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 Span
Long Span =
3.1
4.6 = 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
u
B =
0.988
3.1 = 0.319
v
L =
3.72
4.6 = 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
v
L =
3.72
4.6 = 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 =
1.062
3.1 = 0.343
v
L =
3.72
4.6 = 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 =
2.67
3.1 = 0.861
v
L =
1.546
4.6 = 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 =
1.73
3.1 = 0.558
v
L =
0.856
4.6 = 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
v
L = 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
=
1.73
3.1 = 0.558
v
L = 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

23
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
v
L = 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
v
L = 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
v
L = 0.186
m1=13.5 × 10-2
m2=11.5 × 10-2

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(v1 + Y) = 1.546 m
u
B = 0.172
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
v
L = 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
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 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 =
350
5.3 = 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 =
37.5
1.25 = 29.95 KN
Figure 2.11: Class AA Wheeled loading arrangement as Case 1 for Shear Force
(B) For W2 Load
x
L ) + bw
= 2.84 x 0.1.031 ( 1 -
1.031
2.9 ) + 0.15 + 2 x 0.056 = 2. 149 m
2.149 + 1.2
2 = 1.674 m
62.5
1.674 = 37.32 KN
(C) For W3 Load
x
L ) + bw
= 2.84 x 0.869 (1 -
0.869
2.9 ) + 0.15 + 2 x 0.056 = 1.99 m
1.99 + 1.2
2 = 1.595 m

49
62.5
1.595 = 39.189 KN
(D) For W4 Load
x
L ) + bw
= 2.84 x 0.269 ( 1 -
0.269
2.9 ) + 0.15 + 2 x 0.056 = 0.955 m
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
x
L ) + bw
= 2.84 x 1.019 ( 1 -
1.019
2.9 ) + 0.15 + 2 x 0.056 = 2.139 m
2.139 +1.2
2 = 1.668 m
37.5
1.66 =22.46 KN
(B) For W2 Load
x
L ) + bw
= 2.84 x 1.281 ( 1 -
1.281
2.9 ) + 0.15 + 2 x 0.056 = 2.293 m
2.293 + 1.2
2 = 1.7465 m
62.5
1.7465 =35.78 KN
(C) For W3 Load
x
L ) + bw
= 2.84 x 0.431 ( 1 -0.431/2.9) + 0.15 + 2 x 0.056 = 1.539 m
1.539 + 1.2
2 =1.3969 m
62.5
1.25 = 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
x
L ) + bw
= 2.84 x 0.531 ( 1 -
0.531
2.9 ) + 0.25 + 2 x 0.056 = 1.5939 m

52
1.539 + 1.2
2 = 1.3969 m
57
1.3969 = 40.80 KN
(B) For Load W2
x
L ) +bw
= 2.84 x0.569 ( 1 -
0.569
2.9 ) + 0.25 + 2 x 0.056 = 1.66 m
1.66 + 1.2
2 = 1.43 m
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 1408
1000 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
ƩIidi
2 . 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 =
ƩP
n ( 1 +
n e X
Ʃ X2 ) =
2W
3 ( 1 +
3 x 2.05 x 3.2
2 x (3.2)2 ) = 1.308 W
RB =
2P
3 ( 1 + 0 ) =
2P
3 =
2W
3 = 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 =
ƩP
n (1 +
neX
ƩX2 ) =
4W
3 ( 1 +
3 x 32.25 x 3.2
2 (3.2)2 ) = 2.74 W
RB =
4W
3 = 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 =
4W
3 ( 1 +
3I
2(I x 3.22
) x 3.2 x 1.650) = 2.36 W
Rb =
4W
3 ( 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’ =
2.867
4.667 (1.28P) =0.786 P

66
RD’ =
1.8
4.667 (1.28/P) =0.494 P
RB’ =
2.867
4.667 (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-(20.21x103
x2.72
)/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=(180x2x4/4x82
x1540)/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 1000
11.6 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 10
239 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

85
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 (
238
10 )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

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 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

87
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

88
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 x10-4
14.192 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 10
308 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

89
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 (
308
10 )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

90
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

91
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

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", 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".

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 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

95
Figure A.1: IRC Class A Tracked and Wheeled Vehicle

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 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

97
Figure A.2: IRC Class A and B Loading Vehicles

98
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.

99
Figure B.1: Impact Percentage for Highway Bridges(IRC 6: 2000)

100
Appendix-C: K in Effective Width
Table C.1: Value of Constant 'K' (IRC 21: 2000)

101
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

102
Figure D.2: Moment Coefficient m1 and m2 for K=0.7

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