Structural Steelwork Eurocodes Development of A Trans-national Approach

Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode Module 7 : Worked Examples Lecture 22 : Design of an unbraced sway frame with rigid joints Summary: NOTE This example is a draft version Pre-requisites: Notes for Tutors: This material comprises one 60 minute lecture. Objectives: To explain the main principles of EC by practical worked example. References: Eurocode : Design of steel structures Part 1.1 General rules and rules for buildings Contents: 15/02/07 2

WORKED EXAMPLE Design of a Sway Frame The frame consists of three storeys and three bays. The frames are at 10 m spacing. The beam span is 6,5 m and the total height is 15 m, each storey being,5 m high. It is assumed that the column foot is pinned at the foundation. Roof beams Floor beams,5 m Internal columns Internal columns,5 m External columns External columns,5 m 6,5 m 6,5 m 6,5 m Figure 1 Typical Cross Section of Frame The structure is assumed to be braced out of its plane and to be unbraced in its plane. In the longitudinal direction of the building, i.e. in the direction perpendicular to the frame plane, a bracing does exist so that the tops of 15/02/07

the columns are held in place. The lateral support for the floor beams is provided by the floor slabs. All the beam-to-column joints are assumed to be perfectly rigid. The connections must be capable of transmitting the forces and moments calculated in design. With these assumptions, the frame is classified as continuous, and the internal forces and moments are determined using a global elastic analysis which assumes the members to be effectively held in position. The steel grade selected for beams, columns and joints is Fe60 (f y = 25 N/mm 2 ). Characteristic Loads Floor: Variable actions, Q k = 1,8 kn/m 2, Permanent actions, G k =,0 kn/m 2 Roof: Variable actions, Q k = 6 kn/m 2, Permanent actions, G k = 2,0 kn/m 2 The wind loads are applied as point loads of 15 kn at roof level and 21 kn at the first and second storey levels. The basic loading cases, shown in Figure 2, have been considered in appropriate combinations. 20 kn/m 0 kn/m 15 kn 21 kn 6.4.2.2() 5.2.2..1 0 kn/m 21 kn Permanent Loading (G) Wind Loading (W) 6 kn/m 6 kn/m 6 kn/m 6 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m 18 kn/m loading case 1 (N1) loading case 2 (N2) loading case (N) Imposed Loading Cases Figure 2 Loading Cases 15/02/07 4

Frame Imperfections Frame imperfections are considered by means of equivalent horizontal loads. The initial sway imperfection is given as: φ = k c k s φ 0 5.2.4.(1) where k = k s c 5+ 1 n c but k c 1, 1 = 2 + but k s 1, and n s 1 φ 0 =. 200 In this case, the number of full height columns per floor, n c, is 4 and the number of storeys in the frame, n s, is. Therefore k c = 5+ 1 = 866, and 4 1 k s = 2 + = 7. Substituting into the above equation: 1 φ = 866 x 7 x 200 = 1 15 The equivalent horizontal load, H, at each storey of the frame is derived from the initial sway, φ, and the total design vertical load, N, in any storey for a given load case. Therefore H = φn. The relevant values are listed in 1 for all the basic loading cases. Basic loading case (see Figure 2) Storey N (kn) φn (kn) G Roof 90 1,24 2nd Floor 585 1,86 1st Floor 585 1,86 N1 Roof 117 7 2nd Floor 51 1,11 1st Floor 51 1,11 N2 Roof 9 12 2nd Floor 24 74 1st Floor 117 7 N Roof 78 25 2nd Floor 117 7 1st Floor 24 74 1 Equivalent Horizontal Forces The equivalent horizontal forces must also be multiplied be the appropriate partial safety factors for actions. 15/02/07 5

Load Combinations It was decided to use the simplified combinations for the ultimate limit state and the serviceability limit state. The basic load cases are combined at the ULS as summarised in 2. Load Case 1 1,5G + 1,5W Load Case 2 1,5G + 1,5N1 Load Case 1,5G + 1,5N2 Load Case 4 1,5G + 1,5N Load Case 5 1,5G + 1,5W + 1,5N1 Load Case 6 1,5G + 1,5W + 1,5N2 Load Case 7 1,5G + 1,5W + 1,5N 2 Load Combination Cases at the Ultimate Limit State The basic load cases are combined at the SLS as summarised in. Load Case 1 1,0G + 1,0W Load Case 2 1,0G + 1,0N1 Load Case 1,0G + 1,0N2 Load Case 4 1,0G + 1,0N Load Case 5 1,0G + 9W + 9N1 Load Case 6 1,0G + 9W + 9N2 Load Case 7 1,0G + 9W + 9N Load Combination Cases at the Serviceability Limit State Partial Safety Factors for Strength The following partial safety factors for strength have been adopted during the design: Resistance of Class 1,2 or cross-section, γ M0 = 1,1 Resistance of member to buckling, γ M1 = 1,1 Resistance of bolted connections, γ Mb = 1,25 Trial Sections In order for a global elastic analysis of the structure to be carried out, initial section sizes must be assumed and allocated to the structural members. The analysis must then be carried out and the members checked for the relevant failure modes. The sections will then need to be modified and the structure re-analysed. This can be a long winded iterative process. The engineer may have his own method of selecting initial section sizes. As a guideline, for this example, columns were selected by assuming an average stress of approximately 100 N/mm 2 under axial forces. Axial forces can be estimated by approximating the floor area supported by that column. Generally, the bending moments withing the beams are critical. Simple bending moment diagrams can be constructed, assuming fixed end moments, and the maximum bending moment can then be estimated. An initial section size can then be determined. The trial member sizes for this example are: Inner columns: HEB 260 Outer columns: HEB 220 2...1(5) and 2..4(5) 2...2(1) 5.1.1(2) 5.1.1(2) 6.1.1(2) 15/02/07 6

Floor beams: IPE 450 Roof beams: IPE 60 Roof beams IPE 60 Floor beams IPE 450,5 m Internal columns HEB 260 Internal columns HEB 260,5 m External columns HEB 220 External columns HEB 220,5 m 6,5 m 6,5 m 6,5 m Figure Trial Member Sizes Determination of Design Moments and Forces This worked example uses the amplified sway moments method of analysis. An alternative method is to calculate the sway-mode buckling lengths of members, and then carry out a first order linear elastic analysis. However, it may be possible to directly use a second order elastic analysis. A global linear elastic analysis is carried out on the sway frame in order to determine the moments, axial forces and shear forces in each member. Props are then applied to the structure in order to prevent any horizontal movement (i.e. as if the structure was now braced), and the analysis carried out again. The moments induced in the members from the propped case are then subtracted from the moments determined from the sway case. The resultant moments are those induced by pure sway. The pure sway moments are then amplified to take account of second order effects ignored by the linear elastic analysis. The amplified pure sway moments are then added to the moments obtained from analysis of the propped structure. These are the design moments which each member must be able to resist. The members must also be checked for the axial and shear forces determined from the initial analysis of the sway frame, and in certain cases the interaction of the design moments with the axial forces and/or shear forces must also be checked. This needs to be carried out for every load case so that the critical conditions for each individual member can be identified. 5.2.6.2(1) 5.2.6.2(1) 5.2.6.2(1) 15/02/07 7

Original Sway Case Moments Minus Propped Case Moments Equals "Pure Sway" Moments (To Be Amplified) 15/02/07 8

Figure 4 Determination of Pure Sway Moments Calculation of Amplification Factors The sway moments should be increased by multiplying them by the ratio: 1 1 VSd / Vcr where V Sd is the design value of the total vertical load, and V cr is its elastic critical value for failure in a sway mode. Instead of determining V Sd / V cr directly, the following approximation may be used: VSd V = V δ h H cr where δ is the horizontal displacement at the top of the storey, relative to the bottom of the storey, h is the storey height, V is the total vertical reaction at the bottom of the storey, and H is the total horizontal reaction at the bottom of the storey. This approximation may not be used if V Sd / V cr is greater than 25. If V Sd / V cr is greater than 25, then the frame may be more susceptible to buckling. It is therefore necessary to carry out analysis using a direct second order analysis. It may also be necessary to stiffen the frame - for example increasing the column sizes. Alternatively, if the value of V Sd / V cr is less than 1 then the structure is classified as non-sway. The amplification factor will be different for each storey of the structure. The maximum factor should be used to multiply the moments at all levels of the structure. This is essentially a conservative method as it corresponds to the critical elastic load of the whole structure. The amplification factor for load case 5 of this example was determined as follows: 5.2.6.2() 5.2.6.2(6) 5.2.5.2(4) 5.2.6.2(4) 5.2.5.2() 15/02/07 9

δ 2 δ = δ2 - δ = 15,6 mm 1 h = 500 mm V = 296 kn H = 81 kn h H V δ1 Figure 5 Determination of Amplification Factor Therefore V Sd V 15,6 x 296 = V δ cr h = = 18 H 500 x 81 1 1 The amplification factor = = = 1, 22 1 VSd / Vcr 1 18 All the pure sway moments for load case 5 were amplified by a factor of 1,22. The global linear elastic analysis and amplification of sway moments was carried out for all seven load cases. s 4a and 4b shows the maximum forces in each member. N 19 P 20 Q 21 R 15 16 17 18 12 K 1 L 14 J M 8 9 10 11 5 F 6 G 7 E H 9 = Element G = Node 1 2 4 A B C D Figure 6 Labelling of Members within the Structure 15/02/07 40

Membe r Node Tension (kn) Compressio n (kn) Shear (kn) Bending Moment (knm) 0 Maximum forces 1 A - 517,6 C2-1,1 C4 E - -514, C2 1,1 C4 44,7 C4 2 B - 1196,4 C2 5,9 C7 0 Axial ld + BM F - -1192,1 C2-5,9-142,1 C7 internal col C7 C - 1196, C2 8 C6 0 G - -1192,0 C2-8 -124,0 C6 C6 4 D - 529,0 C5 27,6 C7 0 Axial ld + BM H - -525,7 C5-27,6-105,5 C7 external col C7 8 E - 17,9 C2-4,9 79,8 C2 C2 J - -14,6 C2 4,9 C2 7,4 C2 9 F - 722,8 C2 25,8 C5 58,6 C7 K - -718,5 C2-25,8-69,2 C6 C5 10 G - 722,8 C2 11,5 C1 1,2 C4 L - -718,5 C2-11,5-45,9 C7 C1 11 H - 20 C2 51, C5-91,7 C5 M - -16,7 C2-51, -9,7 C5 C5 15 J - 11,0 C4-4,0 7,5 C C2 N - -109,7 C4 4,0 C2 77, C2 16 K - 25,9 C2 15,2 C6-41,9 C6 P - -249,6 C2-15,2-6,7 C6 C6 17 L - 25,9 C2-9,8 C 4,6 C Q - -249,6 C2 9,8 C 24,5 C4 18 M - 11,5 C4 44,8 C5-76,0 C5 R - -111 C4-44,8 C5-82,4 C5 4a Maximum Axial Forces, Shear Force and Bending Moment in Each Column Shaded boxes indicate the critical case that each section type will be designed to. For example, internal columns are the same section for every storey, therefore the worst case will obviously be one of the base columns. 15/02/07 41

Membe r Node Tension (kn) Compressio n (kn) Shear (kn) Bending Moment (knm) 5 E -28,9 C2 8,5 C1 201,8 C4 F 28,9 C2-8,5 C1 255,2 C5 6 F -2,4 1,4 C1 228-241,2 C2 C2 C G 2,4 C2-1,4 C1 21,6 17 C5 C5 7 G -1,5-244,2 C2 C2 H 1,5 C2-21, C7 12 J -6 G 28,1 C5 207,1-146,8 C2 C K 6 G -28,1 C5 242,7 288,6 C5 C2 1 K -1, C4 15,6 C5 222,0-242,6 C2 C4 L 1, C4-15,6 C5 22,6 261, C5 C2 14 L -1,9 G 6,5 C5 241,0-271,5 C2 C2 M 1,9 G -6,5 C5 208,5 169,7 C5 C 19 N - 55,8 C5 109,7 C4 P - -55,8 C5 14 C2 20 P - 42,4 C5 119, -11,7 C2 C Q - -42,4 C5 119,7 15,8 C5 C2 21 Q - 44,8 C5 129,9 C2 R - -44,8 C5 111 C4 4b Maximum Axial Force, Shear Force and Bending Moment in Each Beam Maximum forces -118,5 C4 Internal BM 41,6 C5 on floor beam -271,7 C2 External BM 194,4 C5 on floor beam -77, C2 Internal BM 152 C5 on roof beam -146,7 C2 External BM 82,4 C5 on roof beam 15/02/07 42

Shaded boxes indicate the critical case that each section type will be designed to. Generally, the beams will be designed to the maximum bending moment, and the columns will be designed for maximum axial load. However, it may also be necessary to check the interaction of the bending moments and axial forces for certain load cases. Floor Beam - Fully Restrained The maximum moment any floor beam has to resist is 41,6 knm at node F, under load case 5. The beam is in tension for this load case, therefore the interaction of the bending moment and axial force will not need to be checked. The maximum compression force any floor beam has to resist is 28,1 kn (member 12) under load case 5. The beam also has to resist a bending moment of 288,6 knm but as the axial force is low it is unlikely that the interaction will be critical. The maximum shear force any floor beam has to resist is 255,2 kn at node F, under load case 5. The maximum tensile force any floor beam has to resist is 1,5 kn (member 7) under load case 2. This will not be the critical condition for the floor beam design. IPE 450 Section properties: Depth, h = 450 mm, Width, b = 190 mm Web thickness, t w = 9,4 mm Flange thickness, t f = 14,6 mm Plastic modulus, W pl = 1702 cm This notation conforms with Figure 1.1 in Eurocode : Part 1.1. Classification of Cross-section Figure 7 shows a typical cross-section for an IPE. 5.4.5.1 15/02/07 4

t t w f Figure 7 A Typical Cross- Section c d Flange Buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 25 / f y and f y = 25 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half t f the width of the flange = 95 mm, and t f is the flange thickness = 14,6 mm (if the flange is tapered, t f should be taken as the average thickness). c 95 = = 6, 5 t f 14, 6 Web Buckling Class 1 limiting value of d/t w for a web of a rolled section under bending is 72ε. ε = 25 / f y and f y = 25 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 78,8 mm t w and t w is the web thickness = 9,4 mm. d 78, 8 = = 4 t w 9, 4 c < 10ε and d < 72ε t t f w Section is Class 1. The beam shown in Figure 8 is fixed at both ends and is fully restrained along its length. For the critical load cases given above, check the beam, assuming it is grade Fe6 and that it is carrying plaster, or a similar brittle finish. Load varies depending on appropriate load case 5..1 (Sheet ) 5..1 (Sheet 1) 5..1 (Sheets 1 and ) 6,5 m Figure 8 Loading on Fully Restrained Floor Beam Check Moment Capacity Design moment, M Sd = 41,6 knm (load case 5 - member 5) The design bending moment, M Sd, must be less than or equal to the design moment resistance of the cross section, M c.rd : M Sd M c.rd M c.rd = M pl.y.rd = W f pl y γ M0 5.4.5.1(1) 15/02/07 44

where W pl is the plastic section modulus = 1702 cm, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. Therefore, M pl.y.rd = W f pl y 1702x10 x 25 = = 64 knm γ 1,1 M0 M Sd = 41,6 knm M pl.y.rd = 64 knm Therefore the section is satisfactory. Check Interaction of Maximum Axial Force and Bending Moment Interaction case: Design moment, M Sd = 41,6 knm and design axial force, N Sd = 28,1 kn (load case 5 - member 12) Lateral support to the floor beams is provided by the floor slabs, therefore there is no need to check for failure due to flexural or lateral torsional buckling. A class 1 member subject to moment about the major axis only should satisfy the following: 2 MSd NSd + 1 Mpl.Rd Npl.Rd Applied bending moment, M Sd, = 41,6 knm, design bending moment resistance of the section, M pl.rd = 64 knm, applied axial load, N Sd, = 28,1 kn, and.1 5.1.1(2) 5.4.8.1(2) design compression resistance of the cross-section, N pl.rd = Af y γ M0 where A is the cross-sectional area of the section = 9880 mm 2, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 9880 x 25 Therefore, N pl.rd = = 2111 kn 11, Substituting into the above equation: 2.1 5.1.1(2) 6 41, 6x10 28, 1x10 + 94 6 64x10 2111x10 = Therefore the section is satisfactory under combined axial load and bending moment Lateral Torsional Buckling Although lateral restraint is provided by the floor slabs, it is necessary to check for lateral torsional buckling over the length of the beam which is in hogging (approximately 1,7 m). This is because the floor beams are restrained only by the upper flange, while the lower flange which is under compression, has no lateral restraint. As this is over such a short length of the beam, it is unlikely that this will be a critical condition. For completeness, however, the check is included in this worked example. A class 1 section should satisfy the following: 5.5.4(2) 15/02/07 45

N k M Sd LT y.sd + 1, 0 χ Af / γ χ W f / γ z y M1 LT pl.y y M1 Applied axial force, N Sd = 28,1 kn Determination of χ z Slenderness, λ z = l/i z where l is taken as the length of beam in hogging, and i is the radius of gyration about the appropriate axis. Slenderness, λ z = 1700 / 41,2 = 41, λ Non-dimensional slenderness, λ z z = β where λ 1 = 9,9ε = 9,9 x 1 = 9,9, and β A = 1 for class 1 members 41, Non-dimensional slenderness, λ z = = 44 9, 9 From 5.5.2, using buckling curve b, reduction factor, χ z = 91 Cross-sectional area, A = 9880 mm 2, Yield strength of the steel, f y = 25 N/ mm 2, and Partial material safety factor for buckling resistance, γ M1 = 1,1. i a Lt Lt λ 1 25 5 A 5.5.1.4 5.5.1.5(2) 5.5.1.2(1) 5.5.1.1(1) 5.5. 5.5.2.1 5.1.1(2) 6 9 IzIw x10 x 791x10 = 25 mm W = 16, 76 = 46, F.2.2() 2 2 ply ( 1702x10 ) 9 Iw x10 = 5 mm 4 = 5 I 791 F.2.2(1) = 1087 66,9x10 t 15/02/07 46

k LT To Calculate k LT LTNSd = 1 µ but k LT 1,0. χ Af z y where µ = 15λ β 15 but µ 9 LT z M.LT LT λ z = 44 β M.LT = 1,8 (ψ = 0) µ LT = (15 x 44 x 1,8) - 15 = -0 = ( 0) x 28,1x10 k LT 1 = 1, 0 91 x 9880 x 25 Applied moment, M y.sd = 41,6 knm To Calculate Reduction Factor, χ LT The value of χ LT can be determined from 5.5.2 for the appropriate value of the non-dimensional slenderness, λ LT. λ λ LT LT λ β 5 = w 1 where λ 1 = 9,9 ε = 9,9 x 1 = 9,9, β w = 1 for class 1 sections, and L / ilt λ LT = 2 25 ( L / alt) 5 C1 1 + 25, 66 where L is the length = 1700 mm, i LT = 46, mm (from section properties), ψ = k = 1, therefore C 1 = 1,879, and a LT = 1087 mm (from section properties). Substituting into the above equation: 1700 / 46, λ LT = = 26, 2 2 25 ( ) 5 1700 / 1087 1, 879 1 + 25, 66 λ λ LT LT = = = λ β 5 26, 2 0 5 w 9 9 1 0,,, 0, 28 1 Where the non-dimensional slenderness λ LT 4, no allowance for lateral torsional buckling is necessary. Therefore, this section is satisfactory for lateral torsional buckling. Shear on Web Design shear force, V Sd = 255,2 kn (load case 5 - member 5) The shear resistance of the web must be checked. The design shear force, V Sd, must be less than or equal to the design plastic shear resistance, V pl.rd : V Sd V pl.rd where V pl.rd is given by A f / y v γ M0 5.5.4(2) 5.5.4(7) and Figure 5.5. 5.5.2(4) 5.5.2(5) 5.5.2(1) Equation F.15 5.5.2(5) 5.5.2(7) 5.4.6 15/02/07 47

For rolled I and H sections loaded parallel to the web, shear area, A v = 1,04 h t w, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 1,04htwfy Vpl.Rd = = 1, 04 x 450 x 9,4 x 25 = 54 kn xγ M0 x 1,1 This is greater than the shear on the section (255,2 kn). The section is satisfactory under shear. A further check is sometimes required, especially when there are significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximum moment and shear are not coincident and this check is not required for beams without web openings. Check Tension Resistance of Member Maximum applied tensile force, N Sd = 1,5 kn (load case 2 - member 7) For members in axial tension, the design value of the tensile force, N Sd must be less than the design tension resistance of the cross-section, N t.rd. N t.rd = N pl.rd = A f y γ M0 where A is the cross-sectional area of the section = 9880 mm 2, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 9880 x 25 Therefore N pl.rd = = 2110 kn 11, The design tension resistance of the cross-section, N t.rd = 2110 kn, is greater than the applied tensile force, N Sd = 1,5 kn, therefore the section is OK. Deflection Check For a plaster or similar brittle finish, the deflection limit is L/250 for δ max. Deflection checks are based on the serviceability loading. The maximum deflection for any floor beam is 6,2 mm which occurs under serviceability load case 4 at member 5. L 6500 Deflection limit for δ max = = = 26 mm 250 250 The actual deflection is less than the allowable deflection: 6,2 mm < 26 mm, therefore the section is OK. The calculated deflections are less than the limits, so no pre-camber is required. 5.4.6(4).1 5.1.1(2) 5.4.7() 5.4.(1).1 5.1.1(2) 4.1 Figure 4.1 4.1 4..2(2) 15/02/07 48

Summary The trial section IPE 450 is satisfactory. Roof Beam The maximum moment any roof beam has to resist is 152 knm at node P (member 19), under load case 5. The maximum compression force any roof beam has to resist is 55,8 kn (member 19) also under load case 5. The interaction of the axial force and bending moment will therefore be critical. The maximum shear force any roof beam has to resist is 14 kn at node P, under load case 2. The roof beams experience no tensile force. IPE 60 Section properties: Depth, h = 60 mm, Width, b = 170 mm Web thickness, t w = 8,0 mm Flange thickness, t f = 12,7 mm Plastic modulus, W pl = 1019 cm This notation conforms with Figure 1.1 in Eurocode : Part1.1. Classification of Cross-Section Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 25 / f y and f y = 25 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half the width of the flange = 85 mm, t f and t f is the flange thickness = 12,7 mm (if the flange is tapered, t f should be taken as the average thickness). c 85 = = 6, 7 t f 12, 7 Web buckling Class 1 limiting value of d/t w for a web of a rolled section under bending is 72ε. ε = 25 / f y and f y = 25 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 298,6 mm t w and t w is the web thickness = 8,0 mm. d 298, 6 = = 7, t w 8, 0 c < 10ε and d < 72ε t t f w Section is Class 1. For the critical load cases given above, check the beam, assuming it is grade Fe6 and that it is for a general roof. Check Interaction of Maximum Axial Force and Bending Moment Interaction case: Design moment, M Sd = 152 knm and design axial 5.4.5.1 5..1 (Sheet ) 5..1 (Sheet 1) 5..1 (Sheets 1 and ) 15/02/07 49

force, N Sd = 55,8 kn (load case 5 - member 19). There is no lateral support provided for the roof beams, therefore they must be checked for failure due to flexural and lateral torsional buckling. Flexural Buckling A class 1 member subject to moment about the major axis only should satisfy the following: N k M sd y y.sd + 1,0 χ Af / γ W f / γ min y M1 pl.y y M1 Applied axial force, N Sd, = 55,8 kn To Calculate Reduction Factor, χ min The reduction factor χ min is the lesser of χ y and χ z, where χ y and χ z are the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively. Determination of χ y The reduction factor χ y depends on the slenderness about the y-y axis. Assuming that the connections at the beam ends are rigid, then the slenderness about the y-y axis is: λ y = l/i y where l is taken as the system length, and i is the radius of gyration about the appropriate axis. Slenderness, λ y = 6500 / 150 = 4, λ 1 = 9,9ε = 9,9 x 1,0 = 9,9 λ Non-dimensional slenderness, λ y y λ β 5 = A where β A = 1 for class 1 members 4, Non-dimensional slenderness, λ y = = 46 9, 9 From 5.5.2, using buckling curve a, the reduction factor, χ y = 95 Determination of χ z Slenderness, λ z = l/i z where l is taken as the system length, and i is the radius of gyration about the appropriate axis. Slenderness, λ z = 6500 / 7,9 = 171,5 λ Non-dimensional slenderness, λ z z λ β 5 = A where β A = 1 for class 1 members 171, 5 Non-dimensional slenderness, λ z = = 1, 8 9, 9 From 5.5.2, using buckling curve b, reduction factor, χ z = 2521 Reduction factor, χ min = χ z = 2521 A is the cross-sectional area = 7270 mm 2, f y is the yield strength = 25 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. 1 1 5.5.4(1) 5.5.1.5 5.5.1.4 5.5.1.5(2) 5.5.1.2(1) 5.5.1.1(1) 5.5. 5.5.2 5.5.1.4 5.5.1.5(2) 5.5.1.2(1) 5.5.1.1(1) 5.5. 5.5.2.1 5.1.1(2) 15/02/07 50

To Calculate k y yn sd k y = 1 µ but k y 1,5 χ y Af y where Wpl.y W µ y = λ y (2β MY 4) + W el.y el.y but µ y 90 β My is an equivalent uniform moment factor for flexural buckling β My = β M,ψ + M Q MQ M ( β β ) M, ψ where β M,ψ = 1,8-7ψ, and ψ = 47 Therefore β M,ψ = 1,8 - (7 x 47) = 1,47, M Q = 187 knm, M = 21 knm, and β M,Q = 1,. Therefore β My = 1,47 + (187/21) x (1, - 1,47) = 1, 1019-904 µ y = 46( 2 x 1, - 4) + = 49 904 (-49) x 55,8x10 k y = 1 = 1, 02 0, 95 x 7270 x 25 M y.sd is the design applied moment = 152 knm, and W pl.y is the plastic section modulus = 1019 x 10 mm. N k M sd y y.sd + 1,0 χ Af / γ W f / γ min y M1 pl.y y M1 6 55,8x10 1, 02 x 152x10 + = 85 2521 x 7270 x 25 / 1,1 1019x10 x 25 / 1,1 Therefore this section is satisfactory for flexural buckling Lateral Torsional Buckling A class 1 section should satisfy the following: N k M Sd LT y.sd + 1, 0 χ Af / γ χ W f / γ z y M1 LT pl.y y M1 Applied axial force, N Sd = 55,8 kn, Reduction factor, χ z = 2521, Cross-sectional area, A = 7270 mm 2, Yield strength of the steel, f y = 25 N/ mm 2, and Partial material safety factor for buckling resistance, γ M1 = 1,1. i a Lt Lt I I = W z w 2 ply 25 6 9 x10 x 14x10 = 1 4 2 ( 1019x10 ) 9 Iw x10 = mm 4 = I 14 = 917, 5 7,x10 t 5 5 25 = 42, 1 mm 5.5.4(1) 5.5.4(7) and Figure 5.5. 5.5.4(2).1 5.1.1(2) F.2.2() F.2.2(1) 15/02/07 51

k LT To Calculate k LT LTNSd = 1 µ but k LT 1,0. χ Af z y where µ = 15λ β 15 but µ 9 LT z M.LT LT λ z = 1,8 β M.LT = β M,ψ + (M Q / M)(β M,Q - β M,ψ ), where β M,ψ = 1,8-7ψ, and ψ = 47, therefore β M,ψ = 1,8 - (7 x 47) = 1,47, M Q = 187 knm, M = 21 knm, and β M,Q = 1,. Therefore β M.LT = 1,47 + (187/21) (1, - 1,47) = 1,. µ LT = (15 x 1,8 x 1,) - 15 = 21. 21 x 55,8x10 k LT = 1 = 97 2521 x 7270 x 25 Applied moment, M y.sd = 151 knm To Calculate Reduction Factor, χ LT The value of χ LT can be determined from 5.5.2 for the appropriate value of the non-dimensional slenderness, λ LT. λ λ LT LT λ β 5 = w 1 where λ 1 = 9,9 ε = 9,9 x 1 = 9,9, β w = 1 for class 1 sections, and L / ilt λ LT = 2 25 ( L / alt) 5 C1 1 + 25, 66 where L is the length = 6500 mm, i LT = 42,1 mm (from section properties), ψ = 47, k = 1, therefore C 1 = 1,107, and a LT = 917,5 mm (from section properties). Substituting into the above equation: 6500 / 42, 1 λ LT = = 111, 9 2 25 ( 6500 917 5) 5 /, 1107, 1 + 25, 66 λ λ LT LT = = = λ β 5 111, 9 0 5 w 1 9 9 1 0,,, 1, 2 Therefore, from 5.5.2, using buckling curve a (for rolled sections), the reduction factor, χ LT = 5 The plastic section modulus, W pl.y = 1019 x 10 mm. 5.5.4(2) 5.5.4(7) and Figure 5.5. 5.5.2(4) 5.5.2(5) 5.5.2(1) Equation F.15 F.1.2(6) 5.5.2(5) 5.5.(4), 5.5. 5.5.2 15/02/07 52

N k M Sd LT y.sd + 1, 0 χ Af / γ χ W f / γ z y M1 LT pl.y y M1 5.5.4(2) 6 55, 8x10 97 x 152x10 + = 1, 41 1 2521 x 7270 x 25 / 1,1 5 x 1019x10 x 25 / 1,1 Therefore, this section is not satisfactory for lateral torsional buckling. It is highly unusual to have, in practice, unrestrained roof beams. It is, therefore, asssumed that beams/purlins are present at quarter points along the beam, reducing the unrestrained length to 1,625 m. The maximum change in moment occurs between the rd restraint and node P 5.5.2(7) on member 19. When the check is carried out again, λ LT 4, therefore no allowance for lateral torsional buckling is necessary. Shear on Web Design shear force, V Sd = 14 kn (load case 2 - member 19) The shear resistance of the web must be checked. The design shear 5.4.6 force, V Sd, must be less than or equal to the design plastic shear resistance, V pl.rd : V Sd V pl.rd where V pl.rd is given by A f / y v γ M0 For rolled I and H sections loaded parallel to the web, 5.4.6(4) shear area, A v = 1,04 h t w, f y is the yield strength = 25 N/mm 2, and.1 γ M0 is the partial material safety factor = 1,1. 5.1.1(2) 1,04htwfy Vpl.Rd = = 1, 04 x 60 x 8,0 x 25 = 69 kn xγ M0 x 1,1 This is greater than the shear on the section (14 kn), therefore this section is satisfactory under shear. A further check is sometimes required, especially when there are 5.4.7() significant point loads, cantilevers or continuity, to ensure that the shear will not have a significant effect on the moment resistance. This check is carried out for the moment and shear at the same point. The moment resistance of the web is reduced if the shear is more than 50% of the shear resistance of the section. With a uniform load, the maximum moment and shear are not coincident and this check is not required for beams without web openings. Deflection Check For a roof generally the deflection limit is L/200 for δ max. Deflection checks 4.1 are based on the serviceability loading. Figure 4.1 The maximum deflection for any roof beam is 6,5 mm which occurs under serviceability load case 4 at member 19. L 6500 4.1 Deflection limit for δ max = = = 2, 5 mm 200 200 The actual deflection is less than the allowable deflection: 6,5 mm < 2,5 mm therefore the section is OK. The calculated deflections are less than the limits, so no pre-camber is 4..2(2) 15/02/07 5

required. Summary The trial section IPE 60 is satisfactory. Columns External Column The maximum compression force any external column has to resist is 529 kn (member 4) under load case 5. The maximum moment any external column has to resist is 106 knm at node H, under load case 7. The column is also subject to an axial load of 458,4 kn under this load case. However, the critical interaction case is likely to be under load case 5, with a maximum axial force of 529 kn and a bending moment of 10 knm. It is this case that is likely to be critical. The maximum shear force any external column has to resist is 51, kn at member 11 under load case 5. Section Properties All external columns are HE 220 B grade Fe60 h = 220 mm b = 220 mm t w = 9,5 mm t f = 16 mm d/t w = 16,0 c/t f = 6,9 A = 9100 mm 2 I y = 891 x 10 6 mm 4 I w = 295 x 10 9 mm 6 I z = 28,4 x 10 6 mm 4 I t = 76,6 x 10 4 mm 4 W pl.y = 827 x 10 mm W el.y = 76 x 10 mm i y = 94, mm i z = 55,9 mm i Lt = I I z w 2 ply W 25 6 9 x10 x 754x10 = 51, 4 2 ( 128x10 ) 5 5 25 = 69, 7 mm 9 Iw x10 alt = mm 4 = It 754 = 779, 8 124x10 All the above properties can be obtained from section property tables. Classification of Cross Section This section is designed to withstand moments in addition to axial force. (Note that the section is always in compression.) Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 25 / f y where f y = 25 N/mm 2, ε = 1. 10ε = 10 x 1 = 10 From section properties, c/t f = 6,9 Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is ε. ε = x 1 = From section properties, d/t w = 16 c/t f 10ε and d/t w ε Therefore the section is Class 1. F.2.2() F.2.2(1) 15/02/07 54 5. 5..1 (Sheet ) 5..1 (Sheet 1)

Resistance of Cross-Section 5.4.4 It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. For members in axial compression, the design value of the compressive 5.4.4(1) force, N Sd, at each cross-section shall satisfy N Sd N c.rd For a class 1 cross-section, the design compression resistance of the 5.4.4(2) cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 9100 mm 2, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 9100 x 25 N c.rd = = 1944 kn 11, x10 N Sd = 529 kn, therefore N sd < N c.rd. The section can resist the applied axial load..1 5.1.1(2) 15/02/07 55

Buckling Resistance of Member A class 1 member subject to combined bending and axial compression should be checked for the following modes of failure: Flexural buckling (Clause 5.5.4(1)), and Lateral Torsional Buckling (Clause 5.5.4(2)). Flexural Buckling A class 1 member subject to moment about the major axis only should satisfy the following: N k M Sd y y.sd + 1, 0 χ A f / γ W f / γ min y M1 pl.y y M1 Applied axial force, N Sd, = 529 kn To Calculate Reduction Factor, χ min The reduction factor χ min is the lesser of χ y and χ z, where χ y and χ z are the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively. Determination of χ y The reduction factor χ y depends on the slenderness about the y-y axis. The connections at the column base are effectively pinned, then the slenderness about the y-y axis is: λ y = l/i y where l is equal to the system length and i is the radius of gyration about the appropriate axis. (Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1) Slenderness, λ y = 500 / 94, = 7,1 λ 1 = 9,9ε = 9,9 x 1,0 = 9,9 λ Non-dimensional slenderness, λ y y λ β 5 = A where β A = 1 for class 1 members 7, 1 Non-dimensional slenderness, λ y = = 4 9, 9 From 5.5.2, using buckling curve b, the reduction factor, χ y = 9261 Determination of χ z Slenderness, λ z = l/i z where l is equal to the system length and i is the radius of gyration about the appropriate axis. (Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1) Slenderness, λ z = 500 / 55,9 = 62,6 λ Non-dimensional slenderness, λ z z λ β 5 = A where β A = 1 for class 1 members 1 1 5.5.4(1) 5.5.1.5 5.5.1.4 5.5.1.5(1) 5.5.1.2(1) 5.5.1.1(1) 5.5. 5.5.2 5.5.1.4 5.5.1.5(2) 5.5.1.2(1) 5.5.1.1(1) 15/02/07 56

62, 6 Non-dimensional slenderness, λ z = = 67 9, 9 From 5.5.2, using buckling curve c, reduction factor, χ z = 7 Reduction factor, χ min = χ z = 7 A is the cross-sectional area = 9100 mm 2, f y is the yield strength = 25 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. 5.5. 5.5.2.1 5.1.1(2) 15/02/07 57

To Calculate k y yn sd k y = 1 µ but k y 1,5 χ y Af y where Wpl.y Wel.y µ y = λ y( 2β My 4) + but µ y 90 W el.y β My is an equivalent uniform moment factor for flexural buckling (ψ = 0) = 1,8-0 = 1,8. 76 µ y = 4( 2 x 1,8-4) + 827x10 x10 = 04 76x10 ( ) = 04 x 529x10 k y 1 = 1, 01 9261 x9100 x 25 M y.sd is the design applied moment = 10 knm, and W pl.y is the plastic section modulus = 827 x 10 mm. N k M Sd y y.sd + 1, 0 χ A f / γ W f / γ min y M1 pl.y y M1 6 529x10 1, 01 x 10x10 + = 96 7 x 9100 x 25 / 1,1 827x10 x 25 / 1,1 Therefore the section is satisfactory for flexural buckling Lateral Torsional Buckling A class 1 section should satisfy the following: N k M Sd LT y.sd + 1, 0 χ Af / γ χ W f / γ z y M1 LT pl.y y M1 Applied axial force, N Sd = 529 kn, Reduction factor, χ z = 7, Cross-sectional area, A = 9100 mm 2, Yield strength of the steel, f y = 25 N/ mm 2, and Partial material safety factor for buckling resistance, γ M1 = 1,1. To Calculate k LT LTNSd k LT = 1 µ but k LT 1,0. χ Af z y where µ = 15λ β 15 but µ 9 LT z M.LT LT λ z = 67 β M.LT = 1,8 (ψ = 0) µ LT = (15 x 67 x 1,8) - 15 = 0 0 x 529x10 k LT = 1 = 99 7 x 9100 x 25 Applied moment, M y.sd = 102,8 knm To Calculate Reduction Factor, χ LT The value of χ LT can be determined from 5.5.2 for the appropriate value of the non-dimensional slenderness, λ LT. 5.5.4(1) 5.5.4(7) and Figure 5.5. 5.5.4(2).1 5.1.1(2) 5.5.4(2) 5.5.4(7) and Figure 5.5. 5.5.2(4) 15/02/07 58

λ LT λ = λ LT 1 β 5 w where λ 1 = 9,9 ε = 9,9 x 1 = 9,9, β w = 1 for class 1 sections, and L / ilt λ LT = 2 25 ( L / alt) 5 C1 1 + 25, 66 where L is the length = 500 mm, i LT = 69,7 mm (from section properties), ψ = k = 1, therefore C 1 = 1,879, and a LT = 779,8 mm (from section properties). Substituting into the above equation: 500 / 69, 7 λ LT = = 4 1 2 25 ( 500 779 8) 5 /, 1879, 1 + 25, 66 λ λ LT LT = = = λ β 5 4 1 0 5 w 1 9 9 1 0,,, 0, 4 Therefore, from 5.5.2, using buckling curve a (for rolled sections), the reduction factor, χ LT = 94 The plastic section modulus, W pl.y = 827 x 10 mm. N k M Sd LT y.sd + 1, 0 χ Af / γ χ W f / γ z y M1 LT pl.y y M1 6 529x10 99 x 102,8x10 + = 98 7 x 9100 x 25 / 1,1 94 x 827x10 x 25 / 1,1 Therefore, this section is satisfactory for lateral torsional buckling. This section is satisfactory for both flexural buckling and lateral torsional buckling. Shear on Web Generally, a shear check on a column web would only be carried out if the structure is to be subjected to considerable seismic loading. For completeness, the check will be included in this worked example. Design shear force, V Sd = 51, kn (load case 5 - member 11) The shear resistance of the web must be checked. The design shear force, V Sd, must be less than or equal to the design plastic shear resistance, V pl.rd : V Sd V pl.rd where V pl.rd is given by A f / y v γ M0 For rolled I and H sections loaded parallel to the web, shear area, A v = 1,04 h t w, 5.5.2(5) 5.5.2(1) Equation F.15 5.5.2(5) 5.5.(4), 5.5. 5.5.2 5.5.4(2) 5.4.6 5.4.6(4) 15/02/07 59

f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 1,04htwfy Vpl.Rd = = 1, 04 x 220 x 9,5 x 25 = 268, 1 kn xγ M0 x 1,1 This is greater than the shear on the section (51, kn). The section is satisfactory under shear. The column is satisfactory under all loading conditions. Internal Column The maximum compression force any internal column has to resist is 1196 kn (member 2) under load case 2. The maximum moment any internal column has to resist is 142 knm at node F, under load case 7. The column is also subject to an axial load of 961 kn under this load case. However, the critical interaction case is likely to be under load case 5, with an axial force of 1149,7 kn and a bending moment of 12,9 knm. It is this case that is likely to be critical. The maximum shear force any internal column has to resist is 5,9 kn at member 2 under load case 7. Section Properties All internal columns are HE 260 B grade Fe60 h = 260 mm b = 260 mm t w = 10 mm t f = 17,5 mm d/t w = 17,7 c/t f = 7,4 A = 1180 mm 2 I y = 149,2 x 10 6 mm 4 I w = 754 x 10 9 mm 6 I z = 51,4 x 10 6 mm 4 I t = 124 x 10 4 mm 4 W pl.y = 128 x 10 mm W el.y = 1148 x 10 mm i y = 112 mm i z = 65,8 mm i Lt = I I z w 2 ply W 25 6 9 x10 x 754x10 = 51, 4 2 ( 128x10 ) 5 5 25 = 69, 7 mm 9 Iw x10 alt = mm 4 = It 754 = 779, 8 124x10 All the above properties can be obtained from section property tables..1 5.1.1(2) F.2.2() F.2.2(1) 15/02/07 60

Classification of Cross Section 5. Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. 5..1 ε = 25 / f y where f y = 25 N/mm 2 (Sheet ), therefore ε = 1. From section properties, c/t f = 7,4 Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is ε. From section properties, d/t w = 17,7 c/t f 10ε and d/t w ε Therefore the section is Class 1. 5..1 (Sheet 1).5.1 (Sheets 1 and ) Resistance of Cross-Section 5.4.4 It is highly unlikely that the resistance of the cross-section will be the critical case - it is generally the buckling resistance that governs the suitability of a cross-section. For the sake of completeness, the check is included in this worked example. For members in axial compression, the design value of the compressive 5.4.4(1) force, N Sd, at each cross-section shall satisfy N Sd N c.rd For a class 1 cross-section, the design compression resistance of the 5.4.4(2) cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 11800 mm 2, f y is the yield strength = 25 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 11800 x 25 N c.rd = = 2521 kn 11, x10 N Sd = 1196 kn, therefore N sd < N c.rd. The section can resist the applied axial load..1 5.1.1(2) 15/02/07 61

Buckling Resistance of Member A class 1 member subject to combined bending and axial compression should be checked for the following modes of failure: Flexural buckling (Clause 5.5.4(1)), and Lateral Torsional Buckling (Clause 5.5.4(2)). Flexural Buckling A class 1 member subject to moment about the major axis only should satisfy the following: N k M Sd y y.sd + 1, 0 χ A f / γ W f / γ min y M1 pl.y y M1 Applied axial force, N Sd, = 1149,7 kn To Calculate Reduction Factor, χ min The reduction factor χ min is the lesser of χ y and χ z, where χ y and χ z are the reduction factors from clause 5.5.1 for the y-y and z-z axes respectively. Determination of χ y The reduction factor χ y depends on the slenderness about the y-y axis. The connections at the column base are effectively pinned, then the slenderness about the y-y axis is: λ y = l/i y where l is equal to the system length and i is the radius of gyration about the appropriate axis. (Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1) Slenderness, λ y = 500 / 112 = 1, λ 1 = 9,9ε = 9,9 x 1,0 = 9,9 λ Non-dimensional slenderness, λ y y λ β 5 = A where β A = 1 for class 1 members 1, Non-dimensional slenderness, λ y = = 9, 9 From 5.5.2, using buckling curve b, the reduction factor, χ y = 97 Determination of χ z Slenderness, λ z = l/i z where l is equal to the system length and i is the radius of gyration about the appropriate axis. (Use the system length as using the amplified sway moment method. If using the effective length method, this is where the effective lengths would be greater than 1) Slenderness, λ z = 500 / 65,8 = 5,2 λ Non-dimensional slenderness, λ z z λ β 5 = A where β A = 1 for class 1 members 1 1 5.5.4(1) 5.5.1.5 5.5.1.4 5.5.1.5(1) 5.5.1.2(1) 5.5.1.1(1) 5.5. 5.5.2 5.5.1.4 5.5.1.5(2) 5.5.1.2(1) 5.5.1.1(1) 15/02/07 62

5, 2 Non-dimensional slenderness, λ z = = 57 9, 9 From 5.5.2, using buckling curve c, reduction factor, χ z = 80 Reduction factor, χ min = χ z = 80 A is the cross-sectional area = 11800 mm 2, f y is the yield strength = 25 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. To Calculate k y µ ynsd k y = 1 but k y 1,5 χ A f y y where Wpl.y Wel.y µ y = λ y( 2β My 4) + but µ y 90 W el.y β My is an equivalent uniform moment factor for flexural buckling (ψ = 0) = 1,8-0 = 1,8. 1148 µ y = ( 2 x 1,8-4) + 128x10 x10 = 01 1148x10 ( ) = 01 x 1154x10 k y 1 = 1, 0 97 x 11800 x 25 M y.sd is the design applied moment = 12,9 knm, and W pl.y is the plastic section modulus = 128 x 10 mm. N k M Sd y y.sd + 1, 0 χ A f / γ W f / γ min y M1 pl.y y M1 6 1149, 7x10 1, 0 x 12,9x10 + = 1, 02 80 x 11800 x 25 / 1,1 128x10 x 25 / 1,1 Therefore the section fails. The internal column section needs to be increased and the analysis and checks carried out on the modified structure. 5.5. 5.5.2.1 5.1.1(2) 5.5.4(1) 5.5.4(7) and Figure 5.5. 15/02/07 6

1. Joint characterisation 1.1. General An important step when designing a frame consists of the characterisation of the rotational response of the joints, i.e. the evaluation of the mechanical properties in terms of stiffness, strength and ductility. Three main approaches may be followed : experimental numerical analytical. The only practical option for the designer is the analytical approach. Analytical procedures have been developed which enable a prediction of the joint response based on the knowledge of the mechanical and geometrical properties of the joint components. In this section a general analytical procedure, termed component method, is introduced. It applies to any type of steel or composite joints, whatever the geometrical configuration, the type of loading (axial force and/or bending moment,...) and the type of member sections. The method is used in the Lecture on Practical procedures for the characterisation of the response of moment resisting joints where the mechanical properties of joints subjected to bending moment and shear force are computed. 1.2. Introduction to the component method The component method considers any joint as a set of individual basic components. For the particular joint shown in Figure 2.b. (joint with an extended end-plate connection subject to bending), the relevant components are the following : J.1.5 Compression in zone : column web in compression; beam flange and web in compression; Tension zone : column web in tension; column flange in bending; bolts in tension; end-plate in bending; beam web in tension; Shear zone : column web panel in shear. Each of these basic components possesses its own strength and stiffness either in tension or in compression or in shear. The column web is subject to coincident compression, tension and shear. This coexistence of several components within the same joint element can obviously lead to stress interactions that are likely to decrease the 15/02/07 64

resistance of the individual basic components. The application of the component method requires the following steps : a) identification of the active components in the joint being considered; b) evaluation of the stiffness and/or resistance characteristics for each individual basic component (specific characteristics - initial stiffness, design resistance,... - or the whole deformability curve) ; c) assembly of all the constituent components and evaluation of the stiffness and/or resistance characteristics of the whole joint (specific characteristics - initial stiffness, design resistance,... - or the whole deformability curve). In Figure 1, the principles of the component method are illustrated in the specific case of a beam-to-column joint with a welded connection. COMPONENT METHOD Three steps F M=Fz F First step: Identification of the components Column web Column web Column web in shear in compression in tension Second step: Response of the components F Ek 1 Ek 2 F F F F Rd1 F Rd2 Rd Ek 1 1 1 Stiffness coefficient k i of each component Resistance F Rdi of each component Third step: Assembly of the components M M Rd S j,ini φ Stiffness of the joint S j,ini = Ez²/Σk i Resistance of the joint M Rd = min(f Rdi).z Figure 1 Application of the component method to a welded joint The assembly procedure consists in deriving the mechanical properties of the whole joint from those of all the individual constituent 15/02/07 65

components. This requires a preliminary distribution of the forces acting on the joint into internal forces acting on the components in a way that satisfies equilibrium. In Eurocode Annex J, the analytical assembly procedures are described for the evaluation of the initial stiffness and the design moment resistance of the joint. These two properties enable the designer to determine the design joint moment-rotation characteristic whatever the type of analysis (Figures 4 to Figure 6). In Annex A of the present lecture, information is provided on how the stiffness and strength assembly is carried out. The application of the component method requires a sufficient knowledge of the behaviour of the basic components. Those covered by Eurocode are listed in 1. The combination of these components allows one to cover a wide range of joint configurations, which should be sufficient to satisfy the needs of practitioners as far as beam-to-column joints and beam splices in bending are concerned. Examples of such joints are given in Figure 2. Some fields of application can also be contemplated : Joints subject to bending moment (and shear) and axial force; Column bases subject to coincident bending moment, shear force and axial force where the components such as : - concrete foundation in compression; - end-plates with specific geometries; - anchorages in tension; - contact between soil and foundation, will be activated. These situations are however not yet covered, or only partially covered, by Eurocode. 15/02/07 66

N 1 Column web panel in shear Component V Sd 2 Column web in compression V Sd F c.sd Beam flange and web in compression F c.sd 4 Column flange in bending F t.sd J.1 5 Column web in tension F t.sd 6 End-plate in bending F t.sd 7 Beam web in tension F t.sd 8 Flange cleat in bending F t.sd 15/02/07 67

9 Bolts in tension F t.sd 10 Bolts in shear F v.sd 11 Bolts in bearing (on beam flange, column flange, end-plate or cleat) F b.sd 12 Plate in tension or compression F t.sd F c.sd 1 List of components covered by Eurocode 15/02/07 68

(a) Welded joint (b) Bolted joint with extended end-plate (c) Two joints with flush end-plates (Double-sided configuration) (d) Joint with flush end-plate (e) End-plate type beam splice (f) Cover-joint type beam splice Figure J.4 (g) Bolted joint with angle flange cleats (h) Two beam-to-beam joints (Double-sided configuration) Figure 2 Examples of joints covered by Eurocode 2. Joint idealisation 15/02/07 69

The non-linear behaviour of the isolated flexural spring which characterises the actual joint response does not lend itself towards everyday design practice. However the moment-rotation characteristic curve may be idealised without significant loss of accuracy. One of the most simple idealisations possible is the elastic-perfectly plastic relationship (Figure.a). This modelling has the advantage of being quite similar to that used traditionally for the modelling of member crosssections subject to bending (Figure.b). The moment M j,rd that corresponds to the yield plateau is termed the design moment resistance in Eurocode. It may be considered as the pseudo-plastic moment resistance of the joint. Strain-hardening effects and possible membrane effects are henceforth neglected, which explains the difference in Figure between the actual M-φ characteristic and the yield plateau of the idealisation. M j M b, M c M j,rd M pl,rd S j,ini /η EI/L (a) Joint φ (b) Member φ Actual M-φ characteristic Idealised M-φ characteristic Figure Bi-linearisation of moment-rotation curves The value of the constant stiffness S j.ini /η is discussed below. In fact there are different possible ways to idealise a joint M- φ characteristic. The choice of one of them is dependent upon the type of frame analysis which is contemplated: - Elastic idealisation for an elastic analysis (Figure 4) : The principal joint characteristic is the constant rotational stiffness. 15/02/07 70