Analysis of portal frame building In accordance to EN (2005)

Transcription

1 nalsis of portal frame building In accordance to E (005) 1 description / The portal frame is the main structural element of the building. The frame is designed for the following loads Roof loads such as workmen, snow or hail Wind loads Wind loads can be positive as on B or negative (suction)as on BC,CD and DE. Roof loads are positive and up to down direction B/ If The joints at B,C and D are not rigid,the will open up and the frame will be unstable

2 C/ 1) Vertical loading on the frame results in and E tending to be pushed outwards.if the foundation cannot resist this horizontal push,outward movement will occur,and the frame will loose structural strength ) Wind subjects the portal frame to uplift forces(the roof tends to fl-off)like an plane wing,to overturning forces on the sides and ends of the building, These destabilizing forces are resisted essentiall b the weight of the building,and in this regard,the foundations contribute significantl to this weight. Generall speaking it is a fact that portal frame buildings of this kind are light weight structures, and as such the tend to collapse sideward and upwards rather than do wnwards. The effect of wind on a light building cannot be overemphasized. The destabilization it causes is a major design consideration, and in this context, foundations can be regarded as the building s anchors D/ the rafter of the portal frame is a slender structural element,and it is restrained it will buckled when loaded.

3 In a braced roof this restraint is provided b the purlins acting together with a braced ba.the purlins provide the restraining force for the rafters,and the braced ba acts as a buttress wich absorbs these purlin restraining forces. While this sstem is effective in restraining the top flange of the rafter I-beam,the bottom flange remains relativel unrestrained, and to achieve the requisite restraint,short lengths of angle iron are connected at intervals between the bottom flange of the I-beam and the purlins.this simple and necessar anti-buckling feature is sometimes neglected in the design of the portal frames. E/

4 building frame subjected to wind forces along its length will tend to collapse as shown above,while a building with a braced side ba as shown below will be stable,since the braced ba will functions as a buttress to resist the wind forces, and transform them to the foundations portal frame design.1 Basic data Total length b= 70 m Ba width d= 5 m Spacing s= 7 m E=10000/mm G=80770/mm Steel:S5 Height h= 7.5 m Roof slope α= 5 Purlin spacing sp=1.5 m Cladding rail spacing Sp =.0m

5 rticulated purlin purlin cladding rail column Internal portal frame door 4*5m α= 5 7.5m 6.41m 5.0m Internal portal frame. Loads..1 Permanent loads Self-weight of the beam Roofing with purlins G= 0.5 K/m For an internal frame G=0.5*7=.45 K/m G=-.45 K/m α= m 7.5 m 5.0 m

6 .. Construction loads Q=0.5 K/m E clause 4.11(005) For an internal portal frame Q=0.5*7=.5 K/m.. Wind loads Take from the document treated wind actions to E (005) as a values described below -.74 (J) (G) -8.88(F) -.74 (H) -.74 (I) w1 +.4(D) -.4 (E) pproximation calculations 1/ wind forces applied to duopitch roofs and partial variables live loads These actions are ver small in ccomparison with the wind actions on vertical walls(0.5% to 1.%). In this case the will be neglected for calculations.

7 / wind forces (up-to fl) The actions applied to duopitch roofs are oriented as described above (perpendicular to rafters).for simplifications we admit that these forces will be oriented verticall as gravit forces. / Forces transmitted b purlins The forces transmitted to rafters b purlins, (are ponctual forces and must be applied in calculations of rafters),will be converted to linear forces.the error caused b this simplification is 0.5%,and conduct to increase the moments at B and D 4/ Stiffness at B and D To conduct manuall calculations we consider that the inertia of the column and the rafter are equals I c =I R The coefficient of stiffness k I h S I R will be C h k s This simplification,justified b the presence of the haunchs,conduct to increase the moment at C and decrease moments at B and D.It will be compensated b the simplification applied to purlin calculations,wich act in opposite sens.. Simple cases 1/Case 1 Vertical actions dead (G) variable (Q) loads

8 Y C s I R f= B D - h= H E H E x V V E Rigidit coefficient at B and D stiffness. of. rafter IR h K stiffness. of. comumn S I In application of Castigliano théorem and with the structure smetr C d ds 0 BCDE where H is the horizontal force EI dh Displacement 1 in B column. In an ordinate point,, of the column B, the moment is H.,then d dh and h h H d H d EI 0 R EI R EI 0 R Hh Displasment in BC rafter The moment expression at abscissa,x,is: x cos H h xsin q Vx cos d h xsin and dh s x cos H h xsin q Vx cos h xsin dx 0

9 We have l f cos and sin then s s 1 f s 5 1. H. h. s h. f. s q l. f. s hl s EI R 96 1 Then the equation 1 0 give the result H ql s 5f 8h IR h h s f f C R I I h h and ma be reduced if we use the rigidit k h s in place of the real expression k IR h s I C We obtain the simplified expression H ql 5f 8h h k f h f Conclusion Hh B D ql C H h f 8 H H H V E ql VE ql 5f 8h h k f h f /Case Vertical actions( wind up to fl) Hh B ql C H h f 8 D H H H E ql 5f 8h ql V VE h k f h f

10 Y q C B + D H E H E x V V E /Case Horizontal actions( wind 1 pressure) B qh HE. h H E qh 5kh 6h f 16 h k f h f D. E H h H q. h H E

11 qh C H Eh f 4 Y V E qh V l C B + D + q - H E H E x V V E 4/ Case 4 Horizontal actions( wind 1 succions) D qh H. h H qh 5kh 6h f 16 h k f h f B. H h H q. h H E qh C H h f 4 V E qh V l Y C B D + q - H E H E x V V E

12 Calculation of the rafter in bending Dead loads G=.45K/m W1 wind in long span (internal surpressure) W c, W c,1 W c, wc,1.4 K / m see fig above wc,.4 K / m wc,.74 K / m W wind in long span (internal depressure) W c, W c,1 W c, Take cpi, 0. E (005) (7..9 (6)note ) wc,1 cp, e cp, i qps * K / m wc, 0 K / m We have choose the max value of G zone for wind calculation but not the better wc, cp, e cp, i qps *7 4.1 K / m nd For the zones H;I,and J this value is

13 wc, cp, e cp, i qps *7 1.4 K / m W wind gear ( with internal surpressure) W c, W c,1 W W c, We take a middle value of the zones G,H and I as described in wind actions to E (005) We take also a middle value of the zones,b and C then we will have wc,1 wc, 0.668* K / m wc, 1.0*7 7 K / m Calculus actions It is to determinate: --the support reactions H ; H E ;V and V E -- the max bending moments B ; C and D These forces are obtained from the actions mentioned in tables above Values for calculations: S=1.55; f=1.09 ; h=6.41; k= *5 5*1.09 8* * H H K E.45* 5 V VE 0.65K C.45* Km * Km B D

14 actions case q(k/m) H (K) H E (K) V (K) V E (K) B (Km) C (Km) D (Km) G Q W 1 ;W c, W 1 ; W c, W 1 W c, Total W ;W c, W ;W c, W ;W c, Total W ;W c, W ;W c, W ;W c, Total Load combinations Partial factor Gmax 1.5 Gmin 1.0 Q 1.50 permanent loads permanent loads variable loads When there is more then one variable action acting,requiring the actions to be combined, the expression is SLS GK, j 0.9 Q, ULS : g, jgk, j 0.9 Q, iqk, i j i1 j i1 K i

15 These combinations are obtained from the DF (French,national annex ) the coefficient 1. applied for wind will be omitted if we use combinations above ULS combination combination Reactions (K) Bending moments (Km) H H E V V E B C D G 1.5Q G 1.5W1 useless G 1.5W G 1.5W G1.5W G1.5W G1.5W G 1.8W G 1.8W G 1.8W The maximum values are collected in the table Reactions (K) Bending moments (Km) H H E V V E B C D / Rafter 4.1/Resistance The maximum moment in: - pex connection : B = D = Km - Eave connection : C =+9. Km The expression Rd must be verified for bending With Rd W pl. f 0 Wpl. f.we have 0 then W pl. 0 f

16 For apex connection Wpl eave connection Wpl In apex connection Wpl cm IPE 60+(1/) IPE 60 - In eave connection Wpl cm IPE 60 The 1.5 IPE60 section is considered as welded beam. the table below show it s characteristics Caractéristiques du profil P.R.S. Caractéristiques géométriques Caractéristiques mécaniques xe neutre élastique xe neutre plastique h = 540 mm h w = 514,6 mm t w = 8 mm b f = 170 mm t f = 1,7 mm g = 66,1 kg/m = 84,5 cm I = 9105,7 cm 4 W el. = 1448,4 cm W pl. = 1668,1 cm i = 1,5 cm I z = 104,1 cm 4 W el.z = 1,6 cm W pl.z = 191,7 cm i z =,5 cm I t = cm 4 I w = 7861 cm 6 Z ane = 70 mm Y ane = 85 mm Z anp = 70 mm Y anp = 85 mm

17 This choice is preliminar and will be completed b others 4./ Vertical deflection Vertical deflection of the rafter The vertical deflection will be calculated under G The moment in a section is ql q x B x x B integration of the equation d dx We have EI Q l l 1 dx B x x dx 0 0 d ql q dx EI EI l x For we have d dx 0 then l 1 ql q l ql B. x x x B. dx EI For x=0 we have =0 then 1 max 5ql 48. l 84EI 4 B E=10000 Pa=10000/mm =.1x10 8 K/m I=1670 cm 4 q= G Q =.45+.5=5.95K/m L=5.1m B = =55.15 Km For IPE 60 the vertical deflection is 4 5*5.95* * 55.15* 5.1 max 0.119m 1.cm *.1*10 *1670*10 In this case we must upgrade to IPE 500 and we obtain a limit value but less because we haven t consider the presence of apex

18 4 5*5.95* * 55.15* 5.1 max 0.105m 10.5cm *.1*10 * 4800*10 f adm l cm CRCTERISTIQUES GEOETRIQUES IPE 500 CRCTERISTIQUES ECIQUES h = 500 mm b = 00 mm t w = 10, mm t f = 16 mm r = 1 mm d = 46 mm g = 90,70 kg/m = 116,00 cm I = 48 00,00 cm 4 W el. = 1 98,00 cm W pl. = 194,00 cm i = 0,4 cm vz = 59,87 cm I z = 14,00 cm 4 W el.z = 14,0 cm W pl.z = 5,90 cm i z = 4,1 cm I t = 89,9 cm 4 We remark that IPE 500 is ver suffisant to resist under positif and negative bending moment 4./Classification The section is class 1 as a similar (but not the same) verification for the column (see 5) 4.4/Buckling resistance This figure shows different Sections categories and buckling modes

19 Lateral torsional buckling check using the simplified assessment methods for beams with restraints in buildings: 8*1.5m Lateral restraints (purlins) IPE 500 Lateral restraints 4.19m (bracing sstem) *4.18m

20 Bracing sstem In buildings, members with discrete lateral restraint to the compression flange are not susceptible to lateral-torsional buckling if the length L c between restraints or the resulting equivalent compression flange slenderness f Where k clc i f, z 1 f c,0 satisfies: c, Rd, [6...4], is the maximum design value of the bending moment within the restraint spacing k c is a slenderness correction factor for moment distribution between restraints, see E Table 6.6; i f, z is the radius of gration of the compression flange including 1/ of the compressed part of the web area, about the minor axis of the section; c,0 is the slenderness parameter of the above compression element: c,0 LT, then, LT,0 0.4 E and f c f 5 mm [6...] I f, z I z d t w * * 1 then

21 I f, z * * 1 1 d f, z * * t w then cm 4 I i f, z f, z * * cm I f, z f, z cm W Wpl, 194cm E f c, Rd W f 194* 5* Km Combination 1.5G 1.5Q B = =66.87 Km We consider that the coefficient is the same if the rafter is unrestraint then C B Then K C *0.655 table 6.6 But between restraints in the centre of the rafter where the moment are maximum, the moment distribution ma be considered as constant :K C =1.0 table 6.6 KCLC 1.0*150 f 0. i 4.96*9.9 f, z 1 The maximum bending moment is at the origin B of the rafter then the lateral torsional buckling ma be also in the origin 66.87Km,

22 c, Rd c,0, f, z * KCLC 1.0*150 f 0. i 4.96* G 1.8W Combination B = 88.7Km, c, Rd c,0, f, z * KCLC 1.0* 418 f i 4.96*9.9 ot verified It s necessar to add other bracing sstems each m spacing then L C =m 88.7Km, c, Rd c,0, f, z * KCLC 1.0*00 f i 4.96* Then the lateral torsional buckling is satisfactor detailed procedure to do verification for the rafter is sho wn below as for column When the above procedure is not satisfactor. OT The real comportement of the rafter is shown in the figure

23 1 tension flange elastic section plastic stable length 4 plastic stable length 5 elastic section 6 plastic hinge 4.5/ the haunch verification 7 restraints 8 bending moment diagram 9 Compression flange 10 plastic stable length 11 plastic stable length 1 elastic section nnex C + O - D F S x the equation of the bending moment curve is a parabolic form Y ax

24 the point F is considered the limit of elastic moment el Wel f 198*10 * Km X 0 m S=1.55 Y C =4.15 D + C = =60.5 Y 60.5 Then a 4 X The bending moment curve equation will be For X 1.55 F then 0 Y 4X Wel f 198*10 * 5 Y el 45.8Km 1.0 Then F then F 100.4F176 0 Conclusion F 1.78 m Length of the rafter F=m The same verification for buckling 1/about as for column in section 5 ma be used 5/COLU /about zz /lateral torsional buckling The verification of the column is carried out for the combination G 1.5 V K 57.K Km 5.1/Classification of the section Web: the web slenderness is d t f 10.* 5 w (assumed to be constant along the column) (assumed to be constant along the column) (at the top of the column) c t dw d d * 46 w Then the limit for the class is w (tab5.) 10. Q

25 96 96* * Until the web is class 1 Flange: the flange slenderness is c t f b tw r (tab5.) t 16 The limit of the class is 9 9*1.0 9 f Until the flange is class 1 So the section is Class 1. The verification of the member will be based on the plastic resistance of the cross-section. 5. /Resistance Verification for shear force Shear area max bt t r t ; h t.6..6 V f w f w w max *00* *1 *16;1.0*46*10. V V max 605.; 445. V 605.mm V pl, Rd V V pl, Rd f V * () K The effect of the shear force on the moment resistance ma be neglected Verification to axial force Pl, Rd f 11600*5 *10 76 K K * K Pl, Rd (4)equ 6.

26 0.5hwtw f 0.5*46*10.* K 1.0* (4)equ 6.4 Since and The effect of the axial force on the moment resistance ma be neglected Verification to bending moment pl, Rd W Pl, Rd 0 f 194* 5 1.0* Km pl, Rd Ok! serviabilit limit state Horizontal Deflection Horizontal Deflection at the top of the column must be verified for two combinations and Combination 01: G+Q combination 0:G+W 1 Combination 01 G+Q The moment at a point x in the column is x. H x G+Q x H

27 B introducing a virtual force P at the summit of the column B This effort generate the following forces k I h h s I s R C f h k R P 1 R P R E B C Ph Ph 1 1 V Ph l VE D Ph For an IPE 500 column we obtain : k Then we have the results R 0.54 P R P E The moment in the point is X 0.54Px The resultant moment under the two actions is H x 0.54Px X the internal potential energ of the column is: h 1 W H 0.54 x Px dx EI 0

28 h 1 W H 0.54P x dx EI EI 0.54 W x H P h W H 0.54P 6EI dw P dp EI 0 C h H 1.07*6.41 * *.1*10 *4800 l cm Since OK! Combination 0 G+W H K h 0 cm B application a similar resolution as the above x X H x q 0.54Px the internal potential energ of the column is: h 1 x W H x q 0.54Px dx EI 0 Using a similar calculation we have h 1 x W H x q 0.54Px dx EI W x qx H 0.14P H 0.54P 0.05q x EI W h qh H 0.14P H 0.54P 0.05q h EI 4

29 dw dp P 1 EI qh H h *4*641 * *40.6* cm 6.1*10 *4800 P B H q qx / x R 5.4 Buckling Resistance The buckling resistance of the column is sufficient if the following conditions are fulfilled (no bending about the weak axis, Z, =0): 6.. k RK LT, RK 1 1 k z RK LT, RK 1 1 Buckling about L CR, 6.41m 1 1 equation 6.61 equation 6.6 h b 00 t f mm mm buckling curve :a(α =0.1) table 6. 4 EI 10000*4800* K cr, Lcr, 6410 *10 f 11600* *10 cr,

30 Buckling about zz Buckling curve :b (α z =0.4) EI *10000*14* K 4 z cr, z Lcr, z 6410 f 11600* 5 z *10 cr, z z 0.5 1z z 0. z z z z z z

31 1 tension flange plastic stable length elastic section 4 plastic hinge 5 restraints 6 bending moment diagram 7 compression flange 8 plastic with tension flange restraint, 9 elastic with tension flange Column with restraints b cladding rail along long span nnex Lateral torsional Buckling h b 00 then buckling curve c(α LT =0.49) oment diagram with linear variation : 0 then C The simplification of critical moment ma be used: C EI z Iw cr, LT cr 1 L I EI cr, LT z z L GI t nnex cr 10000*14* * *80770*89.9 * *10 14* * 14* cr 5.1* * *8077* Km * 140

32 W f pl, 194*10 * 5 LT 6 cr 676.*10 LT 0.5 1LT LT LT, LT * With a values of LT,0 0.4 and 0.75 LT *0.871 LT LT LT For 0 then K c table 6.6 Bending moment diagram and the coefficient f K c LT f LT LT,mod f Calculation of the factor K K C C m mlt 1 cr, 1 C annex C C 1C m m,0 m,0 1 a LT a annex LT C m, LT C m a LT 1 1 cr, z cr, T 1 annex

33 1 1 cr, cr, annex C w C C n b el, 1 1 m max m max pl LT w w Wpl, Calculation of W w W pl, W el, Critical axial force in the torsional buckling mode EI w cr, T GIt I 0 Lcr, T For a doubl smmetrical section I0 I I z 0 z cm *.1*10 *14.9*10 *10 504*10 * cr, T 80770*89.9*10 4, K cr T C L GI L I EI EI z Iw cr, LT t cr,0 1 cr, LT z z cr,0 is the critical moment foe the calculation of 0 for uniform bending moment as specified in annex. Then we have C 1 =1 cr *14* * *80770*89.9*10 1* 8.1Km 6410 *10 14* *14*10,

34 W f pl, 194*10 * cr,0 8.1* C 1 1 0,lim 4 1 cr, z cr, TF For doubl smmetrical section cr, TF cr, T 0,lim Then 0 0,lim Calculation of C m C C 1C m m,0 m,0, 66.87* W el, * a LT a LT a LT It I 149 w Calculation of C m,0 C table m,0 With a value 0 then Cm, cr, 0.65 *0.98 Cm *0.98 C m, LT C m a LT 1 1 cr, z cr, T 1

35 0.98 Cm, LT Then, 1 C m LT Calculation of C C w C C n b el, 1 1 m max m max pl LT w w Wpl, max max ; z z z, 0 blt 0 n pl C W Rk 11600* * *1.588 * *1.588 * el, W pl, W Since Ok! Calculation of K K C C m mlt 1 cr, 1 C K *1* Verification with interaction formula k RK LT, RK 1 1 1

36 Compiled b T.RangaRajan.