Plastic Design of Portal frame to Eurocode 3

Transcription

1 Department of Civil and Structural Engineering Plastic Design of Portal frame to Eurocode 3 Worked Example University of Sheffield

2 Contents 1 GEOMETRY DESIGN BRIEF DETERMINING LOADING ON FRAME Combination factors ψ Snow loading Self weight of steel members INITIAL SIZING OF MEMBERS LOAD COMBINATION (MAX VERTICAL LOAD) (DEAD + SNOW) Frame imperfections equivalent horizontal forces Partial safety factors and second order effects Sway buckling mode Stability ( αcr,s,est) Snap through buckling stability (αcr,r,est ) Accounting Second Order effects MEMBER CHECKS Purlins Column (UB 610 x 229 x 101) Classification Cross section resistance Stability against lateral and torsional buckling (EN : 2005 (E) Sec BB3.2.1): Rafter (UB457 x 191 x 89) Section Classification Cross section Resistance Check rafter buckling in apex region Stability check for lower bending moments Haunch (UB 457 x 191 x 89) Classification Haunch Stability Cross section resistance COMPARISON BETWEEN DIFFERENT CODES APPENDIX... 44

3 Geometry of the Frame 2 1 Geometry

4 Client brief 3 2 Design Brief A client requires a single-storey building, having a clear floor area 30 m x 80 m, with a clear height to the underside of the roof steelwork of 5 m. The slope of the roof member is to be at least 6 o. Figure 1 Plan view of the frame Figure 2 3 Dimensional view of the building (Plum, 1996)

5 Determining load on the frame 4 3 Determining loading on frame EN :2002 (E) Annex A 1 See Supporting Notes Sec Combination factors ψ The combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NAD. Note that because most portal frame designs are governed by gravity (dead + snow) loading, so in this worked example only maximum vertical load combination is considered. Therefore, the combination factor ψ is never applied in this example, but for full analysis the following load combination should be considered 1) Maximum gravity loads without wind, causing maximum sagging moment in the rafter and maximum hogging moments in the haunches. 2) Maximum wind loading with minimum gravity loads, causing maximum reversal of moment compared with case 1. The worst wind case might be from either transverse wind or longitudinal wind so both must be checked. Figure 3 Frame spacing (SX016, Matthias Oppe) Basic data : Total length: b = 72 m Spacing: s = 7.2 m Bay width: d = 30 m Height (max): h = 7.577m Roof slope: α = 6 o

6 Determining loading on the frame 5 3.2Snow loading General EN Sec Eq.5.1 Snow loading in the roof should be determined as follow µ Where: µ is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for relevant altitude. EN Sec 5.3 Table 5.1 See Appendix A Table A1 Roof shape coefficient Shape coefficients are needed for an adjustment of the ground snow load to a snow load on the roof taking into account effects caused by non-drifted and drifted snow loading. The roof shape coefficient depends on the roof angle so 030 µ =0.8 Snow load on the ground EN Annex C See Appendix A Table A2 For the snow load on the ground; the characteristic value depends on the climatic region; for site in the UK the following expression is relevant S k =0.140z-0.1+(A/501) Where: Z is the( zone number /9 ) depending on the snow load on sea level here in Sheffield z=3 A is the altitude above sea level A=175m

7 Determining loading on the frame 6 Snow load on the roof S k = 0.8 x 1 x = 0.54 KN/m 2 Spacing = 7.2 m For internal frame UDL by snow = 0.54 x 7.2 = 3.89 m Figure 4 Distributed load due to snow per meter span (SX016, Matthias Oppe) 3.3Self weight of steel members Self-weight estimated needed to be checked at the end Assume the following weight by members, Roofing = 0.2 KN/m 2 Services = 0.2 KN/m 2 Rafter and column self weight = 0.25 KN/m 2 Total self weight 0.65 KN/m 2

8 Initial sizing if members 7 TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3 4 Initial Sizing of members See Appendix B for the method Figure 5 Dimensions of portal (The institutionof Structural Engineers, TP/08/43 EC3/08/16) a) L/h = 30/6 = 5 r/l = 1.577/30= b) Loading 1) Gravity loading Snow loading = 0.54 x 7.2 = 3.80KN/m Self weight = 0.65 x 7.2 = 4.68 KN/m 2) Factored load w= (4.68 x 1.35 ) + (3.80 x 1.5 ) = 12.0 KN/m c) Finding Mp for the sections 1) Total load on the frame (wl)= 12.0 x 30 = 360.5KN 2) Parameter wl 2 = 12.0 x 30 2 = KNm 3) From Graphs (Figure B2) obtain horizontal force ratio (0.36) H= 0.36 x = KN 4) From Graphs (Figure B3) obtain rafter Mp ratio (0.034) M rafter,,rd = x = KNm

9 Initial sizing if members 8 5) From Graphs (Figure B4) obtained column Mp ratio (0.063) M column, Rd = x = KNm. 6) Selecting members a) W pl (rafter),required = (367.7 x 10 6 ) / 275 = 1337 x 10 3 cm 3 Try UB 457x152x74 b) W pl(column),required = (681.4 x 10 6 )/275= 2478 x 10 3 cm 3 Try UB 533 x 210 x 109 Section Tables of Universal Beams Properties Rafter Section UB 457x152x74 G=74.2 Kg/m h= 462mm b=154.4mm t w =9.6mm t f =17mm A=94.48 x 10 2 mm 2 d=428mm I y = x 10 4 mm 4 W pl,y =1627 x 10 3 mm 3 i y =186 mm i z = 33,3 mm I z = 1047 x 10 4 mm 4 W pl,z = x 10 3 mm 3 I t = x 10 4 mm 4 I w = x 10 6 mm 6 EN : 2005 (E) Table 3.1 Properties Column Section UB 533x210x109 G=109 Kg/m h= 539.5mm b=210.8mm t w =11.6mm t f =18.8mm A=138.9 x10 2 mm 2 d=510.9mm I y = x 10 4 mm 4 W pl,y =2828 x 10 3 mm 3 i y =218.7 mm i z = 45.7 mm I z = 2692 x 10 4 mm 4 W pl,z = x 10 3 mm 3 I t = x 10 4 mm 4 I w = 1811 x 10 6 mm 6 Steel grade is S275 Assume Sections Class1, then check

10 Ultimate Limit State Analysis 9 5 Load Combination (Max vertical Load) (Dead + Snow) 5.1Frame imperfections equivalent horizontal forces EN :2005 (E) Sec See Supporting Notes Section 9 ØØ Ø = Ø = 3.54 x 10-3 The column loads could be calculated by a frame analysis, but a simple calculation based on plan areas is suitable for single storey portals (i) Permanent loads ( un-factored ): Rafter = (74.5 x 15 x 9.8) / 10 3 = 11 KN Roofing = (15 x 0.2 x 7.2) = 21.6 KN Services = (15 x 0.2 x 7.2) = 21.6 KN Total = 54.2 KN (ii) Variable loads ( un-factored ) Snow load = 15 x 0.54 x 7.2 = 58.3 KN Thus the un-factored equivalent horizontal forces are given by: (i) (ii) Permanent/column = 3.54 x 10-3 x 54.2 = 0.19 KN Variable/column = 3.54 x 10-3 x 58.3 = 0.21 KN Note EC3 requires that all loads that could occur at the same time are considered together, so the frame imperfection forces and wind loads should be considered as additive to permanent loads and variable loads with the appropriate load factors.

11 Ultimate Limit State Analysis 10 Figure 6 Frame imperfections equivalent horizontal forces For Second Order effects See Supporting Notes Section 7.1 & Section Partial safety factors and second order effects Second order effects increases not only the deflections but also the moments and forces beyond those calculated by the first order. Second-order analysis is the term used to describe analysis method in which the effects of increasing deflections under increasing load are considered explicitly in the solution method. The effects of the deformed geometry are assessed in EN by calculating alpha crit (α crit ) factor. The limitations to the use of the first-order analysis are defined in EN Section (3) as αcrit 15 for plastic analysis. When a second order analysis is required there are two main methods to proceed: 1) Rigorous 2 nd order analysis (i.e. using appropriate second order software). 2) Approximate 2 nd order analysis (i.e. hand calculation using first order analysis with magnification factors). Although the modifications involve approximations, they are sufficiently accurate within the limits given by EN

12 Ultimate Limit State Analysis 11 See Supporting Notes Section 7.3 Carrying first order analysis to obtain first order moments and member forces using partial safety factors (γ G =1.35) and (γ Q =1.5) with loading calculated above. Then Checks if second order effects are relevant by calculating the following α cr,est =min ( α cr,s,est, α cr,r,est ) where α cr,s,est = estimated of α cr for sway buckling mode α cr,r,est =estimated of α cr for rafter snap-through buckling mode. Figure 7 Bending moment diagram for first order analysis (Burgess, 20/01/1990)

13 Ultimate Limit State Analysis 12 Load Factor Hinge number Member Hinge status RHC Formed LHR Formed Table 1 Position of Hinges and Load factors Figure 8 Member forces (Burgess, 20/01/1990)

14 Ultimate Limit State Analysis 13 See Supporting Notes Section Sway buckling mode Stability ( αcr,s,est). α cr,s,est = 0.8 1,,,,,, is the axial force in rafter {see figure 8 (150.8KN)}, is the Euler load of rafter full span, Where is the in-plan second moment of area of rafter L is the full span length., = 752 KN,,, is the minimum value for column 1 to n, is the horizontal deflection for top of column as indicated in Appendix, is the axial force in columns {see figure 8 (207.5KN, 208.1KN)} See Figure 9 As can be seen that, is the lateral deflection at the top of each column subjected to an arbitrary lateral load H EHF then here an arbitrary load H EHF can be chosen and using analysis software the deflection at top of each column can be obtained. 1) Arbitrary load H EHF =50KN 2), = 98mm, = 98mm

15 Ultimate Limit State Analysis 14 Figure 9 Sway mode check (Burgess, 20/01/1990) So either. OR. Min,,, = min(14.75, ) = Thus α cr,s,est =

16 Ultimate Limit State Analysis 15 See Supporting Notes Section & Section Snap through buckling stability (αcr,r,est ) α cr,r,est =. tan2 D cross-section depth of rafter (462mm). L span of the bay (30m). h mean height of the column (6m). in-plane second moment of area of column (66820 x 10 4 mm 4 ) in-plane second moment of area of rafter (32670 x 10 4 mm 4 ) nominal yield strength of the rafter (275 N/mm 2 ) roof slope if roof is symmetrical (6 o ) F r /F o the ratio of the arching effect of the frame where F r = factored vertical load on the rafter ( 432 KN see section 3) F 0 = maximum uniformly distributed load for plastic failure of the rafter treated as a fixed end beam of span L,, = 1.81 Thus α cr,r,est =.. tan 26 α cr,r,est = 6.2 Hence α cr,est =min ( α cr,s,est, α cr,r,est ) = 6.2

17 Ultimate Limit State Analysis 16 Although the snap-through failure mode is critical mode as shown in calculation above, but because this example is for designing single bay portal frames, the snapthrough mode of failure is irrelevant but included to show complete design steps for simple portal frame design. Snap-through failure mode can be critical mode in three or more spans, as internal bay snap-through may occur because of the spread of the columns inversion of the rafter (The institutionof Structural Engineers, TP/08/43 EC3/08/16) see figure 10. See Supporting Notes Figure 10 Snap through failure mode critical for 3 bay or more 5.3.2Accounting Second Order effects To account for second order effects the partial safety factors can be modified by the following criteria 1) γ G 2) γ Q = 1.50 = 1.68

18 Ultimate Limit State Analysis 17 See Supporting Notes Re-analyze the first order problem with the modified safety factors using same initial sized sections gives the following results, Load Factor Hinge number Member Hinge status RHC Formed LHR Formed Table 1 Hinges obtained from analysis It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 is suitable, although hinge 1 occurs at a load factors 1, a mechanism is not formed until the second hinge is formed. Therefore this combination of section sizes is suitable Hence size of member initially estimated is suitable and can withstand second-order effects. Note that if the load factors in positions 1 and 2 were less than 1, then the members size needs to be increased to sustain second order effects as the initially sized members cannot sustain second order effects. Figure 11 Bending moment diagram for first order analysis (Burgess, 20/01/1990)

19 Ultimate Limit State Analysis 18 Figure 12 Member forces for first order analysis (Burgess, 20/01/1990)

20 Member checks Purlins 19 6 Member checks See Supporting notes section 10.4 Note. Here the safety factors are used as indicated in King span load table 6.1 Purlins Today the design of the secondary members is dominated by cold formed sections. The design of cold formed members consists of looking up the relevant table for the chosen range of sections. The choice of a particular manufacture s products is dependent on clients or designer s experiences and preferences. Table (Appendix C) illustrates a typical purlin load table based on information from manufacture s catalogue (King span) for double span conditions. As the overall distance between columns is 30 meters, which is assumed to be divided to 18 equal portions would gives purlin centers 1.67 meters (on the slope). The gravity loading (dead (cladding Load plus snow load) is w= (0.1x 1.4) + (0.54 x 1.6) = KN/m 2. From the Table (Appendix C), knowing the purlin length of 7.2 m, purlin spacing of 1.25m and the gravity load to be supported by purlin 1.004KN/m 2, the M section seems adequate. See Appendix C Purlin Rafter Figure 13 Connection between rafter section and purlins Figure 14 Purlin cross section (Kingspan)

21 Member checks Column 20 See Figure Column (UB 610 x 229 x 101) - M Ed = KNm - V Ed = KN - N Ed = KN EN : 2005 (E) Section Classification Web ( Bending + Axial ) ε = 275/235 =1.08 actual (d/t w ) = ε Class 1. Flanges ( Axial Compressive ) actual (c/t f )=.... So the column sections are overall class ε Class Cross section resistance The frame analysis assumed that there is no reduction in the plastic moment resistance from interaction with shear force or axial force. This assumption must be checked; See supporting notes section Shear force effects of Plastic moment resistance (EN : 2005 (E) Sec 6.2.6) V Ed < 0.5 V pl,rd Av = 1.04 h t w = 1.04 x x 11.6 = mm 2 V pl,rd = / 3 /γ V pl,rd = / 3 / KN 0.5 V pl,rd = KN V Ed < 0.5 V pl,rd so the plastic moment of resistance is not reduced by the coexistence of axial force

22 Member checks Column Axial force effects of Plastic moment resistance (EN : 2005 (E) Sec 6.2.9) See supporting notes section 12.4 Check i. If, N Ed < < < ii. If N Ed < 0.25 N pl,rd N Ed < 0.25 plastic tensile resisitance of the section N Ed < < < See supporting notes section 13.4 Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN : Stability against lateral and torsional buckling (EN : 2005 (E) Sec BB3.2.1):. The design of the frame assumes hinge forms at the top of the column member, immediately below the haunch level. The plastic hinge position must be torsionally restraint in position by diagonal stays. With the hinge position restraint, the hinge stability is ensured by EC3 by limiting distance between hinge and the next lateral restraint to L m. 38 /

23 Member checks - Column /..... = 1.53 m Thus there must be a lateral restraint at a distance from the hinge not exceeding (1.53m). Figure 15 Column member stability (Plum, 1996) Therefore if 1.5 meters spacing assumed, this would ensure the stability between the intermediate restraints at the top of the column where maximum bending moment occurs, then the spacing of 1.8 meters is OK for sheeting rails below 2.4 meters from the top of the column, where the moment is lower.

24 Member checks - Column 23 It must be checked that the column buckling resistance is sufficient, so that the column is stable between the tensional restraint at S2 and the base. This part of the column would be checked using slenderness calculated. See supporting notes section 13.4 Different countries have different procedure to calculate the slenderness of the column and check the susceptibility of this part to lateral tensional buckling. Thus the designer must refer to the national Annex. In this example the procedure used in for assessing the significance of the mode of failure is taken from (King, Technical Report P164). Figure 16 Column between tensional restraints (King, Technical Report P164) (a) Calculate slenderness λ and λ LT Assume side rail depth = 200 mm Figure 17 Column/ Sheeting rails cross section (King, Technical Report P164)

25 Member checks - Column 24 (King, Technical Report P164) Distance from column shear center to center of the side rail, a a = 607.3/ /2 = mm i 2 s = i 2 y + i 2 z + a 2 i 2 s = = mm 2 Distance between shear center of flanges h s = h t f = = 520.7mm α = using the simplification for doubly symmetrical I sections I w = I z ( h s / 2 ) 2 α = α =.. = The slenderness of the column is given by: =

26 Member checks - Column 25.., Appendix D Figure D1 Where: m t is moment factor obtained from appendix D. Because loads combination considered there is no lateral loads applied to the walls, so there are no intermediate loads ψ = 0 / = 0 y = / ( L t / i z ) = / (4000 / 45.7 ) = m t = 0.53 c =1 for uniform depth members = 42.1 (b) Calculate buckling resistance for axial force EN :2005 Sec /ФФ. Ф h/b = 539.5/210.8 = 2.56, / EN :2005 Table 6.3 curve b for hot rolled I sections α= 0.34 / /

27 Member checks - Column 26 X z = 1/ = Ф , / = (0.741 x x 10 2 x 275) / (1.1 x 10 3 )= KN (c) Calculate buckling resistance for bending EN :2005 Sec EN :2005 Table 6.3 M b,rd,y =, /86.8 = 42.1/ 86.8 = Ф Ф /ФФ. 1/ = 0.92 M b,rd,y =

28 Member checks Column 27 (King, Technical Report P164) (d) Calculate buckling resistance to combined axial and bending Ψ = 0 β M,LT = Ψ = (0) = 1.8,,,,, μ LT 0.15 β M.LT 0.15 but μ LT 0.9 μ LT but μ LT 0.9 μ LT μ LT = 0.91 The column is OK and stable over the section considered (between restraint S o and S 2 ).

29 Member checks - Rafter Rafter (UB457 x 191 x 89) 6.3.1Section Classification EN : 2005 (E) Section 5.5 Ensure the section is class 1 to accommodate plastic hinge formation. ε = 275/235 =1.08 Web ( combined axial and bending ) actual (d/t w ) = ε Class 1 Flanges ( Axial Compressive ) actual (c/t f )=. The rafter section is Class ε Class Cross section Resistance. See Figure11 (Burgess, 20/01/1990) The frame analysis assumed that there is no reduction in the plastic moment resistance from interaction with shear force or axial force. This assumption must be checked because it is more onerous than that the cross-sectional resistance is sufficient. - Max. shear force V Ed = KN at haunch tip - Max. axial force N Ed = KN at haunch tip

30 Member checks - Rafter Shear force effects of Plastic moment resistance (EN : 2005 (E) Sec 6.2.6) See supporting notes section 12.3 V Ed < 0.5 V pl,rd Av = 1.04 h t w = 1.04 x 462 x 9.6 = 4613 mm 2 V pl,rd = / 3 /γ V pl,rd = / 3 / KN 0.5 V pl,rd = 333 KN V Ed < 0.5 V pl,rd so the plastic moment of resistance is not reduced by the coexistence of axial force Axial force effects of Plastic moment resistance (EN : 2005 (E) Sec 6.2.9) See supporting notes section 12.4 Check i) If, N Ed < < < ii) If N Ed < 0.25 N pl,rd N Ed < 0.25 plastic tensile resistance of the section N Ed < < < Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN : 2005.

31 Member checks Rafter Check rafter buckling in apex region See supporting notes section 13.2 Another highly stressed region is the length of rafter in which the apex hinge occur see fig below. Under (dead + snow) loading, the outstand flange is in tension, while compression flange is restrained by purlin/rafter connection. Figure 18 Member stability apex region (Plum, 1996) Therefore, the buckling resistance of the rafter member between purlins in the apex region needs to be checked. Because the apex hinge is the last to form in order to produce a mechanism (which is true for low pitched portal frame under dead + snow loading), then adequate rotation capacity is not a design requirement, i.e. hinge is required only to develop M p not to rotate. It is set by EC3 EN : 2005 that if the value of L m (as defined in BB.3.1.1) is not exceeded by restraint lateral torsional buckling can be ignored. So assuming that the purlins act as restraint because of their direct attachment to the compression flanges in the apex hinge region, then the purlin spacing should not exceed EN : 2005 (E) Sec BB / / See section 6.1 L m = 1221 mm = 1.221m As purlin spacing is 1.67m (on slope), thus because L m has been smaller than 1.67m then the purlin spacing would have to be reduced in the apex region to 1.2m.

32 Member checks Rafter Stability check for lower bending moments Where bending moment is lower, the purlin spacing can be increased: 31 Figure 19 Rafter under lower bending moments See Figure11 (Burgess, 20/01/1990) Next critical case is in right hand rafter. Try purlin spacing at 1670 mm centres Check for lateral torsional buckling between purlins: o M Edmax.y = knm o N Ed.max = kn at haunch tip at haunch tip (a) Calculate buckling resistance to axial force EN :2005 Sec EN :2005 Table 6.3 L =1670 mm λ z = L/i z = 1670/33.3 = λ z = λ z / 86.8 = / 86.8 = Ф = 0.5 [ 1 + α ( λ 0.2 ) + λ 2 ] =0.5 [ ( ) + (0.578) 2 ] = X z = Ф Ф =.. = N b,rd,z = =.. = KN

33 Member checks Rafter 32 EN :2005 Sec (b) Calculate buckling resistance to bending moment, Take C1 = 1 (conservative) λ LT =, λ LT =. = EN :2005 Table 6.3 Ф LT = 0.5 [ 1 + α ( λ 0.2 ) + λ 2 ] Ф LT = 0.5 [ ( ) ] = X LT = X LT = Ф Ф = M b,rd,y =, M b,rd,y =.. = 325.4KNm

34 Member checks Rafter 33 (c) Calculate buckling resistance to combined axial and bending Check that,, + k LT,, 1 Take k LT = 1 ( conservative ). + 1 x The value is slightly greater than one but due to the conservative assumption of K LT =1 the rafter can be assumed to be stable between intermediate restraint (purlin/sheeting rails) and purlin spacing could be increased to 1.67m between apex and hunch region shown in figure 19. Otherwise if the value was significantly greater than 1 the purlin spacing (1.67m) should be reduced.

35 Member checks - Rafter Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic rafter sections, as it is the normal practice to use a UB cutting of the same serial size as that of the rafter section for the haunch, which is welded to the underside of the basic rafter (UB 457x191x 89). Therefore it is assumed that the haunch has an overall depth at connection is 0.90 m. EN : 2005 (E) Section Classification ε = 275/235 =1.08 Web The web can be divided into two, and classified according to stress and geometry of each. actual (d/t w ) = web 1 ( bending ) ε Class 1 web 2 ( Compressive) ε Class 2 Figure 20 Haunch region cross section classification (King, Technical Report P164) Flanges ( Axial Compressive actual (c/t f )=... Thus the haunch section is a class ε Class1

36 Member checks - Haunch Haunch Stability See Supporting notes section 13.1 & 13.3 First, check the stability of the haunched portion of the rafter ( from eaves connection to the haunch/ rafter intersection) as this represents one of the most highly stressed lengths, and with its outstand flange (inner) in compression, this part of the rafter is the region most likely to fail due to instability. As it has already decided to stay the inside corner of the column/haunch intersection (column hinge position), assume that the haunch/rafter intersection is also effectively torsionally restrained be diagonal braces, giving an effective length of 3m as indicated in Fig below. Figure 21 Member stability haunch rafter region (Plum, 1996)

37 Member checks Haunch 36 EN : 2005 (E) clause BB (3)B It would appear that clause BB (3)B is the most appropriate creation to check the stability of the haunched portion, as there is three flanged haunch, so the distance between rotational restraint should be limited to Where: L k is length limit specified where lateral torsional buckling effects can be ignored where the length L of the segment of a member between restraint section at a plastic hinge location and adjacent torsional restraint. L k.. L k... = 3738 mm L k = 3.738m EN : 2005 (E) section BB See supporting notes section Appendix B c is the taper factor (shape factor) which accounts for the haunching of the restraint length (BB.3.3.3) 1 1 = 1.15

38 Member checks - Haunch 37 Figure 22 Dimensions defining taper factor (BS EN :2005) Is the modification factor for non-linear moment gradient (BB.3.3.2). The plastic moduli are determined for five cross-sections indicated on the figure below, the actual cross-section considered are taken as being normal to the axis of the basic rafter (unhaunched member). The plastic moduli together with the relevant information regarding the evaluation of the ratios N i /M i are given in the following table. The worst stress condition at the hunch/rafter intersection (location 5) is taken.

39 Member checks Haunch 38 Figure 23 Member stability haunch region (Plum, 1996) Position on haunch (FIG ) Distance from the eaves (m) Depth of bottom web (mm) Factored moment (M y,ed ) (KNm) Factored axial force (N Ed ) (KN) Moment capacity (KNm) Plastic modulus (cm 3 ) Ratio (N/M) a Value for (R) calculation (mm) R Value Table 2 Member forces at locations indicated in figure 18

40 Member checks - Haunch 39 The modification factor is determined by the form; See supporting notes section Appendix A in which R 1 to R 5 are the values of R according to equation below at the ends, quarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2) In addition, only positive values of ( ) should be included where, - R E is the greater of R 1 and R 5 - R s is the maximum value of R anywhere ( R 1 to R 5 ) -,, Where (a) is the distance between the centroid of the member and the centroid of restraining members (such as purlins restraining rafter). Here for simplicity a conservative value of (a) is found by conservative method of ignoring the middle flange as shown if figure (19). Figure 24 Simplification for distance between centriod of rafter and purlin sections

41 Member checks - Haunch Thus this portion of the rafter is stable over the assumed restrained length of 3 m (haunch length), as L s is around 3m. If the value was found to be less than the haunch length then a torsion restraint should be provided in the haunch region as shown below See supporting notes section Cross section resistance Shear force effects of Plastic moment resistance The shear in the rafter has been checked above, showing that V Ed < 0.5 V pl,rd. In the haunch, the shear area A v increases more than the applied shear V Ed, so the shear force has no effect on the plastic moment capacity of the haunch.

42 Member checks - Haunch Axial force effects of Plastic moment resistance, See supporting notes section 12.4 The tables provided below gives the axial and moment resistances of the haunch section at points 1to 5 shown in figures 18. A series of checks is carried out to determine whether the cross-sectional moment resistance M N,Rd is reduced by coexistence of axial force. Position Distance (mm) N Ed (KN) A (mm 2 ) N pl,rd (KN) A web (mm 2 ) (A web, f y )/y mo (KN) N pl,rd = A f y / y mo and f y =275N/mm 2 Table 3 Axial force at positions indicated in figure 18 for haunch Position Distance (mm) M Ed (KNm) 0.5 A web f y /y mo Is N Ed > 0.25 N pl,rd Does Axial force affect plastic bending resistance No No No No No No No No No No No No No No No Table 4 Checking the significance of axial force on plastic moment of resistance Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN : 2005.

43 Comparison between outcomes of different codes Comparison between Different Codes As the dimensions of portal frame designed in this worked example were deliberately chosen to be exactly the same as worked-example in (King, Technical Report P147) a comparison was done between the carried out worked example to (BS EN :2005), (BS9590-1:2000) and (ENV :1992). The following is a summary of different outcomes and source of design, Design no Design Code Column Section size Rafter Section Size Haunch Length (m) Purlin Spacing (m) Design Source 1 BS EN : x210 UB x152UB Worked - example 2 BS9590-1: ENV : x229 UB x229 UB x210 UB x191UB (King, Technical Report P164) (King, Technical Report P164) Table 5 Comparison between different code outcome

44 Appendix A 43 8Appendix EN :2003 (E) Section A.1 Roof shape coefficient The values given in table A1 apply when the snow is not prevented from sliding off the roof. Where the snow fences or other obstruction exists or where the lower edge of the roof is terminated with a parapet, then the snow load shape coefficient should not be reduced below 0.8. Table A1- Snow load shape coefficients (BS EN :2003) EN :2003 (E) Table C1 A.2 Snow load relationships The snow load on ground; the characteristic value depends on the climatic region; the following table gives different expressions for different regions, Table A2- Altitude-Snow load relationships (BS EN :2003)

45 Appendix A 44 Where: S k is the characteristic snow load on the ground (KN/m 2 ) A is the site altitude above the sea level (m) Z is the zone number given on the map ( see fig A1 ) The following maps gives the zone number Z for UK and republic of Ireland if other Z values for regions mentioned in Table A2 refer to EN Annex C pages ( 41 to 52 ). EN :2003 (E) Figure C.4 Figure A1 UK, Republic of Ireland : snow loads at sea level (BS EN :2003)

46 Appendix B 45 TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3 B1 Initial sizing using Weller s charts The method described relies for its simplicity on a series of three charts developed by Alan Weller. The chart has been constructed with the following assumptions and which leads to reasonably economic solution (Note. This is not a rigorous design method; it is a set of rules to arrive at initial size). 1) The rafter depth is approximately span / 55 2) The hunch length is approximately span /10 3) The rafter slope lies between 0 o and 20 o. 4) The ratio of span to eaves height is between 2 and 5. 5) The hinges in the mechanism are formed at the level of the underside of the haunch in the column and close to the apex. Each chart requires a knowledge of the geometry of the frame and the design loading as input data in order to determine approximate sizes for the column and rafter members Using of charts Figure B1 Dimensions of portal (The institutionof Structural Engineers, TP/08/43 EC3/08/16) a) Calculate the span/height to eaves ratio = L/h b) Calculate the rise/span ratio = r/l c) Calculate the total design load FL on the frame and then calculate FL 2, where F is the load per unit length on plan of span L (e.g. F =qs, where q is the total factored load per m 2 and s is the bay spacing). d) From figure B2 obtain the horizontal force ratio H FR at the base from r/l and L/h e) Calculate the horizontal force at the base of span H=H FR W L.

47 Appendix B 46 TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3 f) From figure B3 obtain the rafter M p ratio M PR from r/l and L/h. g) Calculate the M p required in the rafter from M p (rafter) = M PR x W L 2. h) From figure B4 obtain the column M p ratio M PL from r/l and r/h. i) Calculate the M p required in the rafter from M p (rafter) = M PL x W L 2. j) Determine the plastic moduli for the rafter W pl,y,r and the column W pl,y,c from W pl,y,r =M p,(rafter) /f y W pl,y,c = M p,(column) /f y Where f y is the yield strength. Using the plastic moduli, the rafter and column sections may be chosen from the range of plastic sections as so defined in the section books. 0.2 Span to eaves height Rise/span H/wL Figure B2- Horizontal force at the base (The institutionof Structural Engineers, TP/08/43 EC3/08/16)

48 Appendix B 47 TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3 Rise/span Span to eaves height M pr / wl² Figure B3- M p ratio required for the rafter (The institutionof Structural Engineers, TP/08/43 EC3/08/16) Span to eaves height Rise/span M pl / wl² Figure B4- M p ratio required for the column (The institutionof Structural Engineers, TP/08/43 EC3/08/16)

49 Appendix C 48 King Span Website Link ngspanstructur al.com/multibe am/rp/load_ta bles.htm C1 King Span Multi beam Purlin (Load tables) Loading Load Factor Dead load 1.4 Dead load restraining uplift or overturning 1.0 Dead load acting with wind and imposed loads combined 1.2 Imposed load 1.6 Imposed load acting with wind load 1.2 Wind load 1.4 Wind load acting with wind and imposed load 1.2 Forces due to temperature effects 1.2

50 Appendix C 49

51 Appendix D 50 (King, Technical Report P164) D1 Equivalent uniform moment factor mt for all other cases This formula, derived by (Sinhgh, 1969), is applicable in all cases, especially when the bending moment diagram is not a straight line between the tensional restraints defining the ends of the element. Figure D1- Moment factors (King, Technical Report P164)

52 Appendix D 51 (King, Technical Report P164) M Ed1 to M Ed5 are the values of the applied moments at the ends, the quarter points and mid- length of the length between effective torsional restraints, as shown in Figure D.2. Only positive values of M Ed should be included. M Ed is positive when it produces compression in the unrestrained flange. Figure D2- Intermediate moment (King, Technical Report P164) D2 Equivalent section factor c For uniform depth members, c = 1,0.

53 s BS EN :2003 Eurocode 1 Actions on structures Part 1 3: General actions Snow loads [Book]. 389 Chiswick High Road, London, W4 4AL : Standards, Institution British. BS EN :2005 Eurocode 3: Design of steel structures BS EN :2005 [Book]. 389 Chiswick High Road, London, W4 4AL : Standards, Institution British. Burgess Ian PLT Portal frame design Software. 20/01/1990. Vol. Ver 1.3. King C M Design of Steel Portal for Europe [Book]. [s.l.] : The steel construction Institute, Technical Report P164. Kingspan [Online] // February 10, Plum L J Morris & D R Structural Steelwrok Design to BS5950 2nd Edition [Book]. [s.l.] : Harlow : Longman, Sinhgh K.P. Ultimate behaviour of laterally supported beams [Book]. University of Manchester : [s.n.], SX016, Matthias Oppe Determination of loads on a building envelope [Online] // steel.com. Access Steel. October 20, steel.com/discovery/resourcepreview.aspx?id=j6oslkashmche7ubkvezgw==. The institutionof Structural Engineers Manual for the design of steelwork building structures to Eurcode 3 [Book]. TP/08/43 EC3/08/16.