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Dissertation Titled “COMPARATIVE STUDY OF IS: 800 (DRAFT) CODE & EUROCODE 3 ENV: 1993-1-1 (PART 1-1 GENERAL RULES AND RULES FOR BUILDING)” Submitted by Swapnil B.Kharmale (Roll No. CD-051061) M. Tech. (Civil Engineering with Specialization in Structural Engineering) 2005 - 2007 Guided by Mr. B.A. Naik 2006-2007 Department Of Structural Engineering Veermata Jijabai Technological Institute. (Autonomous Institute Affiliated to University of Mumbai) EStelar

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Page 1: 123 Unlocked

Dissertation Titled

“COMPARATIVE STUDY OF IS: 800 (DRAFT) CODE

& EUROCODE 3 ENV: 1993-1-1

(PART 1-1 GENERAL RULES AND RULES FOR BUILDING)”

Submitted by

Swapnil B.Kharmale

(Roll No. CD-051061)

M. Tech. (Civil Engineering with Specialization in Structural Engineering)

2005 - 2007

Guided by

Mr. B.A. Naik

2006-2007

Department Of Structural Engineering

Veermata Jijabai Technological Institute.

(Autonomous Institute Affiliated to University of Mumbai)

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Statement by the candidate

I wish to state that work embodied in this dissertation titled “COMPARATIVE

STUDY OF IS: 800 (DRAFT) CODE & EUROCODE 3 ENV: 1993-1-1(PART 1-1

GENERAL RULES AND RULES FOR BUILDING)” forms my own contribution to

the work carried out under the Guidance of Mr. B. A. Naik at the Veermata Jijabai

Technological Institute. This work has not been submitted for any other Degree or

Diploma of any University/ Institute. Wherever references have been made to

previous works of others, it has been clearly indicated.

Signature of Candidate

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ACKNOWLEDGEMENT

It give me immense pleasure to present this report entitled

“COMPARATIVE STUDY OF IS: 800 (DRAFT) CODE & EUROCODE 3

ENV: 1993-1-1 (PART 1-1 GENERAL RULES AND RULES FOR

BUILDING)”

I wish to acknowledge with deep sense of gratitude, my indebtedness to my

guide Prof.B.A.Naik for his valuable guidance. In spite of his busy schedule, he

spared time, took keen interest, reviewed my work, discussed at length and gave me

constant encouragement to complete this dissertation.

I am also thankful to Dr. K. G. Narayankhedkar, Director, V.J.T.I., Mumbai

and Prof. M. G. Gadgil, Head, Structural Engineering Department V.J.T.I., Mumbai

for extending relevant facilities during this work.

On many occasions, Prof K.K.Sangle and Prof N.M.Damle took part in

discussion and enlightened about the current practice in the field. I am thankful for

their helpful suggestions and practical hints.

Last but not least, I am deeply grateful to my family members, all my friends

and well-wishers for encouraging and helping me directly or indirectly, throughout

my project work.

SWAPNIL B. KHARMALE

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Swapnil B.Kharmale Comparative study of IS: 800 (Draft) & EC3 CD-051061

i

List of tables

A.1.1 Countries and their design format

A.2.1 Sections of IS: 800 (Draft)

A.2.2 Appendix of IS: 800(Draft)

A.2.3 General comparison between IS: 800-1984 and IS: 800 (Draft)

A.3.1 Sections/chapters of Eurocode 3 Part 1.1

A.3.2 Annexure of Eurocode 3 Part 1.1

A.4.1 List of symbols used in both codes

A.4.2 Conventions for member axis as per IS: 800 (Draft)

A.4.3 Conventions for member axis as per IS: 800 (Draft)

B.1.1 Limit states

B.1.2 Partial safety factors for loads ( f) for different limit states as per IS: 800

(Draft)

B.1.3 Partial safety factors for actions ( f) for persistent and transient design

situation as per Eurocode 3

B.1.4 Design values of actions as per Eurocode 3

B.1.5 Partial safety factor for material property ( m) as per IS: 800 (Draft)

B.1.6 Partial safety factor for material property ( M) as per Eurocode 3

B.1.7 Classification of cross section by both codes

B.1.8 Limiting width to thickness ratios for outstand flange element

B.1.9 Limiting width to thickness ratios for internal flange element subjected to

axial compression

B.1.910 Limiting width to thickness ratios for internal flange element subjected to

bending compression

B.1.11 Limiting width to thickness ratios for web element subjected to axial

compression

B.1.12 Limiting width to thickness ratios for web element subjected to bending

compression

B.1.13 Limiting width to thickness ratios for angles

B.1.14 Limiting width to thickness ratios for tubular sections

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Swapnil B.Kharmale Comparative study of IS: 800 (Draft) & EC3 CD-051061

iii

B.6.4 Shear area Av

B.7.1 Constant 1 ( ) and 2 ( �B.7.2 Design reduced flexural strength for plastic and compact class without bolt

hole (Approximate formulae as per IS code and Eurocode)

B.7.3 Equivalent uniform moment factor

B.8.1 Minimum edge and end distances of fasteners

B.8.2 Slip factor µf

B.8.3 &RUUHODWLRQ�IDFWRU� w

B.8.4 Strength of weld per unit length of weld as per both codes

C.1.1 Analysis results and load combination for design by both codes

C.2.1 Comparison of design capacity of various elements of FOB by both codes

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Swapnil B.Kharmale Comparative study of IS: 800 (Draft) & EC3 CD-051061 v

B.5.10 Shear and moments in batten

B.6.1 Laterally supported beam

B.6.2 Laterally unsupported cantilever undergoing Lateral Torsional Buckling

B.6.3 Behaviour of simply supported beam

B.6.4 Transition from elastic to plastic stage of cross section in bending

B.6.5 Behaviour of continuous beam

B.6.6 Effect of shear force on Mp (Full plastic moment capacity)

B.6.4 Deflection of simply supported beam undergoing Lateral Torsional

Buckling

B.7.1 Elastic behaviour of cross sections in compression and bending

B.7.2 Full plasticity under axial load and moment

B.7.3 Full plasticity interaction major axis bending of ISHB 450

B.8.1 Connections in multistoryed building

B.8.2 Simple connections

B.8.3 Rigid beam to column connection

B.8.4 Symbols for spacing of fasteners

B.8.5 Long joint

B.8.6 Bolted connection with four bolts

B.8.7 Distribution of forces in case of long joint

B.8.8 Prying force

C.1.1 Geometry of FOB

C.1.2 Modeling of FOB in STAAD PRO

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Swapnil B. Kharmale 2 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

This example shows the improvements in understanding of structural behaviour and

subsequently developments in analysis and design from 1950 to till today. Thus to

avail the benefits of these developments the design code of particular country should

incorporate the modern analysis and design technique

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Swapnil B. Kharmale 4 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

maintaining Allowable Stress Design (ASD) as a transition alternative, which will

help the designers to understand both the design methods and utilize the most

advantageous one, and only recently the Indian Standards Institution has taken up

the job of revising IS 800 to the Limit state method design which is at present at an

advanced stage, with a purpose of evolving a code which will be understandable,

easy to use and based on good and widely practiced structural theory to deal

properly with elastic instability, dynamic loads and fatigue.

A.1.2.2 Countries and their Design Formats

Almost all advanced countries are now taking advantage of efficient code

stipulations, and current practice all over the world is based on either Limit State

Method (LSM) or Load and Resistance Factor Design (LRFD).

Following table shows the various major countries and their Design Format

Table A.1.1:- Countries and their Design Formats

Countries

Design Formats (For Steel Structure)

Australia , Canada , China , Europe ,Japan

United Kingdom (UK)

Limit State Method (LSM)

U.S.A. Load And Resistance Factor Design

(LRFD)

India a) IS: 800-(1984)

b) IS: 800 (Draft)

Allowable Stress Design (ASD)

Limit State Method

A.1.2.3 IS: 800 (Draft)

The total Draft is prepared is based on the stipulations of International

Standards as applicable and Teaching Resource for Structural Steel Design of

INSDAG (a committee comprising experts from IIT, SERC)

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Swapnil B. Kharmale 6 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

3) Comparing similarities as well as differences between both codes and also

examining the efficient way of designing and if possible finding how best we

can incorporate it in our code.

4) Searching limitations of both codes and if possible trying to overcome it

through detailed study.

5) To document step-by step procedure for designing different types of structural

elements, clearly highlighting different methodology adopted in two different

countries so that it may be helpful for undergraduate student as well as

practicing engineer.

6) To study economy achieved by designing through both code.

A.1.4 Scope of the present work

IS: 800 (draft) consists of total 17 section and 9 appendices where as

Eurocode 3 (ENV1993-1-1, General rules and rules for building) contains 9 sections

and 10 annexure covering the specifications, standards and rule for design off steel

structure. It is very vast to cover all sections in details for comparison purpose.

Hence it is considered cover the basic and elementary section for in detail study

purpose. The study work is broadly divided in to following three parts

Part 1:- Comparative Study of Basic Sections by Both Codes through

• Basis Of Design

• Section Classification

• Tension Member

• Compression Member

• Member Subjected To Bending

• Member Subjected To Combined Forces

• Connections

This consists of studying the basis of clauses (for above mentioned sections)

mentioned in both codes followed by illustrated examples by corresponding codes.

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Swapnil B. Kharmale 8 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

Section A:- Introduction A.2 About IS: 800 (Draft)

A.2.1 Content of IS: 800 (Draft) Following tables (Table A.2.1 and A.2.2) represent the content of IS:800

(Draft)

Table A.2.1:-Sections of IS: 800 (Draft)

Sections /Chapters Name Of Sections/Chapters

Section 1 General

Section 2 Materials Section 3 General design requirements Section 4 Methods of structural analysis

Section 5 Limit state design

Section 6 Design of tension member Section 7 Design of compression member Section 8 Design of member subjected to bending Section 9 Member subjected to combined forces Section10 Connection Section11 Working load design format Section12 Design and detailing for earthquake load Section13 Fatigue Section14 Design assisted by testing Section15 Durability Section16 Fire resistance Section17 Fabrication and erection

Table A.2.2:-Appendix of IS: 800 (Draft)

Appendix

Name Of Appendix

Appendix A Chart showing highest maximum temperature Appendix B Chart showing lowest minimum temperature Appendix C Advanced method of analysis and design Appendix D Design against floor vibration Appendix E Methods for determining effective length of columns in

frame Appendix F Lateral torsional buckling Appendix G Connections Appendix H General recommendations for steelwork tenders and

contract Appendix I Plastic properties of beams

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Swapnil B. Kharmale 10 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

a) Elastic analysis

b) Plastic analysis

c) Advanced analysis

The assumptions, requirements and application of each above method have

been discussed in detail in this section.

In addition to the above method of analysis, for the purpose of analysis and design

the Classification of structural frames, Forms of constructions assumed for analysis are

described.

2) SECTION -5 LIMIT STATE DESIGN

In this section; basis for limit state design, two limit state viz Limit state of strength

and Limit state of serviceability are discussed

The actions (Load), classification of actions, design action, strength, design

strength, ultimate strength, and partial� VDIHW\� IDFWRUV� IRU� ORDGV� � f) and for material

VWUHQJWK�� m) are described in detail.

The Sections – 6, 7, 8, 9, 10 (considering Design of -Tension member,

Compression member, Members subjected to bending, Member subjected to combined

forces) deals with Limit State Design format.

3) SECTION-11 WORKING LOAD DESIGN FORMAT

This section deals with working load design format In old code design is based on

working stress method which is modified and presented under Working Load Design

Format in the Draft Code This section deals with design criteria for

a) Tension member

b) Compression member

c) Member subjected to bending

d) Member subjected to combined stresses

4) SECTION-12 DESIGN AND DETAILING FOR EARTHQUAKE LOADS

This section covers the requirements for designing and detailing of steel frames so

as to give them adequate strength, stability and ductility to resist sever earthquake in all

zones of IS:1893 without collapse.

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Swapnil B. Kharmale 12 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

8) SECTION-16 FIRE RESISTANCE

This newly introduced section applies to steel building elements designed to exhibit

a required fire resistance level as per given specification

This section include definition of related terms, different fire exposure condition, fire

resistance level, periods of structural adequacy as well as the variation of mechanical

properties of steel with temperature (i.e. variation of yield stress fy and modulus of

elasticity Es)

Appendices:-

Following are 3(three) newly introduced appendices:-

1) APPENDIX -C ANALYSIS AND DESIGN METHODS (ADVANCED STRUCTURAL

ANALYSIS AND DESIGN)

This appendix gives advanced structural analysis and design methods for a frame

comprising members of compact section with full lateral restraints (i.e. laterally

supported members) and Second Order Elastic and Design

2) APPENDIX- D DESIGN AGAINST FLOOR VIBRATION

This section applicable for design of floors with longer spans and of lighter section

and less damping as these structure are more sensitive to vibrations under normal

human activities.

The appendix gives the determination of floor frequency, peak acceleration and

table for critical damping which required for dynamic analysis.

3) APPENDIX-G CONNECTIONS

In this appendix requirements for design of splice (Beam splice, Column splice)

and beams to column connections as well as recommendations for their design are

discussed

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Swapnil B. Kharmale 14 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

a) To comply with essential requirements of Construction Products

Directive (CPD)

b) To frame harmonized technical specification for construction products

A.3.3 Scope of Eurocode 3

1) Eurocode 3 applies to the design of buildings and civil engineering works in

steel. It is subdivided into various separate parts, such as 1.1.2 and 1.1.3.

2) This Eurocode is only concerned with the requirements for resistance,

serviceability and durability of structures. Other requirements, e.g. concerning

thermal or sound insulation are not considered.

3) Eurocode 3 does not cover the special requirements of seismic design. Rules

related to such requirements are provided in ENV: 1998 Eurocode 8 "Design

of structures for earthquake resistance" which complements or adapts the

rules of Eurocode 3 specifically for this purpose.

4) Numerical values of the actions on buildings and civil engineering works to be

taken into account in the design are not given in Eurocode 3. They are

provided in ENV: 1991 Eurocode 1 "Basis of design and actions on

structures" which is applicable to all types of construction.

A.3.3.1 Scope of Part 1.1 of Eurocode 3

1) Part 1 .l of Eurocode 3 gives a general basis for the design of buildings and

civil engineering works in steel.

2) In addition, Part 1.1 gives detailed rules which are mainly applicable to

ordinary buildings. The applicability of these rules may be limited, for practical

reasons or due to simplifications; their use and any limits of applicability are

explained in the text where necessary.

3) The following subjects are dealt with in this initial version of Eurocode 3: Part

1.1

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Swapnil B. Kharmale 16 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

Further Parts of Eurocode 3 are as follows x Part 1.2 Fire resistances x Part 1.3 Cold formed thin gauge members and sheeting x Part 2 Bridges and plated structures x Part 3 Towers, masts and chimneys x Part 4 Tanks, silos and pipelines x Part 5 Piling x Part 6 Crane structures x Part 7 Marine and maritime structures x Part 8 Agricultural structures

A.3.4 National Application Documents (NAD) for Eurocodes

There are 18 countries following the Eurocode. The design parameter

(materials, sections, climatic conditions) may vary from country to country. Therefore

NAD is introduced to express the national choices depending upon the

corresponding design situations.

Alternatively, National Application Documents refer to other publications that

provide the information or guidance to enable the Eurocode to be used for design for

design of a structure in particular country.

In this Dissertation the Eurocode 3 (ENV: 1993-1-1) together with United

Kingdom National Application Document are used.

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Swapnil B. Kharmale 18 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

-- --- G.sup Upper value of partial safety factor for permanent action

-- --- G.inf Lower value of partial safety factor for permanent action

-- --- Q

Partial safety factor for variable action

-- --- APartial safety factor for accidental action

f

Partial safety factor for load for fatigue

FF Partial safety factor for action for fatigue

Qd Design action Fd Design action Sd Design material strength Xd Design material strength Sm Characteristic value of

material strength XkCharacteristic value of material strength

mPartial safety factor for material M

Partial Safety factor for material property

m0

Partial safety factor for material when resistance is governed by yielding or buckling

M

Partial safety factor for material property for particular governing mode of failure

m1 Partial safety factor for material when resistance governed by Ultimate stress.

Mf Partial safety factor for material property for fatigue

mft Partial safety factor for material for fatigue

-- ---

nf Partial safety factor for material strength for bolts-friction type

-- ---

mb Partial safety factor for material strength for bolts-bearing type

-- ---

Yield Stress ratio=¥�����Iy) Yield Stress ratio=¥�����Iy)-- --- cr Critical Plate buckling stress -- --- k Plate buckling factor -- --- S

Plate slenderness

-- --- E�G Effective dimensions of Class 4 section

-- --- eN

Shift of centroidal axis when Class 4 section subjected to compression

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Swapnil B. Kharmale 20 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

D) For chapter “ Design Of Compression Member”

Pd Design compressive strength of member

NSd Design value of compressive force

-- --- Nc.Rd Design compression

resistance of the cross section

-- --- No.Rd Design local buckling resistance of gross section

-- --- Nb.Rd Design buckling resistance of compression member

Ae Effective sectional area Aeff Effective sectional area �-Ratio of effective area in compression to gross area

fcd Design stress in compression -- --- Non dimensional effective slenderness ratio

Non dimensional effective slenderness ratio

KL Appropriate effective length l Buckling length r Appropriate radius of gyration i Appropriate radius of

gyration (KL/r) Effective slenderness ratio �O�L� Effective slenderness ratio

fcc Euler buckling stress Ncr Elastic critical force

Imperfection factor corresponding to appropriate buckling curve

Imperfection factor corresponding to appropriate buckling curve

Stress reduction factor Stress reduction factor w Uniform soil pressure ts Minimum thickness of slab

base t Minimum thickness of slab

base a , b Larger and smaller projection

of slab base beyond rectangle circumscribing the column

ar , br Larger and smaller projection of slab base beyond rectangle circumscribing the column

e Equivalent non dimensional slenderness ratio for single angle strut loaded through one leg

HIIEquivalent non dimensional slenderness ratio for single angle strut loaded through one leg

rvv Radius of gyration about minor axis

ivv Radius of gyration about minor axis

ryy Radius of gyration about y-y axis

iyy Radius of gyration about y-y axis

rzz Radius of gyration about z-z axis

izz Radius of gyration about z-z axis

-- --- Ieff Effective in-plane second

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Swapnil B. Kharmale 22 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

Iyy Second moment of area of section about minor axis (y-y) Izz

Second moment of area of section about minor axis (z-z)

Iw Warping constant Iw Warping constant It Torsion constant It Torsion constant

IsSecond moment of area intermediate transverse stiffener

IsSecond moment of area intermediate transverse stiffener

Ifc Second moment of area of compression flange about minor axis

Ifc Second moment of area of compression flange about minor axis

Ift Second moment of area of tension flange about minor axis

Ift Second moment of area of tension flange about minor axis

LT Non dimensional effective slenderness ratio for Lateral Torsional Buckling

/7

Non dimensional effective slenderness ratio for Lateral Torsional Buckling

(KL/r) Slenderness ratio for Lateral Torsional Buckling LT Slenderness ratio for Lateral

Torsional Buckling=(l/i)z.

fbd Design bending compressive stress

LT Imperfection factor for Lateral Torsional Buckling LT Imperfection factor for

Lateral Torsional Buckling

LT Stress reduction factor for Lateral Torsional Buckling LT Stress reduction factor for

Lateral Torsional Buckling Mcr Elastic critical moment Mcr Elastic critical moment V Factored shear force VSd Design value of shear force

VnNominal plastic shear resistance under pure shear

-- ---

VpDesign shear strength of section

Vpl.Rd Design plastic resistance of cross section

Av Shear area Av Shear area

wNon-dimensional web slenderness ratio :

Web slenderness

cr Elastic critical shear strength cr Elastic critical shear strength kv Buckling factor for shear kt Buckling factor for shear

c Clear spacing between transverse stiffeners

a Clear spacing between transverse stiffeners

d Depth of web (For plate Girder)

d Depth of web (For plate Girder)

Vcr Design shear buckling strength of section by Simple post- critical method

Vba.Rd Design shear buckling resistance of section by Simple post- critical method

bBuckling strength of web

ba Simple post-critical shear strength

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Swapnil B. Kharmale 24 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

diagram between lateral bracing points in the appropriate plane of bending

diagram between lateral bracing points in the appropriate plane of bending

G)For Chapter “ Connection”

d Nominal diameter of the fastener

doNominal diameter of the fastener

e Edge distance

e Edge distance

-- --- Fv.Sd Design shear force per bolt

Vnsb Nominal shear capacity of a bolt

Fv.Rd Design shear resistance per bolt

nn

Number of shear planes with threads intercepting the shear plane

-- ---

ns

Number of shear planes without threads intercepting the shear plane

-- ---

Asb Nominal plain shank area of the bolt

-- ---

Anb Net tensile area at threads

-- ---

tpk Thickness of the thicker packing

tpThickness of plate

Vnpb Bearing strength of a bolt

-- ---

Tnb Nominal tensile capacity of the bolt

-- ---

fub Ultimate tensile stress of the bolt

-- ---

fyb Yield stress of the bolt

-- ---

Asb Shank area of the bolt

-- ---

Vnsf Nominal shear capacity of a bolt as governed by slip for friction type connection

Fs.Rd Design slip resistance

µfCoefficient of friction (slip factor)

µ Coefficient of friction (slip factor

ne

Number of effective interfaces offering frictional resistance to slip

nNumber of effective interfaces offering frictional resistance to slip

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Swapnil B. Kharmale 26 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

Unless otherwise specified convention used for member axes is as follows:

Table A.4.2:- Convention for member axes as per IS: 800 (Draft)

Axis Description

x-x Along the member

y-y An axis of the cross section

x Axis perpendicular to the flanges

x Axis perpendicular to smaller leg in angle section

z-z An axis of the cross section

x Axis parallel to flanges

x Axis parallel to smaller leg in angle section

u-u Major axis (when it does not coincide with y-y or z-z axis)

v-v Minor axis (when it does not coincide with y-y or z-z axis)

As per Eurocode 3

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Swapnil B. Kharmale 28 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

A.4.4 Notes: -

1. In this dissertation whenever the term both codes appears it indicate that IS:

800 (Draft) and Eurocode 3

2. EC 3 is a short form of Eurocode 3

3. In this dissertation while discussing the common theory under different

chapter the term in bracket () in front of the symbols or terms as per IS: 800

(Draft) means the equivalent symbols or terms as per Eurocode 3

For e.g.:- Plastic (Class1):- Here Plastic is term used for cross section type as

per IS: 800 (Draft) and Class 1 is equivalent term as per Eurocode 3

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Swapnil B Kharmale 30 Comparative study of IS: 800 (Draft) &EC3 CD-051061

overlap, shaded area, the effect of the loads is greater than the resistance of the

element, and the element will fail. shown by the shaded area, the effect of the loads

is greater than the resistance of the element, and the element will fail.

.

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Swapnil B Kharmale 32 Comparative study of IS: 800 (Draft) &EC3 CD-051061

x Failure by excessive deformation, rupture of the structure or any of its parts or

components.

x Fracture due to fatigue.

The limit state of serviceability include:-(See figure B.1.4 below)

x Deformation and deflections, which may adversely affect the appearance or,

effective, use of the structure or may cause improper functioning of

equipment or services or may cause damages to finishes and non-structural

members.

x Vibrations in the structure or any of its components causing discomfort to

people, damages to the structure, its contents or which may limit its functional

effectiveness. Special consideration shall be given to floor vibration systems

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Swapnil B Kharmale 34 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Design value of action:-

As per IS: 800 (Draft) (clause 5.3.1)

G IN FNN

4 4 (I.B.1.01)

As per Eurocode 3 (clause 2.2.2.4. (1))

G I N) ) (E.B.1.01)

Where, Qd=Design value of action as per IS:800 (Draft)

Jfk = Partial safety factor for loads

Qck=Characteristic actions

Fd= Design value of action as per Eurocode 3

� f = Partial safety factor for loads

Fk= Characteristic actions

Design value of material property:-

As per IS: 800 (Draft) (clause 5.4.1)

XG

P

66 � (I.B.1.02)

As per Eurocode 3(clause 2.2.3.1.(1))

XG

P

;; (E.B.1.02)

Where Sd= Design value of material property as per IS:800 (Draft)

m =Partial safety factor for material strength

Su=Characteristic material strength

Xd= Design value of material property as per Eurocode 3

m= Partial safety factor for material strength

Xu= Characteristic material strength

Partial safety factors ( ):-

The LSM approach is reliability based design criteria in which partial safety

factors ( ) are to be evaluated for given target (Safety index) Both the partial

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Swapnil B Kharmale 36 Comparative study of IS: 800 (Draft) &EC3 CD-051061

As per Eurocode 3 (Table 2.2 of Eurocode 3)

Table B.1.3:-Partial safety factor for actions on structures for persistent and

transient design situation as per Eurocode 3

For Variable Action( Q)Effect For Permanent Action

( G) Leading variable

action

Accompanying

variable action

Favourable effect

F.inf

1.0 - -

Unfavourable effect

F.sup

1.35 1.5 1.5

Combination of actions as per Eurocode 3(Clause 2.3.2.2)

Table B.1.4:-Design values of action as per Eurocode 3

Variable Action QdDesign

situation

Permanent

Action Gd Leading variable

action

Accompanying

variable action

Accidental

Action Ad

Persistent and

Transient

GGk QQk 0 QQk -

Accidental GAGk 1Qk 2Qk AAk

The design values given in above table shall be combined using the following rules

Persistent and transient design situation

*�M N�M 4�� N�� *�L N�LM L!�

* � 4 � 4 (E.B.1.03)

Accidental design situation

*$�M N� M G ��� N�� ��L N�LM L!�

* �$ � 4 � 4 (E.B.1.03)

where,

Gk.j = the characteristic values of the permanent actions

Qk.1 = the characteristic value of one of the variable actions

Qk. i = the characteristic values of other of the variable actions

Ad = the design value (specified value) of the accidental action

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Swapnil B Kharmale 38 Comparative study of IS: 800 (Draft) &EC3 CD-051061

x Moreover as per Eurocode 3 for Permanent actions where the coefficient

of variation is large or where the actions are likely to vary during the life of

the structure (e.g. for some superimposed permanent loads), two

characteristic values are distinguished, an upper (Gk.sup) and a lower

(Gk,inf) by using respective partial safety factor G.sup and� G.inf (Clause

2.2.2.2.(2) of Eurocode 3)

Partial safety factor for material property�� m)

As per IS: 800 (Draft) (Table 5.2 of IS:800 (Draft))

Table B.1.5:- Partial safety factor for material property�� m) as per IS:800 (Draft)

Sl. No Definition Partial Safety Factor

1 Resistance, governed by yielding m0 1.10

2 Resistance of member to buckling m0 1.10

3 Resistance, governed by ultimate stress m1

1.25

Shop Fabrications

Field Fabrications

4 Resistance of connection m1

i. Bolts-Friction Type, mf ii. Bolts-Bearing Type, mb iii. Rivets iv. Welds

1.25 1.25 1.25 1.25

1.25 1.25 1.25 1.50

As per Eurocode 3 (Table 1 of NAD)Table B.1.6:- Partial safety factor for material property�� m) as per Eurocode 3

Definition Symbol Condition Value

Partial safety factor for steel M0

M1

M1

M2

Resistance of Class 1,2,3

section

Resistance of Class 4 section

Resistance of member to

buckling

Resistance of net section at

bolt hole

1.10

1.10

1.10

1.25

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Swapnil B Kharmale 40 Comparative study of IS: 800 (Draft) &EC3 CD-051061

B.1.3.1 Classification of Cross Section by both codes:-

Following table shows the classification of cross section by IS: 800 (Draft)

and Eurocode 3. The classification is same just the nomenclature is different.

Table B.1.7:-Classification of cross section by both codes (Refer Fig B.1.1)

Section Classification as per

IS:800 (Draft) Eurocode 3 Definitions

Plastic Class 1 Cross section which can develop plastic hinge

and have the sufficient rotation capacity required

for failure of structure by formation of plastic

mechanism

Compact Class 2 Cross section which can develop plastic moment

of resistance but have inadequate plastic hinge

rotation capacity for formation of a plastic

mechanism.

Semi-compact Class 3 Cross section in which extreme fiber in

compression can reach yield stress but can’t

develop the plastic moment of resistance due to

local buckling.

Slender Class 4 Cross section in which element buckle locally

even before reaching yield stress.

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Swapnil B Kharmale 42 Comparative study of IS: 800 (Draft) &EC3 CD-051061

How the limiting width to thickness ratios are decided for particular class of sectionfor that behaviour of plate element in compression should be reviewed

B.1.3.4 Behaviour of plate elements in compression

For a plate element with an aspect ratio, a = a/b (length-to-width), greater than about

0.8, the elastic critical buckling stress (Euler buckling stress) is given by:

��

�FU

( W N �� �� E *2 (B.1.02)

Where k =The plate buckling factor It depends on

o a)Edge condition

o b) Aspect ratio a/b of plate element

o c) Nature of loading

Poisson’s coefficient,

E =Young’s modulus.

The critical buckling stress is proportional to (t/b) 2 and, therefore, is inversely

proportional to (b/t)2. The plate slenderness, or width-to-thickness ratio (b/t), thus

plays a similar role to the slenderness ratio (KL/r or l/i) for column buckling.

To avoid the local buckling phenomenon

cr � fy (B.1.03)

Now from equation (B.1.02) and (B.1.03) we can write

��

� \

( WN I�� �� E*2 From equation (9.7) of “Theory of Elastic stability” By:-S.P.Timoshenko &J.M.Gere

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Swapnil B Kharmale 44 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.1.8 Limiting width to thickness ratio for outstand flange element

Compression Element:-“Outstand Flange” subjected to axial and or bending compression

Code Stress ratio

Section Plastic / Class 1

Compact / Class2

Semi-compact /Class 3

Rolled b/tf���� b/tf����� b/tf�����IS:800(Draft) =¥(250/fy) Welded b/tf���� b/tf���� b/tf�����Rolled c/tf��� c/tf��� c/tf� ��Eurocode 3 =¥(235/fy)

Welded c/tf�9 c/tf��� c/tf���Table B.1. 9 Limiting width to thickness ratio for internal flange element

subjected to axial compression

Compression Element:-“Internal Flange Element” subjected to axial compression

Code Stress ratio

Section Plastic / Class 1

Compact / Class2

Semi-compact /Class 3

Rolled N.A. N.A. b/tf���IS:800(Draft) =¥(250/fy) Welded N.A. N.A. b/tf���Rolled (c-3tf)/tf��� (c-3tf)/tf��� (c-3tf)/tf���Eurocode 3 =¥(235/fy)

Welded c/tf��� c/tf��� c/tf���

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Swapnil B Kharmale 46 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.1.12 Limiting width to thickness ratio for web element subjected to bending compression

Compression Element:-“Web (internal)” subjected to bending compression

Code Yield Stress ratio

Section Plastic / Class 1

Compact / Class2

Semi-compact /Class 3

IS:800(Draft) =¥(250/fy) Any d/tw����� d/tw������ d/tw������

Eurocode 3 =¥(235/fy) Any d/tw��� d/tw��� d/tw����

Table B.1.13 Limiting width to thickness ratio for Angles

Compression Element:-“Angles” subjected to axial compression

Code Yield Stress ratio

Section Plastic / Class 1

Compact / Class2

Semi-compact /Class 3

IS:800(Draft) =¥(250/fy) Rolled N.A. N.A. b/t�����

d/t�����(b+d)/t���

Eurocode 3 =¥(235/fy) Rolled N.A. N.A. h/t���

(b+h)/2t�����

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Swapnil B Kharmale 48 Comparative study of IS: 800 (Draft) &EC3 CD-051061

As per Eurocode 3 (Clause 5.3.5)

The effective cross-section properties of Class 4 cross-sections shall be

based on the effective widths of the compression elements (see 5.3.2 (2)).

The effective widths of flat compression elements should be obtained using Table

B.1.15 (As per Table 5.3.2 and 53.3 of Eurocode 3) for internal elements and

outstand elements respectively.

As an approximation, the reduction factor may be obtained as follows:

S

�S

ZKHQ� ������������ �S�����ZKHQ� ! ������������ S

S

\

FU

ZKHUH� 3ODWH�VOHQGHUQHVVI E � W ���� N

(E.B.1.08)

In which

t =The relevant thickness

cr=The critical plate buckling stress

k 7KH�EXFNOLQJ�IDFWRU�FRUUHVSRQGLQJ�WR�WKH�VWUHVV�UDWLR� IURP� table 5.3.2 or

table 5.3.3 of Eurocode 3 as appropriate

And

E LV�WKH�DSSURSULDWH�ZLGWK�E �G�IRU�ZHEVE �F�IRU�LQWHUQDO�IODQJH�HOHPHQWV�E �E����W�IRU�IODQJHV�RI�5+6E �F�IRU�RXWVWDQG�IODQJHVE E �K � ��IRU�HTXDO � OHJ�DQJOHVE K�RU E �K � ��IRU�XQHTXDO � OHJ�DQJOHV

To determine the effective widths of flange elements, the stress ratio used from

table B.1.15 (i.e. table 5.3.2 or table 5.3.3 of Eurocode 3) may be based on the

properties of the gross cross-section.

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Swapnil B Kharmale 50 Comparative study of IS: 800 (Draft) &EC3 CD-051061

(Continued)

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Swapnil B Kharmale 52 Comparative study of IS: 800 (Draft) &EC3 CD-051061

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Designa c/s Width to thick. Type of class as Type of class Remarks on wholetion Area bf tf h tw h2 ratios for per IS :800(Draft) as per EC 3 section as per

mm2mm mm mm mm mm flange web flange web flange web IS:800 (Draft) EC 3

ISLB 75 7.71 50 5 75 3.7 11.7 5 13.973 Plastic Plastic Class1 Class1 Plastic Class1ISLB 100 10.21 50 6.4 100 4 13.5 3.906 18.25 Plastic Plastic Class1 Class1 Plastic Class1ISLB 125 15.12 75 6.5 125 4.4 14.8 5.769 21.682 Plastic Plastic Class1 Class1 Plastic Class1ISLB 150 18.08 80 6.8 150 4.8 16.6 5.882 24.354 Plastic Plastic Class1 Class1 Plastic Class1ISLB 175 21.3 90 6.9 175 5.1 16.7 6.522 27.765 Plastic Plastic Class1 Class1 Plastic Class1ISLB 200 25.27 100 7.3 200 5.4 17.2 6.849 30.685 Plastic Plastic Class1 Class1 Plastic Class1ISLB 225 29.92 100 8.6 225 5.8 22.4 5.814 31.086 Plastic Plastic Class1 Class1 Plastic Class1ISLB 250 35.53 125 8.2 250 6.1 23.7 7.622 33.213 Plastic Plastic Class1 Class1 Plastic Class1ISLB 275 42.02 140 8.8 275 6.4 25.7 7.955 34.953 Plastic Plastic Class1 Class1 Plastic Class1ISLB 300 48.08 150 9.4 300 6.7 27.5 7.979 36.582 Plastic Plastic Class1 Class1 Plastic Class1ISLB 325 54.9 165 9.8 325 7 29.3 8.418 38.071 Plastic Plastic Class1 Class1 Plastic Class1ISLB 350 63.01 165 11 350 7.4 30.9 7.237 38.959 Plastic Plastic Class1 Class1 Plastic Class1ISLB 400 72.43 165 13 400 8 31.9 6.6 42.025 Plastic Plastic Class1 Class1 Plastic Class1ISLB 450 83.14 170 13 450 8.6 33 6.343 44.651 Plastic Plastic Class1 Class1 Plastic Class1ISLB 500 95.5 180 14 500 9.2 34.9 6.383 46.761 Plastic Plastic Class1 Class1 Plastic Class1ISLB 550 110 190 15 550 9.9 37 6.333 48.091 Plastic Plastic Class1 Class1 Plastic Class1ISLB 600 126.7 210 16 600 11 39.9 6.774 49.543 Plastic Plastic Class1 Class1 Plastic Class1

ISMB 100 14.6 75 7.2 100 4 17.5 5.208 16.25 Plastic Plastic Class1 Class1 Plastic Class1ISMB 125 16.6 75 7.6 125 4.4 17.9 4.934 20.273 Plastic Plastic Class1 Class1 Plastic Class1ISMB 150 19 80 7.6 150 4.8 18.1 5.263 23.729 Plastic Plastic Class1 Class1 Plastic Class1ISMB 175 24.62 90 8.6 175 5.5 20.3 5.233 24.455 Plastic Plastic Class1 Class1 Plastic Class1ISMB 200 32.33 100 11 200 5.7 23.7 4.63 26.789 Plastic Plastic Class1 Class1 Plastic Class1ISMB 225 39.72 110 12 225 6.5 25.9 4.661 26.662 Plastic Plastic Class1 Class1 Plastic Class1ISMB 250 47.55 125 13 250 6.9 28 5 28.13 Plastic Plastic Class1 Class1 Plastic Class1ISMB 300 56.26 140 12 300 7.5 29.3 5.645 32.2 Plastic Plastic Class1 Class1 Plastic Class1ISMB 350 66.71 140 14 350 8.1 31 4.93 35.556 Plastic Plastic Class1 Class1 Plastic Class1ISMB 400 78.46 140 16 400 8.9 32.8 4.375 37.573 Plastic Plastic Class1 Class1 Plastic Class1ISMB 450 92.27 150 17 450 9.4 35.4 4.31 40.34 Plastic Plastic Class1 Class1 Plastic Class1

Sectional dimension

Swapnil B. Kharmale CD-051061 53 Comparative study of IS:800 (Draft) and EC3ESte

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Designa c/s Width to thick. Type of class as Type of class Remarks on wholetion Area bf tf h tw h2 ratios for per IS :800(Draft) as per EC 3 section as per

mm2mm mm mm mm mm flange web flange web flange web IS:800 (Draft) EC 3

ISHB 350 85.91 250 12 350 8.3 27 10.78 35.663 Semicompact Plastic Class3 Class1 Semicompact Class3ISHB 350 92.21 250 12 350 10 27 10.78 29.307 Semicompact Plastic Class3 Class1 Semicompact Class3ISHB 400 98.66 250 13 400 9.1 29.9 9.843 37.385 Compact Plastic Class2 Class1 Compact Class2ISHB 400 104.7 250 13 400 11 29.9 9.843 32.094 Compact Plastic Class2 Class1 Compact Class2ISHB 450 111.1 250 14 450 9.8 31.9 9.124 39.408 Plastic Plastic Class1 Class1 Plastic Class1ISHB 450 117.9 250 14 450 11 31.9 9.124 34.177 Plastic Plastic Class1 Class1 Plastic Class1

Sectional dimension

Swapnil B. Kharmale CD-051061 55 Comparative study of IS:800 (Draft) and EC3ESte

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Designation Cross Width to thickness Section Classification Effective c/sdimensions Eff. Cross sectional area Aeff in ISA(d x b x t) Sect. ratios for as per for Slender /Class 4 axial compression and ISA(h x b x t) Area Long Short Comb Sections as per UDWLR�� A = (Aeff/A)

leg leg of both IS:800 EC3 As per (d/t) (b/t) (d+b)/t (Draft) deff beff IS:800(Draft) Eurocode 3

cm2 (h/t) (b/t) (h+b)/t mm mm mm mm Aeff cm2

A Aeff cm2A

ISA 45x30x6 4.16 7.5 5.0 12.5 Semi-compact Class 3 4.16 1.00 4.16 1.00ISA 50x30x3 2.34 16.7 10.0 26.7 Slender Class 4 47.1 27.9 42.1 30 2.16 0.92 2.07 0.89ISA 50x30x4 3.07 12.5 7.5 20.0 Semi-compact Class 3 3.07 1.00 3.07 1.00ISA 50x30x5 3.78 10.0 6.0 16.0 Semi-compact Class 3 3.78 1.00 3.78 1.00ISA 50x30x6 4.47 8.3 5.0 13.3 Semi-compact Class 3 4.47 1.00 4.47 1.00ISA 60x40x5 4.76 12.0 8.0 20.0 Semi-compact Class 3 4.76 1.00 4.76 1.00ISA 60x40x6 5.65 10.0 6.7 16.7 Semi-compact Class 3 5.65 1.00 5.65 1.00ISA 60x40x8 7.37 7.5 5.0 12.5 Semi-compact Class 3 7.37 1.00 7.37 1.00ISA 65x45x5 5.26 13.0 9.0 22.0 Semi-compact Class 3 5.26 1.00 5.26 1.00ISA 65x45x6 6.25 10.8 7.5 18.3 Semi-compact Class 3 6.25 1.00 6.25 1.00ISA 65x45x8 8.17 8.1 5.6 13.8 Semi-compact Class 3 8.17 1.00 8.17 1.00ISA 70x45x5 5.52 14.0 9.0 23.0 Semi-compact Class 4 66.2 45 5.52 1.00 5.31 0.96ISA 70x45x6 6.56 11.7 7.5 19.2 Semi-compact Class 3 6.56 1.00 6.56 1.00ISA 70x45x8 8.58 8.8 5.6 14.4 Semi-compact Class 3 8.58 1.00 8.58 1.00ISA 70x45x10 10.52 7.0 4.5 11.5 Semi-compact Class 3 10.52 1.00 10.52 1.00ISA 75x50x5 6.02 15.0 10.0 25.0 Semi-compact Class 4 66.5 50 6.02 1.00 5.57 0.93ISA 75x50x6 7.16 12.5 8.3 20.8 Semi-compact Class 3 7.16 1.00 7.16 1.00ISA 75x50x8 9.38 9.4 6.3 15.6 Semi-compact Class 3 9.38 1.00 9.38 1.00ISA 75x50x10 11.52 7.5 5.0 12.5 Semi-compact Class 3 11.52 1.00 11.52 1.00ISA 80x50x5 6.27 16.0 10.0 26.0 Slender Class 4 78.5 46.5 68 50 6.00 0.96 5.65 0.90ISA 80x50x6 7.46 13.3 8.3 21.7 Semi-compact Class 3 7.46 1.00 7.46 1.00ISA 80x50x8 9.78 10.0 6.3 16.3 Semi-compact Class 3 9.78 1.00 9.78 1.00ISA 80x50x10 12.02 8.0 5.0 13.0 Semi-compact Class 3 12.02 1.00 12.02 1.00ISA 90x60x6 8.65 15.0 10.0 25.0 Semi-compact Class 4 79.7 60 8.65 1.00 8.02 0.93ISA 90x60x8 11.37 11.3 7.5 18.8 Semi-compact Class 3 11.37 1.00 11.37 1.00ISA 90x60x10 14.01 9.0 6.0 15.0 Semi-compact Class 3 14.01 1.00 14.01 1.00ISA 100x65x6 9.55 16.7 10.8 27.5 Slender Class 4 94.2 55.8 92.3 65 8.64 0.90 9.08 0.95

IS:800(Draft) EC3

EK

Swapnil B Kharmale CD-051061 57 Comparative study of IS:800 (Draft) and EC3ESte

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Designation Cross Width to thickness Section Classification Effective c/sdimensions Eff. Cross sectional area Aeff in ISA(d x b x t) Sect. ratios for as per for Slender /Class 4 axial compression and ISA(h x b x t) Area Long Short Comb Sections as per UDWLR�� A = (Aeff/A)

leg leg of both IS:800 EC3 As per (d/t) (b/t) (d+b)/t (Draft) deff beff IS:800(Draft) Eurocode 3

cm2 (h/t) (b/t) (h+b)/t mm mm mm mm Aeff cm2

A Aeff cm2A

Equal Leg Angle Section:-ISA 20x20x3 1.11 6.7 6.7 13.3 Semi-compact Class3 1.11 1.00 1.11 1.00ISA 20x20x4 1.44 5.0 5.0 10.0 Semi-compact Class3 1.44 1.00 1.44 1.00ISA 25x25x3 1.41 8.3 8.3 16.7 Semi-compact Class3 1.41 1.00 1.41 1.00ISA 25x25x4 1.84 6.3 6.3 12.5 Semi-compact Class3 1.84 1.00 1.84 1.00ISA 25x25x5 2.25 5.0 5.0 10.0 Semi-compact Class3 2.25 1.00 2.25 1.00ISA 30x30x3 1.71 10.0 10.0 20.0 Semi-compact Class3 1.71 1.00 1.71 1.00ISA 30x30x4 2.24 7.5 7.5 15.0 Semi-compact Class3 2.24 1.00 2.24 1.00ISA 30x30x5 2.75 6.0 6.0 12.0 Semi-compact Class3 2.75 1.00 2.75 1.00ISA 35x35x3 2.01 11.7 11.7 23.3 Semi-compact Class4 33.5 33.5 2.01 1.00 1.92 0.96ISA 35x35x4 2.64 8.8 8.8 17.5 Semi-compact Class3 2.64 1.00 2.64 1.00ISA 35x35x5 3.25 7.0 7.0 14.0 Semi-compact Class3 3.25 1.00 3.25 1.00ISA 35x35x6 3.84 5.8 5.8 11.7 Semi-compact Class3 3.84 1.00 3.84 1.00ISA 40x40x3 2.31 13.3 13.3 26.7 Slender Class4 37.5 37.5 38 38 2.16 0.94 2.19 0.95ISA 40x40x4 3.04 10.0 10.0 20.0 Semi-compact Class3 3.04 1.00 3.04 1.00ISA 40x40x5 3.75 8.0 8.0 16.0 Semi-compact Class3 3.75 1.00 3.75 1.00ISA 40x40x6 4.44 6.7 6.7 13.3 Semi-compact Class3 4.44 1.00 4.44 1.00ISA 45x45x3 2.61 15.0 15.0 30.0 Slender Class4 37.5 37.5 39.9 39.9 2.16 0.83 2.30 0.88ISA 45x45x4 3.44 11.3 11.3 22.5 Semi-compact Class4 44.6 44.6 3.44 1.00 3.41 0.99ISA 45x45x5 4.25 9.0 9.0 18.0 Semi-compact Class3 4.25 1.00 4.25 1.00ISA 45x45x6 5.04 7.5 7.5 15.0 Semi-compact Class3 5.04 1.00 5.04 1.00ISA 50x50x3 2.91 16.7 16.7 33.3 Slender Class4 37.5 37.5 41.3 41.3 2.16 0.74 2.39 0.82ISA 50x50x4 3.84 12.5 12.5 25.0 Semi-compact Class4 49.4 49.4 3.84 1.00 3.79 0.99ISA 50x50x5 4.75 10.0 10.0 20.0 Semi-compact Class3 4.75 1.00 4.75 1.00ISA 50x50x6 5.64 8.3 8.3 16.7 Semi-compact Class3 5.64 1.00 5.64 1.00ISA 55x55x5 5.25 11.0 11.0 22.0 Semi-compact Class3 5.25 1.00 5.25 1.00ISA 55x55x6 6.24 9.2 9.2 18.3 Semi-compact Class3 6.24 1.00 6.24 1.00

IS:800(Draft) EC3

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Designation Cross Width to thickness Section Classification Effective c/sdimensions Eff. Cross sectional area Aeff in ISA(d x b x t) Sect. ratios for as per for Slender /Class 4 axial compression and ISA(h x b x t) Area Long Short Comb Sections as per UDWLR�� A = (Aeff/A)

leg leg of both IS:800 EC3 As per (d/t) (b/t) (d+b)/t (Draft) deff beff IS:800(Draft) Eurocode 3

cm2 (h/t) (b/t) (h+b)/t mm mm mm mm Aeff cm2

A Aeff cm2A

ISA100x100x8 15.36 12.5 12.5 25.0 Semi-compact Class4 99.1 99.1 15.36 1.00 15.22 0.99ISA100x100x10 19.00 10.0 10.0 20.0 Semi-compact Class3 19.00 1.00 19.00 1.00ISA100x100x12 22.56 8.3 8.3 16.7 Semi-compact Class3 22.56 1.00 22.56 1.00ISA110x110x8 16.96 13.8 13.8 27.5 Slender Class4 100 100 102.8 102.8 15.36 0.91 15.81 0.93ISA110x110x10 21.00 11.0 11.0 22.0 Semi-compact Class3 21.00 1.00 21.00 1.00ISA110x110x12 24.96 9.2 9.2 18.3 Semi-compact Class3 24.96 1.00 24.96 1.00ISA110x110x15 30.75 7.3 7.3 14.7 Semi-compact Class3 30.75 1.00 30.75 1.00ISA130x130x8 20.16 16.3 16.3 32.5 Slender Class4 100 100 109.2 109.2 15.36 0.76 16.83 0.83ISA130x130x10 25.00 13.0 13.0 26.0 Slender Class4 125 125 125.6 125.6 24.00 0.96 24.12 0.96ISA130x130x12 29.76 10.8 10.8 21.7 Semi-compact Class3 29.76 1.00 29.76 1.00ISA130x130x15 36.75 8.7 8.7 17.3 Semi-compact Class3 36.75 1.00 36.75 1.00ISA150x150x10 29.00 15.0 15.0 30.0 Slender Class4 125 125 132.9 132.9 24.00 0.83 25.58 0.88ISA150x150x12 34.56 12.5 12.5 25.0 Semi-compact Class4 148.3 148.3 34.56 1.00 34.15 0.99ISA150x150x15 42.75 10.0 10.0 20.0 Semi-compact Class3 42.75 1.00 42.75 1.00ISA150x150x18 50.76 8.3 8.3 16.7 Semi-compact Class3 50.76 1.00 50.76 1.00ISA200x200x10 39.00 20.0 20.0 40.0 Slender Class4 125 125 144.8 144.8 24.00 0.62 27.96 0.72ISA200x200x12 46.56 16.7 16.7 33.3 Slender Class4 150 150 165.3 165.3 34.56 0.74 38.23 0.82ISA200x200x15 57.75 13.3 13.3 26.7 Slender Class4 187.5 187.5 190.3 190.3 54.00 0.94 54.84 0.95ISA200x200x18 68.76 11.1 11.1 22.2 Semi-compact Class3 68.76 1.00 68.76 1.00

Remarks:-1)From table it is observed that the angle section of Slender class /Class4 are not fully utilising its gross sectional area������������������KHQFH� A<1) in axial compression. The thickness of such section should be increased to make it semicompact /Class 3

2) IS:800 (Draft) dose not clearly specifies how to calculate the effective dimensions for slender class.In this table the dimensions are calculated considering the limiting width-to-thickness ratio for semi-compact class On the other hand Eurocode 3 clearly specifies the procedure for calculating the effective dimensions for Class 4 (See Clause 5.3.5 of EC3)

IS:800(Draft) EC3

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Swapnil B.Kharmale 63 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.2.1 Physical properties of structural steel (With reference to

clause 2.2.4.1 of IS: 800 (Draft) and Clause 3.2.5 of Eurocode 3

Physical Properties As per IS:800 (Draft) As per Eurocode 3

Unit mass of steel =7850 kg/m3 =7850 kg/m3

Modulus of elasticity E=2x105N/mm2 E=2.1x105N/mm2

Poisson ratio 0.3 0.3 Modulus of rigidity G=E/[2(1+� )]

=0.769 x105 N/mm2G=E/[2(1+� )] =0.810 x105 N/mm2

Coefficient of thermal expansion

t=12 u 10-6 /o C t=12 u 10-6 /o C(For T<100o C)

Mechanical Properties

The principal mechanical properties of the structural steels important in

design, are the yield stress, fy, the tensile or ultimate stress, fu, the maximum percent

elongation on a standard gauge length and notch toughness. Except the notch

toughness others are determined by conducting tensile tests on samples cut from

the plates, sections etc. These properties for the common steel products of different

specifications are summarized in following tables

As per IS: 800 (Draft):-

Table 2.1 of IS: 800 (Draft) give the mechanical properties of structural steel

From which here only those specifications are mentioned which are required for

general structural purpose.

As per Eurocode 3

This standard specifies the requirements for long products (such as sections

and bars) and flat products (such as plate, sheet and strip) of hot-rolled non-alloy

general purpose (base) and quality steels. These steels are intended for use in

welded, bolted and riveted structures for service at ambient temperature.

Designation of the Steels in Eurocode 3

The designation consists of:

x The number of the European standard (EN 10025).

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Swapnil B. Kharmale 65 Comparative study of IS: 800 (Draft) &EC3 CD-051061

TABLE B. 2.3 Tensile properties of structural steel by IS: 800 (Draft) for general purpose

(TABLE 2.1 Section 2.2.4.2 of IS: 800 (Draft))

Properties SR. NO.

*Specifications No.

Particulars

Grade / Classificatio

n Yield Stress, MPa (Min) Ultimate

Tensile Stress, MPa, (Min)

Elongation Percent (Min)

d or t < 20 20<d or t<40

d or t > 40

1IS:2062-

1999

Specification of steel for general structural purposes

A/ Fe410WA B/ Fe410WB C/ Fe410WC

250 250 250

240 240 240

230 230 230

410 410 410

23 23 23

TABLE B. 2.4 Tensile properties of structural steel by IS: 800 (Draft) for fasteners (TABLE 2.1 Section 2.2.4.2 of IS: 800 (Draft))

Properties SR.

No.

*Specification No.

Particulars

Grade / Classificatio

n Yield Stress, MPa (Min) Ultimate

Tensile Stress, MPa, (Min)

Elongation Percent

(Min)

1IS: 1367-1991 (ISO 898)

Specifications of fasteners-threaded steel for technical supply conditions

3.6 4.6 4.8 5.6 5.8 6.8 8.8 9.8

10.9 12.9

180 240 320 300 400 480 640 720 900

1080

300 400 400 500 500 600 800 900

1000 1200

25 22 14 20 10 8

12 10 98

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Swapnil B.Kharmale 67 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.2.5:-The nominal values of yield strength fy and ultimate tensile stress

fu for hot rolled steel (Table 3.1 of Eurocode 3)

(Continued)

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Swapnil B.Kharmale 69 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Section B: - Study of Both Code B.3 Analysis and Design Requirements

B.3.1 Analysis (Calculation of internal forces, moment or Action effects)

The section 4 of IS: 800 (Draft) and section 5.2 of Eurocode 3 specify the

method analysis. As both codes use LSM approach of design (which utilizes reserve

strength in plastic region) therefore method of analysis mentioned in both codes are

same. Here these methods are discussed in short

B.3.2 Method of analysis

i. The internal forces and moments in a statically determinate structure shall be

obtained using static’s.

ii. The internal forces and moments in a statically indeterminate structure may

generally be determined using either:

x Elastic global analysis (as per 4.4 of IS:800 (Draft) &5.2.1.3 of

Eurocode 3)

x Plastic global analysis (as per 4.5 of IS:800 (Draft) &5.2.1.4 of

Eurocode 3)

iii. Elastic global analysis may be used in all cases.

iv. Plastic global analysis may be used only where the member cross-sections

satisfy the requirements (specified in 4.5.2. of IS: 800 (Draft) & 5.2.7 and

5.3.3 of Eurocode 3) and the steel material satisfies the requirements

(specified in 4.5.2. of IS:800 (Draft) & 3.2.2.2of Eurocode 3.

v. When the global analysis is carried out by applying the loads in a series of

increments, it may be assumed to be sufficient, in the case of building

structures, to adopt simultaneous proportional increases of all loads.

Effects of deformations

i. The internal forces and moments may generally be determined using either:

x First order theory, using the initial geometry of the structure.

x Second order theory, taking into account the influence of the

deformation of the structure.

ii. First order theory may be used for the global analysis in the following cases:

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Swapnil B.Kharmale 71 Comparative study of IS: 800 (Draft) &EC3 CD-051061

When a plastic method of analysis is used, all of the following conditions of this

section shall be satisfied, unless adequate ductility of the structure and plastic

rotation capacity of its members and connections are established for the design

loading conditions by other means of evaluation:

a) The yield stress for the grade of the steel used shall not exceed 450 MPa as

per IS:800 (Draft) (Here Eurocode 3 doesn’t specify a significant value but

mention that steel grades listed in Table 3.1 may be accepted for plastic

analysis in which maximum Grade had yield stress 460 Mpa.)

b) The stress-strain characteristics of the steel shall not be significantly

different from those obtained from Standard Test Result, and shall be such

as to ensure moment redistribution.

(i) The stress strain diagram has a plateau at the yield stress, extending for

at least six times the yield strain;

(ii) The ratio of the tensile strength to the yield stress specified for the grade

of the steel is not less than 1.2

(iii) The elongation on a gauge length complying with IS: 2062 is not less

than 15%; and

(iv) The steel exhibits strain-hardening capability.

c) The members used shall be hot-rolled or fabricated using hot-rolled plates

and section.

d) The cross section of members not containing plastic hinges should be

compact unless the members meet the strength requirements from elastic

analysis.

e) Where plastic hinges occur in a member, the proportions of its cross section

should not exceed the limiting values for plastic section

f) The cross section should be symmetrical about its axis perpendicular to the

axis of the plastic hinge rotation.

g) The members shall not be subject to impact loading, requiring fracture

assessment or fluctuating loading, requiring a fatigue assessment.

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Swapnil B.Kharmale 73 Comparative study of IS: 800 (Draft) &EC3 CD-051061

SI.No. Member

Maximum eff.

slenderness

ratio(KL/r)

6Member always under tension* (Other than

pretensioned member) 400

* Tension member, such as bracing’s, pretensioned to avoid sag, need not satisfy the maximum

slenderness ratios limit.

As per Eurocode 3

Eurocode 3 doesn’t specify such maximum slenderness ratio for structural

member. But United Kingdom National Application Document (NAD)** which used

along with Eurocode 3 gives the guideline about maximum slenderness ratio.

Table B.3.2 Maximum effective slenderness ratios as per NAD (UK) of

Eurocode 3

SI.No. Member

Maximum eff.

slenderness ratio

1 For members resisting loads other than wind loads 180

2 For members resisting self weight and wind loads 250

3For any member normally acting as a tie but subject

to reversal stress resulting from the action of wind 350

** NAD of UK takes the references of BS: 5950:Part1 here the above table is as per reference of BS:

5950:Part1:1990 and BS: 5950:Part1:2000 omitted the maximum slenderness ratio requirements.

About maximum values of effective slenderness ratios of structural member

The restrictions or limitations on maximum effective slenderness ratios are

meant for transportation, erection and fabrication feasibility of structural member.

There is reduction in capacity of compression member with increase in slenderness

also lesser slenderness or limited slenderness result in unnecessary large cross

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Swapnil B.Kharmale 75 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.3.3. Deflection limits other than for pitched roof portal frame (Table 5.3

of IS: 800 (Draft))

Type of building Deflection Design Load Member Supporting Maximum

Deflection Live load Live load

Purlin Simple span

Roof cladding Brittle cladding

Span / 150 Span / 240

Live load Cantilever Brittle cladding Span / 120

Live load Simple span Elastic cladding

Span / 180

Live load Cantilever Elastic cladding

Span / 90

Live load Simple span Floor Span / 300

Live load Cantilever Floor Span / 150 Crane load (Manual operation)

Gantry Crane Span / 500

Crane load (Electric operation up to 50 t)

Gantry Crane Span / 750

Crane load (Electric operation over 50 t)

Gantry Crane Span / 1000

Crane (Vertical)

+ Roof load Gantry Crane Inward –12 mm Outward –25 mm

Ver

tical

Moving load (Charge cars, etc.)

Gantry Crane Span / 600

No cranes Column Elastic cladding Height / 150

No cranes Column Masonry/brittle cladding Height / 240

Crane Span / 400

Crane Gantry

(lateral) Relative between rails 10 mm

Indu

stria

l bui

ldin

g

Late

ral C

rane

+ w

ind

Crane

Column/frame

Column/frame

Gantry (pendent operated) Gantry (cab operated)

Height / 100

Height / 240

Live load Floors & roofs Not susceptible to cracking Span / 300

Oth

er

Bui

ldin

gs

Ver

tical

Live load Floor & Roof Susceptible to cracking Span / 360

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Swapnil B.Kharmale 77 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.3.4. Recommended limiting values of vertical deflections (Table 4.1 of Eurocode 3)

Limits

Condition max 2

Roofs generally L/200 L/250

Roofs frequently carrying personnel other than

for maintenance L/250 L/300

Floors generally L/250 L/300

Floors and roofs supporting plaster or other

brittle finish or non-flexible partitions L/250 L/350

Floors supporting columns (unless the

deflection has been included in the global

analysis for the ultimate limit state)

L/400 L/500

Where max can impair the appearance of the

building L/250 ---

Comment on the maximum limit of maximum vertical deflection of floor by

both code

Referring to limits on vertical deflections of floor by both code (see underlined

cell in Table B.3.3 and Table B.3.4)) it observed the IS:800 (Draft) code limits are

more stringent than Eurocode 3 this is because Eurocode 3 include provision of pre-

FDPEHU� 1 LQ� FDOFXODWLRQ� RI� PD[LPXP� YHUWLFDO� GHIOHFWLRQ� max and also value of

E(Modulus of Elasticity) of steel is higher (i.e. Es=2.1 x105 N/mm2) .The IS:

800(Draft) also have provision for pre-camber but doesn’t specify the limit on

maximum vertical deflection considering pre-camber.

Horizontal deflection

For buildings the recommended limits for horizontal deflections at the tops of the

columns are:

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Swapnil B.Kharmale 79 Comparative study of IS:800 (Draft) &EC3 CD-051061

Section B: - Study of Both Code B.4 Design of Tension Member

B.4.1 General

Steel tension member are probably the most common and efficient member.

These are efficient because the entire cross section is subjected to almost uniform

stress*1 (In other word whole cross sectional area is utilized). Tension members are

linear members in which axial forces act so as to elongate (stretch) the member. A

rope, for example, is a tension member. Unlike compression members, they do not

fail by buckling. Hence their design is not affected by classification of cross section.

The strength of these members is influenced by several factors such as

length of connection, size and spacing of fasteners, net area of cross section, and

type of fabrication, connection eccentricity, and shear lag at the end connection.

B.4.2 Cross section of tension member

*1It is generally assumed that the distribution of stresses in cross-sections of members subjected to axial tensile forces is uniform. However, there are some parameters like residual stresses and connection which result in a non-uniform distribution of stresses

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Swapnil B.Kharmale 81 Comparative study of IS:800 (Draft) &EC3 CD-051061

B.4.3.2 Connections (Effect of holes on tension capacity*2)

Connections are generally made either by bolting or welding. When several

members have to be connected, additional plates must be used which introduce

secondary effects due to the moments developed. Sometimes it is possible to

reduce these local eccentricities by varying the weld lengths or the bolt distribution.

In addition, the holes that are needed to fix the bolt significantly distort the ideal

behaviour of the cross-section. Firstly, there is an area reduction that has to be

taken into account and also a distortion in the stress distribution that induces a non-

uniformity in the strain; the effect of the holes is to increase the stresses locally

around them (Figure B.4.3). For a plate of infinite width the distribution is given by:

For xtR

� �R

� 5 � 5 >�� � @� [ � [ (B.4.1)

*2 “Effect of circular holes on stress distribution in plates” from “Theory of Elasticity” By-S.P.Timoshenko and J.N.Goodier

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Swapnil B.Kharmale 83 Comparative study of IS:800 (Draft) &EC3 CD-051061

x Gross section yielding

x Net section rupture

x Block shear failure

Gross section yielding

Generally a tension member without bolt hole can resists loads up to the

ultimate load without failure. But such a member will deform in longitudinal direction

considerably (nearly 10%-15% of its original length) before failure. At such a large

deformation a structure will become unserviceable. Hence in limit state design, in

addition to net section failure and block shear failure, yielding of the gross section

must also be considered, so as to prevent excessive deformation of the member.()

Net section rupture

A tension member is often connected to main or other member by bolt or

welds. When connected using bolts, tension members have holes and hence

reduced cross section, being referred to as the net area. Holes in the member

causes stresses concentration (as discussed earlier under ‘effect of hole on tension

capacity’)

Block shear failure

Block shear failure commonly refers to the tearing of a block of material, and

it presumes a combination of tension rupture and shears yield or a combination of

shear rupture and tension yield. Although the first failure mode is quite common, the

latter failure mode is uncommon because of the small ductility in tension as

compared with shear. Block shear failure is usually associated with bolted details

because a reduced area is present in that case, but in principle it can also be

present in welded details. Design rules in various codes base block shear failure

calculation on a combination of yield and rupture strength of the net or gross areas

in shear and tension on the potential failure plane

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Swapnil B.Kharmale 85 Comparative study of IS:800 (Draft) &EC3 CD-051061

IS:800 (Draft) Eurocode 3

Design strength due to yielding of gross

section (Tdg):- (Clause 6.2)

\ JGJ

P�

I $7 where

fy =Yield strength of the material in

MPa

Ag =Gross area of cross section in mm2

m0 =Partial safety factor for failure in

tension by yielding

Design plastic resistance of gross section (Npl.Rd):-(Clause 5.4.3.1 (a))

\SO�5G

P�

$ I �1

where fy =Yield strength of the material in

MPa

A =Gross area of cross section in mm2

m0 =Partial safety factor for failure in

tension by yielding

Design strength due to rupture of critical

section (Tdn):- (Clause 6.3.1)

For plates

X QGQ

P�

����I �$7 where

An =Net effective area of member

fu =Ultimate stress of material

m1=Partial safety factor for failure in

tension by rupture

Ultimate resistance of net cross section

(Nu.Rd):- (Clause 5.4.3.1 (a))

For plates

QHW XX�5G

P�

�����$ �I1 where

An =Net area of section

fu =Ultimate stress of material

m2=Partial safety factor for résistance of

cross section at bolt holes

About net area :-

According to both codes: "the net area of a cross-section or element section shall be

taken as its gross area less appropriate deductions for all holes and other openings.

Provided that the fastener holes are not staggered the total area to be deducted

shall be the maximum sum of the sectional areas of the holes in any cross-section

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Swapnil B.Kharmale 87 Comparative study of IS:800 (Draft) &EC3 CD-051061

When the member reaches the yield condition and the connection the failure

condition, the increases in length would be:

&RQQHFWLRQ \ 7RWDO \ 7RWDO

0HPEHU \ 7RWDO \ 7RWDO

&RQQHFWLRQ

0HPEHU

�&RQQHFWLRQ�]RQH��� / / ��� /�����0HPEHU�]RQH��������� / / ���� /���

/L�H�� ���� /

This means that the elongation within the connection zone is much smaller than that

of the entire bar.

This is why both codes (IS:800 (Draft) and Eurocode 3) allows the exceedance of

the yield strength in the connection zone up to the ultimate tensile strength, fu; that

is, it is implicitly assumed that the failure of an member can be described by its

deformation.

Both codes include a reduction coefficient to take account of the unavoidable

eccentricities, stress concentrations, etc. The reduction is taken as 10%, so that the

recommended formula

X Q

PL

����I �$7HQVLOH�VWUHQJWK�RU�UHVLVWDQFH�DW�QHW�VHFWLRQ IS:800 (Draft) Eurocode 3

For angles (Clause 6.3.3)

With bolted & welded connection

JR \X QFGQ

P� P�

XGQ Q

P�

$ [�I����[�I [�$7 � [IRU�7 $ [

where, D = 0.6 for one or two bolts, 0.7 for

For angles

With bolted connection(Clause 6.5.2.3)

� R XX�5G

P�

� QHW XX�5G

P�

� QHW XX�5G

P�

� H ������G � W I ��1 � �IRU���EROW$ I1 � ��IRU���EROW$ I1 � ��IRU���EROW

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Swapnil B.Kharmale 89 Comparative study of IS:800 (Draft) &EC3 CD-051061

About angle connected by one leg

In many cases, angles are connected to gusset plate (which in turn connected to

other members of structure) by welding or bolting only through one of the two legs.

This kind of connection results in eccentric loading (Fig B.4.9), causing non-uniform

distribution of stress over the cross section. Further, since the load is applied by

connecting one leg of member there is a shear lag at end connection.

Effect of shear lag:-

The force is transferred to a tension member (angles, channels, or T-section) by a

gusset or the adjacent member connected to one leg either by bolting or by welding.

The force thus transferred to one leg by the end connection locally gets transferred

as tensile stress over the entire cross section by shear. Hence, the tensile stress on

the section from the first bolt up to the last bolt will not be uniform. The connected

leg will have higher stresses at failure even of the order of ultimate stress while the

outstanding leg stresses may be even below yield stress. Thus transfer of force from

connected leg to outstanding leg will be by shear (Figure B.4.10) and because one

part ‘lags’ behind the other the phenomenon is referred to as ‘Shear Lag’ However ,

at the section away from the end connection , the stress distribution becomes more

uniform. Hence shear lag effects reduces with increase in connection length

Therefore to account for eccentric loading, effect of shear lag etc the reduction

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Swapnil B.Kharmale 91 Comparative study of IS:800 (Draft) &EC3 CD-051061

shear along a line of the

transmitted force, respectively

(1-2 and 4 –3 as shown in Fig

(B.4.3) and 1-2 as shown in

Fig (B.4.4)

Atg, Atn = Minimum gross and net area in

tension from the hole to the toe

of the angle or next last row of

bolt in plates, perpendicular to

the line of force, respectively (2-

3) as shown in Fig (B.4.3) and

Fig (B.4.4)

fu, fy = Ultimate and Yield stress of the

material respectively

comparative purpose:-

Design shear rupture resistance Veff.Rd

(Block Shear failure):-

Y�HII \HII �5G

P�

Y�HII Y�HII

Y�HII Y � � Y�HII �

� � �

X� � R�W

\

� Y � �

� Y � � R�Y

$ [I9 �[ZKHUH�$ / [W� / / � / � / �EXW�/ /

/ D �EXW�/ �GI/ D � N[G [ I

/ / � D � D �� EXW�/ / � D � D � Q[G X

\

I[ I

where

Av.eff = Effective shear area.

Lv.eff =Effective length of

connection in shear.

Lv, a1, a2, a3 =Indicated in figure. (B.4.12)

do.v and do.t =Hole size for shear face

and tension face

respectively normally it is

doh

n =Number of fastener holes

on the shear face

t =Thickness of section EStela

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PROBLEMS ON TENSION MEMBER BY IS:800 (DRAFT)

A. Analysis Problem

A single unequal angle 125 mm x 75 mm x 8mm is connected to 12 mm thick gusset plate at ends with 6 no 16 mm diameter rivets of Grade 4.6 to transfer tension as shown in figure below. Determine the Tension capacity of an angle section if a) If longler leg is connected to gusset plate.

b) If shorter leg is connected to gusset plate Use fy=250 MPa

Sectional Properties

A= 1538 mm2

b= 125mmd= 75mmt= 8mm

g = 75mm

Nominal Dia of rivet= dn=16mm

Effective Dia of rivet = d = 16 +2 =18 mm

ANALYSIS STEPS REFERENCES

Swapnil B.Kharmale CD-051061 93 Comparative study of IS:800 (Draft) and EC 3

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Alternatively

Hence take Tdn lower of 382.66 kN and 365.26 kN

Therefore Tdn =365.26kN

iii) Design strength due to Block Shear Tdb Clause 6.4.2

where Avg &Avn :=Minimum gross and net area in shear along a line of

transmitted force respectivelyAtg & Atn :=Minimum gross and net area in tension from hole to toe of

an angle or next last row of bolts in plate

Here Avg= Lvgxt

Avg= (5 x 50 + 50 ) x 8 :=2400 mm2

Avn= ( 5 x 50 + 50 - 5.5 x 18 ) x 8 :=1608 mm2

Atg = Ltgxt

Atg = ( 50 x 8 ) :=400 mm2

Atn= (50 - 0.5 x 18 ) x 8 :=328 mm2

Therefore

Considering lower value for T db i.e. Tdb = 395.42 kN

Design Tensile strength of ISA125x75x8 if longer leg connected Clause 6.1to gusset plate

Td= Least of Tdg ,Tdn, Tdb

:= 349.5 kN

GQ

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����

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P� P�P� P�

$ [I $ [I$ [I $ [I7 � RU �

�[ �[" "

GE

GE

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Swapnil B.Kharmale CD-051061 95 Comparative study of IS:800 (Draft) and EC 3

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Here Avg= Lvgxt

Avg= (5 x 50 + 50 ) x 8 :=2400 mm2

Avn= ( 5 x 50 + 50 - 5.5 x 18 ) x 8 :=1608 mm2

Atg = Ltgxt

Atg = ( 35 x 8 ) :=280 mm2

Atn= (35 - 0.5 x 18 ) x 8 :=208 mm2

Therefore

Considering lower value for T db i.e. Tdb = 383.14 kN

Design Tensile strength of ISA125x75x8 if shorter leg connected Clause 6.1to gusset plate

Td= Least of Tdg ,Tdn, Tdb

:= 329.77 kN

Conclusion from problemThe Design Tensile strength capacity of an unequal angle section willbe more if longer leg is connected to gusset plate than if shorter legconnected to gusset plate

GE

GE

����[��� ���[��� ����[��� ���[���7 � RU �

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Swapnil B.Kharmale CD-051061 97 Comparative study of IS:800 (Draft) and EC 3

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Note:-Eurocode 3 dosent specify directly calculation of design tensile strength governed by Block Shear ( As per EC3 "Block Shear faliure at a group of fasterner holes near the end of beam web or bracket orplate shall be prevented by appropriate hole spacing)Therefore for given problem we have to check given connecton for Block Sheariii) Design shear rupture resistance Veff.Rd (Block Shear faliure) Clause 6.5.2.2

where

here Lv , a1, a2 , a3,are indicated in figure

d is nominal diameterdo.v and do.t are hole size for shear face and tension face

respectively normally it is d oh.

n is number of fastener holes on the shear facet is thickness of section

Y�HII \

HII�5GP�

$ [I9

�[

Y�HII Y�HII

Y�HII Y � � Y�HII �

� � �

X� � R�W

\

X� Y � � � Y � � R�Y

\

$ / [W

/ / �/ �/ EXW / /

/ D EXW / �G

I/ D �N[G [

I

I/ / �D �D EXW / / �D �D � Q[G [

I

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Swapnil B.Kharmale CD-051061 99 Comparative study of IS:800 (Draft) and EC 3

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Therefore equivalent equal -leg angle section = ISA 75x75 x8

A=1058 mm2

Therefore Anet:= A- Area due to hole

:=1058 - 18 x8=914 mm2

3= 0.5 (as 5xdo<pitch =50mm >2.5 x do ) Table 6.5.1 of

Therefore,

iii) Design shear rupture resistance Veff.Rd (Block Shear faliure) Clause 6.5.2.2

For our caseLv= 5x 50=250mm

L1= 50 mm< 5 x 16mm

L2 = (35-0.5x18) x(410/250)=42.64 mm

k= 0.5 for single row of rivetL3= 250 + 50 + 50 =350mm < (250+50+50-6x18)x(410/250)

Therefore L3= 350 mm

ThereforeLv.eff=250+50 +42.64 =342.64 mm < L3

Hence Lv.eff = 342.64mm

Design Tension Resistance of ISA125x75x8 if longer leg connected to gusset plate Clause 5.4.3 (1)

Nt.Rd= Least of N pl.Rd ,Nu.Rd and Veff.Rd

Hence Nt.Rd= 149.90 kN

X�5G

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����

Y � H I I \

H I I �5 GP �

$ [ I9

� [

�Y�HII

HII�5G

$ ������[� �������PP

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Swapnil B.Kharmale CD-051061 101 Comparative study of IS:800 (Draft) and EC 3

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3.Connection design

Diameter of rivet (To avoid faliure of rivet in bearing)

Nominal dia = d= 18 mm , Effective dia = 19.5mm Rivet value = Shear capacity of rivet in sigle shear =

Clause 10.2.3.2

Therefore number of rivet required =

Provide edge distance = 40 mm > 30 mm for 18 mm dia rivet Table 10.10 of IS:Pitch (p) :- 800 (Draft)For tension member max. pitch =16 x t or 200 mm whichever is less Clause 10.2.2 minimum pitch = 2.5 x d Clause 10.2.1 Hence ,provide p= 60 mm Therefore length of end connection

L = 360 mm

4. Tension capacity of section(Check forTension capacity)

Anc= Net area of connected leg= (150-(8/2)-20) x 8 :=1008 mm2

Ago= Gross area of outstanding leg=(75-(8/2)) x 8 :=568 mm2

Ag= Gross area of whole setion :=1742 mm2

An= Net area of total cross section = Anc + Ago :=1576 mm3

i)Design strength due to Yielding of Gross Section T dg Clause 6.2

ii) Design strength due to Rupture of Critical Section Tdn Clause 6.3.3

����IRU�RQH� or two rivets =0.7 for three rivets =0.8 for four or more rivets

G ����[ W

G ����[ � �����PP ��PP

�QV

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� �[����

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J \GJ

P�

$ [I ����[���7 ������N1

���

JR \X QF XGQ GQ Q

P� P� P�

\ V V

X

$ [I���[I [$ I7 � [ RU 7 $ [

ZKHUH

I E EZ ���� � �����[ [ [ ��� � ����[

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Swapnil B.Kharmale CD-051061 103 Comparative study of IS:800 (Draft) and EC 3

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PROBLEMS ON TENSION MEMBER BY Eurocode 3

B.Design Problem

Design a single angle tie member to carry the design axial tension of 375 kN. With rivetted connections . Use f y =250 Mpa (Provide rivet prefebarly in single row)

1.Data P = 375 kN

Rivetted connections

2. Trial section

Here if we use same section (ISA 150 x 75 x 8 ) as used in design by

IS:800 (Draft) with same connection details it found to be failed by

Eurocode

Therefore let us try ISA 150 x 115 x 8 @0.159kN/m with longer leg connected to 12mm thick gusset plateSectional Properties

A=2058 mm2 ,b=150mm,d=115mm,t=8mm,g=75 mm

3.Connection design Let us provide 7no 18 mm dia.rivet of 4.6 Grade(d=18 mm,do =20mm)

Provide edge distance = 40 mm > 1.5xdo i.e. 30 mm Clause 6.5.1.3 (1)Pitch (p) :- For tension member max. pitch =14 x t or 200 mm whichever is less Clause 6.5.1.7 (1) minimum pitch = 2.2 x do Clause 6.5.1.5 (1)

DESIGN STEPS REFERENCES

�P� �

J UHT\

3 [ ���[ � � [ � �� �$ � ��� P P

I � ��

Swapnil B.Kharmale CD-051061 105 Comparative study of IS:800 (Draft) and EC 3

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Therefore,

Design Tension Resistance of ISA150x115x8 Clause 5.4.3 (1)

Nt.Rd= Least of N pl.Rd ,Nu.Rd and Veff.Rd

Hence Nt.Rd= 435 kN > 375 kNHence safe

"Design of single angle tension member connected by single row of rivets by both code"

Section for tensile ISA 125 x 75 x 8 ISA 150 x 75 x 8 forceP=375 kN at 0.134 kN/m at 0.153 kN/m

Length of end 7 no 18 mm dia 7 no 18 mm dia connection of rivets @ p=60 of rivets @ p=90

mm c/c mm c/cHence L= 360 mm Hence L=540 mm

Faliure mode Yielding of gross Rupture of net section cross section at

holes for fastneri.e. T= Tdg i.e. Nt.Rd = Nu.Rd

IS :800 (Draft)Points Eurocode 3

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HII�5G

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Swapnil B.Kharmale CD-051061 107 Comparative study of IS:800 (Draft) and EC 3

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Swapnil B.Kharmale 109 Comparative study of IS: 800 (Draft) &EC3 CD-051061

the main components but it is also possible to use I -sections; they are laced or

battened together with simple elements (bars or angles or smaller channel sections)

Figure B.5.1 illustrates all the shapes mentioned above

It should be noted that:

x The type of connection is important in the design of simple compression

members because it defines the effective length to be taken into account in

the evaluation of buckling. Circular sections do not represent the optimum

solution if the effective length is not the same in the two principal directions; in

this case, non symmetrical shapes are preferable.

x Members are frequently subjected to bending moments in addition to axial

load; in these conditions ISMB-sections can be preferable to ISWB-sections

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Swapnil B.Kharmale 111 Comparative study of IS: 800 (Draft) &EC3 CD-051061

For these compression member the Euler formula, predicts the strength of

long compression member very well, where the axial buckling stress remain below

the proportional limit. Such compression member buckles elastically.

x Intermediate length compression member (Medium slenderness)

For intermediate length compression member, some fibers would have

yielded and some fiber will still be elastic. These compression members will fail both

by yielding and buckling and their behaviour is said to be inelastic

The detailed behaviour of long and medium length compression member is

discussed in next article ‘Stability of slender steel column’

B.5.4 Stability of slender steel columns

Depending on their slenderness, columns exhibit two different types of behaviour:

those with high slenderness present a quasi elastic buckling behaviour whereas

those of medium slenderness are very sensitive to the effects of imperfections.

x Euler Critical Stress

If leff is the effective length (critical length), the Euler critical load Pcr (Ncr) is equal to:

�FU FUHII

(,3 1 O (B.5.1)

and it is possible to define the Euler critical stress cr as:

�FU FU

�FUHII

3 1 (, $ O $ (B.5.2)

By introducing the radius of gyration r (i) =,$ , and the slenderness �./�U�RU � = leff/i,

for the relevant buckling mode, Equation (B.5.2) becomes

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Swapnil B.Kharmale 113 Comparative study of IS: 800 (Draft) &EC3 CD-051061

x Buckling of Real Columns

The real behaviour of steel columns is rather different from that described in

the previous section and columns generally fail by inelastic buckling before reaching

the Euler buckling load. The difference in real and theoretical behaviour is due to

various imperfections in the "real" element: initial out-of-straightness, residual

stresses, eccentricity of axial applied loads and strain-hardening. The imperfections

all affect buckling and will; therefore, all influence the ultimate strength of the

column. Experimental studies of real columns give results as shown in Figure B.5.4.

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Swapnil B.Kharmale 115 Comparative study of IS: 800 (Draft) &EC3 CD-051061

ZKHUH �� 6OHQGHUQHVV�RI�FRPSUHVVLRQ�PHPEHU�

1=Euler’s slenderness

� A=(A/Aeff) =Ratio of effective area in compression to gross area

B.5.5.1 Basis of the ECCS Buckling Curves

From 1960 onwards, an international experimental programme was carried

out by the ECCS to study the behaviour of standard columns. More than 1000

buckling tests, on various types of members (I, H, T, U, circular and square hollow

sections), with different values of slenderness (between 55 and 160) were studied. A

probabilistic approach, using the experimental strength, associated with a theoretical

analysis, showed that it was possible to draw some curves describing column

strength as a function of the QRQ� GLPHQVLRQDO� VOHQGHUQHVV� �reference

slenderness ). The imperfections which have been taken into account are: a half

sine-wave geometric imperfection of magnitude equal to 1/1000 of the length of the

column; and the effect of residual stresses relative to each kind of cross-section. The

European buckling curves (a, b, c or d) are shown in Figure B.5.5

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Swapnil B.Kharmale 117 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.5.2* Buckling curves for a cross section (From Table 7.2 of IS: 800 (Draft))

Cross Section Limits Buckling about axis

Buckling Curve

h/b > 1.2 : tf d 40 mm

40 mm<tfd100 mm

z-z y-y

z-z y-y

ab

bc

Rolled I-Sections

h/b < 1.2 : tf d 100 mm

tf >100 mm

z-z y-y

z-z y-y

bc

dd

Welded I-Section

tf <40 mm

tf <40 mm

z-z y-y

z-z y-y

bc

cd

Hot rolled Any a Hollow Section

Cold formed Any b

Generally (Except as below)

Any b Welded Box Section

Thick welds and b/tf < 30

d/tw < 30

z-z

y-y

c

c

Channel, Angle, T and Solid Sections

Any

c

dh tw

by

y

zz

tf

h

y

tw

ttfy

b

zz h

y

tw

ttfy

b

zz

y

tw

y

z z

b

tf

h

y

y

zz

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Swapnil B.Kharmale 119 Comparative study of IS: 800 (Draft) &EC3 CD-051061

effective length KL ,to appropriate

radius of gyration r

m0= Partial safety factor for material

strength

M1=Partial safety factor for material

strength

Section Classification for axial

compression:-

For members in axial compression the

’limiting width to thickness ratios” for

Plastic and Compact class is not

applicable. Here we have to just check

that the section dose not fall in ‘Slender

class' so that whole cross sectional area

is effective in compression.

Effective Area Ae:-

For Semi-compact section

Ae=Gross Area A

g

For Slender Class

Ae

= Based on effective c/s

dimension

IS code doesn't give procedure how to

calculate the effective dimensions of

Slender section

Section Classification for axial

compression:-

For member in axial compression Euro

Code3 specifies the ‘Limiting width to

thickness ratios’ for Class 1, Class 2

Class 3 & Class 4 cross sections.

Effective Area Aeff:-

$V� A=Aeff/A

For Class 1,Class 2 & Class 3 section

A= (A

eff/A)=1

For Class 4 section

Aeff = Calculated on the basis of effective

c/s dimension

A= (A

eff/A)<1

Euro code 3 gives procedure to calculate

the effective c/s dimension of Class 4

section (see Chapter B.1 of Dissertation)

B.5.7 Single angle discontinuous strut

In roof trusses ,the single angle web members are often connected by one leg

(thus introducing eccentricity with respect to the centroid of the cross sections) on

one side of chords (Fig B.5.6) and sometimes alternatively on opposite side of T-

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Swapnil B.Kharmale 121 Comparative study of IS: 800 (Draft) &EC3 CD-051061

where L = Laterally unsupported length of

the member

rvv =Radius of gyration about the

minor axis

b1, b2 =Width of the two legs of the angle

t = thickness of the leg

= yield stress ratio ( 250/fy)0.5

Table B.5.3 Constant k1,k2,k3

(As per Table 7.6 of IS:800 (Draft))

No of the

bolt at end

connection

Gusset or

connecting

member

fixity *

k1 k2 k3

Fixed 0.20 0.35 20 > 2

Hinged 0.70 0.60 5

Fixed 0.75 0.35 20 1Hinged 1.25 0.50 60

*Stiffness of in-plane rotational restraint provided to the

JXVVHW�FRQQHFWLQJ�PHPEHU��)RU�SDUWLDO�UHVWUDLQW��WKH� e

FDQ�EH�LQWHUSRODWHG�EHWZHHQ�WKH� e results for fixed and

hinged cases

Buckling curve to be used= ‘c’ curve

Imperfection factor = 0.49for curve c

Y ���HII�Y Y Y $

] ���HII�] ] ] $

\ ���HII�\ \ \ $

��������[ ZKHUH� [)RU�EXFNOLQJ�DERXW�]�]�D[LV

��������[ ZKHUH� [)RU�EXFNOLQJ�DERXW�\�\D[LV

��������[ ZKHUH� [

where 1 = >(�Iy]1/2=93.9 � ¥������Iy)v� z� y= Effective slenderness ratios

“ratios of buckling length l ,to

appropriate radius of gyration i

Buckling curve to be used =’c’ curve

Imperfection factor = 0.49for curve c

Knowing the non dimensional effective slenderness ratio e( HII ) and imperfection

factor � WKH� VWUHVV� UHGXFWLRQ� IDFWRU� DQG� VXEVHTXHQWO\� WKH� VWUHQJWK�RI� VLQJOH� DQJOH�strut can be evaluated as equation given in article “Codal provisions for designing

compression member”

Refer the worked example solved by both code.

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Swapnil B.Kharmale 123 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Here Eurocode 3 doesn’t not give any guidance about the effective length (buckling

length) factor K for different end conditions in plane buckling which required to be

multiplied to system length L to calculate the effective length KL (buckling length l)..

B.5.7 Built-up compression member

For large loads and for effective use of material, built-up columns are often

used. They are generally made up of two or more individual sections such as

angles, channels or I-section properly connected along their length by lacing or

battening so that they act together as a single unit.

Here the design provisions and construction details of laced and battened

compression member by both codes is discussed.

o Laced column

As per IS: 800 (Draft) (Clause 7.6)

a) Spacing between column section members comprising two main components

laced S:- The spacing S should be such that the built-up beam as a whole as

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Swapnil B.Kharmale 125 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Table B.5.5:-Width of lacing bars

Nominal bolt/rivet dia (mm) Minimum Width of Lacing

Bars (mm)

22 65

20 60

18 55

16 50

h) Thickness of lacing bar:-The thickness of flat lacing bars shall not be less

than one-fortieth of its effective length for single lacings and one-sixtieth of

the effective length for double lacings.

i) End Tie Plates � Laced compression members shall be provided with tie

plates at the ends of lacing systems, at intersection with other members and

at points where the lacing systems are interrupted

j) Lacing forces:-The lacing shall be proportioned to resist a total transverse

shear, Vt, at any point in the member, equal to at least 2.5 percent of the axial

force in the member and shall be divided equally among all transverse lacing

systems in parallel planes.

As per Eurocode 3 (Clause 5.9)

a) Basis

x Built-up compression members consisting of two or more main components

connected together at intervals to form a single compound member shall be

designed incorporating an equivalent geometric imperfection comprising an initial

bow eo, of not less than l/500.

x The deformation of the compound member shall be taken into account in

determining the internal forces and moments in the main components, internal

connections and any subsidiary components such as lacings or battens.

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Swapnil B.Kharmale 127 Comparative study of IS: 800 (Draft) &EC3 CD-051061

x The design procedure given here is for a design compressive force NSd applied

to a built-up member consisting of two similar parallel chords of uniform cross

section, with a fully triangulated system of lacing which is uniform throughout the

length of the member.

x Second moment of area of laced column

The effective second moment of area lo* of a laced compression member with two

main components should be taken as:

Ieff = 0.5 ho2Af

where Af = The cross-sectional area of one chord

ho = The distance between centroid of chords.

x Chord forces at mid length of laced compression member

VI�6G 6G

R

01 ���[1 � Kwhere

�6G R HII

�V R FU6G 6G

FU Y

1 [H (,O0 ���H ���1 1 1 ��� O>�� � @1 6

and Sv =The shear stiffness of the laced compression member

The dimension of lacing element and spacing of lacing should be so

Selected the shear stiffness is large enough to nullify the effect of Ms (in

other word the shear deformations should be small).

The values of Sv for various lacing system are given in Table B.5.8

x Buckling resistance of chords (Nb.Rd). The buckling length of a chord in the plane of a lacing system should be taken as the

system length a between lacing connections.

x Lacing forces

The lacing forces adjacent to the ends of the member should be derived from the internal shear force Vs, taken as

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Swapnil B.Kharmale 129 Comparative study of IS: 800 (Draft) &EC3 CD-051061

x The number of battens shall be such that the member is divided into not less

than three bays within its actual length from centre to centre of end connections.

b) Design

x Spacing of Battens:- The spacing of battens centre-to-centre of its end fastenings

shall be such that the slenderness ratio (KL/r) of any component over that distance

shall be not greater than 50 or greater than 0.7 time the slenderness ratio of the

member as a whole about its z-z (axis parallel to the battens).

x Size: – Refer figure B.5.9

For end batten d end batten t (S+ 2Cyy)

For intermediate batten dint batten t(3/4) th of (S+ 2Cyy) or

t 2Bf

The thickness of batten or the tie plates shall be not less than one fiftieth of the

distance between the innermost connecting lines of rivets or welds, perpendicular

to the main member i.e. t t(1/50)th of a

x Batten forces

Battens shall be designed to carry the bending moments and shear forces arising

from transverse shear force Vt

Vt= 2.5 % Total axial force on column

Shear force in batten WE

9&9 16Bending moment in batten W9&0 �1where

Vt = The transverse shear force as defined above

C = The distance between centre-to-centre of battens, longitudinally

N = The number of parallel planes of battens

S =The minimum transverse distance between the centroids of the rivet

group/welding connecting the batten to the main member

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Swapnil B.Kharmale 131 Comparative study of IS: 800 (Draft) &EC3 CD-051061

taken as laterally restrained in the plane of the battens. As far as possible, the

intermediate battens should be spaced and proportioned uniformly throughout

the length of the member.

b) Design x The design procedure given here is for a design compressive force NSd applied

to a built-up member consisting of two similar parallel chords of uniform cross-

section, spaced apart and inter-connected by means of battens, which are

rigidly connected to the chords and uniformly spaced throughout the length of

the member.

x When Sv, is evaluated disregarding the flexibility of the batten plates

For end batten D end batten tho

For intermediate batten Dint batten t0.5 ho

where ho is the distance between the centroids of the chords.

When Sv, is evaluated considering the flexibility of the batten plates the batten

should satisfy

E I

R

Q, ,��K Dwhere,

Ib =The in-plane second moment of area of one batten

I =The in-plane second moment of area of one chord

ho =The distance between centroids of chords

a =The system length between centerlines of battens

n =The number of planes of battens.

The batten size should so adjusted that above conditions should satisfied

x Second moment of area

The effective in-plane second moment of area leff of a battened compression

member with two main components should be taken as:

HII R I I, ���K $ �� ,

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Swapnil B.Kharmale 133 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Vs= 0s/l

For the purpose of this check, the axial force in each chord may be taken as

0.5Nsd even when there are only three panels in the length of the member.

Comment on designing of laced and battened column by both codesx The bearing capacity of built-up columns is largely affected by the shear

deformations. Because of shear deformations the initial lack of straightness of

column is strongly amplified. The design by Eurocode 3 is based on above

approach. On the other hand in design process by IS: 800 (Draft) doesn’t directly

consider shear deformation of laced or battened compression member (The

shear deformations are indirectly accounted by increasing the effective length by

5% for laced column and 10% for battened column.)

x Eurocode 3 doesn’t give formulation for fixing the size of lacing or battens (as IS:

800 (Draft) gives). The size of lacings or battens are decided such the shear

stiffness are large to nullify the effect of Ms.

B.5.8 Worked examples for Compression member

The worked example contain Analysis problem on single angle strut and

design problem on built-up column by IS: 800 (Draft) and Eurocode 3

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For our caseAssuming the fixity as partial, hence taking the value k1,k2 and k3 Table 7.6 of IS:800as average of same mentioned for fixed and hinged connection (Draft)

Therefore k1=0.45

k2=0.475

k3=12.5

Also

Therefore

Now Clause 7.1.2.1

+HUH� e & for angle section buckling curve ’c’ is used irrespective

RI�D[LV�EHQGLQJ�KHQFH� ,PSHUIHFWLRQ�IDFWRU� ����=1.82

Stress reduction factor

Design compressive stress

Design Compressive Strength Clause 7.1.2

Section ClassificationZKHUH� ¥�����Iy)=1

Hence Slender class Table 3.1 of IS:800- (Draft)

Hence Slender class

Hence Slender class

YY � � � �

���[����� ������� ������ �����

[�[�� [�[���[ �[ [�[�

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� > � @ ���� � >���� ����� @

\ \

FGP� P�

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I II

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Swapnil B.Kharmale CD-051061 135 Compartive study of IS:800 (Draft) and EC3

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Stress reduction factor

Design compressive stress

Design Compressive Strength Table 3.1 of IS:800-

Section Classification (Draft)

Hence not Slender class

Hence not Slender class

Hence not Slender class

Hence whole section is of Non Slender class and gross cross sectional area is effective in compressionTherefore Ag=Ae=1257mm2

Capacity of ISA 100 x 65 x 8

Conclusion from problem :- While designing the compression member avoid to design the slender cross section because whole cross section area (A g ) will not be effective in compression and it

will lead wastage of material

� � ��� � � ���

� � �����

�> � @ ����� �>����� ���� @

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W �E �G ��� ���

����� � ��W �

G3 ����[����� �����N1

Swapnil B.Kharmale CD-051061 137 Compartive study of IS:800 (Draft) and EC3

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Calculation of effective section properties and effective area Clause 5.3.3 and Table 5.3.3 of Eurocode3

���ZKHUH� 1�DQG� 2 stresses at tip of outstanding element

)RU�XQLIRUP�FRPSUHVVLRQ�� 2 1�KHQFH� ��DQG�N ����

For our caseEffective section properties

Effective area of cross section

And

(IIHFWLYH�6OHQGHUQHVV�5DWLR� eff Clause 5.8.3

For single angle strut the effective slenderness ratio are as follows

For buckling @ v-v axis:= where

For buckling@ z-z axis:= where

HII IRUK

HII IRUE

S IRUKIRUK S IRUK S IRUK�

S IRUK

S IRUEIRUE S IRUE S IRUE�

S IRUE

K S [K

E S [E

� ���� K � WS ! �����

����[ [ N

� ���� E � WS ! �����

����[ [ N

N ���� IRU �

' "

' "

" "'

S IRUK

IRUK S IRUK�

S IRUE

IRUE S IRUE�

HII

HII

��� �� ���������[����[ ����

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'

'

�HII HII HII

HII$

W W$ K � [W � E � [W ���� � � [� � ���� � � [� �����PP

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����$ ����

HII�Y Y ���� ����[

HII�] ] ��������[

Y ���Y $

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Swapnil B.Kharmale CD-051061 139 Compartive study of IS:800 (Draft) and EC3

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Buckling strength (Capacity) of ISA 125 x 95 x 6 Clause 5.5.1.1

b) When ISA 100 x 65 x 8 is usedSectional Properties

A= 1257 mm2

h= 100mmb= 65mmt= 8mm

izz= 18.3mm

iyy= 31.6mm

ivv= 13.9mm

Section Classification Table 5.3.1 of Eurocode3

Hence Class 3 section

Hence Class 3 section

Hence whole section is of Class 3 and Aeff=A and

(IIHFWLYH�6OHQGHUQHVV�5DWLR� eff Clause 5.8.3For single angle strut the effective slenderness ratio are as followsNow Taking l=L=2.7m Annex E , E.1

And

\E�5G PLQ $

P�

E�5G

E�5G

I1 [ [$[

���1 �����[����[����[

���1 �����N1

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I ���

K ��� ���� ���

W �K �E ������

�����������W �[�

$ �

���Y

���]

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������ [ � �����

�����������

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Swapnil B.Kharmale CD-051061 141 Compartive study of IS:800 (Draft) and EC3

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Effective non dimensional ���� H ���� � PLQ �����Slenderness ratio for strut of ISA 125x95x6 (L= 2.7 m)

Capacity of ISA 125x95x6 Pd=66.67 kN N b.Rd=89.75 kN

(Slender Class)

Effective non dimensional ���� H ��� � PLQ �����Slenderness ratio for strutof ISA 100x65x8 (L=2.7m)

Capacity of ISA100 x65x8 Pd=73.42 kN N b.Rd=67.4 kN

(Non-slender class)

Points IS :800 (Draft) Eurocode 3

"Analysis of single angle discontinuous strut by by both code"

Swapnil B.Kharmale CD-051061 143 Compartive study of IS:800 (Draft) and EC3

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Sectional PropertiesA= 5366mm2

h= 350mmb= 100mmtf= 13.5mmtw= 8.1mm

cyy= 24.4mmIzz= 10008x104mm4

Iyy= 430.6x104mm4

rxx= 136.6mmryy= 28.3mmh2= 30.9mmg= 60mmd= h-2h2 = 288.2mm

4.To find S (spacing between channels placed back to back)

Spacing between the channel section should be such that built -up column as a whole have nearly same strength against both axis buckling For that

Also (ryy) built-up column >(rzz)built-up column Clause7.6.1.1

Therefore S= 218.4mm

say 220mm so that ryy>rzz

rmin= (rzz) built-up column =136.6mm 5.Capacity of Section Now Minimum slenderness ratio

:=36.9 say 37

S4 4 22(10008x10 )=2[430.6x10 +5366( +24.4) ]2

KL 5040=r 136.6

( ) ( )zz builtup column yy built upcolumnI I− −=

Swapnil B.Kharmale CD-051061 145 Comparative study of IS:800 (Draft) and EC3

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8.Design Of Lacing :-

Single Lacing System

8.1Angle of inclination of Lacing bar Clause7.6.5Let angle of inclination of lacing bar ө =45o

8.2 Spacing of lacing bar a1

a1=2xaxCotө Where a =S+2g=220+2x60=340mma1=2x340xCot(45o)a1=680 mm

Checks for a1:- Clause 7.6.6.1

whichever small

where, r1=min.radius of gyration of individual elements

1

1

50 0.7a KLorr r≤

Swapnil B.Kharmale CD-051061 147 Comparative study of IS:800 (Draft) and EC3

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8.7 Design force on Lacing bar

Transverse shear at any point in member = VT= 2.5% of Pd

Therefore,Transverse shear in each panel= (VT/2)=22.5 kN

Design Compressive force Cd/Tensile force Td in lacing bar

8.8 Capacity of Lacing bar

For Compression

As lacing is solid flat plate therefore we have to use buckling curve 'c' Table 7.2 of IS:800-(Draft)

For fy=250 and (KL/r)lacing =111 we have Table 7.4c of IS:800

fbd=94.6-((111-110) x(94.6-83.7)/(120 (KL/r) fbd(N/mm2) (Draft)fcd=93.51 N/mm2 110 94.6

120 83.7

As Hence not slender class

Therefore, Ae=Ag=b x t = 70x15 =1050 mm2

Compression capacity of lacing C= Ae x fcd

:= 1050 x 93.51 :=98.185 x103 N :=98.185 kN>31.82kNHence safe

For TensionTension capacity of lacing T= [(b-d) x t x fu] /γm1 Clause 6.3

(Considering rupture of net c/s)

2.5 (1800) 45100TV kN= =

Td d o

V 45C =T = = =31.82kN2Sinθ 2Sin45

lacingKL 480.8( ) = =111<145

15r12

lacingKL 480.8( ) = =32.05<42ε

t 15

Swapnil B.Kharmale CD-051061 149 Comparative study of IS:800 (Draft) and EC3

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9.5 Design force on End Tie plate

Longitudinal shear where,Vt = Transverse shear =45kNC = The distance between c/c of batten

Take it same as spacing of lacing=680mm

MomentN = The no of parallel planes on which End-Tie plates are provided =2S =The minimum transverse distance between the centroids of rivet groupsS = 220 + 2 x 60 =340 mm

9.6 Capacity of End Tie Plate

Shear Vd= Avx(fy/√3 xγm0)=(335 x 8) x(250 /√3 xγm0)=351.65 kN> 45 kN

Hence safe

Moment Considering end tie plate of semi -compact section

Md = βbZp x(fy/γm0) where βb=(Ze/Zp) for semi-compact sectionMd = Ze x(fy/γm0)Md=(8 x 3352/6) x(250/1.1)=34 kNm>7.65kNm Hence safe

9.7 Connection for End Tie plates

The connection should be designed to transmit both shear and bending moment

Assume 20 mm diameter rivets of Grade 4.6 are used (dn=20mm & d= 21.5mm)

Therefore , Rivet Value Rv =Strength of rivet in single shear = (152.44/2) = 76.22 kN

TV CVbNS

=

2TV C

MN

=

33

36

45x10 x680Vb= =45x10 N2x340

45x10 x680M= =7.65x10 Nmm2x2

Swapnil B.Kharmale CD-051061 151 Comparative study of IS:800 (Draft) and EC3

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DESIGN OF BUILT-UP COLUMN BY Eurocode 3

B. Design ProblemDesign a built-up column to carry an axial load of 1200 kN (working).The length of column is 6.0m.The column is effectively held in positon at both ends and restrained against rotation at one end.Use preferably channel sections. Design appropriate lacing system and connections. Use Fy= 250

1.Data Axial Load = 1200 kNLength of column= 6 mEnd condition = Fixed at one end and other end is pinned

2.Loading Considering partial safety factors corresponding to leading action as γF.sup=1.5 under unfavourable effectsTherefore NSd= Design value of compression force= 1.5 x 1200=1800kN

3.Buckling length of column (l)

Length Of Column L= 6mBuckling length of built -up compression member shall be taken Clause 5.8.2 (1)as equal to system length L unless a smaller value justified by analysis Therefore l = 6m

4.Trial Section

For comparasion point of view let us try same section with samelacing connection detail as we have used in design as per IS : 800( Draft)Let us try 2 ISMC 350 placed back to backat a spacing of 220 mm between them

Sectional Properties of Single ISMC 350 Af=5366mm2 tw= 8.1mmh=350mm czz= 24.4mmc=100mm Iyy= 10008x104mm4

tf=13.5mm Izz= 430.6x104mm4

DESIGN STEPS REFERENCES

Swapnil B.Kharmale CD-051061 153 Comparative study of IS:800 (Draft) and EC3

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6. Design compression resistance of individual channel Nfc.Rd

For Class 2 section =Nfc.Rd= Af x fy /γmo

Nfc.Rd = 5366 x250/1.1Nfc.Rd = 1219.5 kN

7.Lacing Details Let us try 70 mm x 15 mm solid plate lacing bar with an angle of inclination ө =45o

Therefore Ad=1050mm2, a= 2 x cot 45o x (S+2g)=680mm

d =(a/2) x sec 45o =480.83 mm

Therefore , Shear stifness of lacing S v

Clause 5.9.2.4 & figure 5.9.3

8. Second Moment Of area(For laced compression member) Clause 5.9.2.3 The effective second moment of area Ieff of a laced compression & eqn (5.85)member should be taken as

Ieff= 0.5 x Af x ho2= 0.5 x 5366 x 268.82

Ieff= 19385.59 x 104 mm4

9.Chord forces at mid-length Nf.Sd (Check for capacity of the Clause 5.9.2.4 section in axial compression) & eqn (5.86)

2d o

v 3

5 2

v 3

nEA axhS = where n = no of plane containing lacing2d

2x2x10 x1050x680x268.8S = =92813.70kN2x480.83

sf.Sd Sd

o2

Sd o effs o cr 2

Sd Sd

cr v

o

2 5 4

cr 2

v

MN =0.5N +h

N e π EIlwhere M = , e = ,N =N N 500 l[1- - ]N S

6For our case, e = =0.012m,500π x2x10 x19385.59x10 N = =10629.34kN

6000 S =92813.70kN

Sd

s

f.Sd

N =1800kN1800x0.012 M = =26.63kNm1800 1800[1n n ]

10629.34 92813.726.63N =0.5x1800+ =999.07kN

0.2688∴

Swapnil B.Kharmale CD-051061 155 Comparative study of IS:800 (Draft) and EC3

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15. Shear Stifness Sv of End Tie Panel Clause 5.9.3.2 Sv is evaluated disregarding the flexbility of batten plates themsel-vesFor that a) Width of batten plate > h o here 335 mm >268.8 mm hence o.k.

a =System length between centrelines of batten i.e. Spacing of batten If=In-plane moment of inertia of one chord Ib=In-plane moment of inertia of one batten n= No of planes on which battens are provided

Here taking a = 680 mm (same as that of lacing)

As a=680mm > 97.575 mm , Sv is calculated disregarding the effect of flexibility of battens

ThereforeClause 5.9.3.4 eqn (5.92)

16.Second Moment Of area(For battened compression member)

Ieff= 0.5 x ho2 x Af +2xμxIf Taking μ= 1Ieff= 0.5 x 268.82 x 5366 +2x1x10008x104

Ieff= 39401x104 mm4

17.Chord forces at mid-length Nf.Sd (Check for capacity of the Clause 5.9.3.4 section in axial compression) eqn (5.91)

f o

b

10I hb)a³nI

3

3

10x10008x10 x268.8(680) 12x( x440x335 )12

97.575mm

2 2 5 4f

v 2 2

2π EI 2xπ x2x10 x10008x10S = = =854455kNa 680

s o ff.Sd Sd

eff2

Sd o effs o cr 2

Sd Sd

cr v

o

2 5 4

cr 2

v

M h AN =0.5(N + )I

N xe π EIlwhere M = , e = ,N =N N 500 l[1- - ]N S

6For our case, e = =0.012m,500π x2x10 x39401x10 N = =21604kN

6000 S =854455k

Sd

N N =1800kN

Swapnil B.Kharmale CD-051061 157 Comparative study of IS:800 (Draft) and EC3

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For our case for one chord NSd = 900 kN , χmin =0.9977 A =Af=5366 mm2 ,Wpl.y =640.12x104 mm4, Wel.y =571.9 x104mm4

λy = (a/iyy)=(680/136.6)=4.97 , χy =1 , Also βMy =1.8 Figure 5.5.3Hece μy=4.97 x (2 x1.8 -4) +((640.12-571.9)/571.9) =-1.867 And ky =1- (( -1.867 x 900 x103)/(1 x 5366 x 250) =2.25 >1.5Therefore take ky =1.5Hence

"Design Problem on built-up column"by IS:800 (Draft) and Eurocode 3

1.Design Approach As per IS:800 (Draft) the The bearing capacity of built-upbuilt-up columns are design- columns is largely affected by ed and proportioned accor- the shear deformations.Beca- ding to emperical formula use of shear deformations themost of which are releated to initial lack of straightness of local buckling requirements column is strongly amplified

The design by Eurocode 3 is based on above approach.

1.Effective/Buckling length Effective Length =KL Buckling Length =I =L = 1.05 (k1L)

where k1 depends on end No increase in buckling lengthconditions &1.05 indicate for laced columnthe eff.length is increased by 5% for laced column to account shear deformations

2.Trial section's capacity (P/Pd)=(1800/2166.79) (N/Nd)= (2xNf.Sd)/(2xNb.Rd) ratio :- = 0.83 =(2x999.07/2x1216.740i.e(Design Action/Strength) =0.82

Points IS:800 (Draft) Euro Code 3

y sdy y

y y

pl.y el.yy y My y

el.y

μ xNwhere k =10 but k 1.5

χ xAxf

W -W μ =λ (2β -4)+ [ ] but μ £0.90

W

3 6

4

900x10 1.5x2.06x10+250 2500.9977x5366x 640.12x10 x1.10 1.10

=0.74+0.002=0.742<1Hence safe

Swapnil B.Kharmale CD-051061 159 Comparative study of IS:800 (Draft) and EC3

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Swapnil B.Kharmale 161 Comparative study of IS: 800 & EC3 CD-051061

iii. Closely spaced, discrete bracing is provided so that the weak axis

slenderness (KL/r)y [(l/i)z] of the beams is low.

x Laterally unsupported beams (Unrestrained beams)

The beam is considered laterally unsupported when

i) Compression flange of beam is not restrained laterally against the

lateral buckling

ii) Bending take place in weaker direction

For hot rolled beams and channel section which have very small moment of

inertia about minor axis as compared to that about major axis, this make section

relatively weak against torsion and bending about weaker axis, and if not held or

supported in the direction normal to weak axis, it will bend in the weaker direction

even if the load is normal to axis, it will be weakened further if a horizontal load

were to be applied normal to y-y axis. This bending is usually accompanied by

twisting. This phenomenon of bending in the weaker direction and twisting may be

called as “Lateral Torsional Instability”

Refer following Fig B.6.2 for “Lateral Torsional Buckling”

Figure B.6.2 shows the response of a slender cantilever beam to a vertical

end load

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Swapnil B.Kharmale 163 Comparative study of IS: 800 & EC3 CD-051061

x Statically Determinate Beams Figure B.6.3 presents the relationship between applied load and central

deflection that would be obtained from a test on a simply supported steel beam of

Compact (Class 2) proportions. Three distinct phases may be identified:

ESte

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Swapnil B.Kharmale 165 Comparative study of IS: 800 & EC3 CD-051061

B.6.4 Shape factor S

The ratio of fully plastic moment capacity Mp to yield moment My is called as shape

factor

S S \ S SO \ SO

\ H \ H HO \ HO

0 = I = : I :6 ���L�H��6 ����������6 0 = I = : I :$V�SHU�,6 � FRGH������$V�SHU (XURFRGH

B.6.5 Codal provision for design of laterally supported and laterally

unsupported beam

As per IS:800 (Draft) As per Eurocode 3

The factored design moment, M at any

section, in a beam due to external

actions shall satisfy (Clause 8.2)

M � 0d

A)Laterally Supported Beam-

(Clause 8.2.1)

WhenV����9d

Md

= Eb

Zp

fy �m0

d 1.2 Ze

fy �m0

where

Eb =1.0 for plastic and compact section

Eb = Ze/ Zp for semi-compact section

Zp = Plastic section modulus

Ze =Elastic section modulus

fy = Yield stress of the material

m0 = Partial safety factor

The design value of the bending moment

MSd

at each cross-section shall satisfy

(Clause 5.4.5.1)

MSd

� 0c.Rd

A)Laterally Supported Beam-

WhenVSd����9pl.Rd

Class 1 or 2 cross-sections: M

c.Rd = Wplfy/

M0 Class 3 cross-sections:

Mc.Rd = W

elfy/

M0 Class 4 cross-sections:

Mc.Rd = W

efffy/

M1 where

Wpl = Plastic section modulus

Wel = Elastic section modulus

Weff= Eff. section modulus for Class 4

section as per effective dimensions

fy = Yield stress of the material

m1 = Partial safety factor

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Swapnil B.Kharmale 167 Comparative study of IS: 800 & EC3 CD-051061

its web to resists the shear force as well as to assists the flanges in resisting

moment. Thus a cross-section subject to co-existing bending and high shear has a

reduced moment resistance in presence of high shear. The interaction between

moment and shear is shown

HHoolleess iinn tteennssiioonn zzoonnee ((CCllaauussee88..22..44..11))

a) The effect of holes in the tension

flange, on the design bending strength

need not be considered if

(Anf / Agf) t (fy/fu) ( m1 / m0 ) / 0.9

where

Anf /Agf =Ratio of net to gross area of

the flange

fy/fu =Ratio of yield and ultimate

strength of the material

m1/ m0= Ratio of partial safety factors

against ultimate to yield stress

b) When the Anf /Agf does not satisfy the

above requirement, the reduced flange

area Anf satisfying the above equation

HHoolleess ffoorr ffaasstteenneerrss ((CCllaauussee 55..44..55..33))

a) Fastener holes in the tension flange

need not be allowed for, provided that for

the tension

0.9 (Af.net / Af) t (fy/fu) ( m2 / m0 )

where

Af.net /Af =Ratio of net to gross area of

the flange

fy/fu =Ratio of yield and ultimate

strength of the material

m2/ m0= Ratio of partial safety factors

against ultimate to yield stress

b) When Af,net/Af is less than this limit, a

reduced flange area may be assumed

which satisfies the limit.

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Swapnil B.Kharmale 169 Comparative study of IS: 800 & EC3 CD-051061

eeffffeeccttiivvee wwiiddtthh ooff ffllaannggee ffoorr ddeessiiggnn

ssttrreennggtthh mmaayy bbee ccaallccuullaatteedd uussiinngg

ssppeecciiaalliisstt lliitteerraattuurree,, oorr ccoonnsseerrvvaattiivveellyy

ttaakkeenn aass tthhee vvaalluuee ssaattiissffyyiinngg tthhee lliimmiitt

ggiivveenn aabboovvee..

The calculation of effective breadths of

flanges is covered in ENV 1993-1-3

Eurocode 3:Part 1.3) and ENV 1993-2

Eurocode 3: Part 2)

B)Laterally Unsupported Beam-

Effect of Lateral Torsional Buckling (LTB)

on flexural strength need not be

FRQVLGHUHG�LI�LT

d 0.4 (Clause 8.2.2)

where,

�LT

=Non-dimensional slenderness ratio

for lateral torsional buckling

Then

Md

= Eb

Zp

fbd

fbd = Design bending compressive stress,

obtained as follow

\EG /7

PR

/7

� � ���/7/7 /7 /7

�/7 /7 /7 /7

/7

II � [ZKHUH��� 6WUHVV�UHGXFWLRQ�IDFWRU�IRU�/7%

� �> � � @ ����>��� � � � ��� � @ ,PSHUIHFWLRQ�IDFWRUV�

� �����IRU�UROOHG� VHF WLRQ��� ��

E S \

/7FU

FU

� �\ Z� �FU W

���IRU�ZHOGHG�VHFWLRQ[ = �[�I 0

ZKHUH�0 (ODVWLF�FULWLFDO�PRPHQW(, (,0 > @>*, � @�./ ./

B)Laterally Unsupported Beam-

The design buckling resistance moment

of a laterally unrestrained beam shall be

WDNHQ�DV��LT

d 0.4) (Clause 5.5.2)

Mb.Rd

=LT w

Wpl.y fy/ M1

where

w=1 for Class 1 or Class 2

cross-sections

w=W

el.y /W

Pl.y for Class 3 cross

section

w=W

eff.y /W

pl.y for Class 4 cross

section

LT= The reduction factor for lateral-

torsional buckling.

�/7 � ���/7/7 /7

�/7 /7/7 /7

/7

E S \/7

FU

FU

� �> � � @ ����>��� � � � ��� � @ ,PSHUIHFWLRQ�IDFWRUV�

� �����IRU �UROOHG� VHF WLRQ��� �����IRU�ZHOGHG�VHFWLRQ

[ = �[�I 0ZKHUH�0 (ODVWLF �FU

/7 ���/7 :

/W���� \ /7

]

\

LWLFDO�PRPHQWWR�EH�FDOFXODWHG�IURP�IRPXODH�JLYHQLQ�LQIRUPDWLYH�$QQH[�)RU ��DOWHUQDWLYHO\�

O ( � I ���� L��� <LHOG�VWUHVV�UDWLR I

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Swapnil B.Kharmale 171 Comparative study of IS: 800 & EC3 CD-051061

x Here by considering LT

= 0.4 we can find the laterally unsupported length up

to which LTB is not governing mode.Refer Table B.6.2 which give the

unsupported length corresponding to LT

=0.4 for Indian Standard I-section

by both codes

About Lateral Torsional Buckling

Beams experiencing the bending about the major axis and not restrained

against the lateral buckling of compression flange may fail by lateral torsional

buckling before material fails

Consider a simply supported beam with a uniform bending moment M as shown in figure B.6.6

Notations used in derivation

x Iz=Second moment of area @ major axis (X-axis)

x Iy=Second moment of area @ minor axis (Y-axis)

x It=St. Venant’s constant =� >bt3]/3

x E=Young’s modulus

x =Poisson’s ratio

x G=Shear modulus E/(2*(1+ ))

x =Angle of twist

x u ,v=Displacement along X-axis and Y-axis respectively

x Mz=Bending moment @z-axis (Major axis)

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(Continued)Design- Section Classification Sect. Modulii Moment ation Elastic Plastic Capacity

As per As per (@ major (@ major By IS:800 By EC3IS:800 (Draft) Eurocode3 axis) axis) ( Mpor My) (Draft)

cm3 cm3 x106Nmm mm mm

ISMB 100 Plastic Class1 51.5 57.4 13.1 830.0 601.8ISMB 125 Plastic Class1 71.8 80.2 18.2 786.2 589.9ISMB 150 Plastic Class1 96.9 108.4 24.6 791.5 604.5ISMB 175 Plastic Class1 145.4 163.0 37.1 882.7 677.3ISMB 200 Plastic Class1 223.5 249.7 56.7 1024.5 782.9ISMB 225 Plastic Class1 305.9 342.6 77.9 1115.1 852.1ISMB 250 Plastic Class1 410.5 458.4 104.2 1249.6 965.0ISMB 300 Plastic Class1 573.6 641.3 145.7 1323.1 1034.2ISMB 350 Plastic Class1 778.9 877.0 199.3 1328.6 1034.2ISMB 400 Plastic Class1 1022.9 1161.5 264.0 1326.7 1026.9ISMB 450 Plastic Class1 1350.7 1534.2 348.7 1414.0 1096.1ISMB 500 Plastic Class1 1808.7 2047.5 465.4 1640.3 1281.8ISMB 550 Plastic Class1 2359.8 2678.4 608.7 1742.7 1358.3ISMB 600 Plastic Class1 3060.4 3465.4 787.6 1922.5 1500.3

ISWB 150 Plastic Class1 111.9 125.1 28.4 978.9 761.1ISWB 175 Plastic Class1 172.5 192.2 43.7 1190.6 943.2ISWB 200 Plastic Class1 262.5 291.2 66.2 1372.5 1088.8ISWB 225 Plastic Class1 348.5 386.8 87.9 1471.9 1172.6ISWB 250 Semicompact Class3 475.4 108.0 1734.7 1408.3ISWB 300 Compact Class2 654.8 725.0 164.8 1808.9 1463.9ISWB 350 Plastic Class1 887 986.1 224.1 1824.2 1467.6ISWB 400 Plastic Class1 1171.3 1306.9 297.0 1840.8 1471.2ISWB 450 Plastic Class1 1558.1 1742.7 396.1 1880.3 1496.7ISWB 500 Plastic Class1 2091.6 2331.6 529.9 2250.1 1806.2ISWB 550 Plastic Class1 2723.9 3038.2 690.5 2322.0 1860.9ISWB 600 Plastic Class1 3540 3951.5 898.1 2398.2 1911.8ISWB 600 Plastic Class1 3854.2 4302.2 977.8 2456.5 1948.3 (Continued)

KLSection Vs Laterally Unsupported

Length

0 500 1000 1500 2000 2500 3000

ISMB 100

ISMB 125

ISMB 150

ISMB 175

ISMB 200

ISMB 225

ISMB 250

ISMB 300

ISMB 350

ISMB 400

ISMB 450

ISMB 500

ISMB 550

ISMB 600

ISWB 150

ISWB 175

ISWB 200

ISWB 225

ISWB 250

ISWB 300

ISWB 350

ISWB 400

ISWB 450

ISWB 500

ISWB 550

ISWB 600

ISWB 600

Sec

tio

nKL

As per Eurocode 3 As per IS:800 (Draft)

Swapnil B.Kharmale CD-051061 173 Comparative stud of IS:800 (Draft) and EC3ESte

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Swapnil B.Kharmale 175 Comparative study of IS: 800 & EC3 CD-051061

x My=Bending moment @y-axis (Minor axis)

x KL= Effective length for a s/s beam =L

With the angle of twist ; the lateral displacement u and vertical displacement we

can say that

Mz=Bending moment @ major axis

My=Bending moment @ minor axis is M

Then Moment Curvature Relation

�]

G Y(, 0G] (B.6.1)

�\

G X(, 0G] (B.6.2)

As the section is non circular this torsion arising from angle of twist is

accompanied by warping Therefore equation of non-uniform torsion is given by,

�W Z

G G*, �(, 7G] G] (B.6.3)

where T= Induced torque from external loading =-M(du/dz)

Iw=Warping constant= (Iyh2/4) for a I-section; h= Overall depth of I-section.

Now putting for T in equation (B.6.3) we get

�W Z

�W Z

� � �

� � �W Z

G G GX*, �(, �0G] G] G]G G GX*, �(, �0 �G] G] G]

'LII �Z�U�W��]�ZH�JHWG G G X*, �(, �0 ������������������� %����G] G] G]

But from equation (B.6.2), �

�\

G X(, 0G]Thus we get governing differential equation as follow

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Swapnil B.Kharmale 177 Comparative study of IS: 800 & EC3 CD-051061

Where the ends of the beam are not restrained against torsion, or where the

load is applied to the compression flange and both the load and flange are free to

move laterally, the above values of the effective length shall be increased by 20

percent.

Table B.6.3:- Effective length KL for cantilever of length L

(As per Table 8.1of IS:800 (Draft)

Note: If there is a degree of fixity at the fee end, the effective length shall be multiplied by (0.50/0.85)

in (b) and (c), and by (0.75/0.85) in (d), (e) and (f) above.

As per Eurocode 3

x The effective length factors k vary from 0.5 for full fixity to 1.0 for no fixity,

with 0.7 for one end fixed and one end free.

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Swapnil B.Kharmale 179 Comparative study of IS: 800 & EC3 CD-051061

shear failure

Table B.6.4 Shear area Av

(As per Clause 8.4.1.1 of IS:800 (Draft) and Clause 5.4.6 of Eurocode 3)

Shear buckling (Clause 8.4.2)

Resistance to shear buckling of web

Resistance to shear buckling of web

shall be considered if following situation

exists

xZ

G ! ��W for an unstiffened webs

x YZ

G ! �� NW for stiffened webs

where

kv=Shear buckling coefficient

¥�����Iy)Shear buckling design methods

The nominal shear strength, Vn, of webs

with or without intermediate stiffeners as

governed by buckling may be evaluated

Shear buckling (Clause 5.6)

Shear buckling resistance of web

Web shall be checked for the shear

buckling if following conditions exists.

xZ

G ! ��W for an unstiffened webs

x 7Z

G ! �� NW for stiffened webs

where

kT=Buckling factor for shear

¥�����Iy)Shear buckling design methods

For webs without intermediate transverse

stiffeners and for webs with transverse

stiffeners only, the shear buckling

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Problem 1.Laterally supported beamThe secondary beam B1 of a floor beam system is shown in figure below & is simply supported at both ends.The beam carries working dead load =12 kN/m and working live load =12.5 kN/m from slab.The compression flange of beam is fully embedded in R.C.C.slab. Design the floor beam and apply usual checks. Take fy=250 MPa

1. Loading on Beam B1

Permanent Action (Qp) or D.L.From Slab = 12 kN/mSelf weight of beam = 1 kN/m

Total Permanent Action Qp = 13 kN/m

Variable Action (Qv) or L.L.From Slab = 12.5 kN/m

Total Variable Action Qv = 12.5 kN/m

2. Partial Safety Factors Table 5.1of IS:800For D.L. (i.e. Permanent Action ) fp = 1.5 (Draft)

For L.L. (i.e. Variable Action ) fv = 1.5

DESIGN PROBLEM ON BEAMS BY IS:800 (DRAFT)

DESIGN STEPS REFERENCES

Swapnil B. KharmaleCD-051061 181 Comparative study of IS:800 (Draft) and EC3

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Web :-d/tw � ����������� ������������

whered:=Distance between

Hence web is plastic fillet:= h-2h2 := 450-2x35.40 := 379.2 mm

As both flange and web are plastic hence whole section is plastic6.Shear Capacity Clause 8.4Design Shear Strength of cross section Vd=Vp

As (d/tw����� �IRU�XQVWLIIHQHG�WhereVp = Plastic shear resistance Clause 8.2.1.2

= Av x fy

¥��[� mo

= (hxtw) x fy

¥��[� mo

= ( 450 x 9.4) x 250¥��[����

= 555 x 103 N

= 555 kN As V (Design shear force ) < 0.6 Vd , the effect of shear force

on Plastic moment capacity Mp.

7. Moment Capacity Clause 8.2.1Since the section is plastic and there is no effect of shear force on plastic moment capacity Threrfore, design bending strength of cross section

Md = b x Zp x (fy�� mo )

= 1 x 1553.36x103 x (250/1.10)

= 348.5 x106 N.mm = 348.5 kNm > 269 kNm (M= Design B.M. )

Hence safe

8.Checks 8.1. Deflection Check:- Deflections are to be checked for most Clause 5.6.1 and adverse but realistic combination of service load and their Table 5.3 of arrangements by elastic analysis using load factor 1. IS: 800 (Draft)

Now δmax = 5 x w x L4384 E xIzz

Here w := 1.0 x13 +1.0 x 12.5

Swapnil B. KharmaleCD-051061 183 Comparative study of IS:800 (Draft) and EC3

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And = Imperfection factor corresponding to buckling curve= Non dimensional effective slenderness ratio

=

KL= Effective length of web =0.7 x d =0.7 x 379.2 =265.44 mm

rzz= 2.71mm

Therefore ,

For solid web section we have to use buckling curve c irrespective of axis of bendng

Therefore = 0.49

Therefore

Hence ,the buckling resistance of web

Hence safe

8.3 Web Bearing Or Web Crippling Checks :- Clause 8.7.4The crippling resistance of web is given by

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Swapnil B. KharmaleCD-051061 185 Comparative study of IS:800 (Draft) and EC3

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Problem 1.Laterally supported beamThe secondary beam B1 of a floor beam system is shown in figure below & is simply supported at both ends.The beam carries working dead load =12 kN/m and working live load =12.5 kN/m from slab.The compression flange of beam is fully embedded in R.C.C.slab. Design the floor beam and apply usual checks. Take fy=250 Mpa

1. Loading on Beam B1

Permanent Action (Gk) or D.L.From Slab = 12 kN/mSelf weight of beam = 1 kN/m

Total Permanent Action Gk = 13 kN/m

Variable Action (Qk) or L.L.From Slab = 12.5 kN/m

Total Variable Action Qk = 12.5 kN/m

2. Partial Safety Factors Table 2.2 of EuroFor D.L. (i.e. Permanent Action ) G.sup = 1.35 code 3

For L.L. (i.e. Variable Action ) Q = 1.5

DESIGN STEPS REFERENCES

DESIGN PROBLEM ON BEAMS BY EUROCODE 3

Swapnil B. KharmaleCD-051061 187 Comparative study of IS:800 (Draft) and EC3

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Web :-d/tw � ����������� ����������

whered:=Distance between

Hence web is of Class 1 fillet:= h-2h2 := 400-2x32.80 := 334.4 mm

As both flange and web are of Class 1 hence whole section is of Class 1.

6.Shear Capacity Clause 5.4.6Design Shear Strength of cross section =Vpl.Rd

As (d/tw����� �IRU�XQVWLIIHQHG�WhereVpl.Rd = Plastic shear resistance

= Av x fy

¥��[� mo

= (hxtw) x fy

¥��[� mo

= ( 400 x 8.9) x 250¥��[����

= 467.13 x 103 N

=467.13 kN As VSd (Design shear force ) < 0.5 Vpl.Rd,the effect of shear force

on Plastic moment capacity Mp.

7. Moment Capacity Clause 5.4.5.1Since the section is plastic and there is no effect of shear force on plastic moment capacity Threrfore, design bending strength of cross section of class 1

Mc.Rd= Wplzfy/ M0

= 1161.48x103x250/1.10

= 264 x106 NmmMc.Rd= 264 kNm >256 kNm (MSd)

Hence safe

8.Checks 8.1. Deflection Check:- Eurocode requires that the deflections Clause 4.2.2of the beam to be checked under following seviceability loading conditions :- i)Variable action ii)Permanent action

Swapnil B. KharmaleCD-051061 189 Comparative study of IS:800 (Draft) and EC3

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8.2 Web Buckling Check :- Clause 5.7.5The buckling resistance is determined by taking a length of web as a strut.

The length of web is taken as per Eurocode 3 which in this case gives

For our case a =0 s= Width of stiff bearing plate=Let us take it 75 mm h= Depth of web =400 mm

The height of web for buckling should be taken as l = 0.7 x Distance between fillet = 0.7 xd =0.7 x334.4= 234mm

Radius of gyration of web i= tw/¥12= 8.9/¥12 =2.6 mmTherefore slenderness =(l/i)=90

Hence safe

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Swapnil B. KharmaleCD-051061 191 Comparative study of IS:800 (Draft) and EC3

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Problem 2.Laterally unsupported beamThe primary beam B4 of a floor beam system is shown in figure below & is simply supported at both ends.The beam carries wall of 230 mm thick and 3.55 m clear height and in additon to this it support the secondary beam B1 as shown in figure.The beam is restrained only at ends and at point load Design the floor beam and apply usual checks Take fy=250 Mpa

1. Loading on Beam B1

Permanent Action (Qp) or D.L.Wall load := 0.230x (4-0.450)x20 = 16.33 kN/mSelf weight of beam = 1 kN/m

Total Permanent Action Qp = 17.33 kN/m

And two point loads (working) of 191.33 kN as reactions from B1 acts on beam.

Variable Action (Qv) or L.L. = 0kN/m

2. Partial Safety Factors Table 5.1of IS:800For D.L. (i.e. Permanent Action ) fp = 1.5 (Draft)

For L.L. (i.e. Variable Action ) fv = 1.5

DESIGN PROBLEM ON BEAMS BY IS:800 (DRAFT)

DESIGN STEPS REFERENCES

Swapnil B. KharmaleCD-051061 193 Comparative study of IS:800 (Draft) and EC3

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Sectional properties

A= 13211 mm2

h= 550 mmB= 190 mmb= 95 mmtw= 11.2 mm

tf = 19.3 mmd= 467.5 mmIzz= 64893.6x104 mm4

Iyy= 1833.8x104 mm4

ryy= 37.3 mm

Zez= 2359.8x103 mm3

Zpz = 2678.36x103mm3

Bp= 280 mm

5. To find tp:-

But ,

But provide tp =22 mm so that check for plastic section will be satisfied

6.Check for Section:- Clause 3.7.3 ¥�����Iy)=1

Flange Cover Plate

Flange of Rolled Beam

And

Web

p cov erplate p cov erplate p required p ISMB500

3 3p cov erplate

3 3p cov erplate

(Z ) (Z ) (Z ) (Z )

(Z ) 4583.4x10 2678.36x10

(Z ) 1905.04x10 mm

� �

pp cov erplate p p

p3p

p

h t(Z ) 2xB xt x( )

2550 t

1905.04x10 2x280xt x( )2

t 12.2mm

� �

?

S

E �� ���� � ��� 3ODVWLFW ��

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S

% ��� ���� � ��� 3ODVWLFW ��

Z

G ����� ����� � ���� 3ODVWLFW ����

Swapnil B. KharmaleCD-051061 195 Comparative study of IS:800 (Draft) and EC3

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αT= Imperfection factor =0.21 for rolled section =0.49 for welded section

If λLT < 0.4 then effect of lateral torsional buckling is ignored.

LT :=1.2 for plastic and compact section:=1 for semi-compact section

KL :=Effective laterally unsupported length of memberh :=Over all depth of section

Iyy :=Moment of inertia @ minor axis

ryy :=Radius of gyration of@minor axis

tf :=Thickness of flange of section

While calculating Mcr the sectional properties (h,Iyy,ryy, tf etc) of rolled section are considered though it is built-up as this leadto conservative design.

For our case

As λLT =0.847>0.4 the effect of lateral torsional buckling is considerd

b p yLT

cr

xZ Xf

M

EO

2 2LT yy yy

cr 2 2f

x xEXI xh (KL / r )1M x [1 x ]

202x(KL) (h / t )

E S �

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Swapnil B. KharmaleCD-051061 197 Comparative study of IS:800 (Draft) and EC3

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Now

Where,S= Weld size and S< (3/4)th of tmin or tmin-1.5 whichever Clause 10.5.8.1

is minimum. and 10.5.8.2a= Effective length of intermittent fillet welda> 4xtp or 40 mm whichever is moreb= Clear spacing between effective length of weld

< 12t or 200mm whichever less

mo= 1.25 for shop weld

fuw= Ultimate stress of parent material or weld material whichever less

For our caseTake a= 4x22mm or 40 mm whichever is more say 90 mm

b= 12x17.92 mm or 200 mm whichever is less say 200mm

mo= 1.25

fuw= 410 N/mm2

Therefore provide intermittent fillet weld of size 8mm on either side @ a pitch p=290 mm c/c

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Swapnil B. KharmaleCD-051061 199 Comparative study of IS:800 (Draft) and EC3

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Design ActionsFd := G.supGk+ QQk Clause 2.3.2.2

Fd := 1.35 x17.33 +1.5 x 0:=23.5kN/m

And, Factored Point Load = 1.35x97.5+1.5X93.75 = 272.25≅272.5 kN

Design MomentM at midspan =Mmax= 360.6 x3.75-23.5 x (3.752/2)-272.5 x (3.75-2.5)

MSdmax = 846.5 kNm

Munder point load = 360.6 x 2.5 -23.5 x (2.52/2)

= 828 kNm

Design Shear Force

4. Trial section

As design moment MSdmax=846.5 kNm is large and beam is laterally supported only at ends and under point loads , single rolled section is not sufficient. Therefore try a built-up beam

Let us assume built- up beam is of Class 1

As beam is laterally unsupported hence increase above value of (Wp) req by say 45%

Let us select an ISMB [email protected] kN/m with cover plates 280 mm wide on either sides of flange of rolled beam.

3 3p required(W ) 4583.40x10 mm?

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Swapnil B. KharmaleCD-051061 201 Comparative study of IS:800 (Draft) and EC3

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7.Shear Capacity Clause 5.4.6Design Shear Strength of cross section =Vpl.Rd

As (d/tw����� �IRU�XQVWLIIHQHG�WhereVpl.Rd = Plastic shear resistance

= Av x fy

¥��[� mo

= (hxtw) x fy

¥��[� mo

= ( 550 x 11.2) x 250¥��[����

= 808.29 x 103 N =808.29 kN

As VSd (Design shear force ) < 0.5 Vpl.Rd,the effect of shear force

on Plastic moment capacity Mp.

8.Moment Capacity Clause 5.5.2Mb.Rd�� LT w Wpl.y fy� M1

where

w=1 for Class 1 or Class 2 cross-sections

w=W el.y /W Pl.y for Class 3 cross-section

w=W eff.y /W pl.y for Class 4 cross-section

LT= The reduction factor for lateral-torsional buckling.

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Swapnil B. KharmaleCD-051061 203 Comparative study of IS:800 (Draft) and EC3

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Iyy= =

L= 7.5ma= 2.5m

For permanent action W= 97.5 kN w= 17.33kN/m

1= 4.40+2.15=6.55mmFor variable action W= 93.75 kN/m

w= 0kN/m

2= 4.22mm <L/350 Hence safe

Also max= 1� 2= 10.77mm <L/250 Hence safe

10. Connection between coverplate and flange plate

Shear stress at the junction of cover plate and flange of rolled beam.

Now

Clause 6.6.5.3.(4)

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Swapnil B. KharmaleCD-051061 205 Comparative study of IS:800 (Draft) and EC3

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Swapnil B. Kharmale 207 Comparative study of IS: 800 (Draft) & EC3 CD-051061

Section B:- Study Of Both Code B.7 Design of Member Subjected To Combine Forces

B.7.1 Introduction

Generally, in frame structures the particular cases of beams (F = 0) and

columns (M = 0) simply being the two extremes. In practice the structural members

are generally subjected to various type of combination of forces.

Depending upon the external actions over the members in structural framing

system, the combined forces may be broadly categorized as

i. Combined shear and bending,

ii. Combined axial load and bending

a) Combined axial tension and bending and

b) Combined axial compression and bending i.e. Beam-Column*

The “Combined shear and bending” case is discussed in chapter B.6. Here

more emphasis is given on “Combined axial load and bending”

B.7.2 Cross-sectional behaviour for combined axial load and bending

Figure B.7.1 shows a point somewhere along the length of an I-section column

where the applied compression and moment about the z-axis produce the uniform

and varying stress distributions

*Beam-columns are defined as members subject to combined bending and compression. In principle,

all members in frame structures are actually beam-columns.

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Swapnil B. Kharmale 209 Comparative study of IS: 800 (Draft) & EC3 CD-051061

IQ

K��W)RU�\ � ! ��1HXWUDO�D[LV�OLHV�LQ�IODQJHV�0 \ Z I I Q

1 \ Q Q

K1 I >W K � �W � �E>W � � \ @�K0 I E> � \ @>K � \ @�

(B.7.2)

Figure B.7.3 compares Equations (B.7.2) and (B.7.33) with the approximation used

in Eurocode 3 as well as in IS: 800 (Draft) which is as follow

1\�5G SO�5G QG] G]

6G

SO�5G

>�� Q@ >�� Q@0 0 ������0 0 �>�� ���D@ >�� ���D@�$V SHU�(XURFRGH��������������$V�SHU�,6���� 'UDIWZKHUH�Q 5DWLR�RI �D[LDO�ORDG�WR�VTXDVK�ORDG

1 1 $V�SHU�(XURFRGH��� $V�SHU�,1 1GI

6���� 'UDIW$��EWD ���$

Figure B.7.3 shows the theoretical interaction curve and design interaction curve by

formulae mentioned by both codes for a ISHB 450 section. The design interaction

curve is lower bound when plastic netural axis lies in flange (i.e. when the applied

moment is large as compare to applied axial force) and is upper bound when plastic

netural axis lies in web (i.e. when the applied moment is smaller as compare to

applied axial force)

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Swapnil B. Kharmale 211 Comparative study of IS: 800 (Draft) & EC3 CD-051061

D1, D2 = Constants as given in

Table B.7.1 below.

n = N / Nd

D, = Constants as given in

Table B.7.1 below.

n = N / Nd

Table B.7.1 :-Constant 1( ) and 2� � As per

Table 9.1 of IS:800 (Draft) and Clause 5.4.8.(11) of Eurocode3

Table B.7.2:- Design reduced flexural strength for plastic and compact class

without bolt hole (Approximate formulae as per IS code)*

x Same table is applicable for Eurocode with corresponding symbols

Section D1 ( )* D2( )*

I and Channel 5n t 1 2

Circular tubes 2 2

Rectangular tubes

1.66/(1-1.13n2)< 6 1.66/(1-1.13n2) < 6

Solid rectangles 1.73+1.8n3 1.73+1.8n3

Section Mnd Mndz Mndy

Plates Md(1-n2) -- --

Welded I or H-

section

--

G]

>�� Q@0 >�� ���D@ �Mdz

where

I$��EWD ���$

�G\ G\

Q � D0 >�� @ 0�� Dwhere

I$��EWD ���$For standard I-

section and H-

section --

1.11Mdz(1-n) �Mdz For n�2,

Mdy

For n>2,

1.56(1-n)(n+0.6)

Circular hollow

tubes

1.04Md (1-n1.7)

d Md

-- --

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Swapnil B. Kharmale 213 Comparative study of IS: 800 (Draft) & EC3 CD-051061

Member shall be checked for lateral

torsional buckling under reduced

moment Meff due to tension and bending

Meff=[M- TZec/A]�Md

where M, T = Factored applied moment and

tension, respectively

A = Area of cross section

Zec = Elastic section modulus of the

section with respect to extreme

compression fiber

= 0.8 if T and M can vary

independently

= 1.0 otherwise.

Member shall be checked for lateral

torsional buckling under reduced moment

Meff.Sd due to tension and bending

Meff.Sd=Wcom com.Ed

where

com.Ed = MSd/ Wcom - vec Nt.Sd/ A Wcom =Elastic section modulus for extreme

fiber in compression

Nt.Rd =Design value of axial tension

vec = Reduction factor for vectorial effect

=0.8 if T and M can vary

independently

= 1.0 otherwise

b)Bending and axial compression

(Clause 9.3.2.2)

Members subjected to combined axial

compression and biaxial bending shall

satisfy the following interaction

relationship

\ \ ] ]

G G\ G]

. 03 . 0� � ���3 0 0

Ky, Kz = Moment amplification factor

about minor and major axis

respectively

P = Factored applied axial

compression

My, Mz =Maximum factored applied

bending moments about y and

z axis of the member,

b)Bending and axial compression

(Clause 5.5.3.2)

b.1)Members with Class 1 and Class 2

cross-sections subject to combined bending

and axial compression shall satisfy:

\ \�6G6G ] ]�6G

\ \ \PLQ SO�\ SO�]

0� 0� 0�

N 01 N 0� � �I I I$ : :in which

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Swapnil B. Kharmale 215 Comparative study of IS: 800 (Draft) & EC3 CD-051061

axis and minor axis

About major axis

G] G0 0Md =Design flexural strength about z axis

given by Section 8.2.1of IS:800

(Draft) when lateral torsional

buckling is prevented and by Section

8.2.2, of IS:800 (Draft) where lateral

torsional buckling governs

About minor axis

G\ G0 0Md = Design flexural strength about y

axis calculated using plastic section

modulus for plastic and compact

sections and elastic section

modulus for semi-compact sections

For all above case EMy, EMz, EMLT= Equivalent uniform

moment factor obtained

from Table B.7.3,

Factor: Moment about axis: EMy y-y EMz z-z EMLT y-y

Comment :-

x Eurocode 3 give separate interaction formulae for each class of section and for

different mode of failure (i.e. flexural buckling and lateral torsional buckling ) On the

other hand the IS :800 (Draft) mention one interaction formula and it covers Plastic

Compact and Semi-compact class and both failure mode (i.e. flexural buckling and

lateral torsional buckling )

x For Class 4 (Slender) section, Eurocode 3 considers the Lateral Torsional Buckling

as a mode of failure and gives the interaction equation (b.3) is something strange

because Class 4 section shall be avoided as far as possible in design of member

subjected to axial compression and bending. As such design of compression

member and flexural member is based on section classification

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Swapnil B. Kharmale 217 Comparative study of IS: 800 (Draft) & EC3 CD-051061

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3.Section Classification Table 3.1 IS:800Stress ratio =¥(250/fy)=1 (Draft)For bending compression

Therefore whole c/s is compact from bending point of view For axial compression

Therefore whole c/s is non slender from axial compression point of view and Aeff=Ag

4.Check for resistance of cross section to combined effects for Clause 9.3.1 yielding.For Plastic and Compact Section

Clause 8.2.1

Clause 8.2.2

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Swapnil B.KharmaleCD-051061 218 Comparative study of IS:800 (Draft) and EC3

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DESIGN OF MEMBER SUBJECTED TO COMBINED FORCE (BEAM-COLUMN)

A non – sway intermediate column in a building frame with flexible joints is 4.0 m high The column is subjected to following load:Factored axial load = 500 kN and My Mz

Bottom 0 kNm 0kNmTop 15kNm 0.75 kNm

Take Fe 250 grade of steelAssume effective length of the column as 3.4 m along both the axes.

1.DataNSd= Factored applied axial force =500 kN

My.Sd=Factored applied moment @ major axis of c/s=15 kNm at top

Mz.Sd=Factored applied moment @ minor axis of c/s=0.75 kNm at top

Length of column L= 4m Effective length along both axis (KL) z = (KL)y=3.74m

2.Trial Section As predominant force is axial force hence let us chosse the section on the basis of axial force (say compressive nature) Assume fcd = 80MPa

Therefore

Let us try ISWB 300 @ 0.471 kN/m Sectional propertiesA=6133mm2

Wel.y=654.8x103mm3

b=200mm Wel.z=99.0x103mm3

h=300mm Wpl.y=725x103mm3

h2=24.95mm Wpl.z=103.8x103mm3

tf=10mm d= h-2h2=250.1mm

tw=7.4mm

Iyy=9821.6x104 mm4

Izz=990.1x104mm4

iyy=126.6mm

izz=40.2mm

DESIGN STEPS REFERENCES

BY EURO CODE 3

�G �

UHTFG

1 ���[��$ ����PP

I ��

Swapnil B.KharmaleCD-051061 220 Comparative study of IS:800 (Draft) and EC3

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Swapnil B.KharmaleCD-051061 222 Comparative study of IS:800 (Draft) and EC3

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Swapnil B.Kharmale 223 Comparative study of IS: 800 (Draft) &EC3 CD-051061

Section B: - Study of Both Code B.8 Connections

B.8.1 Introduction

Steel frame buildings consist of a number of different types of structural

elements, each of which has to be properly attached to the neighbouring parts of the

structure. This will involve the use of several forms of connection. The main classes

of connection are:

i) Where a change of direction occurs, e.g. beam-to-column connections, beam-to-

beam connections and connections between different members in trusses.

ii) To ensure manageable sizes of steelwork for transportation and erection e.g.

columns are normally spliced every two or three storeys.

iii) Where a change of component occurs, including connection of the steelwork to

other parts of the building, e.g. column bases, connections to concrete cores and

connections with walls, floors and roofs.

Connections are important parts of every steel structure. The mechanical properties

of the connections are of great influence on the strength, stiffness and stability of the

whole structure.

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Swapnil B.Kharmale 225 Comparative study of IS: 800 (Draft) &EC3 CD-051061

c) Semi-rigid connection

A connection which does not meet the criteria for a rigid connection or a

simple pinned connection as discussed above shall be classified as a semi-rigid

connection. In actual practice simple connections do have some degree of rotational

rigidity the developments in the semi-rigid connections. Similarly rigid connections

do experience some degree of joint deformation and this can be utilized to reduce

the joint design moments.

x Classification by components used for connection

a) Riveted.

b) Bolted.

c) Welded

Based on above classification the following combination can be achieved

i. Riveted or bolted shear connections.

ii. Riveted or bolted Moment connections.

iii. Welded shear connections.

iv. Welded moment connections

This chapter specifically deals with design of bolted and welded shear or simple

connections as per both codes

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Swapnil B.Kharmale 227 Comparative study of IS: 800 (Draft) &EC3 CD-051061

c) Maximum edge and end distance of fasteners (Clause 10.2.4)

� For member exposed to weather

e1or e2<40 mm+4tmin

� In general

e1or e2<12tminor 150 mm

whichever is more

� For unstiffened part

e1or e2<12t ZKHUH� ¥�����Iy)

d) Spacing (Pitch p1)

(Clause 10.2.1&2)

Minimum spacing

p1t2.5d

Maximum spacing for compression member

p1����W�RU����PP�ZKLFKHYHU�LV�OHVV

Maximum spacing for tension member

p1����W�RU����PP�ZKLFKHYHU�LV�OHVV

Tacking fasteners

a)In general

p1���Wminor 300 mm whichever is less

Not exposed weather

p1����W�RU����PP�ZKLFKHYHU�LV�OHVV

Exposed weather

b)In compression

p1���U’min of individual component or

c) Maximum edge and end distance of fasteners (Clause 6.5.1.4)

� For member exposed to weather

e1or e2<40 mm+4tmin

� In general

e1or e2<12tminor 150 mm

whichever is more

� For outstand element

e1 and e2 should not exceed the

maximum to satisfy local buckling

requirements for an outstand element.

d) Spacing (Pitch p1)

(Clause6.5.1.5&6&7)

Minimum spacing

p1t2.2do

Maximum spacing for compression member

p1��4 t or200 mm whichever is less

Maximum spacing for tension member

p1i�28 t or400 mm whichever is less

p1o����W�RU����PP�ZKLFKHYHU�LV�OHVV

where

p1i,p1o=Pitch in inner and outer row

respectively

Tacking fasteners

a)In general

p1���Wminor 300 mm whichever is less

Not exposed weather

p1����W�RU����PP�ZKLFKHYHU�LV�OHVV

Exposed weather

b)In compression

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Swapnil B.Kharmale 229 Comparative study of IS: 800 (Draft) &EC3 CD-051061

containing more than two bolts (i.e. the

distance between the first and last rows

of bolts in the joint, measured in the

direction of the load transfer) exceeds

15d in the direction of load, the nominal

shear capacity Vns, shall be reduced by

the factor, Elj, given by

Elj = 1.075 – lj / (200 d)

but 0.75 < Elj < 1.0� Large Grip Lengths

When the grip length, lg (equal to

the total thickness of the connected

plates) exceeds 5d, of the bolts, the

design shear capacity shall be reduced

by a factor Elg, given by

Elg = 8 d /(3d+lg)

but Elg <Elj and Ig<d

� Packing Plates

The design shear capacity of bolts

carrying calculate shear through a

packing plate in excess of 6 mm shall be

decreased by a factor, Epk given by

Epk = (1 - 0.0125 tpk)where tpk=Thickness of packing plate

nominal diameter of the bolts or rivets,

the design shear resistance Fv.Rd of all

the fasteners calculated shall be reduced

by multiplying it by a reduction factor Lf

given by:

M

/I

/ ���G �� ���Gbut 0.75 < Elj < 1.0

� Fasteners through packing Where bolts or rivets transmitting

load in shear and bearing pass through

packing of total thickness tp greater than

one-third of the nominal diameter d, the

design shear resistance Fv.Rd calculated

shall be reduced by multiplying it by a

reduction factor� p, given by:

S SS

�G EXW� ��G��W

Distribution of forces for long connections (joint)

When several bolts are placed in a row, as is indicated in Figure B.8.6, then

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Swapnil B.Kharmale 231 Comparative study of IS: 800 (Draft) &EC3 CD-051061

of plate between a pair of bolts therefore has the same length, the same strain and

consequently also the same stress. In the example of Figure 2, this means that the

forces in the plates between bolt 1 and bolt 2 are: 0,5 F, 1,0 F and 0,5 F. But this

also applies to the plates between bolts 2 and 3 and between bolts 3 and 4.

Conclusion: the bolts 1 and 4 transmit the full load F. The other bolts are not

loaded, see Figure B.8.7.a.

b. Assume the plates are infinitely stiff and the bolts are weak

The plates between the bolts do not deform. In other words, every bolt has the same

deformation and therefore is loaded to the same extent. Every bolt carries 0,5 F, i.e.

0,25 F per shear area.

The real distribution of forces is between these two extremes, as is indicated by the

solid line ("elastic") in Figure B.8.7c.

The difference between the forces in the outer bolts and the inner bolts is greater

when the stiffness of the plates is low.

This situation occurs when the connection is longer (more bolts) and the plate

thickness compared to the bolt diameter is small Therefore both code specify the

reduction factor lj � Lf) on design shear capacity of bolt to account the uneven

distribution of forces for long joint

Bearing capacity of bolt

A bolt bearing on any plate subjected

to a factored shear force (Vsb) shall

satisfy

Vsb < Vnpb��� mb

Vnpb = Bearing strength of a bolt, calculated as follows

Vnpb = 2.5 d t fufu= Smaller of the ultimate tensile stress

Bearing resistance of boltX

E�5G0E

XE� �

R R X

��� I GW) ZKHUH� LV�VPDOOHU�RI�

IH S �� � � ���G �G � I

fub=Ultimate tensile stress of bolt

fu= Ultimate tensile stress of plate

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Swapnil B.Kharmale 233 Comparative study of IS: 800 (Draft) &EC3 CD-051061

offering frictional resistance to slip

Kh = Factor depending upon clearance of

holes

=1for holes with standard clearance

=0.85 for oversize hole

=0.7 for long slotted holes

Fo =Minimum bolt tension (Proof load)

= 0.7 fub An

=0.7 for long slotted holes

Fo =Design pre loading force

= 0.7 fub As

Table B.8.2:- Slip factor µf

(From Table 10.2 of IS:800 (Draft) &Clause 6.5.8.3 of Eurocode 3)

Bolt subjected to combined shear and

tension

Following interaction equation shall be

satisfied

� �

6G G

9 7> @ � > @ �9 7

Bolt subjected to combined shear and

tension

Following interaction equation shall be

satisfied

Y�6G W�6G

Y�5G W�5G

) )� �) ���)

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Swapnil B.Kharmale 235 Comparative study of IS: 800 (Draft) &EC3 CD-051061

B.8.4 Design strength of welds by both codes.

IS:800 (Draft) Eurocode 3

x Fillet weld

Size of fillet weld (S) (Clause 10.5.2)

St3mm

S�tmin-1.5

S �(3/4) tmin

Throat thickness (a) (Clause 10.5.3)

at3mm

a�0.7 t to 1t

Effective length of weld

(Clause10.5.4)

Effective lengtht40 mm or 4 S

whichever is greater

Strength of weld per unit length

(Clause 10.5.7)

Design shear strength of weld fwd

fwd=fwn/ mw in which fwn=fu/¥3

where

fu = Smaller of the ultimate stress of the

weld and the parent metal

mw= Partial safety factor

Therefore,

strength of weld =Throat area x fwd

= 0.707 Sx1x fwd

X

PZ

I D � x1

x Fillet weld

Size of fillet weld (S) (Clause 6.6.5.1)

St3mm

S�tmin-2

Throat thickness (a) (Clause6.6.6.5.2)

at3mm

Effective length of weld

(Clause6.6.5.1 (2))

Effective lengtht40 mm or 4 a

whichever is greater

Resistance of weld per unit length

(Clause 6.6.5.3)

The design resistance per unit length

Fw.Rd shall be determined from

Fvw.D=fvw.dxa

where

fvw.d=The shear strength of weld

fvw.d X

: PZ

I �fu=Nominal ultimate tensile stress of

weaker part of joint

� W=Appropriate correlation factor

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Swapnil B.Kharmale 237 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

Section C: - Project Problem

C.1Problem Data and Analysis

This project problem consists of analysis of a railway foot over bridge by

STAAD-PRO and then design by both codes. The design example covers design of

all basic structural members (i.e. axial, flexural and combined).

C1.1 Design data of railway foot over bridge.

Clear span = 34m

Walk way width = 2.44m

Clear height above top of rail = 6.5m

Type of truss configuration = N type

Depth of truss girder = 2.3m

Lateral restrain for compression chord = X bracing

C.1.2 Assumption: - (For design by both codes)

1) Dead load and live loads IS 875 part I & II

2) Earthquake loading IS 1893-2002

3) Wind loading IS 875 part III

4) Grade of steel (fy) 250 MPa

5) Cross sections used Indian standard hot rolled sections

6) While designing care is taken that strength of system is governed by member and

not by the connections. Therefore connection details are not covered.

7) Comparison is based on interaction ratio (IR =Action/Strength) of cross sections.

C.1.3 Loading calculation as per IS: 875 (Part I to V)

1) Dead Load:-

a) RCC slab 110 mm thick.

Contributory width of floor beams = 2.266 m

Density of concrete = 25 kN/m3

UDL on beams (602 to 617) = 0.11 x 25 x 2.266 = 6.3 kN/m

UDL on beams (601 & 618) = 3.2 MTon/m Say

b) Vertical Cladding @ 15 kg/m2

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Swapnil B.Kharmale 239 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

3) Wind Load (IS 875-III):-

Design wind pressure (N/m2) is given by: �

]S ����9] Where Vz is design wind speed in (m/s) ] E � � �9 9 N N NHere,

Vb = Basic wind speed in m/s for given region.

= 44 m/s (Mumbai)

k1 = probability factor (risk coefficient) (see 5.3.1)

= 1.0 (Table 1 All general buildings and structure)

k2 = terrain, height and structure size factor (see 5.3.2)

= 0.98 (Terrain category 2 & class B i.e. max dimension is greater than 20m)

k3 = topography factor (see 5.3.3).

= 1.0

]9 ��[�[����[� �����P� V�

]

���[������S ����07RQ�P����However we will design the FOB considering minimum Pz = 0.15 MTon/m2

Considering Side cladding of height 2.3 mts

Uniform wind load applied on structure (top and bottom chord)

= 2.3 /2 x 0.15

=0.17 MTon/m

Therefore,

Joint loads to be applied = 0.17 x 2.27 = 0.38 MTon.

(Joints 1 to 14, 110,111, 34 to 46, 104, 114 and 127)

At end joints = 0.38 x 0.5 = 0.19 MTon

(Joints 64, 105, 116 and 128)

UDL applied on column = 0.15 x 2.27 / 2 = 0.17 MTon/m

(Members 401, 402, 404, 405, 407, 408, 410 and 411)

Note: - Applied in Z direction only, not considered in X direction as projection area in

X is less

4) Earth Quake Loading (IS 1893-2002)

Z = 0.16 (Zone factor for Mumbai)

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Swapnil B.Kharmale 241 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

50 2.266 2.3 2.44; 51 4.532 2.3 2.44; 52 6.798 2.3 2.44; 53 9.064 2.3 2.44; 54 11.33 2.3 2.44; 55 13.596 2.3 2.44; 56 15.862 2.3 2.44; 57 18.128 2.3 2.44; 58 20.394 2.3 2.44; 59 22.66 2.3 2.44; 60 24.926 2.3 2.44; 61 27.192 2.3 2.44; 62 29.458 2.3 2.44; 64 -2.266 0 0; 65 -2.266 0 2.44; 66 0 -6.5 0; 67 0 -6.5 2.44; 68 -2.266 -6.5 0; 69 -2.266 -6.5 2.44; 82 -2.266 -3.25 2.44; 83 0 -3.25 2.44; 84 0 -3.25 0; 85 -2.266 -3.25 0; 103 0 2.3 2.44; 104 0 2.3 0; 105 -2.266 2.3 0; 106 -2.266 2.3 2.44; 110 33.99 0 0; 111 31.724 0 0; 112 33.99 0 2.44; 113 31.724 0 2.44; 114 31.724 2.3 0; 115 31.724 2.3 2.44; 116 36.256 0 0; 117 36.256 0 2.44; 118 33.99 -6.5 0; 119 33.99 -6.5 2.44; 120 36.256 -6.5 0; 121 36.256 -6.5 2.44; 122 36.256 -3.25 2.44; 123 33.99 -3.25 2.44; 124 33.99 -3.25 0; 125 36.256 -3.25 0; 126 33.99 2.3 2.44; 127 33.99 2.3 0; 128 36.256 2.3 0; 129 36.256 2.3 2.44; MEMBER INCIDENCES 1 104 105; 2 34 104; 3 34 35; 4 35 36; 5 36 37; 6 37 38; 7 38 39; 8 39 40; 9 40 41; 10 41 42; 11 42 43; 12 43 44; 13 44 45; 14 45 46; 15 114 46; 16 114 127; 17 127 128; 18 1 2; 19 2 3; 20 3 4; 21 4 5; 22 5 6; 23 6 7; 24 7 8; 25 8 9; 26 9 10; 27 10 11; 28 11 12; 29 12 13; 30 13 14; 31 111 14; 32 110 111; 33 2 34; 34 3 35; 35 4 36; 36 5 37; 37 6 38; 38 7 39; 39 8 40; 40 9 41; 41 10 42; 42 11 43; 43 12 44; 44 13 45; 45 14 46; 46 111 114; 47 104 2; 48 34 3; 49 35 4; 50 36 5; 51 37 6; 52 38 7; 53 39 8; 54 8 41; 55 40 9; 56 42 9; 57 43 10; 58 44 11; 59 45 12; 60 46 13; 61 114 14; 62 127 111; 63 103 106; 64 50 103; 65 50 51; 66 51 52; 67 52 53; 68 53 54; 69 54 55; 70 55 56; 71 56 57; 72 57 58; 73 58 59; 74 59 60; 75 60 61; 76 61 62; 77 115 62; 78 115 126; 79 126 129; 80 17 18; 81 18 19; 82 19 20; 83 20 21; 84 21 22; 85 22 23; 86 23 24; 87 24 25; 88 25 26; 89 26 27; 90 27 28; 91 28 29; 92 29 30; 93 112 113; 94 113 30; 95 117 112; 96 50 19; 97 24 57; 98 56 25; 99 51 20; 100 52 21; 101 53 22; 102 54 23; 103 55 24; 104 25 58; 105 26 59; 106 27 60; 107 28 61; 108 29 62; 109 103 18; 110 115 30; 111 126 113; 112 18 50; 113 19 51; 114 20 52; 115 21 53; 116 22 54; 117 23 55; 118 24 56; 119 25 57; 120 26 58; 121 27 59; 122 28 60; 123 29 61; 124 30 62; 125 113 115; 301 64 84; 302 85 66; 303 116 124; 304 125 118; 305 64 1; 306 84 85; 307 116 110; 308 124 125; 309 65 17; 310 82 83; 311 82 67; 312 65 83; 313 122 123; 314 122 119; 315 117 123; 316 83 84; 317 85 82; 318 82 68; 319 17 84; 320 65 85; 321 66 83; 322 64 106; 323 123 124; 324 125 122; 325 122 120; 326 112 124; 327 117 125; 328 118 123; 329 116 129; 401 68 85; 402 85 64; 403 64 105; 404 66 84; 405 84 1; 406 1 104; 407 118 124; 408 124 110; 409 110 127; 410 120 125; 411 125 116; 412 116 128; 413 69 82; 414 82 65; 415 65 106; 416 67 83; 417 83 17; 418 17 103; 419 119 123; 420 123 112; 421 112 126; 422 121 122; 423 122 117; 424 117 129; 501 105 106;

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Swapnil B.Kharmale 243 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

523 532 TABLE SD ISA60X60X6 SP 0.008 519 TO 522 533 TO 536 TABLE SD ISA65X65X8 SP 0.008 508 511 TABLE SD ISA60X60X6 SP 0.008 504 TO 507 512 TO 515 TABLE SD ISA75X75X8 SP 0.008 501 TO 503 516 TO 518 TABLE SD ISA100X100X8 SP 0.008 404 TO 409 416 TO 421 TABLE ST ISMB500 *78 92 122 TABLE TB ISMB300 WP 0.17 TH 0.025 SUPPORTS 66 TO 69 118 TO 121 FIXED *66 TO 69 118 TO 121 PINNED MEMBER RELEASE 601 TO 618 START MZ 601 TO 618 END MZ MEMBER TRUSS 2 TO 16 18 TO 62 64 TO 78 80 TO 94 96 TO 125 MEMBER TRUSS 306 308 310 313 MEMBER TRUSS 316 317 323 324 MEMBER TRUSS 18 TO 32 80 TO 95 305 307 309 619 TO 636 MEMBER TRUSS 1 TO 17 63 TO 79 501 TO 536 MEMBER TENSION 301 TO 304 311 312 314 315 639 TO 642 646 647 650 651 MEMBER TENSION 318 TO 322 325 TO 329 637 638 643 TO 645 648 649 652 TO 654 LOAD 1 DEAD LOAD (DL) SELFWEIGHT Y -1.05 *5 % EXTRA WEIGHT ADDED TO TAKE CARE OF CONNECTIONS MEMBER LOAD 602 TO 617 UNI GY -0.63 601 618 UNI GY -0.32 *THICKNESS OF SLAB 125MM, WEIGHT = 2.5*0.11 = 0.275 T/SQM *CW = 2.266 M, HENCE UDL = 0.275*2.266 = 0.63 T/M SAY *CLADDING AT 15 KG/SQM W = 0.015*2.3 = 0.0345 18 TO 32 80 TO 94 305 307 601 618 UNI GY -0.0345 *STAIRCASE LOAD *Total rise to climb = 6.5 + 0.3+0.11 = 6.9 m *Rise say = 200 mm *No of Risers = 6.9/0.2 = 34.5 say 35 *Hence No of treads = 35-1 = 34 *Say Tread of 300 mm, hence Total going = 0.3x34 = 10.2 m *Let there be central landing of 1.2m *Hence Total Plan length of staircase = 10.2+1.2=11.4m *Provide central column, *Hence contributory span on Bridge structure = 11.4/4 =2.85 *Staircase width = 2.3 m *Hence plan area for one column = 2.85x2.3/2 = 3.2 sqm *DL @ 0.2 T/sqm = 3.2 x 0.2 = 0.64 MTon *LL @ 0.5 T/Sqm = 3.2 x 0.5 = 1.6 MTon JOINT LOAD 17 65 112 117 FY -0.64 LOAD 2 LIVE LOAD (LL) MEMBER LOAD 602 TO 617 UNI GY -1.14

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Swapnil B.Kharmale 245 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

Fig C.1.1:- Geometry of FOB

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Swapnil B.Kharmale 247 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

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Swapnil B.Kharmale 249 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

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Swapnil B.Kharmale 251 Comparative study of IS: 800 (Draft) & EC3 M.Tech (Str) CD-051061

Fig C.1.2:- Modeling of FOB in STAAD PRO 2006

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DL LL WLz Eqx DL LL WLz Eqx Fx Mz Fx Mz Fx Mz Fx Mz Fx Mz Fx My Fx My Fx My Fx My Fx MyFx

Comp(+)Fx

Tens(-).case

# MzFx

Comp(+)Fx

Tens(-).case

# My

1.35(DL+LL+EQ) # 10

1.2(DL+LL+EQ) #5 Design Critical forces

1.35 DL+1.5LL #6

1.35 DL+1.5WL #7

1.35(DL+LL+WL) # 8

1.35 DL+1.5EQ #9

1.5(DL+EQ) # 4

1.5(DL+LL) # 1

1.5(DL+WL) # 2SR

NOMEMB #

Axial Force Fx (Mton) (+, Comp & -, Ten) Design Critical forces

1.2(DL+LL+WL) # 3

BM-Mz (Mton-m)

56 56 -1.9 -2 0 0 -5.9 0.0 -2.9 0.0 -4.7 0.0 -2.9 0.0 -4.7 0.0 -5.6 0.0 -2.6 0.0 -5.3 0.0 -2.6 0.0 -5.3 0.0 - -5.9 1 - - -5.6 6 -57 57 -3.7 -3.9 0 0 -11.4 0.0 -5.6 0.0 -9.1 0.0 -5.6 0.0 -9.1 0.0 -10.8 0.0 -5.0 0.0 -10.3 0.0 -5.0 0.0 -10.3 0.0 - -11.4 1 - - -10.8 6 -58 58 -5.6 -5.9 0 0 -17.3 0.0 -8.4 0.0 -13.8 0.0 -8.4 0.0 -13.8 0.0 -16.4 0.0 -7.6 0.0 -15.5 0.0 -7.6 0.0 -15.5 0.0 - -17.3 1 - - -16.4 6 -59 59 -7.5 -7.8 0 0 -23.0 0.0 -11.3 0.0 -18.4 0.0 -11.3 0.0 -18.4 0.0 -21.8 0.0 -10.1 0.0 -20.7 0.0 -10.1 0.0 -20.7 0.0 - -23.0 1 - - -21.8 6 -60 60 -9.4 -9.8 0 0 -28.8 0.0 -14.1 0.0 -23.0 0.0 -14.1 0.0 -23.0 0.0 -27.4 0.0 -12.7 0.0 -25.9 0.0 -12.7 0.0 -25.9 0.0 - -28.8 1 - - -27.4 6 -61 61 -11.3 -11.7 0 0 -34.5 0.0 -17.0 0.0 -27.6 0.0 -17.0 0.0 -27.6 0.0 -32.8 0.0 -15.3 0.0 -31.1 0.0 -15.3 0.0 -31.1 0.0 - -34.5 1 - - -32.8 6 -62 62 -13.3 -13.7 0 0 -40.5 0.0 -20.0 0.0 -32.4 0.0 -20.0 0.0 -32.4 0.0 -38.5 0.0 -18.0 0.0 -36.5 0.0 -18.0 0.0 -36.5 0.0 - -40.5 1 - - -38.5 6 -63 33 8.4 8.4 0 0 25.2 0.0 12.6 0.0 20.2 0.0 12.6 0.0 20.2 0.0 23.9 0.0 11.3 0.0 22.7 0.0 11.3 0.0 22.7 0.0 25.2 - 1 - 23.9 - 6 -64 34 7 7 0 0 21.0 0.0 10.5 0.0 16.8 0.0 10.5 0.0 16.8 0.0 20.0 0.0 9.5 0.0 18.9 0.0 9.5 0.0 18.9 0.0 21.0 - 1 - 20.0 - 6 -65 35 5.6 5.6 0 0 16.8 0.0 8.4 0.0 13.4 0.0 8.4 0.0 13.4 0.0 16.0 0.0 7.6 0.0 15.1 0.0 7.6 0.0 15.1 0.0 16.8 - 1 - 16.0 - 6 -66 36 4.3 4.3 0 0 12.9 0.0 6.5 0.0 10.3 0.0 6.5 0.0 10.3 0.0 12.3 0.0 5.8 0.0 11.6 0.0 5.8 0.0 11.6 0.0 12.9 - 1 - 12.3 - 6 -67 37 2.9 2.9 0 0 8.7 0.0 4.4 0.0 7.0 0.0 4.4 0.0 7.0 0.0 8.3 0.0 3.9 0.0 7.8 0.0 3.9 0.0 7.8 0.0 8.7 - 1 - 8.3 - 6 -68 38 1.6 1.6 0 0 4.8 0.0 2.4 0.0 3.8 0.0 2.4 0.0 3.8 0.0 4.6 0.0 2.2 0.0 4.3 0.0 2.2 0.0 4.3 0.0 4.8 - 1 - 4.6 - 6 -69 39 0.1 0.1 0 0 0.3 0.0 0.2 0.0 0.2 0.0 0.2 0.0 0.2 0.0 0.3 0.0 0.1 0.0 0.3 0.0 0.1 0.0 0.3 0.0 0.3 - 1 - 0.3 - 6 -70 40 0.1 0.1 0 0 0.3 0.0 0.2 0.0 0.2 0.0 0.2 0.0 0.2 0.0 0.3 0.0 0.1 0.0 0.3 0.0 0.1 0.0 0.3 0.0 0.3 - 1 - 0.3 - 6 -71 41 1.6 1.6 0 0 4.8 0.0 2.4 0.0 3.8 0.0 2.4 0.0 3.8 0.0 4.6 0.0 2.2 0.0 4.3 0.0 2.2 0.0 4.3 0.0 4.8 - 1 - 4.6 - 6 -72 42 2.9 2.9 0 0 8.7 0.0 4.4 0.0 7.0 0.0 4.4 0.0 7.0 0.0 8.3 0.0 3.9 0.0 7.8 0.0 3.9 0.0 7.8 0.0 8.7 - 1 - 8.3 - 6 -73 43 4.3 4.3 0 0 12.9 0.0 6.5 0.0 10.3 0.0 6.5 0.0 10.3 0.0 12.3 0.0 5.8 0.0 11.6 0.0 5.8 0.0 11.6 0.0 12.9 - 1 - 12.3 - 6 -74 44 5.6 5.6 0 0 16.8 0.0 8.4 0.0 13.4 0.0 8.4 0.0 13.4 0.0 16.0 0.0 7.6 0.0 15.1 0.0 7.6 0.0 15.1 0.0 16.8 - 1 - 16.0 - 6 -75 45 7 7 0 0 21.0 0.0 10.5 0.0 16.8 0.0 10.5 0.0 16.8 0.0 20.0 0.0 9.5 0.0 18.9 0.0 9.5 0.0 18.9 0.0 21.0 - 1 - 20.0 - 6 -76 46 8.4 8.4 0 0 25.2 0.0 12.6 0.0 20.2 0.0 12.6 0.0 20.2 0.0 23.9 0.0 11.3 0.0 22.7 0.0 11.3 0.0 22.7 0.0 25.2 - 1 - 23.9 - 6 -77 301 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - - - 5 -78 302 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - - - 5 -79 303 0 0 0 -2.8 0.0 0.0 0.0 0.0 0.0 0.0 -4.2 0.0 -3.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.2 0.0 -3.8 0.0 - -4.2 4 - - -4.2 9 -80 304 0 0 0 -3.8 0.0 0.0 0.0 0.0 0.0 0.0 -5.7 0.0 -4.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -5.7 0.0 -5.1 0.0 - -5.7 4 - - -5.7 9 -81 306 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - - - 5 -82 308 0 0 0 2 0.0 0.0 0.0 0.0 0.0 0.0 3.0 0.0 2.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.0 0.0 2.7 0.0 3.0 - 4 - 3.0 - 9 -83 63 0 0 -2.8 0 0.0 0.0 -4.2 0.0 -3.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.2 0.0 -3.8 0.0 0.0 0.0 0.0 0.0 - -4.2 2 - - -4.2 7 -84 64 6.3 6.5 -5.2 0 19.2 0.0 1.7 0.0 9.1 0.0 9.5 0.0 15.4 0.0 18.3 0.0 0.7 0.0 10.3 0.0 8.5 0.0 17.3 0.0 19.2 - 1 - 18.3 - 6 -85 65 14.2 14.7 -7.4 0 43.4 0.0 10.2 0.0 25.8 0.0 21.3 0.0 34.7 0.0 41.2 0.0 8.1 0.0 29.0 0.0 19.2 0.0 39.0 0.0 43.4 - 1 - 41.2 - 6 -86 66 20.8 21.5 -9.1 0 63.5 0.0 17.6 0.0 39.8 0.0 31.2 0.0 50.8 0.0 60.3 0.0 14.4 0.0 44.8 0.0 28.1 0.0 57.1 0.0 63.5 - 1 - 60.3 - 6 -87 67 26 27.1 -10.5 0 79.7 0.0 23.3 0.0 51.1 0.0 39.0 0.0 63.7 0.0 75.8 0.0 19.4 0.0 57.5 0.0 35.1 0.0 71.7 0.0 79.7 - 1 - 75.8 - 6 -88 68 29.9 31.2 -11.6 0 91.7 0.0 27.5 0.0 59.4 0.0 44.9 0.0 73.3 0.0 87.2 0.0 23.0 0.0 66.8 0.0 40.4 0.0 82.5 0.0 91.7 - 1 - 87.2 - 6 -89 69 32.5 33.9 -12.3 0 99.6 0.0 30.3 0.0 64.9 0.0 48.8 0.0 79.7 0.0 94.7 0.0 25.4 0.0 73.0 0.0 43.9 0.0 89.6 0.0 99.6 - 1 - 94.7 - 6 -90 70 33.8 35.3 -12.7 0 103.7 0.0 31.7 0.0 67.7 0.0 50.7 0.0 82.9 0.0 98.6 0.0 26.6 0.0 76.1 0.0 45.6 0.0 93.3 0.0 103.7 - 1 - 98.6 - 6 -91 71 32.8 34.2 -12.3 0 100.5 0.0 30.8 0.0 65.6 0.0 49.2 0.0 80.4 0.0 95.6 0.0 25.8 0.0 73.8 0.0 44.3 0.0 90.5 0.0 100.5 - 1 - 95.6 - 6 -92 72 33.8 35.3 -12.7 0 103.7 0.0 31.7 0.0 67.7 0.0 50.7 0.0 82.9 0.0 98.6 0.0 26.6 0.0 76.1 0.0 45.6 0.0 93.3 0.0 103.7 - 1 - 98.6 - 6 -93 73 32.5 33.9 -12.3 0 99.6 0.0 30.3 0.0 64.9 0.0 48.8 0.0 79.7 0.0 94.7 0.0 25.4 0.0 73.0 0.0 43.9 0.0 89.6 0.0 99.6 - 1 - 94.7 - 6 -94 74 29.9 31.2 -11.6 0 91.7 0.0 27.5 0.0 59.4 0.0 44.9 0.0 73.3 0.0 87.2 0.0 23.0 0.0 66.8 0.0 40.4 0.0 82.5 0.0 91.7 - 1 - 87.2 - 6 -95 75 26 27.1 -10.5 0 79.7 0.0 23.3 0.0 51.1 0.0 39.0 0.0 63.7 0.0 75.8 0.0 19.4 0.0 57.5 0.0 35.1 0.0 71.7 0.0 79.7 - 1 - 75.8 - 6 -96 76 20.8 21.5 -9.1 0 63.5 0.0 17.6 0.0 39.8 0.0 31.2 0.0 50.8 0.0 60.3 0.0 14.4 0.0 44.8 0.0 28.1 0.0 57.1 0.0 63.5 - 1 - 60.3 - 6 -97 77 14.2 14.7 -7.4 0 43.4 0.0 10.2 0.0 25.8 0.0 21.3 0.0 34.7 0.0 41.2 0.0 8.1 0.0 29.0 0.0 19.2 0.0 39.0 0.0 43.4 - 1 - 41.2 - 6 -98 78 6.3 6.5 -5.2 0 19.2 0.0 1.7 0.0 9.1 0.0 9.5 0.0 15.4 0.0 18.3 0.0 0.7 0.0 10.3 0.0 8.5 0.0 17.3 0.0 19.2 - 1 - 18.3 - 6 -99 79 0 0 -2.8 0 0.0 0.0 -4.2 0.0 -3.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.2 0.0 -3.8 0.0 0.0 0.0 0.0 0.0 - -4.2 1 - - -4.2 6 -

100 309 0.66 0.68 0.8 0 2.0 0.0 2.2 0.0 2.6 0.0 1.0 0.0 1.6 0.0 1.9 0.0 2.1 0.0 2.9 0.0 0.9 0.0 1.8 0.0 2.6 - 3 - 2.9 - 8 -101 80 4.1 4.2 -1.8 0 12.5 0.0 3.5 0.0 7.8 0.0 6.2 0.0 10.0 0.0 11.8 0.0 2.8 0.0 8.8 0.0 5.5 0.0 11.2 0.0 12.5 - 1 - 11.8 - 6 -102 81 -5.23 -4.2 -3.9 0 -14.1 0.0 -13.7 0.0 -16.0 0.0 -7.8 0.0 -11.3 0.0 -13.4 0.0 -12.9 0.0 -18.0 0.0 -7.1 0.0 -12.7 0.0 - -16.0 3 - - -18.0 8 -103 82 -13.2 -13.6 -5.6 0 -40.1 0.0 -28.1 0.0 -38.8 0.0 -19.7 0.0 -32.1 0.0 -38.2 0.0 -26.2 0.0 -43.7 0.0 -17.8 0.0 -36.1 0.0 - -40.1 1 - - -43.7 6 -104 83 -19.7 -20.5 -7.1 0 -60.3 0.0 -40.2 0.0 -56.8 0.0 -29.6 0.0 -48.2 0.0 -57.3 0.0 -37.2 0.0 -63.9 0.0 -26.6 0.0 -54.3 0.0 - -60.3 1 - - -63.9 6 -105 84 -25 -25.9 -8.1 0 -76.4 0.0 -49.7 0.0 -70.8 0.0 -37.5 0.0 -61.1 0.0 -72.6 0.0 -45.9 0.0 -79.7 0.0 -33.8 0.0 -68.7 0.0 - -76.4 1 - - -79.7 6 -106 85 -28.8 -30.1 -8.8 0 -88.4 0.0 -56.4 0.0 -81.2 0.0 -43.2 0.0 -70.7 0.0 -84.0 0.0 -52.1 0.0 -91.4 0.0 -38.9 0.0 -79.5 0.0 - -88.4 1 - - -91.4 6 -107 86 -31.5 -32.8 -9.2 0 -96.5 0.0 -61.1 0.0 -88.2 0.0 -47.3 0.0 -77.2 0.0 -91.7 0.0 -56.3 0.0 -99.2 0.0 -42.5 0.0 -86.8 0.0 - -96.5 1 - - -99.2 6 -108 87 -31.7 -33.1 -8.8 0 -97.2 0.0 -60.8 0.0 -88.3 0.0 -47.6 0.0 -77.8 0.0 -92.4 0.0 -56.0 0.0 -99.4 0.0 -42.8 0.0 -87.5 0.0 - -97.2 1 - - -99.4 6 -109 88 -31.5 -32.8 -9.2 0 -96.5 0.0 -61.1 0.0 -88.2 0.0 -47.3 0.0 -77.2 0.0 -91.7 0.0 -56.3 0.0 -99.2 0.0 -42.5 0.0 -86.8 0.0 - -96.5 1 - - -99.2 6 -110 89 -28.8 -30.1 -8.8 0 -88.4 0.0 -56.4 0.0 -81.2 0.0 -43.2 0.0 -70.7 0.0 -84.0 0.0 -52.1 0.0 -91.4 0.0 -38.9 0.0 -79.5 0.0 - -88.4 1 - - -91.4 6 -111 90 -25 -25.9 -8.1 0 -76.4 0.0 -49.7 0.0 -70.8 0.0 -37.5 0.0 -61.1 0.0 -72.6 0.0 -45.9 0.0 -79.7 0.0 -33.8 0.0 -68.7 0.0 - -76.4 1 - - -79.7 6 -112 91 -19.7 -20.5 -7.1 0 -60.3 0.0 -40.2 0.0 -56.8 0.0 -29.6 0.0 -48.2 0.0 -57.3 0.0 -37.2 0.0 -63.9 0.0 -26.6 0.0 -54.3 0.0 - -60.3 1 - - -63.9 6 -

Swapnil B. KharmaleCD-051061 253 Comparative study of IS:800 (Draft) and EC3ESte

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DL LL WLz Eqx DL LL WLz Eqx Fx Mz Fx Mz Fx Mz Fx Mz Fx Mz Fx My Fx My Fx My Fx My Fx MyFx

Comp(+)Fx

Tens(-).case

# MzFx

Comp(+)Fx

Tens(-).case

# My

1.35(DL+LL+EQ) # 10

1.2(DL+LL+EQ) #5 Design Critical forces

1.35 DL+1.5LL #6

1.35 DL+1.5WL #7

1.35(DL+LL+WL) # 8

1.35 DL+1.5EQ #9

1.5(DL+EQ) # 4

1.5(DL+LL) # 1

1.5(DL+WL) # 2SR

NOMEMB #

Axial Force Fx (Mton) (+, Comp & -, Ten) Design Critical forces

1.2(DL+LL+WL) # 3

BM-Mz (Mton-m)

168 604 0 0 2.3 0 0.5 0.9 0 0 0.0 2.1 3.5 0.8 2.8 1.7 0.0 0.8 0.0 1.7 0.0 2.0 3.5 0.7 3.1 1.9 0.0 0.7 0.0 1.9 3.5 - 2 2.1 3.5 - 7 2.0169 605 0 0 1.9 0 0.5 0.9 0 0 0.0 2.1 2.9 0.8 2.3 1.7 0.0 0.8 0.0 1.7 0.0 2.0 2.9 0.7 2.6 1.9 0.0 0.7 0.0 1.9 2.9 - 2 2.1 2.9 - 7 2.0170 606 0 0 1.5 0 0.5 0.9 0 0 0.0 2.1 2.3 0.8 1.8 1.7 0.0 0.8 0.0 1.7 0.0 2.0 2.3 0.7 2.0 1.9 0.0 0.7 0.0 1.9 2.3 - 2 2.1 2.3 - 7 2.0171 607 0 0 1.1 0 0.5 0.9 0 0 0.0 2.1 1.7 0.8 1.3 1.7 0.0 0.8 0.0 1.7 0.0 2.0 1.7 0.7 1.5 1.9 0.0 0.7 0.0 1.9 1.7 - 2 2.1 1.7 - 7 2.0172 608 0 0 0.8 0 0.5 0.9 0 0 0.0 2.1 1.2 0.8 1.0 1.7 0.0 0.8 0.0 1.7 0.0 2.0 1.2 0.7 1.1 1.9 0.0 0.7 0.0 1.9 1.2 - 2 2.1 1.2 - 7 2.0173 609 0 0 0.3 0 0.5 0.9 0 0 0.0 2.1 0.5 0.8 0.4 1.7 0.0 0.8 0.0 1.7 0.0 2.0 0.5 0.7 0.4 1.9 0.0 0.7 0.0 1.9 0.5 - 2 2.1 0.5 - 7 2.0174 610 0 0 0.3 0 0.5 0.9 0 0 0.0 2.1 0.5 0.8 0.4 1.7 0.0 0.8 0.0 1.7 0.0 2.0 0.5 0.7 0.4 1.9 0.0 0.7 0.0 1.9 0.5 - 2 2.1 0.5 - 7 2.0175 611 0 0 0.8 0 0.5 0.9 0 0 0.0 2.1 1.2 0.8 1.0 1.7 0.0 0.8 0.0 1.7 0.0 2.0 1.2 0.7 1.1 1.9 0.0 0.7 0.0 1.9 1.2 - 2 2.1 1.2 - 7 2.0176 612 0 0 1.1 0 0.5 0.9 0 0 0.0 2.1 1.7 0.8 1.3 1.7 0.0 0.8 0.0 1.7 0.0 2.0 1.7 0.7 1.5 1.9 0.0 0.7 0.0 1.9 1.7 - 2 2.1 1.7 - 7 2.0177 613 0 0 1.5 0 0.5 0.9 0 0 0.0 2.1 2.3 0.8 1.8 1.7 0.0 0.8 0.0 1.7 0.0 2.0 2.3 0.7 2.0 1.9 0.0 0.7 0.0 1.9 2.3 - 2 2.1 2.3 - 7 2.0178 614 0 0 1.9 0 0.5 0.9 0 0 0.0 2.1 2.9 0.8 2.3 1.7 0.0 0.8 0.0 1.7 0.0 2.0 2.9 0.7 2.6 1.9 0.0 0.7 0.0 1.9 2.9 - 2 2.1 2.9 - 7 2.0179 615 0 0 2.3 0 0.5 0.9 0 0 0.0 2.1 3.5 0.8 2.8 1.7 0.0 0.8 0.0 1.7 0.0 2.0 3.5 0.7 3.1 1.9 0.0 0.7 0.0 1.9 3.5 - 2 2.1 3.5 - 7 2.0180 616 0 0 2.7 0 0.5 0.9 0 0 0.0 2.1 4.1 0.8 3.2 1.7 0.0 0.8 0.0 1.7 0.0 2.0 4.1 0.7 3.6 1.9 0.0 0.7 0.0 1.9 4.1 - 2 2.1 4.1 - 7 2.0181 617 0 0 3.3 0 0.5 0.9 0 0 0.0 2.1 5.0 0.8 4.0 1.7 0.0 0.8 0.0 1.7 0.0 2.0 5.0 0.7 4.5 1.9 0.0 0.7 0.0 1.9 5.0 - 2 2.1 5.0 - 7 2.0182 618 0 0 2.8 0 0.3 0.5 0 0 0.0 1.2 4.2 0.5 3.4 1.0 0.0 0.5 0.0 1.0 0.0 1.2 4.2 0.4 3.8 1.1 0.0 0.4 0.0 1.1 4.2 - 2 1.2 4.2 - 7 1.2183 633 0 0 1.3 0 0.0 0.0 2.0 0.0 1.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 0.0 1.8 0.0 0.0 0.0 0.0 0.0 2.0 - 2 - 2.0 - 7 -184 619 0 0 -3.7 0 0.0 0.0 -5.6 0.0 -4.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -5.6 0.0 -5.0 0.0 0.0 0.0 0.0 0.0 - -5.6 2 - - -5.6 7 -185 620 0 0 -3.2 0 0.0 0.0 -4.8 0.0 -3.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.8 0.0 -4.3 0.0 0.0 0.0 0.0 0.0 - -4.8 2 - - -4.8 7 -186 621 0 0 -2.6 0 0.0 0.0 -3.9 0.0 -3.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.9 0.0 -3.5 0.0 0.0 0.0 0.0 0.0 - -3.9 2 - - -3.9 7 -187 622 0 0 -2.1 0 0.0 0.0 -3.2 0.0 -2.5 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.2 0.0 -2.8 0.0 0.0 0.0 0.0 0.0 - -3.2 2 - - -3.2 7 -188 623 0 0 -1.6 0 0.0 0.0 -2.4 0.0 -1.9 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2.4 0.0 -2.2 0.0 0.0 0.0 0.0 0.0 - -2.4 2 - - -2.4 7 -189 624 0 0 -1.1 0 0.0 0.0 -1.7 0.0 -1.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -1.7 0.0 -1.5 0.0 0.0 0.0 0.0 0.0 - -1.7 2 - - -1.7 7 -190 625 0 0 -0.6 0 0.0 0.0 -0.9 0.0 -0.7 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -0.9 0.0 -0.8 0.0 0.0 0.0 0.0 0.0 - -0.9 2 - - -0.9 7 -191 626 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - - - 5 -192 627 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - - - 5 -193 628 0 0 -0.6 0 0.0 0.0 -0.9 0.0 -0.7 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -0.9 0.0 -0.8 0.0 0.0 0.0 0.0 0.0 - -0.9 2 - - -0.9 7 -194 629 0 0 -1.1 0 0.0 0.0 -1.7 0.0 -1.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -1.7 0.0 -1.5 0.0 0.0 0.0 0.0 0.0 - -1.7 2 - - -1.7 7 -195 630 0 0 -1.6 0 0.0 0.0 -2.4 0.0 -1.9 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2.4 0.0 -2.2 0.0 0.0 0.0 0.0 0.0 - -2.4 2 - - -2.4 7 -196 631 0 0 -2.1 0 0.0 0.0 -3.2 0.0 -2.5 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.2 0.0 -2.8 0.0 0.0 0.0 0.0 0.0 - -3.2 2 - - -3.2 7 -197 632 0 0 -2.6 0 0.0 0.0 -3.9 0.0 -3.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.9 0.0 -3.5 0.0 0.0 0.0 0.0 0.0 - -3.9 2 - - -3.9 7 -198 634 0 0 -3.2 0 0.0 0.0 -4.8 0.0 -3.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.8 0.0 -4.3 0.0 0.0 0.0 0.0 0.0 - -4.8 2 - - -4.8 7 -199 635 0 0 -3.7 0 0.0 0.0 -5.6 0.0 -4.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -5.6 0.0 -5.0 0.0 0.0 0.0 0.0 0.0 - -5.6 2 - - -5.6 7 -200 636 0 0 1.3 0 0.0 0.0 2.0 0.0 1.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.0 0.0 1.8 0.0 0.0 0.0 0.0 0.0 2.0 - 2 - 2.0 - 7 -201 501 0 0 3.2 0 0.0 0.0 4.8 0.0 3.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.8 0.0 4.3 0.0 0.0 0.0 0.0 0.0 4.8 - 2 - 4.8 - 7 -202 502 0 0 3 0 0.0 0.0 4.5 0.0 3.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.5 0.0 4.1 0.0 0.0 0.0 0.0 0.0 4.5 - 2 - 4.5 - 7 -203 503 0 0 2.7 0 0.0 0.0 4.1 0.0 3.2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.1 0.0 3.6 0.0 0.0 0.0 0.0 0.0 4.1 - 2 - 4.1 - 7 -204 504 0 0 2.3 0 0.0 0.0 3.5 0.0 2.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.5 0.0 3.1 0.0 0.0 0.0 0.0 0.0 3.5 - 2 - 3.5 - 7 -205 505 0 0 1.9 0 0.0 0.0 2.9 0.0 2.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.9 0.0 2.6 0.0 0.0 0.0 0.0 0.0 2.9 - 2 - 2.9 - 7 -206 506 0 0 1.5 0 0.0 0.0 2.3 0.0 1.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.3 0.0 2.0 0.0 0.0 0.0 0.0 0.0 2.3 - 2 - 2.3 - 7 -207 507 0 0 1.1 0 0.0 0.0 1.7 0.0 1.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.7 0.0 1.5 0.0 0.0 0.0 0.0 0.0 1.7 - 2 - 1.7 - 7 -208 508 0 0 0.8 0 0.0 0.0 1.2 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 0.0 1.1 0.0 0.0 0.0 0.0 0.0 1.2 - 2 - 1.2 - 7 -209 509 0 0 0.3 0 0.0 0.0 0.5 0.0 0.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.5 0.0 0.4 0.0 0.0 0.0 0.0 0.0 0.5 - 2 - 0.5 - 7 -210 510 0 0 0.3 0 0.0 0.0 0.5 0.0 0.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.5 0.0 0.4 0.0 0.0 0.0 0.0 0.0 0.5 - 2 - 0.5 - 7 -211 511 0 0 0.8 0 0.0 0.0 1.2 0.0 1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.2 0.0 1.1 0.0 0.0 0.0 0.0 0.0 1.2 - 2 - 1.2 - 7 -212 512 0 0 1.1 0 0.0 0.0 1.7 0.0 1.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 1.7 0.0 1.5 0.0 0.0 0.0 0.0 0.0 1.7 - 2 - 1.7 - 7 -213 513 0 0 1.5 0 0.0 0.0 2.3 0.0 1.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.3 0.0 2.0 0.0 0.0 0.0 0.0 0.0 2.3 - 2 - 2.3 - 7 -214 514 0 0 1.9 0 0.0 0.0 2.9 0.0 2.3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2.9 0.0 2.6 0.0 0.0 0.0 0.0 0.0 2.9 - 2 - 2.9 - 7 -215 515 0 0 2.3 0 0.0 0.0 3.5 0.0 2.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3.5 0.0 3.1 0.0 0.0 0.0 0.0 0.0 3.5 - 2 - 3.5 - 7 -216 516 0 0 2.7 0 0.0 0.0 4.1 0.0 3.2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.1 0.0 3.6 0.0 0.0 0.0 0.0 0.0 4.1 - 2 - 4.1 - 7 -217 517 0 0 3 0 0.0 0.0 4.5 0.0 3.6 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.5 0.0 4.1 0.0 0.0 0.0 0.0 0.0 4.5 - 2 - 4.5 - 7 -218 518 0 0 3.2 0 0.0 0.0 4.8 0.0 3.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4.8 0.0 4.3 0.0 0.0 0.0 0.0 0.0 4.8 - 2 - 4.8 - 7 -219 519 0 0 -4.2 0 0.0 0.0 -6.3 0.0 -5.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -6.3 0.0 -5.7 0.0 0.0 0.0 0.0 0.0 - -6.3 2 - - -6.3 7 -220 520 0 0 -3.7 0 0.0 0.0 -5.6 0.0 -4.4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -5.6 0.0 -5.0 0.0 0.0 0.0 0.0 0.0 - -5.6 2 - - -5.6 7 -221 521 0 0 -3.2 0 0.0 0.0 -4.8 0.0 -3.8 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -4.8 0.0 -4.3 0.0 0.0 0.0 0.0 0.0 - -4.8 2 - - -4.8 7 -222 522 0 0 -2.6 0 0.0 0.0 -3.9 0.0 -3.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.9 0.0 -3.5 0.0 0.0 0.0 0.0 0.0 - -3.9 2 - - -3.9 7 -223 523 0 0 -2.1 0 0.0 0.0 -3.2 0.0 -2.5 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -3.2 0.0 -2.8 0.0 0.0 0.0 0.0 0.0 - -3.2 2 - - -3.2 7 -224 524 0 0 -1.6 0 0.0 0.0 -2.4 0.0 -1.9 0.0 0.0 0.0 0.0 0.0 0.0 0.0 -2.4 0.0 -2.2 0.0 0.0 0.0 0.0 0.0 - -2.4 2 - - -2.4 7 -

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DL LL WLz Eqx DL LL WLz Eqx Fx Mz Fx Mz Fx Mz Fx Mz Fx Mz Fx My Fx My Fx My Fx My Fx MyFx

Comp(+)Fx

Tens(-).case

# MzFx

Comp(+)Fx

Tens(-).case

# My

1.35(DL+LL+EQ) # 10

1.2(DL+LL+EQ) #5 Design Critical forces

1.35 DL+1.5LL #6

1.35 DL+1.5WL #7

1.35(DL+LL+WL) # 8

1.35 DL+1.5EQ #9

1.5(DL+EQ) # 4

1.5(DL+LL) # 1

1.5(DL+WL) # 2SR

NOMEMB #

Axial Force Fx (Mton) (+, Comp & -, Ten) Design Critical forces

1.2(DL+LL+WL) # 3

BM-Mz (Mton-m)

278 528 0 0 0 0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 - - - 0.0 - - - 0.0279 529 -2.5 -4.9 -0.6 0 -7.4 0.0 -4.7 0.0 -6.6 0.0 0.0 0.0 0.0 0.0 -7.0 0.0 -4.3 0.0 -7.4 0.0 0.0 0.0 0.0 0.0 - -7.4 1 0.0 - -7.4 8 0.0280 530 -4.9 -9.8 -1.1 0 -14.7 0.0 -9.0 0.0 -13.1 0.0 0.0 0.0 0.0 0.0 -14.0 0.0 -8.3 0.0 -14.7 0.0 0.0 0.0 0.0 0.0 - -14.7 1 0.0 - -14.7 8 0.0281 531 -7.2 -14.3 -1.6 0 -21.5 0.0 -13.2 0.0 -19.1 0.0 0.0 0.0 0.0 0.0 -20.4 0.0 -12.1 0.0 -21.5 0.0 0.0 0.0 0.0 0.0 - -21.5 1 0.0 - -21.5 8 0.0282 532 -9.3 -18.4 -2.1 0 -27.6 0.0 -17.1 0.0 -24.6 0.0 0.0 0.0 0.0 0.0 -26.2 0.0 -15.7 0.0 -27.7 0.0 0.0 0.0 0.0 0.0 - -27.6 1 0.0 - -27.7 8 0.0283 533 -11.1 -21.8 -2.6 0 -32.7 0.0 -20.6 0.0 -29.3 0.0 0.0 0.0 0.0 0.0 -31.0 0.0 -18.9 0.0 -32.9 0.0 0.0 0.0 0.0 0.0 - -32.7 1 0.0 - -32.9 8 0.0284 534 -12.4 -24.4 -3.2 0 -36.6 0.0 -23.4 0.0 -33.1 0.0 0.0 0.0 0.0 0.0 -34.7 0.0 -21.5 0.0 -37.3 0.0 0.0 0.0 0.0 0.0 - -36.6 1 0.0 - -37.3 8 0.0285 535 -13.2 -26.1 -3.7 0 -39.2 0.0 -25.4 0.0 -35.8 0.0 0.0 0.0 0.0 0.0 -37.2 0.0 -23.4 0.0 -40.2 0.0 0.0 0.0 0.0 0.0 - -39.2 1 0.0 - -40.2 8 0.0286 536 -13.6 -27.2 -4.2 0 -40.8 0.0 -26.7 0.0 -37.7 0.0 0.0 0.0 0.0 0.0 -38.8 0.0 -24.7 0.0 -42.4 0.0 0.0 0.0 0.0 0.0 - -40.8 1 0.0 - -42.4 8 0.0

Truss members*Considering fixed support at base

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ii)Design Strength due to Rupture of Critical Clause 6.3.3 ii) Design ultimate resistance of the net Clause 5.4.3- Section (Tdn) cross section (Nu.Rd) (1)-b

To find Tdn we require the connection details

Here connections (welded) is designed such a manner that Tdgis preferably less than T dn In angles connected by one leg the eccentricity Clause 6.6.10(This ensures strength of system is governed of welded end connection may be allowed by -(1)by member) adopting an effective cross sectional area and

then treating member as concentrically loadea

For equal-angles the effective area may be Clause 6.6.10taken as gross area -(2)

Therefore Anet = (2x2106)mm2

3.Design Tension resistance of cross Clause 5.4.3section (Nt.Rd) eqn (5.13)

Nt.Rd = Smaller of Npl.Rd and Nu.Rd

= 957 kN >NSd=924 kN

Hence safeConnection

Section:- 2-ISA 110x110x10 (A=2x2106mm2)Interaction Ratio (Action/Strength)=0.97

Here let us take T= Tdg so that Tdn < Tdg

Let us use 6 mm size fillet weld (i.e. S=6mm)

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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode3T= 17.3 MTon NSd= 16.4 Mton

= 173 kN (Tension) = 164 kN (Tension) Load Case =1.5DL+1.5LL Load Case=1.35 DL+1.5LL

1.Trial section 1.Trial section

Increase it by 5 % so that (Ag) req =800mm2 Increase it by 5 % so that A req =758mm2

Let us try ISA 70x70x6 connected to the Let us try ISA 70x70x6 connected to the of gusset plate 8 mm thick gusset plate 8 mm thick

Sectional properties of 1 ISA 70x70x6 Sectional properties of 1 ISA 70x70x6

A=806 mm2,b= 70mm,d=70mm,t=6mm A=806 mm2,b= 70mm,d=70mm,t=6mmczz=19.4 mm czz=19.4 mm

Check for slenderness = Check for slenderness = kL/rzz = 1.0x 3230 / 13.6 = 237.5 < 400 ok l/izz = 1.0x 3230 / 13.6 = 237.5 < 350 ok

i)Design Strength due to yielding of gross Clause 6.2 i)Design Plastic resistance of gross Clause 5.4.3- section (Tdg) cross section (Npl.Rd) (1)-a

DESIGN STEPS REFERENCES DESIGN STEPS REFERENCES

2)Girder 1/2-L/R Inclined Members 51 to 58 and 101 to 106, 98 and 99

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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode3T= 5.7 MTon NSd= 5.7 Mton

= 57 kN (Tension) As per IS:800 (Draft) = 57kN (Tension) Load Case =1.5DL+1.5EQ Load Case=1.35 DL+1.5EQ

1.Trial section 1.Trial section

Increase it by 5 % so that (Ag) req =265mm2 Increase it by 5 % so that A req =758mm2

Let us try ISA 55x55x5 connected to the Let us try 2-ISA 70x70x6 connected to the of gusset plate 8 mm thick gusset plate 8 mm thick

Sectional properties of 1 ISA 55x55x5 Sectional properties of 1 ISA 70x70x6

A=527 mm2,b= 55mm,d=55mm,t=5mm A=806 mm2,b= 70mm,d=70mm,t=6mmczz=15.3 mm czz=19.4 mm

Check for slenderness = Check for slenderness = kL/rzz = 1.0x 4000 /10.6 =377.4 < 400 ok l/izz = 1.0x 4000 / 10.6 = 377.4 >350

In Eurocode there is no strigent requirement i)Design Strength due to yielding of gross Clause 6.2 for slenderness of Tension member section (Tdg) i)Design Plastic resistance of gross Clause 5.4.3-

cross section (Npl.Rd) (1)-a

DESIGN STEPS REFERENCES DESIGN STEPS REFERENCES

3)Girder 1/2-L/R Inclined Members 302, 301, 303, 304 and 311, 312, 314, 315

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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode3P= 25.2 MTon NSd= 23.9Mton

= 252 kN (Compression) = 239kN (Compression) Load Case =1.5DL+1.5LL Load Case=1.35 DL+1.5LL

Length= 2.3m Length= 2.3m

1.Trial Section 1.Trial SectionLet us assume fcd =130 MPa Asssume design buckling stress =130 Mpa

Let us try 2-ISA 70x70x8 connected on either Let us try 2-ISA 70x70x8 connected on either side of gusset plate 10 mm thick side of gusset plate 10 mm thick

Sectional Properties of 2-ISA 70x70x8 Sectional Properties of 2-ISA 70x70x8

A= 2116mm2 A= 2116mm2

b=d = 70mm c=d = 70mmt= 8mm t= 8mm

rzz= 21.2mm iyy= 21.2mm

ryy= 32.9mm izz= 32.9mm

rvv’ = 13.5mm ivv’ = 13.5mmL= 2300mm L= 2300mm

czz= 20.2mm cyy= 20.2mm2.Section Classification Table 3.1 of 2.Section Classification Table 5.3.1 ofStress ratio IS:800 (Draft) Stress ratio EC3

4) Girder 1/2-L/R Vertical struts, Members 33 to 37 and 42 to 46, 112 to 116 and 121 to 125

DESIGN STEPS REFERENCES DESIGN STEPS REFERENCES

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Interaction Ratio (Action/Strength)=0.96 For =0.95 , buckling curve ’c’ and fy =250 Table 5.5.2 of Stress reduction factor =0.57 EC3

7) ConnectionLet us provide 6mm fillet weld Clause 5.5.1.1Strength of weld= 0.707x S x fu/((¥3)x m1) Clause10.5.7

= 0.803 kN/mHence safe

Interaction Ratio (Action/Strength)=0.877) Connection

And L weld required per angle =158 mm Let us provide 6mm fillet weld

Now L1+ L2 = 158 -70 =88mm Strength of weld= 0.707x S x fu/((¥3)x w Mw) Clause 6.6.5.3Taking moment of weld force about the vertical = 0.973 kN/mface of angleL1 x 20.2 = L2 x (70-20.20)

Hence L2=26mm and L1=62mm

And L weld required per angle =123 mm

Tack weld Now L1+ L2 = 123 -70 =53mm

Provide 3mm fillet weld Taking moment of weld force about the vertical face of angle

Pitch of tack weld < i) 600 mm Clause L1 x 20.2 = L2 x (70-20.20)

ii)40xrvv’ =40x13.5=540mm 10.2.4.5 Hence L2=16mm and L1=37mm

iii)0.6 x(KL/r)xrvv’ =0.6x86.8x13.5=703 mm Tack weld Clause 5.9.4

Provide tack weld @ 540 mm c/c Pitch of tack weld < i) 15 ximin of combined

< i.e. 15 x21.2 =318 mm

Provide tack weld @ 315 mm c/c

Section :-2- ISA 70x70x8 (A= 2116mm2) Section:- 2-ISA 70x70x8 (A=2116mm2)Interaction Ratio (Action/Strength)=0.96 Interaction Ratio (Action/Strength)=0.87

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2.Equivalent slenderness ratio ��(IIHFWLYH�6OHQGHUQHVV�5DWLR� eff Clause 5.8.3For single angle strut For single angle effective slenderness ratios

are as follows

For buckling @ v-v axis:=where wherek1,k2,k3= constant depending upon endcondition

For buckling@ z-z axis:= and where

For buckling @ y-y axis:= where

For our caseAssuming the fixity as partial, hence taking the Table 7.6 of Now L= 2.3 m & taking l=Lvalue k1,k2 and k3 as average of same IS:800 (Draft)mentioned for fixed and hinged connection Therefore k1=0.45 ,k2 =0.475 and k3 =12.5

Also

Therefore

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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode3P= 18.3 MTon NSd= 18.2Mton

= 183 kN (Compression) = 182kN (Compression) Load Case =1.5DL+1.5WL Load Case=1.35 DL+1.5WL

Length= 3.25m Length= 3.25m

1.Trial Section 1.Trial SectionLet us assume fcd =60 MPa Asssume design buckling stress =60 Mpa

Let us try an ISMB 200 @ 0.249kN/m Let us try an ISMB 200 @ 0.249kN/mSectional properties Sectional properties

A= 3233mm2 A= 3233mm2

h= 200mm h= 200mmB= 100mm B= 100mmtf= 10.8mm tf= 10.8mm

tw= 5.7mm tw= 5.7mm

rzz= 83.2mm iyy= 83.2mm

ryy= 21.5mm izz= 21.5mm

h2= 23.65mm h2= 23.65mmd= 152.7mm d= 152.7mm

2.Section Classification Table 3.1 of 2.Section Classification Table 5.3.1 ofStress ratio IS:800 (Draft) Stress ratio EC3

DESIGN STEPS REFERENCES

6) Column Member 401 to 403 & 413 to 415, 410 to 412, 422 to 424

DESIGN STEPS REFERENCES

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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode 3P= Nil NSd= Nil

Mz= 2.1MTon m =21kNm My= 2.0MTon m =20kNm Load Case:- 1.5 (DL+LL) Load Case:-1.35DL+1.5LL

P= 5MTon =50kN NSd= 5MTon =50kN

Mz= 0.8Mton m =8 kNm My= 0.7Mton m =7 kNm Load Case:- 1.5 (DL+WL) Load Case:-1.35DL+1.5WL

1.Trial section 1.Trial sectionLet us try an ISMB 175 @ 0.189kN/m Let us try an ISMB 200 @ 0.249kN/mSectional properties Sectional propertiesA=2462mm2

Izz= 1272 x104 mm4 A=3233mm2Iyy= 2235.4 x104 mm4

h=175mm Iyy= 85x104 mm4h=200mm Izz= 150x104 mm4

B=90mm Zez= 145.4x103 mm3B=100mm Wey= 223.5x103 mm3

tf=8.6mm Zey= 18.9x103 mm3tf=10.8mm Wez= 30x103 mm3

tw=5.5mm Zpz= 163x103 mm3tw=5.7mm Wpy= 249.7x103 mm3

rzz=71.9mm Zpy= 19.85x103 mm3iyy=83.2mm Wpz= 31.5x103 mm3

ryy=18.6mm h2= 20.25mm izz=21.5mm h2= 23.65mmd= 134.5mm d= 152.7mm

2.Section classification Table 3.1 of 2.Section classification Table 5.3.1 ofStress ratio =¥(250/fy)=1 IS:800 (Draft) Stress ratio =¥(235/fy)=0.97 EC3

Hence flange is of plastic class from both axial Hence flange is of Class1 from both axial and bending compression point of view and bending compression point of view

DESIGN STEPS REFERENCES DESIGN STEPS REFERENCES

7) Bottom girder main Beam members 601 to 618

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4.Compression capacity Clause 7.1.2 4.Buckling resistance Clause 5.5.1.1Effective length (KL)y=1x2.44 =2.44 m say 2.5m Effective length (l) z=1x2.44 =2.44 m say 2.5m

ryy=18.6mm and izz=21.5mm

Therefore slenderness ratio (KL/r) y =134.40 Therefore slenderness ratio (l/i) z = =116.28

For buckling about y-y axis buckling curve ’b’ is For buckling about z-zaxis buckling curve ’b’ isused used For (KL/r)y=134.40and buckling curve ’b’ the Table 7.4b ofvalue of fcd=75.16N/mm2 IS:800 (Draft)

For =1.28 and for buckling curve ’b’ value of stress reduction factor =0.35

5.Check for resistance of cross section to Clause 9.3.2.2combined effect of buckling

5.Check for resistance of cross section to Clause 5.5.4axial compression and bending

Here P=50 KN ,Mz=0.8 kNm

For given section ISMB 175 Pd =185kN and Mdz=28.67kNm

Here NSd= 50 kN and My.Sd=0.7 kNm

Note that though interaction ratio for combined forces is very low such section selecton is required to take care of moment action underDL+LL combination (For that case IR=0.73)

Section :-ISMB175 (A=2462mm2) Section :-ISMB200 (A=3233mm2)IR =(Action/Strength)=0.73 For 1.5(DL+LL) IR =(Action/Strength)=0.78 For 1.35DL+1.5LLIR =(Action/Strength)=0.55 For 1.5(DL+WL) IR =(Action/Strength)=0.49 For 1.35DL+1.5WL

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10

33 to 37 and 42 to 46,

112 to 116 and 121 to

125

Comp 2ISA 70x70x8 2ISA 70x70x8 252 264.5 32.9 0.95 239 274 32.9 0.87

11

38 to 41and 117 to 120, 317, 316,

323 and 324

Comp ISA 70x70x8 ISA 70x70x8 52.9 74 16.4 0.71 46 60 16.4 0.77

12306, 310, 313 and 308, 305, 95,309, 307

Comp ISA 60x60x6 ISA 60x60x6 30 38.85 7.44 0.77 28 31 7.44 0.90

135 to 13, and

67 to 75.Comp 2ISA 110x110x15 2ISA 110x110x15 1037 1090.6 51.7 0.95 970 1181 51.7 0.82

14

1 to 4 and 63 to 66,

14 to 17 and 76 to 79.

Comp 2ISA 90x90x12 2ISA 90x90x12 635 636 43.1 1.00 610 715 43.1 0.85

15501 to 505, 514 to 518

Comp 2ISA 70x70x10 2ISA 70x70x10 299 302 35 0.99 278 367 35 0.76

16 506 to 513 Comp 2ISA 65x65x8 2ISA 65x65x8 180 206 30.9 0.87 168 259 30.9 0.65

17

401 to 403 & 413 to 415, 410 to 412, 422 to 424

Comp ISMB 200 ISMB 200 183 204 0.90 182 211 0.86

18 601 to 618 Flexure ISMB175 ISMB 200 21 28.53 24.62 0.74 20 26.93 32.33 0.74

Flexure 474 1442.6 450 1636

Comp 179 399.6 167 330.64 110.7

Flexure 369 1442.6 351 1636Comp 158 399.6 147 330.64 110.7

* Considering fixed support at base

ISMB 500404 to 409 & 416 to 421

19 0.78

19*404 to 409 & 416 to 421

ISMB 500 ISMB 500 0.65

ISMB 500

0.66

110.7

110.7

0.78

Swapnil B. KharmaleCD-051061 277 Comparative study of IS:800 (Draft) and EC3ESte

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Swapnil B.Kharmale 279 Comparative study of IS: 800 (Draft) &EC3 CD-051061

� Design of compression member

Both codes use the Perry-Robertson approach to evaluate the compression

capacity of member.

In case of single angle strut the effective slenderness ratios are calculated on

the basis of test results and researches carried out by respective standards. When

we compare the capacities of single angle discontinuous strut (of non slender class)

with partial fixity at the end connection it founds that IS: 800 (Draft) gives higher

capacity as compare to Eurocode 3.

� Design of member subjected to bending

Plastic design strength of flexural member by both codes is same

If member governs by flexural-torsional buckling mode of failure IS (800) Draft

code may prove economical because Eurocode 3 more conservative equation to

calculate non dimensional slenderness ratio for lateral torsional buckling (LTB)

� Design of member subjected to combined forces

The interaction equations specified by both codes for design of member

subjected to combined forces are same.

From Section B “Project problem”

In project problem the interaction ratios (IR=Action/Strength) are calculated

for various structural elements of FOB designed by both codes.

From load combinations by both code (Table C.1.1) and design calculation it

founds that

x When DL + LL combination is governing then design action by IS:

800 (Draft) are slightly higher than that by Eurocode 3 and in such

case Eurocode is slightly economical.

x When DL +LL+ WL (EL) combination is governing then design action

by Eurocode 3 are slightly higher than that by IS: 800 (Draft) and in

such case IS: 800 (Draft) is slightly economical.

This is because:-

IS: 800 specify the same partial safety factor for Permanent action as well as

Variable action. Eurocode 3 specify the different partial safety factor for permanent

and variable action (more value for Variable action as variation are large and smaller

EStela

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Swapnil B. Kharmale 281 Comparative study of IS: 800 (Draft) & EC 3 CD-051061

References

1. Design standards

Indian Standards

� IS: 800-1984 Code of practice for general construction in steel

� IS:875-1975 (Part I to V)Code of practice for design loads

� IS: 1893-2002 Criteria for earthquake resistant design of structure

� IS: 808-1989 Dimensions for hot rolled steel beam, column, channel

and angle sections

� IS:801 -1975 Code of practice for use of cold formed light gauge steel

structure member in general construction

� IS: 814-1991 Covered electrodes for manual metal arc welding of

carbon and carbon manganese steel

� IS: 4000-1992 High strength bolts in steel structures.

Eurocodes

� ENV: 1991 Eurocode 1 Basis of design and actions on structures

� ENV: 1998 Eurocode 8 Design of structure for earthquake resistance

British Standard

� BS: 5950:Part1:2000:-Structural Use of Steelwork in Buildings:

Part1Code of practice for design in simple and continuous construction

2. Books

� Behaviour and Design of Steel Structure -By N.S. Trahair

� Design of Steel Structure -By B.S. Krishnamachar and D.Ajitha Simha

� Steel Structures Design and Behaviour (1971 edition) - By-C.G.

Salmon, J.E.Johnson

� Reliability based design approach By R.Rangnathan

� Theory of Elasticity - By-S.P.Timoshenko and J.N.Goodier

� Theory of Elastic stability By:-S.P.Timoshenko &J.M.Gere

EStela

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