Structural Steelwork Eurocodes Development of A Trans-national Approach

Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode 3 Module 7 : Worked Examples Lecture 20 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads 1.2 Design Loads F d = γ F F k 1.3 Partial Safety Factors for Strength 2. Floor Beam - Fully Restrained 2.1 Classification of Cross-section 2.1.1Flange buckling 2.1.2 Web buckling 2.2 Shear on Web 2.3 Deflection Check 2.4 Additional Checks if Section is on Seating Cleats 2.5 Crushing Resistance 2.6 Crippling Resistance 2.7 Buckling Resistance 2.8 Summary 3. Roof Beam Restrained at Load Points 3.1 Initial Section Selection 3.2 Classification of Cross Section 3.2.1 Flange buckling 3.2.2 Web buckling 3.3 Design Buckling Resistance Moment 3.4 Shear on Web 3.5 Deflection Check 3.6 Crushing, Crippling and Buckling 3.7 Summary 4. Internal Column 4.1 Loadings 4.2 Section properties SSEDTA 2001

4.3 Classification of Cross-Section 4.3.1 Flange (subject to compression) 4.3.2 Web (subject to compression) 4.4 Resistance of Cross-Section 4.5 Buckling Resistance of Member 4.6 Determination of Reduction Factor χy 4.7 Determination of Reduction Factor χz 5. External Column 5.1 Loadings 5.2 Section properties 5.3 Classifcation of Cross-Section 5.3.1 Flange (subject to compression) 5.3.2 Web (subject to compression) 5.4 Resistance of Cross-Section 5.5 Buckling Resistance of Member 5.6 Determination of Reduction factor χy 5.7 Determination of Reduction factor χz 6. Design of Cross-Bracing 6.1 Section Properties 6.2 Classification of Cross-Section 6.3 Design of Compression Member 6.3.1 Resistance of Cross-section 6.3.2 Design Buckling Resistance 6.3.3 Determination of Reduction Factor χ? 6.4 Design of Tension Member 6.4.1Resistance of Cross-Section 7. Concluding Summary 23/02/07 2

1. Simple Braced Frame The frame consists of two storeys and two bays. The frames are at 5 m spacing. The beam span is 7,2 m. The height from column foot to the beam at floor level is 4,5 m and the height from floor to roof is 4,2 m. It is assumed that the column foot is pinned at the foundation. Roof Beam External Column Internal Column 4,2 m Floor Beam 4,5 m 7,2 m 7,2 m Figure 1 Typical Cross Section of Frame It is assumed that resistance to lateral wind loads is provided by a system of localised cross-bracing, and that the main steel frame is designed to support gravity loads only. The connections are designed to transmit vertical shear, and it is also 6.4.2.1(2) assumed that the connections offer little, if any, resistance to free rotation of the beam ends. 5.2.2.2 With these assumptions, the frame is classified as simple, and the internal forces and moments are determined using a global analysis which assumes the members to be effectively pin-connected. 1.1 Characteristic Loads Floor: Variable load, Q k = 3,5 kn/m 2 Permanent load, G k = 8,11 kn/m 2 Roof: Variable load, Q k = 0,75 kn/m 2 Permanent load, G k = 7,17 kn/m 2 1.2 Design Loads F d = γ F F k 2.2.2.4(1) Floor: G d = γ G G k. At ultimate limit state γ G = 1,35 (unfavourable) G d = 1,35 x 8,11 = 10,95 kn/m 2 Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x 3,5 = 5,25 kn/m 2 2.2.2.4(2) 2.2 2.2.2.4(2) 2.2 23/02/07 3

Roof: G d = γ G G k. At ultimate limit state γ G = 1,35 (unfavourable) G d = 1,35 x 7,17 = 9,68 kn/m 2 Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x 0,75 = 1,125 kn/m 2 The steel grade selected for beams, columns and joints is Fe360. (f y = 235 N/mm 2 ) 1.3 Partial Safety Factors for Strength The following partial safety factors for strength have been adopted during the design: Resistance of Class 1,2 or 3 cross-section, γ M0 = 1,1 Resistance of member to buckling, γ M1 = 1,1 Resistance of bolted connections, γ Mb = 1,25 The following load case, corresponding to permanent and variable actions (no horizontal loads) is found to be critical. 2. Floor Beam - Fully Restrained The beam shown in Figure 2 is simply supported at both ends and is fully restrained along its length. For the loading shown, design the beam in grade Fe360, assuming that it is carrying plaster, or a similar brittle finish. F d = γ G G k + γ Q Q k Design load, F d = (5 x 1,35 x 8,11) + (5 x 1,5 x 3,5) = 81 kn/m 81 kn/m 2.2.2.4(2) 2.2 2.2.2.4(2) 2.2 3.1 2.3.3.2(1) 5.1.1(2) 5.1.1(2) 6.1.1(2) 2.1 7,2 m Figure 2 Loading on Fully Restrained Floor Beam 2 FL d Design moment, MSd = 8 Where M Sd is the design moment in beam span, F d is the design load = 81 kn/m, and L is the beam span = 7,2m. 2 81x7,2 MSd = = 525 knm 8 FL d 81x7,2 Design shear force, VSd = = = 292 kn 2 2 To determine the section size it is assumed that the flange thickness is less than 40 mm so that the design strength is 235 N/mm 2, and that the section is class 1 or 2. 3.1 23/02/07 4

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 = Wf pl y γ M0 Where W pl is the plastic section modulus (to be determined), f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. Therefore, rearranging: 3 Msdγ M0 525x10 x1,1 3 Wpl.required = = = 2457 cm f 235 Try IPE 550 Section properties: Depth, h = 550 mm, Web thickness, t w = 11,1 mm y Width, b = 210 mm Flange thickness, t f = 17,2 mm Plastic modulus, W pl = 2787 cm 3 This notation conforms with Figure 1.1 in Eurocode 3: Part1.1. 2.1 Classification of Cross-section As a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance. Figure 3 shows a typical cross-section for an IPE. IPE sections have been used in this example to reflect the European nature of the training pack. 5.4.5.1(1) 3.1 5.1.1(2) 5.4.5.1 5.3 5.3.2 and 5.3.1 t t w f c d Figure 3 A Typical Cross-Section 2.1.1 Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half t f the width of the flange = 105 mm, and t f is the flange thickness = 17,2 mm (if the flange is tapered, t f should be taken as the average thickness). c 105 = = 6,10 t 17,2 f 5.3.1 (Sheet 3) 23/02/07 5

2.1.2 Web buckling Class 1 limiting value of d/t w for a web subject to bending is 72ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 467,6 t w mm and t w is the web thickness = 11,1 mm. d 467,6 = = 42,1 t w 11,1 c < 10ε and d < 72ε t t f w Section is Class 1 and is capable of developing plastic moment. 2.2 Shear on Web 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 : 5.3.1 (Sheet 1) 5.3.1 (Sheets 1 and 3) 5.4.6 V Sd V pl.rd Where V pl.rd is given by A f / 3 y v γ M0 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 = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 5.4.6(4) 3.1 5.1.1(2) 1,04ht f Vpl.Rd = 3xγ w y M0 1,04 x 550 x 11,1 x 235 = = 783 kn 3 3 x 1,1x10 This is greater than the shear on the section (292 kn). The shear on the beam web is OK. If the beam has partial depth end-plates, a local shear check is required on the web of the beam where it is welded to the end-plate. V A f / 3 y pl.rd = v γ M0 where A v = t w d, and d is the depth of end-plate = (for example) 300 mm. 11,1 x 300 x 235 Vpl.Rd = = 411 kn 3 3 x 1,1 x 10 This is greater than the shear on the section (292 kn). The local shear on the beam web is OK. 23/02/07 6

Other simple joints may be used instead, e.g. web cleat joints or fin plate joints. 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. 2.3 Deflection Check Eurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions: Variable actions, and Permanent and variable actions. Figure 4 shows the vertical deflections to be considered. 5.4.7(3) 4.2 δ δ 1 2 δ δ 0 max L Figure 4 Vertical Deflections δ 0 is the pre-camber (if present), δ 1 is the deflection due to permanent actions, δ 2 is the deflection caused by variable loading, and δ max is the sagging in the final state relative to the straight line joining the supports. For a plaster or similar brittle finish, the deflection limits are L/250 for δ max and L/350 for δ 2. Deflection checks are based on the serviceability loading. For a uniform load δ = 5 384 x FL 3 k EIy where F k is the total load = Q k or (G k + Q k ) as appropriate, L is the span = 7,2 m, E is the modulus of elasticity (210 000 N/mm 2 ), and I y is the second moment of area about the major axis = 67120 x 10 4 mm 4. 4.1 Figure 4.1 3.2.5 23/02/07 7

For permanent actions, F k = 5 x 8,11 x 7,2 = 292 kn. Therefore, deflection due to permanent actions, 3 3 5 x 292x10 x 7200 δ 1 = = 10,1 mm 4 384 x 210 000 x 67120x10 For variable actions, F k = 5 x 3,5 x 7,2 = 126 kn. Therefore, deflection due to variable actions, 3 3 5 x 126x10 x 7200 δ 2 = = 4,3 mm 4 384 x 210 000 x 67120x10 The maximum deflection, δ max = δ 1 + δ 2 = 10,1+ 4,3 = 14,4 mm L 7200 Deflection limit for δ 2 = = = 20,6 mm 350 350 The actual deflection is less than the allowable deflection: 4,3 mm < 20,6 mm L 7200 Deflection limit for δ max = = = 28,8 mm 250 250 4.1 4.1 The actual deflection is less than the allowable deflection: 14,4 mm < 28,8 mm OK. The calculated deflections are less than the limits, so no pre-camber is required. It should be noted that if the structure is open to the public, there is a limit of 28 mm for the total deflection of δ 1 + δ 2 (neglecting any pre-camber) under the frequent combination, to control vibration. This is based on a single degree of freedom, lumped mass approach. For the frequent combination the variable action is multiplied by ψ, which has a value of 0,6 for offices. 2.4 Additional Checks if Section is on Seating Cleats There are cases where the beams may be supported on seating cleats, or on other materials such as concrete pads. A similar situation arises when a beam supports a concentrated load applied through the flanges. In these cases the following checks are required: Crushing of the web Crippling of the web Buckling of the web Generally, these checks need only be carried out for short beams or beams with concentrated loads. For the sake of completeness, these checks are included in this worked example. The following detailed checks are for an 85 mm stiff bearing. (The actual length of stiff bearing will depend on the detail of the connection - see Figure 5.7.2) 4.3.2(2) Lecture 3, section 6.2 2.3.4(2) 5.7.3 5.7.4 5.7.5 23/02/07 8

2.5 Crushing Resistance 5.7.3 The crushing resistance is given by (ss + s y)twfyw Ry.Rd = γ M1 where s s is the length of stiff bearing = 85 mm, t w is the web thickness = 11,1 mm, f yw is the yield strength of the web = 235 N/mm 2, γ M1 is the partial material safety factor for buckling resistance = 1,1, and s y is the length over which the effect takes place, based on the section geometry and the longitudinal stresses in the flange. s y = 2t f (b f /t w ) 0,5 (f yf /f yw ) 0,5 [1 - (σ f.ed /f yf ) 2 ] 0,5 At the support, the stress in the beam flange, σ f.ed, is zero, f yf = f yw but the value of s y is halved at the end of the member. 3.1 5.1.1(2) 5.7.3(1) 5.7.3(3) 0,5 2 x 17,2 x (210 / 11,1) sy = = 75 mm 2 (85 + 75) x 11,1 x 235 Crushing resistance, R y.rd = = 379,4 kn 3 1,1x10 This is greater than the reaction (292 kn). The crushing resistance is OK 2.6 Crippling Resistance The crippling resistance is given by 5.7.4(1) R a.rd = 2 w 0,5 0,5 0,5t (Ef ) [(t / t ) + 3(t / t )(s / d)] yw f γ w M1 w f s where t w is the thickness of the web = 11,1 mm, E is the modulus of elasticity = 210 000 N/mm 2, f yw is the yield strength of the web = 235 N/mm 2, t f is the flange thickness = 17,2 mm, s s is the length of stiff bearing = 85 mm, but s s is limited to a maximum of 0,2d (467,6 mm x 0,2 = 93,5 mm), d is the depth between root radii = 467,6 mm, and γ M1 is the partial material safety factor buckling resistance = 1,1. 3.2.5 3.1 5.7.4(1) 5.1.1(2) 2 0,5 0,5 0,5 x 11,1 (210000 x 235) [(17,2 / 11,1) + 3(11,1/ 17,2)(85 / 467,6)] Ra,Rd = 3 1,1x10 = 626 kn This is greater than the reaction (292 kn). The crippling resistance is OK. 23/02/07 9

2.7 Buckling Resistance The buckling resistance is determined by taking a length of web as a strut. The length of web is taken from Eurocode 3, which in this case, gives a length: 2 2 0,5 ss beff = 0,5(h + s s ) + a+ but [h 2 + s 2 s ] 0,5 2 where h is the height of the section = 550 mm, s s is the length of stiff bearing = 85 mm, and a is the distance to the end of the beam = 0 mm. 2 2 0,5 85 beff = 0,5(550 + 85 ) + = 320,5 mm 2 Provided that the construction is such that the top flange is held by a slab and the bottom by seating cleats, against rotation and displacement, the effective height of the web for buckling should be taken as 0,7 x distance between fillets. l = 467,6 mm x 0,7 = 327 mm t w 11,1 Radius of gyration for web, i = = = 32, 12 12 l 327 Slenderness of the web, λ = = = 102 i 3,2 λ Non-dimensional slenderness of the web, λ = λ β 0,5 A 1 Where λ 1 = 93,9 ε = 93,9 x 1 = 93,9, and β A = 1 102 Non - dimensional slenderness of the web, λ = = 109, 93, 9 Using buckling curve c, the value of the reduction factor, χ may be determined from 5.5.2. Reduction factor, χ = 0,49 AAfy Buckling resistance of a compression member, Nb.Rd = χβ γ M1 A is the cross-sectional area = b eff t w, f y is the yield strength = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. 0,49 x 1 x 320,5 x 11,1 x 235 N b.rd = = 372,4 kn 3 1,1x10 This is greater than the reaction (292 kn). The buckling resistance is OK. 2.8 Summary The trial section IPE 550 is satisfactory if the section is on a stiff bearing 85 mm long. If it is supported by web cleats or welded end plates, the web checks, except for shear, are not required and the section is again satisfactory. The beam is satisfactory. 5.7.5 Figure 5.7.3 5.7.5(4) 5.5.1.4(3) 5.5.1.2 5.7.5(3) 5.7.5(3) and 5.5.2 5.5.1.1(1) 3.1 5.1.1(2) 23/02/07 10

3. Roof Beam - Restrained at Load Points The roof beam shown in Figure 5 is laterally restrained at the ends and at the points of application of load. The load is applied through purlins at 1,8 m spacings. Internal point load = 1,8 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 97,2 kn External point load = 0,9 [(5 x 1,35 x 7,17) + (5 x 1,5 x 0,75)] = 48,6 kn It is assumed that the external point loads will be applied at the end of the beams, and will contribute to the maximum shear force applied to the end of the beam, and the moment induced in the column due to the eccentricity of connection. For the loading shown, design the beam in grade Fe360 steel. 48,6 kn 97,2 kn 97,2 kn 97,2 kn 48,6 kn A B C D E 1,8 m 1,8 m 1,8 m 1,8 m 7,2 m Figure 5 Beam Restrained at Load Points Reactions V Sd (at supports) = [(2 x 48,6) + (3 x 97,2)] / 2 = 194,4 kn Design Moment Figure 6 shows the distribution of bending moments. 1,8 m 1,8 m 1,8 m 1,8 m 0 0 262,4 knm 262,4 knm 349,9 knm Figure 6 Bending Moment Diagram Moment at mid-span (maximum) M Sd = [(194,4-48,6) x 3,6] - (97,2 x 1,8) = 349,9 knm 23/02/07 11

3.1 Initial Section Selection Assume that a rolled I beam will be used and that the flanges will be less than 40 mm thick. For grade Fe360 steel, f y = 235 N/mm 2. Because the beam is unrestrained between load points, the design resistance, M c.rd, of the section will be reduced by lateral torsional buckling. The final section, allowing for the buckling resistance moment being less than the full resistance moment of the section, would have to be determined from experience. Try IPE O 450 Section properties: Depth, h = 456 mm, Width, b = 192 mm Web thickness, t w = 11 mm Flange thickness, t f = 17,6 mm Plastic modulus, W pl = 2046 cm 3 This notation conforms with Figure 1.1 in Eurocode 3: Part1.1. 3.2 Classification of Cross-Section As a simply supported beam is not required to have any plastic rotation capacity (only one hinge required), it is sufficient to ensure that the section is at least class 2 to develop the plastic moment resistance. 3.2.1 Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio c, where c is half the width of the flange = 96 mm, t f and t f is the flange thickness = 17,6 mm (if the flange is tapered, t f should be taken as the average thickness). c 96 = = 5,5 t 17,6 f 3.2.2 Web buckling Class 1 limiting value of d/t w for a web subject to bending is 72ε. ε = 235 / f y and f y = 235 N/mm 2, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 378,8 t w mm and t w is the web thickness = 11,0 mm. d 378,8 = = 34,4 t w 11,0 c < 10ε and d < 72ε t t f Section is Class 1. w 3.1 5.4.5.1 5.3 5.3.2 and 5.3.1 5.3.1 (Sheet 3) 5.3.1 (Sheet 1) 5.3.1 (Sheets 1 and 3) 23/02/07 12

3.3 Design Buckling Resistance Moment The design buckling resistance moment of a laterally unrestrained beam is given by: χ M = b.rd β W γ f LT W pl.y y M1 in which χ LT is the reduction factor for lateral torsional buckling, from 5.5.2, for the appropriate value of λ LT, using curve a for rolled sections. 5.5.2 5.5.2 5.5.2(4) In this case full lateral restraint is provided at the supports and at the load points B, C and D. In general, all segments need to be checked, but in this case they are all of equal length. The segments B - C and C - D are subject to the most severe condition, but with symmetrical loading only one segment needs to be checked. Segment B - C The value of λ LT can be determined using Annex F. For segment B - C it is assumed that the purlins at B and C provide the following conditions: restraint against lateral movement, restraint against rotation about the longitudinal axis (i.e. torsional/twisting restraint), and freedom to rotate in plan. i.e. k = k w = 1,0 For this example, the general formula for λ LT has been used, as the section is doubly symmetric and end-moment loading is present. The following formula for λ LT may be used: λ LT = C 05, 1 1 + L/i LT ( L/a ) LT 25, 66 2 025, where L is the length between B and C = 1800 mm, I z is the second moment of area about the z - z axis = 2085 x 10 4 mm 4, I w is the warping constant = 998 x 10 9 mm 6, W pl.y is the plastic modulus about the y - y axis = 2046 x 10 3 mm 3, and I t is the torsion constant = 89,3 x 10 4 mm 4. i a LT LT = II W z w 2 pl.y 0,25 0,25 4 9 2085x10 x 998x10 = 47,2 mm 3 2 = (2046x10 ) 9 Iw 998x10 = 957 mm 4 = I = 109x10 t 0,5 0,5 Annex F F.1.2(2) Equation F.15 F.2.2(3) F.2.2(1) 23/02/07 13

Note: These properties will probably be tabulated in handbooks. See also appendix A at the end of this example. C 1 is the correction factor for the effects of any change of moment along the length, L. ψ = 262,4/349,9 = 0,75, k = 1, therefore C 1 = 1,141 Substituting into the above equation: 1800 / 47, 2 λ LT = = 34, 6 2 025, ( ) 05, 1800 / 957 1141, 1 + 25, 66 λ Non-dimensional slenderness, λ LT LT = λ β w 1 Where λ 1 = 93,9 ε = 93,9 x 1 = 93,9, β w = 1 for class 1 sections. 34, 6 0,5 Therefore, λ LT = (,) 10 = 037, 93, 9 For rolled I sections, buckling curve a should be used. From 5.5.2, the reduction factor, χ LT = 0,96. (This represents a 4% strength reduction due to moment) W pl.y is the plastic modulus about the y - y axis = 2046 x 10 3 mm 3, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The design buckling resistance moment for segment B - C is: χ M = b.rd 3 β W f 096, x 1 x 2046x10 x 235 = = 419, 6 knm 6 γ 1,1 x 10 LT W pl.y y M1 0,5 F.1.1 5.5.2(4) 5.5.2(5) 5.5.3 5.5.2 3.1 5.1.1(2) 5.5.2 M b.rd = 419,6 knm > M Sd = 349,9 knm, therefore the section is satisfactory. 23/02/07 14

3.4 Shear on Web The maximum shear occurs at the supports, V Sd = 194,4 kn. The design shear resistance for a rolled I section is: V pl.rd 1,04htw = γ ( fy / 3) M0 where h is the height of the section = 456 mm, t w is the web thickness = 11 mm, f y is the yield strength of the steel = 235 N/mm 2, and γ M0 is the partial material safety factor for the resistance of the crosssection = 1,1. 5.4.6(1) 5.4.6(4) 3.1 5.1.1(2) V 1,04 x 456 x 11 x 235 = = 643 kn 3 x 1,1 x 10 pl.rd 3 V Sd = 194,4 kn < V pl.rd = 643 kn, therefore the section is satisfactory. Inspection shows that V Sd < (V pl.rd / 2), so there is no reduction in moment resistance due to the shear in the web. 3.5 Deflection Check Eurocode 3 requires that the deflections of the beam be checked under the following serviceability loading conditions: 5.4.7(2) 4.2 Variable actions, and Permanent and variable actions. For a general roof, the deflection limits are L/200 for δ max and L/250 for δ 2. Deflection checks are based on the serviceability loading. Consider the deflection from the permanent loading. For a point load, distance a from the end of the beam: 4.1 Figure 4.1 Central deflection, δ = 2 2 Fa k L a EIy 16 12 3.2.5(1) where F k is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kn, E is the modulus of elasticity = 210 000 N/mm 2, I y is the second moment of area about the major axis = 40920 x 10 4 mm 4, L is the span of the beam = 7,2 m, and a is the distance from the support to the adjacent load = 1,8 m. Central deflection, δ = 3 64,5x10 x 1800 210 000 x 40920x10 4 2 2 7200 1800 = 16 12 40, mm 23/02/07 15

For a central point load: Central deflection, δ = FL 3 k 48EI y where F k is the value of one point load = (7,17 x 5 x 1,8) = 64,5 kn, L is the span of the beam = 7,2 m, E is the modulus of elasticity = 210 000 N/mm 2, and I y is the second moment of area about the major axis = 40920 x 10 4 mm 4. 3.2.5(1) 3 3 64,5x10 x 7200 Central deflection, δ = = 58, mm 4 48 x 210 000 x 40920x10 Total deflection due to permanent loading, δ 1 = 5,8 + (2 x 4,0) = 13,8 mm Consider the deflection from the variable loading. For a point load, distance a from the end of the beam: 2 2 Fa k L a Central deflection, δ = EIy 16 12 where F k is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kn, E is the modulus of elasticity = 210 000 N/mm 2, I y is the second moment of area about the major axis = 40920 x 10 4 mm 4, L is the span of the beam = 7,2 m, and a is the distance from the support to the adjacent load = 1,8 m. 3 2 2 6,75x10 x 1800 7200 1800 Central deflection, δ = 210 000 x 40920x10 = 04, mm 4 16 12 3.2.5(1) For a central point load: Central deflection, δ = FL 3 k 48EI y where F k is the value of one point load = (0,75 x 5 x 1,8) = 6,75 kn, L is the span of the beam = 7,2 m, E is the modulus of elasticity = 210 000 N/mm 2, and I y is the second moment of area about the major axis = 40920 x 10 4 mm 4. 3.2.5(1) 3 3 6,75x10 x 7200 Central deflection, δ = = 06, mm 4 48 x 210 000 x 40920x10 23/02/07 16

Total deflection due to variable loading, δ 2 = 0,6 + (2 x 0,4) = 1,4 mm Therefore, the total central deflection, δ max = δ 1 + δ 2 = 13,8 + 1,4 = 15,2 mm. The limit for δ 2 is L/250 = 7200/250 = 28,8 mm. The limit for δ max = L/200 = 7200/200 = 36 mm. 13,8 mm < 28,8 mm and 15,5 mm < 36 mm, therefore the deflections are within limits and no pre-camber of the beam is required. 3.6 Crushing, Crippling and Buckling If the beam is supported on seating cleats, the checks for web crushing, crippling and buckling must be made. To satisfy the assumptions made in the design, both flanges must be held in place laterally, relative to each other. If seating cleats are used then the top flange must be held laterally. There is no requirement to prevent the flanges from rotating in plan, as k has been taken as 1,0. 3.7 Summary All Eurocode recommendations are satisfied, therefore this beam is satisfactory. The beam is satisfactory. 4.1 and Figure 4.1 5.7.1 4. Internal Column The internal column shown in Figure 7 is subject to loads from the roof and one floor. Design the column for the given loading, in grade Fe360 steel, as a member in simple framing. 23/02/07 17

4.1 Loadings (54 x 7,2) At roof level, the applied axial load = 2 x = 389 kn 2 (81 x 7,2) At first floor level, the applied axial load = 2 x = 583 kn 2 Maximum load, from the first floor to the base, = 389 + 583 kn = 972 kn Roof Internal Column Floor 4,2 m 4,5 m Figure 7 Internal Column Consider the column from ground floor to first floor. The size of the column must be determined from experience and then checked for compliance with the Eurocode rules. 4.2 Section Properties Try an HE 240 A Grade Fe360 (terminology in accordance with EC3) h = 230 mm b = 240 mm t w = 7,5 mm t f = 12 mm d/t w = 21,9 c/t f = 10 A = 7680 mm 2 I y = 77,63 x 10 6 mm 4 I w = 328 x 10 9 mm 6 I z = 27,7 x 10 6 mm 4 I t = 41,6 x 10 4 mm 4 W pl.y = 745 x 10 3 mm 3 W el.y = 675 x 10 3 mm 3 i y = 101 mm i z = 60 mm i Lt = II W z w 2 pl.y 0,5 0,25 0,25 6 9 27,7x10 x 328x10 = 63,6 mm 3 2 = (745x10 ) 0,5 9 Iw 328x10 alt = 888 mm 4 = It = 41,6x10 All the above properties can be obtained from section property tables. F.2.2(3) F.2.2(1) 23/02/07 18

4.3 Classification of Cross Section This section is designed to withstand axial force only. No moment is applied as the connecting beams are equally loaded. 4.3.1 Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y where f y = 235 N/mm 2, therefore ε = 1. From section properties, c/t f = 10 4.3.2 Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is 33ε. From section properties, d/t w = 21,9 c/t f 10ε and d/t w 33ε Class 1 section. 5.3 5.3.1 (Sheet 3) 5.3.1 (Sheet 1) 3.5.1 (Sheets 1 and 3) Class 1 section 4.4 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. The resistance of the cross-section will only be critical if a short, stocky column is used. For members in axial compression, the design value of the compressive 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 cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 7680 mm 2, f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. N 7680 x 235 = = 1641 kn 1,1x10 c.rd 3 N Sd = 972 kn, therefore N sd N c.rd. The section can resist the applied axial load. 5.4.4(1) 5.4.4(2) 3.1 5.1.1(2) 23/02/07 19

4.5 Buckling Resistance of Member A class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since M y = M z = 0, only failure due to flexural buckling needs to be checked. The design buckling resistance of a compression member shall be taken as: N b.rd AAf = χβ γ M1 y where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 7680 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ 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. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from 5.5.2. λ Non-dimensional slenderness, λ = λ β 0,5 A Where the slenderness, λ = l i l is the column buckling length, and i is the radius of gyration about the relevant axis. The braced frame is designed as a simple pinned structure. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length. 0,5 E λ1 = π 93,9ε f = y 4.6 Determination of Reduction Factor, χ y Slenderness, λ y = l/i y = 4500/101 = 44,6 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ y = y λ β 0,5 44,6 0,5 A = 1 93,9 x 1 = 0,47 From 5.5.2, using buckling curve b, the reduction factor, χ y = 0,89 1 5.5.1.1(1) 5.5.1.1(1) 3.1 5.1.1.(2) 5.5.1.2(1) 5.5.1.4(3) 5.5.1.5(2) 5.5.1.2(1) 5.5.1.4(3) 5.5.3 5.5.2 23/02/07 20

4.7 Determination of Reduction Factor, χ z Slenderness, λ z = l/i z = 4500/60 = 75 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = z λ β 0,5 75 0,5 y A = 1 93,9 x 1 = 0,8 From 5.5.2, using buckling curve c,the reduction factor, χ z = 0,6622 Therefore χ = χ z = 0,6622. Design buckling resistance of member: χβ AAfy 0,6622 x 1 x 7680 x 235 Nb.Rd = = = 1086 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (972 kn), therefore the column is satisfactory. The column is OK. 5.5.1.4(3) 5.5.3 5.5.2 5.5.1.1(1) 5. External Column The external column shown in Figure 8 is subject to loads from the roof and one floor. Design the column for the loading given below, in grade Fe360 steel, as a member in simple framing. 5.1 Loadings (54 x 7,2) At roof level, the applied axial load = = 194 kn 2 (81 x 7,2) At first floor level, the applied axial load = = 292 kn 2 Maximum load, from first floor to base, = 194 + 292 kn = 486 kn The beams in the frame are designed to span from column centre to column centre, therefore all axial load is applied at the mid-point of the column. No moment due to eccentricity of applied load is therefore applied to the column. See Annex H 23/02/07 21

Roof 4,2 m First Floor 4,5 m Figure 8 External Column Consider the column from ground floor to first floor. The size of the column must be determined from experience and then checked for compliance with the Eurocode rules. 5.2 Section Properties Try an HE 200 A Grade Fe360 h = 190 mm b = 200 mm t w = 6,5 mm t f = 10 mm d/t w = 20,6 c/t f = 10 A = 5380 mm 2 I y = 36,92 x 10 6 mm 4 I w = 108 x 10 9 mm 6 I z = 13,36 x 10 6 mm 4 I t = 21,0 x 10 4 mm 4 W pl.y = 429 x 10 3 mm 3 W el.y = 389 x 10 3 mm 3 i y = 82,8 mm i z = 49,8 mm i Lt = II W z w 2 pl.y 0,5 0,25 0,25 6 9 13,36x10 x 108x10 = 52,9 mm 3 2 = (429x10 ) 0,5 9 Iw 108x10 alt = 717 mm 4 = It = 21,0x10 All the above properties can be obtained from section property tables. F.2.2(3) F.2.2(1) 23/02/07 22

5.3 Classification of Cross Section This section is designed to withstand axial force only. 5.3.1 Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 10ε = 10 x 1 = 10 From section properties, c/t f = 10 5.3.2 Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is 33ε. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 33ε = 33 x 1 = 33 From section properties, d/t w = 20,6 c/t f 10ε and d/t w 33ε Class 1 section. 5.3 5.3.1 (Sheet 3) 5.3.1 (Sheet 1) 5.3.1 (Sheets 1 and 3) 5.4 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. The resistance of the cross-section will only be critical if a short stocky column is used. For members in axial compression, the design value of the compressive 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 cross-section, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is cross-sectional area = 5380 mm 2, f y is the yield strength = 235 N/mm 2, and γ M0 is the partial material safety factor = 1,1. 5380 x 235 Nc.Rd = = 1149 kn 3 1,1x10 N Sd = 486 kn, therefore N sd N c.rd. The section can resist the applied axial load. 5.4.4(1) 5.4.4(2) 3.1 5.1.1(2) 23/02/07 23

5.5 Buckling Resistance of Member A class 1 member should be checked for failure due to flexural and lateral torsional buckling. Here, since M y = M z = 0, only failure due to flexural buckling needs to be checked. The design buckling resistance of a compression member shall be taken as: AAfy Nb.Rd = χβ γ M1 where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 5380 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. The magnitude of the reduction factor, χ depends on the reduced slenderness of the columns. χ 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. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from 5.5.2. λ Non-dimensional slenderness, λ = λ β A Where the slenderness, λ = l i l is the column buckling length, and i is the radius of gyration about the relevant axis. The braced frame is designed as a simple pinned structure. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length. 0,5 E λ1 = π 93,9ε f = y 5.6 Determination of Reduction Factor, χ y Slenderness, λ y = l/i y = 4500/82,8 = 54,3 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ y = y λ β 0,5 54,3 0,5 A = 1 93,9 x 1 = 0,58 From 5.5.2, using buckling curve b, the reduction factor, χ y = 0,84 1 0,5 5.5.1.1(1) 5.5.1.1(1) 3.1 5.1.1.(2) 5.5.1.2(1) 5.5.1.4(3) 5.5.1.5(2) Annex E Figure E.2.1 5.5.1.2(1) 5.5.1.4(3) 5.5.3 5.5.2 23/02/07 24

5.7 Determination of Reduction Factor, χ z Slenderness, λ z = l/i z = 4500/49,8 = 90,4 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = z λ β 0,5 90,4 0,5 y A = 1 93,9 x 1 = 0,96 From 5.5.2, using buckling curve c,the reduction factor, χ z = 0,55 Therefore χ = χ z = 0,55. Design buckling resistance of member: χβ AAfy 0,55 x 1 x 5380 x 235 N b.rd = = = 632,2 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (486 kn), therefore the column is satisfactory. The column is OK. 6. Design of Cross-Bracing All horizontal loading will be resisted by bracing. For the purpose of illustration assume this will be present on every other frame (i.e. at 10 m spacing). It is more likely that bracing will be located at each end of the building or perhaps in a stair/lift well. The forces may therefore be greater than here but the principles would remain the same. For the loading shown, design the bracing members in grade Fe360 steel. 5.5.1.4(3) 5.5.3 5.5.2 5.5.1.1(1) 2.2.2.4(2) 2.2 Characteristic wind load, Q k = 1,6 kn/m 2. Design wind load, Q d = γ Q Q k At ultimate limit state, γ Q = 1,5 (unfavourable), Q d = 1,5 x 1,6 = 2,4 kn/m 2. Therefore, the total load (per m height of frame) = 10 x 2,4 = 24 kn/m. 23/02/07 25

24 kn/m 4,2 m 4,5 m 7,2 m Figure 9 Wind Load on Frame It is assumed that the uniformly distributed load acts as two point loads on the frame. Top load = (24 x 2,1) = 50,4 kn. Middle load = (24 x 2,1) + (24 x 2,25) = 104,4 kn. Assume that all horizontal load is resisted by the bracing only. Therefore, the load in the top brace = 50,4 / cos 30,3º = 58,4 kn, and the load in the bottom brace = (104,4 +50,4)/ cos 32º = 182,5 kn. 23/02/07 26

50,4 kn 58,4 kn 4,2 m 104,4 kn 182,5 kn 4,5 m 7,2 m Figure 10 Equivalent Point Wind Loads and Loads Within Bracing Design the bottom brace, as this carries the greater load. Try a CHS 175 x 5,0 6.1 Section Properties Depth of section, d = 175 mm, Thickness, t = 5 mm Area of section, A = 2670 mm 2, Ratio for local buckling, d/t = 35, Radius of gyration, i = 60,1 mm. 6.2 Classification of Cross-Section As the bracing is axially loaded, check the section classification is at least class 1, 2 or 3. Figure 11 shows a typical CHS cross-section. 5.4.4(1)a t d Figure 11 Typical CHS Cross-Section 23/02/07 27

Class 1 limiting value of d/t for a tubular section is 50ε 2. ε = 235 / f y where f y = 235 N/mm 2, ε = 1. 50ε 2 = 50 x 1 = 50 From section properties, d/t =35, therefore the section is Class 1. 6.3 Design of Compression Member The bracing members need to be checked as axially loaded. 6.3.1 Resistance of Cross-Section 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. The resistance of the cross-section will only be critical if a short, stocky column is used. The applied axial load, N Sd, must be less than the design compressive resistance of the cross-section, N c.rd. 5.3.1 (Sheet 4) 5.4.4(1) Applied axial load, N Sd = 182,5 kn. Design compressive resistance of cross-section, N c.rd = N Where A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/ mm 2, and γ M0 is the partial material safety factor = 1,1. N 2670 x 235 = = 570, 4 kn 11, x 10 pl.rd 3 pl.rd Af = γ y M0 3.1 5.1.1(2) The design compressive resistance of the cross-section, N c.rd = 570,4 kn, is greater than the applied axial load, N Sd = 182,5 kn. Therefore the section is satisfactory. 6.3.2 Design Buckling Resistance A class 1 member subject to axial compression should be checked for failure due to buckling. The design buckling resistance of a compression member shall be taken as: AAfy Nb.Rd = χβ γ M1 where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 cross-section, A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/mm 2, and γ M1 is the partial material safety factor for buckling resistance = 1,1. Values of χ for the appropriate value of non-dimensional slenderness, λ may be obtained from 5.5.2. 5.5.1.1(1) 5.5.1.1(1) 3.1 5.1.1.(2) 23/02/07 28

λ Non-dimensional slenderness, λ = λ β A Where the slenderness, λ = l i l is the member buckling length, and i is the radius of gyration. The bracing is designed as a simple pinned member. Therefore, the buckling length ratio l/l is equal to 1 - the buckling length is equal to the system length. 2 2 4, 5 + 7, 2 = 8500 mm Length of member = ( ) 0,5 E λ1 = π 93,9ε f = y 6.3.3 Determination Of Reduction Factor, χ Slenderness, λ = l/i = 8500/60,1 = 141 λ 1 = 93,9ε = 93,9 x 1 = 93,9 λ Non-dimensional slenderness, λ = λ β 0,5 = 141 0,5 A = 1 93,9 x 1 1,50 From 5.5.2, using buckling curve b, the reduction factor, χ = 0,3422. Design buckling resistance of member: χβ AAfy 0,3422 x 1 x 2670 x 235 N b.rd = = = 195,2 kn 3 γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (182,5 kn), therefore the bracing is satisfactory. 1 0,5 5.5.1.2(1) 5.5.1.4(3) 5.5.1.5(2) Annex E Figure E.2.1 5.5.1.2(1) 5.5.1.4(3) 5.5.3 5.5.2 5.5.1.1(1) 23/02/07 29

6.4 Design of Tension Member When the wind load is applied in the opposite direction, the bracing will be loaded in tension. The section therefore needs to be checked, for the same magnitude of loading, to ensure it is also satisfactory in tension. 50,4 kn 4,2 m 58,4 kn 104,4 kn 4,5 m 182,5 kn 7,2 m Figure 12 Equivalent Point Wind Loads and Loads Within Bracing 6.4.1 Resistance of Cross-Section The applied axial load, N Sd, must be less than the design tension resistance of the cross-section, N t.rd. Applied axial load, N Sd = 182,5 kn. Afy Design tension resistance of the cross-section, N t.rd = N pl.rd = γ M0 where A is the cross-sectional area = 2670 mm 2, f y is the yield strength of the steel = 235 N/ mm 2, and γ M0 is the partial material safety factor = 1,1. 2670 x 235 N pl.rd = = 570, 4 kn 3 11, x 10 The design tension resistance of the cross-section, N t.rd = 570,4 kn, is greater than the applied axial load, N Sd = 182,5 kn. Therefore the section is satisfactory. 5.4.3(1) 3.1 5.1.1(2) The bracing fulfils all the Eurocode requirements for members in tension and in compression, and is therefore satisfactory. The frame is satisfactory for all EC3 checks 23/02/07 30

7. Concluding Summary 23/02/07 31

Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode 3 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 EC3 by practical worked example. References: Eurocode 3: Design of steel structures Part 1.1 General rules and rules for buildings Contents: 23/02/07 32

WORKED EXAMPLE 3 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 10,5 m, each storey being 3,5 m high. It is assumed that the column foot is pinned at the foundation. Roof beams Floor beams 3,5 m Internal columns Internal columns 3,5 m External columns External columns 3,5 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 23/02/07 33

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 Fe360 (f y = 235 N/mm 2 ). Characteristic Loads Floor: Variable actions, Q k = 1,8 kn/m 2, Permanent actions, G k = 3,0 kn/m 2 Roof: Variable actions, Q k = 0,6 kn/m 2, Permanent actions, G k = 2,0 kn/m 2 The wind loads are applied as point loads of 10,5 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 30 kn/m 10,5 kn 21 kn 6.4.2.2(3) 5.2.2.3 3.1 30 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 3 (N3) Imposed Loading Cases Figure 2 Loading Cases 23/02/07 34

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 5.2.4.3(1) where k = k s c 0,5+ 1 n c but k c 1,0, 1 = 02, + but k s 1,0, 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 3. Therefore k c = 05, + 1 = 0, 866, and 4 1 k s = 02, + = 073,. 3 Substituting into the above equation: 1 φ = 0,866 x 0,73 x 200 = 1 315 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 390 1,24 2nd Floor 585 1,86 1st Floor 585 1,86 N1 Roof 117 0,37 2nd Floor 351 1,11 1st Floor 351 1,11 N2 Roof 39 0,12 2nd Floor 234 0,74 1st Floor 117 0,37 N3 Roof 78 0,25 2nd Floor 117 0,37 1st Floor 234 0,74 1 Equivalent Horizontal Forces The equivalent horizontal forces must also be multiplied be the appropriate partial safety factors for actions. 23/02/07 35

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 1,35G + 1,5W Load Case 2 1,35G + 1,5N1 Load Case 3 1,35G + 1,5N2 Load Case 4 1,35G + 1,5N3 Load Case 5 1,35G + 1,35W + 1,35N1 Load Case 6 1,35G + 1,35W + 1,35N2 Load Case 7 1,35G + 1,35W + 1,35N3 2 Load Combination Cases at the Ultimate Limit State The basic load cases are combined at the SLS as summarised in 3. Load Case 1 1,0G + 1,0W Load Case 2 1,0G + 1,0N1 Load Case 3 1,0G + 1,0N2 Load Case 4 1,0G + 1,0N3 Load Case 5 1,0G + 0,9W + 0,9N1 Load Case 6 1,0G + 0,9W + 0,9N2 Load Case 7 1,0G + 0,9W + 0,9N3 3 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 3 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 2.3.3.1(5) and 2.3.4(5) 2.3.3.2(1) 5.1.1(2) 5.1.1(2) 6.1.1(2) 23/02/07 36

Floor beams: IPE 450 Roof beams: IPE 360 Roof beams IPE 360 Floor beams IPE 450 3,5 m Internal columns HEB 260 Internal columns HEB 260 3,5 m External columns HEB 220 External columns HEB 220 3,5 m 6,5 m 6,5 m 6,5 m Figure 3 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) 23/02/07 37

Original Sway Case Moments Minus Propped Case Moments Equals "Pure Sway" Moments (To Be Amplified) 23/02/07 38

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 cr h H 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 0,25. If V Sd / V cr is greater than 0,25, 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 0,1 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(3) 5.2.6.2(6) 5.2.5.2(4) 5.2.6.2(4) 5.2.5.2(3) 23/02/07 39

δ 2 δ = δ2 - δ = 15,6 mm 1 h = 3500 mm V = 3296 kn H = 81 kn h H V δ1 Figure 5 Determination of Amplification Factor Therefore V Sd V 15,6 x 3296 = V δ cr h = = 018, H 3500 x 81 1 1 The amplification factor = = = 122, 1 VSd / Vcr 1 018, 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 13 L 14 J M 8 9 10 11 5 F 6 G 7 E H 9 = Element G = Node 1 2 3 4 A B C D Figure 6 Labelling of Members within the Structure 23/02/07 40