analysis of deck slab and tee beam of a bridge

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


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



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



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



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



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.



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.



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 )



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



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)



= 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



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



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


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


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



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



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



M1 = 14.3x10-2 and M2= 9x10-2

Multiply by xy = 0.17

M1= 0.024KN-M and M2 = 0.02KN-M


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


M1= 0.024KN-M and M2 = 0.02KN-M


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



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


M1= 0.07KN-M and M2 = 0.0377KN-M


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


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


v = 2(u1+y) = 2(0.45+0.45) = 1.8


M1 = 9.0x10-2 and M2= 4.5x10-2


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



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


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



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


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


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


v = 2(u1+y) = 2(0.3+1.045) = 2.69


M1 = 7.0x10-2 and M2= 2.67x10-2


M1= 0.077KN-M and M2 = 0.03KN-M


v = 2(y) = 2(1.045) = 2.09


M1 = 12.0x10-2 and M2= 4.5x10-2




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


M1 = 8.0x10-2 and M2= 3.9x10-2


M1= 0.069KN-M and M2 = 0.034KN-M


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


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


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


M1= 1.0KN-M and M2 = 0.038KN-M


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



M1= 0.07KN-M and M2 = 0.0243KN-


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


M1= 0.09KN-M and M2 = 0.05KN-M


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


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




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



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 )



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



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



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)}



= 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



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:



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




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



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



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 ]



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



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



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



Fig 5.2c: ILD for BM at L/8


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


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



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


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




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


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



W 700KN

3.77+4.672

=4.22 M2954KN-M


W 700KN

4.4+3.562

=3.98 M2786KN-M



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


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



W 700KN

3.51+4.6782

=4.094 M2866KN-M


Fig 5.2d: ILD for BM at L/8


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



W 700KN

2.05+1.6562

=1.853 M1298KN-M


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



∑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



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



Shear Force at 3 /8 thspan=12 [ 5

8+ 5

8×

8.0911.69 ]×700

¿ 12 [ 1

2+0.44 ]×700

¿371 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



5.4Dead Load Bending Moment and Shear Force in Girder:

5.5.1 Live Load from Cantilever



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


∴ 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




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


- (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



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



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.



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



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 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



¿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



∴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



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



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:



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.



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.



6) RCC Designs(Reinforced Concrete Structures) by Dr. B.C.Punmia, Ashok Kumar Jain,

Arun Kumar Jain, Tenth Edition, Laxmi Publications.


analysis of deck slab and tee beam of a bridge

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