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ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
CHAPTER 1
INTRODUCTION
1. General
A Bridge is a structure providing passage over an obstacle without closing the way
beneath. The required passage may be for a road, a railway, pedestrians, a canal or a pipeline.
The obstacle to be crossed may be a river, a road, railway or a valley.
Bridges range in length from a few metre to several kilometre. They are among the
largest structures built by man. The demands on design and on materials are very high. A
bridge must be strong enough to support its own weight as well as the weight of the people
and vehicles that use it. The structure also must resist various natural occurrences, including
earthquakes, strong winds, and changes in temperature. Most bridges have a concrete, steel,
or wood framework and an asphalt or concrete road way on which people and vehicles travel.
The T-beam Bridge is by far the Most commonly adopted type in the span range of 10 to 25
M. The structure is so named because the main longitudinal girders are designed as T-beams
integral with part of the deck slab, which is cast monolithically with the girders. Simply
supported T-beam span of over 30 M are rare as the dead load then becomes too heavy.
1.1 Main Components of a Bridge
The Superstructure consists of the following components:
i. Deck slab
ii. Cantilever slab portion
iii. Footpaths, if provided, kerb and handrails or crash barriers.
iv. Longitudinal girders, considered in design to be of T-section
v. Cross beams or diaphragms, intermediate and end ones.
vi. Wearing coat
DEPT. OF CIVIL ENGG; UVCE Page 1
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
The Substructure consists of the following structures:
i) Abutments at the extreme ends of the bridge.
ii) Piers at intermediate supports in case of multiple span bridges.
iii) Bearings and pedestals for the decking.
iv) Foundations for both abutments and piers may be of the type open, well, pile, etc.
Apart from the above, river training works and the approaches to a bridge also form a part
of a bridge works.
1.2 Types of Bridges
i) Girder Bridge
ii) Truss Bridge
iii) Arch Bridge
iv) Cantilever Bridge
v) Suspension Bridge
vi) Cable-stayed Bridge
vii) Movable Bridge
viii) Slab Bridge
1.2.1Girder Bridges
There are two main types of girder bridges. In one type, called a box girder bridge, each
girder looks like a long box that lies between the piers or abutments. The top surface of the
bridge is the roadway. Box girder bridges are built of steel or concrete. In the other type of
girder bridge, the end view of each girder looks like an I or a T. Two or more girders support
the roadway. This type of bridge is called a plate girder bridge when made of steel, a
DEPT. OF CIVIL ENGG; UVCE Page 2
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
reinforced or prestressed concrete girder bridge when made of concrete, and a wood girder
bridge when made of wood.
1.3 Parameters governing choice of Superstructure:
The basic function of a bridge superstructure is to permit uninterrupted smooth passage
of traffic over it and to transmit the loads and to transmit the load and forces to the
substructure safely through the bearings. Although it is difficult to stipulate the aesthetic
requirements, it should, however, be ensured that the type of superstructure adopted is
simple, pleasing to the eye, and blends with the environment. No hard and fast rules can be
laid regarding the economy in cost. The designer should, however, be able to evolve the most
economical type of superstructure based on his judgment and experience given the particular
conditions prevailing at the particular site at the particular time.
The following factors are to be considered while selecting the type of a bridge superstructure.
i. The nature of river or streams
ii. Nature of foundation / founding strata available
iii. The amount and type of traffic
iv. Whether used for navigation purposes
v. Climatic conditions
vi. Hydraulic data
vii. Type of available construction material
viii. Labour available
ix. The available facilities for erections
x. Maintenance provisional
xi. The availability of funds
xii. Time available for construction
DEPT. OF CIVIL ENGG; UVCE Page 3
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
xiii. Strategic consideration
xiv. Economic consideration
xv. Aesthetic consideration
1.4 General guidelines for analysis and Design of a Bridge Structure
Procedure for preparation of General Arrangement Drawing of a Bridge:
I. First of all the required formation level is found out. On knowing this the permissible
structural depth is established. This is done after taking into account the following
two things : ( i ) Minimum vertical clearance required taking into account the
difference between the affluxed high flood level and the soffit of the deck. ( ii )
Thickness of wearing coat required below the formation level.
II. Considering the depth of foundations, the height of deck above the bed level and low
water level, average depth of water during construction season, the type of bridge,
span lengths, type of foundations, cross section of the deck, method of construction
and loading sequence.
III. Trial cross sections of the deck, sizes of various elements of the substructure and
superstructure are decided upon and drawn to arrive at the preliminary general
arrangement of the bridge. Various trials lead to a structural form with optimum
placements of its load masses. Relative proportions and sizes of certain members as
well as their shapes are decided upon and drawn to a certain scale on this drawing.
The type of bearing to be used along with their locations depending the support
system is also established. The main basis of the general arrangement drawing of a
bridge structure is a quick preliminary analysis and design of the member sections.
This is essential for forming the basis of the detailed to be carried later on depending
upon the requirements of the project.
1.5 General Procedure for Design of Superstructure of a Bridge:
i) Analyse and design the transverse-deck-slab and its cantilever portions, unless the
superstructure is purely longitudinally reinforced solid slab with no cantilevering
DEPT. OF CIVIL ENGG; UVCE Page 4
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
portions. This is necessitated so as to decide the top flange thickness of the deck
section which is essential to work out the deck section properties for the subsequent
longitudinal design.
ii) Compute the dead load and live load bending moments at each critical section.
iii) In order to determine the maximum and minimum live load effects that a particular
longitudinal can receive, carry out the transverse load distribution for live load placed
in various lanes.
iv) This may be done by Courbon's method, Little and Morice's method, Hendry and
Jaeger methods.
v) Alternatively, use may be made to the Plane-Grid method which involves using one
of the many standard computer programs (.e.g. STAAD program). The Plan Grid
method is basically a finite element method. Though time consuming in writing the
input data, it is nevertheless very useful for the purpose of analysis. For wide and
multi-cell boxes and transverse live load distribution may be studied by the finite
element method but it is time consuming.
vi) Design against bending of critical sections, in reinforced or in prestressed concrete as
the case may be.
vii)Work out dead load and live load shear forces at each critical section in the
longitudinals of the deck and design the sections and reinforcements for effects of
torsion and shear, if required.
1.6 Transverse Distribution of Loads
Analysis based on the elastic theory is recommended to find the distribution in the transverse
direction of the bending Moment in the direction movement in the direction of the span. For
the analysis, the structure May be idealized in one of the following ways:
i. a system of interconnected beams forming a rigid
ii. an orthotropic plate
iii. an assemblage of thin plate elements or thin plate elements and beams
DEPT. OF CIVIL ENGG; UVCE Page 5
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
For the computation of the bending Moment due to live load, the distribution of the live
loads between longitudinals has to be determined. When there are only two longitudinal
girders, the reactions on the longitudinals can be found by assuming supports of the deck slab
as unyielding. With three or more longitudinal girders, the load distribution is estimated
using any one of the above rational methods.
By using any one of the above Methods, the Maximum reactions factors for
intermediate and end longitudinal girders are obtained. The bending Moments and shears are
then computed for these critical values of reaction factors. The above three Methods make
simplifying assumptions relating to the structure and loading. These assumptions introduce
errors but Make these Methods amenable to calculators and graphs. In relative comparison to
this the grillage Method of analysis, pioneered by Lightfoot and Sawko requires lesser
simplifying assumptions.
1.7 About the Project
The Project is an ongoing work across Palar river near Thangalakuppa on road joining
Kangandlahalli-Ramasagara Road and K-V Road in Bangarpet Taluk. The Superstructure for
19.34M effective span is proposed with Reinforced Concrete Deck slab and cast-in-situ three
Reinforced girders which are supported over four cross girders with a total height of the
girders 1.950M at the centre of the span and 1.890M at the end with two end cross girders
supported on the piers. The spacing of R.C. longitudinal girders is 2.5M c/c. The spacing of
the cross girders is 3.742M c/c. The panel size is 2.1M x 3.442M. The deck consists of two
cantilever slabs of 1.750M length from the centre of the end girder. There are two Crash
Barriers at the end of the deck slab. The design of the superstructure is done by the Working
stress method and involves the following procedure:
1. Deck Slab Design
2. Design of Longitudinal Girders and Cross Girders.
DEPT. OF CIVIL ENGG; UVCE Page 6
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
CHAPTER 2
LOADS
The various loads to which the bridge is subjected to are
i) Dead loads
ii) Live loads
iii) Wind loads
iv) Seismic loads
i) Dead Loads: Unit weight for Dead Loads has been considered by adopting unit
weights as per IRC 6:2000 (Standard Specifications and Code of Practice for Road
Bridges, Section II-Loads and Stress)
ii) Super Imposed Dead load: Wearing coat and Crash Barrier loads are taken as
2KN/M2 and 7.75KN/M.
iii)Vehicular Live Load: As per IRC:6 deck the superstructure is analysed for the
following vehicles and whichever produces the severest effect has been considered in
the design. Following combinations are adopted.
1) One Lane of Class AA loading or
2) Two Lanes of Class A loading
iv) Durability and Maintenance Requirements:
Concrete Grades and Reinforcement
1. Concrete: For RC Deck Slab(M-30)
2. Reinforcement: HYSD bars(Fe-415) conforming to IS:1786
3. Clear Cover: Minimum clear cover of 40MM to reinforcement has been adopted
4. Drainage Provision: Deck slab is provided with 2.5% camber and drainage
spouts with 5m c/c are adopted.
DEPT. OF CIVIL ENGG; UVCE Page 7
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
CHAPTER 3
Design of Superstructure
3. Preliminary Design Details
Clear Roadway = 7.5M Concrete Grade = M30
Three T-beams at 2.5M intervals Steel Fe 415
Five Cross beams at 3.742 M intervals
3.1 Deck Slab
The Slab is supported on four sides by beams
Thickness of Slab, H = 225MM
Thickness of Wearing Coat, D = 75MM
Span in the transverse direction = 2.5M
Effective span in the Transverse direction = 2.5 - 0.4 = 2.1M
Span in the Longitudinal direction = 3.742M
Effective span in the longitudinal direction = 3.742 - 0.3 = 3.442M
i) Maximum Bending Moment due to Dead Load
a) Weight of Deck Slab = 0.225 X 24 = 5.4 KN/M2
b) Weight of Wearing Course = 0.075 X 22 = 1.65 KN/M2
c) Total Weight = 7.05 KN/M2 (say 7.1 KN/M2 )
DEPT. OF CIVIL ENGG; UVCE Page 8
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
3.2. Analysis of Inner Panels
a) Slab Dead Load Bending Moment
Since the slab is supported on all four sides and is continuous, Pigeaud’s Curves will be used
to get influence coefficients to compute Moments,
Ratio K, = Short span/Long span
= 2.1/3.44
Therefore, K = 0.61
M1 = 8.2 X 10-2 and M2 = 3.12 x 10-2
Total Dead Weight = 7.1 x 2.1 x 3.44 = 51.3KN
Dead Load Bending Moment along Short span
MB = W (M1 + 0.15 X M2) Fig:3 Position of wheel load for max BM
= 51.3 (0.082 + 0.15 X 0.0312)
MB = 4.45 KN-M
Dead Load Bending Moment along Long span
ML = W (M2 + 0.15 X M1)
= 51.3 (0.0312 + 0.15 X 0.082)
ML= 2.23 KN-M
b) Dead Load Shear Force
DEPT. OF CIVIL ENGG; UVCE Page 9
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Dead Load Shear Force = WL/2
= (7.1 x 2.1) / 2 = 7.45KN
3.3 LOADS DUE TO IRC CLASS AA TRACKED VEHICLE
3.3.1 Live Load Bending Moment due to IRC Class AA Tracked Vehicle
Size of the panel = 2.5 M x 3.742 M
One Track of the Tracked Vehicle is placed symmetrically on the panel as shown in
the figure after the dispersion through the wearing coat of 75MM thickness the
dispersed dimensions are:
u = b + 2t and v = l + 2t
where, u = short span width of the track contact area
b = width of track wheel contact area
l = length of track wheel contact area
t = thickness of wearing coat
Therefore, u = (0.85 + 2 x 0.075) = 1M
v = (3.6 + 2 x 0.075) = 3.75 M
uB =
12.5 = 0.4 and
VL =
3.753.742 = 1
K = 2.5
3.742 = 0.67
Referring to Pigeaud’s Curves for K = 0.67 and interpolating
M1 = 0.082 and M2 = 0.030
Therefore, Live Load Bending Moment along Short span
MB = W (M1 + 0.15 x M2)
DEPT. OF CIVIL ENGG; UVCE Page 10
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
= 350(0.082 + 0.15 x 0.030)
= 30.30 KN-M
Taking Continuity Factor of 0.8 and Impact Factor as 1.25
MB = 1.25 X 0.8 X 30.30
MB = 30.30KN-M
Live Load Bending Moment along Long span
ML = W (M2 + 0.15 X M1)
= 350 (0.030 + 0.15 X 0.082)
= 14.81 KN-M
Taking Continuity Factor of 0.8 and Impact Factor as 1.25
ML = 1.25 X 0.8 X 14.81
ML= 14.81 KN-M
Design Moments due to Dead Load and IRC Class AA Tracked vehicle (Live Load)
Dead Load, MB = 4.45 KN-M
ML = 2.23 KN-M
Live Load, MB = 30.30KN-M
3.3.2 Live Load Shear Force due to IRC Class AA Tracked Vehicle
Shear Force is calculated by effective width method, considering the panel to be fixed on all
the four edges. Hence effective size of panel will be 2.1M x 3.44M. For maximum Shear
Force the load will be so placed that its spread up to slab bottom reaches up to the face of the
rib i.e. the load is kept at 1.452.0
= 0.705M .
Dispersion in the direction of span
DEPT. OF CIVIL ENGG; UVCE Page 11
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
V = x + 2 (D+H)
= 0.85 + 2 (0.075 + 0.225)
= 1.45M
Effective width of slab = K x A x [1- AL
] + bw
BL
= 3.442
2.1
= 1.64
From Table 9 of IRC 21-2000, K for Continuous slab is obtained as K=2.536
Effective width of slab = 2.536 x 0.705 x [1 – (0.705/2.1)] + [3.6 + (2 x 0.075)]
= 4.94M
Load per meter width = 350/4.94
= 70.85KN
Shear Force = 70.85 x (2.1 – 0.705) / 2.1
= 47 KN
Shear Force with Impact = 1.25 x 47 = 58.75KN
4. LOADS DUE TO IRC CLASS AA WHEELED VEHICLE
4.1 Live Load Bending Moment due to IRC
Class AA Wheeled Vehicle
Case 1)
x=0.152
; y=0.92
x=0.075 M ; y=0.45 M
DEPT. OF CIVIL ENGG; UVCE Page 12
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
u 1=0.45 M ;v 1=0.3 M
Step 1: Find M1 and M2
u¿2 (u1+x )=2 (0.45+0.075 )= 1.05M
v¿2 (v 1+ y )=2 (0.3+0.45 )= 1.5M
uB
=1.052.1
=0.5
Fig 4: Position of wheel load for Class AA wheeled vehicle
vL= 1.5
3.442=0.44
BL= 2.1
3.442=0.61
From Pigeaud’s Curves by interpolation
k=0.61,uB
=0.5,vL=0.44
k=0.6, M1= 12x10-2, M2= 5.5x10-2
k=0.61, M1= 12x10-2, M2= 5.6x10-2
k=0.707, M1= 11.8x10-2, M2= 6.5x10-2 ∴ M1= 0.12x10-2 M2= 0.056x10-2
Multiply these by (u1+x)(v1+y) = (0.45+0.075) + (0.3+0.45)=0.4∴M1= 0.12x10-2 M2= 0.056x10-2
Step 2: Find M1 and M2¿2 x=2 (0.075 )U= 0.15M V¿2 y ( 0.45 )= 0.9M
uB
=0.152.1
=0.072
vL= 0.9
3.442=0.26
BL= 2.1
3.442=0.61
From Pigeaud’s Curves by interpolationDEPT. OF CIVIL ENGG; UVCE Page 13
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
k=0.61, uB
=0.5, vL=0.44
k=0.6, M1= 21x10-2, M2= 9x10-2 k=0.61, M1= 21x10-2, M2= 9.2x10-2 k=0.707, M1= 22x10-2, M2= 11x10-2 ∴ M1= 21x10-2 M2= 9.2x10-2Multiply these by xy=0.075x0.45=0.034 ∴M1= 0.714x10-2 M2= 0.32x10-2
Step 3: Find M1 and M2U=2(u 1+ x)=2 (0.45+0.075 ) U= 1.05M V¿2 y= 0.9MuB
=1.052.1
=0.5
vL= 0.9
3.442=0.26
BL= 2.1
3.442=0.61
From Pigeaud’s Curves by interpolation k=0.61, u
B=0.5, v
L=0.44
k=0.6, M1= 13x10-2, M2= 8.6x10-2 k=0.61, M1= 13x10-2, M2= 8.5x10-2 k=0.707, M1= 12.6x10-2, M2= 8.6x10-2 ∴ M1= 13x10-2 M2= 8.6x10-2Multiply these by y(u1+x)=0.45(0.45+0.075)=0.24∴M1= 13x10-2 , M2= 2.07x10-2
Step 4: Find M1 and M2 U¿2 ( x )=2 (0.075 )= 0.15M V¿2 (v 1+ y )=2 (0.3+0.45 )= 1.5MDEPT. OF CIVIL ENGG; UVCE Page 14
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
uB
=0.152.1
=0.072
vL= 1.5
3.442=0.44
BL= 2.1
3.442=0.61
From Pigeaud’s Curves by interpolation k=0.61, u
B=0.5, v
L=0.44
k=0.6, M1= 16.5x10-2, M2= 5.5x10-2
k=0.61, M1= 16.7x10-2, M2= 6.0x10-2
k=0.707, M1= 18x10-2, M2= 6.0x10-2 ∴ M1= 16.7x10-2 M2= 6.0x10-2
Multiply these by x(u1+y) = 0.075(0.3+0.45) =0.057∴M1= 0.96x10-2 M2= 0.035x10-2
Design M1 = 0.048+0.00714-0.0312-0.0096
M1 = 0.01434 KN-M
Design M2 = 0.0224+0.0032-0.0207-0.0035
M1 = 0.0014 KN-MMB due to single load = W
u1 × v1¿M1+0.15 M2)
=22KN-MML due to single load = W
u1 × v1¿M1+0.15 M2)
=5.26KN-MApplying the effect of continuity and Impact
Final MB=22x0.8x1.25=22KN-M and ML= 5.26 x
1.25 x 0.8 =5.26KN-M
Case 2)
i) Effect of wheel no 2 of both the axles: When
wheel no 2 of both axles are centrally placed with
DEPT. OF CIVIL ENGG; UVCE Page 15
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
respect to y axis. The effect of these loads can be found as a difference of two centrally
placed loads on area (1.5x0.45) and (0.9x0.45)
Step 1: Find M1 and M2
uB
=0.452.1
=0.22
vL= 1.5
3.442=0.44
BL= 2.1
3.442=0.61
Fig 4.1: Position of wheel load for Class AA wheeled vehicle
From Pigeaud’s Curves by interpolation
k=0.61,uB
=0.5,vL=0.44
k=0.6, M1= 12.8x10-2, M2= 4.4x10-2
k=0.61, M1= 12.9x10-2, M2= 4.6x10-2
k=0.707, M1= 13.8x10-2, M2= 5.9x10-2 ∴ M1= 13x10-2 M2= 4.6x10-2
MB due to single load = W
u1 × v1¿M1+0.15 M2)
=43KN-M
ML due to single load = W
u1 × v1¿M1+0.15 M2)
=20.5KN-M
For larger load,
uB
=0.452.1
=0.22
DEPT. OF CIVIL ENGG; UVCE Page 16
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
vL= 1.5
3.442=0.26
BL= 2.1
3.442=0.61
k=0.61,uB
=0.22,vL=0.26
k=0.6, M1= 16.0x10-2, M2= 7.0x10-2
k=0.61, M1= 16.2x10-2, M2= 7.4x10-2
k=0.707, M1= 18.0x10-2, M2= 10.5x10-2 ∴ M1= 16.2x10-2 M2= 7.4x10-2
MB due to single load = W
u1 × v1¿M1+0.15 M2)
=32.5KN-M
ML due to single load = W
u1 × v1¿M1+0.15 M2)
=18.5KN-M
Net Moment MB1 = 43-32.5=10.5KN-M
ML1 = 20.5-18.5 = 2KN-M
i) Effect of wheel no.1 of both axles : wheel no1 are not centrally placed on any of the axes
hence their effect will be analysed by treating each load as eccentrically placed.
u1= 0.45M, v1=0.3M, x = 0.375M, y = 0.45M
Step1: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.375) = 1.65
v = 2(v1+y) = 2(0.3+0.45) = 1.5
By Interpolation of values for k, u/B and v/L, we get
M1 = 9.0x10-2 and M2= 4.6x10-2
Multiply by (u1+x)(v1+y) = 0.62
M1= 0.056KN-M and M2 = 0.03KN-M
Step2: Find M1 and M2 for u=2(x) = 2(0.375) = 0.75
v = 2(y) = 2(0.45) = 0.9
By Interpolation of values for k=0.61, u/B=0.36 and v/L=0.26, we get
DEPT. OF CIVIL ENGG; UVCE Page 17
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
M1 = 14.3x10-2 and M2= 9x10-2
Multiply by xy = 0.17
M1= 0.024KN-M and M2 = 0.02KN-M
Step3: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.375) = 1.65
v = 2(y) = 2(0.45) = 0.9
By Interpolation of values for k = 0.61, u/B = 0.8 and v/L = 0.26, we get
M1 = 14.3x10-2 and M2= 9x10-2
Multiply by xy = 0.17
M1= 0.024KN-M and M2 = 0.02KN-M
Step4: Find M1 and M2 for u=2(x) = 2(0.375) = 0.75
v = 2(v1+y) = 2(0.3+0.45) = 1.5
By Interpolation of values for k = 0.61, u/B = 0.36 and v/L = 0.44, we get
M1 = 12.5x10-2 and M2= 5.5x10-2
Multiply by x(v1+y) = 0.28
M1= 0.036KN-M and M2 = 0.015KN-M
Design MB = 0.056+0.024-0.037-0.036
= 0.008KN-M
ML = 0.03+0.02-0.026-0.015
= 0.01KN-M
MB2 due to single load = 2 W
u1 × v1¿M1+0.15 M2)
=5.28KN-M
ML2 due to single load = 2 W
u1 × v1¿M1+0.15 M2)
=6.23KN-M
ii) Effect of wheel no 3 of both axles
u1=0.45, v1= 0.3, x=0.775, y=0.45
Step1: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.775) = 2.45M
DEPT. OF CIVIL ENGG; UVCE Page 18
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
v = 2(v1+y) = 2(0.3+0.45) = 1.5
By Interpolation of values for k =,0.61 u/B = 1 and v/L = 0.44, we get
M1 = 7.5x10-2 and M2= 4.0x10-2
Multiply by (u1+x)(v1+y) = 0.92
M1= 0.07KN-M and M2 = 0.0377KN-M
Step2: Find M1 and M2 for u=2(x) = 2(0.775) = 1.55
v = 2(y) = 2(0.45) = 0.9
By Interpolation of values for k =,0.61 u/B = 0.74 and v/L = 0.26, we get
M1 = 9.9x10-2 and M2= 7.1x10-2
Multiply by (xy) = 0.348
M1= 0.035KN-M and M2 = 0.025KN-M
Step3: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.775) = 2.45
v = 2(y) = 2(0.45) = 0.9
By Interpolation of values for k =,0.61 u/B = 1 and v/L = 0.26, we get
M1 = 6.63x10-2 and M2= 5.3x10-2
Multiply by y(u1+x) = 0.55
M1= 0.0364KN-M and M2 = 0.63KN-M
Step4: Find M1 and M2 for u=2(x) = 2(0.775) = 1.55
v = 2(u1+y) = 2(0.45+0.45) = 1.8
By Interpolation of values for k =,0.61 u/B = 0.74 and v/L = 0.53, we get
M1 = 9.0x10-2 and M2= 4.5x10-2
Multiply by x(v1+y) = 0.58
M1= 0.053KN-M and M2 = 0.0261KN-M
Design MB = 0.07+0.035-0.0364-0.053
= 0.0156KN-M
ML = 0.0377+0.025-0.03-0.0.261
= 0.0066KN-M
MB3 due to single load = 2 W
u1 × v1¿M1+0.15 M2)
=15.36KN-M
DEPT. OF CIVIL ENGG; UVCE Page 19
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
ML3 due to single load = 2 W
u1 × v1¿M1+0.15 M2)
=8.3KN-M
Final bending Moment for case ii
MB= MB1+ MB2+ MB3 ML = ML1+ ML2+ ML3
MB = 31.14KN-M ML = 16.53KN-M
Case 3: Loads placed as per figure:
i) Effect of wheel 2 of axle 1:
u= 0.45, v = 0.3, u/B= 0.45/2.1,
v/L=0.3/3.442 = 0.087
for k = 0.61, by interpolation M1 = 0.2
and M2 = 0.15
MB1 = 14KN-M and ML1=11.6KN-M
ii) Effect of wheel 1 of axle 1:
For larger load, u/B=0.8, v/L=0.09, k =
0.61
MB = 16.33KN-M
ML = 16KN-M
For smaller load, u/B=0.36, v/L=0.09, k = 0.61
Fig 4.3: Position of wheel load for Class AA wheeled vehicle
MB = 11.8KN-M
ML = 10.81KN-M
Net Moments,
MB2 = ½(16.33-11.8) = 2.27KN-M and ML2 = ½(16-10.81) = 2.6KN-M
iii) Effect of wheel load 3 of axle 1:
For larger load,
u/B= 2.45/2.1= 1.2; v/L= 0.3/3.442 = 0.09
DEPT. OF CIVIL ENGG; UVCE Page 20
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
For k = 0.61, by interpolation M1 = 0.085 and M2 = 0.081
MB = ML= 33.14KN-M
For smaller load,
u/B= 1.55/2.1= 0.74, v/L=0.3/3.442 = 0.09
For k = 0.61 by interpolation M1 = 0.108 and M2 = 0.101
MB = 26.35KN-M and ML= 25.00KN-M
Net Moments,
MB3 = ½(33.14-26.35) = 3.4KN-M and ML3 = ½(33.14-25.00) = 4.07 KN-M
iv) Effect of wheel 2 of axle 2:
For larger load, u/B=0.45/2.1= 0.22; v/L= 2.71/3.442 = 0.8;
For k=0.61 by interpolation M1 = 0.13 and M2 = 0.0331
MB = 76.2KN-M and ML= 30.00KN-M
For smaller load, u/B = 0.45/2.1=0.22; v/L = 2.09/3.442 = 0.61
For k=0.61 by interpolation M1 = 0.13 and M2 = 0.045
MB = 60.00KN-M and ML= 28.00KN-M
Net Moments,
MB4 = ½(76.2-60) = 3.4KN-M and ML4 = ½(30-28.00) = 1.0 KN-M
v) Effect of wheel load 1 of axle 2:
u1= 0.45; v1= 0.30; x = 0.37; y = 1.045
Step1: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.775) = 1.64
v = 2(u1+y) = 2(0.3+1.045) = 2.69
By Interpolation of values for k =,0.61 u/B = 0.8 and v/L = 0.8, we get
M1 = 7.0x10-2 and M2= 2.67x10-2
Multiply by (u1+x)(v1+y) = 1.11
M1= 0.077KN-M and M2 = 0.03KN-M
Step2: Find M1 and M2 for u=2(x) = 2(0.775) = 0.74
v = 2(y) = 2(1.045) = 2.09
By Interpolation of values for k =,0.61 u/B = 0.35 and v/L = 0.61, we get
M1 = 12.0x10-2 and M2= 4.5x10-2
Multiply by xy = 0.39
DEPT. OF CIVIL ENGG; UVCE Page 21
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
M1= 0.0468KN-M and M2 = 0.02KN-M
Step3: Find M1 and M2 for u = 2(u1+x) = 2(0.45+0.775) = 1.64
v = 2(y) = 2(1.045) = 2.09
By Interpolation of values for k =,0.61 u/B = 0.8 and v/L = 0.61, we get
M1 = 8.0x10-2 and M2= 3.9x10-2
Multiply by y(u1+x) = 0.86
M1= 0.069KN-M and M2 = 0.034KN-M
Step4: Find M1 and M2 for u=2(x) = 2(0.775) = 0.74
v = 2(v1+y) = 2(0.3+1.045) = 2.69
By Interpolation of values for k =,0.61; u/B = 0.35 and v/L = 0.8, we get
M1 = 10.4x10-2 and M2= 3.3x10-2
Multiply by x(v1+y) = 0.5
M1= 0.052KN-M and M2 = 0.0165KN-M
Design M1 = 0.077+0.0468-0.069-0.052 = 0.0028
M2 = 0.03+0.02-0.034-0.0165 = 0
MB5 = 0.78KN-M and ML5 = 0.17 KN-M
vi) Effect of wheel load 3 of axle 2:
u1= 0.45; v1=0.3; x=0.77; y=1.045
Step1: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.775) = 2.44
v = 2(v1+y) = 2(0.3+1.045) = 2.69
By Interpolation of values for k =,0.61; u/B = 1 and v/L = 0.8, we get
M1 = 5.8x10-2 and M2= 2.3x10-2
Multiply by (u1+x)(v1+y) = 1.641
M1= 1.0KN-M and M2 = 0.038KN-M
Step2: Find M1 and M2 for u=2(x) = 2(0.775) = 1.54
v = 2(y) = 2(1.045) = 2.09
By Interpolation of values for k =,0.61; u/B = 0.74; and v/L = 0.61, we get
M1 = 8.5x10-2 and M2= 3.0x10-2
Multiply by (xy) = 0.81
DEPT. OF CIVIL ENGG; UVCE Page 22
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
M1= 0.07KN-M and M2 = 0.0243KN-
Step3: Find M1 and M2 for u=2(u1+x) = 2(0.45+0.775) = 2.44
v = 2(y) = 2(1.045) = 2.09
By Interpolation of values for k =,0.61; u/B = 1; and v/L = 0.61, we get
M1 = 6.8x10-2 and M2= 3.8x10-2
Multiply by y(u1+x) = 1.28
M1= 0.09KN-M and M2 = 0.05KN-M
Step4: Find M1 and M2 for u=2(x) = 2(0.775) = 1.54
v = 2(v1+y) = 2(0.3+1.045) = 2.69
By Interpolation of values for k =,0.61; u/B = 0.74; and v/L = 0.78, we get
M1 = 7.4x10-2 and M2= 3.9x10-2
Multiply by x(v1+y) = 1.036
M1= 0.0767KN-M and M2 = 0.041KN-M
MB6 = 1.48KN-M and ML6 = 0.22KN-M
Final Moments for Case iii
MB = MB1+ MBB2+ MB3+ MB4+ MB5+ MB6
= 14 + 2.27 + 3.4 + 8.1 + 078 + 1.48
= 30.03 KN-M
ML = ML1+ ML2+ ML3+ ML4+ ML5+ ML6
= 11.6 + 2.6 + 4.07 + 1 + 0.17 + 0.22
= 19.66KN-M
Final Bending Moments, after applying continuity and Impact factor
MB = 1.25 x 0.8 x 30 = 30.03KN-M
ML = 1.25 x 0.8 x 19.66 = 19.66 KN-M
4.2 Live Load Shear Force due to IRC Class AA Wheeled Vehicle
DEPT. OF CIVIL ENGG; UVCE Page 23
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 4.4: Position of wheel load for Class AA wheeled vehicle
Shear Force is computed by effective width Method. The effective size of the panel is taken
as 2.1Mx3.742M. The load is arranged as shown in figure.
Fig 4.4a: Position of wheel load for Class AA wheeled vehicle
Reactions at A and B can be calculated by using the below equations
RA=W2
( α 3−2 ×α 2+2 )
RA=20.84
2( 0.24 3−2 ×0.24 2+2 )
RA=20.24 KN
RA=W2
(2× α2−α 3 )
RB=20.84
2(2× 0.24 2+0.24 3 )
RB=1.05 KN
DEPT. OF CIVIL ENGG; UVCE Page 24
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 4.4b: Position of wheel load for Class AA wheeled vehicle
RA=W (2× β 3−3× β 2+ α2 β2
−α 2
4+1)
RB=62.5(2× 0.313−3×0.31 2+ 0.42× 0.312
−0.42
4+1)
RA=47.1 KN
RB=W −RA
RA=62.5−47.1=15.4 KN
Fig 4.4c: Position of wheel load for Class AA wheeled vehicle
RB=W2
( α 3−2 ×α 2+2 )
RB=62.5
2(0.43 3−2× 0.43 2+2 )
DEPT. OF CIVIL ENGG; UVCE Page 25
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
RB=53.43 KN
RA=W2
(2 × α 2−α 3 )
RA=62.5
2(2×0.43 2−0.433 )
RA=9.1 KN
The values of effective width, reactions and shear force are tabulated below
Load (KN) X(M) e(M) RA(KN) RB(KN) F A(KN) FB(KN)
W1=30.52 0.25 0.86 20.29 1.05 23.53 1.22
W2=62.5 0.65 1.319 47.1 15.4 35.71 11.68
W3=62.5 0.45 1.2 53.43 9.1 44.53 7.59
Total 103.77 20.48
e1= k x X1x{1-(0.25/2.1)} + W W=0.15+(2 x 0.075) = 0.3M
=2.536 x 0.25x {1-(0.25/2.1)} + 0.3 l/L = 3.442/2.1= 1.64
=0.86M hence k=2.536
e2= 2.536 x 0.65x{1-(0.65/2.1)} + 0.3
= 1.438M. However since c/c distance between wheels in the longitudinal direction is >
than 1.2M so take the average i.e. 1.2+1.438/2 =1.319M for each wheel.
e3 = 2.536 x 0.45x {1-0.45/2.1)}+0.3
= 1.2M
Taking into account Impact factor, Design Shear = 1.25 x 103.77
= 130KN
DEPT. OF CIVIL ENGG; UVCE Page 26
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Final Shear Forces due to
Dead Load = 7.45KN
Live Load = 130KN
Total = 137.45KN, say 138KN
5. LOADS DUE TO IRC CLASS A VEHICLE
5.1 Live Load Bending Moment due to IRC Class A Loading:
u = 0.65M; v = 0.4M
i) Effect of wheel load 1 of axle 1:
u = 0.65M; v = 0.4M; u/B=0.65/2.1= 0.31;
v/L=0.4/3.442 = 0.12
By interpolation, For k = 0.61; M1= 0.175KN-M;
M2=0.15KN-M
MB1 = 11.26KN-M and ML1= 10.05KN-M
DEPT. OF CIVIL ENGG; UVCE Page 27
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5: Position of wheel load for Class Avehicle
ii) Effect of load 1 of axle2:
For the bigger load, u=0.65 and v=2.8; u/B=0.31; v/L=0.82
By interpolation, For k = 0.61; M1= 0.105KN-M; M2= 0.032KN-M
MB = 44KN-M and ML = 19.45KN-M
For smaller load, u= 0.65 and v=2; u/B 0.31; v/L = 0.58
By interpolation, For k = 0.61; u/B = 0.31; v/L = 0.58
MB = 37.55KN-M and ML = 18.73KN-M
Net Moments,
MB = ½( 44 - 37.55) = 3.225KN-M and ML = ½(19.53-18.73) = 0.4KN-M
5.2 Shear Force for Class A Loading
Fig 5: Position of wheel load for Shear force for Class A vehicle
Dispersion in the direction of span = 0.5 + 2 x (0.3) = 1.1M
X= 0.55; k=2.536; W=0.25+(2 x 0.075) = 0.4M
∴ e = 2.536 x 0.55 x {1- (0.55/2.1)}
DEPT. OF CIVIL ENGG; UVCE Page 28
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
= 1.43M > 1.2M, Hence e = (1.2+1.43)/2 = 1.315M
RA=W2
( α 3−2 ×α 2+2 )
RA=572
(0.524 3−2 ×0.524 2+2 ) = 45.54KN
RB=W2
(2 × α 2−α 3 )
RB=W2
(2 × 0.524 2−0.524 3 ) = 11.5KN
Shear Force taking intoImpact factor=1.5 x45.451.315
= 51.84KN
5.3 Final Moments:
Live load Moments due all Class of loads
1. Class AA Tracked, MB = 30.30KN-M and ML=14.81KN-M
2. Class AA Wheeled Vehicle
a. Case i Moments, MB = 22.00KN-M and ML = 5.26KN-M
b. Case ii Moments, MB = 31.14KN-M and ML = 16.53KN-M
c. Case iii Moments, MB = 22.00KN-M and ML = 19.66KN-M
3. Class A Loading, MB = 3.225KN-M and ML = 0.4KN-M
Hence take MB = 31.14KN-M and ML = 19.66KN-M for design. Since the moment due to
IRC Class AA Wheeled Vehicle is severe adopt it for design. Hence
MB , Dead Load = 4.45KN-M
Live Load = 31.14KN-M
Total = 36.00KN-M
DEPT. OF CIVIL ENGG; UVCE Page 29
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
5.4 Final Shear Force
Dead Load = 7.45KN
Live Load = 130KN
Total = 138KN
CHAPTER 4
DESIGN OF DECK SLAB
Grade of concrete = M30
Grade of Steel = Fe415
σ cbc=10 N /mm2
σ st=200 N /mm2 , Modular ratio, M = 10
4.1 Constants:
DEPT. OF CIVIL ENGG; UVCE Page 30
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
n= 1
1+σ st
m .σ cbc
= 1
1+200
10 ×10
=0.34
j=(1−n3 )=(1−0.34
3 )=0.89
Q=0.5× σ cbc ×n × j=0.5 × 10× 0.34 × 0.89=1.513
4.2 Check for depth (MB):
d provided=√ MQ ×b
=√ 36 × 106
1.513 ×1000=154.25 mm
d provided=225 MM
d provided>drequired
Provide 225MM overall depth using 40MM
drequired=225−40−8=177 MM
Areaof steelrequired= Mσ st jd
= 36 × 106
200×0.89 ×225=890 MM2along M B
A rea of steel required= Mσ st jd
= 22 ×106
200 × 0.89 ×225=550 MM2 along M L
Sv=1000 ×201
890
Sv=225 mmc /c
Hence provide 16MM diameter bars at 150MM c/c
Sv=1000 ×113
550
Sv=205 mmc /c
Hence provide 12MM diameter bars at 150MM c/c
DEPT. OF CIVIL ENGG; UVCE Page 31
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
CHAPTER 5
LONGITUDINAL GIRDER AND CROSS GIRDER DESIGN
5.1 Reaction Factor Bending Moment in Longitudinal Girders by
Courbons’s Method
5.1.1 Class AA Tracked Vehicle
DEPT. OF CIVIL ENGG; UVCE Page 32
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5: Position of Class AA Tracked Vehicle for obtaining reaction factors
Minimum Clearance Distance: 1.2 + 0.85/2 = 1.625M
e=1.1 m , P=w2
∑ x2=(2.5)2+ (0)2+(2.5)2=2 (2.5)2=12.5 m For outer girder, x=2.6 m, for inner girder x=0
Therefore, RA=
∑ P
n [1+nex
∑ x2 ]RA=
4 P3 [1+ 3×1.1 ×2.5
2(2.5)2 ]RA=0.5536 W and RB= 0.3333W
5.1.2 Class A Loading
DEPT. OF CIVIL ENGG; UVCE Page 33
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.1: Position of Class A Vehicle for obtaining reaction factors
Minimum Clearance Distance: 0.15 + 0.25 = 0.4M
e=0.8 m , P=w2
∑ x2=(2.5)2+ (0)2+(2.5)2=2 (2.5)2=12.5 m For outer girder, x=2.6 m, for inner girder x=0
Therefore, RA=
∑ P
n [1+nex
∑ x2 ] RA=
4 P3 [1+ 3× 0.8× 2.5
2(2.5)2 ]¿ 4 P
3[1+0.48 ]
¿ 4 P3
[1.48 ]
¿1.9734 P
¿0.9867 w
RB=4 P3
[1+0 ]
DEPT. OF CIVIL ENGG; UVCE Page 34
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
RB=4 P3
=43
×w2
=0.67 w
Therefore RB=0.67 w
Impact factor ¿ 4.56+L
= 4.56+18.71
=0.182
5.2Maximum Live load bending Moment for Class A Loading
For reaction factors we considered the moment or shifting of loads in transverse
direction. For finding the maximum B.M., however we have to consider the movement
of the loads along the span. For maximum B.M. at a given section: The maximum B.M.
at any section of a simply supported beam due to a given system of point loads crossing
the beam occurs when the average loading on the portion left is equal to the average
loading to the right of it, when section divides the load in the same ratio as it divides the
span. To get the maximum B.M. at a given section, one of the wheel loads should be
placed at the section. We shall try these rules for both Class A loading as well as Class
AA Tracked loading.
5.2.1 Class A Loading:
5.2.2 Maximum Live Load Bending moment at the mid span i.e. L/2: The below
figure shows the Influence Line Diagram.
Fig 5.2a: ILD for BM at L/2
DEPT. OF CIVIL ENGG; UVCE Page 35
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Load No. Load Value Ordinate Moment W1 27kN 3.885
9.355× 4.678=1.93 m 52.11KN-M
W2 27kN 4.9559.355
× 4.678=2.48 m 67KN-M W3 114kN 8.155
9.355× 4.678=4.1 m 468KN-M
W4 114kN 4 m 456KN-M W5 68kN 5.055
9.355× 4.678=2.53 m 173KN-M
W6 68kN 2.0559.355
× 4.678=1.1m 75KN-M Total1291.11KN-MBending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1291.11× 1.182× 0.9867=1506 KN −M
Inner girder=1291.11× 1.182× 0.67=1023 KN −M
5.1.2 Maximum Live load bending Moment at 3L/8: The below figure shows the Influence
Line diagram
Fig 5.2b: ILD for BM at 3L/8
DEPT. OF CIVIL ENGG; UVCE Page 36
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Load No. Load Value Ordinate Moment
W1 27kN
2.716257.01625
× 4.39=1.7 m46KN-M
W2 27KN
3.816257.01625
× 4.39=2.39 m64.53KN-M
W3 114KN4.39 m
501KN-M
W4 114KN
10.511.7
× 4.39=4.0 m456KN-M
W5 68KN
6.1937511.7
× 4.39=2.33 m159KN-M
W6 68KN
3.211.7
×4.39=1.2 m82KN-M
W7 68KN
0.1937511.7
× 4.39=0.073 m5KN-M
Total=1314KN-M
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.182 ×0.9867 × 1314=1532.5 KN −M
Inner girder=1.182 × 0.67×1314=1041 KN −M
5.1.3 Maximum Live load bending Moment at L/8: The below figure shows the Influence
Line diagram
DEPT. OF CIVIL ENGG; UVCE Page 37
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.2c: ILD for BM at L/8
Load No. Load Value Ordinate Moment
W3 114KN2.05 m
234.00KN-M
W4 114KN
15.1716.37
× 2.05=1.9 m217.00KN-M
W5 68KN10.8716.37
× 3.05=2.025 m 138.00KN-M
W6 68KN7.87
16.37× 2.05=0.99 m 67.32 KN-M
W7 68KN4.87
16.37× 2.05=0.61m 42.00KN-M
W8 68KN
1.8716.37
× 2.05=0.24 m16.40KN-M
TOTAL 715.00KN-M
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.182 ×0.9867 × 715=834.00 KN−M
Inner girder=1.182 × 0.67×715=566.24 KN−M
5.2 Absolute Maximum Bending Moment
Absolute B.M. occurs at under that heavier wheel load which is nearer to the C.G. of the load
system that can possibly be accommodated on the span of 18.71M. The placement should be
DEPT. OF CIVIL ENGG; UVCE Page 38
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
such that the centre of span is mid-way between the wheel load and the C.G. of the load
system. This position is shown below.
X = 6.42M , C.G of Load = 6.42-(1.1+3.2+1.2) = 0.92M from fourth load
Load No. Load Value Ordinate Moment
W1 27kN3.3958.895
× 4.67=1.782 M48.20KN-M
W2 27KN
4.4958.895
× 4.67=2.39 M
64.00KN-M
W3 114KN
7.6958.895
× 4.67=4.04 M
461KN-M
W4 114KN4.67 M
533KN-M
W5 68KN
4.5959.815
× 4.67=2.2 M
150KN-M
W6 68KN
1.5959.815
× 4.67=0.76 M
52KN-M
Total
=1308.2KN-
M
Bending Moment, BM, including Impact Factor and Reaction factor for
DEPT. OF CIVIL ENGG; UVCE Page 39
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Outer girder=1.182 ×1.82 ×1308.2=834.00KN-M
Inner girder=1.182 × 0.67×1308.2=1036 KN-M
5.3 CLASS AA-TRACKED VEHICLE
5.3.1 Bending Moment at centre of the span
Fig 5.3a: ILD for BM at L/2
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.10 × 0.5336 ×2954=1800.00KN-M
Inner girder=1.10 × 0.3333 ×2954=1083 KN-M
5.3.2 Bending Moment at 3L/8 of the span
Fig 5.2b: ILD for BM at 3L/8
DEPT. OF CIVIL ENGG; UVCE Page 40
Load No. Load Value Ordinate Moment
W 700KN
3.77+4.672
=4.22 M2954KN-M
Load No. Load Value Ordinate Moment
W 700KN
4.4+3.562
=3.98 M2786KN-M
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.10 × 0.5336 ×2786=1697.00KN-M
Inner girder=1.10 × 0.3333 ×2786=1022 KN-M
5.3.3 Bending Moment at L/4 of the span
Fig 5.2c: ILD for BM at L/4
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.10 × 0.5336 ×2866=1746.00KN-M
Inner girder=1.10 × 0.3333 ×2866=1051 KN-M
5.3.4 Bending Moment at L/8 of the span
DEPT. OF CIVIL ENGG; UVCE Page 41
Load No. Load Value Ordinate Moment
W 700KN
3.51+4.6782
=4.094 M2866KN-M
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.2d: ILD for BM at L/8
Bending Moment, BM, including Impact Factor and Reaction factor for
Outer girder=1.10 × 0.5336 ×1298=791.00KN-M
Inner girder=1.10 × 0.3333 ×1298=476 KN-M
5.3Live Load Shear Force
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 girder and shear force will be found by the reaction factors derived below. For
intermediate section, same reaction factors will be used as derived for bending Moment.
5.3.1 Shear at the end of girder
Since the length of the track is 3.6M Maximum shear will occur when the C.G. of load is
1.8M away from support A of the girder. The load will be confined between the end and the
first stiffener. Along width of the bridge, the track will be so placed that it maintains a
maximum clearance of 1.2M. Hence distance of C.G. of load from kerb = 1.2+0.425 =
1.625M
DEPT. OF CIVIL ENGG; UVCE Page 42
Load No. Load Value Ordinate Moment
W 700KN
2.05+1.6562
=1.853 M1298KN-M
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.3: Class AA tracked Wheel load position for Live loaf shear force
P 1=1.6752.1
P=0.8 P
P 2=0.4252.1
+ 1.7252.1
=1.03 P
P 3=0.3752.1
P=0.18 P
Reactions at end of each Longitudinal Girder due to transfer of these loads at 1.8M from left support
RA' = 0.374P RD' = 0.347P
RB' = 0.535P RE' = 0.495P.
RC' = 0.093P RF' = 0.087P.
The load RD’, RE’, RF’ should be transferred to the cross girders as per Courbon’s Theory
∑W =0.347 P+0.495 P+0.087 P
∑W =0.929 P
DEPT. OF CIVIL ENGG; UVCE Page 43
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
∑W2= 22.6 ×2.6
RD=∑ P
n [1+ nex
∑ x2 ]RD=
0.929 P3 [1+
3 ×0.5877 × 2.5
2(2.1)2 ]C=0.464 P
RE=∑ P
n [1+ nex
∑ x2 ]RE=
0.929 P3
[ 1+0 ]
RE=0.31 P
These reactions RDand RE act as point loads on outer girder and inner girders at their 1/5th
points of total span.
RA × 18.71=RD ×14.968 RA × 18.71=RA ×14.968 RA=0.8 RD RB=0.8 RE
RA=0.8 ×0.464 RD RB=0.8 × 0.31 P
RA=0.3712 P RB=0.3712 P Hence shear at A= RA' + RA = 0.3712 P+¿ 0.3712 P B= RB' + RB = 0.535 P+¿ 0.248 P , But P = 350KNHence Shear Force at outer girder ¿1.10 ×0.7452 ×350=284 KN
Hence Shear Force at inner girder ¿1.10 ×0.783 × 350=302 KN
DEPT. OF CIVIL ENGG; UVCE Page 44
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
5.4 Shear Force Intermediate Points: Shear at other points will be found on the basis
of the same reaction coefficients as found for B.M. Thus for Class AA Tracked loading,
reaction coefficient for outer girder = 0.5536W and for inner girder = 0.3333W
5.4.1 Shear Force at Mid span:
Fig 5.4a: ILD for SF at L/2
Shear Force at Mid span=12 [ 1
2+ 1
2×
5.7559.355 ]×700
¿ 12 [ 1
2+0.31]× 700
¿284 KN
Shear Force at outer girder ¿1.10 ×0.5536 × 284=173 KN
Shear Force at inner girder ¿1.10 ×0.3333 × 284=105 KN
5.4.2 Shear Force at 3/8th span:
Fig 5.4b: ILD for SF at 3/8
DEPT. OF CIVIL ENGG; UVCE Page 45
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Shear Force at 3 /8 thspan=12 [ 5
8+ 5
8×
8.0911.69 ]×700
¿ 12 [ 1
2+0.44 ]×700
¿371 KN
Shear Force at outer girder ¿1.10 ×0.5536 × 371=226 KN
Shear Force at inner girder ¿1.10 ×0.3333 × 371=136 KN
5.4.3 Shear Force at 1/4th span:
Fig 5.4c: ILD for SF at 1/4th span
Shear Force at 1 /4 th span=12 [ 3
4+ 3
4×
10.43214.032 ]×700
¿ 12 [ 1
2+2.23 ]× 700
¿458 KN
Shear Force at outer girder ¿1.10 ×0.5536 × 458=279 KN
Shear Force at inner girder ¿1.10 ×0.3333 × 458=168 KN
5.4Dead Load Bending Moment and Shear Force in Girder:
5.5.1 Live Load from Cantilever
DEPT. OF CIVIL ENGG; UVCE Page 46
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
i) Class AA Loading: The minimum distance of Class AA loading from the kerb is to be 1.2M. Since the available clear length of cantilever is only 0.75M, class AA loading will not come on the cantilever portion.ii) Class A Loading:
Fig 5.4: Position of Class A loading for max BM
Distance of C.G of wheel from the edge of Cantilever support = 1.05 – (0.15+0.2)= 0.65MDispersed width of load = 0.5 + (2 × 0.275) = 1.05MLoad onCantilever=57 ×[ 0.875
1.05 ]Effective Widthof Cantilever , e=1.2 × x+W Where, x= distance of C.G. of dispersed load in cantilever portion=0.875/2 W= width of concentration area of load perpendicular to span = B+ 2×thicKNess of Wearing Coat = 0.25+(2× 0.075¿ = 0.4MDEPT. OF CIVIL ENGG; UVCE Page 47
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
∴ x=[ 0.8751.05 ]=0.4375
∴ e= (1.2 × 0.4375 )+0.4
¿0.925 MImpact Factor = 1.5 Hence Bending Moment due¿ Live load=(1.5 ×
47.50.925 )+ 0.875
2
Dead Load and Dead load bending MomentComponent DL/M run(KN) C.G from edge of cantilever(M)
Moment(KN-M)Crash Barrier (0.5 ×1 ×1×24 )=12 KN 1.5 18.00Wearing Coat (0.075 ×1.05 × 22 )=1.74 KN 0.525 0.924Slab (1.55 ×0.2 ×24 )=7.44 KN 0.775 5.80Total 21.18KN 25.00KN-M
∴Design Bending Moment = 25 + 33.7 = 58.7KN-M
Design Shear Force = 21.18 + 77 = 98.18KN
Check for depth (MB):
d provided=√ MQ ×b
=√ 58.7× 106
1.513 ×1000=154.25 mm
d provided=196 MM
d provided>drequired
Provide 200MM overall depth using 40MMdrequired=200−40−8=152 MM
Areaof steelrequired= Mσ st jd
= 58.7 × 106
200 ×0.89 ×225=1466 MM 2
Sv=1000 ×201
1466Sv=137 MM
Hence provide 16MM diameter bars at 100MM c/c
DEPT. OF CIVIL ENGG; UVCE Page 48
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
5.5Dead Load Bending Moment and Shear Force in Girders
Fig 5.5: Position of loads for Max BM and SF in Girders
1. Dead load ¿eachCantilever slab=21.18 KN2. Dead load ¿ Slab∧wearing coat=(0.075 ×22 )+(0.225 × 24 )
¿7.05 KN / M M 2
3. Total Dead load from deck = (2×2.18 )+(2 ×2.5+0.4 )×7.05 = 42.43KN / M M 2
4. Assumingthis load ¿beequally shared by 3 girders=42.433
=14.15 KN /M M 2
5. Let the depth of rib = 1.725M6. Weight of rib = 1× 0.4 ×1.725 ×24=16.56 KN7. Total UDL on girders =16.56+14.15=30.71 KN / M8. Let depth of cross girders = 1.5M9. Weight of rib of cross girders = 1 ×0.3 ×1.5 × 24=10.8 KN / M10. Length rib of cross girders = 2×2.1=2.4 M11. Assuming weight of cross girder to be equally shared by all the three longitudinal girders point load on each girder = 1/3( 4.2 × 10.8) = 15.12KNReaction, RA=RB=
12
(4 ×15.12+18.71× 30.71 )
= 317.53KNBending MomentBending Moment at Mid Span
= (317.53× 9.355¿−15.12 ×(3.7422
) –[15.12(3.742+1.871)]DEPT. OF CIVIL ENGG; UVCE Page 49
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
- (30.71× 9.355× 9.355¿/2= 1513KN-MBending Moment at Quarter Span = [317.53× (3.74+1.871 ) ¿−¿]= 1271KN-M Shear ForceShear Force at support, RC=317.53 KN
Shear Force at quarter span = 317.53−15.12−(30.71 ×5.613 ) =131KN Shear Force at 3/8th span = 317.53−15.12−(30.71 ×7.02 ) =87KN Shear Force at Mid span = 317.53−15.12−15.12−(30.71 × 9.355 ) =0KN5.6Design of Outer Girder
1. Total Bending Moment at centre of span = 1513+1800=3313 KN−M
2. Total Bending Moment at quarter span = 1271+1746=3017 KN−M
3. Total Shear at support = 317.53+287=605 KN
4. Total Shear at ¼ span = 131+279=410 KN
5. Total Shear at 3/8 span = 87+226=313 KN
6. Total Shear at Mid span = 173+0=173 KN
For beams, M30 concrete will be used and the Outer girder will be designed as T-beam
having a depth of rib = 1.725M
Total depth = 1.725 + 0.225 = 1.95M
Let us assume an effective depth = 1.725-0.120 = 1.605M
A st=M
σ st × j ×d= 3313 ×106
200× 0.9 ×1605=11467 M M 2
Provide 12 bars of 32MM diameter, having total A st=9651 M M 2 and
DEPT. OF CIVIL ENGG; UVCE Page 50
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Provide 4bars of 25MM diameter, having total A st=1964 M M 2
Arrange these bars in 4 layers with spacing between bars equal to largest diameter bar used
i.e.32MM
Clear cover = 40MM
Height of C.G of bars from bottom of bars = (40+12+32X2)
= 148MM
∴ d= 1725-148=1577MM <1605MM, hence ok
5.7Check for stresses
1. Depth of Neutral Axis: Flange width will be the least of the following
a. 12ds+ br = 12 x 225 + 400 = 3100MM
b. c/c spacing = 2500MM
c. Span/3 = 18.71/3 = 6236.66MM
Hence Flange width = B=2500MM
2. Let depth of Neutral axis be N lying in web
∴ B× ds×(n−ds2 )=m× A st(d−n)
2500 ×225 (n−112.5 )=10 ×11,467 (1577−n )∴ n = 361MM
Critical Neutral axis depth ,n= d
1+σ st
m . σcbc
= 1577
1+200
10 × 10
=526 MM
Actual Neutral axis falls above the critical neutral axis therefore, the stress in the steel
reaches the Maximum value first hence σ st=200 N /M M 2
Corresponding stress in the concrete at the outer fibre is given by
c=σ st
M×
n(d−n)
¿ 20010
×361
(1577−361)
¿5.94 N / M M 2
DEPT. OF CIVIL ENGG; UVCE Page 51
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Similarly,
c1=n−D f
n× c
c1=361−225
361×5.94
= 0.38N / M M2
y=c+2 c1
c+c1( Df
3 )y=5.94+2× 0.38
5.94+0.38 (2253 )=79.51 MM
Lever arM ,a=d− y=1577−79.51=1497.49 MM
M r=σst × A st ×a
M r=¿ 200 ×11467 ×1497049
M r=3434.34 KN−M <3313 KN−M
Area of steel required
Areaof steelrequired= 3017 × 106
200 ×11467× 1497.49=10,443 mm2
No. of 32MM diameter bars = 10443/305 = 12.9∼13 bars
Check for local bond stress as per IRC code
Assume effective depth = 1950 - 60 = 1890MM
required= 605× 103
0.9 ×1890
No . of 32 MM bars= 355.673.142× 32
Hence atleast 4 bars are to be taken straight
Check for shear
1. NoMinal Shear stress at support= 605× 103
400 ×1890=0. 8 KN / M M 2, hence shear
reinforcement is necessary.
DEPT. OF CIVIL ENGG; UVCE Page 52
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
2. Nominal Shear stress at support= 173 ×103
400 × 1700=0.25 K N / M M 2, hence shear
reinforcement is not necessary.
3. Nominal Shear stress at support= 313 ×103
400 × 1700=0.46 K N / M M 2, hence provide shear
reinforcement.
Approximate distance from support at which shear stress is 0.5 N / M M2
= ½(9.455+7)=8.23M
Let us bend up 2 bars at a time at a spacing of 0.707a i.e. = 0.707x0.9x1605=1021.26MM
∴ Bend 2 bars at a time, spacing = 1020MM
If 5 bars are bent up, effective distance = 5 x 1.020 = 5.1M from support
Shear taken up by 4bent bars of 32MM = 4 x 805 x 200 x Sin(45)
= 455KN
∴ Net remaining shear at support for which shear reinforcement has to be provided
= 605-450=150KN
Spacing, Sv=2 ×78.5 ×200 × 1890
150 ×103=392.64 MM
Hence provide 10MM diameter at 180MM c/c at support i.e. up to 4.08M. After 4.08M only
2 bars will be effective.
At quarter span, remaining shear=410 ×103−455 ×103
2
¿182.5 kN
Spacing of two legged stirrups, 10MM diameter ¿2× 78.5× 200 ×1700
182.5 ×103
¿292.49 mm
DEPT. OF CIVIL ENGG; UVCE Page 53
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Hence provide 2L-10MM diameter bars at 200MM c/c from 4.08M to 5.1M
Beyond 5.1M no bent up bars are available. Therefore, shear at 3/8 span
i.e., 38
× 18.71=7.02=313 kN
Therefore spacing of 10MM diameter bars 2L ¿2× 78.5× 200 ×1700
313 ×103
¿170.54 mm
Therefore provide 2L 10MM diameter bars at 150MM c/c
From 7.02M to 8.02M Provide 2L 10MM diameter at 180MM c/c
For remaining distance provide 22L 10MM diameter at 300MM c/c
Summary:
Provide 10MM 2L diameter at 180MM c/c from support upto 4.08M
Provide 10MM 2L diameter at 200MM c/c from 4.08M to 5.1M
Provide 10MM 2L diameter at 180MM c/c from 5.1M to 7.02M
Provide 10MM 2L diameter at 180MM c/c from 7.02M to 8.02M
Provide 10MM 2L diameter at 180MM c/c for remaining length.
5.8Design of Inner Girder
Adopt,
400 × 1500× 2100 mm
i .e . B=2500 mm , b=400 mm, D=1500 mm ,
M=2596 × 106
∴ A st=2596 ×106
200 ×0.9 × 1605
DEPT. OF CIVIL ENGG; UVCE Page 54
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
¿8986mm2
Use 12 bars of 32MM diameter A st=9651 mm2
A st at quarter span= 2322 ×106
200× 0.9 ×1605=8039 mm2
Check for bond stress as per IRC
Shear at support ¿620 kN . Let effective depth be 1665MM
∴Shear= 620 ×103
1 ×0.9 ×1665=414
∴No .of 32 MM diameter barsrequired= 414π ×32
=4.1
However take 6 bars straight up to support.
Check for Shear
τ v at support= 620 ×103
300×1665=1.24 N / MM2 (Shear reinforcement needed)
τ v at centre= 105 × 103
300 ×1605=0.21 N / MM2 (No shear reinforcement)
τ v at38
span= 223 × 103
300 ×1605=0.46 N /MM 2 (However provide shear reinforcement)
Bend bars at 1021.26MM i.e. 1.02M = a
4 bars are effective at every section, hence ¿4 ×805 ×200 × 0.707
¿455 KN
Shear taken by 4 bars of 32 mm=455 KN
∴ Net remaining shear at support for which shear requirement is necessary
¿620−455=165 kN
DEPT. OF CIVIL ENGG; UVCE Page 55
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
∴Sv=2× 78.5× 200 ×1665
165 ×103=316.85 mm
Hence provide 10MM-2L bars at 180MM c/c from support up to 4 ×1.02=4.08 m from
support after 4.08M
At quarter span, ¿300−4552
=72.5 kN
∴Sv=2× 78.5× 200 ×1605
72.5 ×103=695.13 mm
Therefore, provide 2L-10MM diameter at 300 c/c from 4.08 to 5.1M.
Therefore, shear at 3/8 span=223KN.
∴Sv=2× 78.5× 200×1605
223 ×103=225.99 mm
Therefore, provide 2L-10MM diameter at 200 c/c up to 7.02.
From 7.02 to 8.02, 2L-10MM diameter at 300 c/c
For remaining 2L-10MM diameter at 300 c/c
Support to 4.8M→10MM-2L diameter at 180MM c/c
4.8M to 5.1M→10MM-2L diameter at 300MM c/c
5.1M to 7.02M→10MM-2L diameter at 200MM c/c
7.02 to 8.02 and remaining→10MM-2L diameter at 300MM c/c
5.9Design of cross-girder
DEPT. OF CIVIL ENGG; UVCE Page 56
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.9a: Load distribution on each girder Fig 5.9b: DL on Cross girder
i) Dead load
Cross girder are placed at 3.742M c/c
Dimension¿3 ×1275
∴Weight of rib¿0.3 ×1.275 m ×24=9.18 kN /m
Dead weight from slab and wearing coat¿7.05 kN /m2
∴Dead load on each cross girder¿2 ×[2.5 ×1.25 ×12 ]
¿3.125 ×7.05
¿22.1kN
Assuming this to be uniformly distributed, dead load per meter run of girder
¿ 22.12.5
=8.84 kN /m
Therefore, Total weight = 9.18+8.84 = 18.02KN / M
Assuming cross girders to be rigid, reaction on each longitudinal girder = 18.02× 5
3=31 KN
i) Live Load: Maximum bending Moment and shear force due to Class AA-
Tracked Loading
DEPT. OF CIVIL ENGG; UVCE Page 57
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Fig 5.9c: LL on the span
R=700 ×2.8424
=498 KN
Assuming cross girders to be rigid, reaction on each longitudinal girder = 498
3=166 KN
∴ Maximum bending Moment under the wheel load
Fig 5.9d: Max LL Cross girder
M=4983
×1.475=245 KN −M
Taking Impact factor = 1.1x245 =270KN-M
Dead Load bending Moment from 4.75M of support
¿316 ×1.475−18.02 × (1.475 × 1.475 )
2
¿45.73−19.61
= 26.12KN-M
Total Bending Moment = LL BM +DL BM
¿270+26.12=297 KN−M
Live Load Shear=4983
×1.1=183 KN
Section Design:
DEPT. OF CIVIL ENGG; UVCE Page 58
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
Total depth = 1500MM, Effective Depth = 1440MM
A st=297 × 106
200 ×0.9 ×1440=1146 MM2
Hence provide 3nos of 25MM Diameter bars, ∴Provide A st=1473 MM2
Shear Design:Nominal Shear τ v=
vbd = 214 × 103
300× 1440
¿0.49 N / M M 2<¿ τ max∴ok
But τ c = 0.34N
M M 2, hence provide shear reinforcement.
∴Sv=2× 78.5× 200 ×1440
214 × 103=211.28 MM
∴Provide 2 L−10 MM diameter barsat 200 MM C /C both at intermediate∧ends
CHAPTER 6
CONCLUSION
The analysis and design of Deck slab and T-Beam of a Bridge has been carried out manually
as per IRC guidelines and the following results have been noted.
1. Live Load due to Class AA Wheeled Vehicle produces the severest effect.
2. Shear Force due to Class AA Wheeled Vehicle is very high.
3. Bending Moment in the Inner girder is lesser than the Outer girder hence lesser
reinforcement in inner girder when compared to outer girder.
DEPT. OF CIVIL ENGG; UVCE Page 59
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
4. The design of the deck slab and T- beam has been manually done keeping in view the
above results.
REFERENCES
1) Indian Road congress, IRC: 6-2000, Standard Specifications and Code of Practice for
Road Bridges Section: II, Loads and Stresses, 4th revision.
2) Indian Road congress, IRC: 21-2000, Standard Specifications and Code of Practice for
Road Bridges Section: III, cement concrete (plain and reinforced), 3rd revision.
3) Dr. D. Johnson Victor, Essentials of Bridge Engineering, Oxford and IBH Publishing Co.
4) Dr. N. Krishna Raju, Design of Bridges ,Oxford and IBH Publishing Co. Pvt. Ltd
5) Mr. T.R. Jagadeesh and M.A. Jayaram, Design of Bridge Structures, Prentice Hall of
India Pvt. Ltd.
DEPT. OF CIVIL ENGG; UVCE Page 60
ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE
6) RCC Designs(Reinforced Concrete Structures) by Dr. B.C.Punmia, Ashok Kumar Jain,
Arun Kumar Jain, Tenth Edition, Laxmi Publications.
DEPT. OF CIVIL ENGG; UVCE Page 61
Recommended