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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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.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
-- --- 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
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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
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
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
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
Swapnil B Kharmale CD-051061 59 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
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
EK
Swapnil B Kharmale CD-051061 61 Comparative study of IS:800 (Draft) and EC3ESte
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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
+ 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
� 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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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
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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 �
� � �
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X� Y � � � Y � � R�Y
\
$ / [W
/ / �/ �/ EXW / /
/ D EXW / �G
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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
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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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
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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
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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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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
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
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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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
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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
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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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:= 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
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.
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
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
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
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
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
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.
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
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S
E �� ���� � ��� 3ODVWLFW ��
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% ��� ���� � ��� 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
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
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.
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
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Swapnil B.KharmaleCD-051061 220 Comparative study of IS:800 (Draft) and EC3
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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
Swapnil B. KharmaleCD-051061 257 Comparative study of IS:800 (Draft) and EC3ESte
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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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Swapnil B. KharmaleCD-051061 259 Comparative study of IS:800(Draft) and EC3ESte
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Design forces:- As per IS:800 (Draft) Design forces:- As per Eurocode3T= 17.3 MTon NSd= 16.4 Mton
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
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
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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