Final Report Rob Final-libre

8/10/2019 Final Report Rob Final-libre

1/109

DESIGN OF T-BEAM RAIL-OVERBRIDGE

SUBMITTED IN PARTIAL FULFILMENTOF THE REQUIREMENTS FORBACHELOR OF TECHNOLOGY

IN

CIVIL ENGINEERING

BYVASAV DUBEY (109526) MADHURESH SHRIVASTAV(109833)

LOKESH KUMAR (109822) AMAN AGARWAL (109814)

SHOBHIT DEORI (109281) KULDEEP MEENA(109821)

PRADEEP KUMAR (109695) ANIMESH AGARWAL(109788)

MANJEET GOYAT(109597)

BATCH OF 2009-2013

UNDER THE GUIDANCE OFDR. H. K. SHARMA

NATIONAL INSTITUTE OF TECHNOLOGY

KURUKSHETRA MAY 2013


2/109


3/109

4 Design of Longitudinal Girders 60

4.1 Analysis Longitudinal Girder by Courbon's Method 60

4.2 Shear Force in L-girders 65

4.3 Design Of Section 69

5 Design Of Cross Girders 73

5.1 Analysis of Cross Girder 73

5.2 Design of Section 79

6 Design of Bearings 82

6.1 Design Of Outer Bearings 82

6.2 Design Of Inner Bearings 85 7 Conclusion 90

7.1 Deck Slab 90

7.2 Cantilever Slab 90

7.3 Longitudinal Girders 90

7.4 Cross Girders 91

7.5 Bearings 91

References 93

Appendix-A : IRC Loadings 94

Appendix-B: Impact Factors 98

Appendix-C: K in Effective Width 100

Appendix-D: Pigeaud's Curve 101


4/109

i

Acknowledgement

We wish to record our deep sense of gratitude to Dr. H.K. Sharma, Professor,Department of Civil Engineering, National Institute of Technology, Kurukshetra forhis able guidance and immense help and also the valuable technical discussionsthroughout the period which really helped us in completing this project andenriching our technical knowledge.

We also acknowledge our gratefulness to Dr. D.K. Soni, Head of Department,Department of Civil Engineering, National Institute of Technology, Kurukshetra fortimely help and untiring encouragement during the preparation of this

dissertation.


5/109

ii

List of Figures

Figure 1.1: Cutaway view of a typical concrete beam bridge.

Figure 2.1: Plan of Bridge Deck

Figure 2.2: Section X-X of Bridge Deck Plan

Figure 2.3: Section Y-Y of Bridge Deck Plan

Figure 2.4: Class AA Track located for Maximum Moment on Deck Slab

Figure 2.5: Both Class AA Track located for Maximum Moment on Deck Slab

Figure 2.6: Disposition of Class AA Wheeled Vehicle as Case 1 for MaximumMoment



Figure 2.9: Disposition of Class A Train of Load for Maximum Moment

Figure 2.10: Class AA Tracked loading arrangement for calculation of Shear Force

Figure 2.11: Class AA Wheeled loading arrangement as Case 1 for Shear Force

Figure 2.12: Disposition of Class AA wheeled vehicle as Case 2 for Shear Force

Figure 2.13: Disposition of Class A Train of load for Maximum Shear

Figure 3.1: Cantilever Slab with Class A WheelFigure 3.2: Reinforcement Details in Cross Section of Deck Slab

Figure 3.3: Reinforcement Details in Longitudinal Section of Deck Slab


6/109

iii

Figure 4.1: Class AA Tracked loading arrangement for the calculation of reactionfactors for L-girders

Figure 4.2: Influence Line Diagram for Moment at mid span

Figure 4.3: Class AA Wheeled loading arrangement for the calculation ofreaction factors for L-girders

Figure 4.4: Computation of Bending Moment for Class AA wheeled Loading

Figure 4.5: Class A loading arrangement for reaction factors for L-girder

Figure 4.6: Computation of Bending Moment for Class A Loading

Figure 4.7: Class AA tracked loading for calculation of shear force at supports

Figure 4.8: Dead Load on L-girder

Figure 4.9: Reinforcement Details of Outer Longitudinal Girder

Figure 4.10: Reinforcement Details of Inner Longitudinal Girder

Figure 5.1: Triangular load from each side of slab

Figure 5.2: Dead Load reaction on each longitudinal girder

Figure 5.3: Position of class AA tracked loading in longitudinal direction

Figure 5.4: Plan of position of class AA tracked loading in longitudinal direction

Figure 5.5: Reaction on longitudinal girder due to class AA tracked vehicle

Figure 5.6: Position of class AA wheeled loading in longitudinal direction

Figure 5.7: Plan of position of class AA wheeled loading in longitudinal direction

Figure 5.8: Reaction on longitudinal girder due to class AA wheeled loading


Figure 5.10: Reaction on longitudinal girder due to class A loading


7/109

iv

FIgure 5.11: Reinforcement Details of Cross Girder


8/109

1

1 Introduction

1.1 General

A bridge is a structure that crosses over a river, bay, or other obstruction,permitting the smooth and safe passage of vehicles, trains, and pedestrians. Anelevation view of a typical bridge is A bridge structure is divided into an upperpart (the superstructure), which consists of the slab, the floor system, and themain truss or girders, and a lower part (the substructure), which are columns,piers, towers, footings, piles, and abutments. The superstructure provideshorizontal spans such as deck and girders and carries traffic loads directly. Thesubstructure supports the horizontal spans, elevating above the ground surface.

1.2 Classification of Bridges

1.2.1 Classification by Materials

Steel Bridges steel bridge may use a wide variety of structural steel componentsand systems: girders, frames, trusses, arches, and suspension cables.

Concrete Bridges: There are two primary types of concrete bridges: reinforcedand pre-stressed.

Timber Bridges: Wooden bridges are used when the span is relatively short.

Metal Alloy Bridges : Metal alloys such as aluminum alloy and stainless steel arealso used in bridge construction.

Composite Bridges: Bridges using both steel and concrete as structural materials.

1.2.2 Classification by Objectives

Highway Bridges: Bridges on highways.


9/109

2

Railway Bridges: Bridges on railroads.

Combined Bridges: Bridges carrying vehicles and trains.

Pedestrian Bridges : Bridges carrying pedestrian traffic.

Aqueduct Bridges: Bridges supporting pipes with channeled water flow. Bridgescan alternatively be classified into movable (for ships to pass the river) or fixedand permanent or temporary categories.

1.2.3 Classification by Structural System (Superstructures)

Plate Girder Bridges: The main girders consist of a plate assemblage of upper andlower flanges and a web. H or I-cross-sections effectively resist bending andshear.

Box Girder Bridges: The single (or multiple) main girder consists of a box beamfabricated from steel plates or formed from concrete, which resists not onlybending and shear but also torsion effectively.

T-Beam Bridges: A number of reinforced concrete T-beams are placed side byside to support the live load.

Composite Girder Bridges: The concrete deck slab works in conjunction with thesteel girders to support loads as a united beam. The steel girder takes mainlytension, while the concrete slab takes the compression component of the bendingmoment.

Grillage Girder Bridges: The main girders are connected transversely by floorbeams to form a grid pattern which shares the loads with the main girders.

Truss Bridges: Truss bar members are theoretically considered to be connectedwith pins at their ends to form triangles. Each member resists an axial force,either in compression or tension.


10/109

3

Arch Bridges: The arch is a structure that resists load mainly in axial compression.In ancient times stone was the most common material used to constructmagnificent arch bridges.

Cable-Stayed Bridges : The girders are supported by highly strengthened cables(often composed of tightly bound steel strands) which stem directly from thetower. These are most suited to bridge long distances.

Suspension Bridges: The girders are suspended by hangers tied to the main cableswhich hang from the towers. The load is transmitted mainly by tension in cable

1.2.4 Classification by Design Life

Permanent Bridges

Temporary Bridges

1.2.5 Classification by Span Length

Culverts: Bridges having length less than 8 m.

Minor Bridges: Bridges having length 8-30 m.

Major bridges: Bridges having length greater than 30 m.

Long span bridges: Bridges having length greater than 120 m.

1.3 T-Beam Bridges

Beam and slab bridges are probably the most common form of concrete bridge inthe UK today, thanks to the success of standard precast prestressed concretebeams developed originally by the Prestressed Concrete Development Group(Cement & Concrete Association) supplemented later by alternative designs by


11/109

4

others, culminating in the Y-beam introduced by the Prestressed ConcreteAssociation in the late 1980s.

They have the virtue of simplicity, economy, wide availability of the standardsections, and speed of erection.

The precast beams are placed on the supporting piers or abutments, usually onrubber bearings which are maintenance free. An in-situ reinforced concrete deckslab is then cast on permanent shuttering which spans between the beams.

The precast beams can be joined together at the supports to form continuousbeams which are structurally more efficient. However, this is not normally donebecause the costs involved are not justified by the increased efficiency.

Simply supported concrete beams and slab bridges are now giving way to integralbridges which offer the advantages of less cost and lower maintenance due to theelimination of expansion joints and bearings.

1.4 Background

Nearly 590,000 roadway bridges span waterways, dry land depressions, otherroads, and railroads throughout the United States. The most dramatic bridges usecomplex systems like arches, cables, or triangle-filled trusses to carry the roadwaybetween majestic columns or towers. However, the work-horse of the highwaybridge system is the relatively simple and inexpensive concrete beam bridge.

Also known as a girder bridge, a beam bridge consists of a horizontal slabsupported at each end. Because all of the weight of the slab (and any objects onthe slab) is transferred vertically to the support columns, the columns can be lessmassive than supports for arch or suspension bridges, which transfer part of theweight horizontally.

A simple beam bridge is generally used to span a distance of 250 ft (76.2 m) or

less. Longer distances can be spanned by connecting a series of simple beambridges into what is known as a continuous span. In fact, the world's longestbridge, the Lake Pontchartrain Causeway in Louisiana, is a pair of parallel, two-lane continuous span bridges almost 24 mi (38.4 km) long. The first of the twobridges was completed in 1956 and consists of more than 2,000 individual spans.The sister bridge (now carrying the north-bound traffic) was completed 13 yearslater; although it is 228 ft longer than the first bridge, it contains only 1,500 spans.


12/109

5

A bridge has three main elements. First, the substructure (foundation) transfersthe loaded weight of the bridge to the ground; it consists of components such ascolumns (also called piers) and abutments. An abutment is the connectionbetween the end of the bridge and the earth; it provides support for the end

sections of the bridge. Second, the superstructure of the bridge is the horizontalplatform that spans the space between columns. Finally, the deck of the bridge isthe traffic-carrying surface added to the superstructure.

1.5 History

Prehistoric man began building bridges by imitating nature. Finding it useful towalk on a tree that had fallen across a stream, he started to place tree trunks orstone slabs where he wanted to cross streams. When he wanted to bridge a widerstream, he figured out how to pile stones in the water and lay beams of wood orstone between these columns and the bank.

The first bridge to be documented was described by Herodotus in 484 B.C. Itconsisted of timbers supported by stone columns, and it had been built across theEuphrates River some 300 years earlier.

Most famous for their arch bridges of stone and concrete, the Romans also built

beam bridges. In fact, the earliest known Roman bridge, constructed across theTiber River in 620 B.C. , was called the Pons Sublicius because it was made ofwooden beams (sublicae). Roman bridge building techniques included the use ofcofferdams while constructing columns. They did this by driving a circulararrangement of wooden poles into the ground around the intended columnlocation. After lining the wooden ring with clay to make it watertight, theypumped the water out of the enclosure. This allowed them to pour the concretefor the column base.

Bridge building began the transition from art to science in 1717 when Frenchengineer Hubert Gautier wrote a treatise on bridge building. In 1847, an Americannamed Squire Whipple wrote A Work on Bridge Building, which contained the firstanalytical methods for calculating the stresses and strains in a bridge. "Consultingbridge engineering" was established as a specialty within civil engineering in the1880s.


13/109

6

Further advances in beam bridge construction would come primarily fromimprovements in building materials.

1.6 Construction Materials and Their Development

Most highway beam bridges are built of concrete and steel. The Romans usedconcrete made of lime and pozzalana (a red, volcanic powder) in their bridges.This material set quickly, even under water, and it was strong and waterproof.During the Middle Ages in Europe, lime mortar was used instead, but it was watersoluble. Today's popular Portland cement, a particular mixture of limestone andclay, was invented in 1824 by an English bricklayer named Joseph Aspdin, but itwas not widely used as a foundation material until the early 1900s.

Concrete has good strength to withstand compression (pressing force), but is notas strong under tension (pulling force). There were several attempts in Europeand the United States during the nineteenth century to strengthen concrete byembedding tension-resisting iron in it. A superior version was developed in Franceduring the 1880s by Francois Hennebique, who used reinforcing bars made ofsteel. The first significant use of reinforced concrete in a bridge in the UnitedStates was in the Alvord Lake Bridge in San Francisco's Golden Gate Park;completed in 1889 and still in use today, it was built with reinforcing bars oftwisted steel devised by designer Ernest L. Ransome.

The next significant advance in concrete construction was the development ofprestressing. A concrete beam is prestressed by pulling on steel rods runningthrough the beam and then anchoring the ends of the rods to the ends of thebeam. This exerts a compressive force on the concrete, offsetting tensile forcesthat are exerted on the beam when a load is placed on it. (A weight pressingdown on a horizontal beam tends to bend the beam downward in the middle,creating compressive forces along the top of the beam and tensile forces alongthe bottom of the beam.)

Prestressing can be applied to a concrete beam that is precast at a factory,brought to the construction site, and lifted into place by a crane; or it can beapplied to cast-in-place concrete that is poured in the beam's final location.Tension can be applied to the steel wires or rods before the concrete is poured(pretensioning), or the concrete can be poured around tubes containing


14/109

7

untensioned steel to which tension is applied after the concrete has hardened(postensioning).

1.7 Design

Each bridge must be designed individually before it is built. The designer musttake into account a number of factors, including the local topography, watercurrents, river ice formation possibilities, wind patterns, earthquake potential, soilconditions, projected traffic volumes, esthetics, and cost limitations.

Figure 1.1: Cutaway view of a typical concrete beam bridge.

In addition, the bridge must be designed to be structurally sound. This involvesanalyzing the forces that will act on each component of the completed bridge.

Three types of loads contribute to these forces. Dead load refers to the weight ofthe bridge itself. Live load refers to the weight of the traffic the bridge will carry.Environmental load refers to other external forces such as wind, possibleearthquake action, and potential traffic collisions with bridge supports. Theanalysis is carried out for the static (stationary) forces of the dead load and thedynamic (moving) forces of the live and environmental loads.


15/109

8

Since the late 1960s, the value of redundancy in design has been widely accepted.This means that a bridge is designed so the failure of any one member will notcause an immediate collapse of the entire structure. This is accomplished bymaking other members strong enough to compensate for a damaged member.

1.8 Construction Procedure

Because each bridge is uniquely designed for a specific site and function, theconstruction process also varies from one bridge to another. The processdescribed below represents the major steps in constructing a fairly typical

reinforced concrete bridge spanning a shallow river, with intermediate concretecolumn supports located in the river.

Example sizes for many of the bridge components are included in the followingdescription as an aid to visualization. Some have been taken from suppliers'brochures or industry standard specifications. Others are details of a freewaybridge that was built across the Rio Grande in Albuquerque, New Mexico, in 1993.The 1,245-ft long, 10-lane wide bridge is supported by 88 columns. It contains11,456 cubic yards of concrete in the structure and an additional 8,000 cubic

yards in the pavement. It also contains 6.2 million pounds of reinforcing steel.

1.8.1 Substructure

1 A cofferdam is constructed around each column location in the riverbed,and the water is pumped from inside the enclosure. One method of setting

the foundation is to drill shafts through the riverbed, down to bedrock. Asan auger brings soil up from the shaft, a clay slurry is pumped into the holeto replace the soil and keep the shaft from collapsing. When the properdepth is reached (e.g., about 80 ft or 24.4 m), a cylindrical cage ofreinforcing steel (rebar) is lowered into the slurry-filled shaft (e.g., 72 in or2 m in diameter). Concrete is pumped to the bottom of the shaft. As the


16/109

9

shaft fills with concrete, the slurry is forced out of the top of the shaft,where it is collected and cleaned so it can be reused. The abovegroundportion of each column can either be formed and cast in place, or beprecast and lifted into place and attached to the foundation.

2 Bridge abutments are prepared on the riverbank where the bridge endwill rest. A concrete backwall is formed and poured between the top of thebank and the riverbed; this is a retaining wall for the soil beyond the end ofthe bridge. A ledge (seat) for the bridge end to rest on is formed in the topof the backwall. Wing walls may also be needed, extending outward fromthe back-wall along the riverbank to retain fill dirt for the bridgeapproaches.

1.8.2 Superstructure

4 A crane is used to set steel or prestressed concrete girders betweenconsecutive sets of columns throughout the length of the bridge. Thegirders are bolted to the column caps. For the Albuquerque freeway bridge,each girder is 6 ft (1.8 m) tall and up to 130 ft (40 m) long, weighing asmuch as 54 tons.

5 Steel panels or precast concrete slabs are laid across the girders to form asolid platform, completing the bridge superstructure. One manufactureroffers a 4.5 in (11.43 cm) deep corrugated panel of heavy (7-or 9-gauge)

steel, for example. Another alternative is a stay-in-place steel form for theconcrete deck that will be poured later.

1.8.3 Deck

6 A moisture barrier is placed atop the superstructure platform. Hot-applied polymer-modified asphalt might be used, for example.

7 A grid of reinforcing steel bars is constructed atop the moisture barrier;this grid will subsequently be encased in a concrete slab. The grid is three-dimensional, with a layer of rebar near the bottom of the slab and anothernear the top.


17/109

10

8 Concrete pavement is poured. A thickness of 8-12 in (20.32-30.5 cm) ofconcrete pavement is appropriate for a highway. If stay-in-place formswere used as the superstructure platform, concrete is poured into them. Ifforms were not used, the concrete can be applied with a slipform paving

machine that spreads, consolidates, and smooths the concrete in onecontinuous operation. In either case, a skid-resistant texture is placed onthe fresh concrete slab by manually or mechanically scoring the surfacewith a brush or rough material like burlap. Lateral joints are providedapproximately every 15 ft (5 m) to discourage cracking of the pavement;these are either added to the forms before pouring concrete or cut after aslipformed slab has hardened. A flexible sealant is used to seal the joint.

1.9 Problem Statement

A reinforced concrete bridge was to be constructed over a railway line. It wasrequired to Design the bridge superstructure and to sketch the layout of plan,elevation and reinforcement details of various components for the following data:

Width of carriage way = 7.5 m

Effective span = 14 m

Centre to centre spacing of longitudinal girders = 3.2 m

Number of longitudinal girders = 3

No. of cross girders = 4

Thickness of wearing coat = 56 mm

Material for construction = M-35 grade concrete and Fe-415 steel conforming toIS 1786.

Loading = IRC class A-A and IRC class A ,which given worst effect

Footpath = 1.7 m on left hand side of the bridge.


18/109

11

Total width of road = 10.3 m.

Design the bridge superstructure and sketch the layout of plan, elevation andreinforcement details of various components.


19/109

12

2 Deck Slab

2.1 Structural Details

Let us assume slab thickness of 225 mm.

For cantilever slab, thickness at junction = 350 mm and 100 mm at the end.

Providing vehicle crash barriers (for without footpath) on one side of carriage wayand vehicle crash barrier and pedestrian railing on the other side of thecarriageway.

2.2 Effective Span Size of Panel for Bending Moment Calculation

Let us provide longitudinal beam c/c spaced 3.2 m and with rib width 300 mm.

4 cross girders provided with c/c spaced 4.67 m and rib width 250 mm.

Effective depth of slab = 225 - 25 - 8 = 192 mm

Span in transverse direction = 3.2 m

Effective span in transverse direction = 3.2 - 0.3 + 0.192 = 3.092 m 3.1 m

Span in longitudinal direction = 4.67 - 0.25 + 0.192 = 4.6 m

Effective size of panel = 3.1 m x 4.6 m

2.3 Effective Span Size of Panel for Shear Force CalculationEffective span in transverse direction = 3.2 - 0.3 = 2.9 m

Span in longitudinal direction = 4.67 - 0.25 = 4.42 m

Effective size of panel = 2.9 m x 4.42 m


20/109

13

Figure 2.1: Plan of Bridge Deck


21/109

14

Figure 2.2: Section X-X of Bridge Deck Plan


22/109

15

Figure 2.3: Section Y-Y of Bridge Deck Plan


23/109

16

2.4 Moment due to Dead Load

Effective size of panel = 3.1m x 4.6 m

Self Wt. of deck slab = 0.225 x 24= 5.4 KN/m 2

Wt. of wearing course = 0.056 x 22 = 1.23 KN/m 2

Total = 6.63 KN/m 2

Ratio K =Short SpanLong Span =

3.14.6 = 0.674

1 K = 1.48

From Pigeaud's curve, we get by interpolation

m 1 = 4.8 x 10-2

m 2 = 1.9 x 10-2

Total dead wt. = 6.63 x 3.1 x 4.6 = 94.54 KN

Moment along short span = (0.048 + 0.15 x 0.019) x 94.54 = 4.81 KN-m

Moment along long span = (0.019 + 0.15 x 0.048) x 94.54 = 2.38 KN-m

2.5 Moment due to Live Load

2.5.1 Live load BM due to IRC Class AA Tracked VehicleSince the effective width of panel is 3.1 m, two possibilities should be considered

for finding maximum bending moment in the panel due to Class AA trackedvehicle. In the first possibility one of the track of 35t will be placed centrally(figure 2.4) on the panel. In second possibility both track of 35t each will beplaced symmetrically as shown in figure 2.5.

Case 1: Class AA Track located as in figure 2.4 for Maximum Moment


24/109

17

Figure 2.4: Class AA Track located for Maximum Moment on Deck Slab

Impact factor = 25%

u = = 0.988mv = = 3.72mK = 0.674

uB =

0.9883.1 = 0.319

vL =

3.724.6 = 0.809


m 1 = 10.5 x 10-2

m 2 = 4.1 x 10-2


25/109

18

Total load per track including impact = 1.25 x 350 = 437.5 KN

Moment along short span = (10.5 + 0.15 x 4.1) x 10 -2 x 437.5 = 48.63 KN-m

Moment along long span = (4.1 + 0.15 x 10.5) x 10 -2 x 437.5 = 24.83 KN-m

Final Moment after applying effect of continuity

MB = 48.63 x 0.8 = 38.9 KN-m

ML = 24.83 x 0.8 = 19.86 KN-m

Case 2: Both track of 35t each symmetrically as shown in figure 2.5.

Figure 2.5: Both Class AA Track located for Maximum Moment on Deck Slab

X = 0.531 m u 1 =0.988 m v = 3.72 m


26/109

19

i) u = 2( u 1 +X) = 3.038 m v = 3.72m

u B =

3.038 3.1

vL =

3.724.6 = 0.809


m 1 = 5.5 x 10-2

m 2 = 2.5 x 10-2

M1 = (5.5 + 0.15 x 2.5) x 10-2 x 1.519 = 0.0892

M2 = (2.5 + 0.15 x 5.5) x 10-2 x 1.519 = 0.051

ii) u = 2X = 1.062 v =3.72

u B =

1.0623.1 = 0.343

vL =

3.724.6 = 0.809


m 1 = 10.5 x 10-2

m 2 = 4.0 x 10-2

M1 = (10.5 + 0.15 x 4.0) x 10-2 x 0.531 = 0.0589

M2 = (4.0 + 0.15 x 10.5) x 10-2 x 0.531 = 0.0296

Final moment applying effect of continuity and impact

MB = (0.0892 - 0.0589) x 2 x 350 x 1.25 x 0.8/0.988 = 21.468 KN-m

ML = (0.051 - 0.0296) x 2 x 350 x 1.25 x 0.8/0.988 = 15.16 KN-m

2.5.2 Live Load BM due to IRC Class AA Wheeled VehicleSince the effective width is 3.1 m, all four wheels of the axle can beaccommodated on the panel for finding maximum bending moment in the paneldue to Class AA wheeled vehicle. In the first possibility four loads of 37.5 KN and


27/109

20

four loads 62.5 KN are placed symmetrical to both the axis as shown in figure 2.6.In second possibility all four loads of first axle is place symmetrically with all fourwheels of second axle following it as shown in figure 2.7. A third possibility shouldalso be tried in which four wheel loads of the first axle are so placed that themiddle 62.5KN wheel load is placed centrally, with the four wheel loads of secondaxle following it as shown in figure 2.8.

Case 1: All four loads of 37.5 KN and four loads 62.5 KN are placed symmetrical toboth the axis as shown in figure 2.6.

Impact factor = 25%

u1 =

= 0.469 m

v1 = = 0.345 m(A) For Load W 1 of Both Axles

X = 0.865 m Y= 0.428 m

i) u = 2(u 1 +X) = 2 x 1.335 = 2.67 mv = 2(v1 + Y) = 2 x 0.773 = 1.546 m

u B = 2.673.1 = 0.861 vL = 1.5464.6 = 0.336


m 1 = 8.5x 10-2

m 2= 6.0 x 10-2

M1 = (8.5 + 0.15 x 6.0) x 10-2 x 1.335 x 1.546/2 = 0 .097

M2= (6.0 + 0.15 x 8.5) x 10 -2 x 1.335 x 1.546/2 = 0.075


28/109

21

Figure 2.6: Disposition of Class AA Wheeled Vehicle as Case 1 for Maximum

Moment

ii) u = 2X = 1.73 m v = 2Y = 0.856 m

u B =

1.733.1 = 0.558

vL =

0.8564.6 = 0.186


m 1 = 12.0 x 10 -2

m 2 = 10.0 x 10-2

M1 = (12.0 + 0.15 x 10.0) x 10-2 x 0.866 x 0.428 = 0.050

M2= (10 + 0.15 x 12.0) x 10-2 x 0.866 x 0.428 = 0.049


29/109

22

iii) u = 2(u 1 +X) = 2.67 m v = 2Y = 0.856 m

u B = 0.861

vL = 0.186


m 1 = 8.5 x 10-2

m 2 = 7.5 x 10-2

M1 = (8.5 + 0.15 x 7.5) x 10-2 x 1.335 x 0.428 = 0.055

M2 = (7.5 + 0.15 x 8.5) x 10-2 x 1.335 x 0.428 = 0.050

iv) u = 2X = 1.73 m v = 2(v 1+Y) = 1.546 m

=1.733.1 = 0.558

vL = 0.336


m 1 = 11.5 x 10-2

m 2 = 7.5 x 10-2

M1 = (11.5 + 0.15 x 7.5) x 10-2 x 0.866 x 0.773 = 0.0845

M2 = (7.5 + 0.15 x 11.5) x 10-2 x 0.866 x 0.773 = 0.0618

Final M 1 = (0.097+ 0.05 - 0.055 - 0.0845) = 0.0075

Final M 2 = (0.075 + 0.044 - 0.050 - 0.0618) = 0.0072

(Mw 1)B =.0075x4x37.5

.469x.345 = 6.95 KN-m

(Mw 1)L=.0072x4x37.5

.469x.345 = 6.67 KN-m

(B) For Load W 2 of Both AxlesX = 0.266 m Y = 0.428 m


30/109


31/109

24

M1=(13.5 + 0.15 11.5) 10-2 0.428 0.735 = 0.0478

M2=(11.5 + 0.15 13.5) 10-2 0.428 0.735 = 0.0426

iv) u= 2X = 0.532 m v = 2(v 1 + Y) = 1.546 m

u B = 0.172

vL = 0.336


m 1=19.0 x 10-2

m 2=9.5 x 10-2

M1=(19 + 0.15 x 9.5) x 10 -2 x 0.266 x 0.773 = 0.42

M2=(9.5 + 0.15 x 19) x10-2 x 0.266 x 0.773 = 0.025

Final M 1 = (0.0778 + 0.025 - 0.0478 - 0.042) = 0.013

Final M 2 = (0.0561+ 0.018 - 0.0426 - 0.025) = 0.0065

(Mw2)B=.013 x 4 x 62.5 .469 x 0.345 = 20.09 KN-m

(Mw2)L =.0065 x 4 x 62.5

.469 x 0.345 = 10.04 KN-m


MB= (20.09 + 6.95) x 1.25 x 0.8 = 27.04 KN-m

ML= (10.04 + 6.64) x 1.25 x 0.8 = 16.71 KN-m

Case 2: All four loads of first axle is place symmetrically with all four wheels ofsecond axle following it as shown in figure 2.7.


32/109

25

Figure 2.7: Disposition of Class AA Wheeled Vehicle as Case 2 for Maximum

Moment

(A) For Load W 1 of Axle IX= 0.866m

i) u =2(u 1+ X) = 2 x 1.335 = 2.67 m v= 0.345 m

u

B = 0.861

v

L= 0.075


m 1=10.2 x 10-2

m 2=9.8 x 10-2


33/109

26

M1=(10.2 + 0.15 x 9.8) x 10-2 x 1.335 = 0.156

M2=(9.8 + 0.15 x 10.2) x 10-2 x 1.335 = 0.151

ii) u = 2X = 1.732 m v = 0.345 m

u B = 0.559

vL = 0.075


m 1=12.5 x 10-2

m 2=13.5 x 10-2

M1=(12.5 + 0.15 x 13.5) x 10 -2 x 0.866 = 0.126

M2=(13.5 + 0.15 x 12.5) x10-2 x 0.886 = 0.133

Final M 1 = (0.156 - 0.126) = 0.03

Final M 2= (0.151 - 0.133) = 0.018

(MB)W1 =.03 x 2 x 37.5

.469 = 4.797 KN-m

(ML)W1 =.018 x 2 x 37.5

.469 =2.88 KN-m

(B) For Load W 2 of Axle IX = 0.266 m

i) u = 2(u 1+ X) = 0.735 x 2 = 1.47 m v = 0.345m

u B = 0.474

vL = 0.075


m 1=14.0 x 10-2


34/109

27

m 2=13.0 x 10-2

M1=(14.0 + 0.15 x 13.0) x 10-2 x 0.735 = 0.117

M2=(13.0 + 0.15 x 14.0) x 10-2 x 0.735 = 0.111

ii) u = 2X = 0.532 m v = 0.345 m

u B = 0.172

vL = 0.075


m 1=23.0 x 10-2

m 2=20.0 x 10 -2

M1=(23.0 + 0.15 x 20.0) x 10-2 x 0.266 = 0.069

M2=(20.0 + 0.15 x 23.0) x 10-2 x 0.266 = 0.064

Final M 1 = (0.117 - 0.069) = 0.048

Final M 2= (0.111 - 0.069) = 0.042

(MB)W2 = .048 x 2 x 62.5 .469 = 12.79 KN-m

(ML)W2=.042 x 2 x 62.5

.469 = 12.52 KN-m

(C) For Load W 3 of Axle IIX = 0.866 m Y = 1.028 m

i) u = 2(u 1+ X) = 2 x 1.335 = 2.67 mv = 2(v1 + Y) = 2 x 1.373 = 2.746 m

u B = 0.861

vL = 0.597


35/109

28


m 1=7.5 x 10-2

m 2=4.2 x 10-2

M1 = (7.5 + 0.15 x 4.2) x 10-2 x 1.335 x 1.373 = 0.149

M2 = (4.2 + 0.15 x 7.5) x 10-2 x 1.335 x 1.373 = 0.0976

ii) u = 2X = 1.732 m v = 2Y = 1.028 x 2 = 2.056 m

u B = 0.559

vL = 0.447


m 1=11.2 x 10-2

m 2=6.5 x 10-2

M1=(11.2 + 0.15 x 6.5) x 10-2 x 0.866 x 1.028 = 0.108

M2=(6.5 + 0.15 x 11.2) x 10-2 x 0.866 x 1.028 = 0.073

iii) u = 2(u 1+ X) = 2.67 m v = 2Y = 2.050 mu B = 0.861

vL = 0.447


m 1 = 8.2 x 10-2

m 2 = 5.5 x 10-2

M1 = (8.2 + 0.15 x 5.5) x 10-2 x 1.028 x 1.335 = 0.124

M2 = (5.5 + 0.15 x 8.2) x 10-2 x 1.028 x 1.335 = 0.092

iv) u = 2X = 1.732 m v = 2(v 1 + Y) = 2.746 m


36/109

29

u B = 0.559

vL = 0.597


m 1 = 10.2 x 10 -2

m 2 = 5.2 x 10-2

M1 = (10.2 + 0.15 x 5.2) x10-2 x 0.866 x 1.373 = 0.131

M2 = (5.2 + 0.15 x 10.2) x 10-2 x 0.866 x 1.373 = 0.08

Final M 1 = (0.149 + 0.108 - 0.124 - 0.131) = 0.002

Design M 2 = (0.0976 + 0.073 - 0.092 - 0.08) = 0.0006

(MB)W3 =.002 x 2 x 37.5

.469 x .345 = 0.927 KN-m

(ML)W3 =.0006 x 2 x 37.5

.469 x .345 = 0.278 KN-m

(D) For W 4 of Axle IIX = 0.266 m Y = 1.028 m

i) u = 2(u 1+ X) = 2 x 0.735 = 1.47 mv = 2(v1 + Y) = 2 x 1.373 = 2.746 mu B = 0.474

vL = 0.597


m 1 = 11.0 x 10-2

m 2 = 5.2 x 10-2

M1 = (11.0 + 0.15 x 5.2) x 10-2 x 0.735 x 1.373 = 0.119


37/109

30

M2 = (5.2 + 0.15 x 11.0) x 10-2 x 0.735 x 1.373 = 0.069

ii) u = 2X = 0.532 m v = 2Y = 2.056 mu

B = 0.172

v

L= 0.447


m 1 = 16.0 x 10-2

m 2 = 7.5 x 10-2

M1 = (16.0 + 0.15 x 7.5) x 10-2 x 0.266 x 1.028 = 0.047

M2 = (7.5 + 0.15 x 16.0) x 10-2 x 0.266 x 1.028 = 0.027

iii) u = (u 1+ X) = 1.47 m v = 2Y = 2.056 m

u B = 0.474

vL = 0.447


m 1 = 12.2 x 10-2

m 2 = 6.8 x 10 -2

M1 = (12.2 + 0.15 x 6.8) x 10-2 x 1.028 x 0.735 = 0.0998

M2 = (6.8 + 0.15 x 12.2) x 10-2 x 1.028 x 0.735 = 0.065

iv) u = 2X = 0.532 m v = 2(v 1 + Y) = 2.746 mu B = 0.172

vL = 0.597


m 1 = 14.5 x 10-2

m 2 = 5.8 x 10-2

M1 = (14.5 + 0.15 x 5.8) x 10-2 x 0.266 x 1.373 = 0.056


38/109

31

M2 = (5.8 + 0.15 x 14.5) x 10-2 x 0.266 x 1.373 = 0.029

Final M 1 = (0.119 + 0.047 - 0.0998 - 0.056) = 0.0102

Final M 2 = (0.069 + 0.027 - 0.065 - 0.029) = 0.002

(MB)W4 =.0102 x 2 x 62.5

.469 x .345 = 7.87 KN-m

(ML)W4 =.002 x 2 x 62.5

.469 x .345 = 1.55 KN-m


MB = (4.797 + 12.79 + 0.927 + 7.87) x 1.25 x 0.8 = 26.38 KN-m

ML= (2.88 + 12.52 + 0.278 + 1.55) x 1.25 x 0.8 = 17.23 KN-m

Case3: four wheel loads of the first axle are so placed that the middle 62.5KNwheel load is placed centrally, with the four wheel loads of second axle followingit as shown in figure 2.8.

u1 = = 0.469 mv1 = = 0.345 m(A) For LoadW 1 of Axle I

X = 0.366 m

i) u = 2(u 1+ X) = 1.67 m v = v 1 = 0.345 m

u B = 0.538

vL = 0.075


m 1 = 12.8 x 10-2

m 2 = 13.8 x 10-2


39/109

32

M1 = (12.8 + 0.15 x 13.8) x 10-2 x 0.835 = 0.124 KN-m

M2 = (13.8 + 0.15 x 12.8) x 10-2 x 0.835 = 0.131 KN-m


ii) u = 2X = 0.732 m v = v 1= 0.345 m

u B = 0.236

vL = 0.075


m 1 = 19 x 10-2


40/109

33

m 2 = 17.5 x 10-2

M1 = (19.0 + 0.15 x 17.5) x 10-2 x 0.366 = 0.07915 KN-m

M2 = (17.5 + 0.15 x 19.0) x 10-2 x 0.366 = 0.0744 KN-m

Final M 1 = (0.124 - 0.07915) = 0.044

Final M 2 = (0.131 - 0.0744) = 0.0556

(MB) w1 =.044 x 37.5

.469 = 3.518 KN-m

(ML) w2 =.0556 x 37.5

.469 =4.53 KN-m

(B) For Load W 2 of Axle Iu B = 0.151m

vL = 0.075m


m 1 = 24.0 x 10-2

m2

= 22.0 x 10 -2

(MB)W2 = (24.0 + 0.15 x22.0) x 10-2 x 62.5 x 1.25 x 0.8 = 17.06 KN-m

(ML)W2 = (22.0 + 0.15 x 24.0) x 10-2 x 62.5 x 1.25 x 0.8 = 16.00 KN-m

(C) For Load W 3 of Axle IX = 0.776 m

i) u = 2(u 1+ X) = 2.469 m v = v 1= 0.345 m

u B = 0.794

vL = 0.075


m 1 = 9.8 x 10-2


41/109

34

m 2 = 10.8 x 10-2

M1 = (9.8 + 0.15 x 10.8) x 10-2 x 1.2345 = 0.140

M2 = (10.8 + 0.15 x 9.8) x 10-2 x 1.2345 = 0.151

ii) u = 2X = 1.594 m v = v 1 = 0.345 m

u B = 0.492

vL = 0.075


m 1 = 13.5 x 10-2

m 2 = 14.0 x 10 -2

M1 = (13.5 + 0.15 x 14) x 10-2 x 0.766 = 0.119

M2 = (14.0 + 0.15 x 13.5) x 10-2 x 0.766 = 0.122

Final M 1 = (0.140 - 0.119) = 0.021

Final M 2 = (0.151 - 0.122) = 0.029

(MB)W3 = .021 x 62.5 .469 = 2.79 KN-m

(ML)W3 =.029 x 62.5

.469 = 3.864 KN-m

(D) For Load W 4 of Axle IX = 1.366 m

i) u = 2(u 1 +X) = 3.670 m v = v 1 = 0.345 m

u B = 1.183 1

vL = 0.345


m 1 = 8.0 x 10-2


42/109

35

m 2 = 9.0 x 10-2

M1 = (8.0 + 0.15 x 9.0) x 10-2 x 1.835 = 0.171 KN-m

M2 = (9.0 + 0.15 x 8.0) x 10-2 x 1.835 = 0.181 KN-m

ii) u = 2X = 2.732 m v = v 1 = 0.345 m

u B = 0.88

vL = 0.075


m 1 = 9.0 x 10-2

m 2 = 9.8 x 10 -2

M1 = (9.0 + 0.15 x 9.8) x 10-2 x 1.366 = 0.143

M2 = (9.8 + 0.15 x 9.0) x 10-2 x 1.366 = 0.152

Final M 1 = (0.171 0.143) = 0.028

Final M 2 = (0.187 0.152) = 0.035

Since right most wheels of 37.5 KN are extending over the panel so loadcontributed by these wheels will be

W =37.5 x 0.429 x 0.345

0.469 x 0.345 =33.5 KN

(MB)W4 =0.028 x 33.5

0.469 = 3.33 KN-m

(ML)W4 =0.035 x 33.5

0.469 = 4.16 KN-m

(E) For Load W 1 of Axle IIX = 0.366 m Y = 1.028 m

i) u = 2(u 1 +X) = 1.67 m v = 2(v 1 + Y) = 2.746 m


43/109

36

u B = 0.54

vL = 0.6


m 1 = 10.1 x 10 -2

m 2 = 5.2 x 10-2

M1 = (10.1 + 0.15 x 5.2) x 10-2 x 0.835 x 1.373 = 0.12

M2 = (5.2 + 0.15 x 10.1) x 10-2 x 0.835 x 1.373 = 0.076

ii) u = 2X = 0.732 m v = 2Y = 2.053 m

u B = 0.236

vL = 0.45


m 1 = 15.0 x 10-2

m 2 = 7.4 x 10-2

M1 = (15.0 + 0.15 x 7.4) x 10-2 x 0.366 x 1.020 = 0.06

M2 = (7.4 + 0.15 x 15.0) x 10-2 x 0.366 x 1.020 = 0.036

iii) u = 2(u 1 +X) = 1.670 m v = 2Y = 2.056 m

u B = 0.54

vL = 0.45


m 1 = 11.0 x 10-2

m 2 = 6.5 x 10-2

M1 = (11.0 + 0.15 x 6.5) x 10-2 x 1.028 x 0.855 = 0.103

M2 = (6.5 + 0.15 x 11.0) x 10-2 x 1.028 x 0.855 = 0.069


44/109

37

iv) u = 2X = 0.73 m v = 2(v 1 + Y) = 2.746 m

u B = 0.236

vL = 0.596


m 1 = 13.2 x 10-2

m 2 = 5.8 x 10-2

M1 = (13.2 + 0.15 x 5.8) x 10-2 x 0.366 x 1.373 = 0.070

M2 = (5.8 + 0.15 x 13.2) x 10-2 x 0.366 x 1.373 = 0.039

Final M 1 = (0.12 + 0.06 - 0.103 - 0.070) = 0.007

Final M 2 = (0.076 + 0.036 - 0.069 - 0.039) = 0.004

(MB)W1 =.007 x 37.5 .469 x .345 = 1.62 KN-m

(ML)W1 =.002 x 37.5 .469 x .345 = 0.93 KN-m

(F) For Load W 2 of Axle IIY = 1.020 m

i) u = 0.469 m v = 2(v 1 + Y) = 2.746m

u B = 0.151

vL = 0.596


m 1 = 14 x 10-2

m 2 = 5.8 x 10-2

M1 = (14 + 0.15 x 5.8) x 10-2 x 1.373 = 0.204


45/109

38

M2 = (5.8 + 0.15 x 14) x 10-2 x 1.373 = 0.108

ii) u = 0.469 m v = 2(v 1 + Y) = 2.746 m

u B = 0.151

vL = 0.45


m 1 = 16 x 10-2

m 2 = 7.5 x 10-2

M1 = (16 + 0.15 x 7.5) x 10-2 x 1.028= 0.18

M2= (7.5 + 0.15 x 16) x 10 -2 x 1.028 = 0.100

Design M 1 = (0.204 - 0.018) = 0.186

Design M 2 = (0.108 0.10) = 0.008

(MB)W2 =.186 x 62.5

0.345 = 4.34 KN-m

(ML)W2 =

.008 x 62.5

0.345 = 1.45 KN-m

(G) For Load W 3 of Axle IIX = 0.766 m Y = 1.020 m

i) u = 2(u 1 + X) = 2.47 m v = 2(v 1 + Y) = 2.746 m

u B = 0.79

vL = 0.596


m 1 = 8.0 x 10-2

m 2 = 4.4 x 10-2


46/109

39

M1 = (8.0 + 0.15 x 4.4) x 10-2 x 1.373 x 1.235 = 0.147

M2 = (4.4 + 0.15 x 8.0) x 10-2 x 1.373 x 1.235 = 0.095

ii) u = 2X = 1.532 m v = 2Y = 2.056 m

u B = 0.49

vL = 0.45


m 1 = 11.5 x 10-2

m 2 = 6.5 x 10-2

M1 = (11.5 + 0.15 x 6.5) x 10 -2 x 1.028 x 0.766 = 0.098

M2= (6.5 + 0.15 x 11.5) x 10-2 x 1.028 x 0.766 = 0.065

iii) u =2(u 1 +X) =2.47 m v = 2Y = 2.056 m

u B = 0.79

vL = 0.45


m 1 = 8.8 x 10-2

m 2 = 5.0 x 10-2

M1 = (8.8 + 0.15 x 5.0) x 10-2 x 1.028 x 1.235 = 0.0.12

M2= (5.0 + 0.15 x 8.8) x 10-2 x 1.028 x 1.235 = 0.08

iv) u = 2X = 1.532 m v = 2(v 1 + Y) = 2.746 m

u B = 0.49

vL = 0.596


m 1 = 11.0 x 10-2


47/109

40

m 2 = 5.2 x 10-2

M1 = (11.0 + 0.15 x 5.2) x 10-2 x 1.373 x 0.766 = 0.123

M2= (5.2 + 0.15 x 11.0) x 10-2 x 1.373 x 0.766 = 0.070

Final M 1= (0.147 + 0.098 - 0.121 - 0.123) = 0.001

Final M 2 = (0.095 + 0.065 - 0.08 - 0.07) = 0.01

(MB)W3 =.001 x 62.5

0.469 x 0.345 = 0.380 KN-m

(ML)W3 =.01 x 62.5

0.469 x 0.345 = 3.86 KN-m

(H) For Load W 4 of Axle IIX = 1.366 m Y = 1.028 m

i) u = 2(u 1 + X) = 3.67 m v = 2(v 1 + Y) = 2.746 m

u B = 1.18 1

vL = 0.596


m 1 = 6.6 x 10-2

m 2 = 3.8 x 10-2

M1 = (6.6 + 0.15 x 3.8) x 10-2 x 1.366 x 1.835 = 0.181

M2 = (3.8+ 0.15 x 6.6) x 10-2 x 1.366 x 1.835 = 0.121

ii) u = 2X = 2.732 m v = 2Y = 2.056 m

u B = 0.88

vL = 0.45



48/109

41

m 1 = 8.0 x 10-2

m 2 = 5.0 x 10-2

m 1 = (8.0 + 0.15 x 5.0) x 10-2 x 1.028 x 1.366 = 0.123

m 2 = (5.0 + 0.15 x 8.0) x 10-2 x 1.028 x 1.366 = 0.098

iii) u = 2(u 1 +X) = 3.67 m v = 2Y = 2.056 m

u B = 1.18 1

vL = 0.45


m 1 = 7.3 x 10 -2

m 2 = 4.5 x 10-2

M1 = (7.3 + 0.15 x 4.5) x 10-2 x 1.028 x 1.835 = 0.15

M2 = (4.5+ 0.15 x 7.3) x 10-2 x 1.028 x 1.835 = 0.11

iv) u = 2X = 2.732 m v = 2(v 1 + Y) = 2.746 m

u B = 0.88 vL = 0.596


m 1 = 7.4 x 10-2

m 2 = 4.0 x 10-2

M1 = (7.4 + 0.15 x 4.0) x 10-2 x 1.373 x 1.366 = 0.15

M2 = (4.0 + 0.15 x 7.4) x 10-2 x 1.373 x 1.373 = 0.096

Final M 1= (0.181 + 0.123 - 0.15 - 0.15) = 0.004

Final M 2 = (0.121 + 0.087 - 0.11 - 0.096) = 0.002


49/109

42

(MB)W4 =.004 x 33.5

0.469 x 0.345 = 1.38 KN-m

(ML)W4 =.002 x 33.5

0.469 x 0.345 = 0.69 KN-m

Final Moments applying effect of continuity and impact

MB = (3.518 + 17.06 + 2.79 + 3.3 + 1.62 + 4.34 +0.386 + 1.38) x 1.25 x0.08

= 34.39 KN-m

ML = (4.53 + 16.6 + 3.864 + 4.16 + 0.93 + 1.45 + 3.86 + 0.69) x 1.25 x0.08

= 35.48 KN-m

2.5.3 Live Load BM due to IRC Class A LoadingFigure 2.9 shows the placement of loading for maximum B.M. in which wheel ofaxle 1 is placed centrally with wheel of axle 2 behind it, each of 57 KN.

u = = 0.65 mv = = 0.43 m(A) For Load W 1 of Axle I

u = 0.65 m v = 0.43 m

u B = 0.21

vL = 0.09


m 1 = 20.5 x 10-2

m 2 = 16 x 10 -2

(Mw 1)B = (20.5 + 0.15 x 16) x 10-2 x 57.0 = 13.053 KN-m

(Mw 1)L = (16 + 0.15 x 20.5) x 10-2 x 57.0 = 10.87 KN-m


50/109

43

Figure 2.9: Disposition of Class A Train of Load for Maximum Moment

(B) For Load W 2 of Axle IIY = 0.985 m

i) u = u 1 = 0.65 m v = 2(v 1 + Y) = 2.83 m

u B = 0.21

vL = 0.09


m 1 = 13.9 x 10-2

m 2 = 5.8 x 10-2

M1= (13.9 + 0.15 x 5.8) x 10-2 x 1.415 = 0.209


51/109


52/109

45

2.6 Design of Inner Panel

Depth of deck slab = d = cbc = 11.5 MPa st = 230 MPa j = 0.9 Q =1.1

d = = 187.32 mmDepth provided =225 mm

d = 225 - 25 - 8 = 192 mm

Area of steel (along short span) =38.9 x 10 6

230 x 0.9 x 192 = 1126 mm2

Provide 16 mm Dia @ 140 mm c/c (1408 mm 2)

Area of steel (along long span) =35.48 x 10 6

230 x 0.9 x 192 = 1026 mm2

Provide 16 mm Dia @ 140 mm c/c (1408 mm 2)

2.7 Shear Force In Deck Slab

2.7.1 For Class AA Tracked VehicleShear Force is calculated by effective width method for effective size of panel 2.9m X 4.42 m. For maximum Shear Force, the load will be so placed that its spreadup to bottom reaches the face of the rib as shown in figure 2.10.

Dispersion in direction of span or between longitudinal girder

= 0.85 + 2(0.056 + 0.225)

= 1.412 m


53/109

46

Dispersion along width (b e) = K.x (1-x

L ) + bw

B

L =

4.42

2.9 = 1.524

From the table for effective width method

K = 2.84

For Maximum shear, load is kept in such a manner that dispersion lies in span ordispersion length should end at edge.

Load should be kept at

1.412

2 = 0.706 m

be = 2.84 x 0.706 ( 1-0.706

2.9 ) + 3.6 + 2 x 0.056 = 5.3 m

Load per meter width =3505.3 = 66.04 KN

Figure 2.10: Class AA Tracked loading arrangement for calculation of Shear Force


54/109

47

So shear force at edge =66.04 x ( 2.9 - 0.706)

2.9 = 49.962 KN

2.7.2 For Class AA Wheeled VehicleShear Force is calculated by effective width method for effective size of panel 2.9m X 4.42 m. For maximum Shear Force, the loading is to be arranged by trial anderror, keeping in mind the following two points.

i) Wheel 1 is 1.2 m from kerb.ii) The outer line of third wheel from left should be as near to the face of right

hand support as possible.

So there can be two possibilities for placing the loads for Shear Forcecomputation. In first possibility, left most wheel is placed such that its spread upto bottom reaches the face of the rib as shown in figure 2.11. In secondpossibility, third wheel from left is placed as near to the face of right hand supportas possible as shown in figure 2.12.

Case 1: Left most wheel is placed such that its spread up to bottom reaches theface of the rib as shown in figure 2.11.

Dispersion in direction of span = 0.30 + 2(0.056 + 0.225) = 0.862 m

(A) For W 1 Load

Effective width = b e = K.x (1-x

L ) + bw

= 2.84 x 0.431 ( 1 -0.431

2.9 ) + 0.15 + 2 x 0.056 = 1.304 m

Average effective width =1.304+ 1.2

2 = 1.25 m


55/109

48

So, Load per meter width =37.51.25 = 29.95 KN

Figure 2.11: Class AA Wheeled loading arrangement as Case 1 for Shear Force

(B) For W 2 Load


L ) + bw

= 2.84 x 0.1.031 ( 1 -1.031

2.9 ) + 0.15 + 2 x 0.056 = 2. 149 m

Average effective width =2.149 + 1.2

2 = 1.674 m

So, Load per meter width =62.5

1.674 = 37.32 KN

(C) For W 3 Load


L ) + bw

= 2.84 x 0.869 (1 -0.869

2.9 ) + 0.15 + 2 x 0.056 = 1.99 m


2 = 1.595 m


56/109

49

So, Load per meter width =62.5

1.595 = 39.189 KN

(D) For W 4 Load


L ) + bw

= 2.84 x 0.269 ( 1 -0.269

2.9 ) + 0.15 + 2 x 0.056 = 0.955 m

So, Load per meter width =37.5 x 0.550

0.862 x 0.955 = 25.056 KN

So shear force at edge due to all loadings

=29.95(2.9 - 0.431)

2.9 +37.32(2.9 - 1.031)

2.9 +39.189 x 0.869

2.9 +25.056 x 0.269

2.9

= 63.61 KN

Shear force at other edge = 29.95 + 37.32 + 39.189 + 25.056 - 63.61 = 67.905 KN

Shear force with impact = 67.905 x 1.25 = 84.88 KN

Case 2: Third wheel from left is placed as near to the face of right hand support aspossible as shown in figure 2.12.

Figure 2.12: Disposition of Class AA wheeled vehicle as Case 2 for MaximumShear Force


57/109


58/109

51

Shear Force due to all loading at edge

=22.46(2.9 - 1.09)

2.9 +35.78 x 1.281

2.9 +49.92 x 0.431

2.9

= 14.56 + 15.804 + 7.419 = 37.78 KN

S.F. at other edge = 49.92 + 35.78 + 22.46 37.78 = 70. 38 KN

S.F. with impact = 70.38 x 1.25 = 87.915 KN

So, Shear Force due to Class AA wheeled vehicle = 87.975 KN

2.7.3 For Class A LoadingShear Force will be maximum when dispersed edge of the load touches the faceof the support as shown in Figure 2.13.

Dispersion along span = 0.50 + 2 ( 0.056 + 0.225 ) = 1.062 m

Figure 2.13: Disposition of Class A Train of load for Maximum Shear

(A) For Load W 1


L ) + bw

= 2.84 x 0.531 ( 1 -0.531

2.9 ) + 0.25 + 2 x 0.056 = 1.5939 m


59/109

52


2 = 1.3969 m

So, Load per meter width =57

1.3969 = 40.80 KN

(B) For Load W 2


L ) +bw

= 2.84 x0.569 ( 1 -0.569

2.9 ) + 0.25 + 2 x 0.056 = 1.66 m


2 = 1.43 m

So, Load per meter width =57

1.43 = 39.84 KN

So Shear Force at edge due to combined loading

=39.84 x 0.5969

2.9 +40.80 x 2.369

2.9 = 7.74 + 3.32 = 41 .069 KN

Shear force with impact factor = 41.061 x 1.22 = 50.10418 KN

2.7.4 For Dead LoadDead load of panel = 6.63 KN/ m 2

So dead load shear force =6.63 x 2.9

2 = 9.6135 KN

2.7.5 Summary

Live Load Shear Force = 87.975 KN (Class AA Wheeled )

Dead Load Shear Force = 9.6135 KN

Design Shear Force = 87.975 + 9.6135 = 97.5885 KN


60/109


61/109

54

3 Cantilever Slab

3.1 Moment due to Dead Load

The total maximum moment due to the dead load per meter width of cantileverslab is computed as following table 3.1.

Figure 3.1: Cantilever Slab with Class A Wheel

Table 3.1: Computation of Dead Load Bending Moment due to Cantilever Slab

S.No.

Components Dead Load (KN/unitm run)

LeverArm (m)

BendingMoment (KN-m)

1 Vehicle Crash Barrier 0.275x 24= 6.6 1.65 10.892 Slab(rectangular) 1.8x0.1x24=4.32 0.9 3.893 Slab (triangular) .25x1.8x.5x24 = 5.4 0.6 3.24

4 Wearing Coat .056x22x1.3=1.6 0.65 1.04Total 17.92 19.06


62/109

55

3.2 Moment due to Live Load

Effective width of dispersion b e is computed by equation

be = 1.2 x +bw

x = 0.9 m b w = 0.25 + 2 x 0.056 = 0.362 m

be = 1.2 x 0.9 + 0.362 =1.442 m

Impact factor =4.5

6+ Leff

Leff = 14+ 0.192 =14.192m

I.F. =4.5

6+ 14.192 = 2.2

Live Load per meter width including impact =57x1.22

1.442 = 48.23 KN

Maximum Moment = 48.23 x 0.9=43.41 KN-m

Shear Force = 48.23 KN

Design Moment (Dead Load B.M. + Live Load B.M.) = 19.06 +43.41= 62.47 KN-m

Design Shear Force =17.92 + 48.23 =66.15 KN

3.3 Design of Cantilever Slab

Using M -35 concrete

m = 8.11 cbc =11.5 MPa st = 230 MPa

kc = 0.289 j c = 0.904 R = 1.502

d = = 203.94 mm


63/109

56

Effective depth provided = 350 - 40 - 8 = 302 mm

Area of main reinforcement required

Ast =62.47x10 6

230 x 0.9 x 302 = 1149.19 mm2

Spacing of 16 mm bars =100x201.1

1149.19 = 174.99 mm

Provide 16mm Dia bars @ 150 c/c, Area of steel provided = 1340.67 mm 2

Distribution steel provided for

B.M = 0.3 LL BM + 0.2 DL BM

= 0.3 x43.41 + 0.2 x19.06 = 16.84 KN-m

Ast =16.84x10 6

230 x 0.9 x 302 = 309.79 mm2

Spacing of 8 mm Dia bar =100x50.3 309.79 = 162.37mm

Provide 8 mm Dia bars @ 150 c/c, Area of steel provided = 335.33 mm 2

Shear stress ( v) =66.15x10 3

1000 x 302 = 0.22 MPa

P=100 A st

bd =100x 1340.67

100 x 302 = 0.44

c = K1. K2. ca

ca = 0.23+ 0.31- 0.230.25 x 0.19 = 0.291 MPa

d = 0.302 m

K1 = 1.14- (0.7x.302) = 0.929


64/109

57

K2 = 0.5 + 0.25 p= 0.5+ .25x .44 = .61>1

c = 0.929 x1 .291 = 0.27 MPa

c.> v safe


65/109

58


66/109

59


67/109

60

4 Design of Longitudinal Girders

4.1 Analysis Longitudinal Girder by Courbon's MethodThe reaction factors will be maximum if eccentricity of the C.G. of loads withrespect to the axis of the bridge is maximum.

According to Courbon's method, reaction factor R i is given by

Ri=PIi

Ii ( 1 +

IiIid i

2 . e d i)

P = total live load

II = moment of inertia of longitudinal girder i

e = eccentricity of the live load

d i = distance of girder I from the axis of the bridge

Effective span = 14.00 m

Slab thickness = 225 mm

Width of rib = 300 mm

Spacing of main girder = 3200 mm

Overall depth = 1600 mm

4.1.1 Class AA Tracked Vehicle

Minimum clearance = 1.2 + 0.85 / 2 = 1.625 m ( up to centre of track)

e = 2.05 m

X2 = (3.2) 2 + (0)2 + (3.2) 2 = 2 x (3.2) 2

For outer girder, X = 3.2 m


68/109

61

Figure 4.1: Class AA Tracked loading arrangement for the calculation of reactionfactors for L-girders

For inner girder, X = 0

RA =P

n ( 1 +n e X X2 ) =

2W3 ( 1 +

3 x 2.05 x 3.2 2 x (3.2) 2 ) = 1.308 W

RB =2P3 ( 1 + 0 ) =

2P3 =

2W3 = 0.666 W

Figure 4.2: Influence Line Diagram for Moment at mid span


69/109

62

Impact factor for class AA loading = 10%

M = 350 (2.6 + 3.5

2 ) = 1067 KN-m

B.M for outer girder = ( 1.1 x 1.308 )x 1067 = 1535 KN- m

B.M for inner girder = ( 1.1 x 0.666 ) x 1067 = 781.68 KN- m

4.1.2 Class AA Wheeled Vehicle

Figure 4.3: Class AA Wheeled loading arrangement for the calculation ofreaction factors for L-girders

Min clearance = 1.2 + 300 / 2 = 1.2 + 1.5 = 1.35 m

e = 2.250 m

X2 = 2 (3.2) 2

RA =P

n (1 +neX

X2 ) =4W3 ( 1 +

3 x 32.25 x 3.22 (3.2) 2 ) = 2.74 W

RB =4W3 = 1.33 W

Impact factor = 25%

C.G of wheel will be 500 mm from first axle


70/109

63

RA x 14 = 62.5 (6.25 + 7.25)

RA = 60.27 KN

Figure 4.4: Computation of Bending Moment for Class AA wheeled Loading

M = 60.27 x 6.75 = 406.81 KN-m

B.M for outer girder = 406.81 x 1.25 x 2.74 = 1393.32 KN-m

B.M for inner girder = 406.81 x 1.25 x 1.33 = 676.3KN-m

4.1.3 Class A Loading

Here , P= 4 W n= 3 e = 1.650 m

Ra =4W3 ( 1 +

3I2(I x 3.2 2) x 3.2 x 1.650) = 2.36 W

Rb =4W3 ( 1+ 0 ) = 1.33 W

Rc = 4w ( 2.36 + 1.33 ) W

RA + RB = 340.54 KN

In the longitudinal direction the first six loads of class A train can beaccommodated on the span. The centre of gravity of this load system will befound to be located at a distance of 6.42 m from the first wheel.


71/109

64

Figure 4.5: Class A loading arrangement for calculation of reaction factors forL-girder

The loads are arranged on the span such that the max. Moment will occur underthe fourth load from the left. The loads shown in figure are corresponding Class Atrain load multiplied by 1.33 (reaction factor at intermediate beam ) and furthermultiplied by impact factor of 1.225. For example:- the first load of 22 KN, if theproduct of first train load of 13.5 KN and the factor 1.33 and 1.225.

RA + RB = ( 22 x 2 ) + (92.87 9 x 2 ) + ( 55.4 x 2 ) = 340.54 KN

Figure 4.6: Computation of Bending Moment for Class A LoadingTaking moment about A


72/109

65

RB x 14 = ( 22 x 1.040 ) + ( 22x 2.140 ) + ( 92.87 x 5.34 ) + ( 92.37 x 6.54) + ( 55.4 x10.84 ) + (55.4 x 13.84) = 181.47 KN

Max. B.M will occur under 4 th load from left

Max .BM = R B x 7.46 55.4 ( 7.3 + 4.3 ) = ( 181.47 x 7.46 ) 55.4 ( 7.3 + 4.3 )

= 711.13 KN m ~ 712 KN- m

B.M due to D.L. = 915 KN-m

Rea tio fa tor for e d ea a ordi g to Cour o s ethod = .

Hence maximum B.M. due to L.L. =712 x 2.36

1.33 = 1263.4 KN -m

4.2 Shear Force in L-girders

4.2.1 Shear Force due to Live LoadShear Force will be maximum due to class AA tracked vehicle. For maximum shearforce at the ends of the girder, the load will be placed between the support andthe first intermediate.

C.G of load from kerb = 1.2 + 0.4258 = 1.625 m

P1 =3.075P

3.2 +1.025P

3.2 = 1.28 P

P2 =0.125P

3.2 +2.175P

3.2 = 0719 P

The reaction at the end of each longitudinal girder due to transfer of these loadsat 1.8 m from left support

RA =2.8674.667 (1.28P) =0.786 P


73/109

66

RD =1.8

4.667 (1.28/P) =0.494 P

RB =2.8674.667 (0.719P) = 0.442 P

RE =1.8

4.667 (0.719P) = 0.227 P

The load R D, RE a d R F are tra sferred at the ross girde r should be distributeda ordi g to Cour o s theor

W = 0.494 p + +0.227 P =0.721P

If X is the e tre e dista e of C.G fro D, e ha e

X = / . P . P . = .

E = 3.2 8.2 = 2.38

Figure 4.7: Class AA tracked loading for calculation of shear force at supports


74/109

67

These reactions R D and R E act as point loads on the outer and inner longitudinalgirder and their quarter points of total span. Hence reaction at A and B due tothese will be

RA = 3/ 4 R D = 0.301 P

RE = 3/ 4R E = 0.180 P

Hence shear at A = R A +RA = ( 0.381 + 0.786 )P =1.167 P

Shear at B = R B +RB =90.180 + 0.442)P = 0.622 P

Taking into account of impact maximum shear force at support of outer girder =1.1 x 1.167 x 350 = 449.3 KN

Maximum shear force at support of inner girder = 1.1x 0.622 x 350 = 239.47KN

4.2.2 Dead Load B.M. and Shear Force

Dead Load from slab for girder Dead Load KN/m

Vehicle crash barrier 0.275 x 24 =6.6Slab 9.72Wearing coat 1.6Total 17.92

Total load of deck = 2 x 17.92 + 0.056 x 22 x 6.7 + 0.225 x 24 x 6.7 = 80.27 Kn/m

It is assumed that dead load is shared equally by all girders.-

Dead load/girder = 80.27/3 = 26.76 KN/m

Overall depth of girder = 1600 mm

Depth of rib = 1600 225 =1375 mm


75/109

68

Width = 0.3 m

Weight of rib / m = 1 0.3 x 1.375 x 24 = 9.9 KN/m

Cross girder is assumed to have same rib depth and rib width = 0.25 m

Weight of cross girder = 0.25 x 1.375 x 24 = 8.28 KN/m

Reaction on min girder = 8.25 x 3.2 = 26.4 KN

reaction from deck slab on each girder = 26.76

Total dead load on girder = 26.76 +9.9 = 36.66 KN/m

Reaction from deck slab on each girder= 26.76 KN/m

Total dead load on girder = 26.76 + 9.9 = 36.66 KN/m

Figure 4.8: Dead Load on L-girder

RA + RB = (4 x 26.4) + (36.66 x 14) KN

RA = 309.42 KN

Mmax = ( 309.42 x 7) ( 26.4 x 2.33 +26.4 x 7 + 36.66 x7 x 7/2)

= 2165.94 1144.48 = 1021.46 KN-m

Dead load shear at support = 309.42 KN


76/109

69

4.3 Design Of Section

Girder Max. D.L. B.M. Max. L.L.B.M. Total B.M. UnitsOuter girder 1021.46 1535.2 2556.66 KN-mInner girder 1021.46 781.68 1803.14 KN-m

Max. D.L.S.F. Max. L.L.S.F. Total S.F. UnitsOuter girder 309.42 449.3 758.72 KNInner girder 309.42 239.47 548.89 KN

4.3.1 Design of Outer Longitudinal GirderAssuming depth as = 1360 mm, Since the heavy reinforcement will be provided in

four layers.

Ast =2556.6 x 10 6

200 x 0.9 x1360 = 10443.87 mm2

Provide 16 bars of 32 mm dia in four rows, A st p = = 12873.14 mm2

Shear reinforcements are designed to resist the max shear at supports.

v =v

bd =758 x 10 3

300 x 1360 = 1.86 MPa

100 A sbd =

12873.14 x 100300x 1360 = 3.16%

c = 0.62 MPa

Shear reinforcement for v - c = 1.86 0.62 = 1.24 N/mm2

Vs =1.24 x 300 x 1360

1000 = 505.92 KN

Using 10 mm Dia. 4 legged stirrup.

505.92 x 10 3 =200 x 4 x 78.5 x1360

Sv


77/109


78/109

71


79/109

72


80/109

73

5 Design Of Cross Girders

5.1 Analysis of Cross Girder

5.1.1 Dead LoadCross girders are spaced @ 4.667 m c/c

Assuming X-section of X-girder same as that of longitudinal girder(1600 mm)except the width, which is 250 mm.

Self wt. of cross girder = 24000 x 1.4 x .250

= 9600 N/m = 9.6 KN/m

Dead wt. of slab and wearing coat = (0.056 x 22) + (0.225 x 24) = 6.632 KN/m 2

Each X-girder will get the triangular load from each side of the slab as shown infigure 5.1.

Figure 5.1: Triangular load from each side of slab


81/109

74

Hence, Dead load on each X-girder from the slab

= 2(0.5 x 3.2 x 1.6) x 6.632 = 33.956 KN

Assuming this to be uniformly distributed,

Dead load per meter run of girder = 33.956/3.2 = 10.61 KN/m

Total w = 9.6 + 10.61 = 20.21 KN/m

Figure 5.2: Dead Load reaction on each longitudinal girder

Assuming the cross girder to be rigid,reaction on each longitudinal girder

= (20.21 x 10 3 x 3.2 x 2)/3 = 43114.67 N = 43.11 KN

5.1.2 Live Load

(A) Class AA Tracked LoadingFigure 5.3 and 5.4 shows the position of loading for maximum B.M. in the girderdue to Class AA tracked loading. For maximum load transferred to X-girder, theposition of load in the longitudinal direction is shown in figure 5.3.


82/109

75

Figure 5.3: Position of class AA tracked loading in longitudinal direction

Figure 5.4: Plan of position of class AA tracked loading in longitudinal direction

R = = 565009.64 N

Figure 5.5: Reaction on longitudinal girder due to class AA tracked vehicle

Assuming X-girder to be rigid,reaction on each longitudinal girder

= 565009.64/3 = 88336.55 N

Live load B.M. occurs under wheel load(figure 5.5)

M = (565009.64 x 2.175)/3 = 409631.99 N-m

Now taking into account the impact factor


83/109

76

M = 1.25 x 409631.99 = 512039.98 N-m

Dead load BM from 2.175 m support

= (43114.67 x 2.175) - (20.21 x10 3 x (2.175) 2)/2

=45971.44 N-m

Total B.M. = 45971.44 + (409631.99 x 1.25)

= 558011.42 N-m

Live load shear including I.F. = 1.25 x 565009.64/3

= 235420.68 N

Dead load shear = 43114.67 N

Total Force = 235420.68 + 43114.67

= 278535.35 N

(B) Live Load due to class AAa wheeled loading

Position of load for maximum B.M. in girder is shown in figure 5.6 and 5.7



84/109

77

Figure 5.7: Plan of position of class AA wheeled loading in longitudinal direction

R=(200x1000x4.067)/4.667=174.287 kN

Assuming X-girder to be rigid, reaction on each longitudinal girder as shown infigure 5.8

Figure 5.8: Reaction on longitudinal girder due to class AA wheeled loading

= (174.287x1000)/3 = 58095.67 N

Max. live load BM, M=58095.67 x 2.7 = 156.85 KN-m

B.M. with I.F. = 1.25 x 156.85 = 196.0625 KN-m

Dead load BM from 2.7m support=43114.67x2.7-(20.21x10 3x2.7 2)/2

=42.744 kN-m


85/109

78

Total B.M. = 196.0625 + 42.744 = 238.8065 KN-m

Live load shear including I.F. = 1.25 x 58095.67 = 72619.59 N

Dead load shear = 43114.67 N

Total load = 72619.59 + 43114.67 = 115734.28 N

(C) Live Load due to class A loadingPosition of load for maximum B.M. in girder is shown in figure 5.9.


Assuming X-girder to be rigid, reaction on each longitudinal girder

= 198687.81/3 NMaximum live load B.M. = 198687.81x2.6/3

=172.196 kN-m

Now Moment including I.F. = 1.22 x 172.196 = 210.079 KN-m

Dead Load B.M. from 2.6 m support

= 43114.67 x 2.6 - (20.21 x 10 3 x 2.62)/2

= 43.788 KN-m

Total B.M. = 210.079 + 43.788 = 253.867 KN-m


86/109

79

Figure 5.10: Reaction on longitudinal girder due to class A loading

Live load shear including I.F. = 1.21 x 198687.81/3

=80799.71 N

Dead Load shear=43114.67 N

Total Shear Force = 80799.71+43114.67

=123.914 kN

5.2 Design of Section

Total depth = 1.6 m

Effective depth = 1.540 m

Ast = (558011.42x103)/(180x.9x1540) = 2236.7 mm 2

Provide 5 bars of 25mm dia.

Area provided=2454.37mm2

Shear stress = (278535.35)/(250x.9x1540) = 0.8 N/mm 2

p=(100xAs)/bd = (100x2454.37)/(250x1540) = 0.6374

c=0.34 N/mm2


87/109

80

Net shear = V - cbd =278535.35-(.34 x 250 x 1540)

=147635.35 N

Provide 2-legged 8mm stirrups

S=(180x2x4/4x8 2x1540)/147635.35 = 188.75 mm

Thus, provide 2-L, 8mm stirrups @160mm c/c.


88/109

81


89/109


90/109

83

Wind force = 0.91 144.2 = 131.222 KN

Wind load per bearing =131.222

6 = 21.87 KN

Total longitudinal force per bearing = 1.67 + 37.936 + 21.87 = 72 KN

Rotation at bearing = 0.0025 radian


Maximum vertical load on bearing = N ma = 310 + 450 = 760 KN

Minimum vertical load on bearing = N min = 310 KN

Try plan dimensions 250 500 mm and thickness 40 mm

Loaded area A 2 = 11.6 104 mm 2

From clause 307.1 of IRC 21

Allowable contact pressure = 0.25 f c = 0.25 MPaA1 / A2 >2

Allowable contact pressure = 0.25 35 2 = 12.37 MPa

Effective area of bearing pressure =760 X 1000

12.37 = 61.438 103 mm 2

m =760 x 100011.6 x 10 4 = 6.55 MPa

Thickness of individual elastomer h i = 10 mm

Thickness of outer layer h e = 5 mm

Thickness o steel laminates h s = 3 mm

Adapt 2 internal layers and 3 laminates


91/109

84

Overall thickness of bearing = 40 mm

Total thickness of elastomer in bearing = 40 (33) = 31 mm

Side cover = 6 mm

Shear modulus assumed = 1.0 N/mm 2

Shear strain due to creep, shrinkage and temperature is assumed as 5 10 -4 andthis is distributed to 2 bearings.

Shear strain per bearing due to creep, shrinkage and temperature

=5x10 -4 x 14.192 x 10 3

2 x 31 = 0.114

Shear strain due to longitudinal force =72 x10 3

11.6x10 4 = 0.58

Shear strain due to translation = 0.114 + 0.058 = 0.694

(D) Rotation

bi max =0.56 mh i

bS2

S =(a-2c)(b-2c)

2 x(a+b-4c)h i

Here a = 500mm, b = 250mm , c = 6mm , h i = 10 mm

S =11.6 x10 4

2 x ( 238 +488) x 10 = 7.99 > 6

bi max =0.5 x 10 x 10239 x 7.99 2 = 0.0033 radian

= m/10 = 6.55/10 = 0.655

Permissible rotation = n x bi max = 0.655 x 2 x 0.0033

= 4.323 x10 -3 radian > 2.5 x10 -3 radian


92/109


93/109

86

6.2.1 Longitudinal forces

(A) Braking Effect

For two lane bridge, braking effect is computed as 20% of the first train load +plus 10 % of the loads in succeeding trains.

20 % of first train load =20

100 (54 + 228) = 56.4 KN

10 % of the loads in succeeding trains. = 136 x10

100 = 13.6 KN

Total = 70 KN

Longitudinal force /bearing = 70/6 = 11.67 KN

(B) Friction Force( D.L + LL reaction at bearing ) coeff. Of friction = (309.42 + 240) 0.3

= 164.83 KN

Friction Per bearing =164.83

6 = 27.47 KN

(C) Wind LoadAssuming 10 m height

Wind pressure = 0.91 KN/m 2

Plan area of bridge span = 14 10.3 = 144.2 m 2

Wind force = 0.91 144.2 = 131.222 KN

Wind load per bearing =131.222

6 = 21.87 KN

Total longitudinal force per bearing = 11.67 + 27.47 + 21.87 = 61 KN

Rotation at bearing = 0.0025 radian


94/109

87


Maximum vertical load on bearing = N ma = 310 +240 = 550 KN

Minimum vertical load on bearing = N min = 240 KN

Try plan dimensions 320 500 mm and thickness 45 mm

Loaded area A 2 = 15 104 mm 2

From clause 307.1 of IRC: 21

Allowable contact pressure = 0.25 f c = 0.25 MPa

A1 / A2 >2Allowable contact pressure = 0.25 35 2 = 12.37 MPa

Effective area of bearing pressure =550 x 1000

12.37 = 44.462 103 mm 2

m =500 x10 3

15 x10 4 = 3.67 MPa

Thickness of individual elastomer h i = 10 mm

Thickness of outer layer h e = 5 mm

Thickness o steel laminates h s = 3 mm

Adapt 2 internal layers and 3 laminates

Overall thickness of bearing = 39 mm

Total thickness of elastomer in bearing = 39 (33) = 30 mmSide cover = 6 mm

Shear modulus assumed = 1.0 N/mm 2


95/109

88

Shear strain due to creep, shrinkage and temperature is assumed as 5 10 -4 andthis is distributed to 2 bearings.

Shear strain per bearing due to creep, shrinkage and temperature

=5 x10 -414.192 x10 3

2x30 = 0.12

Shear strain due to longitudinal force =61 x10 3

15 x10 4= 0.407

Shear strain due to translation = 0.12 + 0.407 = 0.527 < 0.7

(D) Rotation

bi max = 0.56mh i

bS2

S =(a-2c)(b-2c)

2 x(a+b-4c)h i

Here a = 500mm, b = 320mm , c = 6 mm , h i = 10 mm

Therefore, S =488 x 308

2 x 10 x (796) = 12> 9.44 > 6

Assuming m,max = 10MPa

bi max =0.5 x 10x 10308 x 9.44 2 = 0.0031 radian

= m/10 = 3.67/10 = 0.367

Permissible rotation = n x bi max = 0.367 x 2 x 0.0031

= 2.27 x10 -3 > 2.5 x10 -3 actual shear strain = 0.527 as calculated

0.2 + 0.1 m= 0.2 + 0.1 x 3.67 = 0.567 > 0.527

Also m satisfies 10 MPa > m > 2 MPa


96/109

89

6.2.2 Shear StressShear stress due to compression = 1.5 m / S = 1.5 x 3.67/ 9.44 = 0.58 MPa

Shear stress due to horizontal deformation = 0.527 x 1 =0.527 MPa

Shear stress due to rotation = 0.5 ( b/ h i )2 bi = 0.5 (

30810 )

2 X 0.0025 = 1.18 MPa

Total shear stress = 0.58 + 0.527 + 1.18 = 2.287 MPa < 5 MPa

The elastomeric pad bearing has the characteristics;

Plan dimensions = 320mm x 500 mm

Overall thickness = 39 mm

Thickness of individual layer = 10 mm

Number of internal elastomer layers = 2

Number of laminates = 3

Thickness of each laminate = 3 mm

Thickness or top or bottom cover = 5 mm


97/109

90

7 Conclusion

7.1 Deck Slab

Overall Depth = 225 mm

Reinforcement 16 mm Dia @140 mm c/c (1408 mm 2) along shorter span.

Reinforcement 16 mm Dia @140 mm c/c (1408 mm 2) along longer span.

7.2 Cantilever Slab

Depth at support = 350 mm

Depth at cantilever side = 100 mm

Main Steel Provide 16mm Dia bars @150 c/c (A st = 1340.67 mm2)

Distribution Steel Provide 8 mm Dia bars @150 c/c (A st = 335.33 mm2)

7.3 Longitudinal Girders



Overall Depth = 1600 mm

Outer Longitudinal Girder

Main reinforcement of 16 bars of 32 mm Dia in 4 rows(A st = 12873.14 mm2)

Shear reinforcement of 10 mm Dia 4 legged stirrups @150 c/c

Inner Longitudinal Girder


98/109

91

Main reinforcement of 12 bars of 32 mm Dia in 3 rows(A st = 12873.14 mm2)


7.4 Cross Girders



Overall depth = 1600 mm

Main reinforcement of 5 bars of 25 mm Dia (A st = 12873.14 mm2)


7.5 Bearings

7.5.1 Outer Bearings

Plan dimensions = 250 mm x 500 mm







7.5.2 Inner Bearings

Plan dimensions = 320 mm x 500 mm


99/109

92








100/109

93

References

1. IRC : 5 - 1998, "Standard Specifications and Code of Practice for Road Bridges,Section I General Features of Design", The Indian Road Congress.

2. IRC : 6 - 2000, "Standard Specifications and Code of Practice for Road Bridges,Section II - Loads and Stresses", The Indian Road Congress.

3. IS : 456 - 2000, "Plain and Reinforcement Concrete - Code of Practice", Bureauof Indian Standards, New Delhi, 2000.

4. Krishna Raju, N., "Design of Bridges".

5. Victor, D.J., "Essential of Bridge Engineering".

6. Punmia, B.C. and Jain, A.K., "R.C.C. Designs".


101/109

94

Appendix-A : IRC Loadings

IRC Class AA Loading

General:1. The nose to tail spacing between two successive vehicles shall not less than

90m.2. For multi- lane bridges and culverts, one train of class AA tracked or

wheeled vehicles whichever creates severer conditions shall be consideredfor every two traffic lane width.

3. No other live load shall be considered on any part of the two-lane width

carriageway of the bridge when the above mentioned train of vehicle iscrossing the bridge.

4. The maximum loads for the wheeled vehicles shall be 20 tonnes for a singleaxle or 40 tonnes for a bogie of two axles spaced not more than 1.2mcenters.

5. The maximum clearance between the road face of the kerb and the outeredge of the wheel or tack , C , shall be as under :

(a) Single lane Bridges

Carriage way width Minimum value of C

3.8 m and above 0.3 m

(b) Multi lane Bridges

Less than 5.5 m 0.6 m

5.5 m or above 1.2 m


102/109

95

Figure A.1: IRC Class A Tracked and Wheeled Vehicle


103/109

96

IRC Class A Loading

General:1. The nose to tail distance between successive trains shall not be less than

18.4 m.2. No other live load shall cover any part of the carriage way when a train of

vehicles (or trains of vehicles in multi- Lane Bridge) is crossing the bridge.3. The ground contact area of the wheel shall be as under :

Axle load (tones)Ground contact area

(B) (mm) (W) (mm)

11.4 250 500

6.8 200 380

2.7 150 200

4. The minimum clearance f , between outer edge of the wheel and the

roadway face of the kerb , and the minimum clearance g , between theouter edges of passing or crossing vehicles on multi-lane bridges shall be asgiven below (figure A.2)

Clear carriageway width g f

5.5 m to 7.5 m

Above 7.5 m

Uniformly increasing from0.4 m to 1.2 m1.2 m

150 mm for all carriagewayvehicles


104/109

97

Figure A.2: IRC Class A and B Loading Vehicles


105/109

98

Appendix-B: Impact Factors

Provision for impact or dynamic action shall be made by an increment oflive load by an impact allowance expressed as a fraction or a percentage ofapplied live load.

Class A or Class B LoadingIn the members of any bridge designed either for class A or class B loading, theimpact percentage shall be determined from the curves indicated in figure B.1

The impact factor shall be determined from the following equations which areapplicable for spans 3 m and 45 m.

Impact factor for R.C. bridges, I.F. = 4.5/ (6+L)

where L is the length of the span in meters.

Class AA LoadingThe value of the impact percentage shall be taken as follows:

For spans less than 9 mFor tracked vehicles: 25% for spans up to 5 m, linearly reducing to 10% for spans9 m.

For wheeled vehicles: 25%.

For span of 9 m or moreTracked vehicles: 10% up to a span of 40 m and in accordance with the curve for

spans in excess of 40 m.Wheeled vehicles: 25% for spans up to 12 m and in accordance with the curve forspans in excess of 23 m.


106/109


107/109

100

Appendix-C: K in Effective Width

Table C.1: Value of Constant 'K' (IRC 21: 2000)


108/109

101

Appendix-D: Pigeaud's Curve

Figure D.1: Moment Coefficient for Slabs Completely Loaded with UniformlyDistributed Load, Coefficients are m 1 for K and m 2 for 1/K

Figure D.2: Moment Coefficient m 1 and m 2 for K=0.6


109/109

Figure D.2: Moment Coefficient m 1 and m 2 for K=0.7

Final Report Rob Final-libre

Documents