Steel Design to Eurocode 3 Introduction • EN 1993-5 Piling • EN 1993-6 Crane supporting structures Eurocode 3 Part 1 has 12 sub-parts: Development of Eurocode 3 • EN 1993-1-1 General Rules Aim: to create a common structural language • EN 1993-1-2 Fire • EN 1993-1-3 Cold-formed thin gauge • EN 1993-1-4 Stainless steel • EN 1993-1-5 Plated elements National Annex • EN 1993-1-6 Shells • • EN 1993-1-7 Plates transversely loaded • EN 1993-1-8 Joints • EN 1993-1-9 Fatigue • EN 1993-1-10 Fracture Toughness • EN 1993-1-11 Cables • EN 1993-1-12 High strength steels – And make allowances for National Choice through the use of a National Annex • Eurocode 3 allows some parameters and design methods to be determined at a national level. Where a national choice is allowed, this is indicated in the Eurocodes under the relevant clause. – values or methods to be used in a particular country are given in the National Annex. Nationally Determined Parameters (NDPs) • The recommended values of the parameters and design methods are collectively referred to as Nationally Determined Parameters (NDPs). • NDPs determine various aspects of design but perhaps most importantly the level of safety of structures during construction and service. Key Differences between EC3 and BS 5950 There are several differences between EC3 and BS 5950: BS 5950 Structure EC3 Structure Separate sections for different elements types Sub-parts are based on structural phenomena Structure of Eurocode 3 e.g. Beams, Eurocode 3 is broken into 6 parts: Plate Girders, e.g. Tension, Compression, Bending, Shear • EN 1993-1 Generic rules • EN 1993-2 Bridges • EN 1993-3 Towers, masts and chimneys • EN 1993-4 Silos, tanks and pipelines Compression members... Sub-parts can be applied to any element The arrangement of the sub-parts means less duplication of rules 25 Fracture Figure 1 (Source: Arya (2009) Design of Structural Elements Pg.377) Different Wording ‘Action’ – force or imposed displacement Permanent action (Dead Load) Variable action (Live Load) ‘Effect’ – internal force or moment. deflections Omissions ‘Verification’ –check Notable omissions: ‘Resistance’ – capacity • Effective lengths – Use BS 5950 effective lengths • Different Symbols Formulae for Mcr – Use SN003 NCCI Document BS 5950 EC3 BS 5950 EC3 BS 5950 EC3 A A P N py fy Z Wel Mx My pb χLTfy S Wpl V V pc χfy Ix Iy H Iw r i Iy Iz J It • Deflection limits – Refer to National Annex .Different Axes Informative subscripts BS 5950 Along the member Major Axis Minor Axis Eurocode 3 ‘Ed’ means design effect X ‘Rd’ means design resistance X Y Therefore: Y Z NEd is an design axial force NRd is the design resistance to the axial force Gamma Factors Partial factor γM UK NA value Application γM0 1.00 Member Buckling γM2 1.00 Cross-sections γM1 1. 3 0 0 0 .1Qk.7 0.7 0.10b Method: Get the factors from Tables 1 and 2 FAT: fatigue failure and substitute them into the equation you are using.6 0.1 “+” ΣγQ.1 “+” ΣγQ.7 0.35 1.0 0.5 0. check for a range of different loading combinations and take the least favourable Actions result.10a and 6.1 “+” ΣγQ.5 Category C : congregation areas 0.3.0) Equation 6.6 0.jGk.i Qk ψ0Qk Combination value ψ1Qk Frequent value ψ2Qk Quasi-permanent value Characteristic value (ψ = 1. Category A : domestic/residential areas 0.1 EQU: static equilibrium Combinations of Actions STR: strength/buckling etc Can use either: GEO: Failure of excessive deformation of ground Equation 6. G (Dead loads) Equation 6.5 Category B : office areas 0.7 0 Category H : roofs Snow (sites up to 1000m) 0.j “+” γPP “+” γQ.jGk.10a Partial Factors γG ΣγG.10 Less favourable of 6.7 Category G : traffic area.1ψ0.j “+” γPP “+” γQ.10 Variable actions .i ξj is 0.3 0.8 0.7 0.9 Category F : traffic area.1Qk.iQk.5 0.j “+” γPP “+” γQ.jGk.5 kN 0.Combination Factors ψ Loading Introduction to EN 1990 Covers the „Basis of Structural Design‟ Use with the other Eurocodes Gives safety factors needed for ULS and SLS verifications partial factors (see Table 1) combination factors (See Table 2) ULS Checks Action ψ0 ψ1 Imposed loads in buildings.iψ0.3 0. 30– 160 0.iQk.7 Category D : shopping areas 0.iψ0.5 0 Table 1: Partial Factor values from the UK NA Equation 6.7 Category E : storage areas 1.i Unfavourable Favourable 1.iQk.0 γQ 1.iQk.2 Table 2: Extract from Table NA.2 Wind 0.10b ΣξjγG.2) ψ2 0.7 0. < 30kN 0. Q (Live loads) ΣγG.2.925 (From NA 2.7 0.iψ0.A1. Permanent actions .6 0. . 2. HEd is the horizontal reaction at the bottom of the storey VEd is the total vertical load at the bottom of the storey δH.2 deals with joint modelling For portal frames (with shallow roof slopes less than 26°) and beam and column plane frames: Eurocode 3 recognises the same three types of joint. Joints Clause 5. • if they increase the action effects significantly • or modify significantly the structural behaviour Table 1: Summary of Analysis Types First-Order Analysis A first-order analysis may be used if the following criteria is satisfied: αcr ≥ 10 for elastic analysis αcr ≥ 15 for plastic analysis αcr =Fcr/FEd Figure 1: Load-Deformation graph for different analysis types (Source: Designer’s Guide to EN 1993-1-1 Page 21) αcr is the factor by which the design loading would have to increased to cause elastic instability in a global mode (λcr in BS 5950-1) FEd is the design loading on the structure Fcr is the elastic critical buckling load for global instability based on initial elastic stiffness.and second. Figure 2: Joint stiffness effects (Source: SCI CPD Course Material) h is the storey height.1.Ed is the horizontal deflection at the top of the storey under consideration relative to the bottom of the storey.1(2) states that second order effects shall be considered: Second-order elastic First-order plastic Deformed Geometry Analysis Type Initial Geometry the internal forces and moments.Steel Design to Eurocode 3 Structural Analysis The choice between a first. in terms of their effect on the behaviour of the frame structure. . with all horizontal loads applied to the structure. as BS 5950: Part 1. the extent to which ignoring Analysis Types second-order effects might lead to an unsafe There are four types of global analysis: approach due to underestimation of some of First-order elastic Second-order elastic Non-linear material behaviour Linear material behaviour Clause 5.order analysis should be based on: the flexibility of the structure in particular. We can represent initial sway imperfections by using Equivalent Horizontal Forces (EHFs) which are based on 1/200 of the factored vertical load.Amplifier (h is the height of the structure in metres) If 10 > αcr ≥ 3. Figure3:Replacing initial sway imperfections with equivalent horizontal forces EHF = φ x Vertical Forces φ = φ0αhαm φ0 = 1/200 = 0.0 α m is the reduction factor for columns Increase all lateral loads by the amplifier: Limits on αcr αcr >10 Action First order Analysis First order analysis plus amplification 10>αcr >3 or effective length method αcr ≤ 3 Second order analysis Table 2: Actions to be taken once αcr has been calculated (m is the number of columns contributing to the effect on the bracing system) Summary 1) Model the Frame Imperfections 2) Put all the loads on the frame (Including the EHFs) 3) Calculate αcr 4) Check to see if second-order effects are significant 5) If necessary use the amplifier Figure 2: Typical Imperfections that will be present when designing a structure Frame imperfections appear in (almost) every load case.005 α h is the reduction factor for height: . with reduction factors. it is recommended that you use Published Document PD 6695 instead. so fy(t) = fy. at a given temperature.Charpy value of 27 J can be obtained at +20°C • S275 J0 . It should be considered where there are tensile stresses. The service temperature is lowered i.25 (t) . PD 6695-1-10 • Published Document is much Simpler to use – Internal Tmd is -5°C (Table 2) – External is Tmd -15°C (Table 3) NOTE: Can only use this document for design in the UK .nom – 0.25 (t/t0) but t0 = 1mm.Charpy value of 27J can be obtained at 0°C • S275 J2 . below is an extract from that table. The Charpy test measures how much energy is absorbed by a steel sample.nom – 0.1 of the Eurocodes so determine the steel sub grade. fy(t) fy(t) = fy. • S275 JR . it becomes a reference temperature.Steel Design to Eurocode 3 Brittle Fracture Steel sub-grade selection Brittle failure is most likely to occur at very low temperatures.Charpy value of 27J can be obtained at -20°C EN 1993-1-10 The method given in the Eurocodes can be quite complex to use. Refer to table 2. It can be avoided by choosing a steel with sufficient fracture toughness Failure mainly dependent on: • Steel strength grade • Thickness • Lowest service temperature • Material toughness • Tensile Stress • Notches or defects in the element Steel toughness Steel toughness is measured by Charpy V-notch value.e. PD 6695-1-10 Tables Table 2 Maximum thicknesses for internal steelwork in buildings for T md = -5°C Table 3 Maximum thicknesses for external steelwork in buildings for T md = -15°C . ε Factor BS 5950 EC3 Plastic Class 1 Compact Class 2 Semi-compact Class 3 Slender Class 4 BS 5950 EC3 ε = (275/py)0. Extract from EN 10025-2 .5 Values of ε are given at the bottom of Table 5. S 355 355 345 335 325 The classification of a section will depend mainly on: The material yield strength.75 0. but local buckling is liable to prevent development of the plastic moment resistance.5 ε = (235/fy)0. Limits The limits between the classes depend on the ε factor which is calculated using fy.org/new/pdfs/Chapter8. where the element will fail before the design strength is reached. Classes Class 2 cross-sections are those which can develop their plastic moment resistance. t (mm) 16< t ≤ 40 40 < t ≤ 63 63 < t ≤ 80 Steel Grade t≤16 Similarly to BS 5950.pdf The UK National Annex says that material properties should be taken from the product standards.2: S 275 275 265 255 245 Class 1 cross-sections are those which can form a plastic hinge with the rotation capacity required from plastic analysis without reduction of the resistance.5 covers the cross section classification • Clauses 6.fy (yield strength) values for hot rolled steel: 2 fy (N/mm ) nominal thickness of element.5.00 0. Class 1 is the least susceptible to local buckling and class 4 is the most susceptible.2 Class 1 Class 2 Class 3 Class 4 Image Source: http://www. Class 3 cross-sections are those in which the stress in the extreme compression fibre of the steel member assuming an elastic distribution of stresses can reach the yield strength.81 0. the yield strength of the steel.2 covers the cross-sectional resistance Sections with slender webs or flanges will be more susceptible to local buckling. fy c/t ratio EN 10025-2 (Table 7) .92 0.71 EN 1993-1-1 Table 5.1 and 6. Eurocode 3 takes into account the effects of local through the process of cross section classification. Class 4 cross-sections are those in which local buckling will occur before the attainment of yield stress in one or more parts of the cross-section. but have limited rotation capacity because of local buckling. cross sections will be placed into one of four behaviour classes.2: fy 235 275 355 420 460 ε 1.Steel Design to Eurocode 3 Local Buckling and CrossSection Classification In Eurocode 3 you will need to refer to the following clauses when classifying a section and determining the cross-sectional resistance: • Clause 5.steel-insdag. fy Yield Strength Eurocode 3 defines the classes in Clause 5. 2) b/T = < 15 ε c/tf = < 14 ε d/t = < 120 ε d/tw = < 142 ε d/tw = < 42 ε Class 4: Slender Outstand Flange b = B/2 c = (b – tw – 2 r)/2 Internal Compressio n Part d= D–2T–2r c= h – 2 tf – 2 r Appropriate values of c and t are defined at the top of Table 5. Sheet 3 – Angles and Tubular Sections Summary Cross-section Classification 1.2) b/T = < 10 ε c/tf = < 10 ε d/t = < 100 ε d/tw = < 83 ε d/tw = < 38 ε Class 3 Class 4 Web in bending limiting value. Table 5.2 An element that doesn’t meet the class 3 limits should be taken as a class 4 section.c/t Width-to-Thickness Ratio Class 3: Semi-compact BS 5950 EC3 Limits The width-to-thickness ratios differ in EC3 differs from BS 5950: Flange outstand Web in bending Web in compression BS (Table 11) EC3 (Table 5. c/tf 9ε Class 2 10 ε d/tw = < 33 ε Class 2: Compact Flange outstand Web in bending Web in compression 2.5.2 to work out the class of the flange and web Class 1 Flange outstand limiting value.2. The limits are provided in table 5. the section is classified as Class 4 4.2) b/T = < 9 ε c/tf = < 9 ε d/t = < 80 ε d/tw = < 72 ε Limits 3. Effective widths are assigned to Class 4 compression elements to make allowance for the reduction in resistance as a result of local buckling To calculate the effective width of a Class 4 section. Take the least favourable class from the flange and web results .2(6) states that a cross-section is classified according to the highest (least favourable) class of its compression parts. refer to the relevant section in the Eurocodes: Section Type Reference Cold-formed sections EN 1993-1-3 Hot-rolled and fabricated section EN 1993-1-5 CHS EN 1993-1-6 Internal compression parts and outstand flanges are assessed against the limiting width to thickness ratios for each class. Determine ε from Table 5.2 for different types of sections. Determine fy (UK NA recommends you use the product standards) Limits Class 1: Plastic Flange outstand Web in bending Web in compression BS (Table 11) EC3 (Table 5. Substitute the value of ε into the class limits in Table 5.2 is made up of three sheets: Sheet 1 – Internal Compression Parts Sheet 2 – Outstand Flanges Clause 5. Overall Cross-Section Classification Table 5.2 BS (Table 11) EC3 (Table 5. d/tw 72 ε 83 ε 14 ε 124 ε If it does not meet Class 3 requirements. 5 deals with the crosssectional resistance for bending. web in bending and web in compression results to get the overall section class Bending Moment Resistance Summary 1.y Plastic modulus about the minor axis Syy W pl. A beam is considered restrained if: The section is bent about its minor axis Full lateral restraint is provided Closely spaced bracing is provided making the slenderness of the weak axis low The compressive flange is restrained again torsion The section has a high torsional and lateral bending stiffness There are a number of factors to consider when designing a beam.15 for Class 4 sections. Shear Resistance In Eurocode 3: Clause 6. Substitute the value of ε into the class limits in Table 5.z Table 1.4 Equation 6. MEd 2.13 for Class 1 and 2 cross-sections. and they all must be satisfied for the beam design to be adopted: Bending Moment Resistance Shear Resistance Combined Bending and Shear Serviceability Bending Moment Resistance In Eurocode 3: Clause 6.2 to work out the class of the flange and web 4.y Elastic modulus about the minor axis Zyy W el. Determine the design moment. Choose a section and determine the section classification 3.0 Section modulus terminology comparison between BS 5950 and EC3 Cross-section Classification Summary 1.Rd = Mpl. For Class 1 and 2 cross-sections: Mc.Rd) (6.2. W Restrained Beams Subscripts are used to identify whether or not the section modulus is plastic or elastic and the axis about which it acts.13) For Class 3 cross-sections: Mc.minfy/ɣM0 γM0 =1. the section modulus is used.15) BS EC3 5950 Elastic modulus about the major axis Zxx W el.14) For Class 4 cross sections: Mc.6 deals with the crosssectional resistance for shear.6 Equation 6.2.12 is satisfied.17 states that the design shear force (VEd) must be less than the design plastic shear resistance of the crosssection (Vpl.Rd is dependent on the class of the section.Rd = W eff.z Plastic modulus about the major axis Sxx W pl. Get ε from Table 5.2. using equation 6.12 states that the design moment (MEd) must be less than the design cross-sectional moment resistance (Mc. Ensure that the correct value of W. and equation 6.Steel Design to Eurocode 3 Section Modulus. Take the least favourable class from the flange outstand.Rd = Mel.Rd = W plfy/ɣM0(6.0 (6.2 covers the cross-sectional resistance o Clause 6.Rd = W el.1 2. Get fy from Table 3.Rd) (6.12) The equation to calculate Mc.2 covers the cross-sectional resistance o Clause 6. A detailed assessment of cross-section classification can be found in the ‘Local Buckling and Cross-Section Classification’ handout. EN 1993-1-1 Clause 6. 4.minfy/ɣM0 (6.2.17) . Carry out the cross-sectional moment resistance check by ensuring equation 6.2 3.14 for Class 3 cross-sections. EN 1993-1-1 Clause 6. equation 6.Rd. Determine Mc. Figure 1: Differences in shear area calculated using BS 5950 and EC3 Type of member Shear Area. Substitute the value of Av into equation 6.0 Table 3. 3. reference must be made to the National Annex.0: Shear area parameter descriptions Figure 1: Visual definition of the parameters used in the shear area calculation. Av EC3 should provide a slightly larger shear area compared to BS 5950 meaning that the overall resistance will be larger as shown in Figure 1. (Source: Blue Book) Figure 2: Standard case deflections and corresponding maximum deflection equations The maximum deflection calculated must not exceed the deflection limit.18) Shear Resistance Summary 1.0: Shear area formulas Term A b h hw r tf tw Definition Cross-sectional area Overall breadth Overall depth Depth of web Root radius Flange thickness Web thickness (taken as the minimum value is the web is not of constant thickness) Constant which may be conservatively taken η as 1.0 2. Calculate the shear area.0: Vertical Deflection Limits from NA 2.18 to get the design plastic shear resistance Shear Area.2. The deflection limits are not given directly in Eurocode 3. Serviceability Deflection checks should be made against unfactored permanent actions and unfactored variable actions. instead.(6. Av Rolled I and H sections (load parallel to web) Rolled Channel sections (load parallel to web) Rolled PHS of uniform thickness (load parallel to depth) CHS and tubes of uniform thickness Plates and solid bars Av = A – 2btf + (tw + 2r)tf but ≥ ηhwtw Av = A – 2btf + (tw + r)tf Av =Ah/(b+h) Av =2A/π Av =A Table 2.17 is satisfied. Design Situation Deflection limit Cantilever Length/180 Beams carrying plaster of other brittle finish Other beams (except purlins and sheeting rails) Span/360 Span/200 To suit the characteristics of particular cladding Table 4. Carry out the cross-sectional plastic shear resistance check by ensuring equation 6.23 Clause 7.1(1) B Purlins and sheeting rails . Av γM0 =1. and the second case is the ‘Special Case’ which is specifically for rolled sections of standard dimensions. LTB Resistance EN 1993-1-1 Clause 6.2.fy (yield strength) values for hot rolled steel: fy (N/mm2) nominal thickness of element.2.3. For Class 4 cross-sections: W y = W eff. Simplified assessment method (Clause 6.54 states that the design moment (MEd) must be less than the design buckling resistance moment (Mb.3.2 t≤16 S 275 275 265 255 245 S 355 355 345 335 325 Extract from EN 10025-2 (Table 7) Reduction Factor.1 Equation 6.56) The methods for both cases are very similar with the addition of a few extra parameters in the Special Case.55) where γM1 =1. < 0.3) 2. General and Special Cases When using the primary method.2. fy The UK National Annex says that we should obtain the value of the yield strength from the product standards.y 40 < t ≤ 63 For Class 3 cross-sections: 16< t ≤ 40 The section is bent about its minor axis Full lateral restraint is provided Closely spaced bracing is provided making the slenderness of the weak axis low The compressive flange is restrained again torsion The section has a high torsional and lateral bending stiffness The non-dimensional slenderness.0 (from UK NA) Section Modulus Wy For Class 1 and 2 cross-sections: W y = W pl.3. This small amount of extra work for the Special Case is worthwhile as it provides greater resistance of the section. this mode of buckling is called lateral torsional buckling (LTB).4) 3. General method (Clause 6. Extract from EN 10025-2 . The first case is the ‘General Case’ which can be used for all sections. This handout is a continuation of the ’Restrained Beams’ one and covers the design of unrestrained beams that are prone to lateral torsional buckling.Steel Design to Eurocode 3 (6.y Lateral torsional buckling can be discounted when: Eurocode 3 Approach There are three methods for calculating the LTB resistance of a member in Eurocode 3: 1.Rd) where . Primary method (Clauses 6.3.4) Note: This handout will only deal with the primary method.2.3. t (mm) Steel Grade 63 < t ≤ 80 W y = W el. there are two cases which are available for you to use.y Yield Strength. χLT General Case: (6.2 and Clauses 6.54) Unrestrained Beams Beams without continuous lateral restraint are prone to buckling about their major axis. (6. 57) where EN 1993-1-1 Table 6.34 0. determine the buckling curve that you need to use from the table from the National Annex NA.3(1) Buckling curve αLT a b c b c You will need the value of special cases.4 To get αLT.6 UK NA sets β = 0.49 0.0 kc can be obtained from Table 6.0.3 You can use a modified value of χLT in the special case to give some extra resistance: where L is the distance between points of lateral restraint (Lcr) E is the Young’s Modulus = 210000 N/mm2 G is the shear modulus = 80770 N/mm2 Iz is the second moment of area about the weak axis It is the torsion constant Iw is the warping constant .56) Mcr Refer to SN003 document (NCCI) for detailed description of how to get Mcr d c d d 0.34 0.1 finished hollow sections Angles (for moments in the major principal plane) and other hotrolled sections Welded bi-symmetric h/b ≤ 2 sections and coldh/b > 2 formed hollow sections Table from NA.21 0.3(1) and then refer to table 6.6 in the Eurocodes: d 0.3 to get the corresponding value of αLT Crosssection Rolled I sections Welded I sections Other Buckling curve αLT (6.17 Clause 6.2.75 and = 0.49 0.3.8)2] but f ≤1.76 EN 1993-1-1 Table 6.2.3 Special Case (for rolled sections): (6.3. for both the general and (6.0 < h/b ≤ 3.5(1 .4 a b c .21 0.17 Clause 6.4 and then refer to table 6. determine the buckling curve that you need to use from table 6.76 EN 1993-1-1 Table 6.kc)[1-2.58) f= 1.3 to get the corresponding value of αLT Cross-section Limits Buckling Curve Rolled bi-symmetric I h/b ≤ 2 and H sections and hot2.0.2.0( Limits Buckling Curve h/b ≤ 2 a h/b >2 s h/b ≤ 2 c h/b >2 d d EN 1993-1-1 Table 6.2.To get αLT. use expression 6. the value of C1 is obtained from Table 3.3 to get a value of αLT b. refer to the table in the National Annex (NA.54.3 to get the value of αLT 7. Calculate the design buckling resistance Mc. For the general case use expression 6. . Work out χLT a. Carry out the buckling resistance check in expression 6.Rd using equation 6. For the general case use expression 6.2. For the special case.2 in SN003: 1.2 from SN003 (C1 and C2 values for transverse loading) For members with end moments.56 b. Work out ΦLT a. Refer to SN003 document and work out the value of Mcr. Lcr 4. For the special case.56. use expression 6.k is an effective length factor (usually 1. 3.2. Determine fy (UK NA recommends you use the product standards) and calculate the class of the section.56 b. Work out the effective length.3.3(1)) to get the buckling curve and then refer to Table 6. Work out Table 5. Once you know the class of the section then you will know which value of the section modulus you will need to use in the equation 6.55. Determine the values of αLT a. Draw the bending moment diagram to obtain the value of the maximum bending moment.4 to work out the buckling curve and then refer to Table 6.57 8.17 Clause 6.57 9. Table 3.0) kw is an effective length factor (usually 1.55. the critical moment 5.0) zg is the distance between the point of load application and the shear centre. 6. For transverse loading. For the special case. The value will be positive or negative depending on where the load is applied as shown in figure 1.1 in SN003: using expression 6.1 from SN003 Summary Figure 1 (from SN003 document) C1 and C2 are coefficients. MEd 2. For the general case use Table 6.1 from SN003 (Values of C1 for members with end moments) where Figure 3. C1 and C2 are obtained from Table 5. 10. . 11) γM0 =1. For Classes 1. 2 and 3: (6.11 for Class 4 sections. Get ε from Table 5. using equation 6. they are prone to buckling.10) For Class 4 sections: (6. Table 1 shows the checks required for both slender and stocky columns: Slender column > 0.Steel Design to Eurocode 3 Compression Members must be less than the design cross-sectional resistance of the sections to uniform compression force (Nc.0 Resistance checks required for slender and stocky columns Cross-Section Resistance EN 1993-1-1 Clause 6. if they are used with fasteners in connections. Nmax = Npl = Aeff fy Cross-section Classification Summary 1.4 Equation 6. please refer to the ‘Cross-section Classification’ handout.Rd) (6.9) Columns are vertical members used to carry axial compression loads and due to their slender nature. Determine the design compression force 2. 4.9 states that the design value of the Compression force (NEd) For a more detailed description of cross-section classification.Rd Buckling Resistance Check. web in bending and web in compression results to get the overall section class Figure 2: Resistance of columns depends on different factors Eurocode 3 Approach To take into account the various imperfections which the Euler formula does not allow for. .2 Cross-section Resistance check. Nb.10 for Class 1. Effective Area Aeff The effective area of the cross-section used for design of compression members with Class 1. 2 or 3 cross-sections. Substitute the value of ε into the class limits in Table 5. Choose a section and determine the section classification 3. need not be deducted. Determine Nc. Holes. and equation 6. Take the least favourable class from the flange outstand. Carry out the cross-sectional resistance check by ensuring equation 6.2 to work out the class of the flange and web 4.9 is satisfied.2 3.Rd. Get fy from Product Standards 2. Nc. Cross-section Summary Resistance Check 1.0 Figure 1 Behaviour of columns is determined by their slenderness Stocky Columns are not affected by buckling and the strength is related to the material yield stress fy.2 Stocky Column < 0.2 and 3 sections.Rd Table 1. the Eurocode uses the Perry-Robertson approach.2. is calculated on the basis of the gross cross-section using the specified dimensions. This is approach is the similar to that used in BS 5950. The behaviour of a column will depend on its slenderness as shown in Figure 1 Cross-section resistance in compression depends on cross-section classification. (6. Calculate 6.51) or Similarly to cross-section resistance. 2. 7.48) γM1 =1.1.Member Buckling Resistance EN 1993-1-1 Clause 6.4 in the Eurocodes by using and the required buckling curve.76 EN 1993-1-1 Table 6. Calculate χ by substituting in the values of Φ and 9. .47 or 6. a0 a b c d 0. buckling resistance is dependent on the cross-section classification.46) Non-dimensional Slenderness For sections with Classes 1.Rd) (6. Make sure that the conditions of equation 6.13 0. Buckling Resistance Check Summary Pinned Pinned Fixed . 2 and 3: (6.49 0. For sections with Classes 1.48 and substituting in the value of χ 10. Table 6. Determine α by first determining the required buckling curve from Table 6. For Class 4 sections: (6. Determine the design buckling resistance of the member by using equation 6.Fixed Fixed .2 in EN 1993-1-1 provides guidance on a range of sections. χ may be read from Figure 6. 3. Some typical effective lengths are given in Figure 3.49) where Alternatively. EC3 gives no direct guidance on calculating the buckling length.47) Imperfection Factor.34 0.0 is an imperfection factor. first you will need to determine the required buckling curve from Table 6. χ Effective Buckling Lengths The effective length of a member will depend on its end conditions.1 Equation 6.46 states that the design values of the Compression force (NEd) must be less than the buckling resistance of the compression member (Nb.21 0. Determine the design axial load. and E and I which are section properties 5.50) or For Class 4 sections: (6.Pinned Figure 3: Effective Lengths for three types of end conditions Elastic Critical Buckling Load Ncr is the elastic critical buckling load for the relevant buckling mode based on the gross properties of the cross section 1.46 are satisfied. therefore it is acceptable to use those given in BS 5950 Table 13. NEd Choose a section and determine the class Calculate the effective length Lcr Calculate Ncr using the effective length Lcr.2 and then reading off the required value of α from Table 6.1 to get the value of : Buckling Curves Buckling Curve Imperfection Factor Buckling curve selection is dependent on the section geometry. 4. Calculate Φ by substituting in the values of α and 8.1 Reduction Factor.3. 2 and 3: where (6.2 and refer to Table 6. Rd: Anet for Staggered Fasteners: (6. t ≤ 16 Tension Members Resistance of cross-sections Resistance of cross-sections in tension to fracture 63 < t ≤ 80 Steel Design to Eurocode 3 Anet for Non staggered fasteners Anet = A – Σd0t Design Plastic Resistance. Value 1.Rd Design Ultimate Resistance Nu. For a member that is purely in tension. bars or flats.Rd) UK N. γM2 16 < t ≤ 40 As the tensile force increases on a member it will straighten out as the load is increased. and is concerns with the ultimate fracture of the net cross-section.Rd 3 < t ≤ 100 Steel grade fu (N/mm2) t<3 fy (N/mm2) 430580 510680 410560 470630 Extract from Table 7 of EN 10025-2 (6. and is concerned with the yielding of the gross crosssection.Ed) must be less than the design tensile resistance moment (Nt. we do not need to worry about the section classification since it will not buckle locally. Npl.A.7 gives the expression used to calculate Nu.Rd is the design ultimate resistance of the net cross-section.5 states that the design tensile force (Nt.Rd Nu.Rd: The total area to be deducted should be taken as the greater of: a) The maximum sum of the sectional areas of the holes on any line perpendicular to the member axis (6. Tensile Resistance EN 1993-1-1 Clause 6.Rd Npl.0 Characteristic Strengths fy and fu 40 < t ≤ 63 A tension member fails when it reached the ultimate stress and the failure load is independent of the length of the member. N u. Equation 6.2.6 gives the expression used to calculate Npl. Equation 6.25 The UK National Annex says you should get the values of fy and fu from the product standards.7) b) .5) The tensile resistance is limited by the lesser of: 1. which will normally occur at fastener holes.Partial Factors γM γM γM0 S 275 275 265 255 245 S 355 355 345 335 325 Design Plastic Resistance Npl. Tension members are generally designed using rolled section. For hot-rolled sections you can use the table below.3(1) Equation 6.6) Design Ultimate Resistance.Rd is the plastic design resistance. 5) .11) For 2 bolts: For 1 bolt: (3. Refer to EN 1993-1-8.5 d0 ≥ 5. Find fy and fu from the product standards 5. use the required equation to work out N u.11) For 3 or more bolts: For 2 bolts: (3.Tension Member Design Steps Summary where: t is the thickness of the plate 1.13) 6.Rd 7. Determine the design axial load NEd p is the spacing of the centres of the same two holes measured perpendicular to the member axis 2.6) d0 is the diameter of the hole Angles with welded end connections (6.7) Clause 4.Rd (6. Angles Connected by a single row of bolts For angles connected by a single row of bolts. or unequal angle welded along its larger leg. Get the gross area A and the net area Anet n is the number of holes extending in any diagonal or zig-zag line progressively across the section 3. Substitute the values into the equations to work out Npl.13) For 3 or more bolts: (3.Rd and Nu. Carry out the tension check: Values of reduction factors β2 and β3 can be found in Table 3.Rd and Nu. the effective area = gross area. Choose a section s is the staggered pitch of the two consecutive holes 4. The design tensile Resistance is the lesser of the values of Npl.13(2) of EN 1993-1-8 states that for an equal angle.12) (3.0 d0 β2 (for 2 bolts) 0.7 more bolts) Note: For intermediate values of pitch p 1 values of β may be determined by linear interpolation.8: Pitch p1 ≤ 2.Rd from EN 1993-1-8 which will depend on the number of bolts.5 0.8 (6. EN 1993-1-8Table 3.4 0. For 1 bolt: (3.12) (3.7 β3 (for 3 or 0. 2 Clause 6. Second-order effects of the sway system (P-Δ effects) have to be taken into account.3.Rk axis according to 6.g.Ed design values of the compression force and the maximum moments about the y-y and z-z axes along the member. a distinction is made for: members not susceptible to torsional deformation (e.3.3(1) When checking uniform members in bending and axial compression. fully restrained members) members susceptible to torsional deformation NEd My.Steel Design to Eurocode 3 Combined axial compression and bending Clause 6. χLT kyy. kzy.3. whether by the end moments of the member or by means of appropriate buckling lengths respectively.2 interaction factors kij.61 Uniform members in bending and axial compression demonstrate complex structural behaviour Interaction Method Equation 6.2. where: Clause 6.3.3.3.3.3(2) The resistance of the cross-sections at each end of the member should also satisfy the requirements given in Clause 6.3 Χy reduction factors due to flexural buckling χz from clause 6. . Mz.3(3) For members of structural systems the resistance check may be carried out on the basis of the individual single span members regarded as cut out of the system.1 Clause 6. CHS.Rd moments due to the shift of the centroidal ΔMz.9. kzz reduction factor due to lateral torsional buckling from clause 6.Ed respectively ΔMy.3 of EN 1993-1-1. kyz. SHS.62 When using the interaction method you will need to refer to Clause 6. and you will also need to refer to Annex A or B depending on the specific method being used.3(4) Members which are subjected to combined bending and axial compression should satisfy both: Equation 6. Ed NRk = fyAi Mi.Rk = fyW i Class 1 2 3 4 AI A A A Aeff Wy Wpl.z W el. .0 NOTE 2: Nb.y W eff.Ed 0 0 0 eN.Rk and ΔMi.z ΔMY.y W el.yNEd ΔMz.z Wpl.0 depend on the shape of the bending moments diagram and these factors are determined from Table A.Rd = χ NRk/γM1 Interaction Factors kij Interaction factors are obtained from one of two methods: Method 1 (given in Annex A) Method 2 (given in Annex B) Annex A (Method 1) • Use Table A.Table 6. Determine the equivalent uniform moment factors from Table B.Ed 0 0 0 eN.2 of EN 1993-1-1 Annex B (Method 2) • • Use Table B.1 of EN 1993-1-1 for members not susceptible to LTB Use Table B.3 of EN 1993-1-1.7 – Values for NRk.y Wpl.2 of EN 1993-1-1for members that are susceptible to LTB. Mi.z W eff.1 of EN 1993-1-1 Equivalent uniform moment factors Cmi.zNEd NOTE1 : For members not susceptible to torsional deformation χLT would be χLT = 1.y Wz Wpl. 3 (4) .1 Interaction factors kij for interaction formula in clause 6.Annex A Table A.3. 2 – Equivalent uniform moment factors Cmi.0 .Table A. Interaction factors kij for members not susceptible to torsional deformations Table B.Annex B Table B.Interaction factors kij for members susceptible to torsional deformations .1 .2 . 2 .Table B.1 and B.3 Equivalent uniform factors Cm in Tables B. 6 0.6 0.2. γM for joints are given in Table 2.2 ‘Nominally pinned’ joints are capable of transmitting internal forces without developing significant moments.2.5 0.1 of EC 3-8.5 0.5 0.Steel Design to Eurocode 3 EN 1993-1-8 Clause 2.5 .25 Joint Types CL 5.63 for countersunk bolt. CL 5.8 10.4 of EN 1993-1-8 gives the different checks required for individual fasteners subjected to shear and/or tension.6 0.25) Bolt Strength These values should be adopted as characteristic values in design calculations : Bolt classes 4. Bolted Joints – Table 3.2.4 Table 3. Checks need to be carried out for a number of possible failure modes: ‘Semi-rigid’ joints lie somewhere between ‘nominally pinned’ and ‘rigid’.9 Bolted Joints – Shear Shear resistance per shear plane for ordinary bolts where the shear plane passes through the threaded portion of the bolt: where: • As is the tensile stress area of the bolt • fub is the ultimate tensile strength of the bolt • γM2 = 1.6 5.6 8.6 5.Nominal values of fyb and fub for bolts Steel Strength 2 2 fy (N/mm ) Steel grade S 275 S 355 fu (N/mm ) Nominal thickness of element t (mm) t ≤ 16 16 < t ≤ 40 40 < t ≤ 63 63 < t ≤ 80 275 265 255 245 355 345 335 325 Nominal thickness of element t (mm) 3≤t t<3 ≤ 100 430 410 to to 580 560 510 470 to to 680 630 Extract from Table 7 of EN 10025-2 where: • As is the tensile stress area of the bolt • γM2 = 1.2 • Joints Partial safety factors. • Eurocode 3 Part: 1-8 Refer to NA to get the required values of the different partial safety factors • Resistance of bolts and welds.9 fyb (N/mm2) 240 300 640 900 fub (N/mm2) 400 500 800 1000 EN 1993-1-8 Table 3. Eurocode 3 • Principles mostly the same as BS 5950 • Results are similar although EC3 results are slightly more conservative and this is due to the larger • Shear resistance per shear plate • Bearing Resistance • Tension Resistance • Combined shear and tension Bolted Joints – Tension Tension resistance for ordinary bolts: partial safety factor (γM2=1. otherwise k2 = 0.1 .25 • fub is the ultimate tensile strength of the bolt • k2 = 0. γM2 = 1.2.8 10.8 8.8 6.9 αv 0.6 4.25 Bolt classes 4. and capable of accepting the resulting rotations under the design loads.8 5.3 ‘Rigid and full strength’ joints have sufficient rotational stiffness to justify analysis based on full continuity. 1) γM2 = 1.Rd where: • A is the gross cross-section of the bolt • fub is the ultimate tensile strength of the bolt • γM2 = 1.Ed ≤ Fw.25 Image Source: ESDEP Minimum distance Steel grade Correlation factor βw End distance e1 1. end and edge distances (4.Ed Bolted Joints – Bearing Fw.2) is the design value of the weld force per unit length is the design resistance per unit length Bearing resistance for ordinary bolts: Fw.1 from EN 1993-1-8: Values for correlation factor βw .25 Fw.d is the design shear strength of the weld a is the effective throat thickness (see Figure 1) for end bolts • for inner bolts Perpendicular to the direction of load transfer: • for edge bolts • for inner bolts Figure 1 – effective throat thickness Image Source: Design of Structural Elements (Arya.1 EN 1993-1-8) Fw.4) fu is the minimum ultimate tensile strength of the connected parts βw is a correlation factor (See Table 4.Rd (4.3 of EN 1993-1-8 gives the maximum and minimum spacing.Rd = fvw.3 and Table 4.2d0 S275 0.4d0 where d0 =hole diameter Extract from Table 4. 2009) Page 421 Bolted Joints – Position of Holes Table 3.85 Edge distance e2 1.3.da where: • • • • d t γM2 fu is the bolt diameter is the thickness = 1.Shear resistance per shear plane for ordinary bolts where the shear plane passes through the unthreaded portion of the bolt: Welded Joints Simplified method for design resistance of fillet weld (CL 4.2d0 S355 0.2d0 Spacing p2 2.5.25 is the ultimate tensile strength • fvw.90 Spacing p1 2.