Structural Steelwork Eurocodes Development of A Transnational Approach


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1 Structural Steelwork Eurocodes Development of A Transnational Approach Course: Eurocode Module 7 : Worked Examples Lecture 0 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads 1. Design Loads F d = γ F F k 1. Partial Safety Factors for Strength. Floor Beam  Fully Restrained.1 Classification of Crosssection.1.1Flange buckling.1. Web buckling. Shear on Web. Deflection Check.4 Additional Checks if Section is on Seating Cleats.5 Crushing Resistance.6 Crippling Resistance.7 Buckling Resistance.8 Summary. Roof Beam Restrained at Load Points.1 Initial Section Selection. Classification of Cross Section..1 Flange buckling.. Web buckling. Design Buckling Resistance Moment.4 Shear on Web.5 Deflection Check.6 Crushing, Crippling and Buckling.7 Summary 4. Internal Column 4.1 Loadings 4. Section properties SSEDTA 001
2 4. Classification of CrossSection 4..1 Flange (subject to compression) 4.. Web (subject to compression) 4.4 Resistance of CrossSection 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. Section properties 5. Classifcation of CrossSection 5..1 Flange (subject to compression) 5.. Web (subject to compression) 5.4 Resistance of CrossSection 5.5 Buckling Resistance of Member 5.6 Determination of Reduction factor χy 5.7 Determination of Reduction factor χz 6. Design of CrossBracing 6.1 Section Properties 6. Classification of CrossSection 6. Design of Compression Member 6..1 Resistance of Crosssection 6.. Design Buckling Resistance 6.. Determination of Reduction Factor χ? 6.4 Design of Tension Member 6.4.1Resistance of CrossSection 7. Concluding Summary 15/0/07
3 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, 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, m. It is assumed that the column foot is pinned at the foundation. Roof Beam External Column Internal Column 4, m Floor Beam 4,5 m 7, m 7, m Figure 1 Typical Cross Section of Frame It is assumed that resistance to lateral wind loads is provided by a system of localised crossbracing, 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 () assumed that the connections offer little, if any, resistance to free rotation of the beam ends 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 pinconnected. 1.1 Characteristic Loads Floor: Variable load, Q k =,5 kn/m Permanent load, G k = 8,11 kn/m Roof: Variable load, Q k = 0,75 kn/m Permanent load, G k = 7,17 kn/m 1. Design Loads F d = γ F F k...4(1) Floor: G d = γ G G k. At ultimate limit state γ G = 1,5 (unfavourable) G d = 1,5 x 8,11 = 10,95 kn/m Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x,5 = 5,5 kn/m...4()....4(). 15/0/07
4 Roof: G d = γ G G k. At ultimate limit state γ G = 1,5 (unfavourable) G d = 1,5 x 7,17 = 9,68 kn/m Q d = γ Q Q k. At ultimate limit state γ Q = 1,5 (unfavourable) Q d = 1,5 x 0,75 = 1,15 kn/m The steel grade selected for beams, columns and joints is Fe60. (f y = 5 N/mm ) 1. Partial Safety Factors for Strength The following partial safety factors for strength have been adopted during the design: Resistance of Class 1, or crosssection, γ M0 = 1,1 Resistance of member to buckling, γ M1 = 1,1 Resistance of bolted connections, γ Mb = 1,5 The following load case, corresponding to permanent and variable actions (no horizontal loads) is found to be critical.. Floor Beam  Fully Restrained The beam shown in Figure is simply supported at both ends and is fully restrained along its length. For the loading shown, design the beam in grade Fe60, 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 1,5 x 8,11) + (5 x 1,5 x,5) = 81 kn/m 81 kn/m...4()....4()..1...(1) 5.1.1() 5.1.1() 6.1.1().1 7, m Figure Loading on Fully Restrained Floor Beam Fd L 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,m. 81x7, MSd = = 55 knm 8 Fd L 81x7, Design shear force, VSd = = = 9 kn To determine the section size it is assumed that the flange thickness is less than 40 mm so that the design strength is 5 N/mm, and that the section is class 1 or..1 15/0/07 4
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 Where W pl is the plastic section modulus (to be determined), f y is the yield strength = 5 N/mm, and γ M0 is the partial material safety factor = 1,1. Therefore, rearranging: Msdγ M0 55x10 x1,1 Wpl.required = = = 457 cm fy 5 Try IPE 550 Section properties: Depth, h = 550 mm, Width, b = 10 mm Web thickness, t w = 11,1 mm Flange thickness, t f = 17, mm Plastic modulus, W pl = 787 cm This notation conforms with Figure 1.1 in Eurocode : Part Classification of Crosssection 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 to develop the plastic moment resistance. Figure shows a typical crosssection for an IPE. IPE sections have been used in this example to reflect the European nature of the training pack (1) () and 5..1 t t w f c d Figure A Typical CrossSection.1.1 Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 5 / f y and f y = 5 N/mm, 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, mm (if the flange is tapered, t f should be taken as the average thickness). c 105 = = 6,10 t 17, f 5..1 (Sheet ) 15/0/07 5
6 .1. Web buckling Class 1 limiting value of d/t w for a web subject to bending is 7ε. ε = 5 / f y and f y = 5 N/mm, 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 = = 4,1 t w 11,1 c < 10ε and d < 7ε t t f w Section is Class 1 and is capable of developing plastic moment.. 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..1 (Sheet 1) 5..1 (Sheets 1 and ) 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, f y is the yield strength = 5 N/mm, and γ M0 is the partial material safety factor = 1, (4) () 1,04ht f Vpl.Rd = xγ w y M0 1,04 x 550 x 11,1 x 5 = = 78 kn x 1,1x10 This is greater than the shear on the section (9 kn). The shear on the beam web is OK. If the beam has partial depth endplates, a local shear check is required on the web of the beam where it is welded to the endplate. V A f / y pl.rd = v γ M0 where A v = t w d, and d is the depth of endplate = (for example) 00 mm. 11,1 x 00 x 5 Vpl.Rd = = 411 kn x 1,1 x 10 This is greater than the shear on the section (9 kn). The local shear on the beam web is OK. 15/0/07 6
7 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.. Deflection Check Eurocode 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 () 4. δ δ 1 δ δ 0 max L Figure 4 Vertical Deflections δ 0 is the precamber (if present), δ 1 is the deflection due to permanent actions, δ 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/50 for δ max and L/50 for δ. Deflection checks are based on the serviceability loading. For a uniform load δ = 5 84 x F L k EIy where F k is the total load = Q k or (G k + Q k ) as appropriate, L is the span = 7, m, E is the modulus of elasticity ( N/mm ), and I y is the second moment of area about the major axis = 6710 x 10 4 mm Figure /0/07 7
8 For permanent actions, F k = 5 x 8,11 x 7, = 9 kn. Therefore, deflection due to permanent actions, 5 x 9x10 x 700 δ 1 = = 10,1 mm 4 84 x x 6710x10 For variable actions, F k = 5 x,5 x 7, = 16 kn. Therefore, deflection due to variable actions, 5 x 16x10 x 700 δ = = 4, mm 4 84 x x 6710x10 The maximum deflection, δ max = δ 1 + δ = 10,1+ 4, = 14,4 mm L 700 Deflection limit for δ = = = 0,6 mm The actual deflection is less than the allowable deflection: 4, mm < 0,6 mm L 700 Deflection limit for δ max = = = 8,8 mm The actual deflection is less than the allowable deflection: 14,4 mm < 8,8 mm OK. The calculated deflections are less than the limits, so no precamber is required. It should be noted that if the structure is open to the public, there is a limit of 8 mm for the total deflection of δ 1 + δ (neglecting any precamber) 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..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.) 4..() Lecture, section 6...4() /0/07 8
9 .5 Crushing Resistance 5.7. 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 = 5 N/mm, 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 = t f (b f /t w ) (f yf /f yw ) [1  (σ f.ed /f yf ) ] 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 () 5.7.(1) 5.7.() x 17, x (10 / 11,1) sy = = 75 mm ( ) x 11,1 x 5 Crushing resistance, R y.rd = = 79,4 kn 1,1x10 This is greater than the reaction (9 kn). The crushing resistance is OK.6 Crippling Resistance The crippling resistance is given by 5.7.4(1) R a.rd = w t (Ef ) [(t / t ) + (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 = N/mm, f yw is the yield strength of the web = 5 N/mm, t f is the flange thickness = 17, mm, s s is the length of stiff bearing = 85 mm, but s s is limited to a maximum of 0,d (467,6 mm x 0, = 9,5 mm), d is the depth between root radii = 467,6 mm, and γ M1 is the partial material safety factor buckling resistance = 1, (1) 5.1.1() x 11,1 (10000 x 5) [(17, / 11,1) + (11,1 / 17,)(85 / 467,6)] Ra,Rd = 1,1x10 = 66 kn This is greater than the reaction (9 kn). The crippling resistance is OK. 15/0/07 9
10 .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, which in this case, gives a length: ss beff = (h + s s ) + a + but [h + s s ] 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. 85 beff = ( ) + = mm 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 = 7 mm t w 11,1 Radius of gyration for web, i = = =, 1 1 l 7 Slenderness of the web, = = = 10 i, Nondimensional slenderness of the web, β = 1 Where 1 = 9,9 ε = 9,9 x 1 = 9,9, and β A = 1 10 Non  dimensional slenderness of the web, = = 1, 09 9, 9 Using buckling curve c, the value of the reduction factor, χ may be determined from Reduction factor, χ = 0,49 AAfy Buckling resistance of a compression member, N b.rd = χβ γ M1 A is the crosssectional area = b eff t w, f y is the yield strength = 5 N/mm, and γ M1 is the partial material safety factor for buckling resistance = 1,1. 0,49 x 1 x x 11,1 x 5 N b.rd = = 7,4 kn 1,1x10 This is greater than the reaction (9 kn). The buckling resistance is OK..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. A Figure (4) () () 5.7.5() and (1) () 15/0/07 10
11 . 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 1,5 x 7,17) + (5 x 1,5 x 0,75)] = 97, kn External point load = 0,9 [(5 x 1,5 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 Fe60 steel. 48,6 kn 97, kn 97, kn 97, kn 48,6 kn A B C D E 1,8 m 1,8 m 1,8 m 1,8 m 7, m Figure 5 Beam Restrained at Load Points Reactions V Sd (at supports) = [( x 48,6) + ( x 97,)] / = 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 6,4 knm 6,4 knm 49,9 knm Figure 6 Bending Moment Diagram Moment at midspan (maximum) M Sd = [(194,448,6) x,6]  (97, x 1,8) = 49,9 knm 15/0/07 11
12 .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 Fe60 steel, f y = 5 N/mm. 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 = 19 mm Web thickness, t w = 11 mm Flange thickness, t f = 17,6 mm Plastic modulus, W pl = 046 cm This notation conforms with Figure 1.1 in Eurocode : Part1.1.. Classification of CrossSection 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 to develop the plastic moment resistance...1 Flange buckling Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 5 / f y and f y = 5 N/mm, 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.. Web buckling Class 1 limiting value of d/t w for a web subject to bending is 7ε. ε = 5 / f y and f y = 5 N/mm, therefore ε =1. Calculate the ratio d, where d is the depth between root radii = 78,8 t w mm and t w is the web thickness = 11,0 mm. d 78,8 = = 4,4 t w 11,0 c < 10ε and d < 7ε t t f Section is Class 1. w and (Sheet ) 5..1 (Sheet 1) 5..1 (Sheets 1 and ) 15/0/07 1
13 . Design Buckling Resistance Moment The design buckling resistance moment of a laterally unrestrained beam is given by: M = b.rd χ β W f LT W pl.y y γ M1 in which χ LT is the reduction factor for lateral torsional buckling, from 5.5., for the appropriate value of LT, using curve a for rolled sections (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 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 LT has been used, as the section is doubly symmetric and endmoment loading is present. The following formula for LT may be used: LT = C 0, L / i LT ( L / a ) LT 5, 66 0, 5 where L is the length between B and C = 1800 mm, I z is the second moment of area about the z  z axis = 085 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 = 046 x 10 mm, and I t is the torsion constant = 89, x 10 4 mm 4. i a LT LT = I I W z w pl.y x10 x 998x10 = 47, mm = (046x10 ) 9 Iw 998x10 = 957 mm 4 = I = 109x10 t Annex F F.1.() Equation F.15 F..() F..(1) 15/0/07 1
14 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 = 6,4/49,9 = 0,75, k = 1, therefore C 1 = 1,141 Substituting into the above equation: 1800 / 47, LT = = 4, 6 0, 5 ( ) 0, 5 / 1141, 1 + 5, 66 Nondimensional slenderness, LT = Where 1 = 9,9 ε = 9,9 x 1 = 9,9, β w = 1 for class 1 sections. 4, 6 Therefore, LT = ( = 9 9 1,, 0 ) 0, 7 For rolled I sections, buckling curve a should be used. From 5.5., 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 = 046 x 10 mm, f y is the yield strength of the steel = 5 N/mm, and γ M1 is the partial material safety factor for buckling resistance = 1,1. LT 1 β w The design buckling resistance moment for segment B  C is: M = b.rd χ β W f 0, 96 x 1 x 046x10 x 5 = = 419, 6 knm 6 γ 1,1 x 10 LT W pl.y y M1 F (4) 5.5.(5) () 5.5. M b.rd = 419,6 knm > M Sd = 49,9 knm, therefore the section is satisfactory. 15/0/07 14
15 .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 / ) 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 = 5 N/mm, and γ M0 is the partial material safety factor for the resistance of the crosssection = 1, (1) 5.4.6(4) () V 1,04 x 456 x 11 x 5 = = 64 kn x 1,1 x 10 pl.rd V Sd = 194,4 kn < V pl.rd = 64 kn, therefore the section is satisfactory. Inspection shows that V Sd < (V pl.rd / ), so there is no reduction in moment resistance due to the shear in the web..5 Deflection Check Eurocode requires that the deflections of the beam be checked under the following serviceability loading conditions: 5.4.7() 4. Variable actions, and Permanent and variable actions. For a general roof, the deflection limits are L/00 for δ max and L/50 for δ. 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 = F ka L a EIy (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 = N/mm, I y is the second moment of area about the major axis = 4090 x 10 4 mm 4, L is the span of the beam = 7, m, and a is the distance from the support to the adjacent load = 1,8 m. Central deflection, δ = 64,5x10 x x 4090x10 = 4, mm 15/0/07 15
16 For a central point load: Central deflection, δ = F L 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, m, E is the modulus of elasticity = N/mm, and I y is the second moment of area about the major axis = 4090 x 10 4 mm 4...5(1) 64,5x10 x 700 Central deflection, δ = = 5, 8 mm 4 48 x x 4090x10 Total deflection due to permanent loading, δ 1 = 5,8 + ( x 4,0) = 1,8 mm Consider the deflection from the variable loading. For a point load, distance a from the end of the beam: F ka L a Central deflection, δ = EIy 16 1 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 = N/mm, I y is the second moment of area about the major axis = 4090 x 10 4 mm 4, L is the span of the beam = 7, m, and a is the distance from the support to the adjacent load = 1,8 m. 6,75x10 x Central deflection, δ = x 4090x10 = 0, 4 mm (1) For a central point load: Central deflection, δ = F L 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, m, E is the modulus of elasticity = N/mm, and I y is the second moment of area about the major axis = 4090 x 10 4 mm 4...5(1) 6,75x10 x 700 Central deflection, δ = = 0, 6 mm 4 48 x x 4090x10 15/0/07 16
17 Total deflection due to variable loading, δ = 0,6 + ( x 0,4) = 1,4 mm Therefore, the total central deflection, δ max = δ 1 + δ = 1,8 + 1,4 = 15, mm. The limit for δ is L/50 = 700/50 = 8,8 mm. The limit for δ max = L/00 = 700/00 = 6 mm. 1,8 mm < 8,8 mm and 15,5 mm < 6 mm, therefore the deflections are within limits and no precamber of the beam is required..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..7 Summary All Eurocode recommendations are satisfied, therefore this beam is satisfactory. The beam is satisfactory. 4.1 and Figure 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 Fe60 steel, as a member in simple framing. 15/0/07 17
18 4.1 Loadings (54 x 7,) At roof level, the applied axial load = x = 89 kn (81 x 7,) At first floor level, the applied axial load = x = 58 kn Maximum load, from the first floor to the base, = kn = 97 kn Roof Internal Column Floor 4, 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. Section Properties Try an HE 40 A Grade Fe60 (terminology in accordance with EC) h = 0 mm b = 40 mm t w = 7,5 mm t f = 1 mm d/t w = 1,9 c/t f = 10 A = 7680 mm I y = 77,6 x 10 6 mm 4 I w = 8 x 10 9 mm 6 I z = 7,7 x 10 6 mm 4 I t = 41,6 x 10 4 mm 4 W pl.y = 745 x 10 mm W el.y = 675 x 10 mm i y = 101 mm i z = 60 mm i Lt = I I W z w pl.y 6 9 7,7x10 x 8x10 = 6,6 mm = (745x10 ) 9 Iw 8x10 alt = 888 mm 4 = It = 41,6x10 All the above properties can be obtained from section property tables. F..() F..(1) 15/0/07 18
19 4. Classification of Cross Section This section is designed to withstand axial force only. No moment is applied as the connecting beams are equally loaded Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 5 / f y where f y = 5 N/mm, therefore ε = 1. From section properties, c/t f = 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 = 1,9 c/t f 10ε and d/t w ε Class 1 section (Sheet ) 5..1 (Sheet 1).5.1 (Sheets 1 and ) Class 1 section 4.4 Resistance of CrossSection It is highly unlikely that the resistance of the crosssection will be the critical case  it is generally the buckling resistance that governs the suitability of a crosssection. For the sake of completeness, the check is included in this worked example. The resistance of the crosssection will only be critical if a short, stocky column is used. For members in axial compression, the design value of the compressive 5.4.4(1) force, N Sd, at each crosssection shall satisfy N Sd N c.rd For a class 1 crosssection, the design compression resistance of the 5.4.4() crosssection, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is crosssectional area = 7680 mm, f y is the yield strength = 5 N/mm, and γ M0 is the partial material safety factor = 1,1. N 7680 x 5 = = 1641 kn 1,1x10 c.rd N Sd = 97 kn, therefore N sd N c.rd. The section can resist the applied axial load () 15/0/07 19
20 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 = χβ γ A Af M1 y where χ is the reduction factor for the relevant buckling mode, β A = 1 for class 1 crosssection, A is the crosssectional area = 7680 mm, f y is the yield strength of the steel = 5 N/mm, 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 for the yy and zz axes respectively. Values of χ for the appropriate value of nondimensional slenderness, may be obtained from Nondimensional slenderness, = β Where the slenderness, = 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. E 1 = π 9,9ε f = y 4.6 Determination of Reduction Factor, χ y Slenderness, y = l/i y = 4500/101 = 44,6 1 = 9,9ε = 9,9 x 1 = 9,9 Nondimensional slenderness, y = y β 44,6 A = 9,9 x 1 = 0,47 From 5.5., using buckling curve b, the reduction factor, χ y = 0, A (1) (1) () (1) () () (1) () /0/07 0
21 4.7 Determination of Reduction Factor, χ z Slenderness, z = l/i z = 4500/60 = 75 1 = 9,9ε = 9,9 x 1 = 9,9 Nondimensional slenderness, = z β 75 y A = 9,9 x 1 = 0,8 From 5.5., using buckling curve c,the reduction factor, χ z = 0,66 Therefore χ = χ z = 0,66. Design buckling resistance of member: χβ AAfy 0,66 x 1 x 7680 x 5 Nb.Rd = = = 1086 kn γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (97 kn), therefore the column is satisfactory. The column is OK () (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 Fe60 steel, as a member in simple framing. 5.1 Loadings (54 x 7,) At roof level, the applied axial load = = 194 kn (81 x 7,) At first floor level, the applied axial load = = 9 kn Maximum load, from first floor to base, = 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 midpoint of the column. No moment due to eccentricity of applied load is therefore applied to the column. See Annex H 15/0/07 1
22 Roof 4, 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. Section Properties Try an HE 00 A Grade Fe60 h = 190 mm b = 00 mm t w = 6,5 mm t f = 10 mm d/t w = 0,6 c/t f = 10 A = 580 mm I y = 6,9 x 10 6 mm 4 I w = 108 x 10 9 mm 6 I z = 1,6 x 10 6 mm 4 I t = 1,0 x 10 4 mm 4 W pl.y = 49 x 10 mm W el.y = 89 x 10 mm i y = 8,8 mm i z = 49,8 mm i Lt I I = W z w pl.y 6 9 1,6x10 x 108x10 = 5,9 mm = (49x10 ) 9 Iw 108x10 alt = 717 mm 4 = It = 1,0x10 All the above properties can be obtained from section property tables. F..() F..(1) 15/0/07
23 5. Classification of Cross Section This section is designed to withstand axial force only Flange (subject to compression) Class 1 limiting value of c/t f for an outstand of a rolled section is 10ε. ε = 5 / f y where f y = 5 N/mm, ε = 1. 10ε = 10 x 1 = 10 From section properties, c/t f = Web (subject to compression) Class 1 limiting value of d/t w for a web subject to compression only is ε. ε = 5 / f y where f y = 5 N/mm, ε = 1. ε = x 1 = From section properties, d/t w = 0,6 c/t f 10ε and d/t w ε Class 1 section (Sheet ) 5..1 (Sheet 1) 5..1 (Sheets 1 and ) 5.4 Resistance of CrossSection It is highly unlikely that the resistance of the crosssection will be the critical case  it is generally the buckling resistance that governs the suitability of a crosssection. For the sake of completeness, the check is included in this worked example. The resistance of the crosssection will only be critical if a short stocky column is used. For members in axial compression, the design value of the compressive 5.4.4(1) force, N Sd, at each crosssection shall satisfy N Sd N c.rd For a class 1 crosssection, the design compression resistance of the 5.4.4() crosssection, N c.rd, may be determined as: Afy Nc.Rd = γ M0 where A is crosssectional area = 580 mm, f y is the yield strength = 5 N/mm, and γ M0 is the partial material safety factor = 1, x 5 N c.rd = = 1149 kn 1,1x10 N Sd = 486 kn, therefore N sd N c.rd. The section can resist the applied axial load () 15/0/07
24 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 crosssection, A is the crosssectional area = 580 mm, f y is the yield strength of the steel = 5 N/mm, 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 for the yy and zz axes respectively. Values of χ for the appropriate value of nondimensional slenderness, may be obtained from Nondimensional slenderness, = β Where the slenderness, = 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. E 1 = π 9,9ε f = y 5.6 Determination of Reduction Factor, χ y Slenderness, y = l/i y = 4500/8,8 = 54, 1 = 9,9ε = 9,9 x 1 = 9,9 Nondimensional slenderness, y = y β 54, A = 9,9 x 1 = 8 From 5.5., using buckling curve b, the reduction factor, χ y = 0, A (1) (1) () (1) () () Annex E Figure E (1) () /0/07 4
25 5.7 Determination of Reduction Factor, χ z Slenderness, z = l/i z = 4500/49,8 = 90,4 1 = 9,9ε = 9,9 x 1 = 9,9 Nondimensional slenderness, = z β 90,4 y A = 9,9 x 1 = 0,96 From 5.5., using buckling curve c,the reduction factor, χ z = 5 Therefore χ = χ z = 5. Design buckling resistance of member: χβ AAfy 5 x 1 x 580 x 5 N b.rd = = = 6, kn 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 CrossBracing 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 Fe60 steel () (1)...4(). Characteristic wind load, Q k = 1,6 kn/m. 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 =,4 kn/m. Therefore, the total load (per m height of frame) = 10 x,4 = 4 kn/m. 15/0/07 5
26 4 kn/m 4, m 4,5 m 7, 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 = (4 x,1) = 50,4 kn. Middle load = (4 x,1) + (4 x,5) = 104,4 kn. Assume that all horizontal load is resisted by the bracing only. Therefore, the load in the top brace = 50,4 / cos 0,º = 58,4 kn, and the load in the bottom brace = (104,4 +50,4)/ cos º = 18,5 kn. 15/0/07 6
27 50,4 kn 58,4 kn 4, m 104,4 kn 18,5 kn 4,5 m 7, 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 = 670 mm, Ratio for local buckling, d/t = 5, Radius of gyration, i = 60,1 mm. 6. Classification of CrossSection As the bracing is axially loaded, check the section classification is at least class 1, or. Figure 11 shows a typical CHS crosssection (1)a t d Figure 11 Typical CHS CrossSection 15/0/07 7
28 Class 1 limiting value of d/t for a tubular section is 50ε. ε = 5 / f y where f y = 5 N/mm, ε = 1. 50ε = 50 x 1 = 50 From section properties, d/t =5, therefore the section is Class Design of Compression Member The bracing members need to be checked as axially loaded Resistance of CrossSection It is highly unlikely that the resistance of the crosssection will be the critical case  it is generally the buckling resistance that governs the suitability of a crosssection. For the sake of completeness, the check is included in this worked example. The resistance of the crosssection 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 crosssection, N c.rd (Sheet 4) 5.4.4(1) Applied axial load, N Sd = 18,5 kn. Design compressive resistance of crosssection, N c.rd = N Where A is the crosssectional area = 670 mm, f y is the yield strength of the steel = 5 N/ mm, and γ M0 is the partial material safety factor = 1,1. N 670 x 5 = = 570, 4 kn 11, x 10 pl.rd pl.rd Af = γ y M () The design compressive resistance of the crosssection, N c.rd = 570,4 kn, is greater than the applied axial load, N Sd = 18,5 kn. Therefore the section is satisfactory. 6.. 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 crosssection, A is the crosssectional area = 670 mm, f y is the yield strength of the steel = 5 N/mm, and γ M1 is the partial material safety factor for buckling resistance = 1,1. Values of χ for the appropriate value of nondimensional slenderness, may be obtained from (1) (1) () 15/0/07 8
29 Nondimensional slenderness, = β Where the slenderness, = 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. Length of member = (,, ) E 1 = π 9,9ε f = y 1 A = 8500 mm 6.. Determination Of Reduction Factor, χ Slenderness, = l/i = 8500/60,1 = = 9,9ε = 9,9 x 1 = 9,9 Nondimensional slenderness, = β = 141 A = 9,9 x 1 1,50 From 5.5., using buckling curve b, the reduction factor, χ = 0,4. Design buckling resistance of member: χβ AAfy 0,4 x 1 x 670 x 5 N b.rd = = = 195, kn γ M1 1,1x10 The design buckling resistance of the member is greater than the applied load (18,5 kn), therefore the bracing is satisfactory (1) () () Annex E Figure E (1) () (1) 15/0/07 9
30 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, m 58,4 kn 104,4 kn 4,5 m 18,5 kn 7, m Figure 1 Equivalent Point Wind Loads and Loads Within Bracing Resistance of CrossSection The applied axial load, N Sd, must be less than the design tension resistance of the crosssection, N t.rd. Applied axial load, N Sd = 18,5 kn. Afy Design tension resistance of the crosssection, N t.rd = N pl.rd = γ M0 where A is the crosssectional area = 670 mm, f y is the yield strength of the steel = 5 N/ mm, and γ M0 is the partial material safety factor = 1, x 5 N pl.rd = = 570, 4 kn 11, x 10 The design tension resistance of the crosssection, N t.rd = 570,4 kn, is greater than the applied axial load, N Sd = 18,5 kn. Therefore the section is satisfactory. 5.4.(1) () The bracing fulfils all the Eurocode requirements for members in tension and in compression, and is therefore satisfactory. The frame is satisfactory for all EC checks 15/0/07 0
31 7. Concluding Summary 15/0/07 1
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