Lecture20.doc

Structural Steelwork Eurocodes Development of a Trans-National ApproachDesign of MembersSection classificationStructural Steelwork Eurocodes Development ofA Trans-national ApproachCourse: Eurocode 3Module 7 : Worked ExamplesLecture 20 : Simple braced frameContents: 1. Simple Braced Frame1.1 Characteristic Loads1.2 Design Loads Fd = gF Fk1.3 Partial Safety Factors for Strength 2. Floor Beam - Fully Restrained2.1 Classification of Cross-section2.1.1Flange buckling2.1.2 Web buckling2.2 Shear on Web2.3 Deflection Check2.4 Additional Checks if Section is on Seating Cleats2.5 Crushing Resistance2.6 Crippling Resistance2.7 Buckling Resistance2.8 Summary3. Roof Beam Restrained at Load Points3.1 Initial Section Selection3.2 Classification of Cross Section3.2.1 Flange buckling3.2.2 Web buckling3.3 Design Buckling Resistance Moment3.4 Shear on Web3.5 Deflection Check3.6 Crushing, Crippling and Buckling3.7 Summary4. Internal Column4.1 Loadings4.2 Section properties4.3 Classification of Cross-Section4.3.1 Flange (subject to compression)4.3.2 Web (subject to compression)4.4 Resistance of Cross-Section4.5 Buckling Resistance of Member4.6 Determination of Reduction Factor cy4.7 Determination of Reduction Factor cz5. External Column5.1 Loadings5.2 Section properties5.3 Classifcation of Cross-Section5.3.1 Flange (subject to compression)5.3.2 Web (subject to compression)5.4 Resistance of Cross-Section5.5 Buckling Resistance of Member5.6 Determination of Reduction factor cy5.7 Determination of Reduction factor cz6. Design of Cross-Bracing6.1 Section Properties6.2 Classification of Cross-Section6.3 Design of Compression Member6.3.1 Resistance of Cross-section 6.3.2 Design Buckling Resistance6.3.3 Determination of Reduction Factor c?6.4 Design of Tension Member6.4.1Resistance of Cross-Section7. Concluding Summary1. Simple Braced FrameThe 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.Figure 1 Typical Cross Section of FrameIt 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 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 pin-connected.6.4.2.1(2)5.2.2.21.1 Characteristic LoadsFloor: Variable load, Qk = 3,5 kN/m2 Permanent load, Gk = 8,11 kN/m2Roof: Variable load, Qk = 0,75 kN/m2 Permanent load, Gk = 7,17 kN/m21.2 Design Loads Fd = gF Fk2.2.2.4(1)Floor: Gd = gG Gk. At ultimate limit state gG = 1,35 (unfavourable)Gd = 1,35 x 8,11 = 10,95 kN/m2Qd = gQ Qk. At ultimate limit state gQ = 1,5 (unfavourable)Qd = 1,5 x 3,5 = 5,25 kN/m22.2.2.4(2)Table 2.22.2.2.4(2)Table 2.2Roof: Gd = gG Gk. At ultimate limit state gG = 1,35 (unfavourable)Gd = 1,35 x 7,17 = 9,68 kN/m2Qd = gQ Qk. At ultimate limit state gQ = 1,5 (unfavourable)Qd = 1,5 x 0,75 = 1,125 kN/m22.2.2.4(2)Table 2.22.2.2.4(2)Table 2.2The steel grade selected for beams, columns and joints is Fe360.(fy = 235 N/mm2)Table 3.11.3 Partial Safety Factors for StrengthThe following partial safety factors for strength have been adopted during the design: Resistance of Class 1,2 or 3 cross-section, gM0 = 1,1 Resistance of member to buckling, gM1 = 1,1 Resistance of bolted connections, gMb = 1,252.3.3.2(1)5.1.1(2)5.1.1(2)6.1.1(2)The following load case, corresponding to permanent and variable actions (no horizontal loads) is found to be critical.2. Floor Beam - Fully RestrainedThe 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.Fd = gG Gk + gQ QkDesign load, Fd = (5 x 1,35 x 8,11) + (5 x 1,5 x 3,5) = 81 kN/mTable 2.1Figure 2 Loading on Fully Restrained Floor BeamDesign moment, Where MSd is the design moment in beam span,Fd is the design load = 81 kN/m, and L is the beam span = 7,2m.Design shear force, 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/mm2, and that the section is class 1 or 2.Table 3.1The design bending moment, MSd, must be less than or equal to the design moment resistance of the cross section, Mc.Rd:MSd Mc.RdMc.Rd = Mpl.y.Rd Where Wpl is the plastic section modulus (to be determined),fy is the yield strength = 235 N/mm2, andgM0 is the partial material safety factor = 1,1.5.4.5.1(1)Table 3.15.1.1(2)Therefore, rearranging:Try IPE 550Section properties:Depth, h = 550 mm, Width, b = 210 mmWeb thickness, tw = 11,1 mm Flange thickness, tf = 17,2 mmPlastic modulus, Wpl = 2787 cm35.4.5.1This notation conforms with Figure 1.1 in Eurocode 3: Part1.1.2.1 Classification of Cross-sectionAs 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.5.35.3.2 and Table 5.3.1Figure 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.Figure 3 A Typical Cross-Section2.1.1 Flange bucklingClass 1 limiting value of c/tf for an outstand of a rolled section is 10e. and fy = 235 N/mm2, therefore e =1.Calculate the ratio , where c is half the width of the flange = 105 mm, and tf is the flange thickness = 17,2 mm (if the flange is tapered, tf should be taken as the average thickness).Table 5.3.1 (Sheet 3)2.1.2 Web bucklingClass 1 limiting value of d/tw for a web subject to bending is 72e. and fy = 235 N/mm2, therefore e =1.Calculate the ratio, where d is the depth between root radii = 467,6 mm and tw is the web thickness = 11,1 mm. and Section is Class 1 and is capable of developing plastic moment.Table 5.3.1 (Sheet 1)Table 5.3.1 (Sheets 1 and 3)2.2 Shear on WebThe shear resistance of the web must be checked. The design shear force, VSd, must be less than or equal to the design plastic shear resistance, Vpl.Rd:VSd Vpl.RdWhere Vpl.Rd is given by 5.4.6For rolled I and H sections loaded parallel to the web,Shear area, Av = 1,04 h tw,fy is the yield strength = 235 N/mm2, andgM0 is the partial material safety factor = 1,1.5.4.6(4)Table 3.15.1.1(2) 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.where Av = twd, andd is the depth of end-plate = (for example) 300 mm.This is greater than the shear on the section (292 kN).The local shear on the beam web is OK.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.5.4.7(3)2.3 Deflection CheckEurocode 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.4.2Figure 4 Vertical Deflectionsd0 is the pre-camber (if present),d1 is the deflection due to permanent actions,d2 is the deflection caused by variable loading, anddmax 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 dmax and L/350 for d2. Deflection checks are based on the serviceability loading.Table 4.1 Figure 4.1For a uniform load where Fk is the total load = Qk or (Gk + Qk) as appropriate,L is the span = 7,2 m,E is the modulus of elasticity (210 000 N/mm2), andIy is the second moment of area about the major axis = 67120 x 104 mm4.3.2.5For permanent actions, Fk = 5 x 8,11 x 7