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 p ipeline. The obstacle to be crossed cro ssed 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 we ll as the weight of the t he people peo ple 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 Main Components Components of a 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. vi.

Cross beams or diaphragms, intermediate and end ones. Wearing coat

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The Substructure consists of the following structures: i)

Abutments at the extreme ends of the bridge.

ii) ii)

Piers at intermediate supports in case of multiple span bridges.

iii) 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 Types of Bridges 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 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 br idges are built built of steel or o r 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. DEPT. OF CIVIL ENGG; UVCE

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The Substructure consists of the following structures: i)

Abutments at the extreme ends of the bridge.

ii) ii)

Piers at intermediate supports in case of multiple span bridges.

iii) 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 Types of Bridges 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 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 br idges are built built of steel or o r 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. DEPT. OF CIVIL ENGG; UVCE

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1.3 Para Param meters governing overning choi choice of Superstr Superstruc uctture: 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 t he 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 t ype of traffic

iv.

Whether used for navigation purposes

v. vi.

Climatic conditions 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


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xv.

Aesthetic consideration

1.4 General guidelines for for analy analysis and Design of a of a Bridge Struc tructture 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 t he 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 a nd sizes o f certain cert ain members as 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 o f the project.

1.5 General Proced ocedu ure for for Design of Superstr Superstruc uctture of a 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. DEPT. OF CIVIL ENGG; UVCE

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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 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 Trans Transv verse Distribu istributio tion of Loads 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 r igid

ii.

an orthotropic plate

iii.

an assemblage of thin plate e lements 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. DEPT. OF CIVIL ENGG; UVCE

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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 procedu re: 1. Deck Slab Design 2. Design of Longitudinal Girders and Cross Girders.


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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 Maintenan ce R equirements:

Concrete Grades and Reinforcement 1. Concrete: For RC Deck Slab(M-30) 2. R einforcement: 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.


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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 2

a) Weight of Deck Slab

= 0.225 X 24 = 5.4 KN/M

b) Weight of Wearing Course

= 0.075 X 22 = 1.65 KN/M

c) Total Weight

= 7.05 KN/M (say 7.1 KN/M )


2

2

2

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3.2. Analysis of Inner Panels

a) S lab 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 -2

M1 = 8.2 X 10

-2

and M2 = 3.12 x 10

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


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3.3 LOADS DUE TO IRC CLASS AA TRACKED VEHICLE 3.3.1 Live Load Bending Moment due to IRC Class AA Track ed 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

୳= ଵ = ୆ ଶǤହ ଶǤହ = = ଷǤ଻ସଶ

0.4

K

and

୚ = ଷǤ଻ହ ୐ ଷǤ଻ସଶ

=1

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


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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 Track ed 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 Track ed 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 o f the rib i.e. the load is kept at

ଵǤସହ = 0.705M . ଶǤ଴

Dispersion in the direction of span V = x + 2 (D+H) = 0.85 + 2 (0.075 + 0.225) = 1.45M


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୅୐

Effective width of slab = K x A x [1- ] + bw

୆ = ଷǤସସଶ ୐ ଶǤଵ = 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)

ൌ ͲǤʹͳͷ Ǣ ൌ ͲʹǤͻ ൌ ͲǤͲ͹ͷ Ǣ ൌ ͲǤͶͷ ͳ ൌ ͲǤͶͷ Ǣ ͳ ൌ ͲǤ͵ ൌ ʹሺ ͳ൅ ሻ ൌ ʹሺͲǤͶͷ൅ ͲǤͲ͹ͷሻ ሻ ൌ ʹሺͲǤ͵ ൅ ͲǤͶͷሻ ൌ ʹͳǤሺͳ൅          ൌ ʹǤͲͳͷ ൌ ͲǤͷ Step 1: Find M1 and M2

u

=

v

= 1.5M

1.05M


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Fig 4: Position of wheel load for Class AA wheeled vehicle

       ൌ ͵ǤͳǤͶͷͶʹ ൌ ͲǤͶͶ        ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ

From Pigeaud¶s Curves by interpolation

      ୆୳ ൌ ͲǤͷ       ୴୐ ൌ ͲǤͶͶ

k=0.61, k=0.6,

,

M1= 12x10-2, -2

M2= 5.5x10-2 -2

k=0.61,

M1= 12x10 ,

M2= 5.6x10

k=0.707,

M1= 11.8x10-2, M2= 6.5x10-2

M1= 0.12x10-2 M2= 0.056x10-2

ሺͳ൅ሻሺͳ൅ሻൌሺͲǤͶͷ൅ͲǤͲ͹ͷሻ൅ ሺͲǤ͵൅ͲǤͶͷሻൌͲǤͶ ͳൌͲǤͳʹͳͲǦʹ ʹൌͲǤͲͷ͸ͳͲǦʹ ʹǣͳ ʹ        ൌ ʹ ൌ ʹሺͲǤͲ͹ͷሻ ൌͲǤͳͷ ൌ ʹͲǤሺͳͲͷǤͶͷሻൌͲǤͻ  ൌ ʹǤͲͳǤͻ ൌ ͲǤͲ͹ʹ  ൌ ͵ǤͶͶʹ ൌ ͲǤʹ͸  ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ ǯ୳୴ ൌͲǤ͸ͳǡ      ୆ ൌ ͲǤͷǡ      ୐ ൌ ͲǤͶͶ ൌͲǤ͸ǡ      ͳൌʹͳͳͲǦǦʹʹǡ  ʹൌͻͳͲǦʹ Ǧʹ ൌͲǤ͸ͳǡ     ͳൌʹͳͳͲǦʹǡ  ʹൌͻǤʹͳǦͲʹ    ൌͲǤ͹Ͳ͹ǡ    Ǧʹ ͳൌʹʹͳͲ ǡ ǦʹʹൌͳͳͳͲ    ͳൌʹͳͳͲ ʹൌͻǤʹͳͲ ൌͲǤͲ͹ͷͲǤͶͷൌͲǤͲ͵Ͷ       ͳൌͲǤ͹ͳͶͳͲǦʹ ʹൌͲǤ͵ʹͳͲǦʹ DEPT. OF CIVIL ENGG; UVCE

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͵ǣͳ ʹ        ൌʹሺ ͳ൅ ሻ ൌ ʹሺͲǤͶͷ൅ ͲǤͲ͹ͷሻ        ൌͳǤͲͷ        ൌ ʹͳǤͲൌͷͲǤͻ         ൌ ʹǤͳ ൌ ͲǤͷ        ൌ ͵ǤͲͶǤͻͶʹ ൌ ͲǤʹ͸        ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ ǯ୳୴ ൌͲǤ͸ͳǡ      ୆ ൌ ͲǤͷǡ      ୐ ൌ ͲǤͶͶ ൌͲǤ͸ǡ      ͳൌͳ͵ͳͲǦǦʹʹǡ   ʹൌͺǤ͸ͳͲǦǦʹʹ   ൌͲǤ͸ͳǡ     ͳൌͳ͵ͳͲ ǡǦʹ  ʹൌͺǤͷͳͲ Ǧʹ   ൌͲǤ͹Ͳ͹ǡ    Ǧʹ ͳൌͳʹǤ͸ͳͲ Ǧǡʹ  ʹൌͺǤ͸ͳͲ    ͳൌͳ͵ͳͲ ʹൌͺǤ͸ͳͲ ሺͳ൅ሻൌͲǤͶͷሺͲǤͶͷ൅ͲǤͲ͹ͷሻൌͲǤʹͶ ͳൌͳ͵ͳͲǦʹ ǡ ʹൌʹǤͲ͹ͳͲǦʹ Ͷǣͳ ʹ           ൌ ʹሺ ሻ ൌ ʹሺͲǤͲ͹ͷሻൌͲǤͳͷ      ൌ ʹͲሺǤͳͳ൅ͷ ሻ ൌ ʹሺͲǤ͵ ൅ ͲǤͶͷሻൌͳǤͷ       ൌ ʹǤͳ ൌ ͲǤͲ͹ʹ       ൌ ͵ǤͳǤͶͷͶʹ ൌ ͲǤͶͶ       ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ ǯ୳୴ ൌͲǤ͸ͳǡ      ୆ ൌ ͲǤͷǡ      ୐ ൌ ͲǤͶͶ -2

k=0.6,

M1= 16.5x10 ,

k=0.61,

M1= 16.7x10 ,

k=0.707,

M1= 18x10 ,

-2


-2

-2

M2= 5.5x10

-2

M2= 6.0x10

-2

M2= 6.0x10

Page 14

ANALYSIS AND DESIGN OF DECK SLAB AND T-BEAM OF BRIDGE -2

M1= 16.7x10

-2

M2= 6.0x10

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

 ൌͲǤͲͲͳͶǦௐ ൌ௨ൈଵ௩ଵ ሺͳ൅ͲǤͳͷʹሻ                      ൌ ʹʹௐǦ ൌ௨ൈଵ௩ଵ ሺͳ൅ͲǤͳͷʹሻ                  ൌ ͷǤʹ͸Ǧ M1

ML due to single load

Applying 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) Eff ect 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 o f 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

      ୆୳ ൌ ଴ǤଶǤସଵହ ൌ ͲǤʹʹ        ൌ ͵ǤͳǤͶͷͶʹ ൌ ͲǤͶͶ        ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ Fig 4.1: Position of wheel load for Class AA wheeled vehicle


Page 15


From Pigeaud¶s Curves by interpolation

      ୆୳ ൌ ͲǤͷ       ୴୐ ൌ ͲǤͶͶ

k=0.61,

,

k=0.6,

M1= 12.8x10 ,

-2

-2

M2= 4.4x10

-2

-2

k=0.61,

M1= 12.9x10 ,

M2= 4.6x10

k=0.707,

M1= 13.8x10 , M2= 5.9x10

-2

-2

-2

-2

M1= 13x10

M2= 4.6x10

MB due to single load =

ௐ ௨ൈଵ௩ଵ ሺ

M1+0.15 M2)

=43KN-M ML due to single load =


M1+0.15 M2)

=20.5KN-M For larger load,

     ୆୳ ൌ ଴ǤଶǤସଵହ ൌ ͲǤʹʹ       ൌ ͵ǤͳǤͶͷͶʹ ൌ ͲǤʹ͸       ൌ ͵ǤʹǤͶͳͶʹ ൌ ͲǤ͸ͳ       ୆୳ ൌ ͲǤʹʹ       ୴୐ ൌ ͲǤʹ͸ k=0.61,

,

k=0.6,

M1= 16.0x10-2,

M2= 7.0x10-2

k=0.61,

M1= 16.2x10 ,

k=0.707,

M1= 18.0x10-2, M2= 10.5x10-2

-2

-2

M2= 7.4x10

M1= 16.2x10-2 M2= 7.4x10-2 MB due to single load =


M1+0.15 M2)

=32.5KN-M ML due to single load =


M1+0.15 M2)

=18.5KN-M

Net Moment MB1 = 43-32.5=10.5KN-M ML1 = 20.5-18.5 = 2KN-M

i)

Eff ect 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. DEPT. OF CIVIL ENGG; UVCE

Page 16


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 -2

-2

M1 = 9.0x10

and M2= 4.6x10

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 -2

M1 = 14.3x10

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 -2

M1 = 12.5x10

-2

and M2= 5.5x10

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 DEPT. OF CIVIL ENGG; UVCE

Page 17


= 0.008KN-M ML = 0.03+0.02-0.026-0.015 = 0.01KN-M MB2 due to single load =

ଶௐ ሺM +0.15 M ) ௨ൈଵ௩ଵ 1

2

=5.28KN-M ML2 due to single load =

ଶௐ ሺM +0.15 M ) ௨ൈଵ௩ଵ 1

2

=6.23KN-M

ii) Eff ect 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 -2

-2

M1 = 7.5x10

and M2= 4.0x10


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 -2

M1 = 6.63x10

-2

and M2= 5.3x10

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 DEPT. OF CIVIL ENGG; UVCE

Page 18


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 =

ଶௐ ሺM +0.15 M ) ௨ൈଵ௩ଵ 1

2

=15.36KN-M ML3 due to single load =

ଶௐ ሺM +0.15 M ) ௨ൈଵ௩ଵ 1

2

=8.3KN-M

Final bending Moment for case ii MB= MB1+ MB2+ MB3

ML = ML1+ ML2+ ML3

MB

ML = 16.53KN-M

= 31.14KN-M

Case 3: Loads placed as per f igure:

i) Eff ect 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) Eff ect 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


Page 19


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) Eff ect 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) Eff ect 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) Eff ect 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 DEPT. OF CIVIL ENGG; UVCE

Page 20


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 -2

M1 = 12.0x10

-2

and M2= 4.5x10

Multiply by xy = 0.39 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 -2

-2

M1 = 8.0x10 and M2= 3.9x10 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) Eff ect 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 -2

-2

M1 = 5.8x10 and M2= 2.3x10 DEPT. OF CIVIL ENGG; UVCE

Page 21



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 -2

-2

M1 = 8.5x10 and M2= 3.0x10 Multiply by (xy) = 0.81

M1= 0.07KN-M and M2 = 0.0243KNStep3: 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 -2

-2

M1 = 6.8x10 and M2= 3.8x10 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 -2

-2

M1 = 7.4x10 and M2= 3.9x10 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 DEPT. OF CIVIL ENGG; UVCE

Page 22


= 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

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 

୅ ൌ ʹ ሺȽ ଷ െ ʹൈȽ ଶ ൅ ʹሻ


Page 23












ൌ୅ ʹͲʹǤͺͶ ሺͲǤʹͶ ଷ െ ʹൈ ͲǤʹͶ ଶ ൅ ʹሻ ୅ ൌ ʹͲǤʹͶ ൌ୅ ʹ ሺʹൈ Ƚଶ െ Ƚ ଷሻ ʹͲǤͺͶ ୆ ൌ ʹ ሺʹൈ ͲǤʹͶ ଶ ൅ ͲǤʹͶ ଷሻ ୆ ൌ ͳǤͲͷ

Fig 4.4b: Position of wheel load for Class AA wheeled vehicle





  

୅ ൌ ቆ

ଶ Ⱦ Ƚଶ Ƚ ଷ ଶ ʹൈȾ െ ͵ൈȾ ൅ െ ൅ ͳቇ

ʹ Ͷ ଶ ൈ ͲǤ͵ͳ ͲǤͶଶ Ǥ Ͷ  Ͳ ଷ ଶ ୆ ൌ ͸ʹǤͷ ቆʹൈ ͲǤ͵ͳ െ ͵ൈ ͲǤ͵ͳ ൅ ʹ െ Ͷ ൅ ͳቇ ୅ ൌ Ͷ͹Ǥͳ ୆ ൌ  െ ୅ ୅ ൌ ͸ʹǤͷ െ Ͷ͹Ǥͳ ൌ ͳͷǤͶ 


Page 24


Fig 4.4c: Position of wheel load for Class AA wheeled vehicle 











ൌ୆ ʹ ሺȽ ଷ െʹൈȽ ଶ ൅ ʹሻ ൌ୆ ͸ʹǤʹͷ ሺͲǤͶ͵ ଷ െʹൈ ͲǤͶ͵ ଶ ൅ʹሻ ୆ ൌ ͷ͵ǤͶ͵ ୅ ൌ ʹ ሺʹൈȽ ଶ െ Ƚ ଷሻ ୅ ൌ ͸ʹǤʹͷ ሺʹൈ ͲǤͶ͵ ଶ െ ͲǤͶ͵ ଷሻ ୅ ൌ ͻǤͳ

The values of effective width, reactions and shear force are tabulated below Load (KN)

X(M)

e(M)

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


(KN)

(KN)

(KN)

(KN)

W=0.15+(2 x 0.075) = 0.3M

Page 25


=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


Page 26


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) Eff ect 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) Eff ect 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


Page 27


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

ൌ୅ ʹ ሺȽ ଷ െ ʹൈȽ ଶ ൅ʹሻ ୅ ൌ ହ଻ଶ ሺͲǤͷʹͶ ଷ െ ʹൈ ͲǤͷʹͶ ଶ ൅ ʹሻ ൌ୆ ʹ ሺʹൈȽ ଶ െ Ƚ ଷሻ ୆ ൌ ୛ଶ ሺʹൈ ͲǤͷʹͶ ଶ െ ͲǤͷʹͶ ଷሻ  ൌͳǤͷ  ସହǤଵǤଷସଵହହ 

= 45.54KN





= 11.5KN



= 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


Page 28


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 adop t 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


Page 29


CHAPTER 4 DESIGN OF DECK SLAB Grade of concrete = M30 Grade of Steel = Fe415

ൌ ͳͲΤଶ ൌ ʹͲͲΤଶ

௖௕௖

௦௧

, Modular ratio, M = 10

4.1 Constants:

ൌ ͳ൅ ͳǤ ൌ ͳ൅ ͳͲͳʹͲͲൈͳͲ ൌ ͲǤ͵Ͷ ൌ ቀͳെ ͵ቁ ൌ ൬ͳെ ͲǤ͵͵Ͷ൰ ൌ ͲǤͺͻ ൌ ͲǤͷ ൈ ൈ ൈ ൌ ͲǤͷ ൈ ͳͲ ൈ ͲǤ͵Ͷ ൈ ͲǤͺͻ ൌ ͳǤͷͳ͵ ௦௧

௖௕௖

௖௕௖

4.2 Check for depth (MB):

଺ ͵͸ ൈ ͳ Ͳ ୮୰୭୴୧ୢୣୢ ൌ ඨ ൈ ൌ ඨ ͳǤͷͳ͵ ൈ ͳͲͲͲ ൌ ͳͷͶǤʹͷ ୮୰୭୴୧ୢୣୢ ൌ ʹʹͷ

୮୰୭୴୧ୢୣୢ ൐ ୰ୣ୯୳୧୰ୣୢ

Provide 225MM overall depth using 40MM

୰ୣ୯୳୧୰ୣୢ ൌ ʹʹͷെ ͶͲ െ ͺ ൌ ͳ͹͹ ଺  ͵͸ ൈ ͳ Ͳ  ൌ ɐୱ୲ ൌ ʹͲͲ ൈ ͲǤͺͻ ൈ ʹʹͷ ൌ ͺͻͲଶ୆ DEPT. OF CIVIL ENGG; UVCE

Page 30


଺  ʹʹ ൈ ͳ Ͳ  ൌ ɐୱ୲ ൌ ʹͲͲ ൈ ͲǤͺͻ ൈ ʹʹͷ ൌ ͷͷͲଶ୐ ൌ௩ ͳͲͲͲͺͻൈͲ ʹͲͳ ௩ ൌ ʹʹͷ Τ Hence provide 16MM diameter bars at 150MM c/c

ൌ ͳͲͲͲͷͷൈͲ ͳͳ͵ ௩ ൌ ʹ Ͳͷ Τ ௩

Hence provide 12MM diameter bars at 150MM c/c


Page 31


CHAPTER 5 LONGITUDINAL GIRDER AND CROSS GIRDER DESIGN 5.1 R eaction Factor Bending Moment in Longitudinal Girders by Courbons¶s

Method

5.1.1 Class AA Track ed Vehicle

Fig 5: Position of Class AA Track ed Vehicle for obtaining reaction f actors Minimum Clearance Distance: 1.2 + 0.85/2 = 1.625M

 ൌ ͳǤͳ ǡ    ൌʹ σ ଶ ൌ ሺʹǤͷሻଶ൅ሺͲሻଶ ൅ ሺʹǤͷሻଶ ൌ ʹሺʹǤͷሻଶ ൌ ͳʹǤͷ ǡ  ൌ ʹǤ͸ǡ ൌ Ͳ ǡ ൌ஺ σ ൤ͳ൅ σ ଶ൨ ஺ ൌ Ͷ͵ ൤ͳ൅ ͵ ൈͳǤͳൈʹǤͷ ʹሺʹǤͷሻଶ ൨ ஺ ൌ ͲǤͷͷ͵͸ ஻ ൌͲǤ͵͵͵͵ DEPT. OF CIVIL ENGG; UVCE

Page 32


5.1.2 Class A Loading

Fig 5.1: Position of Class A Vehicle for obtaining reaction f actors Minimum Clearance Distance: 0.15 + 0.25 = 0.4M

ൌ ͲǤͺ ǡ ൌ ʹ σ ଶ ൌ ሺʹǤͷሻଶ൅ሺͲሻଶ ൅ ሺʹǤͷሻଶ ൌ ʹሺʹǤͷሻଶ ൌ ͳʹǤͷ ǡ  ൌ ʹǤ͸ǡ ൌ Ͳ ǡ ஺ ൌ σ௡ ቂͳ൅ σ௫௡௘௫మቃ                                                                                  ஺ ൌ Ͷ͵ ൤ͳ൅ ͵ ൈʹሺʹǤͲǤͺͷൈʹǤͷ ሻଶ ൨ ൌ Ͷ͵ ሾͳ൅ ͲǤͶͺሿ ൌ Ͷ͵ ሾͳǤͶͺሿ ൌ ͳǤͻ͹͵Ͷ ൌ ͲǤͻͺ͸͹ ஻ ൌ Ͷ͵ ሾͳ൅ Ͳሿ DEPT. OF CIVIL ENGG; UVCE

Page 33


ൌ஻ Ͷ͵ ൌ Ͷ͵ ൈ ʹ ൌ ͲǤ͸͹  ஻ ൌ ͲǤ͸͹ ൌ ଺ାସǤହ ൌ ଺ାସǤଵ଼Ǥହ଻ଵ ൌ ͲǤͳͺʹ 5.2 Maximum 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


Page 34


    Ǥ       ͳ       ʹ       ͵       Ͷ       ͷ       ͸

 ʹ͹

         ͵Ǥͺͺͷ ൈͶǤ͸͹ͺ ൌ ͳǤͻ͵ ͻǤ͵ͷͷ ͷʹǤͳͳǦ ͶǤͻͷͷ ൈͶǤ͸͹ͺ ൌ ʹǤͶͺ ͻǤ͵ͷͷ ʹ͹ ͸͹Ǧ Ǥͳͷͷ ൈͶǤ͸͹ͺ ൌ ͶǤͳ ͺ ͻǤ͵ͷͷ ͳͳͶ Ͷ͸ͺǦ ͳͳͶ Ͷ Ͷͷ͸Ǧ ͷǤͲͷͷ ൈͶǤ͸͹ͺ ൌ ʹǤͷ͵ ͻǤ͵ͷͷ ͸ͺ ͳ͹͵Ǧ ʹǤͲͷͷ ൈͶǤ͸͹ͺ ൌ ͳǤͳ ͻǤ͵ͷͷ ͸ͺ ͹ͷǦ  ͳʹͻͳǤͳͳǦ ǡǡ   ൌͳʹͻͳǤͳͳൈͳǤͳͺʹൈ ͲǤͻͺ͸͹ ൌ ͳͷͲ͸ െ  ൌͳʹͻͳǤͳͳൈͳǤͳͺʹൈ ͲǤ͸͹ ൌ ͳͲʹ͵ െ 

5.1.2 Maximum Live load bending Moment at 3L/8: The below figure shows the Influence

Line diagram

ͷǤʹǣ͵Ȁͺ DEPT. OF CIVIL ENGG; UVCE

Page 35


Load No.

Load Value

W1

27kN

W2

27KN

W3

114KN

W4

114KN

W5

68KN

W6

68KN

W7

68KN

Ordinate

ʹǤ͹ͳ͸ʹͷ ൈͶǤ͵ͻ ൌ ͳǤ͹ ͹ǤͲͳ͸ʹͷ ͵Ǥͺͳ͸ʹͷ ൈͶǤ͵ͻ ൌ ʹǤ͵ͻ ͹ǤͲͳ͸ʹͷ               ͶǤ͵ͻ ͳͲǤͷ ൈͶǤ͵ͻ ൌ ͶǤͲ ͳͳǤ͹ ͸Ǥͳͻ͵͹ͷ ൈͶǤ͵ͻ ൌ ʹǤ͵͵ ͳͳǤ͹ ͵Ǥʹ ൈͶǤ͵ͻ ൌ ͳǤʹ ͳͳǤ͹ ͲǤͳͳͳǤͻ͵͹͹ͷ ൈͶǤ͵ͻ ൌ ͲǤͲ͹͵

Moment 46KN-M

64.53KN-M

501KN-M

456KN-M

159KN-M

82KN-M

5KN-M =1314KN-M

Bending Moment, BM, including Impact Factor and Reaction factor for

 ൌͳǤͳͺʹൈ ͲǤͻͺ͸͹ ൈͳ͵ͳͶ ൌ ͳͷ͵ʹǤͷ െ   ൌͳǤͳͺʹൈ ͲǤ͸͹ ൈͳ͵ͳͶ ൌ ͳͲͶͳ െ 5.1.3 Maximum Live load bending Moment at L/8: The below figure shows the Influence

Line diagram


Page 36


Fig 5.2c: ILD for BM at L/8

Load No.

Load Value

W3

114KN

W4

114KN

W5

68KN

W6

68KN

W7

68KN

W8

68KN

Ordinate

             ʹǤͲͷ       ͳͷǤͳ͸Ǥͳ͵͹͹ ൈʹǤͲͷ ൌ ͳǤͻ       ͳͳ͸ǤͲǤ͵ͺ͹͹ ൈ͵ǤͲͷ ൌ ʹǤͲʹͷ       ͳ͸Ǥ͹Ǥͺ͵͹͹ ൈʹǤͲͷ ൌ ͲǤͻͻ       ͳ͸ǤͶǤͺ͵͹͹ ൈʹǤͲͷ ൌ ͲǤ͸ͳ       ͳ͸ǤͳǤͺ͵͹͹ ൈʹǤͲͷ ൌ ͲǤʹͶ

Moment 234.00KN-M

217.00KN-M

138.00KN-M 67.32 KN-M 42.00KN-M

16.40KN-M TOTAL 715.00KN-M


 ൌͳǤͳͺʹൈ ͲǤͻͺ͸͹ ൈ ͹ͳͷ ൌ ͺ͵ͶǤͲͲ െ   ൌͳǤͳͺʹൈ ͲǤ͸͹ ൈ ͹ͳͷ ൌ ͷ͸͸ǤʹͶ െ  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 37


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

W1

27kN

W2

27KN

W3

W4

W5

W6

Ordinate

Moment

        ͵ǤͺǤ͵ͺͻͷͻͷ ൈͶǤ͸͹ ൌ ͳǤ͹ͺʹ ͶǤͶͻͷ ൈͶǤ͸͹ ൌ ʹǤ͵ͻ ͺǤͺͻͷ ͹ͺǤǤͺ͸ͻͷͻͷ ൈͶǤ͸͹ ൌ ͶǤͲͶ 

114KN

ͶǤ͸͹

114KN

ͶǤͷͻͷ ൈͶǤ͸͹ ൌ ʹǤʹ ͻǤͺͳͷ

68KN

ͳǤͷͻͷ ൈͶǤ͸͹ ൌ ͲǤ͹͸ ͻǤͺͳͷ

68KN

48.20KN-M

64.00KN-M

461KN-M

533KN-M

150KN-M

52KN-M

=1308.2KNTotal


M

Page 38



 ൌͳǤͳͺʹൈͳǤ ͺʹൈͳ͵ ͲͺǤʹ ൌ ͺ͵ͶǤͲͲ  ൌͳǤͳͺʹൈ ͲǤ͸͹ ൈͳ͵ͲͺǤʹ ൌ ͳͲ͵͸

KN-M

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

Load No.

W

Load Value

700KN

Ordinate

͹         ͵Ǥ͹͹ ൅ͶǤ͸ ʹ ൌ ͶǤʹʹ

Moment

2954KN-M


 ൌͳǤͳͲ ൈ ͲǤͷ͵͵͸ൈʹͻͷͶ ൌ ͳͺͲͲǤͲͲ  ൌͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈʹͻͷͶ ൌ ͳͲͺ͵

KN-M

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 39


Load No.

Load Value

W

700KN

Ordinate

        ͶǤͶ൅͵Ǥͷ͸ ʹ ൌ ͵Ǥͻͺ

Moment

2786KN-M


 ൌͳǤͳͲ ൈ ͲǤͷ͵͵͸ൈʹ͹ͺ͸ ൌ ͳ͸ͻ͹ǤͲͲ  ൌͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈʹ͹ͺ͸ ൌ ͳͲʹʹ

KN-M

KN-M

5.3.3 Bending Moment at L/4 of the span

Fig 5.2c: ILD for BM at L/4

Load No.

Load Value

W

700KN

Ordinate

 ͵Ǥͷͳ൅ʹͶǤ͸͹ͺ ൌ ͶǤͲͻͶ

Moment

2866KN-M


 ൌͳǤͳͲ ൈ ͲǤͷ͵͵͸ൈʹͺ͸͸ ൌ ͳ͹Ͷ͸ǤͲͲ  ൌͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈʹͺ͸͸ ൌ ͳͲͷͳ

KN-M

KN-M


Page 40


5.3.4 Bending Moment at L/8 of the span

Fig 5.2d: ILD for BM at L/8

Load No.

Load Value

W

700KN

Ordinate

ʹǤͲͷ൅ ͳǤ͸ͷ͸ ൌ ͳǤͺͷ͵ ʹ

Moment

1298KN-M


 ൌͳǤͳͲ ൈ ͲǤͷ͵͵͸ൈͳʹͻͺ ൌ ͹ͻͳǤͲͲ  ൌͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈͳʹͻͺ ൌ Ͷ͹͸

KN-M

KN-M

5.3 Live 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 der ived 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


Page 41


Fig 5.3: Class AA tracked Wheel load po sition for Live loaf shear force

ͳ ൌ ͳǤʹǤ͸ͳ͹ͷ  ൌ ͲǤͺ ʹ ൌ ͲʹǤǤͶͳʹͷ ൅ ͳǤʹǤ͹ͳʹͷ ൌ ͳǤͲ͵ ͵ ൌ ͲʹǤǤ͵ͳ͹ͷ  ൌ ͲǤͳͺ Reactions at end of each Longitudinal Girder due to transfer of these loads at 1.8M from left support R A' = 0.374P

R D' = 0.347P

R B' = 0.535P

R E' = 0.495P.

R C' = 0.093P

R F' = 0.087P.

The load R D¶, R E¶, R F¶ should be transferred to the cro ss girders as per Courbon¶s Theory

σ ൌ ͲǤ͵Ͷ͹ ൅ ͲǤͶͻͷ ൅ ͲǤͲͺ͹ σ ൌ ͲǤͻʹͻ DEPT. OF CIVIL ENGG; UVCE

Page 42


ʹʹǤ͸ൈʹǤ͸ ൌ஽ σ ൤ͳ൅ σ ଶ൨ ൌ஽ ͲǤͻ͵ʹͻ ൤ͳ൅ ͵ ൈ ͲʹሺʹǤǤͷͺ͹͹ͳሻଶൈʹǤͷ൨ ൌ ͲǤͶ͸Ͷ ൌ σ ൤ͳ൅ σ ଶ൨ ൌ ͲǤͻ͵ʹͻ ሾͳ൅ Ͳሿ  ൌ ͲǤ͵ͳ 2

W =

These reactions

஽

and

points of total span.



th

act as point loads on outer girder and inner girders at their 1/5

஺ ൈͳͺǤ͹ͳ ൌ ஽ ൈͳͶǤͻ͸ͺ                     ஺ ൈͳͺǤ͹ͳ ൌ ஺ ൈͳͶǤͻ͸ͺ              ஺ ൌ ͲǤͺ ஽                                    ஻ ൌ ͲǤͺ  ஺ ൌ ͲǤͺൈ ͲǤͶ͸Ͷ ஽                           ஻ ൌ ͲǤͺ ൈ ͲǤ͵ͳ ஺ ൌ ͲǤ͵͹ͳʹ                                  ஻ ൌ ͲǤ͵͹ͳʹ          ൌ  ୅ൌ ͲǤ͵͹ͳʹ ൅ ͲǤ͵͹ͳʹ     R A' +



               ൌ  ୆ൌ ͲǤͷ͵ͷ ൅ ͲǤʹͶͺ  ǡൌ͵ͷͲ ൌ ͳǤͳͲ ൈ ͲǤ͹Ͷͷʹൈ͵ͷͲ ൌ ʹͺͶ ൌ ͳǤͳͲ ൈ ͲǤ͹ͺ͵ൈ͵ͷ Ͳ ൌ ͵Ͳʹ R B' +



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


Page 43




5.4.1 Shear Force at Mid span:

Fig 5.4a: ILD for SF at L/2

 ൌͳʹ ൤ͳʹ ൅ ͳʹ ൈ ͷǤͻǤ͵͹ͷͷͷͷ൨ൈ ͹ͲͲ                          ൌ ͳʹ ൤ͳʹ ൅ ͲǤ͵ͳൈ൨ ͹ͲͲ                           ൌ ʹ ͺͶ ൌ ͳǤͳͲ ൈ ͲǤͷͷ͵͸ൈʹͺͶ ൌ ͳ͹͵ ൌ ͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈʹͺͶ ൌ ͳͲͷ 5.4.2 Shear Force at 3/8th span:

Fig 5.4b: ILD for SF at 3/8

͵Ȁͺ ൌͳʹ ൤ͷͺ ൅ ͷͺ ൈ ͳͳǤͺǤͲ͸ͻͻ൨ൈ ͹ͲͲ                          ൌ ͳʹ ൤ͳʹ ൅ ͲǤͶͶൈ൨ ͹ͲͲ DEPT. OF CIVIL ENGG; UVCE

Page 44


                         ൌ ͵͹ͳ ൌ ͳǤͳͲ ൈ ͲǤͷͷ͵͸ൈ͵͹ͳ ൌ ʹʹ͸ ൌ ͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈ͵͹ͳ ൌ ͳ͵͸ 5.4.3 Shear Force at 1/4th span:

th

Fig 5.4c: ILD for SF at 1/4 span

ͳȀͶ ൌ ͳʹ ൤͵Ͷ ൅ ͵Ͷ ൈ ͳͳͶǤͲǤͲͶ͵ʹ͵ʹ൨ൈ ͹ͲͲ                          ൌ ͳʹ ൤ͳʹ ൅ ʹǤʹ͵ൈ൨ ͹ͲͲ                          ൌ Ͷͷͺ ൌ ͳǤͳͲ ൈ ͲǤͷͷ͵͸ൈͶͷͺ ൌ ʹ͹ͻ ൌ ͳǤͳͲ ൈ ͲǤ͵͵͵͵ൈͶͷͺ ൌ ͳ͸ͺ 5.4 Dead Load Bending Moment and Shear Force in Girder: 5.5.1 Live Load f rom Cantile ver

ሻ ǣ  ͳǤʹǤ ͲǤ͹ͷǡ   Ǥ  ሻ  ǣ DEPT. OF CIVIL ENGG; UVCE

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Fig 5.4: Position of Class A loading for max BM

Ǥ  ൌͳǤͲͷȂሺͲǤͳͷ൅ͲǤʹሻ ൌͲǤ͸ͷ ൌͲ Ǥͷ ൅ሺʹൈͲǤʹ͹ͷሻൌͳǤͲͷ  ൌ ͷ ͹ ൈ ൤ͲͳǤǤͺͲ͹ͷͷ൨     ǡ  ൌ ͳǤʹ ൈ ൅  ǡൌǤǤൌͲǤͺ͹ͷȀʹ        ൌ ൌ൅ʹൈ   ൌͲǤʹͷ൅ሺʹൈ ͲǤͲ͹ͷሻ  ൌͲǤͶͲǤͺ͹ͷ ൌ ൤ ͳǤͲͷ ൨ ൌ ͲǤͶ͵͹ͷ ൌ ሺͳǤʹ ൈ ͲǤͶ͵͹ͷሻ ൅ ͲǤͶ        ൌ ͲǤͻʹͷ ൌͳǤͷ  ൌ ൬ͳǤͷ ൈ ͲͶǤͻ͹ʹͷǤͷ ൰൅ ͲǤͺʹ͹ͷ DEPT. OF CIVIL ENGG; UVCE

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  Ȁሺሻ ሺͲǤͷൈͳൈͳൈʹͶ ሻ ൌ ͳʹ ሺͲǤͲ͹ͷൈͳǤ Ͳͷൈʹʹ ሻ ൌ ͳǤ͹Ͷ ሺͳǤͷͷൈ ͲǤʹൈʹͶ ሻ ൌ ͹ǤͶͶ                     ʹͳǤͳͺ Crash Barrier Wearing Coat Slab

Total

Ǥ ሺǦሻ ሺሻ ͳǤͷ ͳͺǤͲͲ ͲǤͷʹͷ ͲǤͻʹͶ ͷǤͺͲ ͲǤ͹͹ͷ ʹͷǤͲͲǦ

Design Bending Moment = 25 + 33.7 = 58.7KN-M Design Shear Force

= 21.18 + 77 = 98.18KN

Check for depth (MB):

଺ ͷ Ǥ ൈ ͳ ͺ ͹  Ͳ ୮୰୭୴୧ୢୣୢ ൌ ඨ ൈ ൌ ඨ ͳǤͷͳ͵ ൈ ͳͲͲͲ ൌ ͳͷͶǤʹͷ ୮୰୭୴୧ୢୣୢ ൌ ͳͻ͸

୮୰୭୴୧ୢୣୢ ൐ ୰ୣ୯୳୧୰ୣୢ ୰ୣ୯୳୧୰ୣୢ ൌ ʹͲͲ െͶͲ െ ͺ ൌͳͷʹ ͷͺǤ͹ ൈ ͳͲ଺  ൌ ɐୱ୲ ൌ ʹͲͲ ൈ ͲǤͺͻ ൈ ʹʹͷ ൌ ͳͶ͸͸ଶ ൌ௩ ͳͲͲͲͳͶ͸͸ൈ ʹͲͳ ௩ ൌ ͳ͵͹ Provide 200MM overall depth using 40MM

Hence provide 16MM diameter bars at 100MM c/c

5.5 Dead Load Bending Moment and Shear Force in Girders

Fig 5.5: Position of loads for Max BM and SF in Girders

ͳǤ  ൌʹͳǤͳͺ ʹǤ ൌ ሺͲǤͲ͹ͷൈʹʹ ሻ ൅ ሺͲǤʹʹͷൈʹͶ ሻ DEPT. OF CIVIL ENGG; UVCE

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                    ൌ ͹ǤͲͷȀଶ ͵Ǥ ൌሺʹൈʹǤͳ ͺሻ ൅ ሺʹൈʹǤͷ൅ ͲǤͶሻ ൈ ͹ǤͲͷ                           ൌ ͶʹǤͶ͵Ȁଶ ସଶǤସଷ ͶǤ ͵ ൌ ଷ ൌ ͳͶǤͳͷȀଶ ͷǤ ൌͳǤ͹ʹͷ ͸Ǥ ൌͳൈ ͲǤͶൈͳǤ ͹ʹͷൈʹͶ ൌ ͳ͸Ǥͷ͸ ͹Ǥ ൌͳ͸Ǥͷ͸൅ͳͶǤͳͷ ൌ ͵ͲǤ͹ͳȀ ͺǤ ൌͳǤͷ ͻǤ ൌ ͳ ൈ ͲǤ͵ൈͳǤͷൈʹͶ ൌ ͳͲǤͺȀ ͳͲǤൌʹൈʹǤͳ ൌ ʹǤͶ ͳͳǤ   ൌ ͳȀ͵ሺͶǤʹ ൈͳͲǤͺሻൌͳͷǤͳʹ ǡ ஺ ൌ ஻ ൌ ଵଶ ሺͶൈͳͷǤͳʹ൅ ͳͺǤ͹ͳൈ͵ͲǤ͹ͳሻ                   ൌ ͵ͳ͹Ǥͷ͵  ൌሺ͵ͳ͹Ǥͷ͵ൈͻǤ͵ͷͷሻ െ ͳͷǤͳʹൈሺଷǤ଻ଶସଶሻȂሾͳͷǤͳʹሺ͵Ǥ͹Ͷʹ൅ͳǤͺ͹ͳሻሿ Ǧ ሺ͵ͲǤ͹ͳൈͻǤ͵ͷͷൈͻǤ͵ͷͷ ሻȀʹ ൌͳͷͳ͵Ǧ ൌሾ͵ͳ͹Ǥͷ͵ൈ ሺ͵Ǥ͹Ͷ൅ ͳǤͺ͹ͳሻሿ െ ሾͳͷǤͳʹൈሺ͵ͲǤ͹ͳൈͷǤ͸ͳ͵ൈͷǤ͸ͳ͵ ሻȀʹሿ ൌͳʹ͹ͳǦ ǡ ஼ ൌ ͵ͳ͹Ǥͷ͵     ൌ ͵ͳ͹Ǥͷ͵െͳͷǤͳʹെ ሺ͵ͲǤ͹ͳൈͷǤ͸ͳ͵ሻ                           ൌ ͳ͵ͳ    ͵Ȁͺ    ൌ͵ͳ͹Ǥͷ͵െ ͳͷǤͳʹെ ሺ͵ͲǤ͹ͳൈ ͹ǤͲʹሻ                         ൌͺ͹ 

Bending Moment at Mid Span

Bending Moment at Quarter Span

Shear Force


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       ൌ͵ͳ͹Ǥͷ͵െ ͳͷǤͳʹെͳͷǤͳʹെ ሺ͵ͲǤ͹ͳൈͻǤ͵ͷͷሻ                      ൌͲ 5.6 Design of Outer Girder 1. Total Bending Moment at centre of span = 2. Total Bending Moment at quarter span

=

3. Total Shear at support

=

4. Total Shear at ¼ span

=

5. Total Shear at 3/8 span

=

6. Total Shear at Mid span

=

ͳͷͳ͵൅ͳͺͲͲ ൌ͵͵ͳ͵ െ ͳʹ͹ͳ൅ ͳ͹Ͷ͸ ൌ ͵Ͳͳ͹ െ ͵ͳ͹Ǥͷ͵൅ʹͺ͹ ൌ ͸ Ͳͷ ͳ͵ͳ൅ ʹ͹ͻ  ൌ ͶͳͲ ͺ͹ ൅ ʹʹ͸   ൌ  ͵ͳ͵ ͳ͹͵൅ Ͳ    ൌ  ͳ͹͵

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

ୱ୲ ൌ

௦௧

଺ ͵͵ͳ͵ൈͳ Ͳ ଶ ൌ ൌ ͹ ͳͳͶ͸  ൈ ൈ ʹ ൈ Ǥͻൈͳ͸ ͷ ͲͲ Ͳ Ͳ  ୱ୲  ൌ ͻ͸ͷͳଶ ୱ୲  ൌ ͳͻ͸Ͷଶ

Provide 12 bars of 32MM diameter, having total

and

Provide 4bars of 25MM diameter, having total

Arrange these bars in 4 layers with spacing between bars equal to largest diameter bar u sed 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

5.7 Check for stresses 1.

Depth of Neutral Axis: Flange width will be the least of the following


Page 49


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

ൈ ൈ ቀ െ ௗଶ ቁ ൌ ൈ ሺ െ ሻ ʹͷͲͲ ൈʹʹͷሺ െͳͳʹǤͷሻ ൌ ͳͲ ൈͳͳǡͶ͸͹ሺͳͷ͹͹ െ B

௦

ds

௦௧

ሻ ǡ  ൌ ͳ൅ Ǥ ൌ ͳ൅ͳͷͳͲ͹͹ʹൈͳͲͲ Ͳ ൌ ͷʹ͸ ɐୱ୲ ൌ ʹͲͲ Ȁଶ  ൌ ɐୱ୲ ൈ ሺ െ ሻ ൌ ʹͳͲͲͲ ൈ ሺͳͷ͹͹͵͸ͳെ͵͸ͳሻ ൌ ͷǤͻͶȀଶ ଵ ൌ  െ ௙ ൈ  ଵ ൌ ͵͸ͳെʹʹͷ ͵͸ͳ ଶൈͷǤͻͶ ൅ʹȀ  ൌ ൅ ଵଵ ൬ ͵௙൰ ͻͶ൅ ʹൈ ͲǤ͵ͺ ൬ʹʹͷ൰ ൌ ͹ͻǤͷͳ ൌ ͷǤͷǤͻͶ൅ ͲǤ͵ͺ ͵ ǡ  ൌ  െ ൌ ͳͷ͹͹ െ ͹ͻǤͷͳ ൌ ͳͶͻ͹ǤͶͻ ௥ൌ ൈ ൈ ௥ ൌ ʹͲͲ ൈͳͳͶ͸͹ ൈͳͶͻ͹ͲͶͻ  ௥ ൌ ͵Ͷ͵ͶǤ͵Ͷ െ ൏ ͵͵ͳ͵ െ n = 361MM

௦௧

௖௕௖

Actual Neutral axis falls above the critical neutral axis therefore, the stress in the steel reaches the Maximum value first hence

Corresponding stress in the concrete at the outer fibre is given by

Similarly,

= 0.38

௦௧

௦௧


Page 50


Area of steel required

଺ Ͳ ͹ Ͳ ͵ ͳ ൈͳ  ൌ ʹͲͲ ൈͳͳͶ͸͹ ൈͳͶͻ͹ǤͶͻ ൌ ͳͲǡͶͶ͵ ͳ͵

ଶ



No. of 32MM diameter bars = 10443/305 = 12.9

Check for local bond stress as per IRC code

Assume effective depth = 1950 - 60 = 1890MM

ଷ ͸ ͷൈͳ Ͳ Ͳ  ൌͲǤͻൈͳ ͺͻͲ ͵ͷͷǤ͸͹ Ǥ ͵ʹ ൌ͵ǤͳͶʹൈ͵ʹ

Hence atleast 4 bars are to be taken straight Check for shear

1.

଺଴ൈହଵ଴య ൌ ͲǤͺȀଶ  ൌ ସ଴଴ൈଵ଼ଽ଴ ଵ଻ൈଷଵ଴య ൌ ͲǤʹͷ Ȁଶ  ൌ ସ଴଴ൈଵ଻଴଴ ଷଵൈଷଵ଴య ൌ ͲǤͶ͸ Ȁଶ  ൌ ସ଴଴ൈଵ଻଴଴ ͲǤͷȀଶ

, hence shear

reinforcement is necessary. 2.

, hence shear

reinforcement is not necessary. 3.

, hence provide shear

reinforcement.

Approximate distance from support at which shear stress is = 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 DEPT. OF CIVIL ENGG; UVCE

Page 51


Net remaining shear at support for which shear reinforcement has to be provided = 605-450=150KN

ǡ ୴ ൌ ʹൈ ͹ͺͳͷǤͷൈʹͲ ൈ ͲͲͳͲൈͳଷ ͺͻͲ ൌ ͵ͻʹǤ͸Ͷ 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,

 ൌͶͳͲ ൈ ͳͲଷ െ ସହൈହଵ଴ଶ య ൌ ͳͺʹǤͷ ൌ ൈଶ଻଼Ǥଵ଼ଶǤൈହଶ଴଴ൈଵ଻଴଴ ൈହଵ଴య ൌ ʹͻʹǤͶͻ

Spacing of two legged stirrups, 10MM diameter

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.,

ଷ ൈͳ ଼

ͺǤ͹ͳ ൌ ͹ǤͲʹ ൌ ͵ͳ͵

Therefore spacing of 10MM diameter bars 2L

ହଶ଴଴ൈଵ଻଴଴ ൌ ൈଶ଻଼Ǥൈଷଵൈଷଵ଴ య

ൌ ͳ͹ͲǤͷͶ

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


Page 52


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.

5.8 Design of Inner Girder Adopt,

ͶͲͲ ൈͳͷͲͲ ൈʹͳͲͲ Ǥ Ǥ ൌ ʹͷͲͲ ǡ ൌ ͶͲͲ ǡ ൌ ͳͷͲͲ ǡ ൌ ʹͷͻ͸ൈͳͲ଺ ଺ ʹͷͻ͸ൈͳ Ͳ ൌ ʹͲͲ ൈ ͲǤͻൈͳ͸ Ͳͷ        ൌ ͺͻͺ͸ ଶ ൌ ͻ͸ͷͳ ଶ ଺ Ͳ ʹ͵ʹʹൈͳ ൌ ʹͲͲ ൈ ͲǤͻൈͳ͸ Ͳͷ ൌ ͺͲ͵ͻ ௦௧

Use 12 bars of 32MM diameter

௦௧

௦௧

ଶ

Check for bond stress as per IRC

ൌ ͸ʹͲ Ǥ ଷ ͸ʹ ൈͳ Ͳ Ͳ  ൌͳൈ ͲǤͻൈͳ͸͸ͷ ൌ ͶͳͶ Ǥ ͵ʹ ൌ ͶͳͶൈ͵ʹ ൌ ͶǤͳ

Shear at support

Let effective depth be 1665MM

However take 6 bars straight up to support. Check for Shear

଺ଶ଴ൈଵ଴଺଺యହ ൌ ͳǤʹͶ Τଶ ɒ୴ ൌଷ଴଴ൈଵ DEPT. OF CIVIL ENGG; UVCE

(Shear reinforcement needed)

Page 53


ଵ଴ൈହଵ଴య ൌ ͲǤʹͳ Τଶ ɒ୴ ൌଷ଴଴ൈଵ ଺଴ହ ଶଶൈଷଵ଴య ൌ ͲǤͶ͸ Τ ଶ ɒ୴ ଷ଼  ൌଷ଴଴ൈଵ ଺଴ହ

(No shear reinforcement)

(However provide shear reinforcement)

Bend bars at 1021.26MM i.e. 1.02M = a

ൌ Ͷൈ ͺͲͷൈʹͲͲ ൈ ͲǤ͹Ͳ͹ ൌ Ͷͷͷ ൌ Ͷͷͷ

4 bars are effective at every section, hence

Shear taken by 4 bars of

͵ʹ

Net remaining shear at support for which shear requirement is necessary

ͳ͸ͷ

Ǥͷൈʹ ͲͲ ൈͳ͸͸ͷ ൌ௩ ʹൈ ͹ͺͳ͸ͷൈͳ Ͳଷ ൌ ͵ͳ͸Ǥͺͷ

Hence provide 10MM-2L bars at 180MM c/c from support up to support after 4.08M

ൌ ͸ʹͲ െͶͷͷ ൌ

ͶൈͳǤ Ͳʹ ൌ ͶǤͲͺ

from

ൌ ͵ͲͲ െ ସହହଶ ൌ ͹ʹǤͷ ͲͲ Ͳ ൈͳ͸ ͷ ൌ௩ ʹൈ ͹ͺ͹Ǥͷൈʹ ʹǤͷൈͳ Ͳଷ ൌ ͸ͻͷǤͳ͵

At quarter span,

Therefore, provide 2L-10MM diameter at 300 c/c from 4.08 to 5.1M. Therefore, shear at 3/8 span=223KN.

௩

Ǥͷൈʹ ͲͲ ൈͳ͸ ͷ Ͳ ൌ ʹൈ ͹ͺʹʹ͵ൈͳ Ͳଷ ൌ ʹʹͷǤͻͻ

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 4.8Mĺ10MM-2L diameter at 180MM c/c DEPT. OF CIVIL ENGG; UVCE

Page 54


4.8M to 5.1Mĺ10MM-2L diameter at 300MM c/c 5.1M to 7.02Mĺ10MM-2L diameter at 200MM c/c 7.02 to 8.02 and remainingĺ10MM-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

ൌ ͵ൈͳʹ ͹ͷ ൌ ͲǤ͵ൈͳǤʹ ͹ͷ ൈʹͶ ൌ ͻǤͳͺ Ȁ ൌ ͹ǤͲͷ Ȁ ଶ ൌ ʹൈቂʹǤͷൈͳǤʹͷൈ ଵଶቃ  ൌ ͵Ǥͳʹͷൈ ͹ǤͲͷ ൌ ʹʹǤͳ

Dimension

Weight of rib

Dead weight from slab and wearing coat Dead load on each cross girder

Assuming this to be uniformly distributed, dead load p er meter run of girder DEPT. OF CIVIL ENGG; UVCE

Page 55


ൌ ʹʹǤʹǤͷͳ ൌ ͺǤͺͶ Ȁ Ȁ

Therefore, Total weight = 9.18+8.84 = 18.02

Assuming cross girders to be rigid, reaction on eac h longitudinal girder = i)

ଵ଼Ǥ଴ൈଶହ ଷ

ൌ ͵ͳ

Live Load: Maximum bending Moment and shear force due to Class AA-

Tracked Loading

Fig 5.9c: LL on the span



ൌ ͹ͲͲ ൈʹǤͶ ͺͶʹ ൌ Ͷͻͺ

Assuming cross girders to be rigid, reaction on each longitudinal girder =

ସଽ଼ ଷ

ൌ ͳ͸͸

Maximum bending Moment under the wheel load

Fig 5.9d: Max LL Cross girder

 ൌ Ͷͻ͵ͺ ൈͳǤͶ͹ͷ ൌ ʹͶͷ െ 

Taking Impact factor = 1.1x245 =270KN-M Dead Load bending Moment from 4.75M of support


Page 56


ൌ ͵ͳ͸ൈͳǤͶ͹ͷെ  ͳͺǤͲʹൈ ሺͳǤʹͶ͹ͷൈͳǤͶ ͹ͷሻ ൌ ͶͷǤ͹͵െ ͳͻǤ͸ͳ = 26.12KN-M Total Bending Moment = LL BM +DL BM

                     ൌ ʹ͹Ͳ ൅ʹ͸Ǥͳʹ ൌ ʹͻ͹ െ   ൌ Ͷͻ͵ͺ ൈͳǤͳ ൌ ͳͺ͵ Section Design:

Total depth = 1500MM, Effective Depth = 1440MM

଺ ʹͻ ൈͳ ͹ Ͳ ୱ୲ ൌ ʹͲͲ ൈ ͲǤͻൈͳͶͶ Ͳ ൌ ͳͳͶ͸ଶ Hence provide 3nos of 25MM Diameter bars,

ǣ ଶଵൈସଵ଴య  ௩ ൌ ௩ௗൌଷ଴଴ൈଵସସ଴                       ൌ ͲǤͶͻȀଶ ൏ ௠௔௫

ୱ୲  ൌ ͳͶ͹͵ଶ

௕

ok

ǡ ʹൈ ͹ͺǤͷൈʹ ͲͲ ൈͳͶͶ Ͳ ௩ൌ ʹͳͶൈͳ Ͳଷ ൌ ʹͳͳǤʹͺ ʹ െͳ ͲʹͲͲȀ

But

௖

= 0.34

୒

୑୑

మ hence provide shear reinforcement.


Page 57


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.


Page 58

Analysis of Deck Slab and Tee Beam of a Bridge

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