BS5950 Structural Steel Design R

S T R U C T U R A L ," A f lT h e S t r u c t u r a l E n g i n e e rr e v i e w i n g t h e f i r s t e d i t i o nT h i s m a r k e t l e a d i n g s t u d e n t t e x t c o v e r s t h e d e s i g n o fs t r u c t u r a l s t e e l w o r k t o 5 5 5 9 5 0 P a r t 1 .t h e s u b j e c t I n t w o p a r t s , t h e f i r s t d e a l s w i t h d e s i g n a ta n e l e m e n t a r y l e v e l f a m l i l a r i s l n g t h e r e a d e r w i t hB S 5 9 5 0 . P a r t t w o t h e n p r o c e e d s t o c o v e r a l l a s p e c t so f t h o d e s i g n o f w h o l e b u i l d i n g s , h i g h l i g h t i n g t h ei n t e g r a t i o n o f ' e l e m e n t s ' t o p r o d u c e e c o n o m i c , s a f es t r u c t u r e s .T h o s e c o n d e d i t i o n h a s b e e n t h o r o u g h l y a n du p d a t e d t o t a k e a c c o u n t o f r e c e n t r e s e a r c h a n d d e s i g nd e v e l o p m e n t s a n d a n e w c h a p t e r o n p l a t e g i r d e r s h a sb e e n a d d e d . T h e r e v i s e d t e x t r e t a i n s a l l t h e p o p u l a rf e a t u r e s o f t h e o r i g i n a l w o r k . I n p a r t i c u l a r , r e a d e r s v . i l lf i n d t h a t t h e o u t h o r s : e x p l a i n c o n c e p t s c l e a r l y u s e a n e x t r e m e l y p r a c t i c a l a p p r o a c h. i n c i u d o n u S i e r o S v o v k S e x a m p l e s a n d r e a ls c e n a r i o s. c o v e r w h o l e s t r u c t u r e d e s i g n - t a k e t h e r e a d e r s t e p - b y . s t e p t h r o u g h t h e B r i t k hS t a n d a r dS t r u c t u r a l S t c e l w a r k D e s i m r t o B S 5 9 5 0 I s a c a r e t e x tf o r c I v i l / s t r u c t u r a l e n g i n e e r i n g d e g r e e a n d B T E C H I W / Dc o u r s e s . I t w i l l a l s o p r o v e u s e f u l t o p r o f e s s I o n a le n g i n e e r s n e e d i n g t o f a m i l l a r i s e t h e m s e l v e s w i t h5 5 5 9 5 0 P a r t l a n d t h e d e s i g n o f c o m p l e t e b u I l d i n g s ,p a r t i c u l a r l y p o r t a l f r a m e s .L i M o r r i s v : a s f o r m e r l y R e a d e r i i i S t r u c h i r a l E n p i n e c r i n ga t t h e U n l v e r : ; i t y r z i f l r . n c h e s t o r .D R P l u m i s L e c t u r e r i n S t r u c t u r a l 3 1 t h t :U n i t t r r . i t y o f1 19 0 0 0 0 )9 7 8 0 5 8 2 2 3 0 8 8 0I S B Nc a w t e s , o f S u t t o N M D C I. 5. . - -- . - - 'S T R U C T U R A LS T E E L W O R KD E S I G Nt o1 3 5 5 9 5 02 n d E D I T I O N,. ""'.: .. ,-STRUCTURAL STEELWORK DESIGN to BS 5950 ..... -_. -- - . '. - . . ~ " . ,. ",'.' 2nd EDITION S T R I J C T L J R A LS T E E L W O R KD E S I G Nt oB S 5 9 5 02 n d E D I T I O NL J M o r r i s0D R P I u mo t P e a r s o n E d u c a t i o nH a r r o w E n g l a n d L o n d o n - N e w Y o r k P n d i n g k i n s i a c- S . n F r a n c i s c o T o r o n s a - D o s s M I l l s . O n l a s l a - h y d s s c yT o k y oS i n g a p o r e - H o n g K o n g - S e o u lT a i p e i - C a p o b o w s M a d r i d - I l e a l c o C l Y rA n o s l n s d a s o s - M a n l c s s - F a s i o - M i l a ni 1 ! STRUCTURAL STEELWORK DESIGN to BS 5950 2nd EDITION L J Morris o R Plum .." ;mprinl cl Pearson Education Harrow, U1gland, lendon. New Yruk . A.",nn,}, Ma ... Fflln A I. Cb) Loading Self weight of frame Cladding (roof and walls) Snow and servIces Wind pressure Q {walls) Wind pressure q (roof) 38.0 m 0.90kN/m 0.09kN/m' 0.75 kN/m' 1.20 kN/m' 1.20kN/m' Wind pressures are based on a baSIC wmd speed of 50m/s for a locatIOn m Scotland. Factors{2} S, and S2 are both taken as 1.0 and factor S3 as 0.88 for a height of IOm. The deSign wmd speed is hence 44 m/so giVing a dynamiC pressure of 1.20kN/m'. 2 2 S T R U C T U R A L S T E E L W O A K D E S I G N T O 0 5 5 9 5 0I L O A D I N G A N D L O A D C O M B I N A T I O N S 2 3I t i s p o s s i b l e t o u s e a l o w e r w i n d p r e s s u r e b e l o w a h e i g h t o f S m b u t t h i sm a k e s t h e a n a l y s i s m o r e c o m p l e x f o r v e r y l i t t l e r e d u c t i o n i n f r a m e m o m e n t s .( c ) P r e s s u r e c o e f f i c i e n t sE x t e r n a l p r e s s u r e c o e f f i c i e n t s ( C , , , ) m a y b e f o u n d a n d a r e s u m m a n z e d i n t h ef o l l o w i n g t a b l e . T h e s e v a l u e s a r e o b t a i n e d f r o m r e f e r e n c e 1 2 ) .C , , , f o r f r a m e m e m b e rA D B C C D D EW i n d o n s i d e 0 . 7 1 . 2 0 . 4 0 2 5W i n d o n e n d 0 . 5 0 . 6 0 . 6 0 . 5D e a d l o a d o n r o o f i s c a l c u l a t e d o n t h e p r o j e c t e d a r e a ( F i g . 2 . 1 0 ) :S e l f w e i g h t = 0 . 9 x 1 9 . 3 = 1 7 . 5 k NC l a d d i n g = 0 0 9 x 1 9 . 3 x 6 . 0 = l 0 . 5 k NT o t a l W d . = 2 8 . O k ND e a d l o a d o n w a l l s ( F i g . 2 . 1 1 ) :S e l f w e i g h t = 0 . 9 x 5 . 5 = 5 . 0 k NCB5 . 5F i g . 2 . 1 0 F i g . 2 . 1 1W i n d l o a d s f o r c a s e ( 1 ) o n s i d e + i n t e r n a l p r e s s u r c :W i n d l o a d o n ' v a i l ( F i g . 2 . 1 3 ) = 0 . 5 x 1 . 2 0 x 6 . 0 x 5 . 5 = l 5 k NW i n d p r e s s u r e o n r o o f i s d i v i d e d i n t o v e r t i c a l a n d h o r i z o n t a l c o m p o n e n t s( F i g . 2 . 1 4 ) ,--C a s e( C , , . C , , , ) f o r t r a m e m e m b e rA B B C C D D EI . W i n d o n s i d e + i n I e m a t 0 . 5 1 . 4 0 . 6 0 . 4 5p r e s s u r e2 . W i n d o n s i d e + n i e m a l 1 . 0 0 . 9 0 . 1 0 , 0 5s u c t i o n3 . W i n d o n e n d + i n t e r n a l 0 . 7 0 . 8 0 . 8 0 . 7p r e s s u r e4 . W i n d o n e n d + i n t e m a l 0 . 2 0 . 3 0 . 3 0 . 2s u c t i o nV e r t i c a l c o m p o n e n t 1 . 4 x 1 . 2 0 x 6 . 0 ' t 1 9 . 0 l 9 2 k Nl - l o n z o n t a l c o m p o n e n t = 1 . 4 x 1 . 2 0 x 6 . 0 x 3 , 4 ' = 3 4 k NSIF i g . 2 . 1 3 F i g . 2 . 1 4( I t IIC1 0 D m1I t m a y b e n o t e d t h a t c a s e 4 i s s i m i l a r t o c a s e 3 , b u t h a s l o w e r v a l u e s a n d m a yb e d i s c a r d e d .M e m b e r l o a d sI n t e r n a l p r e s s u r e c o e f f i c i e n t s ( C , , , ) s h o u l d b e o b t a i n e d t f l , a n d a r e t a k e n i nt h i s e x a m p l e a s + 0 . 2 ( m a x i m u m ) a n d 0 . 3 ( m i n i m u m ) w h i c h a r ec o m b i n e d a l g e b r a i c a l l y w i t h t h e v a l u e s o f a b o v e .F i g u r e 2 . 8 s h o w s t h e i n d i v i d u a l p r e s s u r e c o e f f i c i e n t s , a n d F i g . 2 . 9 s h o w st h e v a n o u s c o m b i n a t i o n c a s e s .a ) W i n d o n s i d e u I W i n d o n e n dF i g . 2 . 8C l a d d i n g = 0 . 0 9 x 5 . 5 x 6 . 0T o t a l= 3 . O l c H= 8 . O k NI c ) I n t e r n a l p r e s s u r e I d ) l n t t r n s l s u c t i o nF i g . 2 . 9C a s e I I n + b I C a s e 2 I n + d l C a s e 3 ) b ' + c I C a s e 4 l b + d lW II I I H ( l ) i ) l l I I lI m p o s e d l o a d o n r o o f g i v e n o n p l a n a r e a ( a n d s e r v i c e s ) ( F i g . 2 . 1 2 ) :S n o w l o a d l F k = 0 . 7 5 x 1 9 . 0 x 6 . 0 = 8 6 k MF I g . 2 . 1 222 STRUCTURAL STEELWORK DESIGN TO 8S 5950 It IS possible 10 use a lower wmd pressure below a height of 5 m but ihis makes the analYSIS more complex for very little reductton m frame moments. (c) Pressure coefficients 1.2 0.4 0.1 " !i1jWind on side Fig. 2.8 1.4 0.6 Case 1 !a+b) Fig. 2.9 External pressure coeffiCIents may be found and are summanzed in the followmg table. These values are o'btained from reference (2). Cp"..for frame member AB BC CD DE Wind on side 0.7 -1.2 -0.4 -0.25 Wind on end -0.5 -0.6 -0.6 -0.5 Internal pressure coeffiCients (Cpl) should be obtamed{l). and are taken m this example as + 0.2 (maximum) and - 0.3 (minimum) which are combined algebraically with the values of above. Figure 2.8 shows the mdividual pressure coeffiCients, and Fig. 2.9 shows the vanous combination cases. 0.6 0.6 . (hi Wind on end le) Internal pressure Idllnterna! suction 0.9 0.1 0.8 O.B 0.3 0.3 Case 2 ( 1.2, the constant 1.2 is replaced by the mtlO (Yo) of factared load/unfaclored load. The limllatlOn -j .2p,.z is therefore purely notIOnal and becomes in practice 'IoP).z If 0.6 of shear capacity IS exceeded, some reductIOn 10 Mc will occur as set out In clause 4.2.6. BEAMS tN BUILDINGS 31 Local buckling can be avoided by applYing a limitatIOn to the width/thickness ratios of elements of the This leads to the classificatIon of discussed in SectIOn 1.7. Where mem'bers are subjected to bending about both a.xes a combinatlOll relationship must be satIsfied: (a) For plastIC and compact sectiollS: For un, UC and. jOist sections p-r,,/Mat + My/(Mc)') ;t ! For RHS, CHS and solid sections {M,,/kla)5/ + My(/lfcy)5!3 ;f. I For channel, angle and all other secltons M,,/Ma+ My/MC) . "f I where }'-Ix> My are applied moments about x and y axes Ala, !I-fey are moment capacities about x and y axes -(b) For slender sections: (and as a Simplified method for compact sections In (a) above) ),;fr/Ma + }dv/AIry ., 1 3.5 BUCKLING RESISTANCE (MEMBER BUCKLING CHECK) Members not provided with full lateral restraint (Sec non 3.2) must be checked for lateral torstonal buckling resistance (Mb) as well as moment capacity. The buckling reSistance depends on the bending strength (Sectton 3.2) and the plastic modulus: Mb =PbS" Where members are subjected to bending about both axes {without aXial load) a combinatiOn relationship must be satisfied: m"M,,/Jl-/1> + my My/Mc)' ., I where nI", my are eqUivalent unifonn moment factors Mt:y IS the moment capacity about the y aXIS but without the restnctlOn of 1.2pyZ (ns In SectIOn 3.04). This IS described as a more exact' approach {clause 4.8.3.3.2) which IS iess conservaUve thlln the 'sunplified' approach (clause 4.8.3.3.1), m which Mcy IS defined as p,Zyo Also, a sllnplified for bending about two axes (without axml load) does not reduce the calculatIOns. 3.b OTHER CONSIDERATIONS In addition to the above reqUirements for moment capacity and buclding reSistance, a member IS usually reqUired to mee;t some deflecllon cntena. These are outlined in SectIOn 1.5 and reference (1). The applicatmn of heavy loads or reactions to a member may produce high locnl stresses and it IS necessary to check that the web beanng and web buckling reqUirements are satisfied. These reqUirements are generally Significant only in beams canylng heavy pomt loads such ns crane girders (Chapter 5) or beams supportmg column members within the span. 3 . 7 E X A M P L E 4 . B E A M S U P P O R T I N G C O N C R E T E F L O O R S L A B( R E S T R A I N E D B E A M )( a ) r - iB

7 . 4 i n . 2= 0 . 9 0 ( c ) B M a n d S F I D 1 0 ) 2 2 3 7 0 . 6 3 2 S T R U C T U R A L S T E E L W O R K D E S I G N T O B S 5 9 5 0 8 . 4 1 8 . 0 m 2 5 7 1 2 3 4 2 9 0

a n d

2 5 0 = 1 0 1 4 7 3 1 6

7 4 7 4 3 . O m S h e a r c a p a c i t yU s i n g 4 . 2 . 3 32 STRUCTURAL STEELWORK DESIGN TO BS 5950 Fig. 3.2 Slnb nnd beams Connections must be provided at JunctIOns between members and must safely transmit the calculated loads from one member to the next. A variety of connectIon details exist for most common situations and are fully described in the Se! bandbook(4). DeSIgn infonnatlOn for connectIOns is gwen m section 6, BS 5950, which Includes some guidance on bolt spacmg and edge distances. Bolt and weld sizes and capacities are given in reference (5). 3.7 EXAMPLE 4. BEAM SUPPORTING CONCRETE FLOOR SLAB (RESTRAINED BEAM) (a) DimenSions (See Fig. 3.2.) Beams centres Span (Simply supported) Concrete slab (spannmg In two directions) Finishing screed H Mam beam H E I " 1 , 7.4 m 6.0m 7.4 m 250 mm thick 40 mm thick 40mm licreed 2> 7 250 mm slab 6.0 m I .1 Typical bay of larger floor area H (b) Loading Concrete slab Screed (40 mm) Imposed load H 23.7 kN/m3 0.9 kN/ml 5.0 kN/m' For prelimmary calculatIOn, an estimated self weIght IS Included. Assume beam to be 533 x 210 x 92 UB (grade 43A). It IS suffiCiently accurate to take beam weight of 92 kglm = 0.92 kN/m. Member size must be finally confinned after all the deSIgn checks have been carned out. 10kN ! j I ! .. " " 11 (c) rectangles 2xl.4x3.0 = 8.4 m2 tnangles 4 x 3.0 x 3.0/2 = 18.0m2 Dead load Wd: on rectangles 6.83 x 8.4 = 57 kN on tnangles 6.83 x 18.0 = 123 kN Imposed load Wj : all rectangles 5.0 x 8.4 42 kN on triaggles 5.0 x 18.0 90 kN Ultimate load (factored) (Fig. 3.4); uniformly distributed on rectangles on triangles 1.4 x 7 = 10 kN 1.4 x 57 + 1.6 x 42 = 147 kN l.4x 123+1.6x90 =316kN BM and SF (Sec Fig. 3.5.) Maxunum ultimate moment M;x = 10 x 7.4/8+232 x 3.0 -158 x 1.7 -74 x 10.35=573 leNm Maxtmum ultimate shear force F;x = 1012 + 232 = 237 kN (d) Shear capacity Using the deSign strength from Table 1.2 for grade 43A steel, notlng that maxImum thickness of section IS 15.6 mm: Py = 275 NJmm2 clause 4.2.3 Shear capacity P" = 0.6 Py A" = 0.6 x 0.275 x 531.1 x 10.2 = 897 leN Shear force F;x IP" = 0.26 Therefore, as F;xIP" < 0.6, there will be no reduction m moment capacity (see clause 4.2.5). (e) Moment capacity The concrete slab provides full restraint to the compressIOn flange (Fig. 3.6), and lateral torsional buckJing is not considered. The chosen VB IS 3 4 S T R U C T U R A L S T E E L W O R I ( D E S I G N T O 8 5 5 9 5 0 1B E A M S I N B U I L D I N G S 3 5 f o r c e i s k N m 0 . 8 8 = I V , L 3 / 6 0 E 1 , 2 9 7 . 0 8 0 . 7 3 9 7 ( 4 . 4 + = c l a u s e 4 . 2 . 5 3 9 . 7S9 ,F o r c e s D i r e c t i o n r e s u l t a n t

00 34 STRUCTURAL-STEELWORK DESIGN TO SS 5950 clause 3.5.2 a plastic sectIOn (blT= 6.7) and at mid-span the shear force is zero. dause 4.2.5 (f) M==pySx=275 x 2370 x 10-3=651 kNm Note that because the umts are Pr (N/mml) and S;. (cm)) then the 10-3 must be mcluded for M= In order to obtam the correct umt of kNm. Alternatively, there IS no need for 10-3 if 0.275 kN/mm1 is used for Pr BUl /ylez;t 1.2 PyZ-" = 1.2 x 275 x 2080 x 10-3 = 686 kNm Note that for I and H sectIons bent about the x axis the expression governs the design. For bending about the y aXIS, however, the expressIOn 1.2 p)Zy governs the design. The factor 1.2 m this expressIOn may be Increased to the ratio factored loadlunfactored load (clause 4.2.5): M;./fr/o; = 5731652 = 0.88 SectIOn IS satlsfaclOlY. Deflection DeflectIOn (which IS a servIceability limit state) must be calculated on the baSIs of the unfactored imposed loads: W.,=90+42=132 kN Assume the load IS approximately tnnnguJar and hence fonnulae are available for deflectIOn calculalJons(6) bx = WxLJI60Elx = 132 x 74003/(60)( 205 )( 55 400)( 104) =8.4 mm BS table 5 DeflectIOn limit = 74001360 = 20.6 mm (g) 2 no. 90x90x 10 ls 400 long 01_ a_50 " :il E 01- BC;J1n , 0 " ,., u 01-M22 bolts Jlso Fig. 3.7 Connection The deSign of connections which are both robust and practicable, yet economiC, IS developed by expenence. Typical examples may be found in references (4, 7). The connectIOn at each end of the beam mllst be able to transmi! the ultimate shear force of 237 kN to the column or other support. The connectIOn forms part of the beam, I.e. the pOint of support IS the column to cleat mterface. DeSign practice assumes that the column bolts support shear force only, while the beam bolts carry shear force, together with a small bending moment. BM=237xO.05=12.i kNm Assume 9 bolts, 22 mm diameter (grade 4.6) as shown In Fig. 3.7. 39.7 BEAMS IN BUILDINGS 35 (i) COLUMN BOLTS VertIcal shear/bolt = 237/6 = 39.5 kN (smgle shear) clause 6.3.2 Shear capacity = Ps where IS the area at the roO! of the bolt thread: P. = 0.160 x 303 =48.5 kN/bolt clause 6.3.3.2 Beanng capacity of bolts Pbb = dtPbb: P,,=22 x IOx 0.460= 101 kN/bolt dause 6.3.3.3 Beanng capaCJly of the angles fdtpfu) IS the same as that for bolts, because Pbs for grade-43A steel has the same value as Pbb for grade 4.6 bolts, I.e. 460 N/mm2 In addition, the beann'g"capaclty tiflhe angles mllst comply with the cnterion, Pbe i erpb112 (e defined in Fig. 3.7). Pbs=50 x 10 x 0.460/2 = 115 kN/bolt Note that the column flange will also reqUlre checking if it IS less than 10 mm :;:tS:i Fo'",kN : , (ii) thick. Column bolt connection IS satisfactory. Capacities of bolts and beanng values may alternatively be obtamed from reference (5). BEMIBOLTS r-c--- _ Direction of BB.4I:N resultant Fig. 3.8 clause 6.3.3.3 (iii) clause 4.2.3 (See Fig. 3.8.) Double shear capacity /boU Pj= 2 )( 0.160 )( 303 = 97.0 kN Vertical shear/bolt = 242/3 = 80.7 kN (dollble shear) Ma:umum honzontal bolt forces = (A/dtn=/ d2) due 10 bending moment are discussed in SectIOn 8.4. Honzontal shear/boit = 11.9 x 0.15/(2 x 0.152) = 39.7 kN Resultant shear/bolt = V(SO.72 + 39.71) = 88.4 kN Beanng capacity of bolt, Pbb = dtPbb =22 x 10.2 x 460=97.6kN ASPbl=Pbb, thcn beanng capacity of the web plate IS the samC as for the bolt. , Also, Pfu"'l- = 89 x 10.2 )( 460/2 = 209 kN ANGLE CLEATS Shear area of cleats {allOWing for 24 mm holes) =0.9(400)( \0)( 2 - 3 x 2 x 10)( 24)'= 5904 n1ln2 Shear capacity P" = 0.6 x 0.275 x 5904 =974kN ' A check for bending may also be carned out but will generally give a high bending capacity relnuve to the applied moment(3) 3 0 S T R U O 1 U R A L S T E E L W O R K D E S I G N T O S I $ 0 5 0P af l u 3 . 8 E X A M P L E 5 . B E A M S U P P O R T I N G P L A N T L O A D S( U N R E S T R A I N E D B E A M )( a ) D i m e n s i o n sM u m B E A M S I N B U I L D I N G S 3 7( c ) B M a n d S F 5 . 4 4 0 . 5 1 . 4 ( 4 . 0 1 . 4 ( 4 . 0 3 9 8 1 9 / 2

F 2 1 9 / 2 4 2 2 5 0 4 3 3 5 1 0 0 5 f o r 2 6 5 5 0 4 / 1 1 5 3 0 . 4 4 0 . 6 C d ) S h e a r c a p a d t y 0 . 6 1 . 4 1 8 . 8 7 . 2 4 . 5 5 4 . 0 5 0 4 ( e ) M o m e n t c a p a c i t y 2 6 5 1 2 1 0 = l 0 0 5 / 1 2 1 0 = 0 . 8 3 36 STRUCTURAL STEELWORK DESIGN TO 8S 5950 Fig. 3.9 Plant loads 6.0 m , I Flg.3.10 3.8 EXAMPLE 5. BEAM SUPPORTING PLANT LOADS (UNRESTRAINED BEAM) (a) Dimensions Mam beam is simply supported and spans 9.0 m (Fig. 3.9). Boilers are supported symrnetncally on secondary beams (A and B) of span 6.0 m. which are at 5.0 m centres. -.--,--1 It I 1.Sm .. l.5m l.5m 1.5m -E o ,.; E " m E o ,,; E Beams B I, I!- I '.Om ,I, '.Om .\ (b) Loading Boiler loading (each) 400 kN Open steel floonng (earned on beams A and B) Imposed load (outside boiler area) 0.3 kN/m1 4.5 kN/m1 The boilers produce reactions of 100 kN at the end of each secondary beam. Allow 4.0 kN for the selfwclght of each secondary beam Inot desIgned here). Assume 610 x 305 x 149 VB (grade 43) for main beam. Self weight (ultimate) = 1.4 x 9.0 x 1.49 = 18.8 kN With reference to Fig. 3.10: Flooring on beam A = 0.3 x 4.0 x 6.0 = 7.2 kN Imposed load on beam A = 4.5 x 3.0 x 4.0 = 54.0 kN With reference to Fig. 3.11: Floonng on beam B = 5.4 kN Imposed load on beam B = 40.5 kN 8EAMS IN BUILDINGS 37 Ultimate pomt load = 1.4(4.0 + 7.2) + L6(200 + 54) =422 kN Ultimate POlDt load rvB= 1.4(4.0 + 5.4) + 1.6(200 HO.5) = 398 kN (c) BM and SF S.Om ,1=:-1 S.Om (d) clause d.2.3 Ultimate shear force F, = 1912 x 422 x 6.0/9.0 + 398 x l.0/9.0 =335 kN Ultimate shear force F, = 19/2 x 398 x 8.0/9.0 + 422 x 3.0/9.0 = 504 kN With reference to Fig. 3. J 2: Ultimate moment MA =335 x 3.0 = 1005 kNm Ultimate moment MD =504 x l.0 = 504 kNm Note that m calculatmg the moments, the small reductlOn due to the self weight is Ignored. Shear- capacity DeSign stTength Py for steel 19.7 mm thick (grade 43A)=265 N/mm2 (see Table 1.2). Shear capacity P" = 0.6 PvA." = 0.6 x 265 x 609.6 x 11.9 x 10-:; = ll53 k.N Shear force FI = 335 ItN FI/P. =335/1 153 =0.29 Shear force F2 = 504 kN F2/P. = 504/1153 =0.44 S 0.6 This maximum coexistent shear force IS present at pomt B, while the maXImum moment occurs at pomt A. (e) Moment capacity clause 3.5.2 The chosen sectIOn IS a plastiC section (blT=7.7). Moment capacity Me = pyS;r = 265 x 4570 x 10-3 = 1210 kNm Moment raho M"dMe = 1005/]210=0.83 < I SectIOn IS sailsfactory. 3 8 S T R U C T U R A L S T E E L W O R K D E S I G N T O I S 5 9 5 0B E A M S I N B U I L D I N G S 3 0 = 0 . 9 4 ( f o r f l a n g e s ) I 0 . 8 8 6 = n u v A . 0 j ( A f ( x ) / E 1 ) d x 9 0 0 0 1 2 0 0 4 5 1 0 0 5 0 4 / 1 0 0 . 3 7 5 1 3 4 2 2 1 2 1 3 3 8 5 . 0 2 . 7CI nP 32 !

L s o n4 . 5 m .1 0 4 5B M d i a g r a m = 2 0 7 0 . 7 6 = o i M A = 7 6 4 / 9 4 6 0 . 8 1 1 6 1 6 6 8

a s a n 2 5 . 3 38 STRUCTURAL STEELWORK DESIGN TO 8S 5950 (I) clallse .;-.3.5 BS table -'4 BS table 13 clause 4.3.7.5 BS table J 1 clause 4.3.7.2, BS rabies 13, 18 Budding resistance The buckling resistance moment of Ihe beam In the part-span AB must be found, and in this part the momenl vanes from 504 kNm 10 1005 kNm. It IS assumed thal the steel floonng does not provide lateral restralDt, but that the secondary beams give posltiona! and rotatIOnal restraint at 5.0 m spacmg. Loading between the restTamts IS of a mlllOT nature (self weight only) and is Ignored for use of SS table 13 as It would affect the moments by less than 10% (see also BS table 16). LE =5.0 m Slenderness..:. = LefTy =5000169.9=72 (both ry and LE given In mm) TorsIOnal index x = 32.; )Jx = 2.2 \' = 0.94 (for N = 0.5, 1.e. equal flanges) 11 = 1.0 (for member nol loaded between restramts) 11 = 0.886 ALT =1JI1vA = J.O x 0.886 x 0.94 x 72 = 60 Bending strength Pb = 207 N/mm1 Buckling resistance A1b =PbSx = 207 x 4;70 x 10-3 = 946 kNm EqUivalent unifonn moment factor IrJ needs to be obtained for a member not loaded between restramts: hence P =504/1005=050 III =0.76 EqUIvalent uniform moment M = mAlA. = 0.76 x lOO; = 764 kNm hence M 1Mb = 764/946 = 0.81 and section is satisfactory. Try a smaller sectIOn (610 x 229 x 140 UB): SE=4150 cm4 " = 99.4 Pb = 161 Nlmml Mb =668 kNm Ji /Mb = I.l4 which IS not satisfactory. This companson mdicales the senSitiVity of the buckling reSistance moment to small changes In sectIOn properties, particularly to a reduchon In flange width. (g) 1045 355 588 kN m BM diagram (areas In bold) Fig. 3.13 BEAMS IN BUILDINGS 39 Deflection Ca kulatlon of the deflectIOn for tile serviceability imposed loading cannot be carned easily by the use of formulae, which become complex for noo* standard cases. Serviceability point load WA=200+54 kN 1J's=200+40.5 =240 kN With reference to Fig. 3.l3, mid-span deflection may be found by the method{8.9): {, J fM(x)I1)dx =(149 x 0.667+ 1045 x 2.75 +355 x 3.333)/ (205 x 124660 x 16.3 mm Calculation by Macaulay's method(IO.II) gives the poml of maxlmwn deflection 4.65 m from support 2 with a value of 17.6 mm. DeflectIOn limIt = 9000/200 = 45 mm An approximate estlmate of deflection IS often obtamed by treatmg the load as an eqUivalent u.d.l. 2 no. 100 x 100 x 12 ls 500 long (h) Connection 8_50 100 " 100 E Beam , 100 .. o 'lOO u I M22 bolts -I WO The connectIOn at support 2 of the mam beam must tmnsmlt uitimate shear force of 504 kN and follows the method given 111 Section 3.7(g): M=504 x 0.05=25.3 kNm Assume 22 mm bolts (grade 8.B) as shown m Fig. Fig. 3.14 ' 'r"-(i) COLUMN BOLTS Verttcal shearlbolt =;04110 50.4 kN clause 6.2.3 Shear capacltylbolt = 0.375 x 303 = 114 kN dause 6.3.3.3 Beanng capacity of angleslbolt Pb:! = 22 x 12 x 0.460 = 121 kN but Pru-/bolt I' 50 x 12 x 0.460/2 = l38 kN Note that the column flange will require checking if less than 12 mm thick. Column bolt connectIOn is satisfactory. l 0 O . 8 1 ( 1 0 0 . 8 2 + 5 0 6 2 ) 2 3 0 3 1 1 . 9 i . 0 3 5 1 1 . 9 0 . 4 6 0 1 2 0

R e f e r e n c e sB o w l i n g P . 3 . , K n o w l e s P . & O w e n s G . W . ( 3 9 8 8 ) fS t r u c t u r a l

4 2 0 5 2 . I n s t a b i l i t y V a n N o s t r a n d R e i n h o l d v o l . L o n g a n a n4 0 S T R U C T U R A L S T E E L W O R K D E S I G N T O 9 5 5 9 5 0 6 . 3 . 3 . 3B e a r i n g c a p a c i t y / b o l t c a p a c i t y o f w e b p l a t e x 5 0 . 6B E A M S I N n U I L D I N G S 4 1 1 1 3 t Nr e s u l l a n t 4 . 2 . 3 h o l e s )= 0 . 9 ( 5 0 0 2 8 2 0 8 S h e a r 8 2 0 8 1 3 5 0 R E F E R E N C E S C o n n e c t i o n s 40 STRUCTURAL STEELWORK DESIGN TO 8S 5950 (ii) BEAM BOLTS clause 6.3.3.3 , C'! ";'1 - , ;2 (iii) '" \ resullant , 'll> . 40! clause 4.2.3 . I I Fig. 3.15 (See Fig. 3.15.) Vertical shearlbolt Horizontal shearlbolt = 504/5 =Mdm,.)l.d' = 100.8 !eN = 25.3 x 0.20/2(0.10' + 0.20') Resultant shearfbolt = .J(l00.82 +50.62) =50.6 !eN =113 !eN =227 !eN =271 !eN. = 120 !eN Shear capacity (double shear) =0.375 x 2 x 303 Beanng capacttylbolt Pbb = 22 x 11.9 x 1.035 Beanng capacity of web plate PbJ = 22 x 11.9 x 0.460 but P" l' 89 x 11.9 x 0.460/2 = 243 !eN Beam bolt connection is satisfactory. ANGLE CLEATS i Shear area of deats (24 mm holes) =0.9(500 x 12 x 2 - 5x 2 x 12 x 24)=8208 mm2 Shear capacity p ... =0.6 x 0.275 x 8208 = 1350 kN F.jP" =50411350=0.37 I Conne'ctlOll cleat IS satisfactory. ! STUDY REFERENCES TopIC j. Lateral restraint BS 5950 2. StIllt behavIOur 3. Strut behavIOur 4. ConnectIOns 5. Bolt details 6. Deflection fonnulne 7. ConnectIOns 8. Moment*llfell method References Dowllng P.J., Knowle.s P. & Owens G.W. (l988) Structural Steel DesIgn. Stetd Construction institute Marshal! W.T. & Nelson H.M. fl990) ElastIc anaiysls. Structure, pp. 420-52. Longman Coates R.c., Coutie M.G. & Kong F.K. (1988) Instability of struts and frameworks, Stntctural Anaiysls, ' pp. 58-71. Van Nostmnd Remhold . (1993) JOintS In Simple Connections val. 1. Steel Canstruchon Insbtute (985) Steelwork DeSIgn vol. I, Section propertles. member capacities. Sleel ConstructlOn Instttute (1992) DeSIgn theory, Steel DesIgners' Manual, pp. lO26-50. Blackwell (1992) Jomts In Simple Connections voL 2. Steel Construction institute Croxton P .L.e. & Martin L.H. (1990) A",,-n,on"" methods of analYSIS. SolVing Problems In Structures VD!. 2. pp. 2'5-47. Longman BEAMS IN BUILDINGS 41 9. method Contes R.c., Coulie M.G. & Kong F.K. (1988) Moment-area methods. Structllrai AnalysIs, pp. 176-81. Van Nostrnnd Retn.hold 10. DeflectIon 11. DeflectIOn Mnrs-ball W.T. & Nel!on H.M. (1990) Singulanty functIOns, Structures, pp. 233-8. Longman Hearn EJ. (1985) Slope and deflectIOn of beams, Mer:lramcr of Malenals vo!. 1, pp. l02-7. Pergamon H I P U R L I N S A N D S i O E R A I L S 4 3 ( b ) L o a d i n gT 9 4 . 2 E X A M P L E 6 . P U R L I N h i I t J L U i i Fig. 4.1 Cold fonned purlins PURLlNS AND SIDE RAilS In the UK purlins and side rails used in the constructlOn of industrial buildings are often fabncated from cold fonned sectlOns. TIlcse sectlOns can be desIgned in accordance with Part 5 of BS 5950, hut the load tables for these sectIOns are frequently bused on test data. The sectIOns are marketed by compames specUllizmg III this field who will nonnally gIVe the appropnate spans and allowed laadings In their catalogues. SectIOns of this kind are commonly of channel or zed form as illustrated in Fig. 4.1. Although theIr design IS not covered in Ihis chapler, tbe selectIOn of cold formed sectIOns IS discussed in Chapter 12. Hot rolled sectIOns may be used as an alternatlve, and in some situations may be preferred 10 cold formed sectIOns. The deSign of angles and hollow sections may be carned out by empirical methods which are covered by clause 4.12.4. The full deSign procedure (i.e. non-emplOcal) IS set out m this chapter. 4.1 DESIGN REQUIREMENTS FOR PURLlNS AND SIDE RAILS The deSign of steelwork In bending IS dependent on the degree of lateral restramt given to the compreSSlOn flange and the' torsIOnal restramt of the beam, and also on the degree of lateral/torslOnai restramt given at the beam supports. These restraints are given In detail in clause 4.3 and have been discussed and demonstrated in Chapter 3. Side rails and purlins may he considered to have lateral restrmnt of the compressIOn flange oWing to the presence of the cladding, based on adequate flxmgs (clause 4.12.1). Loads will be transferred to the steel member Via the cladding (see Fig. 4.2), and the Fig. 4.2 PURLlNS AND SIDE RAILS 43 1.0 {fl dead, Imposed and wmd pressure loads will cause the flange restrained by the cladding to be m compressIOn. Wind suctIOn load can, reverse this arrangement, I.e. the unrestrained flange will be In compression. TorSIOnal restraint to a beam Involves both flanges bemg held in pOSition and for purl ins and side rails this wil! be tnle only at the supports. Side rails are subjected to both vertical loading (cladding) and honzontal loading (Wind pressure/suction), but In general the vertical loading IS considered to be taken by the cladding actmg as a deep girder. Consequently, only moments In the honzontal plane (due to wmd) are considered in deSign. In the deSIgn of new construction where the cladding is penetrated by holes for access, ductwork or conveyors, the deSign engmeer should be satIsfied that t/le cladding and fixmgs are capable of acting In this manner. Sag rods are sometimes used to reduce the effective length of purl ins and side rails, and result In continuous beam deSIgn (see Section 4.4). Where sag rods are used, prov1Sl0n must be made for the end reactIOn on eaves or apex beams. As IS shown in the following examples, there IS no reason why purl ins and side rails should not be deSigned as beams subject to biaxial bending m accordance with the normal deSIgn rules. 4.2 EXAMPLE 6. PURLlN ON SLOPING ROOF (a) Dimensions See Fig. 4.2: purl ins at 2.0 m centres; span 6.0 m Simply supported; rafter slope 20" (b) Loading Dead load (cladding + insulatIOn panels) Imposed load Wind load 0.15 kN/m' 0.75 kNjm2 (on plun) 0.40kNjm2 (suctIOn) Reference should be made to Chapter 2 for the derivatIOn of loads, lhe directIOn In which each will act, and the area appropnate to each load. Wdlor W/or W,.} MaXimum values of bending moment and shear force must be found at the ,sIll DIJI::IITi ]iJiIi Ii j':IUD'::jl ultimate limit state making due allowance for the slope angle and mcluding 6.0 m the 'If factors. a12.0 m centres Assume purlin to be }52 x 76 channel sectIOn, grade 43A steel (see Fig. 4.3 Fig. 4.3). I, L 8 0 6 . 0 x 0 . 1 8 = L O B n o r m a l f a c t o r s a n d S FM a x . 8 . 3 ( d ) S h e a r c a p a c i t yD e s i g n s t r e n g t h p , . i s g i v e n i n T a b l e 1 . 2 a n d f o r t h e s e l e c t e d p u r l i n s e c t i o n i s 9 7 5 1 3 0 x B u c k l i n g 4 4 S T R U C T U R A L S T E E L W O R I ( D E S I G N T O E S 5 9 5 0F i g . 4 . 4 P U R L I N S A N D S I D E R A i L S4 5C e ) M o m e n t c a p a c i t yT h e b e n d i n g 7 4 . 2 . 5A1 1 1 1 1 1 1 1 1 ! F i N t . 1 '1 , 44 STRUCTURAL STEELWORI( DESIGN TO as 5950 ! Fig. 4.4 Cladding Self weight Tota! dead load 2.0 x 6.0 x 0.15 = L80 kN 6.0 x O.IB LOB kN kN Imposed load =2.0 cos 200x 6.0x 0.75 W, B.46 kN Wind load x 6.0 x (-0.40) -4.BO kN The rafter slope of 200 results in purl ins at the same angle. Components of load are used to calculate moments about the x and y axes, I.e. nonnal and tangential to the rafter (Fig. 4.4). As with side rails, it would be possible to Ignore bending in the plane of the cladding, but in practice, biaXial bending is usually considered in purUn deSign. Wch =2.88 cos 200 =2.71 kN Wdy = 2.88 sin 200 = 0.99 kN Wu; = 8.46 cos 200 = 7.95 kN Jt'iy = 8046 sin 200 = 2. 89 kN W,,,,"= -4.80 kN Note that WilY IS zcro as wind pressure IS perpendicular to the surface on which it acts, I.C. nonnal to the rafter. U1timate load W,.= \:,4 x 2.71 + 1.6 x 7.95 = 16.5 kN where lA and 1.6 are the appropnate Xf 'factors (Section 1.7). (c) BM and SF ' Wk" I I I I "1 Fig. 4.5 (d) clause 4.2.3 Max. ultimate moment M,. = 16.5 x 6.0/8 = 12.4 kNm Max. ultimate shear force F,. = 16.5/2 = 8.3 k.N Similarly, Wy = 6.0 kN Shear capacity My=4.SkNm Fy= 3.0 k.N DeSign strength py IS given In Table 1.2 and for the selected purlin section IS 275 N/mm2 Shear area A,.,. = 152.4 x 6.4 =975mm2 Shear capacity P \.'X = 0.6 PyAv,. =O.6x275 x 975 x 10-3 =161kN Shear area A,y = 0.9Ao .. =0.9 x2 x 76.2 x 9.0 = 1234 mm'" Shear capactty Pvy = 0.6 x 275 x 1234 x 10-3 =204kN It may be noted that in purlin deSIgn, shear capacity is usually high relative to shear force. Ce) BS table 7 clause 4.2.5 PURLlNS AND SIDE RAILS 45 Moment capacity The section classification of a c'hannei subject to biuxllli bending depends on bIT and dlt which In this case are 8.47 and 16.5, respectively. The channel is therefore a plastic sectIOn. Hence, the moment capacity but Mo; must not exceed 1.2 PyZz 1.2 x 275 x 112 x 10-' 37.0 kNm Note that 10-1 must be mduded to give Mo; In kNm, when N/mm2 for Py and cm) for Sx- Alternatively, py may be expressed as 0.275 kN/mm2. but this reqUIres care later when aX:Ial forces and stresses arc used. The raho Sy/Zy IS greater than 1.2 and hence the constaut 1.2 IS replaced by the ratIO factored loadlunfactored ioad (6.0/[0.99 + 2.89J = 1.55). Mcy must not exceed 1.55 P0y 1.55 x 0.275 x 9.0 kNm The local capacity check may now be earned out (SectIOn 3.4): Mz/Mo; + M/MC)' 1- i (for a channel section) 12.4/35.B + 4.519.0 0.B5 The local capacity of the section IS therefore adequate. (f) Buckling resistance The buckling resistance moment Mb of the section does not need to be found because the beam IS restrained by the cladding In the .1' plane (Fig. 4.6) and Instability IS not considered for a moment about the mmor axiS (Fig. 4.7) (Scction 3.1). Fig. 4.6 Fig. 4.7 4 6 S T R U C T U R A L S T E E L W O R K D E S I G N T O E S 5 9 5 0P U R L I N S A N D S I D E R A I L S 4 7( g ) s u c t i o n( h ) D e f l e c t i o nT h e f o r a = 3 M 4 . 0 1 4 . 0 1 = 3 . 0 1 i s 0 . 9 9 k N 0 . 7 5 3 0 m m 3 . 0 6 . 0 L n ' r , 6 0 0 0 / 2 2 . 4 2 6 8 l e s s a n d n u v A 1 4 . 5 2 6 8 / 1 4 . 5 3 8 0 . 4 9 0 . 9 0 2 0 . 9 4 l I t e q u a l 3 0 8 i 4 . O k N m 4 . 3 7 . n o t e 46 STRUCTURAL STEELWORK DESIGN TO SS 5950 (g) Wind suction The effect of the wmd suction load has so far not been considered, and in some sltuallons it could be critical. Tn combinatton with ioads Wd and Wi, a lower total load W is clearly produced. Ultimate load W..- = LO x 2.71 + 1.4 x 4.8 = -4.01 kN M, = -4.01 x 6.0/8 = -3.01 kNm Wy = LO x 0.99 0.99 kN My =0,99 x 6.0/8 0.75kNm The value of MI is much lower than the value 12.8kNm used earlier, but the negattve Sign Indicates that the lower flange of the channel is 10 and this flange is not restrained. The buckling resistance Ah must therefore be found. The effectlve length LE of the purlin may be found from BS table 9. clause 4.3.5 LE = 1.0 x 6.0 = 6.0 m Slenderness l = Ldry = 6000/22.4 = 268 (which is less than 350 as requrred by clause 4.7.3.2) where LE and ry are in mm. EqUivalent slenderness lLT allowing for lateral torsional buckling IS given by: TorSional mdex x = 14.5 A/X =268/14.5 = 18 BS table 14 v = 0.49 11 =0.902 BS tabl, 16 n =0.94 (for P=O and y=O) Clause 4.3.7.4 BS table 11 BS table 13 ALT = 0.94 x 0.902 x 0.49 x 268 = III Bending strengthpb may be obtamed: Pb= 108N/mm2 Buckling resistance M/) = puSx = 108 x 130 x 10-3 = 14.0kNm The overall buckling check may now be carned out using an eqUIvalent uniform moment factor (m) equal to 1.0 (member loaded between restra1Ots): mM..-/ Mb + Mcy -; 1 3.01/14.0+0.75/(275 x 41.3 x 10-'):=0.28 The overall buckling of the section IS therefore satisfactory. The diagrams for bending moment and shear force shown m Fig. 4.5 mdicate that maximum values are not COincident and it IS not therefore necessary to check moment capacity III the presence of shear load. Purlin deSign does not normally need a check on web beanng and buckling as the applied concentrated loads are low - nole the low values of shear force. The check for beanng and buckling of the web IS particularly needed where heavy concentrated loads occur, and reference may be made to Chapter 5 for the reievant calculahons. __ 4 __ 7 :I Defieclton limits for purl ins are not specified ill BS table 5 but a limit of span/200 IS commonly adopted. DeflectIOn o;r=5W;rL3/384EI..-where W;r IS the ser.'lceability Imposed load, I.e. 7.95 kN and E is 205 kN/mm2 ox=5 x 7.95 x6oo0J/(384 x 205 x 852 x 104)=12.8 mm b;.=28.5mm DeflectIOn limit = 6000/200 = 30 mm (i) Connections Fig. 4.8 Purlin connection The connectIOn of the pllrlin 10 the rafter may be made by bolting Ilto a cleal as shown m Fig. 4.8. The deSign of these connectlons IS usually nommal due to the low reactIOns at the end of the pUrlins. However, the transfer of forces between the purlin and rafter should be considered. For the channel secuon chosen, Wx and Wy transfer to the rafter tlrrough a cleat. Bolts must be provided but will be nommal due to the low reachons mvolved (8.0 kN and 3.0 kN). Chapter 3 gives calculatIOns for a bolted connectIOn In more detail. Multi-span (contmuous) purlins may be used and mlOor.changes m deSign are considered in SectIOn 4.4. '. Putlin CUI tram 4.3 EXAMPLE 7. DESIGN OF SIDE RAIL (a) (b) Dimensions (See Fig. 4.9): side rails at 2.0 m centres; span 5.0 m simply supported. Loading Dead load (cladding/insulation panels) Wind load (pressure) O.ISkN/m' O.80kN/m' AI A 4 5 S T R U C T U R A L S T E E L W O R K D E S I G N T O E S 5 5 5 0P I J R U N S A N D S I D E R A I L S 4 9M a x i m u m f o r l o a dA s s u m e t h e s i d e

s e c t i o n

C l a d d i n g I = 2 . 5 5 k N t o a d F i g . p ' 4d p s i g n = 2 7 5 i . 4 ( I ) D e f l e c t i o n( c ) C a l c u l a t i o n

u n f a c t o r e d = 5 x 8 . O x 4 . 4 O F M U L T I - S P A N P L J R L I N S h e a r c a p a c i t y-t : . i r - , , . , ; - ; -o r d e r 4 . 2 . 3 o r - .N A48 STRUCTURAL STEELWORK DESIGN TO BS 5950 Side rail X Wind load y X "g-g ~ E c iqi '0 0 " ~ ~ N . ~ G . , o. Wind load Fig. 4.9 c E , ." u . "E .0, ~ ~ ~ (c) MaxlfllUm values of bending moment and shear force must be found allowlOg for the wmd loading (bonzontal) only (Fig. 4.9) and inciuding tbe safety faclor Yf' Assume the side rail to be 125 x 75 x 10 unequal angle, grnde 43 steel. An angle, such as that chosen,- provides greater resistance to bending (higher section properties) about the x axis than the y axIS, compared to that for an equal angle of the same area (weIght). Cladding 2.0 x 5.0 x 0.18 = L80 kN Selfwelgbt 5.0 x 0.15 =0.75 kN Total dead load Wd =2.55k:N Wind load W", = 2.0 x 5.0 x 0.80 = 8.0 k:N The loads W", and Wd act In planes at nght angles producing moments about x and y axes of the steel sectlon, but only moments about x are used in dS=Slgn (as discussed in SectIOn 4.1). :to . Ultimate ~ " ( n ~ . w :1:_ . .0= Il.2kN pI!! & 80 mz BM and SF WkN With reference 10 Fig. 4.10: i ,Ill! 1 ! 1 I ! i 12 Maximum moment AI., = 11.2 x 5.0/8 =7.0kNm , I Fig. 4.10 Maximum shear force Fx = 11.2/2 = 5.6 kN (d) Shear capacity Design strength py IS 275 NJmm2 (Section 1.7). clause 4.2.3 Sbear area Av =0.9 x 125 x ID = 1125 mm2 Shear capacity P" =0.6 x 275 x 1125 x 10-3 = 186kN The shear capacity is clearly very large relatl ve to the shear force. PURLlNS AND SIDE RAILS 49 (e) Moment capacity For slflgle angles, iaternl restraint IS provided by the cladding, which also ensures bending about the x rous, rather than about a weaker aXIS (Fig. 4.11). The moment capaCIty only of the sectton is therefore Checked. The section BS table 7 chosen IS defined as semI-compact havmg b!T= 7.5 and dJT= 12.5 (both < 15), and (b + d)IT= 20 1< 23), hence: llIa =pyZ;r =275 x 36.5 x 10-3= ID.OkNm M, /Ma = 7.0/10/.0 =0.70 SectIOn IS satisfactory. The deSIgn of side rails does not nonnally mclude a check on web beanng and buckling, as discussed in Section 4.2(g). (l) Deflection 4.4 CalculatIOn of deflectIOn IS based on the serviceability condition, I.e. with unfadored loads. Ww=8.0kN ay = 5WwL' /3841, = 5 x 8.0 X 5000' /(384 x 205 x 302 x 104) =21.0mm Although dause 4.12.2 avoids specifYing any value, use a deflection limit of, say, U200 = 25 mm. EXAMPLE B. DESIGN OF MULTI-SPAN PURLlN Contmuity of a structural element over two or more spans may be useful In order to reduce the maximum moments to be resisted, and hence the sectIOn size. and to unprove the buckling resistance of the member. 1. In general, the bending moments In a contmuous beam are iess thun those in simply supported beams of the same span. It should be noted, however, that a two-span beam has the same moment (WL/8) at the middle support as the mid-span moment of a Simply supported beam. 2. The resistance of a member of lateral torsIOnal buckling IS improved by continUity and this IS reflected in BS table 16. Continuity may be achieved by fabricating members oflength equal to two or more spans. Length will, however, be limited by reqUirements for delivery P U R L I N S A N D S I D E f l A I L S 5 15 0 S T R U C T U R A L S T E E L W O R I < D E S I G N T O O S 5 9 5 0a n d f l e x i b i l i t y d u n n g s i t e e r e c t i o n . F o r t h e p u r l i n d e s i g n e d i n S e c t i o n 4 . 2 a ( I ) B u c k l i n g r e s i s t a n c el e n g t h o f n o t m o r e t h a n t w o s p a n s ( 1 2 m ) w o u l d b e a c c e p t a b l e ( t h e y c a n b ed e l i v e r e d b u n d l e d t o g e t h e r t o r e d u c e f l e x i b i l i t y ) . C o n u n r n t y c a n a l s o b e A = = 6 0 0 0 / 2 2 . 4 = 2 6 8a r r a n g e d b y U s e o f s i t e c o n n e c t i o n s c a p a b l e o f t r a n s m i t t i n g b e n d i n g m o m e n t s .T h e f a c t o r a i s o b t a i n e d f r o m E S t a b l e 1 6 f o r / 3 = 0 a n d y = A l / A l 0 . j 0S u c h c o n n e c t i o n s a r e c o s t l y t o f a b n c a t e a n d t o a s s e m b l e a n d a r e r a r e l y u s e da s t h e e n d m o m e n t A l a n d s i m p l y s u p p o r t e d m o m e n t A l 0 a r e e q u a l ( b u ti n s m a l l s t r u c t u r a l e l e m e n t s s u c h a s p u r l i n s a n d s i d e r a i l s . U s i n g t h e s a m eo p p o s i t e s i g n ) , h e n c ee x a m p l e a s i n S e c t i o n 4 . 2 , t h e d e s i g n i s r e p e a t e d f o r a p u r l i n c o n t i n u o u s o v e rt w o s p a n s o f 6 . 0 m . a 0 . 6 6i t t = 1 . 0i t = 0 . 9 0 2( a ) D i m e n s i o n s A / x = 2 6 8 / 1 4 . 5 = 1 8 . 5- w h e r e x i s t h e t o r s i o n a l i n d e x .A s S e c t i o n 4 . 2 ( a ) .E S t a b l e 1 4 = 0 . 4 9= 0 . 6 6 x 0 . 9 0 2 ' < 0 . 4 9 x 2 6 8 = 7 8( b ) L o a d i n gE S t a b l e 1 1 p j , = 1 7 0 N / r n m 2A s S e c t i o n 4 . 2 : a s s u m e p u r l i n t o b e 1 5 2 x 7 6 c h a n n e l , 4 3 s t e e l : = 1 7 0 x 1 3 0 x = 2 2 . 1 k N m= 2 . 8 8 k NI= 2 . 7 11 2 . 4 / 2 2 . 1 + 4 . 5 / ( 2 7 5 x 4 1 . 3 x = 0 . 9 61 1 ' d y = 0 . 9 9 k NU l t i m a t e l o a d r i ' , = 1 6 . 5 k NU l t i m a t e l o a d I V , , = 6 . 0 k W( g ) D e f l e c t i o nF r o m E x a m p l e 6 . t h e i m p o s e d l o a d a t s e r v i c e a b i l i t y l i m i t s t a t e i s( c ) D I I a n d S F= 7 . 9 5 k WW i t h r e f e r e n c e t o F i g . 4 . 1 2 :& = 7 . 9 5 x 6 0 0 & / ( l 8 5 x 2 0 5 x 8 5 2 x = 5 . 3 m mM , k N mM a x i m u m u l t i m a t e m o m e n t ( a t c e n t r a l s u p p o d )= 2 . 8 9 k W\ : > /5 ' = 1 4 . 5 m m= 1 6 . 5 x 6 . 0 / 8 = 1 2 . 4 k N mD e f l e c t i o n l i m i t G 0 0 0 / 2 0 0 = 3 O m m1 0 4NF , = 0 . 6 5 x 1 6 . 5 = 1 0 . 4 k WS i . = 4 . 5 k W mF I g . 4 . 1 2 F , , = 3 . 8 k W( d ) S h e a r c a p a c i t yS h e a r f o r c e i s l e s s t h a n 0 . 6 s h e a r c a p a c i t y , a s S e c t i o n 4 . 1 ( d ) .( e ) M o m e n t c a p a c i t yE S t a b l e 7 T h e 1 5 2 x 7 6 c h a n n e l i s a c o m p a c t s e c t i o n ( b / T = 8 . 4 5 ) , h e n c e5 x 1 3 0 x 1 0 3 = 3 5 . 8 k N m / v ! ' 7= 9 . O k N mI1 2 . 4 / 3 5 . 8 + 4 . 5 / 9 . 0 = 0 . 8 550 STRUCTURAL STEELWORK DESIGN TO 8S 5950 ---=:::c:] Fig. 4.12 and flexibility dunng Site erectIOn. For the purlin designed in SectIOn 4.2 a length of not more than two spans (12 m) would be acceptable (they can be delivered bundled together to reduce flexibility). ConnnUlty can also be arranged by use of site connectIOns capable oftransmlltmg bending moments. Such connecl1ons are costly to fabncate and to assemble and are rarely used in small structural elements such as pUrlins and side rails. Using the same example as m Section 4.2, the design IS repeated for a purlin contmuous over two spans of 6.0 m. (a) Dimensions As Section 4.2(a). (b) Loading As SectIOn 4.2; assume purlin 10 be 152 x 76 channel, grade 43 steel: H'd =2.88kN W,u- =2.71kN H'dy =0.99kN Ultimate load TV .. = 16.5 kN Ultimate load IVy = 6.0 kN (c) BM and Sf With reference iD Fig. 4.12: MaXimum ultimate moment (at central support) },{, 16.5 x 6.0/8 12.4 kNm Fi = 0.65 x 16.5 = 10.4 kN My =4.5kNm Fy (d) Shear capacity Shear force IS less than 0.6 shear capacity, as SectIOn 4.Hd). (e) Moment capacity BS table 7 The 152 x 76 channel IS a compact sectlon (bIT= 8.45), hence Ma. =275 x 130 x 1O-3=35.8kNm Mey =9.0kNm AMM=+My/Moy/1 12.4/35.8 +4.5/9.0 0.S5 .. (I) BS table 14 BS fable 11 PURUNS AND SIDE RAILS 51 Buckling resistance .l = LE/ry = 6000/22.4 = 268 The factor n IS obtamed from BS table 16 for P = 0 and}' = M/Ala = - J .0, as the end moment AI and Simply supported moment Mo are equal (but Opposite sign), hence n In = 1.0 11 =0.902 l/x IS.5 where x IS the torsIOnal index. v =0.49 ALT =0.66 x 0.902 x OA9 x 268 = 78 Pb = 170 N/mm2 Mb = 170 x 130 X 10-3 = 22.1 kNm M,/M,+My/M",/ I 12.4/22.J + 4.5/(275 x 41.3 x 10,-3 = 0.96 (g) Deflection From Example 6, the Imposed lond at serviceability limll state \s Wu: =7.95kN o:r =7.95 x 60001/(185 x 205 x 852 x 10.\)=5.3mm lV,y =2.89kN oy =14.Smm DeflectIOn limit 6000/200 = 30 mm 1 s tc .C R A N E i n d u s t r i a l b u i l d i n g s h o u s e h e a v y i t e m s m o v e d a s s e m b l y ,f a b n e a t i o n C R A N E W H E E L L O A D SP a r t s T h e a n d

h e a v y a n d s u r g e C R A N E G I R D E R S 5 3

w h e e lc a r r i a g e o f 5 . 2 O v e r h e a d 0- cs oVa

a - c

rail __ Crane gantry gIrder {UB and plate welded logetherj Fig. 5.1 Crane gantry glfder 5.1 CRANE GIRDERS Industnal buildings corrunonly house manufactunng processes which involve heavy Hems being moved from one pOint to another dunng assembly, fabncatlOn or plant maintenance. In some cases overhead cranes are the best way of providing a heavy lifting facility covering virtually the whole area of the building. These cranes are usually electrically operated, ami are provided by specialist suppliers. The crane is usually supported on four wheels running on special crane rails. These rails are not considered to have significant bending strength, and each IS supported on a crane beam or gtrder (Fig. 5.1). The design of this girder, but not the rail, IS part of the steelwork designer's bnef. However, the position and attachment of the rail on the crane gtrder must be considered, as a bad detail can led to fatigue problems, particularly for heavy duty cranes, The attachment of the rail should allow future adjustment to be carried out, as continuous movement of the crane can cause lateral movement of the raiL CRANE WHEEL LOADS Parts of a typical overhead crane are shown In Fig. 5.2. The weIght or load associated with each part should be obtamed from the crane supplier's data, and then be combined to give the crane wheel loads. Altemattve wheel loads may be glven directly by the crane manufacturer. Reference may be made to BS 6399: Part Ill) for full details of loading effects. The follOWing notes apply to smgle crane operatlOn only. The crab with the hook load may occupy any posItion on the crane frame up to the mmimum approach shown m Fig. 5.2. Hence the vertical load on the nearer pair of wheels can be calculated, adding an amount for the crane frame, IS usually divided equally between the wheels. Maximum wheel loads are often provided by the crane manufacturer. An allowance for Impact of 25% 15 made for most lightlmedium dUty cranes (classes Ql and Q2), and this IS added to each vertical wheel load. For heavy duty cranes (classes Q3 and Q4) reference should be mllcie to as 2573(2) and to supplier"s data for appropnate Impact values. In addition to the vertical loads transferred from the wheels to the crane rail, hOrIzontal loads can also develop. The first of these IS ealled surge and acts at nght angles (laterally) to the girder and at the level of the raiL This surge load covers the Ilcceleration and braking of the crab when movlOg . 5.2 Overhead crane 5.3 Crane loads Hook: + lift CRANE GIRDERS 53 Wheel carnage along the crane frame, together with the effects of non-vertical lifting. The value of this load IS assessed in BS 6399(1) at 10% of the sum of the crab weight and hook load. It IS divided equally between the four crane wheels when the wheels are double Hanged and can act in either direction. The second hOrizontal load (longitudinal) IS the braking ioad of the whole crane, and in this case acts along the crane girder at the level oflhe top flange. The value of this load is assessed at 5% of each wheel load. and is therefore a maXImum when the wheel load is a maxunum. As before, the braking load covers acceleration as well as non-vertical lifting. The loads are summanzed in Fig. 5.3. 10 addition, gantry girders Intended to carry class Q3 and Q4 cranes (as defined in BS 2573: Part 1) should be designed for the crabbing forces given In clause 4.11.2. 1 1 1 2 c I L ) W L / 4 c / 4 ) 2 / L 1 v 1 2 o ' L )c f r '( t wL / 2 ( c ) W h e e l l o a d s 0 . 7 ) / ( 1 5 . 0 0 . 7 ) / ( 1 5 . 0 2 8 . 6 6 7 . S k N 1 9 1 . 4 2 6 / 4 6 . 5 1 . 4 9 . 1 1 5 . 3 3 4 S T R U C T U R A L S T E E L W O R K D E S I G N T O S S 5 9 5 0 .C R A N E G I R D E R S 3 55 . 2 M A X I M U M L O A D E F F E C T S 5 . 3 ( a ) D i m e n s i o n s CW t 2 c / L I- i k fB e n d i n g 54 STRUCTURAL STEELWORK DESIGN TO as 5950 The safety factor '11 for crane loads (ultimate limit state) is taken as 1.6, Le. as for Imposed loads generally (SectIOn 1.7). Whenever the vertical load and the surge' load are combined in the deSign of a member, the safety factor should however be taken as lA for both loads (BS table 2). Further detailed proVISions for gantry girders are given In clauses 4.11 and 2.4.1.2. 5.2 MAXIMUM LOAD EFFECTS Fig. 504 MaXimum BM. SF and R Movmg loads, such as crane wheels, will result In bending moments and shear forces which vary as the loads travel along the supportmg gtrder. In slmply.supported beams the maXimum shear force will occur munediately adjacent to a support, while the maximum bending moment will occur near, - -but not necessarily at, In general, influence lineiJ,4,5} should be used to find the load positions producmg maximum values of shear force and bending moment. The maximum effects of two moving loads may be found from fonnulae(J) as demonstrated in Sectlon 2.5. For a simply supported beam the load pOSitions shown In Fig. 5,4 gwe maXlmum values: Shear force (max) = IT'(2 - clL) Bending moment (max) = WL/4 or =2W(Ll2 - C/4)2IL The greater of Ihe bending moment values should be adopted. The deSign of the bracket supporting a crane girder uses the value of maximum reaction from adjacent Simply supported beams, as in Fig. 504. Where adjacent spans are equal, the reachon IS equal to the shear force, I.e. ReactIOn (max) = IT'(2 - clL) ZI L Shaar force and reaction ZI Banding moment Bending moment CRANE GIRDERS 55 In all cases the effect of self weight (unifomlly distributed) of the glrder must be added. 5.3 EXAMPLE 9. CRANE GIRDER WITHOUT LATERAL RESTRAINT ALONG SPAN (a) Dimensions (b) (c) Span of crane Wheel centres 15.0 m 3.5 m 0.7 m Minimum hook approach Span of crane girder 6.5 m (Simply supported) Loading Class Q2 (no crabbing forces need be calculated) Hook load 200 kN Weight of crab 60 kN Weight of crane (excluding crab) 270 kN Wheel loads VertJcal wheel load from: 95.3 kN 28.6 kN 67.5 kN hook load 200(15.0 - 0.7)1(15.0 x 2) crab load 60(15.0 - 0.7)/(15.0 x 2) crane load 27014 Total vertll:al load = 191.4kN per wheel VertIcal load Wc (including allowance for Impact and 1'/) = 1.25 x lA x 191.4 =335 kN Where vertical load is considered acting alone then 'V/