Arch Bridges/

byDouglas A. Nettleton

Deputy ¿hief

Assisted byJohn S. Torkelson

Structural Engineer

Bridge DivisionOffice of Engineering

Federal Highway AdministrationU.S. Department of Transportation

Washington, D.C. 20590

II

PREFACE

Considering the extensive coverage in text books and technicalliterature of arch stress analysis and design, it might be thoughtthat little of value could be added by a new publication on thissubject. Emphasis in this paper is on aspects of arch design whichare not covered in many text books, such as wind stress analysis anddeflection, stress amplification due to deflection, consideration ofrib shortening moments, plate stiffening, and calculations forpreliminary design. In order for a designer to safely and economicallydesign any structure, he must have a clear understanding of all aspectsof the structural behavior. An unfortunate fact of most computerprogram usage is that the designer is much less cognizant of the basicaction and assumptions. Chapter Icovers steel arches and Chapter IIcovers concrete arches in all matters where concrete arches differfrom steel. Chapter 111 covers arch construction.

Much of the material herein is from papers published by Engineeringand Scientific Societies of the United States and other nations, andtext books. Credit is given in the text to these sources. The basisfor the equations developed by the author is given in the Appendix.

Every effort has been made to eliminate errors; but should errorsbe found, the author would appreciate notification from the readers.

ACKNOWLEDGEMENTS

The author would like to express thanks to those people who havehelped in the preparation of this publication.

The following are from the Washington Office Bridge Division ofFHWA: Mr. John S. Torkelson, Structural Engineer, made or checked thedesign calculations for steel and concrete arch examples;Mr. Emile G. Paulet, Staff Specialist, assisted by discussions withthe author and by helping to find sources of material; Mrs. Willy Rudolphand Miss Karen Winters typed the drafts and final pages of the manual.

The many people and organizations which had a part in the design orconstruction of the steel and concrete arch bridges used as examples inthis publication have given valuable indirect aid which is appreciatedby the author.

111

TABLE OF CONTENTSPage

PREFACE ii

CHAPTER I- STEEL ARCH DESIGN 2

1.1 Basic Arch Action 21.1.1 Dead Load Action 21.1.2 Live Load Action 4

1.2 Buckling and Moment Magnification 41.2.1 Tied Arch Buckling 9

1.3 Ratio of Rib Depth to Span and Live Load Deflection 101.3.1 Tied Arch Ratio of Rib Depth to Span 13

1.4 Rise to Span Ratio 13

1.5 Stress from Change of Temperature 141.5.1 Tied Arch Stress from Change of Temperature 15

1.6 Rib Shortening, Camber and Arch Rib Closure 151.6.1 Tied Arch Rib Shortening and Tie Lengthening 18

1.7 Effect of Location of Pins with Respect to Arch Axis 18

1.8 Fixed Arches versus Hinged Arches 19

1.9 Allowable Stress and Plate Buckling 201.9.1 Web Buckling 201.9.2 Flange Buckling 231.9.3 Equations for Load Factor Design 231.9.4 Web Buckling from Shear 24

1.10 Wind Stresses and Wind Deflection 251.10.1 Single Lateral System Between Ribs 251.10.2 Two Lateral Systems Between Ribs 271.10.3 Lateral Deflection 291.10,4 Interaction Between Arch Rib and Roadway

Lateral Systems 321.10.5 Unsymmetrical Wind Load 331.10.6 Longitudinal Wind and Forces 33a

IV

1.11 Lateral Buckling and Lateral Moment Magnification 341.12 Wind Vibration 36

1.13 Vertical Interaction Between Rib and Roadway Framing 391.14 Welding and Other Connections 391.15 Equations and Curves for Approximate Design 411.16 Design Example - Vertical Loads 45

1.16.1 Wind Analysis for Double Lateral System 491.16.2 Wind Analysis for Single Lateral System 541.1b.3 Wind Analysis Struts Only Between Ribs fin1.16.4 Lateral Buckling and Moment Magnification fil1.16.5 Distribution of Wind Loads Between Arch

Rib and Roadway Lateral System fó1.16.6 Longitudinal Forces 65a

CHAPTER II - CONCRETE ARCH DESIGN 67

2.1 Basic Arch Action 672.1.1 Dead and Live Load Action 67

2.2 Buckling and Moment Magnification 69

2.3 Ratio of Rib Depth to Span 70

2.4 Rise to Span Ratio 72

2.5 Rib Shortening, Shrinkage, Temperature Change and Camber 722.5.1 Permanent Arch Deflection 722.5.2 Arch Stresses from Rib Shortening, Shrinkage

and Temperature 732.5.3 Reduction of Rib Shortening and Shrinkage

Stress by Construction Methods 75

2.6 Buckling of Elements of Box Cross Sections 76

2.7 Wind Stress and Wind Deflection 77

2.8 Interaction Between Rib and Roadway Framing 77

2.9 Lateral Buckling and Lateral Moment Magnification 80

V

2.10 Load Factor Versus Service Load Design 80

2.11 Minimum Reinforcing Steel and Details 81

2,12 Design Example-

Vertical Loads 812.12,1 Revision of Cross Section 912.12,2 Vertical Crown Deflections 942.12.3 Wind Analysis 952.12.4 Effect of Live Load Lateral Eccentricity 962.12.5 Lateral Buckling and Moment Magnification 97

CHAPTER 111 - ARCH CONSTRUCTION 99

3.1 Steel Arches 993.1.1 Cantilevering From the Abutments by Tie-Backs 993.1.2 Cantilevering From Falsework Bents 1023.1.3 Off-Si te Construction 1053.1.4 Camber for Tied Arches 107

3.2 Concrete Arches 1083.2.1 Freyssinet and Menager-Hinge Construction 1083.2.2 Tie-Back Construction 1123.2.3 Elimination of Rib Shortening, Creep and

Shrinkage Stresses 112

REFERENCES 117

APPENDIX - Derivation and Origin of Equations 120

VI

Illustrations and Design Charts

Figure Page

1. Arch Nomenclature 32. Arch Deflection 53. Moment Magnification at Service Load 74. Moment Magnification for Design Load 85. Deflection Variation with Rib Depth and Stress 116. Suggested Rib Depth Variation with Span 127. Variation of Allowable Axial Stress with ki/r 218. Wind Analysis by Arch Bending Method 269. Wind Stress Analysis by St. Venant Torsional Method 2810. Torsional Stresses and Torsional Constant 3111. Rainbow Bridge - Erection 10012. Glen Canyon Bridge - Erection 10113. Bayonne Bridge -

Erection 10314. Bayonne Bridge -

Shoe Erection 10415. Fremont Bridge 10616. Cowlitz River Bridge 10917. Gladeville Bridge - Construction 11118. Van Staden's Bridge 11319. Hokawazu Bridge - General View 11520. Hokawazu Bridge - Construction 116

Tables

I. Examples and Dimensions of Concrete Arches 7111. Concrete Arch - Equation for X and Torsional Stress 78

2

CHAPTER I- STEEL ARCHES

1.1 Basic Arch Action

The distinguishing characteristics of an arch are the presenceof horizontal reactions at the ends, and the considerable rise of theaxis at the center of span, see Figure 1. Rigid frames and tiedarches are closely related to the arch. However, both of these typeshave characteristics which cause them to act quite differently fromtrue arches. In the case of the rigid frame, no attempt is made toshape the axis for the purpose of minimizing dead load bending moments,thus resulting in bending stresses which are considerably larger thanaxial compressive stress. In the case of the tied arch rib, thehorizontal reactions are internal to the superstructure, the spangenerally having an expansion bearing at one end. As a result thestresses are different, in several respects, for a tied arch as comparedto an arch with abutments receiving horizontal thrust.

In a true arch, the dead load produces mainly axial stress, and mostof the bending stress comes from live load acting over a part of thespan. Live load over the entire span causes very little bending moment.True arches are generally two-hinged, three-hinged or hinge!ess. Thetwo-hinged arch has pins at the end bearings, so that only horizontaland vertical components of force act on the abutment. The hinge!essarch is fixed at the abutments so that moment, also, is transmitted tothe abutment. The three-hinged arch has a hinge at the crown as well asthe abutments, making it statically determinate and eliminating stressesfrom change of temperature and rib shortening.

1.1.1 Dead Load Stress Action

Since dead load extends over the full span and is a fixed load,the arch axis should be shaped to an equilibrium polygon passingthrough the end bearings and the mid-depth of the rib at the crown,for dead load only. Since part of the dead load is generally appliedto the rib as a series of concentrated loads, the equilibrium polygonhas breaks in direction at the load points. The arch rib is usually acontinuous curve in the case of a solid web rib, and this results insome dead load moments. Trussed ribs have breaks at each panel pointsand, if the dead loads are applied at every panel point, the chordstresses will not be affected by the moment effect mentioned for thesolid web ribs on a continuous curve.

Usually a five-centered curve for the arch span can be fairlyclosely fitted to the dead load equilibrium polygon. Due to the greaterdead load in the outer parts of the span, the radii should increase fromthe crown toward the springing, resulting in an axis lying between aparabola and a circular curve of constant radius.

3

FIG. I -ARGH NOMENCLATURE

4

1.1.2 Live Load Stress Action

Although there is some live load axial stress, most of the live loadstress is bending stress. The maximum live load bending stress at thequarter point of the span is produced by live load over approximatelyhalf the span, the moment being positive in the loaded half and negativein the unloaded half of the span. The maximum axial stress is produced byload over the full span. Thus the loaded length for maximum stress isdifferent from the loaded length for the maximum moment alone. Underloading for maximum live load stress, the loaded half of the archdeflects downward and the unloaded half deflects upward. Thisdeflection in combination with the axial thrust from dead and live. loadresults in additional deflection and moment. This will be discussed atmore length under Moment Magnification.

For "L" in the AASHTO Impact Formula use one half the span lengthfor approximate calculations. For exact calculations use the loadedlength as indicated by the influence line for the point in question,for either lane loading or truck loading.

With two arch ribs, the live load should be placed laterally, inaccordance with AASHTO Specifications, to give the maximum load on onerib under the assumption of simple beam action between the two ribs.In the case of four or more ribs, the effect of live load eccentricitycan be distributed in proportion to the squares of the distance fromribs to the center line. Where there arc two lateral systems, a moreexact method of distribution is explained under 2.1.1.

1.2 Buckling and Moment Magnification

An arch has a- tendency to buckle in the plane of the arch due to theaxial compression. This in-plane buckling tendency is usually greaterthan the lateral buckling because an arch bridge generally has two ribsbraced together, and spaced apart a distance related to the width ofthe roadway. The usual in-plane buckling deflection is in the form ofa reverse curve with part of the arch rib going down and the other partgoing up, as shown in Figure 2. Thus the buckling length for a two-hinged arch is approximately equal to L, defined as one-half the lengthof the rib, and a kL/r value of 80 or more is usual for most solid webarch ribs. This buckling tendency should be taken into account in theallowable axial stress, just as it is in other compression members.Note that L as used here is the half-length of the axis, and z is thehorizontal span.

In most arches live load deflection causes an increase in stressover that shown by a classical elastic analysis. This effect has beenknown for a long time, but it is often overlooked or else treated asa secondary stress for which an increased allowable stress is permitted.There is a similar effect in suspension bridges, but there is a vitaldifference in the deflection effect on the two types of structures.In the case of suspension bridges, deflection decreases stresses and,when taken into account, effects an economy. In the case of arches,deflection increases stresses, and reduces the safety factor if neglectedin the analysis and design.

5

FIG. 2 -ARCH DEFLECTION

6

The live load moment in an arch is increased by the product of thetotal dead plus live arch thrust by the deflection from live load, asshown in Figure 2. The major component of the total arch thrust is deadload. Thus dead load interacts with live load in arch momentmagnification. Maximum positive and negative moments are increased inabout the same proportion.

Additional live load deflection is produced by the increase inmoment, and this increase in deflection produces an additional increasein moment. This effect continues in a decreasing series. Anapproximate method of taking this effect into account is to use a momentmagnification factor, Ap.

Aps = ] Eg. 11 "I (for deflection only)

AFc

Ap = moment magnification factor for deflection u/ider service load

T = arch rib thrust at the quarter point(approximately equal to H x secant of slope of line fromspringing to crown).

A = arch rib area at the quarter point

r if Ere x the Euler buckling stresski ¿/— \vr /

L = half the length of the arch rib(approximately equal to a/2 x secant ofslope of line from springing to crown)

r = radius of gyration at the quarter point

k is a factor varying from 0.7 to 1.16, dependingon end restraint, see Figure 3.

Equation 1 gives the moment magnification factor at service load,and it should be used only for figuring service load deflection.Figure 3 can be used to get the value of this factor. For example,if T/A = 6 ksi and kL/r = 80, the factor is 1.155.

F.KS, ,3

7

Fifí4rTTTTTTTrr

8

9

For an overload, the moment magnification factor does not remainthe same but increases considerably. This is due to the fact thatdeflection is not proportional to load in the case of an arch. Totake account of this fact, the following equation for Ap should be usedfor Service Load Design in order to maintain a desired safety factor:

A = 11 " T -7 T (Service Load Design) Eg. 2

Figure 4 can be used to get the value of this design factor. ForT/A = 6 ksi and kL/r = 80, the design Ap is 1.295 as compared to 1.155for service load deflection. The constant of 1.7, used in the Apequation, is less than the corresponding numerical constant of 2.12which is used for moment magnification in the AASHTO equations ofArt. 1.7.17, for combined bending and axial stress in columns. TheAASHTO value of 2.12 is the safety factor for compression. Since thethrust in an arch is mainly from dead load and the moment is mainlyfrom live load, the increase in thrust due to overload is likely tobe less than that which may occur in a column. Since the numericalfactor in the Ap equation is to allow for non-linearity under overload,it is logical to use a smaller value in an arch formula than in acolumn formula. The numerical constant value of 1.7 has been derivedfrom Load Factor Design safety factors which are smaller for dead loadthan for live load.

The AASHTO Load Factor Design safety factor for compression axialdead load is 1.3 f 0.85 = 1.53, and for compression axial live load is5/3 x 1.53 = 2.55. Thus a weighted safety factor, for a case wherelive load thrust is equal to say 15% of total thrust, would be:0.85 x 1.53 + 0.15 x 2.55 = 1.30 + 0.38 = 1.68. A numerical factor of1.7 is used in the equation. The numerical constant for Load FactorDesign is 1 f 0.85 = 1.18 and the equation for Ap, Load Factor Design is

Ap = ]_!„ 1 " 181 (Load Factor Design) Eg. 3

AFc

The effective length factors k, as given in Figure 3, are based onGuide to Stability Design Criteria for Metal Structures - StructuralStability Research Council, Chapter 16. (1)

1.2.1 Tied Arch Buckling and Moment Magnification

Moment magnification should not be used for tied arches. At anypoint in a tied arch the tie and the rib deflect practically the sameamount. Thus the moment arm between the tension in the tie and thehorizontal component of thrust in the rib remains constant for anysection, regardless of deflection. In this respect the tied arch actswery much like a bow string truss, and there is no stress amplificationdue to deflection in either. In a true arch, however, the line of the

10

horizontal component of the reaction is unchanged in position bydeflection. Since the arch rib position does change due to deflection,the moment arm of "H" is changed by deflection in a true arch, andtherefore the net moment, which is the difference between simple beammoment and the moment resisted by the effect of H, is increased bydeflection in a true arch.

Buckling in the plane of a tied arch becomes a matter of bucklingbetween suspenders rather than in a distance kl_. Where wire ropessuspenders are used, the difference in stretch of the suspenders dueto concentrated live load may be sufficient, due to the high allowableunit stress, to cause the buckling length to be somewhat longer thanthe distance between suspenders. (Ref. 5, page 14).

1.3 Ratio of Rib Depth to Span and Live Load Deflection

Moment magnification is quite sensitive to the ratio of rib depthto span. This is shown by Figure 4. At an axial stress of 8 ksi, anincrease of kL/r from 80 to 100 increases AF from 1.44 to 1.91. Atwo-hinged design first studied for the 950 foot span Rainbow Arch atNiagara Falls (2) had an i/d ratio of 66.5 and a kL/r ratio of 99. Thispreliminary design showed a very large moment magnification and was notused. A fixed arch with an £/d of 78 and a kL/r of 75 was used. Inthis case, due to the long span and the use of silicon steel, the axialstress is 11,700 psi. As can be seen from Figure 4, the momentmagnification is still large.

It can be seen from Figure 5 that deflection also is quite sensitiveto the ratio of span to depth. For a magnified bending stress of 10,000psi at service load, the magnified live load deflection is 1/800 of thespan for i/d = 75, and 1/1200 of the span for i/d = 50.

AASHTO Specifications give a maximum value for live load deflectionof 1/800 of the span for simple or continuous span. It is questionablewhether such a high deflection in terms of spans should be permitted foran arch. Maximum deflection for an arch occurs for approximately halfspan loading and, under this loading, about one half the span goes downand the other part of the span goes up. It could be argued, therefore,that the maximum deflection for an arch should be 1/1600 of the span.Very few existing steel arch bridges would meet such a criteria. Somewill barely meet the criteria of 1/800 of the span. We suggest avalue of 1/1200 of the span. An equation i/d = 44 +0.6 /£, for two-hingedsolid web ribs, is plotted in Figure 6. Use of this depth-to-span ratioshould result in arch ribs which can meet the deflection criterion of1/1200 of the span without loss of economy. Several existing archeswere checked in relation to this curve. Most of them have a lesser depththan that shown by the curve. Two hinged trussed arches are generallyabout 25% deeper at the crown than solid-web arches, and increase indepth toward the springing. As a result, live load deflection isconsiderably smaller for trussed arches. Moment magnification is reduced

FIG., , 5 ,

11

Fifi fi

12

13

in a trussed arch for the same reason, and also because the ratio ofr to d is almost 0.5 for a trussed arch and about 0.4, or less, for asolid web arch.

Fixed arches may have about 0.8 the depth of two-hinged arches.With this depth ratio and the same bending stress, fixed arches willhave about 2/3 the live load deflection of two-hinged arches.

See section 1.15 for approximate L.L. deflection equations

1.3.1 Tied Arch-Rib and Tie Depths

Tied arches may be designed with a rib of sufficient depth to takealmost all of the bending moment. In that case the tie may be madethe minimum depth permitted by the Specifications (3) for a tensionmember with an unsupported length equal to the spacing of the suspenders.The tie in this case is designed mainly for the tension produced bythe horizontal component of arch thrust. The tie will receive somemoment due to arch rib deflection under partial live load. This effectcan be approximately allowed for by dividing the total live loadmoment between the rib and tie in proportion to their moments of inertia.

Many tied arches, however, are designed with a deep tie and shallowrib. The tie then takes most of the bending from partial live load orother causes. The rib then serves principally as a compression member totake the arch thrust, and may be made as shallow as consideration ofallowable compressive stress and maximum £/r, based on supports at thesuspender points, will economically allow. The rib depth will be suchthat bending stress as well as axial stress should be considered in itsdesign.

An approximate method of analysis of a tied arch is to calculate thebending moment in the same manner as for a true arch, and then dividethe moment, at any point, between the rib and tie in proportion to theirmoments of inertia. An exact method of analysis would be to treat eachsuspender force as an "unknown." The amount of work would then be suchthat a computer program would probably be needed.

1.4 Rise-to-Span Ratio

The rise-to-span ratio for arches varies widely. A range from 0.12to 0.3 would include almost all bridge arches. Most are in the rangefrom 0.16 to 0.2. The site along with navigation clearances orvehicular clearances and roadway grades may determine, or have apredominent effect on the rise to span ratio used. A tied arch withsuspended roadway over a navigation channel is one case in which thereis full freedom to vary the rise-to-span ratio to suit economy andappearance.

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An increase of rise decreases arch thrust inversely with therise-to-span ratio, reducing the axial stress from dead andlive load and the bending stress from temperature change. The axialtension in a tie, if used, is also decreased in the same way. Off-setting these effects from the standpoint of economy is the increasedlength of the arch rib. This greater length increases the quantityof steel and the dead load. It also increases the buckling lengthin the plane of the arch and the moment magnification factor. Thelengths of the suspenders are increased. The total length of lateralbracing between the ribs is increased, and the wind overturning andstresses are increased. Many existing tied arches have a rise to spanratio of about 0.2.

1.5 Stress from Change of Temperature

An arch responds to temperature change by increasing or decreasingits rise, instead of by increasing or decreasing the span. If the archhas a hinge at the crown and at the ends, no stresses are produced bya change of temperature. In the case of a two-hinged arch, positivemoment is produced by a drop in temperature and negative moment by arise in temperature. For a fixed arch, the moments reverse in theouter portions of the span. The greater the ratio of rise to span,the smaller is the temperature stress. A lesser depth of rib alsoresults in smaller temperature stresses. The following approximateequations are based on an assumed uniform moment of inertia:

Temperature Rise2-Hinged Arch

H = 15Efrtt Bh2 Eg. 4a

Mx= -Ht

" y Eq. 4b

Fixed Arch

H = MWt Eg. 4c

Ms= |Ht

" h Eg. 4d

Mx= "Ht

(y "|h) Eq> 4e

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Where H = horizontal thrust from change of temperatures

M = moment at point xxy = ordinate to arch axis at point xI = assumed uniform moment of inertiaw = temperature expansion coefficientt = change of temperatureh = arch rise

1.5.1 Tied Arch Stress from Change of Temperature

A tied arch will not be stressed by change of temperature. Adifference in temperature between the rib and tie will produce stresses.In fact differences in temperature between parts of most bridgestructures will produce stresses, but these are generally neglected.It may be that this effect in a tied arch should not be neglected dueto the large distance between the rib and tie, and the fact that the tiemay be protected from the direct sun by the roadway slab. It can becalculated by the equation for a two-hinged arch, using the sum ,of theIvalues for rib and tie and the difference in temperature. As for anytied arch, the calculated bending moment may then be divided between therib and tie in proportion of their respective moments of inertia, as anapproximation.

1 .6 Rib Shortening, Camber and Arch Rib Closure

The shortening of the arch axis from axial thrust due to loads iscalled rib shortening. The stress produced by loads may be divided intotwo parts: that resulting from rib shortening, and that resulting fromflexual deformation. The arch axis can be shaped so as to practicallyeliminate other dead load moments, but the shape of the axis does notaffect the moment from dead load rib shortening. This dead load axialdeformation produces negative horizontal reactions and positive momentthroughout the span of a two-hinged arch. This positive moment resultsin a required larger top flange or chord area than that required for thebottom flange or chord. Using a larger area in the top flange than inthe bottom flange will not balance the maximum unit stresses in the twoflanges, because more of the axial thrust will then go to the top flangeor chord.

Therefore the net result of dead load rib shortening is anunderstressed bottom flange or chord, which means that maximum economyis not reached. However, dead load rib shortening stress can beeliminated.

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As an example of the effect of rib shortening on stress, thestress sheet for the 1700 foot arch span over the New River Gorge inWest Virginia (4) shows that, in the central two-thirds of the span,the dead load upper chord stresses in kips are from 35 to 60 percenthigher than the lower chord stresses. This large difference is partlydue to rib shortening and partly due to larger upper chord areas.However, the design unit stresses in the upper chord are very close tothe allowable, whereas the design unit stresses in the lower chord areabout 15 percent below the allowable. This is a typical relationshipfor many arches, and is due to the effect of dead load rib shortening.

It is not necessary to have rib dead load shortening stresses inan arch, if certain methods of member camber and erection are used.If the lengths of the members are cambered for dead load axial stress,there will be no dead load rib shortening stresses. Steinmanrecognized this in 1936 in the design of the Henry Hudson Bridge (29).This is because the closure of the arch must be forced if the memberlengths have been cambered for dead load axial stress but not forcurvature due to moment from rib shortening. The effect of thisforced closure, which can be done with jacks that are needed for erectionin any case, is to produce equal and opposite flexure to that producedby rib shortening. The almost complete elimination of dead load momentsin the Fremont Bridge (5) by the use of length camber, resulting inforced reverse moments under zero load is illustrated in Figure 15.

The method of analysis used, such as a computer program, mayinclude rib shortening stresses. If so, they may be separately figuredand then excluded by use of the following approximate equations forrib shortening. These equations are for a uniform depth. Seeequations 48a - 48h (Concrete Arches) for a varying depth of rib.

2-Hinged Solid Web

Hrs= -15(-7h)2 HDHD.L. Eq _ 4f

Mx= -Hrs y Eq. 4g

Fixed-Solid Webs

H^ = r|o(r/h)2 HDHD.L. Eq< 4h

Mx= -Hrs(y-0.67h)Hrs(y-0.67h) Eq. 4i

2-Hinged-Trussed Rib

Hrs= -0.47 (dc/h)2HD.L. Eq. 4k

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Where H^ = horizontal reaction from dead load

Hrs= horizonal reaction from rib shortening

r = radius of gyration of a solid web ribdc

= depth center to center of chords at the crownh = arch axis rise at crowny = arch axis rise at point x, measured from springing

The method used for analyzing a trussed rib will most likely includerib shortening deformation and stress, and the above equations will beneeded for exclusion of dead load rib shortening stress. The methodfor solid web ribs may or may not include rib shortening. Thisillustrates the importance of a designer knowing the basic assumptionsof the method of analysis he is using.

A trussed arch may be closed on a temporary pin in either the upperor lower chord. After the arch becomes self-supporting, with all of thethrust going through the pin, the gap for the closing chord member isjacked open an amount sufficient for the length of that member. In orderto insure that the stresses are as calculated, it is preferable to havethe connection at one end of the closing member blank. Holes are drilledin the field to fit the opening produced by a precalculated jackingforce. This force is equal to the calculated stress in the member fromthe dead load, including erection equipment, which is on the bridge atthe time of jacking. In order to exclude all dead load rib shorteningstress, the rib shortening stress from dead load to be placed afterclosure should be subtracted (algebraicly) from the calculated jackingforce. Dead load rib shortening stress should be excluded from thedesign stress and camber in all members. The members are cambered, ofcourse, for the stress from dead load thrust, and the stress from deadload moment due to causes other than rib shortening.

An arch may be designed as three-hinged, only for the dead load attime of closure. It would be closed on a pin at the crown, and thenprovided with a full depth moment connection. No jacking is required,and errors of surveying or fabrication would not affect the stresses.The remainder of the dead load would produce rib shortening stressesand these should be included in the design. If it is desired toeliminate these stresses, it could be done by light jacking at closure.If the pin is in the lower chord, jacks in the line of the upper chordwould exert a pull in the upper chord line, and release after closurewould leave a precalculated tension in the upper chord equal to the ribshortening compression form the additional dead load. For a pin atmid-depth of a solid web rib, simultaneous jacking would be required atboth flanges to eliminate all rib shortening stress.

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It is necessary to indicate the assumptions with regard to ribshortening on the plans, and the requirements at closure to insurethe realization of such assumptions. The contractor should havefreedom in choosing the method of erection, but he must be required toachieve the desired stress condition.

1.6.1 Tied Arch Rib Shortening and Tie Lengthenin

The effect of rib shortening in a tied arch is accentuated by theeffect of tie lengthening, due to tension, from load effects. Thefollowing equation can be used for calculating H:

Where 1^ = moment of inertia of tie

Ir= moment of inertia of rib

At= area of tie

Ar= area of rib

Values of Iand A at the crown may be used

Mr +t= Hrt

" y Eq. 4m

Where M■ = sum of moments in rib and tie at point xr+t

MMr+t may be divided between rib and tie in proportion to their respectivevalues of I. Camber will eliminate this moment and that from hanger stretch

1.7 Effect of Location of End Pins with Respect to the Arch Axis

For trussed arch ribs the end pins are often located at the centerof the lower chord instead of on the arch axis. This has the advantageof simpler framing for the arch truss and for the lateral system at theends of the span. It may result in somewhat more steel in the archchords.

Locating the pin in the lower chord, in effect, is equivalent tointroducing a negative dead load and live load truss moment at the pin,instead of zero moment for a pin on the arch axis. For dead load, thisnegative moment reduces to zero in a little more than one quarter of thespan length and a maximum positive moment is produced at the center ofthe span. As a result the lower chord has more compression than theupper chord in the outer parts of the span, and the less in the centralpart of the span. The lowered pin produces similar results for liveload stresses at the ends and center of the span. The areas of thechord members cannot be fully adjusted to meet these large differences

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in maximum stress between upper and lower chords at the same point inthe span. The true arch axis is changed by the change in chord areas.The result is similar to the effect of rib shortening stress. Thedesigner will be forced to use reduced unit stresses in the upperchords at the ends of the span, and in the lower chords in the centralpart of the span. These effects may offset the saving in cost offraming details at the ends of the span for a pin located on thearch axis.

Appearance may also enter into the decision of end pin location.Tapering the end panel almost to a point from a fairly deep truss maygive the appearance of weakness. When the pin is placed in the lowerchord, the abutments are sometimes designed so as to hide the factthat the upper chord does not thrust against the abutment. The pinis always placed on the axis for a solid web arch.

1.8 Fixed Arches Versus Hinged Arches

Unlike concrete arches, most steel arches are not fixed. TheRainbow Arch over Niagara Falls (2) is a notable exception. This archwas originally planned as a two-hinged arch and changed to a fixedarch to reduce deflection and moment magnification. Of course, reduceddeflection and reduced moment magnification could have been obtained byuse of a deeper two-hinged arch. The fixed arch was found to belighter in weight, and was probably also favored because of its slenderappearance.

Abutment costs are higher for a fixed arch because a large momentmust be transmitted to the foundation by anchorage ties or by sufficientspread of foundation bearing to keep the edge pressure within theallowable. Also heavy anchorage details between the steel and concreteare required. Although the designers of the Rainbow Arch found economyin the use of a fixed arch, studies by the designers of the Bayonneand the Glen Canyon arches indicated the opposite.

Erection is more complicated for a fixed arch, particularly if itis by the tie-back method. Complex calculations were made for theerection of the Rainbow Arch.

The fixed arch is statically indeterminate to the third degree.Analysis is generally made by cutting the arch at the center of span,and taking the horizontal thrust, moment and shear at that point asthe "unknowns." Simultaneous equations can be avoided by use of the"elastic center," a point below the crown. However, there may not bea net saving in work by use of this method.

20

21

1.9 Allowable Stress and Plate Buck! in

The allowable axial stress, Fa, is given by the equation for Fain Table 1.7.1 of AASHTO Specifications, as revised by 1974 Interim 7.This equation is:

F = Jü_ h _ (KL/r)2Fy )a 2.12 4,2E

L for a solid web arch rib is one-half the length of the arch axis. Thisequation has been plotted in Figure 7, and a table of k values for archribs is given in Figure 3. As in the AASHTO Specifications, when KL/rexceeds [ZirE/Fy)^^, the following equation for Fa should be used:

F = 135 x 106a (KL/r)z

The allowable bending stress, Fa, for a solid web arch is 0.55Fy.No reduction for lateral buckling is needed for a box section, or fortwo plate girders laced together, as in the case of the Henry HudsonBridge.

■F fThe interaction equation, _§. + Jl < 1, is used to check the assumed

section. Fa

The plates making up the cross-section must meet requirements forlocal buckling. Where the overall design is based on an interactionequation, as is the case of the arch rib, the equations for platebuckling cannot be based on Fy, or on Fa and Ft). They must be basedon fa and fb with a safety factor included in the equation.

The axial stress fa, from arch thrust, produces uniform compressivestress across the width of both the flange and the web plates. Thebending stress f^, from moment, produces uniform stress across theflange plate; but a varying stress from maximum compression at oneedge to maximum tension at the other edge in the web plate. As a result,separate buckling equations are needed for the web and flange plates.

1.9.1 Web Buck!in

For an unstiffened web, the effect of f^ on buckling is very smallcompared to the effect of fa. For a stiffened web, f^ has a veryappreciable effect on the buckling of the individual panels betweenstiffeners. Although f^ does not appear in the following equations,its effect has been taken into account in the numerical coefficientsof the equations. These coefficients are based on a value of f^ at theedge of the web equal to about 1.75 fa. It is very unlikely that anyarch would have a higher ratio of fb to fa than 1.75.

22

The plate buckling equations, which are based on the AASHTOSpecification equations, are quite conservative for the web. In thecase of the web plate, particularly the unstiffened web, the axialstress has the major effect on local buckling, and this stress isgenerally in the range from 0.15 Fy to 0.3 Fy.

The AASHTO plate buckling equations, and the equations given here,cover the full range of unit stress in the plate. Reference to USSSteel Design Manual, pages 74 and 75, will show two equations for platebuckling, one to be used when the critical stress is below 0.5 Fy andthe other for a critical stress above 0.5 Fy. It will be noticed, inFigure 4.1, that the curve, for critical buckling stress below 0.5 Fy,gives too high a stress above 0.5 Fy. In order to use one equationover the entire range, it is necessary to accept a reduced allowablestress for the lower part of the stress range. Since the axial stressin the web plate of a bridge arch rib will be in this lower range, theallowable D/t for the web will be on the conservative side.

The effect of this conservatism on economy is small, because aheavier web will permit lighter flanges.

Web Buckling Equations:No longitudinal stiffener

D/t = 5000/ v/Ta", max. D/t = 60 Eg. 5

One stiffener at mid-depth

D/t = 7500/ \/T, max. D/t = 90 Eg. 6a

Is= 0.75Dt3 Eg- 7

Two stiffeners at the 1/3 points

D/t = 10000//^, max. D/t = 120 Eg. 8

Is= 2.2Dt3 Eg. 9

Outstanding element of stiffener

b'/f = 1625/ \/fa + fb/3, max. b'/f = 12 Eg. 10

Where D = web deptht = web thicknessIs

= moment of inertia about an axis at the base of thestiffener

b 1 = width and t1 = thickness of outstanding stiffener elements

23

Generally two stiffeners should be used in order to get the mosteconomical section. For spans of 450 feet or more, longitudinaldiaphragms across the width of a box section should probably be used.These would act as rigid lines of support for the webs. Such a diaphragmcould be used at mid-depth of the box, and each panel formed therebycould be stiffened at its mid-depth. The b/t value for each panel wouldthen be based on the average stress in the panel rather than on the axialstress for the whole rib. In the case of the Rainbow Arch (2), continuouslongitudinal diaphragms were used at the mid and fourth points of the webdepth. These longitudinal diaphragms were supported by radial diaphragmsabout 20 feet apart. The stress in a longitudinal diaphragm at mid-depthis fa.

Longitudinal Diaphragm Buckling Equation

b/t = 4500//fa~, max. b/t = 54 Eg. 10a

1.9.2 Flange Bucklin

Flange Buckling Equations (Unstiffened)Between webs

b/t = 4250/ \/fa + fb,max. b/t =47 Eg. 11

Overhang

b'/t = 1625/\/fa + fb,max. b'/t =12 Eg. 12

Where b = distance between websb' = flange overhang outside webt = flange thickness

Stiffeners are seldom used on arch rib flanges. A single stiffenerat mid-width of the flange would permit a b/t = 8500/ \/fa + fb,and therequired Is for that value of b/t would be 4bt3. About the only casewhere flange stiffening might be used would be where the two ribs arenot braced together, laterally.

1.9.3 Equation for Load Factor Design

All the above equations in this article are for Service Load Design.The corresponding equations for Load Factor Design Are:

Web PlatesNo longitudinal stiffener

D/t = 6750//f¡ Eq. 13

One longitudinal stiffener

D/t = 10,150/Vfá Eq. 14

24

Two longitudinal stiffeners

D/t = 13,500/Vfg Eg. 15

b'/f = 2200/ \/fa + fb/3fb/3 Eg. 16

Flange PlatesBetween webs

b/t = 5700/\/fa + fb- unstiffened Eg. 17

b/t = 11,500/\/fa+ fb

- one stiffener Eg. 17a

Overhang

b'/f = 2200/ \/fa + fb Eg. 18

Otherwise the equations are the same.

1.9.4 Web Buckling from Shear

Transverse stiffeners on the web are not generally required forstress in either the rib or tie of an arch,J)ecause of the low unitshear stress. The AASHTO equation t = Dyfv (Article 1.7.71) (3),

7500which permits the omission of transverse stiffeners, is almost certainto be met. The dead load produces very little shear since the thrustline follows the axis. Live load shear is small because the shear isonly a component of the thrust and, at any point, is equal to the thrustmultiplied by the sine of the angle between the direction of the thrustand the direction of the axis at that point.

In the case of the tie, the large axial tension more than nullifiesany buckling tendency from shear. The Fremont Bridge (5) tie is 18 feetdeep with 1/2-inch webs, a D/t ratio of 432. Transverse stiffeners areused on these webs midway between floorbeam diaphragms, thus giving webpanels of 5 feet 7 inches. These stiffeners were probably used toprevent distortion from handling in shipping and erection.

Longitudinal stiffeners are not required in an arch tie because ofthe large tension.

Diaphragms are used for arch ribs and ties at points of loading.The Rainbow Arch has diaphragms at the columns and midway between,giving a 20-foot spacing for the 12-foot rib.

25

1.10 Wind Stress and Wind Deflection

For lateral forces such as wind, the arch is a curved member in theplane normal to the direction of loading. The central angle is largeand, therefore, the torsional effects are very important in the windstress analysis. The exact analysis of torsional effects is quitecomplicated because of the interaction of St. Venant torsion withwarping torsion.

If the two arch ribs are connected by a single lateral system atthe mid-depth of the ribs, the St. Venant torsion is of minor importanceand can be neglected in the analysis. The torsional effects are resistedin this case by equal and opposite bending of the arch ribs in thevertical plane. With lateral systems at both the upper and lower flangeor chord levels, St. Venant torsion becomes predominant because theoverall structure is a closed torsional section.

1.10.1 Single Lateral System

The forces acting on the arch are applied at the connections of thelaterals to the ribs, and they act tangentially to the arch curve. Sincethe laterals carry the transverse wind shear, the tangential forces actingon the arch ribs are equal to the transverse shear, in the lateral systempanel, multiplied by the ratio of the panel length to the distance betweenribs. For the symmetrical load case of wind over the full span, the shearis zero at the center of the span, and the shear in any panel can befound by summing the panel loads outward from the center of the span.This procedure is illustrated in Figure 8. These tangential forces acttoward the crown of the windward rib and opposite on the leeward rib.

One method of determining the arch stresses produced by these loadsis to resolve the tangential forces into vertical and horizontalcomponents. The horizontal component of the reaction at the springingis then determined by assuming one end bearing to be on horizontalrollers, figuring the horizontal movement on the rollers, and then thehorizontal force required to reduce the movement to zero. The windwardarch rib will move downward in the outer portions of the span underpositive moment and upward in the central portion of the span undernegative moment. The leeward rib will do the opposite. In other wordsthe ribs rotate clockwise for the outer parts of the span and counter-clockwise for the inner part of the span, as seen from the leftspringing with the wind coming from the right. The wind stress at anysection is composed of two parts, the axial stress and the bending stressin the vertical plane. The stresses can also be obtained by taking the"unknown" as the crown moment instead of as the horizontal reaction

26

FIG. 8 -WIND ANALYSIS BY ARCH BENDING METHOD(single lateral system of mid -depth of rib)

27

as indicated in Figure 8. The statically determinate structure is athree-hinged arch. The moment in the vertical plane, at the crown,is then solved for by making the rotation in the plane of the rib atthe crown equal to zero, in the case of symmetrical loading. Thismethod of analysis is illustrated by an example.

1.10.2 Two Lateral Systems

On long spans, two lateral systems are used between the arch ribs,at the levels of the upper and lower chords or flanges. With twolateral systems, the structure is stiff in St. Venant torsional action,because it is a closed torsional system with one dimension equal to thearch rib depth and the other dimension equal to the distance center tocenter of ribs. There will be some vertical arch rib bending actionbut the St. Venant torsional action will be dominant. For two lateralsystems between the ribs, it is sufficiently accurate to neglect thearch rib bending stress and consider only the St. Venant torsionalstress and the lateral bending stress. The St. Venant torsionalaction produces stresses in the web members of the rib and in thelaterals. The lateral bending stress which accompanies the St. Venanttorsional analysis produces axial stresses in the arch ribs whichcorrespond to the axial stresses in the arch ribs found by the verticalarch bending method of analysis. An example will later be analyzed bythe St. Venant torsional method and by the vertical arch bending methodto give a comparison of results. The St. Venant torsional analysisinvolves cutting the arch at the crown and solving for the unknownlateral bending moment at the crown. Under symmetrical loading thetorsional moment at the crown is equal to zero. Figure 9 illustratesthe stress action. For varying moment of inertia and any shape ofarch axis, the arch axis may be divided into several sections of equallength A|_, and the "unknown" quantity, the lateral moment Mo at thecrown, is found by solving the following equation which was derived byequating the rotation in a horizontal plane at the crown to zero:

z[cose(Mxcos9 + MySino) *I] +|z[sin9(Mxsine - Mycose) iK]e[cos 2929 vl] +IE[sin2e vK] Eq> 19

Where I = average moment of inertia for each sectionX = average torsional constant for each sectionG = shear modulus of elasticity

28

FIG. 9 -WIND ANALYSIS BY ST. VENANT TORSIONAL METHOD(double lateral system - at lop and bottom of ribs)

29

The second terms in the numerator and denominator come from torsionaldeformation. For most steel arches the second terms are so muchlarger than the first terms that the effect of the first terms may beneglected. This is not true for concrete arches. Neglecting thefirst terms and assuming constant cross section, the equation becomes:

zsine(M sine - M cose)Mn

= x y '_ Eg. 20. 9Esin^e

If we assume a single centered arch axis of radius R and constant crosssection, the equation for MQMQ becomes:

M = WRa[[sing - 3/2 - slip] + El. [slng .gcos g _1/2 (g _ $in ecOs3)]}

To/9 + SJn23-i . ÉI r " 1L3/¿ + —4— J +

2GK [3 " sin 3 cos 3] Eq> 21

Neglecting the first terms in the numerator and denominator, Eg. 21becomes:

Mo= WR2 [ 2(sine - BCQSb) _i] Eg. 22

3-siri3cos3W is the wind load per foot, R is the radius and 3 is one-half thecentral angle. R and 3 are obtained by passing a circular curvethrough the crown and springing of the actual arch axis. At any point,M and T are as follows:

M = MoqosoMoqos0 - WR2(l-cose) Eg. 22a

2T = Mosine - WR (0-sine) Eg. 22b

1.10.3 Lateral Deflection - Two Lateral Systems

The lateral deflection at any point may be found by applying a unittransverse force to the cantilevered half arch, and multiplying thelateral bending moments due to this unit load by the actual lateralbending moments. These products are summed up and divided by El to givethe lateral deflection at the point of application of the unit load.The torsional deformation is practically negligible insofar as anyeffect on lateral deflection. The lateral deflection at the crown,assuming a circular axis is:

6C6C= (Rsin3)2 [-WR2(I - 2 )-Ml Eg. 23c 2EI 1 + cos^ °

30

A more approximate formula will give practically the same answer:

66C= ¿ [Wf -Mo ] Eq. 23a

The value of MQMQ for an arch of average rise to span ratio is about sixtenths of the center moment in a fixed-end beam of span equal to therolled out length of the arch axis. Using 0.6 x M^?. for Mo, the crownlateral deflection for the arch is 1.75 times that of a fixed-end beamof equivalent rolled out length. It will be shown further on that thelateral deflection of an arch with only a single lateral system isconsiderably greater. An approximate formula for the deflection atany point x, measured along the arch axis from the crown is:

á x=

Wl Llí< x4- 4L3x + 314) -Mo(L - x)2] Eq. 24

Torsional moment acting on an arch with two lateral systemsproduces shearing forces on the laterals and on the web members of theribs as shown in Figure 10. These shearing forces produce axialstresses in truss type laterals and rib webs, and shearing stresses insolid rib webs. For Equation 19, it is necessary to determine thetorsional stiffness of the arch cross section, corresponding to thebending stiffness El and the axial stress stiffness EA. The totaltorsional stiffness of the cross section is somewhat complicated by thefact that some of the members resisting torsion will have axial stressand others may have shearing stress. The torsional stiffness will bedetermined by the rotation in one panel length produced by a unittorque. Dividing the panel length by the rotation gives the torsionalstiffness at that point on the arch axis. Referring again to Figure 10and using the method described above, the following equation for thetorsional factor X is obtained:

X = : Eg. 25

This equation applies only for a X lateral system with equal size membersfor the upper and lower systems and a solid web rib, with t representingthe combined thickness of two web plates in the case of a box section.For other lateral system configurations and for a trussed rib insteadof a solid web rib the equation for X will be different. The method ofderivation described for the above equation can be used for any type oflateral system and for trussed ribs. There will generally be smallstresses from torsional moments in the flanges or chords, but they arenegligible in relation to the areas of the chords or flanges, from thestandpoint of either stress or deformation.

31

FIG. 10

32

In addition to the lateral deflection from wind, there is alsorotation in a plane normal to the arch axis. This rotation causes theleeward rib to move down and the windward rib to move up at the crown.This movement is of importance in distributing wind loads between thearch lateral system and a lateral system in the plane of the roadwayabove the arch. The rotation angle multiplied by the vertical distancefrom the arch axis to the roadway lateral system adds to the archdeflection to give the total transverse movement, from arch wind loading,at the roadway level. There is a slight reduction of the rotationaleffect due to the bending of any columns at the crown. However, thesecolumns will be very short at the crown so that this bending effect issmall. The rotation angle at the arch crown can be calculated from thefollowing equation:

Crown rotation = [Mqx 1sin232 3 + WR2(j. sin232 3 - cos 3- 3sin3 + 1)]Eg. 26

This equation has been derived on the basis of a circular arch axis ofuniform cross-section. Torsional deformation only has been includedbecause the effect of lateral bending deformation is negligible forrotation. The effect of variable cross section and the actual shape ofthe arch axis can be taken into account by the summation method. By thismethod, a unit couple in the vertical plane is applied to the cantileveredhalf arch at the point where the rotation is desired. The products ofthe actual torsion by the torsion from the unit couple, divided by thetorsional X at each point are.summed to give the desired rotation at thepoint of the unit load.

1.10.4 Interaction Between Arch Rib and Roadway Lateral Systems

Distribution of wind forces between the roadway and arch rib lateralsmay be accomplished by a trial and error process of balancing lateraldeflections and rotations. Lateral bending deflection of the columns isinvolved in this process unless transverse bracing is used between thecolumns. Use of such bracing is unusual and is not recommended becauseof the rotation of the arch.

Quite often lateral struts only are used between the ribs with nolateral diagonals. This type of lateral system can be solved for thethrust and moment in the vertical plane of the arch rib in the samemanner as for a single lateral system. However, since there are nodiagonals to take the lateral shear, lateral bending moments are producedin the arch ribs and in the struts. An approximate solution can be madefor these moments by assuming points of contraflexure in the membersmidway between the panel points and midway in the length of the transversestruts. With this type of bracing, the lateral deflection from shearwill be considerably greater than when diagonals are used. The deflectionfor one panel is:

A = tt£_ r p + 2b i Ea 2724E Li T J q *

33

Where V = panel shearp = panel lengthb = distance center to center of ribsIr = lateral moment of inertia of one ribIs

= lateral moment of inertia of the strut

The total deflection from shear at any point is found by adding thepanel deflections outward from the end of the span.

1.10.5 Unsymmetrical Wind Load

The discussion of wind stress up to this point has been based onsymmetrical wind load only. Wind on the live load would be unsymmetricalfor most cases. The same general methods of analysis, as have beenexplained for symmetrical wind load, can be used for unsymmetrical windload. However, symmetry can no longer be used to eliminate "unknowns"in the solution. In the case of a double lateral system, there will bean unknown lateral shear and an unknown torque at the crown in additionto the unknown lateral moment. It will be necessary to set up threeequations to be solved simultaneously, and the full length of the archwill have to be used in the summation method.

In the solution for unsymmetrical wind with a single lateral system,the tangential forces can no longer be determined from the fact ofsymmetry. An approximate way of handling this is to assume that thetransverse lateral reactions at the ends of the span are the same as fora straight fixed end beam of uniform moment of inertia. In the part ofthe solution involving the determination of the thrust at the crown ofa three-hinged arch, the vertical shear at the crown must be taken intoconsideration. This can be found by statics. The remainder of thesolution is the same as for the symmetrical case, except that the fullarch instead of the half arch must be included in the summation. Afterthe determination of the arch thrust at the crown, the thrust can bemultiplied by the distance between ribs to get Mo. The Mo can be usedto determine the lateral end reactions. If these differ appreciably fromthose assumed in the beginning, the whole process can be repeated.

Obviously the solution for unsymmetrical wind load involvesconsiderable work, and there is a question as to whether it is reallynecessary. The main reason for such a solution would be for design ofthe laterals in the central portion of the arch. The unsymmetrical windload stresses in these members can be approximated by assuming the lengthof the rib as a straight member and getting the end reactions on the basisof a fixed end beam of constant section.

33a

1.10.6 Longitudinal Wind and Forces from Braking and Traction

An arch may have stresses produced by longitudinal forces causedby wind parallel to the roadway or wind at an angle, and by braking andtraction from live load. AASHTO Specifications mention arches under windand link arches with trusses for transverse wind force. Seventy-fivepounds per square foot is specified for trusses and arches, and 50 poundsper square foot for girders and beams. The lower figure for girders andbeams would appear to be based on the shielding effect of a solid webgirder with respect to girders behind it, as compared to the shieldingof a truss. If so, it would seem that the lower figure of 50 pounds persquare foot should have been specified for a solid web arch rib.

AASHTO Specifications for "Substructure Design" give a table oflongitudinal components of wind acting on the superstructure. However,this is intended to be used only for the purpose of obtaining forces onthe substructure from the superstructure.

In the case of an arch, the effect of longitudinal wind is of moreimportance because of the total height of the structure. However, wherethe roadway passes over the crown of the arch, the arch itself wouldgenerally be shielded in the longitudinal direction by high ground ateach end. Where the roadway is suspended below the arch, there wouldbe no shielding above the roadway level.

For the design example in 1.16, the following analysis is made fora 60° wind acting on the deck. It is assumed that all of the longitudinalforce acting on the deck, in the length of the arch span, is transmittedto the arch at the crown.

34

1.11 Lateral Buckling and Lateral Moment Magnification

An exact analysis of lateral buckling and lateral moment magnificationis quite complicated. An approximate analysis is proposed here becausebuckling in the plane of the arch is almost always controlling, andtherefore the effect of lateral buckling is minor in the overall designof most arch bridges.

Arches have overall fixity at the ends for lateral moment and fortorsion. Due to the arch rise, resulting in a combination of torsionand flexure for lateral buckling, the buckling condition will alwaysbe more severe than that for a straight member with a length equal tothe length of the arch axis. The third edition of "Guide to StabilityDesign Criteria for Metal Structures," pages 479 and 480, gives anequation (16.25) and a table (16.7) for calculating the critical verticalload for lateral arch buckling. The values for XL given below apply toa closed torsional system. In the case of steel arches, this wouldrequire two planes of lateral bracing between the ribs. Most concretearches do have a closed torsional system. This equation and table havebeen used to work out the following lateral buckling lengths for use inthe general equations involving KL/r, such as those for allowable unitstress and moment magnification. The following values apply to singlearch ribs having a ratio of torsional stiffness to lateral bendingstiffness between 0.5 and 1.5:

h/£ XL

0.1 1.04L0.2 1.17L0.3 1.34L

Where L = the half length of the arch axis

The above values, in the case of a single rib, would be used to get Fefor use in Equations 2 and 3 for the lateral moment magnificationfactor, and in Figure 7 to get allowable stress based on lateral KL/r.Braced Ribs with One Plane Diagonal Type Lateral System

h/l XL

0.1 1.36 L1-88Lr = half the distance between0.3 2.30Ltwo outer ribs

* 1° *\1OlJ°* etori?ation of the late^ls, multiply XL above by kfrom Eg. 28 (Bleich (28) page 169).

35

k= Vl+V1+ w " pfe (f + ¿)The above equation applies to a bracing system with one diagonal

of length z and area A^, and one strut of length b and area A^ perpanel. Iis the lateral moment of inertia of the braced system and p isthe panel length. Bleich's notation has been changed to correspond tothe notation being used in this paper. Also XL has been used in placeof a,, and Ej. has been assumed equal to E. This last assumption is basedon the fact that the bridge arch thrust would always be appreciablyless than the maximum allowable because of the presence of bending stressdue to partial live load. Therefore, it is reasonable to assume that,when loaded to ultimate strength, the stress from thrust would be withinthe proportional limit of the stress strain curve.

Braced Ribs with One Plane of Struts Only

The Guide to Stability Design refers to theoretical studies for thecase of lateral bracing by transverse struts only, made by Ostlund (25)and Almeida (26); but goes on to state that these methods are moreinvolved than is desirable for use in preliminary design. The Guide thenmentions an approximate method by Wastlund (27) and others in which thearches are assumed to be straightened out so that the ribs and bracingmembers lie in a horizontal plane. The ASCE paper by Wastlund gives asimple equation for the buckling thrust. However, this equation involvesa coefficient "C" which must be obtained from Ostlund 1s paper. Thefollowing method, based on Bleich (28), is similar to the method proposedfor diagonal bracing. The following is a modification of Bleich'sEquation (350) on page 178.

k=\/ + -¿I— (P-L- + *T) Eg. 28a12 (XL)2 2I1 ¡

b

Where I-, = lateral moment of inertia of one rib

I = moment of inertia of the strut about an axisnormal to the arch axis.

As with diagonal bracing, k is to be used as multiplier of k£/rfor use in equations for allowable stress and moment magnification. Themoment magnification factor should be applied to both the axial forcesin the ribs produced by lateral flexure of the arch as a whole and to thebending stresses in the individual ribs and struts produced by the shearaccompanying lateral flexure. KL/r is the same as for braced ribs withone plane diagonal type lateral system.

36

Stress in Laterals Due to Arch Lateral Buckling

The buckling tendancy will produce lateral shears on the arch asa whole. This shear may be taken as 2 percent of the axial thrust. Themaximum shear from buckling will occur at the lateral inflection point,near the arch quarter point. Since the arch is fixed laterally at theends and is symmetrical, the buckling shear will be zero at the ends ofthe span and at the crown» and maximum at about the quarter point.

The lateral wind shear is a maximum at the ends of the span and,since it may be taken as a partial load, the maximum crown shear isabout one fourth of the end shear. Thus the maximum lateral bucklingshear and the maximum lateral wind shear do not add directly. Thelateral wind shear at the quarter point may be assumed as 0.6 times theend shear. It is proposed that the laterals be of constant cross-section throughout the span, and be designed for either the full windshear at the end of the span or for a shear equal to 2 percent of thetotal axial force at the quarter point plus 0.6 times the end wind shear,whichever is larger. The wind shear factor of 0.6 is very conservativebecause it assumes that the wind can blow at maximum velocity over threequarters of the span, with negligible velocity over the remainder of thespan. For this reason, it is recommended that Wind on the live load beneglected, because of its minor effect and the large amount of work inmovable wind load analysis.

1.12 Wind Vibration

The hangers for suspended roadways and the columns for a roadwayabove an arch have had wind vibration difficulties in several arches.This vibration is due to vortex shedding. The wind stream is split bythe member, causing turbulence and downstream vortexes with alternatingtransverse pressures on the member, resulting in vibration in adirection at right angles to the wind direction. This type of vibrationwas noticed in hangers of the Tacony-Palmyra bridge, a tied arch inPhiladelphia, built before 1930. The hangers were H shaped members.Horizontal and diagonal bracing between the hangers, in the plane of thearch rib, was added to stop the vibration. This occurred about 1929,at the time the Bayonne bridge (6) was under design. H shaped hangershad been designed for this bridge, but they were changed to wire ropehangers, after the trouble at Tacony-Palmyra became known. The Bayonnebridge was probably the first arch bridge to use wire rope hangers.They have been used for a number of arches since. The wire rope hangersin the Fremont bridge (5) developed wind vibration during construction.Spreaders have been used between the four ropes at each panel point tochange their vibration characteristics.

37

Several tied arches have developed vibration in their structuralhangers. In one case, the H shaped sections were converted to boxsections by welding plates on the open sides. The long columns at theends of the Glen Canyon bridge developed vibration. Two long columnsat the end of each rib were braced by horizontal struts to the rib.

This type of vibration is caused by a steady wind with a velocitysuch that the frequency of the vortex shedding is in resonance with thenatural frequency of the member. The following equations may be used todetermine this velocity for a given member.

V-C\^T Eg. 29mi

and y = %L-Where fw

= natural frequency of th/e member in cycles per secondV

I = moment of inertia about the axis parallel to the windi = length of memberm = mass per unit length of memberC = 1.57 for pinned ends and 3.57 for fixed endsV = wind velocityd = width of member at right angles to the wind directionS = Strouhal number =0.12 for H shape

0.15 for box shape

The longest column for the Glen Canyon bridge (7) has a length of156 feet with an overall depth in the plane of the arch of 313 1 3 1/2"and at right angles to the plane of the arch of V - 11/4". The momentof inertia about the axis at right angles to the arch plane is 19,850in.^ and the cross-sectional area is 82.7 square inches. Assumingaverage end conditions as half way between fixed and free, the frequencyis:

-F ■9¿7 \ /32.2 x29 x l(fi x 19,850 ooc , ,f = 2.57 \/ ■— 9- = 2.26 cycles per secondv V 281 x 1564 x 144

38

The presence of axial load will modify the above figures by the followingfactor:

\/I+ — —i—) 2 » where fis the axial unit stress. The plus signr is for tension and the minus is for compression.

Assuming f = 2500 psi and k = 0.75, the modifying factor is

/| _ 2.5 ( 0.75 x 156 x 12 j2j2 = 0.96V 3.142 x 29,000 15.5

and the critical wind velocity is 0.96 x 34 = 33 mph.

The bracing strut reduced the maximum column length, vibration inthe plane of the arch, to about 100 feet. This would increase thecritical wind velocity to:

,156x2x 33 = 80 mph.

The above equations are based on vibration transverse to the winddirection. Torsional vibration may occur for an H-shaped section butwould be very unlikely in a box shape. Vibration can probably be avoidedby using a box or double laced section, and designing for a criticalwind velocity above that likely to occur for a steady wind at thestructure site.

Since the suspended roadways of some suspension bridges have sufferedsevere vibration, it might be though that the same type of floor vibrationcould occur for a floor suspended from an arch. We know of no arch bridgewhich has been subjected to such vibration. The probable reason is thatthe rib or the tie of an arch is much stiffer for a given span than thestiffening truss of a suspension bridge. The stiffening member of anarch must be designed to carry the live load moment for a spanapproximately equal to one half the arch span. The stiffening member ofa suspension bridge may be made as light as the designer determines tobe adequate. The George Washington bridge, with a span of 3500 feet,had no stiffening truss in its initial single deck condition, whichlasted for a period of approximately 25 years. During this period, novibration of any seriousness occurred. This was due to the fact thatthe dead load cable tension supplied the necessary stiffening. The archrib thrust does not supply similar stiffening because it is a compressiveinstead of a tensile force, and a compressive force amplifies change ofshape, rather than resisting as does tension.

39

1.13 Interaction Between Rib and Roadway Framing

If the roadway longitudinal members are continuous across thecolumns or the suspenders they will participate in the arch bendingmoments, since they will take the same deflection curve as the archrib. They will have, approximately, a participation bending stressequal to that in the arch rib multiplied by the ratio of the floormember depth to the depth of the rib.

The Hampton Road bridge (8) at Dallas, with a span of 192 feet,has 24 inch stringers and 58 inch deep ribs. With a maximum ribbending stress of about 10,000 psi, the participation stress in thestringers is roughly 24 x 10,000/58 = 4000 psi. This could beconsidered a secondary stress in the stringers. The total moment ofinertia of the stringers is about one twelfth that of the ribs, so thereduction in moment on the ribs due to the action of the stringers wouldbe about 8 percent. This was neglected in the design of the arch ribs.

The effect of deck participation is likely to be less as the archspan gets longer.

1.14 Welding and Other Connections

Welding is now being very extensively used for connecting the platesmaking up the cross-sections of box ribs and ties; and for making theshop splices of these members. The corner welds for these box sectionscarry the longitudinal shear between the flanges and webs of the riband tie. Although full penetration welds are sometimes used at thesecorners, they are not needed for stress. By over!aping the web plateon the edge of the flange plate, a fillet weld, meeting the longitudinalshear requirement and the minimum size requirement as specified inAASHTO 1.7.26, may be used. The fillet weld costs less than the fullpenetration weld and is desirable from the standpoint of lessershrinkage stresses due to its smaller volume.

Because the tie is a tension member, it is much more susceptibleto possible cracking due to improper welding than is the compressionrib. The corner welds receive the same stress as the tie, which is thefull allowable tensile stress for the steel used. A defect in the weldmay start a crack that can rapidly spread across the tie. Failure atany point in one tie of a bridge supported by only two arches is certainto cause complete collapse of the bridge. The tied arch is 9 in effect,a simple span, and one end rests on an expansion bearing. Even if thepier or abutments could take the horizontal arch thrust, which they arenot designed for, the expansion bearing would prevent transfer to thepier of the horizontal force caused by failure of the tie,

40

Transverse and longitudinal stiffeners are not needed on the tiein the completed bridge. If the webs need stiffening for transportand erection, some form of temporary bracing, not welded, should beused. Use of unnecessary members and welding adds to the cost andincreases the probability of occurance of a weld defect.

Due to the 100 percent certainty of complete collapse of the spanfollowing tie failure, the obtaining of perfect welds cannot be reliedupon, A tougher steel that is capable of arresting crack growth isnecessary in a tie that is fabricated by welding. The added cost ofthis tougher steel in the tie is a very small percentage of the totalcost of the bridge, and is well justified on the basis of safety.

Suspender Connections

Where the roadway is suspended below the arch, the connections ofthe hangers should be designed to permit inspection with minimumtrouble, and should avoid loss of cross-section in the arch rib. Fora large truss, such as the Bayonne Arch, the hangers may be connectedto vertical gusset plates extending below the lower chord. For solidweb sections, the hangers are sometimes extended through large holesin the bottom flange of the rib. This gives a good appearance, butinvolves considerable loss of section or considerable reinforcementaround the holes. Slots for gusset plates parallel to the arch ribresult in a minimum loss of section. The upper connection, if insidethe rib, can be inspected by.a man having access through the insideof the box rib, manholes being provided at the diaphragms.

Wire rope hangers are more frequently used now than structuralhangers. They have given some trouble, particularly on suspensionbridges, by corrosion of the wires, due to holding damp dirt at thesockets. The connection of the ropes to the sockets at the lower endshould be easily accessible and visible to a man on the deck.Column Connections

Short stiff columns near the crown of the arch may result inconsiderable interaction between the deck and the rib. This can beminimized by using rocker type connections.

Splices

Since the arch ribs are principally in compression, 50 percent ofthe load in bearing at splices, as permitted by AASHTO Specifications,can be utilized for bol tedsplices. In the case of the Hampton Roadrib, both shop and field splices were welded, Welding of field splicesis generally not desirable for an arch because of locked-in stressesfrom weld shrinkage and from the method of erection,

41

1.15 Equations and Curves for Design

The equations of this section and Figures 3 to 7 are intended forpreliminary design use, to arrive at an arch rib section for moreexact analysis by classical methods or by computer, Some of thecurves» such as those for deflection and moment magnification may besufficiently accurate for final design.

This section should be used with Section 1.16, a design exampleusing the approximation of Section 1,15.

Steel Weight

The first requirement for preliminary design is an assumed deadload. The equations for weight of steel include not only that in thearch ribs and bracing, but also the roadway framing steel and themembers connecting the roadway framing to the arch. The reason forlumping these different sections together is their interdependencewith regard to steel weight, It is easily seen that a greater spacingof suspenders or columns will result in more weight of roadway framingsteel and less weight of suspender or column steel. It is not soobvious, but nevertheless true that suspender or column spacing alsoaffects the weight of arch rib steel, The equations for steel weightare:

2-Hinged ArchSolid Web

No tie Ws = 0.18 a + 20 Eg, 30With tie Ws = 0.23 i+ 20 Eg, 31

Truss type rib(no tie) Ws = 0.15 i+ 20 Eg. 32

Fixed Arch

Solid Web Ws = 0.16 z + 20 Eg. 33Truss Type Ws = 0.14 % + 20 Eg. 34

Where W = steel weight in pounds per sq. ft. of deck

i = span in feet

The above equations are roughly applicable regardless of the gradeof steel, assuming that A36 steel will be used in the shorter spans andhigher strength steel in the arch rib main members of long spans.

42

This steel weight will not be uniformly distributed over the lengthof the span. By assuming it as uniform for preliminary design, thetotal dead load thrust will be overestimated» but probably by not morethan 6 percent. If desired, the weight of steel only could be reducedby a factor of 0.9 in figuring the dead load thrust by the uniformload formula. For a partially suspended deck» such as the BayonneBridge type, the steel weight would be slightly less due to the useof suspenders instead of columns.

Thrust

The horizontal component of dead load thrust in an arch may beapproximately calculated by the following:

H = Wj¿ 2/8h Eq. 35U " A/ "

Where H = horizontal thrust from uniform load over thefull span

w = load per ft.i = arch spanh = arch rise

This equation may also be used for uniformly distributed live load overthe full span.

For a concentrated load, P, the horizontal thrust is approximately:

(P at the crown) H = P£/5h for 2-hinged andP£/4h for fixed arch Eg. 36

(P at the quarter point) H = Pi/7.3 h for either2-hinged or fixed Eg. 37

The thrust at any point is approximately equal to the secant of theangle of slope of the axis times H.

Moment

Live load positive moment in the vicinity of the quarter pointmay be approximately calculated by using simple spans of the followinglengths:

Fixed arch, equiv. simple span = 0.36(1-0.1Is/IcR Ecl> 38

2-Hinged arch, equiv. simple span = 0.36£ Eg. 39

where Ie = crown and Is= springing moments of inertia

43

The above moments, which occur near the quarter point, are themaximum, except at a fixed end. The crown moments will be somewhatless than the quarter point moments. Negative moments in a 2-hingedarch will generally be smaller than positive moments. Positivemoments at the springing of a fixed arch will be of the order of2.5 times the maximum positive quarter point moment, and the maximumnegative springing moment will generally be less than the positivespringing moment.

The dead load moments in any arch will be quite small, providedthe arch axis is closely fitted to the dead load equilibriumpolygon. There will be rib shortening moments from dead load unlessthese are eliminated by the fabricating and erection procedure,

Effect of Curving Arch Axis

As previously mentioned, uniformly distributed loads appliedthrough columns or suspenders will produce additional moments due totheir application as concentrations on a continuously curving axis,A positive moment will be produced at the point of load and will haveapproximately the following value:

M = PAi/12 Eg. 40cone. L

Where M = additional moment due to non-uniform application

AL = distance between columns and suspendersP = load

The above moment, from the roadway D.L. and the uniform live load,will add to the maximum positive moments at the columns. The maximumnegative moment, at points mid-way between columns, will be increasedby one half of the above, and will be from roadway D.L, only, sincelive load will be placed on a different part of the span for negativemoment.

Deflection

Live load deflection may be approximately calculated by thefollowing equations:

2-Hinged Arch

A = M2Eg. 4141El

44

Fixed Arch

A = _MiL Eg, 41a76TT

Where A = live load deflection (maximum)M = live load moment (maximum)£ = arch spanI= moment of inertia at quarter point

The above equations may be used for either steel or concrete arches,by using the appropriate value for E, The moment for computingdeflection should be magnified for service loading only.

Dead load deflection at the crown may be calculated by the followingapproximate equations:

2-Hinged Arch

Fixed Arch

ía " L_2+ !§. " h Eq. 42aE 4h E

Where f = dead load axial unit stress at the crown,a

The above equations assume that the arch rib shape has beendetermined by the dead load equilibrium polygon» For a tied arch, fain the first term should be the sum of the dead load axial unit stressin the tie and in the rib at the center line of the span.

Temperature deflection at the crown may be calculated by thefollowing approximate equations:

2-Hinged Arch

o)t(^+ h) Eg. 42b

Fixed Arch

a)t(^+ h) Eg. 42c

45

I>l6 Design Example - Vertical Loads

s T-, = ,

I 1—i

Sp- — — N?SpecificationsL.L. = 2 Iones HS2O-44maximum L.L +1 deflection =

A Je stee!Dead Load (3O1 roadway)slab + bituminous pavement ~ % * /SOX 52 = 4OOO 7"curbs + railing -

IOO *rstructural steel Cea, 30) - 32 ¡O./8(425)+2oJ= 3O9O

~779O ¥

reduction in steel wt for egu/vo/entuniform loaa1 ~

~OJ / 3O9O =- 3O9*

748O 9Live Load + Impact of quarter pointtraffic lanes distributed to one rib = Z ¿¿^ = 1286eouivoJent simple span (ey. 39)

1 O.36X 425 = ¡531

live load moment Ca.A.S.H.T.O. Appendix A)t.286 X 2560 = 329O1*

% impact 5O±[0.5 (4-25) +12Si = 14.8 %LL +Imoment = 1.148 X 329O = 378O /K

46

dead load vertical reaction = JJ4 X179X 179 X425 = 828*

deac/ hod horizontal reaction Ceo. 35)//= é (7.48X42S)2 * (8noy /2IO«

Thrust at quarter point ~secant of slope x Hdead had ~

ÁO6* /210 = 128O

uniform /one load over ho/f span (ea. 35)ÁO6* L286X0.64X 425* + /6 X7O = 141

concentrated lone had at quarter point (co. 37)1.06 X 1.286 x J6.OX 425 + 7.3 / 70 = 2O

impact-

0./4S (¡4l + 2O) = 241465 X

From -f¡g.*6 ~ d - 1.5'*/d = 425/7.5 =561

From fig *5 given */A ft */d = 56.7fbs =89 KS.i.L ~ secant of slope at o.p. x */2

approximated: = ¿040.06 * 212.5) -7o7o

assume fa =6.0 i<siFrom f/g.*r3jt**4

-fb = 8.9 x ¿28+115 -'9.91.K5.1

From f/p.^7 ~ Fa =/3.7K5.i& j. fk - &O +9JH - 0.44 + Ú.SO = 0.94 <AOFa Fb /J.T 2Ú.0

Web sizereouired °/t for web with two stiffeners (eg. 8 )

/o,000/^/60o¿> =129 ~~ use Í2O mox.try web 90"X3^

-D/t = 90/0.75 = /ao

'Sftffener size stiffeners at (j points feo. 9)required Is =2.2*90Xa153 - 83.5in 4

use I**34 plates--

Is = ais*73/3= 85.1in4

Flange sizeapproximate rib area = ¡465/6.0 - 244-*'approximate f/onge area= 244-(2*90X34)~(4xix 34)-8834)-88auuse 34''xI'/ plates 6/f-34/f.25 = z.72rzq 'd. fytleq./l) = 4Z50/YfQ +fb = + 99/0 = 33.7

47

}. 34 .\ Area II 1 ¿webs 9OX 34 135 9/JOO4stiff. 7X24 21 4,730„ , ¿-flanges 34*lk 85 176.9OO

h t\^--7x34 24Ia"

¿7ZJ3O//1 4-

8! Ir n j/ r =V'272,700/24! = /y

~~~ ¿ A = I.O4X/.06X 212,5'X/2/33.6= 83.7I I 11 5 = E7at 7OO /46.Z5 =5896¡n

/7¿>/?9e = = = 4a50/V6p80i-J013IO=33.2Web b/t = = ,>4//w. =m°%l6080~ = /¿>^

Combined stressesfa = 14-65/24-1 = 6.O& KS/from fig. *IO ~ Fa = ¡3.2Ksifrom f/g.

*4

—AF

= /J47^ = /.34X378OXI2/S896 - IO.31KSifa , A - 6¿V9 , /g?3/ ~ O.46? +O.S/6 =O.977 < 1.0Fa Fb

~¡3.2 +

2O.OService Load deflectionfrom fig.*23

—/4^ =//a

f¿s - (ua/t.34) /o.3/=9.08*/d = 425/7.7/ =SS./from fi$.*5

—<^d =/^^

Dead Lone/ rib shortening stress Ceo. 4-P)Hts. = /.875r33.6/7Ox¡a]2 /e/o = 3.63K

rib shortening moment at quarter point3.63XO.75X7O = !9l iK

rib shortening stress ot quarter pointJ9I/ /2 /S396 = a39 KS\

Additional stress from column concentration Ceo. 40)D.L. = £ r47+CaO2X3a)J32.72//a = 238LL.+L = Z./4-8 X i.2S6X0.64X32S2//2 = c?J

f¿ = J£/ /2 / 5*A96 = O.6S /rsifa lft - O.977 l 0.39+0.65- 1089Ta* Ft"

+2Q.O

48

Alternate cross section with thinner web by usingC horizontal diaphragm

Web sizeove. stress m one half of web=6.OS+ y20a3l+O.39+O.65)

= // 76 ksirequired D/t for web with one sfiffener at mid- depthCea. S) 75OO/V//76O = 63.2use 88X 58 web ; °/t = 44/O.62S = 7O.4Stiffener size Cea. 7)requiredIs = O.75 X 44 X O.6253 4

use 4 J/x 38"

plates Is =O.375 x 43'

/3 = 8.O in4

Diaphragm size Ceo. 10a)required ó/t = 45OÓ/W6O8O =57.7use 3Oy/ x f/ plate

- b/f » 30/a5 = 6o

2 webs 8B"x 58 „ Jio nooo= = / diaphragm 3O"x fe 15

4 stiffeners 4" X 38 6 ¿9O0c§ == r ¿flanges 34"Xl¡¿" J0¿_ ¿Q43OO

q 233a" 278,E00¡rÜ-

1 I \ r = -^278000/233 = 345"b=3(?\ KL/r = 104X106XE12.5X12/345 = 81.5

S ■ 278,000/45.5 ~6UO in3

Combined stressesfa = 1465/233

-£¿\9 ÁT5/

fro/7? fig.* 7 ~ Fo ~ I3A5 KS\-from fig.

**4 ~ Af == /3-J5

/i -1335 X 378OX12 /6/fO =9.9/ KSt

Ú dfb _ £L29,9JIL - O.468 s-O.496 ~ O.964 < l.OFa Fb

" TJ^S"^20.0~

Service load He.flectiohfrom fig.*3 ~

5 =;-/7.7Jr/A5 = C/J73/J.335)9.91^8.71t/c/= 425/7.58 o =S6Jfrom fig. *S

-<*/a = /246>

49

Dead load rih shortening stress Cea. 4f)Hrs = /.875f34.5/7OX/2]2 J2IO = 383 X

rib shortening moment at quarter point3.83 x 0.75 X 7O « ¿oi

'*rib shortening stress at quarter point

20/ X¡2 /6//O = 0.39 MSiAdditional stress -from column concentration Ceq. 40)fh = 32/X/2 /6iio = a63MSifa , f6f6 _ 0.964 J-0.39-f-0.63 - LO/5Fa F6F6

~20.0

Temperature stress Cecj f ")assume ove. I=0.91cj.p.///== /5X29 X/X /Q6X0.9X278.000X6.5X/0~6X606X6O _ 75~20

**8 Í7OX/2)*

Stress at quarter point = {7.52/3.83)0.39 = 0.766 KS\

1.16.1 Wind Analysis :—

Double Lateral System

CKIXDKDKKIXMXIXIXIXIXIX^

Wind hod on rib (W)= 1.5x15 ~563r useO.6*The equation for a single centered circular curveand rib of constant cross- section will be usedto calculate Mo - fSee fig.*8)

50

radius (R) =*?+ ÍL = 4¿5_ z , 70 « 351.545*8h 2 £*7¿

J3(radians) = 0.63643Moment at crown (ey.*2¿)Mo g 0.6X351.52 raising- &COSB) jl =3/74 /

L CJ3 -s/n&cosp) Urib stress at crown = 3114/26/24-1 = o.4lKsimoment at Springing =Mocosß-W/M0cosB-W/?2(/-cosJ3) = /2t

46o'*torsion at Springing z=-i35O //<

torsion at Quarter pt z -sin 2̂)= 583 //K

n'6 stress at springing = /2460/25X241 - 1.85 KSilength of half arch = 351.5 £ =¿21.5'

Due to the fact that wind force is applied asconcentrated loads at the panel points, stress in thelaterals } using the approximate equation, should bebased on the angle from the crown to the centerof the panel in question.

It will be assumed that the arch is dividedinto 14 equal panels along the axis of 32.5' each.

Maximum shear in end panelO = C6.5/7.0)C0.63643) = 0.59091Sin 6> = 0.551)1torsion -For end panel =MosinO-WR (0-sinO)

3174-X0.551J1 -Ú.6X 357.5Z{a59097--asslJl)=-824

torsional shear -for laterals (see fig.^/O)V2d

--824/ 2X1.5 =i55 X

bending shear in one lateral systemf2f2

*0.6X6.5X32.5 = 63 X

Combined shear in upper lateral system =8K11 " "

lower" " =/;$*

Máximum shear at quarter pointO = C3.5/7.0)(O.63643) = 0.3/822torsion at q.p.

- 3114x0.31281-16,100(0.31822-0.31281)-583

"X

51

tors¡onal shear -for lateralsT/2d = 583/2X1.5 = ± 39X

bending shear in one lateral systeml2l2X0.6XZ.5AZ2.5 = 34 X

combined shear in upper lateral system ~73// // // ¡owet // // =5*

Minimum shear -for lacing (A.A.S.H.T.O. / 7. 83),/

- PX1465 /"" _ fOO_ / /.26X221/14 7- 34*~2 systemsv joo i/.¿6j22i+j¿ +3300/3S J~4Z«~ / system

Assuming on X-system with the diagonaldesigned to take tension only, the maximumdiagonal stress is 1.53 xus = 18!

*Use a wr8x35.5

diagonal. The maximum strut stress is 118*requiring a WT 6* 22.5. The upper and lowerstruts will be braced together to act os cross-"frames. In determining rotation oí the crownby the approximate equationJ assume an averagelateral diagonal area of 9.4 ¡nz and an averagelateral strut area of 5.9 inz Assuming only onediagonal of the X- system in action, the equationfor X Is'

¿,- AL (t>d)Z .A ~JLTJd j.

*3 7jl Aid

32.5 X7.5f= 4.09ft4

Crown rotation (eg.P.4xi2f?z%000x4.09 [2174X0./766 + 76J00X-6.ooSBz} ■00597

52

Jotero!I= ¿(241/144)04)* = 656 ft4'

LoteroI. deflection at crown from arch rib axial stress(e^za) 227t5

*r a6x 221.5* r}74lA =

2X29.0OOXI44-X656 L 4 ¿"rj= 0.0434* ~ O.5211

Lateral deffeetion at crown from shear stress ¡nwind bracing ¿ span—-r

> /4 spa. <§> 32.S'^ 455' r \%k JÍL" ~J¿? -J£r 'Jít" «-Oí-' 'J^r 'JiUr

<j^^ 4y^ w^t? $y\? *>w/$i4yp

nt L, LZ L3 L4 LS Lb ,/r¿ of unit hod fo \each lateral system

Unit load stresses

Hi S S"^ Nl?. >^ \J^ N^ / \OS r

N K K K Is-' 1^Oi OS OS Q-i OS OS

Wind shear stresses

A = lAL(u)AL = PL/ABU -unit bod stress

53

mem- L A P AL u I ALCu)bar (in) (in2) (Kips)UÜ, 5/5 9.4 91.0 ,1832 .383 .O7O2L,UZ

'f " 82.0 J549"

.0593LzU3

" "Gil .1268 n .O4S6

LsU4" ll 52.2 .0986

".0318

LiUs" n 37.3 .OÍOS

".0270

LsUb" "

22A .O423 » .0162LeUr

" "1.5 .OI4I

".0054

Li üi 336 S.9 -63.4 -.1245 -.250 .0311LzU2

» "-53.6 -.1053 "

.0263Li Uj

" "-43.9 -.O862

".0216

L+Ü4" "

"34.1 -.0610"

.0/61LsUs

" " -24.4 -.0419 ".OI20

UU6" " - 14.6 -.0287 " .0012

LyU7" " - 9.8 -.O/9! » .0048

.3842"

Since we hove considered only !2 of oneIoterol system- A= 4/O.3842 - 154-"This deflection may be determined by the followingobbreviated method also:dia9onol stress from unit lood = k X '2*1.53 - .383Strut a a " " = 'z X '2 = .250ove. shear from wind = '2X0.6X221.5 « 68.2ave. diagonal wind stress = 2X68.2X.383 = 52.2ove. strut « " ~¿X68.2X.25 =34.!Since there ore 28 panels, A =■ ZQ(pl/ae)u

a - pa rS2.2X28Xl.S3X.383 , 3A-.IX28X.25 7a " L 94X29,000 5.9X29,000J- 1.53total deflection at crown =0.52+1.53 - 2.05"

54

// the roodwoy lateral system is five -feet abovethe center of the arch rib at the crown, totallateral deflect/on from wind on the arch rib is2.OS + 5X1¿x. 00591 = 2AI" at the roadway lateralsystem level.

For a straight member with -fixed ends andconstantI, the lateral deflection at the center ofspan due to moment from a uniform wind loadiS WJ/*/384E1. This would give "

0.6C455)*/384X29000X144-X655 -O.OZA-S''-O.30"

as compared to 0.5211 for the arch.

/"IS.¿ Wind Analysis-Single Lateral System

/2spo. <§> 37.925= 455' (arc length) ,

-^~H-(

425' — H~^

For analysis,divide the half arch into six equalsegtnents. Transverse wine shear in the lateralsystem may be calculated by summing the windforces outward -from the crawn for symmetricalWind loading. Tronsverse shears result in tangentiallongitudinal forces CAL WRO/b) acting on the archrib at each point of connection of the diagonalto the rib . For simplification of analysis it willbe assumed that tangential -forces are applied otthe center of the panels.

55

For this analysis , a constant cross-sectionis assumed. The arch rib moments will be defermined■for o 3-hinged condition 4 converted fo Z-hinged

Ff - tangential -force -for one panel = WRO Cal)AL - length of panel b

A/It* -moment at Q of tangential forces= Z*FtD-cosce-eF)jRHj ~ thrust at crown -for 3-hinged condition= I*Ft [/-cosC&-eF)]RM ~ moment for 3-hinged condition= Mt,-r H3R('-cose) R(l-cose)m - moment from unit crown moment

- > ~jj~Mv = ¿-hinged crown moment- -¿-Mm/Lm 2-Hz

= thrust at crown for ¿-hinged condition= Hj +hAv/h6 5 4 3 2 1

I , J^i>*^ Í4Y ~>-fiO-COSO)

\Ft€^\( \ len9th of half a/is^E1.553

y^\^ \ " "!panel(Al)=37.925f

\ \ -&= O.63643 radiansN^- \ angle change for

\£, \ ¡panel =0.106013 rod.N^ \ vertical /=273,000 in4

56

Determine Hj

/ I 2 \ 3 \ 4 \ 5Tpanel 9 Ft Rp-cospÁ 3X4

1 .05304 ¡5.4 59.1 9102 .15911 46.2 4O.O 18483 26518 71.0 24,3 18714 .37125 ¡07.9 12.5 ¡3495 .47732 ¡38.7 4.52 6276 | S8340\ ¡69.5 \ 0.503 \ 85

¡40.9 6690H3

~669O/7O = 95.57* —tension in leeward rib

Determine M** values - feeword rib~ponel \Z\3\4\S\6A<9=O-&r JOSO7 .21214- .31822 .42429 .53O36r[I-cosó9] z.Ql 8.02 17.95 3I.7O 49.12

Ft \ \ t \ \¡5.4 31.0 ¡23.5 488 15646.2 92.9 311 82977.0 GIS

107.9 217Z38.7

~Mtf I 31 \ 216 I 802 I 2152 \ 4741

Determine Mv

1 / 1 Z I 3 1 I I Ipane! m^ ¡-coso fcRfhcose) ¡y\ ~ ¡+3 m M-m mz1 o .001406 -48 -48 .993 -48 .9852 31 .O/263I -432 -401 .935 -375 .8755 216 .O34953 ~I192 -975 .822 -80/ .6754 802 .O68O76 -2325 -¡523 .653 -995 .4215 2152 .1/1765 -3820 -¡668 .429 -1/6 .1846 4747 J654I6 -5655 -9O9 .1543 -I4O .023

Springing 6690 .195783 -6694 O O O OMy

~ 3075/3169 = 97O.3 (leeward rib) -3O75 3169Hz s 95.6+C97O.3/7O) = ¡09.5(tension in leeward rib)Mo

-¡O9.5X28 = 3070 JK

57

Determine arch rib moment and thrust

panel 1 7 1 ¿ 1 /7^ 1 J 1 4 I 5*4A± Mvm moment H2 cos& '^(/-coso) thrust

/-

48 963 915 IO9.4 -3.8 105.62 ~ 4OI 907 506 IO8.2 -34:6 13.63 -97S 191 -118 IO5.9 -95.1 IO.Z4 -I5Z3 633 -690 IOZ.Z -186.1 -8455 -1668 416 -IZ5Z 97.5 -306 -¿096 -9O9 50 -859 91.5 -453 -36Z

sprinohoX O I 0 I O 1 88.1 ~ 536 \ -443

stress at crown - /o9.s/za\ ~ 97oxjz/s896 s Z.42 ks'i" "point5 = -2O9/Z4I --/Z52X/Z/S896 =~3AZks\

LateraI deflection

The de-flection may be considered as consistingof three ports '-

A {rom arch rib oxial stressA from orch rib -flexure in the vertical planeA from shear stress in the lateralsThe axial stress effect may be calculated by

eojuation a as in the doable lateral system.6r

~ 2Z1.S rO.6X2Z7.52-

3O7O1 - 0.0443*~O.SS"c ZXZ9OOOXI4A-X656 L 4 J

Determine deflection from arch rib flexure/. Apply unit lateral load at crown and find tangential

■forces at each pointZ. Calculate moments,Mu, in 3-hinged system from unit load.3. Multiply the actual wind moments by the unit load

momenis. (If is a well Known fact that deflectionin a statically indeterminate system con be calculatedby using the unit load moments in the determinatesystemJ

4 Deflection at crown-

ZM-Mu

58

tangential forces from unit load -O.SAL/Z8

- 0.0/786AL Cull points)H -from tangential -forces =0.0/786A1R F /-cos(#~&r)

J ° TO

&-

O.OS6ALMv /AL =[O.01186 RII-COSÍO-Or)] 'HRQ-COSO)° AAT

panel 1 / 1 * I 3 I 4 I 5 1 771RR zeO/-cos(o~z

eO/-cos(o~e

0/-cos(o~eF) O.Q/786X/ dKO^soi Mu/A¡f^3 M

**$

/O O 0.0/S/ -0.0/81 - 915 16.6Z 2.0/ 0.0359 0./636

-0./261 '506 64.S

3 /0.03 0./79 0.450 -0.211 + 118- 48.2

4 ¿1.98 0.500 0.818 -0.318 +690 -2615" 59.68 /.066 1.440 -0.314 + /252 -4686 /08.80 Á943 2.130 "OJB7 +859 ~161

Y-MMu~ -851.5AL

deflection of crown - 4X 31.9 x144x858 -0.08971~ hoe"Z9.000X273.000

The -factor (4) accounts for the 4 hoIf ribs in the system.The lateral deflection -from chord action is 1.08 +0.53

-1.6/" which is 1.6/ -^O.3-531 times that for o straight

beam. The total lateral deflection - 1.61 +¡,53 =3.14", assumingthe some bending shear deflection as for the double Jotero!system. The ratio of lateral deflection for a single tothat for a double lateral system - 3.14/'2.05'= 1.5

To find the lateral deflection from vertical momentot other points in the span, apply unit lateral forcessymmetrically at the points where deflection is desiredand go through the some process as above. A factor of'c must be applied in the final eouotion since two unitloads are applied.

The lateral deflection at any point from oxiot stressmay be obtained from the approximate equation involvingMo given for the double lateral system.

59

Vertical deflectionA = (AL/£l)lMm/. Apply a unit vertical load to the 3-hinged arch at

the crown2. Multiply ¿ L/El by the sum of the products of the

unit load moments Cm) and actual wind moments (A/I)in the vertical plane to get the deflection.

H due to unit load = */Ahm

" " " » ~[(**Mh)(l-coso)l-(R/z)sin<9

pane/ -Rj-71-r*

— ~ m= 1-2 M M-m9% (¡-cose) % sine/ O.76 9.47

- 8.11 ~+ 915 -1910¿ 6.84 28.3 -21.5 + 506 -IO88O3 J8.94 46.8 -27.9

-118 +4960

4 36.9 64.8 ~ 27.9 - 690 +192505 60.6 8Z.O -21.4

-J252 +26800

6 88.5 98.5 -JO.O - 859 + 8590

+4OJ502X40750X37.9X144 , tldeflection at crown ~29,ooo x 273,000 " O.O56

~ a61

The factor (2) accounts for the 2 half ribs in the system,rotation angle - O.6l/C/2X14) - O.OO4

The total lateral deflection at slab level fromwind on the arch is 3.14 +(o.oo4x60) - 3.38".

To find the vertical deflection at other points inthe span, apply symmetrical unit loads at those points andgo through the same procedure as above.

60

1163 Wind analysis - Strui bracing between ribs

I /4 spo.@ 32.5'= 455'(ore length) ,

Assume 14 equal panels in the arch lengfh.Transverse wind shear wili be divided equally betweenthe tuvo ribs. Points of contraflexure will be assumedin the ribs midway between struts . Balancinglongitudino! shear -forces are then required inthe struts of assumed points of contraflexuremidway between the arch ribs.

Design wind shear- 0.6X455/2. * /SIM

Rib moment h end panel- /3lX3Z.5/z*Z - ¡no

'*Strut shear in end panel

-i3l*3Z.5/2xl4 -

fS9 K

Strut moment m end panel= ¡59X /4 =2230'*

y— facing /tr ~x¿ és 4i"x %

\J[ 4- ¿s 9X4X1K lateral T=Iy-y = 4xlZxZOz + li5X413x-k = ¿9,200 in4

\ ! *"— sirut cross section ~ strut depth equals! rib depth

61

% p w p*

p

<^~K--^ _ — ]—

;■

3W %Yi 3/4w ~j*w IW/nef shears

8ending stresses Cfb)Strut ft, « ¿so x !Z*20.5/29,200 = 18.8 KSilaterolIin rib ~

156 X I6Z + 2.5 X34?X '//B = 48JOO in4

rib fb -filOX¡2 X J7/4&JOO

-4.1KSi

Latero! deflectionAs in the previous shear deflection calculotions,

assume on overage V - 68.2*. From eq.&El, thelateral shear deflection for one pone! is *

.se.Z (32.5X12? f 32.5X12 j. 2X88X12 7 s O. 461"

24-x 29,200 L 4&joo 29,ZOOJtotal shear deflection at crown - 7XO.46I - 3.23"This compares with 1.53" shear deflection withdiagonals , Total íateroí deflection of crown ecjuols0.52 + 1.08 +3.23 = 4.83" f/Deflection at slab level ~ 4.83 +■ 0.004 X6O

~ 5.01 /J

I.IS.4Lotera! BucKHno and Moment MagníficationFor diagonal type lateral system ' (Single lotero])

assume d'iogonol oreo - ZZ ¡n.z" Sf-rut " s¡s tn*XL =I7X ¿27.5 : t JB7^overol! lateral I- 2X241X 14* =■ 94 500 ¡nz ft1from ec¡.fZa_X = -,/;+ Tr^xS^soa / r 42.9 3 , 28 3 1A V / 273* 38.5XZ8ZL 22

+71J= LSI

62

HX ~ = 1.87 X *-& = St.7r ¡4

Vertical y=- governs for allowable compressivestress since it /s 83.7.For lateral moment magnification :Fe = 7T 2X 29,000 / S/.7 £ = 107 j<st

lateral moment magnifier = ! ¿ ///- /7 X 29Z(0

4SZ X IQ7

For struts only between ribs'-XL = 387

'overall lateral I= 94:500 fnl-ftzfrom eg.

*Z& a

X - //+ -rr*tg4sOO f 32.5%144- + 32.5*28X144-1V /a y 213* L 2*4-8,WO 29,200 J= 2JI

X x XL/r= 2JJ X 3§7 = 74.9

14Fe- rfz X 29,000 /74.9 z = 5/ ksí¡otero/ moment magnifier -

m / _ / zs/_ ¿7 X2926

481x SIWind stresses in the rib for the two systems-

(stress at point 5) diagonal typestrut bracing sinole lateral

vertical flexure 2.56 2.56lateral flexure i.isxo.ei^ ¡.09 /.// x0.87= .97local bending t.25x4.6 x5.IS O

9.40 3.53

63

1.16.5 Distribution of Wind Load Between Arch Rib and Decklateral Systems

The wind stress analyses made in 1,16,1 and 1,16,4 are for thetwo arch ribs, braced together, and subjected to the lateral forcesfrom wind blowing directly against the arch. There will also bewind lateral forces acting directly on the roadway deck. Since theroadway deck and the arch are not free to deflect laterally,independently of each other, there will be lateral force transfersbetween them through the columns or suspenders.

The roadway deck will probably have a lateral system of its ownand the roadway slab may also act as a lateral system. If there is nodiagonal bracing between the columns within the arch span, the mainlateral force transfer would be through the short columns at oradjacent to the center of span. The bents at the abutments may havediagonal bracing between the columns, in which case they will act asvertical truss cantilevers to take the wind shear from the deck downto the arch abutments. A rough method of analyzing the lateral loadtransfer is to assume that the deck lateral system and the archlateral system have the same lateral deflection at the center of thearch span, but are free to deflect independently at the other columnswithin the arch span. The following example will illustrate a methodof solving for the value of the lateral force required at the centerof the span, between the arch and deck, to equalize the deflection.

The arch single lateral system will be used in this example. Forthis case, the arch was calculated to have a lateral deflection at thecenter of span of 3.38 inches, for the wind directly against it. Itwill be assumed that the deck system, acting alone, has a deflectionof 2.3 inches from the wind loads directly on it and that a unitlateral force of 1 kip at the center of span will deflect the deck25 x 10"3 inches. These values, of course, depend on the deck lateralsystem, the roadway slab and the end support, and would have to becalculated.

The arch lateral deflection of 3.38 inches from wind loads actingdirectly on the arch, was calculated in Section 1.16.2, The lateraldeflection of the arch for a 1 kip lateral load at the crown will becalculated in a similar manner. This lateral deflection from verticalarch bending is:

óvb= em^ al/ci

64

where Mo is the moment at any point, in the 2-hinged arch, produced bya unit horizontal load at the crown. M~ is calculated by the followingequation:

Mo= M + M m2 v v

where MvMv is the moment at any point, in the 3-hinged arch, producedby a unit horizontal load at the crown; m is the moment at any pointdue to a unit crown moment; and Mv is the 2-hinged crown moment,produced by a unit lateral load at the crown. Mv is found by thefollowing equation:

Mv= £mM v ;Em2

To get the total lateral deflection, the lateral deflection fromshear and from axial stress must be added to the lateral deflectionfrom vertical bending.

The shear from a unit lateral load of 1 kip at the crown is 0.5kips and the shear deflection is calculated in the example by multiplyingthe shear deflection from wind loads by the ratio of 0.5 to the averageshear from the wind loads.

The axial stress deflection may be approximately calculated bymultiplying the wind load deflection by the ratio of one to the totalwind load on the span and by the ratio of the fixed end beam deflectionsconstants of 1/192 and 1/384 for concentrated load at the center anduniformly distributed load, respectively.

An equation to solve for the lateral force transfer at the archcrown is set up in the example in terms of the independent deflections,the unit load deflections and the unknown force W[_. The followingexample illustrates the procedure described above.

65

Rib deflection for IKip lateral load at the crown(Single diagonal system)

Values for nij Zm* t Mu/al ore found in Sec. 1-16.2/ ¿ 5 4 Z+4=S S

¿ =6panel m mu/al MMu/áL Mv/r?/AL MyAL Mfaf

/ .993 -.0/3/ - .o/8O'

.249 .23! .05342 .935 - J27 -.119 .235 .108 .01113 .822 - .211 - .223 .206

- -065 .00424 .653 - .378 - .241 .164 - .2/4 .04585 .429 -.314 - J6O .108 -.266 .01086 1 .154 V J87 I- .02891 .039 V .146 .0219

-.196 .2018

Mv^ 0.796 XAL/3.169 =0.251ALFor / Kip lateral loadlateral deflection from vertical bendingr _ 0.208 X 3793X 4 x144 , 0.824-X/Or* f{~ ~ 9.9 XIO"J/h

2B.O0O X 273,000

lateral deflection from shear (with diagonals)¿£ X f.53 = I4.6x/O~3tn.

lateral deflection from axial stress (approximate).¿74F5 X W * ***:■-*l *k>*'<"-

total lateral deflection =Í99+14.6+41)x/o~s * 2&6x/o~s/n.

Wl ~ lateral force tranfer from arch to deck atcenter of span.

3.38 -28.6 X !O~3

'WL

= 2.3 + 25XIO~5 WL

WL = 3.38-2.3 = 20.2*(28.6+25.0)I0'5

65a

Consideration of Longitudinal Forces for Design Example in 1.16

Moments -from 60° wind acting on the arch "

total longitudinal component - 0.6 X J| X 425 =91K

H= 37+2 = 48.5K

[/= ~91xlOxO.75* 485 =-IZ K Cassuming parabolic axis)

M of quarter point'= 48.5xo.isxio-IEXIO6- 1 1̂^-■ a55O-/27O-64O'-r 640'"

M of Crown = 48.5X10-I2*212.5- 48.5X7OX0.2S-34OO- 255O-85O

Moments -From 6O° wind acting on the roadway "'assume longitudinal component = 6 X 0.OI9X 425 -48.5assume this force to oct 5 -Feet above the arch atthe crown, and to be transmitted to the arch atthe crown.

s 48.5*«01 A

"

65b

H = k x 48.5 " 24.25X

V ' -46.5 X75 ± 485 =-<&6*Mat quarter point = 24.25x0.75x70-8.6X4ss *■4= 1210-9/O= 360 '«

M at crown = '2X48.5x5= ISI'*

Total moment at quarter point from longitudinal windM = é (640 +360) - J^^ //r

Wind on live load (WL) moment at quarter pt -half span loadec

M= O.O3S*J>,2 X 3£O_ = 30<«per r/b

Moment -From L.L. longitudinal -force■force = 0.05x2 (0.64- x 425+18) = E9X

quarter point moment = x * ?̂=?= /O7/k

For Group HT loading with 6O° windMa.p. = 101 -f- 0.3x500 +30 = 287'"longitudinal -fcj.p. = 237 x 12 +5836 = 0.564 ksí

lateral fs [see poqe 57)= 3.4-2. xo.3 x ¥n = 0.349 «sir3 0.933 ks\

Total stress from 60° wind=3.42xg + (s°sqq^ '2- 2. Z4 ks¡

Total stress from 90° wind - 3.4-2 ksi

90° wind governs over 60° wind for this exomple.

67

CHAPTER II - CONCRETE ARCHES

2.1 Basic Arch Action

Much of what has been said in Chapter Iwith regard to steel archesapplies to concrete arches also. It is the intent to discuss in thischapter those features in which the action and design of concrete archesdiffer from that of steel arches.

Concrete arches are generally fixed at the springing, and generallyhave a varying rather than a uniform depth. They may be of the barreltype, in which a single rib is used, with a width approximately equal tothat of the deck, or they may have individual ribs braced together bylateral struts. In recent years a number of hollow box-section concretearches have been built. These are generally a single or a multiple cellbox with an overall width somewhat less than the width of the roadway.For short spans many spandrel -filled concrete arches have been built.These have retaining walls at the edges of the barrel to hold an earthfill on which the roadway slab is placed. Concrete arches are frequentlybuilt in multiple spans, so that the dead load horizontal reactions ofadjacent spans balance each other, and the longitudinal force on thepiers is from live load only. Thus high piers with the arch springingwell above the foundation may be used.

Concrete arches are subjected to appreciable stresses from shrinkageand creep of the concrete under dead load stresses. Stress interactionbetween the rib and the deck is likely to be greater for concrete archesthan for steel arches. This is due to the greater degree of monolithicconstruction used in concrete and the fact that the concrete columns anddeck members are generally larger in relation to the arch rib than is thecase for steel arches. The method used in the construction of concretearches does not generally result in the elimination of dead load ribshortening stresses, as is the case for steel arches. Dead load isgreater for a concrete arch than for a steel arch, resulting in a higherpercentage of the stress being axial and, therefore, a greater pre-dominance of compressive stress over tensile stress.

2.1.1 Dead and Live Load Action

The comments in Chapter Iwith regard to dead and live load actionapply, in general, to concrete. There is, however, one fundamentaldifference in the case of many concrete arches. These are arches whichhave a single, wide rib of either the barrel or box type. Live loadlocated laterally eccentric to the rib center line applies twistingmoments as well as vertical loads to the rib, resulting in torsionalshearing and lateral flexural stresses. This is quite different fromarches with two or more ribs, where live load eccentricity is carried byan increase in vertical load to the ribs on the side of the eccentricityand a decrease to the other ribs.

68

The single rib or barrel arch under live load w with eccentricitye is subjected toa twisting moment, w-e, in addition to vertical load.A stress analysis similar to that used for wind load may be used. Thefollowing equations apply to a uniform load w with eccentricity e overthe full span, and a circular axis.

M = transverse bending moment at crownEC = w-e-RsineMQ

= Mocose-w-e-Rsin^e Eq. 43T = Mosin0 + w-e-Rsinecose Eq. 44

Assuming constant cross-section, and using the method of cutting thearch at the crown and solving for statical unknown, Mo

" (To= zero, by

symmetry).16 13

M = H" /o MmRde +W f0 f-fR'deo -i 1El fQ m2R.de + GK 'o t2-R-de

Where M = moment due to external forces on cut structureT = torsion due to external forces on cut structurem = moment due to unit Mo

= cosQt = torsion due to unit M = sin0G = shearing modulus of elasticity = E * 2. 3

69

we-R 3 w-e-R 3M = El -^0 -sin29coseRd9 + GK sin26coseRd0

El 0 cos^eßde +GK 0 sin^-eßde

? ,EI= 2R w-e-siVg {W - 1) Eq> 45

3 [ (3 + singcos3) + (3 - SÍII3COS3) ~ ]

2.2 Buckling and Moment Magnification

A concrete arch will generally have a greater dead load thrust thana steel arch of the same span. However, at service load the stiffness,El, of the concrete arch is generally greater than that of the steelarch. Although the dead load may be twice as great for the concretearch, the stiffness at service load may be 1.5 times as great as thatof the steel arch. In addition, since concrete arches are generallyfixed and steel arches are generally two-hinged, the k factor forbuckling length of the concrete arch may be about seven tenths that forsteel arch. For a concrete arch, therefore, the ratio T/AFc,used inmagnification Equation 1, may be 2 x 0.7^:- 1.5 = (approximately) 0.65that of a steel arch. Thus moment magnification is generally less atservice load for a concrete arch. The combination of more stiffness atservice load with the effect of fixity causes the fixed concrete and tohave about 1/3 the live load deflection of a two-hinged steel arch.

At ultimate load, however, the picture changes completely and theconcrete arch becomes much more flexible. This is due to the wery largedownward curve, from a straight line, of the concrete stress-straincurve as ultimate strength is approached. The AASHTO SpecificationsArticle 1.5.34 give a value of concrete El = (EcIc/5) + ESE SIS forcompression member magnification of live load moment. Since an archmay have as little as 0.5 percent reinforcing in each face, the aboveequation may reduce the El of an arch section to 1/4 the value atservice load. The net result is that a concrete arch will generallyhave considerably more moment magnification than a steel arch, for designpurposes.

AASHTO Article 1.5.33, Compression Members With or Without Flexure,and Article 1.5.34, Slenderness Effects in Compression Members may beused for the design of the concrete arch cross-sections. These articlesare under Load Factor Design. Service Load Design makes use of theseLoad Factor Design Articles by the application of certain factors.

70

Since ultimate strength, rather than working stress, is therefore thebasis of both design methods, it is recommended that load factor designbe used for arches. It is also recommended that a minimum of 1 percenttotal reinforcement be used for concrete arches. More reinforcementmay be needed because of tensile requirements, or to reduce momentmagnification.

There is a question of the value of cj> to be used in AASHTOEquation (6-15). Although <fc should be taken as 0.7 (except forP < 0.1fcAq) for determining the section resistance to combined momentand axial load as outlined in Article 1.5.33, it is recommended that <j>be taken as 0.85 in Equation (6-15) for the determination of momentmagnification. The reason for this recommendation is that momentmagnification applies to flexural action only. This <f> value is alsoconsistent with the numerical constant of 1.18 used in Equation 3 forLoad Factor Design of steel arches.

Moment magnification should not be applied to either stresses ordeflection produced by rib shortening, shrinkage or temperature. Thereason is that this deflection, unlike partial live load deflection, isin the same direction (either up or down) over the entire span. As aresult of this type of deflection the horizontal reaction and theposition of the thrust line can, and does, adjust to this deflection.The calculated amount of H is correct for the final arch position,because of camber.

2.3 Ratio of Rib Depth to Span

As explained in the previous section, live load deflection atservice load is quite small for a fixed concrete arch and, therefore,does not govern the depth of rib to be used. Buckling in the planeof the arch and moment magnification are the important factors indetermining the rib depth. Table Igives dimensions and data forexisting concrete arches. Six of the eight have a ratio of span torib depth at the crown between 70 and 80. A ratio of 75 is a goodaverage figure for a fixed-concrete arch. The Sando Arch (9) inSweden has a ratio of 99. Radious of gyration, rather than overalldepth, is a measure of resistance to buckling and moment magnification.The radius of gyration of a solid section is about 3/10 of the depth,and about 4/10 of the depth for a box section. On this basis the boxsection might have 4/3 the span-to-depth ratio of a solid section.However, there would appear to be no advantage, from the standpoint ofeconomy, in using the smaller depth for a box section. The width ofrib and thickness of slabs and webs can be such that the area ofcross-section can be made a minimum consistent with stress.

Excluding the Hokawazu Arch (10), which is two hinged, the otherfour box sections have a ratio of springing depth to crown depth varyingfrom 1.55 to 1.72. A value of about 1.7 is suggested for the ratio.From the crown to the springing the depth may vary approximately linearlyalong the arch axis.

71

TABLE I - EXAMPLES OF CONCRETE ARCHESGladesvi 11Australia (12)le Sando Sweden (9)

Port ElizabetSouth Africa(13)

okawazu Japan(10)Cowlitz RiverWashington St.(14)

MississippiMinnesota (11)

MemorialWash. D.C.(11)

Arroyo SecCalifornia (11)Deck-Width(B d ) 84' 42' 85.3 33' 31' 57' 94- 32'Span (1)Rise (h)

1000' 134

866'130 1

656'145' 557' 87'

520'148.6 1

300' 80'

180'27.4" 105' 45'Width Rib 4@20l 31 ' 47. 8181 26. V 27' 2@1 2 ' 86' 2@3. 5151

Rib Depth -Crown 14' 8.75' 9.02' 7. 87" 7171 4.0 1 2. 25' 2.0 1

Rib Depth-Springing 23' 14.75' 14.0' 9.84' 12' 10.00' 5.86' 3.5 1

No. of Cells 4 3 3 2 3Slab b/t 14.4 9 19.6 11.3 12.6Web b/t (max) 19.5 12.75 T4.1 5.5 16Rise/Span 0.134 0.15 0.22 0.156 0.286 0.267 0.139 0.43Span/Crown DepthSpan/ Rib Width 71.512. 5151

9928

72.813.7 70.821.2

74.3 20

7525

801J92 5321f'c in psi 6000 5750 5700 5700 4000 2500? 3900 2500?Spring Depth/Crown Depth 1.64 1.69 1.55 1.25 1.72 2.5 2.6 1.75Rib Area-Sq. Ft(at crown) 292 89 100 72 51.5 96 193 14

All of the above are fixed except Hokawazu which is 2-hinged

1 The 20-foot wide ribs are separated by1-foot gaps, and connected by prestressed diaphragms at50-foot intervals. The ribs are precast in lengths of about 10 feet and have no continuous

72

The span-to-crown depth ratio of 75 appears to be applicable tosolid as well as box section. The lower value of 53, used for theArroye Seco Bridge (11) may have been for architectural reasons,possibly to obtain a narrow rib in relation to depth in this shortspan, high rise arch. We see no reason for a larger ratio of springingto crown depth for a solid as compared to a box section and, therefore,recommend that 1.7 be used for both.

2.4 Rise-to-Span Ratio

Just as with steel arches, the rise-to-span ratios for concretearches vary over a wide range. The ratios for the arches listed inTable Ivary from 0.14 to 0.43. The site in combination with requiredclearances and roadway grades generally control the rise and minimumspan. For a bridge over a deep valley either the span or the rise, orboth, may be increased for reasons of economy. In a single span over acanyon the rise may be reduced and the span increased by raising theabutments, to effect economy.

A high rise-to-span ratio for a given span reduces the dead loadthrust and the moments from rib shortening, shrinkage and temperaturechange, but the wind stresses are increased. The reduction in dead loadthrust reduces the live load moment magnification. The resulting addedlength of rib, of course, adds material cost and construction cost.

2.5 Rib Shortening, Shrinkage, Temperature Effects and Camber

Rib shortening, as used here, refers to the shortening of the archaxis due to axial stress and the stresses produced by that shortening.Shrinkage refers to arch axis shortening and resultant stresses due tothe drying out of the concrete after setting. Temperature involves thelengthening or shortening of the arch axis due to rise and fall of theaverage concrete temperature from the temperature at the time of closure.These effects are modified by the plastic quality of concrete resultingin creep and stress relaxation. Changes in length of the axis and theresultant deflections which determine camber will be discussed first.

2.5.1 Permanent Arch Deflections

Since dead load is a permanent load, plastic flow or creep increasesthe initial deflection over a long period of time, at a decreasing rate.The following equations can be used to calculate ultimate crowndeflection, approximately:

73

Elxari ArchInitial Ars

= fa/Ec (h + £ 2/4h) Eg. 46a

Creep Ars= 2.4 Initial Ars (Additional ) Eg. 46b

As= 0.0003 (h + ü 2/4h) Eq. 46c

At= cot (h + £ 2/4h) Eg. 46d

Two-Hinged Arch

Initial Ars= fa/Ec(h + ¿ 2/sh) Eg. 47a

Creep Ars =2.4 Initial Ars(Additional ) Eg. 47b

As= 0.0003 (h + £ 2/sh) Eg. 47c

A t= wt (h + ¿ 2/sh) Eg. 47d

Where f_ = axial unit stress at the crowna

Ars= crown deflection from dead load rib shortening

As= final crown deflection from shrinkage

A = crown deflection from temperature change

a)= coefficient of expansion = 0.000006

t = temperature change in degrees Farenheit

2.5.2 Arch Stresses from Rib Shortening, Shrinkage and Temperature Change

Stress relaxation does not reduce deflection but has a considerableeffect in reducing the stress effects of creep and shrinkage. The initialrib shortening deflection and stress are elastic effects of load. Theeffect of creep is to increase initial deflection and stress. At thesame time that creep is increasing stress, relaxation is reducing stressbut not deflection. There is a residual net stress effect from thecombined action of creep and stress relaxation. This residual stressfactor will be taken as 0.38. Therefore, the final stress from deadload rib shortening will be 1.38 times the initial elastic stress dueto rib shortening. In other words, the initial elastic rib shorteningstress is a load stress which in itself remains unchanged, but thedeformation effects of creep and relaxation add 38 percent to this initialload stress, in a period of time.

74

Shrinkage is entirely a deformation effect, having no connectionwith any load. Shrinkage stress, therefore., is subject to the fulleffect of stress relaxation, The shrinkage factor of 0.0003 should,therefore, be multiplied by the relaxation residual factor of 0.38,giving a net shrinkage stress factor of 0.00012.

In line with the above discussion, the following equations shouldbe used in calculation of reactions and stresses:

Fixed Arch

Hrs = -*> [0-5 (rs +, re) 2 x Eq _°h c c

Hs= r|o [Q. 5(rs + rC,)-|2rC,)-|2 x QOOOU EAc Eg. 48b

H (drop) = :ü¿ [0.5 (rs + rc)j2 xct £A Eq> 48ct 8 h c

M = -H [y - h(l - 0.33 /dc/ds )] Eg. 48dx

2-Hinged Arch

Hrs=

"g- [-^1 x 1.38 fcAc Eg. 48e

H = 2 0.00012 EAr Eg. 48fs o h c

Ht(drop)

-zl| £re.]. x ct EAQ Eg. 48g

Mx= -Hy Eg. 48h

rs= radius of gyration at the springing

rerc= radius of gyration at the crown

y = vertical distance from springing to point x

f & A = unit stress and area of cross-section at the crownc cMc

= crown moment

M = springing moment

75

2.5.3 Reduction of Rib Shortening and Shrinkage Stresses by"Construction Methods

The majority of concrete arch designs allow for full ribshortening and shrinkage stresses, and only minor steps are takenin construction to reduce these stresses. One generally used methodis to pour the arch rib in separated sections along the axis. Key-ways are left between these sections and reinforcing is lapped in thekeyways. The keyways are poured last. This will result in somereduction of shrinkage stress, due to some shrinkage having occurredin the time period between the start of concrete placement and thefinal pour. The reduction effect is greater in long span archesbecause of the greater length of time involved in concrete placementas compared to short span arches.

Stresses opposite to rib shortening and shrinkage stresses can bejacked into the rib before final rib closure. This is similar to themethod in steel arches. Freyssinet developed one method. The twohalves of the rib may be separated at the crown by an opening. Pairsof jacks are used at the introdos and extrados at the crown to furtherseparate and raise the two halves of the rib. The jacking forces areprecalculated to force initial stresses in the arch for the purpose ofcounteracting rib shortening and shrinkage stresses. The sum of thejacking forces must be wery close to the crown thrust from thevertical load at that time. By making the lower jacking forces largerthan the upper forces, negative moment (compression in the bottom ofthe rib)is forced into the arch. The jacking results in separating thetwo halves at the crown, and lifting the entire rib from the falsework,Concrete is then placed in the opening created by jacking, and thejacking forces are transferred to the crown key section after it hasattained sufficient strength. The crown jacking produces rotation, sothat the key section is wedge shaped. The calculated stresses to bejacked are opposite in sign and considerably larger in magnitude thanthe rib shortening and shrinkage stresses, because of stress relaxationUsing the previously mentioned residual factor of 0.38 for the combinedeffect of creep and stress relaxation, the calculated total, final ribshortening and shrinkage forces at the crown must be multiplied by1 t 0.38 = 2.6 to get the jacking forces.

Other methods of producing counteracting stresses, just before thearch becomes self supporting, will be discussed under arch construction

76

2.6 Buckling of Elements of Box Cross-Sections

The slabs and walls of concrete arch box cross-sections should bechecked for buckling, just as are the flanges and webs of steel archboxes. The slabs are more critical than the walls because the averagestress in the slab, at any cross-section, is practically equal to thesum of the axial compression and the bending compression, whereas theaverage stress in the wall is equal to the axial compression.

The following equations are based on two independent sets of tests,one set by S. E. Swartz and V. H. Rosebraugh, reported in the October 1976ASCE Proceedings; and the other set by G. C. Ernst, reported in theDecember 1952 Journal of the American Concrete Institute. The equationsgive results about midway between the two sources.

For slabsfa + fbb/t = 80(1 - 0.85 ), maximum = 20 Eg. 49

allowable fc

For wallsf

b/t = 80(1 - 0.85— —,, — — ), maximum =20 Eg. 49aallowable fc

Where fa= computed axial compressive stress

fjj = computed bending stress

allowable fc = maximum allowable stress for the concrete,without consideration of buckling

b = clear width of slab or wall

t = thickness of slab or wall

The above equations will permit full stress in the slab, with noreduction for buckling, for b/t = 12 or less, and a maximum reductionof stress of 12 percent is required for b/t = 20.

The same is true for the walls, except that the rib bending stressis not considered, and only the rib axial stress is used. Generallyb/t for the walls can be made 20, except in some cases where the wallmay become too thin for proper placement of the concrete.

77

The slabs should have a minimum transverse reinforcement of0.5 percent, and this reinforcement should be equally divided betweenthe top and bottom. The transverse reinforcement should extend to theexterior faces of the outside walls and be anchored by standard 90degree hooks. The webs, or walls, should have the same minimumpercentage of reinforcement as the slabs.

The longitudinal reinforcement should be a minimum of 1 percentfor both slabs and walls, and should be divided equally between thetwo faces. This minimum longitudinal reinforcement is desirable forreduction of dead load axial creep and of live load momentmagnification.

Table I shows the ratio of b/t for the slabs of 5 box-sectionarches. The only one which considerably exceeds 12 is the PortElizabeth Bridge in South Africa. The b/t ratio for the slabs of thiscross-section is 19.6. This ratio, by equation 49, would require astress reduction of 11 percent.

The proposed equation may be somewhat conservative. However, inmost arches, the value of b/t = 12 can be easily met by changing therelationship of overall box width to slab thickness, with no loss ofeconomy.

2,7 Wind Stress and Wind Deflection

The equations and methods for wind stress analysis and winddeflection of steel arches apply to concrete arches as well. Concretearches consisting of a single rib should be analyzed by the method ofArticle 1.13.2. Where two or more ribs are braced together by transversestruts, Article 1.13.1 and Design Example 1.22 should be used.

Table 11, on the following page, gives formulas for the torsionalstiffness constant X and for torsional shear stress.

2.8 Interaction Between Rib and Roadway Framin

Interaction between the roadway framing and the arch rib is likelyto be greater for a concrete arch than for a steel arch. This is mainlydue to the relatively deeper members used for concrete framing and tothe greater rigidity at the intersections. A model arch was tested atthe University of Illinois,and the results are described in theASCE 1935 Transactions, page 1429, "Concrete and Reinforced ConcreteArches - Final Report of the Special Committee."

78

Table JL ~ Equations for X and Torsionol Stress

The equations -for a single cell box may beused for a fhree cell box with negligible errorin K . The effect of the two 'interior walls is toincrease the stress at the center of the longside by about 137* .

Section Tbrsional Stress

eg. 5O a eg. 50db

5

XT3.1 T

For Yd ¿ 10

v. & ey 50 b eg. 50c

td —-Jl f

'td2b2dZ T

+% Zbd t

Use t -for sideH^>of desired stress

eg. 50 c eg. 50f

4.czdzt(3C+sd) T(C+2d)Z(c+d)z-dz ZcdtCJc*sd)

79

This test arch had a springing depth of twice the crown depth,with nine column spaces, and the deck framing was against the rib forfor the length of the center space. The moment of inertia of thedeck was approximately 1.5 times that of the rib at the crown, and thecolumn moment of inertia was 0.4 that of the rib at the crown.

A horizontal force of 2540 pounds was required to reduce the spanby 0.1 inch and this produced a moment at the springing of 131,300 inchpounds. Calculations, for the arch rib acting alone, show a horizontalforce of 950 pounds and a springing moment of 59,000 inch pounds for aspan reduction of 0,1 inch. Thus the effect of the interaction was toincrease the springing moment due to shortening of the rib fromshrinkage, axial stress, and temperature by a ratio of 2.2.

The total moment from these causes, at the mid-point of the span,is increased in a ratio of 4.1 by interaction. The springing momentwould be carried by the arch rib alone, but the moment at the centerline of span would be carried by the combined action of the deck andrib. This is just one example, of course, but it gives an idea of theeffect of interaction on stresses from shrinkage, rib shortening,and temperature.

The effect of interaction on live load stress would be a reductionat the springing and at all points in the rib, The combination of rib,columns, and deck framing would act as a Vierendeel truss to resistlive load moments. The net effect of interaction would be to reducethe total stress from all causes in all parts of the rib.

If the rib is designed to act alone under all loads, it should beadequate. Some stresses from arch action will be induced in the columnsand deck framing, but these can be considered as being in the categoryof secondary stresses in a truss. Just as in a truss the secondarystresses will be minimized by using slender members, for the columnsand deck framing.

Some arches, particularly in Europe, have very slender ribs,designed to take only axial load, and deep deck members are designed totake the moment. This is similar in action to a tied arch in which adeep tie is designed to take the moment, and a slender rib is used.

In some cases, expansion joints have been introduced in the deck,within the arch span, to reduce interaction. This is not desirablebecause expansion joints should be kept to a minimum and because of apossible high stress produced in the rib in the vicinity of the joints.

The 1938 ASCE Transactions, page 62, has a very good paper byNathan Newmark, "Interaction Between Rib and Superstructure in ConcreteArch Bridges."

80

2.9 Lateral Buckling and Lateral Moment Magnification

Arches of the single barrel or multiple box type are appreciablystiffer laterally than vertically so that lateral buckling and lateralmoment magnification are of minor consequence. When two or moreindividual ribs are used, lateral stiffness may play a more importantpart in the design, Referring to Table I, the Sando multiple box archhas a i/b ratio of 28, and the Cowlitz River multiple box has a i/bratio of 20, The Minnesota multiple rib arch has an individual rib spanto width ratio of 25, and the two ribs are braced together by severaltransverse struts. The Arroyo Seco multiple rib arch has two ribs withindividual span-to-width ratios of 21, and the ribs are braced togetherat each column with a spacing of 10 feet 6 inches.

Thus the Sando Bridge is the most flexible laterally of thisgroup of bridges. A multiple box section has a transverse radius ofgyration approximately equal to 0.32b. Referring to Section 1.11, andinterpolating in the table for XL; the Sando Arch, with a rise to spanratio of 0.15, has a ki/r ratio of 1.11 x 1.05 x 433 * 0.32 x 28 = 56.4.In the plane of the arch, the Sando Bridge has a ratio of crown depthto span of 1/99 which would give a k£/r value of approximately0.55 x99 x 0.7 * 0.4 = 86. Thus, for this arch, lateral buckling isconsideratly less critical than in-plane buckling. The span-to-ribwidth ratio of 28 is entirely satisfactory.

The equations of AASHTO Article 1.5.34 may be used for Load FactorDesign Moment Magnification, As with in-plane moment magnification,a <f> factor of 0.85 may be used and Cm should be taken as 1, Use lateral

For moment magnification to determine service load unit stresses,the elastic value of El should be used and <j> should be taken as 1.

A concrete arch with narrow ribs braced together by lateral strutsmay be investigated for lateral buckling and moment magnification by themethods of Section 1.11.

2.10 Load Factor Versus Service Load Design

An arch rib is a slender compression member with flexure, AASHTOSpecifications for Service Load Design of such members refer to thepertinent Articles under Load Factor Design, requiring a factor of 0.35to be applied to capacity and a factor of 2.5 to be applied to theaxial load for use in moment magnification. In effect, UltimateStrength Design is factored for use in Service Load Design, It is moredirect to simply use Load Factor Design. A desirable feature of ServiceLoad Design, the calculation of unit stresses, is not used in AASHTOSpecifications for compression members, A designer using Load FactorDesign may wish to calculate unit stresses, as well as deflection, atservice load to gain a better insight into the member action underservice load.

81

Load Factor Design for compression members subjected to flexureinvolves the construction of an Interaction Diagram. Such diagramsare generally available for solid sections. For box sections theInteraction Diagram can be constructed according to AASHTO Article1.5.31 - Design Assumptions. Load Factor Design for Reinforced ConcreteBridge Structures (29) by P.C.A, gives examples of calculations forsuch construction.

2.11 Minimum Reinforcing Steel Requirements and Other Details

A minimum of 1 percent of reinforcing steel should be used in aconcrete arch rib. For a wide barrel type rib this reinforcementshould be divided half and half between an upper and lower layer. Fora box section this minimum should be distributed uniformly over thecross section. Two layers of steel should be used in both slabs andwebs. The use of this steel reduces creep and gives a minimum tensilestrength. This steel may be very highly stressed in compression dueto creep. This is counted on for stiffness in moment magnification.

Additional reinforcement may be required because of tensile stress.As required in the AASHTO Specifications, a design dead load reductionby 0.75 should be used in investigating tension.

For box sections, diaphragms should be used at columns. Columnsmay be used in pairs, or a single wide column at each panel point maybe used. In addition to the diaphragm at columns, diaphragms arefrequently used at the mid-panel points between columns.

2.12 Design Example

As an example, a concrete arch for the same conditions as thoseassumed for the steel arch design example of Chapter 1 will be designed.The initial preliminary design for the concrete arch will not be givensince it is similar to the method used for the steel arch. The moreexact analysis is used, and it is revised to bring the assumed cross-section closer to the required cross-section. The followingassumptions in regard to the cross-section to be analyzed are made:

Use cellular type box cross-sections

Rib depth at crown = 4p. = 5.67, try s' -9"75

Rib depth at springing = 1.7 x 5.75 x 9.78, try lO'-O1

Rib width = 4.25 f 20 = 21.2, try 20-0"

82

Table Igives examples of the dimensions of a wide variety oftypes and spans of concrete arches. The last line in the tablegives the area of the rib cross-section at the crown. The CowlitzRiver Bridge is the closest to the design example. It has a cross-sectional area at the crown of 51.5 square feet. The design examplehas about the same roadway width but a span about 8/10 that ofCowlitz River. A two-celled section with 10 inch webs and 8 inchslabs will be assumed. The crown area A is:

C

20 x 5.75 - 17.5 x 4.42 = 37.7 sq. ft.

The ratio of b/t for the slabs = 1^ = 13.1

Assume 35 #6 bars, top and bottom, in slabs

p for slabs =2x 35 x .441 ,fio,

H 240 x 8'- b/o

Assume 6 #6 bars, each face, in webs

p for webs = 2x6 x .441 = -j M53 x 10

The weight of the rib can now be calculated. The deck weight willsimply be assumed for the purpose of this example. Normally, of course,it would be designed first in order to more accurately get thesuperimposed dead load on the arch.

83

Concentrated CoJumn Loads ÍC)C, = 0AC5.0XZ2.-r)+80.0 +Í2.0x 66.89) = 219Ct= 1.1( " )+ " +(20X47.90) = 356

h- /3 gy $pq§> 3269 . c*= l.o( " )+"

+(2.0X32.03) = 308C4= " + " +(2.0x 19.36) -282

1 I T~TI^ I I I Cs= " + " -I-(2.0* 9.88) = 263V^^ - ~~~ ;̂::i^J C*~ " + " + (2O* 3.52)-

251V<>^ S C7= ".'■."* " +(2.0X0.38) ■= 244

./</^ JL / * 1983*N< :±± - Arch Rib Loads (P)„ i- 4E5' I I load I sec.<j> I AS I dx I Aa \ kips

ft 1.n04 36.26 9.45 46.95 269.4/? 1. 1211 36.65 8.45 44.4-6 244.4

2O' % 1.0791 35.28 1.60 42.33 224.0

HS2O-44 t-— ~ H 1.0452 34.17 6.85 40.46 207.4

f'c = 4.oks¡ op ps ¿oeo3 33.35 6.30 39.08 195.5

fr =6o.OHs¡ auio rp i r 1 1 p* 10051 32.86 5.95 38.21 188.3rdwy. deck ~ 5X/< -' to" /o" ñ 1.0003 Z6.36 5.11 31.16 92.1bent cap ~ 8O« N [ I i TT Tf *

ft, dx 4Ag at segment center 1421.1Column ~ 2 /' ' o. s-reinforcement: | *

r Cs C,6 \webs- IZ-7>'s ea. web S: 0. S r L-^-l-r-top and bott. slabs~ § b r ,3,

3 12Ployers-*6's*7" Í^P^S \\l^J_c'e'- 75»

i^Cp '^ |. /6,36' - 355^---'I/a L . 3¿7/

'' 6/53"^ ' [.49.QS¿ -

123122 |.65.40[ -

1¿786

V |, 81.14' , -21498

96.Q9'. - 20344©r?\ n\ (T\ r^\ r^\ i^í\ r t

** /14.43' -32269(¿) (3) (£) (¿) (6) (2) *— Column , /30.7a'

' _01 "O)j -Q rvi V) X»| "dpi m 147.IB1 - 45313

,$ <ti Ob N \(j Vtí ifjI 163.47' - 39952SOA

— — —jr\ 179.81

- 640/e

P/ 2/S.SO' 7 +6642/1v_ S.<9;- J-^ *" ""

Í3226159.0' J^ 11- 32Z675 _ 4609.6*<¿ °{ y ,„<' 1 7c

"

/4r<:A /?/¿ Depth (dx) Determine Dead Load Eaui/ibrium Polygon

A Y = s/?ecrr 4XX,/ /iAssume parabolic arch axis: load shear ¿X .A y lAy

C 0.00

f? -269.4Sec 0=V/^f3f/\ z 2856.3 "

I0.13l7y Ct -356 21.21250Q3 a 8.87

\* \ \ \ \ JPi '

2fstt . cooliJJliJJr^*"4^ Cs 3of 38.08Ar^Z^-^~~^ 19479 "6.91Jl^gy P* -224.0

\ Ci -282 51.10\ 1 1441.9 i< 5.1Íy /? -207.4

Column length - /'+ -d*sec d 1234S " 4.38¿ I Cs -263 60.59

ri*

i +, 971.5 a 3A41 Sb9' Ps -'95S2 4190' 776° " 27S% «I", U '251 66.183 32.03 5250 „ ,86,864 19.36' p ,oP7Í. %%%, 336.1 a 1.19% %5¿ c7c7 -244 69.84' Uóti 92.1 8.18 0.16

R -92.1o.o "

0.00crown \ | I I10.00

84

Concentrated column loads applied to the Assume the thrust Une willarch produce an angular . equilibrium be a distance d Co.18') above thepo/yyo/7. The arch axis is a smooth curve. Keep arch axis at the springing anddead /oad moments to o m/rrimum hy a distance d /z (o.Oi') helow thepassing the axis a distance (d) below the arch axis at the crown. A newthrust line at columns and half this distance horizonfal thrust results '■

( d /2) above the thrust line at midpoints 322675 32.692^ t-65335'between columns. H = 70.00-0.2 5 ~ 4626.2* A.

A .UI4.4Z3,.Reca/cu/ote the thrust line I9.Q6' '-\

—= ——7 —^-CT^^.coordinates and construct the 29.73' ~^P^\ *"- —^-I2^\^§arch axis to pass through those 2I.IS 1 \/\ \

X /*~<./ÍN/g points. \ \ \ / /7thrust "a \£t \ / / /

fa 4 line ~\Jl—r \\ \ / //v^Ai J-^^^ % a * /s ~^^^^^ \

!oad \ shear \ AX 1 Ay \ IAy f Ay -~j\ \\/z\^LZUl \^^Sc\><^~~~ 3125.1 16.345 II.O4 Five Centered Curve Jk~A'\"^T/^'^y \ 2SS6.3 " 1O.O9 I

i

¿2 21.32 2115 x > x "AX J 25OO.3 " 8.83i4.pt P? A 3/

= V Cll4. 423) z-f

(R, -19. 06)* -352.988'ZZ55.9 " 7.97 Ci = -^0/4.423)2 +(l9.O6) z = 1/6.OOO'

If a Circular arch axis of constant radius 0?J fs 3&/¿ 379J £, = 2 sin' 1 (58. 000/352.938) = /8.9145 oassumed-

f

1947.9 " 6 88 X' = (Rz-Ri) Sin fr = O.3Z4151 ]

R¿- 1/4.4-23'/?= -VC272.S) Z -i-Cf?-7O.O) z = 358 p^

y' = (R z -R,)cas£, -O.946O03Rz-333.SZB'

"5j= l/V ft Ss = Al 1723.9 " 6.09 R* ~ -Vci79.8oe + X')z + (304.138 1- y' ) z = 369.473'

H AÁ

AX C4 SI.O9 5O.94 X' = 5.344'd¡= (Sa-Sb) ~2~ 1441.9 " 5.09 Y ~ I5595

p 4 Cz = -VC65.385)* +(29.79) z = 7/55^'0= Sin' ('2AS/R) /2345 „ 436 J3¿ = 2 Sin' 1

(35.926/369.473) =11.1600°

d z = R- O -Cose) Cs

■ 60.55 60 41 X" = (#3-Rz) Sin (£,+Pz) = 0.5O//aSR 3 -I8S.ISZ 1

d - d z -d, g ll5 11 34.3 ' Y" = (R3-Rz) COS Cfi +J3z) = O.865375 -R 3 -3l9J33'1-5 p s ' " R 3 = -j(2/2.5O+X'+X")*- + (282.988 + y'+ y") ¿ = 394.918'kcdfion \ Vi fI S*\~S T~^ T~^ \ d z \ d I 776.0 " 2.14 K~ '2.152'hot*» V> 4/ g S A S s fm s (fit)

m Ce 6612 66.58 X* = 22.0/9'

1 3O58 .6634 .6488 .24 3.O6 .51 .18 525.0 " 185 Cj = -VC 32.692) Z + (2I.I5O) Z = 38.937'

2 2439 .529/ .516/ .21 2.93 .47 .17 p ' -/3 3 = Z Sin' 1

(Z9.468/ 394.9/6) = 5.6514°

3 1392 .4IO4 .3383 .20 2.82 .43 .IS 356J » ¡, g IP = 35.7259°

4 I39O .3015 .2903 .18 2.74 .41 .15 q7

' 5577 69.S3 ___T _____^^___^_^_^___^__5 923 .2002 .1894 .18 2.61 .39 .14 92 7 8.18 0.16 \cohmi\ / I 2~\ 3~] 4~\ 5~\ 6~\ 7~

6 478 .1037 .0936 .17 2.63 .38 .14 p 7 X 212.5 179.81 147.12 114.42 81.73 49.O4 ¡6.35

7 1 46 \.0!00\.0000\ J6 \ 2.62 \ .37 \ .14 \ 0.0 " O.O 69.93 7O.OO \ Y \ 0.0 \ 2l.l5\ 37.97\ 5O.94\ &QA1 \66.58\69.62

85

Properties of Arch Rib

s I , 2|

/ radians ft ft. ft ft* ft ft ft*S <V J> r yÍX-- ¡;

\—J I 1 yg I Zo I A3 I dx | ■/« I Í5 I I \AS/l\Z.AS/j\ X \ Y I A 3

~,

I T^Tl^g^í^ I O236 69.90 16.65 5.11 192.8 2O.8 213.6.0779 5448 8.3Z -I6.O8 37.16

/■ i

o^-^i^-^n^^ /¿~-T . 2 .O7O7 ¿,9.12 " 5-88 ZOI.7 21.1 ZZ3.4-.O14S5.I5O 24.35- 15.30 38. O3

■*■

]Q^^<f\^^^^ Afr^A <.

3 " 13 61SS " 6 ° 3 214 -3 Z3 ° 237.3.0702 4.739 4-1. 5Z -13.73 38.4-1

C,

\j^\^éB^^^ . '/ 4 J6SJ ¿>S -2 ° " 6.2O 2 29.Í 24.5 Z53.6.O6Sé4.28O 58.0O- 11.38 38.83\|JZ^C^^^^ N 5 -2 -12 -2 - 6208 " *> AO 2 47 ¿ 2 £ -3 2 13.5. 06O9 3.779 J4.35-8. 2G 39.33%)^^<{íK^^ 6 .2594 58.19 " ¿.68 273.3 28.9 3O2.8 .0550 3.193 SO.S4-431 4-O.O3\£&^^ 7 .3065 53.55 " 1.03 3O9.4 32.4 341.8 .O4S7 2.6O8 /O6. 52 O.28 40.9I"TÜt^ <S .3545 41.92 17.93 n.4O 349.7 36.3 386.0 .0466 2.234 /22.9O 5.9O 41.83Z5j259Vy 3 .4032 41-21 " 7.80 396.5 4O.8 431.3 .O4/J ¿698 139.61 12.55 42.83~""i x IO .4519 33.8/ * 8.23 450.6 45.9 496.5 .0362 ¿225 155.98 Zo.oi 43.9111 .5006 25.56 " 8.68 511.6 51.6 563.2 .03 1 9 0.817 171.97 28.Z6 45.O3Zo = /?, cos-0- 282.988' (seg. hi) X = R, Sin £

#

/ 2 .5496 16.17 19.48 9.18 584.7 58.2 642.9 .0303 O.490 188.17 31.65 46.28Z. - /?z cos J9- 238.583' ( "8-11) X -Rz Sin $-5.344 \ g [,5^3 1-5:5^1 » \ 9.73 \e>7l.8\ 66.O \737.8\.O264\O.I48 \ZO4.S3\ 4-8.231 41.66Zo = /? 3

cosJ3-320.602' C " 12413) X = /? 3 sm^ ~'8.O96' -6654 ?>5.ftl3

A3 = 2.5 d,

+23.333 z AS/IAT = A 3

+ (Zrt-i) Co.53&) for dead loadAr = A o + (n-i) (O.538) for ¡¡ye hodn = 8 fjz ft* ft-K; P t

Jc - 20 dr -/7.5(d x -/333) 3

I \A-(--ÜAr(---.)\ X' I V \^^l\ y^I^'^ll V¿¥l\ M

|M¿S/j|M¿Sfr?jM¿S>/j

r S2 \Z n I 45.83 41.52 é9 258.6 O.649 -J.253 5.4 2O.I6- O o o O

J = Cti-/)j O.4Z8 í d x - O.667) + C¿>OO9/7)(d x -I.Z33)¡ 2 4-6.10 41.80 6Z1 234.0 1.859 -/. 140 46.4 17.44- 3667 -213 -68/8 4I8OL ' 2 / J 3 46.48 4Z.I1 1724 I88.S 2.913 -0.963 120.9 Z3.22 -/O846 -161 -3159/ IO446

4 46.90 42.60 3364 129.5 3.807-0.141 2 20. 8 8.50 -2/8/6 -1432 -83O56 16299

5 47.40 43.10 5528 68.2 4.526 -0.503 336.5 4.15 '36159- 22OI -Z63644 18182

6 48.10 43.8O 8/97 19.1 4.911-0.240 450.6 1.05 -54259 -2983 -¿70066 13042

7 48.98 44.67 11346 O.l 5.188 0.0/3 552.6 0.00 -75651 -3685 -392470- 1013

C 8 49.90 45.60 I5IO4 34.8 5.128 0.275 7O4.0 1.62 -10/632 -4737 -¿92/59-21948y^~ -~^\ 3 50.90 46.6O ¡9491 157.6 5.144 O.5I6 6OI.9 6.48 -/32306 -5443 -153342 -68330■jV^ IO 5L98 47.67 24330 4OO.4 5.652 0.725 88/6 14.51 -/66978-6O5! -343187 -121079A II 53./O 48.80 29574 198.5 5.493 0.903 944.7 25.51 -204389-6548 -IIZ6O83 -18503112 5435 5O.O5 35409 1417.8 5.101 IJ4I IO72.7 4Z.95-248460-7527 -U16357 -283416

1 13 1 55. 13 1 51.42 1 4I83O ¡2326.31 5.399 \/.

273 \IIO4. 2 I 61.41 \-231353\'7849\-l6OSl(S\.3785S4Dead Load Reactions at flosf/c Center Nej/ecfmj Rib Shortening- 1242 3 Z17.OI -49490 '1381336 -1,003258

_fMAS ,„M

e = * J - -2^-49,490) = 74,374I J\E- 2 (0. 6 '654) Dead Load Momen t ¡ n Cant/levered Half -Arch (m)

' MVAS 1 Item \/\C 7 \2\3\Ct\4\S\Cs\6\ 7_y iv] m^ (-mm?") 4f>?T D* Vertical load (Kips) 94.16 244.0 94.97 95.91 251.0 96.97 98.22 263.0 99.96 102.15H. = —g / = P i P/7 O/Y = Z- " " " 94.28 338.Z6433.25 529.16 78O.I6 877.13 975.35 I23S.4 1338.3 1440.5% rj ' y Moment arm (ft) 8.O24 8.604 16.571 7.5/7 8.962 16.35 7.38/ 8.804- 15.984 7.904Moment -(ft-Kipa) °O 756 3667 10846 14324 218/6 36/57 43356 54259 75651re M xAS 3 K/XAS\ lfern C+ 6 9 C 3

IO I' C* 12 I

13 C>-| j " f ') I -V-138I336)'(-1 381 336)\ n" Vert/co/ lood 282.0 1/3.89 HS.59 308.0 118.5 121.53 356.0 13521 139.23 279.0Vo ' A yt AS - "— ' 2(7242 3)' - ° X " " 1722.5 1835.4 1950.9 2258.9 2377.4 2499.02855.0 299O.2 3129.4 3408.4t A f ' y Moment arm 8.474 16.714 7.504 8.867 15.988 1.838 8.366 16.351 7.975 O.O

I Moment \87036\/O1632\l32308\l4£94dJ66978\204S8é¿2457&24846O\i973S3\3ZZ3l0

86

Unit load at C z I Unit load at C3 I Unit load at C*- I Unit load at Cs I Unit load at Ce I Unit load at Crsin<f> cosh AS sinfiAS IV fllcosfñSNSinfAS N Ncosfi&S NsinfAS N HcosÓAS NsinfAS N HCosfiAS N 'Sin fas N vcostAS NSinéAS A/

NcqsóaS NsiniiSAt(l.l.) AtÚ-l) At At At At At At At At At At At At1 .023SS .4OO68 .OX>232 .O7O68 .39627 .00/99 .07O63.02808.OO/99

3 .11763 .38926 .OO546 .11763.046// .OO5464 .16431 .38O23 .0/OSS .16431 .06334 .0IO55 .16431 .O6334 .OIO555 .21063.36911 .0/7/4 .2IO63 .07953 .O/7I4 .21063 .07953 .0/7/46 .25648 .35508 .02500 .25648 .O9422 .02500 .25648.09428 .02500 .2S648 .09422 .O2SO0

7 .30/76 ■33870.03393 .30/76 JO78I .03393 .30/76 .10721 .O3393 .30176 JO72I .033938 .34709 .347O2 .04753 .347O9 .12843 .04753 .347O9 .12843 .O47S3 .347O9 .12843 .04753 .34-709 .12843 .047S39 .39233 .32666.05943 .39233 J3932 .OS943 .39233 J3932 .O5943 .39233 .13932 .O5943 .39233 J393Z .0594310 .43664 .30543 .07/95 .43664 ./

46 '24 .O7I95 .43664 .14824 .07/95 .43664 .14824 .07/sS .43664 .14824- .07/95 .43664. Z4824 .07195// .47991 .28377 .0849/ .47991 .15523 .08491.4799/ J55 23 .08491 .4799/ .15523.0849/ .4799/ J5523 .0849/ .4199/ .15523 .08491¡2 .52231 .28298 .106/6 .52231-/7332 .106/6 .52231 .17332 .106/6 .52231 .17332 .10616 .52231 .17332 JO6I6 .52231 .I733Z JO6/6 .52231 .17332 JÓ6/613 .S637l\.25838\.IZ035 .5637l\.l7634\.l2035 .5637l\.l7634\.l203S .5637/1./7634 1/2035 .5637/ \.

17634 1./2Q35 .S637l\.l7634\.l2O3S .5637l\.l7634\.IZO3S

1 443356 .58463 .34967 .2265/ .65313 .38337 .92088 .49O33 !.I213\ .54926 1.26518 .57695 ¿33937.58440Ho .09 1 8 .3345 .6669 1.0195 /.3150 14845

Vo

I I .0125 I .0505 I .1140 \ .2021 \ .3113 .4353Reactions Including Rib Shortening (R)

N= dead load Shear X i ¡tie U =% l ¿ =*- L ' V£ ¿ °ad ' ¿ Shorfenin 9Fffect° f Y Z AS ,f COS 2 ó AS , . .a I a At ¿/Wf load at:

¿ Cz C3 £V Cs C& Cr

I /V I AS \cos<¿ \Ncasi&&ae\tas\ -\ fMXAS _v MsinéAs\ H ° ne 3 lectin 3 R .0945 .3428 .6826 1.0429 1.3447 1.5/79AtCo-l) Árlol.) ÁTtei.) U -La I A At J H° mdudiho R .0918 .3345 .6669 I.019S 1.3150 1.4845

I O .3633 .9S972 O .363O f X l AS -ff Sin* ¿AS Negative thrust iromR-. 0027 -.0083 -.0/57 -.0234 -.0297 -.03342 23.9 .3611 .99750 8.6/ .3593 k I a At

3 5I.O 3582 99306 18 13 3532 N 7 =fi/e 9 .thrust*Cost -.OO27 -.0083 -.0157 -.0234 -.0297 -.0334

4- 1282 35499864/ 44 88 ' 3453 Mo is not affected M 7 = " »x(y) .0427.1311 .2481 .3697 .4693 .5277

5

%% fsil 97757 gi% 335*6 N s -.OO26 -.008, -.0,53 -.0228 -.0289 -.0325

6 317.6 3461 96655 10624 3233 Load Rib Shortening Effect M* .0178 .0547 JOSS .1542 .1957 .2201

7 4039 3399 95336 I3O86 3089 N -° 0¿( > --0079 --0149 --0221 --028 ' --0316

8 5979' 3605' 93783 202 14' 3/71 H o neglecting R = 46ZZ.O* M* -.0073 -.0233 -.o*52 -.0674 -.O656 -.09629 7201 '3534 '91983 23409 '299O Ho Including R 4521.6* N\ -.0022 -.OO67 -.O/27 -O/90 -.O24I -.027/10 986.3 3461 89964 3O7.I4 28O1 NegoUe fUust from P= 95.4" XML \-.l453\-.4467\-.845o\-1.2594\-1.5985\-/.7976\11 1I41.O .3388 .87732 333.13 .2608 ond M ° are * ot effected by a symmetrical12 1491.2 .3583 .85276 455.65 .26O6 hading. The rib shortening effect on Vo due to13 \I685.6\.3495 \.82S97\486.57\ . 2887\ Thrust (Al)' -95.4 COS <£ /¡ ve load is so small that 'it is neglected2396.86 40447 fiAomentÍM)= -95.4 x distance vertically in this case., to e/asfic center.H= - (-10O3.258 -r 2396.86) - j rrr r* K 4 *Ho 217.Ol-r4.O45 ~ 45Z7.6 , = _g53 * _9O2 *Z' -- 1507.'" M* = -270'"

A/f = -92.8" N' = -77.4"M s = 629." M'=-5I34'*

87

~ Unit load at d 1 Unit load at C3 I Unit load at C4 1 Unit load at Cs I Unit load at Ct I Unit had at ¿V-2 -8.60- .64O9- /6.O 10.31

3 -25.18 -1.7664 - 73.5 24^54 -8.96 -.5881-34.11 6.69 -41. 6S -2.1339- 158. 6 31.125 -25.31 -I.S4O6-H4.S5 12.73 -58.OO-3.53OÍ- 262.5 29.176 -8.81 -.4844 -43.9 2.12 -41.50 -2.Z816 -2O6.6 9.97 -14. 19 -4.0787 -369.3 17.83

7 - 24.79 -I.2O74 -IZ8.6 - O.33 -57.48 -2.7995-298.2 -O.77 -9O./7 -4.39/6 -461.8 - I.ZI8 -8.47 -.3948- 48.5 - 233 -41.17 -1.9/89 -235.8 -11.32 - 73.86 -3.4425-423.1 -2O.3I -I06.SS -4.9662 -6/O.3 -29.309 -25.19-1.0363- 144.1 -13.01 '- 51.88 -2.3812 -332.4 -29.89-90.58 -3.7266-520.3 -46.18-/23.27 -5.O1I5 -708.0-63.66/O -8.87 -32/4 -5O.I -6.43 -41.51 -1.5063 - 235.O -30.14 -14.25 -2.6905 -419.1 -53.84-/06.94-3.875/ -604.4 -71.54-/39.64-5.060Ú-189.3 -101.2611 -24.85-7938 -136.5 -22.43-57.55-1.8384-316.1 -51.95-90.24 -2.8890-496.8 -81.64 -122.93-3.926$ -675.3 -110.97 -155.62 -4.91/1 -8549 -140.4112 -8.31 -.2536 -47.1 -9.55 -41.06 -1.2439 -234.1 -46.84 -13.15 -2.2342-420.4 -84.13 -IO6.4i-3.224S-6O6.8-/2l.41 -139.14 -4.2151 -193.2 -158.72 -171.83 -5.2054-919.5 -196.0013 -24.12 \- .65 25\-/33.5 I- 31.47 -57.41 [-1.5155 \-309.9 \-73.09 -9O.IOV23784V486A 1 114.11 -/22.7S\-3.24/3\-662.9\- 156.33 -1'55AÍ-4.lO4S\-839.5\-l97.97 -/88./8\-4. 96741-1016. 0 \-239.5SZ -.906/ -181.2 -41.02 -3.8745 -730.7 -148.79 -9.3884 -1651.2 -29621 -18.0372 -2926.9 -452.65 -30.5004 -4509.2 -583.66 -47.3836 -63O5.4 -658.82M. O.68O8 2.9113 -7.0545 13.5532 22 9181 35.6042Ho O.O945 0.3428 O.6826 /.O429 1.3447 15179Vo I

I

O.O/2 5 I O.O5O4 I 0.1140 I O.2O2I I 0. 3113 I 0.4353_f MAS/jMo = i AS/ Me -Mo-YoNo f~ N / A - M 9

TI ñ/ T Moment (M) =Mc+ YHo+XVo+m' _Thrust CN)Lt- Heos <fc

-V sin 0 I Column*! I Column^5 I Column* 4 1 Column */~

c

" (N)*t= Hcos<£ +Vs/n& A I C I /" A I C I 7" A I C I I A\C\1-K MVA^jr fy,' -moment -From unit load 4l.60\34.8"\¿ '16.2 43.3s\33.o'\z83.7~4S.lo\41.2'\z6l8 S2.IO\60.o" 787.3H ° ~y z AS/ ° n can^ ileYerecl half -arch f u ¿, f u fi fi ñ fi, ft"a Ci -.0409 .0723 JOO2 -.07/6 .1778 -.1516 -.3832 1.1574Cs -.0998 .2134 ,43/2 -.3280 .7153 -.6205 -1.3582 1

.5802_r MXAS/r C4 ~ o6/o 2S6S 10467 -.8425 1.6183-1.4309-1.1829 1.46852

. /v\Ano/j 1 Unit load At: I Cs .2077 . 1369 2.000S -I.69O5 .6746- .29/2 -.6252 .9750Vo ~ 2Y Z AS/r Cz 1 C3 1 C+\ Cs \ Ct I Cr Ce .8516 -.4080 .7766 -.3064 .0246 .4358 .1100 .28822

.

* ¿S / 1 Cr /.39S2-/4S58 -.077/ .5921 -.3238 .8224 .8100 -.3936Me -.8482 -2.63SZ-3.9SOO -3.3ZO9 1.1609 11.0446 Cs .676/ -.1631 -.5409 1.0463 -.4365 .9221 1.2917 -.8955Mi -.6O79-/.6809 -1.8667 .3797 6.7616 18.7386 Cs -.OS/8 .5450 -.6847 1.1269 -.3986 .8214 1.4548 -1.1154, -> ■** r* t a a m 11 *> a u Ms -10/67 -3.3290 -5S345 -6.2289 -3.4179 4.5042 ¿*> --4047 .7557 -.6/26 .9526 -.2918 .6158 1.3017 -I.O505+ X ' I I \ IS-*'I

? -¿Xj III 1 -X Nj .0938 3401 .6766/0324 1.3289 1.4961 Ca -.4064 .6358 -.4273 .6485 -.1729 .3831 .9457 -.7803]>^ V» .0950 .3447 6872 1.0512 1.3577 1.5365 <m -.2526 .3676 -.2205 .331/ -.0767.1815 .5107 -.4287"\^ \ i .y

I Cis \-.O788\ .0947 \-.O6/S I .O767\-.0/86\ .O474\ .1484 \- JZ6OMs 1.O797 477/4 ¡1.8734 23.1981 6.8O6& -4.2061Mm -.9636 -3.4669-6.761/ -9.8371 -//.3860 -9.9758 2.0-, .y f u 2.0-, . /" f"

v 7 = y 8 = O.38 1 Ns .O89O .3218 .6376 .9678 1.4611 1.6015 if /Y

X r = -X s = /6 35' Al/o .O948 .3452 .6904 I.O614 /.38O3 1.5775 Z£ /.O- I \ 1.0- Á \ /TT^>^M4 2.3832 9.6653 22.06426.9888 -2.9148 -&2S4I -j

O- -aTjUP^J [KJ P>- o—*£A \/\A P>-ys = y"> = 5 55>' Mn -.4773 -1.8682 -4.0235 -6.5675 -8.8281 -9.8313 \ / / XJ_L>^^

X r = -X /fl = 81.73' N* .0853.3080.6087 1.2452 Z.4953 1.6189 \l f * -L0 ~ \L—ft

<ps = p'° = 13.388° Mu .O93S .3406 .6827 I.

O5 2 1 1.3730 1.5770 V \f'^.^¿jj Column 1 . -2.0-1 Column 5

7* = Y" = I9.O6' Mi -24.2693-33.3/38 -30.05SS-I8.I40S-2.0Z04 +13.6451 g.o-\ , 2.O-,X 4 = -X" - 114.42' Mi* 3.1/06 10.6508/9.5670 26.735$ 29.1387 24.7964 S A/" /T~K yf»fl* = f = 18.914° IV' .6531 .8328 1.07/4 /.3125 1.4938 I.562O X L0 - l\ J^- r ~^ t.0~ ( \ /| PvNi* .0840 .3077 .6207 .9646 1.2135 1.4865 A V-/Tli>^ / \\ A \v

f ' f* = 35.726° \f*-ft -1.0- \|J/ N M_-U ><^fi-2.0-^ Column * 4 -2.O-1 Column */

88

+fu -compressive stress in extreme upper hendinq fiber of arch ribf-£= r,, // // // /o

rwet //

*ii " n

Lane load to produce'- Truck load to produce:maximum +-fb at column 7 ,*",,,*

/ , r> 64 oh-6.5_

ii. /SOS 6.5 if iK—

tt* h-^"-""—

M= 1869 a a M=2272, r36r36 N* 305* 16" N- 21/"

313 9I3 N/A+MN/A+Mtfr = 225ps! 9I 31 ffA+Me/l =¿47psi.'. trucK controls

maximum +-ft at column*

7, /JB.B ,, . /89.S1 .' "1 M= -1141 h i *1 /14= -860

N= 285- I A/ = /44

\%íiii it SU! dI 1 dI 9lane load controls

maximum +fu at column**5I M= 214-9 M*2917

, i , N= 215 M= 134

maximum +fj at column**5~

. 2485' .M= -2296 M= -1590M= 3JS i A/= 2OZ

itíiiiiIi 9 % '%lane load controls

maximum +-fu at column

I M * 2462 M= 2733I N = 148 N■ = 84

oal * * * * * * * * ' * V\j * * * * " * * * *qfl P| {^1 § 51 trucK load controls

maximum +-f-¡ at column 4v, 281.2' .

M= -2OBS i M= -1367N= 406 I N = 227

'sl\ Í9 ll ?l §1 |1 I t% Ílane load controls

maximum +fu at column*I

M=6362 I M=4060M= 341 I N= ie-7

l?U*1§! il SÍIt aiiV '''

lane load controlsmaximum +-F. at column**!. 176.5' ,

I M = -5747 M= -46461 N-

25/ N= 125

xittii^ ' ' ■ $1 1a ■" '

lane load controls

89

Increase dead load rib shortening stress 3s'/, to.account for creep.

Dead Load Moment 4 Thrust r , # 1.38 Pot. Ratal=i.3Bßdl tl-fíuMt>- Mo+HoV+'contllever moment Loi +fu +fx +fu +fjtNo = HoCOSp + Vs¡h<p I3BMn(O.L) 8080 2080 2167 2193

7 « Nit i -132 -132 -138 -138~7~~~\ #1 I VZ r~¿- I r I "M* i 863 868 892 946Column I L i L 5 »N* * .IZB

-/2g

-/3/ .¡36

M0OK)M0OK) 575 552 652 874 "Mr t -373 -573 -3&0 -408NAK) 4634 4784 4932 5743

- ""* " '124 ~!2(> "/33II— /V° 'I WJ I *'°* — \_"±£ 1 oi'** 1 HM, » -7085 -7085 -1561 -1306'I »/fo 'I -107 I -107 I -114 1 -lIP

Live Load Moment ( ThrustMoment Magnification Factor (S) Temperature Change Moment 4 Thrust(based on E,l,<t N at col. *4) <*tL

Igross * ¿OX 7.O3/*-

¡7.55.673//2 =306 Ft4H° = f /*AS + f COSCAS "= 6* /0

~

Jreinf. = 2*3J*38Z +15.9X68*/I2 =95,100¡n4n4-4.62 ft4£ I

XAE

A.KSM.J.O. +17 4," I'HU' ZEl=[3600X306/5 +29000x4.62]¡44=510X10* K-fiZ Mt= HofA.A.S.H.T.O. 6-16 Nr = HoCOS^p - 3.14sx s/o'/O6 _ _ 20250" I/-./ # I "fJ

° I fJ?°c '(0.7XJ06XZ1é.5) z " C Mr I Nt Mr I Nrp.

-aC49* + % (WJ.sssS I zl//5 :!if :'ÍH fo¡

r- / IC9

* -387 -127 30/ 996 "I- 6594

= Ib¿ I / \-7228\-109\562Z 850.85(20,250)

Impocf (I) Shrinkage Moment $ Thrustr _ S0 -CL1 -

L+IES Ho- f yzAS ;b COSJ^J C- 0.000/zL

-length of lane loading AEI *

A E- _ era a*

Col.* 1%1 for+fu \%T for +h I M

_ „ v

I 7 i M I U I y^\ & I %^\5 394-58ML+t

-SIM 4 -¡i¿ -57

NNL.i -IN I / \-3218 1 -49Col. * I +fu I *fx

7 Mlh = 4347 -2052Nl+i = Z49 316

5 Mlh= 5463 -42IBNlh= ISS 364

a Mlh- 5193 -3793Nlh= 98 456

I Mlh« 11636 -10856WW 385 I Z9Z I Design Moment ¿ Thrust

GroupI-

B[D+s3a.+I)JRib Shortening Moment 4 Thrust GroupET=/.3CD+(

'L+I)+R+S-*T]Mr = Mk(0.1.) +IMr(l.l.)Nr = N* (d.1.) y-lA/.?q.U

,___--_, „_, I #\l 1 \7l\ I fu-Al 1 fu+AT I fjt-AT 1 fi+ATCol

*+fu +fx Co/ 9"Wp—^f\ N M\ N ~M\ N 'M \ N

7 MitU.L.) 74/02 1 ¡0/68 6564 10168 5564 -3699 6709 -3699 6709N*

" -5 -7 / JSL 13203 5916 8299 6225 4918 6001 14 6310S M* » 21 48 i/2554 6555 12554 6555 -8422 7051 -8422 1051

M* " -3 -7 J jy 10642 6005 8597 6308 -¡813 6296 -3918 65994 Mr " -6 -25 j ¡2099 6624 12099 6624 -7371 1400 -757/ 1400

Nk " -2 -8 q JJT 6378_ 6136 1212 6430 -5340 6592 -4446 6886I M* » -422 -190 t 26348 8300 ¿6346 8300 -22386 8099 -22386 8099

W* "1-61-3 I I 'IJT1-7/46 I76/5 1 95591 7865 \-36054\ 1491\-19349\1149

90

Inferaction Diagrams

*£ .. ® co/umn *^7 ® column

i -$ ■

—\ ■'■ \

""s, to- \ io- _\

11— L— | , 1 1^ I 11

— L—j 1 , 1

Moment(in-Kips */03)

@ column 4- @ co/umn J

\l %% \ iIiI§

91

2.12.1 Revision of Cross-Section

The cross-section which has been analyzed is understressed.There are three possible ways of reducing the section, a reduction inoverall width, a reduction in overall depth, and a reduction inthickness of the members of the box. Each of these will reduce themoment of inertia, resulting in an increase in the moment magnificationfactor with a resulting increase in live load moment. This will bepartially offset by a decrease in dead load thrust due to reducedweight of the arch rib, A reduction in depth would have the biggesteffect on the reduction in moment of inertia and the least effect ondecrease of dead load. Therefore, depth reduction is ruled out, andthe change will be confined to thinning the slab and reduction ofoverall width to give a slab b/t of 12 so as to prevent possiblebuckling control. The walls could be reduced in thickness, but thiswill not be done because of difficulty of concrete placement betweenvertical forms.

The ratio of force to section strength is about 0,85 at thespringing. A 7 inch slab with a 17 foot overall width would give ab/t ratio for the slab of 87 7 7 = 12.4, and roughly a reduction ofsection moment of inertia of about 20 percent. The live load momentwill increase but the thrust will decrease, resulting in only a slightchange in force on the cross-section. The spacing of reinforcingbars in the slabs will be increased from 7 to 8 inches to keep the samepercentage of roughly 1.5 percent in the slabs.

Since the variation of moment of inertia and area along the archaxis will remain practically the same, a new analysis is not necessary,The thrusts and moments can be modified by simple calculations, and thencompared with a new interaction diagram.

The span-to-width ratio will be 425 7 17 = 25. Wind stresses,live load lateral eccentricity and lateral buckling and lateral momentmagnification will be investigated.

92

Ao.p. = 204- XB4--/74-X7O * 4950 inZ

Icj.p. = <^¿?4 X843~/74 x 7O3)+/2= 5./OX iO6in4

Previous Acf.p. ** ¿40 X 84-210 X6B - SBBO/h2

Previous Ic¡.F. ~ (Z4OxB43-ZIO*683)+¡a 6.35x/06

Obtain the previous thrust at the quarterpoint for GroupT /oac/ing by interpolaTion fromthe design moment and thrust tabulption.Prevhus A/q.p. ~ 6607X

Decrease in Nyp. = /.Q6 CSBSO -49Sd)Ca/So)UZS)/4¿ CBX7O)= 331*

A/9.p.9.p. =6607-/3X33Í

= 6/77aRoffo change in rib shortening moment - 6607-33/X/.3

66070.935

Ratio change ¡n shrinkage $ temp, moment= 4-950SBBO

= 0.842

lof reinforcement^ Zx 22.9 (38.sf+/5.9(70f+f2= 74, 4-00/h*ET- (3600XSJOX/O (¿9OOO X744O0)= S63ox/ocF¿

-rrz XSB3Ox/o 6 _ /6jo7Z

(0.7XÍ.06 y 2J2,5)Z144Z 144r IO

-/ 6177

" i-831 0.85X/6,072

93

Revised secfion at sprinping :

Group JZ~ foodtnqnew thrust - 7497-/J X ~ x 331= 1497-481 6"

7OtO*

¡83new moment = /3T874-/§§* ¡QBs6 -0.935* 7306-0.842X1Q4-4-6

« 35JEO IX

Group X loadingnew thrust * 8300-487~ 78/3new momerft= /.3C874 +ss (¿H x //636)J= 29,62O//c

l0 '-

—i— X-— i

—i—

i

<\l M^ *O

Moment (/n.-zr/ps x /O3)

Revised Interact/on Diagram ot Springing

94

2.12.2 Vertical Crown Deflection(dead /oad, creep, / shrinkage)Dead load r/h shortening Heflertinntransformed area at crown (Ac)

Ac= 204x69-/74x55 + 7X61.6 = 4940in2

riDL= 4623-33/ = 4292 X

€ D.i.= 4292/4950 = 0.567xsi

a* = %%(*>+}&)= 7 = 2.06" initially

Creep¿} c

= 2.4 x 2.06" = /7

Ac= aooo3 (115) - 0.21S''= 7

deflection -for comber ot crown CZAC)

£ACA C" Z.06 +4.94+2.58 ="9.58"

Defiection from change in temperature (45°F drop)Crown deflection = 6x/o"6x 45 (7/s)= 0./93 f = //

Z/Vg mod+ /mpact crown a'efiectionMoment magnification ot service foad

II9.p? 5.10 X IO6 ¥■ 7X 14,400 =562 X ¡O6O 6 in4

//*/*.= 6/77+ A3 = 4752*p = 7rzx3e00x562 */o6 - 55,800

*c " (0.7X1.06X212.5)Z 1446 =

/- 4-7SZ = /¿**

/I/. ,- MJtZ- 4347X/.09X425Z XI7EB =1.43c 7651 76X3eOOX^7S/7,0)2x562x/0€' (X/360Ó)

95

Z.l?3\AJind Analysisuse wind load = 7X75 - 525 **/ft of arch axislateralI?.p. = (7*I73-5.83x /5.333)+l¿ = ///6-ft4

/<Q.p. = 2 X J6.172X 6.4PZ - eae ff+/6J7 , 6.42

0.5&3 ~f~ 0.833£ /¿

-2.3

I£ = 1116 X?3 _ 4.¿¿A G 60S

transverse moment at crown (A/lo)) using eauat/on 2/iA sin2^ m- n fi-smPcos&T

Mn - fw/?2) 5Jr)-J9 " 2^4- +X6Lsin£-#coS'&- J< nvr J g sin¿¿i jrf r¿ s/nficostfT. +—} tK6 ¿- -2 J-fi = a63643 roc/lons /? ■» 357.545' ff(

A/L = (67/OO) 'SS4:*2~ 31822-.23899+4.22Í.OQ32&J ¿ 3828.31822+.23899 i-4.22C-07923Jtors/'on ot springincp CTs)

Ts = Mo S//7-J3 ~ WPZ fe-s/nyß)- 3828 X.59432-

67100X .042H '= -551

'*

transformed lateralIat crown (Ic)t

_ A/67X/73XO* 7X.0/6Q) , ¿667 X 458 XB.OS^Í/'7X0096)=

/063ft 4

lateral bending stress at crown (fc)f = 38PS y8.5x1000 - 213 f=>slc /063XJ44

lateral shear stress at springing CH)lateral react/on - 0.525 x/.06x /.O6 x 212.5

-U8M

bending shear ~ //e,ooo/204 x14 = 41 ps/

torsional shear = ff/JÍ2"° = 22 f>s¡

¿X//JX/&4XT 63PSÍ

96

lateral deflection from wind fee), -Prom eauotlon ¿3a

r L2L 2 r wi2 .„ 7= JEJ L ~4~~ M*J

= 2252 f.525X2251 „ ropol2X3500X1063X144 l 4 J&¿&J

= OJ29*= /.SS"This compares with a ioteral de-fiection

of 2.4-1" -for the steel arch with two pionesof Zotera/ bracing, 3.38" for the stee) archwith one p/one of lotera/ bracing, and 2.23

"for the deck.

The concrete arch is Jess -F/exible laterallythan the roadway -floor. This results in atransfer of wind load from the roadwaySlab to the arch. The opposite is true -forthe steel arch.

Z.2A Effect of //ve iood lateral eccentricity

Assume 2. - 12. foot ¡ones at one edge ofthe 3O foot roadway with vehicles shifted/ foot in the ¡ones. Resulting eccentricity fe)is 4 feetfrom eojuotion 45'"*ia - 2X3SZSX¿28X4XJS53 Í4.22-lj _

A 7/7 //c/VJg ~3f/.//44 -h .15846x4.22]

~ 4JU

T5T5 = -430X.55432 + 357.5X/.28*4*A7797 -r 2*26*4-827*

97

Bending stress at crown ("£)fc

= 430 X8.5X /000/ /063* J44 =24 PSÍ

lateral shear stress of springing B?7x/2000shear stress from eccentric LL =2x 113x¡94X7~32 ps!

shear stress from wind ~63 f>siTotal shear stress in transverse slabs -95'ps'i

Due to the large axial compressive stress,diagonal tension is not involved with this shearstress . The minimum requirement for transversereinforcement in the slobs must still he met.

¿ ¡ZSLateral buck/¡no ond moment magnification

At service food-lotero/ r

- V-^4 = ~V ///6/34.4 = 5~.7 /

¡Otero/ *L/r = /./2X/.06X2/2.5/S.7 = 7T

___777/5 compares with a vertical ML/r z=-o.ix/.06x212.5/<J2+G/344lateral Pc

-(USX /.06X 212.5)2 200

lateral moment magnifier -/„ 4752

~ 106as200

vertical moment magnifier ~ f.09 (for comparison)At ultimate hod'Elj.p. = C///6X3600+S -f- .0/24*1116X29000)144 = Z73.5*/06

loteroI/? -a/3 x ¿06X212.5)^- 25.400

lateral moment magnifier -/_ 1.3x4752

= /J8/ aesx 2g,400

vertical moment magnifier - ¡.83 (for comparison)

99

CHAPTER 111 - ARCH CONSTRUCTION

3.1 Steel Arches

It is important for the designer to consider the probable methodof construction of the arch. This may have an effect on the choiceof type and on the design. The principal methods of construction are:full falsework support; falsework bents with increasing cantilevers toeach succeeding bent; cantilevering from each abutment by the use oftiebacks; and off-site construction for tied arches, followed byflotation to the site and vertical lifting of the full tied arch bymeans of cables.

As pointed out in earlier chapters, erection procedure may beused to eliminate stress from rib shortening, shrinkage and creep.

As is the case for all long span bridges, the designer shouldcheck and follow the erection procedure. This is necessary to insurethat the actual structure is equivalent to the one assumed in design;and, in the case of arches, to see that the necessary closure andjacking procedure is followed.

3.1.1 Cantiievering from the Abutments by Tie-Backs

The Eads Bridge over the Mississippi at St, Louis started in 1868,was the first to make use of this method of erection. Temporarywooden towers were built on the piers to support steel tie bars» Sincethis is a multiple span bridge, balanced cantilever erection, avoidingthe use of anchorages, could be used for all of the bridge except one-half of each end span.

No adequate method was provided for adjustment at closure. As aresult, difficulty was encountered in the insertion of the closingsections. The arch chords are steel tubes and an improvised closingsection, made adjustable by screw threads at the ends, was used.

Since then many long span steel arches have been erected by thecantilever tieback method. Some of the notable ones are the Hell GateBridge over the East River in New York City, built in 1916; theRainbow Bridge at Niagara Falls, Figure 11, completed in 1941; theGlen Canyon Bridge, Figure 12, over the Colorado in 1958; the Lewiston-Queens ton Bridge over the Niagara River; the Cold Spring Canyon Bridgein California; the Snake River Bridge in Idaho; and the New River GorgeBridge in West Virginia, see cover.

100

FROM: "Rainbow Arch Bridge"Erection of Steel Superstructure, by E. L.DurkeeASCE Proceedings, October 1943, page 1240Fig. 11

101

GLEM CA?!YON BRIDGE - ERECTICÍIFrom "Technical Record of Design and Construction, " Nov. 1959

U S Dept. of the Interior - Bureau ofReclamation - Fig. 12

102

For the tiebacks of the Rainbow Bridge, prestrained wire bridgestrands were used. These were provided with adjustable links adjacentto their connections to the arch rib. It was necessary to adjust thesestrands to keep the arch profile close to the theoretical. Jacks wereprovided on both flanges at the crown to permit insertion of the closingkey section of the arch rib, 11 inches long and wedge shaped. This keysection was actually fabricated 13 inches long before milling. Ifadjustment were found necessary to get the design crown moment, it wouldhave been possible to mill the section to a length lesser or greaterthan the theoretical 11 inches and to change the angle of wedge.

After the arch was self-supporting, the jacking forces were measuredMaking allowance for the dead load and erection load on the arch atthat time, it was decided to use the theoretical size of the keysection. The moment and thrust did not check out exactly, but thedifference was considered to be within the limits of possible accuracyof steel weight calculations.

The Glen Canyon Arch, see Figure 12, a truss type rib, was closedon a pin in the upper chord. Jacks inside the lower chord were used togive a calculated force of 525 kips at that stage. This force had beencalculated on the basis of the loads, including erection equipment, onthe arch at that time and on the temperature at that time. The chordwas then shimmed in that position, and the holes match-marked in theblind connection, The holes were drilled and the jacks were then usedto remove the shims and allow the placement of drift pins and bolts.The connection at this point, of course, did not provide any bearingand the rivets were designed to take the full maximum stress. Unlikethe Rainbow Arch where an erection traveler on the top flanges was usedfor erection, a cableway was used at Glen Canyon to deliver the steelmembers to their final position.

The horizontal part of the tie backs for the New River GorgeBridge consisted of hollow oil well drill stems. Two of these failedduring construction. Fortunately, the remaining intact ties were strongenough to take the additional load and impact,

3.1.2 Cantilevering over Falsework Bents

The Bayonne Bridge with a span of 1652 feet over the Kill Van Kullwas completed in 1931. As shown in Figure 13 this arch was built bycantilevering over falsework bents. Rock foundation for the bents wasat a shallow depth and plate girders for the approaches were used in thebents. The arch was designed to act as 3-hinged for its own weight onthe basis of closure on a pin in the lower chord at mid-span, The archwas to then be converted to 2-hinged for the remainder of the dead loadand live load by connecting the top chord over the pin in a stress!ess

103

BAYONNE BRIDGE OVER THE KILL VAN ¡CULL - ERECTIONCourtesy American Bridge Division of United States Steel CorporationFig. 13

104

BAYONNE BRIDGE - SHOE ERECTIONCourtesy American Bridge Division of United States Steel CorporationFig. 14

105

condition. However, the off-center location of the ship channel wouldhave required a very long cantilever on one side. It was, therefore,decided to close the arch at a point about 250 feet from the center ofthe span with a cantilever overhang, from the temporary bent,of 413 feet.This required a temporary toggle arrangement over the bent to reinforcethe arch for the large cantilever moment,

Since the arch was closed on a lower chord pin 250 feet from thecenter of the span, calculated jacking at closure was required, Thestress in the closing top chord member at this point was calculatedon the basis of a pin at the center of the span and the dead load attime of closure. The vertical force, acting at the closest bent,required to produce that stress in the 2-hinge was calculated. Thisjacking force was applied at the bent to the 3-hinged arch, Theclosing top chord member was then drilled at the blank connection tofit the opening under this load. The connection was riveted and thejacks released. This produced the required compression in the closingmember and zero stress in the top chord member at the crown as the archwas designed.

Figure 14 shows temporary supports of the arch shoe required duringerection. Since the reaction on the pin was vertical until closure,temporary connections to the shoe were provided to prevent overturning.These consisted of a temporary extension of the shoe in front to providebearing at the edge of the abutment and tension connections at the backof the shoe to the anchor frame in the concrete. Another temporaryconnection was required because of wind. Due to the extremely longcantilevered arch rib, a wind blowing at the right angles to the spanwould tend to move the windward rib away from the pin, due to negativewind moment, It was decided to relieve this condition by allowing thepin to act as a roller, as shown in the enlarged detail at the pin,Figure 14. This, however, introduced another problem. There was apossibility of an off-shore wind rolling the pin up the 10 percentslopes and off the shoes. To offset this possibility, an eyebar tieon the center line of bridge was anchored to the abutment as shown inFigure 14. This eyebar chain was strong enough to take the off-shorewind, but flexible enough to allow the transverse wind negative momentrotation without overstress of the eyebar chain.

3.1.3 Off-Site Construction

Since tied arches form a self-contained structural system,requiring only vertical support, they can be assembled at a convenientshore point, and floated to the site for bridges over navigable water.They are then raised to their final position. The center section ofthe main span of the Fremont Bridge in Portland, Oregon, is, in effect,a tied arch with a span of 902 feet. This section, weighing 6,000 tonswas raised by means of jacks and rods mounted on the cantileveredsections of the span, Figure 15. This is quite similar to theerection of the suspended span of cantilever bridges. Two 535 ft,

FREMONT BRIDGE - ELIMINATION OF DEAD IOAD MOMENTS BY' CAMBERFrom ASCE April 1970 Portland, Oregon Meeting Preprint 1210

by A. Hedefine and L. G. SilanoFig. 15 106

107

tied arch spans over the Tennessee River near Paducah, Kentucky, werefloated to the site on barges and raised immediately adjacent to theline of the bridge on falsework towers by means of jacks and rods.The spans of the parallel twin bridges were then skidded sideways ontothe permanent concrete piers. Falsework in the span opening wouldprobably have been less expensive, but could not be used because ofnavigation requirements.

3.1.4 Camber for Tied Arches

cornbered position■fabricated long^^^^r-"~~"~ \ "

L~ 11— — — -^J_ -final position

/C^"^ i iN

Short

Camber of the rib and tie sections is for length only. Eachsection is fabricated to the radius of curvature that the rib or tieis intended to have under final dead load. Both rib and tie will beforced to assume sharper curvature while supported on falsework, inorder to connect the stressless, cambered members. This forcinginduces flexural stresses in the rib and tie equal and opposite to ribshortening stresses. The arch is erected by use of blocking underthe tie and struts between the tie and rib. After the members areconnected and the blocking and struts are removed, so that only endsupport is provided, the temporary, forced flexural stresses arecounteracted by rib shortening so that zero rib shortening stress,under full dead, results. The vertical and horizontal position ofthe points of falsework support can be precalculated, and this willgive a check on the accuracy of the fabrication and erection. In thecase of the Fremont Bridge, all the design moments for the tie girderincluded an allowance of + 5,000 ft. kips, to allow for possibleinaccuracies in fabrication and erection.

108

3.2 Concrete Arches

Dating back to the early stone arches and continuing to the presenttime, the most used method of building masonry or concrete arches hasbeen on timber falsework. The Cowlitz River Bridge (14) in the Stateof Washington, Figure 16, is an example of the use of full falsework,This arch has a span of 520 feet and the crown is 220 feet above theground. In order to avoid flexural stresses from uneven support duringremoval of falsework, some means of uniform lowering of the centeringover the full span should be used. This may be done by the use ofjacks, sand boxes, or even wedges in the case of short spans,

In setting the elevation of the soffit forms, the deformation ofthe timber supports, including the local deformations due to postbearing normal to the grain of the wood, must be taken into account.The actual movement during concrete placement should be monitored andadjustments made by the jacks if needed.

Where openings for traffic, stream flow or navigation are required,steel falsework or a combination of steel and timber falsework may beused.

The Sando Arch overthe Angerman River in Sweden with a span of866 feet, completed in 1943, was built on multiple timber falseworkin the river. The first attempt at building this arch resulted incollapse in 1939. This support was a lattice type timber arch.ofthe same span as the final arch. Collapse occurred after only a smallamount of concrete had been placed. The rib was jacked at the crownwith a force of 6,700 tons.

Truss type steel arch centering of the same span as the finalconcrete arch has been frequently used, A step from this was the useof a structural steel arch as reinforcing inside the concrete, Theforms were suspended from the reinforcing. An example of this type ofconstruction is an arch bridge over the Connecticut River at Spring-field, built in the early twenties.

3.2.1 Freyssinet and Menager Hinge Construction

Stresses from rib shortening, creep and shrinkage in concretearches are much more important than the rib shortening stresses insteel arches. They can be practically eliminated by the method ofconstruction.

109

COWLITZ RIVER BRIDGE - WASHINGTON STATEReproduced from lASSE Bulletin 25 - 1969 9 pg. 54

Fig. 16

110

One of these methods discussed in 2.5.3 is that developed byFreyssinet before 1930. This method involves jacking at the crown,or the quarter point. The effect of jacking also raises the archrib off the centering. This eliminates the need for a number ofvertical jacks for lowering the centering. Spans as great as 1,000feet have been built by the Freyssinet Method. The necessary camberis produced by the jacking.

Another method for eliminating "parasitical stresses" is by theuse of temporary hinges which cause the arch to act as 3-hinged for aspecified length of time. The 3-hinged arch, being staticallydeterminate, does not develop these stresses. The Menager hinge is awell known type of hinge for this purpose. It consists of rein-forcing bars bent in the vertical plane so as to cross each other at anangle, at mid-depth of the rib. These bars have sufficient strengthto take the stress from arch thrust, and the gap in the concrete is justwide enough to permit rotation of the concrete. Jacks to produce quickrotation can also be used in conjunction with this hinge for eliminationof stress.

The Gladesville Bridge at Sydney, Australia, Figure 17, with theworld's longest concrete arch span of 1,000 feet, was built on steelfalsework with columns at varying intervals. The arch consists offour parallel ribs, each made up of precast single-cell unreinforcedunits 20 feet wide by approximately 10 feet long. The units were raisedto the top of the falsework at the center of the span, and then moveddown the falsework. Three inch joints between the units were pouredin place. Freyssinet flat jacks were placed between the units at thequarter points of the span.

After closure at the crown, the jacks were used to open the ribs3 1/2 inches at each quarter point. This was for the purpose of liftingthe rib from the falsework and for reducing stresses from rib shortening,creep and shrinkage.

The ribs were set with a 1 foot wide opening between them trans-versely, and diaphragms inside the ribs and between the ribs were usedat 50 foot intervals. These diaphragms were pos t- tensioned transverseto the ribs. The 1 foot space between the ribs was continuously filledwith concrete of a thickness equal to that of the top slab of the cells.

Use of jacks at the quarter points instead of the crown permittedjacking for axial force only, since the quarter point is practicallyon the line of the elastic center of the rib. Jacks at the crownwould have been required to jack in a large moment as well as an axialforce.

111

GLADESVILLE BRIDGE, AUSTRALIA - CONSTRUCTIONFrom March 1965 Proceedings - The Institution of Civil Engineers, page 489by J.W. Baxter, A.F. Gee and H.B. JamesFig. 17

112

The use of unreinforced units required a larger cross-sectionalarea; and the resulting increased dead load on such a long span, inturn, further increased the area. Reinforced units with projectingbars and sufficient space between the units for lap of the bars inthe poured-in-place concrete might be more economical.3,2,2 Tieback Construction

Construction of concrete arches by means of anchored tie^backs isbecoming increasingly popular. An arch bridge with a span of 656 feetwas built near Port Elizabeth in South Africa about 1970, Figure 18.The Hokawazu Creek Bridge in Japan, Figure 19, with a span of 560 feetwas built about 1973; and a 315 foot span was recently built inAustria. All of these bridges have hollow box cross-sections and theconcrete was poured in place in about 10 to 20 foot lengths onmovable forms cantilevered from the completed sections.

The South African bridge used cable tie-backs passing over atemporary tower built on top of the land pier adjacent to the archabutment, and anchored in the rock slope. Anchorage jacks were used foradjustment of these ties.

The Hokawazu Bridge used 76 prestressing rods for tie-backs, SeeFigure 20, These rods were anchored in the permanent bridge abutments,designed for this purpose. The horizontal rods were laid on theroadway deck which was constructed along with the arch rib. Temporarydiagonal rods from the deck to the arch rib were used between thecolumns. The arch is two-hinged and the columns are pinned at bothends. There are also joints in the deck over the crown and at thebridge abutments. The deck is a voided slab with a depth of only 26inches. The arch rib has a depth of about 8 feet at the crown and10 feet at the springing. The arch rib is flared laterally at thespringing for earthquake resistance. The column hinges, shallow deckand joints minimize deck participation with the arch rib, The columnsare heavier than normally used, because they must take the verticalcomponent of the temporary diagonals. The temporary rod diagonals,are protected from the sunshine by tubes of vesicated styrene. Theneat would effect tension in the bars. The horizontal bars on thedeck are protected by thick wood planks.

3.2.3 Elimination of Rib Shortening, Creep and Shrinkage Stress

Bridges constructed by the tie-back method also permit thereduction or elimination of stresses from rib shortening, creep andshrinkage. The tension in the several ties can be adjusted by jacks sothat their release, after closure, results in stresses opposite in signto the "parasitical stresses," Due to the effect of stress relaxation,

113 VANSTADEN'SBRIDGE

-SOUTHAFRICA

From"L'lndustria

Italianodel

Cemento,"n.

12,Dec.71

Designers:M.d'Aragona,

G.

Filippazzo-Contractor:

DipentaAfrica

Construction Fig.18

114

larger reverse moments are required than the moments actually producedby rib shortening, creep and shrinkage, see 2.5.3.

An article on the Hokawazu Bridge, in "Annual Report of Roads,1974" of the Japan Road Association, gives a table of stresses. Ananalysis of these stresses indicate that reverse stresses were intro-duced in this bridge. The effect of creep and shrinkage results in alarge change in dead load stress after the time at which the full deadload is in place. In the case of the Hokawazu Bridge, the immediatemaximum dead load stress is 1,700 psi, which compares to a maximum deadload stress after ending of creep and shrinkage of 1,230 psi. Theinitial stresses fall off rapidly in the beginning, so this initialcondition is only for a short time.

The arch rib should be constructed high to allow for the subsequentdeflection, after closure, from dead load rib shortening, creep andshrinkage. This is true regardless of whether or not measures are takenin construction to eliminate stresses from rib shortening, creep andshrinkage. The same ultimate deflection will still occur, becausethis deflection is due to the shortening of the arch axis from deadload thrust and shrinkage.

If the transfer of load is from a rib supported by tie backs to aself-supporting rib after placing the key, there is a considerablechange of stress and position of the rib at release of the tie-backs.The maximum stress in the Hokawazu Arch rib during erection was about2,200 psi near column P3, Figure 19, even though the section area usedat this point was about 1.5 times the section area used at the crown.The maximum stress in the arch rib under design load is about 1,400 psiat the same point. However, this stress is at the intradós, whereasthe 2,200 psi erection stressis at the extrados. The effect ofchange in stress and position of the rib in going from the cantileveredcondition to the arch condition must be taken into account in determiningthe position of the two cantilevers at the time of making the closingpour at the crown.

115

From Japan Road AssociationAnnual Report of Roads, 1974, page 14"Design Construction of the Hokawazu Bridge" by Y.

Inaue, Y.

MiyazakiFig. 19

HOKAWAZU BRIDGEFrom Japan Road AssociationAnnual Report of Roads, 1974, page 14"Design Construction of the Hokawazu Bridge" by Y.

Inaue, Y.

MiyazakiFig- 20

116

117

REFERENCES

1. "Guide to Stability Design Criteria for Metal Structures," 3rdEdition,Edited by B. G. Johnston, Structural Stability ResearchCouncil,

2. "Rainbow Arch Bridge-Design" by S. Hardesty, J. M. Garrelts, andI. G. Hedrick, Jr«, ASCE Proceedings, October 1943.

3. "Standard Specifications for Highway Bridges," AASHTO 1973 and1974, 1975 Interim Specifications.

4. "The New River Gorge Bridge," by C. V. Knudson, Military EngineerSeptember-October 1974.,

5. "Design of the Fremont Bridge," by A. Hedefine and L. G. SilanoASCE Meeting, Preprint 1210, April 1970.

6. The Kill Van Kull Bridge (Bayonne), "Design, Materials andConstruction," by L. S, Moisseiff, Journal of the FranklinInstitute, Vol. 213, No. 5, May 1932. "A Description of theErection," by H. W. Troelsch, American Bridge Co., 1931,

7. "Glen Canyon Bridge-Technical Record of Design and Construction,"United States Department of the Interior - Bureau of Reclamation,1959.

8. "Hampton Road All-Welded Steel Arch Bridge," by W. L. Powell,D. A. Nettleton, M. E. Eliot and J. C. Bridgefarmer, WeldingJournal, April 1959. "Box Girders for Arch," by J. C. Bridgefarmer,D. A. Nettleton and W. L. Powell, Welded Interstate HighwayBridges, Lincoln Arc Welding Foundation.

9. The Sando Bridge - Sweden, Engineering News Record, January 17,1946, p. 90.

10. "Design and Construction of the Hokawazu Bridge," by Y. Inoue andY. Miyozaki, Annual Report of Roads - Japan Road Association,1974.

11. "Reinforced Concrete Construction," - Hool,Vol. 11l - Bridges andCulverts, 1928.

12. "Gladesville Bridge - Australia," by Baxter, Gee and James,British Institution of Civil Engineers Proceedings, March 1965.

118

13. "Van Staden's Bridge - Port Elizabeth, South Africa/ 1 byM. d'Aragons ? L' lndustria Italiana del Cemento t December 1971.

14. Cowlitz River Bridge by Howard, Needles9 Tammen & Bergendoff,,International Association for Bridge and Structural Engineering,Bulletin 25 ? 1969.

15. Structural Design for Dynamic Loads-

1959 by M, J. Hoi ley, Jr.,Chapter 20 - Dynamic Effects of Wind Loads.

16. Creep Shrinkage Temperature in Concrete Structures, ACT.Publication SP-27, 1971,

17. Local Buckling of Long Span Folded Plates» Journal of theStructural Division, ASCE, October 1976.

18. Buckling Tests on Rectangular Concrete Panels by S. E. Swartz,V. H. Rosebraugh and M. Y. Berman, ACI Journal, January 1974,

19. Stability of Thin-Shelled Structures by G. 0, Ernst, ACI Journal,December 1952.

20. Concrete and Reinforced Concrete Arches, Final Report of theSpecial Committee, ASCE Transactions, 1935»

21. Interaction Between Rib and Superstructures by N. M. Newmark,ASCE Transactions, 1938.

22. Load Factor Design for Reinforced Concrete Bridge Structures,Portland Cement Association, 1974,

23. Fifty Year Development: Construction of Steel Arch Bridges byW. F, Hollingsworth, Modern Steel Construction, AISC, Vol. XV,No. 2.

24. Analysis of Statically Indeterminate Structures, Parcel andMoorman, 1955.

25. Lateral Stability of Bridge Arches Braced with Transverse Barsby L. Ostlund, Trans. Royal Institute Technology, Stockholm,Sweden, No. 84, 1954.

26. U. S. S. Steel Design Manual, R. L. Brockenbrough andB. J. Johnston, U. S. Steel Corporation, 1974.

119

27. Lateral Buckling of Twin Arch Ribs with Transverse Bars byP. N, Aimerdia, Dissertations Ohio State University,Columbus, Ohio, 1970.

28. Stability Problems of Compressed Steel Members and Arch Bridgesby G. Wastlund» ASCE Journal Structural Div 9, June 1960.

29. Buckling Strength of Metal Structures-

Bleich, 1955,pages 175-182.

30. Henry Hudson Bridge by D. B. Steinman and C. H. Gro.nquist,Engineering News Record, August 13, 1936.

120

APPENDIX - DERIVATION AND ORIGIN OF EQUATIONS

Equations 2 and 3 are the same as given in the 1977 Interim AASHTOEg. 1, Specification for Bridge, #20, replacing Article 1.7.90, and explained2 & 3 in the Commentary.

Equation 1 is for use in calculating live load deflection underservice load, and is therefore the same as equations 2 and 3 with thenumerical overload factor omitted.

X values, Fig. 3, are from Third Edition, Col. Research CouncilGuide, Chapter 16.

i 600,000 dFig. 5 -.*

—2 x - for 2-hinged steel arch

and A Tbs l

Eg. 41&41a Where i = arch span

d = depth of arch rib

A = live load deflection

f|DS= live load bending stress for service condition

Maximum live load deflection for a 2-hinged arch occurs approximatelyat the quarter point under a load over half the span. The loaded half ofthe span deflects downward and the unloaded half moves upward with apoint of zero deflection at approximately the center of the span.

The approximate equation for live load deflection is based on theassumption of a simple span equal to one half the arch span and aparabolic shape to the moment curve over this simple span. Using thewell known method for deflection whereby the moment curve is treated asa load and the moment due to this "load", divided by El, is equal to thedeflection:

Mí, 5£ 1 5Mil2A= ~6 xl2x U =

192E1

MdThe bending stress, f^, is equal to --=-, where d = rib depth; therefore:

M 2fbs

Id A = 5Ü_ " =192 x 29,000 d 557,000 d

121

rounding off:fbs¿ 2

600,000 dI 600,000 . d

and - - —j TA fbs £

Mdsubstituting for fks and transposing, gives Eg. 41

A 1,200,0001 41EI

Maximum live load deflection for a fixed arch occurs at the crown.Under a load over the central part of the span, the crown moves downward,and the outer parts of the span move upward, the points of zero deflectionare assumed 0.35& apart. This is based on the X value of 0.7Lforbuckling of a fixed arch. Equation 41a was derived in a similar mannerto equation 41, but using the span as 0.35&.

Fig.6 i= 44 + 0.6/1d

This equation for the l/á ratio is based partly on the ratios usedfor existing arches, and partly on the ratio required to meet a live loaddeflection-to-span ratio of 1/1200, as determined by the equation ofFigure 5. The z/á curve of Figure 6 will meet this requirement forfks =9 ksi up to a span of 400 feet, and for fbs

= 8 ksi up to a spanof 900 feet. As the span gets longer the dead load axial stress becomeslarger, resulting in less allowable stress available for bending. Theuse of higher strength steels for longer spans will counteract thislowering of the available allowable stress for bending. The net effectis usage of a somewhat smaller ratio of depth to span as the span getslonger.

Eg. 4a- These equations for the horizontal reactions due to change of4e temperature were obtained by integrating the equations given in many

text books for arches, assuming constant Iand a parabolic arch axis.

Eg. 4f- These equations are based on the same assumptions as for Eg. 4a4g 4e, and the assumption that the change in span length from dead load

axial stress, if unrestrained, would be:

HqL¿ v ACE, where Ac= rib area at the crown

IeI cand r2r 2 = —Ac

drFor the trussed rib, -^- is substituted for r.

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Eg. 4£ These equations are obtained in a similar manner to 4f and&4m 4g, using the sum of the axial change in length of the rib and

tie = Hp|_/Ar + HQ|_/A t and by adding Ir to ít in the equation forHrf

Eg. 5 These are as adopted in the 1977 AASHTO Interim Specificationsto 18 and explained in the Commentary.

Eg. 19 The derivation of these equations is explained in the text,to 27

Eg. 28 The origin of these equations is explained in the text.& 28a

Eg. 29 This equation and the equation on the following page are givenin various forms and rotation in text books on vibration. "EngineeringVibrations" by Jacobsen and Ayre is one.

Eg. 30- These equations are based on weights in actual bridges. Since34 they include both the roadway framing and the arch steel, different

strength steels may be included in a single bridge. In a generalway, the "£" term includes both the effect of more steel per footin a longer span and the use of higher strength steels as the spansget longer.

Eg. 35 H = simple beam moment from uniform load divided by h = W¿ 2/Bh.This assumes that the arch axis is so shaped as to eliminate dead loadbending moment.

Eg. 36 These are based on typical influence lines for arches.& 37

Eg. 38 These equations are based partly on typical influence lines for&39 arches. If Is is placed equal to zero in equation 38, this equation

becomes the same as equation 39. Placing Is equal to zero might beconsidered equivalent to hinged ends. Equation 38 goes from 0.324£for constant Ito 0.222& for Is

= 4IC.The following is proof of equation 38 for use in preliminary

design. Assume a parabolic axis and uniform live load.

Ordinate to arch axis at quarter point = 0.75h

Full span loaded: H = 1/2 W¿ 2/8h and V= W£ 2

Left half of span loaded: H = 1/2 W£ 2/8h = W¿ 2/16hVLVL

= 3Wa/8

123

""■►■Í.-Í-¥*'■&"■V32V 32 32 32 64

Wi,2 1~~o* = 77^2>2 > where &c = equiy. simple span

!2! 2 =ilZ

iQ

* 0.354¿

therefore, use equivalent simple span = 0.36£

A similar approach for a concentrated load at the quarter pointgives approximately the same equivalent simple span.

Eg. 40 The effect of applying uniform load through uniformly spacedcolumns or hangers to the arch rib is to add moment to the momentwhich would occur for directly applied uniform load. This added momentis equivalent to the moment occurring in a uniformly loaded continuousbeam on equally spaced supports, but opposite in sign. Thus, for acolumn spacing of A. , the added rib moment is +WA?/12 = +PA./12 underthe columns, and -PA|_/24 midway between columns.

Eg. 41 See under derivation for Figure 5.& 41a

Eg. 42 These equations assume no dead load bending moment in the arch42a except dead load rib shortening moment, and uniform dead load axial42b stress equal to fa. The second term of each equation is the downward42c deflection at the crown due to shortening of the arch axis from dead

load axial stress, assuming the arch to be unrestrained horizontallyat one abutment. The second term represents the effect of the bendingmoment in the rib produced by the horizontal restraint at the abutments.The second term can be arrived at from the horizontal reactions producedat the abutments by a unit vertical load at the crown. The equationsassume constant fa and constant temperature along the arch axis.

Eg. 43, These equations are the effect of lateral eccentricity of the live44 &45 load on a wide arch barrel. Their derivation is explained in the text.

Eg. 46a- These equations are similar to equations 42 - 42c for steel arches.47d The effect of creep has been assumed as 2.4 times the initial rib

shortening. The shrinkage coefficient has been taken as 0.0003.

Eg. 48a- These equations differ from the corresponding ones for steel arches48h to allow for a non-uniform depth of rib.

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Eg. 49a The basis for these equations is explained in the text.& 49b

Table II Equations 50a and 50b, with variable numerical constants, are givenEg. 50a in a number of texts on torsion. The numerical constants given here areto 50f close enough within the limit of b/d _> 10.

Equations 50b to 50f were derived by a method given in "Torsion inStructures," by C. F. Koelbrunner and K. Basler.