deflection and prestress uplift are equal vaiues of opposite sense.

3.2.6 Shear

Due te the complicated nature of the prestress shear c a l d t i o n , as introduced h the 1977 AASHTO, this topic is campletely covered under 3.4.12 "Design of a 2-Span Continuous Cast-In-Place Prestressed Bax Girder - Shear . "

The upward deflection due to the unífornload andend eccentrícily of thepost-tension force is calculated by the equa tion:

P LZ * =ziz (5q + 6e,) when e, í s below girder center of gravity.

P L2 p = - (5ez - 6e,) when e, is above @der center of graviv.

48 El

CG Girder

The upward deflection due to the concentrated bads at the hrping points and end eccentricity of fhe pretension force ic d d a t e d by the equation:

P L2 Ap = - ( 2 3 q + 27e,) when e, i s be l~w girder CG.

216El

Y LL Ap = - (Be2 - 27e,) when q i s above guder CG.

216 EI

For defiection calculations assume:

E, = w : d 5 x 3 3 f i

Page 3-30 Section 3 - Precfressed Concrete

E* + Bridge Design Practice - March 1993 m

Deflections due to dead bads are:

At the time of stsessing the midspan of the girder will rise by an amount equal to (APi - ADLg,,). Because the bottom fiber ismow highly stressed than the top fiber, the girder will continue to rice with the passage of time (tk is tme even though the prectress f ~ r c e decreases}. T h e total arnount ttm t the girder will rice is d ependent on its age at the time the unif orm ioad of the slab concrete is applied.

Mowances canbemade for dis situation using the following values for caefficimt of creep.

Time Coefficient of Creep 2 weeks 1.25 9 weeks 1 .so

20 weeks 1.75 It is conmon for to assume in camber calculatiom, that t he coefficient of creep is approximately 1.5 and that the prestress force has reIaxed to a valile of about 0.89 P;.

3.3.0 Design of an "I'"Girder Continuuus for Live Load Mos t rndti-span structurec using precast prestrecsed girders are designed for con tinuit y under live Ioading. This elirninates bearings and expansion joint details at the bents and provides riding qualitzes superior to those of simple span ~onctmction There are alsa the added advantagec of reduction of presWssing force, ~eductim of required concrete strengfhc, and some overali economies redting from the simplification of details.

Assume the use af the same typical &m and standard girders ac were used in the previous m p l e (9ction 3.20).

Using a constant mommt of inertia, üw frame shown (whm loaded to produce m u m ~onditíons) yields the foliowing I i ve loa& plus impact moment S and DL,, rnoments (se Table 3-4).

Section 3 - Prestressed Concrete Page 3-3 1

E* Bridge Design Practice - March 1 993 m

Reinforcing for negative moment is pJaced in the top slab in structures of this -e. Jhe gir derr will resist dead load stressesas simple spanmembers, bukwill act ascontinuous frame members when loaded by the rail, AC sur face, hive and impact loading.

Table 3-5 indicates the decign mwientc for these girderc.

The girder design procedure Is the same as shownpreviwsly and iherefore willnotberepea ted in thís example.

Table 3-5

The remajning element of design for thh type of stnicture is ta detemiine the amount of negative reinforcing over the bents as required by the loads for whch the shcture will act as a cmtinuous frame.

Assume that the designof the girders requires f', te be4000 psi. We will designbyultimate strength theory using load factors of 1.3 .t 5/3 (L + l)H] or 1.3 ID + (L + I)J whichever produces the greater rnoments iri order to be consiclent with the design of the gírders.

For this example assume that the "conpression flange" sectionof the I girder hasan effect ive depth or h, = 8.75 in. so that the analysis may be based an a rectangular section condition.

Page 3-32 Sedion 3 - Prestressed Concrete


n-ie ultimate resisting moment of rectangular sections with tension reinforcement only is calnilated by:

where

a = ky As , $ = 0.90, d and a are in inches and A, is in square inches.

o. 85 f; b

= 1.3 (173 + 1188) = 1769 kip ft-controls

M, = 1769 kip ft

d = 46.5 in.

b =19in.

b, =7in.

h, = 8.75 h.

f ', = 4000 psi (PCFS Girder Concrete) (MTD 11-31

Note: A, can also be found by solving a qwdratic equation formed from the above reiationships between A, and a.


E* Bridge Design Practice - March 1993

From reinforccd concrete design,

O. 85f,"(b -b,)hf A,,, = 0.0314 b,d + 0.75 4, = 0.0214 b,,d + 0.75 x

C

0 .85~4 .0~ (19 -7 )~8 .75 km, = 0.0214 (7 ~46.5) + 0.75 x = 11 -43 > 9.32 in? - OK 60

Assume additiod #9 bars are to be used as mild steel, in coqjunction with the ctintinueus top ma t longi tudinal slab bars shown on page 8-30 o f B r i d p Design Detnils.

Try 9- #9 'S in additíon to the cmtimtous slab bars far ea& girder.

A,=9~1.0+0.80=9.80>8.30in.~-OK

M,, = 225 x [(2 ~ 4 6 . 5 ) - 9.101 x 9.80 = 1850 > 1769 kip ft - OK

Therefore, add 9- e ' s to slab reinforcement far ultimate mornent. At least of negative reinforcement shdl be extended pastthepoínt of inBection, 9.32/3 = 3.11 in2. Make 3- m's con~uuous and place over ea& girder.

Cee Figure 3-6 fwrnegativeultimate moments and reinforánglayout. 'Bar cutoffs wodd be govemed by the specifications far reinforced concrete, including extensions beyond the point of inftection,

Page 3-34 Sectíon 3 - Prestressed concrete

c* \: Bridge Design Practice - March 1993

Bent Syrnrnetrical about

7-%9 nnrl nthpr c~ntinuous bars -31 , :Sprextension

ec. Article 8.24.2.1)

I 850 kip )t l t i i 14-

9-#9 and other continuous bars

Negative Moments (Ultirnat e) 1"=4400 kipft

Figure 3-5

Secfion 3 - Prestressed Concrete P age 3-35

Bridge Design Practiee - March 1993 m

No attempt is made in this; section to cover aU steps of the design. For those not covered, the analysis wíll be s d a r to the illustrations in Cection 3.2 "Design of a Simple Cpan Prestressed Precast I Girder Bridge". Arnong thece items are: Girder Design, Deflections, Shear, Ultimate Mommt Design.

3.4 Design of a 2-span Continuous Cast-ln-Place Prestressed Box Girder Superstructure

The designof continuous post-tensioned concretemembers differ from the design of precast members in that secondary moments are induced in the members due to prestressing. Precast meabers are generally desiped to be símply supported for dead load, and continuouc for additional dead load and bve load, whereas the cact-in-place box gsrder is designed as continuous for aU loadings. Also, the cable pathis usuallymuchlonger, and the effectc of friction Iosc due to the angle change play a role in the design process. Certain design Cnteria and specifications also differ between t he two types of design.

This section proceeds by hand similar to the way you would i n p t jnlo the BDC computer progmm. 13ridge geometryisfound, s t ruc t~e load in~ is calcu1ated.a cablepathic jntroduced, prestress losses, stresses, and cable forre are fhen calculated, and ultimate moment and shex are checked. The reader s M d always hand check some of the calculations produced by BDS, and use the plotting aptiom to graphically check results.

TIit! desigier usuallyhas limited infomation when tk design proccss begins. For purposes of thic example problem, ttie following infomation ís given:

* Continuous 2-spm cast-in-place box girder cupermcture.

+ Span len,$hs o£ 162 and 150 feet.

' EOD to EOD width of 37' - 6"

* Columri longitudinal rnoment of inertia given a 92 ft4.

The elwationview of this frame is shownin Figure S. Civcn this preliminary infomation, i t is the designer's j ob to develop tfie typical and E ongjtudinal sections.


E* 1 Bridge Design Practice - March 1993 m

Etevation View

Figure 3-7

The deipth of a superstmcture is a function of the span length. The dep~- to -van ratio equates the depth of the superstnictue to the length of the span. The AhCHTO r e c o m d e d depth-tospanratio for CIP post-tensioned box girderc is 0.045 for single span superstmthires and 0.044) for continuous spans. ?'he suggested stmcture depth of the example problern S:

D = r * L Vllhere: D = Stnrcture depth (ft)

r = Dep th-to-span ratio

D = 0.04 * 1623 = 6.48ft

Use D=6'-6"

L = Span Length (ft)

When span lengltis are of M a r length on the same stmcture, is usually a good idea to use the same depth for the entire sSnicture.

Sectron 3 - Presfressed Concrete Page 3-37

E* ,e Bridge Design Practice - March 1993 m

3.4.2.2 Girder Spacing

?'he spacingcif girdersis a funtion of the depth of the stnicture. ForaCIPpect-tensianed box structure, it is suggested that the girders be spaced not more than 2 times the simchire depth. The suggested rnaximum gitder spacing for the example problern k:

Sc2"D Where: C = girder spacing (C/L to C / t)

D = Structure Depth (ft)

C< 2 ' 6.5ít = 53 ft. Maximum

Using an overhang width of 4'- O", the C/L to C/L dictance bei-ween the exterior girders is :

37-5 ft - 2 (4.0 ft) - 2 (0.5 ft) = 28.5 ft.

T ~ i n g 3 girders,2bays,weneed28.5 ft/2bays= 14.25 ft,gidder spacing, whirhisgreater than ihe suggestedmaxímumvalue. If we try 4 girders, 3 bays, we have a girdex spacing of 25.5 ft/3 bays = 9.50 fit which is within the suggested range.

3.4.2.3 Typical Sectian

The width of t he overhang í s 4 ' 0 " from the face of &e exterior girder to the EOD. The thichess of SLiC otierhang shaU be 7 in. at he EOD, flaring to 12 in, at the face o£ the exterior gkder. AU girders sMl be 1'4" wide, with the exterior girders flanng tc? 18in a t tfie anchorages. The length of this p d e r flare shall be 16ft, and the stsuctural effects of the 6in. increase in web width wiU be neglected in future analysis.

The de& and soffit thichess is a function of the dear distance behwen N o adjacent guders- The dear distance for the example problem is 9'4" - (2'-6") = 8'-6". The Bridge Design Detaik rnanual cantains a table showing the design thíckness of the deck and so ffit as a function of dear spacing between girders. These thirknesses me a product of the AASHTO specifications. It should be noted that mud-r of the steel contained in the superstnicture is also designed as a function of girder spacing, and ic shown in the tabk mntioned above. The de& and wffit thichesc corresponding 20 a dear distance of 8' - 5" are 8.125 in and 6.373 in. respectively, Fow-hch fillets are to be located between perpendiculaf: surfaces except for thoce located adjojning the saffit. Figure 3-8 shows the typicd section based on th~ above design proredure.


E* Bridge Design Practice - March 1993 W

Typical Section

fype 25 Concrete Barrier

3" Future A.C. Overlay

Figure 3-8

3.4.3 Loads

After the development of the elevationview and typical section, &ese isenoughinf cimation to analyze the frameunder the effects of gravity loads. There are several way; to accomplish thisr but the most comnody used method rnirrmtly used at CaItrans íc the use of the 3DS frame d y s i s program. Section andmwiber properties f or the example problem have been input into this program based on the elevation view and typical section, and h e output is incIuded in the following aleveral pages. The BDS program is well suited to analyze bsth gravity loads and the effects of prestressing. However, it is important to h o w how to do the prestressing analysis by hmd for two reasons. First, hand anaiysis leads to a thorough mderstandmg of prestressing mechanics, and second,hand analysis is a great method for chedcing computer results.

3.4.3.1 Dead Load

The twm "dead load" refew to the weight of the box girder, which includes the deck, girders and soffit. BDS computes output relating to dead load is referred to as "Trhl O". Tfie mperstmcture weight used ta develop the section prapertiesis 150 lbs. /fL3 Page 3 of the BDS outpu t contains the relevad section properties ass ocíated with dea d load, and they are summarized as follows:

Seciion 3 - Prestressed Concrete Page 3-39

E* I Bridge Design Psactice - March 1993 m

Loca tion

Property Midspan Face of Bent Area (ft 2, 63.48 75.42

Y @ ) (fi) 3.70 3.24 1 (ft4> 400.29 487.09

The term y @, Is defined as the distance from the center of gravity to the bottom fiher of the niperstsucture. Horizontal member moments, shears, and stresses are tabulated cin

Page 9 of the BDS output.

3.4.3.2 Additional Dead Load

Additiod dead Ioad refers to dead laad that is appLied to the initial section. h h e e m p l e prablem, "ADL" refers to the weight af the Suture A.C. overlay and the Type 25 bmia rail. Most new bridges are designed baced on a future AC. overlay weight of 35 lbs. Jfth2 Th~s value equates ta no overlay thihecc of 3 inches, using a unit weight of AC. of 140 lbs /P. Tkie unifomi weight of Lhe future A.C. overhy is d c d a t e d as follows:

o (A.C. Owrlay) = (De& width - Barrier width) * 35 lbs./fi2

(A-, ,,,,y) = 1-19 klf

Type 25 barrier rail has a weight af 0.392 klf. Therefore w = 0.392 klfq2 - 0.984 kIf. BDS computer autput rehting to additional dead load is seferred to as 'Trial 1". Additional deadload ísnot loadresMant,andhasno structural s i p f t m c e whatsoever.

3.4.3.3 t ive tead

The tem "live l o a r refers €o AAS31TO design -trucks which reprecwt worst case loadúig condi tions. The two load cases that are used in the design af the box girder are Case 1 and Case N, which are the H S 2 W and Pemiit tnick,respec tively . The BDS computer program caiculates momentc, shem and stresces using a Zive load gmerator, whidz utilizes h e influence line concept. Tlie only rnforrnatron that BDSneds is the number of live load lanes. R e líve load h e mlculation for ttie box grrder is shown below:

# LL lanes = Deck widfli / 11 ft.

= 375ft / 14 ft.

# LL bnes = 2-66 lanes

T h e came number of lanes is used for both the HS2U4-4 and pemit truck loading cases. BDS computer autput relating to the HSZO-44 and the pemit tmck loading cases are rtlfersed to as "LL No. 1" and "LL No. 4"' recpectively.


E* Bridge Design Practice - Match 1993 m

At thic pociint, the items of interest regarding the longitudha1 section are the soffit flare and cable path. The Eongitudinal section is mduded as Figure 3-9.

Bent 2

Figure 3-9

3.4.4.1 Soffit Flare

The soffitis S.picaIly h e d to 1' - 0 fhickatthe face ofabmt cap ina CTPpost-tensioned box offíce stnicture, the main reason for this flare is to Iower the center of gravity of the superctmchrre, and thus, bcrease the eccentriüty h t w ~ n the CG of the P/S steel and the C.G. of t he superstnichire. Usualiy, h g h tensile ctresses are present at the top fjbers of the superstnichire near the bent .cap. h increase in eccenhcity in this area allows for more ~ i l e s ~ tobe 'iuiloaded" by theprimqmornent due& Prestre~~ing, ~hichissim~1~:

M,=Pixe Where: Mp = Primary moment due tci prestressing

Pi = Total jackhg force

e = Cable path eccentricity

Figure 3-9 shows graphically the effects of soffit &re. In the example problem, the flare lengths are assumed to begin at the 0.91. point of Span 1, and the 0,lL point of Span 2.

Secibn 3 - Prestressed Cancrere Page 3-41

Bridge Design Praetice - March 1993 m

34-42 Cable Path

GeneraUy, the maxhum eccentriuties (vertical distance between the C.G. of the superstructure and the C.G. o f the P/S steel) should occur at the points of maximum gravity moment. mese points are usually at the bent cap, and for our example, dose to the 0.4L and 0.6L points of Spans 1 and 2, respectively. The m h u m eccwtricity that is physically possible depends on the amount of prestressuig cteel requhed and the geometric limitations created by the hmsverse &el k the deck and soffit. There must be enough clearance to accommodate a wide range of currently used prestressing systems.

The first step in determining the cable path is to get a good preliminaq estimate of the Pi. h order to estimate Pi, it is necessq tol e s b a t e tlie pounds of prestressing steel per cquare foot of deck area.

Usjng the chart shown in BDA 11-61 for 2- Span Box Girder Steel in Lbs. per deck area, with a d / S ratia of 0.4 andan average span Iength of 156 feet, the diart reads m estimate of 4.2 pounds of P/S steel per sq. ft. In order to estimate the total pounds o£ P/S steel needed for the entire stsucSure, we simply multiply the square foat estima te x deckarea as shown below:

# P J S steel = 4.1 lb/ft2 x (37.5ft x 31 2 f t ) - 47,970 Ibs

BDA 11-11 cantains anequation that isused to estimate Pi or #P/S steel, @ven ane w the other. The equation is as fallows:

#F/S steel = x h n g h x 3.4 (O. 75$(270)

Page 342 Section 3 - Prectressed Concrete

E4 Bridge Design Practice - March 1993 m

Now that we have a pxeIiminary estimate of Pj, we can find the mhimum distance between extweme tensiie fibers in the superstructuxe and the C.G. of the P/S steel. Referring to h K D 11-28, this distance = "Y + "D", "Y values have been assumed and are shownúiMTD 11-28. Theminimum distancesbetween~eextreme tensile fibers and the C.G. of the P/S steel are as calcdated:

At Bent: Mh. Dist, = "r + "D" = 63jn. + 7 - h. = 14 in. 8 2 8

Use 15 in.

Figure 3-9 shows the above sesults for the vertical diiension of tIiecabIe path at the 0.4Land centerline Bent 2 locations. The final step in defining the cable path is to fmd the veriical positíon of the t w o inflerkian points, located at the 0.9L and 0.1L points of Spans 1 and 2 respectiveIy. By definition the position of "a" in Figure 3-9 is located where the straight h e between the two adjacent intersects horizontal location of the point of Mection.

By similar Eriangles,

"a" 4.25ft +,a,, -=- = 3.54 ft 81 f t 97.2 fi

"b" 4.25 ft +,lb,, - -- -- - J. 54 ft 75fi 90f t

The forre applied to the prestressing strands, as a function of A', is: Pi = (0.75)(270 ksi)(A',) is: Pj = (0.75) (270 ksi)(A',}. However, becauce losses of forceinthestrands occurs, the actual amountof forre inthe P/Spa&willbesomewhatless thanthejacbgforce. The three types of losses addressed jn the design of the CIP post-tensioned box girder bridge are as follows:

l. Frictional losses.

2. Anchor set lasses

3. Long term lasses

Because we hst have to calculate losses before we c m solve for P,, alZ losses are f o d as fractions of t h jacking force.

Secffon 3 - Prestressed Concrete Page 3-43


3.4.5.1 Frictional Losces

Due to the frictional forces that develop from the curvature of the presíress tendon and the devia tion of the tendon within the semíngid duct. (Friction and w obble coe fficients).

where e is the vertical distame of the parabola between control

pe jnt S and L is h e horizontal distance between control points

By similar kiangles:

a = ''.o' (6.5-2.25)=3.542ft 81.Oft + 16.m

b = 75'0" (6.5-225)=3.Wft 75.Oft +15.0ft

2. Fricticin loss coefficients

2e 2(2.70) a=- + FT B+o:=-=O.O833RAD

L M 8

2I3.54) PT C + a= - = 0,0874 IZAD

81 O

I T D--+a=-- 2'0-"' - 0.0874 RAD 16.2



F.C.I. = e- p* + m) where: p = 0.2

Spec. 9.16.1 gives the e-quation for caldation af fiiction losses as:

The K value Lndicates fnction wobble per foot of prestresshg steel and is given a value of O for the galvanized rigid duct most widely used m cast-S-place post-tensioned brldg~s.

This value T, is also called the initial force coefficient F.C.i wKch illustrate losses that occut immediately d-g the strecsing operationbefore tfte s t r d ¡S cet and m&or set and long term losses occur.

The F.C.i values calcuhted between segments can be linearly interpolated along / 1 Q

points and are @ven in Table 3-6.


r* - Bridge Design Practice - Mareh 1 993 m

3.4.5.2 Anchar Set Lossec

The xecommended method of detailing requises jacking the tendons to 0.75 f ', ,,, and 3 -

anchoringat a stress resultingfromamjnimum anchor set of - m. (stress afterandior 8

set not to exceed 0.70 f ',l.

The effect af anchor sef on the cable stress is approximately as follows:

f jack

X t anchor

figure 3-1 0

Aí = diange in s b s s due to m&or set (ksi)

x = lengfh influenced by anchor set (ft)

d = friction Ioss in length L (ksi)

L = length to a point where loss is h o w n (ft)

AL = anchor set (h)

E = Moddus of Elastiúty (ksi}

Assume E = 27 x 103 ksi

average unit stress = E x (unit strain)

Page 346 Section 3 - Prestressed Concrete


3.4.5.3 Long Term Losces

Assume a Iow Relaxation Strand is used,Spec. Artide 9.16.2.2 aUows 20 ksi ar 0.0988 P,.

Final force coefficients (F.C.0 can be found algebraicaliv by subtcacting the anchor set and 1% t em losses as shown in Table 3-6 horn the initi i forre coeffiOents due to frictional Josses. The last c a l m show the ratio of inikial to final forces.


Location e- F.C.iJF.C.f

O.OOL1

0.1 L l

0.2u

0.3Ll

0.4L1

0.5L-1

0.6L1:

0.7L-F 0.8L1

0.9kl

1 .O

0.0L2

0.1 12

0.2l-2 0.3L2

0.4L2

0.5L.2

0.6L2 0-7L2

0.8L2

0.9L2

1 .OL2

F.C.i. ASET F.C. PJ

1.000

0.983

0.937

0.941

0.945

0.949

0.952

0.956

0.960

O. 964

0.967

0.96ü

1 .O00

0.996

0.992

0.987

0.983

0.981)

0.976

O. 838

0.842 0.846

0.850

0.853

0.857 0.861

0.865

0.868

0.867 0.966

0.950

0.932

0.91 4

0.898

j.193

'1.183

1.173

1.161

1.152

1.1 44

1.134

-1.125

1.116

-1.1 14

O, 933 0.969

O. 966

0.950

0.950

0.932

0.928

0.935

0.921

0.918

0.91 4

0.9'1 0

0.906

0.902

0.898

0.851

0.851

0.833

0.829

0.820

0.822

0.819

0.81 5

0.811

0.807

0.803

0.799

1.1 16

1.116

1.119

1.119

f -120

1.120

1.121

1.121

1.122

f .123

1.123 3.124

E* - Bridge Decign Ptactice - March 1993 m


E* - Bridge Design Practice - March 1993 m

3.4.7 Secondary Mornents h addition to the "Pe" ( p r h q ) moments, secondary moments (MI} are introduced becauce of resic tance to beam distostions as theprestressingis applied. The conjugate beam nethod ic a general method for f i n h g the magr-tihrde of the secondary moments. This method Lnvolves the followhg procedures:

1) Each span ic considered to be a simple span so h t the endc can rotate freely. The applied mement S are equal to a rela tive value of "Fer'.

2) The angle o£ rotation at the ends of these simple spam caused by the "Pe" leads is found using h e conjugate beam method.

3) Morn~ntts are applied at the ends of the beam which will rotate the angle of rotation back to zera. Thew are the f i e d end secondary moments due to prestressing.

4) The fina1 secondary moments over the supportc are found by distributing these fixed end rnornehts.

5 ) The total moment in the rnember due to prestrecsing in terms of Pj is the algebraic sum of the "Fe" value and the secondary rnoment. We cal1 t h i s the Mament Coefficient. This moment coefficient divided by the relative P can be shought of as an effectjve eccentricity, e'. A h e throughthese e' ordinates is called the thrust b e , pressure line or c- line in some textbooks.

The dead and live loadmoments were computed neglecting the flare in thebottomdab. The *condary moment will be computed in the same manner. No appreciable ertor is intscl- duced wGh t h i c prosedure since the increased "e" and increased "1" tend ta.compensate for each other. T h i s ic more accurate than measuring "e" hom the neutral axis of the flared section with a minimum "J". Tñe relative values of M/I at the bent are shown below:

Actual Ordinate

Using efl,, And I,,



The flare pr~perties wZ, of course, be used to find stresses.

By the conjugate beam method, the rotation that would occur at the bent due io the "Pe" moments (8 thebeam was free to rotate) is equal to the reaction of the conjugate beam at that point. Take moments about point Z M, = 0.

Actual Beam

Figure 3-1 2

Conjugate Bearn Z M A = O

r M p E = R ~ X L

Secfion 3 - Prestressed Concrete Page 3-51


Secandary Moments

Span I


1

2

3

4

5

6

7

8 1

9

10

No. Asea (PíLIEl)

O - 0.0497 - 0.0994

- 0.0362 - 0.1717

- 0.0217

- 0.21 51 - 0.0076 - 0.21 74 - 0,0055 - 0.1 834

- 0.01 80 - 0.1237 - 0.0299

- 0,0373 - 0.0452

- 0.0063 0.0240

0.0728

0.04915

-1.1200

Section APM

0.050

0.067

O. 7 50

0.167

0.250

0.267

0.350

0.367

0.450

0.433

0.550

0.533

0.650

0.633

0.750

0.733

0.81 1

0.878

0.950

0.967

1

2

3 4

5

6

7

8

9

1 0

11

12

13

14

15 16 17

18

19

20

~oment (PjL2/EI)

O - 0.00333 - 0.01 491

- 0.00604

- 0.04293 - 0.00579 - 0.07529 - 0.00279 - 0.09873 - 0.00236 - 0.10087 - 0.0~959 - 0.08041 - 0.01 890 - O. 02798 - 0.031 67 - 0.00514 0.021 09

0.0691 6

0.04753

-0.38895

P)(o.l) (0.5>(- 0.994)(0.1 ) (- 0.994)(0.1)

(0.5)(- 0.723)(0.1)

(- t4717)(0.1

{OS)(- 0.434)(0.1)

(- 2.151)(0.1)

(OS)(- O . 152)(0.1)

(- 2.1 94)(0.1)

(0.5)(- 0.109)(0.1)

(- 1.834)(0.1)

(0.5)(- 0.360)(0.1)

(- 1.237)(0.1)

(OS)(- 0.597)(0.1) (- 0.3?3)(0.1) (0.5)(- 0.8W(O.l) (0.5)(- 0.373)(0.034)

(O .S) (O -728) (O. 066) (0.728}(0.1)

(0.5)(0+9~)[a. 1 )


Span 2

-2.201

Section 3 - Pmtressed Cmcreire Page 3-53

Area (PjUEl)

O

- O. 0474

- 0.0945 - o. 0345 -0.1638

- 0.02007 - 0.2052

- 0.0075

- 0.2097 - 0.0052

- 0.1751

- 0.0173 -0.118

- 0.0285 - 0.0356 - 0.0413 - P.0061 0.0231

0.0700

0.0506

- 1.0650

APM Moment (PJWEI) Section

(0)(0-1) (0.5){- 0.945)(0.1)

(- 0.948)(0.1)

(a.s)(- o.sgo)(o.~) (- 1.638)(0.1)

( O S ) { - 0.41 4)(0.1) (- 2.052)(0.1)

(0.5)(- 0.1 49)(0.1)

(- 22.97)(0.1) (0.5)(-0.104)(0.1)

(- I -751 )(O.I) (OS)(-0.346)(Q.1)

(-1.18~)(0-1)

(OS)(-0.570)(0.1)

(- 0.356)(Q.I)

(OS)(-0.825)(0.1)

(OS)(-0.356)(0.034) ( 0 . 5 ) ( 0 . 7 0 0 ) ( 0 . ~ )

(0.700)(0.1)

(0.5)(1 .O1 l ) (O . I )

No.

1

2

3

4

5

6

7

0

9

10

0.050

O. 067 0.150

0.167

0.250

0267

0.350

1

2

3 4

5

6

7

8

9

10

7 1

12

13

14

15

16

17 18

19

20

O

- 0-0031 S

- 0.01 422 - 0.00576 - 0.04095 - 0.00553 - 0,071 82

0.367

0.450

0.433

0.550

0.533

0.650

0.633

0.750

0.733

0.81 1

0.878

0.950

0.9

- 0.0027 3

- 0.09437 - 0.00225 - 0.09631 - 0.00922 - 0.07677 - O.QI804

- 0.02870 - 0.03027 - 0.00491 0,02028

0.06650

0.04885

- 0.36737

E* Bridge Design Practiee - March 1993

Secondary Mements Span 1

PiL2 PjL Conjugate reaction at D = rotation =4.38845 -+ L = 4.38895 -

EI E1

= 0.38895Pi M, = 3(0.38895) Pi 3

Ms = 1.1669 Ti + M,, = 1-1669

Check:

1,1775 -1.1669 % error = 1.1775 - = 0.9% :. OK

1.1669

Pagc 3-54 Section 3 - Presiressed Concrete

E* I Bridge Design Practice - Mareh 1993

Secondary Moments Span 2

PjL? PíL Conjugare reaction at D = Rotation = - 0.36737 - = - 0.36737 -

E1 ET

- P-L 4.36737 1 =O

3EI EI

3 = 0.36737 pj + ?!& = 3 (0.36737) P, 3

= 1.1T)Zll + Mc,, = 1.10211 Pi

1.753 Check; 1

Seciion 3 - Presfrecsed Concrete - Page 3-55

E-t: ! Bridge Design Practice - March 1993 m

Pistfibuted End Moments

Cummary of fixed end moments (conjugate beam):

L = 162ft C ='150fi E = 3600 ksf E = 3600 ksi l = 400.3 fF F = 1 F = 1 1 = 400.3 ft4 R = 8 8 9 6 K = 26688 K = 28821 R = 9607

D = 0.272 D = (3.293

4271 6 a,, = -=e.435 98225

figure 3-1 3

Page 3-56 Seciion 3 - Prestressed Concrere

E* Bridge Design Practice - Mareh 1993

Eccentrlcitiec at each 10th Point:

eEg = 0-84 ft

e,,o = 0.84 + 0.71 + 0.46 =LO1 ft

%y symmetry eccenhicities in Cpan 2 are a mirror image tboce in Span 1.



Determine Moment Coefficient (M.C.)

F.C., (0.099) 201202.5

1 LT. Loss M.C.

F.C.i

F.C.,

Page 3-58 Secfion 3 - Prectressed Concrete

E* + Bridge Design Practice - March 1993 m

Figure 3-i4

3.4.8 Prestress Force

Check for control point at places of maximum eccentricity, 0.4 Span 1,0.6 Span2, and at Bent.

Pj (F. C.) Pj (M. C.)C S S 0 .1 , L 1 + ~,,I,LL,*~

-t 6 & = 0 A 1

:. P. = 1 (F.C.) (M.C.)c +

A 1

Secíion 3 - Prestressed Concrete Pag e 3-59

E4 -1 Bridge Design Pracfice - March 1993 m

0.4 Point Span 1 :

Where: C = 3.70 ft

~ $ . l ) ( f ) 1 = 400.3 ft3 DL Tfid o,, + = (F. C.) (M.C.)=

F.C. =0.853 4-

A 1 A = 63.48 ft2 M.C. =1840

pi = 6558 kipc

S h , O,l , l l .#I . ) Pj =Pi = (F.C.) . [M.C.)c

1 .O Point Span 1 : Face of Bent:

75.42 ' 467.1

Pj = 7561 kips

Page 3-60 Section 3 - Presfresced Concrete


q = 7491 kips

Where: C =6.5 A - 3.24 ft = 3.26 ft

1 = 487.1 ft4

F.C. =0.85ñ

A =75.42 ft' M-C. = 2.870

M T ~ =28443 k - f t

M T ~ ~ = 6M4 k - ft Moments taken from BDS Ou tput Section 3.4.13

MTehl ~ 7 . 8 4 2 k-ft 1 0.6 Point Span 2:

= 54.6 ksf

pj = 5318 kips

Sectiorn 3 - Prestressed Concrete Page 3-67

E* Bridge Design Practice - March f 993 m

Where: C =3.70 ft

1 = 400.3 ft2 F.C. = 0.815

A = 63-48 ft2

M.C. =1.750

MTridO = 13937 k - f t

IdTdd, = 2942 k - fi

M,,, = 5721 k.ft

6 4 m = 379 psi (E) IIXX)

= 54- 6 ksf

:. Pi = 7613 pound up to nearest 10 kips = 7570 kips

Controlling condition: DL + ADL, left face o£ bent.

Page 3-62 Section 3 - Prestressed Concreie

E* Bridge Design Practice - March 1993 M

3.4.9 Stresses

Top Fiber Stresses

Dead Service Final Initial Total Total Load Load fl DOO/ 1 4.4) (1 00011 4-4) PIS PIS Enitial Final

Location Stress Stress Pj (F.C.) /A Pj (M.C.)~/ i Stress Stress Stress Stress

O.OL O O 694 O 694 828 828 694

0.11 400 600 697 - 304 393 465 865 1015

Stress values reported in Psi Pi = 7570 A = 63.48 M2 (Mid) 1 = 400.3 Ft4 (Mid) C = 6.50' - 3.70' = 2.80 Ft (Mid) A = 73.42 FY' (Bent) 1 = 487.1 Ft4@nt) C = 6.50' - 3.24' = 3.26 Ft (Ben t }

Table 3-8

Seclion 3 - Presiressed Concrete Page 3-63

r-tr Bridge Design Practice - March 1993 m

m Q 0 m CI X C S :: m "7 :: O - D 0 ::

Figure 3-15



Battorn Fiber Stresses

Stress valuec xeported in Psi Pj = 7570 A =63.48FYZ(Mid) 1=4003 F9 (Md) C=3.70Ft(Mid) A=75.42FY2 (Bent) I=457.1 Ft4@ent) C= 3.24Ft(Bent)

Table 3-9

Location

0.OL

0.1L


(1 000/14.4) Pj (F.C.) /A

694

697

Dead Load

Stress

O

-528

(1 00011 4.4) Pi (M.c.)~/

D

402

Service Load

Stress

D

- 822

Final PIS

Stress

694

1099

lnitial PIS

Stress

828

7300

Total lnit ¡al

Stress

828

772

I Total Fina 1

Stress

69 4

277


STRESSES ( PSZ ) FOR S D T P ; I W FI3w

m m m CASE i PS - : nb + ADL h---i 1 UL . LSL - P U * , L 1L r AtL - NLL

C 15 r) a O O Q c z Y1

m m X -2 0

m a .. 0 - 5

Figure 3-1 6

Page 3-66 Sectian 3 - Presfressed Concrete

c* Bridge Design Practice - March 1993

3.4.10 Concrete Sffength Required

l . ' -i (choose 3500 psi unIess cales show greater strmgth xequjxed:

f,, + fF' /, 921 psi f ' . = - -

€ 1 = 3636 psi c 40QO O. 55 O. 55

:. f 2 = 3500 psi

2 f', 4 choose 4000 psi unless calcs show greater strength requjred:

fn,- + ADL+ LL# X + fP - 1455 psi f', = - = 3638 pci < 4000

O. 40 O.#

f > = 4000 psi

3.4.1 7 Ultimate Moment Prestressed rnembers are designed a t senrice load IeveI. Af ul timate conditionc, the member is cubject to loads iri excess of t he Design Loads. Therefore, the ultimate flexura1 ctrength of them.ernbermust be checked at aitical sections. These critica1 cections inclnide points of maximum positive and negative moment and changes úi cross-cectianc.

Jf the dtirnate applied momenf is less then the capacity of the section, no additional mild reinforcement ic needed. If, however, the ultimate applied moment is greater then thc capacity of the section, additiml mild reinforcement may be added.

For example:

Tf M, 5 bM,, then, no additional mild reinforcement i c needed

Lf M, > @Nr k, additional mild reinforcement is needed.

3.4.1 1.1 Ultimate Applied Mornent, M,

A. STEP 1:

The ultjmate applied moment at any section can be cdcdated by the greater of the two equations.

or

M, = 1-3Ch + M,,,) + M,

Take note, the member is at the leve1 of incipient failure, therefore, the beam is full y loaded, where,

M, = maximum dead load moment due to Trail ) + Triat 1

Seciion 3 - Prestresced Concrete Page 3-67

E* Bridge Design Practice - March 1993 =

MLL#I = maxímum live load moment due to LL#l (HSZO)

MLL#I = maximum live load moment due to LM4 (P13)

MS = secondary moment where MX = (h4Cs)Pj

3.4.1 1.2 Capacity, bM, (without mild reinforcement)

A. STEP 2:

Fhst calculate the rapacity of the section with na additional mild reinforcernent te d e t e h e jf mild reinf orcement is needed if M,>@M,, then calcula te the area of mild reinforcement required .

Three basic assurnptions are made in cdculating the capacity of t he section:

1. no additional naild reinforcemen t at thic time

2. sec tion is rectangular sec€ion (a < h,}

3. section is under- reinforced (w 1 0.30)

At ultimate conditions, prestressed concrete acts similar to reinforced concrete, therefore,many of theprinciples of reinforced concrete areutilized in caiculating the u! tima te capacity o£ ttie section. The parabolic stress shape may be assumed to be an equivalent rectangular stress block wifti an average compressive skess equal to 0.85f', over a camprecsion block depth a,

Figure 3-1 7

The compressive force equalc the average stress, @.85f',), times the arpa onwhich it acts, @a). And, the tensile force equals the stress in theprestressing steel at dtimate conditions, (P ,), times the area of prestrescing steel, {A*). Epdibrium muct be maintauied, therefore:

Page 5-68 - Secfioh 3 - Preslressed Concrete

r* Bridge Design Practice - March 1 993 m

(.E5 f ,) b, = (P,,) A*, where P,, = Fs [ 1 - 0.5- p;:':] Cpec 9.17-4.1

(Equation 1)

Curnmulg mornents about fie compressive force yields:

M,, = r ) (z* ) , where

d* = distance from wtermost compression fiber to the center of gxavity e£ prestressing steel

BY substitution and multiplying both sides o f the equation by m,

(Equation 2)

Substiuting Equation 1 into Equatien 2 yields:

So naw, algebraically manipula te qua tion to look like equation in specifications.

Note:

l. Check assump tions

a - a l h, b. w S 0.30

2, If & IBM, 'Ef M, > $M,

Then, must calculate area of mild reinforcement requked

(Equation 3)

Section 3 - Predressed Concrete Page 3-69


3.4.1 1.3 Capacity, QM, {With Mild Reinforcernent)

A. STEP 3:

If M, > OM,, add addi tional mlld reinforcement to increase the resisting moment.

%ee basic assumptions are rnade in c a l c u b ~ the capacity of the section:

1. No additiond mild reinforcement at thís time

2. Section is rectangular section (a 1 h+)

3. Sectionisunder-rreinforced (~50.30)

The mild reinfarcement produces anadditional tensile force e q m l to the stress inthe steel, f, = f,, times the area of steel, &.

----- ----- J' = (f ,") A* CG af mild -----. -__----- .......................

reinforcemen t

Figure 3-18

Again, equilibrium must be maintained, therefore:

C = T * + T

By subs titu tion,

(0.85 f', $ b a = (P,, A"+ ($)A, where f+,, = f',

m i n g moments about the compressive fmce yields:

@M,, = fl*)(z*} -t p)(z), where:

Z* = moment arm d -- ( Z ) d* =distame from outemost compression fiber to the center of gravity of

prestsessing steel

Page 3- 70 Sedion 3 - Prestressed Concrete


d = distante from outemast compression fiber to the centes af gravity of mild reinforcement

Bv substitution and mdtiplyirig both sides of the equation by 4,

(Equation 5 )

Having twe equatíons and W o unloi0wni, the area of miId reinforcement can be determined by first substituting Equation 4 into Equation 5. Then, solve the cpadratic equatien for 4.

Note:

1. Check assumptions

a. a 1 b

b. w 10.30

3.4.1 1 "4 0.4 Point

This procedure will be demonstra ted on the 0.4L, poink.

STEP 1: Calculate M,

M, = 1.3(Md, 1.6mLxi) + M,

= 1.3[17,931 + 3,668 + (1.67)(6,147)] + (0.463)(7,570)

M, = 44,929 k*ft

M"= 1.3FI, +M,,,)=M,

= 1.3[17,931+ 3.668 + 17iQ141 + (0.463)(7,570)

M, = 54,222 k-f t t CONTROLS

Section 3 - Presfressed Concrete Page 3-71

E* i I Bridge Design Practice - March 1993 m

STEP 2: Caldate @ M, (without mZld reinforcement)

Equation 1:

a = 6.32in5(hf= 8.25in)

Equation 3:

= 0.95 ((258.5)(37.38)(5.5)[1- (0.60)@.00126)(~~~h)] )

@M, = 48.021 k.ft

& > @M" (54,222 > 48.021)

.-. Additional mild reinforcemeni is needed. So now ca lda te 9. Note:

l. Check as- tions

a. a l h f

6.62 in 5 8.25 in

b. w 10.30

(O. K.)

Page 3- 72 Sectiun 3 - Prestressed Concrete

E* 8 Bridge Design Practiee - March 1993 m

STEP 3: CaicuIate A,

Equation 4:

a 9662.17+60A5 -= z 2244.. O

Equation 5: (set M, = @M,,)

Multiply both sides of the equation by 2244.0 and divide by 9,

(54,222) (12)(2244) / (0.95) = (258.5)(37.38) [(66)(2244F 966273- 60 A,]

Algebraically manipulate equation ylelding:

(&y -3,494.09 A, + 66,058.0 = O

Colve rfuadratic equation,

A, = 19.0 h2

Note:

1. Check assumptions

a. a 5hf

Substitute A, jnto equation 4,

7.06 in 5 8.25 (O. K.)

b. w < 0.30

SecfJon 3 - Prestressed Concrete Page 3- 73

rZ 8ridge Design Practice - March 1993 m

As in the previous design considerations of this exarnple, only Spm I will be analyzed. Agah we mention that in actual design where ther~ is an abscnce of symrrietry, thex i llustrated procedures should be applied throughout the structure lerig th,

The distortion of the member from ~rectressing causes a change in the reactions wlzich in hirn will affect the shear values. The vanous seclion properties, concrete strength, dead Ioad moments and shears and othet live load mornents and shearc have heen calculated. To calculate P , = 7S70k fos this example.

We wi l l use khe method in the Spec. for shear design. Since this method involves lengthy calcula tions, we wjll demonstra te the procedure ody on the 0.9L point.

V, equals the greater of the two equations. "P" loads wil1 conirol for this exarnple,

From BDS outpii t:

Dead Load Sheais (Trial O)

Dead Lead Shears (Trial 1)

Live Load #1 and Impact Shears:

Live Load #4 Impact shears

(Spe~. 3.22.1A)

Abut 1 0.4L 0.9L Bent 2

~ 8 5 . 2 ~ - 31.8 - 803.1 - 971.6

120.6" 7.3 - 1673 - 199.2

214.7" 92.8,- 202.8 - 250.0 - 285.1 5 9 4 9 192.1,-168.1 -695.4 -780.8

Shears induced by M,:

These shears are equal lo reactions a t supports ko resist M,.

@C) P, - 0 . 1 5 9 ) ~ 5 7 0 ~ ) -

L (1 62 ft)

Note: VMs = A R :. VMs = 54.2K

E* I Bridge Desigrii Practice - Match 1993 m

S m a r y of factored shears, Spanl:

Abut 1 0.4L 0.9L Ben t 2

VUPL = 1 .3VDL 917.5 - 50.0 - 50.8 -1261.4 -1522.0

V U , ~ L A I = ~ - ~ ~ L , t 7726 249.7 - 2445 - 904.0 - 1015.0

Km 54.ZK 54.2 54.2 54.2 54.2

Total V,: 1 7Mk 253 -2111 - 2483k

Note: Skewed bridges require modification 05 design sheafi. Consult Meino fo DesCynm 1s- 1.

Resisting shears:

Members subject to shear shall be designed such that:

V, I Q (V,+V,) where I$ = 0.80 (Cpec. 9.20.1.3, eq- 9- 26)

Sheas resistence o f concrete, V, : (Spec. 9.20.2)

The shear resistence of concrete, V,, is taken as the lesser of the values

Vti and V, brit not less then Vrii ,,,, (Spec. 9.20.21& 9.20.2.2)

where Vci,, .13 (&) b' d

@ a9t :

Vd = Due to unfactored self weight shear (trial O)

-803.1

vci : (Spec. 9.20.21)

(Spec. 9.20.2.2, eq. 9- 27)

Vci 2 1.7flc b'd

b' =width of web = 4 x 1ft = 4 ft

d = effective depth to prestrecsing steel or 0.8H, whchever is greater

V, = associated factored shear to Mm,

=t .3 (1 67.2 + 537.6) = 91 6.ZK

Seclion 3 - Prestressed Concrete - Page 3-75

r* Bridge Decign Practice - March 1 993 i

&, =maKimum factored moment due to extemaLly arplled loads.

=1.3(3403 + 8832) = 15,905 k-ft

(Br. Des. Spw. 9.20.2.2, eq. 9-28)

fp, = compressive stress in concrete dueto effective prestresses at extreme fibes where tensile stresses are produced by externally appiied loads

= 1369 psi

fd = stress due to unfactored self wt. at extreme fiber where tensile stresses ase produced by extemaliy applied loads.

= 771 Psi

dieck minimum value of VCi: (Spec. 9.20.22)

Y,

V, = (3.5& + O. 3&)bPd + vp (Spec. 9.20,23, eq 9-29)

d =5.2ft

b' =4.Dft

f, =compressivestress~conc~eteatN.A.Due toeffectiveprestress

Page 3-76 Sectíon 3 - Presffessed Concrete

Bridge Design Practice - March 1993 m

Y, = ver tical romponent of cable f o ~ e (Spec 9 -7-1 )

V, = (F.C.)Pj$

= angle change hom horizontal axis to prestressing cable .-. for 0.9 point

$I = r r ~ ~ = ~ = O . O 8 7 4 r a d i a n s

V, = @.867)(7570)(0.0874) = 57dk

SO, V, = [ 3 . 5 4 z E [ ~ )+~3 (103 .39 ) (4 ) (1 .2 )+574 1 =1882k

cince 1882 i 2076, Y, = 1 882k

Ascume # 5 stimps in each of the four girder webs.

so total A, = (4 @dms)(Z Iegs/girder) (o-:::2 1

- 2 48(60)(5.2)(12) = 20 h. (z- 1882)

(Spec. 9.203.1, eq. 9-30)

Checking mhimum steel, minimum web width, and maximum spaüng:

Min. steel:

0.80 < 248, O.K.

(Spec 9.20.3.3, eq 9-31)

S e c t i ~ ~ 3 - Prestressed Concrete Page 3-77

r-É - Bridge Design Practice - Match 1993

Min, web width:

(Spec. 9.20.3.1)

Avfy d 2.48(60000) :. b'& = - - B& S adGE(20)

14.7 in since 14,7 in 5 48 In, ok web width

Max. spacing:

(Spec. 9.20.3.21

S,,= mirümum of: 0.75 (6.5(12)) = 58.5 h. or 24 h.

S =20ik<24in, O.K.

For the rest of the span, the spacings are:

Lacation S

0.4L 24 in.

The spacing shown are corre& and comply with the Spec. requirements. However, it ha s b-shown thalinpracticalsihiations, it is pmdmt to limitthe stimp spacing to an 18 in maximum and to 12 in. naximum within about 8 feet of the supports.

3.4.13 BDS Computer Output

The following pages are selected parks of the BDS Vession 3.0 romputer output for the exampleproblem. The reades is encouraged to compare these results to thehand calculations. BDS calculates mommts and stresses e~ery ' /~~thpo"t on the span and provides output at the ]/,,th point. These extemive calculations might not match exactly with the hand calcubtions.

Page 3-78 Seciion 3 - Prestfessed Concrete

E* + Bridge Design Praetice - March 1993 m

0 0 D 0 0 0 0 0 0 0 0 a Q " + ~ d 0 0 0 o O O w O o ~ 0 0 D 0 ~ 0 0 0 0 0 a Q O Q 0 0 ~ 0 0 a o O o O 4 ~ ~ ~ ~ m m ~ ~ ~ ~ m ~ ~ m m m m m r n e ~ ~ p i r n u )

O w O O a O O

0 O 0 rl TV t

0 0 0 0 0 0 0

N r U O O O O O O O O O o o o , - l , - ,

~ P 0 0 0 0 0 0 0 0 O

0 0 0 - = = b i d i

O O D N ~ ~ ~ ~ ~ ~ ~ ~ ~ * P r( ~ ~ g a ~ ~ O ~ w ~ ~ O a r( R A

;I Ll O Q O O O O O O O

vi v ) E p: a r x w w o o o g , O

a o o d p: m

0 0 0 0 0 0 0 0 0 0 E

O B O R ' 4 U c . J m m - .

~ ~ ~ 0 0 0 0 0 0 0 4 4 a 4 O o w ? J N z z 0

0 0 0 0 0 0 0 0 0 0 0 U O tl O O O D O O C a O O D D O O ~

m W W S Z L P I H H > > 1 1 m P E m m a Q

0 0 0 O O O O O O Q O

m m m 0 0 0 ° 0 0 - d d r i o u o o o m a o o O O 5 O 0 ~ 0 N f l O 0 0 0 0 0 0 0 w

A d N O ~ 0 0 0 0 0 0 0 0 0 O r y m o o o w 4 rl 0 0 0 0 0 0 0 m r v o O O

O

m O D O O O m o O o Q D '2

w N 0 0 0 0 0 0 0 0 0 0 O

74 N C V O O O r] ,+

O O

C * * o O ~ w m m o o o o o o o

4 C o 0.0 0 o m -k U Ln Ul N $: * O O O k 3 W C i o a o o o - l o * * 0 0 0 D 0 0 0 N O r U O * +: tn ln o Z t m w L n ~ ~ ~ ~ ~ o r J + r- r- r r d r l l r l UI

m m 0 0 0 0 0 0 0 m

I 2 I o o 0 0 9 0 0 0 0 0 0 0 ~ 0 0 ~ a : Y : N

o - ~ O O O O O O O o O O " * W O C O O O * E n * W L 1 N

S i ~ 3 P o o o o r i m * H - C 4 L O O O * W O O ~ ~ W ~ O N + + z z N m r n m m w ~ ~ w ~ ~ ~

S 0 0 0 i F t - r l r l N N N h 1 0 0 : !* ~ 0 0 0 0 0 0 0 0 0 0 4 rl

m m m

* 2 N * , ~ ( T v Q ~ ( N ( U W N ? I A O O O Q r l L D .

* Z c O O O O C O O ~ O O K! m CQ 03 m o N cu e o m + W * d N 9 1 m w w u l e m w 4 Ln

m C d C d d r i W A N r l r U r U N C V N d

r'a r( 4 ~ ~ + d ~ ~ i O ~ d ~ d . ~ m ~ N ~ o m r n m ~ ~ ~ r i m r u m ~

+ F + -k * a X d + C m * O + a * w x m i O # X rl * E * N * 3 % * 0 % * > c rl

d

-

Section 3 - PresFressed Concrete Page 3-79


Page 3-80 Secfion 3 - Presfressed Concrete

E4 Bridge Design Practice - March 1 993 m

m " , E =

A

o:

n ; p J

3 4 \

L n W O

1 O

Y m G W E d

0 3

9 " U

O 3 H

L o a a : E

m rn n m L ,

2 0 2 S z 5 O

Lnoa P O N N O - . . - 0 0 0

o m o 040

0 0 6 o m m

W o o v i U3 w N m m m

I

Section 3 - Presttessed Concrete Page 3-81


L J


E* - Bridge Design Psactice - Match 1993

E-G Bridge Design Practice - March 1993 E

Sectíon 3 - Prestressed Concrete Page 3-85

r* 1 Bridge Design Practice - March 1993 m

- - . -


E* ' Bridge Design Psactice - March 1993 m

Z A

e'd * P P 4

U N O 0 0 m m H . . a . . M E d n i N O 0 0 X

I E

;I E - P W r - m a i m m m w w d W

r u c o P ~ O C O C E I Z m . . . . . E - m e a m o o

O cr: b+

E w m m m a i m m nin t o r i P

m m qin m m W r - a - - . . W

q K l r i N O O L

1


c* ; Bridge Design Practice - March 1993 m

A u1 m

5 r; ~1 r- L l % N i _ r , m Ln S i e w Ln w

E.! a l i]

w E"

m m m E m m N m m Ul m m rl O m m rl

r n l d 4

- N O

. E o e m e I 8 " " " d Ln

$ 2 w n m rl

E co 4

b 4 - H rn . [L . -

m 4 r l U ' m 4 - ~ m m m w e rn " " a " 0 4 W r- rl

w . c m Ln

m 0 J 3

O I . LC: J P W I D O c n r i J D i N I T i J r n d i

2 N Q' A w r l r l J N

A w a : . c3 w a: z = b Lri D: 3 4 - W

E 4 o F e w Z H o : a , L n r n r , N r n

W w m A r i r i A u m m ri ri

51 - tu

d 3 W m

3 H - h

2 b b b d N b W b a i m r - h r n m

K! C) m e 2 Z R Pa

"; EI) > > E L Z J

w o O . o . E N = w m

V ) D 4 e = ' U ) a i L n UJ m m w m h l m d id a . a 6 P

V) C i H M u

E - a : Ul P N a i - W

ln m H m m m ú r ) ~ m

U N rl N > m . m

En E w trl

W W 2 I*; n:

a . E - - r- U) tu

ri E r i n ' g r n m W r i -a I w 4 m . m 17 r

W d. z

Cri d . d

Ff 4 b o m a : o m

Z .--! pl i i a ri b m m 3

Z W W ~ S W i n 0 3 t O 4

U J N N

U H H E

W

4 E

E 0 0 O E Z Z d N Z d N

Page 3-90 Sectbn 3 - Prestressed Concrete

E* -1 Bridge Design Practice - March 1993 m

- E - m

m x r i m r- u-

l-4 Ln m U, H e , m Lrl

W a 1 4

u 2

E m m m C L ~ m m a r p m w m m rl O m m

d Q l l d !+

N O

. X ! % E o d N m di d ln

Cil e m m ,- z m rl

L 4 - E H en E a, . --.

d E m r i r i 9 ' m 2 3 a m m t n l a e -. rn m * y m 0 4 W r- rl

V i .

T r J O wl W m m cI1 tO O 1 d g .1 , , 3 : 0 w m rf

r : W % d d J N * w L m Wi E c3

E. 4 Z E L, D:

5 A. . W - E r i o F e w

rrai ln cn b, N sy W e m A r i r i

m P l L n d rl H - rr:

z L W m

S E Z O . L , - t-

Z B FEi d ru F m b b w r - h m o

I w O Ln rt N m e 4 ri R

U) h > G C 4 cI1

J

3 Z m " " O . Z E w o w m

2 t n P r % ? w I R w ' m VJ m m W I m

F. = Wrn ri W z - E

O 3 H

El W

5 W . n:

V1 m m H Cr) m m l l l r l ~

W N rl W > U ) . tr) x m En H E W W

S a: !x 0 . M *

m N ri E r i + - r g m m

W d 5 4 l W d g m m z 5 2 U r; C4 a . J *

E- Z ri & L c l : b o m S o m m ri m m 3

ZW. w 1 s

w u' O

O ;I 4

U J N N

U 5 2 H

4 rZ; O 0 O

~ X Z f i r u Z r i N

Page 3-90 Section 3 - Preslressed Concrete

CALTRANS BDS-VERCION 3 . 0 0 REL-10 05 /10 /92 , FEB. 02, 1993 PAGE 3 6

D I V I S I O N O F S T R U C T U R E S - C A L T R A N S

TRIAG 1 FRAME 1. P ATH HORIZONTAL MEMBER MOMLHTS DUE TO P/S MGM

NO LEFT . 3 PT . S PT . 3 PT - 4 PT .5 PT .6 PT .'I PT . 8 PT . 9 PT RIGWT i -5. -6482. -10914. -13268. -13513.-11765. -8216, -2830. 4 4 1 5 - 13530, 21780, 2 21725. 13307. 4 4 6 3 . -2545. -7763 . -11208. -12902. -12703. -10463. -6219. -5.

VERTICAL MRMBER MOMENTS DUE TO P/S MEM f.10 LEFT .I PT .S PT .3 PT . 4 PT . 5 PT - 6 PT . 7 PT . 8 PT .9 PT RIGHT 3 -55. - 5 5 . -55. -55. -55. -55. -55. - 5 5 . -55. - 5 5 . -55.

TANGENTIAL ROTATIONS - RADIAN2 - CLOCKWISE POSITIVE SPAN LT. END R T . END SPAN L T . E N D R T . E N D SPAW LT. END RT. END 1 -0.003139 0.000036 2 0 .000036 0 .002770 3 O. 0 0 0 0 0 0 O. 000036

HORIZONTAL MEMBER DEFLECTIONS IN FEET AT 1/4 POINTS FROM LEFT END - DOWNWARD POSITIVE M E ~ E R 1 E= 3 6 0 0 . o. o00 -0.220 -0.373 -0.169 o . a o o

VERTICAL MEMRER DEFLECTIONS IN FEET A T 1/ 4 POINTS F R O M LEFT END.

MEMBER DEFLECTIONS HAVE BEEN MlTLTIPLIED BY A CREEP FACTOR OF 3 . * * * * *

r-G 1- Bridge Design Practice - March 1993 m

E . . m m

m m rl U 3 . N

w I

2 z O W O x z z r l w



l 1

Secfion 3 - Prestressed Concrete Page 3-93


Section 3 - Presiressed Concrete Page 3-95

E* 1 Bridge Design Practice - March 1993 m

Page 3-96 Section 3 - Presfrecced Concrete

E* Bridge Design PracZice - March 1993

Sectjon 3 - P resiressed Concrete Page 3-97

ACCOUNT: CALTRANS BDS-VERSION 3.00 REL-10 05/10/92, FEB. 02, 1993 PAGE 4 4

D I V I S I O N O F S T R U C T L ' R E S - C A L T R A N S

LONG TERM LOSCGS TOTAL LOSS (KSI) = SH + ES + CRC + CRS MEM NO LEFT .1 PT . S PT . 3 PT - 4 PT .5 PT .6 PT - 7 PT - 8 PT .9 PT RIGHT 1 21.1 2 5.7 1 9 . 8 18.9 18.6 18,9 19.9 21.2 22 .0 21.3 15.6 2 15.9 2 0 . 2 21.4 21.5 21.1 20.6 2 0 . 3 20.2 20.2 2 0 . 3 2 0 . 3 * * * * THE AVEMGE LONG TERM PRESTRECS LOSSES IS 20.0 KSI. * * * *

SHEAR DBSIGN - MSHTO 1981

LEFT . 1 PT .S PT - 3 PT - 4 PT .5 PT .6 PT .7 PT .8 PT .9 FT RIGHT MEMBER: 1 V- CABLE 530. 3 9 0 . 261. 131. 1, 111. 2 2 3 . 336, 449. 5 2 0 . 6 9 . SECONDARY 5 6 . 56. 5 6 . 5 6 . 56. 56. 56. 56. 56. 56. 56. W 1747. 1354. 9 7 2 . 605. 255. 595. 969. 1355. 1739. 2109.2481. VC 1774. 1433. 725. 381. 322. 447. 771, 1375. 1428. 187. 1257. REQD WEB 48. 48 . 48 . 4 8 . 48. 4 8 . 4 8 , 48. 48. 4 8 . 4 8 .

MEMBER: 2 V-CABLE75. 5 4 3 . 468. 350. 232 116 1, 137 272 405. 520 . SECONDARY - 6 O a -60. - 6 0 . - 5 0 . -60 0 - 6 0 . 6 O -60 -60. -60. VU 2332. 2014. 1659. 203. 929 584 254. 542 887 1 2 4 7 . 1619. VC 3262, 1632. 1 2 2 2 . 162D- 885 516 322. 397 3 6 1 1507. 1772. REDD WEB

4 8 . 48. 4 8 . 4 8 , 4 8 . 4 8 . 4 8 . 4 8 , 4 8 , 48. 48. AS (IN) /FT

4 . 4 0 1 . 9 4 1.99 0.48" 0.48) 0.48* 0 . 4 8 ' 0.66 0 . 7 2 0.48* 0.48

NOTE: + AFTER R E 2 D WEB INDICATES ADDITIONAL WEB WIDTH REQD. * AFTER AS{IN}/FT I N D I C A T E S MINIMUM

E* -1 Bridge Design Practice - March 1993

I k

Section 3 - PresfresSed Concrete Page 3-99

ACCOUNT: CACTKANS BDS-VERSTON 3.00 REL-10 Q5/10/92, FEB. 02 , 1 9 9 3 PRGE 46

D I V I S I O N O F S T R U C T U R E S - C A L T R A N S

TENDON ELONGATION % JACK FY AS AVE STRESS TENDON LENGTH ELONGATION

(KLPS) EKSI 1 (sQ IN) (Kslj (FT) * (m) 7 4 2 0 . 7 5 . 270. 102.92 316.00 26,13

NOTE: TENDON LENGTH INCLUDES 4 FEET FOR JACKS.

MODULUC USED FOR P/S STEEL IS 2 8 0 0 0 . KSI.

ACCOUMT : CALTPANS BDS-VERSLON 3.00 HEL-IO 0!5/10/92, FED. 0 2 , 1993 PAGE 47

D X V I S I O N O F S T R U C T W R E S - C A L T R A N S

' AP PROXIMATE QUANTITY ' -------+------------ ---------A----------

* * * * * CONCRETE SUPER 740 C . Y , * * * * * * * * * * CONCRETE SUB 26 C.Y. ***** * * * * * PIS TRIAL 38888 LBS- ' ***'*

THE SUPERSTRUCTURE CONCRETE QUANl'IW I S BASED ON THE UNXT WEXGTH O F CONCRETE SUPPLIED ON ,THE FRaME 1 DESCRIPTION CARD. TT ASSUMES THAT ALL THE DEAn LORD TS GTVEN IN TRTAL O.

THE CONCRETE SUBSTRUCTURE QUANTITY LS BASED ON TRIAL O ONLY.

TEiE P/S QUANTLTIES FQR STRAND O N W ARE FOR EACH TRIAL THAT WAS ENTERED AND IN THAT ORDER. STRAND USE IS BASED ON TKE LENGTH FROM ANCHOR SO ANCHOR.

EMD OF JOB - 022086 L

r* Bridge Decign Practice - March 1993 =



Page 3- 102 Secfion 3 - Prestressed Concrete

E* Bridge Design Pradice - June 1994

3.5 Design of a CSpan Continuouc Cast-in-Place Prestressed Box Girder

m example builds upon experience gained in the 2span example and highlights the special considerations needed far multi-span h e s . The BDSprogram serves as the main design toa1 with simple hand check verdying output. As is the practice m design, only selec t points of control are hand checked agains t the computer output.

Only the point of control for qack determinaticm will be hand verified. Shear design will be ;ddresced in subsequent versionc as the LRFD specification is irnplemented.

3-52 Bridge Configuration and Companent Propefiies 4-Span contmuous

Span Lengths IZO', 130: IZO', 70' ( a s m e physical constraints res hict bent location)

EOD to EOD width 108'

Note: Sofft flares at bents, length = 12'-0'

(a) Elevation Figure 3-1 9

3.5.2.1 Strvcture Depth

0.04 x 130" 5 . 2 U s e 5'-3"

Use 5'-3" throughout stnrcture for consistency

Sedion 3 - Prestressed Cancrefe Page 3-7 03

E* W Bridge Design Pradice - June 1994 m

3.5.2.2 Girder Spacing

Assume 4'4" overhang

10810" - (2 x 4'4'') - (2 x 6") = 99'

With 2 x D as a guide, use 11 '0" spacing

Therefore, 99/11 = 9 spaces, 10 girders

3.5.2.3 Typical Section

Based on 11 '-0" girder spacing, Btidge Design Defails suggests,

Top Slab t = 85'' 8

Bnttom Slab 2 = 7 l r ' 2

Note: Typical cectím is syrnrnetricai about bfidge.

(6) Typicai Section Figure 3-20

Page 3-104 Sectian 3 - Presfressed Concrete

r* Bridge Design Practice - June 1994 m

3.5.3 Loads

35.3.1 Pead Load

Superstmctme Dead Load is calcdated from BDC using m c s sectiorial properties. These preperties can be checked by hand or by using the PROPC compu ter progam.

Midspan Face at %nt

Area (ft, 2) 1811.25 213.63

3.5.3.2 Additional Dead Load

Future AC overlay and weight of bamers

AC = (weight of AL) x (width between barríers)

= 0.035 (10B-2-2~1.75) = 3.59 k Jft

Barriel = 2 x 0.392 + 0.41 1 = 120 k/ft

ADL Totd - UnifDrm Loading = 4.79 k/ ft

3.5.3.3 Live Load

# LZ k n ~ s = Deck width / wheeI loads

(Spec. Table 3.23-1)

= 108.0 / 2 x 7 = 7.714 lanec

Use this number far both the HS2Q-44 and Permit truck.

The cable pathis most important when ini tiall y develeping the Tongitudinal girder section. Once established, other pameters su& as soffit &res and post tmsioned anchorage requirements can be checked.

Sectjon 3 - Precfressed Concrete Page 3-105

E* I Bridge Design Practice - June 1994

' Cable path stope at d g e of suppod related to the mornent and shear force.

Figure 3-23

3.5.5 Losses Ffiction, Anchor Set, and Long Term losses wili all be explored.

It should be noted shat only one point in the s h c t u r e w i l l be hand checkd, the 1.0 pt. Spm 1, whichisthesame asthe faceofBent2. IniIialBDSrunc identified this as thepoint of Pjack design, therefore it Is fitting to hand check this crj tical location.

3.5.5.1 Frictional Losses

Due to fictional forres that develop from the curvature of t he prestress tendon and the deviation of the tendon wiffün the semi-rigid duct.

The enginew must start at the beginning of the structure and i n m e n t d y sum up frictioml losses lncurred unti1 the point of interest, in this case 1 .O pt. Span 1

Pt. yb Ae L a=2eSL Z F.C.i

O -4 1.0 1.62 46 0.068 0.068 0.986

Frictional Coefficimt (F.C.) = e - (pa+ KL)

= e - (0.2)(0.239) = 8.953

Page 3-1 08 Section 3 - Prestressed Concrete

Bridge Design Pradice - June 1994 m

At midspan: Clearance to mild reinforcement 4.5'"

Dictanm "D" 6.3'"

10.8" , say 12"

Thickness of Top Slab 8.63"

Clearance to mild reinforcement -1.0

Dxstance '9'' 6.3"

Check clearances at And-iorages

M m o to Dtxipers 21-28 recommends height and width requirements for anchorage devices. APjackper girder of 1860 kips requires aheight of 5Pt',leavingonly a9"possible depth for a joint sed blockout. Withthis tight clearance and recent failures due to dep blockouts required for joint seal assemblies, the mgíneer should attempt a more in depth investigation of the anchorage *ea,

Aftw cmulting wíth the jaint sed and prestress speaalist, check space requirements based on nanufacttrrers actual equipment dimensions. (See Figure 3-22]

ANCHOR

STRAN D C.G. - c - , J -

End Diaphragrn Figure 3-22


E* -1 Bridge Pesign Praclice - June 1 994 m

Abut 1 Bsnt 2 *cm Bent 3 b n t 4 Abut 5

" Cabie path clope at edge of support relatd to the mornent and shear force,

Figure 3-23

3.5.5 Losses

Friction, Anchor Cet, and b n g Tem losses will all be explored.

It should be no ted tha t only one poi nt in the stmcture will be hand checked, the 1 .O p t, Span 1, wkch is the same as the face of Bent 2. hitial BDS nins identilied this as the point of Pjack design, therefore it is fitting to hand check this critica1 EocaHon.

3.5.5.1 FrictionaI Losses

h e to f r i c t i d forces that develop from Lhe curvature of the prestress tendon and the deviation of the tendon within the semi-tigid duct.

The engineer must start at the beg-hdng oi the structure and inuementally sum up frictional losses incurred until the point of interest, in this case 1.0 pt. Span 1

Pt. yb de L a=2eiZ J: F.CI

O .4 1 .O 1.62 48 0.068 0.068 0.986

0.9 3.57 2.57 60 0.086 0.154 0.970

2 .O 4.08 0.51 22 0.085 0.239 0.953

Frictional Coefficient (EC.) = e - (va+ KL)

= e - {0.2)(0.239) = 0.953

Page 3-1 08 Section 3 - Prestresed Concrele

rZ Bridge Pesign Practice - June 1994

Stressing may take place at one or both ends of the structure. Jacking frorn both ends redu res the canb-ibutory 1engt.h for frictional forres, there fore effiaently w orking the prestress strand. However, the labor involved with this procedrire might negat~ an y economy denved from efficient stiand.

Studies conducted in 1971 showed that an increase in Pjack of 3% or less when jacking from one end only wouEd be economicaI. Thic has been h e his toric "break+ven" poht ever since.

Check Qne End versus Two End Stressing

Two separate runs of BDS were rnade in the prehüm'y stage with the following resdts:

one end stressing Pjack = 15,700 k

two end stressing Pjack = 15,880 k

Difference is toughly f % wkch is less than 3%, therefore one end stressing is more economical.

3.5.5.2 Anchor Set Losces

Anchor Set losses oulur at a localized area at the post-tensioned anchorage zone. The s b d wedges seat themselves in the anchor head when released by the hydraulic jacking ram. The anchor set los5 assumed in design is 3/8".

okay 12(1- 0.953)2025

Note ffiat the l e n e of influence foi the anchor cet loss is lo?, there fore the f .O pt. Span 1 is not effected.

Section 3 - Presiressed Concrete Page 3-109

C-tí Bridge Design Practice - June 1994

SPAN 1

Anchor Cet Figure 3-24

3.5.5.3 Long Term Losses

U= Low Relaxation Strand with a 20 ksi lmg tenn Iosc

(5pec. Table 9.16.2.2)

20 ksi / 202.5 ksi = 0.0988 Pjack

3.5.6 Force Coefficien t

Take the Frictional Force Coefficimt (F.C.) and subtract anchor set and Jong tenn losses

F-C. = 0.953

Anchor Set = 0.0

Total = 0.854

Page 3-170 Section 3 - Presfrecsed Concrete

E* Bridge Design Practiee-June 1994 m

3.5.7 Seeondary JMoments

Seccindary moments can be calcdated by hand using the satne procedure as the BDS p r o p . BDS calculates at 1 /M span intervals, so batid calnila tions will obviwsl y no t be as accurate. A 10% 2 allowance is reaconable with h d verification.

The PE diagxmrnust be developed to soive for secondqmoments. Force coefficients and eccwtsicities from the 23DS progran are used.

SPAN 1 P e Pe

Secondary Moment (PE) Figure 3-25

Take mornents about Point A by smmming the areas under the PE mrve.

Section 3 - Prectressed Concrete Page 3-7 11

r-G Bridge Design Practice - June 1 994

Areas Figure 3-26

Therefore, Ms = 0.621

BDC had 0,695, m r g h of error within 10%.

3.5.7.2 Distri buted End M.ornents

Fixed End Moment (FEM) Convention Figure 3-27

Page 3-5 52 Secfion 3 - Prestressed Concrete

E* Bridge Design Practice - June 1994 m

A B A BC BD CB

FEM O 0.095 - 0.461 - 0.487 0.426

A 0.234 0.061

BALANCE 0 -.(-0-082 - 0. 102 ,- 0.025 -- 0.024 COM O & - 0 - 0.025 - 0.01 2 &--

/

A 0,037 -

BALANCE - 0.01 3 - 0.016 DEM 0.59

Figure 3-28

Add the Distributed End Moment 0.59 to the original Pe moment at she 1 .O Pt. Span 1.

Pe = 1.36

DEM = 0.59

1.95

(BDS had 1.9445)

Sedion 5 - P restressed Concrete Page 3-1 13

rZ I Bridge Design Practice- June 1994 m

3.5.8 Prestress Sh ortening

Appkation of a Prestress forre at the ends of a f m e shorten the shuchiTe by an amomt equal to PL / AE. The"L" term in this egua tion is taken from the point of concern ta the poht of zero stnicture movement due to shortening. This point of zero structure movement is ~alculated using individual column stiffnesses. Stiffer coliunns attract the point of no mavement.

Memo fe Designms Chap ter 7 contaim an example for dculatíng the point of no m ovement This procedure is the sarne as what BDS uses.

For example, check the movement of Eent 2 due t o the Pjack force. Fhd the average Force Coefficient (F.C.) behveen %t 2 and the poht of no movement, whichis located 62' to the right af Bent 3. (approx. 0.5 pt.)

F.C. Wt 2 = 0.858

F.C. 0.5 Span 3 = 0.763

The length from k t 2 tci the point of no movment is 192'. Use an area of 180.25 ftz (no bo t tm silab fiare) to approxima te the average area between the two pointc-

The Distrhted End Moment PEM) due to prestress superstnicture shorterUng is 4.341. This decreases the overali prestress moment cogfficient as dmlated in Section 3.5.7.2 from 1.9445 fo 1.799. Using the equation in Cection 3.5.9, total prestress force required therefore increases by 5%.

The effectc of prestress shortening on the required Pjack and on the appiied c o l m moments are especially important for longer structures when detemünhg hinge locations .

Page 3- 1 14 Section 3 - Prestressed Concrete

E* I Bridge Design Practice - June 1994 =

Point of na movement for stnleture shoriening.

@ PMnt of no movement for prestresc ctrand.

Static Diagrarn Figure 3-29

3-59 Prestress Force

Knowing the point of control for Pjack design as the 1.0 pt. Span 1 Top fiber, hand verify-

D t e ADL + P/S

= 15,422 k

BDS had 15,420.



ln C

- d

c

w r n O E Z >

1 2 C1

VI E O C

u -

t m 0 c W m - E .r

m C tr, m Z 0

u' ron 0 C -- = a + 8 u 4 K L

0 m z

t' E c 5 E 3 4 * IL

CU fLc

@Y o - o = O v m - m

W E " o - . - u q 8 6

L-

0 w O ' O CL

+ C' I r z W

- 11) u cu o; M ~ G

LC - - - - - C S - =

t9 :'= & = m E e 0

E : P C O g E E I ir - - g:;: 3 - x

o m E F - m = ' ' '=- - . - 3 - ; - =;z 5 = -

x- , ,.=a=.= O

" m r2 - g 7 : 5 i g & -

N + z W m

d

- 1' Eí [9, = N & + M ~ ' < G ~ o ~ N & + & - C . - - -

-- -- -

Page 3- 1 16 Secfion 3 - P restressed Concrete

E* I Bridge Design Practice - June 1 994 m


E* Bridge Design Practice - June 1 994

Page 3- 1 18 Secfion 3 - Prestrecsed Conerefe

E* Bridge Design Practice - June 1994

Section 3 - Prectressed Concrete Page 3-179

ri. Bridge Design Practice - June 1994 m

Page 3- 120 Seciion 3 - Prestressed Concrete

c* Bridge Design Practice - June 1994 m

Sectian 3 - Prectressed Concrete Page 3- 12 1

I A ~ C O U ~ ~ I CAtTRANSBDLI-VERSiON3~00 REL-10 05 /10 /92# J W . (15,1991PhGE 1

D X V X 8 I Q N O F 8 T R W C T U R 1 8 - C A L F R A N S M B R l D O g DEYXQN PRACTICB ZXAMPLB

1

* * * I * ~ * * I + C I + I I * I . t C * e I m 1 + * 1 ~ * 1 i ~ ~ 1

NOTES a0 USERS / DESIdNgRB * . * . t * + * * * * * * * . * * 4 ~ + 1 * * 4 * * . I C 4 * * * m *

l BUS VBRSION o a . 1 m s R ~ L E A ~ B D ON DEC. os, 1987 nrm MINOR MODIPICATIONS W E BY THB DIVISXON OP STRUCTURBS.

1. THK DBPAPLTB CDMCRBTH ATRgNCITH POR PRBSTRBSSED BRXZiOBB HAS BEEN CHANOBD TO 4 , 0 0 K81.

2, THB PINSROLLBR OR ROLLER/PIN PROBLEH FOR BIMPLY SUPPQRTBD SB1DOB9 HhS BEEN (IORRBCTED.

3 , THE REDDCBD MOmWTB hRB REPORTBP hT TH$ PACE OF THE SUPPOII'P AS PBR SPECIFICATION 0 . 8 . 2 .

4 . aTIRRUP DBSIONB MM VAtID FOR PRB8TRESSED CONCRETE BTRUCTUILgS ONLY. THB PShTWRBI WST NOT USED l ü R

COWENTLDNALLY RBINFORCEII BTRUCTVRBS.

3 , IP A P R ~ S T R F ~ S LQAD COHBINATION ( 5 5 0 1 c m xs WM BUPPL~KD, ALL PR~STKESSINO OUTPUT WYLL BE BABBB ON THB

KBSVLTS PROA LIVg LO- ?#Om 1 $VEN IP OTHBR LIVR L O M S ARE I N P U T .

6 . WITH OR WXTHOUT THE tIVB LOAü CQMBINATION ( 5 5 0 ) C m , USBRS MllY CHOOSB 'TQ POUM MHStPrAND OTHBR LIm LOmINgS

C O W U W N T L Y TQ DEBION PRHSTRBBBBD CONCRETE B R I W P B .

7 . T H ~ LONO TERH tassss MR maum PRgaTRsaama BTEBL ~ N O A LOW-m) WILL BE DEFAWLTBD TO 32 XSI A U T O ~ T I C A L L Y .

0 . THR DEPAULTBD HOPYLUB OP HL~BTXCIPY, BC, POR THB BVBSTRUCTURB HAS BBBN R E ~ S B D FROM 3600 KSY TO 3250 XSI.

9 . U N t E S S S P E C I F I E D BY THB USIRB QTHBRWIQBi THB DBPLECTIQNS UflBD FOR CAMBlRG XKE RBPORTSD AT THHL QUARTER PQINTB.

1P. FiNY 6UOOBSTfONS OR M C O W N D h T I O N S POR BDB UPORhDBS SHOULD 88 DIMCTIID TO TIBN LBB AY 324-9239 .

. -

1

~ 1 ~ I 4 i ~ + ~ * i i b i i t r ~ + ~ n r + c ~ ~ a ~ ~ ~ b 1 1 1 + 1 ~ ~ ~ ~ ~ ~ * * $ n *

UPORADB XEHXNDLR POR BD9 M R S I O N 3 .O6 4 + ~ c ~ ~ I I ~ u ~ X R ~ ~ I ~ ~ B ~ ~ ~ ~ C + C C C ~ ~ C ~ C ~ ~ ~ ~ + ~ C L ~ ~ ~ ~ ~

BDS M R S I O N 3 ~ 0 0 DOS-01 WAa RBLIABBD OH D T / 1 1 t I 9 ~ 0 MYTR IHB POLLOWIWO MODIPICATIONB :

1. A MINOR "BUO" IWXCH MPBCTED THB RHBhR BTTKRUP DBSIIIN hT CBRTAIN POXNTS I U B BEBN CORRgCTRD.

2 . THB OVTPUT NOPI comIme UNITS POR SECONDARY M O M B ~ S ms eaum DISP~ACEMENTS,

3 . THB BDS TOP PfBBR ANü BOTFOH PfDER STR88B PLQT9 MVB BBBN REARRANOED ON THB PLOT OUTPUT BHBBT.

d . THB DIBTRIBUTSD BBCOND)LRY COLVHN MOMBNTS DUB TO EACH PMSFWBS CMLE PATH ARB WORTED.

5 , THB "POINT QP NO MOVEMHTw DUE TO PRICTXON LO88 Hh8 BEEN RRWOR3BD AS THE "POIETZL 01 NO HOVEKENT POA

PRBGI'FRtfBSfWd FiND THB "POINT OP NO MOVBMENT" WWE TO RIaIDITY HAS 8 8 K N RBWORDBD A 8 THBl "POLNT OP NO

MOVKEZBNT POR IHORTBINLN~".

6, m a mPoaTen PJACIC NO# AOUNDBD TO THB m m s ~ 10 KIPS (MBMO TO D E S I C I ~ R 1.1-4).

7 . THB mfm m x a OF SPANB I ~ L L O ~ P ~ I D POR LIVE LO= IDALYSIS IB LIM~TKD TO a o , H O ~ V B R , A SPECIAL RVN CM ~g

ARRaMOBD POR EITRUCTUKSS WITH $PAN& MOHORF THkN 3 0 .

8 . THB WARNING MESShOI ON THE ULTIMATE MOCIKBHT OUTPUT EHlIBT WILL BB PRINTBD ONtr WHBN THB CUMBINBD R B I N P O R C & ~ . I B N T

INBPIK "Rp" f 8 QREATFIR T!W 0.30.

9 . PARTIAL PRltCTRgSSLNa DESXUN 15 ROW AVASWLE POR TESTINU, THOSK DBBIRfNQ TD V8E THTB OPTXON SHQULD CONTACT

TIEN LEE AT 0-9239 POR AS8fBTñNCB.

10. DOS-01 D T E b % / 2 3 / 8 8 1 A BU0 WHfCH CAUBBU THI f N T E R N l t FORCE TENRIQN (T I NOT BQWIiL COMPRXSSION (E) AT THE

ULTIMATE MOKBI4T aTAo3 HAS BEEN CORRBCTEDA 11. DOS-oa DATZD a / r > / e u , !ras R~QUIREP CONCRETE B T R B K ~ T A AT 2 0 DAYS WILL m 8 ROUNL~ED Pa THE W B ~ ~ R B S T CDO S S X .

IP RBQVTRPID. ( W M O 11-16 DATED 3/2/1972)

r-t. -1 Bridge Design Practice - June 1.994 m

Z r a O 4

H . m r * r z .

: P f l : - m - - 0 - a - *

i 2 ; : 5 : r a:. r u . I * * E 4 * o * c k , r

i E r * m # : S:

H . r E * : 2 :

Page 3- 124 Section 3 - Prestresswi Concrete

E* I Bridge Design Practice - June 1994 M


E* I Bridge Pesign Practice - June t 994 m

0 0 ~ 0 0 0 0 g,, .o=non, . . * . . . . 0 0 ODOOPOO

€4 * w 52H ~ ~ Z ~ Z Z ~ 5 &00A&Q&


E* Bridge Design Praclice - June 1994 m

UZO & O oonl zoocfl Zi

m 9 vi2 00 OOOD

- I m I

m z m u u m M XinininHZW N H m-4rPma H

2 :Y;U;Y;2,2 4 d d m m L1

D a a ; : 2 &-a O

dd onqo* -0oCh-I Pi - u.-. .. Óioz oomo at o

0-0 il OrYd€.O 4 r(rlu d r l Y F H

w u z o 9zF;zo d d d w zddd d m z B LOE

11a ~ 0 0 0 0 0 0

Section 3 - Prstressed Concrete P age 3- 127

EZ Bridge Design Practice - June 1994 1

- DOOU J a000

N N

Page 3- 328 Section 3 - Presfressed Concrete

E* i Bridge Design Practice - June 1994

N.rr O tan

2 0 Hd l3

O " a

. .

u m 0 3 1 H Z * O N m i n

Y 1 m 1

m z m w n m m EDE.PM N CI uoa H E - - m w d W O e i O L í ' iTI

=El n O &E32 8 &no U a 0:

oe. -0m o, - v

Az 80 T z c1 O J EDE J H

$20 E E a o YiY 3- J W Z m 2 B urg

O 0 OQOOO


E* I Bridge Design Ptactice - June 1994 m

Page 3- 130 Section 3 - Prestresced Concrete

E* Bridge Design Practice - June 1 994

Section 3 - Prestressed Cancrete Page 3-731

732 - Seetjon 3 - Prestressed Concrete

r-t: Bridge Design Practice - June 1994


1ACCOUNT 1 CALTRANS BDS-VERSION 3 . 0 0 WL-10 0 5 / 1 0 / 9 2 , JAN. 05, 1 9 9 4 PAGE 10

D I V L S I O N O F S T R U C T U R E S - C A L T R A N S AN BRfaGE DE81M PRACFICE EXAMPLE

OgIDESWAY DIA-STICS

D m m O R S FLñlND

RESPLTS OF 1 m H SWAY M THE RIQrn

VERTICAL SKEAñ (KIPS) MCWN'l'S (PT-KLPS) m E R LT RT

BkSED ON E 3250. RSI. 5 1037.8 -19038. 37522. 6 874.7 -17645. 1 7 3 4 4 . 7 4407.8 -52098. 44174 .


. . . m . . - m - m -

R m m O F u m a a m e ?.- m ri 9 t- m r u m b m m m ~ m r * m w e n m m N C Y W

1 1 1 rl l r l O P - V I F l r l t l

1 I I

4 , . * . . . . . - ( Y W P Poi W P d m m ~ m m m d d m m i n

0: m 4 m ni oi N A F ~ 0 d w 4 uidn 0 - m a wi m m ~d d~ fl O ~ N m m m E m m I ru vi I I rl ~ r l n m m - I d i 1 3 1

4 . - * * . - . - . . m

U E + & y; ; 0; O ~ D D R W O ~ m n w e C L i C Y 0i V) 0i m d m w m d m m = m 4 ~ m i n

m V1 r l m n ~ d d 4 w - i R ~ W a m ~ I r - m m d rri L I I 1 I d D W V I

ri r l l F 1

VJ

w . . * . * . . . . - m w w ~ r n rrr n N m m m ~ m u r * - m m e c

E p r m o m m m - 0 - N V ~ N m q m H W Q m u m m q n m d 0 - -

3 m G l W LL) VI i $ 1 I I d - - ( Y . A rl rl I I I

Er

C1 . . . * . . , . . A

3 bt- * io 0 ' - - o m m d , - Y W * b m m w - - = * m m m o w i - ~ r n m n p w - m d U O w w - r l m m r ? w II r l w w

r n w r- r- 9 I r I I rl n * N rl rl 1

H a W

O H - - H rl rl

. i . . . . . . . . , + m r n d w m o o n n o a c i m m m - m m m m m m o n m r - N

n W bi m O w d d O m d d L m d d r - W m H P . 0

.P. W I? 1 1 1 rl N C C P

V l V i U N a O w VI 4 o

H a * Y H > m * M a m , , . A i G ; é ; ; ~ i 6 ; n ; ; a i ; ; ~ . . - C L ~ ? * E O m m r m n r - m m o u w m ~ d w r ( - m 3 m i , i . y g n n z F % Z E m i U l d O W t l *

m

m s m w m m - rrl t z N S IO E I r i a m - i ?

O ' - " . - " - N O E O

H N H

N w E H a d ! 3 # n d , o d n n - o d o n r

o x

O Q 0 ~ 0 ~ 0 ~ ~ 0 0 0 D 0 0 0 0 ~ 0


E* Bridge Design Practice - June 1 994 1

Sectiun 3 - P restressed Concrete Page 3-137

E* -- Bridge Design Practice - June 1994 m

Page 3- 138 Secfion 3 - Prestressed Concrete


Section 3 - Prestressed Concreta Page 3-139

r-É Bridge Design Practlce - June 1994

m W & W E.' . * . P I O w m r i

t- 1 m 1


IACCOUNT : C b L W S BDS-WRSION 3.00 REL-20 05/1O/PZ, JAN. 0 5 , 1994 PAGe 17

D I V f S X O N O F S T R U C T U R E S - C A L T R A N S AN BRIDGE DESIm PRACTICE WAMPLE

OTRIAL 1 OTANGENTIAL RDTATIONS - RADIANS - CtDtKWISE POSITIVE

CPAN LT. END RT. END EPAN LT. W ü RT* SPAEI LTm EHD RT. END O 1 0.000456 -0,000o73 a -0.oo0073 -o.oeooes a - A . 000005 -P. OOOOBO O 1 -0. 060080 -0.OD0042 S 0.000000 -0.000073 6 Q .000000 -0.OOOOQ5 O 7 0.00000~ -0.000080 OHORfZONTAL MEMgER DEFLECTIONS IN FEET AT 1/ 4 P o m a FROM LEF"F "F - W M D POSITIVE: O MEMBER 1 E-3600, 0 .000 0 , 0 3 5 0 . 0 4 2 0,022 O ,000 O MEMBER 2 ~ ~ 3 6 0 0 . 0.000 e. a i o O.OZZ 0.013 o ,000 O ~ E R 3 EW 3600. o , o00 o. 011 e.oza o. 0 1 4 0.000 O MP~BER 4 E- 3600. 0 . 0 0 0 - 0 . 0 5 1 0.001 a . ooa 0 . 0 0 0 OVERTfCAt MEMBER DEFLFCTIONG I N FEET AT 1/ d POfNTB m LEPT END. O -EA 5 En3250. 0.000 0.(100 0.000 -0.001 -0,003 O 6 Es 3 2 5 0 . 0.000 t' 0.000 - 0 . 0 0 1 -0.002 -0.003 O MEMBER 7 E= 3250. 0.000 0.000 -0,001 -0.001 -0.a03

MEMBEñ DEFLECTIONS HAVE BEEN MTLTIPLTED BY A CREEP FACTOR OF 3 . * *

I A C C O W : CALTRANS BDS-VEREION 3,00 RCL-10 05/10/92, ZAN. 0 5 , 1991 PAGE 10

D I V f S I O N O P E T R U C T Q R E S - C A L T R A N S AH M I W E DESIW PRACTICE W L E

OLIW XlOm DIAC1;X)STTCS

O NO ERRORS POVNI) OSOFERSTRUCT[iRE LIVE MAD HS20 LiVE LOADING

1

I O NüM3ER OF LIVE LOAD LANES RESISTlNd HWENT OF PMT PtOT IWLIU- MM SZTPERSTRUCTURE SWSTRUCTETRE UNIT BTEEL M S SCALE ENCE NO. LT-END R T . W LT.ESiD RT-END WSITIYE UEGAl'lfrE W . LINES GEN

1 7.714 7.714 1.0 1.0 O . 6 . O O NO NO 2 7.714 7.714 1.0 1.0 O . D. 3 7.714 7.714 1 .0 l. O O . O . 4 7 . 7 1 4 7 . 7 1 4 1,O l. O O. 0 .

OLIVE 4 U C K - - LANC-- MJS. LfvE: LOAD P1 D1 P2 D2 P3 UNIm3Iu3 W W . SKml Wi LnAI, NO. RIDER RIDER IHPACT LNC. SIDESWAY

l

1. B.O 1 4 . 0 32.0 i a . 0 3 2 . 0 0.640 18.0 26-13 YES 0.00 m

COMMENTSr HS20-44 AASHTC) M M I N O t WITHOUT AL-Tm

AA6Wiü INPhCT FACMRS C A X ü i A T E D BY PIIWFAM O MEM IWACT

NO %

1 ao 44 2 19.6 3 2 0 * 4 4 25.6


-



. . . r l o m m N t- W N N I ? V I C U d ( r l rl w ui m 1 Vi i i r H l 4 W m

1 I Cr] V1 m m W

4 m o r n t m m a r n w ~ ~ ~ n 10 m 01 m m r c u i n m w m

U z O 4 n m d d O M W M O W N W

i. u U N


1ACCMfirP: C A L ~ 3 B D S - V E R S ~ O N 3 3 . 0 0 REL-10 05 /10 /92 , JAN. 05, 1 9 9 d P A G E 21

D I V I S I O N O F S T R V C T U R E S - C A L T R A N S AN BREWE TJESIm PRACTICE EXAMPLE

OLL N. 1. POCITIW LIYE ZOAD mllENT ENVErAlPE AND ASGOCIATED SKERAS OMEM IrEFT . 1 P T . 2 P T . 3 P T . d P T . 5 P T . 6 P T . 7 P T . 8 P T . 9 P T m 1 0. 6362. 10293. 1 2 5 9 5 . 13387. 12881. 11172. B36B. 4854 . 1707.

SHISAR 0.0 513.5 4 2 8 . 9 5 0 4 . 5 226 .3 -310 .4 - 3 8 1 . 5 - 4 4 5 . 3 - 5 0 0 . 5 -213 .9

2 1565. 1981. 5566. 0700. 10652. 11194, 1 0 5 5 4 , 8514. 5338. 2 3 9 8 '

SHEAR -49.3 196.6 459.6 3 8 6 . 5 307.3 -232.1 -313.6 -392.2 - d 6 4 . 2 - 1 7 4 . 6 3 1765. 2395. 5176. 7989. 9637. 9 9 8 0 . 9204 . 7 1 6 4 . 4164. 2001.

CHEhR -24 .6 2 0 4 , 9 453 .3 377.1. 2 9 5 . 4 -242.9 -325 .7 -404.6 -475.5 -158 .3

4 1417. 1639. 275s. 4ua3. 6377. 7 2 0 2 . 7422 . 6 9 3 8 . 5735, 3 4 5 5 . S- -20 .2 197.6 532.3 43342 365,l -329.9 - 2 1 6 . 3 - 3 0 4 . 5 -409 ,7 - 4 9 3 . 5 OHORSZONTAL M E R STRESSEB Lb MAX POS BOTTQM FIBER O 1 O . -154 . -25'7. -314. -334 . -321. -279 . -209 . -121. -43. O 2 -30. - 4 9 . - 1 3 9 . -217. -266. - 2 7 9 . -263. -212. -133. - 60 . O 3 -34 . -60, -129, -199. -240. - 2 4 9 . -230. -179. -104 . -50 . 0 4 -27. -37. -69. -120 , - 1 5 9 . - 1 B O . -185. -1'73. -143. -86. OHORIZONThL MEMBER STRESSEB LL MAX WS 'fWP FTIER 0 1 0. 133. 222. 272. 289. 2 7 9 . 2 1 1 . 101, 105. 37. o 2 33 . 4 3 . rao. IBB. 230. 242. 2 2 0 . 184. 115. 52. O 3 38. 5 2 . 1 1 2 . 173. 208. 216. 199. 1 5 5 . 90 . 43 . 0 4 30. 35. 5 9 . 104. 130, 156. 160, 150. 124. 75.

C-t. Bridge Design Practice - June 1 994

= - . . * . . * P o d o b t c 0 1 0 o m m m r L w Pi ID * m c - m

1 1 )

- . . . 1 - . .

P W P I - m a c s w C Y A r n O N P W W

,-Id,-, I d 4 N I I I

Page 3- 1 4 6 Secfion 3 - Presfresced Concrete

E* Bridge Design Praetice - June 1994 m

E: - Y . ? n m y i w *

w I I m Y L 9 m . C Y - m r l w

m I r -

? . ? .=? O r n b W C c r - W I * O ~ N P E T r n I r l

a i a '4 rl

m . . .

O W P i r l

:-,-Y;$ m 1 m

ta

Section 3 - Presfressed Concrete Page 3-747

E* Bridge Design Practice - Sune 1994 m

g.n E ? ? g q w Q-!? w m d O V I N o m m 0 - n ~ F P ~ m m H ~ P H O P & * a ErnU E r . * c i m m

N I ? rl N "4 TY h M


E* , Bridge Design Practice - June 1994

o q o - m 0 0 . 4

Sectjan 3 - Presfressed Concrete Page 3- 149

el* Bridge Design Practice - June 1994 m

o o o o q q q W t V O W O d O O O W O N O

- . . . w q r l d N m m w r n h P 4 Q 4 a i & D l

0 0 0 0 0 0 0 0 0 0 0 - 4 - r C - m .al . m - 4 -

m r n , - ~ o ~ o m o ~ o n o P d c l C) a CI n x 0

- -

Page 3-150 Section 3 - Prectressed Concrete


O D O O O - 4 . p . m . m . v i 9 d 9 r c 9 n 9

o m o m ~ w o ~ o ~ ~ m ~ m n m ~ C ) n Q a n E 4 C I P

0 0 0 0 0 0 ., . b ., .e, . m . , + ? t , 9 , 9 ~ u i ~ u i o w o w o ~ o m o m o a o

& h & P i & r n & h

Sectíon 3 - Prestressed Concrete Page 3-151

r-G Bridge Design Practice - June 4994 m

- 0 1 -

E r i i w

H P i r l d r i l

Page 3- 152 Section 3 - Presiressed Concrete


. s . . * . . I . . .

b i C H r n t - m m r l m d * m N m m W m l l i l l > Y I

, - í r l w - o p w i m w a I r n , - I D d . - rt 1 1 1

- 1 7 N 1 4 1

Z -

6i : O E s - 0.



Q U Y W . . . . . r ) r l Y I ~ U I I l )

W m m m

la 7 1 -

E 0 0 w W O m W 0 o * n n O N O , - I

Page 3- 154 Sectfon 3 - Prestressed Concrete

r-G Bridge Design Practice - June 1994

w - d 1 1 1 E m . . F . . . . m . . . * 4 m m o m m m D E m o m H P m m n n - m d D m m - 0 - W O m N - a m m ~ rl e, ti r n d r i l t m r l - n ri ~ i i

Section 3 - Presfressed Concrete Page 3- 155

E* Bridge Design Praetice - June 1994

E W . Y . ? E ? . ? . Y k m , i q - 7 E ? . ? . n W W O ~ D ~ w r l m w m t - W ~ m v i v i * W O l r r O W U d w ri m d m m n n r i G I h l O U l O d d w - w w r n

rl rl m . e c w , - ~ m m V I V I I W P - - !--,m I N - * ,-Id I 47 N d r n i * + * r n d O d F I W a d e Q d N CY rl

Page 3- 156 Sectlon 3 - PresEressed Concrete

OLL NO. 4 .

OMEMBER 1 LEFT OPOS, V 2335.3 mG. V 1330 .0

OLL NI]. 4 .

OMEMñER 1 LEET OPOS. V 3 9 5 7 . 3

NEQ. V 2039.8 OLL NO. 4 ,

OMPreER 3 LEPC OPOS. V 3647.5

NEQ, V 1983.2 DLL m. 4 .

OMCKBEñ 4 Liem OPOS. V 2 9 5 5 . 1 ?iTC, V 1504 . O

CALTRANS BDS-VERSION 3.00 RFL-10 05110/92, JAN. 05, 1 4 9 4 PAGE fZ

D X V X E 1 O N O F S T R W C T U R E E : - C A L T R A N E AN BRIWE DESIGN PRACTICE EXMGLL

DEltIl MAJ3 PLVB LIVE WhD SHEAñl ENYELiOPE *** SPECLALTRUCKWITH 1 A X L E S W A S R E W S T E D T H T S L W L O A D * * *

.I PT . Z PT , I PT .U PT . 5 PT . 5 PT .7 PT . 8 PT . 9 PT 2252,6 1597 .9 974,7 384.7 -171.1 -692.7 -1101.5 -1567,6 -1949 -6

956.0 556.2 5 9 . 4 -492 .2 -1066.9 -1661.5 -2302.0 - 2951 .7 -3618.9 DEhD UJAD PLVS LIVE mAD smm ENYEIBPF

*+* SPECIAL TFtUCK WITH 7 AXüE9 WRS REUUJ3STFD THIS LIVE MRD *** . 1 P T . 2 P T . 3 P T . 4 P T . 5 P T . G P T . 7 P T . B P T . 9 P T

3216.6 2501.0 1800.2 1124,O 4B1.3 -122.2 - 5 6 2 . 5 -976 .4 -1390.3 1595.4 1101.5 767.6 2 8 3 . 5 -330.5 -981.7 -1663.8 -236B.O -3082 .4

DEAü PLVS LIW IiQAD $HEAR ENYEiOPR *+* SPECLAL TRUCK WLTH 7 AXLES W A S REQUESTED THIG LIW LOAD ***

. 1 P T S Z P T * 3 P T - 4 P T m59T m 6 P T . 7 P T . 8 P T . 9 P T 2994 .8 2310.8 1669.5 1050.1 467.8 - 7 2 . 5 -158.4 - 8 4 0 . 5 -1222.5 1571~1 11~9.1 B Q T . O a s ~ . g - 2 ~ 3 . 5 -097.6 -1542.1 - 2 z i d . a - z o e a . ó

DEAD MAD PLUS LIVE 113kD SKEM EPPaEiDPE *** B P E C U TRUCK WIm 7 AXLES W A S REQUESTED THIS LIVE MRD ***

. 1 P T . 2 P T . 3 P T . d P T . 5 P T . 6 P T . 7 P T . B P T , 9 P T 2516.6 ai11.7 1712.6 ia91.o 9 o s . a s1r.s 200.1 -14.e -237.7 1 2 5 7 , l 1028.9 806.0 583 .2 312.2 - 8 0 . 6 -492.8 -990.6 -1293.6

l

1 ACCOUNT 1 -9 BDS-VERSION 3.00 REt-10 0 5 / 1 6 / 9 2 , JAN. 05, 1 3 9 4 PAGE 33

D I V I E I O N O F S T R U C T U R E S - C A L T R A N S AN BRIDGE DERXW PRACTICE EXAEiPLE

+** S P E I k L TRWK W I T H 3 AXLEB W A S REQUESTED THIS LIVE IrOkD *** OLL Eñi- 4 . LiVE UJAD CVPPORT RESULTS O MAX. AXIAL I a A D MAX, WNQITIJDINAL MDMPPT

MIAL - MIAL -- LOAD TOP BOT. MhD TOP BoT .

OGUPPORT JT. 1 POSITIVE: 190.4 a . o. o. o 0. o . NEOATIYE: -1607 0 . O. 0 . 0 0 . 0 ,

OMEMBER S POSITIVE 334,l 2 4 2 . -121, 2 3 7 . 5 1324. -662. NBGATnrE -20.6 214. -107, 2 1 4 . 9 -1233. 616.

O W E R 6 POSITIVE 324 .3 144 . -72. 2 2 7 . 7 1057. -529. NEGATIVE: -30.8 -240 . 124. 194 .a -875. 437 .

OMPIBE3t T POSITIVE 317.0 520. -260, 210.7 1228. -614 , N'EGATlxE -34.8 -321. 161. 141.7 -556. 278.

OSUPPORT üT. 5 PORIT~SC 132.1 o. o , o. a o. o , NBGATWE -30.1 5 , O . O . O O . 0 ,

O Ti-iEFlATIOOFGVBS'SRUCTüRE/ SVPERSTRiETIIRELOADIEFGfS0.130

IACCMINT: CAIiTRANS BDS-VERSION 3.00 REL-10 05/10/92, JAN. 0 5 , 1994 PhGE 34

P I V I S I O N O F S T R U C T U R E B - C A L T R h N S AN BRIDGE DESIW PRACTICE EXAMPLE

O NO PRESTRESS COMBINATION DATA G I W SO DEFAULTS WERE VSED. O LIVE LDAD NUMBER 'i* RESULTE USED m e/s DESIGN RND OTHER LIVE MADS, IF P R E S ~ D ,

ALSO WILL BE CHECKED M DETERMINE THE VLTXMATE: MQhENT CAPhCfTY.

0 m FOL- VAT.WES ARE B R m USED IN THE CRtCtTIrATIOW OF mMENT br SHEAR REQUIRE3ENTS, O D.L. mñD FhCMRr 1.30

O LL NO. 1 ULTIMATE MMENT APPLIED - 1.30 X (DLtADL) t 2.17 X (Lt+I) + 1.00 X (P/R SEC. MDMEWF) O LL NO. 4 U L T M T E MübEMT APPLIED - 1,30 X {DL+ADL) + 1.30 X [LLtI] + 1.00 X (P/8 SEC. MDbDZil') 0 Lk mi. 1 QLTIMATE SHEAR APPLIED t. 1.30 X (DL+ADL) + 2.17 X (LL+I) + 1.00 X (P/S SEC. SHEAR) O LL NO. 4 ULTIMATE S- APPLTED - 1.30 X (DL+ADLl + r.30 X (LL+I) + 1.00 X ( P J B $E. SHERR)

c* Bridge Design Practice - June 1994

Page 3-160 Sedion 3 - Presi'ressed Concrete

E4 Bridge Design Pradiee - June 1994 m

E,,,, nrcridrq H . . . a r( d ri Fi

IP f i n n n m * F C - o m w m m

z H m w m m m m m s m ~ m o

u , . . . . m,, , ,,,, 8 % .

.. . . . e . . . .

' Q * W N

Section 3 - Presfsessed Concrete Page 3-16?

E* i Bridge Pesign Practice - June 1994

iY F m w w m o r n o p .

q m l l * r r r m m m n

L . . . . O O O C i I L I I

Page 3-162 Seciion 3 - Prestrecced Concrete

1

1

IhCCbüHT : CALTRANS BDS-WRSION 3.011 REL-10 05 /10 /92 , JAN. (15, 3994 PAGE 38

D I V I S I O N O F C T R U C T U R E S - F A L T f l A N 8 AN BIlIWE DESIW PRACTICE EXAMFLE

OFEMS DELTAS XN COLUMNS DWE M SHDRTmING - P J A C K m l OTRIAL m 1 0 l?FAME 11 PATH m l A O m FM PEM DEmA TOP OF COL,

NO LT. END RT. END (POSITIVE M RIGIPT) -mIT = FT O 5 -0 .43319823 6.43110823 P.00000162

l

1

O 6 -0 .10541139 0.10541427 0.0[1000049 O 7 0,33825767 -0.33825749 -0 .00000040

l

z

E* -1 Bridge Qesign Practice - June 1994 m

Page 3-164 Secfion 3 - Pnestressed Concre fe

E* Bridge Design Practice - Jvne 1 994

Sectíon 3 - Prestressed Concrete Page 3- 165

D I V I S I O N O F S T R V C T V R E B - C A L T R A N S AN BRIIIGE DESIm P W T I C E EKAMPLE

DTRIAG 10 ImmE 1 PATH 1A QHORIZONT& MEMBER MOMENTS DUE TO P/s m NO LEFT . 1 P T . 2 P T . 3 P T , 4 P T m 5 P T - 6 P T s 7 P T . B P T * 9 P T RXGHT

O 1 - 2 4 8 3 . -10941, -16853. -20106. -20909. -18952. -14311. -6931. 3221. 16174. 27713. 0 2 32714 . 19635. 4980. -5558. -11047, - 14551 . -13161. -7875. 1 2 1 6 , 14040. 24637. O 3 26927. 15115, 1873. -7547. -13217. -15201. -13590. -8381. 331. 12484 . 22536. O 4 17442. 11792. 4256. -128. -3737. -5685. -62e9. -5990. -5190. -3890. -2094. OVERTICAL m E R MOMEWfS DV& M P/S MPI

NO LEET ,1PT . 2 P T . 3 D T , 4 P T . 5 P T . 6 P T . 7 P T . 9 P T . 9 P T RIGIPP Q S -6165. -5051. - 3 9 3 0 , - 2824 . -1711. -597. 516. 1630 . 2743. 3 0 5 7 . 4971, O 6 -2245. -1792. -133&. - 0 8 5 . -431. 2 a , 476. 930, 1383. 1837 2 2 9 0 , O 7 4 2 0 5 . 3275. 2 3 4 5 . 1415. 495 . -444. -1374. -2304. - 3 2 3 4 . -4164. -5094, OTANGENTIRL ROTATIOEIS - RADIANS - CU1CKWISE POSITIVE

SPAN LT. END RT. END BPAN LT. END RT. END ElPRN LT, m RT, END O 1 -0.001975 O . 000374 2 0.000374 -0.000016 3 -0.OOPOl6 O , 000170 O 4 0.000170 0,000330 5 O.ODii000 0.000374 6 D . Q O O O 0 0 -0 .000016 O 7 0.000000 0.000170 CIHORiZONThL MEMBER DEFLECTEONS m FEET hT 1/ 4 POINTS FROM LEFT END - m h R D POSITIVE O M E M B ~ 1 E- 3600. 0 . 0 0 0 -0.167 -0.178 - 0 . 092 0.000 O 13EMBER 2 E - 3 6 0 0 . 0 .000 -0.045 -0.100 -0.05B [1.000 O m E ñ 3 E-3600. 0 . 0 0 0 -0 .056 -0.102 - 0 . 0 6 0 0.000 O -m 4 En3600. 0 . 0 0 0 -0 .OOd -0.014 -0.013 0.000 OVERTICAL HEMBER DEFLECTIONS IN FEET AT 1/ d POINTS FROK LEPG END. O MEMBER 5 E= 3250. 0.000 O . O 1 1 0 . 0 3 6 0.064 O . 000 O MEMBW 6 E- 3250. O . 000 O . 005 0 . 0 1 6 0.026 O . 000 O &E!bfBm T E- 3250 . 0 . 0 0 0 -0.003 -0 .008 -0.013 O ,000

UEMBW DE%"i.JECTIONS HAVE BEEN MXTIPLIED BY A CREEP FACMR OF 3 . * +

1 ACCOVEfi r CALTRANS BDb-VERSTON 3.00 REL-10 05/10/92, JAN. 0 5 , 1 9 9 4 PAGE 42

D X V X B I O N O F S T R U C T U R E I G - C A L T R A N S m BRIDGE DESI& PRPICTICE -LE

OTRIAL 10 FRAME 1 OHORIZONTAL MEMBER CTRESSES M R ALL PIS PATHS BEWRE m108988 B O T W FIBBR {PSX) MPI NO LEFT .i PT . S PT . 3 PT . I PT . 5 PT .E PT .7 PT . O PT . s PT RIQHT

O 1 6 2 2 . 860. 1026. 1120. 1141, lOB7 m 954 + 755. 473. 115. -164.

O 2 -163, 20. 624. 714. 891. 9 5 7 . 915. 7 6 3 . 505. 142. -129,

0 3 -148. 97 . 467 . 730, Be&. 942, 896. 7 4 7 . 500, 155. -44 .

O 4 -18. 145 + 358 . 490, 5 8 3 . 638. 654 . 646. 623. 586 . 5 3 5 . DHORIZONTAL MEMBER GTRESSES FOR Alrh PJ9 PATflS BEFORE WSSES M P FIBER IPCI) 0 1 49.3. 231 151 71. 59 . 109. 223, 404, 651. 965. 1192. 0 2 l l B 5 , 1028 . 673. 417. 258 , 196. 2 2 s - 355. 5 7 4 . 8 8 4 - 1082. O 3 1104, 8 9 1 m 566 . 333 . 1916 139. 176. 300, 509. 803. 9 2 3 . O 4 894 746. 5 9 5 , 477. 3 9 4 . 363 . 326. 332. 350. 381. 424,

i

IACCOUNT i CALTMNS BDS-VERBION 3.00 REL-10 O S J 1 0 / 9 2 , JAN. 05, 1994 PAGE 43

D I Y T S I O N O F S T R U C T U R E B - C A L T B A N S AW BRIDGE DESIC3 PRACTICE EXAMPLE

OTRIAL l b !mmE 1 OHORLZONTAL i4EMñER STRESBES FUR ALTA PIS PATHS AFTER Att MSSES 80- FIBER (PSI) MPI NQ LEFP , 1 P T - 2 P T . 3 P T . 4 P t - 5 P T . 6 F T . 7 P T . 0 P T , 9 P T

O 1 556. 769 . 919. 1004. 1024. 977. 863. 681. 430 . 109. 0 2 - 2 0 4 . 12. 374 . 6 3 5 . 794. 054. 8 1 7 . 683. 4 5 3 . 131. O 3 -123. €17. 4 1 5 . 647. 7 8 6 . 832. 790. 658. 438 . 132. O d 2 9 . 133. 320 . 4 3 5 . 316. 563. 576. 5 6 8 . 548. 515 . O HORXZOEPTAL MEMBER STRESSEB f?OR ALL P/S PATHS AFPER &L MSSES MI? FIBER (PSI) U 1 4 4 1 . 260 134 . 64, 51. 95. 197. 350. 579. 061, O 2 1127. 926 606 . 376. 233. 176. 205. 316. 510. 7 0 4 . 0 3 9 7 2 . 791. 502. 296. 17 O. 125 . 1 S B . 2 6 0 . 4 5 3 . 713. O 4 7 3 8 . 651. 510. 4 1 5 . 342. 298 , 281. 290 3 0 6 . 334 .


m , . V . . . . h O ' ~ m ~ m m o m a i ~ r i m o r d - ~ n v i

* m Q i l L F * w I D l n m * m H Ir

Sectian 3 - Prestesed Concrete Page 3-169


u n + + + J . . . . + . . . . . . . . m m m r n a n r n r a c o m p a K n 8 d

!4 H h w d - m ~ W P I , + ~ P - O ~ ~ ~ - - P m m w m U r - n t u m m m w - - m

~ L Z vi N m w v1 m = . 2 El W V1 V1 m

5 m LY . . . g . . . . ,-IZ * A & A i É A 8 d W m o d d A - r n

& ~ ~ m O w ~ w O m L l m ~ d ~ w m m n m w m m m m w w m m w m m m n n ~

D: 0:

. . . . . . . . . . . . . . . . 2 m b w m

~ w ~ w ~ d d r n m m w w o n d b m m , a h m ~ a w , , - # , P w a u i o m m a , N m m 4 5 ~ m m m C ~ u N P l N

8 8 G e H d H

a * H H 8 H

c: d EL: g ~ ~ 4 - ~ m - ~ ~ o n - o ~ ~ n - o ~ m m ~ o O

x x 0 0 0 0 Q 0 0 ~ ~ 0 0 ~ ~ 0 ~ 0 ~ 0 0

Page 3-1 70 Section 3 - PresFresced Concrefe

r-G Bridge Design Practice - June 1 994

Seclian 3 - Pmstressed Concrefe Page 3- 171


t . . . n 2 0 0 O D O O 0 0 0

Z o a a e o a a

Page 3- 1 72 Section 3 - Prestressed Concrete

E* Bridge Design Practice- June 1994 m

Seciian 3 - Prestrecsed Concrete Page 3- 1 73


PiPi -40 p . . . . ~ . m m w i n

Page 3- 1 74 Secfion 3 - Prestresed Concref e


Cecfion 3 - Prestsesced Concrete Page 3- 175

c* Bridge Design Practice - June 1994 m

r a m a 191010 N C Y N

r3 rri pr W I D W N N N

Page 3- 1 76 Sectian 3 - Prestressed Concrete

E-G Bridge Design Practice - Jurie 1 994

-*

Fl 1 . . . . . . . . . . . . . . . . . . . . . . m m + r n e a o m e r n d r - ~ n m m m ~ n m n m

ril o d ~ m w ~ w m m d o a d w d ~ n m o ~ a r3 2 t - m * w I . F b w m w r - ~ w u ? ~ w u l m w u l ~ g :

9 m + -

01 . . . . . . . . . . . . . . . . . . . . . . ol d 8 % E " " " " " " " ' " " O " " " " O " " " " " "

H 0 1 . 1

CI1 Vi O

7 O

H m S bZ E'

z 4

ai ~ ~ w c + o ~ o m n ~ b ~ n m m ~ o b r n d n e O

m " * B Z g m o - r n m - m m v l m m - * O . d 3 0 0 H 3 0 d O d d O O Q O O d , - l O O O d d d , - l H f H < y . . . . . . . . . . . O i [ g O O O O D O O O D D O A D & & 8 8 ó o 0 0 . rn

\ U U W

Q a I W1 a d

p: rl I m

m E $ u

p: P

i 2 O z

i 3 m

L)

m m ~ w ~ m m w m n ~ m ~ m ~ m ~ ~ e ~ r n o E r

52 m m ~ q y ? y m m m m m - g m n T q ? y y ? 8 H . . . . 2 E * . . 3 - ~ ~ n m m m m u ~ m u ~ ~ ~ a ~ ~ m m m u - t m m m m m o n

V1 LI

P UI

I D 1 0 Q Q * O ~ W L n W I p W 7 I L n b t 3 9 i r l p l . O W O H r n u - t ~ r u w d m ~ m m m y y y d * w - - q q ~ y o . . . . . . . . . 4 A o ~ m m m r m n w ~ w n a n n n n n a G - W U V W W W W W W W W ! 4 U U 4 m W W W W W W W H W N N W N N ~ ~ N N W W W N N N N N ~ ~ ~ N a . . . . . . . . . . . . . . . . . . . . . .

d n e d w d w d r n m - d d a d o a a d n ~ o o d ~ m w - w ~ m d o o d - + t - r n m o - -

m m 2 p r n i - m , - t m m m m m e t v w ~ r n w m m m r n r n ~ m W $ 2 5 F m I m e P h - m m b r - w u l ~ ~ m m u l m ~

ci n 9 M S3 . . . . . . m . . . . . . . d d ; d A A 4 A ~ d m m ~ ~ m ~ m n ~ m f i ~ ~2 o , - - o a w o r n - - m a w m m m ~ ~ n m w m

~ m ~ ~ m n m ~ w m o m e - w W m W o ~ m b = e E 0 0 1 a m A m 4 r n m 0 t - * d o - d r n r = r n ~ ~ d F $ ~ W W l f H N U U I 9 O N d N N N N d 9 3 . . . . . . . . . . . . . . . . . . . . . .

d W d W d W O U l O ~ 6 W W ~ ~ d ~ Q ~ P Q O w i - m o n n m m m m d m s o m n m m d ~ n U ~ - - - - ~ ~ L C I H - ~ m f l F 1 f l - d d d 9 l F F W W W W U l A W ~ ~ ! I I I 1 1 1 1 -

n - !Z O

G i G & G & k L & k i - & & & i & G G G G & & 2 + h & b h Q P , & & & & $ m m h b h f i Q h a a G

O O o ~ O o o O D D O o O a D O o Q o O O o O O O

-- - - - -



Page 3-1 78 Section 3 - Presfressed Concrete

E* \\ Bridge Design Practice - June 1994

Section 3 - Prestressed Concrete Page 3- 1 79

E* Bridge Becign Practice - June 1 994


4 Design of Welded Steel Plate Girders In accordance with the I992 F8eenf h Edition AASHTO Specificat-ons and Revisions by Caltruns Bridge Design Specifications

Contents

Notations and Abbreviations .............................. .......... ................... 4-1

..................................................................................... 4.0 Introduction 4-9

.................................................... 4.1 General Design Considerations 4- 10

......................................................... ....................... 4.2 Design Loads .- 4-11

........................................................... 4.3 Design for Maximum Loads 4-11 4.3.1 Braced Sections ................................................ .. 4 - 1 1

4.3.2 Unbraced Sections ................................................................................. 4- 13

4.3.3 ShearCapacityandDesign .................................................................... 4-14

........................................................................... 4.4 Composite Girders 4 1 6

................................................................................ 4.5 Fatigue Design 4-1 8 ................................................. 4.5.1 Factors Affecting Fatigue Performance 4-18

4.5.2 Applied Stress Range ............................................................................. &I9

........................................................................ 4.5.3 Allowable Stress Range 4-20

4.5.4 Type of Loading ................................................................................... -4-20

4-55 Stress Category ................................................................................. 4-20

4.5.6 Redundancy .................................................................. ... ................ 4-20

......................... 4.6 Charpy V-Notch Impact Requirements .... ......... 4-21

4.7 Fracture Control Plan (FCP) ............................................................ 4-22

4.8 Design Example Problem ................................................................ 4-23

Design Example Solution

......................................................................................... 4.9 Loading 4-25 ..*........................ .............*..*...*.....*....*....*..**..*...*.**.**..... 4.9.1 Dead Load .. 4-25

4.9.2 Live Load ....................................... .. .................................................... 4-26

4.10 Composite Section Design ................... ... ....................................... -4-29 4.1 0.1 Design Loads ................... .. .............................................................. 4-29

4.10.2 Fatigue Loads ...................................................................................... 4-30

4.10.3 Girder Section ...................................................................................... 4-31

4.10.4 Width to Thicbess Ratios ........................................ ...... ........... -4-37 ..................... ............................................ 4.10.5 Bracing Requirements .. 6 3 7

......................................................................... 4.10.6 Fatigue Requirements -4-38

.................... ............. 4.10.6. I Applied and Allowable Stress Rmgw ... 4-48

4.10.7 Shearksign ................. ..,., .................................................. 4 - 4 9

...................... 4.10.7. P Moment nnd Shear Interaction ................... .... 4-50

........................................................ 4.10.7.2 Trmncrse Stiffener Design 4-51

4.1 1 Non-Composite Section Design ...................................................... 4-52 ......................................................... 4.11.7 DesignLoads ..................... ... 4-52

............................. ................................................ 4.11.2 Girdersection ... 4-53

4.1 1 -3 Width to Thickness Ratios ........................................ ...-. .................. 4-55

4- 1 1.4 Bming Requirements .................. ..... ......................................... 4-56

4.1 1.5 Fatigue Requirements .................................................. .. ............... 4-56

4.11.6 ShearDesigmga ....................................................................................... 4-57

4.11.6.1 MomentandSheorPnreraction ................................................... 4-58

........................................................ 4 . J 1.6.2 Transverse Stifener Design 4-61

4.12 Flange- to- Web Weld ...................................................................... -4-6 3 4.12.1 Weld Design .................................................................................... 4-63

4.12.2 Fatigue Check ...................................................................................... 4-64

............................................................................. 4.13 Shear Connectors 4-65 ..................................................................................... 4.13.1 FatigueDesign 4-65

4.13.2 Ultimate Strength ................................................................................. 4-68

...................................... 4.13.3 Shear Cannectm at Points of Con&exure 4-69

............................................................... 4.14 Bearing Stiffener at Pier 2 4-70

4.15 Splice Plate Connection TO BE ADDED ATA FUTURE DATE .... 4-75

4.16 Bridge Design System Computer Output ............................... 4-A to 4-1

4 Design of Welded Steel Plate Girders

Notations References wirhin parentheses are to Bridge Design Specificatiom, Section 10.

A = areaofcrosssection(Artic1es 10.37.1.1,10.34.4,10.48.1.1.10.48.2,1,10.48.4.2, 10.48.5.3 and 10.55.1)

A = bending moment coefficient (Anicle 1 0.50.1.1.2)

AF = amplification factor (Articles 10.37,l. 1 and 10.55. I) = product of area and yield point for bottom flange of steel section (Article

10.50.1.1.1)

fAFvIt = product of area and yield point of hat pan of ~'einfminp which lies in the compression tone of the slab (Article 10.50.1.1.1)

(AF,Jf = groductof area and yield p i n t for top flange of steel section (ArtjcIe 10.50.1.1. I ) (My), = product of area and yield point for web of steel section (Article 10.50.1.1 .I )

Af = atea of flange (Anicles 10.39.4.4.2, 10.4821, 10.53.1.2, aud 10.56.3)

AI, = area of compression flange (Article 10.48.4.1)

A; = cod area of longitudinal reinforcing steel ar the interior support within the effective flange width {Article 10.38.5.1 2)

A f = total area of longitudinal slab reinforcement steel for each beam over interior support (Mele 10.38.5.1 -3)

A, = a m of steel section (Articles 20.38.5.12, 10.54.1.1, and 10.54.2.1 1

AW = arm of web of beam (Article 10.53.1.2)

Q = distance from cenltr of bolt under consideration to edge of plate in inches (Articles 10.32.3.3.2 and 10.56.2)

a = spacing of uansvwse stiffenen (Atticle 10.39.4.4.2) - o = depth of smss block (Figure 10.50A)

Q = d o of numerically smaller to the larger end moment (Article 10.54.2-2)

B = constant based on the number of smss cycles (Article 10.335.1.1 1 B = constant for stiffeners (Articles 10.34.4.7 and 10.48.5.3)

b = compression flange width (Table 10.32. E A and Article 10.34.2.1.35 li = distance from center of bolt under cwsideration to toe of Wet of connected

part, in. (Articles 10.32.3.3.2 and 10.56.2) b = effective width of slab (Article 10.50.1.1.1) b = effective flange width (Articles 10.38.3 and 10.38.5. E .2)

b = widest flange width ( M c l e 10.15.2.1)

= distance ftom edze of plare or edge af perforation to the point of support (ArtlcIe 10.35.23)

= unsupported distance between points of suppost (Article 10.35.3.7)

= flangewidthbemeen we&s (Articles 10.37.3.1.10.39.4.2.10.51.5.l.and 10.553)

= widthofsciffeners(krticles10.M.5.2,10.346.10.37.2.4, 1Q.39.4.5.11and10.55.2 = wi& of a projecting flange element. q l e , or stiffener (Articles 10.74.22.

1037.3.2. 10.39.4.5.1.10.48.1, 10.48.2, 10.485,3.10.50. 10.5I.5.5.and 10.55.3) = web buckling coefficient (Articles 10.34.4. 10.48.5.3. 10.48.8, and 10.50(e))

= compressive force in the slab (Artick 10.50.1,1,1) = equivalent moment factor (Article 10.54.2.1)

= compressive force in rop portion of steel section (Article 10.50.1.1.1 )

= bending coefficient vable 10.32.1A, Article 10.48.4.1) = column slenderness ratiodividing elastic and inelastic buckling (Table 10.32.1 A)

= coefficient about X-axis (Article 10.36) = coefficient about the Y-axis (Article 10.36) = buckling stress cuefficient (Anicle 10.5 1-52) = clear distance b e t w m flanges, inches ( Arzicle 10.15.2) = clear unsupported distance between flange components (Articles 10.34.3, 10.34.4,10.34.S,10.37.2,10.48.I, 10.4S.2.10A8.5.10.48.6110.48.8,10.49.2, 10.49.3.2, 10.50(d), 10.50.1.1.2, 10.50.2.1, and 10.55.2)

= dear distance between the neutral axis and the compression flange (Articles 10.48.2.1(b), 10.48.4.1, 10.49.2, 10.49.3 and, EO.SO(d))

= moments caused by dead load acting on composite girder (Article 10.50,1.2.2) = distance to the compression flange from the neutral a x i s for plastic bending,

inches (Articles 10.50. I. 1.2 and P 0.50.2. I ) = moments caused by dead load acting on steel girder (Article 10.50.1.2.2)

= bolt diameter (Table 110.32.3B) = diameter of stud. inches (Article 10.38.5.1)

= depth of beam or girder. inches (Article 10.13. Table 10.32.1A. Articles 10.48.2 10.48.4.1, and 10.50.1.1.2)

= diameter of rwker or d e t , inches {Article 10.32.4.2) = beam depth (Article 10.56.3) = column depth (ArtEcle 10.56.3) = spacing of intermediate stiffener(Articles 10.34.4.10.34.5,10.48.5.3,10.48.6.3,

and 10.48.8) = mdulus of elasticity of steel, psi (Table 10.32.2A and Articles 10.15.3.10.36,

10.37, 10.39.4.4.2, 10-54.1. and 10.55.1)

= modulus of elasticity of concrete. psi (Article 10.38.5.1.2) = maximum induced suess in the bottom flange (Article 10.20.2.1) = maximum compressive stress, psi (Article 1 0.41 -4.6) = dfowable axial unit stress (Table 10,32.1 A and Articles 10.36. 10.37.1.2. and

10.55.1 ) = allowable bending unit stress ('Table 10.32.134 and Articles 10.37.1 -2 and f 0.55.1 ) = buckling stress of the compression flange plate or column (Articles 10.51.1.

EQ.51 5, IO.54.I.1, and 10.54.2.1) = compressive bending stress permitted about the X-axis (Article 10.361 = compressive bending smss pemrirted about the Y-axis (Atticle 10.36) = maximum horizontal force (Article 10.20.2.2)

= Euler buckling stress (Articles 10.37.1, 10.54.2.1, and 10.55.1) = Euler s-ss divided by a factor of safety (Anicle 10.36) = computed bearing stress due to design load (Table 10.323B) = limiting bending stress (Article 10.34.3)

= allowable range of stress Uable 10.3.1A) = specified minimum yield point of the reinforcing steel ( Atocies 10.38.5.1.2)

= factor of safety CTable 10.32, I A and Articles 10.32.1 and 10.36) = specified minimurn tensile strwgth (Tables 10.32.IA and 10.32.3B. Article

lo. 18.4)

= tensile saenglth of electrode classification (Table 10.56A and Article 10.32.2)

= allowable shear s m s pables 10.32.1A, 10.32.3B and Articles 10.32.2. 10.32.3, 10.34.4, 10.40.2.2)

= shear smngth of a fastener (Article 10.56.1.3) = combintd tension and shear in bearing-type connections (Article 10.56.1.3)

= specifred minimum yield point of steel (Articles 10.15 -2.1, 10.I5.3,10.16.11, 10.32.1,10.32.4,10.34.10.35,10.37.13,10.385,1039.4,10.40.2.2,10.41.4.6, 10.46, 10.48. 10.49, 10.50, 10.5I.S. and 10.54)

= specifie$minimvmyie~dsmngthoftheflange(Article 10.48.1.l.md 2053.1) = swfred minimum yield strength of the web (Article 10.53.1) - computed axial compression stress (Articles 10.35.2.10.10.36.1037.10.55.2.

and 10.55.3) = computed compressive bending stress (Articles 10.34.2. 10.34.3, 10.34-5.2,

10.37. 10.39. and 10.55) = unit ultimate compressive strength of concrete as determined by cylinder rests

at ageof 28 days,psi (Anicles 10.38.1, 10.38.5.1.2,10.45,3, and 10.50.1.1.1E = top flangecompressivestressdue to noncompositedeadload (Amcle 10.34.2.1,

1 0.34.2.2 and 10.5Nc))

= m g e of stress due to live load plus impact. in the slab reinforcement over the suppon (Article 10.3 8.5.1 3 1

= maximum 1ongitudina.l bending stress in the flange of the panels on either side of the transverse stiffener (Article 10.39.4.4)

= tensile suess due to applied loads (Artides 10.32.3.3-3 and 1 0.56.1.3.21

= unit shear stress (Articles 10.32.3.2.3 and 10.34.4.3)

= computed compressive bending stress about the x axis Chicle 10.36) = computed compressive bending stress about the y axis (Asticle 10.36) = gage between fasteners, inches (Articles 10.16.14 and 10.245)

= height of stud, inches (Arricle 10.38.5.1.1'1

= average flange thickness of the channel flange. inches (Article 10.38.5.I.2)

= moment of inertia (Articles 10.34.4. 10.34.5. 10.385.1.1, 10.48.5.3, and 10.48.6.3)

= moment of inertia of stiffener {Articles 20.37.2. 10.39.4.4. I , and 10.51.5.4)

= moment of inertia of transverse stiffeners (Article 10.39.4.4.2)

= moment of inertia of member about the vertical axisin the plane of the web. in? (Arude 10.48.4.1)

= moment of inertia of compression flange about h e vertical axis in the plane of the web, jnP (Table 10.32.1 A, Artlcle 10.48.4.1)

= required ratio of rigidity of one mansverse stiffener to that of the web plate (Articles 10.34.4.7 and 10.48.5.3)

= in.4 Uable I0.32.1& Article 10.48.4.1 ) St Venanr torsional consmc = effective length factor in plane of buckling (Table 10.32. I A and Articles 10.37.

10.54.1 and 10542) = effective length factor in the plane of bending (krticle 10.36) = constant: 0.75 for rivets; 0.6 for high-strerrgch bolts with thread excluded from

shear plane (Artlcle 1032.3.3.4) = buckling coefficient (Anicles 10.34.4.10.39.4.3, 10.48.8, and 10.5 1.5.4)

= distance from outer face of flange 'to toe of web fillet of member to be stiffened (Article 10.56.3)

= buckling d c i e n t (Article 10.39.4.4)

= distance between bolss in the dinxtion of ehe applied force (Table 10.32.38) = actual unbraced length flable 10.32.IA and Articles 10.7.4. 10.15.3, and

10.55.1)

= 'h of the length of the arch rib (Atticle 10.37.1) = distance between m v e r s e beams (Anicle 10.41.4.6)

= unbraced Iength vable 10.48.2.1 A and Articles 10.36, 10.48.1.1, 10.48.2. I , 10.48.4.1, and 10.53.1.3)

& = length of member between points of suppon. inches (Article 10.54.1.1)

L~ = limiting unbraced length (Article 10.48.4.1)

JL = limiting unbraced length (Anicle 10.48.4.1)

I = member length (Table 10.32.1 A and Article 10,35.1)

M = maximum bending moment (Articles 10.48.8, and 10.53.2.1)

MI = moments at the ends of a member M i & M= = moments at two adjacent braced points (Table 10.32. IA, Anicles 10.36A and

10.48.4.1)

M, = column moment (Article 10.56.3.2)

M~ = full plastic moment of the secrjon {Articles 10.50.1. I 2 and 10.54.2.1 3 Mr = lateral torsional buckhng moment or yield mornen1 (Articles 10.48.4.1 and

10.53.1.3)

M, = elascic pier moment for laadmg producing maximum positive moment in adjacent span (Article 10.50.1.1.2)

MU = maximum bcnding sucn_eth (Articles E0.48,10.5 1.1. 10.53.1. and 10.54.2.1)

N1 & EJT = number of shear connectors (Article 10.385.1.2)

Nc = number of additional connectors for each beam at point of conrraflexure (Article 10.38.5. J .3)

NS = number of slip planes in a slip critical connection {Articles 1032.3.2.1 and 10.57,3.1)

&. = number of roadway design lanes (Article 10.39.2) = ratio of mdlulus of elasticity of steel to that of concrete (Article 20.38.1 ) = nmkroflongimdinal,stiffeners(ArticIes 10,39,d.3,10.39.4.4,md 10.51.5.4)

= allowable compressive axial load on members (Article 10.35.1) = axialcompression onthemember{Articles 10.48.1.1,10.483.l. and 10.54.2.1 )

= force in the slab (Article 10.38.5.1.2)

= allowable slipmistance (Article 10.32 2-2-13 = maximum axial compression capacity (Anicle 10.54.1.1)

= allowable bearing (Article 10,32.4.2) = prying tension per bolt (Artides 10.32.3.3 -2 and 10.56.2) = statical moment about the neultral axis (drticle 10.38.5.1.1)

Q. = ultimate strength of a shear connector (Amcle 10.50.1.1. 1)

R = radius (Article 1 0.15 -2.1 )

R = number of design lames per box girder (Article 1039.2.1 $

= bending capacity reduction factor (Articles 10.48.4.1. and 10.53.1 -3)

= a range of stress involving both tension and compression durinz a stress cycle (Table 10.3.1B)

= vertical force at connecrions of venicd stiffeners to longitudinal stiffeners (Article 10.39.4.4.83

= venical web force (Article 10.39.4.4.7) = radius of gyration, inches (Articles 10.35.1, 10.37.1, 10.41.4.6, 10.48.6.3,

10.54.1.1, 10.54.2.1. and 10.55.1) = radius of gyration in plane of bending (Article 10.36) = radius of gyration with respect to the Y - Y axis (Article 10.48.1.1)

= radiusofgyrationininches of thecompression flangeabouttheaxis inthe plane of the web (Table 10.32.1A, Article 10.48.4.1)

= allowable rivet or bolt unit s m s in shear (Article 10.323.3.4) = sectjonmodulus, in.3 (Articles 10.48.2, 10.51.1, 1053.1,2, and 10.53+1.3)

= pitch of any two S U C C ~ S S ~ W R holes in the chain (Article 10.16.14.2) = range of horizontal shear (Asticle 10.38.5.1.1) = section mdulus of transverse stiffener. in3 (Articles 10.39.4-4 and 10.48.6.3) = section modulus of longimdinal or transverse stiffener, in.3 (Article 10.48.6.3) = ultimate smgth of the shear connector (Asticle 10.38.5.1.2) = section rndulns with respect to the compression flange. in Uable 10.32.1 A.

Article 10.48.4.1)

= computed rivet or bolt unit smss in shcar [Article 10.32.3.3.4) = range in tensile s t m s (Table 10,3,1B) = direct tension per bolt due to external load (Articles 10.32.3 and 1056.2) = arch rib h s t at the quarter point from dead + live + impact loading

(Articles 10.37.1 and 1055.1)

= thickness of the thinner outside plate or shape ( h i d e 10.35.2)

= thichiss of members in compression (Article 10.35.2) = rhicksess of thinnest part connected, inches (Articles 10.32.3.3.2 and 10.56.2) = cornpured rivet or bolt unit stress in tension, including any stress due to prying

action (Article 10.323.3.4)

= thjchess of the wearing d a c e , inches (Adele 10.41.2) = flange thickness, inches {Artides 10.34.2.1, 10.39.4.2, 10.48.1.1, 10.48.2.1.

10.50, and 10.51.5.1) = thichess of a flange angle (Article 10.34.22) = &chess of the web of a channel, in. (Article 10385.1.2) = thickness of stiffener (Article 10.48.5.3)

= thickness of flange delivering concentsated force (Article 10.56.3.2) = h c h e s s of flange of member to be stiffened (Article 10.56.3.2) = thickness of the flange (Articles 10.37.3, 10.55.3 and 10.39.4.3)

= thickness of stiffener (Article 10.37.2 and 10.552)

= slab aichess (Articles 10.38.5.1 -2. 10.50.1.1.1. 10.50.1.1.2)

= web thickness, inches (Articles 10.15.2.1, 10.34.3, 10.34-4. 10.34.5, 10.37.2. 10.48. 10.49.2, 10.49.3, 10.55.2. and 10.56.3)

= ttuckness of top Ran~e (Article 10.50.1.1.1 3 = thickness of outstanding stiffener element (Articles 10.39.3.5.1 and 10,S 1 3 .5 )

= shearing force (Anicies 10.35.1, 10.48.5.3, IQ.48.8, and 10.51.3) = shear yielding strength of the web (Anides 10.48.8 and: 16.53.1.4) = m g e of shear due to live loads and impact, kips (Article 10.38.5.1 -1)

= maKimurn shear force (Articles 10.34.4. 10.48.5.3. 10.48 -8. and 10,53.1.4$

= vertical shear (Amcle 10.39.3.1 ) = design shear for a web (Articles 10.393.1 and 10.51 -3) = length of a channel shear connector, inches (Article 10.38.5.I.2) = roadway width between curtrs in feet or barriers if curbs are not used (Article

16.39.2.1)

= fraction of a wheel load {Article 10.392) = length of achaanel shear connector ininches measured in a m v e r s e direction

on the Range of a girder (krdcle 10.385.1.1)

= unit weight of concrete, lb. pet cu. ft (Article 10.38.5.1.2) = wdth of flange between longitudinal stiffeners (Articles 10.39.4.3, 10.39.4.4.

and 10.51.5.4)

= ratio of web plate yield strength to stiffener plate yield strength (Articles 10.34-4 and 10.48.5.3)

= distance from the neutral axis to the exireme oum film, inches ( M c Z e 10.15.3)

= location of steel sections from n e u d axis (Article 10.50.1. I .l) = plastic section modulus (Articles 10.48.1. 10.53.1.1, and 10.54.2.I) = allowable range of horizontal shear, in pounds on an individual connmor

(Article 10.38.5.1) = constant based on the amber of stress cycles (Amcle f 0.3 8.5.1. f ) = minimum specified yield strength of the web divided by the minimum specified

yield strength of the tension flange (Articles 10.40.2 and 10.40.41

= area of the web divided by the area of the tension flange (Articles 10.40.2 and 10.53.1.2)

= F J F ~ (Amcle 10.53.1.21

= angle of inclination of the web plate to the v e ~ c a l (Articles 10.39.3.1 and 105 1.31

= ratio of [oral cross secrional area to he cross sectional area of both flanges (Article 20.1521

= distance from the outer edge of the tension flange ro the neuud axis divided by ?he depth of the steel section (Araicles 10.40.2 and 10.53.1.2)

= amount of camber, inches (Article 10.15.3) = dead load camber in inches at any point (Article 10.15.31 = maximum value of ADLq inches {Article 10.15.3)

= duction factor (Articles 10.38.5.1.2. 10.56.1.1, and 10.56.1.3) = longitudinal stiffener coefficient (Articles 10.39.4.3 and 10.5 1.5.3) = slip coefficient in a slip-critical joint (Article 1057.3)

Abbreviations BDS = Bridge Design Specifcations manual

4.0 Introduction This section illustrates Load Factor Design (LFD) for a consinuous, welded. structural steel girder highway bridge, composite for positive live load moments accordins ra Section 10 of the Bridge Design Spec$cations (BDS).

En addition to being classified as symmetrical or unsyrnrnetsical as shown in Figure 4-1, steel g~rders can be further categorized as follows: - Compact

Non-compact

* Braced - Unbraced Transversely stiffened

* Longitudinally stiffened

Composik * Nan~ompasite

Hybrid

Symmetrical Unsymmetrical

Figure 4-1 Type of Steel Girders

The steel girders designed by C a l m s arc usually welded plate girders. Typically rbese are non-compact aad transversely stiffened: they can be eirher braced or unbraced. The use of longitudinal stiffeners should be avoided if possible as they lead to complicated details and, when extended into tension zones, become a fatigue consideration. Non-composite girders are generally symmetrical.

A plate girder is a beam built up from plate elements to achieve a mom- efficienr arrangement of material. Plate girders are economical in the span range between 100 to 300 feer Since the 1950's steel girders designed by Calms have been welded plate girders. They are shop welded using rwo flange plates and one wet, plate to make an tshaped moss section as shown in Figure 4-2.

Bearing Stiffeners - f ransverse Stiffener

F i v 4-2 Details of Welded Steel Plate Girder

4.1 General Design Considerations Members designed by the load factor design (LFD) method are proportioned for a number of design loads. They are required to meet three main theoretical load levels:

1. Maximum Design Lmd 2. Overload 3. Service Load

The maximum design laad and overload requirements are based on multiples of the swvice loads with certain other coefficients necessary to insure the required capabilities of the smcture. The maximum desi_m lmd critetia insures h e smcrrrres capability of wirhstand- ing a few passages of exceptionally heavy vehicles.

The overload criteria insures conml of permanent deformation in a member caused by occasional overweight vehicles as specified in BDS. Article 10.57.

Service loads are utilized for the serviceability criteria to limit the live load deflection and provide an adequate fatigue life of a member.

4.2 Design Loads The moments and shears are determined by subjecting the girder to the design loads. Elastic analysis is used to calculate the various straining actions.

The design loads are given by

For HS20: 1.3 [ D + 73(L + I)]

For permit loading: 1. widely spaced 1.3 [ D + ( L + I ) H ~ ~ o + 1.15 (L+I)p13] 2. closely spaced 1.3 [D + (L + I)pi31

Where D = dead load, L = live load (HSZO. P13). I = impact. The factor 1.3 is included to compensate for uncertainties in strength, theory, loading. analysis and material properties. Also, the factors 7 3 and 1.15 are incorporated to allow for variability in overloads.

4.3 Design for Maximum Loads Welded plate girders of normal proportions are not likely to satisfy the requirements for a compact section, which is capable of developing full plastic stress dismbution. Usually welded plate girders arenon-compact braced or unbraced sections. Thenon-compact braced section is a section that can develop yield smngth in thecompression flange before the onset of local buckling, but it cannot resist inelastic deformation required for full plastic stress distribution.

4.3.1 Braced Sections

For non-compact braced section;

Mu= FJ ............................................................................................................. (10-97)

Where F, = yield stress and S = elastic section modulus.

The section modulus consequently must be proportioned so that

F,S > 1.310 + %(L + I h ]

2 1.3[D + (L + I)HS20+1.15(t + I)p13] for widely spaced

2 1.3[D + (L + I)pl3] for closely spaced

For the relationship to be permitted, the following crireria must be satisfied:

1. Width-thickness ratio of the compression flanse:

b where b' = width of projec'ting flanse element = 1

t = flange thickness - 2, Depth - thickness ratio of he web:

Where D, is the depth of tbe web in compression and t , i s the web thickness. However, for a symmetrical section this ratio can be exceeded by providing uansverse stiffeners and meeting

D 36500 -<' ....,,....,........,..w........ --a-~.................. (1@103$ and (10.50(dl) t w - & or for an unsymmetrical section

3. Spacing of lateral bracing of she compression flange :

20.000, (MIAr 4 5 . . . .-..-...-.-....,.. ................................................................... (10-200)

FJ

where Af = cross sectional area of compression flange d = total deprh of girder

3.3.2 Unbraced Sections

When a @sder does not meet h e lateral bracing requirement, the section i s considesed an unbraced sect~on and its dtimate moment capacity is given by:

Where M, is the maximum bending strength,M,is the Iateraltotsional buckling moment. and Rb is a bendins capacity reduction factor.

When the compression region of a bending member does not have adequare lateral suppon, the member may deflect laterally in a torsional mode before the compressive bendin, stress reaches the yield stress. This mode of failure is known as "lateral torsional bucung" or simply "lateral buckling".

The tendency of the comprrssion flange to tw i s t is resist& by a combination of St. Venant and warping torsion. In misnng W buckhg by warping tomon, the compression flange acts as a column swxgible to buckljng in the lateral direction h closed secrions, such as box girders or m k . torsional stiffness is generally very large and lateral buckling is not a concern. However, for open sections, such as plate girders, lakd buckhng rnust be considemi Because of the complexity of the theofftical expressions for f a t d b t l c b g stress that take into account the simultaneous resisrance to h r a l buckling afforded by SL Venanr and warpinz torsion. conservative simplified expressions have lteen developed for design use thaL consider zhe effects separately. Plate girders, u d y deep grders, areconmlJed by warping torsion since h e effect of St Yenant torsion is s m d The ulrimare moment ~apaciy for unbraced section, as used in AASHTO S p m f ~ d o n s pnor to the fifteen edition, is:

This equation treats tbe compression flange as a column. provided tkar the compression flange is not smaller in width than the rension flange. When using the equation, the moment capacity may kin~reased 20% when the ratio of the end moments is less than 0.7, bur cannot exceed F,S. The specifications also limit the stress in the top flange of acomposite &er to 0.6 F, under dead load. However, if the width of she compression flange is smaller rhan the rensi~n flange, then the a b v e equation is unconsewauve and the moment capacity should be cal~ulated using

4.3.3 Shear Capacity and Design

The shear capacity of girder webs with msverse stiffeners is given by:

Thls quation combines the "beam action*' and he "tension field action." The first tern of the equauon represents web buckling under shear and the second term represents the additional post-buckling strength.

V,= pIastic shearcapacity =0.58FyDi ,.........+................................ ............... (10-1 14)

and

web buckling shear stress C=

web shear yield stress

Depending on the value of D/t, the web can be one of three cases which is given in Article 10.48:

I . Yielding:

2. Inelastic buckling: 6.000& D 7 . 5 0 0 4

5 -5 : C= , .................. (1@i 151 6,000dk I. JF;

D 1,500& 4.5xlOJk ...... 3- Elastic buckling: c' JE : C= .................... ,, ... (10-1 16)

where R is the buckling coefficient given by: k = 5 + 5 v.9 m2

Generally. the effect of bending on the shear strength of a girder can be ignored. However. if the bendins M exceeds 0.75 Mu and the shear capacity is calculated from Equation 10-1 13, then the shear at that section should be limited LO:

Spacing of sransverse stiffeners dong a girder should nor exceed d, derermined from Vb formula nor 30. However. for msverseIy stiffened plate girders with D l t , >150, the stiffener spacing shall not exceed

D [ z ] ' to ensure rficicnr handling, fabricauon and e m i o n of rhc girder

At simply supposted ends of girders, h e first stiffener space shall be such that the applied shear will not exceed the plastic or buckling shear force:

and the maximum spacing is limited to 1.SD.

Transverse stiffeners should be proportioned so that the width-thickness ratio shall be

Also, the p s s cross-sectional area of each one-sided stiffener or pair of two-sided stiffeners shall be at leas1

where Y= racio of web yield strength to stiffener yield strength; B = 1.0 for stiffener pairs; B = 2.4 for single plates, and; Cis the value used In computation ~f V,.

In addition, the m q k d moment of inertia of stiffeners with respect ta midplane of web is

2

where J = 2.5($) --2 10.5 .....................~....~.~-~~..-~..~....~~....~.~~..,............. (10-107)

Composite Girders In the non-composite type of sreel girder bridge. the entire dead load and live load of the supersmcture is svpponed by the steel girders alone, with h e deck on1 y msrnitting loads to the girders. However. in composite consmction, the concrete deck is keyed ro the steel girders by mechanical means and may thus be considered a component part of the girder.

-

I i --Shear connectors

Figure 4-3 Details of Composite Steel Girder

Figure 4 3 shows a s e ~ t i o n and elevation view of a typical composite girder. The concrete deck is keyed to the steel girder by shear connectors. therefore, h e deck sentes as additional upper flange area for rhe steel girder.

In accordance withBDS, M d e 10.383.1. the assumed effective width of tbeconcrete deck shall not exceed the following:

la) one-founh of the span length of the gder. (b) the distance center to center of girders. (c) twelve times h e least thichess of slab

Since the modulus of elasticity of h e concrete deck is different from that of the steel girders, the effectiveness of the concrete as flange material is a function of the modular ratio n = E,/Et. The quivalent net composite section is usually obtained by converting the effedve concrete area to an equivalent area of steeL Thus in Figure 4-3 h e equivalent width of concrete, be, equals the effective widrh, b, divided by n. When the concrete deck has been

convened to an quivalmt area of steel, the section may be considered to be a steel girder composed of (1) the original steel girder and (2) an additional rectangular flange of width b, Thecomposite bridge steel girder is usually designed &acomposite for live load and non- composite for dead load. Since intermediate tempomy supports are not normally used during deck place men^ the nee1 girder alone has to carry I t s own weight in addition to the weight of the deck. Once the concrete hardens the girder and deck will act as a composite section. Usually three t y ~ s of loading act on the girder:

1. Dead load (weight of girder and slab) 2. Additional &ad load (rail, AC overlay) 3. Liveload

For design putposes the girder is considered anon-composi te section for dead load and a full composite section for live load. However. for additional dead load (AC overlay + mil) the girder will act as a partially composite section. This is because the additional dead Ioad will cause sustained stress on the concrete s d o n . Due to this sustained smss, the concrete will undergo plastic flow, aod its effectiveness in resisting stress will be reduced, The main reasun of dus plastic flow is the creep of concrete. One conservative way to account for the creep of concrete under sustained loading is to d u c e the elastic modulus E, to Ih Ec whch means increasing n to 3n as in the BDS Article 10.38.1.4.

Figure 4-4 Effect of Creep on Concrete

4.5 Fatigue Design The farigue provisions of the bridge design specifications w m developed from r e s m h and studies of failurn in the fidd with respect to in-plane bending: wt-of-plane bending is not addressed. Details for main load carrying members, such as butt weld at rmion flanys and sriffener welds. are familiar to designers. However, fie effects of connections to the main members ate not as famiharand have been a s o w e of an ~nmasing number of fatigue problems.

Fatige may be defined as the initiation andlor propagation of cracks due to repeated variation of noma1 stresses which include a tensile component. Therefore, fatigue is the prccess of cumulative darnage hat is caused by repeated fluctuating loads.

Fatigue damage far a component hat is subjected to nomally elastic smss fluctuations occurs at regions of -5s raisers. After a certain number of load fluctuations. the accumulated damage causes the initiation and subsequent propagation of a crack or cracks in she plastically damaged regions. This prmess can and in many cases d m cause fracture of components. The more severe the stress concentration, the shoster the time to initiate a fatigue crack for the same smss cycle.

4.5.1 Factors Affecting Fatigue Performance

Many parameters affect the fatigue perfomance of structural components. They include parameters related to stress. geometry and propeEies of. the component, and external environment.

The s m s paramam include s m s range, constant or variable loading and frequency. The geometry and properties of the component include s@ess misers. size. smss gradient and mechanical propedes of the base metal and weldment The external environment parameters include xernperanrre and aggressiveness of the eavitonment. The major facsors that govern fati-me are:

* applied sks s range number of load cycles applied - type of detail

Smchlres are typically designed with a finite fatigue life of fifty years, however, an infinite fatigue life could be designed for with proper consideration to the items listed above. It is important to note that once fatigue cracks develop, it does not imply hat the useful life of the smcrute has ended. Usually with minor repairs the svucturecan still function inrhe same capacity for many years.

4.5.2 Applied Stress Range

The applied stress rangemay bedefined as the algebraic difference between exmrne s e s s e s resulring horn h e passage of load across rhe suucntre. If. as in a compression member. the stress range remains within compressive values there is no fatigue considerations.

1 cycle C-.

Figure 4-5 Constant Amplitude Cycles

The above figure tepresents fie simplest stress history which is the constant-amplitude cyclic-smss fluctuation. The saess range is h e algebraic difference betwew the maximum stress, f,, and the rmnimurn stress, f,,, in the cycle.

The other type of stress history is the variable-amplitude random-sequence stress history as shown in rhe Figure 4-6. This is a very complex histoy and cannot be represented by an analytical function. The truckloading onbridges isaparucular exampleof this stress hstory.

Time

Figure 4-6 Variable Amplitude Cycles

4.5.3 AUowabIe Stress Range

The following items concrol the allowable stress range.

1. type of loading 2. smss category (connection derail)

2. redundancy

4.5.4 Type of Loading

The number of cycles has a significant affect on the fatigue design. G e n e d y . by increasing the number of cycles. the allow&le stress range would decrease.

The number of cycles used for fatiwe design depends w the type of rad and live load For example, "Case T', which is the most used case for freeways (an average daily mck uaffic wbch exceeds 2500). bas tbe following live load cycles to consider for longitudmd m e m h

I . HS20 (multi-truck) ..,.........,......-..me- --.--..-..,. . ............................... 2,000,000 cycles . 2. HS20 (multi-lane) ...............,.................................................... 50Q.000 cycles

................. 3. Single HS20 (mck) ,.........,......... ~.~............. over 2,000,000 cycles

4. PLoading(P13withHS20) .......,...........~~.~~~~.-.~~~.~~.~.~~................ 100.000cycles

4.5 -5 Stress Category

The main sFress categories A. B , C, D. E and Fare desm-bed in Table 10.3.3 B and illustrated in Figure 10.3.1C of the Bridge Design [email protected]. These categories correspond to plates and rolled beams; welds and welded beams and plate girders; stiffener and shon (less than 2") attachments; intermediate (over 2" but less than 4") attachments; long (over 4") attachments and cover plates; and fillet welds in shear, respectiveIy.

The most severe connection details are in category E and E'. These should be avoided as much as pssible because they are regarded as poor details.

4.5.6 Redundancy

Bridge structures are considered non-redundant when the failure of a member or of a ainzle elemenr could cause collapse of the srrucmre (such as a tension chord in a mss bridge). The design specification places increased ~estrictions on non-redundam smrctures by imposing lower allowable stress ranges in almos~ all categories. This reduction to a lower smess range makes details that fall into Category E very uneconomical and. in essence. resuicts their use.

In summary. the fatigue allowabIestress ranges and number of cycles represent a confidence limit for 95-percent survival of all details in a given category. Also. h e stress ranges are governed by details that have the most severe gwmeDrica1 discontinuities andfor imperfec- dons. Iris imponant to note that the fatigue cmcldpropagation is independent of she strength of steel. Therefore. the allowable stress ranges are tndependenr of steel strenrjlfi.

4.6 Charpy V-Notch Impact Requirements Main load carrying member components subjected to tensile stress are required to provide impact properties as shown in the table IbeIow.

These impact requirements vary depending on the type of steel used and the avetage minimum service ternpewre to which the stmctvre may be subjected.

The basis and philosophy used to develop h a e nquirements are given in a paper entitled 'The Development of AASHTO Fracture-Toughness Requirements for Bridge Sreels"' by John M. Barsom, February 1975. available from the American Iron and Steel Institute, Washingon, D.C.

Charpy V-notch (CVN) impact values shall confom to the following minimorn values:

Table 4-1 Fsactwre Toughness Requirements

Weld& or Mechanically

Fastened

Welded

Non-Ftactu~eGiticaI

Fastened

Fracture-Cnticd

Grade .(Y,P. IY.S.)

36

H3150W

Zone 3 Ft-Lbs @ "F

Zone 1 R-Lbs @ "F

Thiclmess (Inches)

r61H l % < r 1 4

t < 1 & l"h<r12 2 ~ 1 1 4

70W

Zone 2 Ft-Lbs Q "F

Zone3 Ft-Lbs @ OF

Be10 25@-10

Zone 1 Ft-Lbs @ "F

25b70 256370

15@70 15@70

, 15@70 15@70 20@70

Zone 2 Ft-Lbs B "F

25B40 2 5 @ 4 0

1 5 @ 1 0 25@-10 30@-10

25@70 25@70 30@70

1 5 @ 4 0 15@4O

30B-10 1 20B50

2 5 Q 4 0 2 5 @ 4 0 30@40

154310 1 5 @ 1 0

20@20 rSlM 20@-10

I 151% lllirill

15 @ 10 llPlO

1 5 @ 1 O 1 5 @ 1 0

20@-I0 20@-10

15@40 15@40 20@40

1"/3<t12'M 2%cr64

25@70 25070

2 5 @ 7 0 2 5 @ 7 O

30@20 30@20

30B20

2 5 @ 1 0 25s-I0

5OI5OW

70W

1 5 @ 10 15@10 2 0 @ 1 0

2 5 @ 4 0 11040

25@40 2 5 @ 4 0

30@20 30@20

3 0 d 2 0

15 ($40 ::::: 1 I5FIO

r l l h l M < r 1 4

ill% 2' / :<t<4

20@-10 25@-10

30@20 3 5 @ 2 0

30@20 3 5 @ 2 0

20@20 1,5@20

30@-30 35@-30

, 25@10 15@70 25C-10 15@70

30@-10 20@50 30@-30 20@50

2 0 @ 5 0 2 5 @ 5 0

1 5 @ 4 0 1 5 @ 4 0

2 0 8 2 0 20@20

The CVN-impact testing shall be "P" plate frequency tesring in accordance with AASHTO T-243 (ASTM A673). For Zone 3 requiremeets only, Charpy impact tests are required on each piare at each end The Charpy test pieces shall be coded wirh respect to heawlate number and that code shall be recorded on the mill-~est seprt of the steel supplier wirh the test result. If requested by rhe Engineer. the broken pieces from each zest (three speclrnens. six halves) shall be packaged and forwarded to the Quality Assumce organiadoo of the State. Use the average of three (3) tests. Lf the energy value for more than one af h e rest specimens is kl0w the minimum average requirements, or if the energy value for one of the three specimens is less than wo-thirds ($5) of the specified minimum average requirements, a retest shall be made and the energy value obtained from each of the three retest specimens shall equal or exceed the specified minimum average requirements.

Zone 1: Minimum Service Temperature 0°F and above.

Zone 2: Minimum Service Temperature from -1°F to -30°F.

Zone 3: Minimum Service Temperature from -3 1 "F to -60°F

Fracture Control The FCP is a plan devised to prevent collapse of steel bridges. Much of the FCP dates to design, weldmg, and material properties. The designer has h e respwsibibty for designating any member or structural component 8s a Fracture Critical Member (FCM) when failure of that member would cause the strucnue to collapse. The FCP requires the FCM be fabricated in a qualified shop and inspected by qualified inspectors: requires Nondemuctive Inspection (NQI) by qualified testers; supplements the current AWS and AASHTO welding specifications; and specifies material toughness.

It is a comprehensive plan whose adoption shwld improve the w e d quality of steel structures from design through fabrication.

For more detailed infomation see AASHTO's Guide Specification for Fracture Critical Nun-Redundanr Steel Bridge Members,

4.8 Design Example Problem To illustrateload factor design, portions of an interior girder of a three-span bridge as shown in Figure 4-7 will be designed. The section in the positive-moment region consists of a welded steel girder acsing compos~tely with the concrete slab. I n the negative moment resion. the section is designed as a nonaompsite section.

Roadway Section: Figure 4-8 Typical Section

Specification: 1992 Fifreenth E;dition AASHTO with fntwims and Revisions by Calbms Loading: 1. Dead Load

2. Live Load: HS20-44 and alternative and permit design load S m c r u d Steel: A709 Grade 50 - assume for web and flanges

A709 Grade 36 - assume for stiffeners. etc.

Concrete: f,' = 3.250 psi. modular ratio n = 9

88 503'-0' EB A -

1'-6" I - 150'-On 200"-0" 150'-0" 1'-6" I t -

ELEVATION

Figure 4 -7 Design Example

F i 4-8 Typical Section

4.9 Loading Since the spacing between girders exceeds 14 feet (BDS Table x23.1). this is a widely spaced girder and should be checked for load combinations IH and Ipw.

1" Group = 1.3[D + %(L + I)Hs~o] IPwGroup = 1.3[D + ( L + I)Hszo + 1.15(L + I)p13]

4.9.1 Dead Load

Trfbutory to Interior Girder 8'-0" - 4 - 8-0' - -

F i 4-9 Interior Girder Cross Section

Concrete Slab: Assume transverse deck design has been completed and a 10%" thick deck has been selected.

Area = (10%112)(16) = 1450 ft2 w = 14.50 (0.150) = 2.18 klft

Steel Girder

w = 0.30 Wft (including bracing and fillet welds) (estimated weight)

Type 25 Concrete Bamers:

w = '/3(2)(2.61)(0.15) = 0.26 Wft

AC Overlay:

w = 0.035 (16) = 0.56 Wft

Dead Load of steel girder and slab = 2.18 + 0.30 = 2.5 Mt

Dead Load of rail and AC overlay = 0.26 + 0.56 = 0.82 Wft

4.9.2 Live Load

For widely spaced girders. the load on each girder wilI be the reaction of the wheel loads assuming the deck between the girders acts as a simpIe beam.

Number of traffic lanes:

width of deck 'between rails = 44 - 2 ( 1.75) = 40.5 ft

From BDS 3,6. a traffic lane is 12 feet wide

40.5 number of traffic lanes = - = 3.38 :. number of design traffic lanes = 3.

12

40.5-een face of rails

I

1 16' between girders

1 I 1 I 12qlane I 12' lane - I 1 1 2' lane i b-

Figure 4-10 Lomtion of 3 T d c h e s far Maximum Load at B @IS203

Live Load contributory reaction at Interior Girder B (3 lanes)

number of Iive load lanes = 1.44

However, according so BDS 3.12.1. for 3 lanes use 90% of maximum.

number of design live load lanes = 0.90 ( 1.441 = I .30 lanes HS20

40.5' between face of rails

12' lane - 12' lane - - - - -,

Figure 4-11 Location af 2 T d c Lanes for Maximum Load at B (HS20)

Live Load contributory reanion at Interior Girder B (2 lanes)

number of design live Ioad lanes = 1-33 h e s HS20 ,130 for 3-lane trial

:. For IH Group, live load djsmibution = 1.38 lanes

I,= 1.3[D+5/3(1.38)(L+ IH= 1.30 3.O(L + I)m20 (*f-MebJOw)

In these expressions the term (t + I ) represents the effects af one lane of rhe desi-mated live Ioad. including impact,

40.5' between face of rai Is I

I

1 6' between girders

7 2' lane 12' lane

Live Load con~butory reaction at hterior Girder B

13 P13: RB = -= 0.81 lanes

16

9 HS20: Rg = - = 0.56 lanes

16

For IPw Group

In = 1 3 [D + 0.56(L+ I)H520 + 1.15(0.81)[L+ I)p13] IPW = 1.3D + 0.73(L+ I)mzn + 1.22(L+ tSecrDomac-)

la tbese expressions the rerm IL + 1 ) represents the effects of one lane of the desi,gated live load including impact.

4.10 Composite Section Design This seclion i l lustrutes the design ofan interior girder of a composite section at 0.4 point of Span I

4.10.1 Design h a & (See Section 4-16, Bridge Design Sysem Computer Outpur)

Load on Steel Girder Only (Non-composite)

Dead toad girder and slab

Moment = 1.3(3,590) = 4,667 k-ft * Shear = 1.3(-152) = -19.8 k

b a d on Partially Composite Section (PI = 3 x 9 = 27)

Dead Load sail and AC overlay

Moment= 1.3(1.221) = 1387 k-ft * S h w = 1.3(-5.2) = 4.8 k

Laad on Composite Secrion fn = 9)

Live Load Group IH

MaximumMoment =3.0(2,424)=7,272k-ft - AssociatedShear =3.0(17.7)=53.1k Maximum+Shcar =3.0(38.7)= 116k Associated Moment = 3 -0 12,3 191 = 6,957 k-ft Maximum - Sheat = 3.0 (-367) = -1 10 k - Assmiated Moment = 3 -0 1 1,367) = 4.101 k-ft

Live Load Group Ipw

a Maximum Moment = 0.7312.424) + 1.22(6.648) = 9,880 k-ft Associated Shear = 0.73(17.7) + 1.22(48.4) = 72.0 k Maximum + Shear = 0.1308.7) + 1.22(79.1) = 125 k Associated Moment = 0.73(2.3 19) + 1 -22t4.748) = 7,485 k-ft - Maximum - Shear = 0.73C-36.7) + 1.22(40.4) = -100 k Associated Moment = 0.73(1,367) + 1.22(3.865) = 5,7 13 k-ft

HSZO (Single Truck) Over 2,000,000 Cycles

HSZO (Multip k Lanes) 2,1000,000 Cycles (Truck Load)

* +LLM= 1.38(2,360}=3.257k-ft -LLM = 1.38 (- 590) = - 8 14 k-ft

HS20 (Multiple h e s ) 500.000 Cycles ( h n e Loud)

+LLM = 1 -38 (2,424) = 3 345 k-ft -W= 1.38(-805) =-1,111 k-ft

PI3 with HS20 J00,OOO Cycles

+LUI = 0.56/2,424) + 0.8 1 (1.15) 6.648 = 7,550 k-ft -LLM = 0.56(-85) + 0.81 (1.15) (-2,219) = -2518 k-ft

4.10.3 Girder Section

Top Flange

Typically, the maximum transported length of a steel plate girder is 120 feet. Due to consuucrion problems. some erectors limit the length of girder shipping pieces to 85 times the flange width. Based on that, for 1 TO foot length. the widdr of the compression flange wiU be about 18 inches, and this dimension can be used for the first trial size.

Try top flange 18" x I "

where the thickness of the Range can be obtained from the following equation.

b' 9 -=-=9r9.84 Okay $ 1

D Depth to Span Ratio: - = 0.04 ........,.......,.....~......~-~.~~.~,~............. (105.1)

5

D = 0.04{200') = 8 ft = % in.

For initial sizing of the web the following equation a n be used

Try web 96" x s/$''

3osr0m Flange

Try bottom flange 18" x l1/i'

Composite Concrete Slab

From ArticlelO.38.3, the effective flange width of the slab shall not exceed:

1. span length = '/4 (150) = 37.5 fr 2. Spacing between girders = 16 ft 3. Twelve times rhe slab thickness = 12 (107/s) = I3 I in. + control

Figure 4-13 Deck Effective Width

Effective concrete area = 131 (10%) = 1,425

For a @ally composite section. n = 3 (91 = 27 ......................................... (1038.1.4)

Calculation of moment of inertia can be done using b e composite girder worksheet or with "COMB" on IBM mainfmme. Both methods are illusaated on the following pages.

Concrete

STAT€ OF CAUKIRHIA- DEPARTMEW OF TRAHSPbRTAIION COMPOSITE WELDED GlRDER WORKSHEET US-09 t 24 ( REV 1191 1

Job+ Shsel Of

m lnmr CRn-mte - G h r Bmnar

sw .

span SP-T oak

Seetion Ibr Sla4 and Girder Loads Ym h~ Y AY W

3@ -i 6iTopFlange =z *== M 98 1,7& S72,n2 ' 9 Web =* x 5/8= 66 49.5 2,970 14 7, Of 5 , D B O L F I ~ ~ S ~ = ~ X & = 17 0.75 10.3 f5.5.9

3 =- 4,754 31 9,902 - YI I, Wcb 5 - 46,080 10, = 365,9= - y T x S = - 215J66

Yi = ficr =

M n Ibr Cub Lorrds, Railhg, U r l I H m En =

= = = 4,754 3f9,902

GI- i = 518 107.3 -- 5.663 #7,&6 $5 7.8 IO,417 927,548

I m W -. M*om I- = 975,62% -TxL4=- 687,771

Yr- I C L xy]

-n brUmbm%(n=I Z A = = 4,754 319,962

~ms: DL = 0.030

- = = 5 . 3 fo7.s fisw 1 ~ ~ 2 ~ 9 3 ~ 263.3 21,743 Sf42.W

lw web = - 46,080 1.587(123(112.75 - 66.0) = O.l Rail = r, = 2,198,920

285,857(27) - j j x a = - t,7y5,3fu

9.88q12)(112.75 - 82.6) = l.b

Qnb-

-I, ; ~ ~ Z ~ . Y I I *

Section Roperties Using "COMP" an 113M Mainframe

* u * = * * * * + * * . l * ~ * * - * * . * * * * Y * * + + C * 1 1 * 1 *

1 I 1

I I I C

I ~ w w v m o . . . . I E " w - * m I b 2 d l cu-

l I O ~ I I e I mlr. a

c E-* maw r u-

: E W E I

44 l - : s q * w m t-Ql l l I

II

I I1 D O I W m N m P - c + U + d 0 H O Q I U v . * . * \ * h b 4 + 1 a

- + & Z o r r u W \ P BW a * " O e g Q 4 " * * S a, 1

W 4 o m + m m * ; B X O a m w o E 1

C 1 Z l 8, O\Clul * P-mmul C * * I !21 g" U 4 N m I - - - I

vrcocn vl l n v m I V) C * ...... v I 2 """ r r a c - 3 1 0 1 + - O W l I * c 1 1 0 1 &O' C

I r.

: c s ; rl I * a 1 .

~ J I t e a L 1 a tl 1 * r. 1 gs I a n w e * * 5% Q *

I < R = > E 1

C m 3 w m ESm? &low - 1 c3nui l€- * + r o d h o ~ k 4 m w m 1 a 8 d 1 & -3 m d m m t * t

1 g I a: a + I

4 I m S I * I s 0 I E O z I H W V I - l d W - 2 1 a 4 o c l m w m 1 4 a w +

C 1

LI.OWC)O I b c C O ...... m * H W F U l I O U 4 c tl 1 em

a W ag.44 I " i d s ? .. -* *- * - I -... +

W W r f I * WE-m%" 4 0 +

O W 2 - 1 * * a rn I 0 0 * * ...... i r a m - c j r - m ~ 1 P

darn ~ n m m % Ht- E LO kl N - W E W O N

m h - lc& . - .durn 8 U O a r l e d 2 C

F' I .L E d o I - W J

0 W L 0 * * ,...... zmeg* cn l o m * + a . O f l G

1 I W 3 S $ G z g l : p d trl I I ~ . w ~ H u u m m I I

I-% - - I I g l hE.u;l -s

W < W & I U S do I *+ -. * m -la I m 1 8 3 * e 8

I

1 W 1 / B d * -a c g C C m - i % E * a I C I ; I - 2 a g 1 5

.a ............ 2 sa;; s * * 8 * 1 - * = ~ * * * . * * * C - * . * . C * 1 * ~ * * * * * * *

4-34 DESIGN OF WELDED STEEL PLATE G~RDERS

Non Yes Check Shear - Compact V, Lb

Calculate Check - vlh d5 A v Lb

I

Check Fatigue ,

Calculate Steel Girder Design XSection

' M o s t economical between d, f, and stiffener sizes?

Yes

Check b' D -and- t fw

Check: 1. Flange-Web

2. Shear

Completed '"" Flow Chart for h a d Factor Design

1 / , Transverse StlRener - %

I Wak I4mncr,arnmk, m t X L m r r d \ e)nP .In A P 0

wrnpressron flange

Transverse Stiffener BDS 10.48.5.3

B = 1 .qstiffener pairs)

Bending

1. BDS 10.482 Mu = F,S

2. 50s 10.48.4

Figure 4-14 Typical TransverseEy Stiffened Non-Compact S tee1 Section

4.10.4 Width to Thickness Ratios

Requirements for braced oon-compact sections

a) Outstanding leg of compressjon flange. non-compact

b) web requirement for m v e r s e l y smened girder (composite section without longitudinal stiffeners)

Figure 4-1 5 Girder Dimensions

4,10.5 Bracing Requirements

The section Is a non-compact btaced section, It is braced since the compression Range (top flange) is embedded in the concrete which provides a continuous lateral supporr However, the smss that is induced on h e compression flange from noncomposite dead load should be checked because the flange is unbsaced for dead load.

& = unbraced length

Since the spacing between she cross-frames will be 20 feet, the section is unbmed for non- composite dead laad, and therefore. a reduction in allowable stresses is q u i d due to buckhng.

Allowable stress

F,, = 0.6 Fv = 30 ksi .....................-.........~..~.~............................ { 10.501hl)

Lb = 20 (12) = 240 in.

b' = 1 8 / 2 = 9

Fv = 50 ksi

E = 2 9 x lO3ksi

Applied DL stress in top flange

fm = 19.8 ksi ........................................................................... f rom page 4-33

since F, = 30 ksi > fDL = 19.8 hi. top flange bracing of 20 feet is okay,

4.1 0.6 Fatigue Requirements

The allowable fatigue s w ranges are dependent on (BDS Table 10.3. I A):

1, redundant or non-redundant load path strpctures 2. stress cycles 3. stress cattgories

The flaw chart on page 4-39 ilIustmtes a procedure for determining the allowable stress range F,for any fatigue detail.

If failure occurred in the interior girder, the load would be redistributed to rhe exterior girders and the bridge probably would not collapse. Therefore, the interior g-lrder of this three girder system is considered redundant.

The bridge is locared on a major highway with average daily truck traffic mter shan 2,500. From BDS Table 10.32A this is a 'Case I" road and the following stress cycles are to k considered:

Connection I Find:

Illustration and Number

I ~ - ~~

Figure 4-16 (10.3.1C)' Illustrative Examples' I

I Illustrated Find: Example I Number.

Read Description and Stress Category

Table 4-5 (1 0.3.1 8)'

I Find: I

Find: Stress Range, F,.

Non-redundant Table 4.4 (10.3.1A)'

Find: Stress Range, F,,

Redundant Table 4.4 (10.3.1A)'

'( ) Bridge Design Specifications

Flow Chart to Find Allowable Stress Range, F ,

Table 4-2 Number of Cycles for Case I

Due to the uncertainty involved in predicting future MIC levels. it is specified thal "Case I" beused f o r d designs. This alsoinsures that pemit vehicles are consided since PI 3 with IS20 (at 100.000 cycles) has a smns influence on the fatigue behavior.

Loa&ag

.PI3 with HS20

HS20 Lane Loading

The most common types of connections found in plate girders are:

1. Transverse stiffeners 2. Butt weld of flange plates 3. Gusset plates for lateral k i n g 4. Range-teweb weld

Cycles

100.OOO

500,000

Theseconnecsim and &rs arc illusmad in F i p 4-1 6 (lllusmtive Examples) and dm5W in Table 4-5. Table 4-5 is vsed to select the caregory which matches the detail being considered

The four connections listed above have been marked on Figure 4- 16 and Table 4-5 and the results summarized below:

HS2O Tmck h d n g

Table 4-3 Common Types of Bridge Connections

2.W.000

HS20 Single Truck Loading over 2.000.00

Type of Connection S m s

T or Rev.

T or Rev.

I

2

3

4

Toe of transverse stiffeners

Butt weld at flanges

Gusset for lateral bracing lbolt gusset to flange) ---- Flange-wweb weld

Category

C

B

r IHusuatioa

6

8. 10

21

9

T or Rev. B

S bear F

The applicable stress ranges are now read from BDS Table 10.3.1A and shown below:

Table 44 Allowable Fatigue Stress Range

Type of Connection

Figure 4-16 Illustrative Examples

442 DFSIGN OF WELDED STEEL ~ T E GTKDERS

Table 4-5 @DS Table 103,XB)

General Condi don Situation

Saess IIFusmtive Ca~egofi Example

Kind of (See Table (See F~gure Stress 10.3.1A) 10.3.ICF

Plain M e m k Base me& with rolled or cleaned surface. Flamecut T or Reva A 1.2 edges with ANSI ~m~orhness of 1.000 or less.

Built-Up Members Base metal and weld m e d ia members of built-up T or Rev B 3,4.5.7 plates or shapes. bithour attachments) connected by contlouous full penemuon groove welds (with backing bars removed) or by con'tinuous filler welds p d e 1 to rbe ihcfion of applied s m s .

Base metal and weld metal in members of built-op Tor Rev B ' 3.4.5.7 plates or shapes (without aaachments) connected by continuous full penemrim p a v e welds with backing bars not removed. or by conanuous partial p e n e d w ~ w : welds parallel to h e Man of applied mess. - Calculated flexural s a s s at the toe of tmnsverse T or Rev C stiffener welds on ,met webs or flanges. 3

- ,..-- uu - -- -.-- < - l * Y - u t - & . . ~ ~ - --->--*--- -rr--*-u +-*^--rrxx -r

Base mtal at eDds of panial length welded coverplms nawwer b n the flange having square or m p d eods, with or without welds k s s the ends, or wider than flange with welds across be ends.

(a) Flange thickness S 0.8 in. Ib) Range thickness > 0.8 in.

T or Rev E 7 + or Rev E' 7

B, metd at the ends of partial length welded T ar R w E' 7 coverflares wider than lh flange without welds mvss the ends.

Groove Welded Connections

Base metal aud weld metal in or adjacent to full T or Rw B pmmtion groove weld splices of rolled or welded semoas having similar profiles when welds are ground

8-10 0 flush with griodmg in the direction of applied mess and weld soundness established by nondesrmctive inspection.

+.- - - -* + . * + ? - .+- , _ - - ..b? . - . > * . -- - -., - Base metal and weld metal in or adjacenr to f l ; ~ T or Rev B 13 penetration groove weld splices with 2 foot radius transitions in width, when welds m p m d f l u b with grindmg m the dmection of applied m s s and weld soundness esrabirshed by nondesmctive inspection.

General Condition

Table 4-5 ( continued)

Situation Kind of Stress

Stress Jilusmrive Category Example

(See Table (See Figure E0.3.1A) 10.3.1C)

Base meml and weld metaI in or adjacent to full peaemion groove weld splices at Wtions in width or Ihickaess, with wekds ground to provide slopes no steeper than I to 2?/;, wish g l l d m g in he dtrection of h e applied stress. and weld soundness establtskd by nondeshctive inspctioa:

(3) A A S m M270 Grad% 1001 1 mwIASTM A7091 base metal

(b) Other basememls

Base metal and weld metal in m adjacent to hll penemxion groove weld splices, with or without mansitions having slopes no prater than 1 to 21h when b e remfmmmmt is not removed aad weM s w e is established by nondesmctive inspecrim.

Groove Welded Base raetal adj-t to &as attached by full or partial Anachmeots - pmtri~dorr m e welds when the detail lengg. L. io Longitudinally tbe direction of smss, is less tban 2 inches. Loaded"

Basemetaladjjacenttodelailsatracbbdbyfullorpmid pemmion grwve welds when the &mil h g g & in ~ ~ w o f s a e s s . i s b e t w e e o 2 i n c b e s a n d 12tims the place thiclmess but lets rhan 4 i n c h .

B a s e m e r a l a d ~ t e d e t d s a m l c b e d b y f o l l u r p ~ penemiw gmive welds w k the detail kmgth, L in kdirectiw of is^^ 12Limes zbeplafe thickness or greater than 4 inches:

(a) Detail t h i c k s 4.0 inebes. (b) Detail thickness 1 1.0 inches,

Bare metal adjacent to details atmbd by fall clr partial penemtion groove wdds with a wansition radius, R, ~egardless of the detail length:

-With the end welds ground smoorh

(a) Transiticm radius 124 inches, (b) 24 inches > Transition sadrus 2 6 inches.

T or Rev

TmRw

T or Rev

T or Rev

T or Rev

T or Rev T or Rev

T ar Rev

General Condition

Table 4-5 (continued)

Situation

Stsess 11Iustranve Category Example

Iiind of (See Table (See Figure Suess 10.3.1A) 10.3.IC)

(cl 6 inches > Transition radius 2 2 inches. (dl 2 inches > Transition radius 2 0 inches.

-For all mansition radii without end weIds ground T or Rev E 16 smooth.

Groove Welded Detail base metal attached by full penelmtion p v e Attachments welds wirh a uaasition radius. R. regardless of the Transversely detail length and with weld soundness mansverse to Loadedb,= the direction of stress established by nondesrmctfve

inspection:

-With equal phte thickness and reiafarmment T w Rev removed

(a) Transition radius 2 24 inches. (b) 24. inches > Transition radius 2 6 iacbes. (c) 6 inches > Transition radius 2 2 inches. ld) 2 inches >Transition radius 2 0 inches. -With equal plate thic- and minforcenmenl not T or Rev removed.

(a) Transition d i u s 2 6 inches- C (b) 6 inches >Transition radius 1 2 inches. (c) 2 inches > Transition rdus L 0 inches. E -With unequal plate thickness and reinforcemeat T or Rev 16 removed.

(a) Ttansition radius 12 inches. D (b3 2 mches > Transition &us 2 0 iocbes. E

- Fm all &tion radii with unequal plate thiekmss T or Rev E 16 and reinforcement not removed.

Fillet Welded Base m e d at derails commmd with m e r s e l y loaded Connections welds, with the welds peqxndicular to the direction

of s-:

(a) Detail thEchess 10.5 inches. (b) Detail thiehess > 05 inches.

T or Rev C 14 T or Rev See Note d

Base metal at intermittent fillet welds. T or Rev E - Shear m e s s on throat of fillet welds. S bear F 0 . " . + I - XI _ - *A" - . . -

General Condi tian

Table 4-5 (continued)

Kind of sms

Stress lllusmcive Careeary Example

(See Table {See Figure 10.3.lA) 10.3.1C)

Fillet WeIded Attachments- Long~tudinally Loaded bc.=

Fi tlet Welded Attachments - Tramverse t y Zoaded with the weld in the direc'tion of principal stress bx

Basemetaladj;tcentmdetaiIsattae.hedbyflletweIds TorRev Iw*, L. lo tbe -on of m e s s , n less than 2 mches and a u d - w shear conneclors.

Base metal adjacent to details anached by met welds T or Rev with leagtb. L. in the direction of stress. between 2 inches md 12 times the plate rhichss bul less than 4 inches.

Base I& djacenr to derails anached by fillet welds; witb length. L. in the direction of stress greater than 12 rimes the plate thickness or greater thao 4 inches:

la) Detail tbickaess r 1.0 inch. (b) Derail th ichess 2 1.0 inch.

Base mtel adjacent m details amcbed by fillet welds with a mmibw &us. R. regardless of the &tail leu*

T or Rev T Or Rev

-With h e end welds ground smootb T w Rev

(a) Transition radius 2 2 inches. (b) 2 inches >Transition radius 10 inch.

-For all aansidon radii without the end welds T or Rev p u n d smooth.

Detail base metal attacbed by M e t welds witb a transition &us. R. regardless of tbe detail length (shear stress on the throat of fillet welds governed by Category F):

-With the end welds ground smooth

(a) Transition d u s 1 2 inches. (b) 2 inches >Transition rad~us 1 0 inch

T or Rev

-For all ltransition radii without the end welds T or Rev ground smooth.

Table 4-5 (conrinued)

I I + . - - -

Base metal at net secnon of high strength bolred T or Rev B 21

Stress Illusuative Carepry Example

Kind of (See Table (See Figure General Condiuon Situation Smss 10.3. IA) 10.3.TC)

karing-type connections.

Mechanically FaSte~ed Connections

Base metal at net section of rivered comecrions. T or Rev D 21

Base mral at p s s section of hlgb saength bolted slip T or Rev B resacant connec~oas. excepr axially loaded jainrs which induce out-of-plane beadtng in connecnng matends.

*'T' signifies tang in tmsile mess only. Tev" si,gnihes arange of stress involving borh tension and compression dm%ng a saess cycle.

b"Long~ru~ndl y Loaded" signifies direcrion of applied stress is paraIkI to the longitudinal axis of the weld. Tmsverstly Loaded" signifies direction of applied s&ss is perpendicular to h e longttu&nal axis of h e weEd

cTranmersely loaded panid penemtion m o v e welds are prohibited. Allowable fatigue smss rang on b o a t of fitlet welds aansversely loaded is a fundon of the

effective b a r and plare Ibickness. (SeeFrankand Fisher. Journal of the Smcnual Ihvision, ASCE. Yol, 105. No. Sn. Sept 1979.)

where S,' is equal m the allow able mess range for Categwy C given in Table 10.3.1 A. This assumes PO pwePafion at tbe weld root

"Gusset pl ares aaached to girder flange surfaces with only transverse fillet welds are prohibited

4.1 0.6.1 Applied and Allowable Stress Ranges

I - HS20 (Multiple Lanes) 2.000.000 cycles (Truck) +LLM = 3.257 k-ft -LLM = -8 14 k-ft

Smss range = 3.257(12) (82.6) +

,393,606 150,717

= 8.20 + 2.94 = 11.1 ksi < 13 ksi < 18 ksi Okay for Category B and C

2. HS20 (MultipIe Lanes) 500.000 (Lane Load) +u = 3.345 k-fr -LLM =-I,] I I k-ft

=8.42+4.01 =I2.4ksi<21 ksi ~ 2 9 k s i Okay forCategory Band C

3. P13 with HS2Q 1 00,000 cycles +W = 7550 k-ft -LLM = -2.5 18 k-ft

= 19.01 + 9.08 = 28.1 hi < 35.5 ksi e 49 hi Okay for Category B and C

4. Single HS20 over 2,000,000 cycles (Truck) +LLM = 1,912 k-ft -LLnn = -478 k-ft

Stress mge = 1.912(12) (82.6) +

393.606 478(12) (45.3) 150,717

Calculations for flange-wweb weld (Category F) axe not shown, see page 4-63 for procedure.

4.10.7 Shear Desip

The shearcapaciry. V,, of the section is dependent on the yield strength and thickness of h e web and the spacing of the transverse stiffener as

where: I f p = plastic sbear capaciy

........................................................................................... = 058F$f, ( lo- 1 1 4) = 0.5 B(50) 96 ( 5 l g ) = 1 -740 k

C = ratio of buckling shear mess to shear yield stress

The stiffeners are usuahly spaced equally between cross kames up to a maximum of 3D as spexified in BDS M c l e 10.48.8.3.

Maximum do= 3(96$ = 288 inches

Try d, = 20 feet = 240 inches = spacing between cross-frames

6 . d - 6.- = 64.6 and 7,5wfi 7 , 5 0 0 f i

- 4 m SO, 000

Applied Shear = Ifrn = -19.8 - 6.8 - 110 = - 137 k

V,, = I37 k < V, = 821 k Okay

Check requirements for handling

:.do D ( z y = 96(-$-r = 275 inches

do = 240 inches 5275 inches Okay

Spacing of transverse stiffeners and cmss-frames is 20 feet.

As might be expected due to low shear demands at the 0.4 point, only minimal stiffeners are r e q u i d . However, as the design check moves closer to the supports, where the shear is higher, the spacing of the stiffeners may 'become much closer.

4.10.7. I Moment a d Shear Interaction

............................................ Moment - shear interaction .......,**-.-h---..-.----..-m. ( 10.48.8.2)

If M > 0.75 Mu then a reduction in the allowable sheat. V, must be made.

Let M = Mu

V M -=2.2-1.6-= 2.2-1.6=0.6 ...... , ....................................................... (1b1 I f ) v u * MM

.= V = OdVu = 0.6(82 I ) = 493 k

Vm=137k<493k Okay

4.10.7.2 Transverse Stifener Design

Moment of inertia requited:

I = d ~ y ? J ...,...............--.... ................................................................................ (10-106)

where:

Area required:

where: Y = Ratio of web plate yield to stiffener yield

B = 1.0 for stiffener pairs

Since area r e q u i d <: 0, then the transverse stiffener must meet only the moment of inertia requirement (1b106) and the width-tu-thickness d o :

The width of stiffener is preferred to be at l a s t 6 inches to allow adequate space for cross- frame connection.

Try 6" x M" stiffeners

b' 6 -=-=I2<13.7 Okay 1 . c

Figure 4-17 = 83.7 in4 r f q m d =29.5 in4 Okay Web and Stiffener Cross Section

Use 6" x Mu stiffeners

4.1 1 Non-Composite Section Design This section illustrates the design of n~n-composite section at Pier 2.

4.1 1.1 Design h a d s (See Sectiondl 6. BridgeDesignSysrem Computer Oupui)

Dead Load Girder and Slab

Moment =1.3{-7,899) =-10,269k-A Shear = 1.3 (250) = 325 k

Dead Load Rail and AC Overlay

Moment = 1.3 (-2,686) = -3.492 k-fr Shear = 1.3 185) = I l l k

Live Loads

I . Live Load Group In Maximum moment = 3.0(-3,292) = -9.876 k-ft Associated shear = 3.0(94.9) = 285 k

Maximum shear = 3,OC 1 103 = 330 k A s ~ a ~ moment = 3.0(-2,634) = -7,902 k-ft

2. Live Load Group Ip, Maximum moment = 0.731-3,292) + 1-22(-5.548) = -9.172 k-ft Associated shear = 0.73(94.9) + 1.22(2 17) = 354 k Maximum shear = 0.73(110) + 1.221279) =427 k Associated moment = 0.73(-2.634) + 1.22i4.569) = -7.497 k-ft

Fatigue Loads (Case I Road)

1. HS20 (single tsuck) over 2,000.000 cycles +U = 0.811322) = 261 k-ft -LLM = 0.81(-1.476) =-1,196 k-ft

2. HSZO (Multiple Panes) 2,000,000 cycles (uuekS + f L M = 1.38(322) = 444 k-ft -U = 1.38(-1,476) = -2,037 k-ft

3. HS20 (Multiple Lanes) 500,000 cycles (Lute) +LLM = 138(365) = 504 k-ft -LLM = 1.38(-3,292) = 4 , 5 4 3 k-ft

4. PI3 with HS20 100.000 cycles +LLM =056(365)+l.f5(0.81)1,TlO = 1,238 k-ft -LLM = Q.56(-3292) + (1.15) 0.8 1 (-5548) = -7,02 1 k-ft

4.11.2 Girder Section

The section over the pier is designed as a non-composite section. It is Caluans pl icy to minimize theuse of shear connectors in negative moment areas to minimize weMng on the tension flange.

One method of minimizing the welding on the tension flange is to add additional studs near the DL point of conrraflexure and addtianal reinforcement is placed in the concrete deck overthe pierto control cracking in thedeck. Referencecan bemade to BDS Article 10.38.4.3 regarding minimum deck reinforcement in negative moment regions.

Another method would be to use shear connectors at the maximum spacing of 24 inches through the negative moment area.

A symmetrical steel section will be used.

Figure 4-18 Equilibrium of Forces

Design Moment Td = F&f

Design Momenc Girder + Slab = - 10269 k-ft RajI + AC = -3,492 k-ft Live Load = - 9,876 k-ft Mdrd = 23,637 k-ft

- 23,637 k-ft

Disrance between c.g. af h e flanges d = % + 2 = 98 inches + (assuming 2-inch thick flanges)

Ta make the fabrication of ehe plate girder easier. the web depth should m a i n constan[ throughout the length of girder. The depth of the web (D = 96") is the same as used in the composite area

57.9 Let b = 18 inclles, t = - = 3.2 inches I8

Try: 18" x 2%" flanges and 96" x %" web

96 Y,= & =-+2% =50.9 inches

2

= 48.3 hi < SO ksi Okay

4.11.3 Width toThichessRatios

a) Outstanding leg of compression flange - n-ompact

b) Web - umsvenei y stiffened girder

96 - = 154 c 163 Okay s/s

4.1 1.4 Bracing Requirements

A! = Area of compression flange = 18(2%) =S1.8

d =96+2(27h)=101.8

20.000.000(5 1.8) "' 50.OO(fi1.8) = 203.5 inches = 17 feet

Let spacing between cross-frames be 15 feet, and no moment reduction due to bracing will be required.

4.1 1.5 Fatigue Requirements

1- RS20 (Multiple Lanes) 2,000.000 cycles (Truck Load) (434 + 2,03'7)50,9

Stress range = (12) = 5.07 ksi < 13 ksi c 18 ksi 299,04 1

Okay for Category 3 and C 2. HS20 (Multiple lanes) 500,000 cycles (Lane Load)

(504 + 4,543$50.9 stress range = 299.041

(12) = 10.3 ksi < 21 ksi < 29 ksi

Okay for Category B and C

3. PI 3 with HS20 100,000 cycles (1.238 +7.011)50.9

Stress range = 299,04 1 (12) = 16.8ks.i < 35.5 hi < 49 ksi


4. HS20 (Single Tmck) over 2,000,000 cycles


Calculations for flange-to-web weld (Category F3 are shown on page 4-63.

4.1 1.6 Shear Design

Maximum Shear capacity, V,

where V, = 0.S8FyDt, ..................................................... (10- I 14)

= 0.58 (50) 96 (5h) = 1.740 k

whim do = spacing between transverse stiffeners

maximum do = 3D = 3(96) = 288 in.. or for handling = 96 f 53.6

try d, = 15 ft = t 80 in. = spacing between cross-frames

5

Design V=325+111 +421 =857k

Since Design V = 857 k c V, = 959 k, spacing of aansverse stiffeners, do = 15 fr, is okay.

4.11.6.1 Moment and Shear lnreraction

1. Maximum Sbear - Associated Moment

Vappli~d =857 k

MaWllrod = - 10,269 - 3.492 - 7,497 = - 2 1358 k-ft

249,041 1 Mu (12) = 24,479 k-ft

0.75 Mu = 0.75 (24.479) = 18.360 k-ft

Since M = 21,258 k-ft > 0.75 Mu= 18.360 k-ft, a reduction in [he allowable shear. V. must be made.

V = V,(0.81) =959 (0.81)= 777 k

Since applied V = 857 k > allowable V = 7Y7 k N.G. The section must be revised,

The section can be revised by one or more of the following:

1. Increase flange site. 2. Encrease web thickness. 3. Reduce stiffener spacing, d,

Try reducing stiffener spacing. $, = 90 in.

allowable V = 0.81 Vu = 0.81 (1,365) = 1,105 k

applied V = 857 k < allowable V = 1 ,I 05 k Okay

IF web size were inc- L = 3/4 in. and the stiffener spacing rerained the same do = 180 in.

V, = 0.58 FyD r, = 058 150) 96 I%) = 2.088 k

k = 6.42 as before

Fv I 308.247 M, =-= (50) = Z* 234 k-ft c 50.9(12)

0.75 Mu= 18.925 k-ft < M = 21,258 k-ft

applied V = 857 k < allowable V = 1,093 k Okay

So either reduce spacing between transverse stiffeners. do = 90 in., or increase the size of the web. r, = 3/4 in.

For this example use:

t, = 3J4 in. and do = 180 in.

2. Maximum moment - associated shear a) IH Group

M = -10,269 - 3,492 - 9,876 = -23,637 k-ft

0.75M, = 18,925 k-ft < M = 23,637 k-ft

V = 0.70V, = 0.70(1.286) = 900 k

applied V = 721 k < allowable V = 900 k O k y

b) IPW Group

M =-10,269 - 3,492 - 9,172 = -22,933 k-ft

V=325+ 1 1 1 +334=770k

Mu = 25,234 k-ft

0.75 M, = 18,925 k-ft < M = 22.933 k-ft

V= 0.75V"= 0.75(1,286) = 964 k

applied V = 770 k < allowable V = 964 k Okay

4.11.6.2 Transverse StifSener Design

Moment of inertia required:

......................................................................................................... I = d o t 3 J w (10-106)

Where:

Use:

Area Required:

Where:

Y = Rario of web plate yidd ta stiffener yield

B = 1.0 for stiffener pairs

Since area required < 0, then the barisverse stiffener must meet only the moment of inertia requirement ( 10-106) and the width to thickness ratio:

Try 6 x ?hH Stiffeners

-=-- b' -12~13.7 Okay r '/z

= 86.6 in? > l,ad = 38.0 in.'

Use 6" x lh" Stiffeners F * i 4-19

Web aad Stiffener Cross Section

4.12.1 Weld Design

@ Pier 2 VWhd = 857 k = Design Shear

Applied shear flow at flange-teweb weld

where: Q = static moment = 18 (271s)49.4 = 2556 in.3 I = 308,257 in."

According to BDS Article 10,236, the minimum size of fillet weld f0~2~/8" plate is '/2", but need not exceed the thickness of the thinnerpart joined. Use M" fillet welds.

mowable shear on h a t of weld Figum 4-20

Guder Dimensions

F, = ultimate strength of base metal or weld metal. whichevm is smder

For A704 Grade 40 Fm = 65 ksi

For weld metal F, = 70 ksi

Use F, = 65 ksi

Fw = 0.45(65) = 29,3 ksi

allowable shear flow on h a t of two welds

= 2(l/2) 0.707(29.3) = 20.7 Win.

applied shear = 7.1 1 Win. < allowable = 20.7 Win. Okay

Use 'A" weld

4.12.2 Fatigue Check

For flange-to-web weld in shear, the allowable ranges of shear. F,, are:

TabFe 4-6

Mowable shear flow for F, for 15 ksi = 2(V2) 0.707(15) = 1Q.6 Win.

For 12 hi ............... 2(V12) 0.707(12) = 8.48 Win.

For 9 ksi ....,....,...... 2(ll1) 0.70719) = 6-36 Win.

For 8 ksi ................. 2(lit3 0.707(8) = 5.66 Win.

Type of Load

P13 with HS20

HS20 h e Lagd

Applied Shear Range (See Section 4- 16. Bridge Design System Computer Ourput):

1. HS20Ch.lu1tipleLanes) 2,000,000cydes (TmckLaad) Shear m g e = V,= 1.38(87.6) = 121 k

HS20 Truck Load 2,W,MX, F 9 @i

HS20 Single Truck over 2.IKK).000 F 8 hi

Cycles

100.000

500.000

s=-- " - 121'215" =l .O Win. < 6.36 Win. Okay 1 308,257

2. HS20 (Multiple h t x ) 500.000 cycles (Lane Load} Shear range = V, = 1.38(119) = I64 k

Category

F

F

S=-- VrQ - 161(2'556' = 1.36 Win. < 8.48 klin. Okay I 308,257

Frr

15 hi

12 ksi

3. PI3 with HS20 100.000 cycles Shear range = V,= 0.56C119) + 1.15(0.8 11304 = 350 k

s=-= vp 350(2'556) = 2M Win. < 10.6 HI. Okay 1 308,257

4. HS20 (Single Truck) over 2,oOo,000 cycles Shear range = V,= 0.81187.6) = 71 k

$=-- - ' 1(2*556) = 059 Win. < 5.66 k/in. Okay 1 308,257

Shear Connectors

4.1 3.1 Fatigue Design

The shear connectors are designed for fatigue and checked for ultimare strength. Maximum spacing equals 24 inches.

where: S, = range of horizontal shear flow V, = range of vertical shear due to (LL+I) (Service Load) Q = stahc moment of zbe mansformed concrete atea I = moment of inenia of the composite section

where: 2, = allowable range for horizontal shear

for welded studs with Hid 2 4,Zr = ads. where d = diameter of stud and a = 1 3,000 for 100,000 cycles

= 10,600 for 500,000 cycles = 7,850 fm 2,000,000 cycles = 5500 for over 2,000,000 cycles

Q - = 0.010 from " C O W ' progsam on IBM mainframe I

1. HS20 (Multiple Lanes) 2,000,000 cycles (Tmck Load) Allowable range of horizontal shear Assume 7/s" diameter studs, 3 per row

a = 7.850

zzr = 7.850( 3/a )' 3 I8

= 18 kh studs + spacing = - 1.000 sr

Table 4-7

2. HS20 (Multiple Lanes) 500,000 cycles ( h e Load)

Span 1

Abut 1

0.4L1

0.7L1

EZ, = l0.600( %j23 24.3

=24.3 kJ3 studs + spacing=- 1.000 sr

Table 4-8

v, 122

101

3. P13 with HS20 I 00.000 cycles

zzr = 29.9 131000( = 29.9 k/3 stuck -r spacing = -

1,000 Sr

Qll

0-010 . I -

0.010

Table 4-9

SF

1-22

1,Of

105

Spacing

11.8

17.8

1.05 0.0 I0 17.1

span 1

Abut I

0.4L1

0.7Ll

S. [ 2.99

1.71

1.99

v, 299

171

Spacing

, 10.0

17.5

15.0

Qir

0010

0.0 10

199 0.010

4. HS2O (Single Truck) over 2.000.000 cycles

zz, = 12.6 5'500''/a'23 =12.6 k/3 studs -r spacing=- 1, OQO S,

Table 4-10

Spacing for Fatigue

Span 1

~ b ~ t 1

0.4L1

0.7L1

I

40' 70' 1 40' 4

Rows 8 10 Rows 8 15 I No Studs or could use max. spacing = 24"

(z Abut 1

Vr

71.6

59 A

61.3

Q Pier 2

F- 4-21 Spacing of Shear Studs

Qfl

0.010 ----- 0.01 0

0.010

Number of smds provided for Span 1:

=(48+56+ l)x3=31Sshrds

These c ~ ~ o n are also applicable to Span 3 because the bridge is symmetrical. Span 2 calculations are similar.

S,

0.7 I 6

0.594

0.613

Spacing

17.6

21.2

20.6

4.1 3.2 Ultimate Strength

The number of studs provided for fatigue must be checked for the ultimate strength of the structure.

where: N1 = number of studs between point of maximum psirive moment and adjacent

end support or point of inflection. = 0.85. a reduction facror

S, = ultimate strength of connector P =ultimateforcecapacity,sm~erof

PI = ATy ....................................................................... (10-61) .................................................................................... P2 = 0.8Sf: brb (10-62)

where:

A, = area of steel section F, = yield point of ssteeI f: = compressive strength of concrete b = effwtive flange widtb of concrete t~ = h c h e s s of concrete

The ultimate strength, S,, of rhe stud comeccot with Rid 4 is:

where:

f: = 3 350 psi. w = 145 pd, E, = (1 45)" 33 63.250' = 3.3 X 1 @ psi. d = 'la in.

A, = area of smd section F, = yield p i n t of stud

therefore P = 3.936 k controls

P number of studs N, = - =

3.936 = 147

@S,, 0.85(31.7)

number of studs requid in compression flange length of Span I = 2 x 147 = 294 smds

9 15 studs provided for fatigue 294 studs required for strength

Fatigue design governs the number of studs in Span 1

4.1 3.3 Shear Cmectors at Points of Contraflexure

If no studs a~ used overthe negative moment m additional sruds are required at the dead load points of contraflexure to anchor the addi ti& deckreinforcement placed overthe pier. The minimum amout of reinforcement is 1 % of the concrete area of whch two-thirds must be placed in the top layer within the effecrive widdi.

Area of concrete = 14.50 ft2 Af = total m of longitudinal slab reinforcemeor over pier

= 0.01 (14.50) = 0.145 ft' = 20.9 in.'

Number of connectors:

where: N = number of additional connectors at points of c o n ~ e x u r e f, =range of stress due to live load plus impact in the slab reinforcemen~f~may

be taken as 10 ksi. Z = allowable m g e of horizontal shear on an individual shear comectot

- - - 12 6 = 4.2 for 7/sU diameter stud at over 2,000,000 cycles 3

These studs must be placed adjacent te h e dead load point of contdlexure witbin adistance equal to one-third the effxtive slab width. The reinferring should extend 40 diameters beyond this group.

4.14 Bearing Stiffener at Pier 2 Reaction at Fjer 2 (See Secrion 4-1 6, Bridge Design S w e m Computes Ouput) :

1. DL (Girder + SIab) = 1.3(240 + 250) = 637 k 2. DL (rail + overlay) = 1.3182 t 85) = 217 k 3. Live load - greater of either

Ei) IH&oup=3.0(185)=555k b) IW Group = 0.73 (1 85) + 1.22 (344) = 555k

According ta BDS Article 10.34.6. bearing suffeners are designed as concentrically loaded columns.

where:

A, = _mss effective area of column

and

where: K = effective length factor

= 0.75 for welded end connections r = radius of gyrauon. Fy = yield of sfeel. E = 29 x 106 psi. F,, = critical bucklinz

stress

The stiffeners are A709 Grade 36 steel. F, = 36 ksi

for short columns assume

P I,4W Artqhd = - = - = 39.1 in? F, 36

Try:

I Q

Pier 2

Figure 4-22 Plan View of Bearing St'Xfeners

Try 8" x 3/4" PL

web:

between stiffeners = 12(3/4) = 9 9."

outside stiffeners ( 1 &r,.) = 1 8 C3/44) 3/4 = 1 0.1 in.?

total area = 55.1 ia'

36 (1 8.0)' = 35.6 ksi 1 PU =0.85AfF, =0.85(55.1)35.6=1,667 k > I , 4 0 9 k Okay

Check bearing on end of stiffeners

bearing smngth = 1.5F, = 1.5656) = 54 ksi

applied bearing (assuming 192 in. cope on bearing stiffeners) . .

- - '"09 = 48.2 ksi Okay 618- 1.5)0.75

:. use 6 PL 8" x 3/4" @ Pier 2

Note: Spacing of bearing stiffeners is normally controlled by the size of the 'bearing pad. Access for welding should also be considered: h e 6 inches shown in Figure 4-22, while adequate for design purposes, will make welding difficult

4.15 Splice Plate Connection Example and details to be completed at a later date and distributed at that time.

4.16 Bridge Design System Computer Output The following pages are selected parts of '"BridgeDesip System" for the example problem.

INPUT FILE: STRUCTURAL STEEL DESIGN EXAMPLE PAGE 1

FRAME DESCRIPTION END SUPPORT CARRY OVER

MEN JT. COND OR DEAD LOAD K FACTORS RECALL NO CT RT LT RT DIR SPAN J HLNGE E UNf SEC CT RT LT RT MEM

/-/ /-/ /-/ / - I / - J l - / I - / /-/ /-/ /-J /- / / - I 1 1 2 R H 150.0 90.00 0.0 3600. 2.500 . 000 0.00 0.00 0.00 0.00 Ci 2 2 3 H 200.0 40.00 0.0 3600. 2.500 .000 0.00 0.00 0.00 0 . 0 0 0 3 3 4 R H 1 5 0 . 0 90.00 0 , O 3600. 2.500.000 0,OO 0.00 0.00 0.00 0 4 5 2 P 20.0 5.00 0.0 3250. 0.000 . D O 0 0.00 0.00 0.00 0.00 0 5 6 3 P 20.0 5 . O Q 0.0 3250. 0.000 .000 0.00 0 . D O 0.00 0.00 0

F W E PROPERTIES END SUPPORT CARRY OVER DISTRIBUTION

MEM JT COFID OR FACTORS FACTORS NO LT RT LT RT DXR SPAN MIN E*T HINGE e LT RT LT FT I - / /-I /-/ /-/ I--/ /-/ /-/ /-/ / / /- / 1 1 2 R H 150.0 0.3240E+06 0.0 3600, 0 .500 0 .000 0.000 0 . 5 0 0 2 2 3 H 2 0 0 . 0 0.3240~+06 6.0 3600. 0.500 0.500 0 . 5 0 0 0 . 5 0 Q 3 3 4 R H 150.0 0,32403406 0.0 3600, 0.000 0.500 0.500 0.000

4 5 2 P 2 0 , O 0.1625B+05 0.0 3250. 0.000 0.500 0.000 0.000 5 6 3 P 20.0 0.1625E+05 0.0 3250. 0.000 0.500 0.000 0 .000

* * * FT(AMB DOES NOT SWAY WITH THIS LOADINGA*** ~ R I Z O M T A F . MFMRER MOMENTS TRIAL Q MEM NO LEFT . 1 P T . 2 P T . 3 PT .Q PT . 5 PT . 6 PT . 7 P T .R PT . 9 PT RIGHT L 0. 1741. 2920. 3536. 3590. 3082. 2010. 377. - 1 R 1 9 , - 4 5 7 8 . - 7 8 9 9 , 2 - 7 8 9 9 . -3399. 101. 2601. 4101. 4601. 4 1 0 1 . 2601. 101. -3399. -7899. 3 -7899. - 4 5 7 8 . -1613. 377. 2010. 3082. 3590. 3536. 2920. 1741. 0.

HORIZONTAL MEMBER SHEARS TRIAL 0 1 134.8 97 -3 5 9 . B 22.3 - 1 5 . 2 - 5 2 . 7 - 9 0 . 2 -127.7 -165.2 -202 .7 -240 .2 2 2 5 0 . 0 200.0 1 5 0 . 0 lDO.0 5 0 . 0 0 .0 - 5 0 . 0 -100.0 -150.0 -200.0 -250.0 3 2 4 0 . 2 202.3 165.2 127,7 90.2 5 2 , 7 15.2 -22.3 - 5 9 . R -97.3 -134.8

-

LOAD DATA TRIM, 1

LOAD FIXED END MOMENTS LINE MEM w OR P COPE A e CF.FT RIGHT DEFT.T COWENTS

1 0,850 U 0 . 0 0.0 0. 0. 0 0 ASSUMED DATA 150.0

a 0 . 8 5 0 u 0 . 0 0.0 a. 0. o o ASSCIMED DATA 2 0 0 . 0

3 a. 850 v 0.0 0.0 0 . 0. o o ASSUMED DATR 150.0

HORIZONTAL MEMBER MOMENTS TRIAL 1 MEM NO LEFT . 1 P T . 2 P T .3PT . 4 P T . 5 P T . 6 P T . 7 P T , B P T - 9 PT R I G H T 1 0 . 592. 993. 1202. 5221. 1 8 4 0 . 684. 128. -619. -1557. -2686. 2 -2686. -1156. 34. P 8 4 . 1394. 1564. 1394. 8 8 4 . 3 4 . -1156. -2686. 3 -2686. -1557. -619. 128. 684, 1 0 4 8 . 1221. 1202. 993. 5 9 2 . 0.

HORIZONTAL MEMBER SHEARS TRIAL 1 Z 4 5 . 8 33.1 20.3 7 . 6 -5.2 -17.9 - 3 0 . 7 - 4 3 . 4 -56.2 - 6 8 . 9 -81.7 2 8 5 . 0 68.0 51.0 34 . O 17.0 0.0 -17.0 - 3 4 . 0 -51.0 -68.0 -85.0 3 81,7 6 8 . 9 5 6 , 2 43.4 30,7 17.9 5.2 -7 - 6 -20.3 -33.1 - 4 5 . 8

LIVE LOAD DIAGNOSTICS

SUPERSTRUCTURE LIVE: LOAD

MEM SUPERSTRUCTURE SUBSTFXJCTURE UNIT STEEL M S SCALE ENCE NO. LT.END RT. END LT,END RT.END POSITIVE NEGATIVE m. LINES GEN

I 1.000 l.oao l o o 1.0 0. 0. o o NO NO 2 i.ooe 1.000 1.0 1.0 a. 0. 3 1.000 1.000 1.0 1.0 0. 0.

L I V E -TRUCK -LANE- NO. LIVE LOAD P1 Dl P2 D2 P3 UNIFORM MOM. SHEAR LL LOAD NO, R I D E R RIDER IMPACT M S . SIDESWAY

1, 8.0 14.0 32.0 14 .0 32.0 0.640 18.0 26.0 YES 0,00 NO COMMENTS: HS20-44 AASHTO LORDING WITHOUT ALTERNATIVE

2 , 0 . 0 0.0 0.0 0.0 0.0 0 . 6 4 0 18.0 26.0 YES 0.00 NO COMMENTS: LRNE LOADING

1 - 8 . 0 1 d . D 32.0 1 4 . 0 32.0 0 . 0 0 0 0 . 0 0.0 YE5 O . f l 0 NO COMMENTS: TRUCK LORDING

. . .

1

I

m m m o dl - 0 & ? .? , . . - - . W . . . . . - U N q f i * P

C) P a . 4 ( U ~ O O i2 z d 2 w o a H m o r l o

5 & m m m m r p i r l w r l F! I..I I N

r Y I N l m P - 4 I C E I I K

~ W W W d m s!.<."1 . . . . . . . . . . . . W L n N W I O V I I I ) W ( B r l F U l l n F w m W 01 W P I d Q I k V l m N P I Q H z T r ? P ? I I l r * l r D - I N 4 I F I N M l U 7 t O I rl

C R N A I ol 4 L - 1 1

E m F N I

vl W O W V1 PJ ct 0 P I . . . . . . . . . . . . . . . . . P e m w F %+ d n m w n ~ % P rnmmmmm m orc - im W b ddU O d W h m m 4 m w I n - m a r l x w ~ r n - x d r n r t - I a t urn d I I a m d 4 4 . I o n E

E z b d r m

- 0 - E d eJ In _ . _ & . . . . . . * - . - . * g h q p p d 4 2~ m n n a e r Z b r n - w - m d r - r l p w r - m ~a odd O A U & r - 0 4 - m n P - r I ~r, 0 W I r n w g3 ~ l ~ i ~ l m 4 V I P 4 I I e m . I ' 3 . d u rC t;p.m m o

- 0 - P r- w rc o h - . - 4 * . . . . . . . . . . . . ~1 r l w w o m

h 0 4 4 0- Llr ~ ~ U I N N - I 4 - ~r I 4 W r v ~ m m W w 1 r n 8 e 1 4 rl & W rl 8 I & W (V N N 0 . I 0 - d il

Err, d P1 d o e N w N V 1 & - " " 5: . . . . . . . . . . . W m r l w o m

w P f - n n o ~ m d ~ - w o - t + a m ~ n u m ~ h orlrl o d IWE ndrrrlnd 0 - N 1 m

g m Z N T z G m X I K : +l N 4 I*r

W . Z z- 0 z r * o * r P - w r y c ' .': .? . . * . m . . . - - - . K w m w f - I n s g m n n o s n B b s-0-40 d - r n d m w *

odd ad %IL ~ t l ~ r y ~ n bl - n r n 0 O o l l n N X N - I

1 I rl d q N N N V) I

W M

U i3 ,,, z 2 4 o g ? .4 .yl . . . . . . ;I . . a h . 2 J m m m m w

J k + n m o r n n p n r l w - v r - m - w m w p & W P * odd od W L L m m ~ l r l - r w W . rl I A

m l m e . + o r - w + % I , 2 , , , d m 2 L - ;I

3 CO e 0 d 29 .r .CY . . . . . . E z 0, . ? . 4 . ? m e c l c l m

n m m w d n o w m m m ~ P ~ V I P I ~ ~ F E ,A, I,, E ~ V l r l m m m ~ l r n 9 - 1 0 3 w t- rl d 4 W r Y l l d N 4 d A il I ct U

W (D N r l W E? my .? . . . . . . . . . . . . V1 QEmVlr l l l I

a E r l l c ~ w r n m w mm[cWintn r l W 4 r l N O D O u y y y k 0 - m n n n Z C v I o - u l q r r l - 0 7 1 - b d N L F cl 0 In U, ~ ~ b d ~ r n r n t r , rl I -I m d d b Z -444 I I K=. +I

i9 0 01 0 a .l Ut t7 *'? .? . . . . . . . . . . . . C) W P J O ~ O I -

d J r n 4 m I - m m m m R Q O N * C u r l E O D V r m m d w I Y 3 1 - . D l CV r 4 .W F1 rl drl

H 4 4 r? r7 d d E 0 I I

O w Z d m m . +z x ul I

. - . .. , - " --

i g = .N -1 5: .r1 . 7 g U m e D f m U 1 O R O F H m.+m.+ H A m m

KJ Cc N h W d E I

r7 I w

I

W C7 E m e E e n F

& . . . . . a . , . . . QINr7I-m l n r l w - I 1

m 0 m m 0 r n 4 N W N r n I p I W d r - l a

I I . . . . * * * .

0 0 0 0 0 0 0 0

F.e 0 m e m u Pa & . . - - - & . . , . .

~ # v r r n e r l r l t l m o m ~3 r m + l N l - m m rnNrnRVP zs.

r W I C - r - I t - b7 W d 48

d Z z o . . . " . . . * z

Z E r , m mg?.? --- , y O h . . . - a I 0 0 0 0 a o P O

r m n n p n ~ m w m m w 2 Z '--P4m-w- H I P

a a ~ l e 13 - 4 1 4 (3 I W d r l t l C Y 2 I * P

5 5 S u b m m rnuerc rc m Q & . . . - . . . . . d z s 0 0 a m 0 0 0 0 w ~ n m d ~ ~ w ~ r n a m = . - . . . . . t o m m - l - o ~ m m m w m ~ r - x o 0 0 0 0 0 0 0 0 4 - 1 4 - m 1 m ct cl

4 N 0

9 V1 0 F 0 4

bE, O o E d w 2 rl W b . . . . . m & . . . . . W m d m d m W W O O l d R W V1 a r n n r - n v ~ & m ~ r n ~ m ~ P: w

2 . I D N d d N . 2 a - T r . * - . . . W UE

- 0 0 0 0 0 0 0 d o z

H -6,, , , 0

& . * . - . p 1 . - . - . S r 0 4

r n r n P 0 P E e o d w m W C ) P 4

, I m ;Ssm:'r"zp <Zr 9:s: W z x r l r l a u wl . . . " * . s . E a T m a oo 0 0 o o u w a I & 3

LZ d n m m a m d ~ F I + I - 4 b

m W i W d W m @ 7 W P d P m to u . - 1 0 w - W I ~ I P; 2 A d 5 ,-I W

a J el 3

LC lo w m , . . . . . kY ,." .r

e ~ m ~ l t l 4 O l W I I P l I C H 4 CW* rnrl vlrl N W N P N 4 U a N l - 4 1 0 - - L n I O - d m

22 m 7 27 27 r r r : w I E'

U 3

Ern In *r E d d N & . . . . , P I . . . . -

a: + Pl-lmNF1 N U I Q m W yl

d m m I o o d m ~ t r m a I - m m 3

I A a w WrtF W I J OMcn I z z 5s i 3 Z h

hE. e E b E E E . 0 E N .? ." t? .Y .r ~ u d n u ~ n u a w n ~

m u m ( 3 U1u m o o w o w r n w m W m ~ n w ~ o w o w a~ O W H J d m I w d A c n I m o C ; & Z 6 P . Z & & C . ' & Z b - *la N - I . w rut-

"" ';" -Iri ? 4 Q q b c s - 3 - W O W 0 w 0

2 Z S S E Z S i J J G O Z r n ' - ' Z E E 2 Z F a u , , m , ~ 5 ~ a a ~ $ ~ .3 L L L a o 4

.azazzz Z ~ Z Z Z 3 6 ' $ 2 5 - .

e 9 ol a m pl D g". .=? .vO .".Y $"rY rrl . . . . . . W . . . . . . u n r m a ~ O m e o s m u e a f u o ~

0 N ~ N I P O O E r n r l r n l p o ~ H m o n o H m + n d H A m m A m m n m X W W ~ d w 4 z w d - 4 & I P. 2 r.7 I r 4 I N

I? I I E l 8 E

P-l 4 m kr t w P-F .I r- a .cTJ - w o r n & . . . . . a * . . - - & - - . - . . . . . . . , . . . . . w m ~ d m ~ m

I 1 7 U 1 e W r f r r l V l r l F d M Q I rn ~ m 4 m rn ~ m a & m d ~ w ~ m k mr7Nr7P- l E e m d r r l P L n r ? t W - t - 1 0 rl ~r I N m ~ m ~ c n i rl r ~ r (

r n N d 4 m I - 1 1

- 0 - N N ** 7 ? 2 4 .? .'7 ;7 .1 .'9 . . . . . . 5 b . . . . . 2 F W m W P m b 7 m - d d d m m W 3: d r t ~ + m m 4 b n r w o m m m t-r-dm w d w t - m m m m n m r n ~ 4 W & m m m m o w m r r l w 1 r - m ~ w z r o ~ m - x 4 1 4 1 r - t 10 r CJY tn -I

m m 4 I I tnm ,+ A A - I p. n U W a, rl F. - 0 - F m u 1 z b m O h . . Y .?Zk? .T . ? Z E - ? . * . . . . * - . * . . . . . . d r n p - ~ m z m m n p n ~ m w m m ~ S e m m n o - n 2~ m - w - r n + 2 p d w m p p w ~ 4 N w - m p N w w u p U & odd 0-1 utb ~ - ~ - r n n 2 . m ~ w c i w O . 4 1 m o e l m w gF 5 1 ~ 1 4 ) w r l w r l r l W P r r N + h k m - I R % . ;T 5 5 C1 - 0 - . * . . . . .- .+ -9 8EZk6daH"Z_A;E"E;;2

400P-P ~ W d m W W P W U N * ~ P P ~ W ~ W ~ N * D r l d 04 ', d010NN3 4 . 4 1 3 4 - 1 4 . - 1 0

W N I C ~ m o ~ 1 0 1 - I 4 4 N rl TU L L ~ W -I I I a w N N rj 0 . I 0 . ;I il

. . . . . . . . . . . w = * fl * N G O & . . W & . . . . . . . " . . - - W V I * I D O r l W m * l D 4 1 D E h 1 0 0 V l + d W E &=r;l"r-z

Zv, N N N W - I:

0 E

P I . . . - .

i 9 ~ - i- n w ~ w m - W P M o r t . . . . . 2: - o w g - = I - & . . . . . * . . - . . . . . . . . r n r n W P 0 E rn rnPOFE W O ! - d t Q b m n n o m n 4 P 1 O r l d 0 - 4 NrlU1Ncrrn - I r7 m I QI W . d I * G m l m N 0 W I N 3: d r l X 4 4 2 r l d d*r 1 I 4 J q FI N (Y Crl U) V1

r bJ W P 24 - 0 - 2 32'9 . Y .:SET . Y . ? ~ E T .":Y . . . . . . . . . . . . n m m w m d m w m s m ~ p d 4 - m d b e m m a m m E n d w d m d r J ~ * ~ ~ m ~ m w ~ d e r n M W P + I ~ F W p , o d d 0 4 W L L m n t l w ~ w U + 6n 1 4 W * n 1u-t W * w I -

W 1 m P" + O F 2 d d 2 4 4 z 4 , z m P, N 4

E ' 1 z - cl cl 3

K CI1 w o w 0 - N FI E? .? .'4 E 0 .4 ."". E'1 - 4 .? . . . . . . . . . . . . 0. m m m d d ~ W F W P

t , f i * W d m NLnlDO-P MmLnBaFt* N F N A - m N P d I W W 0310 1 ~ I r l k m s m v l n m r - 1 - m l o - d m: w t m w I- -4 4 4 +I 1

C U I 1 4 N r l r l d I

B n -$ r- ho tn a e-4 4 N . . . . . . . . . . W ' 5 W 11 0 w h . . . . . & a . * . . * . . . . . . - m m m d m n d m m n w w - m w r l m - m m w ~ w d d ~ m ~ ~ W I O O d m - ~ k w m P r l N U 7 l C ) r C I g m m F c l m m I c U 7 4 P l W r ? . m l - * P I - . + + N U l N I P -( m rn rn I d

4 1 r l m rl I I L

+I-- vl m . . . . . 0 m e 0 - m , E ? . ? . ? k a .Y .? . . . . . . . . . . - . W N D d O P W O W m d G l W I J ~ - ~ * *

E o w * l L I V l r l d m 4 m 3 d m I W A don 1 m 0

m m m m E O O N w N d W I W I I r A m rnd * d w P I 4 - W CY W - W M rl W d r Y N N PJ- TY I

I m-2 n t-i w a

I I - w > > - s

0 E Z a P : 0 a E O W -

W O W w o w Z r $ 2 5 Z x $ 5 5

A - , "

d m 0 2 - . . . . . . . . . . o m N d - - U F m O r n F O m o m D m H N W O m H N W - m W P W P: P l l r ? E P l I r C L F1: I

I I

u a l l E

m N m VL 4 N . . . . . . . . . . . . N o l W W W P P F i ~ Q U I ~ ~ ~ O J W Ul d P 8 Z rn e m P W P l m rn W P W P m ~ r - d m rn W W N F

r.31- . 4 N u l a 4 1 0 I rl

B a n m G m w in P v o T . . . . . . . . . . . r D m m V1 0 - 0 p, & p , . , . . . . . . . . . . . . . . d w m t t - m o m t+tMPlmO(D u~ C W O P rn - ~ P P ~ A N ~ V P - m r r l - - 1 1 - + m + r -

x - ~ r m ~ r - r 07 r l w r l r l v ) 4 - 1 m w rl I I CnQI rl 4 4 - I a W

. . . . . .

Z F 2 Y 7 u 1 . l F l l r l l W A d N

t7 - r m - .I rl H

U B m n F U ~ N w o v E w f- n . . . . . . . . . . . P W N W 0 m p\ ~ b . . . . . . . . r n ~ q ~ c n

W C P t n 4 TU W N N N G W I I I a w N m N 3 . 0 .

E: 2 m m w

E: . . . . . . . . - . . - 5 , - P 3 0 - P b t O N m - P P b W n O V l O P L U ? ~ P N n P

N n W m N I I 0 * N I N O - TT t 0 . W I N B ne-1- il P I N ZVI n m n w U - E: N N

2 - 2:

. . . . . . . . . . 4 m m & . . - - . . * . . . . . . w w m w m m a r n r n P W * S m * d w - a @ o m r - ~ w m o n m m ~ w d - n - ~ r n m ~ g j ~ w ~ y ~ w $ ~ m ~ ~ r n - 4~ n ~ w ae o ~ m

I

0 L O ~ D ~ 0 m - 0 X N N Z r U N X N r + J t I I J V r-4 F1 N v) rn [R

W m P n O I ( L - . . . -

I . . . . . . ;I . - . . - - a u l W W W J W e - d e d 4 m w F V # F. m r l w m m w J m w w N m p m m m d m r - m m w d r - +

fit& v * m rrr tilh W P U + ~ F - U I w . r( 101 01 I t- rd + ra r w 9 w 0 r l m w 2 r r - ; 2 4 ' ' 2 , l b ,

L- * I Y "I A S - A

2 VI PO - e E W m In . . . . . . . . W C U T X L , b . . . . . . . . Z 2 .? .* .O m 0 - p ~ m i o m d m m m e w m F3. t n m * ~ d m 0 m o l m m ~ N M e d N P N W O I W V N W 0 1 0 P b r n I t - 7 CZI . r - l r ? - m r l . d m

C Y - 4 P m d d rl 4 rl 4 N l i 4 N 4 r l - l

I

ca Pa m t V . l r l V I . . . . . . . . . . . . m I T F l - 0 l n W N m O 3 w m ~ t c m m m m r - a r m 4 w m I wt- dr-4 r i-m - I F N I W P

* 0 d . r l 4 - e m d m 1 7 O V l V "-I I

d l 1 4 rl d r

H W m w B m N w P a 4 0 . . . . . . . . . . O I I ~ O N 4 ~ 4 . - . . . LL. . . . . . . . . . . . . ~ m 0 m 0 m W o r n - N P W A - N N W 0 0 1 0 ~ 1 0 m J t - I W I N W d m 0 1 - W

Irr P m t - W I N 1 - m r l r 7 F l W e q .k! II m m 4 r?N I m m I

rl J rl -4 rl J I I

0 U E P C z C C U W W W

J Z O - I z w t m = ;I Z Z v) vl bl

V . EXAMPLE PROBLEMS . . . . . . . . . . . . . . . . . . . 47 . 5.1 S i n g l e Column Bent w / P i l e Foot ing . . . . . . . . 47 . . . . . . . . . 5.2 Two Column Bent w/Pile Footing 65

BIXM;E DESXGN PRACTfCE JAHUARTp 1982 5-1

1-1 - SCOPE

previous issues of "Bridge Design practice" have n o t treated foundation d e s i g n as a separate topic but contain b i t s of information under various headings.

T h i s section is an attempt to concentrate the material on substructures and foundations and to apply the Load Factor D e s i g n method to the ir des ign .

The method consists of applying factored loads (AASKTO Article 1.2, Table 1 . 2 . 2 2 A ) to ultimate capacities of f o u n d a t i o n e l ements w h i e h have been modified by a s t r e n g t h reduct ion factor ( p ) ,

f n applying group loadings to bent foundat ions , a method of applying seismic loading is presented which is c o n s i s t e n t with t h e c u r r e n t philosophy of seismic analysis and des ign .

The substructure is that part of the structure which serves to transmit t h e forces of the superstructure and the forces on the substructure itself onto t h e foundation.

The foundation is that part of a structure which serves to transmit the forces of the s t r u c t u r e onto the natural ground.

f f a stratum of soi l s u i t a b l e for sus ta in ing a structure is located at a relat ive ly shallow depth, the structure may be supported direct ly on i t by a spread foundation. If t h e upper strata are too weak, the Loads are transferred to mete s u i t a b l e material a t greater depth by means of pi le s or piers.

The des ign af the structural elements for foundations, substructures and retaining walls is in accordance w i t h the provisions of AASHTO.

The design of the structural elements i s w e l l codified; t h e so i l mechanics aspect of the design is not oodif i c d to any extent.

The bearing capacities of foundation soils, settlements, the ability of p i l e s to transfer load to the ground, lateral earth pressures and lateral earth resistances are some of t h e iterhs which are determined by evaluation of site Investigations and/or current practice.

In s t a b i l i t y analyses the factors af safety f o r overturning and sliding are not specified in AASHTO. Determination of values to be used is based on accepted pract ice and evaluation of t h e r i s k involved. Part 111 will enumerate values c u r r e n t l y used.

11- STRUCTURE FOUNDAT1 ONS -

2 *I CAPACITY OF SHALLOW FOUNDATIONS

A s h a l l o w foundat ion is a term applied to a footing having a depth to base w i d t h rat io of less than or equal to 1. ( D ~ J B c 1) Where depth Df = the distance from the ground surf ace 6 the

contact surface between the so i l and the base of the foot ing ,

B = width of foot ing.

Two th ings control the capacity of a shallow foundation:

1) the a b i l i t y of the s a i l to support the loads imposed upon it, known as the bearing capacity of the s o i l .

2 ) the amount of total or d i f f e r e n t i a l settlernent that can be to lerated by the s t r u c t u r e being considered.

2.1.1 Ultimate Bearing Capacity of Soil

When a load is applied to a l imited portion of the surface of a s o i l the surface settles. The r e l a t i on between the sett lement and the average load per u n i t area (qd) is represented by a settlement curve ( F i g . 2-11. ff the so i l is dense at s t i f f the curve is s imilar to C1. me absc i s sa qd of the vertical tangent to the curve represents the ultimate bearing capacity of t h e s o i l . If the so i l is loose or f a i r l y s o f t , the settlement curve may be s i m i l a r to C2 and the bearing capacity is not a l w a y s w e l l - d e f i n e d , The bearing capacity of such so i l s is sometimes assumed to be equal to the abscissa g' of the p o i n t at which the s e t t l e m e n t curve becomesdateep and s tra ight . A more conservative value is to use the bearing capacity at the abscissa q n d B at the point where the settlement curve C2 ceases to be l inear .

When the bearing capacity of a real footing is exceeded the so i l f a i l s along a surface of rupture s i m i l a r to fede l f in Fig. 2-2(a). An approximate method of eva luat ing the u 1 tlmate bear ing capacity consists of equating separately the f o l l o w i n g three components: See Fig. 2-2.

I. The cohesion and friction of a w e i g h t l e s s material carrying no surcharge.

2. The friction of a weightless material upon addit ion of a surcharge q on the ground s u r f a c e ,

3. The friction of a material possessing weight and carrying no surcharge.

B R I X E DESfGN PRACTICE JANUARY, 1982 5-3

The approximate equation for bearing capacity of a shallow - foundation is:

Q ' c N c + 7 DfNq + 1/27 BNI

9d = bearing capacity per unit area

c = cohesion

7 = unit weight of s o i l

NC and N are bearing capacity factors w i t h respect to cohesion and surcharge respective1 y.

?II accounts for t h e i n f l u e n c e of the weight of the s o i l ,

ALI t h e bearing capacity factors are dimensionless quantities depend ing only on 8 .

Meyerhof's values for the bearing capacity factors are g iven in Fig. 2-3. Fig. 2-4 is a d i r e c t correlation between the bearing capacity factors and the N - Y ~ ~ u ~ s obtained from Standard Pene t r a tion T e s t s ,

The sol id l i n e s in Meyerhof8s table are to be used w i t h f i r m s o i l s corresponding to load settlement curve C1 in F i g . 2-1.

The dash l i n e s respect ively are for s o i l s which would correspond ta curve C2 in Fig, 2-1 , These soils would n o t f a i l in quite the same manner as t h e firmer s o i l s , and the foot ings would se t t l e before shear became mabilized along the entire surface of F i g , 2-2 .

For this local shear f a i l u r e an approximate solution is to use 2J3 the value for both cohesion and f r i c t i o n , i,e,,

tan P R = 213 tan B

the equation for bearing capacity becomes: = 2/3 c Ncv + 7 D f N q * + 1/27 BljrB

and the bearing capacity factors N,', H and ~~1 a m taken from the dash l i n e s us ing the angle of ghearing resistance 8 ' . ( F i g . 2-31

The bas kc equation for beating capacity relates to a continuous or s t r i p f o o t i n g . Modif ieations of the formula are ava i l ab l e for use w i t h square, circular or footings of other shapes. ( R e f . 13

Tables are also a v a i l a b l e in the references for modifying t h e basic beat ing capacity equation fo r the c o n d i t i o n of a footing on or at the top of a s lope . (Ref. 2 )

BRIDGE DESXGN PRACTICE JANUARY- 1982 5-4

- 2-1.2 mtermination of Allowable Settlement

There are many methods of estimating the settlement of a shallow foundation, Some of these are:

1. Table, ( R e f . 0 ) t i t l e d "Allowable Bearing on Granular Sediments".

Fig-

3. A Reyerhof relation:

Equation I qs = N/8 when B < 4 "

Equation 2 gs = 8/12 1 7 ) when B 2 4

Where qs = allowable bearing in tons per square foot

N = blow count o b t a i n e d from Standard Penetra t ion T e s t

p = settlement in inches

B = w i d t h of the foundation under consideration

Many other methods exist, however, these three a l l relate to the value N obtained from the Standard Penetration T e s t which i s used by TransLab Engineering Geology for determining soil parameters in most cases,

If a footing is underla in by a layer or layers of compressible material, settlement due to the compressibility of the layers must be added to t h e amount of settlement obtained from the procedures noted above.

2.1.3 Oesisn Procedures (General)

B i s terieally the des ign of s h a l l o w spread foundations cons is ted of proportioning the footing to d i s t r i b u t e service loads on the foundation sail such that the maximum bearing pressure d i d n o t exceed an allowable capacity as predetermined by TransLab Engineering Geology.

This allorable bearing capacity was that unit load which it was est imated would produce a maximum differential settlement a5 1/2".

The allowable bearing capacity in no case, however* was to exceed the ultimate bearing capacity reduced by a factor of safety of 3 .

Load factor design of shallow footing foundations w i l l employ the use of t h e u l t i m a t e bearing capacity of the foundat ion so i l .

TransSab Engineering Geology w i l l . n o w be providing values for - u l t i m a t e bearing capacity of spread foundations and inf omat ion as to the nrethod by w h i c h they were determined.

2 * 2 CAPACITY OF DEEP FOmDATX ONS

A pier is a structural member of steel, concrete or masonry that transfers a load through a poor stratum onto a better one. A p i l e i s e s s e n t i a l l y a slender pier that transfers a l oad e i t h e r through its t i p o n t o a f i r m stratum (point bearing pike) or through side f r i c t i o n onto the surrounding s o i l ever some portion of its l e n g t h (friction pile).

Load settlement curves for piers and piles are similar to those for footings, The d e f i n i t i o n of bearing capacity of piers and p i l e s is i d e n t i c a l w i t h that of footings.

Piers founded an f i r m soil beneath layers of more eomoressible material a r t mse l i k e -read footings with surcharge approximately e q u a l to y D f . The bearing capacity qp may be determined using the prowr form of t h e basic hearing capacity equation considering the shape af the pier,

ff the so i l surrounding the pier i s bomoqeneous the shear patterns in the soil at failure are altered and the bearinq capacity formulas no longer amly.

Consider a cy l indr i ca l pier of radius-t, and d e p t h bf. A t f a i l u r e the load is expressed as:

qp = the bearing capacity per unit of area of t h e soil beneath the base.

Ap = the base area.

f , = the average value a t failure of the -Mined effect of adhes ion and friction along the contact surface between pier and soil. The l a t t e r term, earnmanly referred to as - sk in friction'.

The values for adhesion and friction can b determined approximately in the lab, R r ~ r ~ ~ r r the nethod of i n s t a l l a t i o n nf a pier has a marked i n f l u e n c e on the values,

The bearinq capacity of a pier then is mast t e l iab ly detsnnined u s i n g empirical ~ l u e s for qp and fS as selected by someone exper ienced in evaluatinq conditions e x i s t i n g at the s i t e and construction procedures.

Tables containing approximate values of the parameters (fs,qp) for various s o i l s and conditions are available in R e f , 1.

2.2.2 Bearing Capacity of Piles

In general, the beating capacity of a single p i l e is controlled by the structural s trength of t h e p i l e and t h e supporting strength of t h e soil , The smaller of the t w o values is used for design.

F i l e s driven through s o f t material ta p o i n t bearing may be dependent upon t h e structural s trength of the pile for t h e i r bearing capacity.

The supporting s trength of the soil is t h e sum 05 two factors - t h e beating capacity of t h e area beneath the base, and t h e frictional resistance on the contact surface area for t h e length of t h e p i l e .

For point bearing p i l e s the former is of primary significance w h i l e for friction p i l e s the latter is of primary significance,

S t r u c t u r a l sections of pi l e s are to be designed using t h e provisions for t h e material being used and satisfying t h e minimum requirements specified in AASRTO and this section in foundations. R e f . 13 has standard d e s i g n s for 45 ton and 70 ton piles which are 1 laterally supported by soil, Heme to Des igners .3 -3 may be used for design of IS* diameter p i l e extensions,

Displacement of soil during installation of p i l e s creates varying states of stress i n the surrounding soil and makes computation of " s k i n f rictionw unreliable.

Ranges of empirical values for 'skin frictionn in various soils are in R e f . 1. Local experience is of great value in selecting empirical values to be used.

The Engineering News Record Formula is a Dynamic Formula for determining p i l e capacity, The formula has been in use in t h i s country for some 90 years. The R4R formula, as w i t h several similar dynamie formulas (banish, Janbu, etc, ) , estimates bearing known as dynamic resistance from measuring t h e average penetration of t h e p i l e under the l a s t f e w blows of the hammer. While the farmula is theoretically sound the results obta ined from its use are unreliable. R e f , 1 cites load tests on p i l e s driven using ENR which had bearing capacities ranging from 1.2 to 30 times t h e value obtained -by the formula, The formula itself has a b u i l t in factor of safety af 6, R e f . t and many other respectable so i l t e x t s recommend against the use of E?lR for determining pile capacity.

The Wave Equation is a sophisticated Dynamic P i l e Formula which is now be ing used on an experimental basis by t h e TransLab Engineering Geology.

A pile l o a d test is probably the best method available for determining the bearing capacity of an indiv idual p i l e . The tests are quite expensive, however, and on small jobs the cost of their use cannot be j u s t i f i e d ,

&I large jobs where there are many piles to be driven, a p i l e load test is performed under the d i r e c t i o n of the TtransLab Engineering Geology, Detai ls on the performance 05 a pile load test are available in Reference 8,

Design Proeedur@s I General)

P i l e foundations have h i s tarically been designed using service loads. TransLa b Engineering Geology has u n t i l now recommended allowable bearing capacit ies of p i l e s using a factor of safety of 2 aga ins t bearing f a i lure .

The TransLab Engineering Geology w i l l now furnish the ultimate bearing capacity of the p i l e and the method by which i t was determined.

P i l e capacity designations currently used w i l l not be changed an contract documents, standasd plans , etc. For example, the class 7 0 p i l e w i l l be designed using the ult imate capacity of 140 tons (280 k i p s } , but w i l l still be designated a ? O ten p i l e .

2.3 DESIGN OF BENT FOUNDATIONS

Procedures for design of footings for columns. (Note: Procedures also a p p l i e d to f a o t i n g s for pier w a l l s in the l a n g i t u d i n a l d iree tion. I

Columns on f nd i v i d u a l F o o t inga

betermine calmn section requirements based on the mad Factor Design Group madings in AASBTO and using the design strength a£ the member, The design strength of a member or cross section equals the nominal strength modified by the strength reduction factors (8) s p e c i f i d in AASBTO. The nominal strength of a member or cross section equals the strength c a l e u l a t d using the spcified compressive strength of the concrete and the specified y ie ld strength of the reinforcement,

2. mtermine as a minimum the nominal moment strength of the column in the direction of the principal axes of the footing at the locations where p la s t i c h i n g e s may form when the structure response to se i s m i e loading causes inelas tic action in the columns. These nominal moments s t r e n g t h s s h a l l be those associated w i t h the unfaetosed dead load a x i a l force. Current ly there is a TSO program t i t l e d "YIELDn which can be used to develop interaction diagrams for c o l u m n sections. Also in terac t ion diagrams for t h e standard column sections are a v a i l a b l e .

BRIDGE DESIGN PRACTICE JANUARY, 1982 5-8

- 3. Determine the column probable p l a s t i c moments. The column

probable p l a s t i c moments equal the nominal moment s t r e n g t h s increased by a f a c t o r equal t o 1.30.

4. Using the column probable p l a s t i c moments, determine the corresponding column shear fo rces .

Determine the a x i a l fo rces i n the columns due t o .over turning when the probable p l a s t i c column moments a r e developed. Using these column a x i a l fo rces combined with the dead load a x i a l fo rces , determine new column probable p l a s t i c moments. Using these new probable p l a s t i c moments determine the column s h e a r forces . I f the sum of these new column shea r s a r e not reasonably c l o s e (wi th in 10 pe rcen t ) to t h e sum of the prev ious ly determined column shears , r eeva lua t e the column probable p l a s t i c moments and column shears .

5. The u l t imate moments to be used f o r designing t h e foo t ing s h a l l be those t h a t a r e t he l e a s t c r i t i c a l of t h e following two cases:

A. The f i n a l column probable p l a s t i c moments a t the base of t h e column.

B. The column moments a t the base of the column f r m an e l a s t i c seismic a n a l y s i s before any reduct ion f o r d u c t i l i t y ( 2 f a c t o r ) . Two orthogonal d i r e c t i o n s of ear thquake motion s h a l l be considered. The moments which r e s u l t from the a n a l y s i s of earthquake motion in one d i r e c t i o n s h a l l be combined w i t h 30 p e r c e n t of the moments which r e s u l t from the a n a l y s i s of earthquake motion i n the o t h e r d i r e c t i o n . This w i l l r e s u l t i n appl ied moments ac t ing i n two orthogonal d i r e c t i o n s simultaneously. The two poss ib l e combinations of moments s h a l l be considered. See Figure 1.

6. The ho r i zon ta l fo rce induced i n t o the s t r u c t u r e a t the ben t is l imi ted t o t h e column shear f o r c e a s soc i a t ed with the development of t he probable p l a s t i c moments. The l a t e r a l r e s i s t a n c e of t h e foo t ing may be considered adequate provided the ma te r i a l surrounding the foo t ing and upper po r t i on of the p i l e of p i l e foo t ings has a s tandard pene t r a t i on value, N, of a t l e a s t 4. The p i l e s f o r p i l e foo t ings should be designeri to s u s t a i n l a r g e induced curva tures and still maintain t h e i r des ign a x i a l load.

7. The u l t imate v e r t i c a l fo rces tc be used f o r designing the foo t ing s h a l l be the unfactored dead load force combined w i t h t he ax i a l f o r c e s associa ted with the u l t imate moments of S t ep 5.

8. Design a foo t ing t o r e s i s t the u l t imate moments and forces Of S t eps 5 and 7. For r e s i s t i n g the v e r t i c a l f o r c e s and moments use the u l t i m a t e s o i l bearing capac i ty o r the u l t ima te p i l e

- axial capacity using a s trength reduction factor (81 equa l to 1 * 0 .

When determining the flexural capacity of t h e foo t ing , use a s t reng th s e d u c t i o n factor ( $ 1 equal to 1.0 and a y ie ld s t rength of reinforcement equal to 1.0 t ines f

Y

When determing the shear capacity of t h e foot ing , use a strength reduc t ion factor ( 0 ) equal to 0.85 and a y i e l d strength of reinforcement equal to 1.0 times fy,

9. Design t he p i l e s of p i l e footings to susta in large curvatures and t h e des ign a x i a l force.

When determining t h e transverse reinforcement required in the p i k e s , consideration s h a l l be given to conf ining the eare in those regions where plast ic h inges may be expected-to form. In these regxons the minimum volumetric r a t i o s h a l l be:

L x - e P, - 0.12 2 (0.5 + 1 . 2 5 -1 , except P, need not be

4 ' cAg greater than-0.012.

I Generally, where t h e soil st ra ta below the footing increase in s t r e n g t h w i t h depth t h e plastic h i n g e in t h e pikes can be assmed to form at the footing. For this case the confining reinforcement s h a l l extend not less than twice the l onges t cross-sectional dimension of the p i l e or 36 inches , whichever is greater, )

The minimum recommended transverse reinforeement in the top 6 feet Qf p i l e 20- B or less s h a l l be equivalent to a W6.5 sp i ra l at 3 inch pitch* the minimum transverse reinforcement in the remainder af the p i l e s h a l l be equivalent to a W6.5 spiral at 6 inch pitch.

when determining the axial tensile force resistance of the pi lesp use a strength reduction factor ( 0 ) equal to 1.0 and a yie ld strength of reinforcement equal to 1.0 times fym

Uhen u p l i f t capacity of the piles is required, verify with the TransLab Engineering Geolegy that the p i l e length s p e c i f i e d is adequate for the design axial t e n s i l e force,

10. Check the footing design us ing the Load Factor Design Group Loadings in MSHTO, except omit Group VfI, The u l t i m a t e soil bearing capacity shall be m o d i f i e d by a s t r e n g t h reduct ion factor ( $ 1 equal to 0.5 and t h e u l t imate p i l e bearing capacity shall be modified by a s trength reduct ion factor ( $ 1 equal to 0.75.

When checking t h e adequacy of t h e footing sections use t h e design strength of the member specified in AASATO. Revise footing, if required.

BRIDGE DESIGW PRACTICE 5-10

11. Transverse column reinforcement shall be provided for - confinement and shear resistance.

The cores of the collrmn s h a l l be confined by transverse reinforcement in the regions where p l a s t i c hinges are expected to form.

The extent 0 5 these regions s h a l l be assmed to be length not less than (1) t h e maximum dimension of t h e column, ( 2 ) one-sixth of the clear h e i g h t of t h e column, ( 3 ) 24 inches . For the flared end of a flared column t h e extent of t h e plastic hinge region shall be assumed to be a length equal to the flare length p l u s the greater l e n g t h ef (11, ( 2 1 , or ( 3 ) above.

The transverse reinforcement for son fin em^-t within these regions s h a l l provide t h e seater of t h e 'so fellowing volumetric ratios f o r sp ira l ly reinforce& columns:

volmetsic ratios for sp ira l ly reinforced colmns:

me transverse reinforcement for confinement at any laeation w i t h i n t h e eolwmn s h a l l provide t h e following volumetric ratio f o r spirally reinforced columns:

It i s recommended that a l l coltmvls be spirally reinforced.

For those e o l m s reinforced transversely w i t h rectangular hoop reinforcement, refer to SEAOCUs .Recommended Lateral Sorce Requirements and Commentary' for the required confinement reinforcement, and AASHTO.

The des ign colman shear forces determined in Step 6 s h a l l be resisted by concrete and transverse column reinforcement. fn regions where p l a s t i c hinges may form, use the core section of t h e ealumn to reslst the shear force, fn regions other than where p l a s t i c h i n g e s may form, use the gross section of t h e column to resist the shear force.

When determining the shear resistance of the column, use a - strength reduction factor (41 equal to 0.85 and a y i e l d s trength 0 5 reinforcement equal to 1*0 times f ~f me transverse reinforcement required for con f inemgnt is also adequate for shear then ne addit ional transverse reinforcement is required. The reinforcement requirements far confinement and shear are not a d d i t i v e .

2 For columns considered hinged at the top 0 5 the footing, the bottom of co2umn design shear forces and d e s i g n ax ia l tensile forces s h a l l be resisted by concrete area and vertical reinforcement according to the provisions of AASHTO,

The column design axial compressive force s h a l l be resisted by concrete area and vertical .reinforcement according to the prwis ions of AASHTO. Vertical reinforcement provided to resist the column shear forces and the column axial t e n s i l e forces may te u s d in resisting t h e colvmn ax ia l compressive force.

*

Bridge Plan i- Global x-x & Z-Z Axes

(Local Y-Y r Z-z Axes)

Lnng i tud ina l mad Resul t s

Case f

Transverse Load R e s u l t s

Case 2

where % and HZ are about loeel axes.

l om bin at ion* of Orthoqonnl Seismic Forces

figure I

- 13. A t each bent, individual footings may be connected by ties to d i s t r i b u t e the total horizontal force in the plane of the bent to each footing in proportion to its capabi l i ty to resist horizontal forces. The ties s h a l l be capable of resisting in t e n s i o n and compression the d e s i g n u l t imate axia l force requirled to red l s t r i b u t e t h e total horizontal force.

When determining the tensi le capacity of the tie use a s trength reduction factor ( 8 ) equal to 1.0 and a yield s t r e n g t h of reinforcement equal to 1.0 fy.

When determining the compressiw capacity of t h e tie use a strength reduction factor ( $ 1 e ual to 0 . 7 5 , a concrete P strength equal to f \ and a yie d strength of reinforcement equal ko 1.0 f

Y* 14. The bent cap and girders sha l l be capable sf res i s t ing

unfactored dead load forees ad moments combined with seismic forces aird moments.

The u l t i m a t e seismic moments to be considered sha l l be t h o s e that are the least critical of the fel lowing two eases:

A. The moments which r e s u l t from using the f i n a l top of column probable plastic moments as an applied load.

B. The moments which result from an elast ic seismic analysis before any reduct ion for d u c t i l i t y ( 2 factor). Tm, orthogonal directions of earthquake motion shall be considered.

The moments which result frem the analysis of earthquake motion in one d ircetian s h a l l be combined w i t h 30 percent of the moments which result from the analysis of earthquake motion in the other direct ion . The tw possible eombinatiens of moments shall be cons ider&.

The ultimate seismic a x i d and shear forees to be considered s h a l l be those associated with the Beast critical ult imate seismic moments.

When determining the flexural capacity of the members, use a s trength reduction factor (fl) equal to 1.0 and a yield strength of reinforcement equal to 1.0 t i m e s f Y* When determining the shear capacity of t h e members, use a strength reduetian factor ( j 8 ) equal to 0-85 and a y i e l d s trength of reinforcement equal to 1.0 times fy.

-Columns on Combined 4 Common l Footing

1. Determine the eslvmn section requirements based an the Load Factor Design Group madings in AASHTO, and using the d e s i g n strength of the member.

2. For each column determine the column probable p l a s t i c moments and the column axia l forces and shear forces associated with the development of the probable plastic moments,

3. The u l t i m a t e moments to be used as a p p l i d moments for des ign ing the foot ing s h a l l be determined as specif ied in Step 5 of the procedures for "Columns on Individual Footings".

4. The u l t i m a t e horizontal forces and ultimate vertical forces to be used as applied forces for des igning the f o o t i n g s h a l l k determined as specified in Steps 6 and 7 of the procedures fo r "Columns on Xndividual Footings' .

5. fn the transverse d irect ion , assume the footing is a continuous beam on an e last ie supportl either s o i l or p i l e s . In the l o n g i t u d i n a l direction, assme the foot ing is s one- way f o o t i n g . Using the s o i l or pike reactions obtained, des ign the footing sections. Considerat ion should be given to the tensile force applied to the footing due to the difference in column shears.

Design a foot ing to resist the ultimate moments and forces of Steps 3 and 4 . For resisting the ver t i ca l forces and moments use the u l t i m a t e so i l bearing c a p a c i v or the u l t i m a t e p i l e bearing capacity and ultimate p i l e u p l i f t capacity using a strength s e d u c t i o n factor (83 equal to 1.0* Far res i s t ing the l a t e r a l forces use the ult imate capacity of the s o i l or p i l e s using a strength reduction factor ( 8 ) equal t o 1-0.

When determining the f l exural capacity of the footing, use a strength reduct ion factor ( 0 ) equal to 1.0 and a y i e l d strength of reinforcement equal to 1.0 times fy. When determining the shear capacity of the foot ing , use a strength r e d u c t i o n factor ( 0 ) equal to 0.85 and a yie ld strength of reinforcement equal to 1.0 times fy.

6. Design the p i l e s af pile footings ta susta in large curvatures and the d e s i g n ax ia l force. Refer to Step 9 of the proced ares for Columns on -1nd iv i d u a l Footings' .

7 . Cheek the footing design us ing the mad Factor Design Grow madings in RASHTO, except omit Group V f f . The ultimate s o i l bearing capacity s h a l l be m o d i f i e d by a strength reduct ion factor ( 8 ) equal to 0.5 and the u l t i m a t e p i l e bearing capacity shall k modified by a strength reduction factor (83 equal to 0 , 7 5 ,

When checking the adequacy of the foot ing sect ions, use the d e s i g n s trength of the member specified in AASHTO. R e v i s e f o o t i n g , i f required .

8. Design the column transverse reinforcement Zo provide for con£ inernent and shear resistance. Refer to Step 14 of the procedures for oColmns and Indiv idual Footingsn.

9. Design the connection of hinged columns to resist the u l t i m a t e forces determined in Step 4 . Refer to Step 12 of the procedures Zor *Columns on I n d i v i d u a l Footings ".

10- The bent cap and girders s h a l l $e capable of resisting unfactored dead laad forces and moments combined with seismic forces and moments. Refer to Step 14 of the procedures for "Columns en I n d i v i d u a l Footings' .

Columns as Extensions ef P i l e s

1. Using the Load Factor Design Group madings in AaSRTO, determine the required pile embedment and the required column and p i l e sections.

For determining the required p i l e embedment to resist the applied moments and l a t e r a l forces use a l i m i t i n g equilibrium analys i s , For this analysis, use ultimate l a t e r a l soil pressures modified by a strength reduc t ion factor ( 8 ) equal ta 0.5. For determining the column and p i l e section require ment s , use the d e s i g n strength of the members,

Jf the l o c a t i o n of the maximum moment in the pi le d i f f e r s s i g n i f i c a n t l y from the location of p i l e f i x i t y assumed for the frame analysis, considerat ion should be given t~ making a revised frame analysis,

. Determine the column and pi le probable plastic moments at t h e lacations where p l a s t i c hinges may fom. Assume the p l a s t i c hinge in the pi le o c c u r s at the p o i n t af maximm mment determined in Step I, or if the column s e e k i o n probable p l a s t i c moment at the columnJpile connection is less, assume a plas tie hinge occurs at the colmn/pile connect ion .

mtermine t h e column and pi le shear forces and axial forces t h a t are associated w i t h the development of the selected probable .plastic moments. Reevaluate the probable plast ic moments, shear forces, and a x i a l forces for the effects of over turningf if necessary.

3 . using the ult imate moment at the location of the p l a s t i c h i n g e or p o i n t of maximum moments near the ground surface and the associated shear force as loads, check if the p i l e embedment determined in Step 1 is adequate. U s e a l i m i t i n g

- equilibrium analysis using the u l t i m a t e lateral soil pressures modified by a strength reduetion factor (B) equal to 1.0, Increase the pile embedment, if required,

The ult imate moment to & used s h a l l be that which is the l e a s t critical of the following two cases:

A. The final column probable plastic moment,

8. The resultant moment from an elastic seismic analysis before any zleduction for ductility (2 factor 1, TWO orthogonal directions of earthquake mot ion s h a l l be considered. The rrrrments which result from the analysis of earthquake motion in one direction s h a l l be combined w i t h 30 pzrcent of the m e n t s which result from the analysis of earthquake motion in the other direct ion, The two poss ib le resultant moments s h a l l k cons idered ,

4 . Check t h e p i l e and superstructure deflections using the Service Load Group Loading in AASATo except o m i t Group V I f , Use t h e longest of the following p i l e embedment l engths :

1) t h e length determined in Step 1. 2 ) the l ength determined in Step 3. 3 ) the lengfi specified by t h e TransCab ~ n g i n e e r i n g

Geology to resist the ax ia l forces.

U s e an elastic method of analysis, and revise the p i l e section and/or embedment, if required ,

5. D e s i g n t h e column and p i l e transverse reinforcement to provide for cenf inement and shear resistance. Refer to Step 11 of the procedures for 'Columns on Ind iv idua l ~oot ingsm- A

6. ~esign the zonnectim of columns hinged a t the top of pile to resist the applied ultimate forces.

The forces to be used s h a l l be the least critical of the following two cases=

A, The unfaetored dead load forces combined w i t h the forees associated with the development of the probable plast ic moments a t the t o p of the column.

B. The unfaetosed dead load forces combined w i t h the forces from an elastic seismic analysis before any reduction for d u c t i l i w ( 2 factor) . nJo orthogonal direct i o n s of earthquake mt ion s h a l l be c a n s i d e r e d . The forces which r e s u l t frm t h e analys i s of earthquake motion in one direction shall be combined w i t h 30 percent of the forces, which result frm the ana lys i s of earthquake m t i o n in the other direction.

BRIDGE DESfGN PRACTZCE JAAaARYp 19B2 5-17

The two possible sets of r e s u l t a n t forces s h a l l be cons idered . Refer t w Step 12 of t h e procedures for "Colmns on I n d i v i d u a l Footings',

7 . The bent cap and girders shall be capable of resisting uafactored dead load forces and moments combined w i t h seismic forces and moments. Refer to Step 1 4 of the procedures f o r 'Columns on I n d i v i d u a l Footings*.

2.4 DESIGN OF ABUTMENTS

hx> types of abutments are d i s c u s s e d , a seat- type and a diaphragm- type-

besign of S e a t - w e Abutment

Procedcres for the d e s i g n of a seat-type abutment which uses p i l e s at the end of the wingwalls plus the lever arm afforded by the wingwalls to resist overturning moments,

The procedure w i l l be i l l u s t r a t e d us ing an abutment for the 2-span box girder s t r u c t u r e used in the concrete des i g n course.

1. Determine the embedment of the diaphragm and t h e length of wingwalls required to s a t i s f y site requirements.

2. betermine number and s i ze of elastomesic bearing pads required u s i n g service loads from superstructure analysis. Refer to Memo to Designers 7-1.

3. Assume psel iminaq dimensions far wingwallsr abutment diaphragm, hackwall, curta in wall, w i n g w a l l foo t ing , and transverse shear key. See Figures A-2, A-3 and A-4.

4. Determine unfaetared dead load of abument, s t a t i c earth pressure, seismic earth pressure, and horizon taP and v e r t i c a l Eorees transferred from the superstructure. See Figure A-5,

5, using the load factors and group loadings in AASHTO, determine the number of p i l e s required to resist the vert ica l and horizontal forces and overturning moments. Use the u l t i m a t e bearing capacity, ult imate la tera l res istante and ul t h a te tensile capacity of the pi l e s mod if i e d by a s t s t n g th reduction factor ( 0 ) equal to 0.75 for all group loadings except VIf far which use a strength r e d u c t i o n factor I B ) equa l to 1.0. For purposes of analysis assme a pinned support at the j u n c t i o n of abutment diaphragm and the f r o n t row af p i l e s .

6 . Using factored l o a d s , des ign the fol lowing elements using the d e s i g n strength of the member as spec i f i ed in RASHTO:

BRIPGE DESfGt? PRACTICE JAHI3AKP. 1982 5-18

1) abument diaphragm 2 ) abutment backwall

*

33 l o n g i t u d i n a l shear key at base of backwall 4 ) transverse shear key a t end of abutment diaphragm 5) c u r t a i n wall 6 ) w i n g w a l l s 7 ) foot ing at end of wingwall.

Xn des ign ing the abutment diaphragm cons iderat ion should be given to the torsional load created by the eccentricity of the earth pressure .and forces transferred through the elastomeric bearing pads. T h i s torque is resisted by the wingwalls and wingwall p i l e s . The wingwalls should be des igned for a v e r t i c a l moment in the plane of the wingwal l s to resist the torque frm the abutment diaphragm.

I n des ign ing the wingwall, cheek the vertical shear capacity at the m i n i m u m section at the wingwall footing.

In designing the wingwall to baekwall/abutment diaphragm connection, take into account the shear and moment due to earth pressure an the wingwal l and the fact tha t for this type of corner detailr where the moment tends to open the cornet* it is d i f f i c u l t to: maintain the moment capacity of the vingwall section around the j o in t area, Consideration should be given to using a haunch in the corner even for abutments w i t h o u t skew.

Ilesign of Diaphragm Abument'

Procedure for the design of a diaphragm abutment on either p i l e s or can t inuous spread f o o t i n g w i t h cantilevered wingwalls ,

The procedure is i l lus trated using an abutment 5oc the 3-span T-beam structure used in the concrete d e s i g n course.

I. mtermine the dimensions of the diaphraq, wingwall. etc,, to s a t i s f y site requirements.

2. If diaphragm rests on elastomeric bearing pads, determine their s i ze and number us ing semiee loads £ram superstructure analysis,

3. Determine factored a x i a l load. Find number of piles or area af spread footing required using ult imate capacities of soil or p i l e s reduced by appropriate fl factor i .e . (0.5 £or s o i l . 0.75 for p i l e s ) ,

4 Design abutment wall as cantilever us ing the d e s i g n strength of the member as specified in AASHTO. Longitudinal force applied at base of cantilever w i l l be dependent upon the type of support as f ollovs:

On Concrete P i l e s - Longitudinal Farce HX, = Vlgross section) = vc + fts [per p i l e )

v, - A r a/s v Y

Where: Vc = nominal shear strength provided by eoneretee

Vs = nominal shear strength provided by shear rein£ orcement.

A, = area of shear. reinforcement with in a distance s, In square ~ n c h e s .

fy = specified yield strength of non-prestressed re in£ orcement, in psi,

d = dis tance from extreme compression fiber to c e n t r o i d of longitudinal tens ion r e i n £ otcemen t , but need n o t be less than 0.80h for prestressed members, in inches, (For circular' sectionsr d need not be less than the dis tance from extreme compression f iber to centroid of tension reinforcement in opposite hal f of member. 1

s = spacing of shear reinforcement in direction parallel to longitudinal reinforcement, in inches ,

DL = unfaetored dead load.

Ag = gross area of section, in square inch'es.

f ', = specified compressive strength of mncrete, in ps i .

b, = web width, or diameter of circular section, in inches.

On Steel 45T Piles

U s e HL = 30k [pet pile1

an Elastomeric Bearins Pads

use BL = 25 % of unf actorcd dead load,

These va lues are to be used in l i e u of more exact analys i s whrch glves greater va lues . Minimum reinforcement ta he # 5 at l B u in both faces,

5. Wingwalls are des igned us ing Group froading and the - fallowing:

Earth Pressure E = (0.53 R, y h 2

h = h e i g h t of wall at point considered.

L ive Load Regardless of Group used, consider as 2' equivalent earth surcharge.

For convenience the equations in BRIDGE DESIGN AIDS pages 3-6 may be used considering L of w a l l and s = 2' then applying factors:

BRTDGE DESfGR PRACTICE JARUIVIP, 1982 5-21

Load per Unif of Area

Re/crfiun befween h f e n ~ i f y o f /oad and seftkmenf of a /oof/jy on C, dense or sfiff and C2 /OOJ~ OP ~oiDCf JO~/.

Figure 2-1

B R X E E DESLCN PRACTICE JANUARY, 2982 5-23

BRIDGE DESXGR PRACTICE JARUARYr 1982 5-24

-

Anpfc o/ m f c ~ m l f~~chorr , P, &yew

BEAR1,WG CAPACl T Y FACTORS

Figure 2-4

Char i 0 e s f k f h a//uwab/e soil pressure +?or t"oo tiny on sun ~7 on fhe basis o/ rcsu/fs of sfandcrrd penef r~ f /bn f e d .

Figure 2-5

BRIM;E DESIGN PRACTfCE S W A R T , 1982 5-26

- 1x1- EARTH RETAX Uf NG STRUCTURES

3.1 STATES OF STRESS

When the maximum shearing s trength i s f u l l y mobilized along a surface w i t h i n a s o i l mass, a failure condition known as a state of p l a s t i c (or l i m i t i n g ) equilibrium is reached. Rankine's active and passive states of stress result when shear.stressas equal to the maximum shearing s t r e n g t h of the s o i l develop uniformly and unhindered in two major directions throughout a soil mass due to lateral extension or compression.

Where the combinations of shear and normal stress with a soil mass a l l lie below t h e l i m i t i n g envelope {see F i g . 3-11, t h e soil is in a state of elastic equilibrium. A special condi t ion of e l a s t i c equil ibrium is t h e "a t-rest" s t a t e , where the sail is prevented from expanding or compresring l a t e r a l l y w i t h changes in the vertical stress.

The l i m i t i n g eguilibrim theories all require t h a t the maximum shearing strength of the soil i s mobilized. This however, requires deformation in the soil. me deformation Of a supporting s t r u c t u r e has only a local effect on the state of stress in the s a i l , The remainder of the s o i l remains in a state af elast ic equil ibrium. The s t a t e af stress in t h e locally d i s t u r b e d zone and the shape of t h i s zone is dependent on the amount and type of w a l l deformation* This a lso determines the shape of the pressure d i s t r i b u t i o n on t h e w a l l and t h e i n t e n s i t y of the pressure. When a wall moves outward, the s h e a r i n g strength of the retained soil resists t h e corresponding outward movement a£ t h e soil and r e d u ~ e s the earth pressures 0x1 the wall.

The earth pressure calculated for the active state i s the absolute minimum value. When the w a l l movement i s towards the retained soil, the shearing strength of the soil resists the correspanding s o i l movement and increases t h e earth pressure on the wall, The earth pressure (or rcsistanceS c a l c u l a t e d for the passive stare is t h e maximum value that can be developed.

The mount of movement required t a produce the aetive state in t h e s o i l is dependent mainly en t h e type 05 b a c k f i l l mater ia l* Fig. 3-2 gives the outward movement of a wall which is necessary to produce an active s t a t e of stress in the r e t a i n e d s o i l , The movements required to produce full passive resistance are considerably larger, espec~ally in cohesionless materia l . These requirements apply whether the movement is a l a tera l t t a n s l a t i ~ n of the whole wall or a rotation about t h e base. The pressure distributions for full

B R I X E DESIQI PRACTICE JmARP* 1382 5-27

active and pass ive states are basica l ly triangular for constantly s loping ground,

The amount of wall movement which w i l l take place depends mainly upon t h e foundation conditions and t h e flexibility of the wall. The designer must insure that the calculated earth pressures correspond to t h e avai lable w a l l movement. A free- s tanding wall need only be d e s i g n e d for active earth pressure as f a r as s t a b i l i t y i s concerned, since i f it s tarts to s l ~ d e or overturn under higher pressures, the movement w i l l be sufficient to reduce the pressures to act ive , Howevert if it is on s strong foundation or otherwise f i x e d se t h a t adequate stability is provided, t h e stern may be subject to pressures near those for the at-rest state.

3.2.1 The Rankine Earth Pressure Theory

Rankine's equations g ive the earth pressure on a vertical plane which is sometimes called the virtual back of t h e w a l l , The earth pressure an the vertical plane ac ts in a direc t ion paral le l to t h e ground surface and i s d i r e c t l y proportional to t h e vertical d i s t a n c e below the ground sur face (see Figure 3-31 . Provisions f o r Rankinegs conditions in cohesive so i l s w i t h a horizontal ground surface are available,

3,2.2 The Coulomb Earth Pressure Theory

The theory direc t ly gives the resul tant pressure against the back of a retaining s t r u c t u r e for any slope of the wal l and f a r a range of wall friction angles. It assumes that the s o i l s l i d e s on the back of the w a l l and mobilizes the s h e a r i n g resistance between the back of the wall and soil as w e l l as t h a t on t h e failure surface,

The Coulomb equations reduce to those of the Rankine theory sf a vertical w a l l surface w i t h an angle of w a l l friction equal to the backfill slope is used, O t h e r cases of w a l l slope or w a l l friction require curved surfaces of s l i d i n g to satisfy static equil ibrium. The degree of eusvature may be quite marked, especially for passive conditions. However, Caulombls t h e o r y assmes that t h e failure wedge is always bounded by a plane surface, and it is therefore only an approximation for passive conditions. It is usually on the unsafe s i d e if t h e wall f r i c t i o n angle exceeds l / 3 8.

The simplifying assumption also means that static equilibrium is not always completely satisfied. For example, the forces a c t i n g on the soil wedge cannot all be resolved to act through a common point, The error from an exact s o l u t i o n i s proportional to t h e amount by which s t a t i c equilibrium is n o t s a t i s f i e d ,

B R f E E 5ESXGM PRACTICE JA#UARY, 1982 5-29

The b a c k f i l l is divided into wedges by selecting planes through the heel af the wall. The forces act ing on each of t h e s e wedges are combined in a force polygon so that the magnitude of the resultant earth pressure can be obta ined . A force polygon is constructed even though the forces acting on the wedge are often not in moment equilibrium. T h i s method is therefare an approximation v i t h the same assumption as the equations for Coulomb's condi t ions , and for a ground surface v i t h a constant slope will give the same result, If the conditions arc the same as those for Rankine's equations the Trial Wedge earth pressures w i l l corresp~nd ta t h e s e also. The limitations an wall f r i c t i o n and passive pressures mentioned in t h e use of the Rankine and Coulomb equations also apply to t h e T r i a l Wedge Hethod. The adhesion of the soil to the back of the w a l l in cohesive soils is neglected since it increases t h e tension crack depth and hence r e d u c e s the active pressure.

For the active east the maximum value of the earth pressure calculated for the various wedges is required. T h i s is obtained by interpolating between t h e calculated values. For t h e passive case t h e required minimum value is s imilarly obtained,

The direction of t h e r e s u l t a n t earth pressure and t h e force polygons should be obtained from the consideration of Sections 3.2.1 to 3.2.3. For t h e cases where this force acts parallel to the ground surfaces. a substitute cons tant slope s h o u l d be used, as shown on Figure 3-4, for s o i l both v i t h and without cohesion.

For an irregular ground surface t h e pressure d i s t r ibut ion is not triangular. However, if the ground does not depart signif i e a n t l y from a plane surf ace, a l inear pressure d i s t r i b u t i o n may be assumed and t h e constructions given in Figure 3-5 and 3-6 used. A more accurate method is given in Figure 3-7, The l a t te r should be used when there are nonuni form surcharges.

3 . 3 ELASTIC EQUf Lf BRIUn CO??Pf TI O??S

(At -Rest l?rkssures)

The special state of elastic equilibrium known as the at-rest state is u s e f u l as a reference point'for calculation of earth pressures where only small wall movements occur, For t h e case of a vertical w a l l and a horizontal ground surface the coeff ieient of at-rest earth pressure may be taker! as: KO = 1 - s i n 8' for normally consolidated materia ls , T h i s a s s u m e s t h a t the material has no built in overconsolidation stress. For other wall a n g l e s and backfill slopes, it may be assumed the KO varies proportionally to KA. At-rest e a r t h pressures may be assumed to increase linearly w i t h depth from zero at the ground sur face for a11 materia ls .

-The total, at-rest earth pressure farce i s given by:

This acts as 8/3 from the base of the wall (or bottom of t h e key for walls with keys].

For gravity type retaining walls the at-rest pressure should be taken as acting normal to the back Of the w a l l ( i ,e . 8 = 0). For cantilever and counterfort walls it should be calculated an t h e vertical plane threugh the rear of the heel and taken as ac t ing parallel w i t h the ground surface,

Zn eohcsienless soils, f u l l at-rest pressures w i l l occur only w i t h the .mst r i g i d l y supported walls. 1 n highly p l a s t i c clays, pressures approaching at-rest may devlop u n l e s s w a l l movement can m n t i n u e w i th t i m e (creep).

3-4 SEISMIC EAR= PRESSURE

The most frequently used method for the calculation of the seismic soil forces act ing on bridge abutments or retaining walls is t h e s t a t i c approach developed by Mononobe and Okabe. The Mononobe-Okabe analysis is an extension of the s l i d i n g - wedge theory taking i n t o account horizontal and vertical i n e r t i a f orees acting on the soil. The analysis is described in d e t a i l by Seed and Whitman, (Reference 4 1. The fo l lowing assumptians are made:

1. The abutment is f.r- to move s u f f i c i e n t l y so that the s o i l s t r e n g t h will be n m b i l i z e d . 3E the abutment is r i g i d l y 2 ~ x e d and unable to move, the soil forces w i l l k very much higher than those predicted by the Nononobe-Okabe analysis.

2 . The b a c k f i l l is cohesionless w i t h a f rietion angle 0.

3* The b a c k f i l l is unsaturated, so that liquif ieation pronlcms w i l l not arise.

uniform toads

Unif o m surcharge loads may be canvested t o an equivalent h e i g h t of fill and the earth pressures calculated for the mrrespondingly greater height.

Line Loads

Where there is a superimposed l i n e load running a considerable Length parallel to the wall, the w e i g h t pet u n i t l e n g t h of t h ~ s

BRIDGE DESIGN PRACTICE J m A R Y , 1982 5-31

load can be added to the w e i g h t of the particular trial wedge to which it is applied {Fig. 3-8 ) . The increased t o t a l earth pressure will b~ given from the t r i a l wedge procedure but t h e line Load w i l l also change the point of application of t h i s tota l pressure. The method given in Figure 3-7 may be used to g i v e t h e d i s t r i b u t i o n of pressure.

When the l i n e l oad is small in comparison w i t h ac t ive earth pressure, the effect of the l i n e load on its mrn should be determined $y a method based on stresses in an elast ic medium, The pressures thus determined are superimposed on those due to act ive earth pressure and athes effects (Ref. 2 r Sheet 7-10-10).

Point Loads

P o i n t loaas sannot be taken into account by trial wedge procedures, The method based on Boussinesq's q u a t i o n s should be used (Ref. Z r Sheet 7-10-10) - Static Water ~ e v e l

Where part or a l l of the so51 behind t h e wall is submerged below a s t a t i c water level, the earth pressure is changed due to the hydrostatic pore pressures set up in the soil. The water i t se l f also exerts lateral pressure on the wall q u a 1 to the depth k L o w the water table times the density of water.

3.6 STABILTTY OF RETAINING WALLS

3 1 General

The s t a b i l i t y of a freestanding retaining s t r u c t u r e and the soil containing it is determined by computing the factors of sa fe ty or ' s t a b i l i t y factors' which may be d e f i n e d in general tenas as:

moments or farces aiding s t a b i l i t FmSo = moments or forces c a u s i n g i n s t a b i f i t y .

Factors of safety should be calculated for t h e fo l lowing separate modes of failure:

a) S l i d i n g of t h e wall outwards from the retained s o i l . b] Overturning of the retaining w a l l about its toe, e) Foundation bearing failure. d ) S l i p circle fa i lure in the surrounding soil.

when ca lcu lat ing overall s t a b i l i t y of the wall, t h e lateral earth pressure is calculated to the bottom of the footing, or In the case of a footing w i t h a key, t o the bottan af the key.

The vertical component [ i f any) of the resu l tant earth pressure may be added 50 the w e i g h t of the wall system when mmputing s t a b i l i t y factors.

'If t h e passive resistance of the soil in front of the w a l l is i n c l u d e d in calculat ions for s t a b i l i t y , t h e top 12' of t h e s o i l should be neglected, and passive resistance should k ca lcu lated by Ranking theory.

3.6 - 2 Sliding S t a b i l i t y

Factor af safety: sum of t h e forces resisting s l i d i n g sum of the forces c a u s i n g s l i d i n g

should be a t least egud to 1,s for s t a t i c l o a d i n g and at l east 1.2 for seismic loading,

3.6.3 Overturning Stability

Moments calculated about the bottom of the f r o n t of t h e toe must g ive an w e s t u r n i n g factot of safety:

s m of t h e moments resisting overturning FrSm sum of the moments causing o v e r t u r n i n g

The factor of safety fur overturning should be at least 2,0 for static I p a d i n g , For seismic load ing F.S. for sliding is generally exceeded befom overturning is critical.

Sfiesscs /n 501/ in c / w f i c ranqe - (below //m/tinq enue/wc )

Figure 3-1

Novemenf o f wa// necessary fo produce ~ ~ f t v e pre~sures.

Figure 3 2

SOIL WALL YIELD

Co h es~on/cws, dense Cohesron /ess, base

C /uy , f ~ ~ m C / o y , soPf

0.0Q/ H 0.00/ - 0.002 H 0.OJ -0.02 H 0.02-0.05 H

BRf DGE DESfGA PRACTICE JAMUARY, 1982 5-34

cn- F m s 0 +

2 " m 0 t o m

L + m s m E'

3

BRZDCiE DESfGlU PRACTICE SA#OARf, 1982 - 5-36

TRfAL WEDGE METHOD

COHESIONLESS SOIL

CULMANN'S CONSTRUCTION

(FOR STATIC EARTH PRESSURE OPILY)

PROCEDURE

I, Draw line A-G at an acgle of to the hotizcntal for acFire pressurn.

2, bmw t r i a l wedges AtlCDF, 6 8 m , &c, - a minimum of four w i l l usually suf f i oe.

3. Calculate the weights of the wedges - say wl. w etc., and p lo t these tO a suitable scale on A-G, each measured fmm I:

4. Thmugh rl , r2, etc. , draw l ines a t an angl e E , ( see text for d i reb ion of PA and hence E l , to intersect A-1, A-2, etc,, at H, J , ~ T c .

5. Draw a c u m "hrough A, H, J , etc.

6. PA i s obtained by drawing a tangent to the curve, para1 l e l to A-G to touch at f. PA i s the l ine W-T, to t h e sane scale as w, , etc.

7. The fa i lu re plane i s the l i ne through A and T.

Figure 3-5

B R X Z E DESIGN PRACTICE 5-38

POINT OF APPLICATION 05 RESULTANT PRESSURE AND PRESSURE PISTRIBUTFON

surcharge

P S

A A TRIAL WEDGES PRESSURE ON A-8

Uw when thc ground sur i~c t i s m y Irmgulnr or whrn a non-uniforo surcharge i s carried.

PrnQWRE

1. Subdivi* the Ilnt h4 into abtr? 4 aqwl psr)s hl lbslov thc &p+h ye at tclnslon cracking).

2. w u t a +he sctiuc ma* pmsums PI, P , P9, etc.. a i t e m o+ +ha points 7 , 2 , 4, cte., -re thc b a w of ~ f m mall. fhc t r i a l mdgb

i s U K ~ for eraputation.

3. Do*cmlm the pmsurs dis*rlbutlon by wr*tng dan fm p i n 9 1, A l i nssr vari st im of pmsum m y M .sS#d bstrs4n ths poin+l *hsm pressure hss bebn a leu la td .

1. k t a f m i ~ the elmvatlm of the a n t m i d of +h pnssur t diagrsa, ;. T h i s i s ths appmnimp*e Clewation Of 7- p ~ b n t ei applicnTim O f th. rarulr ant saen pmssum, PA.

Figure 37

BRIDGE DESfGN PRACTICE JAFmARY, 1982 5-40

SUBSTRUCTURES AITD FOUtIDATIONS

REFERENCES

"Soil Mechanics i n Engineering Practice, 2nd Edition," by Terzaghi, K. and Peek, R. B. (1967) John Wiley & Sons.

'Design Manual Soil Flechanies, Foundations, and E a r t h Structures ' NAVFAC DM-7 U * S . Department ef the ? l a y (1971 3 .

mFoundation Design,- by Teng, Wayne C. 119621, Prentiee-Hall Inc.

.Design of Earth Retaining Structures for Dynamic L w d s , ' Seed, H. B. and Whitman, R. V. [ 1 9 7 O ) , ASCE Specialty Conference - Lateral Stresses in t h e Ground and Earth Retaining S t r u c t u r e s .

'Retaining Wall lk s ign Notes , ?lew Zealand Minist ry of Works (1973) Design Hanual prepared i n the.Office of the Chief Design Engineer (civil).

*Steel Sheet P i l i n g Design Manual," U-S. S t e e l Cow. (19751.

"Trenching and Shoring M a n ~ n l , ~ State of California Department of Transportation /1977).

'Randbook of Engineering Geology,. State of California Department of Transportation 2 1958 1.

ATC-6, draft copy dated ?¶arch 7, 1979 the Applied Technology Council .

mReewrmended Lateral Force Requirements and Camentaryrw ( 1975 3 by S t r u c t u r a l Engineers Association of California ( SEAOC 1.

"Standard Speeifieations for Highway B r i d g e s , Twelfth Edition,' (19773 by the American Association of State Highway and Transportation Mfie iaks (AASHTOJ.

=Bridge , Hemas to Designers,' State of California Wpartment of Transportation

'Standard Plans," State of California Department of Transportation (?larch 1977 5 .

BRfQGE D E S f a PRACTICE JANUARY, 1982 5-4 1

BRIDGE DESIGR PRACTICE JANUARY, 1982 5-44

-

- Example Problem 1 - Single Column B e n t W / P i l e Footing

Problem: Design the column and a pile footing f o r t h e s i n g l e eelmn b e n t s of t h e 3 span box girder bridge used i n she Reknforced Concrete Section of t h e Br idge Des ign Praetice Manual.

Noter Article nmbers c i t e d within t h i s example problem refer to t h e article numbers used in t h e RASFfTO Standard Specifications for Bighvay Bridges, 12th Edition and inc luding Interim Specif ieations through 1981 and Caltrans Br idge Design Specifications,

Colmn Loads: Par i l l u s t r a t i v e purposes only dead load, l i v e Joad and seismic loads will be cons idered. The X-ax is equals the c e n t e r l i n e of bent and the Y-axis equals the centerline of column.

Mad load [service levell

DL TOP COL = 1 0 5 6 ~ DL = 1154 k BOT COL

DL DL 'X TOP COL = - 5 2 1 ft-kips H x B O T c O L = 216 f t -kips

DL My TOP COL = 0

Live load + impact [service levell; Impact = 22%

Case I

LL+X ' K TOP COL

' = 60 f t-kips = -146 f t - k i p s H, CoL

LL+T , 1 1 2 l f t - k i p s t L + f My TOP COL ny BOT COL = 1121 f t -kips

Case 2

p bL+* , 131k BOT COL

, 5 1 6 f t - k i p s = - 1 2 5 2 f t - k i p s M X D m c o L 'x TOP COL

LL+f My TOP COL = 654 f t -kips LL** = 654 f t -k ips

. l , s a COL

- Live load + impact (for factored level): Impact = 22%

Case 7 ( l , I f x 1 lane P + 1 lane HS

LL+l k L W I k

TOP cot = 456 BOT COL = 456

LL+I Lt+I

n* .OF COL = -330 f t -k ips 'x BOT COL = 136 ft-kips

LWI LL+f

TOP COL = 3672 ft-kips My BOT COL = 3672 f t - k i p s 1

Case 2 (1 -15 x 1 lane P + 1 lane H)

LL+I LL+f

Mx TOP COL 1 -231 1 f t-kips Mx BOT COL = 952 ft-kips

LWI L W I mp COL = 2179 f t -kips BOT COL - 2179 f t - k i p s

Case 3 (1.75 x 7 lane P)

L W f k LWf k

TOP COL = 204 BOT COL = 204

U+I L W I TOP cob - -1685 ft-kips M, B,,,, = 694 f t -k ips

LL+X LL+S TOP COL = 2244 f t-kips My BM COL = 2244 ft-kips

Seismic load IARS forces and moments before application of 2 = 6 factor)

EQ 'x ARS EQ = $ 2 8 = Hy ARS

Case 1

EQ = . S S k TOP - EQ

Mx TOP COL = 8472 ft-kips

Case 2 11.0 EQL + 0 . 3 3 E%l

EQ = + 7905 ft-kips Mx BOT COL -

EQ + + 12624 f t-kips BOT COL -

Case 3 (0.33 E R + 1.00 EQT)

EQ ' x BOT COL = 2 2640 f t -kips

(Horizontal force)

EQ - + 38256 f t -kips n y B O T C O L -

Column: - Geometry - Standard architectural calrmtn type 2R

Clear height = 20*-O* Length of top flare section = 16'-6" t= 5' -6"

Longitudinal reinforcement - determined by using YIELD program with reinforcement placement c o n t r o l l e d by basic section of column.

*s TOP = 36 - a9 = 36.00 sq,in.

As BM = 54 - $9 = 54.00 sq.in.

9 - 19 extend 11' above top of footing

9 - # 9 extend 8 ' above top of footing

Ag TOP COL ACTUAL = 7777.19 sq.in.

Ag TOP COL DESIGfJ = 3421.19 sq.in.

P s = 36.0/3421.19 = 0,0105 > 0-01 ok

Ag BUT COL ACTUAL & DESIGN = 3421.19 sg-in.

p s = 54.0/3421,19 = 0.0158 > 0.01 ok

Article 1.5.111A) ( 2 ) provisions were used for complying w i t h t h e minimum f p requirements of Article 1.5~11EAI(l).

A t the base of t h e column the p = 0,0158 more than s a t i s f i e s t h e minimum P s = 0.005 specified in Article 1,4.6(J3{41(C),

Nominal moment strength - Article 1.5.2(8)

f ', = 3250 psi

fy = 60000 psi

8 , max = 0.003 in . / in .

E, = 29000000 p s i

- Using the output from the YIELD program determine t h e nominal moment strength at sections w i t h i n the f lared p e r t i e n of the column associated with the dead l o a d p l u s seismic a x i a l load,

Top of column

Pe = 1056 + 55 = l l l l k

Hnx = 8190 ft-kips

n = 14721 ft-kips nY

5'-6* fm tap of colmn

Pe = 1094 + 38 = 1132k

"Lx = 7764 ft-kips

U = 9752 ft-kips nY

11'-Qw from tap of eolrnn

Pe = 1121 + 26 = 1147k

*nx = 7075 f t -k ips

H = 7437 ft-kips *Y

Probable plastic moment - Article 1.5.33(D)

Basic section e top of column

Pe = ZUlk

n ~ x = H = 8233 ft-kips PY

Basie section @ bottom of column

Pe = 1154 + 55 = 1209k

r3p, = m = 10641 ft-kips PY

Colmn shear - article 2.51351G)(13

mtermine the maximum coltnun shears eonsidering t h a t t h e nominal moment strengths can be developed in the gross flare sections and that the probable p l a s t i c moment s trengths will be developed in the basic section.

Case 1 {nominal moment C top of column, probable plastic moment @ bottan of calumn)

Case 2 (probable plastic momenta @ top arid bottom of e o l m n )

Case 3 (nominal moment @ 5'-6' from top of column, probable plastic moment @ bottom of column)

F r o m the above results it can be coneluded that i f plast ic h inges form they probably w i l l form at the top and bottom of t h e column about t h e X-axis and at t h e bottom of the colmn about the Y-axis,

Since the co lmn section at the bottom is circular and spiral shear reinforcement w i l r be used, t h e shear in the Y-axis direction will contra1 the design.

Although Article 1,5.35(6) (1) permits using t h e lesser value of 865k f o r the design shear force Y , i t is des ireable to use the design shear forces associate8 with the developnent of t h e column moment s trengths . T h i s is particularly so f o r short structures for which t h e response of the abutment - abutment foundat ion material system and its influence on t h e response of the s t r u c t u r e as a whole d u r i n g an earthquake is so uncertain at this t i m e ,

_Ttansverse reinforcement - determine reinforcement for conf i n a e n t and shear, Articles 1,5.11(81 and 1.5.3SIG).

Confinement, u s i n g sp i ra l reinforcementr the following three volumetric equations should be satisfied :

Equations @ and @ apply only to regions of potential plastic hinging.

The confinement requirements for t h e basic section a t the bottom of t h e column w i l l s a t i s f y t h e requirements for the remainder 05 the eolumn.

f * c = 3250 psi , fy = 60000 psi

I

3250 controls ~ q . @ PS - 0.12 x 0,63592 = 0.00413 c--------

Try I4 spirals @ 3' pitch

spiral - 0.20 x 2a(33 ,0 - 2.0 - 0.283 = 38.60 cu.in.

concrete = 3.0 x 3019.1 = 9057.3 eu . in .

Shear, use core section of b a s i c section

vu = 94 4 Vu - E T = 0.307 ksi

W

W t e m i n e average compressive stress on t h e cote concrete area due to Pe.

f 1209 = 0.400 ksi c avg. = 3019.'r

O . l f m , = 0.325 ksi < 0.400 k s i

2(f1,) 0 5 /. can use v = e 1000 0,114 k s i

0.5 (3250' = 0.456 k s i > 0.193 ksi :. Sect ion s i z e is adequate. 1000

~ r y #5 spiral a t 3' pitch

= 6 0 8 ~ < 761k N.G.

600 x 3.0 = 2.40" =,eq~d* m T r y 16 spiral at 3.5" p i t c h

c 2 % 0.44 x 60.0 % O.S(C6.0 - 4.0 - 0.88) ,73gk < 761k 3.5

3% under. Say ok.

Check for min. per Article 1.5.lO(A)t2)

- Use 46 spiral 3.5" pitch for full l ength of column and extend i n t o b e n t cap and foot ing per Mticles 1.5.35(1) and 1.4.6EJ) respectively. Spiral may be discontinuous at t h e bottom bent cap reinforcement and top footing reinforcement, but should be anchored on each a i d e of these l eve l s of horizontal reinforcement.

P i l e s : U s e standard 70 t o n p i l e s

U l t i m a t e axial bearing capacity = 280k U l t i m a t e ax ia l uplift capacity = 112k U l t i m a t e lateral resistance = 3ok except for

Group VZI 1 40 k

f'= = 3250 psi fy = 60000 psi

Determine t h e pile layout, footing size, and footing reinforcement required to resist the bottom of column forces and moments. U s e the centerline of bent (X -ax is ) and the centerline of column (Y-axis) as the principal axes of the footing.

Minimum footing thickness

2 6 . O O R development length of outer r i n g of column reinforcement

6.00 addi t iona l embedment of inner ring of column re inforcement

3,26 t l l footing reinforcement 6.00 clearance to bottom footing reinforcement

41.26.

Determine a pi le layaut that i s adequate for Group V f T loads and check for other group loads.

Comparison af the dead l o a d p lus e last ic ARS earthquake forces w i t h t h e forces result ing from seismic plast ic hinging ind i ca te s that t h e latter rill be the lesser of the t w u (Article 1.2.26(f)).

Bottom of eofum forces resulting frm plastic hinging

Case 1 (hinging t i t h e r about X-axis or Y-axis)

Pe = 1209 k , M P = 10641 ft-kips

Case 2 [Ringing about axis at 45* from z-ax i s )

Try a 16 p i l e footing 19.0' x 19.0' x 5,O'

r % Column l Y mis 1

DL = (19.0~19.0-23.8) 2 ~ 0 . 1 2 0 = 81k COVER - 351k

+les = 16 piles

(each direction) 8 x m2 = - 128 512 p i l e f t . 2

P i l e reactions - Group ttff

Case 1

- 9 7 . 5 * 1 6 6 . 3 - = ~ 6 3 . 8 ~ r n a x < 2$ok ok

1-68. nk minl <I-112kl ok

Case 2

= 37.5 - + 117.6 2 58.8 = 273.gk max C 280k ok

I-78.9k mini <)-112kI ok

Factored Group z loads Cat bottom of coltmn)

Case 1

x 60 ) = 388 tt-kips nx = 1.3 (216 + 1.22

Im6' x 1121) = 1995 ft-kips n = 1.3 ( 0 + m 2 Y

- Case 2

Hx 1.3 ( 276 + x 5 1 6 ) = 1199 f t - k i p s

1.67 n - 1.3 ( 0 + 7 - ~ 6 5 4 ) = 1164 f t -k ips Y *

Case 3

136 nx = 1.3 (216 + ) = 426 ft-kips

3672 - 1.3 ( 0 - 3913 f t -k ips

Case 4

952 = 1295 f t -k ips M, 1.3 (216 * m1

27 79 n = 1.3 (0 + = 2322 ft-kips Y

Case 5

694 Mx = 1.3 (216 + ,-) 1020 f t -k ips

2244 n, 1 . 3 t o + -1 1. 2391 ft-kips

P i l e Reactions - Group I loads

Case 1

'n pile = 0.75 x 280 = 210k (Article 1.4.6tD)I

Case 2

Case 3

3.3% over, say ok Case 4

I

k = 140.2 + 10.1 + 36.3 = 186.6 < 210k ok

Case 5

Determine footing shear requirements.

Equivalent square column section (Article 1.4,6(F))

From a comparison of p i l e reactions i t can be determined that Group Vfl case 1 loads w i l l control.

Assume # I 1 bars in bottom mat of reinforcement

Shear a t section through foot ing a t distance d from face of column (Article 7,5.35(F) ( 1 3 ( a ) )

Vu = 3 x 263.8 = 731.4

- 'u af lowable = 4 1 0 ( f @ ~ ) O * ~ b p (Articles 1.5.35(B) ( I )

on section and 1.5,35(C) 1 6 ) )

9 v, = 0.85 x 1310 - 1 1 3 9 ~ > ~ 3 9 . 3 ~ c. do not need shear reinforcement

Shear at section concentric with and st a distance,d/2 from the face a£ column (Article 1.3.35(F) ( 1 ) (b)).

use section d l 2 from face of actual circular column section.

Neglect t e n s i l e pire reactions.

'u allowable = 4 6( fv ,10*5 bad : (Article 1.5.35(P) ( 0 ) )

on seetion

' c without = 4 ( f a c ) 0 ' 5 bod : (Article 1.5.35(F)(3)) shear reinforcement

+ vc = 0.85 x 4347 = 369sk > 1 4 8 4 ~ ok do not need shear reinforcement

- rktermine minimum foo t ing thickness that i s adequate for shear without shear reinforcement.

Try 3.5' footing thickness

P i l e reactions are based en the dead load of a 5.0' thick foot ing, therefore use t h i s dead load far reducing the applied moments and shears.

Shear of section through footing a t dis tance d from face of column.

Shear at section concentric with and a t a dis tance d/2 from the face of column.

'ftg. min. 3'-6" ; use minimum shear reinforcement

BRIDGE bESfG?l PRACTXCE JMWARY. 1983 5-62

' betermine f lexural reinforcement sequired (Articles 1.4. ti (G 3 , 1.5.7, 1.5 .32 and 1 . 5 . 3 7 )

B o t t m of f noting f l e x u r a l reinforcement - Section a t face of column

7.07 - - 2.0 x 7.07 x 19.0 x 0.120 x 7 - - 114 - 5072 f t -k ips

'(19.0 x 1.2)(42.0) ' t 3 2 5 0 ) ~ * ~ TS = 1.2 x 7.5 1000 0.5 x 42,O 412 = 2866 f t -k ips

" = 42.0 - 6,O - 1.5 x 1.25 = 34-12' m i n

= 6291 ft-kips > 5072 f t -kips ok

P s = 39.0/119.0 x 12 x 34.32) = 0.005

- Check serviceability requirements of Article 1.5.39

Determine p i l e reactions under service l o a d s Group f

Case 1

60 = 265 f t -kips n, = 216 * 1;22

'''' = 919 f t -k ips M = O + m 2 Y

k controls = 100.6 + 2.1 + 14.4 = 117.1 max <-------- - - Case 2

516 Wx - 216 * - = 639 it-kips 1.22

Y 654 = 536 ft-kips = O + m

= 95.8 I + 10.0 - + 4.2 - 110-ok max. n 3 x 115.0 (8 .0 - 2,135 = 1922

4 x 107.8 (4.0 - 2,433 - 677

- 3.5 x 7.07 x 19.0 X 0.156 x 3.53 0 - 249

f, = 0.86 ksi

fs = 22.1 ksi < 24.0 ksi ok

- Use $ 9 C 6 2 total 39 each d i r e c t i o n

Top af footing f lexural reinforcement (Article 1 . 4 . 6 ( G ) J

U s e 49 @ 12 - + total 20 each direc t ion

Minimwn shear reinforcement

Use # S J @ 12 placed per Article 1 . 4 . 6 ( A ) .

Compare t h e available lateral resistance of t h e soil and foundation system w i t h the horizontal seismic forces at t h e bottom of the eolmn. It is n o t a requirement to provide a bent foundation d e s i g n to resist the horizontal seismic forces, but i t t h e ava i lab le resistance is significantly less than the seismic forces then large permanent displacements of t h e foundation may result a£ t e t a large earthquake.

Soil parameters of soil in which footing is embedded

'n piles - 16 x I D = 640'

Hn foundation - 153 + 640 = 793k < 94qk 16% under seismic horizontal f oree

Increasing the embedment of t h e footing would increase t h e resistance available and reduce t h e seismic force (column shear) so that it would be possible to reduce the likelihood of large permanent bent footing displacements after a large earthquake,

BRfDGE DESIGN PRACPf CE HAY, 1982 5-65

- Example Problem 2 - Two Column Bent W/Pile Footings

Problem: Design t h e columns and p i l e footings for t h e structure shown below, This is t h e same problem which is used as Problem S e t No. 1 of Reinforced Conetete Sect ion of the Off i ce of Structures Q e s i g n correspondence course.

E R X E E DESIGN PRACTICE HAY, 1982 5-66

_Column loads* For illustrative purposes, only dead load, l i v e load and seismic loads will be considered. The X-axis e q u a l s t h e centerline of k n t and the Y - a x i s e q u a l s the centerline of column.

Dead load (service levell

DL TOP COL =

DL = 914 k

BOT COL

DL DL * x TOP COL = -1845 f t -k ips nx BOT COL = IS9 f t -k ips

DL = 1150 Zt-kips DL

TOP COL B M COL = -360 f t -k ips

Live load + impact (service level); Impact = 21%

Case 1

p LL+' = 256k

LL+l = -391 f t - k i p s Mx rnP COL , 1 5 9 ft-kips "x SOT COL

LL+' = 162 f t-kips TOP COL = - 3 f t -k ips ' x BeT C O t

Case 2

P LL+T , U L k

L W f ' x mp eoL LL+z = 756 f t -k ips = -1864 f t -kips M, ,,

LL+X LL+I m, WL = 83 ft-kips ", BOT COL == -2 f t-kips

Case 3

P L W r , 1 3 0 ~

=-234 f t -kips nx TOP cur, = 95.ft-kips nx Dm ,L

- 6 D l f t - k i p s ny TOP EOL Lwz = -188 f t-kips My DOT ML

BRfM=E DESTG# PRACTICE 1982 5-67

- Live load * impact Ifos factored levell: fmpaet = 21%

Case 1 (1.15 x 1 lane P + 1 lane H)

LL+' = -982 f t - k i p s LL+f 'x TOP COL Mx BOT COL = 399 ft-kips

LL+f = 812 f't-kips LL+f TOP COL 'Iy BOT COL = -102 ft-kips

Case 2 (1.15 x 1 lane P + 1 lane B)

LL+Z = 383 k

LL+X LWf 'x TOP COL = -4304 f t -k ips PI, ,, = 1746 ft-kips

LL+I TOP COL = 535 ft-kips = -80 ft-kips BOT COL

Case 3 (1.15 x 1 lane P + 1 lane H)r

LL+I = 450 k

LL+ f ' x TOP COL = -825 f t -k ips 5 3 3 5 f t -k ips Hx BOT COL

= 1251 f t -k ips 2L+I Wy BOT COL = -287 f t -kips

Seismic load (ARS forces and moments before and after application of z = 8 factor)

Transverse earthquake motion ( W S results1

EQ k EQ k = 32 (~orizantal H, TOP COL = 880 By TOP COL forces)

EQ = 921k EO 'x BOT COL Hy BQT COL = 3rlk

Case 1

EQ = +936 k TOP CQfr - EQ

' x TOP Cot = 328 / 8 = 4 1 f t -k ips

EO TOP COL

= 8240 / 8 = 1030 ft-kips

B R T E E DESIGM PRACTICE m y m 1982 5-68

- Case 2

EQ = +932 k E r n COL - EQ

% BOT COL = 334 J 8 = 42 ft-kips

EQ 5 em COL = 9032 / 8 = 1229 i t -kips

IRngitudinal earthquake motion [RHS results)

EQ - 2~~ EQ = 461k (Eorizontal Hx TOP COL By TOP COL f otces l

EQ = 2gk €0 Hx BOT COL By BOT COL = 49zk

Case 3

FQ +202 k TOP COL - EQ

M~ TOP COG - 4597 / 8 r 575 f t -k ips

EQ TOP COL 384 J B a 48 ft-kips

Case 4

EQ 1 +202 k E r n COL -

EQ ' x BOT COL - SOSO / 8 + 631 f t - k i p s

EQ My BOT COL = 191 / 8 = 24 i t -kips

Combined earthquake motion

Case 5 (1.0 EQL + 0.3 EQT)

EQ TOP COL = 202 + 0.3 x 930 = 481k

EQ 'X TOP COL = 4597 + 0-3 x 328 = 4695 / 8 = 587 i t -k ips

EQ ny TOP COL = 384 + 0.3 x 8240 = 2856 / 8 = 357 f t -k ips

BRf DeE DES XGH PRACTf CE HAY, 1982 5-69

- Case 6 ( 1 . 0 EQL + 0 . 3 EQTl

EQ BOT COL = 202 + 0.3 r 932 = 48zk

EO * X 30, COL = 5050 + 5.3 x 334 = 5350 / B = 644 it-kips.

EQ Hy BOT COL = 191 + 0 . 3 x 9832 = 3141 / 8 = 393 ft-kips

Case 1 ( 0 . 3 EQL + 1.0 EQT)

EQ TOP COL = 0.3 x 202 + 930 - 991k EQ

* X TOP CbL = 0 .3 x 4597 + 328 = 1707 / 8 = 213 ft-kips

EQ My TOP COL = 6.3 x 384 + 8240 = 8355 / 8 = 1044 f t -k ips

Case 8 (0.3 EPL + 1.0 EQT)

EQ 'X BdT Cob = 0.3 x 5050 + 334 = 1849 / 8 = 231 f t -kips

EQ BOT COL = 6-3 x 191 + 9832 = 9889 / 8 = 1236 f t - k i p s

Columns : -

Geometry - Standard architectural column Type ZR

Cleat h e i g h t = 20'-0" Length of top flare section = 16'4" t - 5' -6*

Longitudinal reinforcement - determined by using YIELD program w i t h r e i n f otcement placement controlled by bas ie section at bottom of column.

As TOP = 54-#9 = 54.00 sq.in.

3 6 - t 3 in outer sing for f u l l length column

18-#9 in i n n e r s ing for top 2/3 column

*S B O m n 36-39 = 36.0 3q.in.

Ag TOP COL A m = = 7777.19 sg-in.

Ag M P COL DESXGN = 5401.19 sq-in. (Group loads other than Group V f I )

Ag TOP COL DESIGN = 3421,19 sq.in. (Group VfX loads)

Ag BOT COL ACTUAL c DESIGN = 3421.19 s q - i n .

Probable plastic moment strength - fcm = 3250 p s i . e, ,, = 0.003 in./in,

f~

60000 psi r E, 1 29000000 psi

Full Flare Section @ Top of Column

As = 54-19 = 54.0 sq.in.

BRfDeE DESfGW PRACTICE M Y . 1982 5-72

Full Flare Sect ion @ Top of Column

As = 54-19 - 54.0 sq.in.

F u l l Flare Section @ Top of Colmn

As = 54-19 = 54.0 sq.in.

BRIDGE DESfm PRACTICE M Y . I982 5-74

Basic Section @ Top of Column

As = 54-49 = 54.0 sg.in.

Basic Section @ Top of Column

A, = 54-#9 = 54.0 sq-in.

Basic Section @ mttam of Colmn

As = 36-#9 = 36.0 sq.in.

Fasic Sect ion @ Bottom of Colmn

AS = 3 6 - t 9 = 36.0 sq.in.

BRIDGE PESfGR PRACTICE 1983 5-70

- C o l m n Shear

Deternine the maximum colmn shears considering that t h e nominal moment strengths can be developed in the gross flare sections of the columns and that t h e probable p l a s t i c moment s t r e n g t h s c a n be developed in t h e basic column section, C o n s i d e r the se moments developing about the X-axis, Y-axis and on a x i s 4 5 * from t h e X- a x i s . U s e the dead load axial f o r c e ' p l u s the change in axial force due -to t h e deve lopent of the component of moments about t h e Y-axis t i .@. , take i n t o consideration t h e effects of transverse overturning),

noment strengths - Use the interaction curves developed from Y f ELD program output,

Top of colvmn

%x 0. 3 9600 f t - k i p s

'ny 90. = 16700 f t -k ips

H px om = 10000 f t -k ips

n PY g o 0

= 10000 f t - k i p s

Bottom of eolmn

o tnny 90. XIP + npy 90. BOT) (16700 + 7 9 0 0 ) , 1Z30k YUx

= clear column h e i g h t 20 .0

n v = ( px O 0 m p + 'PX om e m = clooob + 7900) , 895k UY clear column h e ~ g h t 20.0

OT My I =Tux col (clear colmn height + yb of superstructure)

= 2460 (20.0 + 3.54) + 57960 f t -k ips

n OT - ZM p " L + !Y py 90- BOT +

57900 - 2 X 7900 +2339 - colrmrn spaclng - 18.0 - I Note: The a b v e determination of axial column f orees associated w i t h the column moment strength is v a l i d only for 2 column bents w i t h equal l e n g t h eolmns, For b e n t s w i t h more than 2 columns and/or for bents w i t h significantly d i f f e r e n t length columns

BRIDGE DESIGN PRACTICE JAROAR?. 1983 5-79

- w i t h i n t h e b e n t , use a n app rop r i a t e a n a l y s i s f o r determining a x i a l column f o r c e s a s soc i a t ed wi th t h e column moment s t r eng ths .

Determine new moment s t r e n g t h s , s h e a r s and a x i a l f o r c e s u s ing

P DL+OT .

'ny 90 ,= 23600 f t - k ip s n

PY 90° = 12400 f t - k i p s

DLMT- 815 - 2339 = -1524~ 'min

'ny 90. = 6600 i t - k i p s 'py 90° = 5000 i t - k i p s

Bottom o f c o l m n

914 + 2339 = 3 2 ~ 3 ~ 'max

M PY 90.

= 10600 f t - k i p s

DL-T= 914 - 2339 = -1425 k

'min

n PY 90.

= 2200 f t - k i p s

P (23600 + 10600) l,lOk 'ux max 20.0

E (6600 + 2200) , 440k Vux min 20.0

' "ux columns = 1710 + 440 = 2150k, - 13% less t h a n prev ious '"ux columns = 2 4 6 0 ~

n OT= 2150 x 23.54 = 50611 f t - k i p s Y I

Determine new moment strengthsJ shears and a x i a l forces us ing rev ised P DLWT

a

Top of eolmn E

'ny 90. = 23000 ft-kips n

PY 9 0 ° = 12300 ft-kips

DLX)T= 815 - 2100 - -1285 k 'min

EI nny 90. = 7700 f t -k ips

PY 90' + Sf00 f t - k i p s

n = 10460 f t - k i p s PY 90'

DL*T= 914 - 2100 - -1186 k 'min

H PY 90°

= 3000 f t - k i p s

t (23000 + 10400) = Ib70k 'tix max 20.0

a (7700 + 3000) _ 535k Y u x min 20.0

k rVux col-s - 1670 + 535 = 2205 ,*2.6% greater than

previous ZVux columns - 2150 k say close enough

-termin@ moment strengths, shears and axial forces due to y i e l d i n g of columns at top arid bottom due to bending about an axis 45O from the X-axis.

'nx 45' m l h y 45. = 0250 f t - k i p s

HpX 45. - n

PY 45. = 7100 f t - k i p s

- mttom of column

P 914 k

Ipx 45. = F I = 5580 i t - k i p s PY 45'

Zvux columns = 2 x 692 = 1384*

M OT= 1384 (20 .0 + 3.54) - 32579 i t - k i p s Y

Determine new moment s t r e n g t h s , shears and axial forces u s i n g

P DL+OT

Mnx 45. = nny 45. = 10400 I t -k ips

H px 4S0

= n PY 45O

= 8200 f t -k ips

DL+oT= 815 - 1190 - -375 k 'min

'nx 45. I nny 4 5 0 + 5800 f t -k ips

Mpx 45. = M PY 4 s 4

= 5600 ft-kips

mttom of column

DLwT= 914 + 1190 = 2 1 0 4 ~ 'max

Hpx 45. = 21 PY 4 5 O

= 6800 f t -k ips

914 - 1190 r -276 k

'min

npx 45. = n

PY usu = 3800 f t - k i p s

- P (10100 + 6800) , 860k

VuX max ' u y max 20.0

D (5800 + 3800) , aaak V ~ x =in 'uy min 20 .0

B V u ~ columns - 860 + 480 = 1340k,'3.2% less than previous

ZVux columns = 1384): say close enough I

Determine column shears using dead load p l u s elastic earthquake results.

npz BOT COL =9100 <136O+ 98891 ft-kips (Probableplastic moments associated

DL+EQ Xpy R O T COL - 2000-(159 + 1849) f t -k ips w i t h P BOT

*The elastic moment value far Mx BBT COL of (360+9889S would not quite be reached but say close enough for shear

determination.

- Comparison of shear force from elastie analysis w i t h t h e shear forces associated with column y i e l d i n g indicates that the shear force fsum t h e elastic earthquake analysis is least cr i t i ca l ,

" V ~ DESIGN = 1 0 2 6 ~

The above comparison of shear forces also i l l u s t r a t e s an undesireable aspect as far as seismic performance i s concerned of t h e use of flared columns at short multi-column bents , which aspect is the potential for a high demand for shear resistance,

Transverse colmn reinforcement

Shear

Associated a x i a l force

k P ~ ~ ~ E ~ O L = 914 - + 993 = 1907 *- max - 7 g E min

Because the column axial force can be a tensile force, t h e total shear should be resisted by shear reinforcement.

Because plastic hinging may occur at t h e bottom of t h e colmnr use the core section of the bas i c column section f o r shear resistance,

core = 66.0 - 4.0 162 .0 '

d core = 0.8 x 62.0 = 49.60"

V" = 1026 u 0.85 x 62.0 x 49.60

- 0,392 k s i

BRXDCe DESIeff PRACTICE M Y , 1982 5-84

- Try concentric 35 spirals at 3" pitch

A f d 4 x 00.3 X 60.0 x 49.60 = 3230k > l lOtk

vs S 3,O

Concrentri~ spirals require revised arrangement of Longitudinal column reinforcement.

Try 18-#9 full length in outer and inner ring p lus 1 8 4 9 for 2 / 3 t h e length of the top portion of the column in t h e outer ring,

Spacing a t the bottom of t h e eolmn

'outer ring = 33.0 - 2.0 - 0.69 - 0.63 29.68.

Z P x 29.68 spacing = 18 = 10.36" > 8' N.G. See Article 2,5,11(AJ

Try CB longitudinal colmn reinforcement

24-13 full length i n outer and inner r i n g plus 24-18 for 2/3 the l e n g t h of the column at the top o f the column in t h e outes ring.

*S TOP COL = ( 4 8 + 24) = 72 - # 8 36.88 sq. in, -- 54.0 sg. in, ok

As B(TP COL + 4 8 - #8 * 37-92: sq, in.^ 36.0 sq, in. ok

Spacing at bottom of eolmn

+outer r ing = 33.0 - 2.0 - 0.69 - 0.57 * 29-7dW

Spacing at top of column of outer ring

= 3.9' < than the preferred Spacing = 7 minimum spacing, therefare bundle part ia l length bars to t h e f u l l length bars in t h e outer ring,

BRIDGE DESIGN PRACTICE MAY, 1.982 5-85

Confinement

S a t i s f y t h e f o l l o w i n g v o l u m e t r i c e q u a t i o n s . Equa t ions @ and @ app ly to r e g i o n s o f p o t e n t i a l p l a s t i c h ing ing only.

The conf inement r e q u i r e m e n t s f o r t h e b a s i c s e c t i o n a t t h e bot tom o f t h e column w i l l satisfy t h e r e q u i r e m e n t s f o r t h e remainder of t h e column.

A9 = T ( 6 6 * 0 ) L = 4 3421.2 sq. in .

7 (62.0) 2

A, = - = 3019.1 sq. in.

f ' c = 3250 psi , f y = 60000 p s i

k 'e max = 1907 (dead load p l u s elastic e a r t h q u a k e r e s u l t s )

3250 x 0.7144 = 0.00464 c o n t r o l s E¶.@ P, = 0-12 <------

Try 2 c o n c e n t r i c 15 spirals a t 3. p i t c h

cu . i n .

VO1* concrete = 3019.1

U s e 2 concentric 1 5 spirals at 3. p i t c h for t h e f u l l length of column and extend into t h e bent cap and footing.

P i l e s :

A preliminary determination of t h e footing s i z e required u s i n g 70 ton p i l e s ind icated that this was not a practical solution considering the 18' eoltrmn spacing. A common footing using 70 t a n p i l e s could present a practical s o l u t i o n , but in order to i l l u s t r a t e a footing d e s i g n for ind iv idua l footing, 100 ton p i l e s w i l l be used.

Ultimate bearing capacity = 400; U l t i m a t e uplift capacity = ZOOk Ultimate lateral resistance = 40 except for

Group V I f = 5sk

Footing:

f o e * 3250 mi. fy = 6 0 0 0 0 psi

mtermine the p i l e layout, footing size, and footing reinforcement required to resist t h e bottom of column forces and moments. U s e the of bent (X-axis) and the E of column (Y-axis) as the principal axes of t h e footing, I M i n i m u m footing thickness

19 .80n development of outer r i n g of li.8 eoLmn reinforcement

6.00" additional embedment of inner r ing of column reinforcement

3-26" # 11 bottom footing ref nforcement 6 ,QOw clearance to bottom footing reinforcement 35,06*

mtermine a p i l e layout that is adequate for Group YII loads and cheek for t h e other group loads,

Group VII Bottom of co lumn loads

Case - 7900 f t -kips --- - 6800 38 00 53139 5309 2008 2008

rc- f t-kips 10400 Yielding of

3000 column 6 8 0 0 3800 3501 DL + EO 3 5 0 1 e las t i c

10249 analysis 10249

Try 15' x 1 5 9 3-5 ' faating w/16 p i l e s

BRXDGE DESXGH PRACTICE . JmARYn I983 5-88

Apiles - 1 6 p i l e s

= 8 ( 6 . 0 ) ; - 288 = 0 (3 .0 ) = - 72

(each direction) 360 p i l e - f t . 2

P i l e reactions - Group Vff loading

Case 1

k = 6 7 . 5 + - 131.7=199.2kmax. -64.2 min.

Case 2

k k = 198.8 2 173.3 - 372.1 max. 25.5 min,

Case 3

P = (-1186 + 166) + 3000 x 6 16 - 360

Case 4

Case 5

k = -6.9 + 63-3 2 63.3 - 119.7 max. k - -133.5 min .

- Case 6

k = 97.6 + 88.5 2 58.4 = 244.sk nax. -49.3 min. - Case 7

k k = 37.4 - + 0 8 - 4 2 58.4 - 184.3 max, -109.5 min .

Case 8

k = 129.6 + 33.5 + 170.8 = 333.9 max. k - - -74.3 min.

Case !3

P = (-" + 166) + 33.5 + 170.0 16 - - k k = 5.4 - + 33-5 2 170,8 + 209.7 max. -198.9 min,

The p i l e layout satisfies load eases for DL + EO E m an elastic a n a l y s i s a n d a l s o s a t i s f i a s l o a d c a s e s f r o m t h e y i e l d i n g o f the 1 columns, For i n t e r n a l footing d e s i g n use only t h e load cases from an e l a s t i c analysis.

Check p i l e layout far other group loads.

Factored Group I loads Fat bottom of colmnl

1 67 nx = 1.3 (159 + A 1591 = 492 f t -k ips 1.21

B R X E E DESXGN PRACTICE HAT, 1982 5-9 1

Pile reactions - Factored Group 1 loading By inspection Case 4 and Case 5 w i l l produce t h e maximum p i l e reaction.

Case 4

k = 126,4 + 10.6 + 9 . 6 = 146.6 max - - < 0 .75 x 400 = 3 0 8 ok

Case 5

Determine f aoting shear requirements

Equivalent square calmn section

From a comparison of p i l e reactions it can be determined t h a t either Group VII Case 8 or Case 9 l o a d i n g w i l l control, because ef the Group V l I loading cases, those from the elastic analysis of DL+EQ are t h e lesser. See Article 1.2.201 PI.

Assume piles art HPl4 x 89 far determination of contribution of a pi le reaction to the shear on a particular section through the foot ing.

Asstzne 39 footing reinforcement

Shear at section through footing at distance d from fact of column and parallel to the Y-axis.

- Group VIf Case 8 loading I

=. need shear reinforcement sr thicker footing

Shear at section through footing at face of column and para l le l to t h e Y-axis.

Group V Z I Case 9 loading

Need to pick up vertical component of diagonal compressiwe force at t h e bottom af t h e footing w i t h vertical reinfoteemerit in order t o transfer this force ta t h e tap of footing so it can be transferred into t h e column area and be picked up by t h e long i tudina l column reinforcement, Assumed this load condition could occur a b u t each a x i s of the footing.

As mq'd C = 19.26 sq. in. 0.9 x 6 0

using # S L 6 rovs @ 12. x 6w spacing, A; 2 x 6 x 6 x 0 .31 = 22.32

sq. in. ok (Y-axis)

3 raws @ 12' x 6' spacing, AS = 2 x 3 x 11 x 0 . 3 1 = 20.46

t X-axis) sq. in. ak

Shear at section through footing at distance d J 2 from perimeter of column.

Group P f Z Case 8 loading

disregard tensile pile reactions and dead load of footing and ewer.

Using 3.5 ' thick footing deternine shear reinforcement required, I Group VII Case 8 loading

try t S @ 12 each way

Group VTf Case 9 loading

#5 C @ I2 each way ok by comparison

Determine s t irrup layout af t e t flexural reinforcement is determined.

Determine f lexural reinforcement required,

B o t t o m of footing flexural reinforcement. Sect ion at face of calrzmn,

Group V 1 I Case 8 loading will contro l

4 x 215.0 [3 .0 - 2.443 = 4 8 2

C

a760 ft-kips

a > 1.2 ncr aa '* *s reqnd. ' *a min.

try A~ - 49 e 6 -, - 31 x 1.0 - 31.0 sq. in.

9% 1 / 2 - 4999 ft-kips = 1.0 x 31.0 x 60.0 (34.12 - 7 > 4760 i t -k ips ok

cn- F m s 0 +

2 " m 0 t o m

L + m s m E'

3

BRfbeE DESXGM PRACTICE S A R U m , 1983 5-96

= 74.2 - + 4.0 2 8.6 = 8 6 . ~ ~ max

Case 2 controls, cheek section at face of eolumn.

n -- 4 x 87.3 (6 .0 - 2.445 1243 4 x 80.8 ( 3 . 0 - 2.445 = 181 -

1 4 2 4 f t -k ips

f = 17-69 ksi < 24.0 ksi ok S

Use #9 @ 6 - + total 31 each direction

Top of footing f lexural reinforcement. Check section at face of column

Group VII Case 9 loading

H, 4 x 165.4 16.0 - 2.44) = 2355 4 x 80 .0 (3 .0 - 2 - 4 4 ] = 179

5.06 x 15-0 x 3.5 x 0.150 x 2.53 - 101 5.Q6 x 1 5 . 0 x 2 . 0 x 0.120~ 2.53 = - 4 6

2681 f t -kips

1.93 +n, - 1.0 x 16.0 x 60 .0 (37.12 - - 2 1 / 12 = 2892 ft-kips I J 2691 f t - k i p s < 2892 ft-kips ok

BR1XK;E DESIGR PRACTICE M Y , 1982 5-97

Use #9 @ I2 2 total 16 each direction

SECTJON A-A

Cak - Bridge Design Practice . December 1992 m

Section 6 . Underground Structures

Contents

Part 1 . Underground Stauetures

A . General ............................................................................. .... ........................... .. ................... 6-1

.................................................................................................................... B . Standard Plans 6-3

..................................................................................................................... C . Overfill Tables 6-3

......................................................................................... D . Special Design Considerations 6-4

Part 2A . Reinforced Concrete Box Culvert. Cast-In-Place

A . General ................................................................................................................................. 6-7

............................................................................................................... B . Caltrans Research 6-8

...........*...............*.*.......*............*.*..**.*... *.......... . .......*.... C Design Method -. ......................- 6-8

...................... . ................................. 1 Dead Load ... ...-................................................ 6-9

. .................................. 2 Earth Loads .. ......................................................................... 6-9

. 3 Live Loads ............................................*...............-...................................+................... 6-9

. ...........................................................................................*.......................*........ 4 Impact 6-9

. ................................................................*....*........................................ 5 Other Loads 6-10

6 . Parapets ............... ......-............... .............-... .... ............ 6-10 ........................................................................................................ .. D . Design Analysis -6-1 1

. .................................~.*....................................*....*..........*...*....................... 1 AASHTO 6-11

. ...................*...........*.....*......**..*..............‘...................*................................ 2 C a l m 6 - 1 1

. ..........................*.*.........**.......................................*.................................. E Design Criteria 6-16

. .................... 1 Loadings .... ..................................................................................... 6-16

...................................................................................................................... 2 Moments 6-17

.............................................................................................. . 3 Live Load Diskribution 6-17

.............................................................. . 4 Moment Envelopes 19

. 5 Shear ..............................-...................................................*........................................ 6-19


c* ! Bridge Design Practice . December 1992 =

Part 2A . Reinforced Concrete Box Culvert. Cast-In-Place . continued

F . Design Example . Double 12 x 12 RCB with 10' Cover ........................ ... ........... 6-20 1 . Loading Cases to Consider ............................................-.......................................... 6-21

..................... ................... ...................... . 2 Loads - Condition 1 (2' Cover) ...... .... 6-22

......................... ..................................... . 3 Loads - Condiiion 2 (10' Cover) .....-........... 6-31

. 4 %ion by Shear .......................................................................................................... 6-39

5 . Reinforcement by Ultimate Moment ........................................................................ 640

...................................................................................... 6 . Distribution Reinforcement $ 6 4

............................................................... 7 . Load Moment Envelopes . ......., .......... ... 5-44

8 . Reinforced Lengths ........................................ ... ...............d............1.*...............*1..*. .. 6 4

..... ...................... ........................................ Appendix .... ............................-............ 6-49

Section 6 . Contents Page 6-ii

c* -1 Bridge Design Practice - December t 992 m

Part 1 - Underground Structures

A. General

Soil-structure interaction systems include both rigid and flexible drainage and highway separation structures. These are usually buried within the roadway embankment, but may also be "At Grade" lie. the grade-top cast-in-place reinforced conmete box culvert).

These underground stnzctutes4rcular pipe, pipe-arch, arch, and box shapes- be either flexible or rigid structures. AASHTO and Bridge Desip S p ~ $ ~ ~ t i m s presently contain design criteria in Sections 1217 and 16 for Soil-Corrugated Metal Structure Interaction Systems, Soil- Reinforced Concrete Structure Interaction Systems, and Soil-Thermoplas tic pipe Interaction Systems respective1 y . Dimension ra tio (DR) is the inside diameter of the culvert (or structure) in inches divided by the wall thickness in inches. Based on Cdtrans culvert research program, this new des ip concep t has been developed in the design of circdar and semi-circular underground shctures . Specifically, the first applications have been to reinforced conmete pipe and reinforced conmete semitircuh arch d e s i p .

Design methods have previously assumed that a pipe was flexible, semi-flexible or rigid depending upm the pipe material. For example, metal pipe dverts were always considered flexible, prestressed concrete pipe culverts were considered semi-rigid, and reinforced conm te pipe culverts were considered as rigid designs.

Research into culvert usage established tke fhct that prestmd cormete and reinforced concrete pipe adverts have hen vsed in all three ranges, ie., flexible, semi-rigid and rigid (Figure 1).

The culvert ma teriaE properties alone do not dictate the s h c ~ a l performance of a c h m k pipe or semS4rcular arch- A key element is the fact that all underground structures are duenced by soil-structure interaction.

In effect, thinner walled, m ore flexible pipes simply deflect to a more uniform loading condition and consequently, moment is diminished as a design parameter, and thrust becomes the primahy design considera tion.

Another direct consequence of the d v e i t research by Caltrans was the discovq of two si@cant design parameters:

1. h the mse of the 10 ft diameter steel stmctutal plate pipe, there was an effective density increase of 50% subsequent to fill completion, based on readings taken 30 months afker installation Recent researchby N o m y (TRB 1231 - 1989) on a 25 ft x 22 f t steel struchrraI plate pipe-arch and a 35 ft x 23 ft steel structural plate horizontal ellipse confirmed the results of research by Caltrans of this effective density increase. Since readtngs were exkended, by Noway, to 7 vears after installation on the 25 fi x 22 ft steel s h c s u r a l plate pipe-arch, it also showed &t the effective density increase took place within 2 years, and then stabilized.

Seetion 6 - Underground Structures Page 6- 1

Ed - Bridge Design Practice - December 1992 m

2. For thick wall reinforced concrete pipe and pipe arch designs, the long term readings of 24 months, after fill camptetim, established the necessity to design for two loadings:

Loading 1 - 140V: 42H Loading 2 - 140V: 140H

Recent researchby Nebraska (TRB1231-1989),onadouble 12xlZcast-in-phceRCBconfirmed the existence of two horizontal design pressure loadings for RCBs.

Seismic forces are normally not considered in soil-structure interaction systems. Observations of all ty-pes of underground stmchms in the 1971 San Femanda earthquake area and in the 1989 San Francisco (Lorna Prie ta) earthquake area, affirmed the cushioning effect the soil b s on the performance of an underground structure during an earthquake. There were no failures due to an increase in soil pressures. Underground structures must move with the surrounding soil during earthquakes and usually will be supported by the interacting earth against mushing or collapse even if the structure joints are strained. If the earth does fault across a culvert, the tremendous forces will shear the submerged structure regardless of how the structure was designed. In special a s e s where underground sbctures are in soft ground (bay mud), considera tion should be given to providing longitudinal structural continuity.

A most siflcant difference between overhead and underground s ~ ~ s i s in the application of Ioads. In the case of overhead structures, the application of vertical extend loads is limited to live load only- In effxt, the increase in loads is h a t . However, in the case of underground structures, there are combinations of vertical earth loads and Bve loads which are not linear, (Figure 2). Ln effect, the least total combined vertical external load is at 3 to 6 feet af overfill. Consequently, Caltrans has used 10 feet as the minimum design ovwfi11; and specifies that all underground sbctures satisfy all loading combinations of dead load, earth bad and live load between inirknun cover of 2 feet, (at grade for CEP RCB), and 10 feet overfill. Am underground structure designed for 5 feet overfill could k mx@, or inadequate structur- ally, if either an add i t id overfiJ of 5 feet was added during the life of the structure; or 3 feet of overfill (or 5 beet for a CIP RCB), was removed during the specified 50 Far service life. As a consequence, no underground structure designed since 1965 has required replacement.

Page 6-2 Secliun 6 - Underground Structures

c* - Bridge Design Practice - December 1992 m

B. Standard Plans (Ca'ltrans)

Underground Structures Standard Plans 079 Precast RCP D80 CIP Single Box RCB D61 CIP Doubre Box RCB D95 Reinforced Concrete Arch (Horseshoe) Bl4-1 SSP Vehicular UC

C, Overfill Tables

Highway Design Manual, Chapter 850

Corrugated Steel Pipe

854.38 2% in x 44 jn Cormga tions - helical 854.3B 3 in, x 1 in. Cormga tions - helical S54.3C 5 in x 1 in. Corngations -helical 854.3D 2% in x % in, Cormgations - annular

Corrugated Steel Pipe Arch

854.3E 2% in x 'Ji in. Corngations - helid or annular

Corrugated Aluminum Pipe

854.4A 2% in x 1 !A in. Cormgations - helical or annular 654.4A 3 in- x f in. Cormgations - helical or annular

Corrugated Aluminum Pipe Arch

854.4C 2% in- x 54 in Cormgations - helical or annular

Steel Spiral Rib Pipe

854.5A ?4/[email protected] 8.5.4.5I3 34 in. x 3/1 in @ 7% in. pitch

Aluminum Spiral Rib Pipe

854.5C 3/c in. x 34 in @ 7% in pitch

Steel Structural Plate Pipe

854.6A 6 in x 2 in. Corrugations

Page 6-3 Section 6 - Underground Structures

- Bridge Design Practice - December 1992 m

Steel Structural Alate Pipe Arch

854.6B 6 in. x 2 in, Cormgations

Aluminum Structural Plate Pipe

854.6C 9Tn.x21/4in.Cormgations

Aluminum Structural Plate Pipe Arch

854.6D 9 in. x 2M in. Corrugations

Cast-In-Ptace Non-Reinf orced Concrete Pipe

854.2

Plastic Pipe (Preliminary)

fi59.X Plastic High Density Polyethylene Pipe- Cormgated, ribbed. Poly Vinyl Chloride Pipe. Ribbed Profile Wall

D. Special Design Considerations

Load Factor design is to be applied to all underground stnxctures. The &vest research by CaItrans has conclusively shown that initially used empirical design for underground structures and the subsequently developed senice load design does not provide structural adequacy for underground strurtures.

In order to determine the type of culvert material te be used, the resistivity and pH for the soil and water shall be determined for each culvert instahtion. Consult Highway Design ManuaS topic 852-Design Service Life.

The h y d r a d a shall also include information concerning the possibility of scour and abrasion at any proposed culvert irrstalla tion.

Cutoff walls should be provided whenwer scorn is a po tmtial problem. Further, headwalls, endwalls or bred end seclions are design features that may be required to assure the culvert stsuctural integrity.

Page 6-4 Section 6 - Underground Structures

Em - Bridge Design Practice - December 1 992

Figure 1. Soil-Circular Structure Interaction Systems Caltrans Effective Densities Research

1- tm -

Cahrans 11-81

Section 6 - Underground Structures Page 6-5

4l* I-: Bridge Design Practice - December 1992 =

Section 6 - Underground Structures Page 6-6

=* ' Bridge Design Ptactice - December 1992 =

Part 2A Reinforced Concrete Box Culvert, Cast-In-Place

A. General For economic reasons, Cast-Wlace RCB culverts in Callram Standard Plans are designed as rigid frames when either the span or height exceeds 8 feet, and the outer corners are designed as pin-ended if both the span and height are 8 feet or less.

Ends of interior walls (for multiple cells) are normally conside& pinned unless the reinforcement has sufficient embedment into the slabs.

Box culverts under high earth covets are probably less economical Lhan other shapes. Other shapes (&&,arch, and elliptical) should always be investigated for earth covers over 20 feet. If an RCB is the culvert type selected for fills over 20 feet, generally a rigid frame is preferable, regardless of span or height.

For significantly non-unifonrt loads, for example, if the RCB nrns along the toe of an embmk- ment, or next to a retahmg wall, design the stmcttlre as a rigid frame.

The bearing mpacity of the supporting medium shall always be considered. The Division of New Technology, Material and Research (DONTMR), Office of Engineering Geology, shall be consulted where footing pressures exceed 1 '/3 tons per square foot, or the span exceeds 10 feet. [See Higkway Design Manrurl, 8W.2(1) Bedding and B a r n , Paragraph 41

Do not place reinforced cont rek box culverts on piles. Other alternatives shall be considered such as moving the lm tion, using alternate types of culverts, or mbexavating and baddilling with suitable material.

Reinforcement is normally placed bamversely, as this is the most efficient span. Howwer, if u n d conditions indimte that placement along a skew is much more economical and practical, the design frame span will then be parallel to the bars . . . with a resultant increase In conmete depth and reinfotcing steel.

Compressive ~einforcement is not consid& in RCB design h u s e just a d deviation in rebar Iceition (in the= relatively thin members) d d result in a big change in ~mpability. However, if compression steel is considered, for analysis of an existing culvert only, it would be limited to half of t h ~ tension n&tforcern~nt.

Although axial Ioad (thrust) is a valid component member design, it has not been considered in the formula developed or applied to the Standard Plans.

For design notes, construction notes, and pertinent information, see m n t RCB Culvert Standard Plans D80, D81, and D82

Section 6 - Underground Stnrdures Page 6-7

4E - Bridge Design Practice - December 1992

53. Carttans Research The research conducted by C a l m of three reinforced concrete (horseshoe) arches (1963 thru 1975) resulted in a si@iont change in the design of reinforced concrete underground stmchms. It was found h a t the lateral pressures can k as much as the vertical p m s . Therefore, the traditional loading wherein the lateral p r e s m is taken as 30% of the vertical p-ra has k e n supplemented with a second loading wherein the lateral pressure is also taken as lOQX of the vertia1 pressure. These two loadings are applied qmxately and the resulting maximum moments are utilized in design.

R m t research by Tadros, of Nebtaska University (TRB I231 - 1989) has affirmed the two bands of loading concept on RCB Culvert Design.

C. Design Method

Caltrans RCB culverts are analyzed and checked by load factor design only. The service Imd (i-e., working stress or elastic design) does not apply to these stmclhves - see Brrdge Dmgn SpeCiFatrotls 17.2

'The derived applicable Ioad faam for Gmup X (dverts) are obtained from

Where D = Dead Load E = E a r t h h d L = Live Load I = Impact

%e page 12 and 13 for formula derivation and values.

Page 6-9 S- 6- U n d w g m StMdures

- - Bridge Design Practice - December 3992 H

Assume a structure weight for concrete plus reinforcement of 150 pounds per cubic foot.

Based onCaltransculvertresearch, it hasbeen determined that the "equivalent soil density" is 140 pd. This "equivalent soil density" is based on the actual maxiplum ifisitu densities observed on Caltrans culvert research projects. The full lateral pressure condition also satisfies the saturated fiIl situation.

For box culverts under highways, only HS20 tmck loads apply. Mtemative loadings, lane loadings, and P-Loads are not used in the design.

When the RCB is at grade, or with a cover equal ko or less than 2 feet, wheel loads are distributed as though they were applied directly to the roof, as in ordinary slab bridges. Wheel live load distribution to the invert is assumed asa uniform load applied transversely amss the width and 7 feet Iongihrdhlly along the length of the RCB. Conenhated live load distribution reinforcement shallbe placed in the roof. AU RCB's wikhcoveregual to and less than 10' shall be designed far two conditions:

a) 2 feet cover with HS2U4 live load b) 30 feet of covet (Se BDS 6.4.4 for live load distribution)

If loaded construction equipment passes across an RCB when the cover is less than 5 feet, temporary cushioning and possibly struts may be required (see Standard Plan D88$, or the roof shall be designed for the construction equipment loading.

Sx the m t A.RE.A. specifications for the design of reinforced concrete box culverts with railway loading.

The?-foot liveload surcharge,formerlyadded on top of thelaleral load tosimubtehjghway live laads, is no longer applied because of the more conservative design resulting from the two bands of earth loading.

4. Impact lr) Apply impact only to the mf slab of R- culverts.

Railroad impact may be much Ia~ger than 30% and is determined by the fomu k givm in A.R.E.A. Spediwtions,

~ E D ~ ~ w M * . 1 inplrt:!

Sectibn 6 - Undergmmd Stmdures Page 6 9

0 s 7 feet

1 5 2 feet

2 5 3 f d wer 3 fW

30% ~P/o

10% - 0%

c* I Bridge Design Practice - December 1992 =

Tramverseexpansion jointsnteusuallyprovided at intmalsin the roof and walk to control s M g e macking and to relieve stresses caused by differential settltlement. 5ee Standard Plans for application Deep or varying fdls may generate tension forms along an RCB as the foundation compacts; therefore, tenslon continuity is maintained in Ihe invert so that it will net pdI apast or displace v e r t i d y and permit xour or flow obstmction.

Culverts with shallow cover in saturated ground, such as storage boxes far pumping p h t s , should be checked for buoyancy.

Xn the m e case whena head of water mexist (as ha s3pbncendtKon), hydraulic pressure inside the cells of rigid frames will oppose the wall moments (due to earth pressure). Where cover is shallow, tension may occur amoss the top of she roof.

S e i c forces are normally not applimb le in s o ~ - s t r Y ~ interaction systems.

In genera]., wind loads,cmtrifuga3 forces, mdlongitrrsinal forces from highway or d o a d traffic need not be considered.

Parapets, projecting above the roof, SFNET as low barriers to resimin loose earth or other deb* from falling onto the chanrtel. See Parapet D e W on Standard Plan D82

On RCBs with skewed ends, parapets also serve as edge beams to support the antilevered end of the transverse moment. The beamdepth indudes the roof thickness. To avoid an unusually high parapet, consider several possibilities:

hgthen the RCBs lwen though the extra deck is not needed for ground conditions.

* ~gnlngLfremah~t~ltobepac~ltQtheskewedendsoftheRCBdvertwill~ the design frame length, the slabs will be thicker, and the m o u n t of reinfozwment wiU inuease. Care is necessary to modify details which will be incompatible with CaIhans standard sections. Skewing rebar should be a finid resort.

NOTE: The area that loads a skewed parapet is shown shaded (wheel loads are generally the major laad):

Tmtrsverse bats 3

E ace

Page 6-10 Sectrectron 6 - Undergmud Smdufes

c* - Bridge Design Practice - December 1992 m

W h m pzrraps serw as vek& barrim, & sure the reinfoment is suffiaentl y anchored into the culvert roof to transfer the impact forces. Ifslirmps are requited, embed them adequately into the mof. Finally, consider the torque applied to the roofend by live load impact. See "Barrier Section" and "Parapet Detail" on Standard Plan D82

When extending existing boxes (with skewed ends), it will be necessary to consider if the parapet is a supporting b e a m 4 not =we the proj- portion without providing replacement support (during construction). Sometimes, the simplest solution to extending a culvert wit ha skewed end, is to leave theexistingparapet inplaceand add acomsponding parapet on the abuning end af the extension. See cdvert extension details on Standard Plan DB2 Negative moment reductions do not appIy.

D. Design Analysis

I . AASHTO (Ref. BUS 3-22]

Group X

y = 13 Gamma Factor & = 1 .O Beta Factor for dead load of conmete & = P .O Beta Factor for earth presure & = 1.67 Beta Factor Live b a d D = Dead Load E ==Load L = Live Load I = h p a a

a) Formula Derivation:

For simplicity of appliotion, in RCB &ert design, the gamma factor has been m f e r r e d to applied loading from the existlng formulas,

AASHTO, modizied, now becomes:

@ is Strength Reduction Factor, also hewn as Capauty Redudion Factor. (Ref. BDS Attide 8.1 6.12 and 17.6.4.5)

+ = 0.9 for flexure i$ = 0.85 for shear

- ' Bridge Design PtaetIe - December 1992 m

Y - 1.3 - - - = 1.45 for flexure @ 0.9

y 1-3 - = - = 153 for shear 0.85

Average = 145+ 1.53 - = 1.49 - UE f 5

L

Group X = 1.511 .OD + 1 .OE + 1,676 + I)] = 1.5D+ 1.5E+25(L + I )

which is the formula shown on Standard Plan D82

b) Dead Load (D)

Concrete Density = 150 pd

c) €am Pressures (E)

140 @is the h4tu soil density as o h m m i in Calttws c a l v e tesearchpmjects. Note that AASHTO values a different.

. .

d) Live Load and Impact (LL + I) HS2044 Truck Load.

e) Structural Analysis

Loading 1 Loading2

Conventional moment-distributionwi1l be applied to the redmgulat uoss-section of a box culvert.

PesignFomulas(SeeBPSbrti~le&.l63)

The design formulas are:

Reaangu;lar section, no compressive reififo-t.

Vertical

140

, 740

(I) Ultimate Concrete Mmeni Strength

kteral 42

140

Page 662 Sectron 6- Undmgmnd Sltudures

- c* I Bridge Design Practice - December 1992 =

where a = fyAs 0.85fc%

Substituting

(Formula 2)

(Formula 3)

As Reinforring Ratio p = - bd (Formula 4)

(2) Sted Reinforcement

(a) Balanced Steel Ratio

Rectawph section with tension reinfoment only

f', = 3.25 ksi fy =€Oksi & = 0.85, for f', 1 4 ksi & =0.80,for4ksicf',15ksi for Caltrans' basic stresses, pb = 0.023

(b) Maximum Allowable Steel Ratio

P- = 0 - 7 5 ~ ~ p, = 0.0174

(c) Minimum Allowable Steel Ratio

(Formula 5)

Section 6 - Undergmund Structures Page 613

- -

Bridge Design Practice - December 1992 =

d) Design Equations

Calbms basic stiesses for cast-in-place reinfmed cmmte boxes are:

f't = 3.25 hi 51- 60 h i

Concrete Design

M, =5&d(1- l l p ) 4 =pbd;andb=12h M, = 60d2 p(l - l l p ) in. foot-kips /foot

Solving for d

Page 6 1 4 Sdim 6 - Udergmund Sbwtures

c* - Br3dge Design Practice - December 1992 =

(1) Steel Reinforcement

Design A,

M " 4 = 0-25- d

(2) Shear

CaE- allowable shear s h s s

vc= 3.5 JS;

Without m p s v- mind=- bVc

With stbups

V" = V"bd = (vm + VJbd

max V, = 4 E + V, = 4 E + 35e= 75E mula la I 21

6- URdwrgmwrd Strudurar P a p S f 5

A44 8 Bridge Design Practice - December 1992

E, Design Criteria

f , Loadings (Note: h e * design pressures not shown)

Symrnefriml about G

16k I 16k

Loading 1 1 - I Loading 2

Candirtion 1 : 2' Cover

Symmetrical about%

Loading 1

Page &76 Section 6 - Undwgmund m u m s

c* Bridge Design Practice - December 1992 =

2. Moments

WE? Slabs: FEM = -

12 Ph2 Aph2

Exterjor Walls: F E S = -+- 12 30

3. Uve Load Distribution

a) Load at Grade (root)

HSZO w h l loads are applied directly upon the conmete smf. Thewheel is co11cwmted in the direction shown and spread wmly along a distance E (wheel distribution on slabs) longitudinally with RCB.

(h + 30% Impact) Load Fador = 1.3x16x2.5=52k

s-ng grwnd is nof shown

The above depicts the usual situation where the direction of traffic msses normal to rulvert. Therefore, the traffic travels roughly p d e I to the main reinforcement.

Sectron 6- Underground Structures Page 617

w I Bridge Design Practice - December 1992 =

b) Load at Grade Qn\nert)

16 k wheel load is "distributed" (causes soil bearing resistance) under the area klow, without impact.

roof

1 invert I

w = Pw xLoadFactor...inksf 7 x width

- ; ~&p?'*, "" . , <

W Width& . '?F, A : >i4

16 k Under 14 fi 5.7 -

width Icsf

32 k 14ft.tolessthan28ft - lL4 ksf width

36k 129 kd

28 ft. minimum - width

c ) Load Distributed Through Fill

When the depth of fill is more than 2 feet, concentrated lirug loads shall be distributed over a square, the sides of which shall equal 1% times the depth of filL ( S t x B E Section 6.4.4). If multiple lanes are encountered each load s h d d be calculated to determine controlling (maximum) load. (See BDP example pages 6-32).

Page 6.18 S&bn 6 - Underground Structures

- Bridge Design Practice - December "592

-

4. Moment Envelopes lkubie RCB (Units in pcf)

(1 ) When LL applies (2) When U i r a p p t i i e (mtianvanes)

Invert m n t s ate maller than Invert m m m B am Larger mi mments, for k w c~ver than mf mments

5. Sheer

D + E pmmm ( c o m e plus earn, B any) w Critiil shear phnes

8 'd d i s w from lace d supprt A

emring pressure (with D + E + L)

Exlerior call has patest strears, usmlly

invert

=* I Bridge Design Practice - December 9992 W

F. Design Example

Double 12 x 12 RCB Culvert with 10 ft cover.

Notes to Designer:

'I. Use the number of dcsigncydes asnecessary to determine the maxbnumpossiblenqptiw distributed end moments and maximum possible positive midspan member moments.

2 When the depth of earth cover is 10 feet or less, all RCB mlvests shall be designed to meet the requirements of Bridgt Design Spm@tions, M d e 6.4.1 for the following two conditiom:

Condition 1,2 f e t earth cover. Condition 2,10 feet earth cover.

3. Use "Load Factor Designp' only.

Page (i-20 Sectr-on 6 - Undergmund Stmdures


I. Loading Cases te Consider

Condition I - Loadings:

Condition 2 - Loadings:

S W o n 6 - Undemwrd S t m d m Page &Zt

c* 1- Bridge Oesign Practice - December 1992 E

2. tmds - Conditibn f - (2 i?. Cover)

Assume Member Thickness:

Check Standard P h 'P-82" sheet and make best guess.

say t, =rain. & ='loin. tExWd = 11 in. t,, = O B h

Factored toads:

P, 1 6 k S w k I = L=-- E - (4 + 0.06(12'))

x 2 5 = 8.48 Wft

Pressures:

ca - Bridge Design Practice - December 1992 1

one wh-1

DL: (2x11+8)x12112 ~ 0 . 1 5 ~ 1 . 5 = 0 . 2 5 k ~ 1 wd(lzn,2+ 12 12

W d = - 0.61 - w,, = 025 + 0.61 = 0.86 bf

Walls:

up6

Member Lengths:

W Bridge Design Practice - December 1992

O.'l40 a kcf

Loading 1 b --%

Loading 2a

Loading 1c

NOTE: Dierent shaped RCB d& wiU regrrire different loading combinations to acquire the maximum design mommt rneEopes.

Page 6 2 4 S d o n 6 - Undwground Shvdm

c* - Bridge Design Practice - December 1992 - -. hy%y* , , , y<&L.. . +> .<I.* - *--. .+ ,,,A ,, _

> , ~;2r;..fy+h,~;~: ;\:+ Tw<++,> . ..- . + y - - - ,. 5 k.. ..: ,,, ,, ' .? . :*:<%<*: -

Roof (L@S/2)

Pt wlZ -+- 8 12

- - 10.17(12.79) + 0.61(12.79)2 8 12

= 16.26 + 832= 245 1

& @ 1/31

4 wcz Lt -(BS)+- 27 12

= I P Z + 832=27.9% - 2 w12

RL -{Pt) +- 27 12 = 9.64 + 832 = 17-95 1

(L6UJ3) Span I:

Lt = Rt above = 17-95 'k

Rt = Lt &toye = 27.59 k

span 2:

4 e t 5 b = -(I279 8) - 14 = 3.05 ft

3 a = 1279-3.05 = 9.74ft

PbZa wlt3 Lt-+-

t2 12

f

10.17(3.03z x9.74 + 832

12.792 = 5.63 a 832 = 13.95 'k

pa3b w12 Rt -+-

C2 12 = 17.99+ 832 = 2630 l c

,, 7? <q;+ ~ y - ~ ; , , , , .',<,

' , ->A, 22'""3M,:. \k

i.

Rook ( L @ 1 / 2 )

Pt wlf -+- 4 8

= 2 CEE%)+ 1 5 rn%) = 3252 + 1248 = 45.00 "k

.(t @ C/3)

2Pt wtZ - +- 9 9

= 28.95 + 11.08 = 40.00 'k

(L d 2/31 Span 1:

Sameas@L/3=40.(30 'k

span 2

Pab 5 / 2 - a - + L 5 ( m ) h - ( 1 LS2 r} - - 10.17 x 9.74 x3.05 +1248

12.79

= 23.62 + 9.07 = 3269 k

c* Bridge Design Practice - December 1992 D ..a

Page 6-26 Section 6 - Undergmund

I < '..'1 < < , ' 'Xrn'' -,,... -.

d %'

walls: (1M)YB)

Aph2 ~ l l ' +_ top: - 12 30

= 0.50(12.83)~ + 2.70('12.83)' 12 30

= 7.00 + 14,81 = 21.81'k

PEL Aph2 b b : -+-

12 20 = 7.0 + 1.5 (14.81) =7.0+2.22=2922%

(30%)

top: 0.3 (7.0) + 03 (14.81) = 21 + 4.44 = 6.54 'k

b m : 2'1 + 03 =Zf +6.67=8.771

h e * (one w M )

W,P wDt2 +- 12 12

- - 0.22(12.79)2 + 0.86(12.79)' 12 12

= 3.m + 11.72 = 14.72 'k

(two wheels)

Z(3.00) + 1 I .R = 17.72 'k

' '<?," :' b'y: .mJ&!>< , ,

; *,;, . . L < - . -A: < ,* ,>,p<qsK$;+i$ I,,: - > ,

,J<': ? 2 . L .

Wails: (lW/a))@ midspan

p,t2 Aph2 - +- 8 16

= 1.5 tFEM,)l+ + 1.5 (FEM,-)

= 1.5 (7.00) + 1 3 (33_37) = 10.50 + 1.25 (2222)

= 10.50 + 27-77 = 38-27 'b

(30%)

03 (10.50) + 03 (27.77) = 3+15 + 833 = 11.48 "k

lnvee (one Wheel)

= 15 (FEMJ + 15 Q?EM13)

= IS (3.00) + 15 (11.7'2)

= 450 + 1758 =-

(two wheels)

2 (430) + 1758

= 9.W + 17.58 = 26.58%

c* - Bridge Design Practice - December 1992 1

Relative S i i f h e ~ )s PmportionaI to t3/I:

walls: 1f3/1283 = 103.741 103.741 57% 575%

Moment Distn'bufion

Loading 2A

SBM = 1 2 W k A -1 6.74%

Section 6 - Undegmud Wctwes Page 627

4 ] -50 .50

8.32 -8.32 5.50 0.00 - 0.00 2.w

1 -94% - rd I

* ~ymmetri~al ~n

2.1 5 - 0.m 0.00 1.07 w 0.47 - 0.00 -

f 6.74 'k 4.35 'k I

1.9 .w,

- 5

CD

I

y: 32Fk

C

LC! -tt.72 1 1 -72 -

N

-7.53 t - 0.00 0.00 -3.76

-1 -65 SBM = 17.58'k -21.57% - 0.m - 0.00 -0.83

Ed I Bridge Design Practice - December 1992'

Loading Ic (unsymmetrical live load)

-1 8.41 'kc -18.Wk

U SBM = 45.Wk F z V) 7

r

Y

9.07 k 7

Page 628 S&on 6 - U n d d p x d Strudure~

c4 I Bridge Design Practice - December 1992 k

Loading 2b (symrnetriml live load) 4

Check distance betwm wheels + d = - (12179 ft) = 17.05 ft 1 14 tt 3

Section 6- U n d e m s1Ndmes Page 529

SBM = 40.0% N

-50 -50 w N

-2 1.81 3 2 9 -3.31

1.89 -265 2 'k

0 140 - 27.59

17.15'k -1 7.95

-2.49 0.00 0.00 -1.24 Symmet-1 h

1.42 0.00 7 G a - z - - 26.52'k -19,Ig'k a

m I

kct r m

-50 .50 eo f

n SBM= 26.41'k 1

2922 -1 7.61 17.61 -15.114k -6.62 -4.99 0.110 -27.89% -7 -65 0.00 -2.50

c* - Bridge Design Practice - December t992 =

Loading 1 b (unsymmetrical l i e load)

Page 6-30 - S d o n 6- Undergmcd STWures

- 1- Bridge Design Practice - December 1992 =

3, L o a d s - Condition 2 - (50 ft Cover)

ksyrnmetr ica~ h u t P

L d i n g 1

Where:

LRF = Lane Reduetion factor (see BDS S W 3.12.1) H = l O f t M C w e r L = UniformLivehd I = Impact, ( w h H > 3.0 ft, I = 0.0) D = DeadLaad

Section 6 - Undwground Sbvdumi Pege 631

=* 1 ' Bridge Design Practice - December 1992 =

Ptesures (factored):

= two h c k s = ldk(2)(2)(2lanes)2.5

= 0.30 ksf conhols (1.75H t. 1 6)(1.75H + 14)

=threetrudEs= 16k(2)(2)(31anes)(2.5)0.9

= 030 ksf (1.75H + 28)(1.75H t 14)

wtmf = wE + wD = 229 ksf

Walls - {outer Q center line of support dabs)

c* - Bridge Design Practice - December 1992 =

Member Lengths:

Loadings:

(max + M mf 4% invert)

Loading I b

0 . 1 4 0 z ~ &0.140 kcf kcf

- . 1 ~ ~ i . m *

kcf kcf

--G

(ClnsymrnetrhE L) (ma + M mf)

Loading t c

( max - M ext. wan)

Loading 2b

Ststion 6 - Underground Smrdums Page 633

c* - Bridge Design Practice - December I992 m

w lZ w,tZ L+.- 12 12

= 4.09 + 3122 = 35.31'k = 6.13 + 46.83 = 5296"k

Walk- llOQO&

P,h2 Aph2 top: -+-

12 30

- - (2.19)(12.83)' 4- (2.70)(12.83)' = 1.5 (FEM ) + 1-25 (EM,_)

I

= 30.04 + 14.81 = 44.86% = 1.5 (30.04) + 1 2 5 (2222) = 45.06 + 27.78 = 7284lC

= 311.04 4 1.5 (14.81) = 3.04 -+ 222.2 = 5226'k

130%1

tap: 03 (30.04) + 0.3(1481) = 9.01 + 4.44 = 13.4Sk - - 03 (45.06) + 03 (U.78)

= 13.52 + 833 = 21.85'k btm: 03 (30.04) + 03 (2222)

= 9.01 + 6.67 = 15.68'k

Invert:

w 1' w,L2 -+- 15 (4.09) + 1.5 (34.63) 12 12

= 6.14 + 51.94 = 58.08'k

= 4.09 + 34.63 = 38.7l.k

Page 634 SeEtron 6- Undergraud Strudures

c* ! Bridge Design Practice - December 1992 =

Distribution Factors:

Relative Stiffness B Proportional to t3 / I:

Roof: 1@/f279 = 78.186 43%

Walls: 113/1283 = 103.742 103.741 57% 5 W n

Mamen t Distribution:

The following load cycles are the mbthwrt number teguired to fmd all of the rontrohg member moments. Use more load cycles if warranted.

Loading Za (symmetrical loading) U Only

c* Bridge Design Practice - December 1992 =

Loading 2 b (symmetrical loading)

, -", p, - -220 I

* - 41.39'k 4- -32.44'k

17.38'k

5226 -38.71 38.71 -7.72 -5.83 0.00 -3522'k

SBM = 58.OB'k - 2.71 0.00 -2.91 4 6 . 1 8'k

-1.54 - -1.17 - 0.00 1.10 0.00

- -0.58

Page 6-36 Sedon 6 - Underground Structures

- c* - Brldge Design Practice - December 1992

loading 1 b {symmetrical loading)

SBM = 52.m b

F 7 -

SBM = Se.Ds*k rri

-25.11% F;I

S d b n 6 - Underground m u m s Page 6-37


Loading 1 c (unsymmetrical live load)

SBM = 46.85k

-

Page 6-38 S&i~n 6 - U d e r g r ~ n d W d ~ e s

=* Bridge Design Practice - December 1992 m

4. Section by Shear Note: Do not check shear in roof for 2 A covet condition, see Bridge Design Spedfic~1tiens 334.4. Shear is controlled by 10 ft earth cover.

Roof and h e r t

'imm YI= 2 + d = 13.5 in.

Rmk V, = [y - WX] + ; DEM = Distributed End Moments 4!

S d o n 6- U d e - Shdufes - Page 639

. \ Bridge Design Practice - December 1992 1 i

Watt:

v 20-63 knd = - - - 8.60 in. > 0.5 in. .-. NG

bv, 12(02) :. Need to inwase wall thickrtess or add shear reinforcement in actual design practice. The required increase in wall thiclness or the addition of shear ~einforcement will be ignored in this design example, which js for illustrative purposes.

5. Rein forcement by Ultimate Moment

(Note: A, units are sq. in. per foot):

walI!roof

walfimvert

invert

invertlwall

2

2

2

2

2b

2b

l b

l b

0 fbp

Q btm

centerline span

8 centerline RCB

2282

41 -39

46.18

45.40

c* I Bridge Design Practice - December t992

Designs:

f; = 3250 psi; f, = 60 ksi

0.2.5MU = d

"d d,, =dLZM. Roof:

mckness is controlled by negative M @ centerline RCB)

:. use: Wanti 4% @ 81hh z ~ x (space 634% ih max 0.c)

(0.6 +Q-44X22) A = = 1.47 > 139 8 5

.-. use #4 and 87 @ 8% h in. ((spa= @ 4% in max Q.c.)

Wall:

chess controlled by negative M B btm)

c* Bridge Design Practice - December 1992 m

.: #7 and #6 @ 8% in. max (space 42 4% in max o-c.)

0*31(12' =0.8!3<0.90 by abut 2.7% Say OK :. % difference c 5% 4.25

:. #7 and #6 @ 0% in max (space 8 4 h in max ox.)

(0.6 t 0.44) k = =L47<L54 by about 4.6% Say OK :.%I difference < 5%

6.5

Invett:

(Tiidmess controUed by negative M, wall)

0.25(22.82) +ve mom. 4-5 = 0.76

7.5

:. use #4 and 67 @ 8M in. mw (space O 4% in. max o.c)

:.#7 and #6 @ 8% in, max (space @ 4M in max o.c.)

AS= '0-6+0-44'(12) =L47 < L54 by 46% say OK :. % difference c 5% 8.5

Page 642 Sdim 6 - Udwground Sdnrdm


Notes To Designers

I. The use of reinforcing bars larger than #8k should be avoided. The bending radius mqujred for a #9 or larger is very large and would require large conaete corner fillets to mahain tk required design "d" .

2 When miember thidmss changes are greater than ?4 of an inch, the design calculations should be secyded.

6. Distribution Reinfomement: (Ref. BBS 324. lo)

100 100 Amount = - =- = 28.9% < 50% max 6 d z i i

(for 2 ft cover) x 0.29

Find: #4 req'd

A . span #4mq1d= a x - .= 7.5 A,, 2

:. use 8 #4 over middle half span

Design Tempesatutelshrin kage Reinforcement

&=%h = 0.125 in. ' PIX 820.1)

- #r2 a 18 in. = 0.20 h2(12 in) 18 in

= 0.133 in. > 0.125 in. 2 OK

Sectrectron 6 - Underground SlrYdums Page &#3

c* ' Bridge Design Practice - December 1992 1

7. Lead Moment Envelopes: Assume:

1. If the moment disbibution was done for d possible live load loadings dong the roof of tk RCB Culvest with 2 feet earth cover or less, the maxhum Live Load moment envelope would be a paraboh.

2 The -urn: positive nomwts are at midspan- 3. The moment envelope for the exterior walls is a pmbola. 4. We uw h e foUowing information from a parabola to plot 2 additional points of the

&um and minimum moment envelopes at the '/r span:

8. Reinforcement Sengths

8ar Extensions: (See BDS 824)

Page iW4 Section 6- Lld- SrWums

1- Brldge Design Practice - December 1992 m

Determine Negative Mornmt Reinformrnent Cut* Lengths

Rool and Invert:

#6 8.5.; M = 5 (0.62) 7.5 (1 - 11 (0.0069)) = 21 -48 fi--kips

rrs a, n c a.s: M = 5 (1.47) 7.5 (1 - 11 t0.0163)) = 4523 ft-ldps

Wlllr:

Negathn Moment Envelope Scale: (Moment) 1 ' = 30 R-lops

S m t t i c a l a h ~ t of RCB -I

d #6 15 bar dla. #7: 15 bar dia. 9'20 #6: 4, 4

Section 6 - Underground StrUchlres Page 6.45

- ' Bridge Design Practice - December 1992 =

Calculate Bar Cutoffs (corners)

Roof:

# 6 8 8.5 in. : 0.13 (L] + Id 0.13 (1279) (12) + 18.52 = 38.47 ih 036 (L) + 15 bar dia 036 (1279) (1 2 ) + 1 1.25 = 66.50 in. t controls

I = 6650 + 55 t -, - clearance = 66.514 + 35 (11) - 2= 70.W in. use1 =5ft-loin.

#[email protected] in.: td=2526h 0.13 (L) + 15 bar dia 0.1 3 (1279) (12) + 13.125 = 33.0 in t conPols

Wall (top):

# 7 @ 8 5 j n : tobemtinu0~5(lapspke45bar&a)

# 6 @ 8 5 h : td=1852in. + c Q W ~ S 0.03 (L) + 15 bat dia 0.03 (1283) (12) + 11.25 = 15-87 jar.

Wall {Softam):

#7@&5in.: toE#mtinuous

#6@85in.: Ld=18.52in. 0.05 (L) + 15 bar dia 0.05 (1283) (12) + 1125 = 18.95 in 5 conlmk

c = r ~ . 9 5 + ~ t,,-c~- = 15.95 .t % (10) - 2 = 21.95 in. use L = 1 f t - l ob .

c* - Bridge Design Practice - DecembeF 1992 m

invert:

#[email protected] in : 0.12(L}+ td 0.12 (1279) (12) + 18.52 = 36.94 in. 0.32 (L) + 15 bar dia 0,32 (1279) (12) + 1125 = 60.36 in, t conuols

L=6036+ Mtd-ciearanee 6036 + !h (11) - 2 = 63.86 in. mI=5fk-4in.

#7@$.5inr fd=z26in. 0.12 (L) + 15 bar dia 0.12 (1279) (12) t 13.125 = 3 1 9 in. t controls

=* I Bridge Design Practice - December 1992 =

SymmetriFal about RCB

Partial Section %. = 1 '-OW

Note: The bar reinforcing details shown in this design exampleis slightly different than the steel details shown on Standard P h D81.

Page &# Section 6 - Underground Strudwes


Appendix

AREA. RailmdLnads

Bridge Design Specifications Manual, (BDS):

NOtations Combinations of h d s Disteibution of Laads and &ign of C m c ~ k Skbs Bearing Capacity of Foundation So& Dead Loads F o o q Distribution of -1 Loads Through Eatttr, Fills Notations Concrete WnfMCerne~t

Stsength Deign Method (Load Factor Design) ~ ~ ~ e m m t of R e d M e m h S M d a g e & Temperature Reinformerit Spa* Limits For ~ ~ e n t Pmbxtion AgainstCoffosion Hooks and k d s ~ e v e ~ o p ~ ~ ~ t of nexural~0-m Development of Deformxi Bars in T d o n Splices of wom2mmt General b a d Factor Design Reinforced Concrete Box, Cast-In-Phe

Highway Design Manual:

Standard Plans: (1992 Ed.)

D80 Cast-In-place M o d Conaete S h g k Box Culvert D81 Cast-h-Place Wc1fied Caurete Double l3ux Culvert D82 Cast-h-Place W o m d Conmete Box Cdvert Miscelhmm Deb 5

c* I Bridge Design Practice . February 1993 m

Section 7 . Bridge Design Aesthetics

Contents

. .............. ...‘............... .. ........ 7.1.0 Preface - ..- .....- " ...........-w..............w.....a....m......m............. 7-1

7.1 . 1 Definition ........................................................................................................................ 7-1

7 . 1.2 Scope ............................................................................................................................... 7-1

7.2.0 Introduction .................................................................. ........................................... 7-1

7.2.1 General ........................................................................................................................... -7-1

............................................................ ..................... 7.23 Psojec t Development Team ,.. 7-1

7.2.3 Design Philosophy ................................... ........ 7.2.4 Method ............................................................................................................................ 7-2

7.25 Results .................................. , .......... .... ........................................................................... 7-2

.. 7.3.0 Bridge Design and Aesthetics ....... -. .......,.. ..-... ......m...,,....n.n...nn.. 7-2

7.3.1 Theory ............................................................................................................................. 7-2 ............................................................................................................ 7-32 Structural T y ~ e 7-3

7.4.0 Type Selection Meeting ................... ......,..,.... .... ,... 7.4.1 Prepamtion for Type Seleic tion Meeting .......................................................... ..-.... ... 7-3

7.42 Public Meetings ............................................................................................................ 74

.................................................................................................................... 7.4.3 Visual Aids -74

7.5.0 AestheticFeatureGuidelines ..- .... .... ............* .....*.... ......... 5

7.5.1 hilings ........................................................................................................................... 76

.................................................................................. ................... 7.5.2 Girders and Decks ... 7-7

7.5.3 Columns .................... .. ............................................................................................ 7-10

Section 7- Cantents Page 7-i

el* . Bridge Design Practice - February 1993

Aesthetics for Seismic Retrofit ................................................... , ........................ 7-15

General ............. ... ....................................................................................................... 7-15 Steel Column Casing .................................................................................................. 7-16

Fabric Wrap Casing ......................... .... ....................................................... 7 - 2 4

..................................... Cable Wrap Casing ............................ .... ...... .. 7 - 2 4

New Replacement Columns ................................... ..- 4

Welding External S tee1 Plates and Tie Rods .......................................................... 7-26

Bibh ography ....................................................................-........................................... 7-26

Section 7 . Con tents Page 7-ii

! Bridge Design Practice - February 1993 =

Bridge Design Aesthetics

7,l.O Preface

7.1.1 Definition AestJretics - The science or study of the qualities of beauty, including surrounding light, shadow, and color not limited to physical f o m .

Benuti@l Bridg~ - A beautiful bridge makes a minlmaI impression on the environment, has good proportions both in its integral parts and in the space outlined bp itsparts. It is composed of one dominant structural systemusing a minimumnumber of bents witha ninimum number of columns per bent. Size, shape, color, and texture on superstructure, colwmns, and abutments are utilized to either call attention to, or play down, the role of these sbctuta l parts.

7.7.2 Scope The inexact science of aestheticu is h i k d to practid application in bridge design where cost, construction, and maintenance compete with public acceptance. Consult t he bibliography for philosophy, history, and ethics.

7.2.1 Genera I

7.2.2 Project Development Team C A L M S has incorparated the design team concept into its n o d project development proc:ess. This team is composed of professionals from all disciplines participatingin the design process. Representatives horn engineering, architecture, environmental stuches, Jandscape architecture, and legal are directly and indirectly incorporated in the project development team. The Division of Sb.uctmes type selectionmeeting members represent the structural engineering function of the project development team.

7.2.3 Design Philosophy Design philosophy from top m g e m e n t to the design team encourages beautiful structural design while following the rule that "form follows function" The end result can be an aesthetically pleasing structure that is also economical.

Section 7- Bridge Design A esthetics Page 7-1

ca -: Bridge Design Practice - February 1993 m

The method utilized in most CALTRANS structure designs consists of modifying the standard shapes required to perform the shctural function. These modifications are achieved by using grooves, offsets, texhw, m d moss sectional changes.

7.2.5 Results This approach results in using a standard endosure which is basically rectangular in shape, both in m s s section and longitudinal &on for girders; and r m d to for columns. Only when we extend thrs k h o l o g y to its upper limit do we consider the need to reducemass. We p e r a l l y accept redundant mass in order to slmpltfy form w ork whch is the basis for our cost e f f e c ~ e design Concrete Is the primnry construction material. Special site or public relations problems requiw the designer to mamy architecture and engineering by using engineering diciates such as moment dia- to provide the basis for a design subject co skid cost/funchon uiteria.

7.3.0 Bridge Design and Aesthetics

Section 7.1 -1 defines a beautiful bridge for the purpose of directing bridge designers. C A L W bridge engfneers attempt to make every bridge a beautiful bridge. They produce advance planning studies, or the equivalent, to discussxskhetics with the Aesthetics m d Modeksection ( A M ) . A&M coordinates with bridge engineers using visual aids to identify areas of aesthetic concern and ta resolve these areas of concern. Interaction of A&M and engineeringpersennel is necessary as early as possible to coordinate Division of Stmchr~es' work with the Project Development team.

7.3.1 Theory Bridges affect their surroundings by their size, shape, and color. CALTRANS has determined bridges shodd be a good neighbor. They shodd be compatible with their environment and blend rather rhan make a strong statement.

There are two basic positions for viewing a bridge: (1) the position of the bridge user and (2) the position of a viewer looking at the bridge from a location to the side of the bridge.

The bridgeuser should be presented witha minhrrm of distractions. Therefore, thebest bridge is one where the user is not able to determine that he is using a bridge. The second position is generally dependent upon the elevation or profile presented by the bridge.

The profileview desaibes ~ s t r u ~ typesuchasarch, buss, girder, sqmsion, orstayedgkder. S o d stmchrral design (I) function, (2) and appropriate aesthetic treatmenf, (3) have been establisM as the order of priority. The first priority must begin with geology to B e t e m where supports may reasonabl y be Em ted. The structural type is therefore dependent on possible support in order to dde termine the span 1engLl-s. T&noIogy has made j t possible to have s ing lqan lengths of over200feetinconmte.Phisbasecanbeexpanddbyusingsteelandadaplingacontinuousspan procedure to extend the limit. Zncreasing the apparent height of the sbuctural tyFe by constructing a truss, combining a truss with an nrch, building towers, and suspending the deck by stays or suspension cables are other methods of inmasing span lengths.

Page 7-2 Section 7 - Bridge Design Aesthetics

c* 'I Bridge Design Practice - February I993 m

The problem of economically producing large numbers of bridges necessitates the i n p l e m e ta tion of a repetitive process. Basic methods and procedures must remain as simple as possible: therefore, the Ieast complicated method and procedure must be the starting point for selecting struc.haral type. Actual physical and monetary conditions modify this beginning toward an increasingly cornpBca ted problem.

A parallel exists in aesthetics. T k second and third priority may modify the choice of structuraE svstern. Bridges constructed to serve transports tion w o r t s are large stmchms. Therefore, the abpropdateness of the s h b u r a l system is the most important factor in bridge aesthetics. This factor can be smn from any position from which the bridge on be viewed. It is the bridge. The mfhitectcan bring out the aesth;.tic @ties of the shctural system but on never change its basic impression

7.3,2 The Strucfural Type

Aesthetic Considerations Each route should be composed of bridges with an appeasance compatible with other bridges on the route. Overcrossings should contah the aesthetic theme for the route. Undercrossings m y v q from the route theme to satisfy I d community requirements.

Rver crossings and viaducts are u d v not within the driver's focused viewing area; theherefore, they may also vary from the route theme. However, close r e d l a n c e to route theme bridges will produce a desirable unified appearance for the entire route.

7.4.6 Type Selection Meeting

The type selection meeting 5 a t t d e d by repmtatives born Spdicatjons, Maintenance, Construc- tion, kihating, Design, and Aesthetics and Models. The purpose is to provide a bridge with @ties which will sa tisfv all members of the Division of Sbuchms. The sedts of this meeting willbe "sold" to CALM. mi k t r i c t and the c o ~ ~ t m t are SO present for EX temdy w ~ e d Projtxts.

7.4.1 Preparation for Type Selection Meeting The engineer is in charge of producing contract plans. Aesthetics and Models acts as the engineer's comultant. Advance Planning Studies prepared by the engineers are reviewed by Aesthetics and Models. These reviews consist of selecting column type, girder edge treatment, and surface treatment. A preliminary architectural sketCh is drawn consisting of section, elevatim, and a rough perspective. Cost estima tes are prepared, and the suitability of structural dsign to architectural features and cost are determined before work progresses.

PIanning and recommendations by Aesthetics and Models begins with research of the area and route to be occupied by the bridge. These recommendations also incorporate directions or information gathered at public meetings.

Section 7 - Bridge Design Aesthetics Page 7-3

c* I Bridge Design Practice - February 1993 E

7 4.2 Public Meetings PubLicmeetingsmay be workshops organized by the District to produce facilities incorporating a style, theme, or artistic feature required by the community. Workshops are occasionally h e d d e d to insure that features required by the community are presented at public hearings. Public hearings are usually designed to present the community with a proposal or alternate proposals in order to secure a freeway agreement.

Public meetings may be scheduled either before or afker typg selection meetings.

7-43 Visual Aids Presenh tion quality ma berial to be displayed at public meetings or Dishkt meetings is available from Aesthetics and Models. Artwork ranging from simple sketches to photo retouches showing the proposed product in its environment is also available from Aesthe tics and Models. These displavs require from one week to two months t~ produce.

Modeb may also be ordered. The equivalent of a sk t& is a styrofoam model. Elaborate models which correspond to photographic retouches require six months to a year to construct.

Design sections may order visual aids by sending a memo ito Aesthetics and Models.

The memo should indude the type, size, and completion date required. Districts usually order the large elaborate models showing interchanges with several bridges within a community. This type of v h l aid, requiring relatively long construction periods, must be ordered by a letter to the Chief of the Division af S tnrctures.

Page 7-4 Seclion 7 - Bridge Design Aesthetics

c* Bridge Design Practice -- February 1 993 1

7.5.0 Aesthetic Feature Guidelines

Fi y r e 1 illustrates the bridge parts normally given aesthetic consideration This work involves scrutinizing the required structural shape with regard to appearance criteria developed by the Aesthetics and Models Unit from public meetings, District requirements, and aesthetic judgement.

Figure 1. The Parts of a Bridge

Secii~n 7 - Bridge Design Aesthetics Page 7-5

c* - Bridge Design PraetZce - Februaw 1993 =

Type 25 and Type 27 form the basis for our work. The only modifiation allowed is fo stain or paint the depressed or grooved area on the outside face of Type 25 (Figme 23,

Type 25 Type 27

figure 2

Texture or shape change ta the outside face requires the sailing to be designated Type 25M or Type 27M (Figure 3). The texture must be added to the basic structural section.

Basic ISectlonI Basic ~TZ-G-I

Type 25M Minimal Texture

Type 27M Minimal Texture

Section 7 - Bridge Design Aesthetics Page 7-6 -

Ed I Bridge Design Practice - February 1993 m

7.5.2 Girders and Decks a) Figure 4 illustrates the basic girder and deck assembly wherein the ghder and deck are

combined to form a sIab. Aesthetic consideration for slab bridges are generally confined to limiting the apparent thickness (t) of the outside edge of the slab. "t" should be approximately equal to the corresponding dimension for box girder bridges (Figure 5).

Figure 4. Slab Deck

Figure 5. Slab Deck with Sloping Edge

b) Bridges comtructed using steel and reinforced c o n g e te rely on the tensile strength of steel. These bridges do not require the massive abutments used for compression force structures. Our tension force structures demonstrate the physical ability of tension structures to be much thinner or require less depth than compression structures.

Bridges that appear to be horizontal constructions appear, from an artistic point of view, stable and graceful. Thin horjzontal bridges with a minimum number of columns are desirable.

Work in Aesthetics has focused upon applying the previous statement. Horizontal lines and shadows are the devices employed by the architect to produce bridge designs which have been labeled dean, functional, and honest.


c* ' Bridge Design Practice - February 1993 m

c) The exterior girder of the box girder system can be arranged to e h c e the perception of small depth or thinness.

Figure 6 . Shadows cast bv the overhang on the exterior girder place the exterior girder in shadow similar to the sh;de always present on the soffit.

Figure 6. Concrete Box Girder with Vertical Sides

The amount or depkh of the exterior girder shadow is dependent upon the overhang length, the sun angle, the exterio~ girder angle, and the relative northsou th orientation of the girder.

Maximum enhancement, thinness, is achiwed when the shade of the sofit merges with the shadow an the girder (Figures 7 and 8). The railing and edge of deck are the only elemmk remaining in sun (Figure 8).



Figure 7. Concrete Box Girder with Sloping Sides

Figure 8. C o n ~ ~ e t e Box Girder with Sloping Sides + (No Overhang)

Section 7 - Bridge Design Aesthetics Page 7- 9


7.5.3 Columns Standard architectural columns have been designed as a series of modular shapes. Cross sections are available in round, octagonal, and hexagonal form. These forms can be expanded to include a rectangle between the basic fom. A round aoss section can become a cross section that is semicircular at the edge with flat sides joining another semicircular edge. This system is used for all the geometrical forms to increase the load bearing ability of a single &lurnn Therefore, a minimum number of c o l ~ s can be used. The edge column directly adjacent to the viewer provides the impression of column width that the viewer normally perceives. This impression is controlled by light reflecting from the column edge. Octagonal columns appear slimmest as a result of the pea test number af surfaces. The viewer sees a large area broken up by several planes. Round columns are affected less and square or rectangular columns are not zdfected. ?'he architect can take advantage of this light reflection by using the principles to slim down a massive c o l m or increase the apparent sjze of a column to offset a massive superstructure. CoIumn proportions, therefore, have a large effect on the aesthetics of bridges.

Columns that appear larger than necessary to support the superstructure are not desirable because attention is directed away from the primary purpose of a bridge, which is to provide free movment. Columns that are obviously needed to support the supersbucture should appear to be of sufficient size to perform their function. Columns that appear thinner than the visual requirement impart the feeling of possible rollapse to the viewer.

The upper part of standard architectural columns is curved, arched, or !dared to visually integrate the columnwith the superstmcture. This spreading outward of the standard architectural column is designed to be cornpahble with the sloped exterior girder of a trapezoidal box girder (Figure 9). Standard, flared architectural columns are not compatible with vertical exterior girder shapes (Figure 10). A bansition between khe c a l m and the superstructure similar to the capital on classic style columns must be introduced (Figure 1 1). This "capital" usual y takes a simplified form involving straight Lines tap2ring in the opposite direr tion of the flare for a distance Iess than the girder depth The capital actually becomes an exposed column cap, or part of an exposed column crap. This treatment is effective only when the extremities of the flare are wider ham the superstructure.

Page 7- 7 0 Section 7 - Bridge Design Aesthetics

c4 I Bridge Design Practice - February 1993

Figure 9. Compatible - Recommended

figure 10, Incompatible - Not Recommended

Seetion 7 - Bridge Design Aesthetics Page 7- 7 f


figure 11. Acceptable

Vertical exterior girders should be used with prismatic columns (Figure 12).

Figure 12. Compatible - Recommended


c* - Bridge Design Practice - February 1993 M

Flared columns as descnied previously are designated as one-way-flare columns (Figure 13). They promote flow perpendicular to the bridge and under the bridge; therefore they are directional.

Figure 13. Standard Architectural Columns with One-Way Flare

Standard architectural columns have also been designed with two-way flare (Figure 14). These columns are nondirectional (they do not direct flow in a particular direction) and are particularly appropriate in situations &volving more than one bridge, such as an interchange. Two- way 5 r e columns are more effective from a visual judgement because the flare is evident from . . any viewing position.

Figure 14. Standard Architectural Columns with Two-Way Flare


wlIi Bridge 3esign Practice - Febmary '993

- . :e :owe: ?orti02 of z coI-;z:.zst co-ec wit:? ezi3, z.?-?-rzce zzierk:, or wzter. S:z-.d?zt zrChi:ecL~-2: cokr--s z3 k v e verticz: lower ? o i i o ~ s . A!::lo;tg:? 52s zzzy :.ot ?yovide 5 e 3es: . - . soikon zest:?eticzliy, it is 5 e o?Ly ?rzcCcz- so--t:o:: for z cob&--7 5 2 : =.-st co3e w-2- Feeat

. . C;?2ges :. .?eAg:?t wr:?i:e -ek-:-:-o - .c ' .:e s?.?le wid?. zt 5.e to?.

,. ^ L. co'.---s .'' . . .,.... .... .?er lower ?ortions 5 wzter rely 0:: t:?e str;c>:rz: cor:ditio?s for 5e:r s?z?e. P ...-- . . . . . . . LO.,.. ..s res5::g on q rezd foot-?.-.gs or :n cr;..ec :o:es czr: z??e?.r io dise??ezr >.to Lke wzter. - - --s 5 cls~:-ct zdvz?;zoe wiL? f..Lc~..-z:-n .- .-. -. .a .d..-., wz:er ;eve:s. .. . Co2;~zsk wzter s:??ortec o r . ~ . e cz?s rec;12e the ?i:e cz? to 3e 22 e:e?.en: L-. L:lek overz- cesigr:. ?Ie 5e?.ts ex?osed?s c o 5 - s - ere zsel o?Ly w5e: low cletrz-ce lics:es t3.e zse of K - 2 sc?erstrcc>ze. -nese si>cz5o?s zs;1?Ly occrrr i? zrezs of restrictee c:ezrzr:ce. - .:ere w X 2:wz.y~ 3e z -eel for -o-stzz??rt co>~z.?s. S?ec>.: s i2 c o ~ l i ~ o n s , slC? zs -. -. .. .. -.-. - - . .,,re -3, rec;',ire zr: overcrossLL?.g :O s?z2 i: c:v:cec r'ig3wzy w i 5 i: ce?ressio: 5. S-e ce2:er . - for xzss Z?zs1Si:. :?zicesr;z:e s?zce for oze k g e CO>AYZ ?reser:tel ::le ?ro>.ern of t;Yo 5-i:

-es.'L:"" :- -. , - - . . .. cO1..-- . , -..-., -. z 5=ee-s??-? str;c>;re. .:e sL;ckre ce>t? - wo--c x v e visz?:.y over- - .. ~owerec 5:e co2;rr:s. .2e resz2 w2s z --zssiveloo:&o s;?ers>:c?:re. Corn35i~g 7- 5 e . . a. ?ro3;ezs s?eciEc to ::?e s:?ers:zc>:re ?rokcec 2 ..".c;';e soklio.? where co>~?zs azc s..? ers.--c'..- . . .. ..- .,.e ???ear io 3e 5 ?ro?or%o:. w;:z ex:-. oz-.er.


7.6.0 Aesthetics For Seismic Retrofit

7.6.1 General The ideal retrofit would result in the retrofitted structure showing no change in appearance.

Real world conditions will probably dictate the use of seismic retrofit technology which will produce a change in the appearance for the retrofitted structure. This change in appearance should be minimized. Some retrofits will require additional structural parts. These functional parts should be integrated into the design of the original structure. Shape, texture, and color should be utilized to accomplish the integration.

All retrofit work affects the appearance of our structures.

The following excerpt from Memo to Designers 21-18 dated July 1989, was written to provide guidance for girder to column retrofits:

"Appearance of structures being retrofitted should be given consideration. "

Generally speaking:

a) If the cables are between girders, above the girder bottom flange, or are attached by means of small fittings, they are least objectionable.

b) If the cables are wvand aromd colmns or other st!-~cturaimenbers visible frorr. 2 position outside the structure, they are more objectionable @,@re 16).

Figure 16

Section 7- Bridge Design Aesthetics Page 7- 15

c* Bridge Design Practice - February 1993

C) If the cables are visible in silhouette and are obviously not a part of the major structural xheme, they are most objectionable.

All of the above is further influenced by the m*onment: sky, background, color, character, e tc.

It might be argued that precast concrete or steel beam stmctural types fall into a group entitled "articulated" and might be expected to contain hardware. This hardware should be minimized in size and prominence to retain its place in the general structural order.

General tidiness in detailing, a little paint, avoiding profile view contamination bv the '"sytem" and utilizing djps or secondary fasteners in lieu of cables everywhere, could heip preserve the appearance of s trucltures retrofitted by narrow-minded efforts of "restrain at all costs."

COMMENT: Place devices on interior girders if aU girders do not need retrofitting and structural concerns can be satisfied.

7.6.2 Steel Column Casing Three types of steel column casing are currently in use. For the benefit of this guide Tvpe 1 - Prismatic Circular (Figure 17). Type 2 - Primatic Elliptical (Figure 18). and ~ & e 3 - Formfitting (Figure 19) will be considered.

Existing Concrete Column Circular Steel Casing

figure 17, Prismatic Circular Steel Casing

Page 7- 16 Section 7 - Bridge Design Aesthelics

I Bridge Design Practice - February 1993 m

Existing Concrete Column Etlipt*wl Steel Casing

Figure 18. Prismatic Elliptical Steel Casing

Existing Concrete Column Form Ftting Steel Casing

figure 19. Form Fitting Steel Casing

Section 7 - Bridge Design Aesthetics Page 7- 1 7


a) Type 1 - Prismatic Circulnr should be installed from soffit to ground continuously, minus structural gap at ends (Figure 20). The idea1 use is on circular prismatic columns. Existing column cross-sections other than arcular will undergo an appearance change when circular-section retrofit is installed (Figure 21). The extent of this appearance change will be greatest when all the existing columns are not retrofitted. This latter scheme is undesirable.

Figure 20. Circular Prismatic Steel Column Casings Extending from Ground to Soffit - Recommended

Page 7-18 Sedion 7 - Bridge Design Aesthetics

el* - Bridge Design Practice - February 1 993 =

Figure 21. Partially Retrofitted Columns - Not Recommended

Section 7 - Bridge Design Aeslhelics Page 7- 19

c* 1- Bridge Design Practice - February 1 993 1

b) Type 2 - Prkmatic E l l i w l should be installed from sofit to ground continuously, minus s t n x ~ gap at ends. The shape is the mult of pmviding restmht withwt using tie mds on recbngdx or oblong column mosssertions. It is difficult to detemine the difference in appearance when this shape is compared to round or oblong columns in a d w (Figure 22).

Figure 22. Elliptical Steel Column Casing - Recommended


c* Elridge Design Practice - February 1993 =

c) Type 3 - Formfitting i s the ideal type for flared columns and columns with cross-sections which are not circular. Examples in use have been hi led to oblong flared column (Figure 23). Difficulty with restraint along the resulting flat sides has required the use of tie rods. Difficulty a f elliptical casing: on Elared sections to meet construction tolerance for transition horn prismatic section to flared section is also a problem. The ideal application would show no change in the appearance of the column after the steel casing is painted, This ideal will require the tie rods to be countetsunkand filled prior topainting (Figures 24). Analternative detail would involve placing reinforcing structure on the inside of the casing (Figure 25). Although formfitting has been determined to be more expensive than some other methods, it does the best job of providing undetectable seismic retrofit, espeaally if the proposed details to tie bolts are utilized.

Figure 23. Form Fitting Steel Casing

Secfion 7 - Bridge Design A esthetics Page 7-21

Bridge Design Practice - February 1993 =

Typicat Section

r bisting Column r Grout

Steel Plate Column Casing

St-! Cover Plate

Section A-A

figure 24. Proposed Detail for Eliminating Exposed Tie Rods

Page 7-22 Section 7 - Bridge Desbn A esthe fics

c* I Bridge Design Practice - February 1993 m

I Typical Section

Steel Casing Casing Steel on inside of Ribs

Section A-A Section 8.8

Figure 25. Proposed Detail for Elirnlnatin g Exposed Ribs

Steel Casing


Ed ' Bridge Design PractIee - February 1993 m

7.6.3 Fabric Wrap Casing Experimental scale modeIs show promise of providing seismic retrofit capabilihr to existing columw which will make appearance change in structural capacity difficult to detect. Addi- tional testing will be required to prove this method, which appears to offer an excellent method to retain the origrnaI appearance of our shuctures.

7.6.4 Cable Wrap Casing Plastic coated steel tightly wound or wrapped around columns in a continuous spiral appears to offer a method for presercring the silhouette shape af architectural columns. There is a viewing distance from which the individual coils will not be noticed. If this- distance is within the range of normal viewing in interchanges, the cable wrap seismic retrofit would be zlndeteaable for all practical purposes; therefore, this method would accomplish our aesthetic requirement. A workable solution is available and should be considered.

7.6.5 New Replacement Columns Generally speaking, asthet idy acceptable soh tions should b considered in situations w b architectural columns have been built. In these situations, retrofitting a single column bent with two new additional columns should be avoided (Figure 26). If this cannot be avoided, the new columns should be from the same archit&urd M y or gmup as h existing columns (Figure 27). Constructing a new s%le column bent and r e m m g the exhting is the best aesthetic solution.


el* - Bridge Design Practice - February 1993 m

Rgure 26 - Not Recommended

Figure 27 - Recommended


c* I Bridge Design Practice - February 1993 1

7.6.6 Welding External Steel Plates and Tie Rods Protection of Weak Colzrmns

Several methods designed to change the structure's dynamic frequency by changing the stifhess of its parts have been proposed by h4. YasWky. Please remember Section 7.6.1 General:

"The ideal retrofit would result in f he ref roJifted sfnictrrre shaoing no change in a ppearnnce. "

Personnel, as well as models, are available in Aesthetics to help you achieve this goal.

7.7.0 Bibliography

Bacow, A. F. and Kruckemeyer, K. E., BRIDGE DESIGN, Aesthetics nnd Developing Technologies, Massachusetts Deparben t of Public Works, Massachusetts Council on the Arts and Hmani - ties.

Burke, Martin P. Jr., P.E., Bridge Aestheti~ Bibliography, published by Burgess and Niple Limited, Engineers and Arcihitects, Massachusetts Department of Public Works,

Rimer, John C., Bridges Produced by nn Architectural Enginmikg Team, Transportation Research Record 1024,

Rimer, John C., Creating A Beautiful Concreie Bridge, Esthetics in Concrete Bridge Design, American Concrete Institute 1990.

Page 7-26 Section 7 - Bridge Design Aeslh~fics

Seismic Analysis of Bridge Structures

Introduction The purpose, of h s course is to give engineers the tools necesw to obtain &quake farces on bridges. A simple method is presented that rndels a bndge as a single degree of freedom system. As the bridge model becomes more complrcared, this simple pmzedure hecomes less accunte. Then. a multimdd dynamic analysis or time hstory computer analysis is recorn ended

There are two basic concepts that are presented in this course. The first is has there is a relationship between a bndgeas mass and stiffness and h e forces and displacements rhat effect the structure during an ezlrthquake. Therefate, if we can calculare the mass and the stiffness for our structure we can obtain the &quake forces acting on it. The second concepr is rhar C a l m ' s bridges are designed to behave nonlinearly for large eathquakes. Therefore. the engineer is requited to make successive estimates of an equivalent linearized stiffness to obtain the seismic forces and displacemenrs of the bridge.

The unics of measurement for this course are in SI, Sufficient information is provided in this section to do the assignment. However, strucmrd dynamics is a complicated subject and enpinews aa encouraged to read books and lake courses to improve rheir understanding.

Basics Ma!x is a measure of a body" mistance rn -1erar.m. It qnim a force d o n e Newton to accelerate one kilogram at arak of one meter per second squared In this course. we will calculate the weight in Newtons and divide by g. the axeleration due to gravity (9-81 rnlsec2] to obtain a bridge" mass in kilograms.

Stiffn- is a measure of a structuff's resistance to displacement. In this course, we define it as the force (in Newtons) required to move a stnrcture one meter. The boundary conditions for the bridge need to be carefully studted to determine the stiffness of the structure. We typically consider the stiffness of columns and abuments in our analysis. IF the stiffness of column footings or the bridge superstmcmw has a large effect on a bridge's seismic behavior, the bridge shouldn't be analyzed using the simple procedure given in this course. Jf the results of an analysis suggest thar a column may rock on its footing, simple seismic rockmg analyses can be performed.

Period is the time, in seconds it takes to complete one cycle of movement. A cycle is the trip from rhe poiah of zero dispIacement to she completion of the structures hnhest left and right excursions and back to h e point of zero displacement.

Natural Period is the time a single degreeof freedom system will vibrate at in the absence of damping orotherforces. Natural period (T) has the following relationship to the system's mass (m) and stiffness (k).

This is themost fundamental relationshipin structural dynamics. We will useit to obtain the earthquake force and displacement on bridge structures.

Frequency is the inverse of period and can be measured as the number of cycles per second ( f ) or the number of radians per second (a) where one cycle equals 2~radians.

Damping (viscous damping) is a measure of a structure's resistance to velocity. Bridges are underdamped structures. This means that the displacement of successive cycles becomes smaller. The damping coefficient (c) is the force required to move a structure at a speed of one meter per second. Critical damping (c,) is the amount of damping that would cause a structure to stop moving after half a cycle. Bridge engineers describe damping using the damping ratio (8 where

A damping ratio of 5% is used for most bridge structures.

Tne Force Equation The force quation for smmral dynamics can be derived from Newton's second law

Thus, all the forces acting on a body are equal to its mass times its acceleration.

,/ Location of zero displacement

- p ( t ) equals

I- fs -

When a structure is acted an by a force, Newton's second law becomes:

Where:

f, = ku h e f o ~ e due to the stiffness of the sthlcture .,.....,.......,.... ............. 15)

fD = a' the force due to damping of the smcture .......................................... (6)

and

.............. P is the external force acting on the structure ....,................ ... (7)

The variables up and u" and are the first and second derivatives of the displacemen1 u, k is the force required for a unit displ~emenr of the strumre. and c is a measure of h e damping in the system.

Thus, equation (4) can be rearranged as shawn in equation (8).

mu" -t cup + ku = p(t) ................................. .........................,............................... (8)

However, for earthquakes, the force is not applied at the mass b u ~ at the ground,

therefore, equation (8) becomes:

.................... mu" + c(ur - 2') + k(u - :) =g ................ ,,., -.-.----.----.-------... (9)

for the relative displacement w = u - z ......-.-...~.......ti.................3.... ..... ( 10)

and the equation of motion. when there is no externaF force p being applied, is:

In equation (I I), the mass m, the damping factor c, and the stiffness k, are all bowa. The support acceleration r" can be obtained from accelmogsam records of pevi ws earthquakes. Equation ( I 1) is a second order differential equation that can be solved to obtain the relative displacement w. the relative velocity w', and the relative accelmuon w" for a bridge structure due to an earthquake.

Calms ' Response Spectra Response spmahave been developed so that engineers don't have to solve adifferential equation reptedEy 10 capture the maximum force or dsplacement of their smcrure for a gven acceleranon record.

A response specfra is a graph d the maximum mpmm (displacement, velocity or acceleration) of different single degree of f&sm systems For a given earthquake record.

0 0.5 1.0 2.0 3.0 4.0

Period

Example Response Spectra

The horizontal axis is the system's period and the vertical axis is tk system's maximum response. A vertical line is drawn from the perid to h e spectm and a connecting horizontal line is drawn to obtain the respame.

Thus. engineers can calculate the suucture's period from its mass and stiffness, and use the appropriate 5% damped spectra to obtain the smcrure's response from the earthquake. If a bndge has a higher damping ratio, response specm at higher damping can be calculated.

The Force Equation showed three reqmmes that can be obtained from a dynamic analysis; displacement, velocity, and acceleration. We can also obtain them using nzsponse spectra The spectral displacement (Sd) and velocity (Svj can tK obtained from the specml acceleration using the following telationshp.

Therefore

C a l m developed response specma using five large California -quake p u n d motions on w k Twentye~ght d r f f m t specm were m e d based on four soil depths and seven pealt, ground accelerarions (PGA). Therefom, engineers can obtain the &quake forces on a bridge by picking the appropriate response specaa b& on PGA and soiI depth at the bridge site and calculating the natural period of therr structure. These response s p m a wn be found in Cal trans' Bridge Design Specflcations and in the Appendix at he back of this section However, C a i r n is moving towards using site specific response specm for many bridge sites,

Nodinear Behavior

Bridge meml m change sWness during earth- V quakes. A co umn's stiffness is reduced when

the concrete cracks in tenslon It is further re- - - - - - - - - ducd as the !eel begins 20 yield and plassic

I h n g a form. me mai stiffness of a bridge I Firs1 yielding , changes in ter ton and compression as expan- I sion joints op I and cIose. The soil behind h e I I abunnent yie s for hrge compressive f o m s f and may not s jnpan tension We must consider I I all changes of - f ines to accumely obtain force I and rlisplacen .nr values for our bridge. I First eradung I Currently. ou aolicy is lo calculate a cracked I I stiffness for 3ridge columns. A value of

I - A Icr = 0.5(JF can be used unless a moment- Au *Y p=- Au curvarure 2 I. 1:sis is wmanted. Also, since *Y hidge colrlm are designed to yield during

large eanhouc ~s. we take rhe column force Nonlinear Column Stiffness obtained from >ur d y s i s , reduce it by a

ductility factcr, and design the columns for this s d e r force. Caltrans is currently using a ducdlity factor &) of abut 5 for designing new columns. Hawewer, a moment-curvature analysis of columns should be done when t h e column's ductility is uncertain.

Since we do net h o w how large a gap will exis1 at an abutment w hinge during aa eanhquake. the engineer should determine the largest and d e s l gap and perform two analyses and use the largest forceand displacement. The example ~roblem will illustrate rhis procedure. An advantage of doing a hand aadysis is h a t it allows us to consider many nonlinearities that are difficult to model when doing a multirnodal dynamic analysis.

Abutment Stiffness ~ongitudinal l~, the soil khind the baekwd is assumed to h v e a ~ ~ 4 s . which is related to h e area of the backwall as shown below:

kips kN 4 = 178'b- = 102 OmbT In

Abubnent Stiffness

Transversely. the stiffness is considered % effective per length of inside wingwd (assuming the wingwall is designed to take the Ioad) and the oudde wingwall is only lh effective per wingwall length for a resultant stiffness shown in equation 15,

An additional stiffness of 7 000 kN/m for each pile is added in both directions.

Therefore. in the longitudinal direction.

...................... KL = (47 000)Wh + (7 000)n (W/rn) ~...~.~.-...~.............. ( 16)

In the oransverse direction.

...................................................... KT = ( 102 000)b + (7 OM3)n (Wlm) (171

More infonation on abutment stiffness can be obtain in Bridge Design Aids,

The abutment suffness is highly nodineat. Bridge abutments are only effecrive in compression. A gap may need to be closed on sear typeabutmen& before the soil stiffness is initiated. The abutment stiffness remains linear until it reaches the ultimace smngrh of 370 kN/m2- This value was confirmed by testing at the University of California at Davis. after reaching its ultimate smngth, the abutmen1 is assumed to have a perfectly plastic behavior. After about a meter of movement it has a negative stjffness. To capture thrs behavior in a linear analysis is, of course, impossible. However.the engineercan calculare the displacement and adjust she secant stiffness undl the change in stiffness is less than 5%. This will be shown in the example problem.

Force Initial stifhess

Nonlinear Abutment Behavior

Parallel and Series Systems

Parallel System Series System

A simplification that dlows engineers to analyze by hand many complicated and statically indeterminant strueturn is theconcept of parallel and series swcmsalsy stems. Fora patallel system. all the elements share the same displacement, while for a series sysrem. h e y share the same force. Also, h e i r stlffnesses are surnrnd differently. By assuming a ripid supersmcture or by making other simplifying assummons, bridge structures can be analyzed as combinations of parallel and series systems. This concept is particularly useful when evaluating the Iongitudinnl displacement of the supersmture.

Code Requirements E a n h q d a me only considered for the h p W loading. %re are two cases. Case No. 1 is for 10Q% of the m v e f s e force and3096 of the longitudinal force. Case No. 2 is for 100% of the longkudmal force a d 30% of the m e r s e force. This is to take care of unceminty as to the earthquake direction and to account for curved and skewed bridga Gth memben that take a vector cornpanem of b t h the longitudinal and m e r s e fme.

Example Problem

Mwable 35 Fixed L - 35 rn

Mwable %, Fixed 35, Movable

€lev. TO30 / i I I 1030 7 lev

. - . . Sand . . . ;.: .' . k: .. Highest Gmnd Level Water. - ' d . . . . . N = h 0 . ,

Elev. . .

0 - -a nP .'n -... q - - , ~ - ~ *-- QQ g P a p e r O 0 e Q$W.~.D c -0- 00 . a,O:~P D ' - r,-- Q , ~ ~s~~~~~~~~ ~ o b , i ~ ! ~ . ~ ~ ~ Q ~ . ~ , - ~ , : ~ ~ e n ~ ~ ~ ~ D , , D ~ b v ~ ~ ~ e ~ ~ ~ , ~ o b ~ D ~ . - ~ ~ , ~ b U , ~ , 5 - ~ O ~ r '

,- 9 m square pile wp, Detail No. 2 r l l m ' Be- 2 and 4

I I - - 1 ,,'. - - - - - - - . * -

1 ;; 1 - t t:' I I I ; ; r I Z I 1 : ; I I I , I I ' ; I lo4 * L * I - 1 3 1 - 4

I ' : "; I,< - 1

A * 9 m square spread M n g -J'

4 - 2 4 Detail No. 1

L e E i 11

1 1.5m Piles at Bents 2 and 4 dd

. , . * . . . . .. . -

. , , . - .I / . . ::- - -

. . I _ . - - - . . + . r . - I.

(Steel ~ d l e r " Bearing at Bent 3) $%EX j o m m

Reinforced Concrete Brim Oeck

,

0.4064 m (1 6 inch) ClDH piles, total 12 - 0 . 75 m , 2 5 m , 2.5m , 2,5m 1'

I

Location of Pin for 7

9 m by 9 m by 1.9 m M n g . (Bent 2 and 4 with piles, Bent 3 withwt p~les.)

Detail No. 1 Detail No. 2

0.4064 m (16") ClDH piles, 12 per a b ~ .

Wan A-A Section 8-0

A) Calculating bngitudinal Seismic Forces

We wilt assume the bent footings are fixed and ignote the stiffness of the benl caps to simplify the analysis. Bent #3 will not be considered in the analysis since it has a roller bearing and conrributes a negligible resistance to the earthquake force. However, engineers need to make a field inspection of existing bridges 10 make sure beariogs are capable of rolling, For a seismic analysis we will use the cracked moment of inen~a of the columns.

The only othet stiffness we need to consider is at &e abutment Since only one abutmat caa act at a t h e (the one the supersmcture is pressing against), and because both abutments are identical. we will only consider one abutment m our analysis. Equation (14) gives rhe follawing stiffness;

L m g i t u ~ y , rhe bridge behaves as a parallel system, therefore the total stiffness is

W, the smcmres dead load is equal to the dead weight of the superstructure plus the dead weight of the bent caps for bents #2 and #4.

The total weight for our structure is 13 300 kN.

Bent 2 Bent 3 Bent 4 I I I

Superstructure Dead Load = (12.8 + 68.4)35 = 2 850 kN

Bent Cap Dead Load = 970 kN

Total Dead Load (per bent) = 3 820 kN

1.7 m

Bent 2 and 4

Bent 3

Homework Problem This twespan bridge is in a highly seismic area with a Peak ~ m d Acceleration of 0.7 g. Calculate the maximum seismic forces per column to be used in design.

Elevation View

9mx9mfbotingwith 8-0.4064 m dm. ClDH p i k

3m

Qof Abut 1 CofBent2

Plan View

Cof Bridge

A -

LO m &

4 I

?A7 ; g : .-:4-.

1 ; I

j i : I

I : . I

I I .-+; I ; : . I t

Caltrans_Bridge Design Practice.pdf

Documents

Transcript of Caltrans_Bridge Design Practice.pdf