Undergraduate major project_-_design_of

DESIGN OF T-BEAM RAIL-OVER BRIDGE

SUBMITTED IN PARTIAL FULFILMENT

OF THE REQUIREMENTS FOR

BACHELOR OF TECHNOLOGY

IN

CIVIL ENGINEERING

BY

VASAV 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 OF

DR. H. K. SHARMA

NATIONAL INSTITUTE OF TECHNOLOGY

KURUKSHETRA

MAY 2013

Contents

Acknowledgement i

List of Figures ii

1 Introduction 1

1.1 General 1

1.2 Classification of Bridges 1

1.3 T-Beam Bridges 3

1.4 Background 4

1.5 History 5

1.6 Construction Materials and Their Development 6

1.7 Design 7

1.8 Construction Procedure 8

1.9 Problem Statement 10

2 Deck Slab 12

2.1 Structural Details 12

2.2 Effective Span Size of Panel for Bending Moment Calculation 12

2.3 Effective Span Size of Panel for Shear Force Calculation 12

2.4 Moment due to Dead Load 16

2.5 Moment due to Live Load 16

2.6 Design of Inner Panel 45

2.7 Shear Force In Deck Slab 45

3 Cantilever Slab 54

3.1 Moment due to Dead Load 54

3.2 Moment due to Live Load 55

3.3 Design of Cantilever Slab 55

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

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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 for

his able guidance and immense help and also the valuable technical discussions

throughout the period which really helped us in completing this project and

enriching 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 for

timely help and untiring encouragement during the preparation of this

dissertation.

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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 Maximum

Moment


Moment


Moment

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 Wheel

Figure 3.2: Reinforcement Details in Cross Section of Deck Slab

Figure 3.3: Reinforcement Details in Longitudinal Section of Deck Slab

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Figure 4.1: Class AA Tracked loading arrangement for the calculation of reaction

factors 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 of

reaction 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

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FIgure 5.11: Reinforcement Details of Cross Girder

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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. An elevation view of a typical bridge is A bridge structure is divided into an upper part (the superstructure), which consists of the slab, the floor system, and the main truss or girders, and a lower part (the substructure), which are columns, piers, towers, footings, piles, and abutments. The superstructure provides horizontal spans such as deck and girders and carries traffic loads directly. The substructure 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 components and systems: girders, frames, trusses, arches, and suspension cables. Concrete Bridges: There are two primary types of concrete bridges: reinforced and 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 are also 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.

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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. Bridges can alternatively be classified into movable (for ships to pass the river) or fixed and 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 and lower flanges and a web. H or I-cross-sections effectively resist bending and shear. Box Girder Bridges: The single (or multiple) main girder consists of a box beam fabricated from steel plates or formed from concrete, which resists not only bending and shear but also torsion effectively. T-Beam Bridges: A number of reinforced concrete T-beams are placed side by side to support the live load. Composite Girder Bridges: The concrete deck slab works in conjunction with the steel girders to support loads as a united beam. The steel girder takes mainly tension, while the concrete slab takes the compression component of the bending moment. Grillage Girder Bridges: The main girders are connected transversely by floor

beams to form a grid pattern which shares the loads with the main girders.

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

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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 construct magnificent arch bridges. Cable-Stayed Bridges: The girders are supported by highly strengthened cables (often composed of tightly bound steel strands) which stem directly from the tower. These are most suited to bridge long distances. Suspension Bridges: The girders are suspended by hangers tied to the main cables which 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 in the UK today, thanks to the success of standard precast prestressed concrete beams developed originally by the Prestressed Concrete Development Group (Cement & Concrete Association) supplemented later by alternative designs by

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

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

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

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

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

1.4 Background

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

Also known as a girder bridge, a beam bridge consists of a horizontal slab supported at each end. Because all of the weight of the slab (and any objects on the slab) is transferred vertically to the support columns, the columns can be less massive than supports for arch or suspension bridges, which transfer part of the weight 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 beam bridges into what is known as a continuous span. In fact, the world's longest bridge, 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 two bridges 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 years later; although it is 228 ft longer than the first bridge, it contains only 1,500 spans.

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A bridge has three main elements. First, the substructure (foundation) transfers the loaded weight of the bridge to the ground; it consists of components such as columns (also called piers) and abutments. An abutment is the connection between 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 horizontal platform that spans the space between columns. Finally, the deck of the bridge is the traffic-carrying surface added to the superstructure.

1.5 History

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

The first bridge to be documented was described by Herodotus in 484 B.C. It consisted of timbers supported by stone columns, and it had been built across the Euphrates 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 the Tiber River in 620 B.C. , was called the Pons Sublicius because it was made of wooden beams (sublicae). Roman bridge building techniques included the use of cofferdams while constructing columns. They did this by driving a circular arrangement of wooden poles into the ground around the intended column location. After lining the wooden ring with clay to make it watertight, they pumped the water out of the enclosure. This allowed them to pour the concrete for the column base.

Bridge building began the transition from art to science in 1717 when French engineer Hubert Gautier wrote a treatise on bridge building. In 1847, an American named Squire Whipple wrote A Work on Bridge Building, which contained the first analytical methods for calculating the stresses and strains in a bridge. "Consulting bridge engineering" was established as a specialty within civil engineering in the 1880s.

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

1.6 Construction Materials and Their Development

Most highway beam bridges are built of concrete and steel. The Romans used concrete 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 water soluble. Today's popular Portland cement, a particular mixture of limestone and clay, was invented in 1824 by an English bricklayer named Joseph Aspdin, but it was not widely used as a foundation material until the early 1900s.

Concrete has good strength to withstand compression (pressing force), but is not as strong under tension (pulling force). There were several attempts in Europe and the United States during the nineteenth century to strengthen concrete by embedding tension-resisting iron in it. A superior version was developed in France during the 1880s by Francois Hennebique, who used reinforcing bars made of steel. The first significant use of reinforced concrete in a bridge in the United States 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 of twisted steel devised by designer Ernest L. Ransome.

The next significant advance in concrete construction was the development of prestressing. A concrete beam is prestressed by pulling on steel rods running through the beam and then anchoring the ends of the rods to the ends of the beam. This exerts a compressive force on the concrete, offsetting tensile forces that are exerted on the beam when a load is placed on it. (A weight pressing down 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 along the 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 be applied 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

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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 must take into account a number of factors, including the local topography, water currents, river ice formation possibilities, wind patterns, earthquake potential, soil conditions, 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 involves analyzing 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 of the 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, possible earthquake action, and potential traffic collisions with bridge supports. The analysis is carried out for the static (stationary) forces of the dead load and the dynamic (moving) forces of the live and environmental loads.

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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 not cause an immediate collapse of the entire structure. This is accomplished by making 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, the construction process also varies from one bridge to another. The process described below represents the major steps in constructing a fairly typical reinforced concrete bridge spanning a shallow river, with intermediate concrete column supports located in the river.

Example sizes for many of the bridge components are included in the following description as an aid to visualization. Some have been taken from suppliers' brochures or industry standard specifications. Others are details of a freeway bridge 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 contains 11,456 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. As an auger brings soil up from the shaft, a clay slurry is pumped into the hole to replace the soil and keep the shaft from collapsing. When the proper depth is reached (e.g., about 80 ft or 24.4 m), a cylindrical cage of reinforcing steel (rebar) is lowered into the slurry-filled shaft (e.g., 72 in or 2 m in diameter). Concrete is pumped to the bottom of the shaft. As the

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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 aboveground portion of each column can either be formed and cast in place, or be precast and lifted into place and attached to the foundation.

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

1.8.2 Superstructure

4 A crane is used to set steel or prestressed concrete girders between consecutive sets of columns throughout the length of the bridge. The girders 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 as much as 54 tons.

5 Steel panels or precast concrete slabs are laid across the girders to form a solid platform, completing the bridge superstructure. One manufacturer offers 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 the concrete 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 another near the top.

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8 Concrete pavement is poured. A thickness of 8-12 in (20.32-30.5 cm) of concrete pavement is appropriate for a highway. If stay-in-place forms were used as the superstructure platform, concrete is poured into them. If forms were not used, the concrete can be applied with a slipform paving machine that spreads, consolidates, and smooths the concrete in one continuous operation. In either case, a skid-resistant texture is placed on the fresh concrete slab by manually or mechanically scoring the surface with a brush or rough material like burlap. Lateral joints are provided approximately every 15 ft (5 m) to discourage cracking of the pavement; these are either added to the forms before pouring concrete or cut after a slipformed 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 was

required 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 to

IS 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.

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Total width of road = 10.3 m.

Design the bridge superstructure and sketch the layout of plan, elevation and

reinforcement details of various components.

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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 way

and vehicle crash barrier and pedestrian railing on the other side of the

carriageway.

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 Calculation

Effective 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

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Figure 2.1: Plan of Bridge Deck

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Figure 2.2: Section X-X of Bridge Deck Plan

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Figure 2.3: Section Y-Y of Bridge Deck Plan

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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/m2

Wt. of wearing course = 0.056 x 22 = 1.23 KN/m2

Total = 6.63 KN/m2

Ratio K = Short SpanLong Span =

3.14.6 = 0.674

1 K = 1.48

From Pigeaud's curve, we get by interpolation

m1 = 4.8 x 10-2

m2 = 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 Vehicle

Since 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 tracked

vehicle. 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 be

placed symmetrically as shown in figure 2.5.

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

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

Impact factor = 25%

u = = 0.988m

v = = 3.72m

K = 0.674

uB =

0.9883.1 = 0.319

vL =

3.724.6 = 0.809


m1 = 10.5 x 10-2

m2 = 4.1 x 10-2

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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 u1 =0.988 m v = 3.72 m

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i) u = 2( u1 +X) = 3.038 m v = 3.72m

u B =

3.038 3.1

vL =

3.724.6 = 0.809


m1 = 5.5 x 10-2

m2 = 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


m1 = 10.5 x 10-2

m2 = 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 Vehicle

Since the effective width is 3.1 m, all four wheels of the axle can be

accommodated on the panel for finding maximum bending moment in the panel

due to Class AA wheeled vehicle. In the first possibility four loads of 37.5 KN and

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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 four

wheels of second axle following it as shown in figure 2.7. A third possibility should

also be tried in which four wheel loads of the first axle are so placed that the

middle 62.5KN wheel load is placed centrally, with the four wheel loads of second

axle 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 to

both the axis as shown in figure 2.6.

Impact factor = 25%

u1 = = 0.469 m

v1 = = 0.345 m

(A) For Load W1 of Both Axles

X = 0.865 m Y= 0.428 m

i) u = 2(u1 +X) = 2 x 1.335 = 2.67 m

v = 2(v1 + Y) = 2 x 0.773 = 1.546 m

u B =

2.673.1 = 0.861

vL =

1.5464.6 = 0.336


m1 = 8.5x 10-2

m2= 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

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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


m1 = 12.0 x 10-2

m2 = 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

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iii) u = 2(u1 +X) = 2.67 m v = 2Y = 0.856 m

u B = 0.861

vL = 0.186


m1 = 8.5 x 10-2

m2 = 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(v1+Y) = 1.546 m

= 1.733.1 = 0.558

vL = 0.336


m1 = 11.5 x 10-2

m2 = 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 M1 = (0.097+ 0.05 - 0.055 - 0.0845) = 0.0075

Final M2 = (0.075 + 0.044 - 0.050 - 0.0618) = 0.0072

(Mw1)B =.0075x4x37.5

.469x.345 = 6.95 KN-m

(Mw1)L= .0072x4x37.5

.469x.345 = 6.67 KN-m

(B) For Load W2 of Both Axles

X = 0.266 m Y = 0.428 m

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i) u = 2(u1+ X) = 2 x 0.735 = 1.47 m

v = 2(v1 + Y) = 2 x 0.773 = 1.546 m

u B = 0.474

vL = 0.336


m1 = 12.5 x 10-2

m2 = 8.0 x 10-2

M1 = (12.5 + 0.15 x 8.0) x 10-2 x 0.735 x 0.773 = 0.0778

M2 = (8.0 + 0.15 x 12.5) x 10-2 x 0.735 x 0.773 = 0.0561

ii) u = 2X = 0.532 m v = 2Y = 0.856 m

u B = 0.172

vL = 0.186


m1=20.0 x 10-2

m2= 13.0 x 10-2

M1=(20 + 0.15 x13) x 10-2 x 0.266 x 0.428 = 0.025

M2 = (13 + 0.15 x 20) x 10-2 x 0.266 x 0.428 = 0.018

iii) u = 1.47m v = 2Y = 0.856m

u B = 0.474

vL = 0.186


m1=13.5 × 10-2

m2=11.5 × 10-2

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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(v1 + Y) = 1.546 m

u B = 0.172

vL = 0.336


m1=19.0 x 10-2

m2=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 M1 = (0.0778 + 0.025 - 0.0478 - 0.042) = 0.013

Final M2 = (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 of

second axle following it as shown in figure 2.7.

25


Moment

(A) For Load W1 of Axle I

X= 0.866m

i) u =2(u1+ X) = 2 x 1.335 = 2.67 m v= 0.345 m

u B = 0.861

vL = 0.075


m1=10.2 x 10-2

m2=9.8 x 10-2

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


m1=12.5 x 10-2

m2=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 M1 = (0.156 - 0.126) = 0.03

Final M2= (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 W2 of Axle I

X = 0.266 m

i) u = 2(u1+ X) = 0.735 x 2 = 1.47 m v = 0.345m

u B = 0.474

vL = 0.075


m1=14.0 x 10-2

27

m2=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


m1=23.0 x 10-2

m2=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 M1 = (0.117 - 0.069) = 0.048

Final M2= (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 W3 of Axle II

X = 0.866 m Y = 1.028 m

i) u = 2(u1+ X) = 2 x 1.335 = 2.67 m

v = 2(v1 + Y) = 2 x 1.373 = 2.746 m

u B = 0.861

vL = 0.597

28


m1=7.5 x 10-2

m2=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


m1=11.2 x 10-2

m2=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(u1+ X) = 2.67 m v = 2Y = 2.050 m

u B = 0.861

vL = 0.447


m1 = 8.2 x 10-2

m2 = 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(v1 + Y) = 2.746 m

29

u B = 0.559

vL = 0.597


m1 = 10.2 x 10-2

m2 = 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 M1 = (0.149 + 0.108 - 0.124 - 0.131) = 0.002

Design M2 = (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 W4 of Axle II

X = 0.266 m Y = 1.028 m

i) u = 2(u1+ X) = 2 x 0.735 = 1.47 m

v = 2(v1 + Y) = 2 x 1.373 = 2.746 m

u B = 0.474

vL = 0.597


m1 = 11.0 x 10-2

m2 = 5.2 x 10-2

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

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 m

u B = 0.172

vL = 0.447


m1 = 16.0 x 10-2

m2 = 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 = (u1+ X) = 1.47 m v = 2Y = 2.056 m

u B = 0.474

vL = 0.447


m1 = 12.2 x 10-2

m2 = 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(v1 + Y) = 2.746 m

u B = 0.172

vL = 0.597


m1 = 14.5 x 10-2

m2 = 5.8 x 10-2

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

31

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

Final M1 = (0.119 + 0.047 - 0.0998 - 0.056) = 0.0102

Final M2 = (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 62.5KN

wheel load is placed centrally, with the four wheel loads of second axle following

it as shown in figure 2.8.

u1 = = 0.469 m

v1 = = 0.345 m

(A) For LoadW1 of Axle I

X = 0.366 m

i) u = 2(u1+ X) = 1.67 m v = v1 = 0.345 m

u B = 0.538

vL = 0.075


m1 = 12.8 x 10-2

m2 = 13.8 x 10-2

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


Moment

ii) u = 2X = 0.732 m v = v1= 0.345 m

u B = 0.236

vL = 0.075


m1 = 19 x 10-2

33

m2 = 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 M1 = (0.124 - 0.07915) = 0.044

Final M2 = (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 W2 of Axle I

u B = 0.151m

vL = 0.075m


m1 = 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 W3 of Axle I

X = 0.776 m

i) u = 2(u1+ X) = 2.469 m v = v1= 0.345 m

u B = 0.794

vL = 0.075


m1 = 9.8 x 10-2

34

m2 = 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 = v1 = 0.345 m

u B = 0.492

vL = 0.075


m1 = 13.5 x 10-2

m2 = 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 M1 = (0.140 - 0.119) = 0.021

Final M2 = (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 W4 of Axle I

X = 1.366 m

i) u = 2(u1 +X) = 3.670 m v = v1 = 0.345 m

u B = 1.183 1

vL = 0.345


m1 = 8.0 x 10-2

35

m2 = 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 = v1 = 0.345 m

u B = 0.88

vL = 0.075


m1 = 9.0 x 10-2

m2 = 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 M1 = (0.171 – 0.143) = 0.028

Final M2 = (0.187 – 0.152) = 0.035

Since right most wheels of 37.5 KN are extending over the panel so load

contributed 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 W1 of Axle II

X = 0.366 m Y = 1.028 m

i) u = 2(u1 +X) = 1.67 m v = 2(v1 + Y) = 2.746 m

36

u B = 0.54

vL = 0.6


m1 = 10.1 x 10-2

m2 = 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


m1 = 15.0 x 10-2

m2 = 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(u1 +X) = 1.670 m v = 2Y = 2.056 m

u B = 0.54

vL = 0.45


m1 = 11.0 x 10-2

m2 = 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

37

iv) u = 2X = 0.73 m v = 2(v1 + Y) = 2.746 m

u B = 0.236

vL = 0.596


m1 = 13.2 x 10-2

m2 = 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 M1 = (0.12 + 0.06 - 0.103 - 0.070) = 0.007

Final M2 = (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 W2 of Axle II

Y = 1.020 m

i) u = 0.469 m v = 2(v1 + Y) = 2.746m

u B = 0.151

vL = 0.596


m1 = 14 x 10-2

m2 = 5.8 x 10-2

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

38

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

ii) u = 0.469 m v = 2(v1 + Y) = 2.746 m

u B = 0.151

vL = 0.45


m1 = 16 x 10-2

m2 = 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 M1 = (0.204 - 0.018) = 0.186

Design M2 = (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 W3 of Axle II

X = 0.766 m Y = 1.020 m

i) u = 2(u1 + X) = 2.47 m v = 2(v1 + Y) = 2.746 m

u B = 0.79

vL = 0.596


m1 = 8.0 x 10-2

m2 = 4.4 x 10-2

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


m1 = 11.5 x 10-2

m2 = 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(u1 +X) =2.47 m v = 2Y = 2.056 m

u B = 0.79

vL = 0.45


m1 = 8.8 x 10-2

m2 = 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(v1 + Y) = 2.746 m

u B = 0.49

vL = 0.596


m1 = 11.0 x 10-2

40

m2 = 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 M1= (0.147 + 0.098 - 0.121 - 0.123) = 0.001

Final M2 = (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 W4 of Axle II

X = 1.366 m Y = 1.028 m

i) u = 2(u1 + X) = 3.67 m v = 2(v1 + Y) = 2.746 m

u B = 1.18 1

vL = 0.596


m1 = 6.6 x 10-2

m2 = 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


41

m1 = 8.0 x 10-2

m2 = 5.0 x 10-2

m1 = (8.0 + 0.15 x 5.0) x 10-2 x 1.028 x 1.366 = 0.123

m2 = (5.0 + 0.15 x 8.0) x 10-2 x 1.028 x 1.366 = 0.098

iii) u = 2(u1 +X) = 3.67 m v = 2Y = 2.056 m

u B = 1.18 1

vL = 0.45


m1 = 7.3 x 10-2

m2 = 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(v1 + Y) = 2.746 m

u B = 0.88

vL = 0.596


m1 = 7.4 x 10-2

m2 = 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 M1= (0.181 + 0.123 - 0.15 - 0.15) = 0.004

Final M2 = (0.121 + 0.087 - 0.11 - 0.096) = 0.002

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 Loading

Figure 2.9 shows the placement of loading for maximum B.M. in which wheel of

axle 1 is placed centrally with wheel of axle 2 behind it, each of 57 KN.

u = = 0.65 m

v = = 0.43 m

(A) For Load W1 of Axle I

u = 0.65 m v = 0.43 m

u B = 0.21

vL = 0.09


m1 = 20.5 x 10-2

m2 = 16 x 10-2

(Mw1)B = (20.5 + 0.15 x 16) x 10-2 x 57.0 = 13.053 KN-m

(Mw1)L = (16 + 0.15 x 20.5) x 10-2 x 57.0 = 10.87 KN-m

43

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

(B) For Load W2 of Axle II

Y = 0.985 m

i) u = u1 = 0.65 m v = 2(v1 + Y) = 2.83 m

u B = 0.21

vL = 0.09


m1 = 13.9 x 10-2

m2 = 5.8 x 10-2

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

44

M2= (5.8 + 0.15 x 13.9) x 10-2 x 1.415 = 0.112

ii) u = u1 = 0.65 m v = 2Y = 1.97 m

u B = 0.21

vL = 0.43


m1 = 15.5 x 10-2

m2 = 7.6 x 10-2

M1= (15.5 + 0.15 x 7.6) x 10-2 x 0.985 = 0.164

M2 = (7.6 + 0.15 x 15.5) x 10-2 x 0.985 = 0.098

(Mw1)B = (0.209-0.164) x 57.0

0.43 = 5.97 KN-m

(Mw1)L = (0.112- 0.098) x 57.0

0.43 = 1.86 KN-m

Applying effect of continuity and impact

MB = (13.05+ 5.97) x 1.25 x 0.8 = 19.02 KN-m

ML = (10.87+ 1.86) x 1.25 x 0.8 = 12.73 KN-m

2.5.4 Summary

Max Live Load B.M. on slab

MB = 38.9 KN-m (IRC Class AA Tracked)

ML = 35.48 KN-m (IRC Class AA Wheeled)

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 mm

Depth provided =225 mm

d = 225 - 25 - 8 = 192 mm

Area of steel (along short span) = 38.9 x 106

230 x 0.9 x 192 = 1126 mm2

Provide 16 mm Dia @ 140 mm c/c (1408 mm2)

Area of steel (along long span) = 35.48 x 106

230 x 0.9 x 192 = 1026 mm2

Provide 16 mm Dia @ 140 mm c/c (1408 mm2)

2.7 Shear Force In Deck Slab

2.7.1 For Class AA Tracked Vehicle

Shear Force is calculated by effective width method for effective size of panel 2.9

m X 4.42 m. For maximum Shear Force, the load will be so placed that its spread

up 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

46

Dispersion along width (be) = K.x (1- x

L ) + bw

BL =

4.422.9 = 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 or

dispersion 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

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 Vehicle

Shear Force is calculated by effective width method for effective size of panel 2.9

m X 4.42 m. For maximum Shear Force, the loading is to be arranged by trial and

error, 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 Force

computation. In first possibility, left most wheel is placed such that its spread up

to bottom reaches the face of the rib as shown in figure 2.11. In second

possibility, third wheel from left is placed as near to the face of right hand support

as possible as shown in figure 2.12.

Case 1: Left most wheel is placed such that its spread up to bottom reaches the

face 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 W1 Load

Effective width = be = 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

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 W2 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 W3 Load


L ) + bw

= 2.84 x 0.869 (1 -0.869

2.9 ) + 0.15 + 2 x 0.056 = 1.99 m

Average effective width =1.99 + 1.2

2 = 1.595 m

49


1.595 = 39.189 KN

(D) For W4 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 as

possible as shown in figure 2.12.

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

Shear Force

50

(A) For W1 Load


L ) + bw

= 2.84 x 1.019 ( 1 - 1.019

2.9 ) + 0.15 + 2 x 0.056 = 2.139 m

Average effective width = 2.139 +1.2

2 = 1.668 m

So, Load per meter width = 37.51.66 =22.46 KN

(B) For W2 Load


L ) + bw

= 2.84 x 1.281 ( 1 - 1.281

2.9 ) + 0.15 + 2 x 0.056 = 2.293 m


2 = 1.7465 m


1.7465 =35.78 KN

(C) For W3 Load


L ) + bw

= 2.84 x 0.431 ( 1 -0.431/2.9) + 0.15 + 2 x 0.056 = 1.539 m


2 =1.3969 m

So, Load per meter width = 62.51.25 = 49.92 KN

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 Loading

Shear Force will be maximum when dispersed edge of the load touches the face

of 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 W1


L ) + bw

= 2.84 x 0.531 ( 1 - 0.531

2.9 ) + 0.25 + 2 x 0.056 = 1.5939 m

52


2 = 1.3969 m

So, Load per meter width = 57

1.3969 = 40.80 KN

(B) For Load W2


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 Load

Dead load of panel = 6.63 KN/ m2

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

53

2.7.6 Check For Shear

v = V

bd = 97.5885 x 1000

1000 x 192 = 0.498 MPa

For M-35 grade concrete

w = 0.50 MPa

K1 = (1.14 – 0.7 x 0.192) > 0.5 = 1.056

K2 = ( 0.5 + 0.25 p ) > 1 ……( p = 100 x 14081000 x 192 = 0.73)

= (0.5 + 0.25 x 0.73) > 1

c = K1.K2.w

= 1.0056 x 1.0 x 0.5 = 0.5028 MPa

v <c , so safe within permissible limits.

54

3 Cantilever Slab

3.1 Moment due to Dead Load

The total maximum moment due to the dead load per meter width of cantilever

slab 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/unit m run)

Lever Arm (m)

Bending Moment (KN-m)

1 Vehicle Crash Barrier 0.275x 24= 6.6 1.65 10.89

2 Slab(rectangular) 1.8x0.1x24=4.32 0.9 3.89

3 Slab (triangular) .25x1.8x.5x24 = 5.4 0.6 3.24

4 Wearing Coat .056x22x1.3=1.6 0.65 1.04

Total 17.92 19.06

55

3.2 Moment due to Live Load

Effective width of dispersion be is computed by equation

be = 1.2 x +bw

x = 0.9 m bw = 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 jc = 0.904 R = 1.502

d =

= 203.94 mm

56

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

Area of main reinforcement required

Ast = 62.47x106

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 mm2

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 = 16.84x106

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 mm2

Shear stress (v) = 66.15x103

1000 x 302 = 0.22 MPa

P= 100 Ast

bd = 100x 1340.67

100 x 302 = 0.44

c = K1. K2.ca

ca = 0.23+ 0.31- 0.23

0.25 x 0.19 = 0.291 MPa

d = 0.302 m

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

57

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

c = 0.929 x1 .291 = 0.27 MPa

c.>v safe

60

4 Design of Longitudinal Girders

4.1 Analysis Longitudinal Girder by Courbon's Method

The reaction factors will be maximum if eccentricity of the C.G. of loads with

respect to the axis of the bridge is maximum.

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

Ri= PIi

ƩIi ( 1 +

ƩIi

ƩIidi2 . e di)

P = total live load

II = moment of inertia of longitudinal girder i

e = eccentricity of the live load

di = 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

ƩX2 = (3.2)2 + (0)2 + (3.2)2 = 2 x (3.2)2

For outer girder, X = 3.2 m

61

Figure 4.1: Class AA Tracked loading arrangement for the calculation of reaction

factors for L-girders

For inner girder, X = 0

RA = ƩPn ( 1 +

n e X Ʃ 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

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 of

reaction factors for L-girders

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

e = 2.250 m

Ʃ X2 = 2 (3.2)2

RA = ƩPn (1 +

neXƩX2 ) =

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

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 3.22) 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 be

accommodated on the span. The centre of gravity of this load system will be

found to be located at a distance of 6.42 m from the first wheel.

64

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

L-girder

The loads are arranged on the span such that the max. Moment will occur under

the fourth load from the left. The loads shown in figure are corresponding Class A

train load multiplied by 1.33 (reaction factor at intermediate beam ) and further

multiplied by impact factor of 1.225. For example:- the first load of 22 KN, if the

product 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 Loading

Taking moment about A

65

RB x 14 = ( 22 x 1.040 ) + ( 22x 2.140 ) + ( 92.87 x 5.34 ) + ( 92.37 x 6.54) + ( 55.4 x

10.84 ) + (55.4 x 13.84) = 181.47 KN

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

Max .BM = RB 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

Reaction factor for end beam according to Courbon’s method = 2.36

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 Load

Shear Force will be maximum due to class AA tracked vehicle. For maximum shear

force at the ends of the girder, the load will be placed between the support and

the 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 loads

at 1.8 m from left support

RA’ = 2.8674.667 (1.28P) =0.786 P

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 RD’, RE’ and RF’ are transferred at the cross girder should be distributed

according to Courbon’s theory

ƩW = 0.494 p + +0.227 P =0.721P

If X’ is the extreme distance of C.G from D, we have

X’ = 1/ 0.721 P (0.227 P x 2.6) = 0.82 m

E = 3.2 – 8.2 = 2.38

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

67

These reactions RD and RE act as point loads on the outer and inner longitudinal

girder and their quarter points of total span. Hence reaction at A and B due to

these will be

RA = 3/ 4 RD = 0.301 P

RE = 3/ 4RE = 0.180 P

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

Shear at B = RB’ +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 =6.6

Slab 9.72 Wearing coat 1.6

Total 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

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

69

4.3 Design Of Section

Girder Max. D.L. B.M. Max. L.L.B.M. Total B.M. Units

Outer girder 1021.46 1535.2 2556.66 KN-m

Inner girder 1021.46 781.68 1803.14 KN-m

Max. D.L.S.F. Max. L.L.S.F. Total S.F. Units

Outer girder 309.42 449.3 758.72 KN

Inner girder 309.42 239.47 548.89 KN

4.3.1 Design of Outer Longitudinal Girder

Assuming depth as = 1360 mm, Since the heavy reinforcement will be provided in

four layers.

Ast = 2556.6 x 106

200 x 0.9 x1360 = 10443.87 mm2

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

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

v = v

bd = 758 x 103

300 x 1360 = 1.86 MPa

100 As

bd = 12873.14 x 100

300x 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 103 = 200 x 4 x 78.5 x1360

Sv

70

Sv = 168.81 mm

Provide 10 mm Dia. 4 legged stirrups at 150 C/C.

4.3.2 Design of Inner Longitudinal Girder

Ast = 1803.14 x 106

200 x 0.9 x 1360 = 7365.77 mm2

Provide 12 bars of 32 mm Dia. in 3 rows Astp = 9654.86 mm2

v = v

bd = 548.89 x 103

300 x 1360 =1.35 MPa

100 AS

bd = 9654.86 x 100

300x 1360 = 2.37%

c = 0.56 + 0.58 -0.56

0.25 x 0.12 = 0.57 MPa

Vs = V - cbd= 548.89 - 0.57 x 300 x 1360

1000 = 316.33 KN

Providing 10 mm 4 legged stirrups

Sv =200 x 78.5x1360

316.33 x 103 = 270.29 mm

Provide 10 mm 4 legged stirrups at 200 C/C

73

5 Design Of Cross Girders

5.1 Analysis of Cross Girder

5.1.1 Dead Load

Cross 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/m2

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

figure 5.1.

Figure 5.1: Triangular load from each side of slab

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 103 x 3.2 x 2)/3 = 43114.67 N = 43.11 KN

5.1.2 Live Load

(A) Class AA Tracked Loading

Figure 5.3 and 5.4 shows the position of loading for maximum B.M. in the girder

due to Class AA tracked loading. For maximum load transferred to X-girder, the

position of load in the longitudinal direction is shown in figure 5.3.

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

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 x103 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


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 in

figure 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-(20.21x103x2.72)/2

=42.744 kN-m

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 loading

Position 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 N

Maximum 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 103 x 2.62)/2

= 43.788 KN-m

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

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 mm2

Provide 5 bars of 25mm dia.

Area provided=2454.37mm2

Shear stress = (278535.35)/(250x.9x1540) = 0.8 N/mm2

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

Ʈc=0.34 N/mm2

80

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

=147635.35 N

Provide 2-legged 8mm stirrups

S=(180x2x4/4x82x1540)/147635.35 = 188.75 mm

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

82

6 Design of Bearings

6.1 Design Of Outer Bearings

Dead load per bearing = 309.42 KN ~ 310 KN

Live load = 449.3 KN =450 KN approx.

6.1.1 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 x 10

100 = 13.6 KN

Total = 56.4 + 13.6 = 70 KN

Longitudinal force /bearing = 70/6 = 11.67 KN

(B) Friction Force

( D.L + L.L reaction at bearing ) × coeff. Of friction = (309.42 + 449.3 ) × 0.3

= 227.616

Friction Per bearing = 227.616

6 = 37.936 KN

(C) Wind Load

Assuming 10 m height

Wind pressure = 0.91 KN/m2

Plan area of bridge span = 14 × 10.3 = 144.2 m2

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 = Nma× = 310 + 450 = 760 KN

Minimum vertical load on bearing = Nmin = 310 KN

Try plan dimensions 250 × 500 mm and thickness 40 mm

Loaded area A2 = 11.6 × 104 mm2

From clause 307.1 of IRC 21

Allowable contact pressure = 0.25 fc

= 0.25 MPa

A1 / 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 mm2

m = 760 x 100011.6 x 104 = 6.55 MPa

Thickness of individual elastomer hi = 10 mm

Thickness of outer layer he = 5 mm

Thickness o steel laminates hs = 3 mm

Adapt 2 internal layers and 3 laminates

84

Overall thickness of bearing = 40 mm

Total thickness of elastomer in bearing = 40 – (3×3) = 31 mm

Side cover = 6 mm

Shear modulus assumed = 1.0 N/mm2

Shear strain due to creep, shrinkage and temperature is assumed as 5× 10-4 and

this is distributed to 2 bearings.

Shear strain per bearing due to creep, shrinkage and temperature

= 5x10-4 x 14.192 x 103

2 x 31 = 0.114

Shear strain due to longitudinal force = 72 x103

11.6x104 = 0.58

Shear strain due to translation = 0.114 + 0.058 = 0.694

(D) Rotation

αbi max = 0.56 mhi

bS2

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

2 x(a+b-4c)hi

Here a = 500mm, b = 250mm , c = 6mm , hi = 10 mm

S = 11.6 x104

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

αbi max = 0.5 x 10 x 10239 x 7.992 = 0.0033 radian

β = m/10 = 6.55/10 = 0.655

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

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

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actual shear strain = 0.694 as calculated

0.2 + 0. m= 0.2 + 0.1 x 6.55 = 0.855 > 0.694

Also m satisfies 10 MPa > m > 2 MPa

6.1.2 Shear Stress

Shear stress due to compression = 1.5 m / S = 1.5 x 6.5/ 7.99 = 1.23 MPa

Shear stress due to horizontal deformation = 0.694 x 1 =0.694 MPa

Shear stress due to rotation = 0.5 ( b/ hi )2 αbi = 0.5 (

23810 )2 X 0.0025 = 0.71 MPa

Total shear stress = 1.23 + 0.694 + 0.71 = 2.634 MPa < 5 MPa

The elastomeric pad bearing has the characteristics:

Plan dimensions = 250 x 500 mm

Overall thickness = 40 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

6.2 Design Of Inner Bearings

Dead load per bearing = 309.42 KN ~ 310 KN

Live load = 239.42 KN =240 KN approx

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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 x 10

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 Load

Assuming 10 m height

Wind pressure = 0.91 KN/m2

Plan area of bridge span = 14 × 10.3 = 144.2 m2

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

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Maximum vertical load on bearing = Nma× = 310 +240 = 550 KN

Minimum vertical load on bearing = Nmin = 240 KN

Try plan dimensions 320 × 500 mm and thickness 45 mm

Loaded area A2 = 15 × 104 mm2

From clause 307.1 of IRC: 21

Allowable contact pressure = 0.25 fc

= 0.25 MPa

A1 / A2 >2

Allowable contact pressure = 0.25 × 35 2 = 12.37 MPa

Effective area of bearing pressure = 550 x 1000

12.37 = 44.462 × 103 mm2

m = 500 x103

15 x104 = 3.67 MPa

Thickness of individual elastomer hi = 10 mm

Thickness of outer layer he = 5 mm

Thickness o steel laminates hs = 3 mm

Adapt 2 internal layers and 3 laminates

Overall thickness of bearing = 39 mm

Total thickness of elastomer in bearing = 39 – (3×3) = 30 mm

Side cover = 6 mm

Shear modulus assumed = 1.0 N/mm2

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Shear strain due to creep, shrinkage and temperature is assumed as 5× 10-4 and

this is distributed to 2 bearings.

Shear strain per bearing due to creep, shrinkage and temperature

= 5 x10-414.192 x103

2x30 = 0.12

Shear strain due to longitudinal force = 61 x103

15 x104= 0.407

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

(D) Rotation

αbi max = 0.56 mhi

bS2

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

2 x(a+b-4c)hi

Here a = 500mm, b = 320mm , c = 6 mm , hi = 10 mm

Therefore, S =488 x 308

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

Assuming m,max = 10MPa

αbi max = 0.5 x 10x 10308 x 9.442 = 0.0031 radian

β = m/10 = 3.67/10 = 0.367

Permissible rotation = β x n x α 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

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6.2.2 Shear Stress

Shear 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/ hi )2 αbi = 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







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7 Conclusion

7.1 Deck Slab

Overall Depth = 225 mm

Reinforcement 16 mm Dia @140 mm c/c (1408 mm2) along shorter span.

Reinforcement 16 mm Dia @140 mm c/c (1408 mm2) 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 (Ast = 1340.67 mm2)

Distribution Steel Provide 8 mm Dia bars @150 c/c (Ast = 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(Ast = 12873.14 mm2)

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

Inner Longitudinal Girder

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


7.4 Cross Girders



Overall depth = 1600 mm

Main reinforcement of 5 bars of 25 mm Dia (Ast = 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

92







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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", Bureau

of 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".

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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 considered

for 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 is

crossing the bridge.

4. The maximum loads for the wheeled vehicles shall be 20 tonnes for a single

axle or 40 tonnes for a bogie of two axles spaced not more than 1.2m

centers.

5. The maximum clearance between the road face of the kerb and the outer

edge 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

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

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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 the

outer edges of passing or crossing vehicles on multi-lane bridges shall be as

given below (figure A.2)

Clear carriageway width

g

f

5.5 m to 7.5 m

Above 7.5 m

Uniformly increasing from

0.4 m to 1.2 m

1.2 m

150 mm for all carriageway

vehicles

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

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Appendix-B: Impact Factors

Provision for impact or dynamic action shall be made by an increment of

live load by an impact allowance expressed as a fraction or a percentage of

applied live load.

Class A or Class B Loading

In the members of any bridge designed either for class A or class B loading, the

impact percentage shall be determined from the curves indicated in figure B.1

The impact factor shall be determined from the following equations which are

applicable 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 Loading

The value of the impact percentage shall be taken as follows:

For spans less than 9 m

For tracked vehicles: 25% for spans up to 5 m, linearly reducing to 10% for spans

9 m.

For wheeled vehicles: 25%.

For span of 9 m or more

Tracked 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 for

spans in excess of 23 m.

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Figure B.1: Impact Percentage for Highway Bridges(IRC 6: 2000)

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Appendix-C: K in Effective Width

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

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Appendix-D: Pigeaud's Curve

Figure D.1: Moment Coefficient for Slabs Completely Loaded with Uniformly

Distributed Load, Coefficients are m1 for K and m2 for 1/K

Figure D.2: Moment Coefficient m1 and m2 for K=0.6

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Figure D.2: Moment Coefficient m1 and m2 for K=0.7

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