Lightweight Steel Structures part.1 by Archineer

Sabah Sabah Shawkat Sabah Shawkat Sabah Shawkat Shawkat

Sabah Shawkat

STRUCTURES STRUCTURES STRUCTURES STRUCTURES

LIGHTWEIGHT STEEL STRUCTURES

STEEL STEEL STEEL STEEL LIGHTWEIGHT LIGHTWEIGHT LIGHTWEIGHT LIGHTWEIGHT

LIGHTWEIGHT LIGHTWEIGHT LIGHTWEIGHT LIGHTWEIGHT LIGHTWEIGHT STEEL STEEL STEEL STEEL STEEL STRUCTURES STRUCTURES STRUCTURES STRUCTURES STRUCTURES

Sabah Shawkat © Sabah Sabah Sabah Sabah Sabah Shawkat Shawkat Shawkat Shawkat Shawkat

Reviewer: Cover Design: Editor: Software Support: Publisher:

Prof. Dipl. Ing. Ján Hudák, PhD. Ing. Peter Novysedlák, PhD, M.Eng. Mgr. art. Peter Nosáľ prof. Ing. arch. Zuzana Pešková, PhD, Mgr.art. Ing. Richard Schlesinger, PhD. asc. Applied Software Consultants, s.r.o., Bratislava, Slovakia Tribun EU, s.r.o., Brno, Czech Republic

Sabah Shawkat © All rights reserved. No part of this book may be reprinted, or reproduced or utilized in any form or by any electronic, mechanical or other means, including photocopying, without permission in writing from the author.

Lightweight Steel Structures ©

Assoc. Prof. Dipl. Ing. Sabah Shawkat, MSc, PhD. 1. Edition, Tribun EU, s.r.o. Brno, Czech republic 2019 ISBN 978-80-263-1458-5

Definitions

Sabah Shawkat © About the Author

Sabah Shawkat Zahawi is the Head of Engineering Room at the Academy of Fine Arts and Design in Bratislava, Slovakia. He teaches students of architecture several structural engineering subjects. Moreover, he regularly organizes workshops for students and exhibitions of their projects and construction models. He also actively practise in projecting and building constructions as well as reconstructions and modernizations of buildings. Sabah Shawkat is also a passionate expert in reinforced concrete, prestressed concrete structures and structural design. He has published numerous articles in professional journals and wrote several books.

Definitions

Preface Why this publication deals with steel construction, because steel is a recyclable material. A structural steelwork is easily demolished and melted down after use and processed in to a new structures. Demolition is easier if bolted connections were used in the original construction, and the elegance of the welding profiles may be used to construct light and adaptable structures. It is possible to produce a profile for every application with as optimal as possible use of material. This publication gives a general overview of structural design and behaviour of steel structures to the structural Euro-code, and includes a set of worked examples showing the design of structural elements. The idea behind the creation of this book came from trying to find the best way to introduce students, designers in architectural building as an introduction, reference on how to design steel structures. Every step in the design of the steel structure is explained in simple terms. Designers interested in learning more about behaviour of steel members will find a series of 3d pictures which I made them during the realization projects on site that will

Sabah Shawkat ©

complement the text therefore demonstrate the concept and building details of the constructions to further the readers understanding.

The publication has been produced with the assistance of structural design lecturers, the set of worked examples present the design of structural members.

This publication contains comprehensive section property data and member resistances for a wide range of steel sections such as double symmetric hybrid beam, hollow cross-section, and box profiles. It is intended to be of particular help in undergraduate teaching, although it will also provide guidance to practising designers who want to become acquainted with design to the steel structures design. The introductory text and the worked examples have all been evaluated using the values of parameters and design options given in the euro-code. The Structural Euro-codes are a set of structural design standards, which cover the design of all types of structures, however, always be used together with the relevant national application documents (NAD). The designer should separately check the relevant requirements for each country in the national application documents. Sabah Shawkat

Definitions

Bratislava, 02 / 2019

Structures as Architecture Structure as architecture presents a comprehensive analysis of the indispensable role of structure in architecture. the structural design is one of the main generators of the architectural language. Here there is a strong correlation between form, loads and materials and in many cases the equation is - “architecture = structure and structure = architecture”. The structural system is one of the most important components of any entity in the universe, which has a physical dimension. Specifically, the main role of this system is to allow any entity to cope with physical loads, and to ensure the entity’s performance. But generally this system can have other important roles – aesthetical, organizational and even cultural. An exploration as well as a celebration of structure, the book draws on a series of design studies and case study examples to illustrate how structure can be employed to realize a wide range of concepts in contemporary architecture by examining design principles that relate to both architecture and structural engineering. During the design of projects in the frame of the subject of construction in architecture, the

Sabah Shawkat © student will create a structural model with a computer application based on the concepts of the

behaviour and loading of the structural member or assemblage. The student will be able to interpret the modelling results and relate the results to the solution obtained by manual calculations and will be able to articulate the physical phenomena, behaviour and design criteria

which influence structural space and form. Then we tried to provide new insights in to

relationship between both the technical and aesthetic aspects of architecture with a number significant enhancements.

Structure as manifest in existing architecture, this new emphasise on design, rather than analysis brings a welcome balance to the book.

Structure is column or planar or a combination of these which a designer can intentionally use to reinforce or realize ideas. As such structure can be used to define space, create units, articulate circulation, suggest movement, or develop composition and modulations.

Definitions

Preface

Buckling

Definitions

Lateral buckling

Terminology

Buckling resistance Buckling resistance of welded profile for fire situation

Limit state design and partial safety factors

Calculate the stress resistance of the roof beam

Conventions for member axes

The calculation of the Buckling

Classification of cross-sections

Calculate the resistance to lateral-torsional buckling

Bending resistance-circular hollow sections

Calculate the compression resistance of the box column

Resistance of hollow sections

Calculate the moment resistance and the resistance to lateral - torsional buckling

Determine the classification of the cross-section of the hollow section

Determine the cross-section of the cantilever steel beam

Calculate the resistance of a hollow section

Calculate the moment resistance of the hybrid beam

Sabah Shawkat © Holow sections subjected to bending moment and axial force

Calculate the dimensions and resistance of steel columns

Calculate the resistance of a circular hollow section

Calculate the effective widths and the effective second moment of area of the

Calculate the tansion resistance of the flange plate joint

compression elements

Dual profiles

Connections in steel structures

Structural actions

Bolt connections

Self-weight and imposed loads

Welded connections

Imposed load reduction

Splices

Snow load

Slip resistance

Wind load

Calculate the tension resistance of the joint

Designing gable beam in the model building

Bearing resistance of splice plates

Calculate the deflection of the steel I beam

Shear resistance of bolts

Designing the stiffering elements in the model building

Resistant the welds Calculate the resistance of the shear joint Shear resistance of bolts Resistance of welds

Calculate the bending resistance of flange plate joint

Fire design / protection

Calcute the bending resistance of the splice with end-plates

Unprotected steel structure

Rotation capacity of the splice

Steelwork insulated by fire protection material

Design the joint

Calculate the resistance of column in fire situation

Column Bracket

Holow section exposed to fire on three sides

Detail of attachment - Steel Beam to Steel Column

Holow section exposed to fire on two oppposite sides

Beam-to-column joint

Holow section exposed to fire on two adjacent sides

Resistance of the joint

Fire in maisonette flats of a residential building

Joint to the foundation

Composite slab

Column-to Foundation connections

Composite steel concrete ceilings

Sabah Shawkat © Calculate the joint resistance of a hollow section

Lightweight Structures

Trusses and retuculated structures

- Definition of lightweight structures

Example

- The art of tensegrity

Design of trusses

- Attachment of the stiffener to the steel beam

Y joint

- Detail of the support cables on the pylon

K joint

- Detail of the support bearing cables on the pylon

T joint

- Detail of the stiffener connection to column on the bar

X joint

- Membrane structures

A gapped K joint

- Table properties of membrane structures

A lowe corner joint in a lattice structure

- Force dentity method

A T or a Y joint

- Mohr circle of stress

A X joint An overlapped K joint Examples

ce(z)

Definitions

topography and height above ground

The principal terms and the equivalent symbols relating to actions are as follows: A

Accidental action

G, g

Permanent action

Q, q E

exposure coefficient, which takes account of terrain roughness,

external wind pressure

internal pressure

Variable action

Area of the shank - nominal area [mm2]

Stress area [mm2]

Effect of actions

Nominal diameter of the bolt (shank) [mm]

Nominal diameter of the hole [mm]

Force

Diameter of the stress area (As) [mm]

Moment

End distance [mm]

Axial force

Edge distance [mm]

Shear force

Applied load [kN]

Shear force [kN]

Partial safety factor applied to characteristic values of actions to

Fv,Rd

Design shear resistance of a bolt [N]

establish values to be used in design calculations

Tensile force [kN]

Ft,Rd

Design tension resistance of a bolt [kN]

Sabah Shawkat © 



Combination factors used to modify partial safety factors for

Fb.Rd

Design bearing resistance [kN]

Ultimate tensile strength of a steel element [MPa]

Specific notation is used for the parts of EC1 dealing with snow and wind loads

fu,b

Nominal ultimate stress of the bolt material [MPa]

as follows:

fy,b

Nominal yield stress of the bolt material [MPa]

Snow loads

p1, p2 Pitches [mm]

actions used in combination

characteristic snow load at ground level

f

is the partial safety factor for loading



snow load shape coefficient,

m

is the partial safety factor for steel

characteristic snow load on the roof

is characteristic value of the force or moment caused by the load

is characteristic value of the resistance

Wind loads vref,

reference wind velocity

G

for permanent loads f = 1.35

qref

reference mean wind velocity pressure

Q

for variable load Q = 1.5



the density of air

Gj

is the partial safety factor for permanent loads

Gkj

is characteristic value of permanent loads

Definitions Definitions

Ql

is the partial safety factor for the leading variable load

QQl

is characteristic value of the leading variable load

Qi

is the partial safety factor for the other variable loads

QQi

is characteristic value of the other variable loads

Partial safety factors for material properties. M0 = 1.1

For Class 1,2,3 cross-section

M1 = 1.1

For Class 4 cross-section

M1 = 1.1

For Buckling resistance

M2 = 1.25

For net cross-section (deduction for the holes in the gross cross-section)

Mb = 1.25

Resistance of bolted connections

Mr = 1.25

Sabah Shawkat © Resistance of welded connections

Msult = 1.25

Slip resistance (ultimate limit state)

Msult = 1.40

Slip resistance (ultimate limit state, oversize or slotted holes parallel to the direction of load transfer)

Msser = 1.1 Mf = 1.0 Mf = 1.25

Slip resistance (serviceability limit state)

Fatigue resistance (accessibility of the structure (normal) – Fail-safe structure)

Fatigue resistance (accessibility of the structure (normal) – Non fail-safe components)



is the combination factor of the load.

Design value of an effect.

MEd

Design bending moment.

Design resistance.

MRd

Design resistance for bending el Elastic property.

Wel

Elastic section modulus.

Wpl

Plastic section modulus.

Definitions Definitions

Terminology Fail-safe structure = failure of the component does not rapidly result in failure of the entire

Plan: A horizontal, geometrical drawing of a building showing the walls, doors, windows,

structure.

stairs, chimneys, columns, and other structural components.

Non fail-safe structure = failure of the component results in rapid failure of the entire structure.

Plaster: A mixture of lime, hair, and sand (or of lime, cement, and sand) used to cover outside and inside wall surfaces.

Actions: Loads, imposed displacements, thermal strain.

Plate: The top, horizontal piece of the walls of a frame building upon which the roof rests.

Effects: internal bending moments, axial forces etc.

Plot plan (site plan): A drawing showing all necessary property lines, contours, building lines,

Resistance: Capacity of a structural element to resist bending moment, axial force, shear, etc.

building locations, existing or new buildings, and utility easements.

Verification: Check.

Basic component (of a joint): Part of a joint that makes a contribution to one or more of its

Execution: Construction – fabrication, erection.

structural properties.

Anchor bolts: Bolts which fasten columns, girders, or other members to concrete or masonry.

Connection: Location at which two or more elements meet. For design purposes it is the

Beam: An inclusive term for joists, girders, rafters, and purlins.

assembly of the basic components required to represent the behaviour during the transfer of the

Braces: Pieces fitted and firmly fastened to two others at any angle; used to strengthen the

relevant internal forces and moments at the connection.

angle thus treated.

Connected member: Any member that is joined to a supporting member or element.

Column: A square, rectangular, or cylindrical support for roofs, ceilings, and so forth.

Joint: Zone where two or more members are interconnected. For design purposes it is the

Dead load: A permanent load on a building or other structure, which includes the weight of its

assembly of all the basic components required to represent the behaviour during the transfer of

structural members and the fixed loads they carry.

the relevant internal forces and moments between the connected members. A beam-to-column

Diagonal: Inclined member of a truss or bracing system used for stiffening and for wind-

joint consists of a web panel and either one connection (single sided joint configuration) or two

bracing.

connections (double sided joint configuration)

Footing: An enlargement at the lower end of a wall, pier, or column that distributes the load.

Nominally pinned joints: A nominally pinned joint shall be capable of transmitting the internal

Gable: The vertical, triangular end of a building from the eaves to the apex of the roof.

forces, without developing significant moments which might adversely affect the members or

Gambrel: A symmetrical roof with two different pitches or slopes on each side.

the structure as a whole.

Sabah Shawkat © Hanger: A vertical-tension member supporting a load.

Rigid joints: Joints classified as rigid may be assumed to have sufficient rotational stiffness to

Lattice: Crossed wood, iron plate, or bars.

justify analysis based on full continuity.

Level: A term describing the position of a line or plane when parallel to the surface of still

Semi-rigid joint: A joint which does not meet the criteria for a rigid joint or a nominally pinned

water; a tool used for testing horizontal and vertical surfaces and for determining differences of

joint should be classified as a semi-rigid joint.

elevation.

Nominally pinned joints: A nominally pinned joint shall be capable of transmitting the internal

Member: A single piece complete in itself, within a structure.

forces, without developing significant moments which might adversely affect the members or

Partition: A permanent interior wall that divides a building into rooms.

the structure as a whole.

Partition, non-bearing: A dividing wall that separates areas of a structure but does not provide

Rotational capacity: The angle through which the joint can rotate without failing.

support for the room, overhead partitions, or floor joists.

Rotational stiffness: The moment required to produce unit rotation in a joint.

Pitch: Inclination or slope, as for roofs or stairs; the rise divided by the span.

Structural properties (of a joint): Resistance to internal forces and moments in the connected

Pitch board: A board sawed to the exact shape formed by the stair tread, riser, and slope of the

members, rotational stiffness and rotation capacity.

stairs and used to lay out the carriage and stringers.

Uniplanar joint: In a lattice structure a uniplanar joint connects members that are situated in a single plane.

Terminology Definitions

Heat release rate: Is a term to show the amount of heat generated by fire. Reaction to fire: Is a term used to classify how a material responds to heat and fire. Fire resistance: Is a term used to measure the ability of a material (or combinations of materials) to resist the passage of fire from one distinct area to another. Char-line: Borderline between the char-layer and the residual cross-section. Effective cross-section: Cross-section of member in a structural fire design based on the reduced cross-section method. It is obtained from residual cross-section by removing the parts of the cross-section with assumed zero strength and stiffness. Failure time of protection: Duration of protection of member against direct fire exposure; (when the fire protective cladding or other protection falls off the timber member, or when a structural member initially protecting the member fails due to collapse, or when the protection from another structural member is no longer effective due to excessive deformation). Faire protection material: Any material or combination of materials applied to a structural

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member or element for the purpose of increasing its fire resistance.

Residual cross-section: Cross-section of the original member reduced by the charring depth.

Protected members: members for which measures are taken to reduce the temperature rise in the member and to prevent or reduce charring due to fire.

Normal temperature design: Ultimate limit state design for ambient temperatures according to EN 1995-1-1.

Terminology Definitions

References [1] ENV 1993-1-1: Eurocode 3: Design of steel structures: Annex K: Hollow section lattice girder connections, 1994 [2] CIDECT: Design guide for rectangular hollow section joints under predominantly static loading, Verlag TUV Rheinland GmbH, Koln 1992 [3] CIDECT: Design guide for circular hollow section joints under predominantly static loading, Verlag TUV Rheinland GmbH, Koln 1992 [4] CIDECT: Design guide for circular hollow section joints under predominantly static loading, Verlag TUV Rheinland GmbH, Koln 1991

[18] Frame design including joint behaviour. Volume 1. 1997. ECSC. 268 p. [19] Frame design including joint behaviour. Volume 2. 1997. ECSC. [20] Design Handbook for Braced or Non Sway Steel Buildings according to EC3. Brussels 1997. ECCS Publication No. 85. [21] ENV 1993-1-1/A2. 1998. Annex J: Joints in building frames. European Committee for Standardization (CEN). 80 p.

[5] ISO/FDIS 12944-2: Paints and varnishes-corrosion protection of steel structures by protective painting systems. Part 2: Classification of enviroments, 1997 [6] ISO/FDIS 12944-3: Paints and varnishes-corrosion protection of steel structures by protective painting systems. Part 3: Design considerations, 1997 [7] CIDECT: Design guide for fabrication, assembly and erection of hollow section structures, 1996

Sabah Shawkat © [8] ENV 1090-1: Execution of steel structures-Part 1: General rules and rules for buildings, 1996

[9] ENV 1991-2-1: Euro-code 1: Basis of design and actions on structures. Part 2-1: Actions on structures, densities, self-weight and imposed load

[10] ENV 1991-2-3: Euro-code 1: Basis of design and actions on structures. Part 2-3: Actions on structures, Snow loads, 1995 [11] ENV 1991-2-4: Euro-code 1: Basis of design and actions on structures. Part 2-4: Actions on structures, Wind loads, 1995 [12] ENV 1991-1: Euro-code 1: Basis of design and actions on structures. Part 1: Basis of design, 1995

[13] ENV 1991-2-2: Euro-code 1: Basis of design and actions on structures. Part 2-2: Actions on structures. Actions on structures exposed to fire, 1995 [14] ENV 1993-1-2: Euro-code 3: Design of steel structures. Part 1-2: General rules. Structural fire design, 1996 [15] ENV 1994-1-1: Design of composite steel and concrete structures: Part 1.2: Structural fire design, 1994 [16] ENV 1994-1-1: Design of composite steel and concrete structures: Part 1.1: General rules and rules for buildings, 1994 [17] ENV 1993-1-1: Euro-code 3: Design of steel structures: Part 1.1: General rules and rules for buildings, 1993

References Definitions

Sabah Shawkat ©

Definitions

Sabah Shawkat ©

Limit state design Limit state design

Limit - state design and parcial safety factors Limit - state design and parcial safety factors Convention of members Convention of members Classification of cross-sections Classification of cross-sections Bending resistance-circular hollow sections Bending resistance-circular hollow sections Resistance of hollow sections of hollow sections Resistance Square hollow sections hollow sections Square Determine the classification of the cross-section of the hollow section Determine the classification of the cross-section of the hollow section CalculateCalculate the resistance of a hollow section the resistance of a hollow section Hollow sections subjected to bending moment and axial force to bending moment and axial force Hollow sections subjected Calculate the resistance of a circular hollow section Calculate the resistance of a circular hollow section Calculate the tension resistance of the flange plate joint Calculate the tension resistance of the flange plate joint Dual profiles Dual profiles

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Limit – state design and partial safety factors

Nominal characteristic values of yield strength fy and ultimate tensile strength fu

The simple theory of bending is based on the assumption that plane sections remain plane after

Steel grade

bending. The resistance of the structure means its capability to bear loads applied on it without and place. They have thus no single absolute values. But their values are distributed with statistical probability. In design, the distributions of the resistance and laoding are taken into account by using partial safety factors. The general design criterion at the ultimate has the form (to combine loads)





 fiSki

 

Rk 



 m

    Ql QQl

 GjGkj

The expression below can be used to calculate the design value of the loading at the ultimate limit state, when the structure is subject to several variable loads.

 

  0.9    QiQQi     GjGkj     Ql Qkl     0i QiQki

 GjGkj

Material thickness 40mm < t ≤ 100mm fu (MPa) fy (MPa) fu (MPa)

fy (MPa) EN 10025 S235 235 360 215 340 S275 275 430 255 410 S355 355 510 335 490 EN 10113 S275N 275 390 255 370 S355N 355 490 335 470 S420N 420 520 390 520 S355M 355 470 3351 4501 S420M 420 500 3901 5001 S460M 460 530 4301 5301 1) Plates and plain steel products: 40mm<t<63mm 2) Values in this table are basic values in Euro-code 3. Requirements for each country have to be checked in the national application Document (= NAD)

failure or too large deformations. The resistance and loading are quantities that vary with time

 

t ≤ 40 mm

Conventions for member axes

Sabah Shawkat © j



( i l)





( i l)

Alternatively, also the more accurate formula can be used at the ultimate limit state



The convention for member axes and symbols for section dimensions used in the Euro-codes are shown below. Major axis y-y

Partial safety coefficients

Partial safety coefficients of the loads and the combination of cases is carried our according to the NADs of various countries. Typical coefficients are presented in the following:

The load combinations are checked in the ultimate limit state according to the Euro-norm (long-term and temporary) and in the serviceability limit state (unusual combination) as follows:

Minor axis z-z

Longitudinal axis of element x-x

In general the convention in member axes is:

x-x – along the member in new norm in new norm, z-z according to old norm

Limit state Combination Permanent Stationary snow Snow span 1 Snow span 1 Snow span .. Wind suction Wind pressure Imposed span 1 Imposed span 2 Imposed span..

Serviceability limit state A B C 1.0/1.0 1.0/1.0 1.0/1.0 1.0/0.0 1*ψ /0.0 1*ψ /0.0 1.0/0.0 1*ψ /0.0 1*ψ /0.0 1.0/0.0 1*ψ /0.0 1*ψ /0.0 1.0/0.0 1*ψ /0.0 1*ψ /0.0 1.0/0.0 1*ψ/0.0 1*ψ /0.0 1.0/0.0 1*ψ/0.0 1*ψ /0.0 1.0 /0.0 1*ψ/0.0 1*ψ /0.0 1.0 /0.0 1*ψ/0.0 1*ψ /0.0 1.0 /0.0 1*ψ/0.0 1*ψ /0.0

Ultimate limit state B C 1.0/1.0 1.35/1.0 1.5*ψ /0.0 1.5*ψ/0.0 1.5*ψ /0.0 1.5*ψ/0.0 1.5*ψ /0.0 1.5*ψ/0.0 1.5*ψ /0.0 1.5*ψ/0.0 1.5/0.0 1.5*ψ/0.0 1.5*ψ/0.0 1.5/0.0 1.5*ψ/0.0 1.5*ψ/0.0 1.5/0.0 1.5*ψ/0.0 1.5*ψ /0.0 1.5/0.0 1.5*ψ/0.0 1.5*ψ /0.0 1.5/0.0 1.5*ψ/0.0 1.5*ψ /0.0 A 1.35/1.0 1.5/0.0 1.5/0.0 1.5/0.0 1.5/0.0

y-y – axis of the cross-section in new norm, x-x according to old norm z-z – axis of the cross-section in new norm, y-y according to old norm

Rolled-steel (I, Z, U, V) shapes different cross-section steel profiles The reduction coefficient varies according to the load types as follows: Load Snow Wind Imposed

ψ 0.6 0.6 0.7

In addition, the partial safety coefficients of resistance are taken into account in the design of sheets as follows: M1= 1.1 And the design of struts: M2= 1.25

Axis convention and symbols for principal dimensions according to old standard

Limit states - examples Limit states - examples

Classification of cross-sections

Laterally stable steel beams can fail only by Flexure, Shear or Bearing, assuming the local buckling of slender components does not occur. These three conditions are the criteria for limit state design of steel beams. Steel beams would also become unserviceable due to excessive deflection and this is classified as a limit state of serviceability. The factored design moment, M at any section, in a beam due to external actions shall satisfy M < Md Cross-sections are classified in four classes. The same structure may have components in different classes. The various elements (such as flanges and webs) of the same cross-section may also have different classes. The class of the cross-section depends on the proportions of the compression elements of the cross-section and on the stress state. The class of the cross-

Dimensions and axes of sections Composite beam according to old standard

section can be different in bending and in compression. Class 1 cross-section: A plastic hinge can form in the cross-section with the rotation capacity

Sabah Shawkat © required for plastic analysis.

Class 2 cross-section: A plastic hinge can form in the cross-section, but without enough rotation capacity for plastic analysis.

Class 3 cross-section: The compression stress can reach yielding strength in some element of

the cross-section. The cross section buckles locally before the internal moment has developed to the value of the plastic moment resistance of the cross-section.

Class 4 cross-section: The cross-section buckles locally before the greatest compression stress

in some element has reached the value of the yield strength. The whole cross-section is normally classified on the basis of the compression element in the least favourable class. The internal forces, moments and resistances can in all classes be calculated using the theory of elasticity, if the effect of local buckling on the resistance of the cross-section is taken into account. The theory of plasticity can be used to calculate the internal forces and moments in Class 1 and resistances in Class 1 and class 2. In practice, the internal forces and moments can be calculated, for the sake of simplicity, according to the least favourable class. For Class 4 profiles, the calculation of the resistance in combined bending and compression is based on the effective cross-section. The resistances in of the cross-section are then calculated

Different cross-section steel profiles

on the basis of the areas of the effective elements only.

Where x-x = y-y and y-y = z-z according to new code

Limit states - examples Limit states - examples

When the beam is adequately supported against lateral buckling, the beam failure occurs by yielding of the material at the point of maximum moment. The beam is thus capable of reaching its plastic moment capacity under the applied loads. Thus the design strength is governed by yield stress and the beam is classified as laterally supported beam. Beams have much greater strength and stiffness while bending about the major axis. Unless they are braced against lateral deflection and twisting, they are vulnerable to failure by lateral torsional buckling prior to the attainment of their full in plane plastic moment capacity. Such beams are classified as laterally supported beam. When the lateral support to the compression flange is adequate, the lateral buckling of the beam is prevented and the section flexural strength of the beam can be developed. The strength of I-sections depends upon the width to thickness ratio of the compression flange. When the width to thickness ratio is sufficiently small, the beam can be fully plastified and reach the plastic moment, such section is classified as compact sections. However, provided the section can also sustain the moment during the additional plastic hinge rotation till the failure mechanism is formed. Such sections are referred to as plastic sections.

Sabah Shawkat © When the compression flange width to thickness ratio is larger, the compression flange may buckle locally before the complete plasticisation of the section occurs and the plastic moment is reached. Such sections are referred to as non-compact sections. When the width to thickness

ratio of the compression flange is sufficiently large, local buckling of compression flange may occur even before extreme fibre yields. Such sections are referred to as slender sections.

The flexural behaviour of such beams is presented in Figure bellow. The section classified as

slender cannot attain the first yield moment, because of a premature local buckling of the web or flange. The next curve represents the beam classified as 'semi-compact' in which, extreme fibre stress in the beam attains yield stress but the beam may fail by local buckling before further plastic redistribution of stress can take place towards the neutral axis of the beam.

The stress distribution in a double – symmetric hybrid beam with Class 1,2,3,4

Flexural member performance using section classification

Limit states - examples Limit states - examples

Limit values for hollow cross-section classes 1,2,3 Cross-section class

Loading method

Crosssection element

355

h/t b/t

36,6

fy N/mm2

Compression

Web and Flange

Bending

Flange

b/t

29,3

33,4

36,6

Bending

Web

h/t

61,1

70,0

103,3

Sabah Shawkat © Compression and / or bending

Entire crosssection

d/t

33,1

46,3

59,6

fy is the yield strength of steel.

For other steel grades, the values in column 235 N/mm2 are multiplied by correction factor  when using square and rectangular hollow sections and by correction factor 2 when using circular hollo sections 

235 MPa fy



235 MPa fy

The design of a hollow section member or framework is easy and quick: the simple geometry can be expressed with few parameters, which makes computer-aided design a feasible option. The weight, resistance and stiffness of structures can be optimized by modifying the wall thickness, without needing to change the external dimensions of the hollow section or the geometry of the structure. Another application for hollow sections is in composite structures, when using a concrete-filled composite column, the properties of steel and concrete can be efficiently utilized, under normal load and in fire situation. Hollow sections can easily be formed into light-weight and stiff frames grid structures, since their torsional stiffness and bending resistance in all directions is high. The torsional stiffness of hollow sections can be utilized also in various console structures and structures with projecting sections.

Design of tables and chairs using hollow steel pipes

Limit states - examples Limit states - examples

Limit of classes for double-symmetric and box profiles Cross-section class

Class of cross-section

Stress state

Compression

Bending

Crosssection element

Web and Flange

Flange

460

bw / tw

23,6

27,2

c / tf

6,4

7,1

b/tf

7,1

fy N/mm2

6,4

c / tf

Sabah Shawkat © b/tf

23,6

27,2

Bending

Web

bw / tw

51,5

100

88,6

Compression and / or bending

Entire crosssection

d/t

33,1

46,3

59,6

fy is the yield strength of steel For other steel grades, the values are calculated according to Table 5.3.1 in EC3 Part 1-1.

The tangential stress reached 0.65 e

The I-profile is the most common and most recommended profile-type. It is normally used in different beams and girders in houses and in industrial buildings. If it has wide flanges, it is also suitable for column. An asymmetric I-profile has applications in crane runway girders, bridge girders of composite structures. The profile is also commonly used in mechanical engineering applications. The box profile is used both as column and beam. It is possible to get the buckling resistance equal with respect to both principal axes. Box columns find their use as heavily loaded columns. If the cross-section is subject to large torque. It is advantageous to choose a box section because the torsional stiffness of a box cross-section is many times greater that of an I-cross-section. Welded I profile

Limit states - examples Limit states - examples

Bending resistance of class 4 circular hollow sections:

is the cross-sectional area of the tension flange

The buckling stress must be calculated for circular hollow sections with class 4 cross-sections.

Afnet

is the net cross-sectional area of the tension flange

is the ultimate strength of the material

 M0

is the partial safety factor for the material

 M2

is the partial safety factor for the net effective cross-section

The design criterion is that the bending moment due to loading is smaller than the bending resistance of the hollow section. Msd  McRd

Where is the design value for the bending moment

Msd  u

Mcd

Wel

Design value for bending resistance

 M1

is the buckling stress of the hollow section u The buckling stress of a circular hollow section is calculated as follows, when 



u

b

Sabah Shawkat ©

1  0 4123 (  ) 1 2 fy d t



 cr

 b  cr

0 605 E

The radius of the wall central axis

is the reduction factor for buckling stress, which is calculated as follows, when  212

b

0 1887 

0 6734

1  0 01 

r t

For the hollow sections for which

 r  212  and or   t 





Effect of holes on bending resistance: The effect of holes need to be taken into account in a flange subjected to tension when the following criterion is satisfied.

0 9 

Afnet Af



fy  M2  fu  M0 Design of a playground using hollow steel pipes

Limit states - examples Limit states - examples

If the criterion is not satisfied, the cross-sectional area of the tension flange assumed in design,

The design criterion for member subjected to bending about one axis is:

must be reduced to such an extent then the criterion is satisfied. This reduced cross-sectional

Msd  McRd Where

area of the tension flange is then used to calculate the bending resistance. The effect of the holes in the tension area of the web need not be considered, if the criterion is met in the entire tension area. the tension area consists of the tension flange and the tension element of the web. In the

Msd

is the design value for bending moment

McRd

is the design value for bending resistance

compression area, the effect of the holes need not be considered, unless the bolt holes are oversize or slotted.

The bending resistance for hollow sections of different classes of cross-section is calculated as

Calculating the effective cross-section:

follows:

The width reduction factor ρ for class 4 cross-sections of square and rectangular hollow sections is calculated as follows: 

where

1 p



p

 0 673

where

p

MplRd

Wpl 

McRd

MelRd

Wel 

 0 673

 0 22 p

McRd

fy  M0

Class 1 and 2 cross-sections

Class 3 cross-sections

Sabah Shawkat © The slenderness of the flange or web subjected to uniform compression can be determined using

McRd

MeffRd

the following formula.

Weff 

 M0

Class 4 cross-sections, for square and rectangular

p

 cr

56 8 

is the thickness of the hollow section wall

 cr

is the buckling stress

b  3 t

flange design width or

h  3 t

web design height 

235 fy

is the nominal yield strength of the material

Resistance of hollow sections subjected to bending moment: Hollow sections are efficient when subjected to bending about one or both principle axis, in addition, the buckling resistance about the minor axis is superior to an equivalent weight I or H profile section and therefore lateral restraints can be placed at greater spacing. Truss steel frames, using square hollow sections

Limit states - examples Limit states - examples

Consider a structural hollow section with dimensions 145x145x5. The steel grade used is

Internal nominal corner radius

S355J2H.

External nominal corner radius

 b  145 mm  h  145 mm  fy  355 MPa

 M0

ri  5mm

t  5 mm

r0  10mm

 1.1

5mm 10mm

By inserting the hollow section dimensions 145x145x5 in the formula above we obtain the following value for section modulus: The plastic section modulus for the cross-section is determined by the following equation: Wpl 

b h ( b  2 t ) ( h  2 t )   4 Az hz  4A h 4 4

Wpl  142455.382 mm

The bending resistance of the hollow section with dimensions 160x160x5 is thus:

M plRd  Wpl 

Flange: b

 29 t Web: h t



fy  M0

Sabah Shawkat © b

compression

bending

29  29.3

compression

29  61.1( bending)

Where the terms induced by corner rounding are:

 



 



Az   1  A   1  h 

MplRd  45.974 m kN

 r 2  i

4

h  2 t 2

 r 2  0

Az  21.46018 mm

4

 10  3   12  3 



h  0.06638 m

 A 5.36505 mm

 r  i 

 hz 

h 2

 10  3   12  3 



 r  0 

hz  0.07027 m

Limit states - examples Limit states - examples

Consider a structural hollow section with dimensions 180x180x5. The steel grade used is S355J2H.  b  180 mm  h  180 mm

 fy  355 MPa

 M0

ri  5mm

t  5 mm

1 1     r 4  0  3 16 3 ( 12  3  ) 

(compression)

Izz  75.45116 mm

1 1     r 4  i  3 16 3 ( 12  3  ) 

 1.1

I  

Flange:

b  36 t

Izz  

r0  10mm

b t

36  36.6

( compression)

 I 4.7157 mm

Internal nominal corner radius

5mm

External nominal corner radius

10mm

By inserting the hollow section dimensions 180x180x5 in the formula above we obtain the Web: h t

following value for section modulus: 

( bending)

36  103.3( bending)

The plastic section modulus for the cross-section is determined by the following equation: Wpl 

b h ( b  2 t ) ( h  2 t )   4 Az hz  4A h 4 4

Wpl  224016.223 mm

Sabah Shawkat ©

 b h 3

Wel  

 12

3  2 ( b  2 t ) ( h  2 t ) 2 2  4  Az hz  Izz  4 A h  I   12  h





 



Wel  192985.121 mm Wpl Wel

 1.1608

The bending resistance of the hollow section with dimensions 180x180x5 is thus: M plRd  Wpl 

Where the terms induced by corner rounding are:

 

Az   1 



 

 r 2  0

A   1 

4

 hz 

 r 2  i

4

 r  0 

h  2 t  10  3   h    ri 2  12  3  

180x180x5: M elRd  Wel 

hz  0.08777 m MplRd

h  0.08388 m

MplRd  72.296 m kN

The following bending resistance is obtained for a hollow section with dimensions

 A 5.36505 mm

 Az 21.46018 mm

h  10  3   2  12  3 



fy  M0

MelRd

fy  M0

 1.1608

Limit states - examples Limit states - examples

MelRd  62.282 m kN

Consider a structural hollow section with dimensions 200x200x8. The steel grade used is S355J2H.

 b  200 mm  h  200 mm fy  355 MPa 

 M0

t  8 mm

ri  5mm

r0  10mm

1 4  4 1      r0 Izz 75.45116mm  3 16 3 ( 12  3  ) 

Izz  

 1.1

1 4  4 I 1    4.7157 mm  r  i  3 16 3 ( 12  3  ) 

I  

internal nominal corner radius

5mm

external nominal corner radius r0 10mm By inserting the hollow section dimensions 200x200x8 in the formula above we obtain the

Flange: b t

Web:

h t

 25

(Compression)

 25

(Bending)

following value for section modulus:

25  36.6

t h

25  103.3

(Compression)

(Bending)

The plastic section modulus for the cross-section is determined by the following equation:

Wpl 

b h ( b  2 t ) ( h  2 t )   4 Az hz  4A h 4 4

 b h 3

Wel  

 12



( b  2 t ) ( h  2 t ) 12



Wpl  436182.037mm

 

 2h

 4  Az hz  Izz  4 A h  I  

Sabah Shawkat © 3

Wel  371707.573mm W pl W el

 1.17345

The bending resistance of the hollow section with dimensions 200x200x8 is thus:

M plRd  Wpl  Where the terms induced by corner rounding are:

 

Az   1 

 r 2  0 4

 



A   1 

2  2 Az 21.46018 mm  ri  4



h 

  r0 

hz  0.09777 m

 10  3   r  i  12  3  

h  0.09088 m

 10  3   12  3 



h  2 t 2



200x200x8:

M elRd  Wel 

A 5.36505mm  h

MplRd  140.768 m kN

The following bending resistance is obtained for a hollow section with dimensions

 hz 

 M0

fy  M0

MelRd  119.96 m kN

MplRd MelRd

 1.17345

Benefits of hollow sections for the design of new architecturally small cell spaces. We can design the containers using square or rectangular hollow sections in creative way for the series of prototype production for people living in festivals, at stadiums such as cabinets for athletes, cabinets for administrative offices, or on site for designers like the office.

Limit states - examples Limit states - examples

Square hollow section with class 4 cross-section: Consider a hollow section with dimensions 220x220x6. The steel grade used is S355J2H.

 b  220 mm  h  220 mm  fy  355 MPa

 M0

t  6 mm

ri  5mm

r0  10mm

 1.1

Flange: b  36.66667 Compression t

Sabah Shawkat © b t

36.66  36.6 Compression Class 4 cross-section

Web: h h  36.66667 Bending 36.66  103.3 Compression Class 3 cross-section t t As the compression flange belongs to class 4, its effective width must be determined. The slenderness of the flange is obtained from the formula: p





 cr

235 MPa fy



 0.81362

is the buckling stress

is the thickness of the hollow section

is the nominal yield strength of the material

b  3t

 b1  b  3t

Container home prototype

Limit states - examples Limit states - examples

flange design width or web design height

h  3t

The width reduction factor ρ for class 4 cross-section of square and rectangular hollow sections is calculated as follows: 

When

p

 0.673

p

 0.673

when



p

 0.22 p

p

 cr



56.8 

 p  0.7285  p  0.673 Now, the effective width of the flange can be calculated from the formula:

beff 

p

 0.22 p

( b  3 t)

beff  0.19354 m

The neutral axis of the effective cross-section is transformed downwards. The effective section modulus of the cross-section is calculated by subtracting the section modulus of the non-

Sabah Shawkat ©

effective element from the section modulus of the entire cross-section. The effective section modulus for a hollow section with dimensions 220x220x6 is obtained as follows:  bnoneff  b  3 t  beff

bnoneff  8.45506 mm

non effective element of the compression web

I  3813 10 mm

A  50.43 10 mm

is the area of the entire cross-section

is the second moment of area of the entire cross-section

To obtain the bending moment resistance value, the effective section modulus is multiplied by the yield strength. bnoneff t ( 0.5 h  0.5 t )

  1.08731 mm A  bnoneff t Transfer of the neutral axis of the cross-section. 



Weff 

I  A 

M  effRd  Weff 

 bnoneff t ( 0.5 h  0.5 t   ) 0.5 h  

fy  M0

Weff  337961.962mm MeffRd  109.07 m kN

Design of structure using hollow rectangular sections

Limit states - examples Limit states - examples

Calculate the compression resistance of a hollow section with dimensions 300x300x6. The steel grade used is S355J2H, and the buckling length is 6m. The member is nominally pinned at both ends.

Determine the classification of the cross-section of the hollow section:

 b  300 mm

 t  6 mm 2

A  69.63 10 mm

b  50 t

L  c  6 m

b  36.6 t

 0.49

fy  355 MPa



i  11.96 10 mm

E  210 GPa

 M1

 1.1

Class4

As the cross-section of the hollow section is class 4, the effective cross-section must be determined. The

Sabah Shawkat © slenderness of the compression elements is calculated using the formula

The width reduction factor ρ for class 4 cross-section of square and rectangular hollow sections is calculated as follows:



235 MPa





p



  0.22

p

  p

b t t

p

56.8 



When

  0.673

p

  0.673

p

When

2



 0.81362

 1.0603

The dimension of the effective and no-effective elements of the cross-section are as follows:

The designed steel structure supports the entire roof structure of the building using rectangular hollow section frames.

beff 

p

 0.22 p

( b  3 t )

 bnoneff  b  3 t  beff

Limit states - examples Limit states - examples

beff  0.21078 m bnoneff  0.07122 m

heff  beff

heff  0.21078 m

Using the effective cross-section, let as determine the effective area and parameter β .A:  A eff  A   4 bnoneff t 

Aeff  0.00525 m

by nominally pinned connections at both ends. The steel grade used is S355J2H, and the loading

Aeff

 A  0.75451 A The local buckling of the cross-section has now been taken into account. Next, consider the A



Calculate the resistance of a hollow section with dimensions 300x300x6 to the loading shown in the adjacent figure. The buckling length of the structure is 6m, and the member is supported

buckling resistance of the hollow section. The cross-section slenderness is determined using

values are: N sd  750 kN 

Mysd  25 kN m

Mzsd  25 kN m

 Mtsd  7.5 kN m

the formula.

 b  300 mm   

Lc i 



fy E

 A



h  300 mm 2

 0.57031

i  11.96 10 mm The reduction factor for buckling is calculated from:

 

   0.5  1   (   0.2)    2

 0.75335

Wt  996.8 10 mm 3

Wpl  764.2 10 mm 

t  6 mm

A  69.63 10 mm

 E  210 GPa

I  9964 10 mm

Lc  6 m

 1.1

 M0

 M1

Wel  664.4 10 mm

 1.1

 fy  355 MPa 

 0.49

Sabah Shawkat © 1

  0.80284 2 2     The buckling resistance of the hollow section is calculated by multiplying the plastic 





 

compression resistance of the effective cross-section by the reduction factor.  N bRd   Aeff 

 M1

N bRd  1361.22371 kN

The Moment Mtsd assumed constant along the entire hollow section.

b  50 t

since

b t

50  36.6

The hollow section is classified as Class 4

As the cross-section of the hollow section is class 4, the effective cross-section must be determined. The slenderness of the compression elements is calculated using the formula

Limit states - examples Limit states - examples

The width reduction factor ρ for class 4 cross-section of square and rectangular hollow sections

The buckling resistance of the hollow section is calculated by multiplying the plastic

is calculated as follows:

compression resistance of the effective cross-section by the reduction factor.



235 MPa





  p  0.22



 0.81362

 When  p  0.673

p

56.8 

 1.0603

 0.22

p

( b  3 t )

 3 

The dimension of the effective and no-effective elements of the cross-section are as follows:

beff 

NbRd4  1361.22371 kN

a hollow section subjected to compression is calculated using buckling curve C.



fy  M1

must be accounted for in the interaction expression. The reduction factor Χ for the buckling of

b t p

 N bRd4   4 Aeff 

But if the bending moment must be determined using elasticity theory. The effect of torsion

  0.673

p

When

2   p



beff  0.21078 m

heff  beff

heff  0.21078 m

i 



fy E

 

2    3  0.5  1     3  0.2   3  3

 3

3

 2

3

 0.65656

3

 0.82739

3

 0.75138

 3

Sabah Shawkat ©  bnoneff  b  3 t  beff

bnoneff  0.07122 m

Determine the compression and bending resistance of the hollow section:

Using the effective cross-section, let as determine the effective area and parameter βA: A  eff  A   4 bnoneff t 

A



Aeff

 A  0.75451 A The local buckling of the cross-section has now been taken into account. Next, consider the

Aeff  0.00525 m

buckling resistance of the hollow section. The cross-section slenderness is determined using the formula.

 4 

Lc i 



fy E

 A

4

 

1 4

N bRd  1688.45828 kN

MyRd  214.42 m kN

 M1

MzRd  MyRd

MzRd  214.42 m kN

as follows:

2    4  0.5  1     4  0.2   4 



M  yRd  Wel 

 M1

The effect of the calculation methods of class 3 and class 4 on this example on the finale results

 0.57031

The reduction factor for buckling is calculated from:

4

 N bRd   3 A 

4

 2

4

 0.75335

4

 0.80284

3 4

 1.15124

N bRd N bRd4

 1.2404

 4

Limit states - examples Limit states - examples

3 4

 1.09828

3 4

 0.9359

Hollow sections subjected to bending moment and axial force (buckling):

 2  Mz  4 

z

 z

y

y

z

 z

Wplz  Welz

The classification of webs subjected to bending and compressions depends on the stress distribution. In practice, the cross-section is more easily determined by the compression element (web or flange). The interaction expression for a structure subjected to compression and bending is as follows: N Sd N bRd

Where

N bRd



ky MySd



MyRd

 min A 

kz MzSd MzRd



Mtsd MtplRd

 1

Wply 

M zRd

 2  My  4  0 9

Class 3 and 4 cross-sections

 2  Mz  4  0 9

Class 3 and 4 cross-sections

is the slenderness determined by the y axis

z

is the slenderness determined by the z axis

MzRd

Wplz

MyRd

Wely 

for class 1 and 2 cross-sections

 1.3

  y   3  2  My  4



ky   1 

 y N sd

y

  

 0.91918

ky  1.37118

ky  1 5

Sabah Shawkat ©  M1

 M1



 Mz

M zRd

Class 1 and 2 cross-sections

The parameters μ and Χ depending on the shape of the moment diagram are as follows:

 M1

is the minimum value of the reduction factor for buckling (about the y or z axis)

MyRd

 0 9

y

 My  min

Welz

W elz 

 M1

W effz 

for class 3 and 4 cross-sections

 M1

Weffy 

MyRd

 M1

 3 A fy

 1.4

  z   3  2  Mz  4

 kz  1 

 M1

z

 zN sd

 0.78787

kz  1.31815

 3 A fy

kz  1 5

for class 4 square and rectangular hollow sections

The calculating method for torsional resistance is determined by the web slenderness:

1

 y N sd  y A fy

 1 5

1

 zN sd  zA fy

h h calculate the plastic torsion resistance  50  59 1 t t Plastic torsional resistance is calculated using the formula:

 1 5

Where y

is the reduction factor for buckling determined about the y-axis

z

is the reduction factor for buckling determined about the z-axis

Wt MtplRd  185.73043 m kN  3  M0 By adding the effect of torsion in the interaction expression, the following result is obtained: M tplRd 

N sd y

y

 2  My  4 

Wply  Wely Wely

 0 9

N bRd



Class 1 and 2 cross-sections N sd N bRd



ky Mysd MyRd ky Mysd MyRd



kz Mzsd MzRd kz Mzsd MzRd

Limit states - examples Limit states - examples



Mtsd MtplRd Mtsd MtplRd

 1

 0.79813

Calculate the resistance of a circular hollow section with dimensions 323.9x12 to the combined loading shown in the adjacent figure. The steel grade used is S355J2H, and the hollow section length is 8m. The moment is assumed constant along the hollow section length. The hollow section is supported by hinges at both ends. L  c  8 m

 R  323.2 mm

r 

R 2

 0.49

 M1

 1.1

Wel  884.2 10 mm

t  12 mm

fy  355 MPa  E  210 GPa 



Wt  1768 10 mm

Wpl  1168 10 mm

A  117.6 10 mm I  14847 10 mm

r  0.1616m

i  11.04 10 mm

Sabah Shawkat © The loading values are:  N sd  750 kN

Compression

 Mysd  35 kN m

Mzsd  35 kN m

 Mtsd  21 kN m

First, the buckling resistance of the hollow section is calculated. Obtain the following value for the parameter α 0 in compression only. 0



0.83 1  0.01 

Elevation and detachment details of the shelter made from Hollow section steel profiles

Limit states - examples Limit states - examples

r t

0

 0.77919

The buckling stress σ in compression only:

  cr  0.605 E

u

 cr

1

 0 cr



1.2

 1  0.4123  1

fy  fy   1  0.222    cr  3  The torsional buckling resistance

 9434.40594 MPa



1

The shear buckling stress is obtained from the:   ba 

 0.21975

M tbRd   ba 

 fy

u

 331.24 MPa

u





 









 0.58998

VbaRd  1031.62 kN

the effect of torsion the following result is obtained:

N sd N bRd



ky Mysd McRd



kz Mzsd McRd



Mtsd MtbRd

 0.83349

Sabah Shawkat © 1



 1.09505

t  M1

The resistance values calculated can be inserted in the interaction expression. Allowing for

 0.91608



MtbRd  299.39 m kN

 M1

 VbaRd   ba  r 

E  i The buckling resistance of the compression member is derived from:

2    0.5  1   (   0.2)   

 186.27 MPa

The resistance to shear buckling is calculated using the shear buckling stress:

The slenderness is obtained for buckling by inserting the buckling stress value.   

 ba



 2

N  bRd   A 

 

N sd

N bRd

u



ky Mysd McRd



kz Mzsd McRd



Mtsd

MtbRd

 1.0

N bRd  2089.295 kN

 M1

The equivalent uniform moment factor for constant moment is  M   1.1 Parameter μ and χ are derived from the formula   y 



 2  M



 4

y

 1.64895

z

  y

z

 1.64895

 y N sd

ky  1.53812 kz  ky kz  1.53812 A   u The bending resistance for hollow section with dimensions 323.9x5 was calculated in example:  ky  1 



M  cRd  Wel 

u

McRd  266.26038 m kN

 M1

First determine the theoretical shear buckling stress from:

 cr

 cr

 0.747 E

t t   Lc  r 

0.75  cr

 864.25 MPa

0.444 fy  157.62 MPa Circular steel stairs using circular hollow section

 0.444 fy

Limit states - examples Limit states - examples

Calculate the tension resistance of the adjacent flange plate joint which is subjected to axial

bred  b  0.5d  t0

force Nsd. The hollow section dimensions are 150x150x10, and the steel grade used is S355J2H. The steel grade used in the flange is S355J2.

K 

The strength grade of the M30 bolts is 10.9. the flange hole position parameters are:  Nsd  1700 kN  a1  150 mm  Mb

 1.25

 d  30 mm

d0  30 mm

n  8

 b1  150 mm

t0  10 mm

 M0

 fy  355 MPa

pmin  3 d0

fub  1000 MPa

pmin  0.09 m

 amin  1.5 d0

amin  0.045 m

amax  150 mm

 4bred  M0  0.9fy p

1

K  0.0000088kN m

tp  16mm

a red  a  0.5d

 fy  345MPa

ared  112.5mm

 1.1

The axial force in one bolt is:

 fu  490 MPa

pmax  200 mm

bred  0.0925m

p 

a 

Ntsd 

pmin  pmax 2

t2 

amin  amax

Nsd 6

K Ntsd

N tsd  283.33333 kN

t1 

t2  49.89151 mm

tp 

K Ntsd 1 

t1  t2 2

t1  37.25838 mm

tp  43.57494 mm

Sabah Shawkat © a  0.0975 m 2 

 A d 

b  a

A  0.00071 m

b  0.0975 m

Ar  0.75 A

The tension resistance of the bolt is:

 p 145 mm

Ar  530.14376 mm

 BtRd  0.9 fub 

 Mb

BtRd  381.70351 kN

is the tension cross-section of the bolt

fub

is the ultimate strength of the bolt

 Mb

is the partial safety factor for bolt joints

Flange resistance: The minimum and maximum values for the flange thickness are obtained by the formula.

K Ntsd 1

  

 d0    p 

1  

 tp 

K Ntsd

Select a flange thickness of 

 0.7931

Limit states - examples Limit states - examples

K Ntsd  37.25838 mm  K Ntsd 49.89151mm 1  tp  43.57494 mm

The joint resistance is determined using the formula: h

  

K BtRd

 tp2 

  a  0.5 d       a  b  t0  

 1 

tp  1    h  n

h

Flange plate joint in circular hollow sections: The required flange thickness for a circular hollow section is calculated from the following

 0.53008

formula:

N Rd 

N Rd  2455.95 kN

N sd  1700 kN

d 0circl  150 mm

e1min  1.2 d0

e1max  0.12 m

e1 

N Rd  N sd OK

e1min  36 mm

e1min  e1max 2

e1max  12 t0

e1  78 mm

Resistance of bolts: d0circl  t0

The resistance of a fillet weld is calculated as shown in EC 3. w

 0.9

 Mw

d0circl  t0  2 e1

 1.25

is the length of the weld

 Lw  2 a1  2 b1

from the graph we obtain f3  3.8

 0.47297

tp 

2 N sd  M0

tp  30.13419 mm

fy  f3

Lw  0.6 m

Sabah Shawkat ©

a w

is the ultimate strength of the weaker joint component

is the shape coefficient of the flange

is the thickness of the weld throat

Nsd

is the design value for the tensile force of the joint

is the partial safety factor (S355 β w = 0.9)

is the yield strength of the flange

is the partial safety factor of the welded joints

 Mw

for the number of bolts, we obtain the following equation:

In the example the hollow section is welded to the flange from all edges. In such cases the required throat thickness is: a 

3  w  Mw N sd

FwRd 

fu Lw fu a Lw 3  w  Mw

 r1  0.5 d0  2 e1

a  11.26717mm



Nsd  1  FwRd  1700 kN

n 

 

1 MPa  fy 0.67 BtRd

The joint resistance is sufficient for an axial force of FwRd , which is also the plastic tension resistance of the 150x150x10 hollow section.

Limit states - examples Limit states - examples

r2  0.5 d0  e1

  r1   f3 ln    r2   1

n  9.50019

we select n=11

Centre of gravity y0:

Dual profiles Calculate the steel stress of the dual profiles shown on figure bellow. To receive the bending moment Mx we determine the moment of inertia of the whole complex cross-section, while for the bending moment My we consider only the part above the middle, because we cannot expect a more significant participation of the lower flange.

H  t  1u  2 

Au eu  AI  y0  a 

We chosen 1 I400 and 1 U300 to composite the dual cross-section with cross-section areas AI,

y0  14.914 cm

Au  AI H  r1u  e 2

e 

 t1u  y0 e  6.0862 cm

a  0.12314 m

Au.

The total moment of inertia of the dual steel cross-section:

To determine the moment of inertia Ix we must first find the centre of gravity.

Ix  IxI  AI e  Iyu  Au a

Data: Bending moment Mx, and My  Mx  210 kN m

My  9.6 kN m

Wx12 

I400 3

Ix  42879.76062 cm

Wx12  2875.1732 cm

 H  t  e   1u 2 

Sabah Shawkat © AI  11.8 10 mm

 IxI  291 10 mm

IyI  11.4 10 mm

U300

 Au  5880 mm  eu  27.0 mm

Ixu  80.3 10 mm

t1u  10.0 mm

r1u  16.0 mm

Iyu  4930 10 mm

Wx34 

 hu  300 mm

Wx34  1643.77188 cm

 H  e   2 

 Iy  Ixu 

IyI

Iy  8600 cm

Wy12 

 hu    2

Maximum stress we obtain as follows:  1 

Dual profiles as a crane beam

Mx Wx12 Mx

3



4

  3

Wx34

Limit states - examples Limit states - examples



My Wy12

1

 89.78326 MPa

3

 127.75495 MPa

4

 127.75495 MPa

Wy12  573.33333 cm

Sabah Shawkat ©

Limit states - examples

Sabah Shawkat © Structural actions Structural actions Structural actions Structural actions Self-weight and imposed loads Self-weight and imposed loads Imposed Load Reduction Imposed Load Reduction Snow load Snow load Wind load Wind load Designing gable beam in the model building Designing gable beam in the model building Calculate the deflection of the steel I beam Calculate the deflection of the steel I beam Designing the stiffening elements in the model building Designing the stiffening elements in the model building

Sabah Shawkat ©

4040

Structural actions Structural actions

Variable actions include: Variable actions include:

actions which structure subjected divided permanent, variable TheThe actions to to which thethe structure is is subjected areare divided intointo permanent, variable andand

Imposed loads a) a)Imposed loads

accidental actions. A design value is obtained multiplying characteristic value of the load accidental actions. A design value is obtained by by multiplying thethe characteristic value of the load

Snow load b) b)Snow load

partial safety factor. Permanent actions include: by by thethe partial safety factor. Permanent actions include:

Wind load c) c)Wind load

Self-weight of the structure a) a)Self-weight of the structure

Self-weight imposed loads Self-weight andand imposed loads

Fixed equipment b) b)Fixed equipment

weight of the partition walls distributed to generate a uniform load. When TheThe weight of the partition walls cancan be be distributed to generate a uniform load. When designing structural floor elements a single storey building. load must taken designing thethe structural floor elements of aofsingle storey building. TheThe load must be be taken in in to account in the weakest area. effect of concentrate load must considered separately. to account in the weakest area. TheThe effect of concentrate load must be be considered separately.

Partial safety factors actions Partial safety factors forfor actions Accompanying Accompanying

Permanent actions Permanent actions

variable action variable action

(G1()G1)

Variable actions Variable actions (Q( ) Q) Leading variable Leading variable

Accompanying Accompanying

action action Q1Q1

variable action variable action Q2Q2

In some cases, imposed loads structural floor a single storey building In some cases, thethe imposed loads on on thethe structural floor in ainsingle storey building cancan be be reduced. reduced. When designing column, loads storeys assumed uniformly distributed. Also in the When designing column, thethe loads on on storeys areare assumed uniformly distributed. Also in the case of multi-storey buildings, imposed loads sometimes reduced. case of multi-storey buildings, imposed loads cancan sometimes be be reduced.

Favourable effect Favourable effect

1,01,0

- -

Unfavourable effect Unfavourable effect

1,35 1,35

1,51,5

characteristic values impose load areas residential, social, commercial TheThe characteristic values of of thethe impose load forfor areas in in residential, social, commercial andand

Fatigue-inducing Fatigue-inducing

1,01,0

administration buildings provided in table below divided categories according to their administration buildings areare provided in table below divided intointo categories according to their

Fire design Fire design

- -

Serviceability limit Serviceability limit

1,01,0

1,01,0 

Sabah Shawkat © action action

specific uses. specific uses.

Building occupancy categories: Building occupancy categories:

state state

Residential (including hospital wards, hotel bedrooms etc.) Residential (including hospital wards, hotel bedrooms etc.)

partial safety factors presented in this table basic values of EC3. TheThe partial safety factors presented in this table areare thethe basic values of EC3.

Office areas B B Office areas weight fixed partitions may considered self-weight represented TheThe weight of of fixed partitions may be be considered as as self-weight andand represented as as an an

Assembly areas (subdivided 5 sections depending likely areas (subdivided intointo 5 sections depending on on likely C C Assembly

equivalent uniformly distributed load.NoNo guidance is given required magnitude equivalent uniformly distributed load. guidance is given as as to to thethe required magnitude of of such equivalent load intensity, designer should a reasonable approach in making such an an equivalent load intensity, andand thethe designer should useuse a reasonable approach in making 2 2 such estimate. However, a minimum value kN/m is typically used offices with is typically used forfor offices with such an an estimate. However, a minimum value of of 1,01,0 kN/m

normal weight partitions storey heights. normal weight partitions andand storey heights.

density of occupation crowding) density of occupation andand crowding) Shopping D D Shopping Storage areas E E Storage areas

Structuralactions actions Structural Structural Structuralactions actions

The characteristic values of the impose load for storage and industrial activities areas

Residential, social, commercial and administration areas: qk Category Areas for domestic and A

residential activities

kN/m

Qk 2

buildings are provided in table below divided into two categories.

[kN]

Areas for storage and industrial activities:

Rooms in residential buildings and houses;

bedrooms and wards in hospitals; bedrooms in hotels and hostels kitchens and toilets. E

Storage

E1: Areas susceptible to

& Industrial

accumulation of goods,

- Floors

2,0

- Stairs

3,0

2,0

including access areas

- Balconies

4,0

2,0

E2: Industrial use

Office areas - Office

3,0

2,0

Areas where

C1: Areas with tables, etc.

3,0

4,0

people may

e.g. areas in schools, cafés, restaurants, dining halls, reading rooms, receptions.

7,5

7,0

5,0

7,0

The characteristic values of the impose load for roofs categorised according to their accessibility are shown in table below.

Roofs:

Sabah Shawkat ©

congregate

Roofs not accessible except for normal maintenance and

4,0

> 40

conference rooms, lecture halls, assembly

5,0

Roofs accessible with occupancy according to categories A to

4,0

administration buildings, 5,0

7,0

0,0

1,5

According use

landing areas

The characteristic values of the line load qk acting at the height of the partition wall or parapets but not higher than 1,20 m should be taken from Table below.

e.g. dance halls, gymnastic rooms, stages. 5,0

4,0

Horizontal loads on partition walls and parapets:

buildings for public events like concert halls,

Loaded areas

sports halls including stands, terraces and

qk [kN/m]

access areas and railway platforms. Shopping areas

1,5

Roofs accessible for special services, such as helicopter

rooms, etc. and access areas in public and

C5: Areas susceptible to large crowds, e.g. in

0,75

people, e.g. areas in museums, exhibition

C4: Areas with possible physical activities,

(linear interpolation/ A=50x50mm)

halls, waiting rooms, railway waiting rooms. C3: Areas without obstacles for moving

[kN/m ] [kN]

< 20o

e.g. areas in churches, theatres or cinemas,

repair.

C2: Areas with fixed seats,

[kN/m2] [kN]

D1: Areas in general retail shops

5,0

4,0

D2: Areas in department stores

5,0

7,0

Structural actions Structural actions

0,5

B, C1

1,0

C2, C3, C4, D

1,5

3,0

4042

Imposedactions Load Reduction: For the design of columns or walls, loaded from several storeys, the Structural

Snow load Variable actions include:

total imposed loads on the floor of each storey should be assumed to be distributed uniformly. The actions to which the structure is subjected are divided into permanent, variable and For categories A to E loads from more than two stories may be reduced by applying a reduction accidental actions. A design value is obtained by multiplying the characteristic value of the load factor αn according to by the partial safety factor. Permanent actions include:

Those described in this chapter are not applicable for the following cases: a) Imposed loads  at more than 1500m above sea level b) Snow load  for impact load caused by the sliding snow from roof c) Wind load  for ice and snow loads caused by the blockage of gutter Self-weight and imposed loads  for areas where snow is present during the whole year

a) Self-weight of the structure b) Fixed equipment n is the number of storeys (> 2) above the loaded structural elements from the same category.

The weight theload partition walls can be distributed to generate a uniform load. When  forofice

ψ0 is coefficient according table below. For other categories and areas αn = 1,0 should be

designing the rain structural floor elements  for load falling on snow of a single storey building. The load must be taken in to account in the weakest area.ofThe effect of concentrate load must be considered separately. Determination of the value snow load:

applied. Partial safety factors for actions Coefficient ψ0 for buildings: Accompanying Permanent actions Action variable action (G1) Imposed Loads, category:

Variable actions (Q) ψ0 Accompanying Leading variable

1,5

variable action Q2 0.7 0.7 1,5

1,0

0.7

action Q1

A – domestic, residual Favourable effect 1,0 Unfavourable effect B – offices 1,35

= i*Ce*Cfloor t*sk In some cases, the imposed loads on thesstructural in a single storey building can be where reduced. i – form factor for snow load When designing column, the loads on storeys are assumed uniformly 2distributed. Also in the sk – characteristic value of snow load on the surface (kN/m ) case of multi-storey buildings, imposed loads can sometimes be reduced. Ce – reduction factor from the effect of the wind, generally 1,0

Sabah Shawkat © C – congregation Fatigue-inducing 1,0areas D – shopping action Fire design

state

0.7

E – storage-

H - roofs1,0

Serviceability limit

1,0

Snow Load

cat. B

qk2

cat. B

qk3

cat. D

qk4

cat. D

administration are provided in table below divided categories according to their The directionbuildings of the snow load is vertical and it refers to theinto horizontal plan of the roof. specific uses. The characteristic value of the snow load on the surface:

1.0

1,0

1,0 

0.6

The partial safetythe factors presented table are the basic values of EC3. Illustration of applying reduction factorinαthis n qk1

The characteristic values of the impose1,0 load for areas in residential, social, commercial and Ct – thermal factor, generally

sk = 0,25*(1+A/100) kN/m2

Building occupancy categories: where A is the height of the ground level above the sea in [m] Residential (including hospital wards, hotel bedrooms etc.) B Office Pitched roofs areas

S1 S2

The weight of fixed partitions may be considered as self-weight and represented as an S3

Form factors of the snow load of pitched roofs are contained by the following graph and table: Assembly areas (subdivided into 5 sections depending on likely C

q cat. E equivalent uniformly distributed load. No guidance is given as to the required magnitude of k5

qk6

density of occupation and crowding)

cat. F

such an equivalent load intensity, and the designer should use a reasonable approach in making 4 (S1+S2+S3+S4)

2 used for offices with such n,ani estimate. a minimum value of 1,0 Cat. However, qki Si ψ0 αn kN/m is typically Loads from n storeys normal weight partitions and storey heights. 1 B 3,0 S1 0.7 1.0 1.00 S1

3,0

0.7

1.0

1.00 (S1 + S2)

5,0

0.7

0.90

0.80 (S1 + S2 + S3)

5,0

0.7

0.85

0.85 (S1 + S2 + S3 +S4)

6,0

1.0

0.85 (S1 + S2 + S3 +S4) + S5

5,0

Shopping

Storage areas

0.85 (S1 + S2 + S3 +S4) + S5 +S6

Structuralactions actions Structural Structural Structuralactions actions

0° <  < 15°

15° <  < 30°

30° <  < 60°

 > 60°

1 form factor

0,8

0,8*(60-)/30

2 form factor

0,8

0,8+0,6*(-15)/30

1,1*(60-)/30

Roof slope

half-sided load arrangement that acts on the more disadvantageous side of the roof with a value of half of the full load intensity (ii)

Among the following load cases the worst case should be considered:

Joined pitched roofs: In case of joined pitched roofs the worse case of the uniform and asymmetrical snow load determined for the pitched roof, and the snow load calculated from snow accumulation given in the figure must be taken into account.

Sabah Shawkat © Single-pitched roofs: Form factor of the snow load of single-pitched roofs is contained by the following table. The

The form factor of the snow load must be determined separately, in case at least one of the

determination of the values is based upon the assumption, that the sliding of the snow from

roof surfaces joining to the valley of the joined pitched roofs has a slope greater than 60°.

the roof is not hindered. If there is a parapet wall, snow barrier or other element that stops the move of the snow at the lower edge of the roof, the form factor of the snow load must be at

Roof slope

0° <  < 15°

30° <  < 60°

least 0,8.

1 form factor

0,8

0,8*(60-)/30

2 form factor

0,8+0,8*/30

1,6

Roof slope

0° <  < 30°

30° <  < 60°

 > 60°

0,8

0,8*(60-)/30

1 form factor

Form factors of the snow load at sudden change of height of roof Two types of load arrangements must be taken into consideration: -

uniform load arrangement (i)

At sudden change of height of roof, the worse case of the uniform and asymmetrical snow load determined for the pitched roof, and the snow load calculated from snow accumulation given in the figure must be taken into account.

Structural actions Structural actions

4044

Note: Ifactions b2 < ls, the form factor on the edge of the lower roof can be determined with linear interpolation between Variable include:

Structural actions

1 and 2, and beyond the edge of the lower roof no snow load is to be assumed.

The actions to which the structure is subjected are divided into permanent, variable and accidental actions. A design value is obtained by multiplying the characteristic value of the load by the partial safety factor. Permanent actions include:

a) Imposed loads b) Snow load Snow accumulation behind protruding parts and obstacles c) Wind load The form factor of the snow load and the length of the snow accumulation must be

a) Self-weight of the structure

determined asimposed follows: loads Self-weight and

b) Fixed equipment

Leading variable

Accompanying

2 = *h/sk 1 = 0,8, The weight of the partition walls can be distributed to generate a uniform load. When with the following restriction: 0,8 < 2 < 2,0 designing the structural floor elements of a single storey building. The load must be taken in where to account in the weakest area. The effect of concentrate load must be considered separately.  is the bulk density of snow, it’s value can be 2 kN/m3 in this case In some cases, the imposed loads on the structural floor in a single storey building can be Length of the snow accumulation: ls = 2*h reduced. with the following restriction: 5 m < ls < 15 m

action Q1

variable action Q2

When designing column, the loads on storeys are assumed uniformly distributed. Also in the

Partial safety factors for actions Accompanying

Permanent actions

variable action

(G1)

Variable actions (Q)

Favourable effect 1,0 On roofs witheffect different levels snow Unfavourable 1,35 accumulation is realized 1,5 as a result of the effect 1,5 of the

case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © wind and the sliding of the snow snow load is Fatigue-inducing 1,0from higher roof parts. 1,0The form factor of the1,0 determined by the following formulas: action 1 =Fire 0,8 design (assuming that the smaller - roof is flat)

2 = s + w limit Serviceability where state s

1,0

1,0 

The characteristic values of the impose load for areas in residential, social, commercial and administration buildings are provided in table below divided into categories according to their specific uses.

Building occupancy categories:

Residential (including hospital wards, hotel bedrooms etc.)

formpartial factorsafety from factors the sliding of the snow The presented in this table are the basic values of EC3.

w

form factor from the effect of the wind

Determination: The weight of fixed partitions may be considered as self-weight and represented as an If  < 15°: s = 0 equivalent uniformly distributed load. No guidance is given as to the required magnitude of If  > 15°: s can be determined from the extra snow load given by the 50% of the greatest such an equivalent load intensity, and the designer should use a reasonable approach in making snow load determined for the pitched roofs of the neighboring higher roof. such an estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with w = weight (b1+b2)/2*h < *h/s normal partitions andk storey heights. with the following restriction:

0,8 < w < 4,0

where 

Office areas

Assembly areas (subdivided into 5 sections depending on likely C Wind load: density of occupation and crowding) Wind pressure calculation in case of simplified procedure (this refers to such structures which areDnot sensitive Shoppingto dynamic excitation and to maximum 200m high buildings): E Storage areas External pressure:

we = qref*ce(ze)*cpe

Internal pressure:

wi = qref*ce(zi)*cpi

is the bulk density of snow, it’s value can be 2 kN/m3 in this case

Length of the snow accumulation:

ls = 2*h

with the following restriction:

5 m < ls < 15 m

where cpe, and cpi are the external and internal pressure factors. The total wind pressure on the wall or structural element is the difference of the pressures acting on the two surfaces. Pressure values must be taken with signs. (Pressure towards the surface is regarded as positive and suction from the surface is regarded as negative.)

Structuralactions actions Structural Structural Structuralactions actions

The reference value of the dynamic pressure must be calculated with the following formula:

Determination of the external pressure factor:

qref = /2*vref2

In case of buildings and parts of buildings the value of the external pressure factor, cpe is the

(in N/m2)

function of the A loaded area. In the tables for the different cases, values for 1m2 and 10m2 (cpe,1 and cpe,10) are given. (The loaded area is the surface of the structure, which is considered

where

for the determination of the wind effect.) In case of different sizes of area, the external pressure factor, cpe can be obtained from the following graph.

- vref is the reference value of the wind speed which is 20 m/sec in Hungary -  is the density of the air, its value can be taken as 1,25 kg/m3 resulting qref = 0,25 kN/m2.

Determination of the location factor ce(z) With the location factor, ce(z) the roughness of the surface, the topography and the height above ground level can be taken into account. The EC distinguishes the following built up area categories: I. Open sea; at least 5 km long lake in the direction of the wind; uniform, plain land without obstacles

Sabah Shawkat ©

II. Agricultural area with fences, scattered with agricultural buildings, houses or trees III. Suburban or industrial zones; forests

IV. Urban zone where there are buildings on at least 15% of the land, with the average height

of 15m. In case of plane surface, the value of the location factor, ce(z) as a function of the

Note: The figure shows the following function:

height above the ground level, z and the built up area category can be obtained from the following graph:

z (m) 200

Expert's opinion

IV III II

100

cpe = cpe,1

if A < 1 m2;

cpe = cpe,1 + (cpe,10 - cpe,1)*log10 A

if 1 m2 < A < 10 m2;

cpe = cpe,10

if 10 m2 < A.

Both vertical walls and roofs must be divided into zones and the factors must be determined with reference to these zones.

Cpe,1 and cpe,10 values for the most loaded zones for different parts of buildings can be found in

the tables bellow. 20 10 5 2

ce(z)

The location factor, ce(z) as a function of the height above the ground level, z and the built up area category, in case of ct = 1

Structural actions Structural actions

4046

Verticalactions walls: Structural

Single-pitched roofs: Variable actions include:

The actions to which the structure is subjected are divided into permanent, variable and

a) Imposed loads

accidental actions. A design value is obtained by multiplying the characteristic value of the load

b) Snow load

by the partial safety factor. Permanent actions include:

c) Wind load

a) Self-weight of the structure

Self-weight and imposed loads

b) Fixed equipment

The weight of the partition walls can be distributed to generate a uniform load. When designing the structural floor elements of a single storey building. The load must be taken in to account in the weakest area. The effect of concentrate load must be considered separately.

Partial safety factors for actions Accompanying

Permanent actions

variable action

(G1)

Zone Favourable effect

In some cases, the imposed loads on the structural floor in a single storey building can be

Variable actions (Q) Leading variable

Accompanying

action Q1

variable action Q2

A 1,0

reduced. When designing column, the loads on storeys are assumed uniformly distributed. Also in the case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © d/h

cpe,10

-1,0

Unfavourable effect Fatigue-inducing

action Flat roofs: Fire design

Serviceability limit

1,35 1,0

cpe,1

cpe,10

-1,3

-0,8

-1,3

-0,8

cpe,1

cpe,10

cpe,1

-1,0

0,8

1,0

-1,0

0,6

1,0

1,5

1,0

1,5

The characteristic values of the impose load for areas in residential, social, commercial and

1,0

administration buildings are provided in table below divided into categories according to their specific uses.

1,0

1,0 

Building occupancy categories:

state

Residential (including hospital wards, hotel bedrooms etc.) Roof

The partial safety factors presented in this table are the basic values of EC3.

slope

Office areas 

The weight of fixed partitions may be considered as self-weight and represented as an

equivalent uniformly distributed load. No guidance is given as to the required magnitude of such an equivalent load intensity, and the designer should use a reasonable approach in making 2

such an estimate. However, a minimum value of 1,0 kN/m is typically used for offices with

F Rectangular eaves Parapet

cpe,10

cpe,1

-1,8

-2,5

hp/h = 0,025

-1,6

-2,2

hp/h = 0,05

-1,4

-2,0

hp/h = 0,10

-1,2

-1,8

F ( = 180°)

cpe,1

cpe,10

cpe,1

F ( = 90°)

cpe,10

cpe,1

Assembly into 5 sections 5° areas (subdivided -1,7 -2,5 -2,3 depending -2,5 on likely -1,6

-2,2

-0,9 and crowding) -2,0 -2,5 density15° of occupation

-2,8

-1,3

-2,0

0,2

D E

normal weight partitions and storey heights.

F ( = 0°)

cpe,10

Structuralactions actions Structural Structural Structuralactions actions

Shopping 30° Storage areas

-0,5

-1,5

-1,1

-2,3

-1,2

-2,0

0,7

45°

0,7

-0,6

-1,3

-1,2

-2,0

60°

0,7

-0,5

-1,0

-1,2

-2,0

75°

0,8

-0,5

-1,0

-1,2

-2,0

Pressure coefficient Cp in vertical wall (Aref > 10m2):

Pitched roofs:

Wall

d/h<1

-1,0

-0,8

-0,5

0,8

-0,3

d/h>4

-1,0

-0,8

-0,5

0,6

-0,3

The intermediate values can be determined by linear interpolation

Sabah Shawkat © Pressure coefficient for the wall

Roof

slope 

F ( = 0°)

cpe,1

cpe,10

-45°

-0,6

F ( = 90°)

cpe,10

cpe,1

-1,4

-2,0

-30°

-1,1

-2,0

-1,5

-2,1

-15°

-2,5

-2,8

-1,9

-2,5

-5°

-2,3

-2,5

-1,8

-2,5

5°

-1,7

-2,5

-1,6

-2,2

15°

-0,9

-2,0

-1,3

-2,0

30°

-0,5

-1,5

-1,1

-1,5

0,2 0,7 45°

0,7

-1,1

-1,5

60°

0,7

-1,1

-1,5

75°

0,8

-1,1

-1,5

Structural actions Structural actions

4048

Determination Structural actions of the internal pressure factor: Values of internal pressure factors, cpi for buildings without partition-walls are given as a The actions to which the structure is subjected are divided into permanent, variable and function of the opening ratio, by the figure bellow. accidental actions. A design value is obtained by multiplying the characteristic value of the load by the partial safety factor. Permanent actions include:  = the total area of the openings on the sheltered side and the sides parallel to the wind / the a) Self-weight of the structure total area of the openings on all sides b) Fixed equipment

Variable N actions 1    include: L when q q  60  L 2500 a) Imposed loads b) Snow load 500   q  0 2 Wind load L

 c)

Self-weight and imposed loads N is the axial compression

L weight is the of compression The of length the partition walls can member be distributed to generate a uniform load. When is the structural horizontalfloor deflection in the system causedThe by the force andtaken external q designing elements of astiffening single storey building. load mustq be in toload. account in the weakest area. The effect of concentrate load must be considered separately.

Partial safety factors for actions Accompanying

Permanent actions

variable action

(G1)

Variable actions (Q) Leading variable

Accompanying

action Q1

variable action Q2

Favourable effect

1,0

Unfavourable effect

1,35

1,5

Fatigue-inducing

1,0

When there are several supported members, the horizontal load is determined as follows: In some cases, the imposed loads on the structural floor in a single storey building can be k r  0 2 L reduced. when q N  q  60  L 2500 When designing column, the loads on storeys are assumed uniformly distributed. Also in the kr   L case multi-storey buildings, imposed when q ofN   qloads  can sometimes be reduced. 60  L 2500

Sabah Shawkat © action

Fire design

Serviceability limit 1,0 1,0 1,0  Extreme values of the internal pressure factor, cpi can be applied in case of closed buildings state with partition-walls, and openable windows. The values are as follows: The partial safety factors presented in this table are the basic values of EC3. cpi = 0,8, and cpi = -0,5

The characteristic values of the impose load for areas in residential, social, commercial and I administration kr 0 2  buildings are provided in table below divided into categories according to their nr specific uses.

Where

 is the sum of the compressive forces of members Building occupancy categories: is the number of members nr Residential (including hospital wards, hotel bedrooms etc.) B Office Combined loads:areas

TheAdditional weight of fixed partitions horizontal Forces: may be considered as self-weight and represented as an

such an estimate. However, a minimum value ofmember 1,0 kN/m is typically to used offices proving with horizontal forces transmitted from structural in compression thefor members

normal weight partitions and storey heights. restraint.

( i 1)

  G.j Gk.jareas   0 9     Q.i Qk.i  Storage

  Gk.j   I.I Qk.I     2.i  Qk.j j

( i 1)

Normal combination

The load by a compression structural member in compression (e.g. the upper chord of the lattice) on restraining members (e.g. horizontal diagonal members) is determined as follows: when

  Gk.j  Qk.I     0.i  Qk.i

Shopping Rare combination

Transverse force due to a compression structural element

 density Q k.I   G.jof Goccupation k.j      and Q.I crowding)   j 



such an equivalent load intensity, and the designer a reasonable approach in making installation tolerances. A further factor to beshould takenuse into account in the design are the

N 500  L

ULS(subdivided into 5 sections depending on likely SLS Assembly areas

equivalent uniformly load. forces No guidance is given are as to the required magnitude ofand In addition to winddistributed load, horizontal in the structure generated by eccentricities

  Gj Gkj   Qj QkI     0.i  Q i Qk.i j



( i 1)

Long-time combination

L q  2500

Structural Structuralactions actions Structural Structuralactions actions



  Gk.j     2.i  Qk.j j

( i 1)

Load determination in the model building Self-weight The self-weight of the purlin trusses and roof is estimated at Gk = 0,5kN /m2.

Snow Load: The characteristic value of the snow load on the ground in the building area is 1,5kN /m2. The roof angle is 1:6 (9,5 o). For a pitched roof, the shape coefficient when s

 s

0.8  1 5

1 2

kN 2

Wind load: In the building area, the wind velocity ν ref. The reference mean velocity pressure qref is determined from the formula: 

1.25 

 ref

30 

m s

d = 15 m

h = 20 m

Sabah Shawkat ©

2

2 kN 2

d / h = 0.75 qref  0.5625 m kN m In terrain category III, the exposure coefficient ce has the following value with a building

qref  0.5   ref 

height of 20 m measured at the eaves level.

cd  1 ce  2.3 Thus the basic value for the wind load is obtained using the formula: qwk  qref ce cd 

qwk  1.29 m

2

kN

2

qwk qref ce cd cp 1.29 cp kN m The following pressure coefficient values are obtained for the model building: Wind parallel to the side wall cpA  0.8

cpE  0.3

qwkA  qref ce cd cpA

qwkA  1.035 m

qwkE  qref ce cd cpE

qwkE  0.38813 m

2

kN 2

kN

Despite the correct static calculations and the dimensioning of the individual supporting elements on the facade of the individual objects, the unlikely realization may not correspond

Errors and failures during the construction hall made of steel structures

to our realizations see the pictures as realized facades of steel thin-walled profiles where the thermal insulation on the large facades behind the trapezoidal sheets was missing.

Structural actions Structural actions

4050

Structural actions

Variable actions include:

The actions to which the structure is subjected are divided into permanent, variable and

a) Imposed loads

accidental actions. A design value is obtained by multiplying the characteristic value of the load

b) Snow load

by the partial safety factor. Permanent actions include:

c) Wind load

a) Self-weight of the structure

Self-weight and imposed loads

b) Fixed equipment

Partial safety factors for actions Accompanying

Permanent actions

variable action

(G1)

In some cases, the imposed loads on the structural floor in a single storey building can be

Variable actions (Q) Leading variable

Accompanying

action Q1

variable action Q2

Favourable effect

1,0

Unfavourable effect

1,35

1,5

Fatigue-inducing

1,0

Fire design

Serviceability limit

1,0

1,0 

reduced. When designing column, the loads on storeys are assumed uniformly distributed. Also in the case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © action

The characteristic values of the impose load for areas in residential, social, commercial and Conception of the wind load on structures administration buildings are provided in table below divided into categories according to their specific uses.

Building occupancy categories:

state

Residential (including hospital wards, hotel bedrooms etc.)

The partial safety factors presented in this table are the basic values of EC3.

The weight of fixed partitions may be considered as self-weight and represented as an

Office areas

Assembly areas (subdivided into 5 sections depending on likely

equivalent uniformly distributed load. No guidance is given as to the required magnitude of such an equivalent load intensity, and the designer should use a reasonable approach in making such an estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with normal weight partitions and storey heights.

density of occupation and crowding) D

Shopping

Storage areas

Structural Structuralactions actions Structural Structuralactions actions

Calculate the horizontal load on individual columns in the industrial hall: Let us determine the type of the wall cassette and the trapezoidal sheet of the roof for the hall

 H

q2v h2

qv 



8 h1  h1  h2 



The geometry of the hall:

 q1v h12  q2v h22   q1v h12  q2v h22

q1v h1  q2v h2 8 h2  h1  h2

 h3 l     2 8

where

structure with the following parameters:

8 h1  h1  h2 

 q1v h1   q2v h2

  H

 q1v h12  q2v h22 

b = 15 m d = 35 m

h = 8,0 m  = 30°

8 h2  h1  h2 

qv 

h2 l  2 4

qv 

h1 l  2 4

Sabah Shawkat © Single-span structure

q1v h1

is wind load

Calculation of the bending moment on the

t [mm] 0,50 0,63 0,75 0,88 1,00

1,00 7,10 C1 11,45 C1 14,39 C1 17,28 C1 19,94 C1

1,25 4,52 C1 7,30 C1 9,18 C1 11,01 C1 12,72 C1

1,50 3,12 C1 5,05 C1 6,34 C1 7,62 C1 8,80 C1

Maximumload (g+q)Sd [kN/m2] with relation to spanwidth L(m) 1,75 2,00 2,25 2,50 2,75 3,00 3,25 2,27 C1 1,73 C1 1,35 C1 1,08 C1 0,89 C1 0,73 C1 0,61 C1 3,69 C1 2,81 C1 2,20 C1 1,77 C1 1,45 C1 1,20 C1 1,02 C1 4,63 C1 3,53 C1 2,77 C1 2,22 C1 1,82 C1 1,52 C1 1,28 C1 5,57 C1 4,24 C1 3,32 C1 2,67 C1 2,19 C1 1,82 C1 1,54 C1 6,43 C1 4,89 C1 3,84 C1 3,09 C1 2,52 C1 2,10 C1 1,78 C1

3,50 0,52 C1 0,87 C1 1,09 C1 1,31 C1 1,51 C1

3,75 0,45 C1 0,74 C1 0,94 C1 1,13 C1 1,30 C1

4,00 0,39 C1 0,64 C1 0,81 C1 0,97 C1 1,13 C1

slope of the beam

Failure modes:

L cos BH

Support- or span moment

Bsin

A cos q



cos 2

span moment capacity

Support reaction

Further circumstances of the construction: The new, pitched-roofed hall will be built in Bratislava, on plain surface in the local industrial

s  2 8 co q

zone. Each side of the hall is closed; there are 6x6 m doors on the shorter sides, and no openings on the longer sides. Side walls are fixed to frames with 6,0 m spacing; purlins on the roof are placed by 3,0 m.

Determination of the loads of the roof sheet:

At first, let us assume the use of RAN 35 trapezoidal sheet with 0,88 mm thickness. Attention! The self-weight of the sheet should not be considered when determining the loads, because the values in the tables are without the self-weight of the sheet as well!

Structural actions Structural actions

4052

- sum ofactions the loads of the different layers (insulation, etc.) and the mechanical loads hanged on Structural the sheet: The actions to which the structure is subjected are divided into permanent, variable and accidental actions. A design value is obtained by multiplying the characteristic value of the load glay = 0,30 kN/m2 by the partial safety factor. Permanent actions include: Determination ofof thethe value of snow load: a) Self-weight structure b) Fixed equipment

s = i*Ce*Ct*sk

where i – form factor for snow load

Ce – reduction factor from the effect of the wind, generally 1,0 Accompanying Permanent actions Variable actions (Q) Ct – thermal factor, generally 1,0 variable action (G1) Accompanying Leading variable A = 300 m

Characteristic value of the snow1,0 load: Favourable effect

variable action Q2

action Q1

b) Snow load Dimensions of the most loaded zone (F) for pitched roofs: c) Wind load e = min (b, 2h) = 15 m Self-weight and imposed loads e/4 = 3,75 m; The weight of the partition walls can be distributed to generate a uniform load. When e/10 = 1,5 m designing the structural floor elements of a single storey building. The load must be taken in to account in the weakest area. The effect of concentrate load must be considered separately.

characteristic value of snow load on the surface (kN/m2) sk –factors Partial safety for actions

Height above the sea level at site:

Determination of the external pressure factor: Variable actions include:  a) Imposed Wind directionloads  = 0°

In some cases, the imposed loads on the structural floor in a single storey building can be reduced. When designing column, the loads on storeys are assumed uniformly distributed. Also in the case of multi-storey buildings, imposed loads can sometimes be reduced. e e A  A 5.63m2 10 4 The characteristic values of the impose load for areas in residential, social, commercial and Cpe1suc  1.5 Cpe10suc 0.5in table Cpe10pressure Cpe10pressure administration buildings are provided below divided0.7 into categories according0.7to their

sk = 0,25*(1+ - 300/100) = 1,0 kN/m-

Sabah Shawkat © Unfavourable effect 1,35 but we provide sk = 1,25 kN/m2 Fatigue-inducing 1,0

Form factor action for pitched roof with  = 30° slope:

Fire design Roof slope Serviceability limit 1 form factor state  form factor

1,5

1,0

 = 0,8

0° <  < 15°

15° <  < 30°

0,8

1,0

1,5

1,0

specific uses. Cpe,1 and cpe,10 values can be found in the table for pitched roofs

30° <  < 60°

 > 60°

1,0 

0,8*(60-)/30

0,8 0 0,8+0,6*(-15)/30 1,1*(60-)/30 The partial safety factors presented in this table are the basic values of EC3. value of the snow load: s = 1,00 kN/m2 2

Wind load: The weight of fixed partitions may be considered as self-weight and represented as an z Expert's The reference value of the dynamic equivalent uniformly distributed load. No guidance is given as to the required magnitude of opinion (m) pressure is as follows: such an equivalent load intensity, and the designer 200 should use a reasonable approach in making 2 IV III qrefan = 0,24 kN/mHowever, such estimate. a minimum value of 1,0 kN/m2 is typically usedII forI offices with 100 normal weight partitions storeyfactor, heights. 50 Determination of the and location ce with the help of graph on page 9: (z

= 8,0 m, built up

area category: III.)

ce = 1,63

5 2 0

ce (z)

1 < A < 10, therefore

Building occupancy categories: a log( A) a  0.7505 Residential (including hospital wards, hotel bedrooms etc.) Cpesuction Cpe1suc ( 1  a)  Cpe10suc a Cpesuction  0.75 B Office areas Cpepressure Cpe10pressure ( 1  a)  Cpe10pressure a Cpepressure  0.7 Assembly areas (subdivided into 5 sections depending on likely C Roof density of occupation and crowding) slope F ( = 0°) F ( = 90°) cpe,10 D  Shopping

-45° -0,6 -30° -1,1 E-15° Storage areas -2,5 -5° -2,3 5° -1,7 15° -0,9 0,2 30° -0,5 0,7 45° 0,7 60° 0,7 75° 0,8

Structural Structuralactions actions Structural Structuralactions actions

cpe,1 -2,0 -2,8 -2,5 -2,5 -2,0

cpe,10 -1,4 -1,5 -1,9 -1,8 -1,6 -1,3

cpe,1 -2,0 -2,1 -2,5 -2,5 -2,2 -2,0

-1,5

-1,1

-1,5

0,7 0,7 0,8

-1,1 -1,1 -1,1

-1,5 -1,5 -1,5

Determination of the internal pressure factor:

Checking of the load bearing capacity of the roof sheet:

Total area of the openings on the sheltered side and the sides parallel to the wind:

For wind pressure (positive placement):

2 O 1 O 1



72m2

Internal pressure factor suction:

Cpisuction

0.5

Cpipressure

qref Ce  Cpesuction  Cpisuction

Wsuction  0.1 kNm-2

qref Ce  Cpe10pressure  Cpisuction

Wpressure

0.8



 1.5Qsnow

in case of more variable effects:

Pmorepressure

Thus the added values of wind load:

Wsuction

1.35G

Ponepressure

 = A1/A2 = 1

Total area of the openings on all sides: A2 = 72 m2

in case of one variable effect:

1.35G

g*Gk + 1,5*Qk1 Ponepressure

2.28kNm-2

g*Gk + 1,35*Qk

 1.35Qsnow  1.35Wpressure

Pmorepressure

2.73kNm-2

The trapezoid sheet must be checked for the latter value:

Wpressure  0.47kNm-2

Wind direction  = 90° External pressure factor:(suction)

Sabah Shawkat ©

1.5

Cpe1090

1.1

Cpe90suction

Cpe190 ( 1  a)  Cpe1090 a

Cpe90suction

1.2

Two-span structure (b>60mm)

Cpe190

t [mm] 0,50 0,63 0,75 0,88 1,00

1,00 4,67 C4 7,64 C4 10,09 C4 12,99 C4 15,85 C4

1,25 3,2 C4 5,25 C4 6,91 C4 8,87 C4 10,81 C4

1,50 2,32 3,82 5,03 6,44 7,83

C4 C4 C4 C4 C4

Maximum load (g+q)Sd [kN/m ] with relation to span width L(m) 1,75 2,00 2,25 2,50 2,75 3,00 3,25 1,76 C4 1,38 C4 1,1 C4 0,9 C4 0,73 C1 0,6 C1 0,51 C1 2,9 C4 2,28 C4 1,83 C4 1,5 C4 1,24 C1 1,03 C1 0,86 C1 3,82 C4 2,99 C4 2,4 C4 1,96 C1 1,6 C1 1,33 C1 1,11 C1 4,88 C4 3,82 C4 3,07 C4 2,48 C1 2,03 C1 1,68 C1 1,41 C1 5,93 C4 4,64 C4 3,72 C4 2,99 C1 2,44 C1 2,03 C1 1,7 C1

3,50 0,42 C1 0,73 C1 0,95 C1 1,2 C1 1,45 C1

3,75 0,36 C1 0,62 C1 0,81 C1 1,03 C1 1,24 C1

4,00 0,31 C1 0,54 C1 0,7 C1 0,89 C1 1,07 C1

Determination of the internal pressure factor:

Maximum load value from the table: 1.68 kN/m2 < 2,15 kN/m2 incorrect

Total area of the openings on the sheltered

We should provide RAN 35 trapezoidal sheet with 0,88 mm thickness with purlins on the roof

side and the sides parallel to the wind: A1

are placed by 3,25 m.

Total area of the openings on all sides: A2

Therefore 3.07 kN/m2 > 2,15 kN/m2 correct A190

O1 O1

A190

Internal pressure factor:

36m2

72m2

 90

A190 A2

 90

0.5

Suction (negative placement): 

pressure: cpi = +0,20

Thus the added values of wind load: Wsuction90

qref Ce  Cpe90suction  Cpi90pressure 

Wsuction90

0.55kNm-2

Note: Design loads are highlighted with bold characters.



considering only the variable effect of the wind suction:



g*Gk + 1,5*Qk1

Structural actions Structural actions

4054

Structural 1 G  1.5Wsuction90 Psuctionactions

Psuction

Determination of the loads of the side walls: Variable actions include:

0.52kNm-2

The actions to which the structure is subjected are divided into permanent, variable and Maximum load value from the table: accidental actions. A design value is obtained by multiplying the characteristic value of the load 2 3.53 kN/m > 0,68 kN/m2 correct by the partial safety factor. Permanent actions include:

Let use Rannila a) us Imposed loads C150 cassettes with 0,75 mm thickness as 6,0 m long single-spanned structures! b) Snow load Casett125,negative placement c) Wind load Casetti125, positive placement

a) Self-weight of the structure t

Two-span structure (b>60mm)

[mm]equipment 1,00 1,25 b) Fixed 0,50 0,63 0,75 0,88 1,00

5,48 8,90 11,61 14,57 17,41

C4 C4 C4 C4 C4

3,81 6,20 8,04 10,03 11,93

C4 C4 C4 C4 C4

1,50 2,79 4,56 5,90 7,32 8,68

C4 C4 C4 C4 C4

Self-weight and imposed loads

Maximum load (g+q)Sd [kN/m ] with relation to span width L(m) 2,50 2,75 3,00 3,25 1,75 2,00 2,25 2,14 C4 1,69 C4 1,36 C4 1,12 C4 0,94 C4 0,79 C4 0,67 C1 3,50 C4 2,76 C4 2,23 C4 1,84 C4 1,54 C4 1,31 C4 1,13 C4 4,51 C4 3,56 C4 2,88 C4 2,37 C4 1,98 C4 1,67 C4 1,42 C1 5,58 C4 4,39 C4 3,53 C4 2,90 C4 2,42 C1 2,01 C1 1,69 C1 6,60 C4 5,17 C4 4,16 C4 3,41 C4 2,79 C1 2,32 C1 1,95 C1

3,50 0,57 C1 0,95 C1 1,20 C1 1,44 C1 1,67 C1

3,75 0,48 C1 0,82 C1 1,04 C1 1,24 C1 1,43 C1

4,00 0,42 C1 0,71 C1 0,90 C1 1,07 C1 1,24 C1

Partial safety factors for actions Accompanying Deflection check:

Permanent actions

designing the structural floor elements of a single storey building. The load must be taken in to account in the weakest area. The effect of concentrate load must be considered separately. In some cases, the imposed loads on the structural floor in a single storey building can be

Variable actions (Q)

variable action (G1) Calculating with the basic values of loads:

Leading variable

Accompanying

action Q1

variable action Q2

P 1G  0.90 Q snow  0.90W1,0 pressure Favourable effect

-2 1.85kN - m

suction The weight of the partition walls can be distributed to generate a uniform load. When pressure

reduced. When winddesigning load: column, the loads on storeys are assumed uniformly distributed. Also in the case of multi-storey buildings, imposed loads can sometimes be reduced. qref = 0,25 kN/m2 ce = 1,63 The characteristic values of the impose load for areas in residential, social, commercial and

Sabah Shawkat © Unfavourable effect 1,35 1,5 Maximum load value from the table for L/250 deflection limit: Fatigue-inducing 1,0 1,0 action 1,96 kN/m2 > 1,46 kN/m2 Fire design

Serviceability limit t

1,0

1,02 1,0  Maximum load (g+q)Sk [kN/m ] with relation to span width L(m)

1,00 1,25 1,50 1,75 2,00 2,25 2,50 2,75 3,00 3,25 3,50 3,75 4,00 L/150 4,67 3,20 2,32 1,76 1,38 1,10 0,90 0,80 0,60 0,47 0,36 0,30 0,23 L/250 4,67 3,20 2,32 1,76 1,28 0,88 0,62 0,46 0,34 0,26 0,20 0,15 0,12 The partial safety presented in this1,06table are 0,52 the basic of EC3. L/300 factors 4,67 3,20 2,32 1,59 0,73 0,37 values 0,28 0,21 0,16 0,12 0,10 0,63 L/150 7,64 5,25 3,82 2,90 2,28 1,83 1,50 1,11 0,84 0,65 0,51 0,40 0,32 L/250 7,64 5,25 3,82 2,67 1,76 1,22 0,88 0,64 0,49 0,37 0,28 0,22 0,17 L/300 7,64 5,25 3,54 2,21 1,46 1,01 0,72 0,53 0,40 0,30 0,22 0,17 0,13 L/150 10,09 6,91 5,03 3,82 2,99 2,40 1,92 1,42 1,08 0,84 0,66 0,52 0,42 0,75 The weight of fixedL/250 partitions be considered as an 10,09 may 6,91 5,03 3,40 2,26 as 1,57 self-weight 1,13 0,83 and 0,62 represented 0,47 0,37 0,28 0,22 L/300 10,09 6,91 4,54 2,83 1,87 1,29 0,93 0,68 0,51 0,38 0,30 0,22 0,17 equivalent uniformly distributed load. No guidance is given as to the required magnitude of 0,88 L/150 12,99 8,87 6,44 4,88 3,82 3,07 2,40 1,79 1,36 1,05 0,83 0,65 0,53 L/250 12,99 8,87 6,44 4,27 2,83 1,96 1,41 1,04 0,79 0,60 0,46 0,36 0,28 such an equivalent load intensity, designer should a reasonable L/300 12,99 and 8,87 the 5,67 3,54 2,35 1,62use1,16 0,86 0,64 approach 0,49 0,37in making 0,28 0,22 L/150 15,85 10,81 7,83 5,93 4,64 3,72 2,87 2,13 1,62 1,26 0,99 0,78 0,64 1,00 L/250 15,85 10,81 7,83 5,09of 3,38 2,34 21,68 1,24 0,94 0,72 0,43with 0,34 such an estimate. However, a minimum value 1,0 kN/m is typically used for 0,55 offices L/300 15,85 10,81 6,76 4,22 2,80 1,94 1,39 1,02 0,76 0,59 0,45 0,35 0,27 Two-span structure

state[mm] 0,50

correct -

1,5

normal weight partitions and storey heights.

administration buildings Wind direction  = 0°are provided in table below divided into categories according to their specific uses. Determination of the external pressure factor:

Dimensions of the most loaded zone (A, at the corners Building occupancy categories: of the hall) for vertical walls: Residential (including hospital wards, hotel bedrooms etc.)

d e eB  3 mOffice areas  4.38  A h 5 5 h Assembly areas 2 (subdivided A  24m 10  A into 5 sections depending on likely hC  8 m

density of occupation and crowding) A is the size of the loaded area A D Shopping 10 < A, therefore cpe = cpe,10 E

Storage areas

Cpe,10 values can be found in the table for side walls: zone A

Cpe10suctionA  1

zone B

Cpe10suctionB  0.8

zone D

Cpe10pressureD  0.8

 Here, in principle, the thickness of the wall cassette could be reduced. Internal pressure factor:

Structural Structuralactions actions Structural Structuralactions actions

suction: cpi = -0,5

Thus the added values of wind load: WwallsuctionA

B Zone D Zone

WwallsuctionB WwallsuctionD

Zone

Windward

Leeward

WwallsuctionA

0.2kNm

qref Ce  Cpe10suctionB  Cpisuction

WwallsuctionB

0.12kNm

-1,0

-1,3

-0,8

-1,0

-0,5

+0,8

+1,0

-0,3

-1,0

-1,3

-0,8

-1,0

-0,5

+0,6

+1,0

-0,3

qref Ce  Cpe10pressureD  Cpisuction

WwallsuctionD

0.51kNm

cpe,10

cpe,1

cpe,10

cpe,1

cpe,10

cpe,1

cpe,10

cpe,1

cpe,10

cpe,1

Note: Design loads are highlighted with bold characters.

Checking of the load bearing capacity of the side wall:

The external pressure factor similarly to the previous case: zone A

zone B

1

qref Ce  Cpe10suctionA  Cpisuction

Wind direction  = 90°

Cpe10suctionA

d/h

Cpe10suctionB

Internal pressure factor:

For pressure: in case of one variable effect:1,5*Qk1

zone D

0.8

Cpe10pressureD

pressure: cpi = +0,15

Thus the added values of wind load:

0.8

Pwallpressure

1.5WwallsuctionD

Pwallpressure  0.76kNm-2

Maximum load value from the table: t [mm] 0,75 0,88 1,00 1,25

Single-span structure

A Zone

0,98 kN/m2 > 76 kN/m2

3,00 2,91 3,90 4,93 7,43

C2 C2 C2 C2

3,25 2,68 3,60 4,55 6,86

Maximum load (g+q)Sd [kN/m ] with relation to span width L(m), support width 100 mm 3,50 3,75 4,00 4,25 4,50 4,75 5,00 5,25 5,50 2,49 C2 2,32 C2 2,18 C2 2,01 C1 1,78 C1 1,59 C1 1,43 C1 1,29 C1 1,17 3,34 C2 3,12 C2 2,93 C2 2,75 C2 2,45 C1 2,18 C1 1,95 C1 1,77 C1 1,60 4,23 C2 3,95 C2 3,70 C2 3,48 C2 3,12 C1 2,78 C1 2,50 C1 2,26 C1 2,05 6,37 C2 5,95 C2 5,56 C1 4,92 C1 4,39 C1 3,94 C1 3,55 C1 3,22 C1 2,93

C2 C2 C2 C2

WwallsuctionA90

qref Ce  Cpe10suctionA  Cpipressure

WwallsuctionA90

0.45kNm

WwallsuctionB90

qref Ce  Cpe10suctionB  Cpipressure

For suction:

WwallsuctionB90

0.37kNm-2

WwallsuctionD90

qref Ce  Cpe10pressureD  Cpipressure

Maximum load value from the table:

WwallsuctionD90

0.25kNm

C1 C1 C1 C1

5,75 1,07 1,45 1,87 2,68

C1 C1 C1 C1

6,00 0,98 1,33 1,70 2,45

C1 C1 C1 C1

Sabah Shawkat © Single-span structure

t [mm] 0,75 0,88 1,00 1,25

Pwallsuction 1.5WwallsuctionA90

1,51 kN/m2 > 0,67 kN/m2

3,00 5,12 C5 7,17 C1 8,13 C1 10,37 C1

Maximumload (g+q)Sd [kN/m ] with relationto span width L(m), support width100 mm 3,50 3,75 4,00 4,25 4,50 4,75 5,00 5,25 5,50 4,38 C5 3,87 C1 3,40 C1 3,01 C1 2,69 C1 2,41 C1 2,18 C1 1,97 C1 1,80 C1 5,27 C1 4,59 C1 4,03 C1 3,57 C1 3,19 C1 2,86 C1 2,58 C1 2,34 C1 2,14 C1 5,97 C1 5,20 C1 4,57 C1 4,05 C1 3,61 C1 3,24 C1 2,92 C1 2,65 C1 2,42 C1 7,62 C1 6,64 C1 5,83 C1 5,17 C1 4,61 C1 4,14 C1 3,74 C1 3,38 C1 3,09 C1

3,25 4,72 C5 6,11 C1 6,92 C1 8,83 C1

Deflection check: P

Pwallsuction 0.67kNm-2

5,75 1,65 C1 1,95 C1 2,22 C1 2,83 C1

6,00 1,51 C1 1,79 C1 2,03 C1 2,60 C1

Calculating with the basic values of loads:

1 WwallsuctionA90

P  0.45kNm-2

Maximum load value from the table for L/150 deflection limit:

Single-span structure

0,98 kN/m2 > 0,45 kN/m2 t [mm] 0,75 0,88 1,00 1,25

L/150 L/150 L/150 L/150

Structural actions Structural actions

3,00 2,91 3,90 4,93 7,43

ok 3,25 2,68 3,60 4,55 6,86

Maximum load (g+q)S k [kN/m 3,50 3,75 4,00 4,25 2,49 2,32 2,18 2,01 3,34 3,12 2,93 2,75 4,23 3,95 3,70 3,48 6,37 5,95 5,56 4,92

] with relation to span width 4,50 4,75 5,00 5,25 1,78 1,59 1,43 1,29 2,45 2,18 1,95 1,77 3,12 2,78 2,50 2,26 4,39 3,94 3,55 3,22

L(m) 5,50 1,17 1,60 2,05 2,93

5,75 1,07 1,45 1,87 2,68

6,00 0,98 1,33 1,70 2,45

4056

Quasi-permanent value of snow variable action for all snow areas of Slovak republic Structural actions calculated from characteristic value of the snow action on ground level decreased by 1,05 2 ThekN/m actions which the structure is subjected are divided variable and it istonecessary to take into account in the loading casesinto withpermanent, long-term acting snow. accidental A design by multiplying characteristic value of allowed the load to and is Ctobtained are generally 1.0. For the simple shape roofs it is Values actions. of coefficients Cevalue by reduce the partial safety factor. Permanent actions include: the value of coefficients Ce as follows:

a)a)forSelf-weight all snow regions α ≤ 25° (or h ≤ 0.13 b) Ce = 0.9 of the and structure For buildings without skylights, with width less than 60 m and height more than 20 m this b) Fixed equipment coefficient is allowed to be reduced by additional 10% (obtaining values from 0,81 up to 0,90). Partial b) forsafety snowfactors regionsfor IVactions and V and α ≤ 6° (or h ≤ 0.05 b) Ce = 0.8

For buildings without skylights, with width less than 60 m and height more than 20 m this Accompanying Permanent actions Variable actions (Q) coefficient is allowed to be reduced by additional 10% (obtaining values from 0,72 up to variable action (G1) Accompanying Leading variable 0,80). Aforementioned rules are not applicable for: variable action Q2 action Q1 a) buildings protected against wind by surrounding buildings, trees or other objects (buildings Favourable effect 1,0 in forests, etc.) Unfavourable effect 1,35 1,5 1,5 For building with inside heat emissions (power stations, greenhouses) coefficient Ct is as Fatigue-inducing 1,0 1,0 1,0 follows: action a) for roofs without thermal insulation and roof slope α > 1,5° is Ct = 0.8. When heat emission Fire design is removed, snow must be taken away. Serviceability limit 1,0 1,0 1,0 b) for roofs with thermal insulation the coefficient C is from 0,80 up to 1,00. 

The average compression coefficient for the pitch and vault roofs is given as follows Variable actions include: Compression Suction  a) Imposed loads Cecr Cetr b) Snow load

0    10

c) Wind load Duopitch o o Self-weight and imposed   40 10 loads

 

2  0.25 



 

 

1.5  0.333 

100 

   2  0.45   100  

state

The weight of the partition walls can be distributed to generate a uniform load. When

o o   The load must  in  0.40 building.  be taken  designing the structural   elements  10 of a single 0 floor 1.8  0.40  1.8 storey   100  100    to account in the weakest area. The effect of concentrate load must be considered separately. min= 0.80 Vault

o o    floorina single   In some cases, the imposed storeybuilding   on 40the structural 10  loads 1.8  0.40can  be  2  0.50  100 100     reduced. max 0.27 Wall column, the loads on storeys are  in 0.8 Cecw  0.8uniformly distributed. Cetw   1.3 When designing assumed Also o the







area of fixed partitions I IV represented as V an TheSnow weight may beIIconsidered asIIIself-weight and Characteristic equivalent uniformly distributed load. No guidance is given as to the required magnitude of value of snow 0.75 1.05 1.50 2.25 >2.25 such an equivalent action kN/m2 load intensity, and the designer should use a reasonable approach in making such an estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with Wind actions: normal weight partitions and storey heights. The characteristic values are taken from the map given in NAD for STN P ENV 1991-2-4. Wind zone

III (altitude 700 1300 m above sea level)

Wind speed νref,0

24 m/s

26 m/s

30 m/s



load buildings, imposed loads Cic can 0.6sometimes  1.8  1.3be  Cit  0.6  1.3  o  0.8  oreduced. case Internal of multi-storey The characteristic values of the impose load for areas in residential, social, commercial and Overall  Cic into categories according Cetw  Citto their Cecwdivided Wall+interior administration buildings are provided in table below specific uses. where

W(kN)occupancy = 1.3 q (kN m-2) H (m) a (m) alebo 1.3 q(kN m-2) H(m) b(m) Building categories: Residential (including hospital wards, hotel bedrooms etc.)

The partial safety factors presented in this table are the basic values of EC3. Characteristics values of snow action in Slovak republic are given for snow areas in the table

NAD 3

 

   0.5  0.60   100  

Sabah Shawkat © t



100 

Office areas

Assembly areas (subdivided into 5 sections depending on likely density of occupation and crowding)

Shopping

Storage areas

mountain areas (more than 1300 m above sea level) 33 m/s

Structural Structuralactions actions Structural Structuralactions actions

Closed construction with duo pitch and open structure with vault roofs

Roof:

Reinforced concrete chimney Preliminary data and calculations

Compression

 25  o  1 

   C ecr  2  0.45  100   

Cecr  0.4

Wall:

Compression Cecw  0.8



Internal load:

Cetr  0.425

Suction Cetw   1.3  o  0.8 Cetw  0.5

Overall:

    Cetr  0.5  0.60   100  

Chimney is assumed in Region II, where h, dd and dh are the dimensions of the object. d d  16.70 m  d h  6.80 m 110 m Evaluate 

 h 

 

Suction



Cic  0.6  1.8  1.3  o Cic  0.3







 Cit  0.6  1.3  o  0.8 Cit  0.3

Cecw   Cic  0.5

Cetw   Cic 0.8

Cecw   Cit  1.1

Cetw   Cit 0.2

 h1 

  14.29

 dd  dh   h 1  d h h 2   2 

20 m

 h2  h  h1

 

h 2  90 m

1.215

From adjacent graph based on we find the value then we calculate the vibration time T



 m   T  2.42 dd  m  The load is parallel to the wind direction, from adjacent figure based on T we find the value  cto  0.55



Ct  0.67

 Ct   cto

 1.8

If h  60 m



 1

T  0.09 

i

   1    i

Sabah Shawkat © In case of normal overload: Static load on level H:

Tsni  0.67 d i  i qHni

Dynamic loads on level H: Tdn  i   i Tsni

Hi  0

Pressure coefficients for the duo pitch and vault roofs

di  16.7

i



i



Tsni 

0.36

1.648

4.11

kN m

1

Tdni  6.78

11.75

0.36

1.648

3.86

6.8

0.345

1.621

2.65

4.30

6.8

0.33

1.594

2.98

4.74

6.8

0.315

1.567

3.7

5.80

6.8

0.3

1.54

4.44

6.83

6.8

0.285

1.513

4.66

7.05

6.8

0.272

1.49

4.86

7.24

6.8

0.26

1.468

5.02

7.37

6.8

0.25

1.45

5.17

7.49

100

6.8

0.24

1.432

5.29

7.58

110

6.8

0.233

1.419

5.4

7.67

Structural actions Structural actions

6.36

kN m

 1.00

4058

Because the calculation Variable actions include: of the force below the critical pressure value is significantly lower than

Structural actions Extremes of overload:

loadtoonwhich level H: Tsn i  1.75 i TheStatic actions the structure is Tse subjected are divided into permanent, variable and Dynamic loads on level H: Tdei  1.75 Tdni accidental actions. A design value is obtained by multiplying the characteristic value of the load i  Tsei  Tdei  by the partial safety factor. Permanent actions include: 1 1 0.7 7.2 11.86 kN m kN m 0.7 6.75 11.13 a) Self-weight of the structure 0.7

4.64

7.52

b) 0.7 Fixed equipment5.21

8.3

0.8

6.47

10.14

0.9

7.76

11.96

0.9

Partial safety factors for8.16 actions 0.9

0.9 Accompanying 0.9

variable action 0.9

9.26

0.9

9.46

Self-weight and imposed loads

designing the structural floor elements of a single storey building. The load must be taken in to account in the weakest area. The effect of concentrate load must be considered separately.

12.67

8.79 Permanent 9.04

b) Snow load c) Wind load

The weight of the partition walls can be distributed to generate a uniform load. When

12.34

8.5

the force of the wind at normal pressure, it is unnecessary to continue the calculation a) Imposed loads

actions

(G1)

12.9 13.11

Variable actions (Q)

Leading variable

13.27

13.42 action Q1

Accompanying variable action Q2

Favourable effect 1,0 The load is perpendicular to the wind direction, determination of critical speeds Unfavourable effect 1,35 1,5 1,5 dh Fatigue-inducing 1,0  13.6 m dh  6.8 m S  0.20 1,0T  2.5 T  2.5 1,0 Vcr  Vcr S T action dd  16.7 m a  16.3 Fire design 2 Vcr qcr  11.35 qcr    0.3 Serviceability limit 1,0 1,0 1,0  2 a m state

Sabah Shawkat ©

Hi kN  2 kN factors  of EC3. The0.5 partial in this tableTcr are  thebasic values  safety  2 presented 10  Lcri   qcr 10  dh  i i Ct  i d i  qcr 2 2  h  m   m  Tcri  Lcri  The weight of fixed  1 partitions may be 0considered  1 as self-weight and represented as an 1.461 kN m kN m 1.028uniformly distributed load.0.117 equivalent No guidance is given as to the required magnitude of 0.234 such an0.585 equivalent load intensity, and the designer should use a reasonable approach in making 0.575

0.351

such an0.646 estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with 0.468 normal0.715 weight partitions and storey heights. 0.585 0.702

0.701

0.691

0.818

0.681

0.935

0.673

1.052

0.665

1.169

0.659

1.286

Building occupancy categories:

Residential (including hospital wards, hotel bedrooms etc.)

Office areas

Assembly areas (subdivided into 5 sections depending on likely density of occupation and crowding)

Shopping

Storage areas

1- Steel structures, 2- Pre-stress structures, 3-Reinforced concrete structures

Structural Structuralactions actions Structural Structuralactions actions

Designing gable beam in the model building Now design the gable beam of the model building. The width of the building is 80m and the column spacing at the end of the building is 10m. Divide the gable beam into four parts which are joined to the end columns on-site using bolted joints. The gable beam is vertically loaded by the support reactions of the purlin trusses. The gable beam transfers the transfers wind load of the hall to the bracing lattices, which means it is also subjected to axial force. Consider a load combination with dominant snow load. The wind load is multiplied by the combination factor ψ 0 = 0.6. The compressive force on the gable beam is assumed constant along the entire length of the section: 0

 0.6

 H1  15 m

 Q2

 1.5

cp1  0.6

H  2  6 m

Lp  6.666 m 

cp2  0.3

L  120 m

B  80 m

Lf  20 m

fyk  355 MPa

qref  0.6 Lcolumn  10 m

Sabah Shawkat © Lp or a

is the purlin spacing

Structural actions Structural actions

or 2L

is the frame spacing

4060

In the building Structural actions area, the wind velocity is ν ref. The reference mean velocity pressure qref is the formula: Thedetermined actions tofrom which the structure is subjected are divided into permanent, variable and accidental actions. A design value is obtained by multiplying the characteristic value of the load m kg 3  ref  30  ref 30     1.25 10  1.25  3 by the partial safety factor. Permanent actions include: s m a) Self-weight of the structure 2 2 kN qref  0.5625 m kN q  0.5   ref  b)refFixed equipment 2 m In terrain category III, the exposure coefficient Ce has the following value measured at the eaves level. Partial safety factors for actions Accompanying

Permanent actions

variable action

(G1)

Variable actions (Q) Leading variable

Accompanying

action Q1

variable action Q2

Favourable effect

1,0

Unfavourable effect

1,35

1,5

Fatigue-inducing

1,0

Variable actions include: With dominant snow load, the following restraining force for the purlin truss is obtained: a) Imposed loads Self-weight: b) Snow load kN  Gk1Wind  0.9load  c) 2 m Self-weight and imposed loads

self  weigh

 G1 

1.35

load: TheSnow weight of the partition walls can be distributed to generate a uniform load. When designing the structural floor elements of a single storey building. The load must be taken in kN   0.8   1.5  s  1.5 load must be considered separately. to account in the weakest area.Q1The effect of concentrate 2 m In some cases, the imposed loads on the structural floor in a single storey building can be 2 2 Qk1 1.2 m kN  sd   s  sd 1.2 m kN Qk1  sd  reduced.

F  0.5   G1Gk1   Q1Qk1 Lp Lf F  200.98 kN Whenysd designing column, the loads on storeys are assumedysduniformly distributed. Also in the case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © action

The self-weight Gk1 can be assumed smaller 0.9kN/m2, since the weight of the primary lattice The characteristic values of the impose load for areas in residential, social, commercial and need not be taken into account. administration buildings are provided in table below divided into categories according to their calculate specific uses.the forces and moments using plastic theory. For vertical loading, the following static

Fire design

Serviceability limit

1,0

1,0 

state

model is obtained:

By usingoccupancy equalizingcategories: the internal and external work, the plastic moment can be determined: Building Residential (including etc.) Lcolumn  2 hospital  wards, hotel bedrooms Msd ( 3   2  ) Fsd  L   Msd  2 Fysd  Msd  267.9732 m kN 3 15   B Office areas

The partial safety factors presented in this table are the basic values of EC3.

Thus the basic value for the wind load is obtained using the formula: The weight of fixed partitions may be considered as self-weight and represented as an

Msd VCsd Assembly areas (subdivided into 5 sections depending on likely

2

equivalent uniformly distributed of  1.38 mmagnitude kN ce  2.45 cd  1 load. Noqguidance cd as to qthe wk required wk  qrefiscegiven such an equivalent load intensity, and the designer should use a reasonable approach in making 2 qd of1.11628 m 2 iskN wind loadfor offices with  0  Q2However, qwk  cp1  cp2a minimum d  suchqan estimate. value 1,0 kN/m typically used normal weight partitions and storey L heights. Nsd  qd  0.375 H1  H2  Nsd  778.60617 kN 2

density of occupation and crowding) 3 D

Shopping

Resistance at the ultimate limit state: E Storage areas Since the hollow section is also subjected to compression load. Consider a hollow section with dimensions 400x200x10 and steel grade S355J2H. The cross-section of the hollow section is

The total portion of the roof is included in the value of the horizontal load, which is

class 1. The resistance values are as follows.

MplyRd  462.7 kN m

conservative. A more accurate result can be obtained by taking into account the effect of the pressure coefficients of the roofs wind load.

Structural Structuralactions actions Structural Structuralactions actions

VplzRd  1398.0 kN

 NplRd  3408 kN

The bending resistance, reduced by the axial force is: MNyRd

Nsd    1.33 MplyRd  1   N plRd  

 max 

Melysd Wely400.200.10



Nsd A400.200.10

 max  fyk

fyk  355 MPa

MNyRd  474.79616 m kN   max 316.603 MPa

MNyRd  MplyRd

MNyRd  MplyRd We determine the deflection of the gable beam using elasticity theory:

In the interaction expression, there is now only one bending load, so the effect of the parameter α is omitted:

 Msd     0.57915  MNyRd 

Msd MNyRd

 1

E  210 1000MPa

I  23003 10 mm

Lp 1 Fsd Lp  Lcolumn  Lp    2 Lcolumn  Lp 6 EI 2

check that the hollow section will not buckle before the mechanism is generated. The buckling







 17.89286 mm

length is 10m. The buckling resistance is as follows:

Sabah Shawkat ©

 NbyRd  2812. kN

NbyRd  Nsd

Nsd  778.61 kN

1



Lcolumn 200

 1 50 mm



 1

The gable beam is restrained laterally supported by the roof, so lateral buckling need not be checked. In this case, the effect of bending moment on buckling need not be taken into account, since vertical buckling leads to the expected failure mechanism. Stress and deflection at the serviceability limit state:

Check that the stresses do not exceed the yield strength of the material with serviceability limit state loads. The partial safety factors for loads are given the formula: Fsd  0.5  Gk1  Qk1 Lp Lf

Fsd  139.986 kN

self-weight and snow load

 Lf   Fsd 9

M  elysd  

Melysd  311.08 m kN

L  Nsd   0 cp1  cp2 qwk  0.375 H1  H2   2  2

A400.200.10  112.6 10 mm

Nsd  519.07078 kN 3

Wely400.200.10  1150 10 mm

Wall section- vertical cladding as a suggestion compared to poor realization

Structural actions Structural actions

4062

Variable actions include: Timber:

Structural actions Calculate the deflection of the steel I beam shown in adjacent figure. TheFirst actions whichthe thedefinition structure ofis asubjected divided of into permanent, variable andthe of alltoremind force andare a moment force to some point. From accidental A design is obtained by multiplying the acharacteristic valueand of the loadand physics,actions. it is known thatvalue the force is a vector, which has sign, direction, value point.factor. The Permanent force is aactions representation by application the partial safety include: of some load, which causes damages (deformations and displacements) of the body on which act. If the force acts at arbitrary a) Self-weight of the structure direction, we may decompose it at the two main directions – horizontal and vertical as it’s b) Fixed equipment shown on a figure. The action of the decomposed force is the same as this of the whole one. 1

 q  4.90 kN m Partial safety factors for actions



 40 deg

LAccompanying – is beam span

Permanent actions

variable action

(G1)

 L  4.8 m

a) section Imposedproperties loads The b) Snow load cross  section c) Wind load 2   w  1.0 kN cm Self-weight and imposed loads

250x250mm or

w

bw  250 mm

 hw  250 mm

 10 MPa

cw  1.4

The weight of the partition walls can be distributed to generate a uniform load. When We define the value of the coefficient c, for cross-sections subjecting to biaxial bending. designing the structural floor elements of a single storey building. The load must be taken in 3weakest area. The effect of concentrate load must be considered separately. 3 to accountbinthe Ix bw hw w hw 3 wx  Ix  wx  2604.16667 cm Iy  12 the imposed loads onhw In some cases, the structural floor in a single storey building can12 be

Variable actions (Q) Leading variable

Accompanying

action Q1

variable action Q2

Favourable effect

1,0

Unfavourable effect

1,35

1,5

Fatigue-inducing

1,0

reduced.

Iy Wx Wx 3 wy  2604.16667 cm wydesigning  c Wy Also in the or distributed. When uniformly bw column, the loads on storeys are assumed W c y

2 case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat ©

action

specific uses.

Fire design

Serviceability limit

1,0

1,0 

state

The partial safety factors presented in this table are the basic values of EC3.

The weight of fixed partitions may be considered as self-weight and represented as an q is Dead load equivalent uniformly distributed load. No guidance is given as to the required magnitude of

such an equivalent load intensity, and the designer should use reasonable members approach of in structures. making Dead load (q) represents the self-weight of construction anda permanent 2 such an estimate. However,direction a minimum value 1,0 kN/m is the typically forperiod. offices with It has constant intensity, sense and of location during whole used service

normal weight partitions and storey heights. 1 qy  3.75362 m kN  qy  q cos (  )

 qx  q sin(  )

qx  3.14966 m

1

kN

We calculate the bending moment on simple supported beam due to uniform load

 Mx 

1 2 qy L 8

Mx My c Mx  c Mz The characteristic values ofWthe impose load for areas in residential, social, commercial and  xneeded Wx Wx  administration buildings are provided in tableadm below divided into categories according to their

Mx  10.81 m kN

 My 

1 2 qxL 8

For rectangular cross-section:

2 Building occupancy categories: b h 6 Residential h(including hospital wards, hotel bedrooms etc.) c 1.2  1.4 b  h b 2  B  Office areas  6 

Assembly areas (subdivided into sections depending likely stress of the crossC Determination of the section modulus of5the cross-section andonmaximum

section density of occupation and crowding) D Shopping Mx  cw My Wxneeded   E Storage areasw Mx My   max   wx wy

Wxneeded  2350.98454 cm  max

 7.63447 MPa

Steel:   e  240 MPa

My  9.07 m kN

Structural Structuralactions actions Structural Structuralactions actions

 adm

 0.6  e

 adm

 144 MPa

y



y



 adm

y

 e

 M1



 fy    A E

 

0.01309

Stress of steel:

1.1

 M1

 maxsteel

2 0.5 1     y  0.2   y 

y



y

1  M1

0.61245  e

 adm



Mx Wxsteel



 maxsteel

Wysteel

 142.74199 MPa

0.61245 Deflection: When the beam is loaded, it deformeds.

2

  adm 14.4 kN cm

Where Mx- is bending moment, Es- is modulus of elasticity, I- is moment of inertia of the crosssection about the neutral axis.

for I steel cross-section

csteel  8

 is an imperfection factor, is a function of buckling curve, for buckling curve a, b, c, d is

Ix300  97.9 10 mm

 fy 

5 MxL  48 Es Ix300

 fx 

5 My L  48 Es Iy300

0.21, 0.34, 0.49,0,76 y is reduction factor

 is slenderness ratio

Iy300  4.49 10 mm

Needed coefficient c:

fy  0.1262 cm

fx  2.30888 cm

Sabah Shawkat © Then the final deflection will be:

Cross-section

I 140 to I220

I 240 to I 600

9-10

U 120 to U 160

U 180 to U 320

7-8

IPB 100 to IPB 320

f 

Mx  csteel My  adm

f  2.31233 cm

fadm 

L 200

fadm  2.4 cm

f  fadm

deformation due to which the axis of structure is deflected from its original position. The deflections of structures are important for ensuring that the designed structure is not excessively flexible. The large deformations in the structures can cause damage. The deflection in beams is dependent on the acting bending moments and its flexural stiffness.

Wxsteel  579.01784 cm

I300

g0 is self-weight of the steel beam Then we propose the cross-section parameters: 3

W ysteel  71.9 cm

Wxsteel  652 cm

g0  A 0.785

kN m

When a structure is subjected to the action of applied loads each member undergoes

The elastic section modulus of the cross-section then will be: Wxsteel 

fx  fy

10  0.54165 m

1

A  6.90 10 mm

kN The convention for member axes according to Euro-code

Structural actions Structural actions

4064

Structural actions limiting values for vertical deflections Recommended

Variable actions Designing the include: stiffening elements in the model building

The actions to which the structure is subjected are divided into permanent, variable and  Limits accidental actions. A design value is obtained by multiplying the characteristic value of the load  max   1   2   0 2 by the partial safety factor. Permanent actions include: [  0 - is the pre-camber [  2 - is the deflection a) Self-weight of the structure (hogging) of the beam in the of the beam due to

unloaded state – state 0;

b) Fixed equipment Conditions

 1 - is the deflection of the

variable loading plus any time dependent deformations due to permanent load – state 2]

beam due to the permanent loads immediately after Partial safety factors for actions loading –state 1] Accompanying Permanent actions Variable actions (Q) Roofs generally Span/200 Span/250 variable action (G1) Accompanying Leading variable Roofs frequently carrying variable action Q2 action  Q1 personnel other than for Span/250 Span/300 maintenanceeffect Favourable 1,0 -

Simple a) Imposed connections loads are defined as joints between members that have not been designed with Snow load b) intention the that they transmit significant moments. Their purpose is to transfer load from the supported into the supporting member in such a way that essentially only direct forces c) Wind member load are involved, e.g. vertical shear in a beam to column or beam to beam connection, axial tension Self-weight and imposed loads or compression in a lattice girder chord splice, column base or column splice connection. They Themay, weight of the partition cansituations be distributed generate abracing uniformis load. When therefore, only be walls used in wheretosufficient present that, when the designing theassumed structural elements of adequate a single storey The load must be takenPopular in joints are to floor function as pins, overallbuilding. structural resistance is present. to account in the weakest area. The effect of concentrate load must be considered separately. arrangements include lattice girders and bracing systems or connections between beams and columns in rectangular in the which lateral floor loadings resisted stiff systems In some cases, the imposedframes loads on structural in a are single storeybybuilding can beof shear walls, cores or braced bays. reduced. In the model building, the lattice stiffening method and the design of transverse When designing column, the loads on storeys are assumed uniformly distributed. Also in the stiffening elements only for the model building is used. The longitudinal stiffening elements case of multi-storey buildings, imposed loads can sometimes be reduced. are designed according to the same principles.

Sabah Shawkat © Unfavourable effect Floors generally

1,35

Span/250 1,5

1,5 Span/300

Fatigue-inducing 1,0 Floors and roofs supporting action plaster or other brittle finish or non-flexible Fire designpartitions -

1,0

Span/250

Span/350

Floors supporting columns Serviceability limit 1,0 1,0 1,0  (unlessstate the deflection has been Span/400 Span/500 included in the global analysis The partial safety factors presented in this table are the basic values of EC3. for the ultimate limit state) Where  max can impair the

Span/250 the building Theappearance weight ofoffixed partitions may be considered as self-weight and represented as an

The characteristic values of the impose load for areas in residential, social, commercial and The structural idealisations suitable for determining the distribution of member forces will be administration buildings are provided in table below divided into categories according to their as shown in figure below, with all lateral loading being resisted by the bracing or shear wall. specific uses. When considering the design of the frame to withstand gravity loading, the assumption of pin connections makes the overall structural analysis particularly straightforward, since loads can

be traced from floors into beams into columns and eventually into the foundations using a Building occupancy categories: simple statically process. Residential (including hospital wards, hotel bedrooms etc.) B

Office areas

Assembly areas (subdivided into 5 sections depending on likely

equivalent uniformly distributed load. No guidance is given as to the required magnitude of such equivalent intensity, and the for designer should use a reasonable approach Foranbuildings theload recommended limits horizontal deflections at the tops of the in making columns are: However, a minimum value of 1,0 kN/m2 is typically used for offices with such an estimate. Portal frames without gantry cranes (height of the column or of the storey)/150 normal weight partitions and storey heights. Other single storey buildings (height of the column or of the storey)/300

In a multi-storey buildings: - in each storey In a multi-storey buildings: - on the structure as a whole

density of occupation and crowding) D

Shopping

Storage areas

(height of the column or of the storey)/300 Idealisations of bracing

(the overall height of the structure)/500

Structural Structuralactions actions Structural Structuralactions actions

Idealisation of shear wall

Lateral stiffening The wind loads in the side wall of the building are transferred to the stiffening lattices in the end wall through the horizontal lattice in the roof parallel to the side walls. However, the horizontal force of the side wall wind columns is transferred to the primary columns through the roof profile. This way, the eaves section of the side wall can be made lighter. The joints of the stiffening profile must be checked for the loads created by the horizontal forces.

Connecting the bracing The following pressure coefficient values are obtained for the model building:

End wall bracing

cp  cpA   cpE

cpA  0.6

cpE  0.3

walls, as the building is symmetrical and the bracing in the end walls is similar. In the pressure

qwkA  qref ce cd cpA

qwkA  0.82688 m

coefficient of the wind load, the effect of negative pressure must also be taken into account, so

qwkE  qref ce cd  cpE

qwkE  0.41344 m

Design the bracing elements of the end wall. The horizontal load is divided evenly in the end

Cp=0.6+0.3=0.9. Thus the following value for the horizontal force of the end wall is obtained:

2

cp  0.9

kN

Sabah Shawkat ©

In the building area, the wind velocity is ν ref. The reference mean velocity pressure qref is determined from the formula: 3

   1.25 10

 ref

 30



1.25 

2

2 kN 2

qref  0.5625 m kN m In terrain category III, the exposure coefficient Ce has the following value with a building

qref  0.5   ref 

height measured at the eaves level. Thus the basic value for the wind load is obtained using the formula: ce  2.45

cd  1

qwk  qref ce cd

 H1  15 m

H  2  5 m

Ltot  110 m

d  5.33333 H1

H1 H2 Ltot

d H1

is the height of the eaves is the height of the roof structure is the length of the building

qwk  1.38 m  Q1

 1.5

2

kN

d  80 m

Roof bracing

qwd1   Q1qwkA  0.375 H1  H2

qwd1  13.17832 m

qwd2   Q1qwkE 0.375 H1  H2

qwd2  6.58916 m

Fsd   Q1cp qref ce  0.375 H1  H2 

Structural actions Structural actions

Ltot 2

1

kN

pressure negative pressure

Fsd  1087.21143 kN

4066

The bracing is made with two members, one subjected to compression and the other to tension. Structural actions The total horizontal force is transferred by the tension lacing. The tensile force is as follows: The actions to which the structure is subjected are divided into permanent, variable and accidental actions. A design value is obtainedFby sd multiplying the characteristic value of the load cos ( 50)  0.96497 Nsd  Nsd  1126.68363 kN by the partial safety factor. Permanent actions include: cos ( 50 ) a) Self-weight ofsection the structure Consider a hollow of dimensions 150x150x8: b) Fixed equipment 2 2 A150.150.8  43.24 10 mm

fy  355 MPa

fy Ntsdsafety  Afactors  actions Partial for 150.150.8

Permanent actions

When the holes are situated as in the example, the bearing resistance of splice plates is as a) Imposed loads b) Snow load follows: c) Wind load  e1  50 mm d0  1.08 dM30

fu  490 MPa

t1  22 mm

n  6

Self-weight and imposed loads e1     0.55556 The weight3 of the partition walls can be distributed to generate a uniform load. When dM30

 1.1

designing the structural floor elements of a single storey building. The load must be taken in 2.5  fu dM30t1 to account concentratekN load must be considered separately. FbRd of  359.33333 FbRd in the weakest area. The effect

Ntsd  1395.47273 kN

 M1

Accompanying

 M1

Variable actions include: Bearing resistance of splice plates:

 Mb

In some cases, the imposed loads on the structural floor in a single storey building can be

Variable actions (Q)

variable action (G1) is made as aLeading variable The bracing-to-column connection single-lap joint through Accompanying a plate. The thickness variable action action Q1grade of the Q2 The of plates is 20mm, and their width is 250mm. The strength M30 bolts is 8.8. Favourable effect 1,0 - through the shear resistance of bolts is calculated assuming that the-shear plane does not pass

reduced. Now there are 6 bolts per plate, so the bearing resistance is: When designing column, the loads on storeys are assumed uniformly distributed. Also in the FbRdtot  n FbRd FbRdtot  2156 kN case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © Unfavourable effect threaded portion of the bolts. 1,35 Fatigue-inducing

action Shear Fireresistance design of bolts:

1,5

1,0

Serviceability limit of bolts is1,0determined by assuming 1,0 that the shear plane 1,0 does  not pass The shear resistance state through the threaded portion of the bolts: The partial safety factors presented in this table are the basic values of EC3.

The characteristic values of the impose load for areas in residential, social, commercial and Block shear failure resistance of splice plates: administration buildings are provided in table below divided into categories according to their First, uses. calculate the effective shear area from the formula: specific

Lv  100 mm 

B Office areas  t1 Lveff Aveff C

k  2.5

 M0

 1.1

Aveff  5658.64789 mm

Assembly areas (subdivided into 5 sections depending on likely

density of occupation and crowding) Obtain the block shear failure resistance by substituting in formula: D

Shopping

E Storage areas fy Aveff VeffRd  3  M0

 Mb

Nsd  1126.68363 kN

a2  e1

Building occupancy categories:  fu  Lveff  2  Lv  a1   a2  k d0   Lveff  257.21127 mm Residential (including hospital hotel bedrooms etc.) y  fwards,

 fub  800 MPa  Mb  1.25 dM30  30 mm n  6 The weight of fixed partitions may be considered as self-weight and represented as an equivalent uniformly 2distributed load. No guidance is given as to the required magnitude of  dM30 2  AM30  706.85835 mm M30 suchAan equivalent 4 load intensity, and the designer should use a reasonable approach in making such an estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with n 0.6 fub AM30 normal weight FvRdheights.  1628.60163 kN FvRd  Nsd FvRd  partitions and storey Ok

a1  e1

Structural Structuralactions actions Structural Structuralactions actions

VeffRd  1054.35706 kN

Resistant the welds

The axial force values in diagonal members D1 and D2 is obtained from the reaction at end of

Design the fillet welds with a throat thickness of 5 mm, the axial force introduced into the weld

the roof bracing:

is assumed equal to the hollow section plastic tension resistance. The plastic tension resistance

Nsd1  qwd1  

of a hollow section with dimensions 150x150x8 is:  Mw

A150.150.8  4324 mm

 1.25

NplRd  A150.150.8

Lw 

fy  M0

3  w  Mw NplRd 4 fu a

a  5 mm

w

 0.9

 Nsd2  qwd2 

0.5 Ltot

Nsd1  946.94841 kN

cos ( 37) 0.5 Ltot

Nsd2  473.4742 kN

cos ( 37)

NplRd  1395.47273 kN

The diagonal is supported at purlin trusses, so the buckling length is: Lw  277.46514 mm

The entire resistance of the joints then determined by the resistance of the net cross-section:

t1  dM30

Lcy 

Lcy  6.20036 mm

Lcz  Lcy

Lcz  0.0062 m

Sabah Shawkat ©

The splice plates can be taken as tension cross-section. Thus, the resistance of a cross-section

Consider a hollow section of dimension 180x180x12.5. The following values for tension and

containing holes can be obtained from:

buckling resistance is obtained:

The tension resistance of the cross-section is the smallest of the following:

hp  220 mm 

 M2

 1.25

 t1  hp  n1 d0 Anet

 Anet 3414.4 mm

A  v  t1 hp

 Av 4840 mm

N tRd  Av 

Ntsd  A160.160.12.5

NtRd  1562 kN

 M0

 NtRd  0.9 Anet 

fu  M2

NtRd  1204.60032 kN

is the net area (the area of the holes subtracted from the gross area)

is the ultimate strength of the material

NtRd

is the design value of tension resistance

 M0

Ntsd  Nsd1

Ntsd  2163.56364 kN

Nsd1  946.94841 kN

 i  5.83 10 mm  3 

Anet

A160.160.12.5  67.04 10 mm

n1  2

Lc i 



E  210000 MPa

3

 

2    3  0.5  1     3  0.2   3  3

 3

3

 2

 3

Structural actions Structural actions

Lc  8 m

 1.79587

3

 2.55588

3

 0.2286

4068

Structural actions Determine the compression and bending resistance of the hollow section: The actions to which the structure is subjected are divided into permanent, variable and fy NbRd  494.58553 kN NbRd  Nsd2 NbRd   3 A160.160.12.5 accidental actions. A design value  M0 is obtained by multiplying the characteristic value of the load by the partial safety factor. Permanent actions include: Nsd2  473.4742 kN a) Self-weight of the structure b) Fixed equipment In framed structures, the stiffness of the joints can be taken into account, and the bending

moments transferred across the joint. This reduces the span moments and therefore produces an efficient design. The following formulae are obtained for the bending moments of a beam Partial safety factors for actions subjected to uniform load and with a semi-rigid joint at both ends, when the supports are Accompanying Variable actions (Q) assumed non-deflecting.Permanent actions variable action 2  c   q l M1    c  2  12 Favourable effect

(G1)

Leading variable

restraint moment action Q1 1,0

Accompanying variable action Q2

1,5

1,0

Variable actions include: SjL Where c EI a) Imposed loads is the rotational stiffness of the joint Sj Snow load b) L is the length of the section c) Wind load I is the second moment of area of the section The rotational stiffnessloads of a welded T joint in square and rectangular hollow sections can be Self-weight and imposed

determined as follows: The weight of the partition walls can be distributed to generate a uniform load. When  3 designing1000 the structural floor elements of a single storey building. The load must be taken in c t0 S j to account in the weakest area. The effect of concentrate load must be considered separately. 52 This formula yields for rotational stiffness an approximation which best corresponds with the In some cases, the imposed loads on the structural floor in a single storey building can be bending moment values of the joint, up to the yield moment (Melc) of the joint. reduced. Where When designing column, the loads on storeys are assumed uniformly distributed. Also in the is the rotational stiffness of the joint (Nm/Rad) j multi-storey buildings, caseSof imposed loads can sometimes be reduced. is the wall thickness of the hollow section (mm) t0 The ccharacteristic values the impose loadfrom for areas residential, 2) social, commercial and is the of constant obtained figuresin(N/mm

Sabah Shawkat © 2 Unfavourable 6  q l  c  effect M0   Fatigue-inducing  c  2  24

1,35 field moment 1,0

administration buildings are provided in table below divided into categories according to their

action

specific uses.

Fire design

Serviceability limit

1,0

1,0 

Building occupancy categories:

state

Residential (including hospital wards, hotel bedrooms etc.)

The partial safety factors presented in this table are the basic values of EC3.

The weight of fixed partitions may be considered as self-weight and represented as an

Office areas

Assembly areas (subdivided into 5 sections depending on likely

density of occupation and crowding) D

Shopping

Storage areas

A beam with semi‐rigid joints

The moment-rotation curve of a semi-rigid joint, and values of constant 1000/Co for T joints square hollow sections, when b1/b0 ≤ 0,7 and b0/t0 ≥ 10. With square hollow sections, the  constant c is taken from figure when b1/b0 is less than or equal to 0,7.

Structural Structuralactions actions Structural Structuralactions actions

Force quantities The forces on the columns are determined simply by the area of the load carried. The resistance must be checked separately for two different load combinations, since at this stage it is not known whether the dominant load is the snow load or the wind load. In case the wind load is dominant: Wind load  Lf  12 m

 H  14 m

qwkA  1.035m

2

B  48 m

 Gk1  0.5 

kN

 a  6 m kN m

 Q1

 1.5

 Q2

 1.5

 G1 

 s  1.5

1.35

kN m

 02

 0.6



 0.8

Sabah Shawkat ©

frame spacing

is the width of the building

Gk1

self-weight

is the height of the column

Snow load

is the distances between the columns

qwka

wind load

is the buckling length

Incorrect realization of the steel structure from the bearing steel members

Structural actions Structural actions

4070

Structural  Q2qwkA a qwd actions

qwd  9.315 m

1

kN

is the wind load on the column

The actions to which the structure is subjected are divided into permanent, variable and 1 A design 2 accidental is obtained by multiplying thebending characteristic value of the is the moment at the endload of  Mysdwactions.  qwd H Mvalue ysdw  228.2175 m kN 8 by the partial safety factor. Permanent actions include: the column a) Self-weight of the structure 5  Vysdw  equipment qwd H Vysdw  81.50625 kN is the shearing force on the column b) Fixed 8

Snow load:

a) Imposed loads b) Snow load c) Wind load Self-weight and imposed loads The weight of the partition walls can be distributed to generate a uniform load. When designing the structural floor elements of a single storey building. The load must be taken in

Partial safety sd   02factors  Q2 for s actions Accompanying  G d   G1Gk1 variable action

Variable actions include:

sd  1.08 m

2

to account in the weakest area. Thethe effect of concentrate must factors be considered separately. For the column in this example, equivalent uniformload moment are obtained as

kN

follows: In some cases, the imposed loads on the structural floor in a single storey building can be

2 Permanent actions Gd  0.675 m kNVariable actions (Q)

(G1)

Nsdw  0.5  Gd  sd  Lf B

Leading variable

Nsdw  505.44action kN Q1

Accompanying variable action Q2

Favourable effect

1,0

Unfavourable effect Wind load: Fatigue-inducing

1,35

1,5

1,0

reduced. 1 2 MQ q H 8 When designing column, the loads on storeys are assumed uniformly distributed. Also in the 9  25  1  buildings, 2 2 case of multi-storey M H can sometimes be reduced.   q H imposedqloads 8 128 128   The characteristic values of the impose load for areas in residential, social, commercial and

Sabah Shawkat ©   02 Q2qwkA a qwd action Fire design

qwd  5.589 m

2 1,0 Serviceability limit Mysdsnow  0.125 qwd H state

1

kN

1,0 Mysdsnow  136.9305 m kN

1,0 

The partial safety factors presented in this table are the basic values of EC3. 5 Vysdsnow  qwd H Vysdsnow  48.90375 kN 8

load:of fixed partitions may be considered as self-weight and represented as an TheSnow weight equivalent uniformly distributed load. No guidance isgiven as to the required magnitude of 2   sd 1.8 m kN sd   Q1 s such an equivalent load intensity, and the designer should use a reasonable approach in making such an estimate. However, a minimum value of 1,0 kN/m2 is typically used for offices with 2 G   G1Gk1  G  0.675 m kN normald weight partitions and storey heights. d Nsdsnow  0.5  Gd  sd  Lf B

Nsdsnow  712.8 kN

administration in transverse table belowload divided into categories according to their uniform  MQ  1.3buildings are provided specific uses.  M   1.8

restraint moment at the lower end of the column

The finale value for the equivalent uniform moment factors is obtained from the following

formula:occupancy categories: Building

Residential (including hospital wards, bedrooms etc.)   1hotel 2     8 q H     MQ  B 1.8      1.8      1.8  0.64  ( 1.3  1.8 )           My M MQ M Office areas MQ  M    25 q H2    128       Assembly areas (subdivided into 5 sections depending on likely C   My  1.8  0.64 ( 1.3  1.8) density of occupation and crowding)  My  1.48 D Shopping Try a hollow section with dimensions 400x200x10 and steel grade S355J2H. The crossE Storage areas section is class 4, as section of the hollow

b  200 mm 

h  400 mm 2

A  112.6 10 mm

Structural Structuralactions actions

Wt  1373.2 10 mm

i  14.30 10 mm

Wel  1150.0 10 mm

Structural Structuralactions actions

 t  10 mm

Wpl  1434.0 10 mm

I  23003 10 mm

 Lc  0.7 H

fy  355 MPa

 E  210 GPa

The local buckling of the cross-section has now been taken into account. Next, consider the buckling resistance of the hollow section. The cross-section slenderness is determined using

 M1

 1.1

 M0

 1.1

Lc  9.8 m

the formula.  4 

The Moment Mtsd assumed constant along the entire hollow section h

since

40  36.6

The hollow section is classified as Class 4

Lc i 



fy E

 A

4

 0.93164

The reduction factor for buckling is calculated from:

As the cross-section of the hollow section is class 4, the effective cross-section must be determined. The slenderness of the compression elements is calculated using the formula. The width reduction factor ρ for class 4 cross-section of square and rectangular hollow sections

 

2    4  0.5  1     4  0.2   4  4

 4

is calculated as follows:

4

 2

4

 1.11323

4

 0.58052

 4

The buckling resistance of the hollow section is calculated by multiplying the plastic





When

Sabah Shawkat ©

235 MPa fy



 0.81362

  0.673

p



1 When

  p  0.673



  0.22 2   p

 NbRd4   4Aeff 

b t

p



p

 0.22 p

( b  3 t )

 bnoneff  b  3 t  beff

beff  0.19223 m

heff  beff

heff  0.19223 m

A

NbRd4  2276.14254 kN

a hollow section subjected to compression is calculated using buckling curve C.

 3 

Lc i 



fy E

 

2    3  0.5  1     3  0.2   3 

bnoneff  0.02223 m

Aeff  0.01215 m

 M1

must be accounted for in the interaction expression. The reduction factor χ for the buckling of

Using the effective cross-section, let as determine the effective area and parameter β.A: A  eff  A   4 bnoneff t 

But if the bending moment must be determined using elasticity theory. The effect of torsion

56.8 

The dimension of the effective and no-effective elements of the cross-section are as follows beff 

compression resistance of the effective cross-section by the reduction factor.

p



Aeff A

A

 1.07897

3

 3

3

 2

3

 0.8969

3

 1.07296

3

 0.60174

 3

Determine the compression and bending resistance of the hollow section:  NbRd   3A  M  yRd  Wel 

fy  M1

Structural actions Structural actions

NbRd  2186.65454 kN MyRd  371.13636 m kN

4072

Structural actions MzRd  MyRd

MzRd  371.13636 m kN

TheThe actions which the structure is of subjected are class divided effect to of the calculation methods class 3 and 4 oninto this permanent, example on variable the finaleand accidental actions. A design value is obtained by multiplying the characteristic value of the load results as follows: by the partial safety factor. Permanent actions include: 3

3

3

 0.96271  0.96382 a) 4 Self-weight of thestructure 4

4

 1.03655

NbRd NbRd4

 0.96068

b) Fixed equipment The parameters μ and χ depending on the shape of the moment diagram are as follows:

In case the wind load is dominant Partial safety factors for actions Accompanying Permanent actions Variable actions (Q)   y   3  2  My  4  y  0.93278 variable action (G1) Accompanying Leading variable action Q1 1,0

b) Snow load N kysnow Mysdsnow c) sdsnow Wind load   0.79691 NbRd MyRd Self-weight and imposed loads

The weight of the partition walls can be distributed to generate a uniform load. When Regarding shear resistance, the wind load is dominant. we need to calculate the plastic shear designing the structural floor elements of a single storey building. The load must be taken in to account in the weakest The effect of concentrate load must be considered separately. resistance, since h/t=40area. < 59,1

 y N sdw   kyw   1 effect   Favourable  3 A fy   Unfavourable effect

Variable Nsdwactions kywinclude: Mysdw   0.96659 MyRd NbRd a) Imposed loads

kyw  1.19601 -

variable action Q2 ky  1 5-

In some cases, the imposed loads on the structural floor in a single storey building can be reduced. VplRd

1398 kN   Vysdw

81.50625 kN 

When designing column, the loads on storeys are assumed uniformly distributed. Also in the Thus the resistance of a 400x200x10 hollow section is sufficient case of multi-storey buildings, imposed loads can sometimes be reduced.

Sabah Shawkat © 1,35

1,5

The characteristic values of the impose load for areas in residential, social, commercial and

Fatigue-inducing 1,0 In case the snow load is dominant action

1,0

administration buildings are provided in table below divided into categories according to their

  yFire  design 3  2  My  4

specific uses.

y

 0.93278

Serviceability limit 1,0 1,0 1,0   y N sdsnow     1  kysnowstate ky  1 5 kysnow  1.27642   3 A fy   The partial safety factors presented in this table are the basic values of EC3. The calculating method for torsional resistance is determined by the web slenderness

h h The weight considered as self-weight and represented as an calculate the plastic torsion resistance  40 of fixed partitions  59 1 may be t t equivalent uniformly distributed load. No guidance is given as to the required magnitude of Plastic torsional load resistance is calculated using the formula: such an equivalent intensity, and the designer should use a reasonable approach in making such an estimate. a minimum value of 1,0 kN/m2 is typically used for offices with fy However, Wt M  partitions  MtplRd  255.86379 m kN tplRd normal weight and storey heights. 3  M0

Building occupancy categories:

Residential (including hospital wards, hotel bedrooms etc.)

Office areas

Assembly areas (subdivided into 5 sections depending on likely density of occupation and crowding)

Shopping

Storage areas

By adding the effect of torsion in the interaction expression, the following result is obtained: Nsd NbRd



ky Mysd MyRd

Steel structure frames using rectangular hollow sections  1

Structural Structuralactions actions Structural Structuralactions actions

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Structural actions

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Buckling

Lateral buckling Buckling resistance Buckling resistance of welded profile for fire situation Calculate the stress resistance of the roof beam The calculation of the Buckling Calculate the resistance to lateral-torsional buckling Calculate the compression resistance of the box column Calculate the moment resistance and the resistance to lateral-torsional buckling Determine the cross-section of the cantilever steel beam Calculate the moment resistance of the hybrid beam Calculate the dimensions and resistance of steel columns Calculate the effective widths and the effective second moment of area of the compression elements

Sabah Shawkat ©

7676 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling Buckling

 

chapter focus steel columns, why? Because strength steel is very high In In thisthis chapter wewe willwill focus on on steel columns, why? Because thethe strength of of steel is very high which leads to smaller cross sectional area a member to resist a particular force as compared which leads to smaller cross sectional area of aofmember to resist a particular force as compared concrete. problem buckling generally arises steel columns. have a very to to concrete. So So thatthat problem of of buckling generally arises in in steel columns. WeWe have a very interesting example understanding problem buckling, I hope interesting example forfor understanding thethe problem of of buckling, andand I hope thatthat youyou willwill getget involved. involved.

Lc Lc i

equation above describes slenderness ratio, limit slenderness column TheThe equation above describes thethe slenderness ratio, andand thethe limit of of slenderness of of column varies material changes. example, steel column is said to be short if the slenderness varies as as thethe material changes. ForFor example, steel column is said to be short if the slenderness ratio is less than is called intermediate column is between ratio is less than 50,50, it isit called an an intermediate column if itif isit between 50 50 andand 200200 andand it isit is called a long column is greater than 200. Long columns very susceptible to elasticity. called a long column if itif isit greater than 200. Long columns areare very susceptible to elasticity. know effective length of the column depends conditions of the column AsAs wewe know thatthat thethe effective length of the column depends on on thethe endend conditions of the column

Have ever seen Charlie Chaplin stick when rests Have youyou ever seen Charlie Chaplin stick when he he rests on on it? it?

support. support.

stick in the picture bellow describes of the most fundamental characteristic a column TheThe stick in the picture bellow describes oneone of the most fundamental characteristic of aofcolumn

Buckling defined sudden, large, lateral deflection a column owing a small Buckling cancan be be defined as as thethe sudden, large, lateral deflection of of a column owing to atosmall

field structural engineering, called "Buckling Column". why it buckle? in in thethe field of of structural engineering, called "Buckling of of Column". ButBut why diddid it buckle?

increase in an existing compressive load. This response leads to instability collapse increase in an existing compressive load. This response leads to instability andand collapse of of thethe

What made stick to bend instead taking load straight down ground? Well, here What made thethe stick to bend instead of of taking thethe load straight down thethe ground? Well, here

member. section shall describe critical, buckling, load welded bolted member. In In thisthis section wewe shall describe thethe critical, or or buckling, load forfor welded andand bolted

to discuss about event. wewe areare to discuss about thethe event.

profiles, compressive load cases instability. profiles, thethe compressive load thatthat cases thethe instability. calculate critical load flexural - torsional buckling, to allow restraint ToTo calculate thethe critical load forfor flexural - torsional buckling, andand to allow forfor thethe endend restraint conditions. compressive force in this context, thought to be applied at the centroid conditions. TheThe compressive force is, is, in this context, thought to be applied at the centroid of of

Sabah Shawkat © cross-section. thethe cross-section.

1757, mathematician Leonhard Euler derived a formula gives maximum axial load In In 1757, mathematician Leonhard Euler derived a formula thatthat gives thethe maximum axial load

Average stress in columns versus slenderness ratio Average stress in columns versus slenderness ratio

a long, slender, ideal column carry without buckling. ideal column is one thatthat a long, slender, ideal column cancan carry without buckling. AnAn ideal column is one thatthat is is perfectly straight, homogeneous, from initial stress. maximum load, sometimes perfectly straight, homogeneous, andand freefree from initial stress. TheThe maximum load, sometimes called critical load, causes column to be a state unstable equilibrium; called thethe critical load, causes thethe column to be in ainstate of of unstable equilibrium; thatthat is, is, thethe introduction of the slightest lateral force cause column to fail buckling. Buckling introduction of the slightest lateral force willwill cause thethe column to fail by by buckling. Buckling is is called instability occurred a structure because of excessive loading. what mean called an an instability occurred in ainstructure because of excessive loading. ButBut what do do wewe mean slender? When longitudinal dimensions member much greater than cross by by slender? When thethe longitudinal dimensions of of thethe member areare much greater than thethe cross section member then is called a slender member remember should section of of thethe member then it isit called a slender member andand remember youyou should useuse thethe word slender while describing compression in column when there is any tension. word slender while describing thethe compression in column notnot when there is any tension.

Buckling Buckling Buckling Buckling

idealized supports shown below seldom occur. Because of uncertainty relative to the fixity TheThe idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns sometimes taken to be pin-ended. buckling of pin-ended columns of the joints, columns areare sometimes taken to be pin-ended. ForFor buckling of pin-ended columns centrically loaded compressive forces at each end. member is assumed to be perfectly centrically loaded by by compressive forces F atF each end. TheThe member is assumed to be perfectly straight constructed a linearly elastic material-that have ideal column. straight andand to to be be constructed of of a linearly elastic material-that is, is, wewe have an an ideal column. When load F has been increased sufficiently cause a small lateral deflection. This When thethe load F has been increased sufficiently to to cause a small lateral deflection. This is ais a condition of neutral equilibrium. corresponding load value of the load is critical load. condition of neutral equilibrium. TheThe corresponding load value of the load is critical load.

77 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

The critical load is dependent upon the end restrains. For columns with various combinations

failure of an intermediate column occurs by inelastic buckling at stress levels exceeding the

of fixed, free, and pinned supports.

proportional limit. Presented below is one practical approach to determination of inelastic

The behaviour of an ideal column is often represented on a plot of average compressive stress

buckling, this approach is known as the tangent-modulus theory.

versus slenderness ratio. Such a representation offers a clear rationale for the classification of compression bars. Tests of columns verify each portion of the curve with reasonable accuracy. The range of is a function of the material under consideration.

Sabah Shawkat © 2

Fcr A

 E 

 E

 L    i

AL

cr

  EI 

  E I

cr

Because

AL

Fcr

  E I 2

we suppose that the value of =10 cr

Fcr A

 E

 Lc     i

Long columns

 Lc     i

 2E pl



Lc i

210 GPa

max

F ( kN)

 2

A  cm

Short columns

Theoretical buckling length Lc of compression members Most structural columns lie in a region between short and long classifications. Such

intermediate-length columns do not fail by direct compression or by elastic instability. The

Buckling Buckling

  E I

  E

  E i

cr

  EI

Fcr

Fcr A

Then

7678 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

As we know that buckling is characterized by a Buckling

sudden sideways failure of a structural member In this chapter we will focus on steel columns, why? Because the strength of steel is very high subjected to high compressive stress. As an applied which leads to smaller cross sectional area of a member to resist a particular force as compared load is increased on a member, such as a column, it to concrete. So that problem of buckling generally arises in steel columns. We have a very will ultimately become large enough to cause the interesting example for understanding the problem of buckling, and I hope that you will get member to become unstable and is said to have involved. buckled. HaveUsually you everbuckling seen Charlie Chaplin stick he rests and instability are when associated to on it?

I1 I1 h FI

I1 h FI

I1  I2



Lc i

The equation above describes the slenderness ratio, and the limit of slenderness of column varies as the material changes. For example, steel column is said to be short if the slenderness I y

ratio Iis1 less 2 than 50, it is called an intermediate column if it is between 50 and 200 and it is called a long column if it is greater than 200. Long columns are very susceptible to elasticity. As we knowIythat 2the effective length of the column depends on the end conditions of the column CM

 h FI

support.

compression, but recently Zaccaria,one Bigoni, The stick in the picture bellow describes of theNoselli most fundamental characteristic of a column and Misseroni (2011) have shown that buckling and of Column". But why did it buckle? in the field of structural engineering, called "Buckling

Buckling can be 2 defined as the sudden, 2 2 large, lateral deflection of a column owing to a small

can to also occur in of elastic Whatinstability made the stick bend instead takingstructures the load straight down the ground? Well, here subject to dead tensile load. we are to discuss about the event.

member. In this section we shall describe the critical, or buckling, load for welded and bolted

Safety factor vs loading case

  E Iy  G IT

2 Iy  h FI



M Kiin an existing compressive  E  increase load. This response leads to instability and collapse of the 2

profiles, the compressive load that cases the instability. 2 2    2  M Ki  E Iy   G IT   E CM 2 critical  2 flexural  To calculate the load for - torsional buckling, and to allow for the end restraint L L   conditions. The compressive force is, in this context, thought to be applied at the centroid of

Sabah Shawkat © Buckling Force

Safety Coefficients

the cross-section. or

And loading case

Euler´s Buckling Force

SFKl=FKFl





SKFI

  E IFI 2

  E I1



 E Iy  G IT 1 

2 E C M





G IT

Buckling according to Euler´s Force

Engeser Buckling Force



Critical Buckling Force



cr

M Ki

Fcr



 

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

Average stress in columns versus slenderness ratio

steel columns subject largewithout compressive forces slenderiscross-section, that When a long,the slender, ideal column cantocarry buckling. Anhave idealacolumn one that is meaning comprising of thin elements in rolled for perfectly straight, homogeneous, and free (thin from web initialand/or stress.thin Theflange maximum load,I-sections sometimes instance), the thin web might to buckling at localized areas, called the critical load,flange causesand/or the column to bebe in subject a state of unstable equilibrium; that is,which the in turn prevents the section from attaining the the cross-section In suchBuckling a case, steel introduction of the slightest lateral force will cause column to capacity. fail by buckling. is column is said tooccurred have undergone localbecause buckling. called an instability in a structure of excessive loading. But what do we mean Lateral When torsional as dimensions the name suggests, is a condition slender steel by slender? the bucking, longitudinal of the member are much wherein greater than the cross columns lateral combined from plane.use the section of theundergoes member then it isdisplacement called a slender memberwith and twisting remember youitsshould

word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load. Steel column from the I profile on the facade of the building with an shelter construction

79 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

The lateral buckling in beams is a typical problem when them there are not an enough lateral stiff-ness or lateral restrictions along of the length which cause the bent about the minor axis.

  2 E I  1  2  2   h FI L  

M Ki

  E Iy  G IT



To define simply, buckling is aphenomena wherein a compression member is subjected to unwanted bending stresses due to unintended or accidental eccentricities of axial compression force and/or due to curvatures of the compression member. This bending stress causes the

SKFI

  E IFI

  E I1

compression member to bend out of the axis leading to further increase in stress causing the member to ultimately fail.

The critical stress at which the member buckles is arrived upon by solving the elastic curve

M Ki

  EIy  GIT 2



2

 SKFI h FI

equation for column. Below this stress the member maintains its straightness by virtue of its stiffness in stable equilibrium. Above this stress level stable equilibrium ceases and condition

Buckling resistance

changes to unstable equilibrium because stiffness is much less than secondary moment and the The picture bellow describes the buckling load in various cases, but what is effective length?

member fails. This critical stress is given by:

Well, in simple works effective length is the length of the column between points of zero 2

cr

  EI 

moments. In case 1, it is pinned support so the distance between point of zero moments will definitely be L. In case 2, total freedom of rotation and side movement – like the top of a

Sabah Shawkat ©

This critical stress fixes the upper limit of load that may be applied axially even though ultimate

flagpole. This is the weakest end condition. In case 3, the column is fixed-fixed, so the point of

axial stress capacity of the member may be very high.

zero moments will be at a distance of L/2, that is the point from where the curvature of the

Yet another related phenomena is the lateral torsional buckling of laterally unrestrained beams

column changes. You just need to replace the value of L in the Euler's formula with the value

or beam columns. When I sections are used as beams or beam columns the compression flange

of effective L and you will get the modified value. I would recommend everyone to once go

is under compressive stress and has a tendency to buckle but it is attached to the tension flange

through the calculations behind this.

which resists the buckling giving rise to torsion within the beam section. This torsion twists and warps the unrestrained part of beam leading to lateral torsional buckling.

M Ki

  E Iy  G IT L b t

 cm



kN m



12 b t

 kN cm2



3 Ix 2 b

 F

Effective lengths of columns for various end conditions

Buckling Buckling

7680 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Although the buckling of a column can be compared with the bending of a beam, there is an Buckling

Lc in buckling is presented in the form: The design criterion for a compression member 

important difference in that the designer can choose the axis about which a beam bends, but In this chapter we will focus on steel columns, why? Because the strength of steel is very high normally the column will take the line of least resistance and buckle in the direction where the which leads to smaller cross sectional area of a member to resist a particular force as compared column has the least lateral unsupported dimension. to concrete. So that problem of buckling generally arises in steel columns. We have a very As the loads on columns are never perfectly axial and the columns are not perfectly straight, interesting example for understanding the problem of buckling, and I hope that you will get there will always be small bending moments induced in the column when it is compressed. involved. There may be parts of the cross-section area where the sum of the compressive stresses caused Have ever on seentheCharlie Chaplin helarger rests on it? the allowable or even the ultimate by you the load column could stick reachwhen values than

Nsd  NbRd The equation above describes the slenderness ratio, and the limit of slenderness of column

material. Thestrength stick inof thethe picture bellow describes one of the most fundamental characteristic of a column For example welded in different can have differentBut types of did buckling modes. in the field of structuralprofiles engineering, called shapes "Buckling of Column". why it buckle?

for Class 1,2 or 3 cross-section 1 A Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small Aeff for Class 4 cross-section  A in an existing increase compressive load. This response leads to instability and collapse of the A member. In this section we shall describe the critical, or buckling, load for welded and bolted A is the area of the cross – section. profiles, the compressive load that cases the instability. is the effective area of the cross – section. Aeff To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint

Among modes flexural, torsional anddown flexural-torsional buckling. What madethe thepossible stick tobuckling bend instead ofare taking the load straight the ground? Well, hereEC 3, part 1.1 gives instructions for calculating flexural buckling. Torsional buckling can occur as we are to discuss about the event. pure torsional buckling, whereby any cross-section rotates about the shear centre, or as flexural – torsional buckling. whereby rotation and lateral displacement take place simultaneously. Torsional buckling does not normally occur in box profiles due to their large torsional stiffness,

varies as the material changes. For example, steel column is said to be short if the slenderness Where ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is fy called a long column NbRd  A A if it is greater than 200. Long columns are very susceptible to elasticity.  M1 As we know that the effective length of the column depends on the end conditions of the column is the reduction factor for the relevant buckling mode.  support.

is the partial safety factor the material  M1 conditions. The compressive force is, in this for context, thoughtintocalculation be appliedofatthe thestability centroid of

Sabah Shawkat © which means that torsional buckling is mainly associated with the design of open profiles.

the cross-section.

Where the cross-section is double-symmetric or symmetric about its centroid and retains its

For constant axial compression is members of constant cross-section, the value of the reduction

shape, there are three possible buckling modes, flexural buckling in the symmetry plan and

factor may be determined from:

flexural-torsional buckling, also called space buckling.













 1

Where 

0.5 1   (   0.2)    2



In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the Different buckling modes word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

is an imperfection factor

 fy      A i  E  Average stress in columns versus slenderness ratio Lc is the buckling length in flexural buckling The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity i is the radius of gyration about the relevant axis, determined using the of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns properties of the gross cross-section centrically loaded by compressive forces F at each end. The member is assumed to be perfectly In the case of pure torsional buckling the relative slenderness is determined from: straight and to be constructed of a linearly elastic material-that is, we have an ideal column. 



A fyF has been increased sufficiently to cause a small lateral deflection. This is a When the load T  A condition of N neutral equilibrium. The corresponding load value of the load is critical load. crT

81 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

The tortional buckling load NcrT is calculated as follows: 2  EI 1   GIv  2 2 i0  LT

NcrT

is the yield strength of the compression flange

fyf

   

Verify that the flange of the cross-section.  hw  1110 mm

Where i0

 tw  8 mm

tf  8 mm

bf  850 mm  Es  210 GPa

 fyf  355 MPa 2

iy  iz  y0

 Aw  hw tw

Y0 is the distance between the shear centre and the centroid of the gross cross-section.

k  0.55

is the shear modulus.

is the geometrical torsional rigidity of the gross cross-section

is the geometrical warping rigidity of the gross cross-section

tw Aw

is the buckling length for torsional buckling.

Afc

class 3 flange

 Es  Aw   fyf  Afc

 k 

Aw  8880 mm

Af  bf tf

Afc  Af

hw tw

Af  6800 mm

 138.75

 Es  Aw  371.8   fyf  Afc

k 

the criterion is satisfied

 1.30588 The intersection point of the

hw tw

and

Aw Afc

lines lies above the curve Class 3

or 4

Imperfection factor

Sabah Shawkat ©

Buckling curve

Imperfection factor

0.21

0.34

0.49

0.76

Flange induced buckling

If the profile web is slender enough, the compression flange may buckle locally in the plan of

the web. This arises from the fact that a slender web is not capable of giving enough support for the flange. The webs of those profiles used as columns are normally strong enough to prevent local buckling. Instead, to design a beam with a slender web, the phenomenon should be paid attention to. To prevent local buckling. The criterion below shall be satisfied. hw tw

 k

E fyf



Aw A fc

is the free distance between the flanges

is 0.3 for class 1 flanges, 0.4 for class 2 flanges, 0.55 for class 3 or 4 flanges

is the area of the web

A fc

is the area of the compression flange

The limit curve for local buckling due to flange deflection (fy = 355 MPa)

Buckling Buckling

7682 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Lc of structural hollow sections can be utilized In frame structures with rigid joints, the benefits

Buckling resistance of welded profile for fire situation Buckling



In The this chapter will focus why? Because strength ofcross-section steel is very high bucklingwe resistance foron thesteel fire columns, situation is calculated as the follows in the classes which 1. leads to smaller cross sectional area of a member to resist a particular force as compared to concrete. So that problem of buckling generally arises in steel columns. We have a very   fi     Afor  kzunderstanding  fy interesting example the problem of buckling, and I hope that you will get  1.2  NbfiRd  Mfi involved.

when determining the column buckling lengthi values. Another factor influencing the buckling Thelength equation aboveis describes slenderness ratio,Aand the limit of slenderness of column in frames the lateral the support of the frame. non-sway structure can be stiffened either varies the material Foritexample, steel column is said to be(ashort the slenderness withaslattices or bychanges. supporting with a rigid structural element lift if shaft or stair well). ratio is less than 50, it aisframe calledstructure an intermediate column ifnon-sway it is between and 200condition and it is is Generally speaking, can be classified if the50 following called met.a long column if it is greater than 200. Long columns are very susceptible to elasticity. As we know that the effective length of the column depends on the end conditions of the column Vsd

Have fiyou ever seen Charlie Chaplin stick when he rests on it? is the reduction factor in flexural buckling for the fire design

support.  01

The Astick in the picture bellow describes one of the most fundamental characteristic of a column is the cross-section area in the field of structural engineering, called "Buckling of Column". But why did it buckle? the reduction factor of fortaking the yield of steel at temperature  Whatkymade the isstick to bend instead the strength load straight down the ground?θaWell, here

Where can be defined as the sudden, large, lateral deflection of a column owing to a small Buckling

Vcr

The modified slenderness needed in calculating the reduction factor is obtained as follows. we are to discuss about the event. 0

lfi

lfi i



increase in an existing compressive load. This response leads to instability and collapse of the is the design value of the vertical total load Vsd member. In this section we shall describe the critical, or buckling, load for welded and bolted is the buckling load frame elasticity theory in case sway buckling mode. Vcr the profiles, compressive loadaccording that casesto thetheinstability.

stiffening of sway structure is based on columns functioning as cantilevers fixed to ToThe calculate the critical load for flexural - torsional buckling, and to allow for the endand restraint foundations rigid joint,force or onis,the of thethought joints. to be applied at the centroid of conditions. Thewith compressive in rigidity this context,

fy ky  E kE

Sabah Shawkat © kE

is the buckling length of the member in the fire situation

theIncross-section. the case of a continuous column, the buckling length can be determined using figures

is the reduction factor for the modulus of elasticity of steel at temperature θa

below. The distribution factors 1 and 2 in the figures are determined below:

The buckling length of a column in the fire situation is normally determined as at normal temperature. The buckling length of a column in non-sway frame can be determined according

to figure bellow, if the fire resistance of the structural members liable to buckle and separating the fire compartments is at least equal to the fire resistance of the column concerned.

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

Average stress in columns versus slenderness ratio

that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called alength slender remember Buckling of member columns and is braced framesyou should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a Figure: Column buckling length values of non-sway frames Lc / L condition of neutral equilibrium. The corresponding load value of the load is critical load.

83 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

The buckling length of columns in rigid joined structures is obtained from figure below for

K11, K21, K12 and K22 are effective stiffness coefficient for corresponding hollow sections.

a non-sway frames and for sway frames. The curve values represent the relation of buckling

The moments in the hollow section are assumed to be elastic

length to the actual column length. M sd 

W el  fy  M0

The hollow section is assumed pinned if its moment exceeds the elastic moment. Lateral-torsional buckling of an I-profile with fixed support at the one end, and effect of the shape of the cross-section. Slender and deep I-profiles are prone to lateral-torsional buckling. Lateral-torsional buckling does not normally determine the moment resistance of a box profile because of its high torsional stiffness. The factors that affect lateral-torsional buckling are the length of the member, its loading, cross-sectional dimensions, its end restraint conditions and the material properties. The greater the ratio Iy /Ix is, the more easily the beam buckles in the lateral-torsional mode. A uniform moment yields the most severe loading case. The design resistance to lateral-torsional buckling is calculated from the formula:

Sabah Shawkat ©  LT w Wply 

MbRd

Figure: Column buckling length values of sway frames Lc / L

1

2

Kc  K1 Kc  K1  K11  K12 Kc  K2

Kc  K2  K21  K22

(upper assembly point)

( lower assembly point)

1 L

I1 L1

w w

w

for Calss1 or Class 2 cross-section

Wely

for Class 3 cross-section

Wply

Weffy

for Class 4 cross-section

Wply

 LT

is the reduction factor for lateral-torsional buckling.

The reduction factor for lateral-torsional buckling is calculated from the formula as follow: 1

 LT  LT 

I2 L2

 M1

where  LT

I, I1 and I2 are values of inertia for corresponding columns parallel to frame. L, L1 and L2 are values of height for corresponding columns

Buckling Buckling

 LT

but 2

 LT

 LT

 1

  LT

2 0.5 1   LT  LT  0.2   LT 

 w Wply fy

Mcr

 LT

0.49 for welded sections

7684 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling Mcr

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Lc is positive for loads acting towards the shear The distance of the loading from the shear centre

is the elastic critical moment for lateral-torsional buckling



where the we relative slenderness no allowance forstrength lateral-torsional LT  0.4why? In this chapter will focus on steelcolumns, Because the of steel isbuckling very highis which leads to The smaller cross sectional area of member to resistbuckling a particular as compared necessary. elastic critical moment foralateral-torsional for force a profile symmetric to concrete. thataxis problem of by buckling generally arises in steel columns. We have a very about its So minor is given the formula as follow: interesting example for understanding the problem of buckling, and I hope that you will get 2   EIz     k   Iw  ( k L) GIv  C z  C z 2  C z  C z Mcr C1  2 g 2 g  2 g 3 j    Charlie 2  Chaplin 2 he rests on it? Have you ever seen Iz when  kw  stick ( k L)  EI

involved.





The stick in the picture bellow describes one of the most fundamental characteristic of a column in the fieldCof structural called "Buckling of Column". But why did it buckle? where are factors depending on the load and end restraint conditions 1, C 2 and C3engineering, WhatImade the is stick bend instead of taking the load straight down the ground? Well, here the to geometrical torsional rigidity v we are to discuss about the event. is the geometrical warping rigidity Iw

is the second moment of area about the minor axis

Iz L

is the length of the profile between points, which have lateral restraints

centre. For example, for loading situated on i the top flange and directed downwards, zg is The equation above describes the slenderness ratio, and the limit of slenderness of column positive. If the same direction acts on the bottom flange, zg is negative. varies as the material changes. For example, steel column is said to be short if the slenderness In cross-section that are symmetric about the minor axis, zj can be approximately calculated ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is using the method described below. the geometrical warping rigidity is calculated from the called a long column if it is greater than 200. Long columns are very susceptible to elasticity. formula below. As we know that the effective length of the column depends on the end conditions of the column Ifc 2 support. Iw  f  1   f  Izhf f where  Ifc  Ifc Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small If.c is the second moment of area of the compression flange about the minor axis of the increase in an existing compressive load. This response leads to instability and collapse of the section member. In this section we shall describe the critical, or buckling, load for welded and bolted Ift is the second moment of area of the tension flange about the minor axis of the section profiles, the compressive load that cases the instability. hf is the distance between the shear centres of the flanges To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint hf conditions. The force is, thought to be applied at the centroid of forin this context, zj 0.8  2 compressive  j  1   f  0.5 2 the cross-section. hf for zj 1.0  2  f  1   f  0.5 2

Sabah Shawkat © (distance between lateral-torsional supports)

k and kw are factors related to the effective length

zs 

 0.5   

 y2  z2 z dA

for double-symmetric profiles, zj = 0. Moreover, the formula for the geometrical warping rigidity of double-symmetric profiles can be presented in the form:

is the co-ordinate of the point of load application

is the co-ordinate of the shear centre

0.25 hj Iz

The geometrical torsional rigidity of the cross-section for I-profiles is obtained from the formula below:

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the the member arethe much greater Keylongitudinal to symbols dimensions for a profileofsymmetric about minor axis than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio 3  1  3 Iv  bi ti  3 supports shown  below seldom occur. Because of uncertainty relative to the fixity The idealized i  1 



of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns

centrically loaded compressive forces F at each end. The member is assumed to be perfectly is thebywidth of the plate element bi straight and to be constructed of a linearly elastic material-that is, we have an ideal column. thickness of the same plate element Whenti the loadis Fthehas been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load.

85 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Values of effective length factors for different support conditions

The picture on next page shows the buckling of different types of supported column. Load F is

Supports at the ends

Cross-section

Cross-sections free

of the member

restrained against

to warp at both ends

Cross-section

applied on the column and from it's straight configuration it moved to a buckled configuration.

restrained against

Now, why did it move? Because till now we have learnt that the axial forces directly transfers

warping at both

warping at one end,

ends

the other end being

the force without any instability. Here's the answer.

free to warp Restrained against

k = 0,5

rotation at both ends

kw = 0,5

kw = 1,0

kw = 0,7

Free the rotate at

k = 1,0

k =1,0

both ends

kw = 0,5

kw = 1,0

kw = 0,7

rotation at one end,

k = 0,7

k = 0,

k = 0,7

the other end being

kw = 0,5

kw = 1,0

kw = 0,7

Both ends perfectly pinned 2

  E I

Fcr

Restrained against

One end fixed other free 2

2.5 E I

Fcr

  E I

  L    2.5 

2 2

free to rotate

  E I

Fcr

( 2 L)



because

3.948

2.5

2 L

Sabah Shawkat ©

The factor k in the table above is related to the rotation of the end in the xy-plane, and the

One end perfectly fixed other pinned

factor kw to the warping of the end cross-sections of the member.

Fcr

Values of factors C1, C2, C3 corresponding to the values of factor k Loading and support

Bending moment

conditions

diagram

20 E I

  E I

 L

Values factors

  E I

( 0.7 L)

Because



0.7  0.49

 0.493

0.7 L

1,0

1,132

0,459

0,525

0,5

0,972

0,304

0,980

1,0

1,285

1,562

0,753

0,5

0,712

0,652

1,070

1,0

1,365

0,553

1,730

0,5

1,070

0,432

3,050

Both ends perfectly fixed

1,565

1,267

2,640

0,5

0,938

0,715

4,800

1,0

1,046

0,430

1,120

0,5

1,010

0,410

1,890



2 2

E I L



E I

( 0.5  L )

Because



1,0



40  E  I

F cr

 0.247

0.5

 0.25

F cr

40  E  I L

 

2 2

E I L

Buckling Buckling



E I

( 0.5  L )

7686 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling

Values at buckling coefficient k as a function Lof c slenderness ratio 



In this chapter we will focus on steel columns, why? Because the strength of steel is very high

Values of k for e = 24 daN/mm2 (240MPa, 24kN/cm2, kg/mm2) The equation above describes the slenderness ratio, and the limit of slenderness of column

which leads to smaller cross sectional area of a member to resist a particular force as compared

 as the material changes. For example, steel column is said to be short if the slenderness varies

to concrete. So that problem of buckling generally arises in steel columns. We have a very interesting example for understanding the problem of buckling, and I hope that you will get involved. Have you ever seen Charlie Chaplin stick when he rests on it? The stick in the picture bellow describes one of the most fundamental characteristic of a column in the field of structural engineering, called "Buckling of Column". But why did it buckle? What made the stick to bend instead of taking the load straight down the ground? Well, here we are to discuss about the event.

0 1 2 3 4 5 6 7 8 9 0 1,0 1,0 1,0 1,0 1,001 1,001 1,001 1,002 1,002 1,003 ratio10 is less than it is called an intermediate column if it is1,009 between1,010 50 and 1,012 200 and1,013 it is 1,004 50, 1,004 1,005 1,006 1,007 1,008 20 1,018 1,019 1,021 1,023 1,025 1,028 1,030 1,032 called a long1,015 column1,016 if it is greater than 200. Long columns are very susceptible to elasticity. 30 1,035 1,037 1,040 1,043 1,046 1,049 1,052 1,056 1,060 1,063 As we that the effective length of1,080 the column depends on the end conditions the column 40 know 1,067 1,071 1,075 1,085 1,090 1,095 1,100 of 1,105 1,111 50 1,117 1,123 1,130 1,137 1,144 1,151 1,159 1,166 1,175 1,183 support. 60 1,192 1,201 1,211 1,221 1,231 1,242 1,253 1,265 1,277 1,289 70 1,302 1,315 1,328 1,342 1,357 1,372 1,387 1,403 1,420 1,436 Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small 80 1,453 1,471 1,489 1,508 1,527 1,547 1,567 1,587 1,608 1,629 increase existing compressive This 1,743 response1,767 leads to1,792 instability and collapse the 90 in an 1,651 1,674 1,696 load. 1,719 1,817 1,842 of1,868 100 1,894 1,921 1,947 1,975 2,003 2,031 2,060 2,089 2,118 2,148 member. In this section we shall describe the critical, or buckling, load for welded and bolted 110 2,178 2,209 2,204 2,271 2,303 2,335 2,367 2,400 2,433 2,467 profiles, compressive that cases the instability. 120 the 2,501 2,535 load2,570 2,605 2,640 2,676 2,712 2,748 2,785 2,822 130 2,860 2,897 2,936 2,974 2,013 2,052 3,091 3,131 3,172 3,212 To calculate the critical load for flexural - torsional to allow for the3,591 end restraint 140 3,253 3,294 3,335 3,377 3,419 buckling, 3,462 and 3,504 3,548 3,635 150 3,679 3,723 3,768 3,813 3,858 3,904 3,950 3,997 3,043 3,090 conditions. The compressive force is, in this context, thought to be applied at the centroid of 160 4,137 4,18 4,23 4,28 4,33 4,38 4,43 4,48 4,53 4,58 the cross-section. 170 4,63 4,68 4,73 4,78 4,83 4,88 4,94 4,99 5,04 5,09 180 5,15 5,20 5,26 5,31 5,36 5,42 5,48 5,53 5,59 5,64 190 5,70 5,75 5,81 5,87 5,93 5,99 6,05 6,11 6,16 6,22 200 6,28 6,34 6,40 6,46 6,53 6,59 6,65 6,71 6,77 6,84 210 6,90 6,96 6,03 7,09 7,15 7,22 7,28 7,35 7,41 7,48 220 7,54 7,61 7,67 7,74 7,81 7,88 7,94 8,01 8,08 8,15 230 8,22 8,29 8,36 8,43 8,49 8,57 8,64 8,71 8,78 8,85 240 8,92 8,99 9,07 9,14 9,21 9,29 9,36 9,43 9,51 9,58 250 9,66 9,74 9,81 9,88 9,96 10,04 10,11 10,19 10,27 10,35 260 10,43 10,50 10,58 10,66 10,74 10,82 10,90 10,98 11,06 11,14 270 11,22 11,30 11,38 11,47 11,55 11,63 11,71 11,80 11,88 11,96 280 12,05 12,13 12,22 12,30 12,39 12,47 12,56 12,64 12,73 12,82 290 12,90 12,99 13,08 13,17 13,26 13,35 13,44 13,52 13,61 13,71 300 13,79

Sabah Shawkat ©

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load Buckling length in some common cases that a long, slender, ideal column can carry without buckling. An ideal column is one that is Every system the all existingand matter space considered as a maximum whole has load, got different stages perfectly straight,inhomogeneous, freeand from initial stress. The sometimes



Lc i



I kAverage stress in columns versus slenderness ratio F kF i A needed A A 

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity

of equilibrium system get deformed the of minimum load. that Consider called the critical and load,the causes thewill column to be in aatstate unstablepossible equilibrium; is, thethe system shown in slightest figure above. assume deflections system are very small, introduction of the lateralWe force will here causethat thethe column to failin bythe buckling. Buckling is

of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns

thatan is instability the columnoccurred has slightly buckled. because Now if you take the equilibrium the do buckled shape called in a structure of excessive loading. Butof what we mean of the column you will find that the load required to buckle the column is less that the load by slender? When the longitudinal dimensions of the member are much greater than the crossthat

straight and to be constructed of a linearly elastic material-that is, we have an ideal column.

can beofactually carried by it theisfull cross sectionalmember area. and remember you should use the section the member then called a slender

condition of neutral equilibrium. The corresponding load value of the load is critical load.

word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

centrically loaded by compressive forces F at each end. The member is assumed to be perfectly

When the load F has been increased sufficiently to cause a small lateral deflection. This is a

87 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Values at buckling coefficient k as a function of slenderness ratio



Values at buckling coefficient k as a function of slenderness ratio

Values of k for e = 30 daN/mm2 (300MPa, 30kN/cm2, kg/mm2)  0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

0 1,0 1,004 1,018 1,045 1,088 1,158 1,266 1,424 1,639 1,910 2,234 2,604 2,019 3,476 3,974 4,512 5,089 5,705 6,359 7,052 7,782 8,551 9,358 10,20 11,09 12,01 12,96 13,96 15,00 16,07 17,18

1 1,0 1,005 1,020 1,048 1,094 1,167 1,279 1,443 1,664 1,940 2,269 2,644 2,063 3,524 4,026 4,568 5,149 5,768 6,427 7,123 7,858 8,630 9,441 10,29 11,18 12,10 13,06 14,06 15,10 16,18

2 1,0 1,006 1,023 1,052 1,100 1,176 1,293 1,462 1,689 1,971 2,304 2,684 3,107 3,573 4,078 4,624 5,209 5,832 6,494 7,195 7,933 8,710 9,524 10,38 11,27 12,20 13,16 14,16 15,21 16,29

3 1,0 1,008 1,025 1,056 1,106 1,186 1,307 1,482 1,715 2,002 2,340 2,724 3,152 3,621 4,131 4,681 5,269 5,897 6,563 7,267 8,009 8,790 9,608 10,46 11,36 12,29 13,26 14,27 15,31 16,40

4 1,001 1,009 1,027 1,060 1,112 1,196 1,322 1,503 1,741 2,034 2,376 2,765 3,197 3,671 4,184 4,738 5,330 5,962 6,631 7,339 8,085 8,870 9,692 10,55 11,45 12,39 13,36 14,37 15,42 16,51

5 1,001 1,010 1,030 1,064 1,119 1,206 1,338 1,524 1,768 2,067 2,413 2,806 3,243 3,720 4,238 4,795 5,392 6,027 6,700 7,412 8,162 8,950 9,776 10,64 11,54 12,48 13,46 14,47 15,53 16,62

6 1,002 1,012 1,032 1,068 1,126 1,217 1,354 1,546 1,795 2,098 2,451 2,848 3,288 3,770 4,292 4,853 5,454 6,093 6,770 7,486 8,239 9,031 9,861 10,73 11,63 12,58 13,56 14,58 15,63 16,73

Values of k for e = 36 daN/mm2 (360MPa, 36kN/cm2, kg/mm2) 7 1,002 1,013 1,035 1,073 1,134 1,229 1,371 1,568 1,823 2,131 2,488 2,890 3,335 3,821 4,346 4,912 5,516 6,159 6,840 7,560 8,317 9,112 9,946 10,82 11,73 12,67 13,66 14,68 15,74 16,84

8 1,003 1,015 1,038 1,078 1,141 1,240 1,388 1,591 1,852 2,165 2,526 2,933 3,382 3,871 4,401 4,970 5,578 6,225 6,910 7,633 8,395 9,194 10,031 10,91 11,82 12,77 13,76 14,79 15,85 16,95

9 1,004 1,017 1,041 1,083 1,149 1,253 1,406 1,614 1,881 2,199 2,256 2,976 3,429 3,923 4,456 5,029 5,641 6,292 6,981 7,708 8,473 9,276 10,12 10,00 11,91 12,87 13,86 14,89 15,96 16,06

 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

0 1,0 1,005 1,022 1,055 1,111 1,204 1,349 1,561 1,841 2,185 2,586 3,041 3,546 4,100 4,701 5,350 6,045 6,79 7,57 8,40 9,28 10,21 11,18 12,19 13,25 14,36 15,51 16,70 17,95 19,23 20,56

1 1,0 1,006 1,025 1,059 1,118 1,246 1,368 1,586 1,873 2,223 2,630 3,089 3,600 4,158 4,764 5,418 6,12 6,86 7,65 8,49 9,37 10,30 11,27 12,29 13,36 14,47 15,62 16,82 17,07 19,36

2 1,0 1,008 1,027 1,064 1,126 1,228 1,386 1,612 1,905 2,261 2,673 3,138 3,653 4,216 4,828 5,485 6,19 6,94 7,73 8,58 9,47 10,40 11,38 12,40 13,47 14,58 15,74 16,95 17,20 19,49

3 1,0 1,009 1,030 1,069 1,134 1,241 1,406 1,638 1,938 2,300 2,713 3,187 3,707 4,275 4,891 5,554 6,26 7,02 7,82 8,66 9,56 10,49 11,48 12,50 13,58 14,70 15,86 17,07 18,33 19,63

Lc i

4 1,001 1,011 1,033 1,074 1,143 1,255 1,426 1,665 1,972 2,339 2,762 3,237 3,762 4,335 4,955 5,622 6,34 7,09 7,90 8,75 9,65 10,59 11,57 12,61 13,69 14,81 15,98 17,19 18,45 19,76

5 1,001 1,012 1,036 1,079 1,152 1,269 1,447 1,693 2,006 2,379 2,807 3,287 3,817 4,395 5,020 5,691 6,41 7,17 7,98 8,84 9,74 10,69 11,68 12,72 13,80 14,92 16,10 17,32 18,58 19,89

6 1,002 1,014 1,040 1,085 1,161 1,284 1,468 1,721 2,040 2,419 2,853 3,338 3,873 4,455 5,085 5,762 6,48 7,25 8,07 8,93 9,83 10,78 11,78 12,82 13,91 15,04 16,22 17,44 18,71 20,02

7 1,003 1,016 1,043 1,091 1,171 1,299 1,490 1,750 2,076 2,460 2,899 3,389 3,929 4,516 5,151 5,832 6,56 7,33 8,15 9,02 9,92 10,88 11,88 12,93 14,02 15,16 16,34 17,57 18,84 20,16

8 1,003 1,018 1,047 1,097 1,181 1,315 1,513 1,780 2,111 2,502 2,946 3,441 3,985 4,578 5,217 5,903 6,63 7,41 8,24 9,10 10,02 10,98 11,98 13,03 14,13 15,27 16,46 17,69 18,97 20,29

9 1,004 1,020 1,051 1,104 1,192 1,332 1,537 1,810 2,148 2,544 2,993 3,493 4,043 4,639 5,283 5,973 6,71 7,49 8,32 9,19 10,11 11,08 12,09 13,14 14,25 15,39 16,58 17,82 19,10 20,43

Sabah Shawkat ©





kF A

A needed

kF 



Lc i

Buckling Buckling



kF A

A needed

kF 

I A

7688 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling Calculation of critical stresses according to slenderness ratio 

Values of k1

E In this chapter we will focus on steel columns, why? Because the strength of steel is very high Values of the critical stress of Euler k (daN/mm2) according to the elongation  2 which leads to smaller cross sectional area of amember to resist a particular force as compared

to concrete. So that problem of buckling generally arises in steel columns. We have a very 0 2 3 problem 4 of buckling, 5 6 I hope 7 that you 8 will get 9  interesting example for1 understanding the and 0

involved.

∞

207262 51816

23029

12954

8290

5757

4230

3238

2559

 1   1.3

   1.3

and of Lc 1

i  1.3

in terms of



The equation above describes the slenderness ratio, and the limit of slenderness of column 

 1



  1.3

 1



  1.3



example, steel column k1 material changes. k1 varies as the For is said to be short if the slenderness   1.3 1,80 column 1,60 3,60 50 2,00 ratio 1,3 is less than if it is between and 200 and it is ∞ 50, it is called ∞ an intermediate ∞ 1,31 31 131 1,54 are very 3,36 1,82 to elasticity. called a long column if it is greater 100 than 200.1,85 Long columns susceptible 1,32

1,90

1,50

3,17

1,67

10 2073 1713 1439 1226 1057 921 810 717 640 574 20 518,16 469,98 428,23 391,80 359,83 331,62 306,60 284,31 264,36 246,45 Have you ever seen Charlie Chaplin stick when he rests on it? 30 230,29 215,67 202,40 190,32 179,29 169,19 159,92 151,40 143,53 136,27 40 129,54 123,30 117,50 112,09 107,06 102,35 97,95 93,83 89,96 86,32 The stick the picture bellow76,65 describes one of the most 68,52 fundamental of a column 50 in 82,90 79,69 73,78 71,08 66,09 characteristic 63,79 61,61 59,54 55,70 53,92 52,22 50,60 49,06 47,58 But 46,17 44,82 43,53 in the60 field 57,57 of structural engineering, called "Buckling of Column". why did it buckle? 70 42,30 41,12 39,98 38,89 37,85 36,35 35,88 34,96 34,07 33,21 What80 made 32,38 the stick31,59 to bend 30,82 instead of taking the load straight down the ground? Well, here 30,09 29,37 28,69 26,02 27,38 26,76 26,17 25,59 about 25,03 24,49 23,96 23,46 22,97 22,49 22,03 21,58 21,15 we are90 to discuss the event. 100 20,73 20,32 19,92 19,54 19,16 18,80 18,45 18,10 17,77 17,44 110 17,13 16,82 16,52 16,23 15,95 15,67 15,40 15,14 14,89 14,64 120 14,39 14,16 13,93 13,70 13,48 13,26 13,06 12,85 12,65 12,45 130 12,26 12,08 11,90 11,72 11,54 11,37 11,21 11,04 10,88 10,73 140 10,57 14,43 10,28 10,14 10,00 9,86 9,72 9,59 9,46 9,34 150 9,21 9,09 8,97 8,85 8,74 8,63 8,52 8,41 8,30 8,20 160 8,10 8,00 7,90 7,80 7,71 7,61 7,52 7,43 7,34 7,26 170 7,17 7,09 7,01 6,93 6,85 6,77 6,69 6,62 6,54 6,47 180 6,40 6,33 6,26 6,19 6,12 6,06 5,99 5,93 5,86 5,80 190 5,74 5,68 5,62 5,56 5,51 5,45 5,40 5,34 5,29 5,23 200 5,18 5,13 5,08 5,03 4,98 4,93 4,88 4,84 4,79 4,74 210 4,70 4,66 4,61 4,57 4,53 4,48 4,44 4,40 4,36 4,32 220 4,28 4,24 4,21 4,17 4,13 4,09 4,06 4,02 3,99 3,95 230 3,92 3,88 3,85 3,82 3,79 3,75 3,72 3,69 3,66 3,63 240 3,60 3,57 3,54 3,51 3,48 3,45 3,42 3,40 3,37 3,34 250 3,32 3,29 3,26 3,24 3,21 3,19 3,16 3,14 3,11 3,09 260 3,07 3,04 3,02 3,00 2,97 2,95 2,93 2,91 2,89 2,86 270 2,84 2,82 2,80 2,78 2,76 2,74 2,72 2,70 2,68 2,66 280 2,64 2,62 2,61 2,59 2,57 2,55 2,53 2,52 2,50 2,48 290 2,46 2,45 2,43 2,41 2,40 2,38 2,37 2,35 2,33 2,32 In 1757, a formula that gives 300 mathematician 2,30 Leonhard Euler derived the maximum axial load

As we know that11 the effective length 33 of the column conditions 1,33 44,33 1,95 depends 1,46 on the end 3,00 1,54 of the column

that a long, slender, ideal column can carry without buckling. An ideal column is one that is

tf1  tto  relative f2the fixity The idealized supports shown below seldom occur. Because of uncertainty h w   Lc I 2 2 2   i  sometimes taken i to be i pin-ended. i i I  of pin-ended columns of the joints, columns are ForCbuckling

1,34 support.

8,5 33,50 25 2,00 1,43 2,86 1,43 1,35 7 27 20 2,10 1,37 2,62 1,25 1,36 can 6,5,29 16,67 large, 2,20 1,33 Buckling be defined 22,67 as the sudden, lateral deflection of 2,44 a column1,11 owing to a small 1,37 4,75 19,57 14,28 2,30 1,30 2,30 1,00 increase existing compressive load. This response1,27 leads to instability and collapse of the 1,38 in an 4,33 17,25 12,50 2,40 2,18 0,91 1,39 In this4 section we 15,44 11,11 the2,50 0,83 and bolted member. shall describe critical, or1,25 buckling, 2,08 load for welded 1,40 3,73 14 10 2,60 1,23 2 0,77 profiles, load that cases instability. 1,20 1,41 the compressive 3,50 12,82 9,09 the 2,80 1,87 0,67 1,42 3,31 11,83 8,33 3,00 1,18 1,76 0,59 To calculate the critical load for flexural allow for0,53 the end restraint 1,43 3,14 11,00 7,69 - torsional 3,20 buckling, 1,16 and to1,68 1,44 3 10,28 7,14 3,40 1,14 1,62 0,48 conditions. The compressive force is, in this context, thought to be applied at the centroid of 1,45 2 9,67 6,67 3,60 1,13 1,57 0,44 the cross-section. 1,46 87 9,13 6,25 3,80 1,12 1,52 0,40 1,47 2,77 8,65 5,88 4,00 1,11 1,48 0,37 1,48 2,67 8,22 5,56 4,50 1,09 1,41 0,31 1,49 2,58 7,84 5,26 5,00 1,08 1,35 0,27 1,50 2,50 7,50 5,00 6 1,06 1,28 0,21 1,52 2,36 6,91 4,55 7 1,05 1,23 0,17 1,54 2,25 6,42 4,17 8 1,01 1,19 0,15 1,56 2,15 6,00 3,85 10 1,03 1,15 0,11 1,58 2,07 5,64 3,57 12 1,03 1,12 0,09 1,60 2,00 5,33 3,33 15 1,02 1,09 0,07 1,62 1,94 5,06 3,12 20 1,02 1,07 0,05 1,65 1,86 4,71 2,86 30 1,01 1,05 0,03 1,70 1,75 4,25 2,50 50 1,01 1,03 0,02 1,75 1,67 3,89 2,22 100 1,00 1,013 0,01

Sabah Shawkat ©

2homogeneous, and free from initial stress. The maximum load, sometimes perfectly straight, F L

c c

Ineeded called the critical 2 load, causes the column to be in a state of unstable equilibrium; that is, the  E

introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross

section of the member then it is called a slender member and remember you should use the

word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio Case is I cross-section

centrically loaded forces F at each end. The member is assumed to be perfectly 3 n by compressive im  t i  b i   Telastic material-that is, we have an ideal column. I T and 1.31to be constructed  3  of a linearly straight C    1  m i  1   0.039  I  When the load F has been increased sufficiently toAcause This is a  L 2a small lateralT deflection.    condition of neutral equilibrium. The corresponding load value of the load is critical load. if  T   then this case is without torsion, and the verification of stress is not necessary



89 Behaviour of Steel Structures 2019



For pure compression    k    

      

 E 1

F A

     

   2 2   E i  A   2 Lc  F  

 E 2 Lc 2 i

F A

Fc Lc

Ineeded

 D  k1



 e

 max



s

 k1

 1.3   cr

 s

  s   cr  1.3  e   e  cr

 k

 e

 

1  1.3

 cr

1.3

and multiplying each term by

e  cr s

, we get

Lc F



e



 cr 

 2 k

Sabah Shawkat © e

E

0  



 E

k



 E





e



 cr  cr



with 2

F A



k

Behaviour of Steel Structures 2019

 0.65 e A

A  1

 cr

  0.25

constant moment kf   1.3 or moment varying linearly   0.18 concentrated load kf   1.3 in the middle c   0.25  1.72   L  kf   1.3 concentrated load at a distance c

  1.3

k1   kf  f  e 6 M 6 F h f W 2 ea  h   0.03 load uniform kf   1.3

 k  1  1.3 

e  e     0.5  0.65     0.5  0.65      cr cr    

e

 cr

We see that this value of k depends only on the ratio

e

 e 

 cr



and that consequently

E

the values of k do not change when the product  e remains constant. so if we calculate the values of k as a function of  for a given value of  e Dutheil method of checking the deflection on members subjected to compression force. Amplification of deflections: F

 A

 cr





Values of k1 and  depending on the slenderness ratio 

Fcr



Buckling Buckling

EI

E

Fcr Fcr  F

 cr  cr





F A

7690 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Lc Deflection representative of structure irregularities

Buckling



In this chapter we will focus on steel columns, why? Because the strength of steel is very high which leads to smaller cross sectional area of a member to resist a particular force as compared to concrete. So that problem of buckling generally arises in steel columns. We have a very interesting example for understanding the problem of buckling, and I hope that you will get involved.

k

1 1 1 l l The fequation of slenderness of column a   above describes the slenderness ratio, and the limit  2 E l  Fcr  cr   ( 1  b ) Fcr column varies as the material changes. For example, steel is said to be short if the slenderness 

ratio is less 1 than 50, it is called an intermediate column if it is between 50 and 200 and it is if a e 240 MPa a b a 0.3 12 column if it is greater than 200. Long columns are very susceptible to elasticity. called a long

for member subjectedlength to pure then theonvalue of conditions deflection of will AsCase we know that the effective of compression the column depends the end thecalculate column as

Have you ever seen Charlie Chaplin stick when he rests on it?

follow: support.

The stick in the picture bellow describes one of the most fundamental characteristic of a column in the field of structural engineering, called "Buckling of Column". But why did it buckle?

Buckling can be defined as the 1 cr sudden, large, lateral deflection of a column owing to a small f1 0.3     Fcr compressive increase in an existing cr  1.3  load. This response leads to instability and collapse of the

What made the stick to bend instead of taking the load straight down the ground? Well, here

member. In this in section we shall describe the critical, or buckling, load for welded and bolted Deformation two perpendicular planes

we are to discuss about the event.

profiles, the compressive load that cases the instability. 2 0.3  I Fcr to allow A  cr for the end restraint M F f1the critical  A f1  - torsional because To calculate load for flexural buckling, and k 1.3   conditions. The compressive force is, in this context, thought to be applied at the centroid of

Sabah Shawkat © the cross-section.  max because

s

s

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender memberon and you should Values of (/‐1.3) and (1/‐1.3) depending theremember slenderness ratio  use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

 cr



 1.3 

e

 max



 e   cr

e

  s   cr  1.3  e   e  cr

1   cr  1.3  e  2



 M



 1.3  



 cr





 1.3 

 s   cr

 

then

1 2   cr  1.3  e   e  cr 4

Average stress in columns versus slenderness ratio

is the member length l The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity is the area of the constant cross-section A of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns F is Normal force centrically loaded by compressive forces F at each end. The member is assumed to be perfectly cr is Euler critic stress straight and to be constructed of a linearly elastic material-that is, we have an ideal column. is Euler critic force Fcr the load F has been increased sufficiently to cause a small lateral deflection. This is a When condition neutral equilibrium. The corresponding load value of the load is critical load. (cr / of ) coefficient of increase cr-



is stress in simple or pure compression

91 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Buckling lengths according to A and B

Buckling length according to A and B, case where the ends A and B are fixed in position The charts are based on the relationships  Lc   L   L A   Lc   L n  L B     L  cotg  L  3 L      L  cotg  L  3 L  c c  c c  

one,

the

   L  sin    L   cos    a L Lc  Lc  c 

Lc   1     L  L  sin    Lc  

and

2 2   a    L n  A  L   1  cos  L  3 L   sin  L c c c  

values of the ratio Lc / L of the length of

A value

L Lc

, approached to less than 0.5% by default and 1.5% overpriced (on the security

side) is given by:

buckling / to the length l, of the bar

6

 6

A 0.8  A A 0.8  A





B



A  B  0.7 A  B  0.48

B

A  B  A  B  0.96

0.8  B

Sabah Shawkat ©

the other, the

An approximate value of a is given by: L

ratio a / L of

the distance a between



0.8  B

0.12  0.3 B

A  B  0.6 A  B  0.48

the

end A and the

nearest point

These expressions are reduced: If A B symmetrical system

of inflection,

0.6  A

1.2  A

to the length

and

0.3

1.2  A

L of the bar. If B 0 perfect fixed in B, then Lc

0.48  0.7 A

0.96  A

and

0.2

0.8  A

If B  Buckling lengths according to a and b where the A and B ends are fixed in position

0.3

0.70  A

1  A

Buckling Buckling

and b = 0

7692 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling Case where one of the ends is free to move laterally

Lc laterally Case where one of the extremities is free to move 

In this chapter we will focus on steel columns, why? Because the strength of steel is very high which leads to smaller cross sectional area of a member to resist a particular force as compared

A and the problem charts below give, depending to General concrete.case So -that of buckling generallyonarises in steel columns. We have a very  interesting example for understanding the problem of buckling, and I hope that you will get one, the value involved. of the ratio Lc /

L you of the Have everlength seen Charlie Chaplin stick when he rests on it? of buckling to The stick in the picture bellow describes one of the most fundamental characteristic of a column the length of the in the field of structural engineering, called "Buckling of Column". But why did it buckle? bar. What made the stick to bend instead of taking the load straight down the ground? Well, here we are to discuss about the event.

The equation above describes the slenderness ratio, and the limit of slenderness of column The charts are based on the relationships varies as the material changes. For example, steel column is said to be short if the slenderness ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is 

3 L



3 L

 L c  a c called ifAit is Bgreater than 200. are very susceptible to elasticity. tg  a long column  tg  andLong columns 2 Lc  L A    L Lc 3 Lc   As we know that the effective length of the column depends on the end conditions of the column A  B      L  support.

Buckling can be defined thethan sudden, large,oflateral is given by:of a column owing to a small An approximate value,as less 2% near, Lc / L deflection increase in an existing compressive load. This response leads to instability and collapse of the member. In this section we shall describe the critical, or buckling, load for welded and bolted Lc  3.2 A  B  4 A  4 B  3.75  profiles, load that casesthe instability. L the compressive A  B  3.75   To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint  2 A  3.75force  4 Bcompressive  a 1 The conditions. is,and in this context, thought to a be  applied at the centroid of  b   1     if we have L 2 A  B  3.75 L   L   the cross-section.

Sabah Shawkat © B >  A

the other, the ratio a / L of the distance

inflection,



 4 A  2 B  3.75    A  B  3.75 

and

a  L

1

b



L

if we have

These expressions are reduced:

A of the bar and point

A >  B

between the end

the

If A

perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

symmetrical system

1  1.6 A

the length L of In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load bar. slender, ideal column can carry without buckling. An ideal column is one that is thatthe a long,

B

0.5

Average stress in columns versus slenderness ratio

If B 0 perfect fixed in B, then The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity L 3.75  4  L L c

1 of theL joints,3.75 columns are sometimes taken For buckling of pin-ended columns  2 L to beLpin-ended. 2 L L

centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight If Bandto be constructed of a linearly elastic material-that is, we have an ideal column. When the load bF has been Lincreased sufficiently to cause a small lateral deflection. This is a c a 1

2 1  0.8 

A L L L condition of neutral equilibrium. The corresponding load value of the load is critical load.

93 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Length of buckling of columns in buildings with fixed nodes

Case of the ends fixed elastically This is the special case where the fixed disappears in B (B = ∞). The Lc / L ratio is given by the graph below. The end of the buckling length is at the free end of the B-bar.

Where

transverse

stability

provided by bracing or shear walls, the ratio Lc / L is given by the

graph

below

hereafter,

according to the coefficients of installation KA and KB at the ends of the member AB of column

considered KA and KB, as well as

Case of the fixed elastically this is the special case where the fixed fail in B

Lc and L, are defined below.

Sabah Shawkat ©

This is also frequently the case with articulated frames

The graph is plotted from the relationship 

L Lc

 tg 

L Lc

1

 3 A

A

An approximate value, less than 2% near, of Lc / L is given by: We can also use the simpler expression

Lc L

2  0.60 A

Lc L

1

 3 B

, which gives an error of less than





3  1.6 KA  KB  0.84 KA  KB

3  KA  KB  0.28 KA  KB





lower than the value (k.which would be reached in the middle of the buckling length.  L It is reduced to  1  ( k  1)  sin   that can be replaced in practice by Lc  

2



2



  1 



k 1

1  0.6 KA

1  0.2KA

if the end B perfectly fixed KB = 1

if the both extremity have the same

embedding (fixed)coefficient KA = KB

In the case of compression without bending, for example, the comparison stress is always

 

1  0.36KA

this formula reduces it to

point of the AB bar is fixed in A.



0.7  0.38 KA

2 1  0.8 A

1.5% as long as a  2 which then places in safety. It should be noted that the most stressed

 1  4 ( k  1) 



1  0.8 A  

3  1.6 KA

3  KA

if the end B is

articulated KB = 0

Buckling Buckling

7694 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling Buckling column length in building with displacement of free ends

Buckling the web of beam

why? Because the strength of steel is very high In this chapter we will focus on steel columns,

which leads to smaller cross sectional area of a member to resist a particular force as compared to concrete. So that problem stability of buckling When the transverse is generally arises in steel columns. We have a very interesting for integration understanding the problem of buckling, and I hope that you will get ensuredexample that by the of the

beams on the columns, the Lc / L ratio involved.



Lc i

The equation above describes the slenderness ratio, and the limit of slenderness of column varies as the material changes. For example, steel column is said to be short if the slenderness ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is called a long column if it is greater than 200. Long columns are very susceptible to elasticity. As we know that the effective length of the column depends on the end conditions of the column

is given by graph below hereafter, Have you ever seen Charlie Chaplin stick when he rests on it? according to the coefficients of Theembedding stick in the KA picture describes one of the most fundamental characteristic of a column andbellow KB at both ends of

support. Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small

in the of structural called "Buckling of Column". But why did it buckle? and KB, as well thefield member AB of KAengineering, What made the stick to bend instead of taking the load straight down the ground? Well, here as Lc and L, are defined below.

increase in an existing compressive load. This response leads to instability and collapse of the

we are to discuss about the event.

profiles, the compressive load that cases the instability.

1 KA

1  2 A

member. In this section we shall describe the critical, or buckling, load for welded and bolted

To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied at the centroid of

Sabah Shawkat © 1

Lc L

the cross-section.

1  2 B

  KA  KB  5.5 KA KB

1.6  2.4 KA  KB  1.1 KA  KB  

this formula reduces it to

if the end B perfectly fixed KB = 1

4  3.5 KA

1  6.5KA

Lc mathematician 1.6  2.4 KA Leonhard Euler derived a formula that gives the maximum axial load In 1757, if the both extremity have the same

Average stress in columns versus slenderness ratio

KA that aL long, slender, ideal column can carry embedding without buckling. An ideal column one that is (fixed)coefficient KA = Kis B perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity

called the critical load, causes the column to be in a state of unstable equilibrium; that is, the if the end B is articulated KB = 0 introduction of the slightest lateral force will Lcause the column to fail by buckling. Buckling is 0.8  0.2 K c

called an instability occurred in a structure because of excessive loading. But what do we mean L K A

by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

1000ea For buckling of pin-ended columns of the1000e joints, taken to be2 pin-ended. a columns are sometimes 2  5.667

F  0.04 E ea

3

F  840 ea

 5.667

h a loaded by compressive forces F at each end. hThe centrically a member is assumed to be perfectly 2

2 2 straight of linearly elastic  material-that is,kgwe have an ideal column.  F to be constructed 1 a1000e a  and  E  21000  2.667 F ( kg) ea ( mm)      1    2  9 2 h When the load F has increased lateral deflection. This is a  0.04  ebeen   asufficiently  to cause a smallmm   E e a  condition of neutral equilibrium. The corresponding load value of the load is critical load.

condition of local non-buckling

95 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Tangential stress 160 MPa 180 MPa 220 MPa 240 MPa 300 MPa 320 MPa

e

0.65  e



1.54    e

VS  0.65  e b Ix

 yz

v

z

  zy2    z2



 W h   2

z

h M   2 a Ix h   2

is the tangential stress

e

is limit of elasticity

Shear stress 3







for rectangular cross-section

2 

V A

for thin circular tube

for full circular section



Sabah Shawkat ©

3 tw

0.006 h w



tf  tw

 0.65 e



M hw 2 I

Buckling Buckling

7696 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Simple supported beam without joints in Maximum Stress: Buckling Qx L Qy  L midspan  max  In this chapter we will focus on steel columns, why? Because theI strength ofI steelisevery high 8

x

8

y

which leads to smaller cross sectional area of a member to resist av xparticularv yforce as compared 3

Qx La very to concrete. So that problem of buckling generally arises in steel columns. We have L fx 0.013  Deflection: 200 interesting example for understanding the problem of buckling, and I hope that youE Iwill get x

x,y – are plastic adaptation coefficient

0.013

Qy  L



E Iy

Have you ever seen Charlie Chaplin stick when he rests on it?

fx  fy

200

Lc i

 1 limit ofslenderness of column The equation above describes the slenderness ratio, and the  inspan 2  k  k  short if the slenderness varies as the material changes. For example, steel column is saidchto be 2  1

ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is 2 Long columns are very susceptible to elasticity. called a long column if it is greater than 200. k  k    4  conditions 2 k sh   of  the  k column ch    As we know that the effective length of the column depends on the end 8 2  2   inspan 2   support. 2 sh k  k ch k  k 



involved.

 1

x,y – are plastic adaptation coefficient The stick in the picture beam bellowwith describes one of theormost fundamental characteristic of a column Simple supported one connection Maximum Stress:







I I What made the stick to bend instead of taking the load straight   x the   y Well, here 8 down 32ground?

Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small  2  k 2 sh k   ( k )  4   the increase in an existing compressive load. This response leadsto2  instability and collapse of  insupport 2 sh k  k ch k  k  member. In this section we shall describe the critical, or buckling, load for welded and bolted

we are to discuss about the event.

profiles, the compressive load that cases the instability.

Qx L But why Qy  L did it buckle? in thejoinit fieldinofmidspan structural engineering, called "Buckling of Column".  max



 e

Deflection:

0.013

Qy L3  2 2

Qx L



E Ix

L 200

To calculate the critical load for flexural - torsional buckling, 24  and tok allow  for the end restraint  inspan  1   at the centroid of 2 conditions. The compressive force is, in this context, thought applied k k  to2 be sh   2   the cross-section.

Sabah Shawkat © x,y – are plastic adaptation coefficient

0.052

E Iy



Simple supported beam with two connection or Maximum Stress: Qx L joinits  max

8

x

8

 insupport

Qy  L



fx  fy

400

y



 e

Deflection:

0.00675

0.013

E Iy

x,y – are plastic adaptation coefficient

Qx L



E Ix

200

 inspan

L  2  L f R

 

400

fx  fy

Maximum Stress: Simple supported beam with two connection In 1757, mathematician Leonhard Euler derived aorformula that gives the maximum axial load Qy L joinits at 1/3 L distance of span beam   0.025 that a long, slender, ideal column can carry without buckling. An ideal column is one that is Q L 3 3

 max   e perfectly straight, homogeneous, and free from initial stress. TheI maximumI load, sometimes 8

x

y

vy vx called the critical load, causes the column to be in a state of unstable equilibrium; that is, the 3

Qx L introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is L  Deflection: fx 0.013  Ix do 200we mean called an instability occurred in a structure because of excessive loading. ButEwhat 3

y  L greater than the cross by slender? When the longitudinal dimensions of the member areQmuch   x,y – are plastic adaptation coefficient 3 3 L f 0.00675   section of the member then it is called a slender member and remember you should use the y

E Iy

k  12   1   2 k k  2 th  

600

word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling



 th  k   2  k   2

Average stress in columns versus slenderness ratio  k 

 th  4 

 of uncertainty  The idealized supports shown below seldomoccur. Because relative to the fixity insupport k 



 of the joints, columns are sometimes taken to be pin-ended. For4 buckling of pin-ended columns k

k

L 2 J e e e e  e F at each end. e forces centrically k assumed 0.878 to be perfectly thx The member is shx loaded by compressive chx k k h Iy 2 2  e e straight and to be constructed of a linearly elastic material-that is, we have an ideal column. Q ytsufficiently L L  insupport to cause a small lateral deflection. This is a When the loadQxFL has beenQyincreased  max 



 y200 The corresponding x200 y200 condition of8neutral load value of the load is critical load.   x equilibrium.  y   y 32  32  I

 vy



97 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the stress resistance of the roof beam, where the length of the beam is 10m, and the

Calculation the loads: Qxr 

simply supported at both ends and the loading is uniform

 qdead  qs  a  pp cos( ) L Qxr  110.62909kN

Qyr   qdead  qs  a  pp sin(  ) L

resistance to lateral-torsional buckling for a beam with the cross-section I500, the beam is

Qyr  22.12595kN

Verification of deformation: the main beam is connected on the perpendicular direction with stiffer beam to avoid the lateral deformation.

QxL

 max

8

I vx

 x

32 

 fx  0.013 

Data:

lp 

 L  10 m  a  2.5 m

L 2

lp  5 m

fR 

h1  0.10 m

Material properties: I500

Iy500  21.4 10 mm 3

h500  500mm

Wy500  214 10 mm Wx500  1930 10 mm 3

Es  210GPa  b500 2

Wx 

 Wx 1928000mm

Ix500 vx

Wy 

Iy500 vy

 tf  16 mm

b500  200mm

Wx500

Wy500

1

Wy 214000mm 

Qyr  L  3   2 2  fy  0.052  fx  0.01421m Es Iy500

fy  0.016m

fR  0.0214m

4 3  Qx   qdead a   pp cos (  )    qs a cos (  )    L 3 2 

Qx  151.79549kN

4 3  Qy   qdead a   pp sin(  )    qs a sin(  )    L 3 2 

Qy  30.35928kN

Dead load: qdead  3.10 kN m

2

Snow load: qs  1.05 kN m 

vx 

h500

Load calculated in sin direction without self-weight of beam

4 3  q   qdead   qs   a sin(  ) 3 2 

 h500   h1  2  

M  t  q 

Loads:

 x 

 e

 y

tw  10.2mm

pp  0.907 kN m

 J500  896 10 mm

 steel 

 Ix500  482 10 mm

vy 

Q xrL

Es Ix500

fx  fy

Qy L



2

 Q yt 

Mt h500

q  2.79876 m

y

 1.185

are plastic adaptation coefficients:

Qyt  19.5913kN

L

Calculate the value of k:

k  0.878     11.31 deg

cos (  )  0.98058 sin(  )  0.19612

Buckling Buckling

L h500

kN

Mt  0.97957kN

240MPa

1.06

1



2 J500 Iy500

k  5.08144

7698 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

k k k Buckling k e e e e chx chk  chk  80.50581 2 In this chapter 2we will focus on steel columns, why? Because the strength of steel is very high

1 sectional 8  cross   which leads to smaller area of a member8 toresist a1particular force as compared  inspan  0.30213  inspan  1  inspan   1    2  2 chk k of buckling generally arises to concrete. Sokthat problem in steel columns. We have a very k   ch    2   2  interesting example for understanding the problem of buckling, and I hope that you will get

involved.

Calculate the stress in bending: Have you ever seen Charlie Chaplin stick when he rests on it? QxL Qy L Qyt L   one  f 58.3529 MPa  most fundamental  inspan max e  The  stick in the picture bellow describes off the characteristic of a column I I Wy500 8   x 32   y 8    y in the field of structural engineering, called "Buckling of Column". But why did it buckle? vx vy 2 What made the stick to bend instead of taking the load straight down the ground? Well, here We are therefore obliged to check the value of this torsion and to limit its value to a maximum we are to discuss about the event. of 2o. 1 G

2.6 Es

G 

2.6

Es 2.6

 maxideg 

 maxi

180

 maxideg



  6.22761 i

 maxideg 

The equation above describes the slenderness ratio, and h500the limit of slenderness of column  tanas    0.06874  17.185 mm  the   0.10912 maxi varies material changes. tg For example, steel column h500 2 is said to be short if the slenderness ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is 2 called a long column if it is greater than 200. Long columns are very susceptible to elasticity. As we know that the effective length of the column depends on the end conditions of the column It is therefore necessary in this case to propose beam in transversal direction at the mid-span support. of the main beam to a void lateral torsion buckling of the structures, thus in this case the Buckling canbetween be defined as the sudden, large, lateral deflection of a column owing to a small distance support should be l/2. increase in anLexisting compressive load. This response leads2toJ instability and collapse of the Lmod 500 Lmod  Lmod  5 m kmod  2.54072 kmod  0.878   member. In this section we shall describe the critical, or 2 h500buckling, Iy500 load for welded and bolted profiles, the compressive load that cases the instability. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint kmod  kmod e e conditions. thought to be applied at the centroid of shkthis  6.30499 shkmod The  compressive force is, in modcontext, 2 the cross-section.

G  80769.23077 MPa



L G J500



  h500  tf 

Es Iy500





th 

2







kmod

th  0.99994

chkmod 

e e

 



e e

 5.24379

 insupport

Calculate load in direction sin without self- weight of beam

qt   qdead  qs  a sin(  )

s a sin(  )  0.49029 m time

qdead a sin(  )  1.51991m

1

qt  2.03472m

1

 insupport 

kN

 perfectly straight, homogeneous,  4 and free from initial stress. The maximum load, sometimes 2    M  L    called the criticaltt load,causes state of180 unstable equilibrium; that is, the 57.2974  maxi  1  the column to be in arad   8  G J 1   500 introduction of the slightest 1  force  lateral  will cause the column to fail by buckling. Buckling is 2

th

 kmod

e

  because of excessive loading. But what do we mean called an instability occurred in a structure 4 

by slender? When the longitudinal dimensions of the member are much greater than the cross we should transform this to deg.  maxi  0.10869  maxi 0.10869radian section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

chkmod  6.3838





 2  k 2 sh k   ( k )  4   2  2  sh k  k ch k  k

kN

  h500 In 1757, mathematician derived a formula that gives the maximum axial load 0.71215kN M   h1Leonhard  Mtt Euler tt  qt  2  that a long, slender, ideal column can carry without buckling. An ideal column is one that is





4 kmod

2 







 2  kmod 2 shkmod   kmod      shkmod  kmod chkmod 

 insupport

0.67513

Verification of stress:Average stress in columns versus slenderness ratio Q yt LBecause Q xL shown below Q y Lseldom occur.   insupport The idealized supports of uncertainty relative to the fixity   max    I I W of the joints, columns  y500taken to be pin-ended. x500 are sometimes y500 For buckling of pin-ended columns  x 32   y   y 8 32  vxby compressive centrically loaded  vyforces F at each end.2The member is assumed to be perfectly

straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased max sufficiently cause a Ok small lateral deflection. This is a   max 162.85 MPa   e to 240MPa condition of neutral equilibrium. The corresponding load value of the load is critical load.

99 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Data Mechanical properties: 6

 k1  3.51  ( 3.51  3.29) 

 Iy  1.42 10  mm Ix  19.4 10  mm  It  69.8 10  mm  W x  194 10  mm  W y  28.5 10  mm 9

Iw  13 10  mm  W w  2.71 10  mm  h  200 mm

G 

ix  82.6 mm

iy  22.4 mm

tf  8.5 mm

 b  100 mm

 iw  25 mm

tw  5.6 mm

 Es  210 GPa

cr 

L  3 m

 2tf    2

Es  Iy G It

N 

 2tf    2

 h´  h  

h´  0.1915m

Es  Iy G It

N 

N  1.14 kN

 

N  4.56 kN

2.6 Iy

h´     It  2 L 

 

  0.013470426

2.6 Iy

h´     It  2 L 

  0.053881704

 1  2  2         9.56 k 

h 2



M D 

 L N

k

 d 

d  413.12MPa 

 Es  Iy G It 1 

 h

2 E I s y

4 L





M D 

G It

cruniform 

tf  b  h´ 24

 55.6

 h



 

  p  3.14

p

 p  103.85

  89.597

 Vcr  1.5  1.1 

Vcr  3.496

cr



 

Vcr  2.319

 p 

cradmissible  30.274MPa 

Vcr



 Jb  4

1 3

tf  b  h´ 24

 glcr 30.86 kN

L glcr h

cr

cradmissible 

2 4

  m C1 2.7275kN

Vcr

 p  192 MPa

cruniform

 C 4.212kN  m

  uniform    p 

 Vcr  1.5  1.1  cradmissible 

1.9 100

cruniform

Vcr

cradmissible  3.908 MPa

 13.9

 k1  4.24  ( 4.24  4) 

  p  3.14

cruniform 

Es  Iy C

 glcr 149.46kN 

L glcr h

cruniform  154.08MPa 

cruniform  p

p

 p  103.85

 p  192 MPa

 uniform  3.14

cruniform

 uniform  115.921

  p  3.14

p

 p  103.85

The safety coefficient we obtain as follows: The safety coefficient we obtain as follows:

gl  cr  k3

cruniform  31.81 MPa

 uniform  255.129

L C

 uniform  3.14

G It

 Jb 52151.821mm    2 b  t f  h´ t w 

C  G Jb

Es  Iy C

cruniform  p

2 E I s y

4 L

  m C1 2.7275kN

   2 b  t f  h´ t w  Jb 52151.821mm 3

 Es  Iy G It 1 

EIy, GIt are the transversal flexion stiffness and elastic torsion stiffness in elastic phase  C1  Es 

p

Es cr

Case for uniform load acting along the span Case for uniform load acting along the span of the beam of the beam 7.6 5.9  k3  31.5    ( 31.5  30.5) k3  31.342  k3  42.6    ( 42.6  36.3) k3  37.954  48  8 

d  673.44MPa 

M D  53.97 kN m

C 4.212kN  m  C  G Jb  L C

 L N

and elastic torsion stiffness in elastic phase

  3.14 

 p  192 MPa

cradmissible  111.228MPa 

gl  cr  k3

EIy, GIt are the transversal flexion stiffness

 Jb 

0.93

 M D 22.97 kN m

 C1  Es 

cr  257.921MPa 

M cr  50.037kN  m

M cr

 p  103.85

cradmissible 

 1  2  2         11.73 k 

0.93

 d 

  p  3.14

cr

Es  Iy C

cr  p

 p  192 MPa

 p 

The critical moment of buckling or discharge will be:  1  2  2    L N       M  M 130.65kN  m

k

 cr 105.827MPa 

 Vcr  1.5  1.1 

The critical moment of buckling or discharge will be:  1  2  2    L N      M  M  80.15 kN m d

M cr  k1

The safety coefficient we obtain as follows: The safety coefficient we obtain as follows:

k1  3.47568

 M cr 20.531kN  m

  139.875

h´  0.1915m

100

cr 

M cr

   3.14  L  6 m

 h´  h  

cr  p

G  80769.23MPa 

2.6

M cr  k1

Es  Iy C

15.6

k1  4.23544

Buckling Buckling

Vcr  8.139

  uniform    p 

 Vcr  1.5  1.1  cradmissible 

cruniform

Vcr

cradmissible  53.674MPa 

Vcr  2.871

76100 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

The calculation of the Buckling Buckling Case for pure bending In this chapter we will focus on steel columns, why? Because the strength of steel is very high which leads to smaller cross sectional area of a member to resist a particular force as compared IPE600 to concrete. So6that4 problem of buckling6 generally arises in steel columns. We have a very 4 3 4

 Iy  33.9 10  mm

 Ix  921 10  mm It  1660 10  mm

interesting example for understanding the problem of buckling, and I hope that you will get 3 3  W x  3070 10  mm involved. 6

 W y  308 10  mm

 Iw  2850 10  mm

 89.0seen 10  mm  ix  243  mmwhen he rests on it? W w  ever Have you Charlie Chaplin stick

iy  46.6 mm 

iw  54.6 mm

h  600 mm

 tf  19 mm

tw  12 mm

Es  210 GPa

 b  220 mm

The stick in the picture bellow describes one of the most fundamental characteristic of a column  L  6 m

in the field of structural engineering, called "Buckling of Column". But why did it buckle? Es  2 t f  down the ground? Well, here WhatGmade the stick toGbend instead of taking  80769.23MPa h´  0.581m   h´ thehload   straight  2.6  2  we are to discuss about the event. 2.6 Iy  h´  2 Es  Iy G It    N  27.14kN   0.124467146 N   It  2 L  2 L

M cr  k1

Es  Iy C L

Mcr  732.682kN m

Lc M cr  cr  i Wx

cr  238.659MPa

cr  p

The equation above describes the slenderness ratio, and the limit ofEsslenderness of column Es if     3.14   3.14   93.143    192 MPa   103.85 varies as thep material changes. For example, steel column ispsaid to bepshort pif the slenderness cr ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is calledThe a long column if it iswe greater 200. Long columns are very susceptible to elasticity. safety coefficient obtainthan as follows: As we know that the effective length of the column depends on the end conditions of the column 2 cr   Vcr  1.5  1.1   cradmissible  Vcr  2.385 cradmissible  100.069MPa support.  V cr  p Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small Case in foranuniform increase existingload compressive load. This response leads to instability and collapse of the Es  Iy Cfor welded and bolted member. In this2.53 section we shall describe the critical, or buckling, load gl  k3  59.578  k3  53    ( 53  42.6) cr  k3 2 4.0   L profiles, the compressive load that cases the instability.

glcr  1453.02kN

To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied at the centroid of (L2C)/C1 0,4 4 8 16 24 32 48 the cross-section. K3 143 53 42,6 36,3 33,8 32,6 31,5

The critical moment of buckling or discharge will be:

k

 d 

 L N

 1  2  2         10.54 k 

Md  1715.57kN m

d  558.82MPa

M D 



0.93

 Es  Iy G It 1 

 h

2 E I s y



4 L

G It

M D  777.5kN m

(L2C)/C1

128

200

280

360

400

30,5

30,1

29,4

28,8

28,6

 cruniform 

h 2

(EIy), (GIt) are the transversal flexion stiffness and elastic torsion

glcr h Wx

p  192 MPa 

cruniform  283.98MPa

 uniform  3.14

cruniform  p

cruniform

 p  3.14  p  103.85 

p

 p  103.85

stiffness in elastic phase

 C1  Es 

tf  b  h´ 24

2 4

C1  597.56061kN m

 Jb 

1 3

  2 b  t f  h´ t w 3

3

The safety coefficient we obtain as follows:



2 cruniform   uniform  cradmissible  Vcr  2.244  Average stress in columns versus slenderness ratio  Vcr p  

 Vcr  1.5  1.1 

4 In 1757, Jb  mathematician 1340642.667mm Leonhard Euler derived a formula that gives the maximum axial load

that a long, slender, ideal column can carry without buckling. An ideal column is one that is 2 L C 2 C  108.283kN m see table below  6.52  5.007 perfectly straight, homogeneous, and freek1from initial stress. The maximum load, sometimes C1

cradmissiblesupports  126.566MPa The idealized shown below seldom occur. Because of uncertainty relative to the fixity

called(Lthe 2 critical load, causes the column to be in a state of unstable equilibrium; that is, the C)/C1 0 0,1 1 2 4 6 8 10 12 introduction slightest31,4 lateral force cause the5,85 column to fail by4,70 buckling.4,43 Buckling is K of the ∞ 10,36will 7,66 5,11 4,24

two cases thatbydoes it mean, the beam should onThe several supports. This case as centrically loaded compressive forces F at each be end. member is assumed to beshow perfectly

called an instability occurred in a structure because of excessive loading. But what do we mean (L2C)/C1 16 20 24 28 32 36 40 100 ∞ by slender? When the longitudinal dimensions of the member are much greater than the cross K1 4,00 3,83 3,73 3,66 3,59 3,55 3,51 3,29 ߨ section of the member then it is called a slender member and remember you should use the

word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

wecolumns obtain from the calculation admissible stresses hasofallow value in the of thewhen joints, are sometimes takenthat to bethe pin-ended. For buckling pin-ended columns thatand the to simple supported beam above two supports is not allowable or sufficient straight be constructed of a linearly elastic material-that is, we have an ideal any column. Whenmore. the load F has been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load.

101 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the cross-section class for the profile WII900-5-14x250 and the effective widths of

The entire flange is effective, calculate the effective width of the web:

the compression elements as well as the effective second moment of area, when the crosssection is subject to bending moment, the steel grade is S355J2G3

fyf  460 MPa  H  900 mm b  250 mm tf  14 mm tw  5 mm a  4 mm  fyf   1.1 fyd  fyd  418.18182 MPa

k  23.9 bw p







c 

 b  tw

 2 a 2  bw  H  2  tf  2 a





 hw  H  2 tf

c  11 4 tf

p

bw tw

 0.81362



28.4   k  0.22

   beff

2  be2  0.6 beff

Class 3

p

 1.524



 0.5615

beff  0.24481 m

be1  0.4 beff

be1  0.09792 m

be2  0.14689 m

The depth of the ineffective region: hw

Web:

235 355

The compression region and the tension region of the web are equal in the gross cross-section:

bw  0.86069 m hw  0.872 m

Flange:

c  8.34594 tf



p

c  0.11684 m



 100.9

Class 4

bw tw

 be1  be2

 A f  tf b

 172.13726

The profile is Class 4.

Aw  tw  H  2 tf  Aw  0.00436 m

bneg  0.19119 m

Af  0.0035 m

A  2 Af  Aw

A  0.01136 m

i  1  4

 b 1  250 mm

H h 2   be2  h 1 2 h 4  tf

z1 

2 z1  0.007 m

 Ai  b i h i

 0.0035    0.00291  2 Ai   m  0.00049     0.0035 

Buckling Buckling

b 2  5 mm

b 3  5 mm  b 4  250 mm

h 2  582.887 mm h 3  be1

h 1  tf

h 3  97.925 mm

h 4  0.014 m

 z2  tf 

2 z2  0.30544 m

 z3  H  tf 

h3 2

z3  0.83704 m

Ri  Ai zi

 0.00002    0.00089  3 Ri   m  0.00041     0.00313 

 Ii 

 z4  H 

2 z4  0.893 m

1 3 b i  h i 12

 57166.66666667    82516877.38571519  4 Ii   mm  391262.93479209     57166.66666667 

76102 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling The depth of the tension side:



 we Ri will focus on steel columns, why? Because the strength of steel is very high In this chapter

  i smaller  to  cross sectional area of a member to resist a particular force as compared which leads et  et  0.42772 m  SoAthat i  problem of buckling generally arises in steel columns. We have a very to concrete.   i  for understanding the problem of buckling, and I hope that you will get interesting example



The depth of the compression side: involved.  ec  0.47228 m ec  H  et Have ever seen Chaplin Theyou centroid axis Charlie shifts down by: stick when he rests on it?

  i

 

 i

Ai 

 

support.

Ieffy  0.00159 m

Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small increase in an existing compressive load. This response leads to instability and collapse of the member. In this section we shall describe the critical, or buckling, load for welded and bolted profiles, the compressive load that cases the instability. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint

if the yield strength of the web were 460 MPa, the moment resistance would be Ieffy

As we know that the effective length of the column depends on the end conditions of the column

The stick inHthe picture bellow describes one of the most fundamental characteristic of a column eM    et eM  0.02228 m in the field 2of structural engineering, called "Buckling of Column". But why did it buckle? The effective second moment of area about the new centroid axis: What made the stick to bend instead of taking the load straight down the ground? Well, here

Aievent. Ieffyto   zi   et  we are discussIiabout the



conditions. The compressive force is, in this context, thought to be applied at the centroid of

MeffRD  1404.1953 m kN

the cross-section.

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

Average stress in columns versus slenderness ratio

that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it depth is called slender member and remember should use the Approximate optimum web of aadouble-symmetric I-profile with ayou class 3 cross-section on slender the basiswhile of thedescribing bending resistance of the cross-section. word the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load. Steel bridge for pedestrians across the stream, realized from the I profile

103 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the resistance to lateral-torsional buckling for a beam with the cross-section I300,

This coefficient is a function of torsion

when the length of the beam is L = 10 m. The beam is simply supported at the both ends.

Correction coefficient for opened profile: Profile

Properties of the member: St37   e  240 MPa

  d  0.60  e

d

 144 MPa  Es  210 GPa

C M  Iy 

t1  10.8 mm  H  300 mm 

t2  16.2 mm

Lc  10 m

 a  H  t2

b  125 mm

h  H  2 t2 h  26.76 cm

a  28.38 cm 3



 1

ip 

ix 

 iy 2.55093 cm 0

ix  11.91151 cm

 cr

 1

2 2 ip  148.391 cm ix  iy  ip 12.182 cm

Lc is the buckling length

1 3 1 3 1 3   b1 t1  b2 t2  b3 t3  3 3 3 



iM  ip iM  12.182 cm





Es

T

A  d k1

 64.32478



 cr

Fallowable 

Fcr 

T

 1  CM   0.039 IT  A L 2  

i is the radius of gyration

c  5511.30605 cm

c  74.23817 cm

k1  1.231

4   IT 61.1325 cm

3 

I B, I PB 1,29

3 h t1 

I300



CM  0.039   Lc IT

c 

iy 

I 1,31



I Bolded 1,15

CM  90408.8889 cm

 b t23

a 4

 IT  1.31  2 

Iy  4.49 10 mm

Ix  97.9 10 mm

TU 1,12

I300

A  6.90 10 mm

L 1,0



 500.91307 MPa

 cr

2

 50.09131 kN cm

Fallowable  807.14866 kN

Fcr  807.14866 kN

 x is slenderness ratio cr is buckling critical stress  x 

Lc ix

Buckling Buckling

x

 83.95242

then

k1x  1.527

76104 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

2 Buckling  Es 2  cr   cr  294.0713 MPa  cr  29.40713 kN cm 2 In this chapter  xwe will focus on steel columns, why? Because the strength of steel is very high

which leads to smaller cross sectional area of a member to resist a particular force as compared

to concrete. A So  d that problem of buckling generally arises in steel columns. FallowableWe have a very Fx  650.69 kN Fx  Fallowable Fx   1.24045 ok interesting example for understanding the problem of buckling, and I hope Fx that you will get k1x involved.

Lc Have yyou when he rests on it?  ever seen Charlie Chaplinstick then k1y  14 y  392.01382 iy The stick in the picture bellow describes one of the most fundamental characteristic of a column 2

Es in the field of structural engineering, called "Buckling of Column". But why did it buckle? 2  cr    13.48703 MPa   1.3487 kN cm 2stick to bend insteadcrof taking the load straight downcr the ground? Well, here What made the  y

we are to discuss about the event. A  d Fy  Fy  70.97143 kN k1y

The most commonly used lattice types are K. LAc K truss is suitable for long-spanned structures 

i where loads be transferred directly to lattice joint locations. In K type trusses, the number of Themembers equationis above describes the slenderness ratio, limit of slenderness column in small and joints are simple. In general,and a Kthe truss is simple and very of affordable

varies as of thefabrication material changes. terms costs. For example, steel column is said to be short if the slenderness ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is Check the buckling resistance of the profile-T80 as shown in the figure below. called a long column if it is greater than 200. Long columns are very susceptible to elasticity. properties: AsSteel we know that the effective length of the column depends on the end conditions of the column support.   e  240 MPa

 0.60  e  adm  144 MPa  Es  210 GPa Buckling be defined asarea. the sudden, large, lateral deflection of a column owing to a small A is thecan cross-sectional  adm

increase in an existing compressive load. This response leads to instability and collapse of the t is the flange and web thickness of the T-profile member. In this section we shall describe the critical, or buckling, load for welded and bolted Cross-section properties andthat classification of the cross-section: profiles, the compressive load cases the instability.

Fy  Fallowable

profilebuckling, T80  F  150the kNcritical L  for100 cm - torsional To calculate load flexural and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied 4at the centroid of 4 2 Iy  37.0 cm Ix  73.7 cm   b  80 mm  h  80 mm  A  13.6 cm  the cross-section. t  9 mm   e  22.2 mm a  6 cm

 11.37287

Buckling due to torsion is not allowed, the calculation is ok

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross Steel staircase using I cross - section section of the member then it is called a slender member and remember you should use the slender while describing the compression in column not when there is any tension. word

Buckling Buckling Buckling Buckling

i is radius of gyration of the cross-section

Average stress in columns versus slenderness ratio

For cross-sections (O, L, T) the value of CM=0 CM  0 Thus The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity t of the columns areysometimes taken to be pin-ended. For buckling of pin-ended columns  yMjoints,  e M  1.77 cm 2 centrically loaded by compressive forces F at each end. The member is assumed to be perfectly Iy be constructed of a linearly elastic material-that is, we Ix have an ideal column. straight and to iy  ix  iy  1.64942 cm ix  2.3279 cm A A When the load F has been increased sufficiently to cause a small lateral deflection. This is a 2 condition of neutral load. 2 2 equilibrium. The corresponding load value2of the load is critical  ip 8.13971 cm ip  ix  iy ip  2.85302 cm

105 Behaviour of Steel Structures 2019

ip  yM

iM 

Behaviour of Steel Structures 2019

iM  3.35747 cm

CM  0.039   Lcy  IT

 iM 11.27261 cm

c 

Lcx is the buckling length cr s critical buckling stress X - direction  Lcx  L  a

 cr





Lcx  0.94 m

Es

 cr

x

Fx 

A  adm k1x

 x 

Lcx ix

 1271.13569 MPa

x

 cr

 40.38

  vi 

k1x  1.067

 63.22221

 cr





y

Es

y



 cr

Lcy

y

 563.87372 MPa

Fy 

Fy  F

 1

Es

0

 60.63

k1y  1.201 2

  cr 56.38737 kN cm

A  adm

 vi

 x

T



F A

 cr

 518.53688 MPa

 cr

d

 134.66912 MPa

d

 vi

 y

 1  CM   0.039 IT  A L 2  

Fy  163.06411 kN

k1y

T

 30.92778



The criterion is satisfied because y is greater than vi

 1

1 3 IT  1.12  t [ b  ( h  t ) ] 3

  d  k1vi 

Geometrical torsional rigidity: 

k1vi  1.221

 vi

y  direction





2

 51.85369 kN cm

From the calculation we see that the force Fy is greater than F





The critical buckling stress was calculated in the simplified way below.

buckling critical stress. The calculation of the buckling length and the slenderness ratio are

 cr

2 c

  2   2 2 2  4 c ip  0.093   1  yM   2        0   1  1  2   2 2 c  iM  

Fx  F

Determining the member Fx, Fy forces at the both direction x respectively y direction from

Lcy  L

c  iM



2

x is slenderness ratio

below.

Lcy

 43.31757 cm c

 127.11357 kN cm

 vi

Fx  183.54264 kN

c  6.58161 cm

IT  41096.16 mm

Buckling Buckling

  adm

and

76106 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Calculate the compression resistance of the box column, where the length of the column is Buckling 15m. Steel S355J2G3 is the material used. In this chapter we will focus on steel columns, why? Because the strength of steel is very high

The cross-section is at least class 3



which leads to smaller sectional when area ofthe a member a particular force compared Classification of thecross cross-section, effect of to theresist welds is not taken intoasaccount:

The equation above describes the slenderness ratio, and the limit of slenderness of column The cross-sectional parameters: area of the web varies as the material changes. For example, steel column is said to be short if the slenderness

to concrete. So that problem of buckling generally arises in steel columns. We have a very tf  25 mm tw  15 mm bf  450 mm  b  370 mm  E  210000 MPa interesting example for understanding the problem of buckling, and I hope that you will get fy  355 MPa  M1  1.1  M0  1.1  h  550 mm  involved.

ratio is less than 50, it is called an intermediate column 2 if it is between 50 and 200 and it is A   Aw 7500 mm w  tw hw called a long column if it is greater than 200. Long columns are very susceptible to elasticity.

Have you ever seen Charlie Chaplin stick when he rests on it? Buckling length:

support.

As wearea know thatweb the effective length ofA the column endAconditions the2 column of the t b depends on the 11250of mm f

The stick bellow describes of a0.34 column  15 picture m Lcz  15one m of the most fundamental  A  1 characteristic Lcy in the   in the field of structural engineering, called "Buckling of Column". But why did it buckle? hw  h  2 tf  hw  0.5 m What made the stick to bend instead of taking the load straight down the ground? Well, here we are to discuss about the event. Flange:

b  14.8 tf

b tf

 26.8

class 1

area of the flange A  2 Aw  2 Af  A 37500 mm Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small 3 compressive 3 2 increase in 2ant existing load. This tresponse leads to instability and collapse of the 2 bf tf h f w hw 4 Iy In this section  we shalldescribe 2 Af  the  critical, Iy for 1864062500 mm  member. or buckling, load welded and bolted 12 12 2 2 profiles, the compressive load that cases the instability. 3 3 2 2 tf bf 2 hw tw  b tw  4 Iz  the critical  load forflexural 2 Aw - torsional   buckling, and to Iallow To calculate for the endmm restraint z  935812500 12 12 2 2  conditions. The compressive force is, in this context, thought to be applied at the centroid of

Web:

hw tw

 34.2

hw tw

 33.333

iy  222.95366 mm

iz 

iz  157.97152 mm

class 3

Resistance based on the strength of the cross-section: NplRd  A 

 M0

NplRd  12102.273 kN

Resistance of the cross-section to the flexural buckling:

bf In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

 18

 36.667

Average stress in columns versus slenderness ratio tw

that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity Buckling about y-axis: of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight andLcy to be constructed of a linearly elastic material-that is, we have an ideal column. fy   y  0.8805  y      A  iy F has When the load  E been  increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load. 2   y  0.5   y  1.00333 1     y  0.2   y 

107 Behaviour of Steel Structures 2019

y

Behaviour of Steel Structures 2019

 

y

N  ybRd   y A 

 y

y

 0.67369

y

 1

NybRd  8153.239 kN

 M1

 

y

2   y  0.5  1     y  0.2   y  y

 

y

N  ybRd   y A 

 y

y

 1.00333

y

 0.67369

y

Aeff

is the area of the cross-section

Aeff

is the effective area of the cross-section

 M1

is the partial safety factor for the material in calculation of the stability

 1 1



NybRd  8153.239 kN



 iz

 fy    A E

 

2   z  0.5  1     z  0.2   z 

z





 1



2 0.5 1   (   0.2)   

is an imperfection factor





Lcz



Buckling about z-axis:

 z 

for Class 4 cross-section



for Class 1,2 or 3 cross-section

A

 0.8805

fy  M1

is the reduction factor for the relevant buckling mode

A

 fy    A E

Lcy  iy

  A A 



Buckling about y-axis:

 y 

NbRd

z

 1.2427

z

 1.44941

z

 0.4555

z

i

is the buckling length in flexural buckling.

is the radius of gyration about the relevant axis, determined using the

z

 1

 z

The compression resistance of the cross-section is: N  zbRd   z A 



 fy    A E

 

properties of the gross cross-section



z

fy  M1

T

NzbRd  5512.565 kN

Nsd  NbRd

Buckling Buckling

A fy NcrT

 A

NcrT

2  EI 1   GIv  2 2 i0  LT

   

iy  iz  y0

76108 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Check the cross-section shown in the figure below, should be used as column with buckling Buckling length Lc, in one steel skeleton construction. It need to be examined how much centric load F In this chapter we will focus on steel columns, why? Because the strength of steel is very high can column receive. which leads to smaller cross sectional area of a member to resist a particular force as compared to Steel concrete. So that problem of buckling generally arises in steel columns. We have a very properties: interesting example for understanding the problem of buckling, and I hope that you will get St37 involved.   e  240 MPa  d  0.60  e  d  144 MPa Es  210 GPa Have you ever seen Charlie Chaplin stick when he rests on it?   3.14159 TheCross-section stick in the picture bellow describes one of the most fundamental characteristic of a column properties: in the field of structural engineering, called "Buckling of Column". But why did it buckle? Lx Ly kN  Lx  instead 8 m ofLtaking m load  Fstraight  the q ground? Ftot Well, 208 kN  q made  13 the stick to bend What down here y  8 the tot  2 2 2 m we are to discuss about the event. t1  t1  1.2 cm  t3  1.2 cm Lc  6.80 m b3  40 cm  a3  b3  2  2

 2 a t  a    b t  t3   1   3  2 2   y0 



Lc i

y0  13.797 cm The equation above 2 a t1describes   b t3 the slenderness ratio, and the limit of slenderness of column varies as the material changes. For example, steel column is said to be short if the slenderness 2 a3   4 200 and it is ratioIyis less between 50 and  Ian a1t1   column if 2 I1than  I350, it is called I1 it is17794.3008 cm 1  intermediate 2 called a long column if it is greater than 200. Long columns are very susceptible to elasticity. 1 4 3 4 As we the column of the column Iy  41429.7088 cm  I3 know that t3  the a3 effective length I3  of 5841.1072 cmdepends Iy on  the 2 I1end  I3conditions 12 support. Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small t3 increase load. This e1  iny0an existing compressive  e1  13.19728 cmresponse leads to instability and collapse of the 2 member. In this section we shall describe the critical, or buckling, load for welded and bolted profiles, the compressive load that cases the instability. 2

 t1a  a  yflexural 2  4 To calculate andIxtoallow for the end Ix  2  the critical  t3aload  for a3  t1 buckling, e1 24351.94776 cmrestraint 0   t3-torsional 12  2  conditions. The compressive force is, in this context, thought to be applied at the centroid of Since there is no buckling stress, let´s put b and b0 it 1 the cross-section. 3

b1  a3

b1  38.8 cm  a1  b1 

a1  39.4 cm

iy 



iy  17.13411 cm

 1

0

ix 

ix  13.13629 cm

 1

Distance to the shear centre from the centre of gravity we calculate below. ip 

ix  iy

ip  21.59027 cm

ip  466.13986 cm

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load Cross-section required static An values that a long, slender, ideal column can carrywith without buckling. ideal column is one that is perfectly homogeneous, andmfree from initialbstress. load,b sometimes  a  astraight,  b3 The  2 tmaximum a  0.4  0.376 m 1  0.5 t3 1 called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is 2 A1an instability  2 a t1 occurredA  t3  b3 because 2 t1 of A  A1 loading. A2  A what 141.12 called in2 a structure excessive But do wecm mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio

a1  a3 t3 c  16.92253 cm c1  The idealized supports shown below1 seldom occur. Because of uncertainty relative to the fixity 4 Iy 2

of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns Then it is centrically loaded by compressive forces F at each end. The member is assumed to be perfectly yM andc1to be  e1 constructed of yaMlinearly  30.11981 straight elasticcm material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a 2 is critical load. 2 2 2 condition of neutral equilibrium. The corresponding load value of the load iM  ip  yM iM  37.05864 cm iM  1373.34269 cm

109 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

According to DIN 4144 for the cross-section the value of CM be: 4

I3  5841.1072 cm

Fallowable 

Iy  41429.7088 cm

I1  17794.3008 cm

Fcr  2

CM  

a1 I1  2 I1I3  3 Iy





 x 

 cr

CM  0.039   Lc IT 2

c 

Lc  6.8 m 2

c  13.69713 cm

c  187.61139 cm

iM  37.05864 cm

iM  1373.34269 cm

iy  17.13411 cm

T



c  iM 2

2 c

  2  2  2  2   1  yM   4 c ip  0.093     2     0   1  1    2 2 2 c  iM  



 1  CM   0.039 IT  A L 2  c 

k1  2.240 



x

ix 2



Es

 cr

x

 51.76502

 773.47512 MPa

then

k1x  1.13

 cr

2

 77.34751 kN cm

 d 144 MPa

 A 141.12 cm

Fallowable  907.2 kN

 vi



T

 112.32135

Es 2

 vi

  vi  k1 

Ftot A

 cr

 164.28366 MPa

 vi

 33.01587 MPa

Fx  Fallowable

 0.50446

y

 cr



 112.32135

Fx  1798.34 kN

k1x

Fallowable

 107.37597

 vi

A  d

Then the relationship gives



2

  cr 16.42837 kN cm

y



Es

y

Fy 

Then the curvature of bending and torsion, proof of stability can be carried out with lvi

 cr



Fx 



Fcr  907.2 kN

  vi 

A  d

Therefore it would be: 2

ip  466.13986 cm

Fallowable  907.2 kN

Determine of the slenderness ratio, to gain k1

CM  6551132.36756 cm

1 3 3 4  2 b1 t1  b3  t3 IT 67.7376 cm 3

 IT 

A  d

 cr

A  d k1y

Fallowable Fy

 39.68691

 1315.90497 MPa

Fy  1904.52484 kN

 0.47634

Exact calculation should not be avoided d

 144 MPa

Buckling Buckling

then

k1y  1.067

2

  cr 131.5905 kN cm

Fy  Fallowable

76110 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Flange:

Calculate the moment resistance and the resistance to lateral-torsional buckling for a beam with Buckling the cross-section WI1200-8-20-400, when the length of the beam L is 14m, the beam is simply In this chapter we will focus on steel columns, why? Because the strength of steel is very high supported at both ends and the loading is uniform. which leads to smaller cross sectional area of a member to resist a particular force as compared

varies as the material changes. For example, steel column is said to be short if the slenderness Web: ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is bw bw Class 4 Thevery profile is Class 4to elasticity.  100.9  143.58579 called a long column if it is greater than 200. Long columns are susceptible tw tw As we know that the effective length of the column depends on the end conditions of the column

involved. throat thickness of the neck weld a  4 mm  Have you ever seen Charlie Chaplin stick when he rests on it?

support. The entire flange is effective. Calculate the effective width of the web:

The in the picture bellow fdescribes one of the most fundamental characteristic of a column  fstick Es  210 GPa  steel  0.29 yf  460 MPa y  355 MPa in the field of structural engineering, called "Buckling of Column". But why did it buckle? Es fyf Gsteel the  ground? Well, here  the stick to bend instead fyd  418.18182 MPa Whatfydmade of taking the load straight down 2  1     steel  we are to discuss about the event.

 b  tw 2



2 a

c  0.19034 m

i c c Class 3  9.51716  11 4 The tequation above describes the slenderness ratio, and the limit of slenderness of column t f f

to concrete. So that problem of buckling generally arises in steel columns. We have a very   1.1  H  1200 mm b  400 mm t  20 mm t  8 mm L  14 m interesting example for understanding fthe problem of wbuckling, and I M1 hope that you will get

c 





 bw  H  2  tf 

2 a



Buckling can be defined as the235 sudden, large, lateral deflection of a column owing to a small k  23.9     0.81362 increase in an existing compressive 355 load. This response leads to instability and collapse of the member. In thisbsection we shall describe the critical, or buckling, load for welded and bolted w tw profiles, the compressive load that cases theinstability. p  0.22  p    2 28.4   k  p buckling, and to allow for the end restraint To calculate the critical load for flexural - torsional

bw  1.14869 m

conditions. The compressive force is, in this context, thought to be applied at the centroid of

the cross-section.  p  1.271

hw  1.16 m



 0.65056

The compression region and the tension region of the web are equal in the gross cross-section:    beff

beff  0.37733 m

 be2  0.6 beff

be1  0.4 beff

be1  0.15093 m

be2  0.2264 m

The depth of the ineffective region: bneg 

perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is Lateral-torsional buckling for a beam with the cross-section WI1200-8-20-400 called an instability occurred in a structure because of excessive loading. But what do we mean

 be1  be2

A  w  tw  H  2 tf 

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load that a long, slender, ideal column can carry without buckling. An ideal column is one that is

bneg  0.20267 m

2 Average stress in columns versus slenderness ratio

Aw  0.00928 m

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity 2  Af  0.008 m A f  tf b of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns 2  A  2loaded Af  Aby A F0.02528 m The member is assumed to be perfectly w compressive forces centrically at each end.

Buckling Buckling Buckling Buckling

straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When i the1 load 4 F has been increased sufficiently to cause a small lateral deflection. This is a condition neutral The corresponding the is critical  b 1  of 400 mm equilibrium. b b 3  8load mmvalue of b 4load  400 mm load. 2  8 mm

111 Behaviour of Steel Structures 2019

h 1  tf

h 2 

H 2

 be2  h 1 h 2  806.396 mm h 3  be1

h 3  150.93 mm

 z2  tf 

z1  0.01 m

 z3  H  tf 

z2  0.4232 m

 A i  b i h i

 z4  H 

z3  1.10453 m

Ri  Ai zi

 0.008    0.00645  2 Ai   m  0.00121     0.008 

Ii  

tf 2

z4  1.19 m 1 12

b i  h i 

 0.00008   266666.66666667      0.00273  3 349585363.13646835  4 Ri   m Ii   mm  0.00133   2292129.19895206       0.00952   266666.66666667 

Ieffy

MeffRD  4312.48025 m kN

The effective section moduli are calculated as follows: Weffyc 

Weffyt 

Ieffy

Weffyc  10312452.779 mm

ec Ieffy

Weffyt  11114552.936 mm

The moment resistance: The smaller of the values calculated above is used as the value of the section modulus.

The depth of the tension side:

   et    

If the yield strength of the web were 460 MPa, the moment resistance would be MeffRD  fyd 

h 4  0.02 m

h 4  tf

z1 

Behaviour of Steel Structures 2019

 Ri i

 i



MyRd  3328.11 m kN

 M1

To calculate the lateral-torsional buckling moment and the resistance moment to lateral-

et  0.57754 m

Ai 

torsional buckling, the geometrical warping rigidity, the geometrical torsional rigidity as well

 

as the second moment of area about the z-axis are needed, in addition to the cross-sectional parameters calculated above.

The second moment area about z-axis:

The depth of the compression side: ec  H  et 

bf  b 1

ec  0.62246 m

Iz 

The centroid axis shifts down by  eM 

eM  0.02246 m

 et

  Ii 

Ai  zi 2  et2   

 i



 H  2 tf  tw3 12

Iz  213382826.67 mm

The geometrical warping rigidity: The distance between the centre lines of the flanges is

The effective second moment of area about the new centroid axis: Ieffy 

2 tf bf

bf  0.4 m

Ai 

 

Ieffy  6419093951.19 mm

hf  H  tf 

hf  1.18 m 2

 Iw  0.25 hf Iz

Buckling Buckling

Iw  7.42786  10 mm

76112 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

for double-symmetric profiles, zj = 0. Moreover, the formula for the geometrical warping Buckling of double-symmetric can be why? presented in thethe form: In rigidity this chapter we will focus on profiles steel columns, Because strength of steel is very high

Lc The reduction factor for lateral-torsional buckling is calculated from the formula as follow: 

which leads to smaller cross sectional area of a member to resist a particular force as compared geometrical to The concrete. So thattorsional problemrigidity: of buckling generally arises in steel columns. We have a very

The equation above describes the slenderness ratio, and the limit of slenderness of column Weffyc Wply fy Weffyc fy Weffyc fy  LTif the 1.81006  LT  LT  varies as the material changes. For example, steel column is said to be short slenderness Wply Mcr Mcr Mcr ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is

interesting example for understanding the problem of buckling, and I hope that you will get 1  3 3 4 involved.  Iv  2 bf tf   H  tf  tw   Iv 2334720 mm 3 Have you ever seen Charlie Chaplin stick when he rests on it?

called a long column if it is greater than 200. Long columns are very susceptible to elasticity. The lateral-torsional buckling shall be calculated  LT  0.4 As we know that the effective length of the column depends on the end conditions of the column

lateral against latera-torsional buckling are 14m apart. Because of thea column loading is TheThe stick in therestraints picture bellow describes one of the most fundamental characteristic top flange. The rotation and warping of the of the But member restrained in applied the fieldatofthestructural engineering, called "Buckling of ends Column". why are didnot it buckle? for instead uniformofloading C2straight . The profile double symmetric, which k = made kw = 1,0. 1 and What the The stickfactor to bend taking Cthe load downisthe ground? Well, here 0. zj = about wemeans are to that discuss the event.  zg  600 mm

kw  1

k  kw

C1  1.132

C2  0.459

zj  0

support. where C1, C2 and C3 are factors depending on the load and end restraint conditions. Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small The distance of the loading from the shear centre is positive for loads acting towards the shear increase in an existing compressive load. This response leads to instability and collapse of the centre. For example, for loading situated on the top flange and directed downwards, zg is member. In this section we shall describe the critical, or buckling, load for welded and bolted positive. if the same direction acts on the bottom flange, zg is negative. profiles, the compressive load that cases the instability.

where the relative slenderness  LT  0.4 no allowance for lateral-torsional buckling is To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint necessary conditions. The compressive force is, in this context, thought to be applied at the centroid of

The lateral -torsional buckling moment:

for welded sections

The elastic critical moment for lateral-torsional buckling for a profile symmetric about its minor axis is given by the formula as follow:

Mcr  C1

2   LT  0.5  1   LT  LT  0.2   LT 

2  Es Iz   k  2 Iw ( k L) Gsteel Iv 2       C  z  C  z     2 g 2 g   2  2  Es Iz ( k L)   kw  Iz 



 LT

Mcr  1117.388 m kN



 LT

 LT 

 LT

 LT

 2.53263

 0.23234

 LT

 1

  LT

is the reduction factor for lateral-torsional buckling

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load Iv is the geometrical torsional rigidity that a long, slender, ideal column can carry without buckling. An ideal column is one that is Iw is the geometrical warping rigidity perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes is the second momenttheofcolumn area about theinminor calledIzthe critical load, causes to be a stateaxis of unstable equilibrium; that is, the

L introduction slightest lateral force will cause thewhich column to lateral fail by restraints buckling. (distance Buckling is is of thethe length of the profile between points, have called an instability occurred in a structure because of excessive loading. But what do we mean between lateral-torsional supports) by slender? When the longitudinal dimensions of the member are much greater than the cross section ofkthe member it is a slender member and remember you should use the factors then related to called the effective length k and w are word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio fy

MbRd  773.2544 m kN MbRd   LTWeffyc  ok  M1 The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load.

113 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Sabah Shawkat © Repair and reconstruction-removal of the load-bearing masonry wall and replacement by a steel frame using rectangular hollow welded sections Repair and reconstruction- design of a shelter above the entrance of the building from steel I profiles

Buckling Buckling

76114 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling



Lc i

Have you ever seen Charlie Chaplin stick when he rests on it?

support.

The stick in the picture bellow describes one of the most fundamental characteristic of a column

Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small

in the field of structural engineering, called "Buckling of Column". But why did it buckle?

increase in an existing compressive load. This response leads to instability and collapse of the

What made the stick to bend instead of taking the load straight down the ground? Well, here

member. In this section we shall describe the critical, or buckling, load for welded and bolted

we are to discuss about the event.

profiles, the compressive load that cases the instability. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied at the centroid of

floor plan of the shelter construction

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

Average stress in columns versus slenderness ratio

of the word slender while describingcross-section the compression in shelter columnconstruction not when there is any tension.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns Graf illustrated shearing forces vs depth of the beam centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load.

115 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Determine the cross-section of the cantilever steel beam loaded by axial force at the free end as

IPE400

shown in figure. Assume that the beam has a I cross-section of constant depth.

 Mmax  F L  0.5 g0 L

Data:

Mmax  92.9835 m kN

 L  3.0 m

 F  30 kN

  e  240 MPa

d

 g  0.5 kN m

 0.6  e

d

1

Es  210 GPa

 144 MPa

h  b1  180 mm

t1  8.6 mm

g0  0.663 kN m

 H  400 mm b2  H  2 t2 

1

b2  0.373 m

 t2  13.5 mm 6

 Ix  231 10 mm  0.29  It  514 10 mm

 steel

 max

 80.50519 MPa

Then the maximum stress at the axis of the beam will be:

We propose profile IPE400

Iy  13.2 10 mm

  max  Mmax

 IT 

H  t2





Mmaxh Ix

  Fl 77.78814 MPa



1 3 3 4  2b1 t1  b2 t2  IT 38.22333 cm 3

Gsteel 



 Fl

Gsteel  81395.35 MPa

2  1   steel 

 hFl   Gsteel IT  2 L  Es Iy





 0.36971

adjacent figure.

k  0.878 

Solution:

The bending moment generally vary along the length of a beam. When designing a beam, it is

  kl 

L 2 It  Iy H

k hFl

2 IxL

k  1.83766

 Es Iy Gsteel IT

 kl

 150.49163 MPa

useful to have available a graphical visualization of the variation when the bending moment plotted. 

The calculation of the bending moment due to uniform load and bending moment due to axial



 kl  max



 1.86934



 1.71

force at the end of the beam. 2

 Mmax  F L  0.5 gL

Wxneeded 

 hFl  H  2 

Mmax d

t2 2

Mmax  92.25 m kN

The ratio of σkl to σmax bigger than 1.7. 3

640.62 cm W xneeded

This is the safety of the I steel beam to avoid lateral buckling. hFl  38.65 cm

Buckling Buckling

76116 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Lc The entire flange is effective. calculate the effective width of the web:

Calculate the moment resistance of the hybrid beam shown in the figure below. Buckling



In this chapter we will focus on steel columns, why? Because the strength of steel is very high tf  40 mm tw  25 mm  H  1100 mm  b  500 mm  a  8 mm which leads to smaller cross sectional area of a member to resist a particular force as compared fyf  460  fyw  of355 MPa generally arises in steel columns. We have a very to  concrete. SoMPa that problem buckling interesting example for understanding fyf the problem of buckling, and I hope that you will get fyfd  418.182 MPa  M0  1.1 fyfd  involved.  M0 Have you ever seen Charlie Chaplin fywstick when he rests on it? fywd  322.727 MPa fywd   M0 one of the most fundamental characteristic of a column The stick in the picture bellow describes in the field engineering, called "Buckling of Column". But why did it buckle?  b of tstructural w bw  H  2  tf  2 a bw  0.99737 m  to2 bend a c  c  0.22619 m  What made the stick instead of taking the load straight down the ground? Well, here 2 we are to discuss about the event.  hw  1.02 m hw  H  2 tf





The equation above describes the slenderness ratio, and the limit of slenderness of column bw

varies as the material changes. For example, steel column is said tto be short if the slenderness w 235  p   23.9 0.40202   0.71475 p  200 ratio kis less than 50, it  is called is between 50and and it is 460 an intermediate column if it28.4   k  called a long column if it is greater than 200. Long columns are very susceptible to elasticity.  p  0.22 As we know the column depends on the end conditions of the column   1.12622  that the effective length of 2 p support. Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small The compression region and the tension region of the web are equal in the gross cross-section: increase in an existing compressive load. This response leads to instability and collapse of the hw member. describe the critical, beff   section we bshall In  this be1orbuckling, 0.4 beff load be1for  welded 0.22975and m bolted eff  0.57437 m 2 profiles, the compressive load that cases the instability. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint be2  The  0.6 bcompressive be2  0.34462 m context, thought to be applied at the centroid of eff conditions. force is, in this

Flange:

c  6.4 tf

the cross-section. The depth of the ineffective region:

Class 1

Web:

bw tw

 51.5

Class 1

bw tw

bneg 

 39.8949

hw 2

 be1  be2

A  f  tf b

The profile is Class 1.

bneg  0.06437 m Af  0.02 m

Aw  tw  H  2 tf  Aw  0.0255 m A  2 Af  Aw

A  0.0655 m

Plastic moment resistance:

MplRD  Af fyfd  hw  tf   Aw fywd  In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

hw 4

MplRD  10963.989 m kN

Average stress in columns versus slenderness ratio

that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a structure because of excessive loading. But what do we mean by slender? When the longitudinal dimensions of the member are much greater than the cross section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

117 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the moment resistance of the hybrid beam shown in the figure below.

tf  25 mm

 H  1000 mm b  500 mm  1.1

 M0

fyfd 

fyf  M0

 fyf  460 MPa

 2 a

 a  5 mm

fyw  355 MPa

 fyfd 418.182 MPa

 b  tw

c 

tw  25 mm

fywd 

c  0.23043m

fyw  M0

fywd  322.727 MPa



 bw  H  2  tf 

2 a



bw  0.93586 m

hw  H  2 tf 

hw  0.95 m

Flange:

 9.21716

 10

Class 3

Web:

bw tw

 88.6

bw tw

 37.43431 Class 3

The entire flange is effective. calculate the effective width of the web

Garage for passenger cars across the stream, realized from the I welded profile

Buckling Buckling

76118 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling  Ai 

335 why? Because the strength of steel is very high In this chapter we will focus on steel columns, k  23.9     0.85338 which leads to smaller cross sectional area 460 of a member to resist a particular force as compared bw









Buckling can be defined 9875000 of a column owing to a small  10000  as the sudden, large, lateral deflection

be2  0.27393 m

ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is  10000   125000  called a long are very  column if it is greater than 200. Long columns  susceptible to elasticity. 4750 1246875     As we know that the effective of the column 2 length of the column depends on the end conditions 3 Ai   2739.31  mm Ri   1744845.837  mm support.     1826.207 1613799.874

Have you ever seen Charlie Chaplin stick when he rests on it? The compression region and the tension region of the web are equal in the gross cross-section: The stick in the picture bellow describes one of the most fundamental characteristic of a column

 be2  0.6 beff

The equation above describes the slenderness ratio, and the limit of slenderness of column  R i  A i zi varies as the material changes. For example, steel column is said to be short if the slenderness

to concrete. So that problem of buckling generally arises in steel columns. We have a very tw p  0.22   0.96116     0.31594   p p interesting example for understanding the problem of buckling, and 2 I hope that you will get 28.4   k p involved.

in the field of structural engineering, called "Buckling of Column". But why did it buckle? hw Whatbeff made ofmtaking bthe load ground?   the  stick to bbend 0.45655 0.4straight beff down be1  the 0.18262 m Well, here eff instead e1  2 we are to discuss about the event.



b i h i

increase in an existing compressive load. This response leads to instability and collapse of the 2  Qi  Ai  zi member. In this section we shall describe the critical, or buckling, load for welded and bolted 1 compressive profiles, the load that cases the instability. 3 Ii   b i  h i  12 To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint  520833.333  force is, in this context, conditions. The1562.5 compressive thought to beapplied at the centroid of





 89309895.833    4 Ii   17129402.914  mm  5075378.641     520833.333 

bneg 

hw 2

 be1  be2

bneg  0.01845 m

Af  0.0125 m

A  f  tf b

Aw  tw  H  2 tf  Aw  0.02375 m A  2 Af  Aw

A  0.04875 m

327304.688 the cross-section.   3 Qi   1111406.587  m mm

 1426098.282     9751562.5 

i  1  5

 b 1  400 mm

h 1  tf

The depth of the tension side:

 b2 

 h2 

10 mm  b3 

H 2

 h1

10 mm

h 3  be2

 b4 

10 mm

h 4  be1

h 5  tf

In 1757, Eulerh derived a formula the maximum h mathematician 0.025 m  h Leonhard 475  mm 273.931 mm that h gives  0.18262 m h  axial 0.025load m 1

   et    

 b 5  400 mm

that a long, slender, ideal column can carry without buckling. An ideal column is one that is tf hw be2 be1 h2 perfectly z1  straight, homogeneous, z3  initial stress. h 1  The maximum z4  hwload,  h 1sometimes   z2  tf  and free from 2 2 2 2 2 called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is z1  0.0125 m z2  0.2625 m z3  0.63697 m z4  0.88369 m called an instability occurred in a structure because of excessive loading. But what do we mean tf z5  H When   0.9875 mdimensions of the member are much greater than the cross by slender? thez5longitudinal 2 section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

 Ri i

 i



Ai 

et  0.49822 m

 

Average stress in columns versus slenderness ratio

TheThe idealized supports shown below depth of the compression side:seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly  ec  H  et ec  0.50178 m H  1m straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When load Faxis hasshifts beendown increased The the centroid by sufficiently to cause a small lateral deflection. This is a condition of neutral equilibrium. The corresponding load value of the load is critical load. H eM 1.78191 mm  eM   et  2

119 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

The effective second moment of area about the new centroid axis:

Calculate the moment resistance of the hybrid beam shown in the figure below

Ieffy 

 Ii   Ai  zi   e   Ai 2



Ieffy  5453756204.927mm



 H  1300 mm  b  450 mm  fyf  460 MPa

The elastic section moduli of the cross-section are calculated from the formulas below:

c 

Weffc 

Ieffy

Wefft 

Ieffy et

Weffc  0.01087m

Wefft  0.01095m

 b  tw 2





tf  25 mm tw  10 mm

 1.1

fyd 

2 a



 bw  H  2  tf 

2 a

Wel 



bw  1.23586 m hw  1.25 m

Flange:

fyw

c  8.51716 tf

 M0

c  10 tf

 MelRD  Weffc fywd

McRD

fyd  418.18182 MPa

Elastic moment resistance: MelRD



c  0.21293 m

 hw  H  2 tf

McRD

fyf

 a  5 mm

MeffRD

MelRD  3507.65113 m kN

Weff 

Web:

 M1

bw tw

 88.6

The profile is Class 4.

Realisation of steel stair case using I profile

Buckling Buckling

Class 4

Class 3

bw tw

 123.58579

76120 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

The entire flange is effective. calculate the effective width of the web: Buckling

 A i  b i h i



w In this chapter we will focus on steel columns, why? Because the strength of steel is very high tw 235 which to smaller cross sectional area of a memberto resist a particular force pascompared  k leads  23.9   1.24536  0.71475 p 460 28.4   k to concrete. So that problem of buckling generally arises in steel columns. We have a very  p  0.22 interesting problem of buckling, and I hope that you will get  0.66113   example for understanding the

involved.

p

Ri  Ai zi

The equation11250 slenderness of column  abovedescribes the slenderness ratio, and the limit  of140625   material    varies as the changes. For example, steel column is said to be short 6250 2109375if the slenderness









 

 

 

 

3 ratioAis less2479.23 is 2called an intermediate column if Rit is between 50 and 200  than 50,itmm  1918828.41  mmand it is i i  1652.82  called a long column if it is greater than 200. Long columns are very susceptible to elasticity. 1970754.671

As we know 11250 that the effective length of the column depends on the end14484375 conditions of the column

Have ever seen Charlie Chaplin stick when heofrests it?are equal in the gross cross-section: Theyou compression region and the tension region the on web

support.

The stick in thehw picture bellow describes one of the most fundamental characteristic of a column beff    beff  0.4132 m be1  0.4 beff be1  0.16528 m in the field of structural engineering, called "Buckling of Column". But why did it buckle? 2

1 3 to a small 2 as the sudden, large, lateral deflection of a column Buckling  can defined  Ii  b i owing h i Qi be A i  zi 12 increase in an existing compressive load. This response leads to instability and collapse of the

What made the stick to bend instead of taking the load straight down the ground? Well, here be2  0.6 beff  be2  0.24792 m we are to discuss about the event.

member. In this section we shall describe the critical, or buckling, load for welded and bolted  1757.813   585937.5  profiles, the load that cases the instability.  compressive   203450520.833  711914.063     3 4 To calculate the critical load flexural - torsional buckling, allow for the end restraint Q   1485099.298 mm I and   to12698988.097  mfor  mm

The depth of the ineffective region: bneg 

 be1  be2

bneg  0.2118 m

Aw  tw  H  2 tf 

Aw  0.0125 m

 force is, in this context, thought tobe3762663.14 conditions. The compressive applied at thecentroid of 2349847.081     the cross-section. 18648632.813 585937.5

 A f  tf b

Af  0.01125 m

A  2 Af  Aw

A  0.035 m

i  1  5

 b 1  450 mm

h 1  tf

h 1  0.025 m

 b 2  10 mm

 h2 

H 2

 h1

h 2  625 mm

b 3  10 mm

h 3  be2

b 4  10 mm

 b 5  450 mm

h 4  be1

h 3  247.923 mm h 4  0.16528 m





The depth of the tension side:

   et    

 Ri i

 i



et  0.62721 m

Ai 

 

h 5  tf

The depth of the compression side:

h 5  0.025 m

ec  H  et  ec  0.67279 m The centroid axis shifts down by:

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load tf hw be2 be1 h2 that za1 long, buckling.  slender, ideal  z2 column tf  can carryz3without   h 1  An ideal z4 column  hw  ish 1one  that is 2 2 2 2 2 perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the z2  0.3375 m z3  0.77396 m z4  1.19236 m z1  0.0125 m introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is tf called an instability occurred in a structure because of excessive loading. But what do we mean z5  1.2875 m z5  H   2 the longitudinal dimensions of the member are much greater than the cross by slender? When section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling



H  et eM  0.02279 m Average stress in columns versus slenderness ratio 2 effective secondshown moment of area about the new centroid axis: TheThe idealized supports below seldom occur. Because of uncertainty relative to the fixity eM  

 i 



10 columns 4 2 2 of the sometimes For of pin-ended  I buckling  1.04828  10 mm I joints,  columns I  areA  z   taken e  to beApin-ended. effy







effy

centrically loaded by compressive forces F i i  ati each end. The member is assumed to be perfectly

If the and yieldtostrength of the web 460 MPa, moment resistance would straight be constructed of awere linearly elasticthe material-that is, we have anbe ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a Ieffy MeffRD  6515.71589 m kN MeffRD  fyd  condition of neutral eequilibrium. The corresponding load value of the load is critical load. c

121 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the dimensions and resistance of steel columns loaded by axial force F = 215 kN axial force

F  215kN

column length

 L  8.20 m 

 95 slenderness ratio

To be evaluate the dimensions of steel columns.  e36 

36 MPa

k136  2.379

  e  30 MPa  e24 

k1  2.067

24 MPa

k124  1.767

if we propose that the slenderness λ be 95

0.6  e24

 36d

 0.6  e36

 14.4 MPa

 36d

 21.6 MPa

 24d   24d

 A36  k136

 24d

A24  263.82292 cm

A36  236.79861 cm

I180

I200

A180  2.79 10 mm

A200  3.34 10 mm

 72 mm ix180

ix200  80 mm

 180   180

L ix180

 200 

 113.89

 200

k1180  2.003 F   24  k1180 A24  24

Reconstruction of the hotel, using I and circular Hollow sections

 36d

 24

 16.32326 MPa   24d

Buckling Buckling

does not

L ix200

 102.5

k1200  1.961   36  k1200

F A36

 36

 17.80479 MPa

 36

  36d

 36d

 21.6 MPa

76122 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling

Lcx  L Lcx  8.2 m Ineeded 0.12 F Lcx 1755 cm In this chapter we will focus on steel columns, why? Because the strength of steel is very high for x

L sectional area of a member to resist a particular force as compared which leads to smaller cross 4 Imeeded 0.12 F Lyc 205 cm for y Lyc  Lyc  2.05 m 4 to concrete. So that problem of buckling generally arises in steel columns. We have a very

interesting example for understanding the problem of buckling, and I hope that you will get   2.5 2 involved. F Lcx 4  Ineeded   Ineeded  1743.76169 cm 2 Have you ever seen Chaplin stick when he rests on it?  ECharlie s

Lc by axial force F (kN) and horizontal force H Design the IPE cross-sectional column loaded 

i (kN). The column is fixed in the transverse plane and hinges placed in a perpendicular plane. TheAgainst equation above describes the slenderness ratio, andcolumn the limit of slenderness of top. column the buckling from the cross-section plane, the is also secured at the

varies as the material changes. For example, steel column is said to be short if the slenderness L  8.20 m

ratio is less than 50, it is called an intermediate column if it is between 50 and 200 and it is Extreme Design Loads: called a long column if it is greater than 200. Long columns are very susceptible to elasticity.  215that the 320column kN kN the effective lengthHof As  weF know depends on the end conditions of the column

support.

The stick in the picture bellow describes one of the most fundamental characteristic of a column

Columncan in the transverse plane is under compressive stress and acts as Buckling be defined as the sudden, large, lateral deflection of bending a columnand owing to aa console. small

in the field of structural engineering, called "Buckling of Column". But why did it buckle?

The buckling length compressive perpendicularload. to the y axis is: leads to instability and collapse of the increase in an existing This response

What made the stick to bend instead of taking the load straight down the ground? Well, here

member. In this section we shall describe the critical, or buckling, load for welded and bolted

we are to discuss about the event.

profiles, the compressive load that cases the instability.

Ly  2 L

To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint For buckling perpendicular to the axis of the buckling length: conditions. The compressive force is, in this context, thought to be applied at the centroid of

Lz  L the cross-section. To assess the critical section in the fifth column, where the max. all internal forces. The cross-

sectional area is estimated based on a deflection assessment that is limited at the top. max 

max  27.333mm 

300

Deflection of the column loaded due to horizontal force will be: 3



H

3 E Iy

section of the member then it is called a slender member and remember you should use the Steel frames structure, using I profiles, columns and beams word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio We calculate the required moment of inertia of column as follow:

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity 3

Iymin   Hcolumns are sometimes  1.025to 10 be  cmpin-ended. For buckling of pin-ended columns Iymin taken of the joints, 3 E max

centrically loaded by compressive forces F at each end. The member is assumed to be perfectly or straight and to be constructed of a linearly elastic material-that is, we have an ideal column.





2 4 When the load has been increased sufficiently to cause a small lateral deflection. This is a 0.12FF L 1724 cm Ineeded

condition of neutral equilibrium. The corresponding load value of the load is critical load.

123 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

As we know that the structural elements that are subjected to axial compressive forces only are

Our first proposal will be HE 220 B

called columns. Columns are subjected to axial loads through the centroid. The load applied at the ends of the member, producing axial compressive stresses. The load is known and column size (cross-sectional area) is unknown, the compressive stress and the cross-section area may be computed as:

is radius of gyration of the column cross section

is cross-sectional area of the column

The radius of gyration is geometric property that is used in the analysis and design of

is axial compressive load acting on the column

 F  700kN 2

is the area where the column has the ability to bearing the equivalent load from the slab

 qeq  13.0kN m

2

is the height of the storey

 df  0.20m

is the depth of slab floor

columns.  is known as the slenderness ratio 3

AI  9.10 10 mm

is the value of equivalent load

 H  3.4 m

over the entire cross-section. Iy,Ix is moment of inertia

Load cases:

 A  35 m

The stress in the column cross-section can be calculated as where, s is assumed to be uniform

Iy  28.4 10 mm 



is the buckling column length



Ix  80.9 10 mm

 ix  94.3 mm

iy  55.9 mm k1  1.265

 57.245

 L  H  df

L  3.2 m

  e  240 MPa

d

Lc  L

 0.6  e

d

Lc  3.2 m

 144 MPa

q AI



 162.16327 MPa

 d



does not

A higher slenderness ratio means a lower critical stress that will cause buckling. Conversely, a lower slenderness ratio results in a higher critical stress (but still within the elastic range of the material).

 cr





Es 2

Column sections with large ixy values are more resistant to buckling

Second proposal will be HE 240 B Weight of HE 240 B is Calculate the entire load, to transfer it to column:  qd  qeq A  q0  0.01  F  qd   q  qd  F  q0

qd  455 kN q0  11.55 kN

is load equivalent from the area of floor is the self-weight of the column-assumption

q  1166.55 kN is the final load on the column

83.2

AI  10.6 10 mm Iy  39.2 10 mm 



Lc iy

Buckling Buckling



 52.632

kg m 6

 Ix  113 10 mm iy  60.8 mm k1  1.135

 ix  103 mm

76124 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling q   d ok   124.90889 MPa    k1  AI In this chapter we will focus on steel columns, why? Because the strength of steel is very high 2

Es which leadsto smaller cross sectional area of a member to resist a particular force as compared  cr  748.21471 MPa  cr  2 that problem of buckling generally arises in steel columns. We have a very to concrete. So  interesting example for understanding the problem of buckling, and I hope that you will get



Lc i

involved. The most efficient column sections for axial loads are those with almost equal ix and iy values.

As we know that the effective length of the column depends on the end conditions of the column

a. Circular and square the most Have you ever pipe seen sections Charlie Chaplin sticktubes when are he rests on it?effective shapes since the radii of

support.

gyration about both axes are the same (ix = iy). The stick in the picture bellow describes one of the most fundamental characteristic of a column b. Circular pipe sections and square tubes are often used as columns for light to moderate loads. in the field of structural engineering, called "Buckling of Column". But why did it buckle?

Buckling can be defined as the sudden, large, lateral deflection of a column owing to a small

What made the stick to bend instead of taking the load straight down the ground? Well,kg here 77.2  Circula cross-section D  273 mm t  12 mm Weight m we are to discuss about the event. 2

iy  92.4 mm

AI  98.39 10 mm

increase in an existing compressive load. This response leads to instability and collapse of the member. In this section we shall describe the critical, or buckling, load for welded and bolted profiles, the compressive load that cases the instability. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied at the centroid of





   k1 

 cr

the cross-section.





q AI 2



Es 2

 34.632



k1  1.046

 124.01782 MPa

 cr



 d

d

 144 MPa

 1728.07675 MPa

In 1757, mathematician Leonhard Euler derived a formula that gives the maximum axial load

Average stress in columns versus slenderness ratio

that a long, slender, ideal column can carry without buckling. An ideal column is one that is perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes called the critical load, causes the column to be in a state of unstable equilibrium; that is, the introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called in a structure because excessive loading. what do mean We an areinstability proposingoccurred and assessing that we have the of right dimension of theBut profiles, butwe sometimes by we slender? Whenathe longitudinal member are muchingreater than the cross are making mistake becausedimensions of the lack of of the professional workers the production and the section of the member then be it isaccurate, called aon slender member andwill remember usesee thethe cross-sections that should the contrary, they be untrueyou andshould unusable word slender while describing the compression in column not when there is any tension. pictures below.

Buckling Buckling Buckling Buckling

The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity of the joints, columns are sometimes taken to be pin-ended. For buckling of pin-ended columns centrically loaded by compressive forces F at each end. The member is assumed to be perfectly straight and to be constructed of a linearly elastic material-that is, we have an ideal column. When the load F has been increased sufficiently to cause a small lateral deflection. This is a condition neutral equilibrium. corresponding is critical load. Theof improper production ofThe steel elements thatload werevalue sent of to the the load site for their realization

125 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

Calculate the cross-section class for the profile WII965-13-26x300 and the effective widths of the compression elements as well as the effective second moment of area, when the crosssection is subject to bending moment. the steel grade is S335J2G3.

 H  965 mm

 bf  300 mm

 Es  210 GPa fyf  460 MPa 

tf  26 mm

tw  13 mm

 1.35

Lc  2.5 m





 1.1

 L  7.5 m

Bending (also known as flexure) characterizes the behaviour of a slender structural element subjected to an external load applied perpendicularly to a longitudinal axis of the element. The structural element is assumed to be such that at least one of its dimensions is a small fraction, typically 1/10 or less, of the other two. When the length is considerably longer than the width and the thickness, the element is called a beam.

Sabah Shawkat © The term bending is ambiguous because bending can occur locally in all objects. Therefore, to make the usage of the term more precise, engineers refer to a specific object such as; the bending of rods, the bending of beams, the bending of plates, the bending of shells and so on.

equivalent load

p  100 kN m 

1

imposed load

concentrated load on the middle of the beam

 q pg

q  102.5 m

M H  max  2   max  Ieffy

The improper production of steel elements that were sent to the site for their realization

Buckling Buckling

1

 max

kN

 g  2.5 kN m

1

P  250 kN

Mmax 

 108.9673 MPa

q L 8

Mmax  720.70313 m kN

76126 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

The profile is Class 3.

Buckling



i The entire flange is effective. Calculate the effective width of the web: The equation above describes the slenderness ratio, and the limit of slenderness of column 235 k asthe 23.9   steel column is said 0.81362 varies material changes. For example, to be short if the slenderness 355 ratio is less thanbw50, it is called an intermediate column if it is between 50 and 200 and it is

tw  0.22  pLong called a long column if it is greater than 200. columns are very susceptible to elasticity.  p    1.0451    p  0.614 2 As we know28.4 thatthe length of the column  keffective   depends on the end conditions of the column

Have you ever seen Charlie Chaplin stick when he rests on it?

support.

The stick in the picture bellow describes one of the most fundamental characteristic of a column

The compression region tension region of the web are of equal in the gross Buckling can be defined asand the the sudden, large, lateral deflection a column owingcross-section: to a small

increase in an hexisting compressive load. This response leads to instability and collapse of the w beff    beff  0.47709 m be1  0.4 beff member. In this2 section we shall describe the critical, or buckling, load for welded and bolted be1  0.19083 m profiles, the compressive load that cases the instability.  be2  0.28625 m be2  0.6 beff To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint

in the field of structural engineering, called "Buckling of Column". But why did it buckle? What made the stick to bend instead of taking the load straight down the ground? Well, here we are to discuss about the event.

fyf

fyd 

fyd  418.18182 MPa

d

 fyd 0.6

d

conditions. The compressive force is, in this context, thought to be applied at the centroid of

 250.91 MPa



 bf

 tw



2 a

hw  H  2 tf 

h1 

hw 5



bw  H  2  tf  c  0.13784 m 

2 a



theThe cross-section. depth of the ineffective region:

bneg 

bw  0.90169 m

h2 

 be1  be2

A  w  tw  H  2 tf 

hw  0.913 m

hw  2 

h2  0.4695 m

h3 

hw  tf  tf 2

Af  0.0078 m

 A  2 Af  Aw

A  0.02747 m

i  1  4  b 1  250 mm

Flange: In 1757, Euler derived a formula that gives the maximum axial load c c mathematician Leonhard Class 3  5.30166  11 4 tf that taf long, slender, ideal column can carry without buckling. An ideal column is one that is

by slender? When the longitudinal dimensions of the member are much greater than the cross

Aw  0.01187 m

 A f  tf bf

h3  0.4825 m

perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes Web: called the critical load, causes the column to be in a state of unstable equilibrium; that is, the bw bw introduction of the slightest lateral the column to fail by buckling. Buckling is Class 3 force will cause  100.9 69.36048 t tw because of excessive loading. But what do we mean w called an instability occurred in a structure

bneg  0.02059 m

b 2  5 mm

b 3  5 mm  b 4  250 mm

h 1  tf

H Average stress in columns versus slenderness ratio h 2  742.752 mm h 3  190.835 mm  be2  h 1 h 3  be1 2 The idealized supports shown below seldom occur. Because of uncertainty relative to the fixity h 2 

tf buckling of pin-ended h 2 columns of the joints, columns are sometimes taken to be pin-ended. For h 4  0.026 m  z2  tf  z1  h 4  tf 2 centrically loaded by compressive forces F at each end. The member is assumed to2be perfectly straight and to be constructed of a linearly tf elastic material-that is, we have an ideal column. h3 z   H  t   z  H  3 f 4 When the load F has to cause a small lateral deflection. This is a 2 been increased sufficiently 2

section of the member then it is called a slender member and remember you should use the word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

condition of neutral equilibrium. The corresponding load value of the load is critical load. z1  0.013 m z2  0.39738 m z3  0.84358 m z4  0.952 m

127 Behaviour of Steel Structures 2019

Behaviour of Steel Structures 2019

A i  b i h i 

Ri  Ai zi

 Ii 

1 12

b i  h i 

Both bending and buckling can be consider as the deformation of beam than its normal position.

The main element that differs in bending and bucking is the direction of the force applied on the beam or structural member.

 0.0065    0.00371  2 Ai   m  0.00095     0.0065 

 0.00008    0.00148  3 Ri   m  0.0008     0.00619 

 366166.66666667    170734102.33857602  4 Ii   mm  2895747.17764175     366166.66666667 

Bending: Deviation of axes of structural member under the action of transverse shear load is known as bending. Buckling: Deviation of axes of structural member under axial compressive load is known as buckling. We take part of the upper beam see the adjacent figure to avoid the lateral buckling.

The depth of the tension side:

   et    

 Ri 



 Ii  17436.21828 cm

Ai 

 

et  0.48411 m

tf bf

 IyG 

 ec  H  et

h1 tw



IyG

iyG

ec  0.48089 m

 yG



H  et 2

IyG  5853.3431 cm





1.14 0.6 fyd

If the yield strength of the web were 460 MPa, the moment resistance would be

 fyd  MeffRD

Ieffy ec

Mmax h2

iyG 

izG

 32.95942

 267.57377 MPa

IyG A

iyG  7.58508 cm

k1  1.069

for lateral buckling

Ieffy

 max

 106.03139 MPa



  max

When the concentrated load is acting on the middle of the beam then the bending moment will be as follow:

 Mc  MeffRD  2775.07606 m kN



Ieffy  319122.58021133 cm



 yG

iyG

 max 

 Ii   Ai  zi   e   Ai

 yG

The maximum stress at the lower part of the beam will be:

The effective second moment of area about the new centroid axis:

c 40



eM  0.00161 m

2



The centroid axis shifts down by

Ieffy 

A  101.738 cm

The depth of the compression side:

eM  

A  bf tf  h1 tw 

P L gL  4 8

Buckling Buckling

Mc  486.32813 m kN

76128 Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Buckling

 max 

Behaviour of Steel Structures 2019 Behaviour of Steel Structures 2019

Mc h3

 max  73.53078 MPa  max   d  d  250.9 MPa I In this chapter effy we will focus on steel columns, why? Because the strength of steel is very high L  yG 98.88 to resist a particularkforce  yG  to smaller cross sectional area which leads ofa member as compared 1  2.544 iyG to concrete. So that problem of buckling generally arises in steel columns. We have a very

0.6 ffor interesting 1.14 example yd understanding the problem of buckling, and I hope that you will get     112.43568 MPa k1 involved.

Have you ever Mc seen h2 Charlie Chaplin stick when he rests on it?  Fl   Fl  71.54964 MPa  Fl   Ok Ieffy The stick in the picture bellow describes one of the most fundamental characteristic of a column in the field of structural engineering, called "Buckling 2 of Column". But why did it buckle?  A  hw tw  2 bf tf A  274.69 cm What made the stick to bend instead of taking the load straight down the ground? Well, here 3 3 we are to discuss hw tw about 2 tf the bf event. Iy  12

Iy  11716.71551 cm

 Kl  Fl

 4.24626

 Kl  Fl

 1.71



Lc i

The equation above describes the slenderness ratio, and the limit of slenderness of column varies as theismaterial Forstructural example,member steel column to be short if the slenderness Bending natural changes. reaction of whenisitsaid is applied to the transverse load. · ratio is less isthan 50,when it is called an intermediate if it is between 50 and 200 and ittoisthe Bending cause the application of loadcolumn is perpendicular (i.e. transverse loading) called a long column it is greater than 200. Long columns are very susceptible to elasticity. length of the beam if (i.e. longitudinal axis) As we know that the effective length of the column depends on the end conditions of the column · Failure is cause due to transverse shearing. support.

· During bending top fibre aresudden, in tension while bottom fibre are compression Buckling can be defined as the large, lateral deflection ofin a column owing to a small increase in an existing compressive load. This response leads to instability and collapse of the Every structural member has its limit to nullify the effect of transverse load, if the load gets member. In this section we shall describe the critical, or buckling, load for welded and bolted beyond that limit, the structure will get bending. For some extend bending can be consider but profiles, the compressive load that cases the instability. if it is crossing the safety limit then it causes threat. To calculate the critical load for flexural - torsional buckling, and to allow for the end restraint conditions. The compressive force is, in this context, thought to be applied at the centroid of

 3 1

 hFl  hw  2 

ix  

 1

the cross-section.



 2 bf tf  hw tw

IT  418.38203 cm

hFl  93.9 cm

Ieffy

ix  34.08453 cm

0

 1



 

CM 

iy 

Iy hFl 4

CM  25827185.29 cm

iy  6.53103 cm

(  L) ITEuler derived a formula that gives 2the maximum axial 2load CM   0.039Leonhard In 1757, mathematician c  c  0.54659 m c  2987.64966 cm Iy column can carry without buckling. An ideal column is one that is that a long, slender, ideal perfectly straight, homogeneous, and free from initial stress. The maximum load, sometimes  Es Iy  4317.2 SKlthe  critical load, causes Sthe called tokN be in a state of unstable equilibrium; that is, the Kl column 2 L introduction of the slightest lateral force will cause the column to fail by buckling. Buckling is called an instability occurred in a2structure because   of excessive loading. But what do we mean  SKl hFl   5 h3  2 5 h3   Kl  When thelongitudinal c   Kl  303.8185   dimensions by slender? are muchMPa greater than the cross of the member 2 Ieffy       section of the member then it is called a slender member and remember you should use the ok word slender while describing the compression in column not when there is any tension.

Buckling Buckling Buckling Buckling

Average stress in columns versus slenderness ratio

External steel staircase using I profile cross-section

Behaviour of Steel Structures 2019

Buckling Buckling

129