Design of Composite Bridge

University of Dar es Salaam

College of Engineering & Technology

DEPARTMENT OF STRUCTURAL AND CONSTRUCTION ENGINEERING (SCE)

SD 677 ADVANCED BRIDGE ENGINEERING

Title

DESIGN OF COMPOSITE BRIDGE

Students: Divecha, Jiten L

Reg. No: 2010-06-01268

Program: MSc in Structural Engineering

Lecturer: Dr. MAKUNZA Year: August 2011

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges TABLE OF CONTENT SD677 August 2011 1/

Calculations by: Divecha, Jiten Checked by:

TABLE OF CONTENT

CHAPTER ONE: GENERAL INFORMATION........................................................................................................4

1.0 INTRODUCTION............................................................................................................................................... 4

1.1 LITERATURE REVIEW ON COMPOSITE ACTION...............................................................................................4

1.2 LITERATURE REVIEW ON SHEAR CONNECTORS.............................................................................................5

1.2.1 DESIGN REQUIREMENTS OF SHEAR CONNECTORS....................................................................................6

1.2.2 TRANSFORMED SECTION......................................................................................................................... 7

CHAPTER TWO: DESIGN METHODS.................................................................................................................. 9

2.0 INTRODUCTION............................................................................................................................................... 9

2.1 DESIGN OF INTERIOR PANEL OF SLAB.............................................................................................................. 9

2.2 DESIGN OF LONGITUDINAL GIRDERS.................................................................................................................9

CHAPTER THREE: SCOPE OF WORK.............................................................................................................. 12

CHAPTER FOUR: DESIGN OF COMPOSITE BRIDGE......................................................................................14

4.0 INTRODUCTION............................................................................................................................................. 14

4.1 DESIGN STEPS............................................................................................................................................. 14

Block 38, Plot 14 , Sukuma Street , Kariakoo area Ilala District. P.O. Box 20341, Dar Es Salaam , Tanzania.Tel: +255 786 794 401.Email: [email protected]

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges SEPERATION SHEET SD 677 August 2011 2/


Description/Calculation


1.0 GENERAL INFORMATION

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges GENERAL INFORMATION SD 677 August 2011 3/


CHAPTER ONE: GENERAL INFORMATION

1.0 Introduction

A Composite bridge is one whose decking system consists of a concrete slab and which in conjunction with steel

girders resists moving loads on the bridge. This type of bridge is found to be economical for spans of 10 to 20 m.

A composite bridge offers the following advantages over other types of bridges

More efficient use of materials, since the size of the steel member can be significantly reduced owing to

incorporation of the deck into the resisting cross section, into the compression zone.

Greater vertical clearance by effecting reduction in beam depth.

Enhanced stiffness, which in turn makes the deck sustain greater vehicle loading.

1.1 Literature Review on Composite Action

It is said that composite construction has its roots in the mid nineteenth century. However, the composite bridge

construction did not take effect until about late 1940’s. To understand how composite construction brings in

economy of materials, we have to look back at the basic strength of materials.

From bending theory, the maximum bending stress in a beam subjected to pure bending is given by

Where

= bending stress in the beam

= bending moment

= distance of extreme fibre from the neutral axis

= moment of inertia of resisting section

The above equation can be modified to

Where , is the section modulus

The section modulus is dependent only on the geometry of the cross section. By Observation, we can see that

the bigger the value of , the smaller the resulting stress. Therfore, it is in the best interest of the designer to

increase the section modulus as much as possible. Composite section provide substantial section modulus with

minimum material and it is here that the principal advantage of composite action comes into play.




1.2 Literature Review on Shear Connectors

The shear connectors are part and parcel of a composite deck system. The need for shear connectors can be

understood by considering the interaction between the slab and the steel beam. If the slab simply rests on the

steel beam, a phenomenon known as slippage occurs. As the loads are placed on the top of the slab, the top of

slab and beam will be in compression while the bottom of the slab and beam will be in tension. Both the slab and

the steel beam behave independently deflecting like a beam. Since the bottom of the slab is in tension( tending

to push outwards) and the top of the beam is in compression ( tending to move inwards ) , the resulting effect is

manifested by extension of the slab over the ends of the beam.

It is possible to some how connect the concrete slab and the steel beam such that they resist the loads like a

common unit. Such a one to one unity between the two units can be achieved by providing shear connectors

between the slab and the beam.

A shear connector is generally a metal element of particular shape, which extends vertically from top flange of

the supporting beam and gets embedded into the slab. Depending upon the magnitude of the shear force at the

interface of the bema and the slab, a number of shear connectors can be placed along the length of the beam.

With shear connectors in place, the slab and beam can now be analyzed as a single unit. The composite section

will now have a higher section modulus that allows the composite beam to resist higher loads. In a nut shell, the

I shaped beam gets replaced more or less by a T shaped beam.




1.2.1 Design Requirements of Shear Connectors

The shear connectors have to be designed to facilitate conjoint action between the RC slab and the steel beam.

Their basic function is:

To transfer the shear force at the interface of the slab and the beam without slip.

To prevent separation of the slab from the steel beam in the perpendicular direction.

There are rigid and flexible shear connectors. The rigid types include channel angles, tee sections while the stud

types of shear connectors come under the flexible type.

The characteristic strengths of stud connectors in normal weight concrete as given by the code for composite

beams are shown in table below. The characteristic values are multiplied by the following reduction factors:

0.8 for positive bending moments

0.6 for negative bending moments

0.9 for light weight concrete

Table 1: Characteristic resistance of headed studs in normal weight concrete

Dimensions of stud shear

connectors

Characteristic strength of concrete ( N/mm2)

Nominal

shank

diameter

(mm)

Nominal

height

(mm)

As- welded

height

(mm)

25

KN

30

KN

35

KN

40

KN

25 100 95 146 154 161 168

22 100 95 119 126 132 139

19 100 95 95 100 104 109

19 75 70 82 87 91 96

16 75 70 70 74 78 82

13 65 60 44 47 49 52





1.2.2 Transformed Section

The composite slab – beam section is converted into a modified section where the concrete slab turns into

equivalent area of steel. This conversion is brought through the use of modular ratio is given by

Where

= modulus of elasticity of steel

= modulus of elasticity of concrete





2.0 DESIGN METHODS

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN METHODS SD 677 August 2011 8/


CHAPTER TWO: DESIGN METHODS

2.0 Introduction

A beam and slab bridge or T beam bridge is constructed when the span is between 10 – 20 m. The bridge deck

essentially consists of a concrete slab monolithically cast over longitudinal girders so that the T beam effect

prevails. To impart transverse stiffness to the deck, cross girders or diaphragms are provided at regular

intervals. The number of longitudinal girders depends on the width of the road. Three girders are normally

provided for a two lane road bridge. A complete design of a T beam deck would involve the design of the interior

panel of the slab, longitudinal girders and cross girders.

2.1 Design of interior panel of slab

In a T- beam bridge deck with cross beams, the slab may be regarded as supported on all the four edges and

continuous over the beams. Many methods are available for analysis of such two way slabs subjected to

concentrate loads. Among them are the following methods:

Rankine-Grashoff method.

Diagonals method.

Westergaards method.

Pigeauds method.

The method used in this design is pigeauds method, the short span and long span bending moment coefficients

are read from curves developed by M. Pigeaud. These curves are used for slab supported along four edges with

restrained corners and subjected to symmetrically placed loads distributed over some well defined area. These

curves were developed for thin plates using the elastic flexural theory. However their use has been extended to

concrete slabs too.

2.2 Design of longitudinal girders

It is known that the bridge loads are transmitted from deck to the superstructure and then to the supporting

substructure elements. It is rather difficult to imagine how these loads get transferred. If a vehicle is moving on

the top of a particular beam, it is reasonable to say that this particular beam is resisting the vehicle or truckload.

However this beam is not alone, it is connected to adajacent members through the slab and cross girders. This

connectivity allows different members to work together in resisting loads, though it is logical to assume that , this

specific beam is carrying most of the load. As a result of being connected to other members, the adajacent


Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN METHODS SD 677 August 2011 9/


Members will also assist in carrying part of the load. The supporting girders share the live load in varying

proportions depending on the flexural stiffness of the deck and the position of the live load on the deck. For

determining the fraction of the load carried by the longitudinal girders, several methods have been suggested.

Among them, the rational ones are :

Guyon Massonet method

Hendry Jaegar Method

Courbon’s Method

In the analysis of longitudinal girder of the current design work, Courbon’s method ha sbeen utilized and

presented.






3.0 SCOPE OF WORK

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges SCOPE OF WORK SD 677 August 2011 11/


CHAPTER THREE: SCOPE OF WORK

The purpose of the work presented in this project is to carry out detailed engineering design of a 24 m span

composite bridge. The bridge has a 7.15m wide carriage way width with 900mm wide raised shoulders on both

sides of the carriage way. The Superstructure of the bridge rests on a two abutments 8.95m wide, the hydrology

of the river and the road geometry suggested minimum of 5.5 m high abutments to be provided.

In Chapter 4, Detailed engineering design of the complete bridge including the deck slab, longitudinal girder,

cross girders, shear connectors, abutments and foundation were carried out and detailed engineering drawings

were presented in the appendix.






4.0 DESIGN OF COMPOSITE BRIDGE

Project/Structure: Subject Project Ref: Date: Sheet No:

Design of Bridges DESIGN OF COMPOSITE BRIDGE SD 677 August 2011 13/


CHAPTER FOUR: DESIGN OF COMPOSITE BRIDGE

4.0 Introduction

A 24 meter span composite bridge is to be provided over a road section with 7.15m wide carriage way width and

900mm wide raised shoulders on both sides of the carriage way. The superstructure of the bridge rest on a 7.5

meter high abutment .The type of bearing used is a elastometric bearing.

4.1 Design steps

Step 1: Obtaining all design data, such as references, proposed size, material properties, load cases and all

assumptions.

Step 2: Approximating the proposed structural framing, and checking the dimensions with applicable codes.

Step 3: Carry out the analysis and design of each bridge element in accordance to relevant methods and codes.

Step 4: Detailed Engineering drawings are produced.


Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges Design Data SD 677 August 2011 14/


Reference Calculation Output


STEP 1: DESIGN DATA

A. References

B. C. Pumia, Ashock Kumar Jain, Arun Kumari Jain: Reinforced concrete structures:Vol I (1997) Laxmi Publications (P) Ltd.

BS 8110:1:1997: Structural use of concrete.

BS 5950:1:2000: Structural use of structural steel.

BS5400:2:1978: Steel, concrete and composite bridges: Specification for loads.

BS 5400:4:1990: Steel, concrete and composite bridges: Code of practice for design of concrete bridge.

BS 648: Weights of building materials

BS8002:1994: Earth retaining structures

Charles Reynolds: Reinforced concrete designer’s hand book: 10th Edition

W.H. Mosley, J.H. Bungey, R. Hulse: Reinforced concrete design: 5th Edition (1999) Macmillan press limited

Choo, B. S. MacGinley, T. J. ( ). Reinforced Concrete Design Theory and Examples.

Jagadeesh, T.R, Jayaram. M.A, Design of bridge structures, 3rd

Edition (2003), Pretence hall of India (Pvt) Limited.

B. Proposed Dimensions

Effective span = 24m

Height of abutment = 5.5m

Total carriage way width = 7.15m

Pedestrian kerb (2 on each side) = 0.9 m

Proposed thickness of deck slab = 0.24m

Surface slope = 3%

Abutment thickness =1.3m

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges Design Data SD 677 August 2011 15/




Footing thickness =1.0m

Heel length =1.6m

Longitudinal beam Spacing = 2.25 m

Cantilever slab = 1.1 m

C. Materials properties and specifications.

Density of concrete: =24 kN/m3

Density of surface course = 20 KN/m3

Density of soil = 17 KN/m3

Density of steel = 7800kg/m3

Concrete grade = C30

Steel reinforcement = Grade 460

D. Loading Given

HB Vehicle = 37.5 units

Max temp difference = 120C

E. Safe Soil bearing Pressure

For all location of the Abutments, a safe soil bearing pressure of

250 kN/m3 is assumed.

F. Assumptions

Concrete cover to slab = 30 mm

Concrete cover to abutments =50 mm

Concrete cover to foundations =75 mm

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges Structural Framing SD 677 August 2011 16/




STEP 2: STRUCTURAL FRAMING

Effective Longitudinal Span = 24 m

Slab thickness =240 mm

Surfacing slope = 3%

Carriage way width = 7150 mm

Kerb width =900 mm

Assuming transverse beam every 3000 mm

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN OF INTERIOR PANEL SD 677 August 2011 17/




BS 5400

Part 2:1978

Table 1

Pigeurd’s

Curve

STEP 3: DESIGN OF INTERIOR PANEL

3.1 Panel Size

Interior panel dimension 2.25m x 3m

3.2 Analysis

Using Pigeaud’s method

3.2.1 Loading

A: Dead Load

S.W of slab = 0.24 x 24 x 1.15 = 6.624 kN/m2

S.W of wearing coarse = 0.08375 x 24 x 1.75

= 3.5175 kN/m2

Total Design Dead load = 10.142 kN/m2

Dead load on panel = 10.142 x 2.25 x 3

= 68.46 kN.

Since dead load spreads uniformly on entire slab, we have;

, and

From Pigeurd’s curve , the coefficients M1 and M2 are

obtained, hence

M1 = 4.9 x 10-2

M2 = 3.5 x 10-2

Short span Bending Moment,

Mb = W (M1 + 0.15 M2) = 68.46 (0.049 + 0.15 (0.035)) =3.714 kN. M

Long span Bending Moment

ML = W (M2 + 0.15 M1) = 68.46 (0.035 + 0.15 (0.049)) =2.899 kN. M





Continuity effect on slab is accounted for by a continuity factor which is taken as 0.8.Therefore, bending moment after continuity factor;

Short span, Mb = 0.8 x 3.714 = 2.9712 kNm

Long span, ML = 0.8 X 2.899 = 2.3192 kNm

Shear force;

Dead load shear force = (10.142 x 2.25) ÷ 2

= 11.41 kN.

B: Imposed Load

Taking H B vehicle as critical

HB Loading

Total wheels: 16 wheels in 4 axles

Load per wheel: 37.5 x 2.5 x 1.3 = 121.875 kN

Assuming one wheel can be accommodated centrally on panel so as to produce maximum Bending moment.

Contact area A: A = L/ P

= 121.875 X 103 / 1.1

A = 110795.45 mm2

Taking square dispersion =

= 332.86 mm

Take square area of 340 x 340 mm





Pigeard’s Curve

M1 ≈ 16 x 10-2

M2 ≈ 12 x 10-2

Short span Bending Moment,

Mb = W (M1 + 0.15 M2) = 121.875 (0.016 + 0.15 (0.014)) = 2.21 kN. M

Long span Bending Moment

ML = W (M2 + 0.15 M1) = 121.875 (0.014 + 0.15 (0.016)) =1.999 kN. M

Taking Impact Factor to be 25%

Continuity factor to be 0.8

Therefore, Actual live bending moment

Short span = 1.25 x 0.8 x 2.21 = 2.21kNm

Long span = 1.25 x 0.8 x 1.999 = 1.999kNm

Total design bending moment

Short span bending moment = 2.9712 + 2.21 = 5.18 kNm

Long span bending moment = 2.3192 + 1.999 = 4.32 kNm

Taking design bending moment = 5.2 kNm

3.3 Design of Slab

Taking bar diameter = 12 mm

Concrete cover = 30 mm

Therefore, Effective depth,

d= 240 - 30 - 6 = 204 mm

( Hence, two way slab )





BS 8110:Part 1:1997

Table 3.25

Table 3.10

Short span bending moment = 5.18 kNm

Long span bending moment = 4.31 kNm

Bending short span

Hence, taking

z = 0.95d = 193.8mm

Check minimum reinforcement,

Hence ,use minimum steel

Provide

Span effective depth ratio

Service stress,

Modification factor = =2.36 ≥ 2.0

Hence Take modification factor = 2.0





BS 8110:Part 1:1997Table 3.9

Limiting = 26 x 2.0 = 56

Actual = = = 11.03

Effective depth, d = 204mm is adequate.

Bending Long span

Hence, taking

z = 193.8mm – 12 = 181.8mm

(since the reinforcement for this span will have a reduced eff depth)

As = = = 54.25mm2/m

Check minimum

As min = =

=312mm2/mHenceProvide minimum Y12-300 c/c (As prov 377mm2/m)

Summary

Y12-300c/c – longitudinally

Y12-300c/c – Transversely


Design of Bridges DESIGN OF CANTILEVER SLAB SD 677 August 2011 22/




BS 5400:Part 2 :1978

STEP 4: DESIGN OF CANTILEVER SLAB

5.1 Analysis

5.1.1 Loading

A: Dead + Super Imposed Dead Load

COMPONENT Dead load/Super Imposed Load

Distance from Edge beam ( m )

Bending Moment( KN.M)

1. Railing 5x1.75 =8.75kN/m

1.10 9.625

2. Footpath 24x0.125x0.9x1.15=3.105kN/m

0.65 2.02

3. Triangular deck

1/2x0.9x0.24x24x1.15=2.9808kN/m

0.50 1.4904

4.Rectasngular deck

24x0.2x0.24x1.15=1.325 kN/m

0.10 0.1325

5.Wearing coarse

20x0.03x0.2x1.15=0.21 kN/m

0.10 0.021

TOTAL 16.371 kN/m 13.289kNm






BS 5400P2:1978:Cl3.2.9.3.1

B: Imposed Load

Loaded length = 24 m

Carriage way = 7.15m

No. of notional lane = 2 numbers

B.1 HB Wheel load

Load/wheel = 2.5x37.5x1.3 = 121.875 kN

Contact Area; A =

A = 110795.45mm2

Square Area 332.86 x 332.86 mm

Hence, take square area of 340 x 340

Effective load carried by cantilever portion

= 121.875 x 332/664 = 60.94 kN






BS 5400P2:1978:Cl 7

Table 1

Effective width

beff = 1.2x + bi

where, x = 332/2 =166mm and

bi = w + 2h

= 0.340 + 2(0.08375)

=0.508m

hence

beff = 1.2(0.166) + 0.508

= 0.7072m

Taking impact factor of 0.5

Bending moment due to HB loading

= 1.5x 60.94/0.707 x 0.332/2

= 21.46kNm

B.2 Pedestrian Live Load

L≤ 30M P = 5kN/m2

Design load = 5 x 1.5 = 7.5kN/m2

Bending moment due to pedestrian

= 7.5 x 0.9 (0.9/2 = 0.2)

= 4.39kNm

Total Design Moment = (13.289+ 21.46 + 4.39) = 39.14 kNm

Total Design Shear = 16.371 + (1.5 x 60.94) + (7.5 x 0.9)= 114.531 KN






BS 8110:Part 1:1997Table 3.25

4.2 Design of Cantilever Slab

Design bending moment = 39.14 kNm

Taking bar diameter = 12 mm

Concrete cover = 30 mm

Therefore, Effective depth,

d= 240 - 30 - 6 = 204 mm

Bending reinforcement

z = 0.964d > 0.95d

Take, z = 0.95d=193.8mm

Check minimum

As min = =

=312mm2/mHenceProvide Y12-200 c/c (As prov =566mm2/m)






BS 8110:P1:1997CL 3.4.5.1

BS 8110:Part 1:1997Table 3.25

Shear reinforcement

Max Shear

v =

v =

v = 0.561N/mm2 < 0.8

= = 0.2775≈ 0.28

vc = 0.6N/mm2

The slab is safe against shear.

Distribution of steel

As min = =

=312mm2/m

Provide Y12 – 300c/c (377 mm2/m)

For raised Kerb.

Provide minimum steel 312mm2/m

Therefore Y12 – 300 c/c ( both way)


Design of Bridges DESIGN OF LONGITUDINAL GIRDER SD 677 August 2011 27/




BS 5400P2:1978:Cl 3.2.9.3.1

STEP 5: DESIGN OF LONGITUDINAL GIRDER

5.1 Analysis

5.1.1 Loading

A. Dead Load

A.1. Loading from cantilever = 2x16.371 = 32.74kN

A.2. Loading from deck = 24x0.24 x 6.75 x 1.15 kN

= 44.712 kN

Total Dead Load = 77.412 kN

B. Superimposed Dead Load

B.1. Surfacing (Assuming 30mm at ends)

A =1/2 x [0.03 + 0.1375] x 3.573

= 0.3 x 2

= 0.6 m2

Total surfacing load = 20 x 0.6 x 1.75= 21 kN

Total permanent load = 77.412 + 21 = 98.412 kN/m

Assuming the total dead load is taken equally by four girders.

Permanent Load per girder = = 24.603 kN/mC: Live Load

Loaded length = 24m

Carriage way width = 7.15m

No. of notional lane = 2






BS 5400:P2: 1978:Cl: 6.2

BS 5400:P2: 1978:Cl: 6.3

BS 5400:P2: 1978:Cl: 7

C.1 HA Alone

UDL: 30kN/m/lane =

KEL: 120kN/lane =

C.2 HA with HB

UDL: 30kN/m/lane =

KEL: 120kN/lane = C.3 HB Alone

Load/wheel = 2.5 x 37.5 x 1.3 = 121.875 kN

Load/axle = 121.875 x 4 =487.5 kN

Total vehicle load = 487.5 x 4 =1950 kN

C.4 Pedestrians loading

=5 x 1.5 = 7.5 kN

D: Loading from girder

D.1. Self Weight. of Girder

Assume (0.2 L + 1) kN/m

= (0.2 x 24 + 1) kN/m = 5.8 kN/m Take 6 kN/m

Design self weight = 6 x 1.05 =6.3 kN/m

Total load (Permanent) on each girder = 24.603 + 6.3

= 30.903 kN/m≈ 31 kN/m

D.2 Cross Girders at 3m center ( Assume Self Weight. 1 kN/m )

Each cross girder = 2.5 x 1 = 2.5 kN/m

Load on main girder =1.25 kN (each side)

Design point load = 1.25 x 1.05 = 1.313 kN






5.1.2 Loading Distribution on Girder

Distribution method – using Courbon’s method

Application check

O.K

No. of cross Girder

n = 24/3 = 8

n > 5 O.K

Hence Courbon’s method applicable.

A: HA alone( on one lane ) Loading Distribution on Girder

Wi =

For UDL

=45.05kN

n = 4

e = 3.575 – 1.7875= 1.7875






= distance of girder A from center

= 3.515- 0.2 = 3.375m

= 3.515- 0.2 – 2.250 = 1.125m

= -1.125m

= -3.375m

= 25.313

Load on beam A

WA =

= 11.263 x 1.9533 = 22 kN/m

Load on beam B

WB =

= 11.263 x 1.318 = 14.845 kN/m

Load on beam C

WC = = 11.263 x 0.6822 = 7.684 kN/m

Load on beam D

WD = = 11.263 x 0.0467 = 0.526 kN/m






For KEL

Load on beam A

WA = = 12.59 x 1.9533

= 24.59 kN

Load on beam B

WB = = 12.59 x 1.318

= 16.59 kN

Load on beam C

WC = = 12.59 x 0.6822

= 8.589 kN

Load on beam D

WD = = 12.59 x 0.0467

= 0.588 kN

Design for Beam A Highly loaded.






B: HB alone ( on one lane ) Loading Distribution on Girder

Wi =

=4 x 121.875 = 487.5kN

n = 4

= 25.313

Load on beam A

WA = = 121.875x 1.973 = 240.46 kN

Load on beam B

WB = = 121.875x 1.324 = 161.36 kN

Load on beam C

WC = = 11.263 x 0.6756 = 82.34 kN

Load on beam D

WD = = 11.263 x 0.0267 = 3.254kN

Design for Beam A Highly loaded.






C: HA on one lane and HB on other Loading Distribution on Girder

This is obtained by superimposing Case A ( HA Alone ) and Case B

( HB Alone )

D: Pedestrian Loading distribution

Assuming pedestrian on both sides, Hence loading carried equally by

all girders.

5x1.5 = 7.5x0.9 = 6.75 kN/m (each kerb)

=2x6.75 = 13.75 kN/m ( two kerbs)

Load on each girder =

5.1.3 Analyzing most loaded Girder

Case 1a: Dead Load + live Load (HA Alone )

Moment

Dead load = 2232kNm

Live Load

HA UDL = 1621.44kNm

HA KEL = 536.22kNm

Pedestrian = 243kNm

Cross girder = 31.51 kNm






Total Design Moment = 4664.17 kNm

Case 1b: Dead Load + live Load (HA Alone ) ( Position of

maximum shear )

Knife edge load on one of the support

Dead load = 372kN

Live Load

HA UDL = 270.24kN

HA KEL = 89.37kN

Pedestrian =39.9kN

Cross girder = 4.595 kN

Total Design Shear = 776.105 kN

Summary case 1.

Max Design Moment = 4664.17 kNm

Max Design Shear = 776.105 kN






Case 2a: Dead Load + live Load (HB Alone ) ( Position of

maximum moment )

Moment

Dead load at 12m = 2232kNm

at 10.8m = 2209 kNm

Live Load

HB at 12m = 3895kNm; RA = 420.805 kN

at 10.8m = 3967.59kNm; RB = 541.053 kN

Pedestrian. at 12m = 243kNm

at 10.8m = 240.57kNm

Cross Girder at 12m =31.512 kNm

at 10.8m = 30.724kNm

Total Design Moment at 12m = 6401.96 kNm

at 10.8m = 6447.88 kNm






Case 2b: Dead Load + live Load (HB Alone ) ( Position of

maximum Shear )

HB loading near one of the support

Shear

Dead load =372 kN

Live load

HB =529.01kN

Pedestrian =39.9 kN

Cross girder =4.595 kN

Total Shear = 945.51 kN

Summary case 2.

Max Design Moment at 12m = 6401.96 kNm

at 10.8m = 6447.88 kNm







Case 3a: Dead Load + live Load (HA on one lane and HB on other

lane ) ( Position of maximum moment )

Moment

Dead load at 12m = 2232kNm

at 10.8m = 2209 kNm

Live Load

HB at 12m = 3895kNm;

at 10.8m = 3967.59kNm;

HA- UDL at 12m =471.6 kNm

At 10.8m =466.88 kNm

HA- KEL at 12m =109.32 kNm

At 10.8m =98.388 kNm

Pedestrian. at 12m = 243kNm

at 10.8m = 240.57kNm

Cross Girder at 12m =31.512 kNm

at 10.8m = 30.724kNm

Total Design Moment at 12m = 6982.88 kNm

at 10.8m = 7013.16 kNm






Case 3b: Dead Load + live Load (HA on one lane and HB on other

lane ) ( Position of maximum shear )

HA-KEL and HB loading near one of the support

Shear

Dead load =372 kN

Live load

HB =529.01kN

HA-UDL =78.6kN

HA-KEL =18.22kN

Pedestrian =39.9 kN

Cross girder =4.595 kN

Total Shear = 1042.33 kN

Summary case 3.


at 10.8m = 7013.16 kNm







BS 5950 P1 2000Table 9

BS 5950 P1 2000CL4.3.7

5.2 Structural Design of longitudinal beam

Taking Case 3 loading as critical and hence


at 10.8m = 7013.16 kNm


A. Initial Sizing

Assuming

Depth of girder d ≈

≈ 1333.33 mm

Hence take depth of Girder as d = 1400 mm

B. Section Sizing

Assuming 16 ≤ T ≤ 40 mm, py = 265 N/mm2

Flange force ≈ 5009.4 kN

Since the flange is not fully restrained a value less than 265 N/mm2 should be

used when estimating the required area.

Moment capacity Assume

Area of flange

Af ≈

≈ 19.27 x 103 mm2

Try flange plate 500mm wide x 40 mm thick

Aprovide = 500 x 40 = 20 x 103 mm2

Assume a 20 mm thick web






Trial section

Section properties

Area of section

Moment of Inertia

Ixx = -

= 1.35 x 1011 – 1.098 x 1011

= 2.534 x 1010 mm4

Iyy = 2 x -

= 834.267 x 106 mm4

Radius of Gyration

ryy = = = 110.6 mm

Plastic Modulus

Sxx = (500 x 40) = 28.8 x 106 mm3

Self weight = 68 x 103 x 1000 x 7.8 x 10-8 = 5.34 kN/m

(this is less than the assumed value) Hence, OK!






BS 5950 P1 2000Table 7

Table 6

Cl 3.6.2

Cl 4.4.4.2

C. Section Classification

C.1 Flanges

T = 40 mm, hence y = 265 N/mm2

b = = 245 mm

C.2 Web

t = 10 mm , hence y = 275 N/mm2

,

web is thin, use clause 4.4.4.2 to determine moment capacity

b) with transverse stiffners only

where stiffner spacing a > d then

where stiffner spacing a d then

assume the more critical case with stiffners then

assume the more critical case with stiffners then

t ≥ = = 4.3 mm

Now, web is adequate with respect to serviceability






BS 5950 P1 2000CL 4.3.7.3

CL 4.4.5.3

Table 21b

Cl 4.4.4.2

Cl 4.4.2.3

Cl 4.5.2.2

D. Moment Capacity

compression flange is fully restrained

= 265 x 28.8x106

= 7632kNm

Mb > Mapplied Section is adequate with respect to bending

E. Shear Capacity

y = 275 N/mm2 , = 70, qcr = 151 n/mm2

t ≥ = = 5.6mm

t ≥ = = 4.3 mm

f

ed ≤ ed

ed = = = 41.84 N/mm2

ed = = 41.84 N/mm2






BS 5950 P1 2000CL 4.2.5

Hence provide minimum.

= 3

a = 3x1400

a ≤ 4200 mm

Provide intermediate stiffness at 1400 mm

F. Deflection

δmax ≤ = = 66.67 mm

δudl = (UDL)

δpoint load = (at center-KEL)

Deflection due to unfactored imposed load

W = 5.04 kN/m2,P1 = 14.015kN ,P2 = 184.97kN

δudl = ,

δudl = 0.17 mm

δpoint load = ,δpoint load = 0.759 mm

δat HB = b1 = 15.3 m, b2 = 13.5m, b3 = 7.5m

δmax1 = 8.99mm b1 = 15.3m

δmax2 = 9.8mm b2 = 13.5m

δmax3 = 8.2mm b3 = 7.5m

δmax4 = 6.71mm b4 = 5.7m

δmax(HB) = 8.99 + 9.8 + 8.2 + 6.71 = 33.7mm

δmax = 33.7 + 0.17 + 0.759 = 35.218mm < 66.67 mm






BS 5950 P1 2000CL 4.4.6CL 4.5.12

BS 5950 P1 2000CL 4.4.6.4

CL 4.5.4.2

CL 4.5.1.2

G. Intermediate stiffnersAssume 8mm thick flats y = 275 N/mm2

Outstand bs ≤ 19 x 8 = 152mm

Maximum flange width available

= 240

Hence Stiffner Outstand adequate

Is ≥ = = 16.8 x 106

bs = 136.6mm say 140mm

adopt 2/stiffness – 140mm x 8mm thick

H. Load bearing stiffners

Contact Area A >

> 3032.24 mm2

Assume 12mm stiffner 12mm thick and allow 20mm fillet for web/flange web.

A = 2(bs – 20) x 20 = 3032.24

bs = 95.8 mm

Try stiffness comprising 2 flats 100 x 12mm thick.

= (13 x 12 x 1) = 156 mm

= (19 x 12 x 1) = 228 mm

Therefore ,Use core section equal to 156 mm






BS 5950 P1 2000CL 4.5.15

CL 4.7.5Table 27c

= + = 36.87 x 106 mm4

A =(332 x 12) + (394 x 20) = 11869 mm2

= = 55.75 mm

Le = (0.7 x 1400) = 980 mm

= 17.6

y = 255 N/mm2 (table 6 value less than 20N/mm2)

c = 254 N/mm2

Buckling resistance = 254 x 11864/1000

= 3013.456 > 1043.33 kN OK!

Bearing capacity ≥ (applied load – )

where, b1 = 0, n2 = 100 mm and t = 20 mm

= 100 x 20 275 x 10-3

= 550 kN

(Applied load – ) = 1042..33 – 550 = 492.33 kN






BS 5950 P1 2000CL 4.5.3

Bearing Capacity

= 275 x 11864 x 103= 3263.6 kN

>> 492.33 kN ( Adequate in bearing )

Welded connection

Tension Capacity,

Applied force

Design weld for 1042.33 kN

Length of stiffner = 1400 mm

Strength of weld

= 0.745 kN/mm

Strength of 6mm fillet weld = 2 x 0.903 = 1.806 kN/m

( Adopt 2 – continous 6mm fillet welds )

Flange to web connection

q =

Q = 1042.33kN

A = 500 x 40 x 740 = 14.8 x 106 mm3

= 2.534 x 1010 mm4

q =

Adopt 6mm fillet weld


Design of Bridges DESIGN OF SHEAR CONNECTORS SD 677 August 2011 47/




STEP 6: DESIGN OF SHEAR CONNECTOR

6.1 Composite section properties

The flange width of the composite section is taken as center to center of

girder.

Modular ratio,

Equivalent Area, Ac = = 41538.5 mm2

Determination of neutral axis of composite section,

we have.

= (41538.5x 1600) + (500 x 40 x 1460) + (1400 x 20 x 740) +

(500 x 40 x 20)

= 116.782 x 106

= 41538.5 + (500 x 40 x 2) + (1400 x 20) = 109538.5 mm2

= 1066.13 mm






Moment of inertia of composite section:

= + - +

+

= 3.528 x 1010

6.2 Shear Connectors

,

= 360.65N/mm

Total horizontal shear force on width

= 360.65 x 500 = 180325 N

Taking shear connector -19mm - nominal height= 100mm

= 1042.33 kN , A= 41538.5 mm2 ,y= 293.87 mm

3.528 x 1010 mm2






= characteristics resistance = 100kN

Design shear capacity qe of each stud is;

where and

but not greater than 0.8

= is breadth of concrete rid in profile decking

= 150 mm

= 50mm (depth of profiled decking) h = 100 mm

take = 0.8

Number of studs required = =2.8 studs

Take 3 studs

Place one stud at centerline of girder

Place other two studs at 150 c/c of girder

Spacing = = 532.4 mm adopt 350mm c/c

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN OF ABUTMENT SD 677 August 2011 50/




STEP 7: DESIGN OF ABUTMENT

7.1 Initial Sizing

Thickness: = 1.5 m

Height from base to bearing =5.5 m

Heel length = 1.6 m

Toe length = 1.5 m

Width = 4.6 m

Footing thickness = 1.0 m

7.2 Structural Framing

5.5m

2.0m

1.0m





BS 5400:P2:1978:Table 1

7.3 Loading

Total width of Abutment

W = 7.15 + (2 x 0.9) =8.95 m

A: Dead load

A:1 Self Weight of abutment

Characteristic Load = (0.5 x 2 x 24) + (1.5 x 4.5 x 24) = 186 kN/m

Design Load = 186 x 1.15 = 213.9 kN/m

A.2: Self Weight of beam

Characteristic Load = x 24 x 0.5 x 4 = 32.18 kN/m

Design Load = 32.18 x 0.5 =33.79 kN/m

A.3: Self Weight of slab

Characteristic Load = 0.24 x 24 x x = 55.22 kN/m

Design Load = 55.22 x 1.15 = 63.503 kN/m

A.4: Self Weight. Cantilever slab

Characteristics Load

=

= 17.3 kN/m

Design Load : = 17.3 x 1.15 = 19.89 kN/m





BS 5400:P2:1978:Table 1

B: Super Imposed Dead load

B.1 Surfacing Load

Surfacing = 0.1073 x 20 = 2.145 kN/m2

Surfacing = 0.03 x 20 = 0.6 kN/m2

Characteristic Load

=(2.145 + 0.6) x 7.15 x 24 x

= 13.16 kN/m

Design Load

= x (2.145 + 0.6) x 7.15 x 24 x x 1.75

= 23.03 kN/m

B.2: Parapet Load

Characteristic Load

= 2 x 5 x 24 x =13.41 kN/m

Design Load

= 13.41 x 1.75 = 23 45 kN/m





BS 5400:P2:1978:Table 1

C: Imposed load

Critical case when HA Loading and HB Loading on both Lanes

C1: HA Loading

Characteristic Load

UDL = = 40.22kN/m

KEL = = 13.41 kN/m

Design Load

UDL = 40.22 x 1.3 = 52.29 kN/m

KEL = 13.41 X 1.3 = 17.433 kN/m

C2: HB Loading

Characteristic Load

Load/wheel = 2.5 x 37.5 = 93.75 KN

Load/ wheel = 93.75 x 4 = 375 KN

= 375 x x

=375 x = 134.08 KN/m

Design Load

= 134.08 x 1.3 = 174.304 KN/m

C3: Pedestrian Loading

Charactaristic Load

=5 x 0.9 x 24 x = 12.067 kN/m

Design Load

=12.067 x 1.5 = 18.1 kN/m





BS 5400:P2:1978:Table 1

D: Longitudinal load

D.1: Braking load

Characteristic Load

Due to HA: = (8 x 24 + 200) = 392 KN

Due to HB, 25 % Total HB = 0.25 X 1500 =375 kN

HA = = 43.799kN/m

HB = = 41.899 kN/m

For, critical HA use 43.799 kN/m

Design Load

Due to HA = 43.799 x 1.25 = 54.75 kN/m

Due to HB = 41.899 x 1.1 = 46.09 kN/m

HA critical = 54.75 kN/m

D.2 Earth Pressure due to back fill

Characteristic Load

P = 17 x 0.271 x 75 = 34.5525 kN/m2 ( triangular)

Pn = x 34.5525 x 7.5 = 129.57 kN/m

Design Load

P = 34.5525 x 1.5 = 51.83 kN/m2

Pn = 129.57 x 1.5 = 194.36 kN/m





BD 31/01P12:cl.3.2.6

BS 5400:P2:1978:Table 1

D: Longitudinal load

D.3: Due to Surcharge

Characteristic Load

For HA Loading =10 kN/m2

For HB Loading = 20 kN/m2

Hence HB critical

Design Load





7.4 Stability Check

Load Combination

A: Case 1: Back fill + construction surcharge

B: Case 2: Back fill + surcharge + Deck dead load

C: Case 3: Back fill + surcharge + Deck dead load + (H A + H B)

Loading+ Braking

A: Case 1: Back fill + construction surcharge

2.0m

5.5m

1.0m





Load type N(kN/m)

V(kN/m)

La

(m) (kNm) (kNm)

DL from abutment 186.0 2.25 418.5DL from foot 110.4 2.3 253.92Backfill -Earth 176.8 3.8 671.84Const. surcharge 19.2 3.8 72.96Earth Backfill 129.57 2.5 323.95Surcharge 20.33 3.75 76.24

1417.2 400.17

Safety against

Overturning

OK!

Sliding

Active force = 129.57 + 20.33 = 149.9 kN/m

Friction force

= 284.3 kN/m

(NOT OK!!)

Hence

Change the dimension of the base, Base width changed from 4.6m to 8.5m

Heel = 5.5 m

Width = 1.5 + 1.5 + 5.5 = 8.5 m





Carry out stability check with revised footing size

Load type N(kN/m)

V(kN/m)

La

(m) (kNm) (kNm)

DL from abutment 186.0 2.25 418.5DL from foot 204.0 4.25 867Backfill -Earth 607.8 5.75 3494.9Const. surcharge 66.0 5.75 379.5Earth Backfill 129.57 2.5 323.95Surcharge 20.33 3.75 76.24

1063.8 149.99 5159.9 400.17

Safety against

Overturning

OK!

Sliding

Active force = 129.57 + 20.33 = 149.9 kN/m

Friction force

OK!





Bearing pressure

P = 1063.8 kN/m

A = 8.5 m2/m

Z = 8.52/6 = 12.042 m3/m

Net moment = 5159.9 – 400.17 = 4759.73 kNm/m

Eccentricity (e) of p about center line.= 4.5 –

= 4.5 – 4.474

=0.026 m

Pressure under the base

Pressure under toe = 127.447 < 250 kN/m2

Pressure under Heel = 122.853

OK! Hence abutment stable for case 1.





B: Case 2: Back fill + surcharge + Deck dead load

Load type N(kN/m)

V(kN/m)

La

(m) (kNm) (kNm)

DL from abutment 186.0 2.25 418.5DL from foot 204.0 4.25 867.0Backfill -Earth 607.8 5.75 3494.9DL from superstructure

104.7 5.99 627.15

Superimposed DL 26.57 5.99 159.15Earth Backfill 129.57 2.5 323.95Surcharge 20.33 3.75 76.24

1129.07 149.9 1417.2 400.17

Safety against

Overturning

OK!

Sliding

Active force = 129.57 + 20.33 = 149.9 kN/m

Friction force

OK!





Bearing pressure

P = 1129.07 kN/m

A = 8.5 m2/m

Z = 8.52/6 = 12.042 m3/m

Net moment = 5566.7 – 400.17 = 5166.53 kNm/m


=0.0759 m


132.83 7.116 < 250 kN/m2






B: Case 3: Back fill + surcharge + Deck dead load +

(H A + H B) Loading+ Braking

Load type N(kN/m)

V(kN/m)

La

(m) (kNm) (kNm)

DL from abutment 186.0 2.25 418.5DL from foot 204.0 4.25 867.0Backfill -Earth 607.8 5.75 3494.9DL from superstructure

104.7 5.99 627.15

Superimposed DL 26.57 5.99 159.15Live Load 199.77 5.99 1196.7Braking Load 43.77 4.5 196.97Earth Backfill 129.57 2.5 323.95Surcharge 20.33 3.75 76.24

1328.84 193.67 6763.4 597.14

Safety against

Overturning

OK!

Sliding

Active force = 129.57 + 20.33 + 43.77 = 193.67 kN/m

Friction force

OK!





Bearing pressure

P = 1328.84 kN/m

A = 8.5 m2/m

Z = 8.52/6 = 12.042 m3/m

Net moment = 6763.37 – 597.14 = 6166.23 kNm/m


=0.14 m


156.24 15.45 < 250 kN/m2


Hence the revised sizing of wall is stable against all three cases





7.5 Structural design of Abutment

7.5.1 Structural Framing of the Wall

7.5.2 Analysis

Taking load case 3 as critical, and taking moment about center line of the wall

Load type N(kN/m)

V(kN/m)

La

(m)M (kNm)

DL from abutment 213 0.00 0.00Superstructure D.L 117.185 0.255 29.882Superimposed D.L 46.48 0.255 11.8524Live load 244.03 0.255 62.22Braking 54.75 4.50 246.38Earth Backfill 194.36 2.167 421.18Surcharge 30.495 3.25 99.11

621.595 279.61 870.624

2.0m

5.5m

1.0m

5.5m 1.5m 1.5m





BS 5400:P4:cl.5.6

cl.5.4.

7.5.3 Reinforcement Design

Design values

Normal force = 621.595kN/m

Shear force = 279.61 kN/m

Bending moment = 870.624 kN/m

Total Axial load = 621.595 kN

Check

0.1fcu Ac = 0.1 x 30 x 103 x 8.95 x 1 = 26850 kN > 621.595kN

Hence design as a slab

Let d = 1000 – 50 - use = 25

=1000- 50 – 12.5 =937.5 mm

Use Y 25 – 150 c/c (AS = 3270 mm2/m)

z =

= 0.94d < 0.95d

Mu = 0.95 fy AS z

=0.95 x 460 x 3270 x 0.94 x 937.5 x 106

= 1259.3 kNm/m > 870.624 KNm/m

For horizontal bar provide minimum

As = x 1500 x 1000 = 1950 mm2/m

Y16 – 100c/c (2010mm2/m)





BS 5400:P4:cl.5.3.3

Table 8

Check Shear

= 0.298 N/mm2 < 0.75 of 4.755 N/mm2

=

= 0.464 N/mm2

Corrected = 0.75 X 0.464 = 0.348 N/mm2

(No shear reinforcement required)

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN OF BASE SD 677 August 2011 67/




STEP 8: DESIGN OF BASE ( FOUNDATION )

8.1 Analysis

Load type N(kN/m)

V(kN/m)

La

(m) (kNm) (kNm)

DL from abutment 214.0 2.25 481.5DL from Base 234.6 4.25 997.05Backfill -Earth 729.4 5.75 4193.8Superstructure DL 117.2 5.99 701.94Live load 244.0 5.99 1461.7Superimposed D.L 46.5 5.99 278.42Braking 54.75 4.5 246.38Earth Backfill 194.36 2.167 421.18Surcharge 30.495 3.75 114.36

1585.66 279.61 8114.5 781.92

Bearing pressure

P = 1585.66 kN/m

A = 8.5 m2/m

Z = 8.52/6 = 12.042 m3/m

Net moment = 8114.5 – 781.92= 7332.55 kNm


=0.374 m






186.55 49.25 < 250 kN/m2

Pressure under the toe = 235.8 kN/m2

Pressure under the heel = 137.3kN/m2

= 137.3 + 63.74

= 201.04kN/m





BS 5400:P4:1978cl.5.7.3

8.2 Design of Heel

Taking moment about the stem center line,

M=531.3 + 2552.76 – 2646.875 – 452.76

M= -15.58 kNm

Use Y25 – 200 c/c (2450 mm2/m)

Effective depth, d = 1000 – 75 – 12.5 = 912.5mm

z = 0. 955d > 0.95d

Change As,

use Y25 – 175 c/c (2810 mm2/m)

z = 0.948d < 0.95d

Check

Mu = 0.95fyAsz

Mu= 0.95 x460 x2810 x0.948 x 912.5

Mu=1062.25 kNm > 15.58kNm

Provide Y25 – 175c/c (2810 mm2/m)





BS 5400:P4:1978cl.5.7.3

8.4 Design of Toe

Taking moment about the stem center line

M =

M = 62.1 – 550.55

M = -468.45 kNm

Try Y25 – 175 c/c (2810mm2/m)

Effective depth, d = 1000 – 75 – 12.5 = 912.5

z = 0.948d > 0.95d

Check

Mu = 0.95fyAsz

Mu = 0.95 x460 x2810 x0.948 x 912.5

=1062.25 kNm > 468.45kNm

Provide Y25 – 175c/c (2810 mm2/m)

Distribution steel for both Toe and Heel

= 1300 mm2/m

Provide Y16 – 150 c/c (1340 mm2/m)

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN OF CURTAIN WALL SD 677 August 2011 71/




STEP 9: DESIGN OF CURTAIN WALL

9.1 Analysis

The wall is designed to be cast onto the top of the abutment

Loading will be applied from the backfill, surcharge and braking loads on

top of the wall.

A: Braking load

A.1 HB critical

25% x 37.5 units x 10 = 93.75 kN

assuming 450 dispersion to the curtain wall and max dispersal width of the

abutment (8.950 meter )

1st axle = = 31.25kN/m

2nd axle = = 14.205 kN/m

3rd & 4th axle = 20.95 kN/m

Maximum load on back of abutment

= 31.25 + 14.205 + 20.95= 66.41 kN/m

Bending and shear at the base of 2m high curtain wall

A.1.1 Horizontal load due to HB surcharge

= 20 x 0.271 x 2 = 10.84 kN/m

A.1.2 Horizontal load due to backfill

= 17 x 0.271 x 2 = 9.214 kN/m2

= = 9.214 kN/m





BS 5400:P2:1978cl.5.7.3

B: Design Moment and Shear

B.1 ULS Moment

1.1x 183.094

201.4kNm/m

B.2 ULS Shear

113.4452kN/m

9.2 Reinforcement Design

Bending reinforcement

Effective depth ,d = 500 – 75 – 12.5 = 412.5mm

Try Y20 – 200 c/c

z = 0.93d > 0.95d

Check

Mu = 0.95fyAsz

Mu = 0.95 x460 x1570 x 0.936 x 412.5

Mu = 264.89 kNm > 201.4 kNm/m

Hence ,Provide Y20 – 200c/c





BS 5400:P4: 1978cl.5.3.3

BS 5400:P4.1978

Table 8

Table 9

BS 8110:P1:Table3.25

Shear reinforcement

or 4.75 N/mm2

= 0.381

From table 8.

vc= 0.4772

depth factor correction = 1.0

= 0.4772

vc> v [no shear reinforcement required]

Distribution steel

Provide minimum

= 536.25 mm2/m

Provide Y16 – 200 c/c (1010 mm2/m)

Project/Structure: Subject Project Ref: Date: Sheet No: Design of Bridges DESIGN OF BEARING SD 677 August 2011 74/




STEP 10: DESIGN OF ELASTOMETRIC BEARING

10.1 Loading

Total Dead Load = 328.49kN/m

Superimposed Dead Load = 46.48kN/m

HA Loading = 69.723kN/m

HB Loading = 134.08kN/m

Total Vertical Load = 578.78kN/m

Total Horizontal load

( Braking ) = 54.75kN/m

10.2Assumption

Modulus of rigidity = 1N/mm2

Friction coefficient ( ) = 0.3

Design based on Indian standard and British standard

10.3Bearing Sizing

Selecting Index NO 6 (bearing) based on IRC 83 1987 part 11

10.4Design of bearing

A: Thickness

Selecting the thickness of the bearing to be

Check

OK!





B: Bearing Check

Area ,A = 250 X 500 =125000mm2

But,

OK!

C: Axial Stress

Check

But

and

Hence DESIGN IS SAFE !





B: Slip Check

Check 1:

But

OK!

Check 2

Hence

OK!

HENCE THE DESIGN IS SAFE !

SELECT INDEX SIZE NO 6, ELASTOMETRIC BEARING


Design of Bridges DESIGN OF EXPANSION JOINT SD 677 August 2011 77/




BS 5400:P2: 1978Figure 7 andFigure 8

Figure 9

Cl 5.4.6

BS 5400:P2: 1978cl.5.4.3Table 10

STEP 11: DESIGN OF EXPANSION JOINT

From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air

temperatures are -19 and +37oC respectively.

For a Group 4 type structure (see fig. 9) the corresponding minimum and

maximum effective bridge temperatures are -11 and +36oC

Hence the temperature range = 11 + 36 = 47oC.

The range of movement at the free end of the 24m span deck

= 47 x 12 x 10-6 x 24 x 103 = 13.5mm.

The ultimate thermal movement in the deck will be

= ±[13.5 x 1.1 x 1.3 /2] = ± 9.6mm

Taking the air temperature range to be -19 to 37 degree centigrade

The bearings to be installed at a shade air temperature of

[(37+19)/2 -19] = 9oC to achieve the ± 9.6mm movement.

hence ,If the bearings are set at a maximum shade air temperature of 12oC

then, by proportion the deck will

Expand

Contract

Provide 10mm expansion gap





5.0 DESIGN DRAWINGS AND SCHEDULES

Documents

Design of Composite Bridge