PRELIMINARY TOOLS FOR ANALYSIS AND DESIGN OF CABLE STAYED BRIDGES

A DISSERTATION

Submitted in partial fulfillment of the requirements for the award of the degree

of

MASTER OF TECHNOLOGY in

CIVIL ENGINEERING nth Specialization In Structural Engineering with diversification to bridge Engineering)

By

DEBARAJ BAiWNG ~ r

P` Y

DEPARTMENT OF CIVIL ENGINEERING :INDIAN INSTITUTE OF TECHNOLOGY ROORKEE

ROORKEE -247 667 (INDIA) JUNE, 2009

CANDIDATE'S DECLARATION

I hereby declare that the work presented in this dissertation entitled

"PRELIMINARY TOOLS FOR ANALYSIS AND DESIGN OF CABLE STAYED BRIDGES" submitted in partial fulfillment of the requirements for the award of the degree of Master of Technology with specialization in Structural Engineering with diversification to Bridge Engineering in the Department of Civil Engineering, Indian

Institute of Technology, Roorkee, is an authentic record of my own work which has been carried out from July 2008 to June 2009 under the guidance of Dr. Vipul Prakash, Associate Professor, Department of Civil Engineering, Indian Institute of Technology Roorkee.

I have not submitted the matter embodied in this report for the award any other degree or diploma

2'eb" tgaAfu61~ Sonoce d. Date: June 30, 2009 (DEB (ARAJ 1 BAILUNG SONOWAL)

CERTIFICATE This is to certify that the above statement made by the candidate is true to the best

of my knowledge and belief.

'&- (Dr. Vipul Prakash) Associate Professor

Department of Civil Engineering Indian Institute of Technology

Roorkee-247667, India

1

ACKNOWLEDGEMENT

I would like to express my deep sense of gratitude to my dissertation guide Dr. Vipul Prakash, Associate Professor, Civil Engineering Department, Indian Institute of

Technology for providing me with precious guidance and help at every step of my dissertation work. I am also thankful to him for his rendered help as and when needed, gave useful suggestions and guided through valuable discussions.

I would also like to record my thanks to my fellow class-mates and my family for their constant support and encouragement.

Deb&tej .4-n sondwa-( Date : June, 2009

Debaraj Bailung Sonowal

Place : Roorkee

TABLE OF CONTENTS

Page

Acknowledgement i

Abstract ii

List of figures iii

Nomenclatures vi

CHAPTER I INTRODUCTION

1.1 General 1 1.2 Historical review 3 1.3. List of longest cable stayed bridges in the world 5 1.4 Components of cable stayed bridges 9

1.4.1 Cable system supporting the deck 9 1.4.2 Positions of the cables in space 10 1.4.3 Cables 11 1.4.4 Tower types 12 1.4.5 Deck types 13 1.4.6 Main girders and trusses 14 1.4.7 Advancement of cable stayed bridges over suspension bridges 15

Chapter H Literature Review

2.1 Basic Concepts 16 2.2 Static Analysis 16

Chapter III Analyses

3.1 When the unit load is at a distance x from support A 18

3.1.1 Claypeyron's Theorem of Three moments 18

3.2 When the unit load is at a distance x from support B 20

3.3 When the unit load is at a distance x from support C 21

3.4 When the unit load is at a distance x from support D 22

3.5 When the unit load is at a distance x from support E 23

3.6 When the unit load is at a distance x from support F 24

3.7 Muller-Breslau's Principle 25

3.8 Problem taken for analysis 51

3.8.1 Calculation ofRA, R3. Rc, RD under dead load 52

3.8.2 Calculation ofRA, RB, Rc, Rn under live load 52

3.8.3 Calculation of reaction by using Courbon's method 53

3.9 Values cable forces 56

3.10 Comparison with software STTAD. Pro 57

3.11 Calculation of bending moment at longitudinal girder 57

2.11.1 Values of bending moments 58 3.12 Compression in tower 58

Chapter IV Designs

4.1 Design of cables 59

4.2 Design of longitudinal girder 61 4.3 Design of deck slab 64

4.4 Design of cross girder 68

4.5 Design of tower 72

Chapter 4 Economic Evaluation

5.1 General 77 5.2 Bar binding Schedule 78 5.3 Comparison with Prestressed concrete bridges 80

5.4 Comparison wit Suspension bridges 81 5.5 Comparison between cable stayed and suspension bridges 85

Chapter 6 Conclusions 86

Chapter 7 References 89

LIST OF FIGURES

Figure.1.1 Slay ropes on Egyptian sailing ships ...................................................3 Figure 1.2 Sutong Bridge .............................................................................6 Figure1.3 Pont de normandie .........................................................................6 Figure 1.4 Second Nanjing Yangtze Bridge .............................................................7 Figure1.5 Stonecutters Bridge .....................................................................6 Figure 1.6 Tatra Ohashi Briedge ....................................................................8 Figure 1.7 Longitudinal cable arrangements ......................................................9 Figure1.8 Stay cable types .........................................................................11 Figure 1.9 Different shape of pylon ..............................................................12 Figure1.10 Girder types ............................................................................13 Figure 3.1 Model of the cable stayed bridge ....................................................17 Figure 3.2 Continuous beam ....................... 18' Figure 3.3 When unit load moves at a distance x from A..................................... 18 Figure 3.4 load diagram due to unit load .........................................................19 Figure 3.5 When unit load moves at a distance x from B......................................20 Figure 3.6 When unit load moves at a distance x from C..................................... 21 Figure 3.7 When unit load moves at a distance x from D...................................... 22 Figure 3.8 When unit load moves at a distance x from E ......................................23 Figure 3.9 When unit load moves at a distance x from F ......................................24 Figure3.10 Continuous beam and elastic curve .................................................25 Figure3.11 Conjugate beam .......................................................................25

Figure 3.12 When unit load is at a distance 2m from A .......................................26 Figure3.13 Loading diagram ......................................................................26 Figure 3.1.4 Free body diagram of part FG .......................................................27 Figure 3.15 Free body diagram of part EF .......................................................28 Figure 3.16 Free body diagram of part DE .......................................................28 Figure 3.17 Free body diagram of part CD ......................................................29

Figure 3.18 Free body diagram of part BC ......................................................30

Figure 3.19 Free body diagram of part AB ......................................................30

iii

Figure.3.20 ILD of reaction RA .....................................................................32 Figure.3.21 ILD of reaction RH ....................................................................33 Figure 3.22 ILD of reaction Rc .....................................................................34 Figure3.23 ILD of reaction RD .....................................................................35 Figure 3.24 ILD of reaction Ra ......................................................................36 Figure 3.25 ILD of reaction RF ....................................................................37 Figure 3.26 ILD of reaction RG ...................................................................38 Figure 3.27 ILD for moment in the span AB .................. ...............................39 Figure 3.28 ILD for moment in the span BC .....................................................40 Figure 3.29 ILD for moment in the span CD ....................................................41 Figure 3.30 ILD for moment in the span DE ......................................................42 Figure 3.31 ILD for moment in the span EF .....................................................43 Figure 3.32 ILD for cable force PA ...............................................................44 Figure 3.33 ILD for cable force Pe ................................................................45 Figure 3.34 ILD for cable force Pc ................................................................46 Figure 3.35 ILD for cable force PE ................................................................47 Figure 3.36 ILD for cable force PF .................................................................48 Figure 3.37 ILD for cable force P0 ...............................................................49 Figure 3.38 ILD for Tower compression ..........................................................50 Figure 3.39Transverse Disposition of Class AA Tracked Vehicle Loading for Determination of Reactions on Longitudinal Beams .............................................52 Figure 3.40 Transverse Disposition of one lane Class A Vehicle Loading for Determination of Reactions on Longitudinal Beams ...........................................53 Figure 3.41 Transverse Disposition of two lane Class A Vehicle Loading for Determination of Reactions on Longitudinal Beams ...........................................54 Figure 4.1 Reinforcement detailing of longitudinal girder .....................................63 Figure 4.2 Reinforcement detailing of deck slab ...............................................67 Figure 4.3 Deck panel showing trapezoidal distribution of dead load ......................69 Figure 4.4 Disposition of Class AA tracked vehicle for maximum bending moment on crossbeam ............................................................................................70 Figure 4.5 Reinforcement detailing of cross girder ..............................................71

iv

NOMINCLATIIRE

Ast = Total cross sectional area of tensile steel of a structural member

D = Overall depth of beam or slab, dimension of a rectangular column in the direction under consideration

d = Effective depth of beam d' = Depth of compression reinforcement from top of beam f& = Characteristics compressive strength of concrete f3,= Characteristics strength of steel Minim = Limiting moment

P„= Factored column axial load Pt = Percentage of tensile steel in singly or doubly reinforced section

Pe = Percentage of compression steel = Shear strength in concrete

tc,ma,; = Maximum shear stress in concrete with shear reinforcement Vu = Factored shear force b = Width of the beam Asv = Area of stirrups

Sv = Spacing of stirrups 0Cbc = Permissible stress in concrete 6s, = Permissible stress in steel k = Neutral axis depth factor J = Lever arm n = no of panel a = length of each panel I = moment of inertia

Leff, effective length

r = Corresponding distance of the element from the line about which the moment of

inertia is required to be found out

vi

CHAPTER I

INTRODUCTION

1.1 General The use of cable stayed system in modem bridge engineering was due to the tendency of

bridge engineers in Europe, primarily Germany, to obtain optimum structural performance from material which was in short supply during the post war years.

Cable bridges are constructed along a structural system which comprises an orthotropic

deck and continuous girders which are supported by stays, i.e. inclined cable passing over or attached to towers located at the main piers.

The cables of a cable stayed structure work solely in tension. The cables must not only

have sufficient capacity to carry the dead load, but must also have enough reserve

capacity to carry the live load. Otherwise, the horizontal surface may suffer serious

deformations. The horizontal surface must be sufficiently stiff to transfer and / or resist the lateral and torsional stresses induced by wind, unbalanced live loads, and the normal force created by upward pull of the stays. The stays are usually attached symmetrically to

the column or tower with an equal number of stays on both sides. This is so that the horizontal force component of the inclined cables will cancel each other out and minimize at the top of the tower or column.

Wide and successful application of cable stayed systems was realized only recently, with the introduction of high strength steels, orthotropic type decks, development of welding

techniques and progress in structural analysis. The development and application of computers opened up new and practically unlimited possibilities for the exact solution of

these highly statically indeterminate systems and for precise statically analysis of their

three dimensional performance.

1

Basically the following important factors helped for the successful; development of cable stayed bridges.

1. The development of orthotropic steel decks.

2. Application of high strength steels, new method of fabrication and erection. 3.- The development of method of structural analysis of highly statically

indeterminate structures and application of electronic computers. 4. Experience with previously built bridges containing basic elements of cable

stayed bridges.

5. The ability to analyse such structure through model analysis

2

as

y ~ a

1

rC

• _ F

ry 1 Y Y J

1.2 Historical review

In ancient times, Egyptians built their boats in the form of cable stayed structures. In

other parts of the world, people built cable stayed rope bridges for pedestrian use. Since,

then, advancements in the cable stayed concept are most notable during two periods:

• from the 1600s to 1950

s from the 1950s onwards

Fig. 1.1 Stay ropes on Egyptian sailing ships

In 1617, Faustus Verantius proposed a bridge system having a timber deck supported by

inclined eye bars.

In 1784, a German carpenter, Immanuel Loscher in fribourge designed a timber bridge.

In 1817, two British engineers, Redpath and Brown, built the Kings Meadows Bridge

using sloping wire stay cables attached to the cast iron towers.

In 1824, Motley built a bridge at Tiverton, England.

In 1843, Clive proposed an original system of cable stayed bridge.

3

The first modem cable stayed bridge being the Stromstund Bridge in Sweden designed by

Dischinger and constructed in the year 1955.

In 1952 Leonhardt designed the cable stayed bridge across the Rhine in Dusseldorf. After

the first two cables stayed bridges of modem design had proved to very stiff under traffic

load, aesthetically appealing, economical and relatively easy to erect, the way was open for farther wide and successful application. The new system became rapidly popular

among German bridge engineers and about ten years later, in several other countries also.

It is now increasingly applied by engineers all round the world.

1.3. List of longest cable stayed bridges in the world

NAME LOCATION COUNTRY LONGEST COMPLETION PYLONS SPAN

SUTONG SUZHOU, CHINA 1088 m 2008 2 BRIDGE NANTONG

STONECUTTERS RAMBLER HONG KONG 1018 m 2009 2 BRIDGE CHANNEL

TATARA SETO JAPAN 890 m 1999 2 BRIDGE INLAND

SEA

PONT DE LE HAVRE FRANCE 856 m 1995 2 NORMANDIE

SECOND NANJING CHINA 628 M 2001 2 NANJING

YANGTZE BRIDGE

5

%

Fig 1.4 Second Nanjing Yangtze Bridge

Fig 1.5 Stonecutters Bridge

Rig 1.6 Tatssa Obashi Bridge

smallest amount of steel. The cables carry the vertical gravity load from the deck

structure while generating a minimum horizontal component of cable tension.

2. Harp or parallel system In this system the cables are connected to the tower at different heights and placed parallel to each other. This system may be preferred from an aesthetic point of view.

However it causes bending moments in the tower. In addition it is necessary to study

whether the support of the lower cables can be fixed at the tower leg or must be made movable in horizontal direction.

3. Fan or intermediate system The fan or intermediate stay cable arrangement represents modification of the harp system. The forces of the stays remain small so that single ropes could be used. All

ropes have fixed connections in the tower.

4. Star system

The star pattern is an aesthetically attractive cable arrangement. However, in

contradicts the principle that the points of attachment of the cables should be distributed as much as possible along the main girder.

1.4.2 Positions of the cables in space

1 Two vertical planes system Two alternative layouts may be adopted when using this system: the cable anchorages

may be situated outside the deck structure, or they may be built inside the main girders.

2. Two inclined plane system This system was first used for the Severin Bridge in Cologne, Germany, where the

cables run from the edges of the bridge deck to a point above the centerline of the bridge on an A-shaped tower.

3. Single plane system Another system is that of bridge with only one vertical plane of stay cable along the

middle longitudinal axis of the superstructure. In this case the cables are located in a

single vertical strip, which is not being used by any form of traffic.

10

1.4.3 Cables

Most existing cable stayed girder bridge has prefabricated high strength strands as stay

cables. In Europe, these strands are almost exclusively lock coil strands with several

layers of Z or S shape wires covering a smooth coil strands with round wires in fig. In United States, however, simple smooth coil strands or ropes, which are composed of several strands, are most commonly used.

The diameter of the strands, for erection and transportation convenience, ranges from 2

in. to about 4 in. They are then grouped together, usually after erection, to form the

required cross section ofthe stay cables by wire wrapping.

Besides Z-lock strands, smooth coil strands, parallel wires or even rolled steel I section have been used for cables. However, a recent design using high strength threaded bars is a revolutionary idea. These high strength threaded bars are threaded over their entire strength so that erection is very simple.

A B

bar♦.r0

♦rrr4•rr♦ ri•~r~r~r

r s

C D

Fig 1.8 Stay cable types : A) Parallel bar cables, B) Parallel wire cables, C) Stranded cables, A) Locked coil cables.

11

1.4.4 Tower types

The various possible types of tower constructed are illustrated in fig which shows that may be taken the form of

1. Trapezoidal portal frame 2. Twin towers 3. A-frames 4. Single towers

Twin Towers A-Shaped

Diamond Single tower Fig. 1.9 Different shapes of pylon

Portal type towers are used in the design of early cable stayed bridges, where it is used to

obtain stiffness against the wind load which the cable transfers to the top of the towers. However, later investigation of cable stayed bridges indicated that the horizontal forces

of the cables were in fact, relatively small, so that the freely standing legs could be used without disadvantage. With single towers or twin towers with no cross member, the tower

is stable in the lateral direction as long as the I'evel of the cable anchorages is situated above the level of the base of the tower.

12

1.4.5 fleck types

For a more efficient bridge deck, a major advance has been made with the development of the orthotropic steel deck.

Cross girders are usually 1.8- 2.5 in apart for decks stiffened by flexible ribs, and 4.6-5.5 apart in the case of decks stiffened by box type ribs possessing a high degree of torsional rigidity.

The orthotropic deck performs as the top chord of the main girders or truss. It may be considered as one of the main structural elements which lead to the successful development of modem cable stayed bridge. The advantage of orthotropic steel deck over that of concrete is the elimination or reduction of creep and shrinkage strains, which are large in concrete. In design of cable stayed bridge the major challenge is perhaps to keep the profile of the bridge as close to

the desired profile as possible. If large deviations from the desired profile occur, then riding comfort shall be reduced and remedies shall be to impose limits as vehicle speed or to adjust the lengths of the cables.

o-a m7

Cc] Cd )

Ce, C)

Fig 1.10: Girder types: (a) twin I-girder, (b) multiple I-girder, (c) rectangular box girder, (dl tranezoidal box girder. (el twin reetan'ular box girder. (fl twin tranezoidal box girder.

13

Trusses may be used instead of girders for aerodynamically reasons. In the case of

combined highway and rail road, when double tracks are used, trusses should be provided as the main members of such bridges.

3. Reinforced or Prestressed'concrete girders

A number of cable stayed bridges have been built with a reinforced or prestressed concrete deck and main girders during the last decades. These girders are economical, posses high stiffness and exhibit relatively small deflections. The damping effect of these monolithic structures is very high and vibrations are very small.

1.4.7 Advancement of cable stayed bridges over suspension bridges

Both types of bridge have two towers and a suspended deck structure. Whether the towers are equivalent may become apparent. There is a difference in the deck structures. The deck of a suspension bridge merely hangs from the suspenders, and has only to resist bending and torsion caused by live loads and aerodynamic forces. The cable-stayed deck is in compression, pulled towards the towers, and has to be stiff at all stages of construction and use.

A great advantage of the cable-stayed bridge is that it is essentially made of cantilevers, and can be constructed by building out from the towers. Not so a suspension bridge. Once the towers have been completed, steel cables have to be strung across the entire length of the bridge.

The deck of a suspension bridge merely hangs from the suspenders, and has only to resist

bending and torsion caused by live loads and aerodynamic forces. The cable-stayed deck is in compression, pulled towards the towers, and has to be stiff at all stages of construction and use.

Because the cable-stayed bridge is well-balanced, the terminal piers have little to do for the bridge except hold the ends in place and balance the live loads, which may be upward

or downward, depending on the positions of the loads.

15

CHAPTER II LITERATURE REVIEW,

2.1 Basic Concept

The application of inclined cables gave a new stimulate to the construction of large bridges. The importance of cable stayed bridges increased rapidly and within only one

decade they have become so successful that they have taken their rightful place among

classical bridge systems. "Modem cable-stayed bridges present a three-dimensional system consisting of stiffening girders, transverse and longitudinal bracings, orthotropic type deck and supporting parts such as towers in compression and inclined cables in

tension. The important characteristics of such a three-dimensional structure are the fall participation of the transverse construction in the work of the main longitudinal structure.

This means a considerable increase in the moment of the construction which permits a reduction in the depth of the girders and economy in steel", (Troitsky, 1977). 2.2 Static Analysis

The successful development in Europe within the last decade of cable stayed bridges, with their great structural advantages for medium and long bridges, has created a need for

a comprehensive presentation of the theory and design of this modem bridge system. The fan type is more aesthetic and as a rule the most economical for a pylon of slenderness

ratio (h/L) < 0.3. For an equal tower height, the average inclination of the cable stays is

lower. The harp system is preferred in a double plane system as it minimizes the intersection of cables. The structural behavior of the tower varies depending upon the type of cable system. The fan type increases buckling problems due to greater effective

length and the harp type increases bending moments. The mixed type represents a

compromise between the extremes of the harp and the fan systems and it is useful when it becomes difficult to accommodate all cables at the top of the tower. The star system may

be preferred due to its aesthetic appearance. A large number of stay cables with smaller

spacing simplifies the anchoring and permits shallower main girder. This shallowness facilitates a favorable cross section for aerodynamic stability and simplifies erection.

CHAPTER III

ANALYSIS

Several methods can be employed to carry out the analysis of cable stayed bridge. The

cable stayed bridges display different structural behaviors. The loads acting on the stiffening girder of a cable stayed bridges are transferred to the cables at connections.

It is possible to choose the statically determinate system as the basic structural scheme, assuming a cross section along the symmetry axis and provides supports in the location of the attachment of the cables (as shown in the fig). The analysis can be simplified by

taking into consideration the symmetry of the bridge system.

The method that here is considered can be achieved as follows: 1. Form a structure by adding supports in the intermediate cable connection as shown in

fig.2.1 below

Fig.3.1 Model of the cable stayed bridge

[7

/ =AA cKA \ i rnA \ CPA \ C1¼A

Fig.3.2 Continuous beam 2. This structure can be now treated as continuous beam. 3. Apply Muller-Breslau principle Influence Line method to determine the influence

lines for reaction, moment, cable forces.

3.1 When the unit load is at a distance x from support A

I

Fig.3.3 When the unit load is at a distance x from support A

Then, Applying three moment equation

3.1.1 Claypeyron's Theorem of Three moments

It states, "If a beam has n supports, the end ones being fixed, then the same number of equations required to determine thee support moments may be obtained from the consecutive pairs of spans i.e. AB-BC, BC-CD, CD-DE and so on."

m

Let, MA= Support moment at A

Mn= Support moment at B

Mc= Support moment at C

MD= Support moment at D ME= Support moment at E

MF= Support moment at F

MG= Support moment at G

Let us consider the beam AB as a simply supported beam. Therefore bending moment at the location of unit load,

Al=1x xx (5-x)/5= (5x-x2)/5

Fig.3.4 load diagram due to the unit load

For the span AB and BC A1X1= (1/2) x (5x-x2)/5x (2/3) x+ (1/2) (5-x) (5x-x2)/5x(x+ (5-x)/3)

A2X2=0 MAL1+2MB (L,+L2) +McLZ= - {(6 A1XI)/L1+ (6 A2X2)/L2}

MA+4MB+Mc= - (5x-x2) (5x+25)/125 MB+4MC+MD=O

Mc+4MD+ME=O

MD+4ME+MF=O ME+4MF+MG=O

RAy (ME+5-x)/5

RB= (Mc+10-x-IORA)15

Rc= (MD+15-x-15RA-I ORB)/5 RG= MF/5

RF= (ME-1ORG)/5

RE= (MD-I5RG-1ORF)/5

RD=1-(RA+RB+Rc+RE+RF+RG)

3.2 When the moving load moves at a distance x from the support B in the span BC

Fig.3.5 When the unit load is at a distance x from support B

MA+4MB+Mc= - (5x-x2) (50-5x)/125 MB+4M12+MD= - (5x-x2) (5x+25)/125

Mc+4MD+ME=O

MD+4ME+MrO

ME+4MF+MG=O

RA= MB/5

RB= (Mc+5-x-10RA)/5

Rc= (MD+10-x-15RA-10RE)/5

Rc= MF/5

RF= (ME-IORG)/5

Ef

RE= (MD-15R0-1ORF)/5

RD=1-(RA+RB+Rc+RE+RF+Rc)

3.3 When the unit load moves at a distance x from support C in the span CD

Fig.3.6 When the unit load is at a distance x from support C

MA+4MB+Mc= 0 MB+4Mc+MD= - (5x-x2) (50-5x)/125

Mc+4MD+ME=- (5x-x2) (5x+25)/125 MD+4ME+MF=O

ME+4MF+MG=O R,4= MS/5

RE = (Mc-IORA)/5

Rc= (M0+5-x-15RA-1 ORB)/5 RG= MF/5

RF= (ME-10RG)/5 R&= (M -I5RG-1 ORF)/5 RD= 1-(RA+RB+Rc+RE+RF+RG)

21

3.4 When the unit load moves at a distance x from the support D in the span DE

Fig.3.7 When the unit load is at a distance x from support D

MA+4MB+Mc= 0

MB+4Mc+MD= 0 Mc+4Mp+ME= - (5x-x2) (50-5x)/125

MD+4ME+MF= - (5x-x2) (5x+25)/125 ME+4MF+MG=O

RA= MH/5

R3= (Mc- I ORA)/5

Rc= (MD-I5RA-IORB)/5 R0= MF/5

RF= (Mr IOR0)/5 RE= (MD-15RG-1ORF+x)15

RD=1-(RA+RB+Rc+RE+Rp+RG)

22

3.5 When the unit load moves at a distance x from the support E in span EF

Fig.3.8 When the unit load is at a distance x from support E

MA+4MB+Mc= 0

MB+4Mc+Mp= 0

Mc+4MD+ME= 0

MD+4ME+MF= - (5x-x2) (50-5x)/125 ME+4MF+MG= - (5x-x2) (5x+25)/125

RA= MB/5

Rg= (Mc-1ORA)/5

Rc= (MD-15R4-I0RB)/5

RG= MF/5

RF=' (Mr 10RG+x)/5

RE= (Mn-15RG-1ORF+x+5)/5

RD=1-(RA+RB+RC+RE+RF+RG)

23

3.6 When the unit load moves at a distance x from the support F in the span FG

Fig.3.9 When the unit load is at a distance x from support F

MA+4MB+Mc= 0 MB+4MC+MD= 0

Mc+4MD+ME= 0

MD+4ME+MF= 0

ME+4MF+MG= - (5x-x2) (50-5x)/125

RA= MB/5

RB= (Mc-IORA)/5

RC— (Mn-15RA-1ORB)/5

RG= (MF +x)/5

R (ME- l ORG+5+x)15

RE= (Mo-15RG-IORF+x+10)/5

RD= 1-Rn+Rs+Rc+RE+RF+Rc)

24

3.7 Muller-Breslau's Principle

The Muller-Breslau principle states that if a reaction (or internal force) acts through an

imposed displacement, the corresponding displaced shape (elastic curve) of the structure is, to some scale the influence line for the particular reaction (or internal fore). The force

and displacement, of course, can be replaced by moment and rotation respectively.

Stepl: replace unknown reaction with unit deflection or load.

1 BRA \/ -.a \ i flA \ . rRl \ , rr.a 'J iKA \

Fig3.10 Continuous beam (top) and elastic curve (bottom) due to unit load applied

atA

Step 2: Convert in conjugate beam

G

hinges

Fig.3.11 Conjugate beam

25

Step 3: Applying three moment equation to get the ordinate of the elastic curve.

Consider that the unit load is moved at a distance 2 m from the support A

Fig.3.12 When the unit load is at a distance 2 m from support A

Then the corresponding loading diagram is as follows

0.121

Fig.3.13 loading diagram

26

iy Considering the free-body diagram of part FG

Fig.3.14 Free body diagram of part FG

Rgx5 = 0.5x0.002x5x5/3 = 0.00167

Rf = 0.5 x 0.002x5 — 0.00167 =0.033

Me = -0.0033x5-0.5xO.95x0.002x (4.05+2/3x0.95)+0.5x0.008x4.05x 1/3x4.05 = -0.0165-0.0044+0.021

=-0.021+0.021

M

Considering the free body diagram EF

0.

Pj

Fig.3.15 Free body diagram of part EF

Re + 0.0033 = -0.5xO.002xO.95+0.5x0.008x4.05 Re + 0.0033 = -9.5 x 10'+0.0162

Re = 0.01195

Md = 0.01195x5+0.5x 1.06x0.008x (3.94+2/3x 1.06)-0.5xO.0323x3.94x 1/3x3.94 = 0.06+0.02-0.08

Considering the free body diagram of DE

['At Tcl

Gl:

Fig.3.16 Free body diagram of part DE

Rd+Re = 0.5x 1.06x0.008-0.5x3.94xO.0323

Rd + 0.01195 = 4.24x 10-3-0.063 Rd = -0.071

Considering the free body diagram CD

Fig.3.17 Free body diagram of part CD

29

Rc+Rd =-0.5x0.0323x1.06+0.5x3.94x0.121 Re + 0.071 = -0.017+0.24 Re — 0.152

Considering the free body diagram BC

Rb+Rc = 0.5x0.121 x 1.06-0.5x3.94x0.45 Rb + 0.152 = 0.06413-0.8865 Rb = -0.974

m Fig.3.18 Free body diagram of part BC

30

Considering the free body diagram AB

i u

Fig.3.19 Free body diagram of part AB

MA = -0.974 x 5-0.5 x .45 x 5 x 2/3x5 = -4.87-3.75 --8.62

Ml =-0.974x3-0.18x3x3/2- 0.5x.27x3x 2/3X3 _ -2.922-0.81-0.81= -4.542

Now the scale of the influence line diagram can be fixed on the following basis. Since a unit loading at A must produce a reaction of unity at A, the deflection obtained for the original beam at A must represent unity, that is, the conjugate beam moment MA = -8.62 must be equated to unity. Therefore, the ordinate of reaction at a distance 2 m from the support can be obtained by dividing the moment value MI by moment value MA. Therefore ordinate value at 1 is,

RI = -4.542/-8.62 = 0.52

This value is nearly same to the value obtained from Influence line method

31

KM 0 a 0 0

N O IV A 01 Oi ~ N

1,26 /

,16 /2,5

15

10

1.3 2,5

13,8 g 15

163 11.6 18,8

20 1M3

2.5 3.8

25

w j I

1;5

8.B

LjVI N

5 1.2

1

0.8

0.6

0.4

0.2

0

-0.2

—Sedes1

M

Fig.3.21 ILD for reaction RB

33

Fig.3.22ILD for reaction Re

34

1.2

1

0.8

0.6

0.4

0.2

0

-0.2

= Series9 Sedes2

Fig.3.23 ILD for reaction Rn

35

RE

1.2

_ /

1 =

0.8.

0.6 ,

0.4, —Series1

—Seoes2 0.2

0• .

Fig3.24 LCD for reaction RE

36

RF

12 Th

1 ;

0.8

0.6 — Senesl

0.4 .

0.2

0 LII EflTflTtiTiRllflH + it rnatl ly.~.. O ~. CV Cr) 10 CD P .r CN P7 LO CO N-GO O

N m ". ~. CO r c- . N CV (4 . N N N '0.2

M

Fig.3.25 ILD for reaction RF

37

1.2

1

0.8

0.6

0.4

0.2

0

-0.2

Fig.3.26 ILD for reaction Rc

m

1.2

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

Rm

ti

fi

'~ a

Fig.3.27 ILD for moment in the span AB

39

MC

0.z

0.1

L11 :111111 u Ill lt?Tku II II JU

iO [A It) LID In 1i) ) l}) co it) C.) IYJ. co:' O c'7 tO 0 LID C?) to .GO. N CV "," N r- ~ a-- CTJ CD 1C GD

-0.1 —Senes1

-0.2

-0.3

-0.4

-0.5

M

Fig.3.28 ILD for moment in the span BC

40

Fig.3.29 ILD for moment in the span CD

41

Fig.3.30 ILD for moment in the span DE

42

MF

Fig.3.31 ILD for moment in the span EF

43

PA

2.5

2

1.5

z 1

0.5

0

-0.5

—Seriesl

M

Fig.3.32 ILD for cable force PA

PB

1.6

9.4 ...

1 L

0.8

0.6 F - — Sedes1

0.4 t

0.2

O i •0.2 u~ u) ara era in ~n In o CO u) m Ln CO o CO IS) aa Sri ~a IS)

! N 1__ N C:J CD 1•-00 a— C'4 (0 N t0 I— aO . «~ r~ -0.4

040104 C'10401 -

M

Fig.3.33 ILD for cable force PB

45

1.4 1.2

PC

0.8 0.6 0.4 0.2

0 -0.2 -0.4

u'] in eA afl 14) O. M V] GO I.CJ OD O CO Y'> . Go 10 ftl aD

~ I~ CV MCO f~ QD r N M [D I— OD r r..r r [V. N (V t'J N CJ

M

Fig.3.34 ILD for cable force Pc

PE

1.4 1.2

1 0.8 0.6 0.4 0.2

0 -0.2 -0.4

—Series

M

Fig.3.35 ILD for cable force PE

47

PF

1 1.4 1.2

1 0.8

Y 0.6 0.4 0.2

0 -0.2 -0.4 P Jar.

M

Fig.3.36 ILD for cable force PF

PG

2.5

2

1.5

Z 1

0.5

0

-0.5

- Cli ao o r4 Cfl lf) (O 1~ CO 0 N O) lf) l0 1~ aD

— —

Fig.3.37 ILD for cable force Pa

Fig.3.38 ILD for Tower compression

50

3.8 Problem taken for analysis

For the analysis, the bridge is taken as follows It is a radial pattern type cable connection connected in two vertical systems Total span 30 m Panel length 5m Assume height of the tower =h Inclination of the cable =30 degree From the geometry, h=nxaxtan 30

=3x5xtan 30

'z9m Assume,

Total length of the bridge = 30 m Width of the deck = 7.5 = two lane No. of cables 11 Thickness of r.c.c. slab = 200 mm

Wearing coat - = 70 mm Cross section of longitudinal girder = 600 x 800 mm Cross section of cross girder = 450x 800 nun Loading class AA TRACKED vehicle

Dead Load Weight of deck slab and wearing coat = 25x0.270x7.5 =50.63 kN/m Weight of longitudinal girders (2 nos.) = 2x25x0.6x0.8 = 24 kN/m Weight of cross girder = (7x0.45x0.8x7.5x25)/30 = 15.75 kN/m Total dead load = 90.38 kN/m Dead load on each girder = 45.19 kN/m

51

3.8.1 Calculation of RA, RB, Rc, RD under dead load

From the ILD diagram for RA, calculating maximum value of RA due to dead load RA = area of ILD x load intensity

=1.97x45.19 = 89.02 kN

RB = area of ILD x load intensity = 5.66 x 45.19 =225kN

Re = area of ILD x load intensity = 4.81 x 45.19 =217.36kN

RD = area of ILD x load intensity = 5.095 x 45.19 = 230.24 kN

3.8.2 Calculation of RA, RB, Rc, Rn under live load The critical loading position for maximum reaction due to I.R.C. class AA TRACKED vehicle loading is Reaction = (700x4.9)/7.5 = 457 kN

Fig.3.39 Transverse Disposition of Class AA Tracked Vehicle Loading for Determination of Reactions on Longitudinal Beams

52

w

oc`~~ o

The critical loading position for maximum reaction due to I.R.C. Class A Wheeled vehicle loading is as follows,

C.G

All dimensions are in meter Fig.3.40 Transverse Disposition of one lane Class A Wheeled Vehicle Loading for

Determination of Reactions on Longitudinal Beams

3.8.3 Calculation of reaction by using Courbon's method

According to Courbon's method, the reaction Ri of the cross beam on any girder i of a typical bridge consisting of multiple parallel beams is computed assuming a linear variation of deflection in the transverse direction. The deflection will be maximum on the exterior girder on the side of the eccentric Ioading (or c.g of loads if there is a system of concentrated loads) and minimum on the other exterior girder.

The reaction Ri is then given by Ri = (PI;/ EIi) (1 + Y, Ii/ E Ii d; (e d;)

Where, P = total live load I; = moment of inertia of longitudinal girder i e = eccentricity of the live load

d; = distance of girder from the axis of the bridge

53

Here, P = (37.5+62.5+62.5+37.5) = 200 kN n = no. of longitudinal girder = 2 e=1.3m

RA = (200xL 2xI) (1 + 2xI1/ 2(Ix 3.75)2(1.3 x3.75)

= 134.67 kN RB = 200- 134.67 = 65.33 kN

The critical loading position for maximum reaction due to two lanes of I.R.C. Class A Train of vehicles loading is as follows,

C.G c~

LADS x

11 11.4 11 k'1 kM kF ~14 N - O 5

All dimensions are in meter Fig.3.41 Transverse Disposition of Two lane Class A Train of Vehicle Loading for Determination of Reactions on Longitudinal Beams

Here, P = 4x114 kN n = no. of longitudinal girder = 2 e = 0.7 m

Ra = (4x 1144/ 2xI) (1 + 2xIi/ 2(Ix 3.75)2X(0.7 x3.75)

= 270.56 kN

54

Rb = 4x 114- 270.56 = 185.44 kN

So, maximum critical condition comes under the IRC Class AA Tracked of vehicle loading.

RA = area of ILD x load intensity = 2.07 x 457/3.75 =252.6kN

RB = area of ILD x load intensity = 3.26 x 457/3.75 = 397.64 kN

Rc = area of ILD x load intensity = 3.231 x 457/3.75 = 393.75 kN

RD = area of ILD x load intensity = 3.228 x 45713.75 =393.39kN

55

3.9 Values cable forces

MAXIMUM

REACTION

D.L

kN

L.L

kN

D.L+L.L

kN

CABLE FORCE

kN

RA 89.02 252.6 341.62 683.24

RB 225 397.64 622.64 930.88

RC 217.36 393.75 611.11 699.12

RD 230.24 393.39 623.63 Tower

R6 89.02 252.6 341.62 683.24

12F 225 397.64 622.64 930.88

RE 217.36 393.75 611.11 699.12

56

a

3.10 Comparison with software STTAD. Pro

Result obtained from manual analysis Result obtained from STAAD. Pro

Cable force PA = 683.24 kN PA =656.57 kN

PB = 930.88 kN PB =907.72 kN PC = 699.12 kN PC =648.26 kN

3.11 Calculation of bending moment at longitudinal girder Due to live load

In the span AB = area of ILD x load intensity = 1.45x457/3.75 = 176.71kNm

In the span BC = area of ILD x load intensity = 1.19x457/3.75 =145kNm

In the span CD = area of ILD x load intensity = 1.178x457/3.75 = 143.55 kNm

Due to dead load

In the span AB = area of ILD x load intensity = 2.63x45.19 = 119.19 kNm

In the span BC = area of ILD x load intensity = 1.9175x45.19 = 86.65 kNm

In the span CD = area of ILD x load intensity = 2.1575 x45.19

= 97.49 kNm

57

3.11.1 Values of bending moments

MOMENT DUE TO D.L kNm

DUT TO L.L

kNm TOTAL MOMENT kNm

AB 119.19 176.71 295.9

BC 86.65 145 231.65

CD 197.49 143.55 241.04

DE 197.49 143.55 241.04

EF 86.65 145 231.65

FO 119.19 176.71 295.9

3.12 Compression in tower

Compression in tower = summation of all the reaction = RA+RB+Rc+RD+RE+RF+RG = 341.62+622.64+611.11+623.63+611.11+622.64+341.62 = 3774.37 kN

W

CHAPTER IV DESIGN

4.1 Design of cables

Maximum cable force in outermost cable = 683.24 kN Using 7 mm dia. tensile wires initially stressed to 1200 N/mm2

Area of each wire = n/4 X 72 = 38.5 mm2

Force in each wire = 38.5 x 1200 N =33.5x 1200/1000 kN = 46.2 kN

No. of wires = 683.24/46.2 = 14.78 = 15 no.

Maximum cable force in middle wire = 930.88 104 Using 7 mm dia. tensile wires initially stressed to 1200 N/mm2

Area of each wire = n/4 X 72 38.5 mm2

Force in each wire = 38.5 x 1200 N =38.5x 1200/1000 kN = 46.2 kN

No. of wires = 930.88/46.2 = 20.14 z 20 no.

Maximum cable force in innermost cable= 699.12 kN Using 7 mm dia. tensile wires initially stressed to 1200 N/mm2 Area of each wire = zz/4 x 72

=38.5mm2 Force in each wire = 38.5 x 1200 N

59

= 38.5 x 1200/1000 kN

= 46.2 kN No. of wires = 699.12/46.2

= 15.13 = 16 no,

4.2 Design of longitudinal girder

Maximum moment =295.9 kNm Provide grade of concrete M40 Fe-415 steel

acbc= 13.33 Mpa 6St = 200 Mpa k= 280/(280+3x 65i)

= 0.318 J= 1-k/3 =0s

R= 0.5 x u b, xjxk = 1.91

Effective depth required='I (M/Rxb) =1] {(295.9X106)/ (1.91X600)} = 508.13

509 mm Effective depth provided assuming 12 mm dia main bars and clear cover 40 mm

= 800-40-6 = 754 mm So, the provision of total depth 800 mm is safe. Area of main reinforcement = (295.9x 1000x 1000)/ (200x0.9x754)

=2180.22mm2 Adopt 20 mm dia bars So, no of bars required = 2180.22/ (m/4 x202)

= 7 nos. Adopt 7 nos. bar giving an area of 2198 mm2

Design for shear Maximum reaction V = 622.64 kN

Vu= 1.5x622.64

61

= 983.96 kN

(983.96 x 103 ) / (600x754) = 2.075 MPA

Which is less than'ta = 4.0 PAPA (for M40 grade concrete) Design for shear strength of concrete Ast = 2198 mmz Pt = (100x2198) /(600x754)

= 0.5 So as per table table 19, IS456:2000 Design shear strength of concrete x. = 0.5 1MPa <'r. = 2.075 MPa Design of vertical stirrups Vus=(t -i,)bd Vus/d = (2.075-0.5 1) x600

= 939 N/mm Assuming, two legged closed stirrups of 8 mm dia. Asv = 2xir,/4x82

100.6mm2 Required spacing sv <_ (0.87fyAsv)/ (Vus/d)

_ (0.87x415x 100.6)/939 = 38.68 mm

Code requirement for maximum spacing Sv = 2.175fyAsv/b = (2.175 x415 x 100.6)1600 = 151.34 mm

Sv= 0.75d = 0.75x754 = 565.6 mm Sv = 300 mm Provide, 8tD two legged stirrups @ 40 mm c/c spacing. Since the depth is more than 750 mm, so side reinforcement is provided @300 mm c/c spacing.

62

40mm dear cover

4-2Odia

mm

Fig.4.lReinforcement detailing of longitudinal girder

63

4.3 Design of deck slab

Maximum bending moment due to dead load

Weight of deck slab = 25 x 0.2 =5 kN(m2

Weight of wearing course = 25 x 0.07

= 1.75 kN/m2 Total dead load= 5+ 1.75

= 6.75 kN/m2 Since the slab is supported on all four sides and is continuous, Pigeauds curves will be

used to get influence coefficient to complete moments. Ratio k = shorter span / longer span

= 4.55/6.9 = 0.655

1/k = 1/0.655 = 1.44

From, Pigeaudes curve, m1 = 0.048

And m2=0.018 Total dead weight = 4.55 x6.9x 6.75

= 211.91 kN

Moment along short span = (ml+ µ m2) x P = (0.048+0.15x0.018) x 211.91

= 10.74 kNm

Moment along long span = (m2+ g ml) x P

= (0.018+0.15x0.048) x 211.91 =5.34kNm

Live load bending moment due to IEC class AA tracked vehicle.

Size of an panel deck slab = 7.5m x 4m

Impact factor fraction = 25 %

m

Width of the load along short span U=/ {(0.85+2x0.07)2+ (0.200)2}

= 1.01 m V=./ {(3.6+2x0.07)2+ (0.200)2}

=335m K= 0.652 UB= 1.01!6.9

= 0.146 V/L= 3.7514.55

= 0.824 Using Pigeauds curve m1= 10 and m2 = 2 Total load including impact = 1.25 x 350

= 437.5 kN

Moment along short span = (m 1+ µ m2) x P _ (10.0+0.15x 2.0) x 437.5x102

= 45.06 kNm Moment along long span = (m2+ Lt ml) x P

= (2.0+0.15x 10.0) x 437.5x 102

= 15.31 kNm The above computations assumed a simply supported condition along the four edges. In fact, the deck slab is continuous. To allow for continuity, the computed moments are multiplied by a factor 0.8. Design bending moment along short span = (10.74+45.06) x 0.8

=44.64kNm Design bending moment along long span = (5.34+15.31) x 0.8

= 16.52 kNm

Provide grade of concrete M40 Fe-415 steel

6cbc 8.3 Mpa 6,t = 200 Mpa

65

k= 280/(280±3x crt)

= 0.318 J= 1-k/3

=0.9

R= 0.5xab,xjxk

=1.1

Effective depth required=V (MJRxb)

=V {(44.64x106)/ (1.91x 1000)} = 152.87

153 mm

Effective depth provided assuming 12 mm dia main bars and clear cover 40 mm = 200-40-6 =154 mm

So, the provision of total depth 200 mm is safe. Area of main reinforcement = (44.64x 1000x 1000)/ (20Dx0.9x 154)

=1610.38mm2 Adopt 12 mm dia bars

So, no of bars required = 1610.38/ (a/4 x122)

= 14.24 nos. Adopt 16 nos. bar giving an area of 1808.64 mm2

Area of longitudinal reinforcement = (I6.52x 1000x 1000)/ (200x0.9x 142)

= 646.32 nun2

So, no of bars required = 646.32/ (7r/4 x 122)

= 5.71 nos. Adopt 6 nos. bar giving an area of 678.24 mm2

it'

40mm clear cover

Fig.4.2 Reinforcement detailing of deck slab

M

4.4 Design of cross girder

7J

Fig.4.3 Deck panel showing trapezoidal distribution of dead load

Weight of deck slab and wearing course = 6.75 kN/m2 Total load on cross beam due to slab by trapezoidal load distribution, = 2x0.5x3.45x6.9x6.75 = 160.68 kN

Self weight of cross beam and weight of wearing course on cross beam = 6.9x0.45x0.8x25+6.9x0.45x0.07x25 = 62.1+ 5.44

= 67.54 kN

Total dead load= 160.68+67.54 = 228.22 kN

Coefficient of maximum positive bending moment = 0.051 Coefficient of maximum negative bending moment = 0.097 Positive bending moment = 0.051 x228.22=11.64 kNm Negative bending moment = 0.097x228.22= 22.14 kNm

Bending moment due to live load

5-1.01=3.987m i.ol

Fig. 4.4 Disposition of Class AA tracked vehicle for maximum bending moment on cross beam

Load on cross beam=2x (350x1.575/3.6x3.987/5)+350x0.45/3.6 = 244.2+43.75

= 287.95 kN

Coefficient of maximum positive bending moment due to concentrated load = 0.151

Coefficient of maximum negative bending moment = 0.146 Positive bending moment including impact = 0.151x6.9x287.95x 1.25=375.02 kNm

Negative bending moment including impact =0. 146x6.9x287.95x 1.25=362.60 kNm

Design positive moment = 11.64 + 375.02 = 386.66 kNm

Effective depth required

Provide grade of concrete M40

Fe-415 steel 6,se= 13.33 Mpa

ast = 200 Mpa

k= 280/(280+3x 65,)

=0.318

J= 1-k/3

= 0.9 R=0.5xcoboxixk

= 1.91 Effective depth required=" (M/Rxb)

=1i {(386.66x106)/ (1.91x 450)}

-- 671 mm

Effective depth provided assuming 12 mm dia main bars and clear cover 40 mm

= 754 mm Area of steel required = (386.66x 1000x 1000)/(200x0.9x754)

= 2848.95 mm'

So, no of bars required = 2848.95/ (it/4 x252) = 6 nos.

Adopt 6 nos. bar giving an area of 2989.04 mm2 Design negative moment = 362.60 kNm

Area of longitudinal reinforcement = (362.60x 1000x 1000)/ (200x0.9x754) = 2671.67 mm2

So, no of bars required = 2671.67/ (2r/4 x252) = 6 nos.

Adopt 6 nos. bar giving an area of 2712.96 mm2

Provision for shear By nominal provisions

Nominal shear reinforcement 2 legged stirrups 8 dia @ 100 mm c/c spacing

FZ:

3-12dda

4o mm clear cover

Fig. 4.5 Reinforcement detailing of cross girder

71

4.5 Design of tower

Total vertical compressive force = 3774.37 kN Maximum moment in tower = maximum horizontal force in tower x height of the tower

= (397.64/sin 41.98) x (cos 41.98) x 9 m = 441.51 x 9 = 3973.5 kNm

Assume the cross section of the tower as I= bD3/12 = Db3/12 A=bD I=A? So,r=d/'112 And, r = bhI l2 Lett/r = 0.5 1/ D/'112 = 0.71/ b/'112 Or b/D = 1.4 Assume D = 0.85 m So,b=1.2m Therefore assume croos section is 850 mmx 1200mm As per IRC 6:2000 Table 4 Tower height 9 m (above deck) + 11 m (below deck) =20m For 20 m height, corresponding, V (velocity of wind) = 128 km/h

P (pressure) = 119 kg/m2

Force = pressure intensity x area =119kg/mz x0.85mx9m =910.35 kg= 9103.5n=9.1035kN

Assume the tower as cantilever in the transverse direction So, the moment at fixed end = Wx L/2

= 9.1035 x 9/2 = 40.96 kNm

72

Therefore, Pux = 1.5x3774.37 = 5661.555 kN M„x = 1.5x3973.5 = 5960.25 kNm M y = 1.5x40.96 = 61.44 kNm P„y = 1.5x9.1035 = 13.655 kN

Height of the tower 9 m Check for short or slender column lx=ly=9000 mm, Dx=850 min

Dy = 1200 mm Slenderness ratio lex/Dx = kx x 9000/850 = 7.5 kx

And ley/ Dy = ky x 9000/1200 = 7.5 ky As the column is fixed in both the directions, effective length ratios kx and ky are both less than unity, and hence the two slenderness ratios are both less than 12. Hence the column is designed as a short column. Minimum Eccentricities Applied eccentricities, ex = 5960.25/5661.555 = 1.05 m

And, ey = 61.44/13.655 = 4.45 m ex, min = 9000/500 + 850/30 = 46.33 mm > 20 mm ey, min = 9000/500 + 1200/30 = 58 mm > 20 mm As, the minimum eccentricities is less than the applied eccentricities, no modification to Mux and Muy, is called for. Trial section: longitudinal reinforcement Design for uniaxial eccentricity with Pu = 1.15x1i(P2 + P2„y)

= 1.15x'I(5661.5552+ 13.6552) = 6510.8 kN

Mu = 1.15 x'd (M2 + M2 ) = 1.15x1 (5960.252+ 61.442) = 6854.65 kNm

Pu/fckx bx D = 6510.8 x 103/(40 x 850 x 1200) = 0.159 Mu/fckx bx D2 = 6854.65 x 106/(40 x 1200 x 12002) = 0.139 Assume a clear cover of 40 mm, 8 mm dia. ties and 32 mm dia main bars.

73

d'=40+8+16=64 --70 mm d'/D=70/ 1200=0.058 Using SP-16, chart 43, for the value d '/D= 0.05

p/fck = 0.08 p = 40 x 0.08

= 3.2

As — pbD/100 = 3.2x850x 1200/100

= 32640 mm2

No. of bars required = 32640/ (7E/4 x 402) = 25.98 26 nos.

As provided, 26x (7x14 x 402) = 32656 mm2>32640 mm2

Uniaxial moment capacities: Muxl, Muyl

Pu/fckx bx D = 0.159 (as calculated above)

Pprovided = (32656x 100)/ (850x 1200)

=3.2

p/fck = 3.2/40 = 0.08

d'= 40+8+16 =64 70 mm

d '/D= 70/ 1200 = 0.05 Using SP-16, chart 43, for the value d'/D= 0.05 p/fck = 0.08 Mu/fckx bx D2= 0.15

Muxl= 0.15x40x850x 12002

= 7344 kNm Muyl= 0.15x40x 1200x8502

= 5202 kNm Which are significantly greater than Mux= 5960.25 kNm and Muy= 6i.44 kNm

Values of Puz and oo, Puz = 0.45fokAg + (0.751y-0.45fck)Asc

= 0.45 x40x 850x 1200 + (0.75x415-0.45 x40) x32656

74

= 27936.372 kN

Pu/Puz = 6510/27936.372 = 0.23(which is lies between 0.2 to 0.8) cc= 1+ (0.23-0.2) x (2.0-1.0)1 (0.8-0.2)

= 1.05

Check safety under biaxial loading (Mux/Mux 1) °̀" + (Muy/Muy 1)m"

= (5960.25/7344)' ° + (6l.44/5202)' ° = 0.803 + 0.0094 = 0.8124<1.0

Hence the trial section is safe under the applied loading. Design of vertical stirrups Tie diameter ❑t = 40/4 = 10 mm

Or ❑t = 6 mm Provide 10 mm dia. Tie spacing Sv = D = 1200 nun Sv = 16x40 = 640 mm Sv = 300 mm Provide 8❑ two-legged stirrups @100 nun c/c spacing

75

4od1a@ tamm dear

mm

Fig. 4.6 Reinforcement detailing of Tower

i

CHAPTER V

ECONOMIC EVALUATION

5.1 General

CONCRETE

Concrete in tower (2 nos.) = 2x (9x.85x 1.2) = 18.36 m3 In longitudinal girder (2 nos.) =2 x (0.6x0.8x30)

=28.8m3 In croos giders (8 nos.) = 8x (0.45x0.8) x7.5

= 21.6 m3 In deck = 30x0.2x6.9

=41.4m3 Total concrete = 18.36+28.8+21.6+41.4

= 110.16 m3 CABLE Prestressed concrete bridge As per standard plans for highway bridges (prestressed and R.0 type of bridge) 42 nos 7.0 mm dia. wire Length of cable = (2x30.47+2x30.47+30.510+30.550)

= 182.94 m

Total length = 2x 182.94 =365.88m

Total volume of cables = 42xic/4x0.0072x365.88 =0.59 coin

Cable stayed bridge Length of cable, LI = -4 (152+92) = 17.49 m

Length of cable, L2 = I (102+92) = 13.49 m Length of cable, L3 = J(52+92) = 10.29 m Total length = 2 x (17.49+13.45+10.29)

= 82.46 m

77

For two tower =2x82.46

= 164.92 m Total volume = 200xz/4X0.0072X 164.92 = 0.4 cum

5.2 Bar binding Schedule

SCHEDULE OF DIA LENGTH NO. TOTAL WEIGHT= REINFORCEMENT OF OF OF LENGT LENGTHx

BAR EACH BARS H UNIT BAR WT/METR

E mm mm m kgs

SHAPE OF BARS LONGITUDINAL GIRDER 700 I700 20 6400 84 537.6 537.6x2.5

5000 12 5000 72 360 360x0.9

5000 =324

8 2864 1500 4296 4296x0.4

-~20

192 sz0 =1718.4

192.

120mm 8 2005.6 1500 3008.4 3008.4x0.4 =1203.36

90.Omm

TOWER

9000mm

140Omm

40 10400 52 540.8 540.8x9.9 = 5353.92

770

300 1120

10 4380 80 350.4 350.4x0.6 = 210.24

300m

770m m t

10 2788 160 446.08 446.08x0.6 = 267.648

324mm

3 Arcen

tt]Omn

10 3204 80 256.32 256.32x0.6

mn

= 153.792

79

5.3 Comparison with Prestressed concrete bridges

CABLE STAYED QUANTITY PRESTRESSED QUANTITY • BRIDGE CONCRETE

BRIDGE SPAN 30M

CABLE 0.4 cum CABLE 0.59 cum

AS PER STANDARD PLANS FOR HIGHWAY BRIDGES

CONCRETE 110.16 cum CONCRETE 214 cum

AS PER STANDARD PLANS FOR HIGHWAY BRIDGES

STEEL 9372 kg STEEL 21839.2 kg AS PER STANDARD PLANS FOR HIGHWAY BRIDGES

:1

5.4 Comparison wit Suspension bridges

For the analysis, the bridge is taken as follows

Fig. 4.1 Suspension bridge Panel length Sm

Height of the tower =9 m

Assume, Total length of the bridge = 30 in Width of the deck = 7.5 = two lane

Thickness of r.c.c. slab = 200 mm

Wearing coat = 70 mm

Cross section of longitudinal girder = 600 x 800 mm Cross section of cross girder = 450x 800 mm

Loading class AA TRACKED vehicle The critical loading position for maximum reaction due to I.R.C. class AA TRACKED

vehicle loading is

m

Fig.5.2 Transverse Disposition of Class AA Tracked Vehicle Loading for Determination of Reactions on Longitudinal Beams

Reaction = (700 X4.9)/7.5= 457 kN Udl = 457/3.6 = 126.944 kN/m

DEAD LOAD

Weight of deck slab and wearing coat = 25x0.270x7.5 =50.63 kN/rn Weight of longitudinal girders (2 nos.) = 2x25x0.6x0.8 = 24 kN/m Weight of cross girder = (7x0.45x0.8x7.5x25)/30 = 15.75 kN/m Total dead load = 90.38 kN/m Dead load on each girder = 45.19 kN/m

82

Fig.5.3 Distribution of forces in the suspension bridge

Taking moment @ A,

Rb x30-45.19x3Ox3O/2+126.944x3ox3o/2 = 0

Rb = (20335.5+57124.8)/30

= 2582.01 kN

Ra = 45.19x30 + 126.944x30-2582.01

= 2582.01 kN

EMc=0

Moment due to given loading + moment due to We =0 RA x15-45.19x15x15/2-126.944x15x15/2-Wex 15x15/2=0

2582.01-338.925-952.08 = Wex 7.5

We = 1291.005/7.5

= 172.134 kN/m

Moment in girder Taking moment @1 M1 =2582.01x5-45.19x5x5/2— 126.944x5x5/2-172.134x5x5/2

= 12910.5-564.88-1586.8-2151.675 = 8607.15 kNm

VA = Wel/2 =(172.134x30)/2 = 2582.01 kN IMc=0 H = We12/8h

_ (172.134x302)/8x3 = 6455.025 kN

Tension in the cable Tmax = 'I (VZA+Hz)

_ I(2582.012+6455.0252)

= 6952.27 kN Assuming that the anchored cable is connected through with a inclination of 30 degree Therefore inclination of the main cable e = cos '(H/Tmax)

= cos(6455.025/6952.27) = 21.80 Vertical force on tower = Tmax(sin30+sin21.80)

= 6952.27x0.871 = 6055.43 kN

Horizontal force on tower = Tmax(cos2l.80-cos3O) = 6952.27x0.062 = 434.24 kN

Moment in tower = 434.24x9 = 3908.16 kNm

5.5 Comparison between cable stayed and suspension bridges

ITEM CABLE STAYED BRIDGE SUSPENSION BRIDGE

MAXIMUM CABLE

FORCE

4626.48 kN 6952.27 kN

COMPRESSIVE FORCE IN

TOWER

3774.37 kN 6055.43 kN

MOMENT IN GIRDER 231.65 kNm 8607.15 kNm

85

CONCLUSION

The evolution of cable-stayed bridges is proceeding rapidly in many different

directions: the development of slender and flexible decks which open many possibilities

for medium spans, including competition with other bridge types; the application of

cable-stayed bridges to multiple spans, which will certainly have some importance for

large projects; and, of course, the rapid increase in span length, in competition with

suspension bridges. A considerable amount of analysis work has been done in the analyze

of cable-stayed bridges. From an in-depth of the review of the analytical works, it has

been seen that lot of analysis of cable-stayed bridge has been worked out in stiffness

method, flexibility method, finite element method and in various software programs

which are very difficult to understand as well as complicated for a new beginner in the

design field of cable stayed bridge. Besides these, it has been also noticed that most of the

analysis, design and construction techniques of the cable stayed bridges are developed by

German. These engineers are very familiar with this type of bridge. In India, the

technique of cable stayed bridges is still very new and we are still depending on the

German designers for the design of cable stayed bridges. Beside these, the Indian Authors

are also silent about the analyses and design techniques of cable stayed bridges in their

books. In the current study, a cable stayed bridge model with two vertical planes of

cables and towers with equal span of 15m on either side are studied. To understand the

analytical behavior of the cable stayed bridge in a simple and easy manner, Influence line

method and Muller —Breslau's principle are adopted. Then ILD of each reaction at cable

connection, cable forces, axial force and bending moment in tower and moment in

longitudinal girders are drawn. Then, design of each components i.e. deck slab,

longitudinal girder, cross girder and towers are carried out by assuming suitable cross

section. For this purpose, some design ideas are taken from existing solved analysis

problem of AKKAR Bridge, Sikkim, India. Finally, an economic evolution has been

carried out between cable stayed bridges and prestressed bridges and with suspension

bridges also. The conclusions drawn from the present study are:

1. The initial profile of a cable stayed bridge is designed when the bridge is

loaded by self weight. All the cables are stressed at this initial profile. The bridge

displaces from this initial profile as it is subjected to other loads such as due to

temperature, traffic loads and wind. In most other structures, it is customary to

assume that the structure is unstressed and unloaded initially. "Gravity turn on", analysis is then easily carried out. However, since this cannot be done in the case

of cable stayed bridges, their analysis is significantly more complicated than that

of the other structures. The initial cable stresses are also a function of the stage-

by-stage construction segmented. Most structural analysis programs do not support stage-by-stage construction/dismantling.

2 In this present work, only a simple analysis for the purpose of preliminary design

has been attempted. The supporting cables have been assumed to inextensible,

and replaced by simple supports. This makes the structure simply supported

continuous beam. Now, by using the Influence line method, the continuous

structure is analyzed and values obtained from this method are compared with

software STAAD.Pro. Since, the values are nearly matching, so one can use this

newly adopted method to understand and feel the structural behavior of cable stayed bridge.

3 The Muller-Breslau principle was tried for obtaining the influence lines. But this

method was found to be too much lengthy and full of calculation in comparison to

the three moment equation method (Clapeyron's teorem) which was finally adopted to obtain the ILD's

4 From the adopted method, it has been seen that the moment at the longitudinal

girder can be evaluated very easily. For the design of other components such as

deck slab, cross girder one can use any by general method such as Pigeaud's method, Courbon's method etc.

5 The adopted method gives quantitative as well as qualitative values of influence line diagram.

6 In short span cable stayed bridges, live load plays the key role while in long span

dead load is expected to play the vital role in designing the cable-stayed bridges.

87

7 To understand the actual behavior of the cable stayed bridge, one should go

through the detailed drawing of an existed cable stayed bridge. It helps the

beginner and designer to get an overall concept and idea about the behavior of

such bridges. In this present work, unavailability of such drawings leads to loss of

valuable time and also, creates difficulty in understanding the structure.

8 The economic evaluation shows that the cable stayed bridge my be more economical than the other two prestressed and suspension bridges for the span of

30 m. It uses less material than the other bridge types.

9 In the present work, the work is limited to a span of 30 m. So, one can also carry

out such studies for longer spans such as 50 to 100 m, which become

uneconomical for prestressed concrete construction.

From the entire study, it has been observed that the analysis of cable stayed bridge is

not so much complicated as initially feared. Further, the advancement of software's

makes it easier and less time consuming. The critical situation arises for the cable

stayed bridge is due to temperature difference during construction, effect of creep and

shrinkage. Because, these factors change the actual profile of the cable stayed bridges

and thereby create problem in construction technology as well as creating discomfort

to the passengers. However, the study presented here would certainly useful for

design engineers basically for the beginners for understanding the behaviors and to

analyse method of cable stayed bridges in very simplified and easy manner.

REFERENCES

1. Trotsky M.S. "Cable Stayed Bridges: Theory and Design", Crossby Lockwood

Staples, 1977.

2. Niles J. Gimsing,"Cable Supported Bridges: concept and design", John Wiley &

Sons, 1983.

3. Raju N. Krishna," Design of Bridges", 31d Edition, 1998.

4. Bangash M Y H, "Prototype Bridge Structures: Analysis and Design", l

published 1999.

5. O'Conner, C, "Design Of Bridge Superstructures", 1971.

6. Victor Johnson, "Essentials of Bridge Engineering", Oxford & IBH Publishing

Co. Pvt.Ltd. Sixth Edition, 2007.

7. Podolony, W, "Construction and Design of Cable-Stayed Bridges", John Willey

& Sons, 1976.

8. Agarwal, T.P, "Cable Stayed Bridges: Parametric study", Ph. D. Thesis, 1979.

9. Bhagwat Madhav, "Investigation of Dynamic Behavior of Cable Stayed Bridge",

M.Tech. Dissertation, lIT Roorkee, 2008.

10. Meena Murarilal, "Soil Structure Interaction in A Cable Stayed Bridge", M.Tech. .

Dissertation, IIT Roorkee, 2008.

11. Bureau of Indian Standards (2000) , "IS 456 : 2000",

12. Standard Specifications And Code of Practice for Road Bridges , "IRC 6 : 2000,

Section : II, Load And Stresses",

13. Standard Specifications And Code of Practice for Road Bridges , "IRC 21 : 2000,

Section : III, Cement Concrete (Plain And Reinforced)",

14. Reddy, C S. "Basic Structural Analysis", Tata McGraw Hill, 2003.

15. Tang M, "Analysis of Cable Stayed Girder Bridges", Journal of the Structural

Engineering, ASCE, May 1971, pp. 1481-1496.

16. Tang M, "Design of Cable Stayed Girder Bridges", Journal of the Structural

Engineering, ASCE, August 1972, pp. 1789-1802.