DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4 TH EDITION

Transcription

1 DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4 TH EDITION

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3 SCI PUBLICATION P413 DESIGN MANUAL FOR STRUCTURAL STAINLESS STEEL 4 TH EDITION i

4 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 Individual & corporate membership Advice Members advisory service Information Publications Education Events & training Consultancy Development Product development Engineering support Sustainability Assessment SCI Assessment Specification Websites Engineering software Front cover credits Top left: Canopy, Napp Pharmaceutical, Cambridge, UK Grade , Courtesy: m-tec Bottom left: Dairy Plant at Cornell University, College of Agriculture and Life Sciences, Grade /7, Courtesy: Stainless Structurals Top right: Skid for offshore regasification plant, Grade , Courtesy: Montanstahl Bottom right: Águilas footbridge, Spain Grade 1.446, Courtesy Acuamed 017 SCI. All rights reserved. Publication Number: SCI P413 ISBN 13: Published by: SCI, Silwood Park, Ascot, Berkshire. SL5 7QN UK T: +44 (0) F: +44 (0) E: sci.com sci.com To report any errors, contact: sci.com 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 publication 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 publishers, SCI. 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, SCI, 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. Publications supplied to the members of the Institute at a discount are not for resale by them. British Library Cataloguing in Publication Data. A catalogue record for this book is available from the British Library. ii

5 PART II - DESIGN EXAMPLES 169

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7 This part of the Design Manual gives fifteen design examples that illustrate the application of the design rules. The examples are: Design example 1 A circular hollow section subject to axial compression. Design example A welded I-beam with a Class 4 cross-section subject to combined axial compression and bending. Design example 3 Trapeziodal roof sheeting with a Class 4 cross-section subject to bending. Design example 4 A welded hollow section joint subject to fatigue loading. Design example 5 A welded joint. Design example 6 A bolted joint. Design example 7 A plate girder with a Class 4 cross-section subject to bending. Shear buckling is critical. Design example 8 A plate girder with a Class 4 cross-section subject to bending. Resistance to transverse loads is critical. Design example 9 A cold formed channel subject to bending with intermediate lateral restraints to the compression flange. Lateral torsional buckling between intermediate lateral restraints is critical. Design example 10 A rectangular hollow section subject to combined axial compression and bending with 30 minutes fire resistance. Design example 11 Trapezoidal roof sheeting with a Class 4 cross-section subject to bending a comparison of designs with cold worked material and annealed material. Design example 1 A lipped channel from cold worked material in an exposed floor subject to bending. Design example 13 A stainless steel lattice girder from cold worked material subject to combined axial compression and bending with 30 minutes fire resistance. Design example 14 The enhanced average yield strength of a cold-rolled square hollow section is determined in accordance with the method in Annex B. 171

8 Design examples Design example 15 The bending resistance of a cold-rolled square hollow section is determined in accordance with the Continuous Strength Method (CSM) given in Annex D. The sheeting in example 3 is from ferritic stainless steel grade The plate girders in examples 7 and 8 are from duplex stainless steel grade The members in the other examples are from austenitic stainless steel grades or The references in the margin of the design examples are to text sections and expressions/equations in this publication, unless specifically noted otherwise. 17

9 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Design Example 1 CHS Column Research Fund for Coal and Steel Sheet 1 of Made by HS Date 07/0 Revised by JBL Date 03/06 Revised by FW Date 05/17 design example 1 chs column The circular hollow section column to be designed is an interior column of a multi-storey building. The column is simply supported at its ends. The inter-storey height is 3,50 m. N sd Ed t l d Structure Simply supported column, length between supports: l = 3,50 m actions Permanent and variable actions result in a vertical design compression force equal to: = 50 kn N Ed cross-section properties Try a cold-formed CHS, austenitic grade Geometric properties d = 159 mm t = 4,0 mm A = 19,5 cm² I = 585,3 cm 4 W el = 73,6 cm 3 W pl = 96,1 cm 3 Material properties Take f y = 0 N/mm (for cold-rolled strip). Table. E = N/mm and G = N/mm Section.3.1 classification of the cross-section = 1,01 Table 5. Section in compression : dd/tt = 159/4 = 39,8 For Class 1, dd/tt 50εε, therefore the section is Class

10 Design Example 1 Sheet of compression resistance of the cross-section Section For a Class 1 cross-section: NN c,rd = AA g ff y /γγ M0 Eq. 5.7 NN c,rd = 19, = 390 kn 1,1 resistance to flexural buckling Section NN b,rd = χaaff y /γγ M1 Eq. 6. = 1 φφ + [φφ λλ ] 0,5 1 Eq. 6.4 φφ = 0,5(1 + αα(λλ λ 0) + λλ ) Eq. 6.5 Calculate the elastic critical buckling load: NN cr = π EEEE LL cr = π , (3, ) 10 3 = 943,1 kn Calculate the flexural buckling slenderness: λλ = 19, , = 0,67 Eq. 6.6 Using an imperfection factor = 0,49 and λ 0 = 0, for a cold-formed austenitic stainless steel CHS: φφ = 0,5 (1 + 0,49 (0,67 0,) + 0,67 ) = 0,84 χχ = 1 0,84 + [0,84 0,67 = 0,74 ] 0,5 Table 6.1 NN b,rd = 0,74 19, ,1 The applied axial load is N Ed = 50 kn. = 88,6 kn Thus the member has adequate resistance to flexural buckling. 174

11 Promotion of new eurocode rules for structural stainless steels (PureSt) Title Sheet 1 of 4 Design Example Welded I-section beam-column with lateral restraints Made by HS Date 07/0 calculation SHeet Client Research Fund for Coal and Steel Revised by JBL Date 03/06 Revised by FW Date 06/17 design example welded I-SectIon BeaM-coLuMn with LateraL restraints The beam-column to be designed is a welded I-section, simply supported at its ends. Minor axis buckling is prevented by lateral restraints. The inter-storey height is equal to 3,50 m. The column is loaded by a vertical single load with an eccentricity to the major axis. N Ed Sd e e Load 6 l Structure Simply supported column, length between supports: l = 3,50 m Eccentricity of the load: e = 00 mm actions Permanent and variable actions result in a vertical design compression force equal to: NN Ed = 10 kn Structural analysis Maximum bending moment occurs at the top of the column: MM y,max Ed = 10 0, 0 = 4 knm cross-section properties Try a doubly-symmetric welded I-section 00 00, thickness = 6,0 mm, austenitic grade Geometric properties b = 00 mm t f = 6,0 mm W el,y = 59,1 cm 3 h w = 188 mm t w = 6,0 mm W pl,y = 85,8 cm 3 a = 3,0 mm (weld thickness) I y = 591,1 cm 4 A g = 35,3 cm² i y = 8,6 cm 175

12 Design Example Sheet of 4 Material properties f y = 0 N/mm (for hot-rolled strip). Table. E = N/mm and G = N/mm Section.3 Classification of the cross-section Table 5. = 1,01 Web subject to compression: ( ) cc/tt = = 30,3 6 For Class 1, cc/tt 33,0ε, therefore web is Class 1. Outstand flange subject to compression: cc/tt = (00/ 6/ 3) 6 = 94/6 = 15,7 For Class 3, cc/tt 14,0ε, therefore outstand flange is Class 4. Therefore, overall classification of cross-section is Class 4. effective section properties Web is fully effective; calculate the reduction factor for welded outstand elements: 1 ρρ = 0,188 but 1 Eq. 5. λλ p λλ p λλ p = bb tt 8,4εε kk σ where bb = cc = 94 mm Eq. 5.3 Assuming a uniform stress distribution within the compression flange: ψψ = σ σ 1 = 1 kk σ = 0,43 Table 5.4 λ p = ρρ 94/6 8,4 1,01 0,43 = 0,833 = 1 0,188 = 1 λ pp 0,833 0,188 0,833 λ pp = 0,93 bb eff = 0,93 94 = 87,4 mm Calculate the effective cross-section for compression only: AA eff = AA g 4(1 ρ)cccc = 35,3 4 (1 0,93) = 33,7 cm Calculate the effective cross-section for major axis bending: AA eff = AA g (1 ρ)cccc = 35,3 (1 0,93) = 34,5 cm 176

13 Design Example Sheet 3 of 4 Taking area moments about the neutral axis of the gross cross-section, calculate the shift in the position of the neutral axis: zz = (1 ρ)cccc(h w + tt f )/ AA eff = (1 0,93) 94 6 ( )/ 34,5 10 =, mm shifted in the direction away from the compression flange Calculate the effective second moment of inertia for major axis bending: II eff,y = II y (1 ρ)cccc [ tt 1 + (h w + tt f ) ] zz AA 4 eff II eff,y = 591,1 (1 0,93) 94 6 [ 6 1 II eff,y = 515,1 cm 4 II eff,y WW eff,y = = h w + tt f + zz 515,1 3 = 46,1 cm 18,8 + 0,6 + 0, ( ) + ] 10 4 (,) 34, resistance to major axis flexural buckling NN b,rd = χaa eff ff y /γγ M1 Eq. 6.3 AA eff = 33,7 cm for Class 4 cross-section subject to compression 1 = φφ + [φφ λλ ] 0,5 1 Eq. 6.4 = 0,5(1 + α(λλ λλ 0) + λλ ) Eq. 6.5 λλ LL cr NN cr = AA effff y NN cr Eq. 6.7 = 350 cm (buckling length is equal to actual length) = π EEEE LL cr = π , = 4175, kn λλ = 33, , 10 3 = 0,41 Using imperfection factor α = 0,49 and λλ 0 = 0, for welded open sections, buckling about the major axis: = 0,5 (1 + 0,49 (0,41 0,) + 0,41 ) = 0,643 1 = 0,643 + [0,643 0,41 = 0,886 ] 0,5 NN b,rd,y = 0,886 33, / 1,1 = 597,3 kn Table

14 Design Example Sheet 4 of 4 resistance to axial compression and uniaxial major axis moment NN Ed MM y,ed + NN Ed ee Ny + kk y 1 Eq (NN b,rd ) ββ min W,y WW pl,y ff y /γγ M1 ββ W,y = WW eff /WW pl,y for a Class 4 cross -section = 46,1/85,8 = 0,861 ee Ny is zero, due to the symmetry of the cross-section kk y = 1,0 + (λλ y 0,5) NN Ed = 1,0 + (0,41 0,5) 10,0 NN b,rd,y 597,3 = 0,968 Eq , + NN Ed NN b,rd,y = 1, ,3 = 1,60 but 1, kk y 1,60 kk y = 1, 10,0 597,3 + 1, 4, ,861 85, /1,1 = 0,786 1 Thus the member has adequate resistance. 178

15 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Design Example 3 Design of a two-span trapezoidal roof sheeting Research Fund for Coal and Steel Sheet 1 of 7 Made by AAT Date 06/0 Revised by JBL Date 04/06 Revised by SJ Date 04/17 design example 3 design of a two-span trapezoidal roof SHeetInG This example considers the design of a two-span trapezoidal sheeting. The material is ferritic grade stainless steel and the material thickness is 0,6 mm. The dimensions of the cross-section are shown below. 4 x 1,5 = The example shows the following design tasks: - determination of effective section properties at the ultimate limit state; - determination of the bending resistance of the section; - determination of the resistance at the intermediate support; - determination of deflections at serviceability limit state. design data Spans L = 900 mm Width supports s s = 100 mm Design load Q = 1,4 kn/m Self-weight G = 0,07 kn/m Design thickness t = 0,6 mm Yield strength f y = 80 N/mm Table. Modulus of elasticity E = N/mm Section.3.1 Partial safety factor M0 = 1,1 Table 4.1 Partial safety factor M1 = 1,1 Table 4.1 Load factor G = 1,35 Section 4.3 Load factor Q = 1,5 Section 4.3 Symbols and detailed dimensions used in calculations are shown in the figure below. The position of the cross-section is given so that in bending at the support the upper flange is compressed. 179

16 Design Example 3 Sheet of 7 h su b u0 / b su / b su0 / h 0 b sl0 / h sl Mid line dimensions: h 0 = 70 mm ww 0 = 1,5 mm bb u0 = 65 mm bb ll0 = 57 mm bb su = 0 mm bb su0 = 8 mm h su = 6 mm bb sll = 0 mm bb s10 = 8 mm h sll = 6 mm rr = mm (internal radius of the corners) b sl / w 0 / b l0 / Angle of the web: h 0 θθ = atan 0,5(ww 0 bb u0 bb ll0 ) = atan 70 0,5 (1, ) = 57,1 effective section properties at the ultimate limit state (uls) Section 5. Check on maximum width to the thickness ratios: h 0 /tt = 70/0,6 = sinθ = 336 Table 5.1 max(bb ll0 /tt ; bb uu0 /tt) = bb uu0 /tt = 65/0,6 = Table 5.1 Angle of the web and corner radius: 45 θθ = 57,1 90 bb p = bb u0 bb su = 65 0 =,5 mm The influence of rounded corners on cross-section resistance may be neglected if the internal radius rr 5tt and rr 0,10bb p rr = mm min(5tt ; 0,1bb pp ) = min(5 0,6; 0,1,5) =,5 mm Section 5.6. The influence of rounded corners on cross-section resistance may be neglected. Location of the centroidal axis when the web is fully effective Calculate reduction factor for effective width of the compressed flange: ρρ = 0,77 0,079 but 1 Section λλ p λλ p Eq. 5.1 where bb tt λλ p = Eq ,4εε kk σσ bb = bb p =,5 mm = 1 kk = 4 Table

17 Design Example 3 Sheet 3 of 7 εε = [ 35 ff y 0,5 EE ] = [ 35 0, ] = 0,894 Table 5.,5/0,6 λλ p = 8,4 0,894 4 = 0,738 Eq. 5.3 ρρ = 0,77 λλ p 0,079 = λλ p 0,77 0,738 0,079 0, 738 = 0,901 1 bb eff,u = ρρbb = 0,901,5 = 0,3 mm Table 5.3 Effective stiffener properties Section tt su = bb h su + ( su bb su0 ) h su tt = 6 + ( 0 8 ) 6 0,6 = 0,849 mm AA s = (bb eff,u + bb su0 )tt + h su tt su = (0,3 + 8) 0, ,849 = 7, mm Fig. 5.3 ee s = bb su0h su tt + h su h su tt su AA s = 8 6 0, ,849 7, =,18 mm II s = (15tt ee s ) + bb su0 tt(h ssss ee s ) + h su tt su ( h su ee s) + ( 15tt4 1 ) + bb su0tt tt 3 suh su 1 II s = (15 0, 6, 18 ) + 8 0,6 (6,18) + 6 0, 849 ( 6,18) Fig. 5.3 bb su bb su0 bb s = h ssss + ( ) 15 0, , 63 0, ( ) + + = 159,5 mm bb su0 = 6 + ( 0 8 ) + 8 = 5,0 mm ll b = 3,07 [II s bb p ( bb 1 4 p + 3bb s tt 3 )] Eq /4 ll b = 3,07 [159,5, 5, ( 0, 6 3 )] = 51 mm ss w = ( ww 0 bb u0 bb ll0 ) + h 0 1, = ( ) + 70 = 83,4 mm Fig. 5.5 bb d = bb p + bb s =,5 + 5 = 70 mm 181

18 Design Example 3 Sheet 4 of 7 kk w0 = ss w + bb d 83, = ss w + 0,5bb d 83,4 + 0,5 70 = 1,37 Eq ll b = 51 ss w 83,4 = 3,01 kk w = kk w0 = 1,37 Eq. 5.8 σσ cr,s = 4,kk wee AA s II s tt 3 4bb p (bb p + 3bb s ) Eq. 5.4 σσ cr,s = 4, 1, , 159,5 0, 6 3 4, 5 (, ) = 503,4 N/mm λλ d = ff y σσ cr,s = ,4 = 0,746 0,65 < λλ d = 0,746 < 1,38 Eq χ d = 1,47 0,73λλ d = 1,47 0,73 0,746 = 0,93 tt red,u = χ d tt = 0,93 0,6 = 0,558 mm The distance of neutral axis from the compressed flange: tt sl = h sl + ( bb sl bb sll0 ) tt = h sl 6 + ( 0 8 ) 6 tt w = tt/sinθθ = 0,6/sin(57,1 ) = 0,714 mm 0,6 = 0,849 mm ee ii [mm] AA ii [mm ] 0 0,5bb eff,u tt = 6,1 0 0,5bb eff,u χχ d tt = 5,66 0,5h su = 3 h su χχ d tt su = 4,74 h su = 6 0,5bb su0 χχ d tt =,3 0,5h 0 = 35 h 0 tt w = 49,98 h 0 = 70 0,5(bb ll0 bb ssss ) tt = 11,1 h 0 0,5h sll = 67 h sll tt sll = 5,09 h 0 h sll = 64 0,5bb sll0 tt =,4 AA tot = AA ii = 87,3 mm ee c = AA iiee ii AA tot = 34,9 mm Effective cross-section of the compression zone of the web EE 00 EN ss eff,1 = ss eff,0 = 0,76tt = 0,76 0,6 γγ M0 σσ com,ed 1, = 11,6 mm Clause (4-5) ss eff,n = 1,5ss eff,0 = 1,5 11,6 = 17,4 mm 18

19 Design Example 3 Sheet 5 of 7 Effective cross-section properties per half corrugation h eff,1 = ss eff,1 sinθθ = 11,6 sin(57,1 ) = 9,74 mm h eff,nn = ss eff,nn sinθθ = 17,4 sin(57,1 ) = 14,61 mm ee eeeeee,ii [mmmm] AA eeeeee,ii [mmmm ] II eeeeee,ii [mmmm 44 ] 0 0,5bb eff,u tt = 6, ,5bb eff,u χχ dd tt = 5,7 0 0,5h su = 3 h su χχ dd tt su = 4,7 χχ dd tt su h su 3 /1 = 14, h su = 6 0,5bb su0 χχ dd tt =, 0 0,5h eff,1 = 4,9 h eff,1 tt w = 7,0 tt w h eff,1 3 /1 = 55,0 h 0 0,5(h 0 ee c + h eff,nn ) = = 45,1 (h 0 ee c + h eff,nn ) tt ww = = 35,5 h 0 = 70 0,5(bb ll0 bb sll ) tt = 11,1 0 (h 0 ee c + h eff,nn ) 3 tt w = 1 = 7308,8 h 0 0,5h sl = 67 h sl tt sl = 5,1 tt sl h sl 3 /1 = 15,3 h 0 h sl = 64 0,5bb sl0 tt =,4 0 AA tot = AA eff,i = 79,8 mm ee c = AA eff,i ee eff,i AA tot = 36,8 mm II tot = II eff,i + AA eff,i (ee c e eff,i ) = 7 393, , = ,5 mm Optionally the effective section properties may also be redefined iteratively based on the location of the effective centroidal axis. EN Bending strength per unit width (1 m) Section II = II 0,5w tot = ,5 = ,5 mm4 0 0,5 1,5 WW u = II ee c = WW l = ,5 36,8 II ,5 = h 0 ee c 70 36,8 = ,0 mm 3 = 16 74,9 mm3 Because WW u < WW l WW eff,min = WW u = ,0 mm 3 MM c,rd = WW eff,min ff y γ M0 = , ,1 = 3,84 knm Eq determination of the resistance at the intermediate support Section Web crippling strength cc 40 mm rr/tt = /0,6 = 3,33 10 EN h w /tt = 70/0,6 = sinθθ = 00sin(57,1 ) = 168 Clause

20 Design Example 3 Sheet 6 of 7 45 θθ = 57,1 90 β V = 0 0, α = 0,15 (category ) ll aa = ss ss = 100 mm RR w,rd = α tt ff y EE (1 0,1 rr tt ) (0,5 + 0,0 ll a tt ) [,4 + ( φφ 90 ) ] EN γ M1 0,5w 0 Eq RR w,rd = 0,15 0, (1 0,1 100 ) (0,5 + 0,0 0,6 0,6 ) [,4 + ( 57,1 90 ) ] 1 1, ,5 1, = 18,4 kn combined bending moment and support reaction Factored actions per unit width (1 m): qq = γγ G GG + γγ Q QQ = 1,35 0,07 + 1,5 1,4 =,19 kn/m MM Ed = qqll 8,19, 9 = 8 =,30 knm FF Ed = 5 4 qqqq = 5,19,9 = 7,94 kn 4 MM Ed =,30 MM c,rd 3,84 = 0,599 1,0 FF Ed = 7,94 RR w,rd 18,4 = 0,43 1,0 MM Ed MM c,rd + FF Ed RR w,rd = 0, ,43 = 1,031 1,5 EN Eq. 6.8a - c Cross-section resistance satisfies the conditions. determination of deflections at serviceability limit state (SLS) Effective cross-section properties For serviceability verification the effective width of compression elements should be based on the compressive stress in element under the SLS loading. A conservative approximation is made for the maximum compressive stress in the effective section at SLS based on W u determined above for ULS. MM y,ed,ser = (GG + QQ)LL 8 σσ com,ed,ser = MM y,ed,ser WW u = = (0,07 + 1,4), 9 = 1,55 knm 8 1, = 10,6 N/mm The effective section properties are determined as before at ULS except that f y is replaced by σσ com,ed,ser and the thickness of the flange stiffener is not reduced. The results of the calculation are: Effective width of the compressed flange: Location of the centroidal axis when the web is fully effective: Effective cross-section of the compression zone of the web: Effective cross-section properties per half corrugation: The flange is fully effective. e c = 34,48 mm The web is fully effective. EN Clause

21 Design Example 3 Sheet 7 of 7 Effective section properties per unit width (1 m): A tot = 88,41 mm e c = 34,48 mm I tot = 63759,0 mm 4 I = ,7 mm 4 W u = 17403,8 mm 4 W l = 16894,3 mm 4 Determination of deflection Secant modulus of elasticity corresponding to maximum value of the bending moment: 1,Ed,ser = MM y,ed,ser WW u =,Ed,ser = MM y,ed,ser WW l = 1, ,8 1, ,3 = 89,06 N/mm = 91,75 N/mm nn = 14 (for ferritic grade stainless steel) Table 6.4 EE 00 EE S,1 = EE 1 + 0,00 ( σσ nn = 1,Ed,ser ) σσ 1,Ed,ser ff 1 + 0, = 00,0 kn/mm y 0,089 (0,089 0,8 ) Eq EE EE S, = EE 1 + 0,00 ( σσ,ed,ser σσ,ed,ser EE S = EE S,1 + EE S, = nn = ) ff y , ,09 (0,09 0,8 ) = 00,0 kn/mm Eq = 00 kn/mm Eq. 6.5 There is no effect of material non-linearity on deflection for the given steel grade and stress level. Check of deflection: For cross-section stiffness properties the influence of rounded corners should be taken into account. The influence is considered by the following approximation: δ = 0,43 nn j=1 rr j φ jj mm bb p,i i=1 90 o = 0,43 94,oo 90 oo 149,3 = 0,019 Eq. 5. II y,r = II (1 - ) = ,7 (1-0,019) = 57781,5 mm 4 Eq. 5.0 For the location of maximum deflection: xx = δδ = δδ = LL = ,9 = 1, m 16 (GG + QQ)LL4 ( xx 48EE S II y,r LL 3 xx3 LL 3 + xx4 LL 4) (0,07 + 1,4) 10 3, ,5 10 δδ = 4,64 mm 1 (1,48,9 1, 483 1, , 93, 9 4 ) The permissible deflection is L/00 = 900/00 = 14,5 mm > 4,64 mm, hence the calculated deflection is acceptable. 185

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23 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of Design Example 4 Fatigue strength of a welded hollow section joint Research Fund for Coal and Steel Made by AAAT Date 06/0 Revised by MEB Date 04/06 Revised by UDE Date 01/17 design example 4 fatigue StrenGtH of a welded HoLLow SectIon JoInt This example considers the fatigue strength of the chord of a welded hollow section joint. Fatigue should be considered in the design of stainless steel structures which are subjected to repeated fluctuations of stresses, e.g. in oil platforms, masts, chimneys, bridges, cranes and transport equipment. Section 9 EN is applicable for estimating the fatigue strength of austenitic and duplex stainless steel structures. The example shows the following design tasks for fatigue assessment: - determination of the fatigue strength curve - determination of secondary bending moments in the joint - determination of the partial safety factor for fatigue strength - fatigue assessment for variable amplitude loading. The chords of the joint are RHS and braces are RHS The material is austenitic grade stainless steel with 0,% proof stress of 10 N/mm. Table. 30x30x x30x 50x50x4 actions The fatigue stress spectra for the chord during the required design life is: Nominal stress range: Number of cycles: Δ 1 = 100 N/mm Δ = 70 N/mm Δ 3 = 40 N/mm n 1 = n = n 3 = Structural analysis The detail category of the joint depends on the dimensions of chord and braces. In this example b 0 = 50 mm, b i = 30 mm, t 0 = 4 mm and t i = mm. Because t 0/t i =, the detail category is 71. Because 0,5(b 0 - b i) = 10 mm, g = 11 mm, 1,1(b 0 - b i) = mm and t 0 = 8 mm, the joint also satisfies the conditions 0,5(b 0 - b i) g 1,1(b 0 - b i) and g t 0. All subsequent references to EN Table

24 Design Example 4 Sheet of effect of secondary bending moments in the joint The effects of secondary bending moments are taken into account by multiplying the stress ranges due to axial member forces by coefficient k 1 = 1,5. Partial factors When it is assumed that the structure is damage tolerant and the consequence of failure is low, the recommended value for the partial factor for fatigue strength is Mf = 1,0. The partial safety factor for loading is Ff = 1,0. fatigue assessment Reference stress range corresponding to 10 6 stress fluctuations for detail category 71 is c = 71 N/mm. The fatigue strength curve for lattice girders has a constant slope m = 5. The number of stress fluctuations corresponding to the nominal stress range Δ i is: N i 10 6 c Mf Ff k1i m Clause 4 (), Table 4. Clause 3 (7), Table 3.1 Figure 7.1 Δ 1 = 100 N/mm Δ = 70 N/mm Δ 3 = 40 N/mm N 1 = 47, N = N 3 = Palmgren-Miner rule of cumulative damage Partial damage because of n i cycles of stress range Δ i: D d,i = n Ei / N Ri Therefore, for Δ 1 = 100 N/mm Δ = 70 N/mm Δ 3 = 40 N/mm D d,1 = 0,1 D d, = 0,35 D d,3 = 0, A.5 (1) The cumulative damage during the design life is: D n n Ei d Dd,i i NRi 0,78 1,0 Because the cumulative damage is less than unity, the calculated design life of the chord exceeds the required design life. Eq. A.1 Clause 8 (4) The procedure described above shall also be repeated for the brace. 188

25 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Design Example 5 Welded joint Research Fund for Coal and Steel Sheet 1 of 7 Made by IR Date 08/0 Revised by MEB Date 04/06 Revised by UDE Date 01/17 design example 5 welded JoInt The joint configuration and its loading are shown in the figure below. Noting that there are two identical plane fillet weld joints of constant throat thickness sharing the applied loading, the required throat thickness for the welds shall be determined. Right angle (equal leg) welds will be used throughout. Fillet welds : throat size a throughout c a 300 n x e yc n z = 300 kn n y = 30 kn e zc = y axis c n c n 1 x axis 175 c : centre of gravity of a weld joint 300 n z Plan n x = - 0 kn y axis for joint n n y = 30 kn y axis for joint n z axis Elevation Material properties Use austenitic grade : f y = 0 N/mm, f u = 530 N/mm, E = N/mm and G = N/mm. It is assumed that the yield and ultimate tensile strength of the weld exceed those of the parent metal. Partial factor The partial factor on weld resistance is M = 1,5. The need to include a reduction factor on the weld resistance to account for its length will also be examined. analysis of welded joints An elastic analysis approach is used here for designing the right-angle equal-leg fillet weld for the load case indicated above which leads to a conservative estimate of the joint resistance. The coordinates (x c, y c,, z c) of a point on the welded joint are taken with reference to a right hand axis system with an origin at the centre of gravity of the welded joint (In the present case the joint is taken to be in the y-z plane so that x c = 0 throughout.). The main purpose of the elastic analysis is to determine the design forces in the weld at the most severely loaded point or points of the welded joint, often referred to as the critical points. For the welded joint being examined the critical point can be taken as being the point furthest from the centre of gravity of the joint. Table. Section.3.1 Section Table 4.1 EN clause.5 189

26 Design Example 5 Sheet of 7 The vectors of the applied force, its eccentricity and the resulting moments acting on a welded joint of general form and centre of gravity C can be expressed as follows: Applied force N N, N, N w,ed x,ed y,ed z,ed Eccentricity of the applied force e e, e, e N xc yc zc (these are the coordinates of the point of application of the force) Applied moments M e N e N xc,ed yc z,ed zc y,ed M e N e N yc,ed zc x,ed xc z,ed M e N e N zc,ed xc y,ed yc x,ed A linear elastic analysis of the joint for a general load case leads to the following force components per unit length of weld at a point with coordinates (x c, y c,, z c), where the throat thickness is denoted by a: F F F wx,ed wy,ed N zm ym a Aw Iyc Izc N xm zm a Aw Izc Ixc N x,ed c yc,ed c zc,ed y,ed c zc,ed c xc,ed z,ed c xc,ed c yc,ed wz,ed a Aw Ixc Iyc ym xm In the above expressions, the resisting sectional throat area and the second moment of area about the principal axes of the welded joint are: Aw adl aii l for a weld of straight segments of length l i and throat thickness a i, I a y z dl xc c c yc c c I a x z dl I a x y dl zc c c Assuming that all welds have equal thickness a: Aw dl li a and since x c = 0: Izc yc dl a Iyc zc dl a I I xc yc Izc yc zcdl a a a 190

27 Design Example 5 Sheet 3 of 7 design of fillet welds Two different design methods are allowed for designing fillet welds and thus to determine the required weld throat thickness at the critical point: The first procedure is based on the simplified and more conservative design shear strength for a fillet weld. The design shear force per unit length of the weld at any point of the joint is defined as the vector sum of the design forces per unit length due to all forces and moments transmitted by the welded joint. This design shear force per unit length should not exceed the design resistance per unit length which is taken as the design shear strength multiplied by the throat thickness. This approach ignores the throat plane orientation to the direction of resultant weld force per unit length. The second procedure is based on comparing the basic design strength of the weaker part joined to the applied design weld stress in the weld throat determined by a von Mises type of formula. This approach is the most precise as it allows for the throat plane orientation to the direction of resultant weld force per unit length. Section Simplified method The design resistance check of the fillet weld is as follows: fu 3 EN Fw.Ed Fwx,Ed Fwy,Ed Fwz,Ed Fw,Rd afvw,d a clause w M where: f vw,d is the design shear strength of the weld, F w,rd is the design (shear) resistance per unit length of weld of throat thickness a. For stainless steel, w is taken as 1,0. Section 7.4. When the design procedure requires that a suitable throat thickness be obtained, the design expression becomes: Fw,Ed a f vw,d. Directional method In the directional method, the forces transmitted by a weld are resolved into normal stresses and shear stresses with respect to the throat section (see Fig. 4.5 in EN ): A normal stress perpendicular to the throat section, A shear stress acting in the throat section parallel to the axis of the weld, A shear stress acting in the throat section transverse to the axis of the weld. The normal stress parallel to the weld axis does not need to be considered. For the combination of stresses,, and, the design requirement is: fu 0,9 fu 3( ) and w M M Eqs and 7.15 For the present case of a plane fillet weld joint with right angle (equal leg) welds this latter check is not critical. However, it may be critical for partial penetration welds in bevelled joints. Instead of having to calculate the stress components in the weld throat the following design check expression may be used for y-z plane joints with right angle (equal leg) welds: F F F F Cos F Sin F F Sin F F Cos w,x w,y w,z w,y w,z w,x w,y w,x w,z f u Fw,yFw,z a wm Sin Cos Note: The subscripts have been shortened: F w,x for F wx,ed etc. 191

28 Design Example 5 Sheet 4 of 7 In the above expression, the angle is that between the y-axis and the axis of the weld as shown below: attached element F w,x Section 1 1 support attached element 1 1 F w,z z F w,y y attached element Fillet weld axis F w,z z F w,y y F w,x attached element support Section The force components at the critical point of the weld are determined in the Appendix to this design example. 1. Simplified method The design shear strength for the simplified design approach is: f u vw,d w f M , 0 1, N/mm² The value of the resultant induced force per unit length in a weld throat of 1 mm is : F F F F w,ed wx,ed wy,ed wz,ed N/mm The required throat thickness is therefore: a F w,ed f vw,d 145 5,0 mm 45 EN : Eq directional method At the point (a), where the angle is 0, the design check expression becomes: f u Fwx,Ed 3Fwy,Ed Fwz,Ed Fwx,EdFwz,Ed a M The required throat thickness is therefore: a ( 43) 3 (747) (966) ( 43) (966) 530 /1,5 4,8 mm Adopt a 5 mm throat thickness and assume that the weld is full size over its entire length. Note: A reduction factor is required for splice joints when the effective length of fillet weld is greater than 150a. The reduction factor would seem to be less relevant for the present type of joint. Nevertheless, by considering safely the full length of the welded joint and a throat thickness of 5 mm one obtains: LW.1 0,Lj 0, 600 1, 1, 1, 04 1, 0 150a 1505 Take LW.1 = 1,0. It is concluded that the use of a reduction factor on the design strength of the weld is not required. EN Eq

29 Design Example 5 Sheet 5 of 7 Appendix Calculation of the force components at the critical point of the weld Geometric properties of the welded joint There are two similar joints, one on each side of the column, resisting the applied loads. Only one of the joints needs to be examined. Throat area and the positions of the centre of gravity and the critical point The throat area (resisting section) of each of the joints made up of straight segments of length L i and constant throat thickness a is for each 1 mm of throat thickness: A a ds A al a a a a w w, i w, i i mm²/m L Distance of the centre of gravity from the vertical side (parallel to the z axis) of the joint with constant throat thickness a: Aw, i yi yl i i 87, y a 51 mm Aw, i L 600 i a 51 y ca = = +14 e yc Load point 15 r c,a a e zc z ca = 15 c y-y 15 d 175 z-z The coordinates of the critical point of the joint (a) relative to the principal axes through the centre of gravity C are: y mm z 15mm ca ca Note: The point (d) might also be chosen as a potential critical point, for which: y ca mm z 15mm ca However, for the load case considered it is evident that the point (a) is the most critical. Second moment of area of the joint resisting section For each of the joints, for each 1 mm of throat thickness: I ,77 10 mm /mm 1 3 yc 6 4 zc ds a Izc a yc ds ,5 51,0110 mm /mm 1 For the torsion moment the relevant inertia per joint is: I a r ds a y ds a z ds I I xc c c c zc yc so that: 193

30 Design Example 5 Sheet 6 of 7 I xc (6,77,01) , mm 4 /mm a Applied forces and moments It is assumed that the applied loads and moments are shared equally by both joints. The applied axial and shear force components per joint are: N x,ed 0 10 kn N y,ed kn N z,ed kn The applied moments are calculated by using the applied force components and their eccentricities. The eccentricities, i.e. the coordinates of the effective load point, are: e xc = 0 as the effective load point is taken to be in the y-z plane of the joint e yc = = + 34 mm e zc = 140 mm As a result the applied moments per joint are: Mxc,Ed eyc Nz,Ed ezc Ny,Ed ,7 knm Myc,Ed ezc Nx,Ed exc Nz,Ed ,4 knm Mzc,Ed exc Ny,Ed eyc Nx,Ed ,4 knm force components at the critical point of the weld For the y-z plane joint the force components per unit length of the weld, at the point (a) are as follows: F F F wx,ed wy,ed wz,ed N z M y M A a I a I a x,ed ca yc,ed ca zc,ed w yc zc N z M A a I a w w y,ed ca xc,ed xc N y M A a/ I a zc,ed ca xc,ed xc The contributions to the weld force components (at all points of the welded joint) from the applied force components are: F F F N N/mm a 600 N x x,ed w,x Aw Ny w,y N z wz, Ny,Ed 15 5 N/mm A a 600 N w z,ed A w a N/mm The various contributions to the weld force components per unit length of weld at the point (a) from the applied moment components are : 194

31 Design Example 5 Sheet 7 of 7 F F F F z 15 M 50, N/mm Mxc c,a 6 w,y xc,ed 6 Ixc a 8,7810 y 14 M 50, N/mm Mxc c,a 6 w,z xc,ed 6 Ixc a 8,7810 z 15 1, N/mm Myc c,a 6 w,x Myc,Ed 6 Iyc a 6,7710 y 14 M 3, N/mm Mzc c,a 6 w,x zc,ed 6 Izc a,0110 Combining the contributions from the forces and the moments at the point (a) one obtains: N M x yc Mzc wx,ed w,x w,x w,x N/mm F F F F Ny Mxc wy,ed w,y w,y N/mm F F F F F F N z M xc wz,ed w,z w,z N/mm These resultant design force components per unit length apply to a welded joint with a weld throat thickness of 1 mm throughout its entire effective length. 195

32 196

33 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Design Example 6 Bolted joint Research Fund for Coal and Steel Sheet 1 of 6 Made by IR Date 10/0 Revised by MEB Date 04/06 Revised by UDE Date 01/17 design example 6 BoLted JoInt An angle ( ) loaded in tension is to be connected to a gusset plate of 10 mm thickness. Stainless steel austenitic grade is used for the angle and the gusset plate. Eight bolts made of austenitic stainless steel, property class 50 with a diameter of 16 mm are used in a staggered line to connect one leg of the angle to the gusset plate. It is required to determine the design resistance of the joint. 100x100x10 angle 10mm thick gusset plate M16 property class 50 bolts 18 mm diameter holes x30 The connection is a Category A: Bearing Type connection. The design ultimate shear load should not exceed the design shear resistance nor the design bearing resistance. EN Clause Material properties Both the angle and the plate are made of stainless steel austenitic grade : Table. f y = 0 N/mm and f u = 530 N/mm Section.3.1 The bolt material is of property class 50: f yb = 10 N/mm and f ub = 500 N/mm. Table.6 Partial factors Partial factor on gross section resistance: M0 = M1 = 1,1 Table 4.1 Partial factor on net section resistance: M = 1,5 Partial factor on bolt resistance in shear and in bearing: M = 1,5 Position and size of holes Section 7..3 For M16 bolts a hole diameter d 0 = 18 mm is required. End distances e 1 = 30 mm and edge distance e = 5 mm. e 1 and e < 4t + 40 = = 80 mm and > 1,d 0 = 1, 18 = 1,6 mm 197

34 Design Example 6 Sheet of 6 For the staggered bolt rows: - spacing p 1 = 60 mm >,d 0 = 39,6 mm - distance between two bolts in staggered row: ,1 mm,4d0 43, mm - therefore, spacing for staggered rows p = 35 mm > 1,d 0 = 1,6 mm Note: For compression loading e and p 1 should be checked so that they satisfy local buckling requirements for an outstand element and an internal element respectively. Checks on both the angle and gusset plate are required. design resistance of the angle gross cross-section in tension Section 7..3 Gross cross-sectional area of the angle A g = 1915 mm Design plastic resistance: N Af kn g y pl,rd 3 M0 1,110 Eq. 7.6 design resistance of the angle net cross-section in tension For staggered holes the net cross-sectional area should be taken as the lesser of the gross area minus the deduction for non-staggered holes or: s Ag t nd0 4 p Deductions for non-staggered holes: Ag td mm Net cross-sectional area through two staggered holes: n =, s = 30 mm and p = 35 mm Section p= 35 x 18 mm holes s = 30 s = 30 s 30 net g ( 18) A A t nd 4 p , mm² Therefore, A net = 1619 mm². Conservatively the reduction factor for an angle connected by one leg with a single row of bolts may be used. By interpolation for more than 3 bolts in one row: 3 = 0,57. Table

35 Design Example 6 Sheet 3 of 6 Design ultimate resistance of the net cross-section of the angle: N A f 0, kn 3 net u u,rd 3 M 1, 510 Section 7..3 Eq design resistance of the angle in block tearing The expressions for block tearing are taken from EN (instead of EN ) since EN explicitly covers angles Design resistance in block tearing considering rows as staggered: V 0,5 fa fa 0,5 530 (60 18) 10 0 ( ) 10 u nt y nv eff,,rd 3 3 M 3 1, 510 M0 3 1, kn Design resistance in block tearing considering rows as if non staggered: V 0,5 fa fa 0,5 530 ( ) 10 0 ( ) 10 u nt y nv eff,,rd 3 3 M 3 1,510 M0 3 1, kn EN Clause 3.10.(3) Eq EN Clause 3.10.(3) Eq design resistance of the gross cross-section of the gusset plate Gross cross-sectional area towards the end of the angle: Section 5.7. A g = 10 ( ) = 400 mm Design plastic resistance N Af g y pl,rd 3 M0 1, kn Eq. 5.3 design resistance of the net cross-section of the gusset plate Net cross-sectional area towards the end of the angle (where the applied load is greatest) through one hole non symmetrically placed on an element of width: Section 5.7. b = = 40 mm A net = A g d 0t = = 0 mm Net cross-sectional area towards the end of the angle through two staggered holes with s = 30 mm and p = 35 mm: st Anet Ag dt p mm² Therefore, A net = 104 mm². 199

36 Design Example 6 Sheet 4 of 6 Design ultimate resistance of the net cross-section of the gusset plate near the end of the angle: N ka f net u u,rd Eq. 5.4 M Take factor k = 1,0 for this example (k = 1,0 for sections with smooth holes) N 1, , 510 u,rd 3 89 kn It is advisable to check the resistance of net cross-sections at intermediate cross-sections along the gusset plate. Cross-section at the 1 st bolt hole near the gusset plate edge (Where b = / = 117,5 mm) A net = A g d 0t = 117, = 995 mm This cross-section must be capable of transmitting the load from one bolt. Design ultimate resistance at the section: kanet fu 1, Nu,Rd 41 kn 3 M 1, 510 It is obvious that there is no need to check any other cross-sections of the gusset plate as the load applied cannot exceed the design resistance of the angle itself, which has been shown to be smaller than the above value. Eq. 5.4 design resistance of the gusset plate in block tearing 3 5 Design resistance to block tearing considering rows as staggered: fa fa u nt y nv Veff,1,Rd M 3 M0 530 (35 9) 10 0 ( ) , , , 398,4 508 kn Design resistance to block tearing considering rows as if non staggered: fa fa u nt y nv Veff,1,Rd M 3 M0 530 (35 9) 10 0 ( ) , ,110 7,1408,8 480 kn 4 0 EN Clause 3.10.() Eq. 3.9 EN Clause 3.10.() Eq

37 Design Example 6 Sheet 5 of 6 design resistance of the bolts in shear Design resistance of class 50 and M16 bolt of sectional area A = A s = 157 mm : F α f A ub v,rd Eq M The value of may be defined in the National Annex. The recommended value is 0,6, which applies if the shear plane passes through the unthreaded or threaded portions of the bolt. F α f A 0, ,7 kn ub v,rd 3 M 1, 510 Design resistance of the bolt group in shear: n b F v,rd = 8 37,7 = 30 kn Section 7..4 design resistance of the bolts/ply in bearing The design resistance for bolted connections susceptible to bearing failure is given by: Section 7..3 F,5α k td f b t u b,rd Eq. 7.1 M Design resistance in bearing on the ply with t = 10 mm for the M16 bolt at the end. End distances e 1 = 30 mm, edge distances e = 5 mm ( > 1,d 0 = 1,6 mm), and bolt spacings p 1 = 60 mm and p = 35 mm. Bolted connections are classified into two groups, based on the thickness of the connected plates. Thick plate connections are those between plates with thicknesses greater than 4 mm, while connections between plates with thicknesses less than or equal to 4 mm are defined as thin plate connections. This example is a thick plate connection with t min = 10 mm and deformation should not be a key design consideration. For the end bolt nearest the ends where e 1 = 30 mm and p 1 = 60 mm the bearing coefficient b in the load transfer direction is determined as follows: 1,0 b min e1 3d 0 1,0 min 30 0,556 0, The bearing coefficient k t in the direction perpendicular to load transfer is determined as follows: e 1,0 for 1,5 d0 kt e 0,8 for 1,5 d 0 e 5 kt 0,8 for 1,39 1,5 d 18 0 Section

38 Design Example 6 Sheet 6 of 6 The design resistance for this bolted connection susceptible to bearing failure for the end bolt is as follows: F,5α k td f,50,556 0, ,44 kn b t u b,rd 3 M 1, 510 Eq. 7.1 Design resistance of the joint in bearing: n b F b,rd = 8 75,44 = 604 kn design resistance of the joint at the ultimate Limit State Design resistance of the angle gross cross-section in tension N pl,rd 383 kn Design resistance of the angle net cross-section in tension N u,rd 391 kn Design resistance of the angle in block tearing (for staggered rows) V eff,,rd 83 kn Design resistance of the angle in block tearing (for non staggered rows) V eff,,rd 74 kn Design resistance of the gusset plate gross cross-section in tension N pl,rd 480 kn Design resistance of the gusset plate net cross-section in tension N u,rd 89 kn Design resistance of the gusset plate net cross-section in tension (at the 1 st bolt hole near the gusset plate edge) N u,rd 41 kn Design resistance of the gusset plate in block tearing (for staggered rows) V eff,1,rd 508 kn Design resistance of the gusset plate in block tearing (for non staggered rows) V eff,1,rd 480 kn Design resistance of the bolts in shear F v,rd 30 kn Design resistance of the bolts/ply in bearing F b,rd 604 kn The smallest design resistance is for the angle in block tearing (for non staggered rows): V eff,,rd = 74 kn Note: The critical mode for all of the bolts in the joint is shear (F v,rd = 30 kn). 0

39 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of 5 Design Example 7 Shear resistance of plate girder Research Fund for Coal and Steel Made by AO Date 06/0 Revised by MEB Date 04/06 Revised by ER/IA Date 04/17 design example 7 SHear resistance of PLate GIrder Design a plate girder with respect to shear resistance. The girder is a simply supported I-section with a span according to the figure below. The top flange is laterally restrained. FEd = 440 kn hw bf Use lean duplex grade f y = 480 N/mm for hot rolled strip E = N/mm Table. Section.3.1 Try a cross section with Flanges: 1 00 mm Web: mm Stiffeners: 1 98 mm Weld throat thickness: 4 mm Structural analysis Maximum shear and bending moment are obtained as FEd 440 VEd 0kN F Ed L 440, 5 MEd 75kNm 4 4 Partial factors M0 = 1,1 Table 4.1 M1 = 1,1 classification of the cross-section Section Table 5. = 0,

40 Design Example 7 Sheet of 5 Web, subject to bending c = ,9 90 therefore the web is Class 4. t 4 0,683 Flange, subject to compression c = ,3 14,0 therefore the compression flange is Class 3 t 1 0,683 Thus, overall classification of cross-section is Class 4. Table 5. Table 5. Shear resistance Section ,3 The shear buckling resistance requires checking when hw / tw k for vertically stiffened webs. a/h w k = = 150/500 =,5 > 1, and since the web is not stiffened, k st=0. Hence, hw 500 5,34 4 5,34 4 5,98 a 150 Eq. 6.6 EN recommended value for 1, Section h w/t w = 500 4,3 15 0,683 5,98 33,8 4 1, Therefore, the shear buckling resistance has to be checked. It is obtained as fyw hw tw 1, V b,rd = V bw,rd + V bf,rd ,6 kn Eq ,1 V bw,rd = f w yw w w 3 h t M1 For non-rigid end posts: hw w = 37,4tw 1,19 w = 0,54 w k M1 500 =,00 0,65 37, 4 4 0,683 5,98 Hence the contribution from the web is obtained as 1,19 0,54,00 0,468 w = M1 Eq. 6.3 Eq. 6.5 for w 0,65 Table 6.3 Table 6.3 f h t V bw,rd = w yw w w 0, = 35,9 kn Eq ,1 The contribution from the flanges may be utilised if the flanges are not fully utilised in withstanding the bending moment. The bending resistance of a cross section consisting of the flanges only is obtained as M f,rd = 100 (500 1) , knm 1,1 M f,rd > M Ed = 75 knm, therefore the flanges can contribute to the shear buckling resistance. Section

41 Design Example 7 Sheet 3 of 5 b f t f f yf V bf,rd = M 1 c M 1 M 3,5bt f c = a 0,17 tw hw fyw Ed f,rd f f yf c but 0, 65 a 3, = 150 0, , 5 mm ,5mm < 0, ,5 mm Eq. 6.9 Eq V bf,rd = 1 7, , 1 536, kn Eq. 6.9 V b,rd = V bw,rd + V bf,rd = 35,8 + 7,4 = 63, kn 604,6 kn Eq. 6. transverse stiffeners Section The transverse stiffeners have to be checked for crushing and flexural buckling using Table 6.1 = 0,49 and 0 = 0,. An effective cross section consisting of the stiffeners and parts of the web is then used. The part of the web included is 11 t w wide, therefore the cross section of the transverse stiffener is Class 3. a / h w 150/500, 5, hence the second moment of area of the intermediate Eq stiffener has to fulfil I 0, 75h t 0, mm Eq st w w 3 (11 0, ) I st = 8, 0010 mm 4, hence fulfilled The crushing resistance is obtained as N c,rd = A g f y/ M0 Eq. 5.7 A g = ( , ) 460, 1 mm N c,rd = 3 460, /1,1 1073,5 kn The flexural buckling resistance is obtained as N b,rd = A f y / M1 Eq. 6. = 1 1 0, 5 =, 5 = 1 Eq Eq L cr 1 i f yw E Eq. 6.6 L cr = 0,75h w = 0, = 375 mm Section = , ,1 6 Eq , 103 0, 48 = 0, 5 1 0, 49 0, 103 0, Eq

42 Design Example 7 Sheet 4 of 5 = 0, 48 0, , 103 0, 5 1, , 0 Since N b,rd = N c,rd =1073,5 kn > N Ed, the transverse stiffeners are sufficient. Eq. 6.4 Interaction shear and bending If the utilization of shear resistance, expressed as the factor 3, exceeds 0,5, the combined effect of bending and shear has to be checked. VEd 3 = 1, 0 V 3 bw,rd 0 35,9 = 0, 933 0, 5, therefore interaction has to be considered. The condition is 1 1, 0 Mf,Rd Mpl, Rd Where: 1 = M M Ed pl, Rd M f,rd = 536, knm (Sheet 3) M pl,rd is the plastic resistance of the cross-section. Section Eq M f,rd for 1 Eq M pl,rd t h f w w y M pl,rd = Mf, Rd 536, 645, 3 knm , 1 10 M0 Eq Evaluate conditions M Ed = 75 knm, hence: 75 1 = 0, 46 1, 0 OK Eq , 3 1 fulfils its condition. Now it remains to check the interaction. 536, 1 0, , , 553 1, 0 M f, Rd M pl,rd 645, 3 It therefore follows that under the conditions given, the resistance of the plate girder is sufficient with respect to shear, bending as well as interaction between shear and bending. Calculation of effective cross-section properties The flanges are Class 3 and hence fully effective. The depth of the web has to be reduced with the reduction factor, welded web. 0, 77 0, 079 = 1 p p Eq. 5.1 p = b / t 8, 4 k where b = d = , 68 mm Eq

43 Design Example 7 Sheet 5 of 5 Assuming linearly varying, symmetric stress distribution within the web, = = 1 1 k = 3,9 Table ,68/ 4 p = 1, 9 8, 4 0, 683 3, 9 Eq , 77 0, 079 = 0, , 9 1, 9 Eq. 5.1 b eff= b c = b / (1-) = 0, ,68/(1 ( 1 )) 134, 76mm Table 5.3 b e1 = 0,4b eff = 0, 4 134,76 53, 9 mm Table 5.3 b e = 0,6b eff = 0, 6 134,76 80, 9 mm Calculation of effective section modulus under bending is taken as positive from the centroid of the upper flange and downwards. e i b e1 b e h w/ Ai bt f f be1 4 tw betw hw / tw 6361,7 mm i A eff = e eff = Ae i i bt 0 bt h t b 4 t 0,5b 4 t A A A f f f f w f e1 w e1 f eff i eff eff I eff = 0.5 / / 0,75 0,5 bt h t b h t h t = 66,4 mm i e w w f e w w w f i i eff i i 3 e1 4 tb / 3 3 bt t f f w b t w e w hw IAe ( e) ,5 4 bt f f eeff bt f f eeff hw tf be1 tw eeff be1 t f betw eeff 0,5 hw tf be hw / tw eeff 0,75hw 0,5tf = 3, mm

44 08

45 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of 5 Design Example 8 Resistance to concentrated loads Research Fund for Coal and Steel Made by AO Date 06/0 Revised by MEB Date 04/06 Revised by ER/IA Date 04/17 design example 8 resistance to concentrated LoadS An existing plate girder, previously subjected to an evenly distributed load, will be refurbished and will be subjected to a concentrated load. Check if the girder can resist the new load applied through a 1 mm thick plate. The girder is a simply supported I-section with a span according to the figure below. The top flange is laterally restrained. F Ed = 110 kn Use duplex grade f y = 460 N/mm for hot rolled strip E = N/mm Table. Section.3.1 Flanges: 1 00 mm Web: mm Stiffeners: 1 98 mm Weld throat thickness: 4 mm Structural analysis Maximum shear and bending moment are obtained as FEd 110 V Ed = 55kN F Ed L 110, 5 M Ed = 68, 75kNm 4 4 Partial safety factors M0 = 1,1 Table 4.1 M1 = 1,1 classification of the cross-section Section = 0, 698 Table

46 Design Example 8 Sheet of 5 Web, subject to bending c = 175, 1 90, therefore the web is Class 4. t 4 0, 698 Flange, subject to compression c = t 1 0, , 0 14, 0, and the compression flange is Class 3. Thus, overall classification of cross-section is Class 4. Table 5. Table 5. resistance to concentrated force Section The design load should not exceed the design resistance, i.e. F Rd = f yw L eff / M 1 Eq t w The effective length L eff is given by L eff = F ly Eq where the reduction function is 0, 5 F = 1. 0 F with the slenderness given by F = l y t w F cr f yw The effective loaded length is given by l y = s t m m s f 1 1 Eq Eq Eq Where s s is the length of the stiff bearing and m 1 and m are dimensionless parameters: m 1 = f f yf yw b t f w hw m = 0, 0 f o r F 0, 5 tf Eq Eq m = 0 f o r F 0, 5 Eq s s is conservatively taken as twice the thickness of the load bearing plate, i.e. 4 mm. Figure m 1 = m = 0, 0 34, 7 1, assuming F 0, 5 Eq Eq l y = , 7 68,9 mm Eq The critical load is obtained as F cr = 3 w t 0, 9 k F E Eq h w where the buckling coefficient is given by the load situation, type a. k F = h 6 w a 500 = 6 6, Figure 6.4

47 Design Example 8 Sheet 3 of 5 F cr = 4 kn Eq ,9 6, ,1 68, ,110 F = 3 1,88 0,5, assumption OK Eq , 5 F = 0, 7 1, 0, OK Eq , 88 L eff = 0,768,9 7,6 mm F Ed = ,6 4 / (1,1 10 ) 11,4 kn Eq Hence the resistance exceeds the load. Interaction between transverse force, bending moment and axial force Interaction between concentrated load and bending moment is checked according to EN :006. 0, 8 Where 1 1, 4 EN , Eq. 7. N Ed M Ed N Ed e N EN , 1 = 1, 0 f A f W / Eq y e f f / M0 y e f f M0 F EN , Ed = 1, 0 f L Eq yw eff t w / M 1 Calculation of effective cross-section properties The flanges are Class 3 and hence fully effective. The depth of the web has to be reduced with the reduction factor, welded web. 0, 77 0, 079 = 1 p p Eq. 5.1 p = b / t 8, 4 k where b = d = , 68 mm Eq. 5.3 Assuming linearly varying symmetric stress distribution within the web, = = 1 1 k = 3,9 Table ,68/ 4 p = 1, 6 8, 4 0, 698 3, 9 0, 77 0, 079 = 0, , 6 1, 6 b eff= b c = b / (1-) = 0,56488,68 / (1 ( 1)) 137,3mm Table 5.3 b e1 = 0,4b eff = 0,4 137,3 54,9 mm b e = 0,6b eff = 0,6 137,3 8,4 mm 11

48 Design Example 8 Sheet 4 of 5 Calculate effective section modulus under bending e i is taken as positive from the centroid of the upper flange and downwards. b e1 b e h w/ Ai bt f f be1 4 tw betw hw / tw 637, mm i A eff = e eff= 0 4 0, Ae bt bt h t b t b t A A A i i f f f f w f e1 w e1 f eff i eff eff 0,5 / / 0,75 0,5 bt h t b h t h t = 66,4 mm e w w f e w w w f 3 e1 4 tb / bt tw b t f f w e w hw I eff = Ii Ae i( eff ei ) i i bt f f eeff 0 bt f f eeff hw tf be1 4 t w eeff 0,5be1 4 tf betw eeff 0,5 hw tf be hw / tw eeff 0,75hw 0,5tf W eff = = 3, mm 4 e eff I eff 0, 5 t f 6 1, 9310 mm ,7510 EN = 0, , 9310 / 1, 1 Eq = 0, ,63 0,8 0,8 0,1930,919 1,01 1,4 1 Therefore, the resistance of the girder to interaction between concentrated load and bending moment is adequate. EN Eq Shear resistance Section The shear buckling resistance requires checking when webs. h w / t w , 0, 698 3, 7 1, h w / t w 56, Therefore the shear buckling resistance has to be checked. It is obtained as for unstiffened Eq

49 Design Example 8 Sheet 5 of 5 V b,rd = V bw,rd fywht w V w bf,rd Eq M1 w f yw h w t w V bw,rd = M1 3 For non-rigid end posts Table 6.3 provides w = w = w = h w 86, 4 t w 1,19 0,54 w 500 =, 07 0, 65 86, 4 4 0, 698 1,19 0,455 0,54,07 Eq. 6.3 Eq. 6.4 for w 0,65 Table 6.3 The contribution from the flanges may be utilised if the flanges are not fully utilised to withstand the bending moment. However, the contribution is small and is conservatively not taken into account, i.e. Vbf,Rd 0. The shear buckling resistance can be calculated as: V b,rd = V bw,rd = 0, ,8 1,1 3 V b,rd = V bw,rd > V Ed = 55 kn The shear resistance of the girder is thus adequate. f ht kn < yw w w 3 M 1 Table , 45 kn Eq. 6.3 Interaction between shear and bending If 3 does not exceed 0,5, the resistance to bending moment and axial force does not need to be reduced to allow for shear. 3 = = V V Ed bw,rd 55 19,8 concluding remarks 1,0 Eq , 5 0,5, therefore interaction need not to be considered. The resistance of the girder exceeds the load imposed. Note that the vertical stiffeners at supports have not been checked. It should be done according to the procedure used in Design Example 7. 13

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51 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Design Example 9 - Beam with unrestrained compression flange Research Fund for Coal and Steel Sheet 1 of 7 Made by SMH Date 09/01 Revised by NRB Date 04/06 Revised by SJR Date 04/17 design example 9 - BeaM with unrestrained compression flange Design a staircase support beam. The beam is a channel, simply supported between columns. The flight of stairs between A and C provides restraint to the top flange of this part of the beam. The top flange is unrestrained between B and C. The overall span of the beam is taken as 4, m. w Beam 3 Down A C B, m R A 1,5 m,7 m restrained unrestrained R B actions Assuming the beam carries the load from the first run of stairs to the landing only: Permanent actions (G): Load on stairs 1,0 kn/m = 1,0, =, kn/m Self-weight of beam 0,13 kn/m Variable actions (Q): Load on stairs 4 kn/m = 4,0, = 8,8 kn/m Load case to be considered (ultimate limit state): G,jGk,j + Q,1 Qk,1 j1 + Q,i0,iQk,i i1 As there is only one variable action (Q k,1 the last term in the above expression does not need to be considered in this example. G, j = 1,35 (unfavourable effects) Q,1 = 1,5 Factored actions Permanent action: Load on stairs = 1,35, =,97 kn/m Self-weight of beam = 1,35 0,13 = 0,18 kn/m Variable action: Load on stairs = 1,5 8,8 = 13, kn/m Structural analysis Reaction at support points: R A + R B = (, ,) 1,5 + 0,18 4, = 5,01 kn 15

52 Design Example 9 Sheet of 7 Taking moments about A: 1, 5 (, 9713, ) 0, 75 0, 18 4, ( 4, / ) R B = = 4,71 kn 4, R A = 5,01 4,71 = 0,30 kn 1,5 Maximum bending moment occurs at a distance: 1,5 1 4, = 1,3 m from A. 1, 3 M Ed,max = 0,30 1, 3 (, 9713, ) Maximum shear occurs at A: F Ed,max = 0,30 kn 1, 3 0, 18 = 1,60 knm Material properties Use austenitic grade ,% proof stress = 40 N/mm (for cold formed steel sheet) Table. f y = 40 N/mm E = N/mm and G = N/mm Section.3.1 Try a channel section, thickness t = 5 mm. cross-section properties I y = 9, mm 4 W el,y = 94, mm 3 I z = 0, mm 4 W pl,y = 11, mm 3 I w = mm 4 A g = 1650 mm I t = 1, mm 4 classification of the cross-section Section 5.3. = 35 E = 0,97 Table 5. f y Assume conservatively that c = h t = 00 5 = 190 mm for web Web subject to bending: c t 5 c For Class 1, 7 69,8, therefore web is Class 1. Table 5. t Outstand flange subject to compression: c t 5 c For Class 3, 14 13,6, therefore outstand flange is Class 4. Table 5. t Therefore, overall classification of cross-section is class 4. 16

53 Design Example 9 Sheet 3 of 7 calculation of effective section properties Section Calculate reduction factor for cold formed outstand elements: 1 0,188 p p but 1 Eq. 5. b/ t p where b = c = 75 mm Eq ,4 k Assuming uniform stress distribution within the compression flange = = 1 k = 0,43 Table 5.4 p 1 75 / 5 0,830 8, 4 0,97 0, ,188 0,93 0,830 0,830 Table 5.4 c eff = c = 0,93 75 = 69,9 A eff = A ct = , mm g 1 Calculate shift of neutral axis of section under bending: Non-effective zone y - y y Centroidal axis of gross cross-section Centroidal axis of effective cross-section y h t 00 5 Ag 1 cth , A 165 eff y 98,44 Shift of neutral axis position, y y = 3 1 I eff,y = ct I 1 ct h t A 1 h 00 y 98, 44 1,56mm y eff y-y 3 10, I eff,y = 9, , ,5 1651,56 1 I eff,y = 9, mm 4 6 W eff,y = I 9, ,69 10 h 00 y-y 1,56 6 eff,y 3 mm 3 17

54 Design Example 9 Sheet 4 of 7 Shear lag Section 5.4. Shear lag may be neglected provided that b 0 L e/50 for outstand elements. L e = 400 mm (distance between points of zero moment) L e/50 = 84 mm, b 0 = 75 mm, therefore shear lag can be neglected. flange curling Section ab EN s u = E t z Clause 5.4 Eq. 5.3a a = 40 N/mm (maximum possible value) b s = 75 5 = 70 mm z = 100,5 = 97,5 mm u = ,5 = 0,08 mm Flange curling can be neglected if u < 0,05 00 = 10 mm Therefore flange curling is negligible. Partial factors The following partial factors are used throughout the design example: M0 = 1,1 and M1 = 1,1 Table 4.1 Moment resistance of cross-section For a class 4 cross-section: M c,rd = Weff,min fy M0 3 90, ,110 M Ed,max = 1,60 knm < M c,rd = 19,79 knm cross-section moment resistance is OK. = 19,79 knm Eq cross-section resistance to shear V pl,rd = 3 A f Eq. 5.3 v y M0 A v = h t = 00 5 = 1000 mm V pl,rd = = 15,97 kn 3 1, F Ed,max = 0,30 knm < V pl,rd = 15,97 knm cross-section shear resistance is OK. 18 Check that shear resistance is not limited by shear buckling: Section Assume that h w = h t = 00 5 = 190 mm h w = 190 t 5 = 38, shear buckling resistance needs to be checked if h w 56, Eq. 6.0 t = 1,0 h w 5 6, = 38 < = t 56, 0, 97 = 45,4 1, 0 shear resistance is not limited by shear buckling.

55 Design Example 9 Sheet 5 of 7 resistance to lateral torsional buckling Section 6.4. Compression flange of beam is laterally unrestrained between B and C. Check this portion of beam for lateral torsional buckling. M b,rd = LTWeff,yfy for a Class 4 cross-section Eq.6.13 M1 3 W eff,y = 90,69 10 mm 3 LT = 1 λ LT LT LT 0,5 LT = 0,5 1 0,4 W f y y LT = M cr 1 LT LT LT Eq.6.14 Eq.6.15 Eq.6.16 Determine the elastic critical moment (M cr): 1/ EI z k I w kl GI t Mcr C 1 Cz g) Cz g kl kw Iz EI z C is simply supported, while B approaches full fixity. Assume most conservative case: k = k w = 1,00 C 1 and C are determined from consideration of bending moment diagram and end conditions. Appendix E Eq. E.1 From bending moment diagram, = 0, C 1 = 1,77 Table E.1 C = 0 (no transverse loading) E.3 M M cr , ,77 1, cr 6 4 1, , , , 00 0, , = 41,9 knm 0,5 LT = 3 90, , = 0,71 Using imperfection factor LT = 0,34 for cold formed sections: Section 6.4. LT = 0,5 1 0,34 0,71 0, 4 0,71 = 0,814 1 LT = = 0,839 0,5 0,814 0,814 0,71 M b,rd = 0,839 90, / 1,1 M b,rd = 16,60 knm < M Ed = 1,0 knm (max moment in unrestrained portion of beam) member has adequate resistance to lateral torsional buckling. 19

56 Design Example 9 Sheet 6 of 7 deflection Section Load case (serviceability limit state): Gk,j Qk,1 0,iQk,i j1 i1 As there is only one variable action (Q k,1), the last term in the above expression does not need to be considered in this example. Secant modulus is used for deflection calculations thus it is necessary to find the maximum stress due to unfactored permanent and variable actions. S1 S The secant modulus E E ES Where E S,i 1 0,00 E E i,ed,ser i,ed,ser f y n and i = 1, From structural analysis calculations the following were found: Maximum moment due to permanent actions= 1,90 knm Maximum moment due to imposed actions Total moment due to unfactored actions = 6,68 knm = 8,58 knm Eq. 6.5 Eq Section is Class 4, therefore W eff is used in the calculations for maximum stress in the member. Assume, conservatively that the stress in the tension and compression flange are approximately equal, i.e. E S1 = E S For austenitic grade stainless steel, n = 7 Table 6.4 Serviceability design stress, M 8, ,6N/mm 6 max i,ed,ser 3 We ff, y 90, ES,i ,6N/mm ,6 10,00 94,6 40 Maximum deflection due to patch loading occurs at a distance of approximately 1,9 m from support A. Deflection at a distance x from support A due to patch load extending a distance a from support A is given by the following formulae: When x a: Where m = x/l and n = a/l wal 4 n 3 m 6 m m (4 n ) n 4aE I S When x = 1,9 m and a = 1,5 m: m = 1,9/4, = 0,45; n = 1,5/4, = 0,357 Steel Designer s Manual (5 th Ed) Patch load (permanent + variable unfactored actions): w = 11,0 kn/m Uniform load (permanent action): w = 0,18 kn/m Deflection due to patch loads at a distance of 1,9 m from support A, 1: , ,6 9, ,357 0, , 45 0, ,357 0,357 = 7,04 mm 0

57 Design Example 9 Sheet 7 of 7 Deflection at midspan due to self weight of beam, 5 ( w LL ) 5 (0,1810 4, ) 400 = EI ,6 9,0610 S Total deflection 1 + = 7,04 + 0,9 = 7,33 mm limiting = L ,8 mm > 7,33 mm deflection is acceptably small. = 0,9 mm (A finite element analysis was carried out on an identical structural arrangement. The total beam deflection at mid-point was 7,307 mm see deformed beam shape with deflections below.) 1

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59 Promotion of new eurocode rules for structural stainless steels (PureSt) Title Sheet 1 of 7 Design Example 10 Axially loaded column in fire Made by SMH Date 08/01 calculation SHeet Client Research Fund for Coal and Steel Revised by MEB Date 04/06 Revised by SA Date 05/17 design example 10 axially Loaded column In fire Design an unprotected cold-formed rectangular hollow section subject to axial load and bending for 30 minutes fire resistance. The column length is,7 m and is subject to axial load from the end reaction of a floor beam at an eccentricity of 90 mm from the narrow face of the column. y h Point of application of load z b z 90 mm y Section A A A A Floor beam Column,7 m actions This eccentricity is taken to be 90 mm + h/, where h is the depth of the section. Thus the beam introduces a bending moment about the column s major axis. The unfactored actions are: Permanent action: 6 kn Variable action: 7 kn The column will initially be checked at the ultimate limit state (LC1) and subsequently at the fire limit state (LC) for fire duration of 30 minutes. The load cases are as follows: 3

60 Design Example 10 Sheet of 7 LC1 (ultimate limit state) G,jGk,j + Q,1 Qk,1 G, j = 1,35 (unfavourable effects) Q,1 = 1,5 LC (fire limit state) GA,jGk,j 1,1Qk,1 GA = 1,0 j j Values for 1, 1 are given in EN 1990 and NA for EN 1990, but for this example conservatively assume 1, 1 = 1,0. design at the ultimate Limit State (Lc1) Loading on the corner column due to shear force at end of beam (LC1): Axial force N Ed = 1, ,5 7 = 18,6 kn Try cold-formed RHS. Major axis bending moment (due to eccentricity of shear force from centroid of column): M y,ed = 18,6 (0,09 + 0,10/) =,60 knm Partial factors The following partial factors are used throughout the design example for LC1: Table 4.1 M0 = 1,10 and M1 = 1,10 Material properties Use austenitic grade f y = 0 N/mm and f u = 530 N/mm (for hot-rolled strip). Table. E = N/mm and G = N/mm Section.3.1 cross-section properties 100 x 50 x 6 mm rhs W el,y = 3, mm 3 i y = 3,9 mm W pl,y = 43, mm 3 i z = 19,1 mm A = 1500 mm t = 6,0 mm cross-section classification Section ,5 0,5 35 E ,01 f y Table 5. For a RHS the compression width c may be taken as h 3t. Table 5. For the web, c = = 8 mm Web subject to compression: ct 8 6 = 13,7 Table 5. Limit for Class 1 web = 33ε = 33,33 Table 5. 33,33 > 13,7 Web is Class 1 By inspection, if the web is Class 1 subject to compression, then the flange will also be Class 1. 4

61 Design Example 10 Sheet 3 of 7 The overall cross-section classification is therefore Class 1 (under pure compression). compression resistance of cross-section Section N N Af y c,rd for Class 1, or 3 cross-sections Eq. 5.7 M0 c,rd ,1 300 kn 300 kn > 18,6 kn acceptable Bending resistance of cross-section Section M M W f pl,y y c,y,rd for Class 1, or 3 cross-sections Eq. 5.9 M0 c,y,rd ,75 knm 1,1 8,75 knm >,60 knm acceptable axial compression and bending resistance of cross-section Section MM y,ed MM N,Rd Eq The following approximation for MM N,y,Rd may be used for RHS: EN MM N,y,Rd = MM pl,y,rd (1 nn)/(1 0,5aa w ) but MM N,y,Rd MM pl,y,rd 1, clause (5) Where AA bbbb aa w = but aa AA w 0, aa w = = 0,6 but aa 1500 w 0,5, therefore aa w = 0,5 nn = NN Ed = 18,6 NN pl,rd 300 = 0,06 MM N,y,Rd = 8,75 ( ,5 0,5 ) = 10,94 MM pl,y,rd = 8,75 Therefore MM N,y,Rd = 8,75 knm, and MM y,ed MM N,Rd Member buckling resistance in compression Section A f for Class 1, or 3 cross-sections Eq. 6. y N b,rd M1 1 0,5 where 0 L cr 0,5 1 1 Eq. 6.4 Eq. 6.5 L i f cr 1 y for Class 1, or 3 cross-sections Eq. 6.6 E = buckling length of column, taken conservatively as 1,0 column length =,7 m 5

62 Design Example 10 Sheet 4 of 7 y z = 0,866 3, = 1,49 19, Buckling curves: major (y-y) axis: For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = 0,49 and = 0, ,5 1 0,49 0,866 0,3 0,866 = 1,014 = 1 y = 0,5 1,014 1,014 0,866 N b,y,rd = 0, ,10 194,70 kn > 18,6 kn acceptable Buckling curves: minor (z-z) axis: = 194,70 kn = 0,649 0,5 1 0,49 1,49 0,3 1,49 = 1,905 1 z = 0,5 1,905 1,905 1, 49 N b,z,rd = 0, ,10 = 97,0 kn 0,34 Table ,0 kn > 18,6 kn acceptable (Resistance to torsional buckling will not be critical for a rectangular hollow section with a h/b ratio of.) Section Member buckling resistance in combined bending and axial compression Section 6.5. N M N e k ( N ) W f / Ed y,ed Ed Ny y b,rd min W,y pl,y y M1 W,y = 1,0 for Class 1 cross-sections NEd y k D D D D D 1 Eq N Ed y 1, Eq Nb,Rd,y Nb,Rd,y From Table 6.6, D 1 =,0 and D = 0,3 and D 3 = 1,3 Table ,6 18,6 ky 1,0 0,866 0,3 1, ,3 0,3 1, ,7 194,7 k y = ,108 0,51 1 acceptable 6 18,6, , 0 1,0 43, / 1,10 6

63 Design Example 10 Sheet 5 of 7 design at the fire Limit State (Lc) For LC, the column is designed for the following axial loads and moments. Axial compressive force N fi,ed = 1, ,0 7 = 13,0 kn Maximum bending moment M y,fi,ed = 13,0 (0,09 + 0,05) = 1,8 knm determine temperature in steel after 30 minutes fire duration Section Assume that the section is unprotected and that there is a uniform temperature distribution within the steel section. The increase in temperature during time interval t is found from: Δθ t = Am V h net,d t Eq c Eq. 8.4 h n e t, d = h net, c h net, r h net,c = cg 8 h = 5, n e t, r res 4 4 g where: g = gas temperature of the environment of the member in fire exposure, given by the nominal temperature time curve: Eq Eq g = log 10(8t + 1) Eq = surface temperature of the member Initial input values for determination of final steel temperature are as follows: A m/v = 00 m -1 c = 5 W/m K Section Initial steel temperature: = 0 C Resultant emissivity: res = 0,4 Section Density of stainless steel: = 8000 kg/m 3 for austenitic grade Table.7 Configuration factor: = 1,0 EN cl. 3.1(7) The specific heat is temperature-dependent and is given by the following expression: c = ,8, , J/kgK Eq t = seconds The above formulae and initial input information were coded in an Excel spreadsheet and the following steel temperature, after a fire duration of 30 minutes, was obtained. = 89 C reduction of mechanical properties at elevated temperature The following reduction factors are required for calculation of resistance at elevated temperatures. Section 8. Young s modulus reduction factor: k E, = E /E Eq ,% proof strength reduction factor: k p0,, = f p0,,/f y Eq. 8.1 Strength at % total strain reduction factor: k, = f,/f y but f, f u, Eq. 8. The values for the reduction factors at 89 C are obtained by linear interpolation: k E, = 0,578 Table 8.1 k p0,, = 0,355 Table 8.1 7

64 Design Example 10 Sheet 6 of 7 k, = 0,430 k u, = 0,97 f, = 0,430 0 = 94,6 and f u, =0, = 157, therefore f, f u, Partial factor Section 8.1 M,fi = 1,0 cross-section classification Section 8.3. Under compression, k y, should be based on f p0,,, i.e. k y, = k p0,, Section 8. 0,5 0,5 ke,θ 0,578 θ k y,θ 0,355 1,01 1, 9 Web subject to compression: ct 8 6 = 13,7 Limit for Class 1 web = 33 εε θ = 4,57 4,57 > 13,7 Web is Class 1 The overall cross-section classification is Class 1 (under pure compression). Eq. 8.6 Member buckling resistance in compression N b,fi,t,rd = Ak fi p0,,θ y M,fi 1 fi = 0,5 θ θ θ where φφ θ = = θ y,θ = z,θ = f for Class 1, and 3 cross-sections Eq θ 0 θ Eq ,5 1 Eq k k p0,,θ E,θ 0,5 0,355 0,866 0,578 0,355 1,49 0,578 0,5 0,5 for all classes of cross-section Eq = 0,679 = 1,169 Buckling curves: major (y-y) axis: For cold-formed austenitic stainless steel hollow sections subject to flexural buckling, = 0,49 and = 0, ,5 1 0,49 0,679 0,3 0,679 = 0,83 φφ θ,y = 1 fi,y = 0,5 0,83 0,83 0,679 N b,y,fi,t,rd = 0,7760, ,0 90,91 kn > 13,0 kn acceptable 0,776 = 90,91 kn Table 6.1 8

65 Design Example 10 Sheet 7 of 7 Buckling curves: minor (z-z) axis: 0,5 1 0,49 1,169 0,3 1,169 = 1,396 φφ θ,z = 1 fi,z = 0,5 1,396 1,396 1,169 N b,z,fi,t,rd = 0,4630, ,0 54,4 kn > 18,6 kn acceptable 0,463 = 54,4 kn Member buckling resistance in combined bending and axial compression min,fi Where N Ak k y = fi,ed p0,,θ km km f M M y y,fi,ed z z,fi,ed y y,fi,θ,rd z,fi,θ,rd M,fi N y fi,ed 1 3 fy y,fiak p0,,θ M,fi y = y, M,y M,y 1 Eq. 8.6 Eq , 3 0,44 0,9 0,8 Eq Assuming the column is pinned at the base, a triangular bending moment distribution occurs and M = 1,8: y = (1, 1,8 3) 0,679 0,441,8 0,9 k y = = 0,068 3 ( 0,068) 13, , ,355 1,0 = 1,010 < 3,0 Table 8.3 M k M M0 y,fi,,rd, Rd M,fi M 1,10 y,fi,,rd 0,4308,75 4,14 1,0 knm for Class 1, or 3 sections Eq ,0 1,010 1,8 0, , ,355 4,14 1,0 Eq. 8.6 Therefore the section has adequate resistance after 30 minutes in a fire. 9

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67 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of 8 Design Example 11 Design of a two-span coldworked trapezoidal roof sheeting Research Fund for Coal and Steel Made by JG/AO Date 0/06 Revised by GZ Date 03/06 Revised by SJ Date 04/17 design example 11 design of a two-span cold-worked trapezoidal roof SHeetInG This example deals with a two-span trapezoidal roof sheeting with a thickness of 0,6 mm from stainless steel austenitic grade CP500, i.e. cold worked with f y = 460 N/mm. Comparisons will be made against the design of identical sheeting of ferritic grade in the annealed condition, i.e. f y = 80 N/mm (see Design Example 3). (There are no differences in the design procedure for ferritic and austenitic sheeting.) The dimensions of the roof sheeting are shown below. 4 x 1,5 = The example shows the following design tasks: - determination of effective section properties at the ultimate limit state; - determination of the bending resistance of the section; - determination of the resistance at the intermediate support; - determination of deflections at serviceability limit state. design data Spans L = 3500 mm Width of supports s s = 100 mm Design load Q = 1,4 kn/m Self-weight G = 0,07 kn/m Design thickness t = 0,6 mm Yield strength f y = 460 N/mm Table.3 Modulus of elasticity E = N/mm Section.3.1 Partial safety factor M0 = 1,1 Table 4.1 Partial safety factor M1 = 1,1 Table 4.1 Load factor G = 1,35 Section 4.3 Load factor Q = 1,5 Section 4.3 A detailed sketch of the roof sheeting is given in the figure below. The upper flange will be in compression over the mid support and therefore this case will be checked in this example. 31

68 Design Example 11 Sheet of 8 h su b u0 / b su / b su0 / h 0 b sl0 / Mid line dimensions: h 0 70 mm w 0 1, 5 mm b u0 65 mm b l0 57 mm b su 0 mm b su0 8 mm h su 6 mm bs l 0 mm b 8 mm s10 h 6 mm sl w 0 / b l0 / b sl / h sl r mm (internal radius of the corners) Angle of the web: h 0 θθ = atan 0,5(ww 0 bb u0 bb ll0 ) = atan 70 0,5 (1, ) = 57,1 effective section properties at the ultimate limit state (uls) Section 5. Check on maximum width to the thickness ratios and angle of web: h 0 /tt = 70/0,6 = sinθ = 336 Table 5.1 Angle of the web and corner radius: max(bb ll0 /tt ; bb uu0 /tt) = bb uu0 /tt = 65/0,6 = θθ = 57,1 90 bb p = bb u0 bb su = 65 0 =,5 mm The influence of rounded corners on cross-section resistance may be neglected if the internal radius rr 5tt and rr 0,10bb p Table 5.1 rr = mm min(5tt ; 0,1bb pp ) = min(5 0,6 ; 0,1,5) =,5 mm Section 5.6. The influence of rounded corners on cross-section resistance may be neglected. Location of the centroidal axis when the web is fully effective Calculate reduction factor for effective width of the compressed flange: ρρ = 0,77 0,079 but 1 λλ p λλ p Section Eq. 5.1 where bb tt,5/0,6 λλ p = = 8,4εε kk σσ 8,4 0,698 4 = 0,946 Eq. 5.3 = 1 kk = 4 Table 5.3 bb = bb p = bb u0 bb su = 65 0 =,5 mm 3

69 Design Example 11 Sheet 3 of 8 εε = [ 35 ff y ρρ = 0,77 λλ p 0,5 EE ] = [ 35 0, ] 0,079 = 0,77 λλ p 0,946 0,079 0, 946 = 0,78 1 = 0,698 Table 5. bb eff,u = ρρbb = 0,78,5 = 16,38 mm Table 5.3 Effective stiffener properties tt su = bb h su + ( su bb su0 ) h su tt = 6 + ( 0 8 ) 6 0,6 = 0,849 mm AA s = (bb eff,u + bb su0 )tt + h su tt su = (16,38 + 8) 0, ,849 = 4,8 mm Fig. 5.3 ee s = bb h su0h su tt + h su su tt su = 8 6 0,6 + 6 AA s 4,8 6 0,849 =,39 mm II s = (15tt ee s ) + bb su0 tt (h su ee s ) + h su tt su ( h su ee s) + ( 15tt4 1 ) + bb su0tt tt 3 suh su 1 II s = (15 0, 6, 39 ) + 8 0,6 (6.39) + 6 0, 849 ( 6,39) bb su bb su0 bb s = h ssss + ( ) 15 0, , 63 0,849 Fig ( ) + + = 159,07 mm bb su0 = 6 + ( 0 8 ) + 8 = 5,0 mm ll b = 3,07 [II s bb p ( bb 1 4 p + 3bb s tt 3 )] = 3,07 [159,07, 5, ( 0, 6 3 )] 1/4 = 51 mm Eq ss w = ( ww 0 bb u0 bb ll0 ) + h 1, = ( ) bb d = bb p + bb s =,5 + 5 = 70 mm + 70 = 83,4 mm Fig. 5.5 kk w0 = ss w + bb d 83, = ss w + 0,5bb d 83,4 + 0,5 70 = 1,37 Eq ll b = 51 ss w 83,4 = 3,01 kk w = kk w0 = 1,37 Eq

70 Design Example 11 Sheet 4 of 8 σσ cr,s = 4,kk wee AA s II s tt 3 4bb p (bb p + 3bb s ) Eq. 5.4 σσ cr,s = 4, 1, ,8 λλ d = ff y σσ cr,s = ,3 = 0, ,07 0, 6 3 4, 5 = 551,3 N/mm (, ) 0,65 < λλ d = 0,913 < 1,38 Eq χ d = 1,47 0,73λλ d = 1,47 0,73 0,913 = 0,81 tt red,u = χ d tt = 0,81 0,6 = 0,486 mm The distance of neutral axis from the compressed flange: tt sl = h sl + ( bb sl bb sll0 ) tt = h sl 6 + ( 0 8 ) 6 tt w = tt/sinθθ = 0,6/sin(57,1 ) = 0,714 mm 0,6 = 0,849 mm ee ii [mm] AA ii [mm ] 0 0,5bb eff,u tt = 4,9 0 0,5bb eff,u χχ dd tt = 3,98 0,5h su = 3 h su χχ dd tt su = 4,13 h su = 6 0,5bb su0 χχ dd tt = 1,94 0,5h 0 = 35 h 0 tt w = 49,98 h 0 = 70 0,5(bb ll0 bb ssss ) tt = 11,1 h 0 0,5h sll = 67 h sll tt sll = 5,09 h 0 h sll = 64 0,5bb sll0 tt =,4 AA tot = AA ii = 83,5 mm ee c = AA iiee ii AA tot = 36,46 mm Effective cross-section of the compression zone of the web EE 00 ss eff,1 = ss eff,0 = 0,76tt = 0,76 0,6 γ M0 σ com,ed 1, = 9,07 mm ss eff,n = 1,5ss eff,0 = 1,5 9,07 = 13,61 mm EN (4-5) Effective cross-section properties per half corrugation h eff,1 = ss eff,1 sinθθ = 9,07 sin(57,1 ) = 7,6 mm h eff,nn = ss eff,nn sinθθ = 13,61 sin(57,1 ) = 11,43 mm 34

71 Design Example 11 Sheet 5 of 8 ee eeeeee,ii [mmmm] AA eeeeee,ii [mmmm ] II eeeeee,ii [mmmm 44 ] 0 0,5bb eff,u tt = 4, ,5bb eff,u χχ dd tt = 4,0 0 0,5h su = 3 h su χχ dd tt su = 4,1 χχ dd tt su h su 3 /1 = 1,4 h su = 6 0,5bb su0 χχ dd tt = 1,9 0 0,5h eff,1 = 3,8 h eff,1 tt w = 5,4 tt w h eff,1 3 /1 = 6,3 h 0 0,5(h 0 ee c + h eff,nn ) = 47,5 (h 0 ee c + h eff,nn ) tt ww = = 3,1 tt w (h 0 ee c + h eff,nn ) 3 h 0 = 70 0,5(bb ll0 bb sll ) tt = 11,1 0 h 0 0,5h sl = 67 h sl tt sl = 5,1 tt sl h sl 3 /1 = 15,3 h 0 h sl = 64 0,5bb sl0 tt =,4 0 AA tot = AA eff,i = 71,0 mm ee c = AA eff,i ee eff,i AA tot = 40,0 mm II tot = II eff,i + AA eff,i (ee c e eff,i ) = 5 465, ,6 = ,7 mm 1 = 5411,1 Optionally the effective section properties may also be redefined iteratively based on the location of the effective centroidal axis. EN Bending strength per unit width (1 m) II = II 0,5w tot = ,7 = ,7 mm4 0 0,5 1,5 WW u = II ee c = WW l = ,7 40 II ,7 = h 0 ee c = 1 114,5 mm 3 = 16 15,7 mm3 Because WW u < WW l WW eff,min = WW uu = 1 114,5 mm 3 MM c,rd = WW eff,minff y γ M0 = 1114, ,1 = 5,07 knm Eq determination of the resistance at the intermediate support Section Web crippling strength cc 40 mm EN rr/tt = /0,6 = 3,33 10 Clause h ww /tt = 70/0,6 = sinθθ = 00sin(57,1 ) = θθ = 57,1 90 β V = 0 0, αα = 0,15 (category ) ll aa = ss ss = 100 mm 35

72 Design Example 11 Sheet 6 of 8 RR w,rd = α tt ff y EE (1 0,1 rr tt ) (0,5 + 0,0 ll a tt ) [,4 + ( φφ 90 ) ] 1 γ M ,5w 0 EN Eq RR w,rd = 0,15 0, (1 0,1 100 ) (0,5 + 0,0 0,6 0,6 ) [,4 + ( 57,1 90 ) ] 1 1, ,5 1, = 3,6 kn combined bending moment and support reaction Factored actions per unit width (1 m): qq = γγ G GG + γγ Q QQ = 1,35 0,07 + 1,5 1,4 =,19 kn/m MM Ed = qqll 8,19 3, 5 = 8 = 3,35 knm FF Ed = 5 4 qqqq = 5,19 3,5 = 9,58 kn 4 MM Ed = 3,35 MM c,rd 5,07 = 0,661 1,0 FF Ed = 9,58 RR w,rd 3,6 = 0,406 1,0 EN Eq. 6.8a - c MM Ed + FF Ed = 0, ,406 = 1,067 1,5 MM c,rd RR w,rd Cross-section resistance satisfies the conditions. determination of deflections at serviceability limit state (SLS) Effective cross-section properties For serviceability verification the effective width of compression elements should be based on the compressive stress in the element under the serviceability limit state loading. Maximum compressive stress in the effective section at SLS. A conservative approximation is made based on W u determined above for ultimate limit state. MM y,ed,ser = (GG + QQ)LL 8 σσ com,ed,ser = MM y,ed,ser WW uu = = (0,07 + 1,4) 3, 5 =,5 knm 8, ,5 = 185,7 N/mm The effective section properties are determined as before in ultimate limit state except that f y is replaced by and the thickness of the flange stiffener is not reduced. The com,ed, ser results of the calculation are: Effective width of the compressed flange: Location of the centroidal axis when the web is fully effective: Effective cross-section of the compression zone of the web: The flange is fully effective. e c = 34,1 mm The web is fully effective. Effective part of the web: s 68mm s eff,1 14, eff, n 1, 4 mm Effective cross-section properties per half corrugation: A tot = 8,44 mm e c = 36,5 mm I tot = 5976,1 mm 4 EN Clause

73 Design Example 11 Sheet 7 of 8 Effective section properties per unit width (1 m): I = 5618,0 mm 4 W u = 15507,0 mm 4 W l = 16655,6 mm 4 Determination of deflection Secant modulus of elasticity corresponding to maximum value of the bending moment: 1,Ed,ser = MM y,ed,ser WW u =, = 145,096 N/mm,Ed,ser = MM y,ed,ser WW l =, ,6 = 135,090 N/mm nn = 7 (for austenitic grade stainless steel) Table 6.4 EE EE S,1 = EE 1 + 0,00 σσ ( σσ 1,Ed,ser 1,Ed,ser EE EE S, = EE 1 + 0,00 σσ ( σσ,ed,ser,ed,ser EE S = EE S,1 + EE S, = nn = ) ff y nn = ) ff y 199, , , ,145 (0,145 0,460 ) , ,135 (0,135 0,460 ) = 199,83 kn/mm = 199,89 kn/mm Eq Eq = 199,86 kn/mm Eq. 6.5 Check of deflection As a conservative simplification, the variation of neglected. Es,ser along the length of the member is For cross-section stiffness properties the influence of rounded corners should be taken into account. The influence is considered by the following approximation: δ = 0,43 nn jj=1 rr j φ j mm bb p,i ii=1 90 o = 0,43 94,o 90 o 149,3 = 0,019 Eq. 5. II y,r = II (1 - ) = 5618,0 (1-0,019) = ,1 mm 4 Eq. 5.0 For the location of maximum deflection: xx = δδ = δδ = LL = ,5 = 1,48 m 16 (GG + QQ)LL4 ( xx 48EE S II y,r LL 3 xx3 LL 3 + xx4 LL 4) (0,07 + 1,4) , , ,1 10 δδ = 11,1 mm 1 (1,48 3,5 1, 483 1, , 53 3, 5 4 ) The permissible deflection is L/00 = 3500/00 = 17,5 mm > 11,1 mm, hence the calculated deflection is acceptable. 37

74 Design Example 11 Sheet 8 of 8 comparison between sheeting in the annealed and cold worked conditions A comparison of the bending resistance per unit width and resistance to local transverse forces of identical sheeting in the annealed condition (f y = 80 N/mm ) and cold worked condition (f y = 460 N/mm ) is given below: f y = 80 N/mm (Design example 3) M c,rd = 3,84 knm and R w,rd = 18,4 kn f y = 460 N/mm (Design example 11) M c,rd = 5,07 knm and R w,rd = 3,6 kn With sheeting in the annealed condition, the span must be reduced to,9 m compared to 3,5 m for material in the cold worked strength condition. Hence, sheeting made from cold worked material enables the span to be increased, meaning that the number of secondary beams or purlins could be reduced, leading to cost reductions. 38

75 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of 7 Design Example 1 Design of a lipped channel subject to bending Research Fund for Coal and Steel Made by ER/EM Date 0/06 Revised by HB Date 03/06 Revised by ER/IA Date 04/17 design example 1 design of a LIPPed channel SuBJect to BendInG Design a lipped channel subject to bending with an unrestrained compression flange from austenitic grade in the cold worked condition CP500. The beam is simply supported with a span, l = 4,0 m. The distance between adjacent beams is 1,0 m. As the load is not applied through the shear centre of the channel, it is necessary to check the interaction between the torsional resistance of the cross-section and the lateral torsional buckling resistance of the member. However, this example only checks the lateral torsional buckling resistance of the member. factors Partial factor M0 = 1,1 and M1 = 1,1 Table 4.1 Load factor G = 1,35 (permanent loads) and Q = 1,5 (variable loads) EN 1991 actions Permanent actions (G): kn/m Variable actions (Q): 3 kn/m Since the distance between adjacent beams is 1m, G k = kn/m Q k = 3 kn/m Load case to be considered at the ultimate limit state: * q G,jGk,j Q,1Qk,1 7, kn/m EN 1991 j Structural analysis Reactions at support points (Design shear force) * q 4 VEd 14,4 kn Design bending moment * q 4 M Ed 14,4 knm 8 Material Properties f y = 460 N/mm Table.3 Modulus of elasticity E = N/mm and shear modulus G = N/mm Section.3.1 cross-section Properties The influence of rounded corners on cross-section resistance may be neglected if the internal radius r 5t and r 0,10b p and the cross section may be assumed to consist of plane elements with sharp corners. For cross-section stiffness properties the influence of rounded corners should always be taken into account. Section

76 Design Example 1 Sheet of 7 z b h y r c t y h = 160 mm b = 15 mm c = 30 mm t = 5 mm r = 5 mm z rm rt 7,5 mm g r r mtan sin, mm b btg 115,6 mm Figure 5.5 p r5 mm 5t5 mm r r 5 mm 0,10b 11,56 mm p The influence of rounded corners on section properties may be taken into account with sufficient accuracy by reducing the properties calculated for an otherwise similar crosssection with sharp corners, using the following approximations: Notional flat width of the flange, bp,f btgr 115,6 mm Notional flat width of the web, bp,w htgr 150,6 mm Notional flat width of the lip, bp,l ct/ gr 5,3 mm A g,sh = t bp,f bp,w b p,l 16 mm I yg,sh = bp,ft bp,ft(0,5h0,5 t) bp,ltbp,lt 0,5 h( cbp,l) 0,5bp,l b t = p,w 9, mm 4 n m j = 0, 43 rj / bp,i 0,0 o 90 j1 i1 Eq. 5. A g = A g,sh (1 ) = 119 mm Eq I g = I g,sh (1 ) = 9, mm 4 Eq. 5.0 classification of the cross-section Section E f y ,5 0,698 Flange: Internal compression parts. Part subjected to compression. c b p,f 115,6 mm and c/t = 3,1 For Class, c/t 354,43, therefore the flanges are Class Web: Internal compression parts. Part subjected to bending. c = b p,w = 150,6 mm and c/t = 30,1 For Class 1, c/t 750,6, therefore the web is Class 1. Table 5. 40

77 Design Example 1 Sheet 3 of 7 Lip: Outstand flanges. Part subjected to compression, tip in compression. c b p,l 5,30 mm and c/t = 5,06 For Class 1, c/t 9 6,8, therefore the lip is Class 1. effects of shear lag Section 5.4. Shear lag in flanges may be neglected if b 0 < L e/50, where b 0 is taken as the flange outstand or half the width of an internal element and L e is the length between points of zero bending moment. For internal elements: b o = (b t)/ = 60 mm The length between points of zero bending moment is: L e Therefore shear lag can be neglected. = 4000 mm, L e /50 = 80 mm flange curling Section Flange curling can be neglected if the curling of the flange towards the neutral axis, u, is less than 5% of the depth of the profile cross-section: u b E t z 4 a s a is mean stress in the flanges calculated with gross area (f y=460 N/mm is assumed) b s = is the distance between webs = b p,f + b p,l = 140,9 mm t = 5 mm z = is the distance of the flange under consideration from neutral axis = 77,5 mm u =,15 mm < 0,05h = 8 mm, therefore flange curling can be neglected. EN , clause 5.4 Eq. 5.3a Stiffened elements. edge stiffeners distortional buckling. Plane elements with edge stiffeners Section and EN , clause Step 1: Initial effective cross-section for the stiffener EN , For flanges (as calculated before) clause b =15 mm and b p = b p,f = 115,6 mm For the lip, the effective width c eff should be calculated using the corresponding buckling factor k and expressions as follows: b p,c = b p,l = 5,30 mm p 41

78 Design Example 1 Sheet 4 of 7 b p,c/b p = 0, < 0,35 then k = 0,5 EN , Eq. 5.13b bt p 0,36 ( b 5,3 mm ) Eq ,4 k σ 1 0,188 Cold formed outstand elements: 1,33 1 then = 1,0 p Eq. 5. p c eff = b p,c = 5,30 mm EN , Eq. 5.13a Step : Reduction factor for distortional buckling Calculation of geometric properties of effective edge stiffener section b e = b p,f = 115,6 mm In this example, since the compressed flange is Class, b e already considers the whole flange and therefore b e1 = 0 is adopted. c eff = b p,l = 5,30 mm A s = (b e + c eff)t = (b b,f+ b b,l)t = 704,5 mm Calculation of linear spring stiffness K 3 Et 1 6, b1hwb1 0,5bb 1 hwkf N/mm EN , Eq. 5.10b b 1 = b y b t/ r = 71,1 mm (the distance from the web-to-flange junction to the gravity centre of the effective area of the edge stiffener, including the effective part of the flange b e). k f = 0 (flange is in tension) h w = h t r = = 140 mm Elastic critical buckling stress for the effective stiffener section, adopting K = K 1 KEI EN , s cr,s 565,8 N/mm Eq As Reduction factor d for distortional buckling d fyb cr,s 0,90 EN , Eq. 5.1d 0,65< d <1,38 then d 1,47 0,73d 0,8 EN , Eq. 5.1b Reduced area and thickness of effective stiffener section, considering that com,ed = f yb/ M0 A fyb γ EN , M0 s,red d As 576,4 mm Eq com,ed t red = ta s,red/a s = 4,1 mm Calculation of effective section properties with distortional buckling effect A g,sh = t bp,f bp,w b p,l t red bp,f b p,l 034,0 mm n m j = 0, 43 rj / bp,i 0,0 o 90 Eq. 5. j1 i1 A g = A g,sh (1 ) = 1993,3 mm Eq The new e eff, adopting distances from the centroid of the web, positive downwards: 4

79 Design Example 1 Sheet 5 of 7 e eff = I y,g,sh= 0,5 0,5 0,5 0,5 0,5 0,5 0,5 b t h t b t h t b t h tg b p,f red red p,f p,l red r p,l bp,lt 0,5h0,5tgr 0,5bp,l bp,l0 4,7 mm A g,sh A g,sh bp,fttred bp,ftred(0,5h0,5 tred eeff ) bp,ltred bp,ltred 0,5h0,5tgr 0,5bp,l eeff bp,ft bp,ft(0,5h0,5 teeff ) bp,ltbp,lt0,5h0,5tgr 0,5bp,l eeff bp,wtbp,wt( eef f ) 8,64 10 mm 1 I y,g = I y,g,sh (1 ) = 8, mm 4 Eq. 5.0 z max = h/ + e eff = 160/ + 4,7 = 84,7 mm (distance from the top fibre to the neutral axis) W y,g = I y,g / z max = 97, mm 3 resistance of cross-section Section 5.7 Cross-section subject to bending moment Section M W f / γ 41, 0 knm Eq. 5.9 c,rd pl y M0 Design bending moment M Ed 14, 4 knm, therefore cross-section moment resistance is OK. Cross-section subject to shear Section A v = 800 mm pl,rd v y M0 V A f 3 /γ 193,15 kn Eq. 5.3 Design shear force VEd 14,4 kn, therefore cross-section shear resistance is OK Cross-section subjected to combination of loads Section V Ed = 14,4 kn > 0,5V pl,rd = 96,57 kn Therefore, there is no need to take into account interaction between bending moment and shear force. flexural members Section 6.4 Lateral-torsional buckling Section 6.4. M Wf Eq b,rd LT y y M1 LT 0,5 LT LT λlt 1 LT LT LT LT 1 Eq ,5 1 0,4 Eq y y LT Eq LT Wf M cr = 0,34 for cold-formed sections 43

80 Design Example 1 Sheet 6 of 7 Determination of the elastic critical moment for lateral-torsional buckling EI t z k k L GI I w Mcr C 1 Cz g Cz g kl kw Iz EI z Eq. E.1 For simply supported beams with uniform distributed load: C 1 = 1,13, and C = 0,454 Table E. Assuming normal conditions of restraint at each end: k = k w = 1 z a is the coordinate of point load application z s is the coordinate of the shear centre z g = z a z s= h/ = 80 mm y G y G = = distance from the central axis of the web to the gravity centre bp,ft( gr 0,5 bp,f ) bp,lt( b0,5 t) 46,4 mm A I z,sh = 4, mm 4 I t,sh = 18,010 3 mm 4 I w,sh = 3, mm 6 I z = I z,sh (1 ) = 4, mm 4 I t = I t,sh (1 ) = 17, mm 4 I w = I w,sh (1 4) = 1, mm 6 Then, s EI z k I kl GI w t Mcr C 1 Cz g Cz g 34,76kNm kl kw Iz EI z W f Eq. E.1 y,g y LT 1,14 (W y,g = 97, mm 3, compression flange) Eq M cr LT LT LT LT 0,5 1 0,4 1,7 Eq M LT 0,5 LT LT λ LT 1 b,rd LT y y M1 0,54 Eq Wf γ,1 knm Eq Design moment M Ed 14,4 knm, therefore lateral torsional buckling resistance OK. Note: As the load is not applied through the shear centre of the channel, it is also necessary to check the interaction between the torsional resistance of the cross-section and the lateral torsional buckling resistance of the member. Shear buckling resistance Section The shear buckling resistance only requires checking when h w / t 56,ε η for an Eq. 6.0 unstiffened web. The recommended value for = 1,0. h w / ( h t r ) / 140 / 5 8,0, 56,ε η 3,67, therefore no further check required. 44

81 Design Example 1 Sheet 7 of 7 deflections Section Deflections should be determined for the load combination at the relevant Serviceability Limit State, with: Load factors G = 1,00 (permanent loads) and Q = 1,00 (variable loads) EN 1991 Permanent actions (G): kn/m and Variable actions (Q): 3 kn/m Load case to be considered at SLS, assuming distance between adjacent beams is 1,0 m : qg,jgk,j Q,1Qk,1 5,0 kn/m EN 1991 j The deflection of elastic beams may be estimated by standard structural theory, except that the secant modulus of elasticity should be used instead of the modulus of elasticity: ES1 ES ES where: E S1 is the secant modulus corresponding to the stress in the tension flange and E S is the secant modulus corresponding to the stress in the compression flange Eq. 6.5 E S1 and E S for the appropriate serviceability design stress can be estimated as follows: E S,i 1 0,00 E E i,ed,ser i,ed,ser f y n and i = 1, where: i,ed,ser is the serviceability design stress in the tension or compression flange Eq n is the Ramberg Osgood parameter; for austenitic stainless steel , n = 7. Table 6.4 The non-linear stainless steel stress-strain relationship means that the modulus of elasticity varies within the cross-section and along the length of a member. As a simplification, the variation of E S along the length of the member may be neglected and the minimum value of E S for that member (corresponding to the maximum values of the stresses σ 1 and σ in the member) may be used throughout its length. The stresses in the tension and compression flanges are the following: Compression flange: M Ed,max Ed,ser,1 10,1 MPa and E S1 = , MPa Eq W y,sup with M Ed,max = 10 knm and W y = 97, mm 3 Tension flange: M Ed,max Ed,ser, 100,8 MPa and E S = ,8 MPa W y,inf with M Ed,max = 10 knm and W y = 99, mm 3 And therefore: E S = ,0 MPa Eq. 6.5 The maximum deflection can be estimated by standard structural theory assuming the secant modulus of elasticity: d max 4 5ql 384EI S y Since I y = 8, mm 4, q = 5,0 kn/m and l = 4,0 m Sheets 1 & 5 d 10,0 mm max 45

82 46

83 Promotion of new eurocode rules for structural stainless steels (PureSt) calculation SHeet Title Client Sheet 1 of 8 Design Example 13 Hollow section lattice girder Research Fund for Coal and Steel Made by PTY/AAT Date 01/06 Checked by MAP Date 0/06 Revised by MIG Date 06/17 design example 13 - HoLLow SectIon LattIce GIrder The lattice girder supports roof glazing and is made of square and rectangular hollow sections of grade stainless steel; a comparison is made between material in two strength levels - the annealed condition ( f y =10 N/mm ) and in the cold worked condition (strength level CP500, f y = 460 N/mm ). Calculations are performed at the ultimate limit state and then at the fire limit state for a fire duration of 30 minutes. For the CP500 material, the reduction factors for the mechanical properties at elevated temperatures are calculated according to Section 8.. The structural analysis was carried out using the FE-program WINRAMI marketed by Finnish Constructional Steelwork Association (FCSA) ( The WINRAMI design environment includes square, rectangular and circular hollow sections for stainless steel structural analysis. WINRAMI solves the member forces, deflections and member resistances for room temperature and structural fire design and also joint resistance at room temperature (it also checks all the geometrical restraints of truss girder joints). In the example, the chord members are modelled as continuous beams and the diagonal members as hinge jointed. According to EN , the buckling lengths for the chord and diagonal members could be taken as 0,9 times and 0,75 times the distance between nodal points respectively, but in this example conservatively the distance between nodal points has been used as the buckling length. The member forces were calculated by using WINRAMI with profile sizes based on the annealed strength condition. These member forces were used for both the annealed and CP500 girders. This example focuses on checking 3 members: mainly axial tension loaded lower chord (member 0), axial compression loaded diagonal (member 31) and combination of axial compression and bending loaded upper chord member (member 5). The weight of the girders is also compared. The welded joints should be designed according to the Section 7.4, which is not included in this example. Annealed : lower chord 100x60x4, upper chord 80x80x5, corner vertical 60x60x5 diagonals from left to middle: 50x50x3, 50x50x3, 40x40x3, 40x40x3, 40x40x3,40x40x3, 40x40x3. CP500 : lower chord 60x40x4, upper chord 70x70x4, corner vertical 60x60x5, all diagonals 40x40x3. Span length 15 m, height in the middle 3,13 m, height at the corner 0,5 m. Weight of girders: Annealed: 407 kg, CP kg. The weight is not fully optimised. 47