• Author / Uploaded
• jirong

• Categories
• Ingeniería
• Ingeniería estructural
• Ingeniero civil
• Ingeniería de Edificación
• Tecnología

Citation preview

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

TECHNICAL REPORT SCI DOCUMENT ED 005

Design of Light Steel Sections to Eurocode 3

A G J Way MEng, CEng, MICE M D Heywood MEng, PhD, CEng, MICE

Issued (in electronic format only) by: The Steel Construction Institute Silwood Park Ascot Berkshire, SL5 7QN 01344 636525 www.steel-sci.org

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

SCI (The Steel Construction Institute) is the leading, independent provider of technical expertise and disseminator of best practice to the steel construction sector. We work in partnership with clients, members and industry peers to help build businesses and provide competitive advantage through the commercial application of our knowledge. We are committed to offering and promoting sustainable and environmentally responsible solutions. Our service spans the following areas: Membership

Consultancy

Individual & corporate membership

Members advisory service

Development Product development Engineering support Sustainability

Information

Assessment SCI Assessment

Publications Education Events & training

Specification Websites Engineering software

Advice

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

SCI Technical Reports Technical Reports are intended for the cost effective dissemination of the latest research results and design guidance, as and when they become available, or as specialist documents for further discussion. A Technical Report may serve as the basis for a future SCI publication or the updating of an existing SCI publication.

 2012 The Steel Construction Institute Apart from any fair dealing for the purposes of research or private study or criticism or review, as permitted under the Copyright Designs and Patents Act, 1988, this document may not be reproduced, stored or transmitted, in any form or by any means, without the prior permission in writing of the publishers, or in the case of reprographic reproduction only in accordance with the terms of the licences issued by the UK Copyright Licensing Agency, or in accordance with the terms of licences issued by the appropriate Reproduction Rights Organisation outside the UK. Enquiries concerning reproduction outside the terms stated here should be sent to The Steel Construction Institute, at the address given on the title page. Although care has been taken to ensure, to the best of our knowledge, that all data and information contained herein are accurate to the extent that they relate to either matters of fact or accepted practice or matters of opinion at the time of publication, The Steel Construction Institute, the authors and the reviewers assume no responsibility for any errors in or misinterpretations of such data and/or information or any loss or damage arising from or related to their use. SCI Document Number: SCI ED 005

ED005V01.doc

ii

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

FOREWORD Light gauge cold-formed steel members (generally referred to as light steel sections) are commonly used in a range of building types as secondary steelwork (e.g. purlins and cladding rails in industrial buildings) and as the primary load-bearing elements in light steel frames (e.g. in residential buildings). They may be used as individual structural members (e.g. floor joists) or as part of a structural frame. In Eurocode 3 Design of steel structures, design rules for such members in the UK are given in several separate Parts, notably BS EN 1993-1-1, BS EN 1993-1-3 and BS EN 1993-1-5, each together its National Annex. This Report has been prepared to aid designers in the application of the various rules in those documents, particularly in the applications in wall panels and floors. The report has been prepared by Andrew Way and Martin Heywood, both of SCI. The worked examples have been independently checked by Stephen Napper of Stephen Napper Associates Ltd, and his involvement is gratefully acknowledged.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The work leading to this Report was funded by Tata Steel Strip Products UK.

ED005V01.doc

iii

Created on This material

23 is

February copyright

2012 all

rights

reserved.

Use

of

this

document

is

subject

to

the

terms

and

conditions

of

the

Steelbiz

Licence

Agreement

ED005

Discuss

Technical me ...

Report: Design

ED005V01.doc

of Light Steel Sections to

iv

Eurocode 3

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Contents

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Page No FOREWORD

iii

SUMMARY

vii

1

INTRODUCTION 1.1 Light steel sections 1.2 Eurocodes 1.3 Design to Eurocode 3

1 1 2 3

2

SECTION PROPERTIES 2.1 Core steel thickness 2.2 Mid-line theory 2.3 Corner radii

5 5 6 6

3

LOCAL BUCKLING 3.1 Effective width concept 3.2 Eurocode calculation procedure for unstiffened plane elements

9 9 10

4

DISTORTIONAL BUCKLING 4.1 Design of stiffened sections 4.2 Eurocode calculation procedure for stiffened elements

12 12 14

5

DESIGN OF COMPRESSION MEMBERS 5.1 Design issues 5.2 Eurocode calculation procedures

17 17 17

6

DESIGN OF MEMBERS IN BENDING 6.1 Laterally restrained members 6.2 Lateral-torsional buckling 6.3 Serviceability

20 20 22 23

7

FRAME DESIGN 7.1 Frame stability 7.2 Structural robustness 7.3 Frame anchorage

25 25 26 27

8

WORKED EXAMPLES 8.1 Gross section properties for a cold-formed lipped C section 8.2 Effective section properties for a lipped C section in compression 8.3 Effective properties for a lipped C section in major axis bending 8.4 Effective properties for a lipped C section in minor axis bending 8.5 Design of a wall stud in a light steel frame building 8.6 Design of a floor joist (restrained) in a light steel frame building 8.7 Design of a floor joist (unrestrained) in a light steel frame building 8.8 Design of a lattice floor truss in a light steel frame building

28 29 38 43 48 52 60 64 67

9

REFERENCES

78

ED005V01.doc

v

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3

Discuss me ...

ED005V01.doc

vi

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

SUMMARY This Technical Report provides guidance in accordance with Eurocode 3 for the design of steel sections as used in light steel framing applications. The guidance includes a brief introduction to Eurocodes and light steel sections, followed by detailed design guidance. Since light steel members are especially prone to local buckling, the design consequences of this behaviour are dealt with in depth, notably the calculation of effective cross section properties. Design guidance for members in compression and members in bending is also given.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Eight worked examples are provided to illustrate the application of the design rules to practical building applications. The examples include the calculation of gross and effective section properties, the design of sections subject to bending and compression and serviceability design of light steel floors.

ED005V01.doc

vii

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3

Discuss me ...

ED005V01.doc

viii

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

1

INTRODUCTION

Light gauge cold-formed steel members (generally referred to as light steel sections) are commonly used in a range of building types as secondary steelwork (e.g. purlins and cladding rails in industrial buildings) and as the primary load-bearing elements in light steel frames (e.g. in residential buildings). They may be used as individual structural members (e.g. floor joists) or as part of a structural frame. Light steel members are often prefabricated off-site to form wall panels, floor cassettes or volumetric modular units, but are equally suited to stick build applications.

1.1

Light steel sections

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

For the purposes of this report, the term ‘light steel’ refers to galvanized cold-formed steel sections with a maximum gauge (thickness) of 4 mm, although gauges from 1.2 mm to 2.0 mm are the most common for light steel framing applications. For purlins and cladding rails, the thickness generally lies in the range 1.4 mm to 3.2 mm. Light steel members are usually cold-formed from hot-dip galvanized strip steel, which is supplied to the section manufacturers as pre-galvanized coil conforming to BS EN 10346: 2009[1]. For light steel framing applications, the commonly used grades of steel are S350, S390 and S450, while purlin and cladding rail products tend to use S390 and S450. Lipped C sections are the most common section shape for light steel framing applications, including wall studs and floor joists. The C section shape is simple to roll and widely manufactured, while the lip provides additional stiffness to the flange and increases the stress at which local buckling occurs in the flange. Depths commonly range from 70 mm to 120 mm for wall studs and from 120 mm to 250 mm for floor joists. Purlins are typically made from Zed or sigma sections with depths ranging from 140 mm to 300 mm. A typical light steel frame is shown in Figure 1.1. The photograph shows a prefabricated volumetric module, but similar framing arrangements are also used in panellised and stick-built construction.

Figure 1.1

Load bearing light steel frame in a volumetric module

1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

1.2

Eurocodes

The Eurocodes are a set of structural design standards, developed by CEN (the European Committee for Standardization) over the last 30 years, to cover the design of all types of structures in steel, concrete, timber, masonry and aluminium. In the UK they are published by BSI under the designations BS EN 1990 to BS EN 1999, each in a number of ‘Parts’. Each Part is accompanied by a National Annex that implements the CEN document and adds certain UK-specific provisions.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

There are ten separate Eurocodes: BS EN 1990

Eurocode: Basis of structural design

BS EN 1991

Eurocode 1: Actions on structures

BS EN 1992

Eurocode 2: Design of concrete structures

BS EN 1993

Eurocode 3: Design of steel structures

BS EN 1994

Eurocode 4: Design of composite steel and concrete structures

BS EN 1995

Eurocode 5: Design of timber structures

BS EN 1996

Eurocode 6: Design of masonry structures

BS EN 1997

Eurocode 7: Geotechnical design

BS EN 1998

Eurocode 8: Design of structures for earthquake resistance

BS EN 1999

Eurocode 9: Design of Aluminium Structures

Each Eurocode is comprised of a number of Parts, which are published as separate documents. The main Eurocodes that may be required for the design of a light steel buildings and elements are: BS EN 1990

Basis of structural design

BS EN 1991

Actions on structures

BS EN 1993

Design of steel structures

BS EN 1990[2] can be considered as the ‘core’ document of the structural Eurocode system because it establishes the principles and requirements for the safety, serviceability and durability of structures. This document specifies the limit states and load combinations that should be considered. Eurocode 1 is used to determine the actions (loads) that structures must be designed to resist. Eurocode 3 comprises twelve parts (BS EN 1993-1-1 to BS EN 1993-1-12). When designing light steel framed buildings, the following parts of BS EN 1993-1 will generally be required: BS EN 1993-1-1[3]

General rules and rules for buildings

BS EN 1993-1-3[4]

Supplementary rules for cold-formed members and sheeting

BS EN 1993-1-5

[5]

BS EN 1993-1-8[6]

Plated structural elements Design of joints

Further guidance on the Eurocode system of design for steel structures and the documents involved is provided in SCI publications P362[7] and P387[8]. 2

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

The focus of this report is the design of light steel elements for which BS EN 1993-1-3 is the main reference. However, for the design of light steel frames it is important to note that other parts of the Eurocodes include essential design requirements, e.g. frame stability in BS EN 1993-1-1 and structural robustness in BS EN 1991-1-7[9] (see Section 7).

1.3

Design to Eurocode 3

Although BS EN 1993-1-3 is only a ‘supplementary’ Part of Eurocode 3 and has many references to the general rules in BS EN 1993-1-1 and to rules in some other Parts, as noted above, it contains most of the rules that will be needed for the design of light steel structural members. BS EN 1993-1-3 has been specifically written for light steel and makes special allowances for the structural behaviour commonly encountered with this product. BS EN 1993-1-3 permits two alternative design routes:  Design by calculation

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

 Design by testing. As the name suggests, in the former route the structural designer follows an analytical procedure laid down in BS EN 1993-1-3 to arrive at a calculated value of section resistance (to compression, bending etc.). The method is fairly complex due to the need to take account of local buckling through the use of effective widths, but can be attempted by hand for relatively simple sections such as lipped C sections. One disadvantage of this approach is that it tends to be conservative due to the assumptions and simplifications on which the method is based. For this reason, this approach is rarely used by purlin and cladding rail manufacturers, as it would place their products at a commercial disadvantage. Furthermore, while the scope of BS EN 1993-1-3 includes a range of section shapes, it is less well suited to some of the more complex sections, especially those with multiple stiffeners and curved webs, flanges or lips. However, despite the apparent limitations of this approach, design by calculation remains the preferred option for designers of light steel frames and floor joists and is the primary focus of this report. Design by testing overcomes the limitations of the calculation approach by permitting design resistances to be obtained accurately for almost any shape of section. However, the benefits must be weighed up against the costs of undertaking a programme of tests followed by the statistical analysis required by BS EN 1993-1-3 to convert raw test data into usable design values. Where testing is undertaken, such as for purlins and cladding rails, the design data are usually tabulated in the form of load span tables. These tables are published by the product manufacturer and are used by potential specifiers to choose the most suitable section for the given span and load. The testing route can be sub-divided into two options:  Design data derived directly from test data.  Design data derived from a numerical model. In the first option, an appropriate number of tests are undertaken on a range of test specimens and the results are analysed statistically to obtain a characteristic resistance for each section size within the range. The rules for statistical analysis are laid out in BS EN 1993-1-3 (or alternatively the method in Annex D of BS EN 1990 may be used) and involve subtracting a prescribed multiple of the standard 3

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

deviation from the mean test result. The multiple is dependent on the number of tests undertaken. This option is the simpler of the two, but has the disadvantage that tests must be performed on multiple samples of every section size within the product range. It is, therefore, not suitable for products with a wide range of dimensional variations, such as purlins and cladding rails, which are normally sold in a wide range of depths and gauges. It is, however, a useful method where the product range is limited, such as purlin cleats or tie wires.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

In the second option, testing is only conducted on a limited number of specimens from within the full range. The test results are then normalised (by comparison with the equivalent theoretical value) and used to derive a numerical model that aims to accurately predict a safe design resistance across the full product range. This approach avoids the need to test the full range of section sizes and even allows for new sections to be introduced at a later date without the need for additional testing. Furthermore, where test results are normalised against a theoretical value, it is often possible to consider results from different section sizes as belonging to the same ‘family’ of data, thereby reducing the multiple of standard deviations to be subtracted. The accuracy of the final resistances will depend on the complexity of the numerical model and the number of tests undertaken. For example, a simple model might be used to calculate a realistic bending resistance of a section, while maintaining the theoretical assumption of a simply supported beam. By comparison, a more complex model might take account of the stiffness in the end connections of the beam, using data from a separate set of tests, in order to increase the beam’s resistance. A typical test on a light steel purlin is shown in Figure 1.2. In this instance, a point load is being applied to a cleat connecting two lengths of purlin at the mid-point of a simply supported span. The purpose of this test was to assess the moment-rotation behaviour of the joint in order to model accurately the behaviour of a complete purlin system. The data from this series of tests were combined with results from two other types of test (gravity and uplift loading on a pair of purlins with sheeting attached) to create a numerical model, which was then used to produce load span data for the full range of purlin sizes.

Figure 1.2

Physical testing of purlin specimen

4

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

2

SECTION PROPERTIES

Before a light steel member’s resistance to bending, compression or other loading type can be calculated, it is necessary to determine the dimensional properties of the section under consideration. To those unfamiliar with light steel, this first step might appear to be a trivial exercise of calculating the cross-sectional area and second moments of area, but in reality it is a complex process that lies at the heart of the Eurocode design procedure for light steel. It is important to distinguish between the following types of section property:  Gross section properties  Effective section properties.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The term ‘effective section properties’ refers to the properties of a fictitious cross section that has been reduced in area to take account of the impact of local buckling on its resistance (see Section 3). Further reductions may also be necessary to allow for distortional buckling (see Section 4). The bending and compression resistances of light steel members are always calculated using the effective properties of the section. The calculation of the effective section properties is dealt with in detail in Section 4. The remainder of this section discusses the determination of gross section properties because these are required for some aspects of light steel design. As the name suggests, the term gross section properties refers to the whole cross section without any reduction for local buckling. The process of calculating the gross section properties is relatively straightforward for most common section shapes as it involves little more than the summation of elemental areas and first and second moments of area (for flanges, web, stiffeners etc.), the calculation of the position of the major and minor centroidal axes and, from these values, the second moment of area for the whole section. A similar process can be repeated for other properties as required. There are, however, three important issues that need to be addressed when considering light steel sections:  Core steel thickness;  Use of mid-line theory;  Impact of corner radii.

2.1

Core steel thickness

The galvanized cold rolled strip steel of the type used in light steel construction is normally delivered pre-coated. Therefore, when specifying the thickness of the steel it is common practice to include the thickness of the coating in the specified value. However, BS EN 1993-1-3 (§3.2.4) requires that all section properties are based on the core thickness of the steel, excluding the coating. The standard zinc coating for construction products is 275 g/m2 (referred to as Z275), which corresponds to a coating thickness of 0.02 mm on each surface. Hence, the nominal (specified) steel thickness should be reduced by 0.04 mm for design purposes. The tolerances to which the steel is manufactured and specified should also be considered when determining the thickness to use in design calculations in accordance with BS EN 1993-1-3 (§3.2.4(3)). The application of this clause is explained in detail in SCI advisory note, AD 358: Design thickness of cold formed members and sheeting[10]. 5

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

2.2

Mid-line theory

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

When calculating the section properties of light steel sections, it is standard practice to measure all dimensions along the mid-lines of the individual elements. Initially, corner radii are ignored (see Section 2.3), resulting in an idealised section consisting of a series of thin rectangular elements. In calculating the lengths of the individual elements, an allowance must be made for the intersection between adjacent elements, to avoid double-counting the over-lapping corner regions. Using the midline theory, this is simply achieved by measuring each element length between the points of intersection of the mid-lines. This results in a reduction in the element length below its nominal value of either t/2 or t, depending on the number of corners. The mid-line dimensions for a lipped C section are shown in Figure 2.1.

Figure 2.1

2.3

Mid-line dimensions for a lipped C section

Corner radii

The use of mid-line theory described in Section 2.2 results in an idealised section that is easy to analyse. However, without modification, the impact of the rounded corners on the section properties, which could be significant, is ignored. The problem is illustrated by Figure 2.2.

Figure 2.2

Rounded corner of a light steel section

6

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

According to mid-line theory, the web and flange intersect at the point X (the intersection of the two mid-lines. The true intersection point is P, a distance gr from X, given by:

      g r  rm  tan   sin     2   2

(Eq.1)

where

rm  r 

t 2

It is apparent that any section properties derived from mid-line theory will not be exact. The important question for designers is whether or not the error is significant. BS EN 1993-1-3 gives some guidance in this respect (in §5.1) stating that the influence of rounded corners on cross section resistance may be neglected provided that both of the following conditions are satisfied:

r  5t

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

r  0.1b p where bp is the width of the element measured between the midpoints of the corners (see Figure 2.2). Consider the following example: A typical lipped C section has a nominal width of 65 mm, corner radii of 3.0 mm and a nominal thickness of 1.5 mm. Allowing for the standard 275 g/m2 zinc coating, the core thickness t = 1.46 mm. The flange width measured between mid-lines = 65 – 1.5 = 63.5 mm. (It is assumed that the nominal width of 65 mm includes the galvanising, so it is appropriate to subtract the nominal thickness when calculating the mid-line dimension.)

rm  r 

t = 3.73 mm 2

      g r  rm  tan   sin    = 1.09 mm  2   2 bp  63.5  2 g r  61.32 mm (which corresponds to a reduction in area of about 3.4%) Checking the corner radii: 5t = 7.3 mm, r = 3.0 mm, therefore r  5t 0.1bp = 6.13 mm, r = 3.0 mm, therefore r  0.1bp

7

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Therefore, in this instance, the influence of the rounded corners may be neglected when calculating the cross section resistance. Note: The influence of rounded corners should always be taken into account when calculating cross section stiffness properties (BS EN 1993-1-3 (§5.1(3)). Where the influence of rounded corners needs to be accounted for, this is achieved by first calculating the section properties assuming sharp corners (i.e. ignoring the corner radii) and then applying reduction factors as follows: For area, Ag  Ag,sh 1   

(Eq.2)

For second moment of area, I g  I g,sh 1  2 

(Eq.3)

For warping, I w  I w,sh 1  4 

(Eq.4)

In these expressions, the subscript ‘sh’ denotes the section property based on sharp corners and δ is a reduction factor given by:

j

n

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

  0.43

j1 r j 90

(Eq.5)

m

i1 bp,i

where:

rj

is the internal radius of curved element j

n

is the number of curved elements (number of corners)

øj

is the angle between two plane elements

bp,i

is the notional flat width of plane element i

m

is the number of plane elements.

The same reduction factors may also be applied to the effective section properties (Aeff, Iy,eff, Iz,eff and Iw,eff) provided that the notional flat widths of the plane elements are measured to the points of intersection of their midlines. BS EN 1993-1-3, 5.1(6) states that where r is greater than 0.04t E / fy then the resistance of the cross section should be determined by physical testing. This situation is unlikely to arise for any of the standard sections used in light steel framing, but designers need to be aware of this limit when dealing with some of the more unusual section shapes that are introduced into the light steel market from time to time.

8

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

3

LOCAL BUCKLING

Light steel sections of the types used in construction are profiled in such a way that they are extremely efficient in terms of their use of material. However, the associated penalty from a designer’s point of view is the need to consider local buckling and its impact on the structural resistance of the cross-section.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Designers of hot-rolled structural steelwork will be familiar with the concept of section classification in which a section’s susceptibility to local buckling is determined by comparing dimensional ratios (e.g. the depth to thickness ratio of the web) to a set of specified limits. This process results in each section being assigned a ‘class’ (which in some cases is dependent on the magnitude of any compression in the member). BS EN 1993-1-1 describes four such classes and prescribes design rules for each class that reflect the impact of local buckling on the resistance of the section. The classifications range from ‘Class 1’, which is defined as being a section capable of sustaining its full plastic moment while accommodating rotation of a plastic hinge, to ‘Class 4’, defined as a section whose moment resistance is limited by local buckling. (The bending resistance of a Class 4 section is less than that corresponding to elastic bending resistance of the gross cross-section.) The approach used in light steel design is quite different. Rather than classifying the cross-section, there is an implied assumption that the section is class 4 (although this term is not used in BS EN 1993-1-3). Having made this assumption, the design process focuses on the calculation of effective section properties, following the same procedures for design resistance as for class 4 sections to BS EN 1993-1-1. The use of effective section properties stems from the need to simplify the complex stress distributions associated with local buckling, in order to minimise the required computational effort, without being over conservative in terms of the cross section resistance.

3.1

Effective width concept

When dealing with local buckling of slender plate elements, the behaviour of the plate may be analysed approximately using the effective width method, as illustrated by Figure 3.1.

Figure 3.1

The effective width concept applied to a plate

In this approach, the actual stress distribution acting over element width b is replaced by simplified equivalent stresses acting over two equal widths of beff/2. The central portion of the plate, the region most affected by local buckling, is assumed to have no stress and is ignored completely. The result is a simple model in which a 9

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

uniform stress equal to the yield strength of the steel is assumed to act over a reduced width of plate. The method adopted by BS EN 1993-1-3 takes the effective width concept illustrated above and applies it to the cross section of a light steel member. The cross section is divided into elements (flanges, web, lips etc.), with each element being treated like the flat plate in Figure 3.1. An effective width beff is calculated for all elements that are subjected to compressive stress (either due to axial compression or bending). The effective area of the element Aeff is then obtained by multiplying beff by the section thickness t. Elements not subjected to compressive stress are not susceptible to local buckling, so the full element width b may be used in the calculation of the effective section properties.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Having obtained the elemental effective widths and areas, the properties of the effective cross section are determined in the usual manner by calculating the position of the neutral axis followed by the first and second moments of area about this axis. The resulting set of ‘effective properties’ should be used when calculating the resistance of the cross section to bending or compression as appropriate. Since the distribution of compressive stress across the section differs between sections subjected to pure axial compression and those subjected to bending, it follows that the effective section properties will differ between these two cases. Furthermore, asymmetric sections subjected to bending may have one set of effective properties when sagging and another when hogging. It is important to use the relevant effective properties for the case under consideration. For sections at the heavier end of the ‘light steel’ range, especially those with relatively stocky webs and flanges, it might not be appropriate to reduce the crosssectional properties to allow for local buckling. Since BS EN 1993-1-3 does not permit the classification of the section in the manner familiar to designers of hotrolled structural steel, the procedure outlined above must be followed even for stocky sections. However, in this case, the calculation procedure will automatically yield effective widths beff equal to the full widths b, resulting in effective section properties equal to the gross section properties.

3.2

Eurocode calculation procedure for unstiffened plane elements

This Section focuses on the calculation of beff for an element subjected to compression. The calculation of beff for plane elements without stiffeners is introduced in §5.5.2 of BS EN 1993-1-3. However, the detail of the method, including the relevant equations can be found in BS EN 1993-1-5. For each element, the effective width is given by:

beff  b

(Eq.6)

where:

b

is the width of the element

ρ

is the reduction factor to allow for local buckling.

10

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

The reduction factor ρ takes account of the slenderness of the element, whether it is an internal or outstand element and the stress distribution within the element. For an internal element, ρ is given by:



 p  0.0553   2

p

 1.0

(Eq.7)

For an outstand element, ρ is given by:



 p  0.188 2

p

 1.0

(Eq.8)

where ψ is the stress ratio between the ends of the element and slenderness of the element given by:

p 

fy

 cr



bt 28.4 k σ

p

is the

(Eq.9)

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

where

fy

is the design strength

σcr

is the elastic critical plate buckling stress

b

is the appropriate width of the compression element

t

is the steel core thickness (i.e. minus the coating)

kσ

is the buckling factor corresponding to the stress ratio ψ and the boundary conditions. Values of kσ should be obtained from Tables 4.1 and 4.2 of BS EN 1993-1-5 for internal and outstand elements respectively.



 235 / f y

From the discussion in Section 3.1, it follows that there must be limits of slenderness below which local buckling does not influence the resistance of the section. These limits correspond to ρ = 1 in equations 7 and 8 and, according to BS EN 1993-1-5, the limits are: Internal compression elements:

 p  0.673 Outstand compression elements:

 p  0.748 Where p is lower than the appropriate limit, ρ should be taken as 1.0 in the calculation of the effective width of that element. This does not necessarily mean that the section is fully effective, since there may be other elements for which ρ < 1.0.

11

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

4

DISTORTIONAL BUCKLING

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

In the discussion on local buckling, it was assumed that the corners of the section remained fixed in position, so that the buckling deformation takes place within the length of the element. This case is represented on the left hand side of Figure 4.1. By contrast, the right hand side of Figure 4.1 shows a situation in which the right hand corners of the flanges are not fixed in position, allowing the flanges to rotate. This is known as distortional buckling.

Figure 4.1

Local and distortional buckling

The susceptibility of a section to distortional buckling depends on the ability of the stiffeners to prevent displacement of the adjacent flange corners. This is dependent on the geometry of the stiffeners relative to the flanges and, in particular, their relative stiffness. BS EN 1993-1-3 presents a detailed procedure for the design of stiffened sections based on a simple spring model. This approach is described in Section 4.1 with a summary of the procedure given in Section 4.2.

4.1

Design of stiffened sections

Due to the susceptibility of light steel sections to local and distortional buckling, it is common practice for manufacturers to roll stiffeners into the sections. It is most common to stiffen the free edge of the flanges by the provision of a lip, or in some cases a double lip, but intermediate flange stiffeners may also be used in order to permit the use of thinner gauge steel. It is less common to stiffen the web in sections used for light steel framing applications, although deeper purlins and floor joists may benefit from this technique. Flange and web stiffeners are often used in trapezoidal deck profiles of the types used for roofing and flooring applications due to the very light gauge steel used in these products. BS EN 1993-1-3 provides guidance to cater for all of the options discussed above. In all cases, the underlying assumption is that the stiffener behaves like a compression member with a continuous partial restraint. This is a reasonable assumption since, whether the member is subjected to pure axial compression or bending, one flange and its stiffener at least will be subjected to a longitudinal compressive stress. In the design model, the stiffener is represented by a linear spring of stiffness K, as shown in Figure 4.2.

12

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Figure 4.2

Linear spring model

The spring stiffness depends on the boundary conditions and the flexural stiffness of the adjacent elements. The spring is assumed to act at the centroid of the effective stiffener section. Figure 4.2 shows two varieties of edge stiffener and one intermediate stiffener. In each case, the ‘effective stiffener section’, which is depicted as a dark solid line, comprises the stiffener itself plus an adjacent length (or lengths) of flange. In order that the stiffener provides sufficient stiffness and to avoid buckling of the stiffener, BS EN 1993-1-3 gives limits for the geometry of the stiffener in relation to the adjacent flange as follows:  For a single or double-lipped section, the length of the lip c, measured perpendicular to the flange width b, should lie in the range 0.2 ≤ c/b ≤ 0.6.  For a double-lipped section, the return length of the lip d, measured parallel to the flange (width b), should lie in the range 0.1 ≤ d/b ≤ 0.3. For the case of the edge stiffeners of lipped C sections and Z sections, the spring stiffness K1 for flange 1 may be obtained from the following equation:

E t3 1 K1   2 2 3 4(1   ) b1 hp  b1  0.5b1b2 hp kf

(Eq.10)

where:

b1 and b2 are the distances from the web-to-flange junction to the centroid of the effective stiffener section for flanges 1 and 2 respectively.

hw

is the depth of the web.

kf

is the ratio of the effective areas of the two edge stiffeners (including the effective portion of the flange).

All other symbols have their usual meanings.

13

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

The term kf can have the following values:  For sections subjected to axial compression, such that flange 1 and flange 2 are both in compression: kf = As2 / As1 (where As1 and As2 are the effective areas of the edge stiffeners).  For sections subjected to bending about the major axis so that flange 1 is in compression and flange 2 is in tension: kf = 0  For symmetric sections in compression: kf = 1 Having calculated the equivalent spring stiffness of the stiffener, the method described in BS EN 1993-1-3 proceeds to determine the elastic critical stress for the stiffener σcr followed by the relative slenderness  d and the reduction factor for the distortional buckling resistance χd. Finally, χd is used to determine the reduced effective area of the stiffener, which is generally represented as a reduced thickness when determining the effective properties of the section. A detailed algorithm for determining χd and hence the reduced thickness of the stiffener is described below.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

4.2

Eurocode calculation procedure for stiffened elements

Detailed procedures for the design of plane elements with edge or intermediate stiffeners are presented in §5.5.3 of BS EN 1993-1-3. The procedures combine the calculation of beff with the calculation of a reduced thickness for the stiffener. The former takes account of local buckling within the length of the element, while the latter makes an allowance for the impact of distortional buckling. The procedure described below is for a flange with an edge stiffener. It is divided into three steps, the last of which involves an optional iteration in order to refine the value of the reduction factor χd. BS EN 1993-1-3 also presents procedures for flanges with intermediate stiffeners, stiffened webs and trapezoidal decking profiles. Step 1: The procedure begins with the calculation of the effective width of the flange beff following the method described in Section 3.2. At this point in the procedure, it is assumed that the edge stiffener is infinitely stiff and, therefore, provides full restraint to the free end of the flange. This corresponds to the left hand side of Figure 4.1, in which the corners of the section are fixed in position and failure is due to local buckling. It is also assumed that the maximum compressive stress in the flange is equal to the design strength of the material, i.e.

 com,Ed  f yb  M0

(Eq.11)

Step 2: In the second step, the edge stiffener is considered in isolation in order to calculate the reduction factor χd for distortional buckling. At this point, the infinitely stiff spring used in Step 1 is replaced by a spring of stiffness K, as illustrated in Figure 4.2. The spring stiffness K may be calculated from Equation 10, using the initial effective cross section of the stiffener determined in Step 1.

14

conditions

ED005

Technical

Report:

Design

of

Light

Steel

Sections

to

Eurocode

3

...

Once K is known, the elastic critical buckling stress σcr,s of the stiffener may be calculated from:

 cr,s 

subject

to

the

and

me

terms

Discuss

2 KEI s

(Eq.12)

As

Is and As are the effective second moment of area and the effective cross-sectional area respectively of the stiffener.

The relative slenderness of the stiffener for distortional buckling d is given by:

d 

f yb

(Eq.13)

 cr,s

this

document

is

where fyb is the basic design strength of the steel. The reduction factor for distortional buckling χd is dependent on the slenderness d as follows: For  d  0.65 ,  d  1.0

Eq.14a)

For 0.65   d  1.38 ,  d  1.47  0.723 d

(Eq.14b)

For  d  1.38 ,  d  0.66  d

(Eq.14c)

2012 -

all

rights

reserved.

Use

of

Step 3: The value of χd may be refined iteratively by returning to Step 1 and calculating a modified effective flange width beff based on a revised compressive stress σcom,Ed. This is achieved by calculating a modified value of ρ (see Section 3.2) using a reduced p given by:

 p,red   p  d

Step 2 may then be repeated for the modified effective section to obtain a new value of χd. Steps 1 and 2 may be repeated until the desired degree of convergence on the value of χd has been achieved. Step 3 is entirely optional and it is perfectly acceptable to use the initial value of χd for the calculation of the reduced area of the stiffener. Where the designer chooses to iterate to obtain an improved value of χd, one or two iterations should suffice. Once χd has been calculated to the desired degree of refinement, the reduced effective area of the stiffener As,red may be calculated using :

February copyright

As,red   d As

23 is

(Eq.15)

f yb /  M 0

(Eq.16)

 com,Ed

where σcom,Ed is the compressive stress at the centreline of the stiffener based on the effective cross section. It is usually more convenient to work in terms of a reduced thickness such that:

t red  t As,red As

Created This

on material

(Eq.17)

15

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

This may be calculated directly using:

t red  t d

(Eq.18)

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Worked Examples 2 and 3 in Section 8 illustrate the use of this procedure for a lipped C section under pure axial compression and in bending, respectively.

16

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

5

DESIGN OF COMPRESSION MEMBERS

5.1

Design issues

Light steel members are often required to carry axial compression loads, such as in the case of studs in a load-bearing wall. As with their hot-rolled counterparts, the failure of light steel compression members is likely to be governed by buckling rather than yielding of the cross section, resulting in a member resistance significantly lower than the squash load of the section. The design method for such a member is, therefore, focused on the calculation of its buckling resistance and is similar in many respects to the design of hot-rolled steel columns. However, the behaviour of light steel wall studs differs in a number of respects from that of hot-rolled columns and these differences must be accounted for in the design procedure.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Unlike columns, which act as independent members within a structural frame, light steel wall studs are used in conjunction with plasterboard, and often some form of sheathing board, to form a load-bearing panel. The presence of the boards will provide a certain degree of lateral restraint in the minor axis of the studs, which may be utilized when calculating the buckling resistance. However, any restraint must be verified by testing, using studs of a representative slenderness range and a similar build-up of boards to that used in practice. While flexural buckling usually governs the behaviour of hot-rolled steel columns, many light steel sections are also susceptible to torsional-flexural buckling. If torsional-flexural buckling occurs at a lower magnitude of load than flexural buckling, this mode of failure will naturally govern the resistance of the member. This is reflected in the Eurocode design rules, in which the elastic critical buckling load used for design is taken to be the smallest of the elastic critical buckling loads for flexural buckling, torsional buckling and torsional-flexural buckling. Finally, as noted in Sections 3 and 4, light steel sections are susceptible to local and distortional buckling, both of which can have an adverse impact on the compression resistance of a member. This should be accounted for by using the effective crosssectional area instead of the area of the gross cross section when calculating the compression resistance.

5.2

Eurocode calculation procedures

The design procedures for light steel compression members are given in §6.2 of BS EN 1993-1-3. However, due to the similarities with the design of hot-rolled columns, designers are referred to §6.3 of BS EN 1993-1-1 for much of the detail, including the buckling curves. The design buckling resistance of a member subjected to axial compression is given by:

N b,Rd 

Aeff f y  M1

(Eq.19)

where

χ

is the reduction factor for flexural buckling

Aeff

is the area of the effective cross section (Sections 3 and 4)

17

con

ED005

Technical me

Report:

Design

of

Light

Steel

Sections

to

Eurocode

3

...

fy

is the design strength

γM1

is the partial safety factor for buckling

Since fy and γM1 are known values (γM1 = 1.0 in the UK NA) and the calculation of Aeff has been dealt with earlier in this chapter, the procedure described below focuses on the calculation of χ. The reduction factor χ is used to quantify the reduction in resistance below the squash load of the section due to buckling. It may be obtained from BS EN 1993-1-1 using the appropriate buckling curve and the value of slenderness corresponding to the critical mode of failure. BS EN 1993-1-1 offers a choice of 5 buckling curves, but this is restricted to 3 curves (a, b and c) for light steel according to §6.2.2 of BS EN 1993-1-3. The appropriate choice of curve for various types of cross section is given in Table 6.3 of BS EN 1993-1-3. The relationship between χ and slenderness for buckling curves a, b and c is shown in Figure 5.1. The squash load of the section corresponds to χ = 1.0.

all

rights

reserved.

Use

of

this

document

is

subject

to

the

terms

and

Discuss

February copyright

2012 -

Figure 5.1

Buckling curves a, b and c

The slenderness is given by:



Aeff f y

(Eq.20)

N cr

23 is

Ncr is the elastic critical buckling load, which for flexural buckling is equal to the Euler load and is given by:

Created This

on material

N cr 

 2 EI Lcr

(Eq.21)

2

18

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

where:

E

is Young’s modulus for the material.

I

is the appropriate second moment of area (for the gross cross section).

Lcr

is the buckling length in the plane considered.

Alternatively, may be obtained from:



Aeff Agr

Lcr i

(Eq.22)

1

where i is the radius of gyration and λ1 is given by

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

 

E fy

(Eq.23)

If the buckling lengths differ between the major and minor axes, for example where a mid-height noggin provides restraint to the minor axis of a wall stud, values of should be obtained for both axes (since major axis flexural buckling might govern in this case). In cases where the critical mode of failure is either torsional buckling or torsional-flexural buckling, should be obtained from Equation 20 using the value of Ncr corresponding to the critical mode of failure (i.e. the elastic critical buckling load for torsional or torsional-flexural buckling). Equations for Ncr for both modes of failure are given in BS EN 1993-1-3. The reduction factor χ may be obtained directly from the buckling curves printed in BS EN 1993-1-1 or from the following equations:



1

   2 2



but ≤ 1.0

  0.5 1     0.2   2

(Eq.24)



(Eq.25)

α is the imperfection factor corresponding to the chosen buckling curve. Values for α are given in Table 6.1 of BS EN 1993-1-1. The design of members subject to axial compression is demonstrated in Worked Examples 5 and 8 in Section 8.

19

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

6

DESIGN OF MEMBERS IN BENDING

Several construction applications require light steel members to carry loads in bending. Examples include floor joists, roof purlins and wall studs (when subjected to wind loading). As with hot-rolled steel beams, it is essential that a distinction is made between members which are laterally restrained and those that are susceptible to lateral-torsional buckling, since the bending resistance will differ significantly between the two cases. While many joists, purlins and beams have some form of attachment providing restraint, designers should be aware of the potential lack of restraint during construction and the risk of load reversal. In addition to checking the member resistance, it may also be necessary to check serviceability limits such as deflections and dynamic response.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

6.1

Laterally restrained members

Beams and similar structural members may be considered to be laterally restrained when their compression flange is held in position to the extent that lateral-torsional buckling is prevented. This is often the case with light steel framing members due to the attachment of sheathing or plasterboard (for walls) or flooring products (e.g. wooden floor boards or concrete slabs). Purlins and cladding rails are also often laterally restrained due to the attachment of cladding. However, as the cladding is only attached to one side of the member, which may be subjected to sagging or hogging bending depending on the direction of the loading, purlins and cladding rails must usually be designed for lateral-torsional buckling for at least one load case. The design of laterally restrained light steel beams is similar to the design of the equivalent hot-rolled members. As such, the following issues need to be considered:  Bending resistance  Shear resistance  Local web failure  Deflections (at SLS). As noted previously, the key difference between light steel and hot-rolled steel is the susceptibility to local and distortional buckling, both of which are dealt with by the use of effective section properties. However, the use of light gauge material has other consequences, such as the increased risk of shear buckling and of crushing, crippling or buckling of the web under local transverse forces. A typical failure of a light steel member subjected to bending is shown in Figure 6.1. The member shown is a Z section purlin, but similar failure modes can be observed in lipped C sections of the type used in framing applications.

20

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Figure 6.1

Bending failure of a light steel purlin

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The resistance of a light steel cross section to bending is considered in §6.1.4 of BS EN 1993-1-3, which states that the moment resistance is given by:

M c,Rd 

Weff f yb

(Eq.26)

M0

Weff is the elastic modulus of the effective cross section, as discussed in Sections 3 and 4 (See Worked Examples 2 and 3 in Section 8). Equation 26 assumes that failure is due to yielding of the compression flange. Where yielding occurs first in the tension flange, plastic reserves in the tension zone may be utilized, as explained in §6.1.4.2 of BS EN 1993-1-3. In this case, the bending moment will be limited by the maximum compressive stress σcom,Ed reaching fyb / γM0. The resistance to shear is considered in §6.1.5 of BS EN 1993-1-3, which states that the design shear resistance is given by:

Vb,Rd 

hw tf bv sin 

(Eq.27)

M0

where:

fbv

is the shear strength allowing for buckling

hw

is the web height between the mid-lines of the flanges



is the slope of the web relative to the flanges.

It is apparent from the use of the term Vb,Rd that this is a buckling resistance rather than a cross section resistance. This is due to the susceptibility of some light steel sections to shear buckling. The risk of failure due to shear buckling is dependent on the slenderness of the web, so deep sections made from very light gauge steel are most at risk. Shear buckling is accounted for in the design procedure by the use of

21

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

fbv, which is a function of the basic yield strength fyb and the relative web slenderness w. Values of fbv may be obtained from Table 6.1 of BS EN 1993-1-3. Local transverse forces are considered in §6.1.7 of BS EN 1993-1-3. Several equations are presented for the local resistance of the web Rw,Rd, taking into account the location of the applied load (e.g. whether close to the end of the member), the number of webs in the cross section and the inclusion or absence of stiffeners.

6.2

Lateral-torsional buckling

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Ideally, light steel members should be incorporated into systems that provide lateral and torsional restraint and, therefore, prevent lateral-torsional buckling. However, there are occasions when this is not possible and the member has to be designed as an unrestrained beam. This situation is considered in §6.2.4 of BS EN 1993-1-3. Alternatively, for purlins and cladding rails, the cladding may provide full lateral restraint when the flange to which it is attached is in compression, but only partial restraint when it is in tension. Design rules for this situation are provided in §10.1.4 of BS EN 1993-1-3. The behaviour of an unrestrained beam loaded in bending about its major axis is analogous to the behaviour of a column under axial load and the buckling curves depicted in Figure 5.1 are equally applicable to lateral-torsional buckling. The essential feature of this mode of failure is that the compression flange becomes unstable and, as it is not restrained, attempts to buckle laterally. However, since it is attached to the tension flange via the web of the beam, it cannot move freely and must pull the tension flange over as it deforms. The tension flange resists, resulting in the classical combination of lateral and torsional deformation, commonly known as lateral-torsional buckling. This type of failure only occurs when a member is bent about its major axis; members bent about their minor axis will always fail by minor axis bending and never by lateral-torsional buckling. The impact of lateral-torsional buckling on the bending resistance of a beam is dependent on a number of factors. Principal among these are the geometry of the cross section and the slenderness of the member. In terms of geometry, sections that have a low minor axis flexural rigidity relative to the flexural rigidity of the major axis are most likely to fail by lateral-torsional buckling. The ease with which a section is able to twist and warp is also important. Light steel sections (e.g. C sections and Z sections) tend to be poor on both counts. By contrast, square hollow sections cannot suffer lateral-torsional buckling. The relationship between slenderness and bending resistance can be represented by a buckling curve of the type shown in Figure 5.1. For very stocky members, the bending resistance will be limited by the resistance of the cross section (given by Equation 26), but as the slenderness increases, the influence of lateral-torsional buckling also increases, resulting in a significant reduction in the bending resistance. As with the case of axial compression, it is convenient to quantify this reduction in terms of a reduction factor, expressed as a proportion of the section capacity. For unrestrained beams, the Eurocode symbol for this reduction factor is χLT. Due to the similarities between the design of light steel unrestrained beams and the equivalent hot-rolled steel members, designers are referred to §6.3 of BS EN 1993-1-1 for the detailed design procedures and the buckling curves. However, important guidance is given in §6.2.4 of BS EN 1993-1-3 regarding the choice of design method and buckling curve.

22

Steelbiz

ED005

Technical me

Report:

Design

of

Light

Steel

Sections

to

Eurocode

3

...

the

Discuss

M b,Rd 

 LTWeff,y f y  M1

(Eq.28)

where

χLT

is the reduction factor for lateral-torsional buckling

Weff,y

is the elastic modulus of the effective cross section (major axis)

fy

is the design strength

γM1

is the partial safety factor for buckling (= 1.0 according to UK NA).

BS EN 1993-1-1 gives two alternative methods for the calculation of χLT. However, only the ‘General case’ given in §6.3.2.2 is permitted for light steel members. This method resembles that used for column buckling and uses the same buckling curves. As such, the equations resemble those given in Section 5.2 with the addition of the subscript ‘LT’. The reduction factor χLT is given by:

 LT 

1 2

 LT   LT  LT

2

but ≤ 1.0



 LT  0.5 1   LT LT  0.2  LT 2



(Eq.30)

αLT is the imperfection factor corresponding to the chosen buckling curve and is

given by Table 6.3 of BS EN 1993-1-1. However, since §6.2.4 of BS EN 1993-1-3 states that buckling curve b should always be used for light steel in bending, αLT will always have the value 0.34. Lateral torsional buckling design checks are demonstrated in Worked Examples 5 and 7 in Section 8.

6.3

Serviceability

In addition to checking the resistance of the member to the applied loading and associated bending moments and shear forces, designers should also check that the member is adequate at the serviceability limit state (SLS). For normal building applications, this involves checking the imposed load deflections against specified limits. Occasionally, the dynamic response of light steel floors will also need to be checked. Guidance for light steel floors is given in P301[12] and specialist guidance for more general consideration of dynamic response is given in P354[11]. For the purpose of calculating deflections, the second moment of area should be calculated using the following equation:

I fic  I gr 

 gr I  I  eff   gr

(Eq. 31)

on material Created This

(Eq.29)

and

23 is

February copyright

2012 -

all

rights

reserved.

Use

of

this

document

is

subject

to

the

terms

and

conditions

of

The design buckling resistance of a member subjected to bending is given by:

23

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

where:

Igr

is the second moment of area of the gross cross section.

σgr

is the maximum compressive bending stress at SLS based on gross cross section properties.

I(σ)eff

is the second moment of area of the effective cross section calculated for a maximum stress σ ≥ σgr.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Serviceability limit state checks are demonstrated in Worked Examples 5, 6 and 8 in Section 8.

24

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

7

FRAME DESIGN

Verifying that each member within a steel frame is capable of withstanding the applied loads is an essential part of the structural design process, but it is not sufficient by itself. Other design considerations such as frame stability and structural robustness must also be considered.

7.1

Frame stability

The stability of a building as a whole must be checked. While the methods used to provide stability differ between hot-rolled and cold-formed steel, the principles remain the same:  A suitable load path must be provided to safely transmit horizontal forces to the foundations.  The system used to provide stability must be sufficiently stiff to avoid excessive lateral deflections.  Some account should be taken of the instability arising from frame imperfections.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

 Where appropriate, account should be taken of second order effects. The four issues outlined above are dealt with by the design rules in BS EN 1993-1-1. These rules cover hot-rolled and cold-formed steel structures and no distinction is made between the two forms of construction. The stability of light steel framing is usually achieved through one of the following methods:  Integral bracing  X bracing  Diaphragm action. Integral bracing consists of C section members placed diagonally between the vertical wall studs and within the depth of the studs. The use of a C section means that the integral bracing members are capable of carrying tension and compression forces. However, careful detailing and connection design are important. X bracing consists of diagonal crossed flat straps attached to the face of the vertical studs. Unlike integral bracing, the flats usually extend across several studs and should be connected to every stud that they cross. Each individual bracing element is only capable of acting in tension (hence the need for the X arrangement). As an alternative to steel bracing members, the frame designer may choose to rely on the racking resistance of the wall itself. In this case, stability is provided by diaphragm action in the plane of the wall due to the attached board or cladding. Board options include:  Plywood  Cement Particle Board  OSB  Plasterboard.

25

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

The racking resistance of a particular board and frame combination should be determined by testing.

7.2

Structural robustness

The term ‘robustness’ when used in the context of building design relates to the ability of the structure to withstand accidental actions without the spread of damage or disproportionate collapse. In this sense, robustness is synonymous with structural integrity. The essential principles of robustness, which apply across all forms of construction and materials, are summarised below:  Robustness relates to the ability of a structure to withstand events such as explosions, impact or the consequences of human error.  The structure does not have to be serviceable after the event. Large deformations and plasticity are permitted. It is expected that the structure will need to be repaired before it can be re-occupied. In some cases, it may need to be demolished. The objectives are:

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

 To restrict the spread of localised damage and to prevent collapse of the structure disproportionate to the original cause.  To ensure the safety of the structure while building occupants make their escape and the emergency services are in attendance. Structural robustness and design to avoid disproportionate collapse is a requirement of Building Regulations. The required level of robustness is dependant on the building class, which is related to the size, type and use of the building. The same building classification system is given in Approved Document A, BS 5950-5 and BS EN 1991-1-7. The requirement to design and construct buildings to have robustness and avoid disproportionate collapse under accidental design situations is given in BS EN 1990. Details of how the requirement should be met are given in BS EN 1991-1-7. The guidance given in BS EN 1991-1-7 is material independent. The only guidance that is specific for light steel framing is found in the UK National Annex, where it states in §NA.3.1 that: In the case of lightweight building structures (e.g. those whose primary structure is timber or cold formed light gauge steel), the values for minimum horizontal tie forces in expression A.1 and A.2 should be taken as 15 kN and 7.5 kN, respectively. This guidance is comparable to that given in BS 5950-5, which was based on the guidance given in SCI publication P301[12]. Light steel multi-storey structures are generally structurally robust because of their construction using a large number of regularly distributed structural elements, with a high degree of connectivity. In most applications, the provision of continuous ties between the components is straightforward because of the multiple inter connections. However, the connection resistances must be checked to ensure the appropriate requirements are satisfied.

26

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

7.3

Frame anchorage

Light steel framed buildings must be adequately held in position by a suitable anchorage system to prevent sliding, overturning or lifting off the foundations. Two types of anchorage are generally employed:  Holding down bolts connecting the stud and track to the ground slab.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

 Steel straps connecting wall studs to the foundations.

27

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

8

WORKED EXAMPLES

There are 8 worked examples which show the design of light steel sections to BS EN 1993-1-3. The examples include the calculation of effective section properties and member capacities. The examples are for cross-sections and member designs that may typically be used in light steel framing.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The examples are: 8.1

Gross section properties for a cold formed lipped C section

8.2

Effective section properties for a lipped C section in compression

8.3

Effective properties for a lipped C section in major axis bending

8.4

Effective properties for a lipped C section in minor axis bending

8.5

Design of a wall stud in a light steel frame building

8.6

Design of a floor joist (restrained) in a light steel frame building

8.7

Design of a floor joist (unrestrained) in a light steel frame building

8.8

Design of a lattice floor truss in a light steel frame building

The marginal references to clauses in various Eurocode Parts are given in the form 3-1-3/ 5.1, meaning clause 5.1 of BS EN 1993-1-3.

28

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 1

Subject

Gross properties of a lipped C section

Client

SCI

CALCULATION SHEET

8.1

Sheet

1

Date

Oct 2010

Checked by

SN

Date

Jan 2011

Section dimensions and material properties = 200 mm = b2 = 65 mm = 25 mm = 3 mm = 2 mm = 1.96 mm b p1

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

b1

c

r

h

r

cp

hp

t

t

b2

Mid-line dimensions Web depth Flange width

hp bpl

= h  t nom  200  2  198 mm = bp2  b1  t nom  65  2  63 mm

Stiffener depth

cp

= c  t nom 2  25  2 2  24 mm

Mean radius Corner

rm gr

= r  t 2  3  1.96 2  3.98 mm = rm tan  / 2  sin  / 2   3.98tan 45  sin 45  1.17 mm

29

Rev

AW

This worked example presents design calculations for the gross section properties of a cold-formed lipped C section.

h b1 c r tnom t

9

Made by

Gross section properties for a cold-formed lipped C section

Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness

of

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

2

of

9

Rev

Flat width dimensions Web

bp,h  h p  2 g r  198  2 1.17  195.7 mm

Flange

bp,b  b p  2 g r  63  2 1.17  60.66 mm

Stiffener

bp,c  c p  g r  24  1.17  22.83 mm

3-1-3/ 5.1

Geometry checks Checks on the geometry of the cross section to ensure that the dimensions are within the scope of 3-1-3/ . bt  65 1.96  33.16  60 OK ct  25 1.96  12.76  50 OK ht  200 1.96  102.0  500 OK Check on the dimensions of the stiffener. c b1  25 65  0.38 0.2  0.38  0.6

OK

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Check to see whether the rounding of the corners may be neglected. r t  3 1.96  1.53  5 OK r bp, h  3 195.7  0.02  0.10 OK r bp, b  3 60.66  0.05  0.10

OK

r bp, c  3 22.83  0.13  0.10

Outside limit

Therefore, the influence of rounded corners should be considered in the calculation of section properties. The influence of rounded corners on section properties may be taken into account by reducing the properties calculated for an otherwise similar cross section with sharp corners, using the following approximations. Ag Ig Iw

 Ag, sharp 1   

 I g, sharp 1  2 

 I w, sharp 1  4 

where n



 0.43



 r j 90j j 1 m

 bp,i i 1

n



 0.43



 r j 90j j 1 m

 bp,i

3-1-3/ 5.2

90 90 = 0.43  0.014 195.7  2  60.66  2  22.83 4  3

i 1

30

3-1-3/ 5.1(3)

3-1-3/ 5.1(4)

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

3

of

9

Rev

Gross section properties (sharp corners) BS 3-1-3/ Annex C has been used to calculate the gross section properties that 3-1-3/ Annex C may be required for structural calculations. The section is divided into flat elements. Each element is defined by the coordinates of its ends points based on the mid-line properties, as shown below. The end points of each element are called nodes. 3

4

5

z 0

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

2

y

Node j 0 1 2 3 4 5

Part i

1 2 3 4 5

1

y coordinate yj mm 63 63 0 0 63 63 Start Node No. y z coord coord yi-1 zi-1 mm mm 0 63 24 1 63 0 2 0 0 3 0 198 4 63 198

z coordinate zj mm 24 0 0 198 198 174 End Node No. y z Thick- Length Area coord coord ness yi zi ti dAi mm mm mm mm mm2 1 63 0 1.96 24 47.0 2 0 0 1.96 63 123.5 3 0 198 1.96 198 388.1 4 63 198 1.96 63 123.5 5 63 174 1.96 24 47.0

Area Agr

= Σ dAi = 729.1 mm2

31

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

4

of

9

Rev

Centroid The z coordinate of the centroid (zgc) defines the position of the y axis of the 3-1-3/ Annex C gross section. The y axis is central to the section as the section is symmetrical about the y axis. Hence, zb1 will be equal to zb2. The calculation of zgc is shown here for completeness. The y coordinate of the centroid defines the position of the major z axis of the gross section. The terms are defined on the figure below. ygc = yb1

z

ylip

zb1 y

y

zb2 = zgc

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

z

Determine the z coordinate of the centroid.

Part i 1 2 3 4 5 Total

3-1-3/ Annex C

(zi + zi-1).dAi / 2 mm3 564.0 0.0 38420 24450 8742 72180

First moment of area for the y axis is the total from above, = 72180 mm3. Sy0 The z coordinate of the centroid is given by: zgc = Sy0 / Agr zgc = 72180 / 729.1 zgc = 99.0 mm zgc

= zb1 = zb2 = 99.0 mm

Determine the y coordinate of the centroid.

Part i 1 2 3 4 5 Total

(yi + yi-1).dAi / 2 mm3 2961 3890 0.0 3890 2961 13700

32

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

5

of

9

Rev

First moment of area for the z axis is the total from above, Sz0 = 13700 mm3. The y coordinate of the centroid is given by: ygc = Sy0 / Agr ygc = 13700 / 729.1 ygc = 18.8 mm ygc

= yb1 = 18.8 mm

ylip ylip ylip

= bp1 - yb1 = 63.0 – 18.8 = 44.2 mm

Second Moment of Area

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Second moment of area about major axis.

Part i 1 2 3 4 5 Total

(zi2 +zi-12 +zi.zi-1).dAi / 3 mm4 9024 0.0 5072000 4842000 1628000 11550000

3-1-3/ Annex C

The second moment of area about the y axis with respect to origin (bottom flange) is the total from above, Iy0 = 11550000 mm4 The second moment of area about the y axis with respect to the centroid is given by: Iy = Iy0 – Agr zgc2 Iy = 11550000 – 729.1  99.02 Iy = 4404000 mm4 Second moment of area about minor axis.

Part i 1 2 3 4 5 Total

(yi2 +yi-12 +yi. yi-1).dAi / 3 mm4 186500 163400 0.0 163400 186500 699800

3-1-3/ Annex C

The second moment of area about the z axis with respect to origin (web) is the total from above, Iz0 = 699800 mm4 The second moment of area about the z axis with respect to centroid is given by: Iz = Iz0 – Agr ygc2 Iz = 699800 – 729.1 × 18.82 Iz = 442100 mm4

33

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

6

of

9

Rev

Radius of Gyration Radius of gyration for major axis iy = (Iy / Agr)0.5 iy = (4404000 / 729.1)0.5 iy = 77.7 mm Radius of gyration for minor axis iz = (Iz / Agr)0.5 iz = (442100 / 729.1)0.5 iz = 24.6 mm

Elastic Section Modulus Elastic section modulus for major axis Wy = Iy / Max{ zb1 zb2} Wy = 4404000 / 99.0 Wy = 44480 mm3

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Elastic section modulus for minor axis Wz = Iz / Max{ ygc ylip} Wz = 442100 / 44.2 Wz = 10000 mm3

Shear Centre The position of the shear centre is calculated with respect to the centroid. The section is symmetrical about the y axis. Hence, the shear centre will be a point on the major axis. The terms are defined on the figure below.

3-1-3/ Annex C

z

Shear Centre y

y ysc zsc

y0

z

Part i 1 2 3 4 5 Total

(2.yi-1.zi-1 + 2.yi.zi + yi-1.zi + yi.zi-1) dAi / 6 mm4 35530 0.0 0.0 770300 550700 1357000

34

3-1-3/ Annex C

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

7

of

9

Rev

Product moment of area with respect to origin is the total from above, Iyz0 = 1357000 mm4. Product moment of area with respect to centroid is given by: Iyz = Iyz0 – (Sy0.Sz0 / Agr) Iyz = 1357000 – (72180 × 13700 / 729.1) Iyz = 716.9 mm4

3-1-3/ Annex C

Sectorial Coordinates Part ω0 i yi-1.zi  yi.zi-1 i 0 na 1 1512 2 0.0 3 0.0 4 12470 5 1512

ωi ωi-1 + ω0i 0 1512 1512 1512 13980 15490

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Sectorial Constants Iyω0 i

1 2 3 4 5 Total

2.24 ×106 5.88 ×106 0.00 3.82 ×107 4.37 ×107 9.00 ×107

Izω0

Iωω0

2.84 ×105 0.00 5.81 ×107 1.89 ×108 1.29 ×108 3.76 ×108

3.58 ×107 2.82 ×108 8.87 ×108 9.02 ×109 1.02 ×1010 2.04 ×1010

Iω Iω

= Σ(ωi-1 + ωi) dAi / 2 = –2458000 mm3

Iyω Iyω Iyω

= Iyω0 – Sz0 × Iω / Agr = –9.00×107 – 13700 × (–2458000) / 729.1 = –43710000 mm4

Izω Izω Izω

= Izω0 – Sy0 × Iω / Agr = –3.76×108 – 72180 × (–2458000) / 729.1 = –132700000 mm4

Iωω Iωω Iωω

= Iωω0 – Iω2 / Agr = 2.04×1010 – (–2458947)2 / 729.1 = 1.211×1010 mm4

(ωi-1 + ωi)dAi / 2 3.553 ×104 1.867 ×105 5.868 ×105 9.566 ×105 6.925 ×105 2.458 ×106

The y coordinate of the shear centre is given by: ysc = (Izω Iz – Iyω Iyz) / (Iy Iz – Iyz2) ysc = (–132700000  442100 – (–43710000)  716.9) / (4404000  442100 – 716.92) ysc = –30.10 mm The distance of the shear centre from the centroid is given by: y0 = –ysc + ygc y0 = 30.10 + 18.8 y0 = 48.9 mm 35

3-1-3/ Annex C

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

8

of

9

Rev

The z coordinate of the shear centre is at mid height. zsc = 99.0 mm

Warping Constant The warping constant is given by:

= Iωω + zsc × Iyω – ysc × Izω = 1.211×1010 + 99.0 × –43710000 – –30.10 × –132700000 = 3.797 ×109 mm6

Iw Iw Iw

3-1-3/ Annex C

Torsion Constant

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Part i 1 2 3 4 5 Total

dAi.(ti)2 / 3 mm4 60.2 158.1 497.0 158.1 60.2 933.6

3-1-3/ Annex C

The torsion constant is given by the total from the table above. It = Σ dAi.(ti)2 / 3 It = 933.6 mm4

Summary of Gross Properties for a section with sharp corners Area centroid from flange centroid from web centroid from lip Second moment of area about major axis Second moment of area about minor axis Radius of gyration for major axis Radius of gyration for minor axis Elastic section modulus for major axis Elastic section modulus for major axis Shear centre from flange Shear centre y from web Shear centre from centroid Warping constant Torsion constant

Agr zgc ygc ylip Iy Iz iy iz Wy Wz zsc ysc y0 Iw It

= 729.1 mm2 = 99.0 mm = 18.8 mm = 44.2 mm = 4404000 mm4 = 442100 mm4 = 77.7 mm = 24.6 mm = 44480 mm3 = 10000 mm3 = 99.0 mm = –30.10 mm = 48.9 mm = 3.797 ×109 mm6 = 933.6 mm4

Gross section properties (rounded corners) The influence of rounded corners on section properties is taken into account by reducing the properties calculated for an otherwise similar cross section with sharp corners, using the following approximations: Ag Ig Iw

 Ag, sharp 1   

 I g, sharp 1  2 

 I w, sharp 1  4 

36

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 1: Gross properties of a lipped C section

Sheet

For properties such as the position of the neutral axes and the position of the shear centre, the values for the section with rounded corners are taken as equal to those calculated for the section with sharp corners. The section properties that are reduced to take account of rounded corners are calculated below. Ag Iy

 Ag, sharp 1     729.1(1  0.014)  718.9 mm2

 I y,sharp 1  2   4404000(1  2  0.014)  4281000 mm4

Iz

 I z,sharp 1  2   442100(1  2  0.014)  429700 mm4

iy

 ( I y / Ag ) 0.5  (4281000 / 718.9) 0.5  77.2 mm

iz

 ( I z / Ag ) 0.5  (429700 / 718.9) 0.5  24.5 mm

Wy

 I y / zgc  4281000 / 99.0  43240 mm3

Wz

 I z / ylip  429700 / 44.2  9724 mm3

Iw

 I w, sharp 1  4   3.797  109 (1  4  0.014)  3.584  109 mm6

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Summary of Gross Properties for a section with rounded corners Area centroid from flange centroid from web centroid from lip Second moment of area about major axis Second moment of area about minor axis Radius of gyration for major axis Radius of gyration for minor axis Elastic section modulus for major axis Elastic section modulus for major axis Shear centre from flange Shear centre y from web Shear centre from centroid Warping constant Torsion constant

Agr zgc ygc ylip Iy Iz iy iz Wy Wz zsc ysc y0 Iw It

= 718.9 mm2 = 99.0 mm = 18.8 mm = 44.2 mm = 4281000 mm4 = 429700 mm4 = 77.2 mm = 24.5 mm = 43240 mm3 = 9724 mm3 = 99.0 mm = –30.10 mm = 48.9 mm = 3.584 ×109 mm6 = 933.6 mm4

37

9

of

9

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

CALCULATION SHEET

8.2

Job No.

BCS 320

Title

Example 2

Subject

Effective properties of a lipped C section in compression

Client

SCI

Sheet

1

of

5

Rev

Made by

AW

Date

Oct 2010

Checked by

SN

Date

Jan 2011

Effective section properties for a lipped C section in compression

This worked example presents design calculations for the effective section properties of a cold-formed lipped C section in compression. The chosen section is identical to that considered in Example 1, so some of the calculations and checks relating to the gross cross section have been omitted. It was shown in Example 1 that the influence of rounded corners must be included for this cross-section. The section properties are initially calculated on the basis that the section has sharp corners. The influence of rounded corners is taken into account by applying reductions to the sharp corner properties.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Section dimensions and material properties h b1 c r tnom t fy E v

M0

= 200 mm = b2 = 65mm = 25 mm = 3 mm = 2 mm = 1.96 mm = 350 N/mm2 = 210000 N/mm2 = 0.3 = 1.00

Web depth Flange width

hp bp1

= h  t nom  200  2  198 mm = b p 2  b1  t nom  65  2  63 mm

Stiffener depth

cp

= c  t nom 2  25  2 2  24 mm

Section depth Flange widt Stiffener depth Corner radius Nominal thickness Core thickness Design strength Young’s modulus Poisson’s ratio Partial safety factor

Mid-line dimensions

Effective section properties (sharp corners) 3-1-3/ 5.5.3.2

Effective properties of the flanges and lips Step 1

3-1-3/ 5.5.2

Effective width of the flanges

For a stress ratio   1 (uniform compression), k σ  4



=

235 f yb

38

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 2: Effective properties of a lipped C section in compression bp1 t







beff

  bp1  0.987  63  62.2 mm

be1

 be2  0.5beff  0.5  62.2  31.1 mm

28.4  k σ

p, b  0.0553    p, b

2



Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

kσ  0.5  0.83 3 bp,c bp1  0.35

bp,c bp1  24 63  0.38 so

k σ  0.5  0.833 0.38  0.35  0.582





cp t 28.4  kσ



p, c  0.188 p, c

but   1 so

2

5

Rev

3-1-3/ 5.5.3.2(5a)

if 0.35  bp,c bp1  0.6 :



of

0.691  0.055  3  1  0.986  1.0 0.6912

Effective width of the edge stiffener The buckling factor is given by: if bp,c bp1  0.35 : kσ  0.5

p, c

2

3-1-5/ 4.4

63 1.96  0.691 28.4  235 350  4

p, b



Sheet

2

2

24 1.96  0.690 28.4  235 350  0.58



3-1-5/ 4.4

0.690  0.188  1.05 0.690 2

 1 3-1-3/ 5.5.3.2(5a) 5.5.3.2(6)

The effective width is given by: ceff   cp  1  24  24 mm The effective area of the edge stiffener is: As  t be2  ceff   1.96  31.1  24   108.0 mm 2 Step 2

The elastic critical buckling stress for the edge stiffener is given by: 2 K E Is  cr,s  As where K is the spring stiffness per unit length and Is is the effective second moment of area of the stiffener.

3-1-3/ 5.5.3.2(7)

3-1-3/ 5.5.3.1(5)

1 E t3  2 2 3 4(1   ) b1 hp  b1  0,5 b1 b2 hp k f

K



b1

 bp1 

b2

 b1  54.22 mm (for a section with equal flanges)

kf



K

 0.421 N mm 2

be2t be2 2 31.1  1.96  31.1 2  63   54.22 mm (be2  ceff )t (31.1  24)  1.96

As2 108   1.0 for a member in axial compression As1 108

39

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 2: Effective properties of a lipped C section in compression 2

Is

  b t3 c 3 t ceff 2  e2  eff  be2 t    ceff 12 12  2be2  ceff 

Sheet

c   ceff 2 t  eff     2  2be2  ceff   

3

of

5

Rev

2

 6101 mm 4 As the section has equal flanges, the spring stiffness K and second moment of area Is are applicable to both edge stiffeners. Had the section been asymmetric, it would have been necessary to repeat the process shown above for the upper and lower edge stiffeners.

 cr,s



d



2  0.421  210000  6101  430.1 N mm 2 108.0 3-1-3/ 5.5.3.1(7)

f y  cr,s  350 430  0.902

Since 0.65  d  1.38 ,  d  1.47  0.723 d d  1.47  0.723  0.902  0.818 Step 3

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

3-1-3/ permits the optional iteration to refine the value of χd. This iteration has not been undertaken for this example, so the initial value must be used. t red

 t d  1.96  0.818  1.60 mm

3-1-3/ 5.5.3.2(12)

Effective properties of the web For uniform compression, the stress ratio   1 and the buckling factor k σ  4 (for an internal compression element). hp t

198 1.96  2.172 28.4  235 350  4

p, h







heff

 hp  0.414  198  82.0 mm

he1

 he2  0.5heff  0.5  82.0  41.0 mm



28.4  kσ

p, h  0.0553    p, h

2



2.171  0.055  3  1  0.414 2.1712

40

3-1-5/ 4.4

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 2: Effective properties of a lipped C section in compression

Sheet

Effective properties of the whole cross section The theoretical effective section is shown below. The effective properties of the whole section are calculated using this theoretical effective section. b e1

b e2

c eff h e1

t red t y

y

t red

h e2

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

c eff

b e1

b e2

Aeff Aeff

 t 2be1  he1  he2  2be2  ceff  d   1.962  31.1  41.0  41.0  231.1  24.0 0.816

Aeff

 459.0 mm 2

Since the section is symmetrical and subjected to pure axial compression, the position of the centroidal axis remains unchanged from that of the gross cross section, i.e. 99.0 mm from either flange. Other effective properties for the section subject to axial load are given below but the calculation of these properties is not shown. Summary of Effective Properties for a section with sharp corners Effective area Effective centroid from flange Effective centroid from web Effective centroid from lip Second moment of area about major axis Second moment of area about minor axis

Aeff zgc ygc ylip Iy Iz

41

= 459.0 mm2 = 99.0 mm = 25.0 mm = 38.0 mm = 3780000 mm4 = 311800 mm4

4

of

5

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 2: Effective properties of a lipped C section in compression

Sheet

Effective section properties (rounded corners) Influence of rounded corners The influence of rounded corners must be included for this section. For properties such as the position of the neutral axes, the values for the section with rounded corners are taken as equal to those calculated for the section with sharp corners. The section properties that are reduced to take account of rounded corners are calculated below. Aeff Iy Iz

 Aeff,sharp 1     459.0(1  0.014)  452.6 mm2

 I y,sharp 1  2   3780000(1  2  0.014)  3675000 mm4  I z, sharp 1  2   311800(1  2  0.014)  303100 mm4

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Summary of Effective Properties for a section with rounded corners Effective area Effective centroid from flange Effective centroid from web Effective centroid from lip Second moment of area about major axis Second moment of area about minor axis

Aeff zgc ygc ylip Iy Iz

42

= 452.6 mm2 = 99.0 mm = 25.0 mm = 38.0 mm = 3675000 mm4 = 303100 mm4

5

of

5

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 3

Subject

Client

CALCULATION SHEET

8.3

Sheet

1

of

5

Effective properties of a lipped C section in bending about the major axis Made by MDH Date Oct 2010 SCI Checked by SN Date Jan 2011

Effective properties for a lipped C section in major axis bending

This worked example presents design calculations for effective section properties of a cold-formed lipped C section in bending about the major axis. It was shown in Example 1 that the influence of rounded corners must be included for this cross-section. The section properties are initially calculated on the basis that the section has sharp corners. The influence of rounded corners is taken into account by apply reductions to the sharp corner properties.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Section dimensions and material properties Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness Design strength Young’s modulus Poisson’s ratio

Rev

h b1 c r tnom t fy E v

= 200 mm = b2 = 65mm = 25 mm = 3 mm = 2 mm = 1.96 mm = 350 N/mm2 = 210000 N/mm2 = 0.3

Web depth Flange width

hp bp1

= h  t nom  200  2  198 mm = b p 2  b1  t nom  65  2  63 mm

Stiffener depth

cp

= c  t nom 2  25  2 2  24 mm

Mid-line dimensions

Effective section properties (sharp corners) The effective cross section for a lipped C section in bending is shown below. Note the ineffective portions of the flange and web and the reduced thickness t d of the stiffener and adjacent section of flange.

43

3-1-3/ 5.5

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 3: Effective properties of a lipped C section in bending about the major axis

Sheet

2

of

5

Rev

The effective properties of the flange and web are determined separately as shown below, after which the effective properties of the whole cross section may be calculated. 3-1-3/ 5.5.3.2

Effective properties of the compression flange and lip

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Step 1 3-1-3/ 5.5.2 and 3-1-5/ 4.4

Effective width of the compression flange

For a stress ratio   1 (uniform compression), kσ  4



 235 f y

p, b







beff

  bp1  0.986  63  62.1 mm

be1

 be2  0.5beff  0.5  62.1  31.1 mm

bp1 t 28.4  k σ



63 1.96  0.691 28.4  235 350  4

p, b  0.0553    p, b 2



0.691  0.055  3  1  0.986  1.0 0.6912

3-1-3/ 5.5.3.2(5a)

Effective width of the edge stiffener The buckling factor is given by: if bp,c bp1  0.35 : kσ  0.5 kσ  0.5  0.83 3 bp,c bp1  0.35

2

if 0.35  bp,c bp1  0.6 :

bp,c bp1  24 63  0.38 so k σ  0.5  0.83

p, c







cp t 28.4  kσ



p, c  0.188 p, c 2

but   1 so

3-1-5/ 4.4

3

0.38  0.352

24 1.96  0.690 28.4  235 350  0.58



 0.582 3-1-5/ 4.4

3-1-5/ 4.4

0.690  0.188  1.05 0.690 2

 1

44

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 3: Effective properties of a lipped C section in bending about the major axis

Sheet

3

of

5

Rev

3-1-3/ 5.5.3.2(5a) 5.5.3.2(6)

The effective width is given by: ceff   cp  1  24  24 mm The effective area of the edge stiffener: As  t be2  ceff   1.96  31.1  24   108.0 mm 2 Step 2 The elastic critical buckling stress for the edge stiffener is given by: 2 K E Is  cr,s  As where K is the spring stiffness per unit length and Is is the effective second moment of area of the stiffener. 

K

3-1-3/ 5.5.3.1(5)

E t3 1  2 2 3 4(1   ) b1 hp  b1  0.5 b1 b2 hp kf

be2t be2 2 31.1  1.96  31.1 2  63   54.22 mm (be2  ceff )t (31.1  24)  1.96  0 (for major axis bending)  0.586 N mm  bp1 

b1 kf K Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

3-1-3/ 5.5.3.2(7)

2

3 2   b t3 c t ceff  e2  eff  be2 t    ceff 12 12  2be2  ceff  

Is

2 c  ceff t  eff   2be2  ceff    2

2

 6101 mm 4

 cr,s



d



2  0.586  210000  6101  507.4 N mm 2 108.0

3-1-3/ 5.5.3.2(7)

3-1-3/ 5.5.3.1(7)

f y  cr,s  350 507.4  0.831

Since 0.65  d  1.38 ,  d  1.47  0.723 d  d  1.47  0.723  0.831  0.869

Step 3 3-1-3/5.5.3.2(3) permits the optional iteration to refine the value of  d . This iteration has not been undertaken for this example, so the initial value of  d and the associated effective properties must be used. The next and, therefore, final step for the flange is the calculation of the reduced thickness for the stiffener. t red  t d  1.96  0.870  1.70 mm

3-1-3/ 5.5.3.2(12)

Effective properties of the web The position of the neutral axis with regard to the flange in compression: 2 2 cp hp  cp 2  bp2 hp  hp 2  ceff  d 2 hc  hc  101.1 mm cp  bp2  hp  be1  be2  ceff  d





45

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 3: Effective properties of a lipped C section in bending about the major axis

Sheet

4

of

5

The stress ratio is given by: hc  hp 101.1  198     0.958 hc 101.1 Referring to EN 1993-1-5, the buckling factor for the web is given by: kσ  7.81  6.29  9.78 2 k σ  22.81 hp t

198 1.96  0.909 28.4  235 350  22.81

p, h







heff he1 he2

  hc  0.965  101.1  97.5 mm  0.4heff  0.4  97.5  39.0 mm  0.6heff  0.6  97.5  58.5 mm



28.4  k σ

p, h  0.0553    p, h

2



0.909  0.055  3  0.958  0.964 0.909 2

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The effective width of the web is divided into two portions as follows: h1  he1  39.0 mm h2  hp  hc  he2   198  101.1  58.5  155.4 mm

Effective properties of the whole cross section Aeff

 t[cp  bp 2  h1  h2  be1  (be 2  ceff )  d ]

Aeff

 1.96  24  63  39  155.4  31.08  31.08  24   0.870

Aeff

 706.3 mm 2

The position of the neutral axis with regard to the compression flange is given by: zc

=

 







2

2

t cp hp  cp 2  bp2 hp  h2 hp  h2 2  h1 2  ceff  d 2

 = 101.7 mm

Aeff The position of the neutral axis with regard to the tension flange is given by: zt =  h p  zc  198  101.7  96.3mm

I eff, y

3 cp 3t be1t 3 be2 (  d t ) 3 ceff 3 (  d t ) h13t h2 3t bp2t         12 12 12 12 12 12 12

cp t ( z t  cp 2) 2  bp 2tz t 2  h2t ( z t  h2 2) 2  h1t ( zc  h1 2) 2  2

2

be1t zc  be2 (  d t ) zc  ceff (  d t )( zc  ceff 2) 2

I eff, y

 4235000 mm 4

Weff, y,c  Weff, y, t 

I eff, y zc I eff, y zt



4235000  41640 mm3 101.7



4235000  43980 mm3 96.3

46

3-1-5/ 4.4

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 3: Effective properties of a lipped C section in bending about the major axis

Sheet

Summary of Effective Properties for a section with sharp corners Effective area Effective centroid from compression flange Effective centroid from tension flange Second moment of area about major axis Section modulus about major axis compression flange Section modulus about major axis tension flange

Aeff zgc zgc Ieff,y Weff,y,c Weff,y,t

= 706.3 mm2 = 101.7 mm = 96.3 mm = 4235000 mm4 = 41640 mm3 = 43980 mm3

Effective section properties (rounded corners) Influence of rounded corners The influence of rounded corners must be included for this section. For properties such as the position of the neutral axes, the values for the section with rounded corners are taken as equal to those calculated for the section with sharp corners. The section properties that are reduced to take account of rounded corners are calculated below.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Aeff Iy

 Aeff,sharp 1     706.3(1  0.014)  696.4 mm2

 I y,sharp 1  2   4235000(1  2  0.014)  4116000 mm4

Weff, y,c  Weff, y, t 

I eff, y zc I eff, y zt



4116000  40480 mm3 101.7



4116000  42740 mm3 96.3

Summary of Effective Properties for a section with rounded corners Effective area Effective centroid from compression flange Effective centroid from tension flange Second moment of area about major axis Section modulus about major axis compression flange Section modulus about major axis tension flange

47

Aeff zgc zgc Ieff,y Weff,y,c Weff,y,t

= 696.4 mm2 = 101.7 mm = 96.3 mm = 4116000 mm4 = 40480 mm3 = 42740 mm3

5

of

5

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 4

Subject

Client

CALCULATION SHEET

8.4

Sheet

1

This worked example presents the design procedure for the calculation of the effective section properties of a cold-formed lipped C section in bending about the minor axis such that the web is in compression. It was shown in Example 1 that the influence of rounded corners must be included for this cross-section. The section properties are initially calculated on the basis that the section has sharp corners. The influence of rounded corners is taken into account by apply reductions to the sharp corner properties.

Section dimensions and material properties

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

4

Rev

Effective properties of a lipped C section in bending about the minor axis Made by MDH Date Oct 2010 SCI Checked by SN Date Jan 2011

Effective properties for a lipped C section in minor axis bending

h b1 c r tnom t fy

= 200 mm = b2 = 65mm = 25 mm = 3 mm = 2 mm = 1.96 mm = 350 N/mm2

Web depth Flange width

hp bp1

= h  t nom  200  2  198 mm = b p 2  b1  t nom  65  2  63 mm

Stiffener depth

cp

= c  t nom 2  25  2 2  24 mm

Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness Design strength

of

Mid-line dimensions

Effective section properties (sharp corners) The effective properties of the flange and web are determined separately as shown below, after which the effective properties of the whole cross section may be calculated.

48

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 4: Effective properties of a lipped C section in bending about the minor axis

Sheet

2

of

4

Rev

y h e1 hc z

z

h e2

h2

y

Effective properties of the web in compression

3-1-3/ 5.5.3.2

Step 1 Effective width of the web in compression

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

For a stress ratio   1 (uniform compression), kσ  4



 235 f y

p, b







heff

 h p  0.414  198  82.0 mm

hp t 28.4  k σ



198 1.96  2.172 28.4  235 350  4

p, b  0.0553    p, b 2

3-1-3/ 5.5.2 and 3-1-5/ 4.4



2.172  0.055  3  1  0.414  1.0 2.172 2

3-1-5/ 4.4

Effective width of the part of the flange in compression The effective area of the section considering only the web as having an ineffective part is calculated.

Aeff Aeff Aeff

= Agr – (hp – heff).t = 729.1 – (198.0 – 82.0) × 1.96 = 501.7 mm2

The position of the effective neutral axis with regard to the web in compression is calculated. hc hc hc

= t (2.bc.cc + 2.bc.(bc/2)) / Aeff = 1.96 (2 × 63.0 × 24.0 + 2 × 63.0 (63.0/2)) / 501.7 = 27.32 mm 3-1-5/ Table 4.1

The stress ratio ψ is given by: ψ = (hc – bc) / hc ψ = (27.32 – 63.0) / 27.32 ψ = –1.31 For ψ between –1 and –3, the buckling factor is given by: kσ = 5.98 (1 – ψ)2 kσ = 5.98 (1 – –1.31)2 kσ = 31.91

49

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 4: Effective properties of a lipped C section in bending about the minor axis

p,b

=



=

bp / t 28.4 k σ

=

63.0 / 1.96 = 0.245 28.4  235 / 350  31.91

p, b  0.0553    2

but   1 so

Sheet

p, b  1



0.245  0.055  3  1.31  2.533 0.2452

The flanges are fully effective. The effective width of the zone in compression of the web is: heff = ρ × hc heff = 1.0 × 27.32 = 27.32 mm The effective width of the flange near the web in compression is: he1 = 0.4 heff = 10.93 mm The effective width of the flange near the lips in tension: h2 = bp – hc + 0.6 heff = 52.07 mm The lips are fully effective because they are in tension.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Effective properties of the whole cross section = 501.7 mm2

Aeff

The position of the neutral axis with regard to the web in compression is given by: yc







t 2 cp bp  2bp bp / 2

 = 27.3 mm

Aeff The position of the neutral axis with regard to the lips in tension is given by: yt  b p  yc  63  27.3  35.7 mm

Other effective properties for the section subject to minor axis bending are given below but the calculation of these properties is not shown. Second moment of area about minor axis Effective elastic modulus about minor axis

Iz,eff = 325800 mm4 Weff,z = 9126 mm3

Summary of Effective Properties for a section with sharp corners Effective area Effective centroid from web Effective centroid from lips Second moment of area about minor axis Effective elastic modulus about minor axis

Aeff yc yt Iz,eff Weff,z

50

= 501.7 mm2 = 27.3 mm = 35.7 mm = 325800 mm4 = 9126 mm3

3

of

4

Rev

3-1-5/ 4.4

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 4: Effective properties of a lipped C section in bending about the minor axis

Sheet

Effective section properties (rounded corners) Influence of rounded corners The influence of rounded corners must be included for this cross-section. For properties such as the position of the neutral axes, the values for the section with rounded corners are taken as equal to those calculated for the section with sharp corners. The section properties that are reduced to take account of rounded corners are calculated below. Aeff Iz

 Aeff,sharp 1     501.7(1  0.014)  494.7 mm2

 I z, sharp 1  2   325800(1  2  0.014)  316700 mm4

Summary of Effective Properties for a section with rounded corners

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Effective area Effective centroid from web Effective centroid from lips Second moment of area about minor axis Effective elastic modulus about minor axis

Aeff yc yt Iz,eff Weff,z

51

= 494.7 mm2 = 27.3 mm = 35.7 mm = 316700 mm4 = 8871 mm3

4

of

4

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 5

Subject

Design of a wall stud

Client

SCI

CALCULATION SHEET

8.5

Sheet

Oct 2010

Checked by

SN

Date

Jan 2011

The wall stud is located in a load bearing external. Therefore, the stud is subjected to wind loads and gravity loads. The wall stud is restrained at mid height by a lateral restraint within the wall construction. L = 2700 mm s = 600 mm NEd = 4.5 kN qk = 1.2 kN/m2 My,Ed = 0.98 kNm

Section dimensions and material properties The wall stud is a lipped C section, manufactured from S350 steel with a Z275 coating to BS EN 10346. Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness

h b c r tn t

= 100 mm = 45 mm = 12.0 mm = 1.5 mm = 1.2 mm = 1.16 mm

Basic yield strength Modulus of elasticity Shear modulus Partial factor Partial factor

fyb E G γM0 γM1

= 350 N/mm2 = 2100000 N/mm2 = 80770 N/mm2 = 1.0 = 1.0

Section Properties The calculation of section properties is not included in this example. See Examples 1 to 4 for procedure of calculation of gross and effective section properties.

52

Rev

Date

Wall details

Actions Design value of axial force Characteristic wind load Maximum design moment due to wind

8

AW

Design of a wall stud in a light steel frame building

Dimensions Stud height Stud spacing

of

Made by

This worked example presents the design of a wall stud in a light steel frame building. Ultimate limit state and serviceability limit state design checks are included.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

Gross Properties Area Radius of gyration about y axis Radius of gyration about z axis Position of y axis from flange Position of z axis from web Position of shear centre with respect to the z axis Position of shear centre with respect to the y axis Torsion constant Warping constant Second moment of area about y axis

Agr = 242.7 mm2 iy = 40.2 mm iz = 16.8 mm yflange = 49.4 mm zweb = 13.9 mm yo = 35.0 mm zo = 0.0 mm It = 108.8 mm4 Iw = 1.41 ×108 mm6 Iy = 392000 mm4

Effective Section Properties Effective area subject to compression Elastic section modulus for bending about y axis Elastic section modulus for bending about z axis Position of y axis from flange (due to compression) Position of z axis from web (due to compression) Second moment of area about y axis (due to bending)

Aeff = 153.6 mm2 Weff,y = 6929 mm3 Weff,z = 2101 mm3 yflange = 49.4 mm zweb = 15.5 mm Iy,eff = 360800 mm4

2

of

8

Rev

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Resistance of Cross-Section Axial Compression Design resistance of cross section for compression is given by: Nc,Rd = Aeff × fyb / γM0 Nc,Rd = 153.6 × 350 ×10-3/ 1.0 = 53.8 kN

3-1-3/ 6.1.3

Bending Design moment resistance for bending about y axis is given by: Mcy,Rd = Weff,y × fyb / γM0 Mcy,Rd = 6929 × 350 ×10-6/ 1.0 = 2.4 kNm

3-1-3/ 6.1.4

Design moment resistance for bending about z axis: Mcz,Rd = Weff,z × fyb / γM0 Mcz,Rd = 2101 × 350 ×10-6/ 1.0 = 0.7 kNm

Combined Compression and Bending The shift in the position of the effective section neutral axis relative to gross section neutral axis causes an additional moment due to axial load which is assumed to be applied at the gross section neutral axis. The shift in y axis due to axial compression is given by: eNy = 49.4 – 49.4 = 0.0 mm The shift in z axis due to axial compression is given by: eNz = 15.5 – 13.9 = 1.6 mm Additional y axis moment due to shift in axis: ΔMy,Ed = eNy × Ned = 0.000 kNm

53

3-1-3/ 6.1.9

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

3

of

8

Rev

Additional z axis moment due to shift in axis: ΔMz,Ed = eNz × Ned = 4.5 × 1.6 ×10-3 = 0.007 kNm M y, Ed  M y, Ed M z, Ed  M z, Ed N Ed   1 N c, Rd M cy,Rd M cz, Rd

3-1-3/ 6.1.9

4.5 0.98  0.0 0  0.007   1 53.8 2.4 0.7 0.08 + 0.41 + 0.01 = 0.50 ≤ 1.0

OK

Buckling Resistance of Member Flexural Buckling about major axis (y axis) From 3-1-3/Table 6.3, the buckling curve for a lipped C section buckling about any axis is buckling curve b.

3-1-3/ 6.2.2

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The buckling length for buckling about the y axis is taken as equal to the member length. Lcr,y

= 2700 mm

1

=

3-1-1/ 6.3.1.3

E 210000 = = 76.95 fy 350

Non dimensional slenderness factor Aeff 153.6 A L 2700 242.7 gr = 0.695  = cr  = i 1 40.2 76.95 For buckling curve b, the imperfection factor α is 0.34



  = 0.51  0.340.695  0.2  0.695  = 0.825



=



=



= 0.5 1     0.2   2

2

1

    2

2

≤ 1.0

1 0.825  0.825 2  0.695 2

= 0.787

Flexural buckling resistance Nb,Rd

=

 Aeff f y  M1

Nb,Rd

=

0.787  153.6  350  10 3 = 42.3 kN 1.0

Validation NEd / Nb,Rd = 4.5 / 42.3 = 0.11 < 1.0

OK

54

3-1-1/ Table 6.1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

4

of

8

Rev

Flexural Buckling about minor axis (z axis) The buckling length for buckling about the z axis is taken as half the member length because there is a mid-height lateral restraint in the wall. Lcr,z

= 2700 / 2 = 1350 mm

1

=

3-1-1/ 6.3.1.3

210000 E = = 76.95 fy 350

Non dimensional slenderness factor Aeff 153.6 Lcr Agr 1350 242.7  =  = = 0.830 1 i 16.8 76.95 For buckling curve b, the imperfection factor α is 0.34



  = 0.51  0.340.830  0.2   0.830 



=



=

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement



= 0.5 1     0.2   2

2

1

   2 2

= 0.952

≤ 1.0

1 0.952  0.952 2  0.830 2

= 0.706

Flexural buckling resistance Nb,Rd

=

 Aeff f y  M1

Nb,Rd

=

0.706  153.6  350  10 3 = 37.9 kN 1.0

Validation NEd / Nb,Rd = 4.5 / 37.9 = 0.12 < 1.0

OK

Torsional Buckling The torsional buckling lengths are: LT,y = 2700 mm LT,z = 1350 mm The polar radius of gyration is calculated as below: io2 = iy2 + iz2 + yo2 + zo2 io2 = 40.22 + 16.82 + 35.02 + 0.02 = 3123 mm2 io = 55.9 mm

55

3-1-1/ Table 6.1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

5

of

8

Rev

The elastic critical force for torsional buckling of a simply supported member is 3-1-3/ 6.2.3(5) given by: Ncr,T

Ncr,T

=

 2E Iw 1   G I t io2  LT2

1 = 55.9 2

   

2 8    80770  108.8    210000  1.41  10  = 54.27 kN   1350 2  

3-1-1/ 6.3.1.4

Non dimensional slenderness factor:

T

=

T

=

Aeff f y N cr

153.6  350  10 3 = 0.995 54.27

For buckling curve b, the imperfection factor α is 0.34

T

  = 0.51  0.340.995  0.2  0.995 

T

=

T

=

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

T

3-1-1/ 6.3.1.2

= 0.5 1     0.2   2

2

1

    2

2

= 1.131

≤ 1.0

1 1.131  1.1312  0.995 2

= 0.600

Torsional buckling resistance Nb,Rd

=

 T Aeff f y  M1

Nb,Rd

=

0.600  153.6  350  10 3 = 32.3 kN 1.0

Validation NEd / Nb,Rd = 4.5 / 32.3 = 0.14 < 1.0

OK

Torsional-Flexural Buckling Ncr,y

=

3-1-3/ 6.2.3(7)

 2E Iy L2cr,y

 2  210000  392000  10 3

Ncr,y

=



y  = 1   o   io 

2700 2 2

= 111.5 kN

2

 35.0   1   = 0.608  55.9 

56

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

The elastic critical force for torsional-flexural buckling is given by: Ncr,TF

2  2  N cr,y  N cr,T N cr,T   yo  N cr,T   4   1  = 1 i  N   2   N cr,y N cr, y  cr, y  o   

Ncr,TF

2 2  111.5  54.27  54.27   35.0  54.27   1  = = 43.41 kN 1   4  2  0.608  111.5  111.5   55.9  111.5   

=

TF

=

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

8

Rev

3-1-3/ 6.2.3(7)

3-1-1/ 6.3.1.4

Aeff f y N cr

153.6  350  10 3 = 1.113 43.41

For buckling curve b, the imperfection factor α is 0.34

TF

  = 0.51  0.341.113  0.2  1.113 

TF

=

TF

=

TF

of

   

Non dimensional slenderness factor:

TF

6

3-1-1/ 6.3.1.2

= 0.5 1     0.2   2

2

1

   2 2

= 1.274

≤ 1.0

1 1.274  1.274 2  1.1132

= 0.528

Torsional-flexural buckling resistance: Nb,Rd

=

 TF Aeff f y  M1

Nb,Rd

=

0.528 153.6  350  10 3 = 28.4 kN 1.0

Validation NEd / Nb,Rd = 4.5 / 28.4 = 0.16 < 1.0

OK

Lateral Torsional Buckling The maximum length of member between points of minor axis lateral restraint is half the member length. L

= 1350 mm 13]

SN003[

Coefficients dependent on loading and end restraint: C1 = 1.127 C2 = 0.454 Effective length factor for end rotation on plan

k

= 1.00

Effective length factor for end warping

kw

= 1.00

57

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

7

of

8

Rev

The distance from the point of load application to the shear centre is taken as half the stud depth. zg = 100 / 2 = 50 mm The factor g is used in the calculation of Mcr, it may conservatively be taken as 1.0 or may be calculated as below. I 68550 = 1  z g = 1 = 0.908 g 392000 Iy

14]

SN002[

The elastic critical moment for lateral torsional buckling is given by: 2   2 π 2 EI z   k  I w kL  GI t  2    2  C2 zg  C2 zg  = C1 Mcr 2   kL  g   k w  I z π EI z   



Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Mcr



= 2.82 kNm

For lateral torsional buckling curve b should be used.

3-1-3/ 6.2.4

Non dimensional slenderness factor:

3-1-1/ 6.3.1.4

LT

=

LT

=

Wy,eff f y M cr 6929  350  10 6 = 0.927 2.82

For buckling curve b, the imperfection factor α is 0.34

LT

  = 0.51  0.340.927  0.2  0.927 

LT

=

LT

=

LT

= 0.5 1     0.2   2

2

1

   2 2

= 1.054

≤ 1.0

1 1.054  1.054 2  0.927 2

= 0.644

Lateral torsional buckling resistance: M b,Rd  M b,Rd

 lT Weff,y f y  M1

0.644  6929  350  10 6 = 1.56 kNm  1.0

Validation MEd / Mb,Rd = 0.98 / 1.56 = 0.63 < 1.0

OK

58

3-1-1/ 6.3.2.1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 5: Design of a wall stud

Sheet

8

of

8

Rev

Combined Bending and Axial Compression

The interaction of bending moment and axial force may be obtained from a second order analysis or by using the formula below.  N Ed    N   b, Rd 

0.8

 M    Ed  M   b, Rd 

3-1-3/ 6.2.5

0.8

 1.0

Nb,Rd is the minimum value from flexural, torsional and torsional-flexural buckling. MEd includes any additional moments due to the shift in the neutral axis.  4.5     28.4 

0.8

 0.98  0.0     1.56 

0.8

 0.23 + 0.69 = 0.92 < 1.0

OK

Serviceability Deflections

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

For cross section stiffness properties the influence of rounded corners should always be taken into account. For this example it is assumed that the maximum stress at serviceability is the design yield strength divided by 1.5. Ific Ific





 1  = I gr    I gr  I eff  1.5   1  = 392000   392000  360800 = 371200 mm4  1.5  n



= 0.43

j 1 m

 bp,i

= 0.43

3-1-3/ 7.1(3)

3-1-3/ 5.1(4)

The corner factor δ is given by:

 rj

3-1-3/ 5.1(3)

4  1.5 = 0.01 (98.8  2  43.8  2  11.4)

i 1

Taking account of rounded corners, the second moment of area for serviceability is given by: ISLS = I (1 – 2δ) = 371200 (1 – 0.02) = 363800 mm4 Total wind load: W = L s qk = 2700 × 600 × 1.2 × 10-6 = 1.94 kN Deflection due to wind is given by; 5 WL3 5 1.94  2700 3 10 3  = 6.5 mm wind = 384 EI SLS 384 210000  363792 Deflection limit is taken as length divided by 360, OK δlimit = 2700 / 360 = 7.5 mm > 6.5 mm

59

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 6

Subject

Design of a floor joist (restrained)

Client

SCI

CALCULATION SHEET

8.6

Sheet

1

Date

Oct 2010

Checked by

SN

Date

Jan 2011

Floor details For this example the floor joist is assumed to be restrained along the length of the top flange by the connection to the timber floor boarding. The unrestrained situation during construction is considered in Example 7.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

= 4800 mm = 600 mm

Actions Permanent action (characteristic) Variable action (characteristic) Factors on actions Permanent Variable

γG γQ

gk qk

= 0.5 kN/m2 = 1.5 kN/m2

= 1.35 = 1.50

Section dimensions and material properties The wall stud is a lipped C section, manufactured from S350 steel with a Z275 coating to BS EN 10346. Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness

h b c r tn t

= 200 mm = 65 mm = 25 mm = 3.0 mm = 2.0 mm = 1.96 mm

Basic yield strength Modulus of elasticity Shear modulus Partial factor Partial factor

fyb E G γM0 γM1

= 350 N/mm2 = 2100000 N/mm2 = 80770 N/mm2 = 1.0 = 1.0

60

Rev

AW

This worked example presents the calculations for a restrained floor joist in a light steel frame building. Ultimate limit state and serviceability limit state design checks are included.

L s

4

Made by

Design of a floor joist (restrained) in a light steel frame building

Dimensions Floor span Joist spacing

of

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 6: Design of a floor joist (restrained)

Sheet

2

of

4

Rev

Section Properties The calculation of section properties is not included in this example. See Examples 1 to 4 for procedure of calculation of gross and effective section properties. Gross Properties Second moment of area about y axis

Iy = 4281000 mm4

Effective Section Properties Second moment of area about y axis Elastic section modulus for bending about y axis

Ieff,y = 4116000 mm4 Weff,y = 40480 mm3

Design Moment Applied design moment is given by, Mcy,Ed = (γG × gk + γQ × qk) s L2 / 8 Mcy,Ed = (1.35 × 0.5 + 1.50 × 1.5) 600 × 48002 × 10-9 / 8 Mcy,Ed = 5.1 kNm

Resistance of Cross-Section

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Bending Moment Design moment resistance for bending about y axis is given by, Mcy,Rd = Weff,y × fyb / γM0 Mcy,Rd = 40480 × 350 ×10-6/ 1.0 = 14.2 kNm Validation My,Ed / Mcy,Rd = 5.1 / 14.2 = 0.36 < 1.0

3-1-3/ 6.1.4

OK

Buckling Resistance of Member Lateral Torsional Buckling For this example the floor joist is assumed to be restrained along the length of the top flange by the connection to the timber floor boarding. Therefore, the joist is not required to be checked for lateral torsional buckling. The joist is unrestrained during construction; this situation is considered in Example 7.

Serviceability Deflections For cross section stiffness properties the influence of rounded corners should always be taken into account. For this example it is assumed that the maximum stress at serviceability is the design yield strength divided by 1.5. Ific Ific





 1  = Igr    I gr  I eff  1.5   1  = 4281000   4281000  4116000 = 4171000 mm4  1.5 

61

3-1-3/ 5.1(3)

3-1-3/ 7.1(3)

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 6: Design of a floor joist (restrained)

Sheet

The influence of rounded corners has been taken account in the calculation of the Igr and Ieff values used to calculate Ific. Therefore, the second moment of area for serviceability is given by; ISLS

= Ific = 4171000 mm4

Serviceability Criteria For light weight steel floors there are four serviceability criteria that should be checked to ensure acceptable performance of the floor in service. The criteria are detailed in SCI publication P301 and in Chapter 6.10 of the NHBC Standards. Criterion 1 Dead load plus imposed load deflection less than span/350 or 15 mm whichever is smaller.

Total load is given by: W = L s (gk + qk) = 4800 × 600 (0.5 + 1.5) × 10-6 = 5.76 kN

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Deflection due to load is given by: 5 WL3 5 5.76  4800 3 10 3   = = 9.5 mm 384 EI SLS 384 210000  4171000

Deflection limit: δlimit = 4800 / 360 = 13.3 mm > 9.5 mm

OK

Criterion 2 Imposed load deflection less than span/450.

Total load: = L s qk = 4800 × 600 × 1.5 × 10-6 = 4.32 kN W Deflection due to load is given by: 5 WL3 5 4.32  4800 3 10 3   = = 7.1 mm 384 EI SLS 384 210000  4171000 Deflection limit: δlimit = 4800 / 450 = 10.7 mm > 7.1 mm

OK

Criterion 3 Natural frequency of the floor not less than 8 Hz.

Total load for this criterion is given by: = L s (gk + 0.2 qk) = 4800 × 600 (0.5 + 0.2 × 1.5) × 10-6 = 2.30 kN W Deflection due to load is given by: 5 WL3 5 2.30  4800 3 10 3   = = 3.8 mm 384 EI SLS 384 210000  4171000 Deflection limit for 8 Hz is; δlimit = 5.0 mm > 3.8 mm

OK

62

3

of

4

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 6: Design of a floor joist (restrained)

Sheet

4

of

4

Rev

Criterion 4 Deflection of floor system less than critical value subject to 1 kN point load.

Total load for this criteria is given by: = 1.0 kN W Table 6.1 of SCIThe number of effective joists depends on the floor construction. For this example the floor construction is taken as chipboard on the floor joists spaced at P301 600 mm. = 2.35 Neff

Deflection due to load is given by: WL3 1 1 1.0  4800 3 10 3   = = 1.1 mm 48 EI SLS N eff 48 210000  4171000  2.35

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The deflection limit for this criterion is dependant on the span of the joist. OK δlimit = 1.37 mm > 1.1 mm

63

Table 6.2 of SCIP301

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 7

Subject

Design of a floor joist (unrestrained)

Client

SCI

CALCULATION SHEET

8.7

Sheet

1

Date

Oct 2010

Checked by

SN

Date

Jan 2011

Floor details

= 4800 mm = 600 mm

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Actions Permanent action (characteristic) gk Variable action (characteristic) qk Factors on actions Permanent Variable

γG γQ

= 0.3 kN/m2 = 0.7 kN/m2

= 1.35 = 1.50

Section dimensions and material properties The wall stud is a lipped C section, manufactured from S350 steel with a Z275 coating to BS EN 10346. Section depth Flange width Stiffener depth Corner radius Nominal thickness Core thickness

h b c r tn t

= 200 mm = 65 mm = 25 mm = 3.0 mm = 2.0 mm = 1.96 mm

Basic yield strength Modulus of elasticity Shear modulus Partial factor Partial factor

fyb E G γM0 γM1

= 350 N/mm2 = 2100000 N/mm2 = 80770 N/mm2 = 1.0 = 1.0

64

Rev

AW

This worked example presents the design calculations for an unrestrained floor joist during the construction stage in a light steel frame building.

L s

3

Made by

Design of a floor joist (unrestrained) in a light steel frame building

Dimensions Floor span Joist spacing

of

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 7: Design of a floor joist (unrestrained)

Sheet

2

of

3

Rev

Section Properties The calculation of section properties is not included in this example. See Examples 1 to 4 for procedure of calculation of gross and effective section properties. Gross Properties Second moment of area about y axis Second moment of area about z axis Torsion constant Warping constant

Iy Iz It Iw

= 4281000 mm4 = 429700 mm4 = 933.6 mm4 = 3.57 × 109 mm6

Effective Section Properties Elastic section modulus for bending about y axis

Weff,y = 40480 mm3

Design Moment Applied design moment is given by, My,Ed = (γG × gk + γQ × qk) s L2 / 8 My,Ed = (1.35 × 0.3 + 1.50 × 0.7) 600 × 48002 × 10-9 / 8 My,Ed = 2.5 kNm

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Resistance of Cross-Section Bending Moment

Design moment resistance for bending about y axis is given by, Mcy,Rd = Weff,y × fyb / γM0 Mcy,Rd = 40480 × 350 ×10-6/ 1.0 = 14.2 kNm Validation My,Ed / Mcy,Rd = 2.5 / 14.2 = 0.18 < 1.0

3-1-3/ 6.1.4

OK

Buckling Resistance of Member Lateral Torsional Buckling

The joist is unrestrained during construction. Therefore, it must be checked for lateral torsional buckling. The maximum length of member between points of minor axis lateral restraint is taken as the whole length of the member. L

= 4800 mm SN003

Coefficients dependent on loading and end restraint; = 1.127 C1 C2 = 0.454 Effective length factor for end rotation on plan Effective length factor for end warping

k kw

= 1.00 = 1.00

The distance from the point of load application to the shear centre is taken as half the stud depth. = 200 / 2 = 100 mm zg

65

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 7: Design of a floor joist (unrestrained)

Sheet

The factor g is used in the calculation of Mcr, it may conservatively be taken as 1.0 or may be calculated as below. I 429700 g = 1  z g = 1 = 0.948 4281000 Iy

3

of

3

Rev

SN002

The elastic critical moment for lateral torsional buckling is given by: 2   2 π 2 EI z   k  I w kL  GI t  2      C z C z Mcr = C1 2 g 2 g 2   2  kL  g   k w  I z π EI z   



Mcr



= 3.14 kNm

For lateral torsional buckling curve b should be used.

3-1-3/ 6.2.4

Non dimensional slenderness factor:

3-1-1/ 6.3.1.4

LT

=

LT

=

Wy,eff f y M cr 40480  350  10 6 = 2.12 3.14

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

For buckling curve b, the imperfection factor α is 0.34

LT

  = 0.51  0.342.12  0.2   2.12 

LT

=

LT

=

LT

= 0.5 1     0.2   2

2

1

   2 2

= 3.07

≤ 1.0

1 3.07  3.07 2  2.12 2

= 0.189

Lateral torsional buckling resistance: Mb,Rd =

 lT Weff,y f y  M1

Mb,Rd =

0.189  40480  350  10 6 = 2.68 kNm 1.0

Validation MEd / Mb,Rd = 2.5 / 2.68 = 0.93 < 1.0

OK

66

3-1-1/ 6.3.2.1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Silwood Park, Ascot, Berks SL5 7QN Telephone: (01344) 636525 Fax: (01344) 636570

Job No.

BCS 320

Title

Example 8

Subject

Design of a lattice floor truss

Client

SCI

CALCULATION SHEET

8.8

Sheet

1

Date

Oct 2010

Checked by

SN

Date

Jan 2011

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Floor details

= 4800 mm = 600 mm = 200 mm = 12 = 400 mm = 45o = 283 mm

Actions Permanent action (characteristic) Variable action (characteristic) Factors on actions Permanent Variable

γG γQ

gk qk

= 0.6 kN/m2 = 1.5 kN/m2

= 1.35 = 1.50

Lattice truss dimensions and material properties The lattice truss is fabricated from lipped C sections, manufactured from S350 steel with a Z275 coating to BS EN 10346. The same size section is used for the top chord, bottom chord and brace members. The lattice truss is shown below.

DT

Lbay

67

Rev

AW

This worked example presents design calculations for a restrained lattice floor truss in a light steel frame building. Ultimate limit state and serviceability limit state design checks are included.

L s DT N Lbay θ Lbrace

11

Made by

Design of a lattice floor truss in a light steel frame building

Dimensions Floor span Joist spacing Depth of truss Number of bays in truss Length of each bay Angle of brace member Length of brace member

of

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

The section dimensions are: Section depth h Flange width b Stiffener depth c Corner radius r Nominal thickness tn Core thickness t

= 75 mm = 40 mm = 10 mm = 2.0 mm = 1.6 mm = 1.56 mm

Basic yield strength Modulus of elasticity Shear modulus Partial factor Partial factor

= 350 N/mm2 = 2100000 N/mm2 = 80770 N/mm2 = 1.0 = 1.0

fyb E G γM0 γM1

In this example the nodal eccentricities of the neutral axes of the brace members with the top and bottom chords is considered in the design. The eccentricity values used in this example are estimated. Nodal eccentricity at the top chord Nodal eccentricity at the bottom chord

etop = 60 mm ebottom = 20 mm

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

nodal eccentricity

nodal eccentricity

Section Properties The calculation of section properties is not included in this example. See Examples 1 to 4 for procedure of calculation of gross and effective section properties. Gross Properties Area Position of y axis from flange Position of z axis from web Radius of gyration for y axis Radius of gyration for z axis Second moment of area about strong axis y-y Second moment of area about weak axis z-z Position of shear centre with respect to the z axis Position of shear centre with respect to the y axis Torsion constant Warping constant

68

Ag = 263.0 mm2 yflange = 36.7 mm zweb = 12.9 mm iy = 30.4 mm iz = 14.8 mm Igry = 242500mm4 Igrz = 57230m4 yo = 31.8 mm zo = 0.00 mm It = 213.4 mm4 Iw = 6.64 ×107 mm6

2

of

11

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

Effective Section Properties Effective area subject to compression Aeff Position of y axis from flange (section in compression) yflange Position of z axis from web (section in compression) zweb Elastic section modulus for bending about y axis Weff,y Modulus for bending about z axis (lips in compression) Weff,z,lip Modulus for bending about z axis (web in compression) Weff,z,web Section modulus for bending about z axis (minimum) Weff,z Second moment of area about weak axis z-z Ieff,z

= 222.3 mm2 = 36.7 mm = 13.2 mm = 6297 mm3 = 2247 mm3 = 2179 mm3 = 2179 mm3 = 51530 mm4

Analysis of forces in lattice truss A cross section of the lattice truss is shown below. Centroid of top chord

Chord centroid from w eb Depth of truss

Centroid of lattice truss

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Chord centroid from w eb

Centroid of bottom chord

Maximum moment is given by: My,Ed = (γG × gk + γQ × qk) s L2 / 8 My,Ed = (1.35 × 0.6 + 1.50 × 1.5) 600 × 48002 × 10-9 / 8 = 5.3 kNm End reaction is given by: R,Ed = (γG × gk + γQ × qk) s L / 2 R,Ed = (1.35 × 0.6 + 1.50 × 1.5) 600 × 4800 × 10-6 / 2 = 4.4 kN Maximum compression in top chord is taken as: Nc,Ed = My,Ed / (DT – 2 × zweb) Nc,Ed = 5.3 × 103 / (200 – 2 × 13.2) = 30.5 kN Maximum tension in bottom chord is taken as: Nt,Ed = My,Ed / (DT – 2 × zweb) Nt,Ed = 5.3 × 103 / (200 – 2 × 13.2) = 30.5 kN Maximum compression in brace member is given by: Nc,Ed = R,Ed × Lbrace / DT Nc,Ed = 4.4 × 283 / 200 = 6.2 kN For the bay in the lattice truss adjacent to the central bay: Compression in top chord Nc,Ed = 29.6 kN Tension in bottom chord Nt,Ed = 29.6 kN Moment from nodal eccentricity in the top chord is taken as: ΔMe,top = ΔNc,Ed × etop Me,top = (30.5 – 29.6) × 60 = 0.05 kNm

69

3

of

11

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

4

of

11

Rev

Moment from nodal eccentricity in the bottom chord is taken as: = ΔNc,Ed × ebottom Me,bottom Me,bottom = (30.5 – 29.6) × 20 = 0.02 kNm Moment from load applied between nodes in the top chord is taken as: Mw,top = (γG × gk + γQ × qk) s Lbay2 / 8 Mw,top = (1.35 × 0.6 + 1.50 × 1.5) 600 × 4002 × 10-9 / 8 = 0.04 kNm Moment from load applied between nodes in the bottom chord is taken as zero. Mw,bottom = 0.0 kNm Total moment in top chord Total moment in bottom chord

Mz,Ed,top = 0.05 + 0.04 = 0.09 kNm Mz,Ed,bottom = 0.02 + 0.0 = 0.02 kNm

Resistance of Cross-Section Axial Tension

The average yield strength may be used to determine tension resistance. f u  f yb k nt2 = f yb  f u  f yb but  fya Agr 2

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement







fya

= 350  420  350

fya

= 368.1 N/mm2



3-1-3/ 3.2.2

420  350 7  4  1.56 2 but  2 263.0

Design resistance of cross section for tension is given by: Nt,Rd = Ag × fya / γM0 Nt,Rd = 263.0 × 368.1 ×10-3/ 1.0 = 96.8 kN Validation (Bottom Chord) Nt,Ed / Nt,Rd = 30.5 / 96.8 = 0.32 < 1.0

3-1-3/ 6.1.2

OK

Axial Compression

Design resistance of cross section for compression is given by: Nc,Rd = Aeff × fyb / γM0 Nc,Rd = 222.3 × 350 ×10-3/ 1.0 = 77.8 kN Validation (Top Chord) Nc,Ed / Nc,Rd = 30.5 / 77.8 = 0.39 < 1.0

OK

Validation (Brace Member) Nc,Ed / Nc,Rd = 6.2 / 77.8 = 0.08 < 1.0

OK

3-1-3/ 6.1.3

Bending Moment

Design moment resistance for bending about y axis is given by: Mcy,Rd = Weff,y × fyb / γM0 Mcy,Rd = 6297 × 350 ×10-6/ 1.0 = 2.2 kNm Design moment resistance for bending about z axis Mcz,Rd = Weff,z × fyb / γM0 Mcz,Rd = 2179 × 350 ×10-6/ 1.0 = 0.8 kNm

70

3-1-3/ 6.1.4

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

Validation (Top Chord) Mz,Ed / Mz,Rd = 0.09 / 0.8 = 0.11 < 1.0

OK

Validation (Bottom Chord) Mz,Ed / Mz,Rd = 0.02 / 0.8 = 0.03 < 1.0

OK

5

of

11

Rev

Combined Compression and Bending

The shift in the position of the effective section neutral axis relative to gross section neutral axis causes an additional moment due to axial load which is assumed to be applied at the gross section neutral axis.

3-1-3/ 6.1.9

The shift in y axis due to axial compression is given by: = 36.7 – 36.7 = 0.0 mm eNy The shift in z axis due to axial compression is given by: = 13.2 – 12.9 = 0.3 mm eNz

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Additional y axis moment due to shift in axis is given by: ΔMy,Ed = eNy × Ned = 0.000 kNm Additional z axis moment due to shift in axis is, ΔMz,Ed = eNz × Ned = 0.3 × 30.5 ×10-3 = 0.009 kNm 3-1-3/ 6.1.9

Validation (Top Chord) M y,Ed  M y,Ed M z,Ed  M z,Ed N Ed   1 N c,Rd M cy,Rd M cz,Rd 30.5 0.0  0.0 0.09  0.009   1 77.8 2.2 0.8 0.39 + 0.00 + 0.12 = 0.51 ≤ 1.0

OK

Combined Tension and Bending 3-1-3/ 6.1.8

Validation (Bottom Chord) M y,Ed M z,Ed N Ed   1 N t,Rd M cy,Rd M cz,Rd 30.5 0.0 0.02   1 96.8 2.2 0.8 0.32 + 0.00 + 0.03 = 0.35 ≤ 1.0

OK

Buckling Resistance of Member The buckling resistance of the top chord in compression is checked in this example. The buckling resistance of the brace members in compression should also be checked but are omitted from this example. Flexural Buckling about major axis (y axis)

From Table 6.3 of BS 3-1-3/ the buckling curve for a lipped C section buckling about any axis is buckling curve b. The buckling length for buckling about the y axis is taken as equal to the truss bay length because the top chord is restrained at the node points. Lcr,y

= 400 mm

71

3-1-3/ 6.2.2

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

1

=

Sheet

6

of

11

Rev

3-1-1/ 6.3.1.3

E 210000 = = 76.95 350 fy

Non dimensional slenderness factor: Aeff 222.3 A L 400 263.0 gr  0.157  = cr   30.4 76.95 1 i For buckling curve b, the imperfection factor α is 0.34



  = 0.51  0.340.157  0.2  0.157  = 0.505



=



=



3-1-1/ Table 6.1

= 0.5 1     0.2   2

2

1

   2 2

≤ 1.0

1 2

0.505  0.505  0.157

2

= 1.02

but ≤ 1.0

= 1.0

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Flexural buckling resistance: Nb,Rd

=

 Aeff f y  M1

Nb,Rd

=

1.0  222.3  350  10 3 = 77.8 kN 1.0

Validation NEd / Nb,Rd = 30.5 / 77.8 = 0.39 < 1.0

OK

Flexural Buckling about minor axis (z axis)

The buckling length for buckling about the z axis is taken as equal to the truss bay length because the top chord is restrained at the node points. Lcr,z

= 400 mm

1

=

3-1-1/ 6.3.1.3

E 210000 = = 76.95 fy 350

Non dimensional slenderness factor: Aeff 222.3 A Lcr 400 263.0 gr    0.324 14.8 76.95 1 i For buckling curve b, the imperfection factor α is 0.34

 

  = 0.51  0.340.324  0.2   0.324  = 0.5 1     0.2   2

2

= 0.574

72

3-1-1/ Table 6.1

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss



=



=

1

   2 2

Sheet

7

of

11

Rev

≤ 1.0

1 0.574  0.574 2  0.324 2

= 0.955

Flexural buckling resistance:  Aeff f y Nb,Rd =

 M1

Nb,Rd

=

0.955 222.3  350  10 3 = 74.3 kN 1.0

Validation NEd / Nb,Rd = 30.5 / 74.3 = 0.41 < 1.0

OK

Torsional Buckling

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The torsional buckling lengths are: LT,y = 400 mm = 400 mm LT,z The polar radius of gyration is calculated as below: = iy2 + iz2 + yo2 + zo2 i o2 2 = 30.42 + 14.82 + 31.82 + 0.02 = 2154 mm2 io io = 46.4 mm The elastic critical force for torsional buckling of a simply supported member is 3-1-3/ 6.2.3(5) given by: Ncr,T

 2E Iw 1  = 2 G It  io  LT2

Ncr,T

 2  210000  6.64  10 7 1   = 80770  213.4  46.4 2  400 2

   

Non dimensional slenderness factor is given by:

T

=

T

=

T

3-1-1/ 6.3.1.4

Aeff f y N cr 222.3  350  10 3 = 0.437 407.5

For buckling curve b, the imperfection factor α is 0.34

T

  = 407.5 kN  

  = 0.51  0.340.437  0.2  0.437  = 0.5 1     0.2   2

2

= 0.636

73

3-1-1/ 6.3.1.2

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

T

=

T

=

1

   2  2

Sheet

8

of

11

Rev

≤ 1.0

1 0.636  0.636 2  0.437 2

= 0.911

Torsional buckling resistance is given by:  T Aeff f y Nb,Rd =

 M1

Nb,Rd

=

0.911 222.3  350  10 3 = 70.9 kN 1.0

Validation NEd / Nb,Rd = 30.5 / 70.9 = 0.43 < 1.0

OK

Torsional-Flexural Buckling

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Ncr,y

=

3-1-3/ 6.2.3(7)

 2E Iy L2cr,y

 2  210000  242500  10 3

Ncr,y

=



y = 1   o  io

400 2 2

= 3142 kN

2

  31.8   = 1    = 0.530 46 . 4   

The elastic critical force for torsional-flexural buckling is given by: 2   2  N cr,T  N cr,y  N cr,T  yo  N cr,T    1 Ncr,TF =  4   1  N cr,y  2  N cr,y  io  N cr,y      2 2  111.5  407.5  31.8  407.5   407.5  1 = 382.5 kN Ncr,TF =  1     4 2  0.530  3142 3142   46.4  3142    

3-1-3/ 6.2.3(7)

Non dimensional slenderness factor is given by: Aeff f y TF = N cr

3-1-1/ 6.3.1.4

TF

=

222.3  350  10 3 = 0.451 382.5

For buckling curve b, the imperfection factor α is 0.34

TF

  = 0.51  0.340.451  0.2  0.451 

TF

=

TF

=

TF

= 0.5 1     0.2   2

2

1

   2 2

= 0.644

≤ 1.0

1 0.644  0.644 2  0.4512

= 0.905

74

3-1-1/ 6.3.1.2

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

9

of

11

Rev

Torsional-flexural buckling resistance is given by:  TF Aeff f y Nb,Rd =

 M1

Nb,Rd

=

0.905 222.3  350  10 3 = 70.4 kN 1.0

Validation NEd / Nb,Rd = 30.5 / 70.4 = 0.43 < 1.0

OK

Lateral Torsional Buckling

Lateral torsional buckling of the top and bottom chords is not considered as there are no major axis moments applied to these sections. Lateral torsional buckling of the whole lattice truss is not considered as the top chord (compression flange) is assumed to be restrained by the timber floor system. During construction the lattice truss will be unrestrained. In practice, the unrestrained capacity of the lattice truss should be check for construction loading but it is omitted from this example.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Serviceability Deflections For cross section stiffness properties the influence of rounded corners should always be taken into account. For this example it is assumed that the maximum stress at serviceability is the design yield strength divided by 1.5.

3-1-3/ 5.1(3)

For minor axis (z axis):

3-1-3/ 7.1(3)





Ific

 1  = I gr    I gr  I eff  1.5 

Ific

 1  = 57230   57230  51530 = 53430 mm4  1.5  3-1-3/ 5.1(4)

The corner factor δ is given by: n



 rj = 0.43

j 1 m

 bp,i

= 0.02

i 1

Taking account of rounded corners, the second moment of area for serviceability is given by: ISLS = I (1 – 2δ) = 53430 (1 – 0.04) = 51290 mm4 The top chord is subject to axial compression therefore ISLS is used which is based on effective properties and includes allowance for rounded corners. The bottom chord is subject to axial tension therefore Igr,round is used which is based on gross properties and includes allowance for rounded corners. Igr,round = I (1 – 2δ) = 57230 (1 – 0.04) = 54940 mm4.

75

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

For this example the effective neutral axis of the truss subject to bending is taken at mid-height of the truss. The moment of inertia for the lattice truss is given by: Ieff,truss = (ISLS + Aeff.y2) + (Igr,round + Agr.y2) Ieff,truss = (51290 + 222.3 ×(100 – 13.2)2) + (54940 + 263.0 ×(100 – 12.9)2) Ieff,truss = 3776000 mm4 : Serviceability Criteria

For light weight steel floors there are four serviceability criteria that should be checked to ensure acceptable performance of the floor in service. The criteria are detailed in SCI publication P301 and in Chapter 6.10 of the NHBC Standards. Criterion 1 Dead load plus imposed load deflection less than span/350 or 15 mm whichever is smaller.

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

Total load: = L s (gk + qk) = 4800 × 600 (0.6 + 1.5) × 10-6 = 6.05 kN: W Deflection due to load is given by: 5 WL3 5 6.05  4800 3  103  = = 11.0 mm  384 EI 384 210000  3776000 Deflection limit: δlimit = 4800 / 360 = 13.3 mm > 11.0 mm

OK

Criterion 2 Imposed load deflection less than span/450.

Total load: = L s qk = 4800 × 600 × 1.5 × 10-6 = 4.32 kN W Deflection due to load is given by; 5 WL3 5 4.32  4800 3  10 3  = = 7.9 mm  384 EI 384 210000  3776000 Deflection limit: δlimit = 4800 / 450 = 10.7 mm > 7.9 mm

OK

Criterion 3 Natural frequency of the floor not less than 8 Hz.

Total load for this criterion is given by: = L s (gk + 0.2 qk) = 4800 × 600 (0.6 + 0.2 × 1.5) × 10-6 = 2.59 kN W Deflection due to load is given by: 5 WL3 5 2.59  4800 3  10 3  = = 4.7 mm  384 EI 384 210000  3776000 Deflection limit for 8 Hz δlimit = 5.0 mm > 4.7 mm

OK

76

10

of

11

Rev

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Example 8: Design of a lattice floor truss

Sheet

11

of

11

Rev

Criterion 4 Deflection of floor system less than critical value subject to 1 kN point load.

Total load for this criterion is given by: = 1.0 kN W The number of effective joists depends on the floor construction. For this example the floor construction is taken as chipboard on the floor joists spaced at 600 mm. = 2.35 Neff

Table 6.1 of SCIP301

Deflection due to load is given by: 1 WL3 1 1.0  4800 3  10 3  = = 1.24 mm  48 E I N eff 48 210000  3776000  2.35

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

The deflection limit for this criterion is dependant on the span of the joist. OK δlimit = 1.37 mm > 1.24 mm

77

Table 6.2 of SCI-P301

ED005 Technical Report: Design of Light Steel Sections to Eurocode 3 Discuss me ...

Created on 23 February 2012 This material is copyright - all rights reserved. Use of this document is subject to the terms and conditions of the Steelbiz Licence Agreement

9

REFERENCES

1.

BS EN 10346: 2009 Continuously hot-dip coated steel flat products. Technical delivery conditions BSI 2009

2.

BS EN 1990: 2002 Eurocode: Basis of structural design. BSI, 2002

3.

BS EN 1993-1-1: 2005 Eurocode 3: Design of steel structures. General rules and rules for buildings (incl. Corrigendum, February 2010) BSI, 2010

4.

BS EN 1993-1-3: 2005 Eurocode 3: Design of steel structures. Supplementary rules for cold-formed members and sheeting (incl. Corrigendum, April 2010) BSI, 2010

5.

BS EN 1993-1-5: 2005 Eurocode 3: Design of steel structures. Plated structural elements (incl. Corrigendum, February 2010) BSI, 2010

6.

BS EN 1993-1-8: 2005 Eurocode 3: Design of steel structures. Design of joints (incl. Corrigendum, August 2010) BSI, 2010

7.

BRETTLE, M. E. and BROWN, D. G. Steel Building Design: Concise Eurocodes (P362) SCI, 2009

8.

BRETTLE, M. E. and BROWN, D. G. Steel Building Design: Worked Examples for Students (P387) SCI, 2009

9.

BS EN 1991-1-7:2006 Eurocode 1: Actions on structures. General actions. Accidental actions (incl. Corrigendum, April 2010) BSI, 2010

10.

AD 358: Design thickness of cold formed members and sheeting New Steel Construction, 2011 (available on www.steelbiz.org)

11.

SMITH, A.L,HICKS, S.J. and DEVINE, P.J. Design of Floors for Vibration: A New Approach (P354) SCI, 2009

12.

GRUBB, P.J, GORGOLEWSKI, M.T. and LAWSON, R.M. Building Design using Cold Formed Steel Sections: Light Steel Framing in Residential Construction (P301) SCI, 2001

13

SN003, NCCI: Elastic critical moment for lateral torsional buckling www.steel-ncci.co.uk

14

SN002 NCCI: Determination of non-dimensional slenderness of I and H sections www.steel-ncci.co.uk

78